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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.01%3A_Introduction_to_Acid_Base_Equilibria
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13.1: Introduction to Acid/Base Equilibria
Make sure you thoroughly understand the following essential concepts:
- Write a chemical equation for the autoprotolysis of a species such as HCO 3 – ; write an equation for the equilibrium constant for this reaction.
- Write equations that express the chemical reactions and equilibrium constant expressions that define the strength of an acid or base in terms of proton transfer processes (Brønsted-Lowry concept).
- Write expressions defining the equilibrium constants K a and K b .
- Given the value of any two of K a , K b , and K w , evaluate the third one. (This requires that you can show that K a × K b = K w .)
- Explain the significance of the value of pK a for a given monoprotic acid-base system — that is, what it tells us about that particular acid and its conjugate base.
- Sketch out a simple proton free energy (PFE) diagrams illustrating the distinction between monoprotic strong and weak acids .
- State the meaning of the leveling effect , and construct a simple PFE diagram that illustrates its action.
- Describe, in your own words, the meaning of PFE , how it relates to pH, and in its significance in terms of the relative concentrations of a conjugate acid-base pair in a solution.
Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. In order to thoroughly understand the material in this unit, you are expected to be familiar with the following topics which were covered in the separate unit Introduction to Acid-Base Chemistry:
- The Arrhenius concept of acids and bases
- The Brønsted-Lowry concept, conjugate acids and bases
- Definition of pH and the pH scale
- Strong vs. weak acids and bases
You should also have some familiarity with the principles of chemical equilibrium, and know how to write and use equilibrium constants.
Proton transfer reactions in acid-base equilibria
As you should recall from your earlier introduction to acids and bases, the +1 electric charge of the tiny proton (a bare hydrogen nucleus) is contained in such a miniscule volume of space that the resulting charge density is far too large to enable its independent existence in solution; it will always attach to, and essentially bury itself in, the non-bonding orbitals of a solvent. Thus in aqueous solution, what we commonly represent as the "hydrogen ion" H + is more accurately described as the hydronium ion H 3 O + .
Acid-base chemistry is a transactional process in which protons are exchanged between two chemical species. A molecule or ion that loses or "donates" a proton is acting as an acid ; a species that receives or "accepts" that proton plays the role of a base .
Defining an acid as a substance that simply "dissociates" into its component parts
\[HA \rightarrow H^+ + AB^–\]
fails to capture the concept outlined in the above box. It is therefore important that you become comfortable with the Brønsted-Lowry model of acid-base behavior that represents the generalized acid-base reaction as a proton transfer process :
\[\underbrace{HA}_{\text{acid}} + \underbrace{B}_{base} \rightarrow \underbrace{A^-}_{\text{conjugate base}} + \underbrace{HB^+}_{\text{conjugate acid}} \]
The actual electric charges of the reaction products will of course depend on the particular nature of the species \(A\), but the base will always have one more negative charge than the acid HA. Although anyone familiar with chemicals tends to regard certain substances as "acids" and others as "bases", it is important to bear in mind that these labels have meaning only in terms of the particular process being considered. For example, in the absence of water, ammonia, which we usually think of as a base, can lose a proton (and thus act as an acid) to a "stronger" base \(B\) to form the amide ion:
\[NH_3 + B → NH_2^– + HB^+\]
Similarly, anhydrous sulfuric acid can accept a proton from a stronger acid AH:
\[H_2SO_4 + AH \rightarrow H_3SO_4^+ + A^–\]
However, this is rather exotic stuff that is beyond the scope of this introductory lesson. The acid-base processes we describe in this set of lessons take place in aqueous solution, so we will always assume that \(H_2O\), which can act either as a proton donor or proton acceptor, plays an active role:
- A substance that acts as an acid in aqueous solution donates a proton to \(H_2O\), yielding the hydronium ion \(H_3O^+\).
\[HA + H_2O → H_3O^+ + A^–\]
- Similarly, a base will remove a proton from \(H_2O\), yielding the hydroxide ion \(OH^–\).
\[B + H_2O → BH^+ + OH^–\]
Autoprotolysis
A small number of acids are themselves amphiprotic, meaning that one molecule can transfer a proton to another molecule of the same kind. The best-known example of this process, known as autoprotolysis , is of course that of water:
\[2 H_2O_{(l)} \rightleftharpoons OH^– + H_3O^+ \label{1-1}\]
Pure liquid ammonia and sulfuric acid are other well-known examples:
\[2 NH_{3(l)} \rightleftharpoons NH_2^– + NH_4^+\]
\[2 H_2SO_{4(l)} \rightleftharpoons HSO_4^– + H_3SO_4^+\]
Autoprotolysis reactions such as these typically proceed to only a very small extent. The equilibrium constant for Equation \(\ref{1-1}\) is commonly written as
\[K_w = [H^+][OH^–] \label{1-2}\]
and is usually known as the dissociation constant or ion product of water. For most purposes in elementary courses, we can use the value 1.0 × 10 –14 for K w . But you should understand that this value is correct only for pure water at 25° C and 1 atm pressure. K w varies considerably with temperature, pressure, and the ionic content of water:
It should be obvious that anyone working in the fields of physiological or oceanographic chemistry cannot routinely use \(1.0 \times 10^{–14}\) for K w !
The equilibrium constants K a and K b
Acids and bases vary greatly in their "strength" — that is, their tendencies to donate or accept protons. As with any chemical reaction, we can define an equilibrium constant \(K_c\) whose numerical value reflects the extent of the reaction — that is, the relative concentrations of the products and reactants when the reaction reaches equilibrium. The strength of an acid \(HA\) in water can be defined by the equilibrium
\[HA + H_2O \rightleftharpoons H_3O^+ + A^–\]
\[K_a =\dfrac{[H_3O^+][A^-]}{[HA]} \label{1-3a}\]
Similarly, the strength of the base \(A^−\) in water is defined by
\[A^− + H_2O \rightleftharpoons HA + OH^–\]
\[K_b = \dfrac{'[HA][OH^-]}{[A^-]} \label{1-3b}\]
Note carefully that reaction \(\ref{1-3a}\) is not the reverse of \(\ref{1-3b}\).
K a and K b are inversely proportional
How are K a and K b related? The answer can found by multiplying the above two expressions for K a and K b :
(1-3)
Alternatively, we can simply add the two corresponding reaction equations:
and — as you should recall from your earlier study of equilibrium, the equilibrium constant of the sum of two reactions is the product of the two K's. So either way you figure it, the \(K_a\) for an acid and \(K_b\) for its conjugate base are related in a very simple way:
\[K_a K_b = K_w \label{1-4}\]
Clearly, as the strength of a series of acids increases, the strengths of their conjugate bases will decrease, this inverse relation between K a and K b is implicit in Equation \(\ref{1-4}\). You will recall that the pH scale serves as a convenient means of compressing a wide range of [H + ]-values into a small range of numbers. Just as we defined the pH as the negative logarithm of the hydrogen ion concentration, we can define
\[pK = −\log K \label{2-5}\]
for any equilibrium constant. Acid and base strengths are very frequently expressed in terms of \(pK_a\) and \(pK_b\). From Equation \(\ref{1-4}\), it should be apparent that
\[pK_a + pK_b = pK_w \;(14.0\; at\; 25^o C) \label{1-5}\]
pK values of some common acids and bases
In order to convert pK a into K a , simply take the negative antilog of pK a . Thus, for acetic acid with
\[pK_a = 4.7\]
\[K_a = 10^{–4.7} = 2.0 \times 10^{–5} \]
The following table gives the p Ka values for a number of commonly-encountered acid-base systems which are listed in order of decreasing acid strength. It's worth taking some time to study this table in some detail.
lease take special note of the following points:
- Acids (1-4) are the major strong acids. All dissociate in water to produce the hydronium ion (5) , the strongest acid that can exist in water.
- (6) H 2 SO 3 has never been detected; "sulfurous acid" and its formula refer to a solution of SO 2 in water.
- The names bisulfate (3, 7) , bisulfite (6, 16) , and bicarbonate (13) are still commonly used by chemists, but the preferred names are now hydrogen sulfate , hydrogen sulfite , and hydrogen carbonate , respectively. Similarly, bisulfide (14, 26) is also known as hydrosulfide .
- The ion (9) as it exists in aqueous solution is often represented as Fe 3+ and is known more commonly as the ferrous or iron(III) ion. The aqueous aluminum ion Al 3+ (12) as well as many transition metal ions have similarly complex structures and give acidic solutions in water because the high positive charge increases the acidity of the attached H 2 O units. For more on these aquo complexes, see this brief discussion, or this much more detailed one by Jim Cook of the UK.
- The molecule H 2 CO 3 (13) exists as a minority species in an aqueous solution of CO 2 and has never been isolated. In most contexts, the formula H 2 CO 3 , the name carbonic acid and the pK a 6.3 refer to "total dissolved CO 2 " in the water. "True" H 2 CO 3 is about a thousand times stronger than the pK a of its equilibrium mixture with CO 2 (aq) .
- The sulfide, amide and oxide ions (26-28) , being stronger bases than OH – , cannot exist in water.
| ref | acid | pK a | conjugate base | pK b | |
|---|---|---|---|---|---|
| 1 | HClO 4 | perchloric acid | ~ –7 | ClO 4 – chlorate | ~ 21 |
| 2 | HCl | hydrochloric acid | ~ –3 | Cl – chloride | ~ 17 |
| 3 | H 2 SO 4 | sulfuric acid | ~ –3 | HSO 4 – bisulfate | ~ 17 |
| 4 | HNO 3 | nitric acid | ~ –1 | NO 3 – nitrate | 15 |
| 5 | H 3 O + | hydronium ion | 0 | H 2 O water | 14.00 |
| 6 | H 2 SO 3 | sulfurous acid | 1.8 | HSO 3 – bisulfite | 12.2 |
| 7 | HSO 4 – | bisulfate ion | 1.9 | SO 4 2– sulfate | 12.1 |
| 8 | H 3 PO 4 | phosphoric acid | 2.12 | H 2 PO 4 – dihydrogen phosphate | 11.88 |
| 9 | [Fe(H 2 O) 6 ] 3+ | hexaaquoiron(III) | 2.10 | [Fe(H 2 O) 5 OH] 2+ | 11.90 |
| 10 | HF | hydrofluoric acid | 3.2 | F – fluoride | 10.8 |
| 11 | CH 3 COOH | acetic acid | 4.76 | CH 3 COO – acetate | 9.3 |
| 12 | [Al(H 2 O) 6 ] 3+ | hexaaquoaluminum(III) | 4.9 | [Al(H 2 O) 5 OH] 2+ | 9.1 |
| 13 | H 2 CO 3 | carbonic acid | 6.3 | HCO 3 – bicarbonate | 7.7 |
| 14 | H 2 S | hydrogen sulfide | 7.04 | HS – bisulfide | 6.96 |
| 15 | H 2 PO 4 – | dihydrogen phosphate | 7.2 | HPO 4 – monohydrogen phosphate | 6.8 |
| 16 | HSO 3 – | bisulfite ion | 7.21 | SO 3 2– sulfite | 6.79 |
| 17 | HOCl | hypochlorous acid | 8.0 | OCl – hypochlorite | 6.0 |
| 18 | HCN | hydrocyanic acid | 9.2 | CN – cyanide | 4.8 |
| 19 | H 3 BO 4 | boric acid | 9.30 | B(OH) 4 – borate | 4.79 |
| 20 | NH 4 + | ammonium ion | 9.25 | NH 3 ammonia | 4.75 |
| 21 | Si(OH) 4 | ortho -silicic acid | 9.50 | SiO(OH) 3 – silicate | 4.50 |
| 22 | HCO 3 – | bicarbonate ion | 10.33 | CO 3 2– carbonate | 3.67 |
| 23 | HPO 4 2– | monohydrogen phosphate ion | 12.32 | PO 4 3– phosphate | 1.67 |
| 24 | SiO(OH) 3 – | silicate ion | 12.6 | SiO 2 (OH) 2 2– | 1.4 |
| 25 | H 2 O | water | 14.00 | OH – hydroxide | 0 |
| 26 | HS – | bisulfide ion | ~ 19 | S 2– sulfide | ~ –5 |
| 27 | NH 3 | ammonia | ~ 23 | NH 2 – amide | ~ -9 |
| 28 | OH – | hydroxide ion | ~ 24 | O 2– oxide | ~ –10 |
The Fall of the proton
The concept of the "potential energy of the proton" (more properly, proton free energy ), is not nearly as intimidating as it might sound; it is easy to grasp, requires no mathematics, and will enable you to achieve a far better understanding of acid-base chemistry.
Most courses (and their textbooks) do a good job of explaining the concept of proton transfer, but few of them provide you with a clear basis for understanding and making useful predictions about the direction and extent this transfer will take. For example, when the gas HCl is added to water, you probably know that the reaction proceeds strongly to the right:
\[HCl + H_2O \rightleftharpoons H_3O^+ + Cl^– \label{2-1}\]
...but what principle determines this, and what is to prevent the hydronium ion (also an acid) from pushing one of its own protons back onto a chloride ion, and thus reversing the reaction?
The concept of the proton free energy level was introduced in R.W. Gurney's classic Ionic Processes in Solution (1953). The best detailed exposition can be found in the first (1970) edition of Aquatic Chemistry by W. Stumm and J.L. Morgan.
The answer is that a proton is never really "pushed"; it will spontaneously seek out the lowest potential energy state (that is, the strongest base) it can find. Think how a macroscopic object such as a book will tend to fall to a location where its [gravitational] potential energy is as low as possible. Of course, the potential energy of a proton has nothing to do with gravity, but rather with how tightly it can bind to a base. This kind of potential energy is known as proton free energy (PFE) — but in the context of first-year chemistry, we can also just call it "proton energy".
You are not sure what free energy is? Don’t worry about it for the time being; just think of it as you would any other form of potential energy: something that falls when chemical reactions take place. See here for the lesson on free energy.
For the present, all you need to know is that a proton will tend to "fall" to the lowest potential energy state it can find. In keeping with this idea, you can think of an acid as a proton source , and a base as a proton sink . Using this terminology, we can depict the generalized process HA + B → AB – + HB + as in this plot, which depicts the proton on HA "falling" to the lower-PFE "sink" provided by the base B, leaving AB – and HB + as products.
We will be making a lot of use of schematics like this farther on, so take a moment to familiarize yourself with information it depicts.
This "source/sink" nomenclature recalls the tendency of water to flow down from a high elevation to a lower one; this tendency (which is related to the amount of energy that can be extracted in the form of electrical work if the water flows through a power station at the bottom of the dam) will be directly proportional to the difference in elevation (difference in potential energy ) between the source (top of the dam) and the sink (bottom of the dam).
Thus you can regard proton free energy as entirely analogous to the gravitational potential energy of the water contained in a dam; in this case, we are dealing with chemical potential energy, more commonly known as free energy . Take particular note of the following:
- The ordinate represents the relative free energy of a mole of protons when attached to a given base; we label it simply "proton energy" for brevity.
- The gray vertical line at the center of the diagram separates the acidic form of each conjugate pair from its base form; the terms "proton sources" and "proton sinks" are there to remind us of the role these forms play in acid-base chemistry.
- The red diagonal arrow represents the actual process in which a mole of protons "falls" from the acid HA to the base B. The fact that the arrow slants downward indicates that this process is spontaneous, meaning that the reaction proceeds to completion. In other words, the equilibrium constant for this reaction is very large.
- The green horizontal arrows depict the changes (that is, the formation of the two new conjugate species) that result from the proton transfer as the protons "fall" (in free energy) from the acid HA to the base B.
Strong acids in water
When you were first introduced to the Brønsted-Lowry model, you learned that the proper way to represent the "dissociation" of a strong acid like hydrochloric is
\[HCl + H_2O \rightarrow H_3O^+ + Cl^- \label{2-1a}\]
Thus water is revealed as an active participant in the behavior of an acid, rather than as an inert spectator or solvent. Although analogous acid-base reactions can take place in other solvents such as liquid ammonia, virtually all of the acid-base chemistry we encounter in daily life is based on water. Since water is acting as the proton acceptor here, it must be fulfilling the role of the base B in the schematic given above. We can therefore construct a similar plot specifically for HCl (or, in fact, for any strong acid).
Recall that, for strong acids, this reaction is essentially complete. This means that a "solution of hydrochloric acid" is in reality a solution of hydronium and chloride ions; except in solutions much more concentrated than 1 M, the species HCl is only present in trace amounts.
Weak acids in water
When acetic acid CH 3 COOH is dissolved in water, the resulting solution displays only very mild acidic properties. For simplicity we will represent acetic acid and the acetate ion CH 3 COO – by HAc and Ac – , respectively:
\[HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{2-2}\]
But although this equation has the same form as (2-1) above, it hides an important qualitative difference: HAc is a weak acid, meaning that the proton transfer takes place to only a tiny extent; the equilibrium strongly favors the left side:
HAc + H 2 O H 3 O + + Ac –
The corresponding proton free energy diagram now shows the H 3 O + -H 2 O system at the top, with the proton-transfer arrow pointing upward, indicating that the protons get kicked up to a higher free-energy level in this process.
Where does this additional energy come from? Simply from the random thermal energy in the solution. As you might expect, this is not a thermodynamically favorable process, so it only happens to a very small extent. In contrast to the HCl example we described previously in which 99.99+ percent of the HCl molecules end up as hydronium ions, almost all of the acetic acid molecules remain unchanged, and only tiny amounts of H 3 O + and Ac – are produced.
Strong and weak acids both donate protons to water, yielding hydrogen ions and thus rendering the solution acidic.
What's different is that virtually all of the protons of the strong acid "fall" to the H 2 O level, whereas only a tiny fraction of the weak acid molecules acquire enough thermal energy to promote their protons up to the H 2 O level.
The fundamental difference between strong acids and weak acids is clearly evident when we combine the preceding two proton-free energy diagrams into one. Strong acids (strong "conjugate-pairs") are always above the PFE of the H 3 O + -H 2 O system, while the weak conjugate pairs are below it.
The other big difference with acetic acid is that water is acting as a base here, rather than as an acid as it did in the HCl example. So it's not surprising that water, the basis of almost all acid-base chemistry, should appear in the middle of these diagrams.
Water can swing both ways
Substances that can both donate and accept protons are said to be amphiprotic . Many oxides and hydroxides behave in this way, and since H 2 O falls into both of these classes, it is not surprising that water is amphiprotic:
- Water ("the-acid") : \[H_2O + B \rightleftharpoons BH^+ + OH^–\]
- Water ("the base") : \[H_2O + AH \rightleftharpoons H_3O^+ + A^–\]
("AH" above represents any acid; B stands for any base)
These two opposing tendencies can be represented by
In the case of water, the contest is a draw; not only does neither side win, but the fraction of
H
2
O molecules that break up in this way is miniscule: In pure water, only about one H
2
O molecule out of 10
9
is “dissociated” at any instant:
H 2 O H + + OH – (2-4)
The actual reaction, of course, is the proton transfer
(2-5)
for which the equilibrium constant
K w = [H 3 O + ][OH – ] (2-6)
is known as the ion product of water . The value of K w at room temperature is 1.008 x 10 –14 . We can also express this in logarithmic terms: pK w = 14.0.
As with any equilibrium constant, the value of K w is affected by the
- temperature : K w undergoes a 10-fold increase between 0°C and 60°C
- pressure : K w is about doubled at 1000 atm
- presence of ionic species in the solution .
Because most practical calculations involving K w refer to ionic solutions rather than to pure water, the common practice of using 10 –14 as if it were a universal constant is unwise; under the conditions commonly encountered in the laboratory, pK w can vary from about 11 to almost 15. In seawater, K w is 6.3 × 10 −12 . (See Stephen J. Hawkes: “ pK w is almost never 14.0 ”, J. Chem. Education 1995: 72(9) 799-802)
Notice that under conditions when K w differs significantly from 1.0 × 10 –14 , the pH of a neutral solution will not be 7.0.
For example, at a pressure of 93 kbar and 527°C, K w = 10 −3.05 , the pH of pure water would be 1.5. Similarly unusual conditions apply to deposits of water in geological formations and in undersea vents.
At 60° C, the ion product of water is \(9.6 \times 10^{-14}\). What is the pH of a neutral solution at this temperature?
Solution:
Under these conditions, [H + ][OH – ] = 9.6E–14. If the solution is neutral, [H + ] = [OH – ] = (9.6E–14) ½ = 3.1E–7, corresponding to a pH of log 3.1E-7 = 6.5 .
The leveling effect
The strengths of strong acids and bases are "leveled" in water. This means that all strong acids and strong bases are equally strong in water . To see this more clearly, notice that the free energies of protons in all of these acids (even nitric) are sufficiently above the H 3 O + /H 2 O level that their "dissociation" is essentially complete. Thus
The strongest acid that can exist in water is H
3
O
+
;
The strongest base that can exist in water is OH
–
Similarly, the hydroxide ion is the strongest base that can exist in water.
So what happens if you add a soluble oxide such as Na
2
O to water? Since O
2
–
is a proton sink to H
2
O, it will react with the solvent, leaving OH
–
as the strongest base present: Na
2
O + H
2
O → 2 OH
–
+ Na
+
.
Thus all bases stronger than OH – appear equally strong in water — simply because they are all converted to OH – .
PFE's are the key to understanding acid-base reactions
In the preceding section, we have employed simple PFE (proton free energy) diagrams to help you better understand the concepts of strong and weak acids, water as an acid or a base, and the leveling effect. But we can gain a more generalized and comprehensive view of acid-base chemistry with the aid of a PFE diagram that shows a whole series of acid-base systems, arranged in order of decreasing acid strengths.
How acid strengths relate to proton free energies
As we explained near the beginning of this lesson, we ordinarily express the strength of an acid by quoting its "dissociation constant" K a , which is the equilibrium constant for transfer of a proton to the base H 2 O.
HA + H 2 O → H 3 O + + A –
We also introduced the concept of PFE as an alternative way of expressing the strength of an acid. If both PFE and K a express the same thing, you may wonder how these two quantities are related.
If you have had an introduction to thermodynamics and specifically to free energy G , you might wish to look at the Appendix section of this lesson to review the relationship between the fall in free energy and the value of pK a .
A thorough answer to this question requires a bit of thermodynamics, which you may not have yet studied, but you don't really need any of the math here. It should suffice to show the results in the form of a plot, which yields a simple straight-line relationship when plotted on a semi-logarithmic scale.
Note that
- The extent of any chemical reaction is proportional to the fall in free energy when it takes place. So negative Δ G values correspond to larger equilibrium constants.
- For acid-base reactions specifically, K a is inversely proportional to Δ G , and thus to the fall in PFE.
- This means that the proton on a strong acid ( K a > 1) has a large negative PFE; it has a "long way to fall" to the H 2 O level.
- Conversely, a proton on a weak acid ( K a <1) can only move up to the H 2 O level when supplied with thermal energy from the surroundings.
- A K value of 10 0 = 1 corresponds to a zero-free energy change, meaning that the reaction is at equilibrum when [HA] = [A – ].
- Proton-free energy changes exceeding –20 kJ mol –1 result in K a values so large that the proton transfer can be considered "complete" for most practical purposes.
Understanding the PFE chart
On the PFE diagram shown below, each acid-base system is represented by a horizontal line connecting an acidic species with its conjugate base. Each line is placed at a height on the diagram that is proportional to its pK a , as measured on the logarithmic scale (labeled "pH") at the right. What we have, then, is a list of acid-base systems arranged in order of decreasing acid strength.
Think of this chart as an alternative view of the table of acid strengths displayed near the top of this page, in which some of the information that is "hidden" in the table is now revealed in an easily-grasped way.
Take particular note of the following points:
- The Δ G scale on the left measures the fall in potential energy when one mole of protons is transferred from a given acid to H 2 O. Negative values correspond to stronger acids.
-
Any acid shown on the left side of the vertical line running down the center of the diagram can donate protons to any base (on the right side of the line) that appears
below
it.
The value of the equilibrium constant for such a reaction will exceed unity; for Δ G separations of more than about 10 kJ/mol, the reaction can be considered to be complete. - The zero free energy assigned to the hydronium ion corresponds to the standard free energy of formation Δ G f ° of this ion.
- Notice how the two conjugate pairs H 3 O + / H 2 O and H 2 O/OH – bracket the 80-kJ/mol region that comprises the weak acids. The weakness of these acids reflects the energetic cost of moving a proton "up-hill" to the H 2 O level.
- The strong acids at the top of the table can all be regarded as totally "dissociated" in water. This means that these acids cannot exist in water, as they are totally transformed into H 3 O + . Of course, they will also react with any bases below the H 2 O level that may be available. (See "leveling effect" below.)
- Similarly, the two strong bases NH 2 – and O 2 – shown at the bottom of the table cannot exist in water. Thus if solid sodium oxide Na 2 O is added to water, the oxide ions are such a low-energy sink that will totally fill up with protons from the H 2 O acid level which defines the top of the strong-base range.
- Notice the pH scale along the right side, and note the range of pH values possible in water. We will say more about the significance of pH in these diagrams farther on.
The PFE chart helps you visualize (and understand!) acid-base chemistry
The virtue of a PFE chart is that you can see, at a glance, the relations between conjugate acid-base pairs that enables you to make quick, qualitative determinations of what is going on even in complex mixtures of acids and bases. The speed and convenience this affords enables you to build a much deeper conceptual understanding of acid-base chemistry.
In order to do the same thing by consulting a simple list of acid K a 's or pK a 's, you also need to be familiar with the principles of chemical equilibrium and thermodynamics — and then take the time to work through the arithmetic.
In the preceding sections we have shown how the PFE-chart approach can clarify the distinction between strong and weak acids. We will now go on to describe a variety of other principles that are easily grasped by consulting a PFE chart.
Significance of the pH scale
You will have noticed the scale labeled "pH" at the left side of the PFE plot.
First, please understand what this scale does NOT do; it does not tell you what the pH of the solution will be when you add an acid to pure water. This pH will depend on both the strength ( K a or pK a ) of the acid, and on its concentration in the resulting solution; you will learn how to carry out pH calculations in later lessons.
All but the very weakest acids can drive the pH down to near the bottom of the scale if their solutions are sufficiently concentrated. The only thing you can be sure of is that the pH of a solution of a solution of an acid in pure water will never be greater than 7.
pH sets the average proton free energy
When pH was first introduced in 1909, it was defined in terms of the "hydrogen ion concentration" [H + ]. This has since been amended to the hydrogen ion activity {H + }. The latter term refers to the "effective" concentration of these ions — that is to the "availability" of protons (regardless of whether they physically exist as H 3 O + units or in other forms) to react with bases. But from the standpoint of thermodynamics, "availability" is just a loose term for "free energy". So we can also say that
pH is a measure of the average PFE in an aqueous solution
The more negative the pH, the higher the proton free energy in the solution. (The inverse relation is a consequence of the negative-logarithm definition of pH.)
The pH has the dual advantages of being both dimensionless and readily observable. More importantly, we can easily alter the pH of a solution by adding some strong acid or base. And when we do this, we are in effect using the pH as a tool for controlling the average PFE in the solution.
More on this in a moment; first, let us introduce an important relation that you may have seen in an earlier chemistry course, and which we will discuss more in a later lesson in this group.
pH can control the relative abundance of conjugate acid-base pairs
Recalling the equilibrium expression for a weak acid
\[K_a = \dfrac{[H^+][A^-]}{[HA]} \label{3-1}\]
We can solve this for \([H^+]\):
\[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{3-2}\]
Re-writing this in terms of negative logarithms, this becomes
(3-3)
or, since pK a = – log K a , we invert the ratio to preserve the positive sign:
We will make extensive use of this equation in a later lesson on acid-base buffering and titration.
(3-4)
This extremely important relation tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system. Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its
pK
a
. According to the above equation, when pH =
pK
a
, the log term becomes zero, so that the ratio
[AB
–
] / [HA] = 10
0
= 1, meaning that [HA] = [AB
–
]. In other words,
This pH adjustment can be easily made by adding appropriate amounts of a strong acid or strong base to the solution.
When the pH of a solution is set to the value of the pK a of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically:
This is the basis for the arrangement of the various acid-base systems depicted in the PFE diagram; each conjugate pair is placed at a location on the vertical pH scale that corresponds to the acid's pK a . It is instructive to consider how sensitively the concentration ratio [A – ] / [HA] depends on the pH.
Hypochlorous acid, HOCl, has a pK a of 8.0.
- Estimate the ratio R = [OCl – ] / [HOCl] in a solution of sodium hypochlorite whose pH has been adjusted to 8.2.
- If the initial concentration of the hypochlorite solution was 0.10 M , what will be the concentrations of the acid and base forms at pH 8.2?
Solution
a)
Substituting in Eq 3-4, the log term becomes (8.2 – 8.0) = 0.2,
so
R
= 10
0.2
= 1.6.
Note that it does not matter whether the initial solution consists of HOCl or NaOCl or some mixture thereof; adjusting the pH forces the system composition to the ratio calculated above.)
b) In the resulting solution, the base form OCl – is the predominant species, the mole ratio R being 1.6/1. The mole fraction of OCl – will therefore be 1.6/2.6 = 0.62, making that of HOCl (1.0 – .62) = 0.38. The actual concentrations will be 0.1 times these values, reflecting the 0.10 M concentration of the solution.
To what pH must we adjust a solution of acetic acid ( pK a 4.7) in order to convert 20 percent of the acid into its base form?
Solution:
In the final solution, the mole fractions will be [AB – ] = .20, [HA] = 0.80; the ratio [AB – ]/[HA] = 0.20/0.80 = 0.25. Substituting into Eq 2-9, we have
\[pH = 4.7 + \log\' 0.25 = 4.7 – 0.6 = 4.1\]
We can explore the relation between the pH and the distribution of conjugate species between the acid and base forms by carrying out calculations similar to those in Problem Example 2 for an acid-base system at different pH values.
The orange bars and blue numbers show the mole fractions of the acid and base forms at pH values of the pK a and the pK a ± 1. The green numbers give the percent dissociation of the acid at each pH.
The apparent symmetry between the two extreme pH values suggests a simple relation between the log of the ratio
[AB
–
] / [HA] and the pH. This is seen very clearly when this ratio is plotted as a function of pH.
Solutions containing many different weak acid-base systems are very commonly encountered in nature, especially in biological fluids or natural waters, including the ocean. When the pH of such solutions is altered by the addition of strong acid or base (as can occur, for example, when wastes from industrial processes or mine drainage, or acidic rain, enter a lake or stream); the distribution of all conjugate acid-base systems will change.
It is important to note that the [AB – ] / [HA] ratios (calculated here from Eq 3-4) yield only a qualitative picture and do not take into account the concentrations of the different acid-base systems or proton transfers between them. The latter processes would cause the lowest proton-vacant levels (that is, the strongest bases) to be completely filled from the bottom up, depending on the number of protons available. An exact calculation would require solving a set of simultaneous equations and is beyond the scope of this lesson.
A . What do we mean by the "strength" of an acid ?
The Brønsted-Lowry model defines the strength of an acid HA as its tendency to donate protons to a base B. But bases vary in their abilities to accept protons, so the tendency for the complete transaction
HA + B → BH + + A – (A-1)
to occur depends on the two processes
HA → H + + A – K 1 (A-2)
B + H + → BH + K 2 (A-3)
K a 's are always expressed on a scale relative to the K b of the solvent
Because it is impossible to study either of these steps independently, what we might call the "absolute" strength of an acid (Eq A-2) cannot be measured. For this reason, we define a "standard" base, usually the solvent, and most commonly, water. Thus we replace Eq A-3 with
H 2 O + H + → H 3 O + K 3 (A-4)
so that Eq A – 1 now expresses the strength of HA relative to the base H 2 O :
HA + H 2 O → H 3 O + + A – (A-5)
which, for convenience, we write in its abbreviated form
HA → H + + A – (A-5)
whose equilibrium constant is
(A-6)
This value of K a is, of course, numerically identical to that for Eq A-5.
B . pH and proton free energy
This section may be of interest to readers who have studied elementary thermodynamics, and have some familiarity with Gibbs free energy . it is not required for understanding and using the concept of PFE.
From elementary thermodynamics, the driving force of a chemical reaction is given by the standard free energy change:
Δ G ° = –RT ln K (B-1)
Solving this equation for K yields
(B-2)
For a weak acid we know that
(B-3)
Substituting Eq B-2 into this expression, we get
(B-4)
Next, we take the negative log of each term, recalling that ln x = 2.3 log x :
(B-5)
or
(B-6)
For the special case in which the concentrations of the conjugate pairs [H + ] and [HA] are equal, pH = pK a , and the rightmost term above disappears, leaving
(B-7)
Thus the pH is a direct measure of the average proton free energy in a solution.
Construct a PFE diagram showing a) H 3 O + , b) acetic acid HAc ( pK a = 4.76), and c) H 2 O.
Solution
Solving Eq A – 13 for ΔG° yields ΔG° = (2.3RT) × pK a , or, at 25°C, ΔG° = (2.3 × 8.314 J K –1 mol –1 ) × pK a = 5698 J K –1 mol –1 × pK a
a) The proton in H 3 O + is already at the level of that in the standard base H 2 O (Eq A – 5), so for this transfer is [by definition] ΔG° = 5698 J K –1 mol –1 × 0 = 0
b) The increase in the free energy required to transfer a proton from its level in HAc to that in H 2 O is ΔG° = 5698 J K –1 mol –1 × 4.76 = 27.1 kJ mol –1
c) To transfer a proton from H 2 O to another H 2 O (autoprotolysis), its free energy must increase by 5698 J K –1 mol –1 × 14.0 = 79.8 kJ mol –1
|
libretexts
|
2025-03-17T19:53:15.774185
| 2016-02-13T03:24:26 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.01%3A_Introduction_to_Acid_Base_Equilibria",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.1: Introduction to Acid/Base Equilibria",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.02%3A_Strong_Monoprotic_Acids_and_Bases
|
13.2: Strong Monoprotic Acids and Bases
Make sure you thoroughly understand the following essential concepts that have been presented above.
- Estimate the pH of a solution of a strong acid or base, given its concentration.
- Explain the distinction between the concentration of a substance and its activity in a solution.
- Describe the origin and effects of ion-pairing in concentrated solutions of strong acids.
- Name and write formulas for the four major strong acids .
To a good approximation, strong acids, in the forms we encounter in the laboratory and in much of the industrial world, have no real existence; they are all really solutions of \(\ce{H3O^{+}}\). So if you think about it, the labels on those reagent bottles you see in the lab are not strictly true! However, if the strong acid is highly diluted, the amount of \(\ce{H3O^{+}}\) it contributes to the solution becomes comparable to that which derives from the autoprotolysis of water. Under these conditions, we need to develop a more systematic way of working out equilibrium concentrations.
At Moderate Concentrations, Forget About Equilibria
A strong acid , you will recall, is one that is stronger than the hydronium ion \(\ce{H3O^{+}}\). This means that in the presence of water, the proton on a strong acid such as HCl will "fall" into the "sink" provided by H 2 O, converting the latter into its conjugate acid \(\ce{H3O^{+}}\). In other words, \(\ce{H3O^{+}}\) is the strongest acid that can exist in aqueous solution .
As we explained in the preceding lesson, all strong acids appear to be equally strong in aqueous solution because there are always plenty of H 2 O molecules to accept their protons. This is called the "leveling effect".
This greatly simplifies our treatment of strong acids because there is no need to deal with equilibria such as for hydrochloric acid
\[\ce{HCl + H_2O → H_3O^{+} + Cl^{–}}\]
The equilibrium constants for such reactions are so overwhelmingly large that we can usually consider the concentrations of acid species such as "HCl" to be indistinguishable from zero. As we will see further on, this is not strictly true for highly-concentrated solutions of strong acids (Figure \(\PageIndex{3}\)).
Over the normal range of concentrations we commonly work with (indicated by the green shading on this plot), the pH of a strong acid solution is given by the negative logarithm of its concentration in mol L –1 . Note that in very dilute solutions, the plot levels off, showing that this simple relation breaks down; there is no way you can make the solution alkaline by diluting an acid!
What will be the pH of a 0.025 mol/l solution of hydrochloric acid?
Solution
If we assume all the hydronium concentration originates from the added acid. So
\[\ce{H3O^{+}}\]
and we just find the negative logarithm of the concentration
\[pH = -\log_{10} [\ce{H3o^{+}}] = –\log_{10} 0.025 = 1.6\]
Since this pH is so far away from 7, our assumption is reasonable.
Major Strong Acids
Mineral acids are those that are totally inorganic. Not all mineral acids are strong; boric and carbonic acids are common examples of very weak ones. However, in ordinary usage, the term often implies one of those described below. With the exception of perchloric acid, which requires special handling, these are all widely used in industry and are almost always found in chemistry laboratories. Most have been known since ancient times
| acid | name | pK a | base | pK b |
|---|---|---|---|---|
| HClO 4 | perchloric acid | ~ –7 | ClO 4 – | ~ 21 |
| HCl | hydrochloric acid | ~ –7 | Cl – | ~ 17 |
| H 2 SO 4 | sulfuric acid | ~ –7 | HSO 4 – | ~ 17 |
| HNO 3 | nitric acid | –1 | NO 3 – | 15 |
| H 3 O + | hydronium ion | H 2 O | 14 | |
You should know the names and formulas of all four of these widely-encountered strong mineral acids.
- Hydrochloric acid \(HCl\): In contrast to the other strong mineral acids, no pure compound "hydrochloric acid" exists. One of the most commonly used acids, hydrochloric acid is usually sold as a constant-boiling 37% solution of hydrogen chloride gas in water, making its concentration about 10 M . Although higher concentrations are possible, the high partial pressures of HCl in equilibrium with the solution makes them difficult to store, ship and work with. Commercial-grade hydrochloric acid is often called muriatic acid .
- Nitric acid \(HNO_3\): Concentrated nitric acid is a 68% constant-boiling solution of HNO 3 in water, corresponding to about 11M. Although the acid itself is colorless, its slow decomposition into NO 2 (especially in the presence of light) often results in a yellow or orange color. More concentrated solutions are sold as "fuming nitric acid"; one form, available as "white fuming nitric acid" or "anhydrous nitricacid" contains 97.5% HNO 3 and only 2% water; the remainder consists of dissolved NO 2 . Pure HNO 3 forms crystals that melt at –42°C. In addition to being a strong acid, nitric acid at high concentrations acts as a powerful oxidizing agent, a property that accounts for its use as a rocket fuel.
- Perchloric acid \(HClO_4\) : This is the strongest of the mineral acids, thanks to the electron-withdrawing action of the oxygen atoms which make it energetically easier for HClO 4 to lose its proton. Its usual commercial form is a constant-boiling 72.5% solution in water. Concentrated perchloric acid is a powerful oxidizing agent that can react explosively with organic materials and certain metals. Its use in the laboratory requires special handling. A 1947 explosion of a 1000-L vat of HClO 4 in a Los Angeles electroplating plant killed 17 people and damaged over 250 homes and other buildings.
- Sulfuric acid \(H_2SO_4\): Sulfuric acid is by far the most industrially important acid, and is also the most concentrated one available. "Concentrated" sulfuric acid contains 98% by weight of H 2 SO 4 ; its density of 1.83 kg/L and oil-like viscosity reflect this high concentration. "100% H 2 SO 4 " (which can be prepared However, is not stable) actually contains a variety of other species, all in equilibrium with H 2 SO 4 . These include H 2 S 2 O 7 , H 2 S 4 O 13 , and H 2 SO 4 's autoprotolysis products H 3 SO 4 – and HSO 4 + .
Super acids
There is a class of super acids that are stronger than some of the common mineral acids. According to the classical definition, a superacid is an acid with an acidity greater than that of 100% pure sulfuric acid. Some, like fluorosulfuric acid \(FSO_3H\) are commercially available. Strong superacids are prepared by the combination of a strong Lewis acid and a strong Brønsted acid. The strongest known super acid is fluoroantimonic acid (\(H_2FSbF_6\)). This acid is so corrosive that the fumes alone will dissolve entire fume hoods, glass and plastic beakers, human skin, bone and most synthetic compounds.
Strong bases
The only strong bases that are commonly encountered are solutions of Group 1 hydroxides, mainly NaOH and KOH. Unlike most metal hydroxides, these solids are highly soluble in water, and can thus yield concentrated solutions of hydroxide ion, the strongest base that can exist in water — the ultimate aquatic proton sink. Sodium hydroxide is by far the most important of these strong bases; its common names "lye", "caustic soda" (or, in industry, often just "caustic"), reflect the diverse uses of NaOH.
Solid NaOH is usually sold in the form of pellets. When exposed to air, they become wet ( deliquescence ) and absorb CO 2 , becoming contaminated with sodium carbonate. NaOH is the most soluble of the Group 1 hydroxides, dissolving in less than its own weight of water (111 g / 100 ml) to form a 2.8 M/L solution at 20°C. However, as with strong acids, the pH of such a solution cannot be reliably calculated from such a high concentration.
Acids at High Concentrations
At higher concentrations, intermolecular interactions and ion-pairing can cause the effective concentration (known as the activity ) of \(\ce{H3O^{+}}\) to deviate from the value corresponding to the nominal or "analytical" concentration of the acid. Activities are important because only these work properly in equilibrium calculations. Also, pH is defined as the negative logarithm of the hydrogen ion activity, not its concentration. The relation between the concentration of a species and its activity is expressed by the activity coefficient \(\gamma\):
\[a = \gamma C\]
As a solution becomes more dilute, \(\gamma\) approaches unity. At ionic concentrations not exceeding about 2 M , concentrations of typical strong acids can generally be used in place of activities without serious error. Note that activities of single ions other than \(\ce{H3O^{+}}\) cannot be determined, so activity coefficients in ionic solutions are always the average, or mean , of those for the ionic species present. This quantity is denoted as \(\gamma_±\).
| molarity | \(\gamma_±\) |
|---|---|
| 0.0005 | 0.975 |
| 0.01 | 0.904 |
| 0.10 | 0.796 |
| 1 | 0.809 |
| 2 | 1.01 |
| 5 | 2.38 |
| 10 | 10.44 |
| 12 | 17.25 |
Because activities of single ions cannot be measured, these mean values are the closest we can get to \(\{H^+\}\) in solutions of strong acids.
Activity is Important for Concentrated Samples
Activity is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater.
In a 12 M solution of hydrochloric acid, for example, the mean ionic activity coefficient is 17.25. This means that under these conditions with [H 3 O + ] = 12 M, the activity {H 3 O + } = 12 × 17.25 = 207, corresponding to a pH of about –2.3, instead of –1.1 as might be predicted if concentrations were being used.
These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl (g) over a concentrated hydrochloric acid solution even though this acid is supposedly 100% dissociated. It turns out that the activity {HCl} (which represents the "escaping tendency" of HCl (g) from the solution) is almost 49,000 for a 10 M solution! The source of this gas is best described as the result of ion pairing.
Similarly, in a solution prepared by adding 0.5 mole of the very strong acid HClO 4 to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 M. This does not mean that the acid is only 80% dissociated; there is no evidence of HClO 4 molecules in the solution. What has happened is that about 20% of the \(\ce{H3O^{+}}\) and ClO 4 – ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces (Figure \(\PageIndex{4}\)).
Ion-pairing reduces effective dissociation at high concentrations. If you have worked with concentrated hydrochloric acid in the lab, you will have noticed that the choking odor of hydrogen chloride gas is very apparent. How can this happen if this strong acid is really "100 percent dissociated" as all strong acids are said to be?
At very high concentrations, too few H 2 O molecules are available to completely fill the extended hydration shells that normally help keep the ions apart, reducing the fraction of "free" \(\ce{H3O^{+}}\) ions capable of acting independently. Under these conditions, the term "dissociation" begins to lose its meaning. Although the concentration of HCl (aq) is never very high, its own activity coefficient can be as great as 2000 (Table \(\PageIndex{2}\)), which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable.
Estimate the pH of a 10.0 M solution of hydrochloric acid in which the mean ionic activity coefficient - is 10.4.
Solution
pH = – log {H + } ≈ – log (10.4 × 10.0) = – log 104 = – 2.0
Compare this result with what you would get by using – log[H + ]
Systematic treatment of strong acids
In this section, we will derive expressions that relate the pH of solution of a strong acid to its concentration in a solution of pure water. We will use hydrochloric acid as an example. When HCl gas is dissolved in water, the resulting solution contains the ions H 3 O + , OH – , and Cl−, However, except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions.
To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them. These relations are obtained by observing that certain conditions must always be true in any solution of HCl. These are:
1. The autoprotolysis equilibrium of water must always be satisfied:
\[[H_3O^+][OH^–] = K_w \label{4-1}\]
2. For any acid-base system, one can write a mass balance equation that relates the concentrations of the various dissociation products of the substance to its “nominal concentration”, which we designate here as C a . For a solution of HCl, this equation would be
\[[HCl] + [Cl^–] = C_a \label{4-2}\]
However, since HCl is a strong acid and therefore no "HCl" exists in the solution, we can neglect the first term, so the mass balance equation becomes simply
\[[Cl^–] = C_a \label{4-3}\]
3. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle .
\[[H_3O^+] = [OH^–] + [Cl^–] \label{4-4}\]
The next step is to combine these three equations into a single expression that relates the hydronium ion concentration to \(C_a\). This is best done by starting with an equation that relates several quantities, such as Equation \(\ref{4-4}\), and substituting the terms that we want to eliminate. Thus we can get rid of the [Cl−] term by substituting Equation \(\ref{4-3}\) into Equation \(\ref{4-4}\) :
\[[H_3O^+] = [OH^–] + C_a \label{4-5}\]
The [OH – ] term can be eliminated by use of Equation \(\ref{4-3}\):
\[[H_3O^+] = C_a + \dfrac{K_w}{[H_3O^+]} \label{4-6}\]
This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. Notice that Equation \(\ref{4-6}\) is a quadratic equation. Recalling that K w = 10 –14 , it is apparent that the final term of the above equation will ordinarily be very small in comparison to the other terms, so it can be ordinarily be dropped, yielding the simple relation
\[[H_3O^+] \approx C_a \label{4-7}\]
Only in extremely dilute solutions, around 10 –6 M or below (where the plot curves), does this approximation become untenable. However, even then, the effect is tiny. After all, the hydronium ion concentration in a solution of a strong acid can never fall below 10 –7 M ; no amount of dilution can make the solution alkaline! So for almost all practical purposes, Equaton \(\ref{4-7}\) is all you will ever need for a solution of a strong acid.
|
libretexts
|
2025-03-17T19:53:15.875700
| 2016-02-13T03:27:03 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.02%3A_Strong_Monoprotic_Acids_and_Bases",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.2: Strong Monoprotic Acids and Bases",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.03%3A_Finding_the_pH_of_weak_Acids_Bases_and_Salts
|
13.3: Finding the pH of weak Acids, Bases, and Salts
The numerical examples and the images in this section have some problems ( source text with comments).
Make sure you thoroughly understand the following essential concepts that have been presented above.
- Explain the difference between the terms C a and [HA] as they relate to an aqueous solution of the acid HA.
- Define degree of dissociation and sketch a plot showing how the values of \(alpha\) for a conjugate pair HA and A – relate to each other and to the pK a .
- Derive the relationship between acid concentration and dissociation constant to the hydrogen ion concentration in a solution of a weak acid.
- Re-write the above relation in polynomial form .
- Derive the approximation that is often used to estimate the pH of a solution of a weak acid in water.
- Calculate the pH of a solution of a weak monoprotic weak acid or base, employing the "five-percent rule" to determine if the approximation 2-4 is justified.
- Predict whether an aqueous solution of a salt will be acidic or alkaline, and explain why by writing an appropriate equation.
Most acids are weak; there are hundreds of thousands of them, whereas there are fewer than a dozen strong acids. We can treat weak acid solutions in much the same general way as we did for strong acids. The only difference is that we must now take into account the incomplete "dissociation"of the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which salts of the acid are present. The latter mixtures are known as buffer solutions and are extremely important in chemistry, physiology, industry and in the environment. To keep our notation as simple as possible, we will refer to “hydrogen ions” and [H + ] for brevity, and, wherever it is practical to do so, will assume that the acid HA "ionizes" or "dissociates" into H + and its conjugate base A−.
1 Aqueous solutions of weak acids or bases
A weak acid (represented here as HA) is one in which the reaction
\[HA \rightleftharpoons A^– + H^+ \label{1-1} \]
is incomplete. This means that if we add 1 mole of the pure acid HA to water and make the total volume 1 L, the equilibrium concentration of the conjugate base A – will be smaller (often much smaller) than 1 M /L, while that of undissociated HA will be only slightly less than 1 M /L. Equation \(\ref{1-1}\) tells us that dissociation of a weak acid HA in pure water yields identical concentrations of its conjugate species. Let us represent these concentrations by x . Then, in our "1 M " solution, the concentration of each species is as shown here:
(1-2)
When dealing with problems involving acids or bases, bear in mind that when we speak of "the concentration", we usually mean the nominal or analytical concentration which is commonly denoted by C a . For example, for a solution made by combining 0.10 mol of pure formic acid HCOOH with sufficient water to make up a volume of 1.0 L, C a = 0.10 M . However, we know that the concentration of the actual species [HCOOH] will be smaller the 0.10 M because some it ends up as the formate ion HCOO – . It will, of course, always be the case that the sum
\[[HCOOH] + [HCOO^–] = C_a \nonumber \]
For the general case of an acid HA, we can write a mass balance equation
\[C_a = [HA] + [A^–] \label{1-3} \]
which reminds us the "A" part of the acid must always be somewhere!
Similarly, for a base B we can write
\[C_b = [B] + [HB^+] \label{1-4} \]
Equilibrium concentrations of the acid and its conjugate base
According to the above equations, the equilibrium concentrations of A – and H + will be identical (as long as the acid is not so weak or dilute that we can neglect the small quantity of H + contributed by the autoprotolysis of H 2 O). Let us represent these concentrations by x . Then, in a solution containing 1 M /L of a weak acid, the concentration of each species is as shown here:
(1-5)
Substituting these values into the equilibrium expression for this reaction, we obtain
\[\dfrac{[A^-][H^+]}{[HA]} = \dfrac{x^2}{1-x} \label{1-6} \]
In order to predict the pH of this solution, we must solve for x . The presence of terms in both x and x 2 here tells us that this is a quadratic equation. In most practical cases in which K a is 10 –4 or smaller, we can assume that x is much smaller than 1 M , allowing us to make the simplifying approximation
\[(1 – x) \approx 1 \label{1-7} \]
so that
\[x^2 \approx K_a \nonumber \]
and thus
\[x = \sqrt{K_a} \label{1-8} \]
This approximation will not generally be valid when the acid is very weak or very dilute.
Solutions of arbitrary concentration
The above development was for a solution made by taking 1 mole of acid and adding sufficient water to make its volume 1.0 L. In such a solution, the nominal concentration of the acid, denoted by C a , is 1 M .
We can easily generalize this to solutions in which C a has any value:
(1-9)
The above relation is known as a "mass balance on A". It expresses the simple fact that the "A" part of the acid must always be somewhere — either attached to the hydrogen, or in the form of the hydrated anion A – .
The corresponding equilibrium expression is
\[\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{x^2}{\left(C_a-x\right)}=K_a \label{1-10} \]
and the approximations (when justified) Equation \ref{1-3a} and \ref{1-3b} become
\[(C_a – x) ≈ C_a \label{1-11} \]
\[x ≈ \sqrt{C_a K_a} \label{1-12} \]
Estimate the pH of a 0.20 M solution of acetic acid, K a = 1.8 × 10 –5 .
Solution
For brevity, we will represent acetic acid CH 3 COOH as HAc, and the acetate ion by Ac – . As before, we set x = [H + ] = [Ac – ], neglecting the tiny quantity of H + that comes from the dissociation of water.
\[\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Ac}^{-}\right]}{[\mathrm{HAc}]}=\frac{x^2}{(1-x)} \nonumber \]
Substitution into the equilibrium expression yields
\[\frac{x^2}{(0.20-x)}=1.8 \times 10^{-5} \nonumber \]
The rather small value of \(K_a\) suggests that we can drop the \(x\) term in the denominator, so that
\[\dfrac{x^2}{ 0.20} \approx 1.8 \times 10^{-5} \nonumber \]
or
\[x ≈ \sqrt{(0.20)(1.8 \times 10^{–5})} = 1.9 \times 10^{-3}~ M \nonumber \]
The pH of the solution is
\[-\log (1.9 \times 10^{–3}) = 2.7 \nonumber \]
Degree of dissociation
Even though we know that the process HA → H + + A – does not correctly describe the transfer of a proton to H 2 O, chemists still find it convenient to use the term "ionization" or "dissociation". The "degree of dissociation" (denoted by \(\alpha\) of a weak acid is just the fraction
\[\alpha = \dfrac{[\ce{A^{-}}]}{C_a} \label{1-13} \]
which is often expressed as a per cent (\(\alpha\) × 100).
Degree of dissociation depends on the concentration
It's important to understand that whereas K a for a given acid is essentially a constant, \(\alpha\) will depend on the concentration of the acid. Note that these equations are also valid for weak bases if K b and C b are used in place of K a and C a .
This can be shown by substituting Eq 5 into the expression for K a :
(1-14)
Solving this for \(\alpha\) results in a quadratic equation, but if the acid is sufficiently weak that we can say (1 – ) ≈ 1, the above relation becomes
(1-15)
the amount of HA that dissociates varies inversely with the square root of the concentration; as C a approaches zero, \(\alpha\) approaches unity and [HA] approaches C a .
This principle is an instance of the Ostwald dilution law which relates the dissociation constant of a weak electrolyte (in this case, a weak acid), its degree of dissociation, and the concentration.
A common, but incorrect explanation of this law in terms of the Le Chatelier principle states that dilution increases the concentration of water in the equation HA + H 2 O → H 3 O + + A – , thereby causing this equilibrium to shift to the right. The error here is that [H 2 O] in most aqueous solutions is so large (55.5 M) that it can be considered constant; this is the reason the [H 2 O] term does not appear in the expression for K a . Another common explanation is that dilution reduces [H 3 O + ] and [A – ], thus shifting the dissociation process to the right. However, dilution similarly reduces [HA], which would shift the process to the left. In fact, these two processes compete, but the former has greater effect because two species are involved.
It is probably more satisfactory to avoid Le Chatelier-type arguments altogether, and regard the dilution law as an entropy effect, a consequence of the greater dispersal of thermal energy throughout the system. This energy is carried by the molecular units within the solution; dissociation of each HA unit produces two new particles which then go their own ways, thus spreading (or "diluting") the thermal energy more extensively and massively increasing the number of energetically-equivalent microscopic states, of which entropy is a measure. (More on this here )
Degree of dissociation depends on the pH
Plots of this kind are discussed in more detail in the next lesson in this set under the heading ionization fractions .
The Le Chatelier principle predicts that the extent of the reaction
\[\ce{HA → H^{+} + A^{–} } \nonumber \]
will be affected by the hydrogen ion concentration, and thus by the pH. This is illustrated here for the ammonium ion. Notice that when the pH is the same as the pK a , the concentrations of the acid- and base forms of the conjugate pair are identical.
A 0.75 M solution of an acid HA has a pH of 1.6. What is its percent dissociation?
Solution
The dissociation stoichiometry HA → H + + AB – tells us the concentrations [H + ] and [A – ] will be identical. Thus [H + ] = 10 –1.6 = 0.025 M = [A – ]. The dissociation fraction
\[α = \dfrac{[\ce{A^{–}}]}{[\ce{HA}]} = \dfrac{0.025}{0.75} = 0.033 \nonumber \]
and thus the acid is 3.3% dissociated at 0.75 M concentration.
Sometimes the percent dissociation is given, and K a must be evaluated.
A weak acid HA is 2 percent dissociated in a 1.00 M solution. Find the value of K a .
Solution
The equilibrium concentration of HA will be 2% smaller than its nominal concentration, so [HA] = 0.98 M, [A – ] = [H + ] = 0.02 M. Substituting these values into the equilibrium expression gives
"Concentration of the acid" and [HA] are not the same
When dealing with problems involving acids or bases, bear in mind that when we speak of "the concentration", we usually mean the nominal or analytical concentration which is commonly denoted by C a .
So for a solution made by combining 0.10 mol of pure formic acid HCOOH with sufficient water to make up a volume of 1.0 L, C a = 0.10 M . However, we know that the concentration of the actual species [HCOOH] will be smaller the 0.10 M because some it ends up as the formate ion HCOO – . However, it will always be the case that the sum
[HCOOH] + [HCOO – ] = C a .
For the general case of an acid HA, we can write a mass balance equation
C a = [HA] + [A – ](1-16)
which reminds us the "A" part of the acid must always be somewhere!
Similarly, for a base B we can write
C b = [B] + [HB + ](1-17)
Degree of dissociation varies inversely with the concentration
If we represent the dissociation of a C a M solution of a weak acid by
(1-18)
then its dissociation constant is given by
(1-19)
Because the C a term is in the denominator here, we see that the amount of HA that dissociates varies inversely with the concentration; as C a approaches zero, [HA] approaches C a .
If we represent the fraction of the acid that is dissociated as
(1-20)
then Eq 8 becomes
(1-21)
If the acid is sufficiently weak that
x
does not exceed 5% of
C
a
,
the
-term in the denominator can be dropped, yielding
K a ≈ C a 2 (1-22)
Note that the above equations are also valid for weak bases if Kb and C b are used in place of K a and C a .
Compare the percent dissociation of 0.10 M and .0010 M solutions of boric acid (\(K_a = 3.8 \times 10^{-10}\)).
Solution
Boric acid is sufficiently weak that we can use the approximation of Eq 1-22 to calculate a: = (5.8E–10 / .1) ½ = 7.5E-5; multiply by 100 to get .0075 % diss. For the more dilute acid, a similar calculation yields 7.6E–4, or 0.76%.
2 Carrying out acid-base calculations
In Problem Example 1, we calculated the pH of a monoprotic acid solution, making use of an approximation in order to avoid the need to solve a quadratic equation. This raises the question: how "exact" must calculations of pH be? It turns out that the relation between pH and the nominal concentration of an acid, base, or salt (and especially arbitrary mixtures of these) can become quite complicated, requiring the solution of sets of simultaneous equations.
However, for almost all practical applications, one can make some approximations that simplify the math without detracting significantly from the accuracy of the results.
Equilibrium constants are rarely exactly known
As we pointed out in the preceding lesson, the "effective" value of an equilibrium constant (the activity ) will generally be different from the value given in tables in all but the most dilute ionic solutions. Even if the acid or base itself is dilute, the presence of other "spectator" ions such as Na + at concentrations much in excess of 0.001 M can introduce error.
The usual advice is to consider K a values to be accurate to ±5 percent at best, and even more uncertain when total ionic concentrations exceed 0.1 M. As a consequence of this uncertainty, there is generally little practical reason to express the results of a pH calculation to more than two significant digits.
Finding the pH of a solution of a weak monoprotic acid
This is by far the most common type of problem you will encounter in a first-year Chemistry class. You are given the concentration of the acid, expressed as C a moles/L, and are asked to find the pH of the solution. The very important first step is to make sure you understand the problem by writing down the equation expressing the concentrations of each species in terms of a single unknown, which we represent here by x :
(2-1)
Substituting these values into the expression for \(K_a\), we obtain
(2-2)
Don't bother to memorize these equations! If you understand the concept of mass balance on "A" expressed in (2-1), and can write the expression for K a , you just substitute the x 's into the latter, and you're off! If you feel the need to memorize stuff you don't need, it is likely that you don't really understand the material — and that should be a real worry!
In order to predict the pH of this solution, we must first find [H + ], that is, x . The presence of terms in both x and x 2 here tells us that this is a quadratic equation. This can be rearranged into x 2 = K a (1 – x ) which, when written in standard polynomial form, becomes the quadratic
\[[\ce{H^{+}}]^2 – C_a [H^{+}] – K_w = 0 \label{2-3} \]
However, don't panic! As we will explain farther on, in most practical cases we can make some simplifying approximations which eliminate the need to solve a quadratic. And when, as occasionally happens, a quadratic is unavoidable, we will show you some relatively painless ways of dealing with it.
How to deal with Quadratic Equations
What you do will depend on what tools you have available. If you are only armed with a simple calculator, then there is always the venerable quadratic formula that you may have learned about in high school, but if at all possible, you should avoid it: its direct use in the present context is somewhat laborious and susceptible to error.
Use of the standard quadratic formula on a computer or programmable calculator can lead to weird results!
The reason for this is that if b 2 >> |4 ac |, one of the roots will require the subtraction of two terms whose values are very close; this can lead to considerable error when carried out by software that has finite precision.
One can get around this by computing the quantity
\[Q = –\dfrac{b + \pm (b) \sqrt{ b^2 – 4ac}}{2} \nonumber \]
from which the roots are x 1 = Q / a and x 2 = c / Q . (See any textbook on numerical computing for more on this and other metnods.)
However, who want's to bother with this stuff in order to solve typical chemistry problems? Better to avoid quadratics altogether if at all possible!
Remember: there are always two values of x (two roots ) that satisfy a quadratic equation. For all acid-base equilibrium calculations that are properly set up, these roots will be real , and only one will be positive; this is the one you take as the answer.
Approximations, judiciously applied, simplify the math
We have already encountered two of these approximations in the examples of the preceding section:
- In all but the most dilute solutions, we can assume that the dissociation of the acid HA is the sole source of H + , with the contribution due to water autoprotolysis being negligible.
- We were able to simplify the equilibrium expressions by assuming that the x -term, representing the quantity of acid dissociated, is so small compared to the nominal concentration of the acid C a that it can be neglected. Thus in Problem Example 1, the term in the denominator that has the form (0.1 - x ) , representing the equilibrium concentration of the undissociated acid, is replaced by 0.1.
Most people working in the field of practical chemistry will never encounter situations in which the first of these approximations is invalid. This is not the case, however, for the second one.
Should I drop the x , or forge ahead with the quadratic form?
If the acid is fairly concentrated (usually with C a > 10 –3 M) and sufficiently weak that most of it remains in its protonated form HA, then the concentration of H + it produces may be sufficiently small that the expression for K a reduces to
K a ≈ [H + ] 2 / C a
so that
[H + ] ≈ ( K a C a ) ½ (2-4)
This can be a great convenience because it avoids the need to solve a quadratic equation. However, it also exposes you to the danger that this approximation may not be justified.
The usual advice is that if this first approximation of x exceeds 5 percent of the value it is being subtracted from (0.10 in the present case), then the approximation is not justified. We will call this the "five percent rule".
Problem Example 5 - pH and degree of dissociation
- Estimate the pH of a 0.20 M solution of acetic acid, K a = 1.8 × 10 –5 .
- What percentage of the acid is dissociated?
Solution
For brevity, we will represent acetic acid CH 3 COOH as HAc, and the acetate ion by Ac – . As before, we set x = [H + ] = [Ac – ], neglecting the tiny quantity of H + that comes from the dissociation of water.
Substitution into the equilibrium expression yields
\[K_a = x^2/(0.20 - x)
Can we simplify this by applying the approximation 0.20 – x ≈ 0.20 ? Looking at the number on the right side of this equation, we note that it is quite small. This means the left side must be equally small, which requires that the denominator be fairly large, so we can probably get away with dropping x. Doing so yields
( x 2 / 0.20) = 1.8E-5 or x = (0.20 × 1.8E–5) ½ = 1.9E-3 M
The "5 per cent rule" requires that the above result be no greater than 5% of 0.20, or 0.010. Because 0.0019 meets this condition, we can set
x
= [H
+
] ≈ 1.9 × 10
–3
M, and the pH will be –log (1.9 × 10
–3
) =
2.7
b) Percent dissociation: 100% × (1.9 × 10 –3 M) / (0.20 M) = 0.95%
This plot shows the combinations of K a and C a that generally yield satisfactory results with the approximation of Eq 4.
Weak bases are treated in an exactly analogous way:
Methylamine CH 3 NH 2 is a gas whose odor is noticed around decaying fish. A solution of CH 3 NH 2 in water acts as a weak base. A 0.10 M solution of this amine in water is found to be 6.4% ionized. Use this information to find \Kb and pK b for methylamine.
Solution: When methylamine "ionizes", it takes up a proton from water, forming the methylaminium ion:
CH 3 NH 2 + H 2 O → CH 3 NH 3 + + OH –
Let x = [CH 3 NH 3 + ] = [OH – ] = .064 × 0.10 = 0.0064
[CH 3 NH 2 ] = (0.10 – .064) = 0.094
Substitute these values into equilibrium expression for \Kb:
\[K_b = \frac{0.0064^2}{0.094} \nonumber \]
To make sure we can stop here, we note that (3.6E4 / .01) = .036; this is smaller than .05, so we pass the 5% rule and can use the approximation and drop the
x
-term in the denominator.
pK b = – log \Kb = – log (4.4 × 10 –10 ) = 3.36
However, one does not always get off so easily!
With a K a of 0.010, HClO 2 is one of the "stronger" weak acids, thanks to the two oxygen atoms whose electronegativity withdraws some negative charge from the chlorine atom, making it easier for the hydrogen to depart as a proton.
Find the pH of a 0.10 M solution of chloric acid in pure water.
\[0.010 = \frac{x^2}{0.10 - x} \nonumber \] (i)
The approximation 0.10 – x ≈ 0.10 gives us
x ≈ ( K a C a ) ½ = (0.010 ×0.10) ½ = (.001) ½ = .032 (ii)
The difficulty, in this case, arises from the numerical value of
K
a
differing from the nominal concentration 0.10 M by only a factor of 10. As a result,
x
/
C
a
= .032 / 0.10 = 0.32 which clearly exceeds the 5% limit; we have no choice but to face the full monte of the quadratic solution. (see Problem Example 8 below).
Successive approximations will get you there with minimal math
In the method of successive approximations , you start with the value of [H + ] (that is, x ) you calculated according to (2-4) , which becomes the first approximation. You then substitute this into (2-2) , which you solve to get a second approximation. This cycle is repeated until differences between successive answers become small enough to ignore.
Estimate the pH of a 0.10 M aqueous solution of HClO 2 , K a = 0.010, using the method of successive approximations.
Solution: The equilibrium condition is
\[K_{\mathrm{a}} = \dfrac{x \cdot x}{c_{\mathrm{a}} - x} \nonumber \]
We solve this for x (neglecting the x i n the denominator), resulting in the first approximation x 1 , and then successively plug each result into the previous equation, yielding approximations x 2 and x 3 :
\[x_1 = \sqrt{K_{\mathrm{a}} \cdot c_\mathrm{a}} =0.032 \nonumber \]
\[x_2 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_1)} =0.026 \nonumber \]
\[x_3 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_2)} =0.027 \nonumber \]
The last two approximations x 2 and x 3 are within 5% of each other.
Note that if we had used x 1 as the answer, the error would have been 18%.
(An exact numeric solution yields the roots 0.027016 and –0.037016)
Use a graphic calculator or computer to find the positive root
The real roots of a polynomial equation can be found simply by plotting its value as a function of one of the variables it contains.
In this example, the pH of a 10
–6
M solution of hypochlorous acid (HOCl,
K
a
= 2.9E–8) was found by plotting the value of
y =
ax
2
+
b
x +
c
, whose roots are the two values of
x
that correspond to
y
= 0.
This method generally requires a bit of informed trial-and-error to make the locations of the roots visible within the scale of the axes.
Be lazy, and use an on-line quadratic equation solver
If you google "quadratic equation solver", you will find numerous on-line sites that offer quick-and-easy "fill-in-the-blanks" solutions.
Unfortunately, few of these will be useful for acid-base problems involving numbers that must be expressed in "E-notation" ( e.g. , 2.7E-11.) Of those that do, the one at the MathIsFun site is highly recommended; others can be found here and at the Quad2Deg site.
If you can access a quad equation solver on your personal electronic device or through the Internet, this is quick and painless. All you need to do is write the equation in polynomial form
ax
2
+
b
x +
c
= 0, insert values for
a, b
, and
c
, and away you go!
This is so easy, that many people prefer to avoid the "5% test" altogether, and go straight to an exact solution. However, make sure can do the 5%-thing for exams where Internet-accessible devices are not permitted!
Estimate the pH of a 0.10 M aqueous solution of HClO 2 , K a = 0.010.
Solution
The reaction equation HClO 2 → H + + ClO 2 – defines the equilibrium expression
\[0.010 = \dfrac{x \cdot x}{0.10 - x} \nonumber \]
Multiplying by the denominator yields
x 2 = 0.010 × (0.10 – x ) = .0010 – .01 x which we arrange into standard polynomial form:
x 2 + 0.01 x – 0.0010 = 0
Entering the coefficients {1 .01 –.001} into an online quad solver yields the roots
.027 and –.037. Taking the positive one, we have [H
+
] = .027 M;
the solution pH is – log .027 =
1.6
.
Note: a common error is to forget to enter the minus sign for the last term; try doing this and watch the program blow up!
Avoid math altogether and make a log-C vs pH plot
This is not only simple to do (all you need is a scrap of paper and a straightedge), but it will give you far more insight into what's going on, especially in polyprotic systems. All explained in Section 3 of the next lesson .
Solutions of salts
Most salts do not form pH-neutral solutions
Salts such as sodium chloride that can be made by combining a strong acid (HCl) with a strong base (NaOH, KOH) have a neutral pH, but these are exceptions to the general rule that solutions of most salts are mildly acidic or alkaline.
Salts of a strong base and a weak acid yield alkaline solutions.
"Hydro-lysis" literally means "water splitting", as exemplified by the reaction A – + H 2 O → HA + OH – . The term describes what was believed to happen prior to the development of the Brønsted-Lowry proton transfer model.
This important property has historically been known as hydrolysis — a term still used by chemists.
Some examples:
- Potassium Cyanide
- KCN can be thought of the salt made by combining the strong base KOH with the weak acid HCN: K + + OH – → KOH. When solid KCN dissolves in water, this process is reversed, yielding a solution of the two ions. K + , being a "strong ion", does not react with water. However, the "weak ion" CN – , being the conjugate base of a weak acid, will have a tendency to abstract protons from water: CN – + H 2 O → HCN + OH – , causing the solution to remain slightly alkaline.
- Sodium bicarbonate
- NaHCO 3 (more properly known as sodium hydrogen carbonate) dissolves in water to yield a solution of the hydrogen carbonate ion HCO 3 – . This ion is an ampholyte — that is, it is both the conjugate base of the weak carbonic acid H 2 CO 3 , as well as the conjugate acid of the carbonate ion CO 3 2 – :
The HCO 3 – ion is therefore amphiprotic : it can both accept and donate protons, so both processes take place:
However, if we compare the K a and K b of HCO 3 – , it is apparent that its basic nature wins out, so a solution of NaHCO 3 will be slightly alkaline. (The value of pK b is found by recalling that K a + K b = 14.)
- Sodium acetate
- A solution of CH 3 COONa serves as a model for the strong-ion salt of any organic acid, since all such acids are weak: CH 3 COO – + H 2 O → CH 3 COOH + OH –
Salts of most cations (positive ions) give acidic solutions
The protons can either come from the cation itself (as with the ammonium ion NH 4 + ), or from waters of hydration that are attached to a metallic ion.
This latter effect happens with virtually all salts of metals beyond Group I; it is especially noticeable for salts of transition ions such as hexaaquoiron(III) ("ferric ion"):
Fe(H 2 O) 6 3+ → Fe(H 2 O) 5 OH 2 + + H + (2-5)
This comes about because the positive field of the metal enhances the ability of H 2 O molecules in the hydration shell to lose protons. In the case of the hexahyrated ion shown above, a whose succession of similar steps can occur, but for most practical purposes only the first step is significant.
Find the pH of a 0.15 M solution of aluminum chloride.
Solution
The aluminum ion exists in water as hexaaquoaluminum Al(H 2 O) 6 3+ , whose pK a = 4.9, K a = 10 –4.9 = 1.3E–5. Setting x = [H + ] = [Al(H 2 O) 5 OH 2+ ], the equilibrium expression is
Using the above approximation, we get
x
≈ (1.96E–6)
½
= 1.4E–3, corresponding to
pH = 2.8
.
Finally, we compute x/ C a = 1.4E–3 ÷ 0.15 = .012 confirming that we are within the "5% rule".
The only commonly-encountered salts in which the proton is donated by the cation itself are those of the ammonium ion :
\[\ce{NH_4^{+}→ NH)3(aq) + H^{+}\lable{2-6} \nonumber \]
Calculate the pH of a 0.15 M solution of NH4Cl. The ammonium ion K a is 5.5E–10.
Solution
According to Eq 6 above, we can set [NH3] = [H + ] = x , obtaining the same equilibrium expression as in the preceding problem.
Because K a is quite small, we can safely use the approximation 0.15 - 1 ≈ .015, which yields pH = –log 0.90E–5 = 5.0
Most salts of weak acids form alkaline solutions
As indicated in the example, such equilibria strongly favor the left side; the stronger the acid HA, the less alkaline the salt solution will be.
Because an ion derived from a weak acid such as HF is the conjugate base of that acid, it should not surprise you that a salt such as NaF forms an alkaline solution, even if the equilibrium greatly favors the left side::
F – + H 2 O HF + OH – (2-7)
Solution of sodium fluoride
Find the pH of a 0.15 M solution of NaF. (HF K a = 6.7E–4)
Solution: The reaction is F- + H 2 O = HF + OH – ; because HF is a weak acid, the equilibrium strongly favors the right side. The usual approximation yields
However, on calculating x/ C a = .01 ÷ 0.15 = .07, we find that this does not meet the "5% rule" for the validity of the approximation. We therefore expand the equilibrium expression
into standard polynomial form x 2 + 6.7E–4 x – 1.0E–4 = 0 and enter the coefficients {1 6.7E–4 –.0001} into a quadratic solver. This yields the positive root x = 0.0099 which turns out to be sufficiently close to the approximation that we could have retained it after all.. perhaps 5% is a bit too restrictive for 2-significant digit calculations!
What about a salt of a weak acid and a weak base?
A salt of a weak acid gives an alkaline solution, while that of a weak base yields an acidic solution. What happens if we dissolve a salt of a weak acid and a weak base in water? Ah, this can get a bit tricky! Nevertheless, this situation arises very frequently in applications as diverse as physiological chemistry and geochemistry.
As an example of how one might approach such a problem, consider a solution of ammonium formate, which contains the ions NH 4 + and HCOO-. Formic acid, the simplest organic acid, has a pK a of 3.7; for NH 4 + , pK a = 9.3.
Three equilibria involving these ions are possible here; in addition to the reactions of the ammonium and formate ions with water, we must also take into account their tendency to react with each other to form the parent neutral species:
| NH 4 + → NH 3 + H + | K 1 = 10 –9.3 |
| HCOO – + H 2 O → HCOOH + OH – | K 2 = (1O –14 /10 –3.7 ) = 10 –10.3 |
| NH 4 + + HCOO – → NH 3 + HCOOH | K 3 |
Inspection reveals that the last equation above is the sum of the first two,plus the reverse of the dissociation of water
| H + + OH – → H 2 O | K 4 = 1/ K w |
The value of K 3 is therefore
(2-8)
A rigorous treatment of this system would require that we solve these equations simultaneously with the charge balance and the two mass balance equations. However, because K 3 is several orders of magnitude greater than K 1 or K 2 , we can greatly simplify things by neglecting the other equilibria and considering only the reaction between the ammonium and formate ions.
Notice that the products of this reaction will tend to suppress the extent of the first two equilibria, reducing their importance even more than the relative values of the equilibrium constants would indicate.
ammonium formate
Estimate the pH of a 0.0100 M solution of ammonium formate in water.
Solution: From the stoichiometry of HCOONH 4 ,
[NH 4 + ] = [HCOO – ] and [NH 3 ] = [HCOOH] (i)
then, from Eq 8 above,
(ii)
in which K b is the base constant of ammonia, K w /10 –9.3 .
From the formic acid dissociation equilibrium we have
(iii)
We now rewrite the expression for K 3
(iv)
which yields
(v)
and thus the pH is 6.5
What is interesting about this last example is that the pH of the solution is apparently independent of the concentration of the salt. If K a = K b , then this is always true and the solution will be neutral (neglecting activity effects in solutions of high ionic strength). Otherwise, it is only an approximation that remains valid as long as the salt concentration is substantially larger than the magnitude of either equilibrium constant. Clearly, the pH of any solution must approach that of pure water as the solution becomes more dilute.
Salts of analyte ions
Polyprotic acids form multiple anions; those that can themselves donate protons, and are thus amphiprotic, are called analytes . The most widely known of these is the bicarbonate (hydrogen carbonate) ion, HCO 3 – , which we commonly know in the form of its sodium salt NaHCO 3 as baking soda .
The other analyte series that is widely encountered, especially in biochemistry, is those derived from phosphoric acid:
\[\mathrm{H}_3 \mathrm{PO}_4 \xrightarrow[10^{-2.12}]{K_1} \mathrm{H}_2 \mathrm{PO}_4^{-} \xrightarrow{10^{-7.2}} \mathrm{KPO}_4{ }^{2-} \frac{K_3}{10^{-12.3}} \mathrm{PO}_4^{3-} \nonumber \]
The solutions of analyte ions we most often need to deal with are the of "strong ions", usually Na + , but sometimes those of Group 2 cations such as Ca 2 + .
The exact treatment of these systems is generally rather complicated, but for the special cases in which the successive K a 's of the parent acid are separated by several orders of magnitude (as in the two systems illustrated above), a series of approximations reduces the problem to the simple expression
\[\left[\mathrm{H}^{+}\right] \approx \sqrt{K_1 K_2} \label{2-9} \]
which, you will notice, as with the salt of a weak acid and a weak base discussed in the preceding subsection predicts that the pH is independent of the salt's concentration. This, of course, is a sure indication that this treatment is incomplete. Fortunately, however, it works reasonably well for most practical purposes, which commonly involve buffer solutions.
When dealing with acid-base systems having very small K a 's, and/or solutions that are extremely dilute, it may be necessary to consider all the species present in the solution, including minor ones such as OH – .
This is almost never required in first-year courses. However, for students in more advanced courses, this "comprehensive approach" (as it is often called) illustrates the important general methodology of dealing with more complex equilibrium problems. It also shows explicitly how making various approximations gradually simplifies the treatment of more complex systems.
Solutions of polyprotic acids
A diprotic acid H 2 A can donate its protons in two steps:
\[\ce{H2A → HA^{–} →HA^{–}} \nonumber \]
and similarly, for a tripotic acid H3A:
\[\ce{H3A → H2A^{–} → HA^{2–} → A^{3–}} \nonumber \]
In general, we can expect K a 2 for the "second ionization" to be smaller than K a 1 for the first step because it is more difficult to remove a proton from a negatively charged species. The magnitude of this difference depends very much on whether the two removable protons are linked to the same atom, or to separate atoms spaced farther apart.
Some polyprotic acids you should know
These acids are listed in the order of decreasing K a1 . The numbers above the arrows show the successive K a 's of each acid.
- Notice how successive K a 's of each acid become smaller, and how their ratios relate to the structures of each acid.
- Sulfuric acid is the only strong polyprotic acid you are likely to encounter.
-
Sulfurous acid
has never been detected; what we refer to as H
2
SO
3
is more properly represented as a hydrated form of sulfur dioxide: which dissociates into the hydrosulfite ion:
SO 2 ·H 2 O → HSO 3 – + H + . The ion HSO 3 – exists only in solution; solid bisulfite salts are not known. - Similarly, pure carbonic acid has never been isolated, but it does exist as a minority species in an aqueous solution of CO 2 : [CO 2 (aq) ] = 650[H 2 CO 3 ]. The formula H 2 CO 3 ordinarily represents the combination CO 2 (aq) and "true" H 2 CO 3 . The latter is actually about a thousand times stronger than is indicated by the pK a of 6.3, which is the weighted average of the equilibrium mixture.
Compare the successive pK a values of sulfuric and oxalic acids (see their structures in the box, above right), and explain why they should be so different.
Solution
The two pK a values of sulfuric acid differ by 3.0 – (–1.9) = 4.9, whereas for oxalic acid the difference is 1.3 – (–4.3) = 3.0. That's a difference of almost 100 between the two K a 's. Removal of a second proton from a molecule that already carries some negative charge is always expected to be less favorable energetically. In sulfuric acid, the two protons come from –OH groups connected to the same sulfur atom, so the negative charge that impedes loss of the second proton is more localized near the site of its removal. In oxalic acid, the two protons are removed from –OH groups attached to separate carbon atoms, so the negative charge of the mono-negative ions will exert less restraint on loss of the second proton.
Solutions of polyprotic acids in pure water
With the exception of sulfuric acid (and some other seldom-encountered strong diprotic acids), most polyprotic acids have sufficiently small K a1 values that their aqueous solutions contain significant concentrations of the free acid as well as of the various dissociated anions.
Thus for a typical diprotic acid H 2 A , we must consider the three coupled equilibria
\[\ce{H_2A → H^+ + HA^{–} } \tag{K1} \]
\[\ce{HA^{–} → H^{+} + HA^{2–}} \tag{K2} \]
\[\ce{H_2O → H^{+} + OH^{–} } \tag{Kw} \]
An exact treatment of such a system of four unknowns \(\ce{[H2A]}\), \(\ce{[HA^{–}]}\), \(\ce{[A^{2–}]}\) and \(\ce{[H^{+}]}\) requires the solution of a quartic equation. If we include \(\ce{[OH^{–}]}\), it's even worse!
In most practical cases, we can make some simplifying approximations:
- Unless the solution is extremely dilute or \(K_1\) (and all the subsequent K 's) are extremely small, we can forget that any hydroxide ions are present at all.
- If \(K_1\) is quite small, and the ratios of succeeding K 's are reasonably large, we may be able, without introducing too much error, to neglect the other K 's and treat the acid as monoprotic.
In addition to the three equilibria listed above, a solution of a polyprotic acid in pure water is subject to the following two conditions:
Material balance : although the distribution of species between the acid form \(\ce{H2A}\) and its base forms \(\ce{HAB^{–}}\) and \(\ce{A^{2–}}\) may vary, their sum (defined as the "total acid concentration" \(C_a\) is a constant for a particular solution:
\[C_a = [H_2A] + [HA^{–}] + [A^{2–}] \nonumber \]
Charge balance : The solution may not possess a net electrical charge:
\[[\ce{H3O^{+}}] = [\ce{OH^{–}}] + [\ce{HA^{–}}] + 2 [\ce{A^{2–}}] \nonumber \]
Why do we multiply \(\ce{[A^{2–}]}\) by 2? Two moles of \(\ce{H3O^{+}}\) are needed in order to balance out the charge of 1 mole of \(\ce{A^{2–}}\).
Simplified treatment of polyprotic acid solutions
The calculations shown in this section are all you need for the polyprotic acid problems encountered in most first-year college chemistry courses. More advanced courses may require the more exact methods in Lesson 7.
Because the successive equilibrium constants for most of the weak polyprotic acids we ordinarily deal with diminish by several orders of magnitude, we can usually get away with considering only the first ionization step. To the extent that this is true, there is nothing really new to learn here. However, without getting into a lot of complicated arithmetic, we can often go farther and estimate the additional quantity of H + produced by the second ionization step.
a) Calculate the pH of a 0.050 M solution of CO 2 in water.
For H 2 CO 3 , K 1 = 10 –6.4 = 4.5E–7, K 2 = 10 –10.3 = 1.0E–14.
b) Estimate the concentration of carbonate ion CO 3 2 – in the solution.
Solution
a) Because K 1 and K 2 differ by almost four orders of magnitude, we will initially neglect the second dissociation step. Because this latter step produces only a tiny additional concentration of H + , we can assume that [H + ] = [HCO 3 – ] = x :
(i)
Can we further simplify this expression by dropping the x in the denominator? Let's try:
x = [0.05 × (4.5E–7)] ½ = 1.5E–4.
Applying the "5-percent test", the quotient x/ C a must not exceed 0.05. Working this out yields (1.5E–4)/(.05) = .003, so we can avoid a quadratic.
x = [H + ] ≈ ( K a C a )½ = [(4.5E–7) × .01] ½ = (.001)½ = 0.032 M,
and the
pH = – log .032 = 1.5 .
b) We now wish to estimate [CO 3 2 – ] ≡ x .
Examining the second dissociation step, it is evident that this will consume x mol/L of HCO 3 – , and produce an equivalent amount of H + which adds to the quantity we calculated in (a) .
(ii)
Owing to the very small value of K 2 compared to K 1 , we can assume that the concentrations of HCO 3 – and H + produced in the first dissociation step will not be significantly altered in this step. This allows us to simplify the equilibrium constant expression and solve directly for [CO 3 2 – ]:
(iii)
It is of course no coincidence that this estimate of [CO 3 2 – ] yields a value identical with K 2 ; this is entirely a consequence of the simplifying assumptions we have made. Nevertheless, as long as K 2 << K 1 and the solution is not highly dilute, the result will be sufficiently accurate for most purposes.
Sulfuric acid: A Special Case
Although this is a strong acid, it is also diprotic, and in its second dissociation step it acts as a weak acid. Because sulfuric acid is so widely employed both in the laboratory and industry, it is useful to examine the result of taking its second dissociation into consideration.
Estimate the pH of a 0.010 M solution of H 2 SO 4 with K 1 = 10 3 and K 2 = 0.012
Solution:
Because K 1 > 1, we can assume that a solution of this acid will be completely dissociated into H 3 O + and bisulfite ions HSO 4 – . Thus the only equilibrium we need to consider is the dissociation of a 0.010 M solution of bisulfite ions.
\[\ce{HSO4^{–} + H2O → SO4^{2–} + H3O^{+}} \nonumber \]
\[K_2=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4{ }^{2-}\right]}{\left[\mathrm{HSO}_4{ }^{-}\right]}=\frac{(x)^2}{.010-x}=0.012 \nonumber \]
Setting \(\ce{[H^{+}] = [SO4^{2–}] = x}\), and dropping \(x\) from the denominator, yields
\[x ≈ \sqrt{(0.010)(0.012)} = \sqrt{1.2 \times 10^{-4}} = 0.0011 \nonumber \]
Applying the "five percent rule", we find that x / C a = .0011/.01 = .11 which is far over the allowable error, so we must proceed with the quadratic form. Rewriting the equilibrium expression in polynomial form gives
\[x^2 + 0.022x – 1.2 \times 10^{–4} = 0 \nonumber \]
Inserting the coefficients {1 .022 .000012} into a quad-solver utility yields the roots \(4.5 \times 10^{–3}\) and \(–0.0027\). Taking the positive root, we obtain
\[pH = –\log (0.0045) = 2.3 \nonumber \]
Thus the second "ionization" of \(\ce{H2SO4}\) has only reduced the pH of the solution by 0.1 unit from the value (2.0) it would theoretically have if only the first step were considered.
Amino acids and Zwitterions
Zwitterions: molecular hermaphrodites
Amino acids, the building blocks of proteins, contain amino groups –NH 2 that can accept protons, and carboxyl groups –COOH that can lose protons. Under certain conditions, these events can occur simultaneously, so that the resulting molecule becomes a “double ion” which goes by its German name Zwitterion .
Glycine: the simplest amino acid
The simplest of the twenty natural amino acids that occur in proteins is glycine H 2 N–CH 2 –COOH, which we use as an example here.
Solutions of glycine are distributed between the acidic-, zwitterion-, and basic species:
In the following development, we use the abbreviations H 2 Gly + (glycinium), HGly (zwitterion), and Gly – (glycinate) to denote the dissolved forms.
The two acidity constants are
\[K_1=\frac{\left[\mathrm{H}^{+}\right][\text {HGly }]}{\left[\mathrm{H}_2 \mathrm{Gly}^{+}\right]}=10^{-2.25} \label{3-2A} \]
and
\[K_2=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Gly}^{-}\right]}{[\mathrm{HGly}]}=10^{-9.78} \label{3-2B} \]
If glycine is dissolved in water, charge balance requires that
\[H_2Gly^+ + [H^+] \rightleftharpoons [Gly^–] + [OH^–] \label{3-3} \]
Substituting the equilibrium constant expressions (including that for the autoprotolysis of water) into the above relation yields
\[\frac{\left[\mathrm{H}^{+}\right][\mathrm{HGly}]}{K_1}+\left[\mathrm{H}^{+}\right]=\frac{K_2[\mathrm{HGly}]}{\left[\mathrm{H}^{+}\right]}+\frac{K_w}{\left[\mathrm{H}^{+}\right]} \label{3-4)} \]
Calculate the pH and the concentrations of the various species in a 0.100 M solution of glycine.
Solution
Substitution into Eq 4 above yields
\[\left[\mathrm{H}^{+}\right]=\sqrt{\frac{10^{-9.8} \times 0.10+10^{-14}}{0.1 /\left(10^{-2.3}+1\right)}}=10^{-6.1} \tag{i} \]
The concentrations of the acid and base forms are found from their respective equilibrium constant expressions (Eqs 2):
\[\left[\mathrm{H}_2 \mathrm{Gly}^{+}\right]=\frac{\left[\mathrm{H}^{+}\right] \times 0.10}{10^{-2.3}}=10^{-4.7} \tag{ii} \]
\[\left[\mathrm{Gly}^{-}\right]=\frac{10^{-9.8} \times 0.10}{\left[\mathrm{H}^{+}\right]}=10^{-4.7} \tag{iii} \]
The small concentrations of these singly-charged species in relation to \(C_a = 0.10\) shows that the zwitterion is the only significant glycine species in the solution.
If Equations ii and iii in this Example are recalculated for a range of pH values, one can plot the concentrations of each species against pH for 0.10 M glycine in water:
|
libretexts
|
2025-03-17T19:53:16.111807
| 2016-02-13T03:27:54 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.03%3A_Finding_the_pH_of_weak_Acids_Bases_and_Salts",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.3: Finding the pH of weak Acids, Bases, and Salts",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.04%3A_Conjugate_Pairs_and_Buffers
|
13.4: Conjugate Pairs and Buffers
Make sure you thoroughly understand the following essential concepts:
- State the values of C a and C b after a weak monoprotic acid or base has been partially neutralized with strong base or acid.
- Define the ionization fraction , and calculate the percent ionization, of a C a M solution of a weak monoprotic acid.
- Sketch out a rough plot showing the distribution of conjugate species in a solution of a weak acid having a given value of K a .
- Define a buffer solution , and explain how it works.
- Calculate the approximate pH of a solution of a weak monoprotic acid having a given pK a that has been partially neutralized by addition of a strong base.
- Specify how to make a given volume of a buffer solution having a certain pH, starting with a weak acid and sodium hydroxide.
- Define the buffer index , and explain its significance and the conditions under which its value is maximized.
- Sketch out a –log C vs pH (Sillén) plot for a weak monoprotic acid/base system having a given nominal concentration and pK a , and use it to estimate the pH of the solution.
We often tend to regard the pH as a quantity that is dependent on other variables such as the concentration and strength of an acid, base or salt. But in much of chemistry (and especially in biochemistry), we find it more useful to treat pH as the "master" variable that controls the relative concentrations of the acid- and base-forms of one or more sets of conjugate acid-base systems. In this lesson, we will explore this approach in some detail, showing its application to the very practical topics of buffer solutions, as well as the use of a simple graphical approach that will enable you to estimate the pH of a weak monoprotic or polyprotic acid or base without doing any arithmetic at all!
When the pH takes control
If we add 0.2 mol of sodium hydroxide to a solution containing 1.0 mol of a weak acid HA, then an equivalent number of moles of that acid will be converted into its base form A – . The resulting solution will contain 0.2 mol of A – and 0.8 mol of HA. Note that because we are discussing stoichiometry here, we are interested in quantities (moles) of reactants, not concentrations of reactants.
The important point to understand here is that we will end up with a “partly neutralized” solution in which both the acid and its conjugate base are present in significant amounts. Solutions of this kind are far more common than those of a pure acid or a pure base, and it is very important that you have a thorough understanding of them.
To a solution containing 0.010 mole of acetic acid (HAc), we add 0.002 mole of sodium hydroxide. If the volume of the final solution is 100 ml, find the values of C a , C b , and the total system concentration C t .
Solution
The added hydroxide ion, being a strong base, reacts completely with the acetic acid, leaving 0.010 – 0.002 = 0.008 mole of HAc and 0.002 mole of acetate ion Ac – . The final concentrations are
\[C_a = \dfrac{0.008 \;mol}{0.10\; L} = 0.08 \; M \nonumber\]
\[C_b = \dfrac{0.002 \; mol}{ 0.10 \; L} = 0.02 M \nonumber\]
\[C_t = \dfrac{0.010 \;mol}{0.10\; L} = 0.10\; M. \nonumber\]
Note that this solution would be indistinguishable from one prepared by combining C a = 0.080 mole of acetic acid with C b = 0.020 mole of sodium acetate and adjusting the volume to 100 ml.
Thus starting with a solution of a pure weak acid or weak base in water, we can add sufficient strong base or strong acid, respectively, to adjust the ratio of the conjugate species — that is, the ratios [HA]/[A – ] in the case of an acid, or [B]/[BH + ] for a base, to any value we want.
Ionization Fractions
To express the relative concentrations of the protonated and deprotonated forms of an acid-base system present in a solution, we could use the simple ratio [HA]/[A] (or its inverse), but this suffers from the drawback of yielding an indeterminate result when the concentration in the denominator is zero. For many purposes it is more convenient to use the ionization fractions
\[\alpha_0 = \dfrac{[HA]}{[HA]+A^-]} = \dfrac{[HA]}{C_a} \label{1-7a}\]
\[\alpha_1 = \dfrac{[A^-]}{[HA]+A^-]} = \dfrac{[A^-]}{C_a} \label{1-7b}\]
The fraction \(\alpha_1\) is also known as the degree of dissociation of the acid. By making appropriate substitutions using the relation
\[[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{1-8}\]
we can express the ionization fractions as functions of the pH:
\[\alpha_0 = \dfrac{[H^+]}{K_a+[H^+]} \label{1-9a}\]
\[\alpha_1 = \dfrac{K_a}{K_a+[H^+]} \label{1-9b}\]
Notice that the the values for both of these functions are close to zero or unity except within the pH range pK a ± 1 (Figure \(\PageIndex{1}\)).
Plots of the \(\alpha\) functions vs. pH for several systems are shown below. Notice the crossing points where [HA] = [A] when [H + ] = K a that corresponds to unit value of the quotient in the Henderson-Hasselbalch approximation .
The ionization fractions of a series of acids over a wide pH range can conveniently be summarized as shown below.
If you know the pK a of an acid, you can easily sketch out its ionization fraction plot. The extreme tops and bottoms can only be estimated, but the rest of the plots are essentially straight lines.
As valuable as these plots are for showing how the distribution of conjugate species varies with the pH, they suffer from two drawbacks:
- Plots of\(\alpha\) cover only a single order of magnitude, but the actual concentrations, which we are often more interested in, vary over a far greater range from pH 0 to 14.
- They are of little help if one wishes to estimate the pH of a solution of the acid or base, or of one of its salts.
Both of these limitations are readily overcome by the use of easily-constructed logarithmic plots which we describe in the following section.
Buffer Solutions
The word buffer has several meanings in English, most of them referring (in its verb form) to cushion, shield, protect, or counteract an adverse effect. In chemistry, it refers specifically to a solution that resists a change in pH when acid or base is added. A buffer (or buffered ) solution is one that resists a change in its pH when H + or OH – ions are added or removed owing to some other reaction taking place in the same solution. Buffer solutions are essential components of all living organisms.
- Our blood is buffered to maintain a pH of 7.4 that must remain unchanged as metabolically-generated CO 2 (carbonic acid) is added and then removed by our lungs.
- Buffers in the oceans, in natural waters such as lakes and streams, and within soils help maintain their environmental stability against acid rain and increases in atmospheric CO 2 .
- Many industrial processes, such as brewing, require buffer control, as do research studies in biochemistry and physiology that involve enzymes, are active only within certain pH ranges.
The essential component of a buffer system is a conjugate acid-base pair whose concentration is fairly high in relation to the concentrations of added H + or OH – it is expected to buffer against. A simple buffer system might be a 0.2 M solution of sodium acetate; the conjugate pair here is acetic acid HAc and its conjugate base, the acetate ion Ac – . The idea is that this conjugate pair "pool" will be available to gobble up any small (≤ 10 –3 M ) addition of H+ or OH – that may result from other processes going on in the solution.
Sketch out a similar diagram showing what happens when you remove H + or OH – .
- Answer
-
When this happens, the ratio [HAc]/[Ac – ], will remain substantially unchanged, as will the pH, as you will see below.
You can also think of the process depicted above in terms of the Le Chatelier principle: addition of H + to the solution suppresses the dissociation of HAc, partially counteracting the effect of the added acid, as illustrated by the equation at the bottom left of the above diagram.
The Henderson-Hasselbalch Approximation
To develop this more quantitatively, we will consider the general case of a weak acid HA to which a quantity C b of strong base has been added; think, for example, of acetic acid which has been partially neutralized by sodium hydroxide, yielding the same conjugate pair described above, although not necessarily identical concentrations. The mass balance for such a system would be
\[[HA] + [A^–] = C_a + C_b = C_T \label {2-1}\]
in which C t denotes the total concentration of all species in the solution. Because we added a strong (completely dissociated) base NaOH to the acid, we also note that
\[C_b = [Na^+] \label{2-2}\]
Recalling the equilibrium expression for a weak acid
\[K_a = \dfrac{[H^+][A^-]}{[HA]} \label{2-3}\]
We can solve this for [H + ]:
\[[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{2-4}\]
Re-writing this in terms of negative logarithms, this becomes
\[-\log [H_3O^+] = -\log K_a - log [HA] + \log[A^-] \label{2-5}\]
or, since \(pK_a = –\log K_a\), we invert the ratio to preserve the positive sign:
\[pH = pK_a + \log \dfrac{[A^-]}{[HA]} \label{2-6}\]
This equation is known as the Henderson-Hasselbalch Approximation . It tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system.
It is interesting to note that the H-H equation was not developed by chemists!
- Lawrence Henderson (1878-1942) was an American medical doctor who taught at Harvard and who studied the acidity of blood and its relation to respiration. In 1908 he worked out the relation shown in Equation \(\ref{2-3}\).
- In 1916, K. Hasselbalch, a physiologist at U. of Copenhagen, derived the logarithmic form in Equation \(\ref{2-6}\).
You may wonder why these two equations, whose derivation we now consider almost trivial, should have immortalized the names of these two scientists. The answer is that the theory of chemical equilibrium was still developing in the early 1900's, and had not yet made its way into chemistry textbooks. Even the concept of pH was unknown until Sørenson's work appeared in 1909. It was the mystery (and medical necessity) of understanding why shortness of breath made the blood more alkaline, and too-rapid breathing made it more acidic, that forced the work of H&H into modern Chemistry.
Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its pK a . According to the above equation, when pH = pK a , the log term becomes zero, so that the ratio [AB – ] / [HA] = 10 0 = 1, meaning that [HA] = [AB – ]. In other words, when the pH of a solution is set to the value of the pK a of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically on a proton-free energy diagram:
This says, in effect, that at when the pH of a solution containing both the acid and it conjugate base is made identical to the acid pK a , the forms HA and A – possess identical free energies, and will therefore be present in equal concentrations.
Note
When the pH of a solution is set to the value of the pK a of an acid-base pair, the concentrations of the acid- and base forms will be identical.
Making and using Buffer Solutions
Buffers are generally most effective when the buffer system pK a is not too far from the target pH. Under these conditions, both [HA] and [A – ] are large enough to compensate for the withdrawal or addition, respectively, of hydrogen ions.
| acid | formulas | pK a | pH range |
|---|---|---|---|
| phosphoric | H 2 PO 4 / HPO 4 – | 2.16 | 1-3 |
| carbonic | H 2 CO 3 * / HCO 3 – | 3.75 | 3-5 |
| acetic | HCOOH / HCOO – | 4.75 | 4-6 |
| dihydrogen phosphate | H 2 PO 4 – / HPO 4 2 – | 7.21 | 6-8 |
| boric | H 3 BO 3 / H 2 BO 3 – | 9.24 | 8-10 |
| ammonium | NH 4 + / NH 3 | 9.25 | 8-10 |
| bicarbonate | HCO 3 – / CO 3 2 – | 10.3 | 9-11 |
| monohydrogen phosphate | HPO 4 2 – / PO 4 3 – | 12.3 | 11-13 |
Equation \(\ref{2-4}\) above and its logarithmic equivalent in Equation \(\ref{2-6}\) are of limited use in calculations because the exact values [A – ] and [HA] are known only for the special case when the pH of the solution is identical to the pK a . Most buffer solutions will be adjusted to other pH's, and of course, once the buffered solution begins doing it work by counteracting the effect of additions of H + or OH – , both [A – ] and [HA] will have changed. These two equations are so widely used in practical chemistry (and especially in biochemistry) that they are worth committing to memory.
\[[H^+] \approx K_a \dfrac{C_a}{C_b} \label{2-7}\]
\[pH \approx pK_a + \log \dfrac{C_b}{C_a} \label{2-8}\]
Although these are approximations, they are usually justified because useful buffer systems are always significantly more concentrated than those responsible for adding or removing hydrogen ions. Also, at these higher ionic concentrations, the K a of the buffer system will seldom be precisely known anyway. Don't expect actual buffer pH's to match calculations to better than 5%.
Compare the effects of adding 1.0 mL of 2.5 M HCl to
- 100 mL of pure water
- 100 mL of 0.20 M sodium dihydrogen phosphate solution.
Solution
a)
- quantity of HCl added: (1.0 mL) × (2.5\ mM/mL) = 2.5 mMol = 0.0025 mol
- concentration of H + in solution: (.0025 mol) ÷ (.101 L) = .025 M
- pH = – log (.025) = 1.60, change in pH = 1.60 – 7.00 = –5.4
b) (From Table \(\PageIndex{1}\), the pK a of H 2 PO 4 – is 7.21)
- Initial C a (H 2 PO 4 – ) = C b (HPO 4 2 – ) = 0.20 M
- Initial pH of buffer solution: pH = pK a + log ( C b / C a ) = 7.21 + 0 = 7.21
-
Added HCl will convert .0025 mol of HPO
4
2
–
to H
2
PO
4
–
,
changing C a to (.20 + .0025) mol ÷ (.101 L) = 2.00 M , and
reducing C b to (.20 – .0025) mol ÷ (.101 L) = 1.95 M - Resulting pH = 7.21 + log (1.95 / 2.00) = 7.21 –.011 = 7.00
- Change in pH = 7.00 – 7.21 = –.21
Note that in both cases the pH is reduced (as it must be if we are adding acid!), but the change is far less in the buffered solution.
How would you prepare 200 mL of a buffer solution whose pH is 9.0?
Solution
The first step would be to select a conjugate acid-base pair whose pK a is close to the desired pH. Looking at the above table of pK a values, either boric acid or ammonium ion would be suitable. For this example, we will select the NH 4 + /NH 3 system, K a = 5.5E–10, pK a = 9.25.
2) We will use Equation \(\ref{2-7}\) to determine the ratio C a / C b required to set [H + ] to the desired value of 1.00 × 10 –9 . To do this, write out the complete equilibrium constant expression for the acid NH 4 + : NH 4 + → NH 3 + H + , and then solve it to find the required concentration ratio:
\[K_a = \dfrac{[NH_3][H^+]}{[NH_4^+]} = 5.5 \times 10^{-10} \nonumber\]
\[ \dfrac{C_a}{C_b} = \dfrac{[NH_3]}{[NH_4^+]} = \dfrac{5.5 \times 10^{-10}}{[H^+]} = \dfrac{5.5 \times 10^{-10}}{1.00 \times 10^{-9}} = \dfrac{5.5 \times 10^{-10}}{10.00 \times 10^{-10}} = 0.55 \nonumber \]
3) The easiest (if not particularly elegant) way to work this out is to initially assume that we will dissolve 1.00 mole of solid NH 4 Cl in water, add sufficient OH – to partially neutralize some of this acid according to the reaction equation NH 4 + + OH – → NH 3 + H 2 O, and then add sufficient water to make the volume 1.00 L.
mass of ammonium chloride:
(1 mol) × (53.5 g M ol) = 53.8 g
let
x
= moles of NH
4
+
that must be converted to NH
3
.
Then
C
a
/
C
b
= (1-
x
)/
x
= .55; solving this gives x = .64, 1–x = .36.
4) So to make 1 L of the buffer, dissolve .36 × 53.8 g = 19.3 g of solid NH 4 Cl in a small quantity of water. Add .64 mole of NaOH (most easily done from a stock solution), and then sufficient water to make 1.00 L.
5) To make 200 mL of the buffer, just multiply each of the above figures by 0.200.
Pitfalls of the Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch Approximation is widely used in practical calculations. What most books do not tell you is that Eqution \(\ref{2-4}\) is no more than an “approximation of an approximation” which can give incorrect and misleading results when applied to situations in which the simplifying assumptions are not valid. An exact treatment of conjugate acid-base pairs, including a correct derivation of the Henderson-Hasselbalch equation, is given in the chapter on Exact Calculations .
The Henderson-Hasselbalch Approximation is only valid for fairly high concentrations
The approximations that lead to the H-H equation limit its reliable use to values of C a and C b that are within an order of magnitude of each other, and are fairly high. Also, the pK a of the acid should be moderate.
The shaded portion of this set of plots indicates the values of C a and K a that yield useful results. Clearly, the smaller the buffer concentration, the narrower the range of useable acid pK a s. Most buffer solutions tend to be fairly concentrated, with C a and C b typically around 0.01 - 0.1 M . Thus a buffer based on a .01 M solution of an acid such as chloric (HClO 3 ) with pK a of 1.9 will fall just outside the "safe" boundary near the upper left part of the diagram.
\[\underbrace{[H^+] = K_a \dfrac{[HA]}{[A^-]}}_{\text{exact}} \label{2-9a}\]
Equation \(\ref{2-9a}\) is simply a re-writing of the equilibrium constant expression, and is therefore always true. Of course, without knowing the actual equilibrium values of [HA] and [A – ], this relation is of little direct use in pH calculations.
\[\underbrace{[H^+] \approx K_a \dfrac{C_a}{C_b}}_{\text{approximate}} \label{2-9b}\]
Equation \(\ref{2-9b}\) is never true, but will yield good results if the acid is sufficiently weak in relation to its concentration to keep the [H + ] from being too high. Otherwise, the high [H + ] will convert a significant fraction of the A – into the acid form HA, so that the ratio [HA]/[A – ] will differ from C a / C b in the above two equations. Consumption of H + by the base will also raise the pH above the predicted value as we saw in the preceding problem example.
Buffer Index
The terms buffer intensity and buffer capacity are commonly employed as synonyms for buffer index, but in some contexts, buffer capacity denotes the quantity of strong acid or strong base which alters the buffer's pH by 1 unit. (see below). How effective is a given buffer system in resisting changes in the pH? The most direct expression of this is the rate of change of the pH as small quantities of strong acid or base are added to the system: Δ C /Δ(pH). Expressed in calculus notation, this is the buffer index , defined as
\[\beta = \left| \dfrac{dC}{d(pH)} \right| \label{2-19}\]
C here refers to the concentration of strong acid or base added to the solution. Because added acid or base affects the pH in opposite ways, we take the absolute value of this function in order to ensure that β is always positive. The value of β can be calculated analytically from C a , C b , K a , K b and [H + ]. By taking the second derivative of β, it can be shown that the buffer index has a maximum value when the pH = pK a .
This buffer index plot for a 0.10 M solution of sodium acetate is typical, and confirms that buffering is most efficient within about ±1 pH unit of pK a . But what about the even greater buffering that apparently occurs at the two extremes of pH (orange shading)? This is due to the buffering associated with the water itself, and will be seen in all aqueous buffer solutions.
Pure water is buffered by the H 3 O + /H 2 O conjugate pair at very low pH, and by H 2 O/OH- at high pH. This is easily understood if you think about adding some strong acid acid to pure water; even one drop of HCl will send the pH shooting down toward 0. If you continue adding acid, the pH will not drop significantly below 0, because there won't be enough free water molecules remaining to hydrate the HCl to produce H 3 O + ions — thus the solution is strongly buffered.
Note that if we were to subtract the effect of the NaAc buffering from the above plot, the remaining plot for water itself would exhibit a minimum at pH 7, where both [H 3 O + ] and [OH – ] share a common minimum value.
This alternative view shows how the distribution fractions of HAc and Ac – relate to the effective buffering range of this conjugate pair, which is conventionally defined as ±1 pH unit of the pK a . The term buffer capacity is an alternative means of expressing the ability of a buffer system to absorb the addition of strong acid or base without causing the pH to deviate by more than one unit from that of the pure buffer.
In this example, the buffer capacities for addition of acid and base will differ because the buffer pH has been adjusted to a value that differs from its pK a .
The Importance of buffer concentration
Buffer systems must be appreciably more concentrated than the concentrations of strong acid or base they are required to absorb while still remaining within the desired pH range. Once the added acid or base has consumed most of one or other of the conjugate species comprising the buffer, the pH will no longer be stabilized. And of course, very small buffer concentrations will approach the pH of pure water.
A more complete view: Log-C vs pH plots
These plots are often referred to as Sillén plots . Lars Gunnar Sillén (1916-1970) was a Swedish chemist who studied the distribution of ionic species in aqueous solutions and especially in the oceans. Because the concentrations of conjugate species can vary over many orders of magnitude, it is far more useful to express them on a logarithmic scale. Since pH is already logarithmic, one can obtain a "bird's-eye view" of an acid-base system in a compact log-log plot.
Even better, these plots are easily constructed without any calculations or arithmetic — or even any graph paper — any scrap paper, even the back of an envelope, will be sufficient. A ruler or other straightedge will, however, give more accurate results. In addition, you can use these plots to estimate the pH of a solution of a monoprotic or polyprotic acid or base without contending with quadratic or higher-order equations — or doing any arithmetic at all! The results will not be as precise as you would get from a proper numerical solution, but given the uncertainties of how equilibrium constants are affected by the presence of other ions in the solution, this is rarely a problem.
Aside from these advantages, the use of log-C vs pH plots will afford you an insight into the chemistry of acid-base systems that cannot be obtained simply by doing numerical calculations. The basic form of the plot, and the starting point for any use of such plots, looks like this:
If you examine this plot carefully, you will see that it is nothing more than a definition of pH and pOH, as well as a definition of a neutral solution at pH 7. Notice, for example, that when the pH is 4.0, [H + ] = 10 –4 M and [OH – ] = 10 –10 M .
A simple example: acetic acid solution
Here is a log C plot for a 10 –3 M solution of acetic acid ("HAc") in water. Although it may look a bit complicated at first, it is really very simple. The heavy maroon line on the left plots the concentration of the acid HAc as a function of pH. The blue line on the right shows how the concentration of the base Ac – depends on pH. The horizontal parts of these lines are aligned with "3" on the –log- C axis, corresponding to the 10 –3 M nominal concentration ( C a ).
How do we know the shapes and placement of the plots for the concentrations of acetic acid [HAc] and the acetate ion [Ac – ] ? Although both of these concentrations will of course vary with the pH, their deviations from 10 –3 M are too small to reveal themselves on a logarithmic plot until the pH approaches the pK a .
- At very low pH, virtually all of the acetate system will be in its acid form (that is, [HAc] = 10 –3 ), and similarly, at high pH, the base form [Ac – ] = 10 –3 M .
- When pH = pK a , the concentrations of the conjugate species are identical, as indicated by the crossing of the lines representing the two concentrations. These concentrations will be half of the total concentration given by C a , so [HAc] = [Ac – ] = ½ C a = 0.5E–3 M . Where does this go on the log-C axis? Well, the logarithm of 0.5 is –0.3 (a useful fact to remember!), so the crossing point at is displaced by 0.3 of a log-C unit below the –3 level on the y -axis.
- Because point defines both the pK a and concentration of a particular acid-base system, it is known as the system point .
-
What about the slopes of the plots when they bend down? It turns out that these slopes are +1 for
[Ac – ] and –1 for [HAc]. Since the slopes of the lines for [H + ] and [OH – ] are ±1, we can use these as guidelines; just make the plots of [Ac – ] and [HA] parallel to them. - The curved portions of the plot that joins the the horizontal and diagonal parts on either side of the system point can be drawn freehand without undue error; logarithmic plots are very forgiving!
Estimating the pH
Of special interest in acid-base chemistry are the pH values of a solution of an acid and of its conjugate base in pure water; as you know, these correspond to the beginning and equivalence points in the titration of an acid with a strong base. Continuing with the example of the acetic acid system, we show below another plot that is just like the one above, but with a few more lines and numbers added.
pH of acetic acid in pure water. Suppose you would like to find the pH of a 0.001 M solution of acetic acid in pure water. From the equation
\[HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{Acetate 2-1} \]
we know that equal numbers of moles of hydronium and acetate ions will be formed, so the concentrations of these species should be about the same:
\[[H^+] \approx [Ac^–] \label{Acetate 2-2}\]
will hold. The equivalence of these two concentrations corresponds to the point labeled on the log C -pH plot; this occurs at a pH of about 4, and this is the pH of a 0.001 M solution of acetic acid in pure water.
pH of sodium acetate in pure water. A 0.001 M solution of NaAc in water corresponds to the composition of a solution of acetic acid that has been titrated to its equivalence point with sodium hydroxide. The acetate ion, being the conjugate base of a weak acid, will undergo hydrolysis according to
\[Ac^– + H_2O \rightleftharpoons HAc + OH^– \label{Acetate 2-3}\]
which establishes the approximate relation
\[[HAc] \approx [OH^–] \label{Acetate 2-4}\]
This condition occurs where the right sloping part of the line representing [HAc] intersects the [OH – ] line at in the plot above. As you would expect for a solution of the conjugate base of a weak acid, this corresponds to an alkaline solution, in this case, at about pH = 7.8.
You will have to admit that estimating a pH in this way is far more painless than an ordinary numerical calculation would be!
It is interesting to show the Sillén plot for a weak acid system with its titration curve, both on the same pH scale. In this example, the titration plot has been turned on its side and reversed in order to illustrate the correspondence of its ƒ=0 (initial), ƒ=0.5 and ƒ=1 (equivalence) points with the corresponding points on the upper plot.
Ammonia - a weak base
Ammonia, unlike most gases, is extremely soluble in water. Solutions of NH 3 in water are properly known as aqueous ammonia , NH 3 (aq) ; the name "ammonium hydroxide" is still often used, even though there is no evidence to suggest that a species NH 4 OH exists. Ammonia is the conjugate base of the ammonium ion NH 4 + . The fact that ammonia is a base has no special significance insofar as the construction and interpretation of the log-C vs pH diagram is concerned; such diagrams always refer to conjugate acid-base systems , rather than to individual acids and bases.
Thus the system point defines the NH 4 + /NH 3 pair ( pK a 9.3) at 25°C and a total concentration
\[C_T = C_a + C_b = 0.0010\; M. \label{NH3 2-5}\]
To estimate the pH of a solution of ammonia in pure water, we make use of the charge balance requirement (known as the proton condition )
\[[NH_4^+] + [H^+] = [OH^-] \label{NH3 2-5.5}\]
which we simplify by assuming that, because NH 4 + is a weak acid, [NH 4 + ] >> [H + ] and we can drop the [H + ] term; thus the approximation
\[[NH_4^+] \approx [OH^-] \label{NH3 2-6}\]
which corresponds to the crossing point and a pH of 10.0. Similarly, at the crossing point , [NH 3 ] ≈ [H + ] . This comes from the proton condition
\[[NH_3] + [H^+] – [OH^–] = 0 \label{NH3 2-7}\]
by dropping the [OH – ] term on the assumption that its value is negligible compared to [H + ].
Equation \(\ref{NH3 2-7}\) is referred to as the "proton condition". Sorry to sneak it in without warning, but we hope that having piqued your curiosity, you might take the time to look at the following discussion of this important tool for estimating the pH of a solution of a pure acid, base, or a salt.
Mass- and charge balance, and the proton condition
In any aqueous solution of an acid or base, certain conservation conditions are strictly observed. Together with concentration and the K a or K b , these put certain constraints on the system that determine the state of the system. The most fundamental of these are conservation of mass and of electric charge, which of course apply to chemical changes of all kinds. We commonly express these as mass balance and charge balance , respectively. (The latter of these is sometimes referred to as the electroneutrality condition .)
Using the above example of the ammonium system as an illustration of this, we are interested in two particular instances of practical importance: what conditions apply to solutions made by dissolving a) pure ammonia, or b) ammonium chloride, in water?
Sillén diagram
Charge balances
- C T , NH 3 = [NH 3 ] + [NH 4 + ] = 10 –2 M
- C T , Ac = [HAc] + [Ac – ] = 10 –2 M
- NH 3 : [H + ] + [NH 4 + ] = [OH – ]
- NaAc: [Na + ] + [H + ] = [Ac – ] + [OH – ]
OK so far? That's really all we need, but it's usually more convenient to combine these with mass balances on the protons alone, so that we have a single equation for each of the two solutions. The resulting equations are known as the proton conditions for the two solutions.
To avoid a lot of algebra, there is a simple short-cut for writing a proton condition equation:
- Identify the substance you are starting with — the pure acid, base, or salt, and also H 2 O itself, that make up the solution whose pH you wish to know at the concentration for which the log C-pH chart is drawn. we call this the proton reference level (if that appellation scares you, it is sometimes referred to as the "basis substance".)
- On the left side of the proton condition, write concentration expressions for all species that possess protons in excess of the reference level.
- On the right side, write concentrations of all species that possess fewer protons the reference level.
Let's try it for our ammonia system.
Solution of NH 3 in water:
The proton reference level (PRL) is defined by NH 3 (aq) and H 2 O.
Proton condition: [H 3 O + ] + [NH 4 + ] = [OH – ]
(Notice that the substances that define the PRL (in case, H 2 O and NH 3 ) never appear in the proton condition equation.)
Solution of NH 4 Cl (or of any other strong-anion ammonium salt) in water:
The PRL is defined by NH 4 + and H 2 O.
Proton condition: [H 3 O + ] = [NH 3 ] + [OH – ]
(The chloride ion has nothing to do with protons, so it does not appear here.)
Good question: the answer is that what seems complicated at first sometimes turns out to be simpler and easier to understand than any alternative. Think of it this way: acid-base reactions involve the transfer of protons: some protons jump up to higher proton-free energy levels (e.g., NH 3 + H 2 O → NH 4 + + OH – ), others drop down to proton-vacant levels (H 3 O + → NH 3 → H 2 O + NH 4 + ). But no matter what happens, the total number of "available" protons (that is, all "dissociable" protons, including those in H 3 O + ) must be conserved — thus the "mass balance on protons". So let's look again at the log C - pH plot for the ammonium system, which we reproduce here for your convenience:
Consider first the solution of ammonium chloride, whose proton condition is given by [H 3 O + ] = [NH 3 ] + [OH – ]. We know that NH 4 Cl, being the salt of a weak base and a strong acid, will give a solution that is slightly acidic. In such a solution, [OH – ] will be quite small so that we can neglect it without too much error. We can then write the proton condition as [H + ] ≈ [NH 3 ]. On the plot above, this corresponds to point where the lines representing these quantities cross. At this pH of around 6.2, [OH – ] is about two orders of magnitude smaller than [NH 3 ], so we are justified in dropping it.
Similarly, for a solution of ammonia in water, the proton condition [H 3 O + ] + [NH 4 + ] = [OH – ] that we worked out above can be simplified to [NH 4 + ] ≈ [OH – ] (point ) with hardly any error at all, since [H + ] is here about six orders of magnitude smaller than [NH 4 + ].
In our discussion of the plot for the acetic acid system, we arrived at the proton conditions [H + ] ≈ [Ac – ] (for HAc in water) and [H + ] ≈ [OH – ] (for NaAc) by simply using the stoichiometries of the reactions. This can work in the simplest systems, but not in the more complicated ones involving polyprotic systems. In general, it is far safer to write out the complete proton condition in order to judge what concentrations, if any, can be dropped.
Ammonium acetate solution: solution of a salt
Having looked at the Sillén diagrams for acetic acid and ammonia, let's examine the log C-pH plot for this salt of acetic acid and ammonia.
You will recall that we dealt with solutions of the salts sodium acetate and ammonium chloride in the two examples described above. What is different here is that both components of the salt CH 3 COONH 4 ("NH 4 Ac") are "weak" in the sense that their conjugate species NH 3 and HAc are also present in significant quantities. This means that we are really dealing with two acid-base systems — the ones shown previously — on the same plot.
Each system retains its own system point, for the acetate system and for the ammonium system. Note that the two plots previously shown were for 10 –3 M solutions, so the non-system points for this more concentrated solution occur at different pH values. The pH's of 10 –2 M solutions of HAc , NaAc , and NH 3 are located by using the same proton conditions as before.
The only new thing here is the point , which corresponds to the ammonium acetate solution in which we are interested. The proton condition for this solution is [H + ] + [HAc] = [NH 3 (aq) ] + [OH – ]. Inspection of the plot reveals that [H + ] << [HAc] and [OH – ] << [NH 3 (aq) ], so the proton condition can be simplified to [HAc] ≈ [NH 3 (aq) ], corresponding to the intersection of these two lines at .
Polyprotic systems
Now that you are able to find your way around log C-pH plots that encompass two acid-base systems, the polyprotic systems that we describe below should be no trouble at all.
The Carbonate System
A thorough understanding of the carbonic acid-carbonate system is essential for understanding the chemistry of natural waters and the biochemistry of respiration in humans and other animals. There are two special things about this system that you need to know:
-
A solution of carbon dioxide in water is mostly in the form of hydrated CO
2
, CO
2
(aq). But a small fraction of dissolved CO
2
reacts with water to form carbonic acid, H
2
CO
3
. Because this acid cannot be isolated, common practice is to add an asterisk to the formula to designate the "total" CO
2
concentration
[CO 2 + H 2 CO 3 ]: thus H 2 CO 3 * . - Because CO 2 is a gas, solutions containing carbonate species of all kinds can equilibrate with the atmosphere, which normally contains small quantities of CO 2 . This means that a solution of sodium bicarbonate that is open to the atmosphere can lose CO 2 to the atmosphere at low pH, and absorb it at high pH. A system of this kind is said to be open . Alternatively, a system can be closed to exchange with the atmosphere. Sillén plots for open and closed carbonate systems have very different appearances.
This Sillén plot for the H 2 CO 3 * closed system is typical of other diprotic acid systems in that there are three conjugate species and two system points. The 10 –3 M concentration plotted here is typical of what is found in groundwaters that are in contact with carbonate-containing sediments.
The pH of a 10 –3 M solution of CO 2 , NaHCO 3 and Na 2 CO 3 are indicted by the blue annotations.
| solute, (reference level) | proton condition | approximation |
|---|---|---|
| carbon dioxide ( H 2 CO 3 *) | [H + ] = [HCO 3 – ] + 2[CO 3 2 – ] + [OH – ] | [H + ] ≈ [HCO 3 – ] |
| sodium bicarbonate ( NaHCO 3 ) | [H + ] + [ H 2 CO 3 ] = [CO 3 2 – ] + [OH – ] | [ H 2 CO 3 ] ≈ [OH – ] |
| sodium carbonate ( Na 2 CO 3 ) | [H + ] + 2[ H 2 CO 3 *] + [HCO 3 – ] = [OH – ] | [HCO 3 – ] ≈ [OH – ] |
You should take a few moments to verify the approximations in the rightmost column. Notice the factors of 2 that multiply some of the concentrations. For example, the "2[CO 3 2 – ]" term in the first line of the table means that the CO 3 2 – species is two steps away from the proton reference level H 2 CO 3 *, so it would take two hydrogen ions to balance a single carbonate ion.
Another new feature introduced here is the doubling of the slopes of the lines representing [H 2 CO 3 *] and [CO 3 2 – ] where they cross the pH values corresponding to the K a s that are one step removed from their own system points and near –log C = 7.
The Phosphate System
This triprotic system is widely used to prepare buffer solutions in biochemical applications. In its fundamental form, there is nothing new here, other than an additional system point corresponding to the pK a of H 2 PO 4 – .
The pH of a 10 –3 M solutions of H 3 PO 4 , NaH 2 PO 4 , Na 2 HPO 4 and Na 3 PO 4 can be estimated by setting up the proton conditions as follows. Unfortunately, these estimates become a bit more complicated owing to the proximity of the three system points. But it's far easier than doing a numerical solution which requires solving a firth-degree polynomial!
| solute, (reference level) | proton condition | approximation |
|---|---|---|
| phosphoric acid (H 3 PO 4 ) | [H + ] = [H 2 PO 4 – ] + 2[HPO 4 2 – ] + 3[PO 4 3 – ] + [OH – ] | [H + ] ≈ [H 2 PO 4 – ] |
| sodium dihydrogen phosphate (NaH 2 PO 4 ) | [H + ] + [H 3 PO 4 ] = [HPO 4 2 – ] + 2[PO 4 3 – ] + [OH – ] |
[H + ] + [H 3 PO 4 ] ≈ [HPO 4 2 – ] |
|
disodium hydrogen phosphate (Na 2 HPO 4 ) |
[H + ] + 2[H 2 PO 4 – ] + [H 2 PO 4 – ] = 3[PO 4 3 – ] + [OH – ] | [H 3 PO 4 -] = [PO 4 3 – ] + [OH – ] |
| trisodium phosphate (Na 3 PO 4 ) | [H + ] + 3[H 3 PO 4 ] +2 [H 2 PO 4 – ] + [HPO 4 2 – ] = [OH – ] | [HPO 4 2 – ] ≈ [OH – ] |
Because the phosphate plot is rather crowded, we show here a modified one in which the details of the pH estimates for the solutions of H 3 PO 4 and its salts are emphasized.
Estimation of the pH of 10 –3 M solutions of phosphoric acid and trisodium phosphate , based on the approximations in the above table are straightforward.
Those for NaH 2 PO 4 and Na 2 HPO 4 , however, are complicated by the presence of two concentration terms on the right side of the respective proton condition approximations.
Unless you are in an advanced course, you may wish to skip the following details. The important thing to know is that graphical estimates of these more problematic systems are fundamentally possible.
Thus for sodium dihydrogen phosphate, if we were to follow the pattern of the other systems, the approximation at the crossing point would be
[H
+
] = [HPO
4
2
–
], corresponding to point
. But inspection of the plot shows that at that pH (4.0), the concentration of H
3
PO
4
is ten times greater than that of H
+
(point
). A rough estimate of the pH can be made by constructing a line (shown in red) that is parallel to those for H
3
PO
4
and H
+
, but is raised up by a factor of 10. This results in the new crossing point
.
Similarly, the pH of a disodium hydrogen phosphate solution does not correspond to the crossing point because of the significant value of [HPO 4 3 – ] at this pH. Again, we construct a line above this crossing point that predicts a pH corresponding to point .
Although these stratagems may seem rather crude, it should be noted that the uncertainties associated with them tend to be minimized on a logarithmic plot. In addition, it should be borne in mind that in a solution as concentrated as 0.1M, the pK a values found in tables are not applicable. The alternative of solving for the pH algebraically, taking into consideration activity coefficients for the high ionic concentrations, is rarely worth the effort.
How to draw Sillén diagrams
The relation between the pH of a solution and the concentrations of weak acids and their conjugate species is algebraically rather complicated. But over most of the pH range, simplifying assumptions can be made that, when expressed in logarithmic form, plot as straight lines having integral slopes of 0, ±1, ±2, etc. It is only within the narrow pH range near the pK a that these simplifying assumptions break down, but on a logarithmic plot, one can draw a smooth curve that covers this range without introducing significant error.
|
Here is your basic stationery — a plot showing the two ions always present in any aqueous solution. In practice, you can usually cut off the bottom part where concentrations smaller than \(10^{–10}\, M\). |
|
|
For any given acid-base system, you need to know the pK a and nominal concentration CT. Find the point at which these two lines intersect. |
|
|
Locate and mark the system point on the pK a line. It will be 0.3 of a log-C unit below the concentration line.
This is where [HA]
|
|
|
For pH < pK a , [HA] plots as a line through the system point with slope = 0.
For pH >
pK
a
, the line assumes a slope of –1. Use the [H
+
] and
|
|
| In the pH interval of ± pK a , the slope changes from 0 to +1. You can approximate this with a sloped line as shown, or "prettify" it with a smooth curve. | |
| Finally, draw the lines for the conjugate form [A – ], using slopes +1 and 0. |
What's all this good for?
Suppose you have a solution of NaA whose pH is 8. Estimate the concentration of the acid HA in the solution. All you need do is look at the plot. The [HA] line crosses the line marking this pH at – log C = 6, or [HA] = 10 –6 . This quantity too small to significantly reduce [A – ], which remains at approximately CT = 10 –6 M .
References
- Boundless outlines on buffer solutions. These summarize the main points and problem examples as presented in major textbooks. They are an excellent aid for pre-exam review. Chemistry: The Central Science (8 subtopics)
- A collection of practice exercises (with solutions) on buffers and titrations from Bryn Mawr College.
- The Henderson-Hasselbalch equation: its history and limitations - Henry N. Po and N.M. Senoza, J.Chem. Ed. 78 (11) 2001 (1499-1503)
- A detailed treatment of log-C vs pH diagrams can be found in the book Acid-base diagrams by Kahlert and Schotz. It is available online in some college libraries.
|
libretexts
|
2025-03-17T19:53:16.327700
| 2016-02-13T03:28:26 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.04%3A_Conjugate_Pairs_and_Buffers",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.4: Conjugate Pairs and Buffers",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid_Base_Titration
|
13.5: Acid/Base Titration
Make sure you thoroughly understand the following essential concepts:
- Sketch out a plot representing the titration of a strong monoprotic acid by a strong base, or of a strong base titrated by a strong acid. Identify the equivalence point and explain its significance.
- On the plot referred to above, draw a similar plot that would correspond to the same system at a higher or lower concentration .
- Sketch out a plot representing the titration of a weak monoprotic acid by a strong base, or of a weak base titrated by a strong acid. Identify the equivalence point and half-equivalence points.
- Explain what an acid-base indicator is, and how it works.
- When solutions of some polyprotic acids are titrated with strong base, not all of the equivalence points can be observed. Explain the most common reasons for this.
- Calculate the molarity of a monoprotic acid HA whose titration endpoint occurs after V ml of strong base of a given concentration has been added.
The objective of an acid-base titration is to determine \(C_a\), the nominal concentration of acid in the solution. In its simplest form, titration is carried out by measuring the volume of the solution of strong base required to complete the reaction
\[\ce{H_nA + n OH- → n A- + n H_2O} \label{0-1}\]
in which \(n\) is the number of replaceable hydrogens in the acid. The point at which this reaction is just complete is known as the equivalence point . This is to be distinguished from the end point , which is the value we observe experimentally. A replaceable hydrogen atom (sometimes called an "acidic" hydrogen) is one that can be donated to a strong base — that is, to an OH – ion. Thus in acetic acid HCOO H , only the hydrogen in the carboxyl group is considered "replaceable".
What we actually measure, of course, is the volume of titrant delivered by the burette. Learning to properly control the stopcock at the bottom of the burette usually requires some instruction and practice, as does the reading of the volume. For highly precise work, the concentration of the titrant itself must be determined in a separate experiment known as "standardization".
Understanding Titration Curves
A plot showing the pH of the solution as a function of the quantity of base added is known as a titration curve . These plots can be constructed by plotting the pH as a function of either the volume of base added, or the equivalent fraction \(ƒ\) which is simply the number of moles of base added per mole of acid present in the solution. In most of the titration curves illustrated in this section, we plot pH as a function of \(ƒ\). It's worth taking some time to thoroughly familiarize yourself with the general form of a titration curve such as the one shown below, in which a weak acid HA is titrated with a strong base, typically sodium hydroxide.
The equivalence point occurs at the pH at which the equivalent fraction ƒ of base added is unity. At this point, the reaction
\[HA + OH^– → AB^– + H_2O \label{1-1}\]
is stoichiometrically complete; a solution initially containing n moles of a monoprotic acid HA will now be identical to one containing the same number of moles of the conjugate base A – . At the half-equivalence point ƒ = 0.5, the concentrations of the conjugate species are identical: [HA] = [A – ]. This, of course corresponds to a buffer solution (hence the relatively flat part of the curve) whose pH is the same as pK a .
As base is added beyond ƒ = 1, the pH begins to level off, suggesting that another buffered system has come into play. In this case it involves the solvent (water) and hydroxide ion: {H 2 O} ≈ {OH-}.
A similar effect is seen at the low-pH side of the curve when a strong acid is titrated, as in the plot for the titration of HCl below. In this case, the buffering is due to {H 3 O + ) ≈ {H2O}.
How can this be? Surely, the concentration of OH – , even when the pH approaches 14, cannot be anything like that of [H 2 O] which will be about 55.5 M in most solutions! This fine point (along with the mention of H 2 O/OH – buffering) is rarely mentioned in elementary courses because the theory behind it involves some rather esoteric elements of solution thermodynamics. However, in case you are curious, note that the curly brackets in {H 2 O} ≈ {OH – } denote activities, not concentrations. And by convention the activity of a pure liquid (H 2 O in this case) is unity. At a pH of around 12, pOH = 2, [OH – ] = .01. At this rather high ion concentration, {OH – } will be somewhat smaller than this, but the two activities will be similar enough to produce the buffering effect we observe.
The pH of the solution at its equivalence point will be 7 if we are titrating a strong acid with strong base, as in HCl + NaOH → H 2 O + NaCl. However, if the acid is weak, as in the above plot, the solution will be alkaline. This pH can be calculated from C b and K b in a manner exactly analogous to that used for calculating the pH of a solution of a weak acid in water.
It is important to understand that the equivalent fraction ƒ of base that must be added to reach the equivalence point is independent of the strength of the acid and of its concentration in the solution. The whole utility of titration as a means of quantitative analysis rests on this independence; we are in all cases measuring only the total number of moles of “acidic” hydrogens in the sample undergoing titration.
Acid and base strengths determine the shape of the curve
Although the strength of an acid has no effect on the location of the equivalence point, it does affect the shape of the titration curve and can be estimated on a plot of the curve.
The weaker the acid being titrated, the higher the initial pH (at ƒ=0), and the smaller will be the vertical height of the plot near the equivalence point. As we shall see later, this can make it difficult to locate the equivalence point if the acid is extremely weak.
Estimating the acid strength
As shown in plot above, the pK a of a weak acid can be estimated by noting the pH that corresponds to the half-titration point ƒ = 0.5. Recalling that the pH is controlled by the ratio of conjugate species concentrations
\[pH = pK_a + \log \dfrac{[A^]}{[HA]} \label{1-2}\]
it will be apparent that this equation reduces to pH = pK a when the titration is half complete (that is, when [HA] = [A – ]), the pH of the solution will be identical to the pK a of the acid. This equation does not work for strong acids owing to the strong buffering that occurs at the very low pH at which ƒ = 0.5.
As indicated here, the buffering has nothing to do with the acid HCl itself (which does not exist as such in water), but rather with its dissociation products H 3 O + and OH – , "the strongest acid and base that can exist in water."
Monoprotic titration curves
The following two principles govern the detailed shape of a titration curve:
- The stronger the acid or base, the greater will be the slope of the curve near the equivalence point;
- The weaker the acid or base, the greater will the deviation of the pH from neutrality at the equivalence point.
It is important to understand the reasons for these two relations. The second is the simplest to explain. Titration of an acid HA with a base such as NaOH results in a solution of NaA; that is, a solution of the conjugate base A – . Being a base, it will react with water to yield an excess of hydroxide ions, leaving a slightly alkaline solution. Titration of a weak base with an acid will have the opposite effect.
The extent of the jump in the pH at the equivalence point is determined by a combination of factors. In the case of a weak acid, for example, the initial pH is likely to be higher, so the titration curve starts higher. Further, the weaker the acid, the stronger will be its conjugate base, so the higher will be the pH at the equivalence point. These two factors raise the bottom part of the titration curve. The upper extent of the curve is of course limited by the concentration and strength of the titrant.
These principles are clearly evident in the above plots for the titrations of acids and bases having various strengths. Notice the blue curves that represent the titration of pure water (a very weak acid) with strong acid or base.
Monoprotic titration curve gallery
When both the titrant and sample are "strong", we get long vertical plots at ƒ = 1. Adding even half a drop of titrant can take us across the equivalence point!
When one of the reactants is weak, the pH changes rapidly at first until buffering sets in.
← In (C), the onset of H 2 O/OH- buffering near ƒ=1 makes the equivalence point more difficult to locate.
"Weak/weak" titrations tend to be problematic as the buffered regions move closer to ƒ=1. The equivalence point pH of 7 in these examples reflects the near-equality of pK a and pK b of the reactants.
Dealing with very weak acids
It can be difficult to reliably detect the equivalence point in the titration of boric acid ( pK a = 9.3) or of other similarly weak acids from the shape of the titration curve*. *An interesting student laboratory experiment that employs an auxiliary reagent (mannitol) to make boric acid stronger and thus more readily titratable was described in J. Chem Ed . 2012, 89, 767-770.
The problem here is that aqueous solutions are buffered against pH change at very low and very high pH ranges. An extreme example occurs in the titration of pure water with a strong acid or base. At these extremes of pH the concentrations of H 3 O + and of OH – are sufficiently great that a competing buffer system (either H 3 O + /H 2 O or H 2 O/OH – , depending on whether the solution is highly acidic or highly alkaline) comes into play.
Why we usually use a "strong" titrant
The above plots clearly show that the most easily-detectable equivalence points occur when an acid with is titrated with a strong base such as sodium hydroxide (or a base is titrated with a strong acid.)
In practice, many of the titrations carried out in research, industry, and clinical practice involve mixtures of more than one acid. Examples include natural waters, physiological fluids, fruit juices, wine making, brewing, and industrial effluents. For titrating these kinds of samples, the use of anything other than a strong titrant presents the possibility that the titrant may be weaker than one or more of the "stronger" components in the sample, in which case it would be incapable of titrating these components to completion.
In terms of proton-free energies, the proton source (the acidic titrant) would be unable to deliver an equivalent quantity of protons to the (stronger) component of the mixture.
Polyprotic acids
There will be as many equivalence points as there are replaceable hydrogens in an acid. Thus in the extremely important carbonate system , equivalence points are seen at both ƒ=1 and ƒ=2:
In general, there are two requirements for a clearly discernible jump in the pH to occur in a polyprotic titration:
- The successive K a 's must differ by several orders of magnitude;
- The pH of the equivalence point must not be very high or very low.
Separation of successive equilibrium constants
The effect of the first point is seen by comparing the titration curves of two diprotic acids, sulfurous and succinic. The appearance of only one equivalence point in the latter is a consequence of the closeness of the first and second acid dissociation constants. The pK a 's of sulfurous acid (below, left) are sufficiently far apart that its titration curve can be regarded as the superposition of those for two independent monoprotic acids having the corresponding K a 's. This reflects the fact that the two acidic –OH groups are connected to the same central atom, so that the local negative charge that remains when HSO 3 – is formed acts to suppress the second dissociation step.
*It can be shown that in the limit of large n , the ratio of K 1 /K 2 for a symmetrical dicarboxylic acid HOOC-(CH 2 ) n - COOH converges to a value of 4.
In succinic acid , the two –COOH groups are physically more separated and thus tend to dissociate independently*. Inspection of the species distribution curves for succinic acid (above, right) reveals that the fraction of the ampholyte HA can never exceed 75 percent. That is, there is no pH at which the reaction H 2 A → HA – + H + can be said to be “complete” while at the same time the second step HA – → A 2– + H + has occurred to only a negligible extent. Thus the rise in the pH that would normally be expected as HA is produced will be prevented by consumption of OH – in the second step which will be well underway at that point; only when all steps are completed and hydroxide ion is no longer being consumed will the pH rise.
Two other examples of polyprotic acids whose titration curves do not reveal all of the equivalence points are sulfuric and phosphoric acids. Owing to the leveling effect, the apparent K a1 of H 2 SO 4 is so close to K a2 = 0.01 that the effect is the same as in succinic acid, so only the second equivalence point is detected.
In phosphoric acid , the third equivalence point (for HPO 4 2 – ) is obscured by H 2 O-OH – buffering as explained previously.
Detection of the equivalence point
Whether or not the equivalence point is revealed by a distinct "break" in the titration curve, it will correspond to a unique hydrogen ion concentration which can be calculated in advance. There are many ways of determining the equivalence point of an acid-base titration.
Indicators
Don't overshoot the equiv point!
The traditional method of detecting the equivalence point has been to employ an indicator dye, which is a second acid-base system in which the protonated and deprotonated forms differ in color, and whose pK a is close to the pH expected at the equivalence point. If the acid being titrated is not a strong one, it is important to keep the indicator concentration as low as possible in order to prevent its own consumption of OH – from distorting the titration curve.
The observed color change of an indicator does not take place sharply, but occurs over a range of about 1.5 to 2 pH units. Indicators are therefore only useful in the titration of acids and bases that are sufficiently strong to show a definite break in the titration curve. Some plants contain coloring agents that can act as natural pH indicators. These include cabbage (shown), beets, and hydrangea flowers.
For a strong acid - strong base titration, almost any indicator can be used, although phenolphthalein is most commonly employed. For titrations involving weak acids or bases, as in the acid titration of sodium carbonate solution shown here, the indicator should have a pK close to that of the substance being titrated.
When titrating a polyprotic acid or base, multiple indicators are required if more than one equivalence point is to be seen. The pK a s of phenolphthalein and methyl orange are 9.3 and 3.7, respectively.
Potentiometry: Use a pH meter
The pH meter detects the voltage produced when the H + ions in the solution displace Na + ions from a thin glass membrane that is dipped into the solution.
A more modern way of finding an equivalence point is to follow the titration by means of a pH meter. Because it involves measuring the electrical potential difference between two electrodes, this method is known as potentiometry . Until around 1980, pH meters were too expensive for regular use in student laboratories, but this has changed; potentiometry is now the standard tool for determining equivalence points.
Improved plotting methods
Plotting the pH after each volume increment of titrant has been added can yield a titration curve as detailed as desired, but there are better ways of locating the equivalence point. The most common of these is to take the first or second derivatives of the plot: d (pH)/d V or d 2 (pH)/d V 2 (of course, for finite increments of pH and volume, these terms would be expressed as Δ(pH)/Δ V and Δ 2 (pH)/Δ V 2 .)
A second-derivative curve locates the inflection point by finding where the rate at which the pH changes is zero.
The differential plot , showing rate-of-change of pH against titrant volume, locates the inflection point which is also the equivalence point
In a standard plot of pH-vs-volume of titrant added, the inflection point is located visually as half-way along the steepest part of the curve.
The idealized plots shown above are unlikely to be seen in practice. When the titration is carried out manually, the titrant is added in increments, so even the simple titration curve must be constructed from points subject to uncertainties in volume measurement and pH (especially if the latter is visually estimated by color change of an indicator.)
If this data is then converted to differential form, these uncertainties add a certain amount of "noise" to the data.
A second-derivative plot uses pH readings on both sides of the equivalence point, making it easier to locate in the presence of noise.
Locating the equivalence point depends very strongly on correct reading of only one or two pH readings near the top of the plot.
A simple curve, plotted from a small number of pH readings, will not always unambiguously locate the equivalence point.
The "noise" in differential plots can usually be minimized by keeping the titrant and analyte concentrations above 10 –3 M .
Some other ways of following a titration
Monitoring the pH by means of an indicator or by potentiometry as described above are the standard ways of detecting the equivalence point of a titration. However, we have already seen that in certain cases involving polyprotic acids or bases, some of the equivalence points are obscured by their close proximity to others, or by the buffering that occurs near the extremes of the pH range. Similar problems can arise when the solution to be titrated contains several different acids, as often happens when fluids connected with industrial processes must be monitored.
Take the temperature!
Acid-base neutralization reactions HA + B → A – + BH + are always exothermic; when protons fall from their level in the acid to that in the base, most of the free energy drop gets released as heat. If the acid and base are both strong (i.e., totally dissociated), the enthalpy of neutralization for the reaction
\[\ce{H3O+ + OH- → 2 H2O}\]
is -68 kJ mol –1 .
See this Wikipedia page for more on thermometric titrations, including many examples. Note also the video on this topic in the "Videos" section near the end of this page.
- Thermometric titrations are not limited to acid-base determinations; they can also be used to follow precipitation-, complex formation-, and oxidation-reduction reactions.
- They can be used with polyprotic acids or bases, and with mixtures containing more than one acid or base.
- They are able to follow acid-base titrations that must be carried out in non-aqueous solvents, where other titration methods are not possible.
A typical thermometric titration curve consists of two branches, beginning with a steep rise in temperature as the titrant being added reacts with the analyte, liberating heat. Once the equivalence point is reached, the rise quickly diminishes as heat production stops. Then, as the mixture begins to cool, the plot assumes a negative slope.
Although a rough indication of the equivalence point can be estimated by extrapolating the linear parts of the curve (blue dashed lines), the differential methods described above are generally preferred.
Follow the electrolytic conductance
Acids and bases are electrolytes , meaning that their solutions conduct electric current. The conductivity of such solutions depends on the concentrations of the ions, and to a lesser extent, on the nature of the particular ions. Any chemical reaction in which there is a change in the total quantity of ions in the solution can usually be followed by monitoring the conductance. Acid-base titrations fall into this category. Consider, for example, the titration of hydrochloric acid with sodium hydroxide. This can be described by the equation
\[\ce{H^{+} + Cl^{–} + Na^{+} + OH^{–} → H2O + Na^{+} + Cl^{–}} \label{2-1}\]
which shows that two of the four species of ions being combined disappear at the equivalence point. During the course of the titration, the conductance of the solution falls as H + and Cl – ions are consumed. At the equivalence point the conductance passes through a minimum, and then rises as continued addition of titrant adds more Na + and OH – ions to the solution.
Each kind of ion makes its own contribution to the solution conductivity. If we could observe the contribution of each ion separately, see that the slopes for H + and OH – are much greater. This reflects the much greater conductivities of these ions owing to their uniquely rapid movement through the solution by hopping across water molecules.
However, because the conductances of individual ions cannot be observed directly, conductance measurements always register the total conductances of all ions in the solution. The change in conductance that is actually observed during the titration of HCl by sodium hydroxide is the sum of the ionic conductances shown above.
For most ordinary acid-base titrations, conductimetry rarely offers any special advantage over regular volumetric analysis using indicators or potentiometry. This is especially true if the acid being titrated is weak; if the pK a is much below 2, the rising salt line (Na + when titrating with NaOH) will overwhelm the fall in the contribution the small amount of H + makes to the conductance, thus preventing any minimum in the total conductance curve from being seen.
However, in some special cases such as those illustrated below, conductimetry is the only method capable of yielding useful results.
These examples illustrate two unique capabilities of conductimetric titrations: (left) Titration of a mixture of two acids and (right) Titration of a strong polyprotic acid →
Because pure H 2 SO 2 is a neutral molecule, it makes no contribution to the conductance, which rises to a maximum at the equivalence point.
Automating the process
In four years of college lab sessions, many Chemistry majors will likely carry out fewer than a dozen titrations. However, in the real world, time is money, and long gone are the days when technicians were employed full time just to titrate multiple samples in such enterprises as breweries, food processing (such as blending of canned orange juice), clinical labs, and biochemical research.
Titration calculations
A titration is carried out by adding a sufficient volume \Vo of the titrant solution to a known volume \Vt of the solution being titrated. This addition continues until the end point is reached. The end point is our experimental approximation of the equivalence point at which the acid-base reaction is stoichiometrically complete (ƒ = 1). The quantity we actually measure at the end point is the volume V_ep of titrant delivered to the solution undergoing titration.
The solution being titrated is often referred to as the analyte (the substance being "analyzed") or, less commonly, as the titrand . We shall employ the latter term in what follows.
In a simple titration of a monoprotic acid HA by a base B, the equivalence point corresponds to completion of the reaction
\[\ce{HA + B → A^{–} + BH^{+} }\label{3-1}\]
where equimolar quantities of HA and B have been combined; that is,
M HA = M B (where M represents the number of moles.)
Recall that the number of moles is given by the product of the volume and concentration:
L × mol L –1 = mol,
mL x mMol ml –1 = mmol.
Because we are measuring the volume of titrant rather than the number of moles, we need to use its concentration to link the two quantities. So if we are titrating the base B with acid HA, the end point is reached when a volume V of HA has been added. The number of moles of HA we have added at the end point is given by the product of its volume and concentration
\[M_{HA} = V_{HA} \times C_{HA} \label{3-2}\]
And because the reaction \ref{3-1} is now complete, M HA = M B ; thus, at the end point of any monoprotic titration,
\[V_{HA} C_{HA} = V_B C_B \label{3-3}\]
Equation \ref{3-3} is i mportant! In any titration, both the volume and the concentration of the titrant are known, so the unknown concentration is easily calculated.
In titrations carried out in the laboratory, the titrant is delivered by a burette that is usually calibrated in milliliters, so it is more convenient to express M HA in millimoles and C HA in millimoles/mL (mMol ml –1 ); note that the latter is numerically the same as moles/L .
50.0 mL of 0.100 M hydrochloric acid is titrated with 0.200 M sodium hydroxide. What volume of NaOH solution will have been added at the equivalence point?
Solution
First, find the number of moles of HCl in the titrand:
\[\begin{align*} M_{\ce{HCl}} &= C_{\ce{HCl}} \times V_{\ce{HCl}} \\[4pt] &= (0.100\, mMol\, mL^{–1}) \times (50.0\, mL) \\[4pt] &= 5.0 \,mMol\,\ce{HCl} \end{align*}\]
This same number of moles of NaOH must be delivered by the burette in order to reach the equivalence point ( i.e. , M NaOH = 5.0 mMol.)
The volume of NaOH solution M NaOH required is
V NaOH = ( M NaOH / C NaOH ) = (5.0 mMol) / (0.200 mMol/mL) = 25 mL
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libretexts
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2025-03-17T19:53:16.456179
| 2016-02-13T03:29:09 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid_Base_Titration",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.5: Acid/Base Titration",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.06%3A_Applications_of_Acid-Base_Equilibria
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13.6: Applications of Acid-Base Equilibria
Make sure you thoroughly understand the following essential concepts:
- Describe the major features of the global carbon cycle , and outline the role of acid-base processes in both the slow and fast parts of the cycle.
- Describe the principal acid-base reactions in aerobic respiration .
- Define acid rain , and explain its origins. Outline its major effects on the environment.
Acid-base reactions pervade every aspect of industrial-, physiological-, and environmental chemistry. In this unit we touch on a few highlights that anyone who studies or practices chemical science should be aware of.
Some applications of Buffers
Buffer solutions and the buffering effect they produce are extremely important in many practical applications of chemistry. The reason for this is that many chemical processes are quite sensitive to the pH; the extent of the reaction, its rate, and even the nature of the products can be altered if the pH is allowed to change. Such a change will tend to occur, for example, when the reaction in question, or an unrelated parallel reaction, consumes or releases hydrogen- or hydroxide ions.
Buffers in biochemistry and physiology
Many reactions that take place in living organisms fall into this category. Most biochemical processes are catalyzed by enzymes whose activities are highly dependent on the pH; if the local pH deviates too far from the optimum value, the enzyme may become permanently deactivated.
- For example, the oxidation of glucose to carbon dioxide and water — the fundamental energy-producing process of aerobic metabolism — releases H + ions. The multiple steps of this process are catalyzed by enzymes whose activity is strongly affected by the pH. If the region of the cell where these reactions take place were not buffered, the very act of living would soon reduce the local pH below the range in which the enzymes are active, halting energy production and killing the cell.
- Blood is strongly buffered (mainly by bicarbonate) to maintain its pH at 7.4±0.3; pH values below 7.0 or above 7.8 cause death within minutes.
- The interiors of most cells are buffered near pH 7.0 (by phosphates and proteins)
Buffers in industry
Buffers are employed in a wide variety of laboratory procedures and industrial processes:
- solutions for the calibration of pH meters
- growth of bacteria in culture media
- fermentation processes, including winemaking and the brewing of beer
- controlling the colors of dyes used in coloring fabrics
- shampoos (maintaining pH at or below 7 to conteract the alkalinity of soaps that can cause irritation)
- other personal care products such as baby lotions (keeping the pH around 6 to inhibit the growth of bacteria)
- pH control in printing inks to assure their proper penetration into the paper
Buffers in the environment
- Natural waters (lakes and streams) able to support aquatic life are buffered by the sediments they are in contact with. This buffering is not always able to offset the effects of acid rain.
- Seawater is buffered mainly by bicarbonate and borates. This allows the oceans to absorb about half of the CO 2 emitted to the atmosphere by human activities.
- Soils, in order to remain productive must maintain a near-neutral pH, not only for plants but also for microorganisms which fix- and recycle nitrogen. Soils rendered highly acidic by acid rain can keep essential nutrients such as phosphates in forms such as insoluble Ca 3 (PO 4 ) 2 that prevents their uptake by plants.
Acid-base chemistry in physiology
Acid-base chemistry plays a crucial role in physiology, both at the level of the individual cell and of the total organism. The reasons for this are twofold:
- Many of the major chemical components of an organism can themselves act as acids and/or bases. Thus proteins contain both acidic and basic groups, so that their shapes and their functional activities are highly dependent on the pH.
- Virtually all important metabolic processes involve the uptake or release of hydrogen ions. The very act of being alive tends to change the surrounding pH (usually reducing it); this will eventually kill the organism in the absence of buDering mechanisms.
About two-thirds of the weight of an adult human consists of water. About two-thirds of this water is located within cells, while the remaining third consists of extracellular water, mostly interstial fluid that bathes the cells, and the blood plasma. The latter, amounting to about 5% of body weight (about 5 L in the adult), serves as a supporting fluid for the blood cells and acts as a means of transporting chemicals between cells and the external environment. It is basically a 0.15 M solution of NaCl containing smaller amounts of other electrolytes, the most important of which are HCO 3 – and protein anions.
Respiration, the most important physiological activity of a cell, is an acid-producing process. Carbohydrate substances are broken down into carbon dioxide, and thus carbonic acid:
\[C(H_2O)_{n} + O_2 → CO_2 + n H_2O\]
Maintenance of acid-base balance
It is remarkable that the pH of most cellular fluids can be kept within such a narrow range, given the large number of processes that tend to upset it. This is due to the exquisite balance between a large number of interlinked processes operating at many different levels. Acid-base balance in the body is maintained by two general mechanisms: selective excretion of acids or bases, and by the buffering action of weak acid-base systems in body fluids.
-
Over a 24-hour period, the adult human eliminates the equivalent of about 20-40 moles of H
+
by way of the lungs in the form of CO
2
. In addition, the kidneys excrete perhaps 5% of this amount of acid, mostly in the form of H
2
PO
4
–
and
NH 4 + . Owing to their electric charges, these two species are closely linked to salt balance with ions such as Na + or K + and Cl – . - The major buffering system in the body is the carbonate system, which exists mainly in the form of HCO 3 – at normal physiological pH. Secondary buffering action comes from phosphates, from proteins and other weak organic acids, and (within blood cells), the hemoglobin.
There is a huge industry aimed at the notoriously science-challenged "alternative health" market that promotes worthless nostrums such as "ionized water" that are claimed to restore or preserve "the body's acid-base balance". They commonly point out that most foods are "acidic", (i.e., they are metabolized to H 2 CO 3 ), but they never explain that most of this acid is almot immediately exhaled in the form of CO 2 . The implication is that our health is being ruined by the resulting "acidification" of the body; some further imply this can be a cause of cancer and other assorted ailments.
People who fall for these expensive scams are in effect paying a tax on scientific ignorance. Those who, like you, have studied Chemistry, should consider that they have a social duty to counter this kind of predatory and deceptive pseudoscience.
Disturbances of acid-base balance
Deviations of the blood plasma pH from its normal value of 7.4 by more than about ±.1 can be very serious. These conditions are known medically as acidosis and alkalosis . They can be caused by metabolic disturbances such as diabetes and by kidney failure in which excretion of H 2 PO 4 – , for example, is inhibited.
Numerous other processes lead to temporary unbalances. Thus hyperventilation, which can result from emotional upset, leads to above-normal loss of CO 2 , and thus to alkalosis. Similarly, hypoventilation can act as a compensatory mechanism for acidosis. On the other hand, retention of CO 2 caused by bronchopneumonia, for example, can give rise to acidosis. Acidosis can also result from diarrhea (loss of alkaline fluid from the intestine), while loss of gastric contents by vomiting promotes alkalosis.
Acid rain
The atmosphere is naturally acidic
The natural, unpolluted atmosphere receives acidic, basic, and neutral substances from natural sources (volcanic emissions, salt spray, windblown dust and microbial metabolism) as well as from pollution (Figure 11.6.X). These react in a kind of gigantic acid-base titration to give a solution in which hydrogen ions must predominate to maintain charge balance (indicated by the identical widths of the bar graphs at the bottom labeled "anions" and "cations").
Carbon dioxide, however, is the major source of acidity in the unpolluted atmosphere. As will be explained farther on, CO 2 makes up 0.032 vol-% of dry air, and dissolves in water to form carbonic acid:
\[CO_{2(g) }+ H_2O(l) \rightleftharpoons H_2CO_{3(aq)}\]
Thus all rain is “acidic” in the sense that it contains dissolved CO 2 which will reduce its pH to 5.7. The term acid rain is therefore taken to mean rain whose pH is controlled by pollutants which can lower the pH into the range of 3-4, resulting in severe damage to the environment.
Origin of acid rain
Acid rain originates from emissions of SO 2 and various oxides of nitrogen (known collectively as "NO x ") that are formed during combustion processes, especially those associated with the burning of coal. Incineration of wastes, industrial operations such as smelting, and forest fires are other combustion-related sources; these also release particulate material into the atmosphere which plays an important role in concentrating and distributing these acid-forming substances.
Origins of NO x
Nitrogen oxides are formed naturally by soil bacteria acting on nitrate ions, an essential plant nutrient and itself a product of natural nitrogen fixation. Ammonia, a product of bacterial decomposition of organic matter, is eventually transformed into NO 3 – by bacteria into nitrogen oxides.
The quantity of fixed nitrogen produced industrially to support intensive agriculture now exceeds the amount formed naturally, and has become the major source of anthropogenic nitrogen oxides.
- Atmospheric nitrogen can be thermally decomposed to its various oxides NO x when subjected to high temperatures; one minor natural source is lightning. More significantly, these oxides are also formed when "clean" fuels such as natural gas (mainly methane, CH 4 ) is burnt in air.
- NO x can also react with other atmospheric pollutants to produce photochemical smog .
Formation of acids
The gases noted above react with atmospheric oxygen, with each other, and with particulate materials to form sulfuric, nitric, and hydrochloric acids.
- Most of the H 2 SO 4 comes from the photooxidation of SO 2 released from the burning of fossil fuels and from industrial operations such as smelting.
- Nitric acid similarly results from photooxidation of nitrogen oxides. Large quantities of these oxides are formed in high-temperature combustion processes in gasoline- and jet engines, and in the turbine engines of fossil-fueled electric power plants.
- Hydrochloric acid is formed naturally through the reaction of sea salt aerosols with atmospheric H 2 SO 4 and HNO 3 . The salt particles form when ocean waves break the surface and some of the water evaporates before the spray settles back to the surface. On a much smaller scale, fossil fuel combustion and waste incineration are believed to the principal human-caused sources of HCl.
Wet and dry deposition
There are two basic mechanisms by which acidic material is transported through the environment. The acid that dissolves in the water droplets that form clouds and precipitation, and is eventually incorporated into fog and snowflakes, undergoes wet deposition ; this is ordinarily what is referred to as "acid rain". Because it is widely dispersed, often high in the atmosphere, it can travel very large distances from the original source.
About 40% of the acidic substances introduced into the atmosphere becomes adsorbed onto particulate matter, which can be soot, smoke, windblown dust, and salt particles formed naturally from sea spray. This spreads less widely (typically around 30 km from the source), and falls onto surfaces by " dry deposition ". Once the acid-laden particles land on surfaces, ordinary rain, fog, and dew release the acidic components, often in considerably more concentrated form than occurs through dry deposition alone.
Effects on soils
Because soils support the growth of plants and of the soil microorganisms that are essential agents in the recycling of dead plant materials, acid rain has a indirect but profound effect on soil health and plant growth.
Soils containing alkaline components (most commonly limestone CaCO 3 and other insoluble carbonates) can neutralize added acid and mitigate its effects. But soils in high mountain regions tend to be thin and unable to provide adequate buffering capacity. The same is true in almost half of Canada, in which granite rock of the Canadian Shield is very close to the surface; the eastern provinces of the country are strongly impacted by acid rain.
- Most soils also contain clay and humic substances that bind and retain ions such as Ca 2 + , Mg 2 + and K + which are essential plant nutrients. Added H + binds even more strongly to these substances, displacing the nutrient ions from the upper layers. The overall effect is to leach these ions to greater depths where they may be inaccessible to plants, or to wash them away into ground water.
- The SO 4 2 – component of acid rain converts some nutrient cations into insoluble sulfates, reducing their availability to plants.
- Clays, which are complex aluminosilicates, gradually break down, releasing aluminum ions which normally form insoluble Al(OH) 3 . Addition of H + dissolves this hydroxide, raising the concentration of Al(H 2 O) 6 3+ to levels that can be toxic to plants.
- Many toxic heavy metals such as Pb, Cd and Cr are present in trace amounts in soils but are tied up in the form of insoluble salts. Acid rain can mobilize the ions of these metals, much as happens with aluminum.
Effects on plants
↓ Click on image to enlarge
↓ Click on image to enlarge
The effects on soils noted above affect plants most strongly. However, direct impingment of acidic rain and fog on leaves has other effects which can be especially serious when air pollutants such as SO 2 are present.
- Acid deposition on leaves and needles tends to weaken them by eroding away their protective waxy coatings. This often results in the development of brown spots that interfere with photosynthesis.
- Forests in mountain regions are frequently bathed in clouds and fog which can be even more acidic than normal precipitation, exacerbating the above problem. In addition, the leaves and needles tend to accumulate the fog aerosol into larger droplets. These eventually fall to the forest floor, magnifying the effects on soils described in the previous section.
- Food crops tend to be less affected by acid rain where good farming practices such as the addition of fertilizers to replace depleted nutrients and the addition of limestone to raise the soil pH.
Effects on lakes and aquatic ecosystems
Lakes are subject not only to wet and dry acidic deposition, but also by the water they receive from streams and surface runoff. Any toxic elements released by the action of acid rain on soils and sediments can thus be conveyed to, and concentrated within lakes and the streams that empty them.
↓ Click on image to enlarge
↓ Click on image to enlarge
Lakes in poorly-buffered areas such as are found in alpine regions (western North America, Colorado and much of Switzerland) or on the Canadian Shield and in the Adirondacs and Appalachians) are highly sensitive to acid deposition.
Severely acidified lakes (such as the one depicted at the above right) can be so devoid of life, including algae, that the water appears to be perfectly clear and bright blue in color.
↓ Click on image to enlarge
Aquatic organisms are generally adapted to "ordinary" pH conditions of 6-8, but vary greatly in their tolerance of low pH. As pH declines below 6, the diversity of aquatic animals, plants, and microorganisms diminishes.
- Some plant species such as sphagnum mosses and certain filamentous algaes that thrive at very low pH can proliferate in acidified lakes, producing thick mats that seal off oxygen and thus inhibit the decay of litter on the lake floor.
- Acid deposited onto winter snow is released during spring melting and can cause rapid drops of pH in poorly-buffered lakes that receive waters from these sources. This "spring acid shock" can seriously affect viability of organisms such as fish, amphibians, and insects that lay their eggs in the water which hatch in the spring. The hatchlings are often unable to adapt to the rapid change, and end up deformed or killed.
-
Even those aquatic species able to survive at in low-pH waters can be indirectly affected if their food supply is pH-limited. For example, mayflies and some other insects, which, which are important food sources for frogs, cannot survive below pH 5.5.
↓ Click on image to enlarge
- Fish are severely affected by the aluminum that is released from the action of acid on sediments; it causes a coating of mucus to form on the gills, impeding absorption of oxygen. This has led to the extinction of some species from affected lakes.
- Low pH increases the solubility of calcium salts, making it more difficult for bone to form in developing embryos of fish and amphibians.
- Low pH can also make the eggs of aquatic organism thicker and more difficult for embryos to penetrate, thus delaying hatching. When the embryos continue to grow within the confined space of the egg, their spines can be deformed, interfering with their viability when they finally hatch.
Effects on buildings and statues
Acid rain is not new!
"It has often been observed that the stones and bricks of buildings, especially under projecting parts, crumble more readily in large towns where coal is burnt... I was led to attribute this effect to the slow but constant action of acid rain." Robert Angus Smith, 1856
Acid deposition most strongly affects heritage buildings made of limestone and similar carbonate-containing stone materials.
The principal chemical process involves the reaction of sulfuric acid with calcium carbonate:
\[\ce{CO3(s) + H2SO4 → CaSO4(s) + CO2}.\]
The acid first erodes and breaks up the surface of the stone. As the hydrated calcium sulfate (gypsum) forms, it picks up iron and other components of the stone and forms an unsightly black coating. Some of this gradually blisters off, exposing yet more stone suface. The gypsum crystals can sometimes grow into the stone, further undermining the surface.
Very old structures such as the Taj Mahal, Notre Dame, the Colosseum and Westminster Abbey have all been affected.
↓ Click on image to enlarge
Statues and monuments, including those made from marble, are also susceptible to erosion and damage from acid deposition.
Modern buildings are less affected, although acid rain can erode some painted surfaces.
The Geochemical Carbonate system
The carbonate system, consisting as it does of a soluble gas CO 2 , soluble ions HCO 3 – and CO 3 2– , and sparingly soluble carbonate salts, spans all four realms of nature: the atmosphere, hydrosphere (mainly the oceans), the lithosphere, and the biosphere. And owing to the acid-base equilibria that govern the transformations between these carbonate species, carbon is readily transported between these geochemical reservoirs. This chapter presents an overview of carbon geochemistry.
Distribution of carbon on Earth
Carbon is the fourth most abundant element in the universe. Within the Earth's crust it ranks 15th, mostly in the form of carbonates in limestones and dolomites. Kerogens, which are fossilized plant-derived carbon mostly in the form of oil shale, constitute another major repository of terrestrial carbon.
| Source | relative to atmosphere | |
|---|---|---|
|
carbonate rock
|
carbonate rock
|
28,500
|
|
fossil carbon
|
572
|
10,600
|
|
land - organic carbon
|
0.065
|
1.22
|
|
ocean
|
3.2
|
61.8
|
|
atmospheric CO
2
|
0.0535
|
1.0
|
The geochemical carbon cycle
The carbonate system encompasses virtually all of the environmental compartments– the atmosphere, hydrosphere, biosphere, and major parts of the lithosphere. The complementary processes of photosynthesis and respiration drive a global cycle in which carbon passes slowly between the atmosphere and the lithosphere, and more rapidly between the atmosphere and the oceans. Thus the "carbon cycle" can be divided into "fast" and "slow" parts, operating roughly on annual and geological time scales.
Figure (\PageIndex{2}\): Image by David Bise, Pennsylvania State University; see here for a full discussion of this diagram.
Another excellent depiction of the carbon cycle, by the U.S. Dept of Energy. ( See description ):
Carbon Cycle
Carbon dioxide in the atmosphere
CO 2 has probably always been present in the atmosphere in the relatively small absolute amounts now observed. Precambrian limestones, possibly formed by reactions with rock-forming silicates, e.g.
\[\ce{CaSiO_3 + CO_2 → CaCO_3 + SiO_2} \label{4-1}\]
have likely had a moderating influence on the CO 2 abundance throughout geological time.
The volume-percent of CO 2 in dry air is 0.032%, leading to a partial pressure of 3 × 10 –4 (10 3.5 ) atm. In a crowded and poorly-ventilated room, P CO2 can be as high as 100 × 10 –4 atm. About 54 × 10 14 moles per year of CO 2 is taken from the atmosphere by photosynthesis, divided about equally between land and sea. Of this, all except 0.05% is returned by respiration (mostly by microorganisms); the remainder leaks into the slow, sedimentary part of the geochemical cycle where it can remain for thousands to millions of years. Trends in the growth of atmospheric CO 2 for the past five years can be seen on this US NOAA page .
Since the advent of large-scale industrialization around 1860, the amount of CO 2 in the atmosphere has been increasing. Most of this has been due to fossil-fuel combustion; in 1966, about 3.6 × 10 15 g of C was released to the atmosphere; this is about 12 times greater than the estimated natural removal of carbon into sediments. The large-scale destruction of tropical forests, which has accelerated greatly in recent years, is believed to exacerbate this effect by removing a temporary sink for CO 2 .
About 30-50% of the CO
2
released into the atmosphere by combustion remains there; the remainder enters the hydrosphere and biosphere. The oceans have a large absorptive capacity for CO
2
by virtue of its transformation into bicarbonate and carbonate in a slightly alkaline aqueous medium, and they contain about 60 times as much inorganic carbon as is in the atmosphere. However, effcient transfer takes place only into the topmost
(100 m) wind-mixed layer, which contains only about one atmosphere equivalent of CO
2
; mixing time into the deeper parts of the ocean is of the order of 1000 years. For this reason, only about ten percent of added CO
2
is taken up by the oceans.
Most of the carbon in the oceans is in the form of bicarbonate, as would be expected from the pH which ranges between 7.8 and 8.2. In addition to atmospheric CO 2 , there is a carbonate input to the ocean from streams. This is mostly in the form of HCO 3 – which derives from the weathering of rocks and terrestrial carbonate sediments, and the acid-base reaction
\[H_2CO_3 + CO_3^{2–} → 2 HCO_3^– \label{4-2}\]
which can be considered to be the source of bicarbonate in seawater. In this sense, the ocean can be considered the site of a gigantic acid-base titration in which atmospheric acids (mainly CO 2 but also SO 2 , HCl, and other acids of volcanic origin) react with bases that originate from rocks and are introduced through carbonate-bearing streams or in windblown dust.
Carbon in the hydrosphere
Dissolution of CO 2 in water
Carbon dioxide is slightly soluble in water:
| °C |
0
|
4
|
10
|
20
|
|---|---|---|---|---|
| mol L –1 |
0.077
|
0.066
|
0.054
|
0.039
|
Henry's law is followed up to a CO 2 pressure of about 5 atm:
\[[\ce{CO2 (aq)}] = 0.032 P_{\ce{CO_2}} \label{4-3}\]
“Dissolved carbon dioxide” consists mostly of the hydrated oxide CO2(aq) together with a small proportion of carbonic acid:
\[[\ce{CO2 (aq)}] = 650 [\ce{H2CO3}] \label{4-4}\]
The acid dissociation constant \(K_{a1}\) that is commonly quoted for "H 2 CO 3 " is really a composite equilibrium constant that includes the above equilibrium. This means that "pure" H 2 CO 3 (which cannot be isolated) is a considerably stronger acid than is usually appreciated.
Distribution of carbonate species in aqueous solutions
Water exposed to the atmosphere with P CO2 = 10 3.5 atm will take up carbon dioxide, which becomes distributed between the three carbonate species CO 2 , HCO 3 – and CO 3 2– in proportions that depend on K 1 and K 2 and on the pH. The "total dissolved carbon" is given by the mass balance
\[C_t = [H_2CO_3] + [HCO_3^–] + [CO_3^{2–}] \label{4-5}\]
The distribution of these species as a function of pH can best be seen by constructing a log C -pH diagram for C t = 10 –5 M .
This Sillén plot is drawn for two different concentrations of the carbonate system. The lower one, in heavier lines, is for a 10 –5 M solution, corresponding roughly to atmospheric CO 2 in equilibrium with pure water. The upper plot, for a 10 –3 M solution, is representative of many natural waters such as lakes and streams where the water is in contact with sediments.
It is important to note that this diagram applies only to a system in which C t is constant. In a solution that is open to the atmosphere, this will not be the case at high pH values where the concentration of CO 2 is appreciable. Under these conditions this ion will react with H 2 CO 3 and the solution will absorb CO 2 from the atmosphere, eventually resulting in the formation of a solid carbonate precipitate.
The lower of the two above plots can be used to predict the pH of 10^{5} M solutions of carbon dioxide, sodium bicarbonate, and sodium carbonate in pure water. The pH values are estimated by using the mass and charge balance conditions for each solution as noted below.
The reasoning leading to these calculations is explained in the discussion of the 10 –3 M carbonate system in a previous chapter.
- Solution of CO 2 or H 2 CO 3
- [H + ] = [OH – ] + [HCO 3 – ] + 2 [CO 3 2– ](4-6)
- which, since the solution will be acidic, can be simplified to
- [H + ] = [HCO 3 – ] (point )(4-7)
- Solution of NaHCO 3
- [H + ] + [H 2 CO 3 ] = [CO 3 2– ] + [OH](4-8)
- or
- [H + ] = [HCO 3 – ] (point ) (4-9)
- Solution of Na 2 CO 3
- [H + ] + 2 [H 2 CO 3 ] + [CO 3 2– ] = [OH – ](4-10)
- or
- [HCO 3 – ] = [OH – ] (point ) (4-11)
Carbon in the oceans
Natural waters aquire carbon from sediments they are in contact with, and of course also from the atmosphere. The pH is an important factor here; CO 2 and solid carbonates are more soluble at high pH, and pH also controls the distribution of carbon species, as is seen in the plot just above.
| rain | river/lake water | sea water | |
|---|---|---|---|
| ppm of carbon |
1.2
|
35
|
140
|
| pH (unpolluted) |
5.6
|
6.5 - 8.5
|
7.5 - 8.4
|
At the slightly alkaline pH of most bodies of water (of which the oceans constitute 97% of the earth's surface waters), bicarbonate is the principal dissolved carbon species. The quantity of organic carbon is fairly small.
| CO 2 (aq) | HCO 3 – | CO 3 2– | dead org. | living org. |
|---|---|---|---|---|
|
0.18
|
2.6
|
0.33
|
0.23
|
0.05
|
|
libretexts
|
2025-03-17T19:53:16.604857
| 2016-02-13T03:29:38 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.06%3A_Applications_of_Acid-Base_Equilibria",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.6: Applications of Acid-Base Equilibria",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.07%3A_Exact_Calculations_and_Approximations
|
13.7: Exact Calculations and Approximations
- Understand the exact equations that are involves in complex acid-base equilibria in aqueous solutions
The methods for dealing with acid-base equilibria that we developed in the earlier units of this series are widely used in ordinary practice. Although many of these involve approximations of various kinds, the results are usually good enough for most purposes. Sometimes, however — for example, in problems involving very dilute solutions, the approximations break down, often because they ignore the small quantities of H + and OH – ions always present in pure water. In this unit, we look at exact, or "comprehensive" treatment of some of the more common kinds of acid-base equilibria problems.
Strong acid-base systems
The usual definition of a “strong” acid or base is one that is completely dissociated in aqueous solution. Hydrochloric acid is a common example of a strong acid. When HCl gas is dissolved in water, the resulting solution contains the ions H 3 O + , OH – , and Cl – , but except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions. To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them.
Limiting Conditions
These relations are obtained by observing that certain conditions must always hold for aqueous solutions:
- The dissociation equilibrium of water must always be satisfied : \[[H^+][OH^–] = K_w \label{1-1}\]
- The undissociated acid and its conjugate base must be in mass balance . The actual concentrations of the acid and its conjugate base can depend on a number of factors, but their sum must be constant, and equal to the “nominal concentration”, which we designate here as \(C_a\). For a solution of HCl, this equation would be \[[HCl] + [Cl^–] = C_a \label{1-2}\] Because a strong acid is by definition completely dissociated, we can neglect the first term and write the mass balance condition as \[[Cl^–] = C_a \label{1-3}\]
- In any ionic solution, the sum of the positive and negative electric charges must be zero ; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle . \[[H^+] = [OH^–] + [Cl^–] \label{1-4}\]
The next step is to combine these three limiting conditions into a single expression that relates the hydronium ion concentration to \(C_a\). This is best done by starting with an equation that relates several quantities and substituting the terms that we want to eliminate. Thus we can get rid of the \([Cl^–]\) term by substituting Equation \(\ref{1-3}\) into Equation \(\ref{1-4}\) :
\[[H^+] = [OH^–] + C_a \label{1-5}\]
The \([OH^–]\) term can be eliminated by the use of Equation \(\ref{1-1}\):
\[[H^+] = C_a + \dfrac{K_w}{[H^+]} \label{1-6}\]
This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. As the acid concentration falls below about 10 –6 M, however, the second term predominates; \([H^+]\) approaches \(\sqrt{K_w}\) or \(10^{–7} M\) at 25 °C. The hydronium ion concentration can of course never fall below this value; no amount of dilution can make the solution alkaline!
No amount of dilution can make the solution of a strong acid alkaline!
Notice that Equation \(\ref{1-6}\) is a quadratic equation; in regular polynomial form it would be rewritten as
\[[H^+]^2 – C_a[H^+] – K_w = 0 \label{1-7}\]
Most practical problems involving strong acids are concerned with more concentrated solutions in which the second term of Equation \(\ref{1-7}\) can be dropped, yielding the simple relation
\[[H^+] \approx [A^–]\]
Activities and Concentrated Solutions of Strong Acids
In more concentrated solutions, interactions between ions cause their “effective” concentrations, known as their activities , to deviate from their “analytical” concentrations. Thus in a solution prepared by adding 0.5 mole of the very strong acid HClO 4 to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 M . This does not mean that the acid is only 80% dissociated; there is no evidence of HClO 4 molecules in the solution. What has happened is that about 20% of the H 3 O + and ClO 4 – ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces. Similarly, in a 0.10 M solution of hydrochloric acid, the activity of H + is 0.81, or only 81% of its concentration. (See the green box below for more on this.)
Activities are important because only these work properly in equilibrium calculations. The relation between the concentration of a species and its activity is expressed by the activity coefficient \(\gamma\):
\[a = \gamma C \label{1-8}\]
As a solution becomes more dilute, \(\gamma\) approaches unity. At ionic concentrations below about 0.001 M , concentrations can generally be used in place of activities with negligible error. Recall that pH is defined as the negative logarithm of the hydrogen ion activity , not its concentration.
Activities of single ions cannot be determined, so activity coefficients in ionic solutions are always the average, or mean , of those for all ionic species present. This quantity is denoted as \(\gamma_{\pm}\).
At very high concentrations, activities can depart wildly from concentrations. This is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater. In a 12 M solution of hydrochloric acid, for example, the mean ionic activity coefficient* is 207. This means that under these conditions with [H + ] = 12, the activity {H + } = 2500, corresponding to a pH of about –3.4, instead of –1.1 as might be predicted if concentrations were being used. These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl over a concentrated hydrochloric acid solution even though this acid is supposedly "100% dissociated".
At these high concentrations, a pair of "dissociated" ions \(H^+\) and \(Cl^–\) will occasionally find themselves so close together that they may momentarily act as an HCl unit; some of these may escape as \(HC l (g)\) before thermal motions break them up again. Under these conditions, “dissociation” begins to lose its meaning so that in effect, dissociation is no longer complete. Although the concentration of \(HCl (aq)\) will always be very small, its own activity coefficient can be as great as 2000, which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable.
Weak monoprotic acids and bases
Most acids are weak; there are hundreds of thousands of them, whereas there are no more than a few dozen strong acids. We can treat weak acid solutions in exactly the same general way as we did for strong acids. The only difference is that we must now include the equilibrium expression for the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which strong cations are present. In this exposition, we will refer to “hydrogen ions” and \([H^+]\) for brevity, and will assume that the acid \(HA\) dissociates into \(H^+\) and its conjugate base \(A^-\).
Pure acid in water
In addition to the species H + , OH – , and A− which we had in the strong-acid case, we now have the undissociated acid HA; four variables, requiring four equations.
- Equilibria
\[ [H^+][OH^–] = K_w \label{2-1}\]
\[ K_a = \dfrac{[H^+][A^–]}{[HA]} \label{2-2}\]
- Mass balance
\[ C_a = [HA] + [A^–] \label{2-3}\]
- Charge balance
\[[H^+] = [OH^–] + [HA^–] \label{2-4}\]
To eliminate [HA] from Equation \(\ref{2-2}\), we solve Equation \(\ref{2-4}\) for this term, and substitute the resulting expression into the numerator:
\[ K_a =\dfrac{[H^+]([H^+] - [OH^-])}{C_a-([H^+] - [OH^-]) } \label{2-5}\]
The latter equation is simplified by multiplying out and replacing [H + ][OH – ] with K w . We then get rid of the [OH – ] term by replacing it with K w /[H + ]
\[[H^+] C_b + [H^+]^2 – [H^+][OH^–] = K_a C_a – K_a [H^+] + K_a [OH^–]\]
\[[H^+]^2 C_b + [H^+]^3 – [H^+] K_w = K_a C_a – K_a [H^+] + \dfrac{K_a K_w}{[H^+]}\]
Rearranged into standard polynomial form, this becomes
\[[H^+]^3 + K_a[H^+]^2 – (K_w + C_aK_a) [H^+] – K_a K_w = 0 \label{2-5a}\]
For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.
Approximation 1: Neglecting Hydroxide Population
Unless the acid is extremely weak or the solution is very dilute, the concentration of OH – can be neglected in comparison to that of [H + ]. If we assume that [OH – ] ≪ [H + ], then Equation \(\ref{2-5a}\) can be simplified to
\[K_a \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{2-6}\]
which is a quadratic equation:
\[[H^+]^2 +K_a[H^+]– K_aC_a \approx 0 \label{2-7}\]
and thus, from the quadratic formula,
\[ [H^+] \approx \dfrac{K_a + \sqrt{K_a + 4K_aC_a}}{2} \label{2-8}\]
Calculate the pH of a 0.0010 M solution of acetic acid, \(K_a = 1.74 \times 10^{–5}\).
Solution
First approximation:
\[[H^+] = \sqrt{(1.0 \times 10^{–3}) × (1.74 \times 10^{–5}} = \sqrt{1.74 \times 10^{–8}} = 1.3 \times 10^{–4}\; M. \nonumber \]
Applying the "5% test",
\[\dfrac{1.3 \times 10^{–4}}{1.0 \times 10^{–3}} = 0.13\nonumber \]
This exceeds 0.05, so we must explicitly solve the quadratic Equation \(\ref{2-7}\) to obtain two roots: \(+1.2 \times 10^{–4}\) and \(–1.4 \times 10^{-4}\). Taking the positive root, we have
\[pH = –\log (1.2 \times 10^{–4}) = 3.9 \nonumber \]
Approximation 2: Very Concentrated Acids
If the acid is fairly concentrated (usually more than 10 –3 M), a further simplification can frequently be achieved by making the assumption that \([H^+] \ll C_a\). This is justified when most of the acid remains in its protonated form [HA], so that relatively little H + is produced. In this event, Equation \(\ref{2-6}\) reduces to
\[ K_a \approx \dfrac{[H^+]^2}{C_a} \label{2-9}\]
or
\[[H^+] \approx \sqrt{K_aC_a} \label{2-10}\]
Calculate the pH and percent ionization of 0.10 M acetic acid "HAc" (CH 3 COOH), \(K_a = 1.74 \times 10^{–5}\).
Solution:
It is usually best to start by using Equation \(\ref{2-9}\) as a first approximation:
\[[H^+] = \sqrt{(0.10)(1.74 \times 10^{–5})} = \sqrt{1.74 \times 10^{–6}} = 1.3 \times 10^{–3}\; M\nonumber \]
This approximation is generally considered valid if [H + ] is less than 5% of C a ; in this case, [H + ]/ C a = 0.013, which is smaller than 0.05 and thus within the limit. This same quantity also corresponds to the ionization fraction, so the percent ionization is 1.3%. The pH of the solution is
\[pH = –\log 1.2 \times 10^{-3} = 2.9\nonumber \]
Approximation 3: Very Weak and Acidic
If the acid is very weak or its concentration is very low, the \(H^+\) produced by its dissociation may be little greater than that due to the ionization of water. However, if the solution is still acidic, it may still be possible to avoid solving the cubic equation \(\ref{2-5a}\) by assuming that the term \(([H^+] - [OH^–]) \ll C_a\) in Equation \(\ref{2-5}\):
\[ K_a = \dfrac{[H^+]^2}{C_a - [H^+]} \label{2-11}\]
This can be rearranged into standard quadratic form
\[[H^+]^2 + K_a [H^+] – K_a C_a = 0 \label{2-12}\]
For dilute solutions of weak acids, an exact treatment may be required. With the aid of a computer or graphic calculator, solving a cubic polynomial is now far less formidable than it used to be. However, round-off errors can cause these computerized cubic solvers to blow up; it is generally safer to use a quadratic approximation.
Boric acid, B(OH) 3 ("H 3 BO 3 ") is a weak acid found in the ocean and in some natural waters. As with many boron compounds, there is some question about its true nature, but for most practical purposes it can be considered to be monoprotic with \(K_a = 7.3 \times 10^{–10}\):
\[Bi(OH)_3 + 2 H_2O \rightleftharpoons Bi(OH)_4^– + H_3O^+\nonumber \]
Find the [H + ] and pH of a 0.00050 M solution of boric acid in pure water.
Solution
Because this acid is quite weak and its concentration low, we will use the quadratic form Equation \(\ref{2-7}\), which yields the positive root \(6.12 \times 10^{–7}\), corresponding to pH = 6.21. Notice that this is only six times the concentration of \(H^+\) present in pure water!
It is instructive to compare this result with what the quadratic approximation would yield, which yield \([H^+] = 6.04 \times 10^{–7}\) so \(pH = 6.22\).
Weak bases
The weak bases most commonly encountered are:
- anions A – of weak acids:
\[A^– + H_2O \rightleftharpoons HA + OH^–\]
\[CO_3^{2–} + H_2O \rightleftharpoons HCO_3^– + OH^–\]
- ammonia
\[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^–\]
- amines, e.g. methylamine
\[CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^++ H_2O\]
Solution of an anion of a weak acid
Note that, in order to maintain electroneutrality, anions must be accompanied by sufficient cations to balance their charges. Thus for a C b M solution of the salt NaA in water, we have the following conditions:
- Species
Na + , A – , HA, H 2 O, H + , OH –
- Equilibria
\[[H^+][OH^–] = K_w \label{2-13}\]
\[ K_b =\dfrac{[HA][OH^-]}{[A^-]} \label{2-14}\]
- Mass balance
\[C_b = [Na^+] = [HA] + [A^–] \label{2-15}\]
- Charge balance
\[[Na^+] + [H^+] = [OH^–] + [A^–] \label{2-16}\]
Replacing the [Na + ] term in Equation \(\ref{2-15}\) by \(C_b\) and combining with \(K_w\) and the mass balance, a relation is obtained that is analogous to that of Equation \(\ref{2-5}\) for weak acids:
\[K_b =\dfrac{[OH^-] ([OH^-] - [H^+])}{C_b - ([OH^-] - [H^+])} \label{2-17}\]
The approximations
\[ K_b \approx \dfrac{[OH^-]^2}{C_b - [OH^-]} \label{2-18}\]
and
\[[OH^–] \approx \sqrt{K_b C_b} \label{2-19}\]
can be derived in a similar manner.
Calculate the pH and the concentrations of all species in a 0.01 M solution of methylamine, CH 3 NH 2 (\(K_b = 4.2 \times 10^{–4}\)).
Solution
We begin by using the simplest approximation Equation \(\ref{2-14}\):
\[[OH^–] = \sqrt{(K_b C_b}- = \sqrt{(4.2 \times 10^{-4})(10^{–2})} = 2.1 \times 10^{–3}\nonumber \]
To see if this approximation is justified, we apply a criterion similar to what we used for a weak acid: [OH – ] must not exceed 5% of C b . In this case,
\[ \dfrac{[OH^–]}{ C_b} = \dfrac{(2.1 \times 10^{-3}} { 10^{–2}} = 0.21\nonumber \]
so we must use the quadratic form Equation \(\ref{2-12}\) that yields the positive root \(1.9 \times 10^{–3}\) which corresponds to \([OH^–]\)
\[[H^+] = \dfrac{K_w}{[OH^–} = \dfrac{1 \times 10^{-14}}{1.9 \times 10^{–3}} = 5.3 \times 10^{-12}\nonumber \]
and
\[pH = –\log 5.3 \times 10^{–12} = 11.3.\nonumber \]
From the charge balance equation, solve for
\[[CH_3NH_2] = [OH^–] – [H^+] \approx [OH^–] = 5.3 \times 10^{–12}\; M. \nonumber \]
For the concentration of the acid form (methylaminium ion CH 3 NH 3 + ), use the mass balance equation:
\[[CH_3NH_3^+] = C_b – [CH_3NH_2] = 0.01 – 0.0019 =0.0081\; M.\nonumber \]
Mixtures of Acids
Many practical problems relating to environmental and physiological chemistry involve solutions containing more than one acid. In this section, we will restrict ourselves to a much simpler case of two acids, with a view toward showing the general method of approaching such problems by starting with charge- and mass-balance equations and making simplifying assumptions when justified. In general, the hydrogen ions produced by the stronger acid will tend to suppress dissociation of the weaker one, and both will tend to suppress the dissociation of water, thus reducing the sources of H + that must be dealt with.
Consider a mixture of two weak acids HX and HY; their respective nominal concentrations and equilibrium constants are denoted by C x , C y , K x and K y ,
Starting with the charge balance expression
\[ [H^+] = [X^–] + [Y^–] + [OH^–] \label{3-1}\]
We use the equilibrium constants to replace the conjugate base concentrations with expressions of the form
\[ [X^-] = K_x \dfrac{[HX]}{[H^+]} \label{3-2}\]
to yield
\[ [H^+] = \dfrac{[HX]}{K_x} + \dfrac{[HY]}{K_y} + K_w \label{3-3}\]
If neither acid is very strong or very dilute, we can replace equilibrium concentrations with nominal concentrations:
\[ [H^+] \approx \sqrt{C_cK_x + C_yK_y K_w} \label{3-4}\]
Estimate the pH of a solution that is 0.10M in acetic acid (\(K_a = 1.8 \times 10^{–5}\)) and 0.01M in formic acid (\(K_a = 1.7 \times 10^{–4}\)).
Solution
Because K w is negligible compared to the C a K a products, we can simplify \Equation \(ref{3-4}\):
\[[H^+] = \sqrt{1.8 \times 10^{–6} + 1.7 \times 10^{-6}} = 0.0019\nonumber \]
Which corresponds to a pH of \(–\log 0.0019 = 2.7\)
Note that the pH of each acid separately at its specified concentration would be around 2.8. However, if 0.001 M chloroacetic acid ( K a = 0.0014) is used in place of formic acid, the above expression becomes
\[ [H^+] \approx \sqrt{ 1.4 \times 10^{-6} + 1.75 \times 10^{-14}} = 0.00188 \label{3-5}\]
which exceeds the concentration of the stronger acid; because the acetic acid makes a negligible contribution to [H + ] here, the simple approximation given above \Equation \(\ref{3-3}\) is clearly invalid. We now use the mass balance expression for the stronger acid
\[[HX] + [X^–] = C_x \label{3-6}\]
to solve for [X – ] which is combined with the equilibrium constant K x to yield
\[[X^-] = C_x - \dfrac{[H^+][X^]}{K_x} \label{3-7}\]
Solving this for [X – ] gives
\[ [X^-] = \dfrac{C_xK_x}{K_x + [H^+]} \label{3-8}\]
The approximation for the weaker acetic acid (HY) is still valid, so we retain it in the substituted electronegativity expression:
\[ [H^+] \dfrac{C_xK_x}{K_x+[H^+]} + \dfrac{C_yK_y}{[H^+]} \label{3-9}\]
which is a cubic equation that can be solved by approximation.
Several methods have been published for calculating the hydrogen ion concentration in solutions containing an arbitrary number of acids and bases. These generally involve iterative calculations carried out by a computer. See, for example, J. Chem. Education 67(6) 501-503 (1990) and 67(12) 1036-1037 (1990).
Equilibria of Polyprotic Acids
Owing to the large number of species involved, exact solutions of problems involving polyprotic acids can become very complicated. Thus for phosphoric acid H 3 PO 4 , the three "dissociation" steps yield three conjugate bases:
Fortunately, it is usually possible to make simplifying assumptions in most practical applications. In the section that follows, we will show how this is done for the less-complicated case of a diprotic acid. A diprotic acid HA can donate its protons in two steps, yielding first a monoprotonated species HA – and then the completely deprotonated form A 2– .
Since there are five unknowns (the concentrations of the acid, of the two conjugate bases and of H + and OH – ), we need five equations to define the relations between these quantities. These are
- Equilibria
\[[H^+][OH^–] = K_w \label{4-1}\]
\[ K_1 = \dfrac{[H^+][HA^-]}{[H_2A]} \label{4-2}\]
\[ K_1 = \dfrac{[H^+][HA^{2-}]}{[HA^-]} \label{4-3}\]
- Mass balance
\[C_a = [H_2A] + [HA^–] + [A^{2–}] \label{4-4}\]
- Charge balance
\[[H^+] = [OH^–] + [HA^–] + 2 [A^{2–}] \label{4-5}\]
(It takes 2 moles of \(H^+\) to balance the charge of 1 mole of \(A^{2–}\))
Solving these five equations simultaneously for \(K_1\) yields the rather intimidating expression
\[ K_1 = \dfrac{[H^+] \left( [H^+] - [OH^-] \dfrac{2K_2[H^+] - [OH^-]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] - [OH^-] \dfrac{K_2 [H^+] -[OH^-]}{[H^+] + 2K_2} \right)} \label{4-6}\]
which is of little practical use except insofar as it provides the starting point for various simplifying approximations.
If the solution is even slightly acidic, then ([H + ] – [OH – ]) ≈ [H + ] and
\[ K_1 = \dfrac{[H^+] \left( [H^+] \dfrac{2K_2[H^+]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] \dfrac{K_2 [H^+]}{[H^+] + 2K_2} \right)} \label{4-7}\]
For any of the common diprotic acids, \(K_2\) is much smaller than \(K_1\). If the solution is sufficiently acidic that \(K_2 \ll [H^+]\), then a further simplification can be made that removes \(K_2\) from Equation \(\ref{4-7}\); this is the starting point for most practical calculations.
\[ K_1 \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{4-8}\]
Finally, if the solution is sufficiently concentrated and \(K_1\) sufficiently small so that \([H^+] \ll C_a\), then Equation \(\ref{4-8}\) reduces to:
\[ [H^+] \approx \sqrt{K_a C_a}\]
Acid with conjugate base: Buffer solutions
Solutions containing a weak acid together with a salt of the acid are collectively known as buffers . When they are employed to control the pH of a solution (such as in a microbial growth medium), a sodium or potassium salt is commonly used and the concentrations are usually high enough for the Henderson-Hasselbalch equation to yield adequate results.
Exact solution
In this section, we will develop an exact analytical treatment of weak acid-salt solutions, and show how the H – H equation arises as an approximation. A typical buffer system is formed by adding a quantity of strong base such as sodium hydroxide to a solution of a weak acid HA. Alternatively, the same system can be made by combining appropriate amounts of a weak acid and its salt NaA. A system of this kind can be treated in much the same way as a weak acid, but now with the parameter C b in addition to C a .
- Species
Na + , A – , HA, H 2 O, H + , OH –
- Equilibria
\[[H^+][OH^–] = K_w \label{5-1}\]
\[K_a = \dfrac{[H^+][A^-]}{[HA]} \label{5-2}\]
- Mass balance
\[C_a + C-b = [HA] + [A^–] \label{5-3}\]
\[C_b = [Na^+] \label{5-4}\]
- Charge balance
\[[Na^+] + [H^+] = [OH^–] + [A^–] \label{5-5}\]
Substituting Equation \(\ref{5-4}\) into Equation \(\ref{5-5}\) yields an expression for [A – ]:
\[[A^–] = C_b + [H^+] – [OH^–] \label{5-6}\]
Inserting this into Equation \(\ref{5-3}\) and solving for [HA] yields
\[[HA] = C_b + [H^+] – [OH^–] \label{5-7)}\]
Finally, we substitute these last two expressions into the equilibrium constant (Equation \(\ref{5-2}\)):
\[ [H^+] = K_a \dfrac{C_a - [H^+] + [OH^-]}{C_b + [H^+] - [OH^-]} \label{5-8}\]
which becomes cubic in [H + ] when [OH – ] is replaced by ( K w / [H + ]).
\[[H^+]^3 +(C_b +K_a)[H^+]^2 – (K_w + C_aK_a) [H^+] – K_aK_w = 0 \label{5-8a}\]
Approximations
In almost all practical cases it is possible to make simplifying assumptions. Thus if the solution is known to be acidic or alkaline, then the [OH – ] or [H + ] terms in Equation \(\ref{5-8}\) can be neglected. In acidic solutions, for example, Equation \(\ref{5-8}\) becomes
\[ [H^+] = K_a \dfrac{C_a - [H^+]}{C_b + [H^+]} \label{5-9}\]
which can be rearranged into a quadratic in standard polynomial form:
\[ [H^+]^2 + (C_b + C_a) [H^+] – K_a C_a = 0 \label{5-10}\]
If the concentrations C a and C b are sufficiently large, it may be possible to neglect the [H + ] terms entirely, leading to the commonly-seen Henderson-Hasselbalch Approximation .
\[ \color{red} [H^+] \approx K_a \dfrac{C_a}{C_b} \label{5-11}\]
It's important to bear in mind that the Henderson-Hasselbalch Approximation is an "approximation of an approximation" that is generally valid only for combinations of K a and concentrations that fall within the colored portion of this plot.
Most buffer solutions tend to be fairly concentrated, with C a and C b typically around 0.01 - 0.1 M . For more dilute buffers and larger K a 's that bring you near the boundary of the colored area, it is safer to start with Equation \(\ref{5-9}\).
Chlorous acid HClO 2 has a pK a of 1.94. Calculate the pH of a solution made by adding 0.01 M/L of sodium hydroxide to a -.02 M/L solution of chloric acid.
Solution
In the resulting solution, C a = C b = 0.01 M . On the plots shown above, the intersection of the log C a = –2 line with the plot for pK a = 2 falls near the left boundary of the colored area, so we will use the quadratic form \(\ref{5-10}\).
Substitution in Equation \(\ref{5-10}\) yields
\[H^+ + 0.02 H^+ – (10^{–1.9} x 10^{–2}) = 0 \nonumber\]
which yields a positive root 0.0047 = [H + ] that corresponds to pH = 2.3 .
Note: Using the Henderson-Hassalbach Approximateion (Equation \(\ref{5-11}\)) would give pH = pK a = 1.9.
|
libretexts
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2025-03-17T19:53:16.808482
| 2016-02-13T03:30:23 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.07%3A_Exact_Calculations_and_Approximations",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "13.7: Exact Calculations and Approximations",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry
|
14: Thermochemistry
All chemical changes are accompanied by the absorption or release of heat. The intimate connection between matter and energy has been a source of wonder and speculation from the most primitive times; it is no accident that fire was considered one of the four basic elements (along with earth, air, and water) as early as the fifth century BCE. In this set of lessons, we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics , which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes. This will build the groundwork for a more ambitious goal: to predict the direction and extent of change itself.
-
- 14.1: Energy, Heat and Work
- All chemical changes are accompanied by the absorption or release of heat. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes.
-
- 14.2: The First Law of Thermodynamics
- "Energy cannot be created or destroyed"— this fundamental law of nature, more properly known as conservation of energy, is familiar to anyone who has studied science. Under its more formal name of the First Law of Thermodynamics, it governs all aspects of energy in science and engineering .
-
- 14.3: Molecules as Energy Carriers and Converters
- All molecules at temperatures above absolue zero possess thermal energy— the randomized kinetic energy associated with the various motions the molecules as a whole, and also the atoms within them, can undergo. Polyatomic molecules also possess potential energy in the form of chemical bonds. Molecules are thus both vehicles for storing and transporting energy, and the means of converting it from one form to another is accompanied by the uptake or release of heat.
-
- 14.4: Thermochemistry and Calorimetry
- The heat that flows across the boundaries of a system undergoing a change is a fundamental property that characterizes the process. It is easily measured, and if the process is a chemical reaction carried out at constant pressure, it can also be predicted from the difference between the enthalpies of the products and reactants. The quantitative study and measurement of heat and enthalpy changes is known as thermochemistry.
-
- 14.5: Calorimetry
- Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.
-
- 14.6: Applications of Thermochemistry
- Virtually all chemical processes involve the absorption or release of heat, and thus changes in the internal energy of the system. In this section, we survey some of the more common chemistry-related applications of enthalpy and the First Law. While the first two sections relate mainly to chemistry, the remaining ones impact the everyday lives of everyone.
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libretexts
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2025-03-17T19:53:16.874565
| 2016-03-05T20:09:21 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14: Thermochemistry",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.01%3A_Energy_Heat_and_Work
|
14.1: Energy, Heat and Work
Make sure you thoroughly understand the following essential ideas:
- The potential energy of an object relates to its location, but there is one additional requirement that must be satisfied for potential energy be present. Explain and give an example.
- Distinguish between the nature of kinetic energy that is associated with macroscopic bodies and that is found in microscopic objects such as atoms and molecules.
- Describe the meaning and origins of "chemical" energy .
- Define the calorie .
- Heat and work are both expressed in energy units, but they differ from "plain" energy in a fundamental way. Explain.
- ... and state the distinction between heat and work.
All chemical changes are accompanied by the absorption or release of heat. The intimate connection between matter and energy has been a source of wonder and speculation from the most primitive times; it is no accident that fire was considered one of the four basic elements (along with earth, air, and water) as early as the fifth century BCE. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics , which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes. This will build the groundwork for a more ambitious goal: to predict the direction and extent of change itself.
What is Energy?
Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it.
The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning. Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". René Descartes (1596-1650) stated it explicitly:
When God created the world, He "caused some of its parts to push others and to transfer their motions to others..." and thus "He conserves motion". *
In the 17th Century, the great mathematician Gottfried Leibniz (1646-1716) suggested the distinction between vis viva ("live force") and vis mortua ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853).
Kinetic Energy and Potential Energy
Whatever energy may be, there are basically two kinds. Kinetic energy is associated with the motion of an object, and its direct consequences are part of everyone's daily experience; the faster the ball you catch in your hand, and the heavier it is, the more you feel it. Quantitatively, a body with a mass m and moving at a velocity v possesses the kinetic energy mv 2 /2.
A rifle shoots a 4.25 g bullet at a velocity of 965 m s –1 . What is its kinetic energy?
Solution
The only additional information you need here is that
1 J = 1 kg m 2 s –2 :
KE = ½ × (.00425 kg) (965 m s –1 ) 2 = 1980 J
Potential energy is energy a body has by virtue of its location . But there is more: the body must be subject to a "restoring force" of some kind that tends to move it to a location of lower potential energy. Think of an arrow that is subjected to the force from a stretched bowstring; the more tightly the arrow is pulled back against the string, the more potential energy it has. More generally, the restoring force comes from what we call a force field — a gravitational, electrostatic, or magnetic field. We observe the consequences of gravitationa l potential energy all the time, such as when we walk, but seldom give it any thought.
If an object of mass m is raised off the floor to a height h , its potential energy increases by mgh , where g is a proportionality constant known as the acceleration of gravity ; its value at the earth's surface is 9.8 m s –2 .
Find the change in potential energy of a 2.6 kg textbook that falls from the 66-cm height of a table top onto the floor.
Solution
PE = m g h = (2.6 kg)(9.8 m s –2 )(0.66 m) = 16.8 kg m 2 s –2 = 16.8 J
Similarly, the potential energy of a particle having an electric charge q depends on its location in an electrostatic field.
"Chemical energy"
Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules.
"Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy.
In an exothermic chemical reaction, the electrons and nuclei within the reactants undergo rearrangment into products possessing lower energies, and the difference is released to the environment in the form of heat.
Interconversion of potential and kinetic energy
Transitions between potential and kinetic energy are such an intimate part of our daily lives that we hardly give them a thought. It happens in walking as the body moves up and down. Our bodies utilize the chemical energy in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of chemical energy to other forms.
Energy is conserved: it can neither be created nor destroyed. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal.
- When drop a book, its potential energy is transformed into kinetic energy. When it strikes the floor, this transformation is complete. What happens to the energy then? The kinetic energy that at the moment of impact was formerly situated exclusively in the moving book, now becomes shared between the book and the floor, and in the form of randomized thermal motions of the molecular units of which they are made; we can observe this effect as a rise in temperature.
- ← Much of the potential energy of falling water can be captured by a water wheel or other device that transforms the kinetic energy of the exit water into kinetic energy. The output of a hydroelectric power is directly proportional to its height above the level of the generator turbines in the valley below. At this point, the kinetic energy of the exit water is transferred to that of the turbine, most of which (up to 90 percent in the largest installations) is then converted into electrical energy.
- Will the temperature of the water at the bottom of a water fall be greater than that at the top? James Joule himself predicted that it would be. It has been calculated that at Niagra falls, that complete conversion of the potential energy of 1 kg of water at the top into kinetic energy when it hits the plunge pool 58 meters below will result in a temperature increase of about 0.14 C°. (But there are lots of complications. For example, some of the water breaks up into tiny droplets as it falls, and water evaporates from droplets quite rapidly, producing a cooling effect.)
- Chemical energy can also be converted, at least partially, into electrical energy: this is what happens in a battery. If a highly exothermic reaction also produces gaseous products, the latter may expand so rapidly that the result is an explosion — a net conversion of chemical energy into kinetic energy (including sound).
Thermal energy
Kinetic energy is associated with motion, but in two different ways. For a macroscopic object such as a book or a ball, or a parcel of flowing water, it is simply given by ½ mv 2 .
But as we mentioned above, when an object is dropped onto the floor, or when an exothermic chemical reaction heats surrounding matter, the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as thermal energy . We observe the effects of this as a rise in the temperature of the surroundings. The temperature of a body is direct measure of the quantity of thermal energy is contains.
Thermal energy is never completely recoverable
Once kinetic energy is thermalized, only a portion of it can be converted back into potential energy. The remainder simply gets dispersed and diluted into the environment, and is effectively lost.
To summarize, then:
- Potential energy can be converted entirely into kinetic energy..
- Potential energy can also be converted, with varying degrees of efficiency,into electrical energy.
- The kinetic energy of macroscopic objects can be transferred between objects (barring the effects of friction).
- Once kinetic energy becomes thermalized, only a portion of it can be converted back into either potential energy or be concentrated back into the kinetic energy of a macroscopic. This limitation, which has nothing to do with technology but is a fundamental property of nature, is the subject of the second law of thermodynamics .
- A device that is intended to accomplish the partial transformation of thermal energy into organized kinetic energy is known as a heat engine .
Energy scales and units
You might at first think that a book sitting on the table has zero kinetic energy since it is not moving. But if you think about it, the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and the sun itself is moving away from the other stars in the general expansion of the universe. Since these motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity of the book is measured with respect to the table; on this so-called laboratory coordinate system , the kinetic energy of the book can be considered zero.
We do the same thing with potential energy. If the book is on the table, its potential energy with respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then push the book off the table, its potential energy will be negative after it reaches the floor.
Energy is measured in terms of its ability to perform work or to transfer heat. Mechanical work is done when a force f displaces an object by a distance d : w = f × d .
The basic unit of energy is the joule . One joule is the amount of work done when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec 2 , so the basic dimensions of the joule are kg m 2 s –2 . The other two units in wide use. the calorie and the BTU (British thermal unit) are defined in terms of the heating effect on water. Because of the many forms that energy can take, there are a correspondingly large number of units in which it can be expressed, a few of which are summarized below.
|
1 calorie will raise the temperature of 1 g of water by 1 C°. The “dietary” calorie is actually 1 kcal. An average young adult expends about 1800 kcal per day just to stay alive. (you should know this definition) |
1 cal = 4.184 J |
| 1 BTU (British Thermal Unit) will raise the temperature of 1 lb of water by 1F°. | 1 BTU = 1055 J |
| The erg is the c.g.s. unit of energy and a very small one; the work done when a 1-dyne force acts over a distance of 1 cm. |
1 J = 10
7
ergs
|
| The electron-volt is even tinier: 1 e-v is the work required to move a unit electric charge (1 C) through a potential difference of 1 volt. | 1 J = 6.24 × 10 18 e-v |
| The watt is a unit of power, which measures the rate of energy flow in J sec –1 . Thus the watt-hour is a unit of energy. An average human consumes energy at a rate of about 100 watts; the brain alone runs at about 5 watts. |
1 J = 2.78 × 10
–4
watt-hr
|
| The liter-atmosphere is a variant of force-displacement work associated with volume changes in gases. | 1 L-atm = 101.325 J |
| The huge quantities of energy consumed by cities and countries are expressed in quads ; the therm is a similar but smaller unit. | 1 quad = 10 15 Btu = 1.05 × 10 18 J |
| If the object is to obliterate cities or countries with nuclear weapons, the energy unit of choice is the ton of TNT equivalent . |
1 ton of TNT = 4.184 GJ
(by definition) |
| In terms of fossil fuels , we have barrel-of-oil equivalent, cubic-meter-of-natural gas equivalent, and ton-of-coal equivalent. |
1 bboe = 6.1 GJ
|
Heat and work
Heat and work are both measured in energy units, so they must both represent energy. How do they differ from each other, and from just plain “energy” itself? In our daily language, we often say that "this object contains a lot of heat", but this kind of talk is a no-no in thermodynamics! It's ok to say that the object is "hot", meaning that its temperature is high. The term "heat" has a special meaning in thermodynamics: it is a process in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a different temperature than its surroundings (the rest of the world).
Heat and work are processes and cannot be stored
Thermal energy can only flow from a higher temperature to a lower temperature. It is this flow that constitutes "heat". Use of the term "flow" of heat recalls the 18th-century notion that heat is an actual substance called “caloric” that could flow like a liquid.
Heat is transferred by conduction or radiation
Transfer of thermal energy can be accomplished by bringing two bodies into physical contact (the kettle on top of the stove, or through an electric heating element inside the kettle). Another mechanism of thermal energy transfer is by radiation; a hot object will convey energy to any body in sight of it via electromagnetic radiation in the infrared part of the spectrum. In many cases, a combination of modes will be active. Thus when you place a can of beer in the refrigerator, both processes are operative: the can radiates heat to the cold surfaces around it, and absorbs it by direct conduction from the ambient air.
So what is work ?
Work refers to the transfer of energy some means that does not depend on temperature difference.
Work, like energy, can take various forms, the most familiar being mechanical and electrical. Mechanical work arises when an object moves a distance Δ x against an opposing force f :
\[w = f Δ x N^{-m}\]
with \(1\, N^{-m}\) = 1\, J.\)
Electrical work is done when a body having a charge q moves through a potential difference Δ V .
Work, like heat, exists only when energy is being transferred.When two bodies are placed in thermal contact and energy flows from the warmer body to the cooler one,we call the process “heat”. A transfer of energy to or from a system by any means other than heat is called “work”.
Interconvertability of heat and work
Work can be completely converted into heat (by friction, for example), but heat can only be partially converted to work. Conversion of heat into work is accomplished by means of a heat engine , the most common example of which is an ordinary gasoline engine. The science of thermodynamics developed out of the need to understand the limitations of steam-driven heat engines at the beginning of the Industrial Age. A fundamental law of Nature, the Second Law of Thermodynamics, states that the complete conversion of heat into work is impossible. Something to think about when you purchase fuel for your car!
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libretexts
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2025-03-17T19:53:16.968625
| 2016-03-05T20:11:38 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.01%3A_Energy_Heat_and_Work",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.1: Energy, Heat and Work",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.02%3A_The_First_Law_of_Thermodynamics
|
14.2: The First Law of Thermodynamics
Make sure you thoroughly understand the following essential concepts:
- Define: system , surroundings , state properties , change of state (how the latter is calculated).
- Write out the equation that defines the First Law , and explain its physical meaning.
- Describe the meaning of a pathway in thermodynamics, and how it relates to state functions .
- Comment on the meaning of pressure-volume work : how is it calculated? What kinds of chemical reactions involve P-V work?
- Define isothermal and adiabatic processes, and give examples of each.
- Using the expansion of a gas as an example, state the fundamental distinction between reversible and irreversible changes in terms of the system + surroundings.
- Explain how enthalpy change relates to internal energy , and how it can be observed experimentally.
- Define heat capacity ; explain its physical significance, and why there is a difference between C p and C v .
"Energy cannot be created or destroyed" — this fundamental law of nature, more properly known as conservation of energy , is familiar to anyone who has studied science. Under its more formal name of the First Law of Thermodynamics , it governs all aspects of energy in science and engineering applications. It's special importance in Chemistry arises from the fact that virtually all chemical reactions are accompanied by the uptake or release of energy. One of the interesting things about thermodynamics is that although it deals with matter, it makes no assumptions about the microscopic nature of that matter. Thermodynamics deals with matter in a macroscopic sense; it would be valid even if the atomic theory of matter were wrong. This is an important quality, because it means that reasoning based on thermodynamics is unlikely to require alteration as new facts about atomic structure and atomic interactions come to light.
A Thermodynamic view of the world
In thermodynamics, we must be very precise in our use of certain words. The two most important of these are system and surroundings . A thermodynamic system is that part of the world to which we are directing our attention. Everything that is not a part of the system constitutes the surroundings . The system and surroundings are separated by a boundary .
If our system is one mole of a gas in a container, then the boundary is simply the inner wall of the container itself. The boundary need not be a physical barrier; for example, if our system is a factory or a forest, then the boundary can be wherever we wish to define it. We can even focus our attention on the dissolved ions in an aqueous solution of a salt, leaving the water molecules as part of the surroundings. The single property that the boundary must have is that it be clearly defined, so we can unambiguously say whether a given part of the world is in our system or in the surroundings.
If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is open. A closed system may still exchange energy with the surroundings unless the system is an isolated one, in which case neither matter nor energy can pass across the boundary. The tea in a closed Thermos bottle approximates a closed system over a short time interval.
The properties of a system are those quantities such as the pressure, volume, temperature, and its composition, which are in principle measurable and capable of assuming definite values. There are of course many properties other than those mentioned above; the density and thermal conductivity are two examples. However, the pressure, volume, and temperature have special significance because they determine the values of all the other properties; they are therefore known as state properties because if their values are known then the system is in a definite state .
In dealing with thermodynamics, we must be able to unambiguously define the change in the state of a system when it undergoes some process. This is done by specifying changes in the values of the different state properties using the symbol Δ ( delta ) as illustrated here for a change in the volume:
\[ΔV = V_{final} – V_{initial} \label{1-1}\]
We can compute similar delta-values for changes in P, V, n i (the number of moles of component i ), and the other state properties we will meet later.
Internal energy is simply the totality of all forms of kinetic and potential energy of the system. Thermodynamics makes no distinction between these two forms of energy and it does not assume the existence of atoms and molecules. But since we are studying thermodynamics in the context of chemistry, we can allow ourselves to depart from “pure” thermodynamics enough to point out that the internal energy is the sum of the kinetic energy of motion of the molecules, and the potential energy represented by the chemical bonds between the atoms and any other intermolecular forces that may be operative.
How can we know how much internal energy a system possesses? The answer is that we cannot, at least not on an absolute basis; all scales of energy are arbitrary. The best we can do is measure changes in energy. However, we are perfectly free to define zero energy as the energy of the system in some arbitrary reference state , and then say that the internal energy of the system in any other state is the difference between the energies of the system in these two different states.
The First Law of Thermodynamics
This law is one of the most fundamental principles of the physical world. Also known as the Law of Conservation of Energy , it states that energy can not be created or destroyed; it can only be redistributed or changed from one form to another. A way of expressing this law that is generally more useful in Chemistry is that any change in the internal energy of a system is given by the sum of the heat q that flows across its boundaries and the work w done on the system by the surroundings.
\[ΔU = q + w \label{2-1}\]
This says that there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a system. Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the world outside the system- in other words, energy cannot be created or destroyed.
There is an important sign convention for heat and work that you are expected to know. If heat flows into a system or the surroundings to do work on it, the internal energy increases and the sign of q or w is positive. Conversely, heat flow out of the system or work done by the system will be at the expense of the internal energy, and will therefore be negative. (Note that this is the opposite of the sign convention that was commonly used in much of the pre-1970 literature.) The full significance of Equation \(\ref{2-1}\) cannot be grasped without understanding that U is a state function . This means that a given change in internal energy Δ U can follow an infinite variety of pathways corresponding to all the possible combinations of q and w that can add up to a given value of Δ U .
As a simple example of how this principle can simplify our understanding of change, consider two identical containers of water initially at the same temperature. We place a flame under one until its temperature has risen by 1°C. The water in the other container is stirred vigorously until its temperature has increased by the same amount. There is now no physical test by which you could determine which sample of water was warmed by performing work on it, by allowing heat to flow into it, or by some combination of the two processes. In other words, there is no basis for saying that one sample of water now contains more “work”, and the other more “heat”. The only thing we can know for certain is that both samples have undergone identical increases in internal energy, and we can determine the value of simply by measuring the increase in the temperature of the water.
Pressure-volume work
The kind of work most frequently associated with chemical change occurs when the volume of the system changes owing to the disappearance or formation of gaseous substances. This is sometimes called expansion work or PV -work, and it can most easily be understood by reference to the simplest form of matter we can deal with, the hypothetical ideal gas .
The figure shows a quantity of gas confined in a cylinder by means of a moveable piston. Weights placed on top of the piston exert a force f over the cross-section area A , producing a pressure P = f / A which is exactly countered by the pressure of the gas, so that the piston remains stationary. Now suppose that we heat the gas slightly; according to Charles’ law, this will cause the gas to expand, so the piston will be forced upward by a distance Δ x . Since this motion is opposed by the force f , a quantity of work f Δ x will be done by the gas on the piston. By convention, work done by the system (in this case, the gas) on the surroundings is negative, so the work is given by
\[w = – f Δx \label{3-1}\]
When dealing with a gas, it is convenient to think in terms of the more relevant quantities pressure and volume rather than force and distance. We can accomplish this by multiplying the second term by A/A which of course leaves it unchanged:
\[ w = -f Δx \dfrac{A}{A} \label{3-2}\]
By grouping the terms differently, but still not changing anything, we obtain
\[ w = -\dfrac{f}{A} Δx A \label{3-3}\]
Since pressure is force per unit area and the product of the length A and the area has the dimensions of volume, this expression becomes
\[w = –P ΔV \label{3-4}\]
It is important to note that although \(P\) and \(V\) are state functions , the work is not (that's why we denote it by a lower-case w .) As is shown farther below, the quantity of work done will depend on whether the same net volume change is realized in a single step (by setting the external pressure to the final pressure P ), or in multiple stages by adjusting the restraining pressure on the gas to successively smaller values approaching the final value of \(P\).
Find the amount of work done on the surroundings when 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature to 10 liters by:
- reducing the external pressure to 1 atm in a single step,
- reducing \(P\) first to 5 atm, and then to 1 atm,
- allowing the gas to expand into an evacuated space so its total volume is 10 liters.
Solution
First, note th at \(ΔV\), which is a state function, is the same for each path:
\[V_2 = (10/1) × (1 L) = 10 L\]
so
\[ΔV = 9\; L\]
For path (a)
w = –(1 atm)× (9 L) = –9 L-atm .
For path (b) , the work is calculated for each stage separately:
w = –(5 atm) × (2–1 L) – (1 atm) × (10–2 L) = –13 L-atm
For path (c) the process would be carried out by removing all weights from the piston in Figure \(\PageIndex{1}\) so th at the gas expands to 10 L against zero external pressure. In this case w = (0 atm) × 9 L = 0 ; that is, no work is done because there is no force to oppose the expansion.
Adiabatic and Isothermal Processes
When a gas expands, it does work on the surroundings; compression of a gas to a smaller volume similarly requires that the surroundings perform work on the gas. If the gas is thermally isolated from the surroundings, then the process is said to occur adiabatically . In an adiabatic change, q = 0, so the First Law becomes Δ U = 0 + w . Since the temperature of the gas changes with its internal energy, it follows that adiabatic compression of a gas will cause it to warm up, while adiabatic expansion will result in cooling.
In contrast to this, consider a gas that is allowed to slowly escape from a container immersed in a constant-temperature bath. As the gas expands, it does work on the surroundings and therefore tends to cool, but the thermal gradient that results causes heat to pass into the gas from the surroundings to exactly compensate for this change. This is called an isothermal expansion. In an isothermal process the internal energy remains constant and we can write the First Law (Equation \(Ref{2-1}\)) as
\[0 = q + w\]
or
\[q = –w\]
This illustrates that the heat flow and work done exactly balance each other.
Because no thermal insulation is perfect, truly adiabatic processes do not occur. However, heat flow does take time, so a compression or expansion that occurs more rapidly than thermal equilibration can be considered adiabatic for practical purposes. If you have ever used a hand pump to inflate a bicycle tire, you may have noticed that the bottom of the pump barrel can get quite warm. Although a small part of this warming may be due to friction, it is mostly a result of the work you (the surroundings) are doing on the system (the gas.)
Adiabatic expansion and contractions are especially important in understanding the behavior of the atmosphere . Although we commonly think of the atmosphere as homogeneous, it is really not, due largely to uneven heating and cooling over localized areas. Because mixing and heat transfer between adjoining parcels of air does not occur rapidly, many common atmospheric phenomena can be considered at least quasi-adiabatic.
Reversible processes
From Example \(\PageIndex{1}\) we see that when a gas expands into a vacuum (\(P_{external} = 0\) the work done is zero. This is the minimum work the gas can do; what is the maximum work the gas can perform on the surroundings? To answer this, notice that more work is done when the process is carried out in two stages than in one stage; a simple calculation will show that even more work can be obtained by increasing the number of stages— that is, by allowing the gas to expand against a series of successively lower external pressures. In order to extract the maximum possible work from the process, the expansion would have to be carried out in an infinite sequence of infinitesimal steps. Each step yields an increment of work P Δ V which can be expressed as ( RT/V ) dV and integrated:
\[ \begin{align} w &= \int_{V_1}^{V_2} \dfrac{RT}{V} dv \\[4pt] &= RT \ln \dfrac{V_2}{V_1} \label{3-5} \end{align}\]
Although such a path (which corresponds to what is called a reversible process ) cannot be realized in practice, it can be approximated as closely as desired.
Even though no real process can take place reversibly (it would take an infinitely long time!), reversible processes play an essential role in thermodynamics. The main reason for this is that q rev and w rev are state functions which are important and are easily calculated. Moreover, many real processes take place sufficiently gradually that they can be treated as approximately reversible processes for easier calculation.
Each expansion-compression cycle leaves the gas unchanged, but in all but the one in the bottom row, the surroundings are forever altered, having expended more work in compressing the gas than was performed on it when the gas expanded. Only when the processes are carried out in an infinite number of steps will the system and the surroundings be restored to their initial states— this is the meaning of thermodynamic reversibility.
Heat changes at constant pressure: the Enthalpy
For a chemical reaction that performs no work on the surroundings, the heat absorbed is the same as the change in internal energy: q = Δ U . But many chemical processes do involve work in one form or another:
- If the total volume of the reaction products exceeds that of the reactants, then the process performs work on the surroundings in the amount P Δ V , in which P is the pressure exerted by the surroundings (usually the atmosphere) on the system.
- A reaction that drives an electrical current through an external circuit performs electrical work on the surroundings.
For an isothermal process, pressure-volume work affects the heat q
We will consider only pressure-volume work in this lesson. If the process takes place at a constant pressure, then the work is given by P Δ V and the change in internal energy will be
\[ΔU = q – PΔV \label{4-1}\]
Bear in mind why \(q\) is so important: the heat flow into or out of the system is directly measurable. Δ U , being "internal" to the system, is not directly observable. Thus the amount of heat that passes between the system and the surroundings is given by
\[q = ΔU + PΔV \label{4-2}\]
This means that if an exothermic reaction is accompanied by a net increase in volume under conditons of constant pressure, some heat additional to Δ U must be absorbed in order to supply the energy expended as work done on the surroundings if the temperature is to remain unchanged (isothermal process.)
For most practical purposes, changes in the volume of the system are only significant if the reaction is accompanied by a difference in the moles of gaseous reactants and products.
For example, in the reaction H 2 (g) + O 2 (g) → ½H 2 O (l) , the total volume of the system decreases from that correponding to 2 moles of gaseous reactants to 0.5 mol of liquid water which occupies only 9 mL — a volume so small in comparison to that of the reactants that it can be neglected without significant error. So all we are really concerned with is the difference in the number of moles of gas Δ n g :
Δ n g = (0 – 2) mol = –2 mol
This corresponds to a net contraction (negative expansion) of the system, meaning that the surroundings perform work on the system.
The molar volume of an ideal gas at 25° C and 1 atm is
(298/273) × (22.4 L mol –1 ) = 24.5 L mol –1
Remember the sign convention: a flow of heat or performance of work that supplies energy is positive; if it consumes energy, it is negative. Thus work performed by the surroundings diminishes the energy of the surroundings ( w surr < 0) and increases the energy of the system ( w sys > 0).
and the work done (by the surroundings on the system) is
(1 atm) (–2 mol)(24.5 L mol –1 ) = –49.0 L-atm.
Using the conversion factor 1 J = 101.3 J, and bearing in mind that work performed on the system supplies energy to the system, the work associated solely with the volume change of the system increases its energy by
(101.3 J/L-atm)(–49.0 L-atm) = 4964 J = 4.06 kJ
The above reaction H 2 (g) + ½ O 2 (g) → H 2 O (l) is carried out at a constant pressure of 1 atm and a constant temperature of 25° C. What quantity of heat q will cross the system boundary (and in which direction?) For this reaction, the change in internal energy is Δ U = –281.8 kJ/mol.
Solution
281.8 kJ of heat. The work performed by the surroundings supplies an additional energy of P Δ V = 4.06 kJ to the system. The total energy change of the system is thusq = (–281.8 + 4.06) k J = –285.8 kJ
(Eq. 4.2 above.) In order to maintain the constant 25° temperature, an equivalent quantity of heat must pass from the system to the surroundings.
This terminology can be somewhat misleading unless you bear in mind that the conditions Δ P and Δ T refer to the differences between the inital and final states of the system — that is, before and after the reaction. During the time the reaction is in progress, the temperature of the mixture will rise or fall, depending on whether the process is exothermic or endothermic. But because Δ T is a state function, its value is independent of what happens "in between" the initial state (reactants) and final state (products). The same is true of Δ V.
Enthalpy hides work and saves it too!
Because most chemical changes we deal with take place at constant pressure, it would be tedious to have to explicitly deal with the pressure-volume work details that were described above. Fortunately, chemists have found a way around this; they have simply defined a new state function that incorporates and thus hides within itself any terms relating to incidental kinds of work ( P-V , electrical, etc.). Since both Δ P and Δ V in Equation \(\ref{4-2}\) are state functions, then \(q_P\), the heat that is absorbed or released when a process takes place at constant pressure, must also be a state function and is known as the enthalpy change Δ H
\[ΔH ≡ q_P = ΔU + PΔV \label{4-3}\]
Since most processes that occur in the laboratory, on the surface of the earth, and in organisms do so under a constant pressure of one atmosphere, Equation \(\ref{4-3}\) is the form of the First Law that is of greatest interest to most of us most of the time.
Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find Δ U for the system when one mole of HCl dissolves in water under these conditions.
Solution
In this process the volume of liquid remains practically unchanged, so Δ V = –24.5 L. The work done is
\[ \begin{align*} w &= –PΔV \\[4pt] &= –(1\; atm)(–24.5\; L) \\[4pt] &= 24.6 \;L-atm \end{align*}\]
(The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = 101.33 J mol –1 and substituting in Equation \ref{4-3} we obtain
\[ \begin{align*} ΔU &= q +PΔV \\[4pt] = –(75,300\; J) + [101.33\; J/L-atm) (24.5\; L-atm)] \\[4pt] &= –72.82\; kJ \end{align*}\]
In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of Δ U to –72.82 k J.
The Heat Capacity
For systems in which no change in composition (chemical reaction) occurs, things are even simpler: to a very good approximation, the enthalpy depends only on the temperature. This means that the temperature of such a system can serve as a direct measure of its enthalpy. The functional relation between the internal energy and the temperature is given by the heat capacity measured at constant pressure:
\[ c_p =\dfrac{dH}{dT} \label{5-1}\]
(or Δ H /Δ T over a finite duration) An analogous quantity relates the heat capacity at constant volume to the internal energy:
\[ c_v =\dfrac{dU}{dT} \label{5-2}\]
The greater the heat capacity of a substance, the smaller will be the effect of a given absorption or loss of heat on its temperature. Heat capacity can be expressed in joules or calories per mole per degree (molar heat capacity), or in joules or calories per gram per degree; the latter is called the specific heat capacity or just the specific heat. The difference between \(c_p\) and \(c_v\) is of importance only when the volume of the system changes significantly— that is, when different numbers of moles of gases appear on either side of the chemical equation. For reactions involving only liquids and solids, \(c_p\) and \(c_v\) are for all practical purposes identical.
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libretexts
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2025-03-17T19:53:17.075409
| 2016-03-05T20:12:20 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.02%3A_The_First_Law_of_Thermodynamics",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.2: The First Law of Thermodynamics",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.03%3A_Molecules_as_Energy_Carriers_and_Converters
|
14.3: Molecules as Energy Carriers and Converters
Make sure you thoroughly understand the following essential concepts:
- Describe the sources of potential energy and kinetic energy contained in a molecule.
- Describe the nature of "thermal" energy , and how it relates to other forms of kinetic energy and to temperature .
- Explain why the simplest molecules (monatomic and diatomic) have smaller heat capacities than polyatomic molecules.
- Similarly, explain why the dependence of heat capacity on the temperature is different for monatomic and polyatomic molecules.
All molecules at temperatures above absolue zero possess thermal energy — the randomized kinetic energy associated with the various motions the molecules as a whole, and also the atoms within them, can undergo. Polyatomic molecules also possess potential energy in the form of chemical bonds. Molecules are thus both vehicles for storing and transporting energy , and the means of converting it from one form to another when the formation, breaking, or rearrangement of the chemical bonds within them is accompanied by the uptake or release of heat .
Chemical Energy: Potential + Kinetic
When you buy a liter of gasoline for your car, a cubic meter of natural gas to heat your home, or a small battery for your flashlight, you are purchasing energy in a chemical form. In each case, some kind of a chemical change will have to occur before this energy can be released and utilized: the fuel must be burned in the presence of oxygen, or the two poles of the battery must be connected through an external circuit (thereby initiating a chemical reaction inside the battery.) And eventually, when each of these reactions is complete, our source of energy will be exhausted; the fuel will be used up, or the battery will be “dead”.
Where did the energy go? It could have gone to raise the temperature of the products, to perform work in expanding any gaseous products or to push electrons through a circuit. The remainder will reside in the chemical potential energy associated with the products of the reaction.
Chemical substances are made of atoms, or more generally, of positively charged nuclei surrounded by negatively charged electrons. A molecule such as dihydrogen, H 2 , is held together by electrostatic attractions mediated by the electrons shared between the two nuclei. The total potential energy of the molecule is the sum of the repulsions between like charges and the attractions between electrons and nuclei:
\[PE_{total} = PE_{electron-electron} + PE_{nucleus-nucleus} + PE_{nucleus-electron} \label{1-1}\]
In other words, the potential energy of a molecule depends on the time-averaged relative locations of its constituent nuclei and electrons . This dependence is expressed by the familiar potential energy curve which serves as an important description of the chemical bond between two atoms.
Translation refers to movement of an object as a complete unit. Translational motions of molecules in solids or liquids are restricted to very short distances comparable to the dimensions of the molecules themselves, whereas in gases the molecules typically travel hundreds of molecular diameters between collisions.
In gaseous hydrogen, for example, the molecules will be moving freely from one location to another; this is called translational motion , and the molecules therefore possess translational kinetic energy
\[KE_{trans} = \dfrac{mv^2}{2}\]
in which \(v\) stands for the average velocity of the molecules; you may recall from your study of gases that \(v\), and therefore \(KE_{trans}\), depends on the temperature.
In addition to translation, molecules composed of two or more atoms can possess other kinds of motion. Because a chemical bond acts as a kind of spring, the two atoms in H 2 will have a natural vibrational frequency . In more complicated molecules, many different modes of vibration become possible, and these all contribute a vibrational term KE vib to the total kinetic energy. Finally, a molecule can undergo rotational motions which give rise to a third term \(KE_{rot}\). Thus the total kinetic energy of a molecule is the sum
\[KE_{total} = KE_{trans} + KE_{vib} + KE_{rot} \label{1-2}\]
The total energy of the molecule (its internal energy U ) is just the sum
\[U = KE_{total} + PE_{total} \label{1-3}\]
Although this formula is simple and straightforward, it cannot take us very far in understanding and predicting the behavior of even one molecule, let alone a large number of them. The reason, of course, is the chaotic and unpredictable nature of molecular motion. Fortunately, the behavior of a large collection of molecules, like that of a large population of people, can be described by statistical methods.
How Molecules Store Thermal Energy
As noted above, the heat capacity of a substance is a measure of how sensitively its temperature is affected by a change in heat content; the greater the heat capacity, the less effect a given flow of heat q will have on the temperature.
Thermal energy is randomized kinetic energy
We also pointed out that temperature is a measure of the average kinetic energy due to translational motions of molecules. If vibrational or rotational motions are also active, these will also accept thermal energy and reduce the amount that goes into translational motions. Because the temperature depends only on the latter, the effect of the other kinds of motions will be to reduce the dependence of the internal energy on the temperature, thus raising the heat capacity of a substance.
| monatomic | diatomic | triatomic | |||
|---|---|---|---|---|---|
| He | 20.5 | CO | 29.3 | H 2 O | 33.5 |
| Ne | 20.5 | N 2 | 29.5 | D 2 O | 34.3 |
| Ar | 20.5 | F 2 | 31.4 | CO 2 | 37.2 |
| Kr | 20.5 | Cl 2 | 33.9 | CS 2 | 45.6 |
Whereas monatomic molecules can only possess translational thermal energy, two additional kinds of motions become possible in polyatomic molecules.
A linear molecule has an axis that defines two perpendicular directions in which rotations can occur; each represents an additional degree of freedom, so the two together contribute a total of ½ R to the heat capacity.
Vibrational and rotational motions are not possible for monatomic species such as the noble gas elements, so these substances have the lowest heat capacities. Moreover, as you can see in the leftmost column of Table 1, their heat capacities are all the same. This reflects the fact that translational motions are the same for all particles; all such motions can be resolved into three directions in space, each contributing one degree of freedom to the molecule and ½ R to its heat capacity. ( R is the gas constant, 8.314 J K – 1 ).
Think of a "degree of freedom" as a kind of motion that adds kinetic energy to a molecule.
For a non-linear molecule, rotations are possible along all three directions of space, so these molecules have a rotational heat capacity of 3/2 R . Finally, the individual atoms within a molecule can move relative to each other, producing a vibrational motion. A molecule consisting of N atoms can vibrate in 3 N –6 different ways or modes. Each vibrational mode contributes R (rather than ½ R ) to the total heat capacity. (These results come from advanced mechanics and will not be proven here.)
| type of motion → |
translation
|
rotation
|
vibration
|
|---|---|---|---|
| monatomic | 3/2 R | 0 | 0 |
| diatomic | 3/2 R | R | R |
| polyatomic | 3/2 R | 3/2 R | 3 N – 6 |
| separation between adjacent levels, (kJ mol –1 ) | 6.0 × 10 –17 J (O 2 ) | 373 J (HCl) | 373 J (HCl) |
Monatomic molecules have the smallest heat capacities
Now we are in a position to understand why more complicated molecules have higher heat capacities. The total kinetic energy of a molecule is the sum of those due to the various kinds of motions:
\[KE_{total} = KE_{trans} + KE_{rot} + KE_{vib} \label{2-1}\]
When a monatomic gas absorbs heat, all of the energy ends up in translational motion, and thus goes to increase its temperature. In a polyatomic gas, by contrast, the absorbed energy is partitioned among the other kinds of motions; since only the translational motions contribute to the temperature, the temperature rise is smaller, and thus the heat capacity is larger. There is one very significant complication, however: classical mechanics predicts that the energy is always partitioned equally between all degrees of freedom. Experiments, however, show that this is observed only at quite high temperatures. The reason is that these motions are all quantized . This means that only certain increments of energy are possible for each mode of motion, and unless a certain minimum amount of energy is available, a given mode will not be active at all and will contribute nothing to the heat capacity.
Translational energy levels are effectively a continuum
The shading indicates the average thermal energy available at 300 K. Only those levels within this range will have significant occupancy as indicated by the thickness of the lines in the two rightmost columns. At 300 K, only the lowest vibrational state and the first few rotational states will be active. Most of the thermal energy will be confined to the translational levels whose minute spacing (10 –17 J) causes them to appear as a continuum.
Heat capacity of dihydrogen as a function of temperature. This plot is typical of those for other polyatomic molecules, and shows the practical consequences of the spacings of the various forms of thermal energy. Thus translational motions are available at virtually all temperatures, but contributions to heat acapacity by rotational or vibrational motions can only develop at temperatures sufficiently large to excite these motions.
It turns out that translational energy levels are spaced so closely that they these motions are active almost down to absolute zero, so all gases possess a heat capacity of at least 3/2 R at all temperatures . Rotational motions do not get started until intermediate temperatures, typically 300-500K, so within this range heat capacities begin to increase with temperature. Finally, at very high temperatures, vibrations begin to make a significant contribution to the heat capacity
The strong intermolecular forces of liquids and many solids allow heat to be channeled into vibrational motions involving more than a single molecule, further increasing heat capacities. One of the well known “anomalous” properties of liquid water is its high heat capacity (75 J mol –1 K –1 ) due to intermolecular hydrogen bonding, which is directly responsible for the moderating influence of large bodies of water on coastal climates.
Heat capacities of metals
Metallic solids are a rather special case. In metals, the atoms oscillate about their equilibrium positions in a rather uniform way which is essentially the same for all metals, so they should all have about the same heat capacity. That this is indeed the case is embodied in the Law of Dulong and Petit . In the 19th century these workers discovered that the molar heat capacities of all the metallic elements they studied were around to 25 J mol –1 K –1 , which is close to what classical physics predicts for crystalline metals. This observation played an important role in characterizing new elements, for it provided a means of estimating their molar masses by a simple heat capacity measurement.
Standard enthalpy change
Under the special conditions in which the pressure is 1 atm and the reactants and products are at a temperature of 298 K, Δ H becomes the standard enthalpy change Δ H° . Chemists usually refer to the "enthalpy change of a reaction" as simply the "enthalpy of reaction", or even more simply as the " heat of reaction ". But students are allowed to employ this latter shortcut only if they are able to prove that they know the meaning of enthalpy.
Since most changes that occur in the laboratory, on the surface of the earth, and in organisms are subjected to an approximately constant pressure of "one atmosphere" and reasonably salubrious temperatures, most reaction heats quoted in the literature refer to Δ H° . But the high pressures and extreme temperatures frequently encountered by chemical engineers, geochemists, and practicioners of chemical oceanography, often preclude the convenience of the "standard" values.
The rearrangement of atoms that occurs in a chemical reaction is virtually always accompanied by the liberation or absorption of heat. If the purpose of the reaction is to serve as a source of heat, such as in the combustion of a fuel, then these heat effects are of direct and obvious interest. We will soon see, however, that a study of the energetics of chemical reactions in general can lead us to a deeper understanding of chemical equilibrium and the basis of chemical change itself.
In chemical thermodynamics, we define the zero of the enthalpy and internal energy as that of the elements as they exist in their stable forms at 298K and 1 atm pressure. Thus the enthalpies H of Xe (g) , O 2 (g) and C (diamond) are all zero, as are those of H 2 and Cl 2 in the reaction
\[H_{2(g)} + Cl_{2(g)} → 2 HCl_{(g)}\]
The enthalpy of two moles of HCl is smaller than that of the reactants, so the difference is released as heat. Such a reaction is said to be exothermic . The reverse of this reaction would absorb the same quantity of heat from the surroundings and be endothermic . In comparing the internal energies and enthalpies of different substances as we have been doing here, it is important to compare equal numbers of moles, because energy is an extensive property of matter. However, heats of reaction are commonly expressed on a molar basis and treated as intensive properties .
Changes in enthalpy and internal energy
We can characterize any chemical reaction by the change in the internal energy or enthalpy:
\[ΔH = H_{final} – H_{initial} \label{3-1}\]
The significance of this can hardly be exaggerated because Δ H , being a state function, is entirely independent of how the system gets from the initial state to the final state. In other words, the value of Δ H or Δ U for a given change in state is independent of the pathway of the process.
Consider, for example, the oxidation of a lump of sugar to carbon dioxide and water:
\[\ce{C12H22O11 + 12 O2(g) → 12 CO2(g) + 11 H2O(l)}\]
This process can be carried out in many ways, for example by burning the sugar in air, or by eating the sugar and letting your body carry out the oxidation. Although the mechanisms of the transformation are completely different for these two pathways, the overall change in the enthalpy of the system (the atoms of carbon, hydrogen and oxygen that were originally in the sugar) will be identical, and can be calculated simply by looking up the standard enthalpies of the reactants and products and calculating the difference
\[ΔH = [12 \times H(\ce{CO2})] + [11 \times H(\ce{H2O})] – H(\ce{C12H22O11}) = –5606\, kJ\]
The same quantity of heat is released whether the sugar is burnt in the air or oxidized in a series of enzyme-catalyzed steps in your body.
Enthalpy increases with temperature
When the temperature of a substance is raised, it absorbs heat. The enthalpy of a system increases with the temperature by the amount \(ΔH = C_p ΔT\). The defining relation
\[ΔH = ΔU + P ΔV\]
tells us that this change is dominated by the internal energy, subject to a slight correction for the work associated with volume change. Heating a substance causes it to expand, making Δ V positive and causing the enthalpy to increase slightly more than the internal energy. Physically, what this means is that if the temperature is increased while holding the pressure constant, some extra energy must be expended to push back the external atmosphere while the system expands. The difference between the dependence of U and H on temperature is only really significant for gases, since the coefficients of thermal expansion of liquids and solids are very small.
Enthalpy of phase changes
A plot of the enthalpy of a system as a function of its temperature is called an enthalpy diagram . The slope of the line is given by C p . The enthalpy diagram of a pure substance such as water shows that this plot is not uniform, but is interrupted by sharp breaks at which the value of C p is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature the water boiling in a kettle can never exceed 100 until all the liquid has evaporated, at which point the temperature of the steam will rise as more heat flows into the system.
A plot of the enthalpy of carbon tetrachloride as a function of its temperature provides a concise view of its thermal behavior. The slope of the line is given by the heat capacity C p . All H -vs.- C plots show sharp breaks at which the value of C p is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change ; you already know that the temperature of the water boiling in a kettle can never exceed 100°C until all the liquid has evaporated, at which point the temperature (of the steam) will rise as more heat flows into the system.
The lowest-temperature discontinuity on the CCl 4 diagram corresponds to a solid-solid phase transition associated with a rearrangement of molecules in the crystalline solid.
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libretexts
|
2025-03-17T19:53:17.172946
| 2016-03-05T20:13:00 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.03%3A_Molecules_as_Energy_Carriers_and_Converters",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.3: Molecules as Energy Carriers and Converters",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.04%3A_Thermochemistry_and_Calorimetry
|
14.4: Thermochemistry and Calorimetry
Make sure you thoroughly understand the following essential concepts:
- Write a balanced thermochemical equation that expresses the enthalpy change for a given chemical reaction. Be sure to correctly specificy the physical state (and, if necessary, the concentration ) of each component.
- Write an equation that defines the standard enthalpy of formation of a given chemical species.
- Define the standard enthalpy change of a chemical reaction. Use a table of standard enthalpies of formation to evaluate ΔH f °.
- State the principle on which Hess' law depends, and explain why this law is so useful.
- Describe a simple calorimeter and explain how it is employed and how its heat capacity is determined.
The heat that flows across the boundaries of a system undergoing a change is a fundamental property that characterizes the process. It is easily measured, and if the process is a chemical reaction carried out at constant pressure, it can also be predicted from the difference between the enthalpies of the products and reactants. The quantitative study and measurement of heat and enthalpy changes is known as thermochemistry .
Thermochemical Equations and Standard States
In order to define the thermochemical properties of a process, it is first necessary to write a thermochemical equation that defines the actual change taking place, both in terms of the formulas of the substances involved and their physical states (temperature, pressure, and whether solid, liquid, or gaseous.
To take a very simple example, here is the complete thermochemical equation for the vaporization of water at its normal boiling point:
\[\ce{H2O(\ell,\, 373 \,K, \,1 \,atm) → H2O(g, \,373 \,K, \,1 \,atm)} \,\,\,ΔH = 40.7\, kJ\, mol^{-1} \nonumber\]
The quantity 40.7 is known as the enthalpy of vaporization (often referred to as “heat of vaporization”) of liquid water. The following points should be kept in mind when writing thermochemical equations:
Any thermodynamic quantity such as \(ΔH\) that is associated with a thermochemical equation always refers to the number of moles of substances explicitly shown in the equation. Thus for the synthesis of water we can write
\[\ce{2 H2(g) + O2(g)→ 2 H2O(l) }\,\,\,ΔH = -572 \,kJ \nonumber\]
or
\[\ce{H2(g) + 1/2 O2(g)→ H2O(l)} \,\,\, ΔH = -286 \,kJ \nonumber\]
The thermochemical equations for reactions taking place in solution must also specify the concentrations of the dissolved species. For example, the enthalpy of neutralization of a strong acid by a strong base is given by
\[\ce{H^{+}(aq,\, 1M,\, 298\, K,\, 1\, atm) + OH^{-} (aq,\, 1M,\, 298 \,K,\, 1 \,atm) → H2O(\ell, 373\, K,\, 1 \,atm) } \,\,\, ΔH = -56.9\, kJ\, mol^{-1} \nonumber\]
in which the abbreviation aq refers to the hydrated ions as they exist in aqueous solution. Since most thermochemical equations are written for the standard conditions of 298 K and 1 atm pressure, we can leave these quantities out if these conditions apply both before and after the reaction. If, under these same conditions, the substance is in its preferred (most stable) physical state, then the substance is said to be in its standard state .
Thus the standard state of water at 1 atm is the solid below 0°C, and the gas above 100°C. A thermochemical quantity such as \(ΔH\) that refers to reactants and products in their standard states is denoted by \(ΔH^°\).
In the case of dissolved substances, the standard state of a solute is that in which the “effective concentration”, known as the activity, is unity. For non-ionic solutes the activity and molarity are usually about the same for concentrations up to about 1M, but for an ionic solute this approximation is generally valid only for solutions more dilute than 0.001-0.01M, depending on electric charge and size of the particular ion.
Standard Enthalpy of Formation
The enthalpy change for a chemical reaction is the difference
\[ΔH = H_{products} - H_{reactants}\]
If the reaction in question represents the formation of one mole of the compound from its elements in their standard states, as in
\[\ce{H2(g) + 1/2O2(g) -> H2O(l)} \;\;\; ΔH = -286\; kJ\]
then we can arbitrarily set the enthalpy of the elements to zero and write
\[\begin{align*} H_f^o &= \sum H_f^o (products) - \sum H_f^o (reactants) \\[4pt] &= -286\; kJ - 0 \\[4pt] &= -268\; kJ \,mol^{-1} \end{align*}\]
which defines the standard enthalpy of formation of water at 298 K. The value H f ° = -268 kJ tells us that when hydrogen and oxygen, each at a pressure of 1 atm and at 298 K (25° C) react to form 1 mole of liquid water also at 25°C and 1 atm pressure, 268 kJ will have passed from the system (the reaction mixture) into the surroundings. The negative sign indicates that the reaction is exothermic: the enthalpy of the product is smaller than that of the reactants. The standard enthalpy of formation is a fundamental property of any compound. Table T1 list H f ° values (usually alongside values of other thermodynamic properties) in their appendices.
The standard enthalpy of formation of a compound is defined as the heat associated with the formation of one mole of the compound from its elements in their standard states.
In general, the standard enthalpy change for a reaction is given by the expression
\[ΔH_f^o= \sum H_f^o (products) - \sum H_f^o (reactants) \label{2-1}\]
in which the \(\sum H_f^o\) terms indicate the sums of the standard enthalpies of formations of all products and reactants. The above definition is one of the most important in chemistry because it allows us to predict the enthalpy change of any reaction without knowing any more than the standard enthalpies of formation of the products and reactants, which are widely available in tables.
The following examples illustrate some important aspects of the standard enthalpy of formation of substances.
The thermochemical equation defining \(H_f^o\) is always written in terms of one mole of the substance in question> For example, the relavante thermodynamic equation for the heat of formation of ammonia (\(\ce{NH3}\)) is:
\[\ce{ 1/2 N2(g) + 3/2 H2(g)→ NH3(g)}\,\,\, ΔH^o = -46.1\, kJ \,(\text{per mole of } \ce{NH3}) \nonumber\]
The standard heat of formation of a compound is always taken in reference to the forms of the elements that are most stable at 25°C and 1 atm pressure.
A number of elements, of which sulfur and carbon are common examples, can exist in more then one solid crystalline form (called allotropes ). For carbon dixoide (\(ce{CO2}\), one can construct two thermodynamics equations:
\[\ce{C(graphite) + O2(g) → CO2(g)}\,\,\, ΔH^o ≡ H_f^o = -393.5\, kJ\, mol^{-1} \nonumber\]
\[\ce{C(diamond) + O2(g) → CO2(g)}\,\,\, ΔH^{o} = -395.8\, kJ\, mol^{-1} \nonumber\]
However for carbon, the graphite form is the more stable form and the correct thermodynamic equations for the heat of formation.
The physical state of the product of the formation reaction must be indicated explicitly if it is not the most stable one at 25°C and 1 atm pressure:
\[\ce{H2(g) + 1/2 O2(g) → H2O(aq)}\,\,\, ΔH^o ≡ H_f^o = -285.8\, kJ\, mol^{-1} \nonumber\]
\[\ce{H2(g) + 1/2 O2(g) → H2O(g)} \,\,\, ΔH^o = -241.8\, kJ\, mol^{-1} \nonumber\]
Notice that the difference between these two \(ΔH^o\) values is just the heat of vaporization of water.
Although the formation of most molecules from their elements is an exothermic process, the formation of some compounds is mildly endothermic:
\[\ce{1/2 N2(g) + O2(g) -> NO2(g)}\,\,\, ΔH^o ≡ H_f^° = +33.2\, kJ\, mol^{-1} \nonumber\]
A positive heat of formation is frequently associated with instability— the tendency of a molecule to decompose into its elements, although it is not in itself a sufficient cause. In many cases, however, the rate of this decomposition is essentially zero, so it is still possible for the substance to exist. In this connection, it is worth noting that all molecules will become unstable at higher temperatures.
The thermochemical reactions that define the heats of formation of most compounds cannot actually take place. For example, the direct synthesis of methane from its elements
\[\ce{C(graphite) + 2 H2(g) → CH4(g)} \nonumber\]
cannot be observed directly owing to the large number of other possible reactions between these two elements. However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained in the next section.
The standard enthalpy of formation of gaseous atoms from the element is known as the heat of atomization .
Heats of atomization are always positive, and are important in the calculation of bond energies.
\[\ce{Fe(s) → Fe(g)}\,\, ΔH° = 417 \,kJ\, mol^{-1} \nonumber\]
The standard enthalpy of formation of an ion dissolved in water is expressed on a separate scale in which that of \(\ce{H^{+}(aq)}\) is defined as zero .
The standard heat of formation of a dissolved ion such as \(\ce{Cl^{-}(aq)}\) (that is, formation of the ion from the element) cannot be measured because it is impossible to have a solution containing a single kind of ion. For this reason, ionic enthalpies are expressed on a separate scale on which \(\ce{H_f^o}\) of the hydrogen ion at unit activity (1 M effective concentration) is defined as zero:
\[\ce{1/2 H2(g) → H^{+}(aq)}\,\,\, ΔH^o ≡ H_f^o = 0\, kJ\, mol^{-1} \nonumber\]
Other ionic enthalpies (as they are commonly known) are found by combining appropriate thermochemical equations (as explained in Section 3 below). For example, H f ° of HCl (aq) is found from the enthalpies of formation and solution of HCl (g) , yielding
\[\ce{1/2 H2(g) + 1/2 Cl2(g) → HCl(aq)}\,\,\,\ ΔH^o ≡ H_f^o = -167\, kJ\, mol^{-1} \nonumber\]
Because H f ° for H + (aq) is zero, this value establishes the standard enthalpy of the chloride ion. The standard enthalpy of formation of Ca 2 + (aq) , given by
\[\ce{Ca(s) + Cl2(g) -> CaCl2(aq)} \nonumber\]
could then be calculated by combining other measurable quantities such as the enthalpies of formation and solution of \(\ce{CaCl2(s)}\) to find H f ° for CaCl 2 (aq) , from which H f ° of \(\ce{Ca^{2+}(aq)}\) is found by difference from that of \(\ce{Cl^{-}(aq)}\). Tables of the resulting ionic enthalpies are widely available ( see here ).
Hess’ Law and Thermochemical Calculations
Two or more chemical equations can be combined algebraically to give a new equation. Even before the science of thermodynamics developed in the late nineteenth century, it was observed that the heats associated with chemical reactions can be combined in the same way to yield the heat of another reaction. For example, the standard enthalpy changes for the oxidation of graphite and diamond can be combined to obtain ΔH° for the transformation between these two forms of solid carbon, a reaction that cannot be studied experimentally.
\[\ce{C(graphite) + O2(g)→ CO2(g)}\,\,\, ΔH^o = -393.51\, kJ\, mol^{-1}\]
\[\ce{C(diamond) + O2(g)→ CO2(g)} \,\,\, ΔH^o = -395.40\, kJ\, mol^{-1}\]
Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields
\[\ce{C(graphite) → C(diamond)}\,\,\, ΔH^{°} = 1.89\, kJ\, mol^{-1}\]
This principle, known as Hess’ law of independent heat summation is a direct consequence of the enthalpy being a state function. Hess’ law is one of the most powerful tools of chemistry, for it allows the change in the enthalpy (and in other thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small base of experimental data.
Germain Henri Hess (1802-1850) was a Swiss-born professor of chemistry at St. Petersburg, Russia. He formulated his famous law, which he discovered empirically, in 1840. Very little appears to be known about his other work in chemistry.
Standard Enthalpies of Combustion
Because most substances cannot be prepared directly from their elements, heats of formation of compounds are seldom determined by direct measurement. Instead, Hess’ law is employed to calculate enthalpies of formation from more accessible data. The most important of these are the standard enthalpies of combustion . Most elements and compounds combine with oxygen, and many of these oxidations are highly exothermic, making the measurement of their heats relatively easy. For example, by combining the heats of combustion of carbon, hydrogen, and methane, we can obtain the standard enthalpy of formation of methane, which as we noted above, cannot be determined directly (Example \(\PageIndex{1}\)).
Use the following heat of formation/combustion information to estimate the standard heat of formation of methane \(\ce{CH4}\).
\[\begin{aligned} \ce{C(graphite)} + \ce{O2(g)} &→ \ce{CO2(g)} \quad &ΔH° = -393\, kJ\, mol^{-1} \label{P1-1} \\[4pt] \ce{H2(g)} + \ce{1/2 O2(g)} &→ \ce{H2O(g)} \quad &ΔH° = -286 kJ mol^{-1} \label{P1-2} \\[4pt] \ce{CH4(g)} + \ce{2O2(g)} &→ \ce{CO2(g)} + \ce{2H2O(g)} \quad &ΔH^o = -890\, kJ\, mol^{-1} \label{P1-3} \end{aligned}\]
Solution
The standard heat of formation of methane is defined by the reaction
\[\ce{C(graphite) + 2H2(g) → CH4(g)} \quad ΔH^o = ? \label{P1-4}\]
Our task is thus to combine the top three equations in such a way that they add up to (4).
Step 1
Begin by noting that (3), the combustion of methane, is the only equation that contains the CH 4 term, so we need to write it in reverse (not forgetting to reverse the sign of Δ H °!) so that CH 4 appears as the product.
\[\ce{CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)} \quad ΔH^o = +890\, kJ\, mol^{-1} \label{P1-3Rev}\]
Step 2
Since H 2 O does not appear in the net reaction (4), add two times (2) to cancel these out. Notice that this also cancels one of the oxygens in (3Rev):
CO
2
(g)
+
2H
2
O
(g)
→
CH
4
(g)
+
2
O
2
(g)
Δ
H
°
= +890 kJ mol
-1
(P1-3Rev)
→
2H
2
O
(g)
Δ
H
°
= -484 kJ mol
-1
(P1-2)
Step 3
Finally, get rid of the remaining O 2 and CO 2 by adding (1); this also adds a needed C:
→ CH 4 (g) + 2O 2 (g) Δ H ° = +890 kJ mol -1 (P1-3Rev)→ 2H 2 O (g) Δ H ° = -572 kJ mol -1 (P1-2)
→ CO 2 (g) Δ H ° = -393 kJ mol -1 (P1-1)
Step 4
So our creative cancelling has eliminated all except the substances that appear in (4). Just add up the enthalpy changes and we are done:
\[\ce{C(graphite) + 2H2(g) → CH4(g)} \]
(The tablulated value is -74.6 kJ mol -1 )
Calorimetry: Measuring Δ H in the laboratory
How are enthalpy changes determined experimentally? First, you must understand that the only thermal quantity that can be observed directly is the heat q that flows into or out of a reaction vessel, and that q is numerically equal to Δ H ° only under the special condition of constant pressure. Moreover, q is equal to the standard enthalpy change only when the reactants and products are both at the same temperature, normally 25°C. The measurement of q is generally known as calorimetry .
The most common types of calorimeters contain a known quantity of water which absorbs the heat released by the reaction. Because the specific heat capacity of water (4.184 J g -1 K -1 ) is known to high precision, a measurement of its temperature rise due to the reaction enables one to calculate the quantity of heat released.
In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by the components of the calorimeter itself. It is therefore necessary to "calibrate" the calorimeter by measuring the temperature change that results from the introduction of a known quantity of heat. The resulting calorimeter constant , expressed in J K -1 , can be regarded as the “heat capacity of the calorimeter”. The known source of heat is usually produced by passing a known quantity of electric current through a resistor within the calorimeter, but it can be measured by other means as described in the following problem example.
In determining the heat capacity of a calorimeter, a student mixes 100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (yhe specific heat of water is 4.184 J g -1 K -1 ). After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at 38.7°C. Calculate the heat capacity of the calorimeter in J/K.
Solution
The hot water loses heat, the cold water gains heat, and the calorimter itself gains heat, so this is essentially a thermal balance problem. Conservation of energy requires that
\[q_{hot} + q_{cold} + q_{cal} = 0 \nonumber\]
We can evaluate the first two terms from the observed temperature changes:
\[q_{hot} = (100\; g) (38.7\;K - 57.0\;K) (4.184\; J \;g^{-1}K^{-1}) = -7,657\; J \nonumber\]
\[q_{cold} = (100\; g) (38.7 \;K - 24.2 \;K) (4.184\; J\; g^{-1} K^{-1}) = 6,067\; J \nonumber\]
So
\[q_{cal} = 7,657 \;J - ,6067\; J = 1,590;\ J \nonumber\]
The calorimeter constant is
\[\dfrac{1590\; J}{38.7\; K - 24.2\; K} = 110\; J\; K^{-1} \nonumber\]
Note : Strictly speaking, there is a fourth thermal balance term that must be considered in a highly accurate calculation: the water in the calorimeter expands as it is heated, performing work on the atmosphere.
For reactions that can be initiated by combining two solutions, the temperature rise of the solution itself can provide an approximate value of the reaction enthalpy if we assume that the heat capacity of the solution is close to that of the pure water — which will be nearly true if the solutions are dilute.
For example, a very simple calorimetric determination of the standard enthalpy of the reaction
\[\ce{H^{+}(aq) + OH^{-}(aq) → H2O(\ell)} \nonumber\]
could be carried out by combining equal volumes of 0.1M solutions of HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat q will be released into the solution. From the temperature rise and the specific heat of water, we obtain the number of joules of heat released into each gram of the solution, and q can then be calculated from the mass of the solution. Since the entire process is carried out at constant pressure, we have Δ H ° = q .
For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly placed within a larger insulated container of water. During the reaction, heat passes between the inner and outer containers until their temperatures become identical. Again, the temperature change of the water is observed, but in this case we need to know the value of the calorimeter constant described above.
The Bomb Calorimeter
\(\Delta H_{combustion}\), since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:
- Steel bomb which contains the reactants
- Water bath in which the bomb is submerged
- Thermometer
- A motorized stirrer
- Wire for ignition
Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as bomb calorimetry . The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.
Another consequence of the constant-volume condition is that the heat released corresponds to q v , and thus to the internal energy change Δ U rather than to Δ H . The enthalpy change is calculated according to the formula
\[ΔH = q_v + Δn_gRT\]
- \(Δn_g\) is the change in the number of moles of gases in the reaction.
A sample of biphenyl (C 6 H 5 ) 2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C 6 H 5 COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol -1 (that is, Δ U = -3226 kJ mol -1 .) Use this information to determine the standard enthalpy of combustion of biphenyl.
Solution
Begin by working out the calorimeter constant:
- Moles of benzoic acid:
\[\dfrac{(0.825 g}{122.1 \;g/mol} = 0.00676\; mol\nonumber\]
- Heat released to calorimeter:
\[(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ\nonumber\]
- Calorimeter constant:
\[\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K\nonumber\]
Now determine \(ΔU_{combustion}\) of the biphenyl ("BP"):
- moles of biphenyl:
\[\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol\nonumber\]
- heat released to calorimeter:
\[(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ\nonumber\]
- heat released per mole of biphenyl:
\[\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol\nonumber\]
\[ΔU_{combustion} (BP) = -6,293\; kJ/mol\nonumber\]
(This is the heat change at constant volume, \(q_v\); the negative sign indicates that the reaction is exothermic, as all combustion reactions are.)
From the reaction equation
\[(C_6H_5)_{2(s)} + \frac{19}{2} O_{2(g)} \rightarrow 12 CO_{2(g)} + 5 H_2O_{(l)} \nonumber\]
we have
\[Δn_g = 12 - \frac{19}{2} = \frac{-5}{2} \nonumber\]
Thus the volume of the system decreases when the reaction takes place. Converting to Δ H , we can write the following equation. Additionally, recall that at constant volume, \(ΔU = q_V\).
\[\begin{align*} ΔH &= q_V + Δn_gRT \\[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \\[4pt] &=(-6,293 \; kJ/mol)-(6,194\; J/mol)=(-6,293-6.2)\;kJ/mol= -6299 \; kJ/mol \end{align*}\]
A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in \(ΔH\) compared to \(ΔU\) reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation.
Determining the Heat of Reaction
The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the heat of reaction . The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents (\(q_{calorimeter}\)) through the combustion reaction.
\[q_{rxn} = -q_{calorimeter} \label{2A}\]
where
\[q_{calorimeter} = q_{bomb} + q_{water} \label{3A}\]
If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula:
\[q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}\]
Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate \(q_{rxn}\) from \(q_{calorimeter}\) calculated by Equation 2. The heat capacity of the calorimeter can be determined by conducting an experiment.
1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, \(C_{12}H_{22}O_{11}\), in kJ per mole of \(C_{12}H_{22}O_{11}\).
Given:
- mass of \(C_{12}H_{22}O_{11}\): 1.150 g
- \(T_{initial}\): 23.42°C
- \(T_{final}\):27.64°C
- Heat Capacity of Calorimeter: 4.90 kJ/°C
Solution
Using Equation 4 to calculate \(q_{calorimeter}\):
\[q_{calorimeter} = (4.90\; kJ/°C) \times (27.64 - 23.42)°C = (4.90 \times 4.22) \;kJ = 20.7\; kJ \]
Plug into Equation 2:
\(q_{rxn} = -q_{calorimeter} = -20.7 \; kJ \;\)
But the question asks for kJ/mol \(C_{12}H_{22}O_{11}\), so this needs to be converted:
\(q_{rxn} = \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} = \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}}\)
Per Mole \(C_{12}H_{22}O_{11}\):
\(q_{rxn} = \dfrac{-18.0 \; kJ}{g \; C_{12}H_{22}O_{11}} \times \dfrac{342.3 \; g \; C_{12}H_{22}O_{11}}{1 \; mol \; C_{12}H_{22}O_{11}} = \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; C_{12}H_{22}O_{11}}\)
Calorimeters
Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms:
The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.
The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C.
|
libretexts
|
2025-03-17T19:53:17.305598
| 2016-03-05T20:13:52 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.04%3A_Thermochemistry_and_Calorimetry",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.4: Thermochemistry and Calorimetry",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.05%3A_Calorimetry
|
14.5: Calorimetry
Make sure you thoroughly understand the following essential concept:
- Describe a simple calorimeter and explain how it is employed and how its heat capacity is determined.
Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.
Introduction
Calorimetry is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a coffee-cup calorimeter is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions.
Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion \(\Delta H_{combustion}\), since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, which accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:
- Steel bomb which contains the reactants
- Water bath in which the bomb is submerged
- Thermometer
- A motorized stirrer
- Wire for ignition
Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as bomb calorimetry . The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.
Another consequence of the constant-volume condition is that the heat released corresponds to \(q_v\), and thus to the internal energy change \(ΔU\) rather than to \(ΔH\). The enthalpy change is calculated according to the formula
\[ΔH = q_v + Δn_gRT\]
where \(Δn_g\) is the change in the number of moles of gases in the reaction.
A sample of biphenyl (\(\ce{(C6H5)2}\)) weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid (\(\ce{C6H5COOH}\)) weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant volume is known to be 3,226 kJ mol –1 (that is, Δ U = –3,226 kJ mol –1 .) Use this information to determine the standard enthalpy of combustion of biphenyl.
Solution
Begin by working out the calorimeter constant:
- Moles of benzoic acid:
\[\dfrac{0.825 g}{122.1 \;g/mol} = 0.00676\; mol \nonumber\]
- Heat released to calorimeter:
\[(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ \nonumber\]
- Calorimeter constant:
\[\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K \nonumber\]
Now determine \(ΔU_{combustion}\) of the biphenyl ("BP"):
- moles of biphenyl:
\[\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol \nonumber\]
- heat released to calorimeter:
\[(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ \nonumber\]
- heat released per mole of biphenyl:
\[\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol \nonumber\]
\[ΔU_{combustion} (BP) = –6,293\; kJ/mol \nonumber\]
This is the heat change at constant volume, \(q_v\); the negative sign indicates that the reaction is exothermic, as all combustion reactions are.
From the balanced reaction equation
\[\ce{(C6H5)2(s) + 29/2 O2(g) \rightarrow 12 CO2(g) + 5 H2O(l)} \nonumber\]
we can calculate the change in the moles of gasses for this reaction
\[Δn_g = 12 - \frac{29}{2} = \frac{-5}{2} \nonumber\]
Thus the volume of the system decreases when the reaction takes place. Converting to \(ΔH\), we can write the following equation. Additionally, recall that at constant volume, \(ΔU = q_V\).
\[ \begin{align*} ΔH &= q_V + Δn_gRT \\[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \\[4pt] &= (-6,293 \; kJ/mol)–(6,194\; J/mol) \\[4pt] &= (-6,293-6.2)\;kJ/mol \\[4pt] &= -6299 \; kJ/mol \end{align*} \]
A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in \(ΔH\) compared to \(ΔU\) reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation.
Determining the Heat of Reaction
The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the heat of reaction . The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents (\(q_{calorimeter}\)) through the combustion reaction.
\[q_{rxn} = -q_{calorimeter} \label{2A}\]
where
\[q_{calorimeter} = q_{bomb} + q_{water} \label{3A}\]
If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula:
\[q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}\]
Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate \(q_{rxn}\) from \(q_{calorimeter}\) calculated by Equation \ref{2A}. The heat capacity of the calorimeter can be determined by conducting an experiment.
1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42 °C to 27.64 °C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, \(\ce{C12H22O11}\) (in kJ per mole of \(\ce{C12H22O11}\)).
Solution
Given:
- mass of \(C_{12}H_{22}O_{11}\): 1.150 g
- \(T_{initial}\): 23.42°C
- \(T_{final}\):27.64°C
- Heat Capacity of Calorimeter: 4.90 kJ/°C
Using Equation \ref{4A} to calculate \(q_{calorimeter}\):
\[\begin{align*} q_{calorimeter} &= (4.90\; kJ/°C) \times (27.64 - 23.42)°C \\[4pt] &= (4.90 \times 4.22) \;kJ = 20.7\; kJ \end{align*}\]
Plug into Equation \ref{2A}:
\[\begin{align*} q_{rxn} &= -q_{calorimeter} \\[4pt] &= -20.7 \; kJ \; \end{align*}\]
But the question asks for kJ/mol \(\ce{C12H22O11}\), so this needs to be converted:
\[\begin{align*}q_{rxn} &= \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} \\[4pt] &= \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}} \end{align*}\]
Convert to per Mole \(\ce{C12H22O11}\):
\[\begin{align*}q_{rxn} &= \dfrac{-18.0 \; kJ}{\cancel{g \; \ce{C12H22O11}}} \times \dfrac{342.3 \; \cancel{ g \; \ce{C12H22O11}}}{1 \; mol \; \ce{C12H22O11}} \\[4pt] &= \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; \ce{C12H22O11}} \end{align*}\]
"Ice Calorimeter"
Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms:
The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.
The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C.
|
libretexts
|
2025-03-17T19:53:17.384601
| 2017-06-09T03:00:43 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.05%3A_Calorimetry",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.5: Calorimetry",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.06%3A_Applications_of_Thermochemistry
|
14.6: Applications of Thermochemistry
Virtually all chemical processes involve the absorption or release of heat, and thus changes in the internal energy of the system. In this section, we survey some of the more common chemistry-related applications of enthalpy and the First Law. While the first two sections relate mainly to chemistry, the remaining ones impact the everyday lives of everyone.
Enthalpy diagrams and their uses
Comparison and interpretation of enthalpy changes is materially aided by a graphical construction in which the relative enthalpies of various substances are represented by horizontal lines on a vertical energy scale. The zero of the scale can be placed anywhere, since energies are always arbitrary; it is generally most useful to locate the elements at zero energy, which reflects the convention that their standard enthlapies of formation are zero.
This very simple enthalpy diagram for carbon and oxygen and its two stable oxides (Figure \(\PageIndex{1}\)) shows the changes in enthalpy associated with the various reactions this system can undergo. Notice how Hess’s law is implicit in this diagram; we can calculate the enthalpy change for the combustion of carbon monoxide to carbon dioxide, for example, by subtraction of the appropriate arrow lengths without writing out the thermochemical equations in a formal way.
The zero-enthalpy reference states refer to graphite, the most stable form of carbon, and gaseous oxygen. All temperatures are 298 K.
This enthalpy diagram for the hydrogen-oxygen system (Figure \(\PageIndex{2}\)) shows the known stable configurations of these two elements. Reaction of gaseous H 2 and O 2 to yield one mole of liquid water releases 285 kJ of heat . If the H 2 O is formed in the gaseous state, the energy release will be smaller. Notice also that...
- The heat of vaporization of water (an endothermic process) is clearly found from the diagram.
- Hydrogen peroxide H 2 O 2 , which spontaneously decomposes into O 2 and H 2 O, releases some heat in this process. Ordinarily this reaction is so slow that the heat is not noticed. But the use of an appropriate catalyst can make the reaction so fast that it has been used to fuel a racing car.
Why you cannot run your car on water
You may have heard the venerable urban legend, probably by now over 80 years old, that some obscure inventor discovered a process to do this, but the invention was secretly bought up by the oil companies in order to preserve their monopoly. The enthalpy diagram for the hydrogen-oxygen system shows why this cannot be true — there is simply no known compound of H and O that resides at a lower enthalpy level.
Enthalpy diagrams are especially useful for comparing groups of substances having some common feature. This one shows the molar enthalpies of species relating to two hydrogen halides, with respect to those of the elements. From this diagram we can see at a glance that the formation of HF from the elements is considerably more exothermic than the corresponding formation of HCl. The upper part of this diagram shows the gaseous atoms at positive enthalpies with respect to the elements. The endothermic processes in which the H 2 and the dihalogen are dissociated into atoms can be imagined as taking place in two stages, also shown. From the enthalpy change associated with the dissociation of H 2 (218 kJ mol –1 ), the dissociation enthalpies of F 2 and Cl 2 can be calculated and placed on the diagram.
Bond Enthalpies vs. Bond Energies
The enthalpy change associated with the reaction
\[\ce{HI(g) → H(g) + I(g)}\]
is the enthalpy of dissociation of the \(\ce{HI}\) molecule; it is also the bond energy of the hydrogen-iodine bond in this molecule. Under the usual standard conditions, it would be expressed either as the bond enthalpy H °(HI,298K) or internal energy U° (HI,298); in this case the two quantities differ from each other by Δ PV = RT . Since this reaction cannot be studied directly, the H–I bond enthalpy is calculated from the appropriate standard enthalpies of formation:
| ½ H 2 (g) → H (g) | + 218 kJ |
| ½ I 2 (g) → I (g) | +107 kJ |
| ½ H 2 (g) + 1/2 I 2 (g) → HI (g) | –36 kJ |
| HI (g) → H (g) + I (g) | +299 kJ |
Bond energies and enthalpies are important properties of chemical bonds, and it is very important to be able to estimate their values from other thermochemical data. The total bond enthalpy of a more complex molecule such as ethane can be found from the following combination of reactions:
| C 2 H 6 (g) → 2C (graphite) + 3 H 2 (g) | 84.7 kJ |
| 3 H 2 (g) → 6 H (g) | 1308 kJ |
| 2 C (graphite) → 2 C (g) | 1430 kJ |
| C 2 H 6 (g) → 2 C (g) + 6H (g) | 2823 kJ |
When a molecule in its ordinary state is broken up into gaseous atoms, the process is known as atomization (Figure \(\PageIndex{4}\)). The standard enthalpy of atomization refers to the transformation of an element into gaseous atoms:
\[\ce{ C_{(graphite)} → C(g)} \;\;\;\; ΔH^o = 716.7\; kJ\]
Atomization is always an endothermic process. Heats of atomization are most commonly used for calculating bond energies. They are usually measured spectroscopically.
Pauling’s rule and Average Bond Energy
Pauling’s Rule
The total bond energy of a molecule can be thought of as the sum of the energies of the individual bonds.
Pauling’s Rule is only an approximation, because the energy of a given type of bond is not really a constant, but depends somewhat on the particular chemical environment of the two atoms. In other words, all we can really talk about is the average energy of a particular kind of bond, such as C–O, for example, the average being taken over a representative sample of compounds containing this type of bond, such as CO, CO 2 , COCl 2 , (CH 3 ) 2 CO, CH 3 COOH, etc.
Despite the lack of strict additivity of bond energies, Pauling’s Rule is extremely useful because it allows one to estimate the heats of formation of compounds that have not been studied, or have not even been prepared. Thus in the foregoing example, if we know the enthalpies of the C–C and C–H bonds from other data, we could estimate the total bond enthalpy of ethane, and then work back to get some other quantity of interest, such as ethane’s enthalpy of formation. By assembling a large amount of experimental information of this kind, a consistent set of average bond energies can be obtained. The energies of double bonds are greater than those of single bonds, and those of triple bonds are higher still (Table \(\PageIndex{1}\)).
| H | C | N | O | F | Cl | Br | I | Si | |
|---|---|---|---|---|---|---|---|---|---|
| H | 436 | 415 | 390 | 464 | 569 | 432 | 370 | 295 | 395 |
| C | 345 | 290 | 350 | 439 | 330 | 275 | 240 | 360 | |
| N | 160 | 200 | 270 | 200 | 270 | ||||
| O | 140 | 185 | 205 | 185 | 200 | 370 | |||
| F | 160 | 255 | 160 | 280 | 540 | ||||
| Cl | 243 | 220 | 210 | 359 | |||||
| Br | 190 | 180 | 290 | ||||||
| I | 150 | 210 | |||||||
| Si | 230 |
Energy Content of Fuels
A fuel is any substance capable of providing useful amounts of energy through a process that can be carried out in a controlled manner at economical cost. For most practical fuels, the process is combustion in air (in which the oxidizing agent O 2 is available at zero cost.) The enthalpy of combustion is obviously an important criterion for a substance’s suitability as a fuel, but it is not the only one; a useful fuel must also be easily ignited, and in the case of a fuel intended for self-powered vehicles, its energy density in terms of both mass (kJ kg –1 ) and volume (kJ m –3 ) must be reasonably large. Thus substances such as methane and propane which are gases at 1 atm must be stored as pressurized liquids for transportation and portable applications.
|
|
|
|---|---|
| wood (dry) | 15 |
| coal (poor) | 15 |
| coal (premium) | 27 |
| ethanol a | 30 |
| petroleum-derived products | 45 |
| methane, liquified natural gas | 54 |
| hydrogen b | 140 |
Notes on the above table
a Ethanol is being strongly promoted as a motor fuel by the U.S. agricultural industry. Note, however, that according to some estimates, it takes 46 MJ of energy to produce 1 kg of ethanol from corn. Some other analyses, which take into account optimal farming practices and the use of by-products arrive at different conclusions; see, for example, this summary with links to several reports.
b Owing to its low molar mass and high heat of combustion, hydrogen possesses an extraordinarily high energy density, and would be an ideal fuel if its critical temperature (33 K, the temperature above which it cannot exist as a liquid) were not so low. The potential benefits of using hydrogen as a fuel have motivated a great deal of research into other methods of getting a large amount of H 2 into a small volume of space. Simply compressing the gas to a very high pressure is not practical because the weight of the heavy-walled steel vessel required to withstand the pressure would increase the effective weight of the fuel to an unacceptably large value. One scheme that has shown some promise exploits the ability of H 2 to “dissolve” in certain transition metals. The hydrogen can be recovered from the resulting solid solution (actually a loosely-bound compound) by heating.
Energy content of foods
What, exactly, is meant by the statement that a particular food “contains 1200 calories” per serving? This simply refers to the standard enthalpy of combustion of the foodstuff, as measured in a bomb calorimeter . Note, however, that in nutritional usage, the calorie is really a kilocalorie (sometimes called “large calorie”), that is, 4184 J. Although this unit is still employed in the popular literature, the SI unit is now commonly used in the scientific and clinical literature, in which energy contents of foods are usually quoted in kJ per unit of weight.
Although the mechanisms of oxidation of a carbohydrate such as glucose to carbon dioxide and water in a bomb calorimeter and in the body are complex and entirely different, the net reaction involves the same initial and final states, and must be the same for any possible pathway :
\[C_6H_{12}O_6 + 6 O_2 → 6 CO_2 + 6 H_2O \;\;\;\; ΔH^o = – 20.8\; kJ \;mol^{–1}\]
Glucose is a sugar , a breakdown product of starch , and is the most important energy source at the cellular level; fats, proteins, and other sugars are readily converted to glucose. By writing balanced equations for the combustion of sugars, fats, and proteins, a comparison of their relative energy contents can be made. The stoichiometry of each reaction gives the amounts of oxygen taken up and released when a given amount of each kind of food is oxidized; these gas volumes are often taken as indirect measures of energy consumption and metabolic activity; a commonly-accepted value that seems to apply to a variety of food sources is 20.1 J (4.8 kcal) per liter of O 2 consumed.
For some components of food, particularly proteins, oxidation may not always be complete in the body, so the energy that is actually available will be smaller than that given by the heat of combustion. Mammals, for example, are unable to break down cellulose (a polymer of sugar) at all; animals that derive a major part of their nutrition from grass and leaves must rely on the action of symbiotic bacteria which colonize their digestive tracts. The amount of energy available from a food can be found by measuring the heat of combustion of the waste products excreted by an organism that has been restricted to a controlled diet, and subtracting this from the heat of combustion of the food (Table \(\PageIndex{3}\)).
| type of food | food |
ΔH° (kJ g –1 ) |
percent availability |
|---|---|---|---|
| Protein | meat | 22.4 | 92 |
| egg | 23.4 | ||
| Fat | butter | 38.2 | |
| animal fat | 39.2 | 95 | |
| Carbohydrate | starch | 17.2 | |
| glucose (sugar) | 15.5 | 99 | |
| ethanol | 29.7 | 100 |
The amount of energy an animal requires depends on the age, sex, surface area of the body, and of course on the amount of physical activity. The rate at which energy is expended is expressed in watts: 1 W = 1 J sec –1 . For humans, this value varies from about 200-800 W. This translates into daily food intakes having energy equivalents of about 10-15 MJ for most working adults. In order to just maintain weight in the absence of any physical activity, about 6 MJ per day is required.
| animal | kJ hr –1 | kJ kg –1 hr –1 |
|---|---|---|
| mouse | 82 | 17 |
| cat | 34 | 6.8 |
| dog | 78 | 3.3 |
| sheep | 193 | 2.2 |
| human | 300 | 2.1 |
| horse | 1430 | 1.1 |
| elephant | 5380 | 0.7 |
The above table is instructive in that although larger animals consume more energy, the energy consumption per unit of body weight decreases with size. This reflects the fact the rate of heat loss to the environment depends largely on the surface area of an animal, which increases with mass at a greater rate than does an animal's volume ("size").
Thermodynamics and the weather
Hydrogen bonds at work
It is common knowledge that large bodies of water have a “moderating” effect on the local weather, reducing the extremes of temperature that occur in other areas. Water temperatures change much more slowly than do those of soil, rock, and vegetation, and this effect tends to affect nearby land masses. This is largely due to the high heat capacity of water in relation to that of land surfaces— and thus ultimately to the effects of hydrogen bonding . The lower efficiency of water as an absorber and radiator of infrared energy also plays a role.
The specific heat capacity of water is about four times greater than that of soil. This has a direct consequence to anyone who lives near the ocean and is familiar with the daily variations in the direction of the winds between the land and the water. Even large lakes can exert a moderating influence on the local weather due to water's relative insensitivity to temperature change.
During the daytime the land and sea receive approximately equal amounts of heat from the Sun, but the much smaller heat capacity of the land causes its temperature to rise more rapidly. This causes the air above the land to heat, reducing its density and causing it to rise. Cooler oceanic air is drawn in to fill the void, thus giving rise to the daytime sea breeze . In the evening, both land and ocean lose heat by radiation to the sky, but the temperature of the water drops less than that of the land, continuing to supply heat to the oceanic air and causing it to rise, thus reversing the direction of air flow and producing the evening land breeze .
Why it gets colder as you go higher: the adiabatic lapse rate
The air receives its heat by absorbing far-infrared radiation from the earth, which of course receives its heat from the sun. The amount of heat radiated to the air immediately above the surface varies with what's on it (forest, fields, water, buildings) and of course on the time and season. When a parcel of air above a particular location happens to be warmed more than the air immediately surrounding it, this air expands and becomes less dense. It therefore rises up through the surrounding air and undergoes further expansion as it encounters lower pressures at greater altitudes.
Whenever a gas expands against an opposing pressure, it does work on the surroundings. According to the First Law Δ U = q + w , if this work is not accompanied by a compensating flow of heat into the system, its internal energy will fall, and so, therefore, will its temperature. It turns out that heat flow and mixing are rather slow processes in the atmosphere in comparison to the convective motion we are describing, so the First Law can be written as Δ U = w (recall that w is negative when a gas expands.) Thus as air rises above the surface of the earth it undergoes adiabatic expansion and cools. The actual rate of temperature decrease with altitude depends on the composition of the air (the main variable being its moisture content) and on its heat capacity. For dry air, this results in an adiabatic lapse rate of 9.8 C° per km of altitude.
Santa Anas and Chinooks
Just the opposite happens when winds develop in high-altitude areas and head downhill. As the air descends, it undergoes compression from the pressure of the air above it. The surroundings are now doing work on the system, and because the process occurs too rapidly for the increased internal energy to be removed as heat, the compression is approximately adiabatic.
The resulting winds are warm (and therefore dry) and are often very irritating to mucous membranes. These are known generically as Föhn winds (which is the name given to those that originate in the Alps). In North America they are often called chinooks (or, in winter, "snow melters") when they originate along the Rocky Mountains. Among the most notorious are the Santa Ana winds of Southern California which pick up extra heat (and dust) as they pass over the Mohave Desert before plunging down into the Los Angeles basin (Figure 13.5.X). Their dryness and high velocities feed many of the disastrous wildfires that afflict the region.
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libretexts
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2025-03-17T19:53:17.498745
| 2016-03-05T20:14:36 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.06%3A_Applications_of_Thermochemistry",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.6: Applications of Thermochemistry",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.0E%3A_14.E%3A_Thermochemistry_(Exercises)
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14.E: Thermochemistry (Exercises)
13.5: Calorimetry
Q13.5.1
After going through combustion in a bomb calorimeter a sample gives off 5,435 cal. The calorimeter experiences an increase of 4.27°C in its temperature. Using this information, determine the heat capacity of the calorimeter in kJ/°C.
Q13.5.2
Referring to the example given above about the heat of combustion, calculate the temperature change that would occur in the combustion of 1.732 g \(C_{12}H_{22}O_{12}\) in a bomb calorimeter that had the heat capacity of 3.87 kJ/°C.
Q13.5.3
Given the following data calculate the heat of combustion in kJ/mol of xylose,\(C_{5}H_{10}O_{5}\)(s), used in a bomb calorimetry experiment: mass of \(C_{5}H_{10}O_{5}\)(s) = 1.250 g, heat capacity of calorimeter = 4.728 kJ/°C, Initial Temperature of the calorimeter = 24.37°C, Final Temperature of calorimeter = 28.29°C.
Q13.5.4
Determine the heat capacity of the bomb calorimeter if 1.714 g of naphthalene, \(C_{10}H_{8}\)(s), experiences an 8.44°C increase in temperature after going through combustion. The heat of combustion of naphthalene is -5156 kJ/mol \(C_{10}H_{8}\).
Q13.5.5
What is the heat capacity of the bomb calorimeter if a 1.232 g sample of benzoic acid causes the temperature to increase by 5.14°C? The heat of combustion of benzoic acid is -26.42 kJ/g.
S13.5.1
Use equation 4 to calculate the heat of capacity:
\(q_{calorimeter} = \; heat \; capicity \; of \; calorimeter \; x \; \Delta{T}\)
5435 cal = heat capacity of calorimeter x 4.27°C
Heat capacity of calorimeter = (5435 cal/ 4.27°C) x (4.184 J/1 cal) x (1kJ/1000J) = 5.32 kJ/°C
S13.5.2
The temperature should increase since bomb calorimetry releases heat in an exothermic combustion reaction.
Change in Temp = (1.732 g \(C_{12}H_{22}O_{11}\)) x (1 mol \(C_{12}H_{22}O_{11}\)/342.3 g \(C_{12}H_{22}O_{11}\)) x (6.61 x 10³ kJ/ 1 mol \(C_{12}H_{22}O_{11}\)) x (1°C/3.87kJ) = 8.64°C
S13.5.3
[(Heat Capacity x Change in Temperature)/mass] =[ ((4.728 kJ/°C) x(28.29 °C – 24.37 °C))/1.250 g] = 14.8 kJ/g xylose
\(q_{rxn}\) = (-14.8 kJ/g xylose) x (150.13 g xylose/ 1 mol xylose) = -2.22x10³ kJ/mol xylose
S13.5.4
Heat Capacity = [(1.714 g \(C_{10}H_{8}\)) x (1 mol \(C_{10}H_{8}\)/128.2 g \(C_{10}H_{8}\)) x (5.156x10³ kJ/1 mol \(C_{10}H_{8}\))]/8.44°C = 8.17 kJ/ °C
S13.5.5
Heat Capacity = [(1.232 g benzoic acid) x (26.42 kJ/1 g benzoic acid)]/5.14°C = 6.31 kJ/ °C
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libretexts
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2025-03-17T19:53:17.568417
| 2017-06-09T02:54:19 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.0E%3A_14.E%3A_Thermochemistry_(Exercises)",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "14.E: Thermochemistry (Exercises)",
"author": "Stephen Lower"
}
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria
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15: Thermodynamics of Chemical Equilibria
Trying to introduce chemical thermodynamics to beginning students is always problematic; to do it "properly" requires a degree of rigor that rarely succeeds for more than a small fraction of the class. Although a full formal development is rarely appropriate at this level, I believe that the value of developing students' understanding of the fundamental concepts is generally underappreciated. This requires some understanding of the ways thermal energy is dispersed in matter— something that is not a part of classical thermodynamics and is not supported by most textbooks, but which is in keeping with the molecular focus of modern chemical science.
The equilibrium value for a reversible reaction is an important quantity that characterizes a chemical reaction, but what factors govern its value? In particular, is there any way that we can predict the value of the equilibrium constant of a reaction solely from information about the products and reactants themselves, without any knowledge at all about the mechanism or other details of the reaction? The answer is yes, and this turns out to be the central purpose of chemical thermodynamics:
The purpose of thermodynamics is to predict the equilibrium composition of a system from the properties of its components.
Don’t let the significance of this pass you by; it means that we can say with complete certainty whether or not a given change is possible, and if it is possible, to what extent it will occur— without the need to study the particular reaction in question. To a large extent, this is what makes chemistry a science, rather than a mere cataloging of facts.
-
- 15.1: Energy Spreading Drives Spontaneous Change
- Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as natural processor spontaneous changes. In many cases our everyday life experiences teach us the direction in which spontaneous change can occur, and anything that runs counter to these expectations is immediately sensed as weird.
-
- 15.2: Entropy Rules
- Previously, we explained how the tendency of thermal energy to disperse as widely as possible is what drives all spontaneous processes, including, of course chemical reactions. We now need to understand how the direction and extent of the spreading and sharing of energy can be related to measurable thermodynamic properties of substances— that is, of reactants and products.
-
- 15.3: The Second Law of Thermodynamics
- The First Law of thermodynamics, expressed as ΔU = q + w, is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process.
-
- 15.4: Free Energy and the Gibbs Function
- In this unit we introduce a new thermodynamic function, the free energy, which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. As we will explain near the bottom of this page, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as the "Gibbs function" or "Gibbs energy."
-
- 15.5: Thermodynamics of Mixing and Dilution
- This lesson goes somewhat beyond what is covered in most first-year courses, and can usually be skipped by students in non-honors beginners' courses. The concepts presented here are not especially complicated, but they don't really become essential until one gets into more advanced courses in chemistry, physiology, and similar subjects.
-
- 15.6: Free Energy and Equilibrium
- Under conditions of constant temperature and pressure, chemical change will tend to occur in whatever direction leads to a decrease in the value of the Gibbs free energy. In this lesson we will see how G varies with the composition of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The equilibrium composition of the mixture is determined by ΔG° which also defines the equilibrium constant K.
-
- 15.7: Some Applications of Entropy and Free Energy
- Thermodynamics may appear at first to be a rather esoteric subject, but when you think about it, almost every chemical (and biological) process is governed by changes in entropy and free energy. Examples such as those given below should help you connect these concepts with the real world.
-
- 15.8: Quantum states, Microstates, and Energy spreading in Reactions
- Entropy (S) is a state function whose value increases with an increase in the number of available microstates.For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases.
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libretexts
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2025-03-17T19:53:17.633826
| 2013-10-03T01:38:12 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15: Thermodynamics of Chemical Equilibria",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.01%3A_Energy_Spreading_Drives_Spontaneous_Change
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15.1: Energy Spreading Drives Spontaneous Change
You are expected to be able to define and explain the significance of terms identified in bold .
- S pontaneous change is that which, once initiated, proceeds on its own until some state of equilibrium (mechanical, thermal, chemical, etc.) is attained.
- All natural processes (those that occur in the world) ultimately involve spontaneous change.
- T hermal energy is kinetic energy associated with the random motions of atoms and molecules. These states of motion are quantized into energy states which can be realized in huge numbers of different ways that we call microstates . will be t hermally accessible and likely to be populated.
- Spontaneous change is driven by the tendency of thermal energy to spread into as many microstates of the system and surroundings as are thermally accessible.
Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly distributed throughout the water. What you will never see is the reverse of this process, in which the tea would be sucked up and re-absorbed by the teabag. The making of tea, like all changes that take place in the world, possesses a “natural” direction.
What is a spontaneous process?
Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as natural process or spontaneous changes. In many cases our everyday life experiences teach us the direction in which spontaneous change can occur, and anything that runs counter to these expectations is immediately sensed as weird. In other cases, including that of most chemical change, we often have no obvious guidelines, and must learn how to apply the laws of thermodynamics which ultimately govern all spontaneous change.
In order to answer the above question, let's begin by thinking about the outcomes of the following four experiments, each of which illustrates a natural process that proceeds spontaneously only in a single direction.
A stack of one hundred coins is thrown into the air. After they have come to rest on the floor, the numbers that land “heads up” and “tails up” are noted.
Net change: ordered coins → randomized coins (roughly equal numbers of heads and tails.)
Energetics: no relevant net change in energy
Why it does not go in reverse: Simple statistics shows that the probability that the coins will land in one particular arrangement of the huge number that are possible is vanishingly small.
Two identical blocks of copper, one at 200°C and the other at 100°C, are brought into contact in a thermally-insulated environment. Eventually the temperatures of both blocks reach 150°C.
Net change: block1 (200°) + block2 (100°) → combined blocks (150°)
Energetics: Heat (randomized molecular kinetic energy) flows from the warmer block to the cooler one until their temperatures are identical.
Why it does not go in reverse: Dispersal of kinetic energy amongst the copper atoms is a random process; the chances that such a process would lead to a non-uniform sharing of the energy are even smaller than in the case of the 100 coins because of the much greater number (around 10 22 ) of particles involved.
A book or some other solid object is held above a table top, and is then allowed to fall.
Net change: book in air book on table top; potential energy organized kinetic energy thermal energy.
Energetics: At the instant just before the end of its fall, the potential energy the object acquired when it was raised will exist entirely as kinetic energy mv 2 /2 in which m is the mass of the object and v is its velocity. Each atom of which the object is composed will of course possess a proportionate fraction of this energy, again with its principal velocity component pointing down. Superimposed on this, however, will be minute thermal displacements that vary randomly in magnitude and direction from one instant to the next. The sum total of these constitutes the thermal energy contained in the object.
When the object strikes the table top, its motion ceases and we say its kinetic energy is zero. Energy is supposed to be conserved, so where did it disappear to? The shock of impact has resulted in its dispersal into greatly augmented thermal motions of the atoms, both of the object itself and of the area of the table top where the impact occurred. In other words, the kinetic energy of organized motion the object had just before its motion stopped has been transformed into kinetic energy of random or disorganized motion (thermal energy) which spreads rapidly away from the point of impact.
Why it does not go in reverse: Once the kinetic energy of the book has been dispersed amongst the molecules of the book and the table top, the probability of these randomized motions reappearing at the surface where the two objects are in contact and then acting in concert to propel the object back into the air) is negligible.
One mole of gas, initially at 300 K and 2 atm pressure, is allowed to expand to double its volume, keeping the temperature constant.
Net change : Increase in volume of gas.
Energetics : No change in energy if the gas behaves ideally.
Why it does not go in reverse: Simple statistics: the probability that N randomly moving objects (flies in a bottle, for example,) will at any time all be located in one half of the container is (1/2) N . For chemically-significant values of N (10 20 , say) this probability is indistinguishable from zero.
All of the changes described above take place spontaneously , meaning that
- Once they are allowed to commence, they will proceed to the finish without any outside intervention.
- It would be inconceivable that any of these changes could occur in the reverse direction (that is, be undone) without changing the conditions or actively disturbing the system in some way.
What determines the direction in which spontaneous change will occur? It is clearly not a fall in the energy , since in most cases cited above the energy of the system did not change. Even in the case of the falling book, in which the potential energy of the system (the book) falls, energy is conserved overall; if there is no net loss of energy when these processes operate in the forward or natural direction, it would not require any expenditure of energy for them to operate in reverse. In other words, contrary to what all too many people appear to believe, the First Law of Thermodynamics cannot predict the direction of a natural process.
The direction of a spontaneous process is not governed by the energy change and thus the First Law of Thermodynamics cannot predict the direction of a natural process
Direction through disorder
In our examination of the processes described above, we saw that although the total energy of the system and the surroundings (and thus, of the world) is unchanged, there is something about the world that has changed, and this is its degree of disorder .
After coins have been tossed or cards shuffled, the final state is invariably one of greater disorder. Similarly, the molecules of a gas can occupy a larger number of possible positions in space if the volume is larger, so the expansion of a gas is similarly accompanied by an increase in randomness
How can we express disorder quantitatively? From the example of the coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins.
Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its members are heads-up constitutes a macroscopic state of the system. Since we do not care which coins are heads-up, there are clearly numerous configurations of the individual coins which can result in this “macrostate”. Each of these configurations specifies a microscopic state of the system. The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate. To see what this means, consider the possible outcomes of a toss of four coins (Table \(\PageIndex{1}\) ):
| macrostate | ways | probability | microstates |
|---|---|---|---|
| 0 heads | 1 | 1/16 | TTTT |
| 1 head | 4 | 4/16 = 1/4 | HTTT THTT TTHT TTTH |
| 2 heads | 6 | 6/16 = 3/8 | HHTT HTHT HTTH THHT TTHH THTH |
| 3 heads | 4 | 4/16 = 1/4 | HHHT HTHH HHTH THHH |
| 4 heads | 1 | 1/16 | HHHH |
A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the table. The second column gives the number of “ways”— that is, the number of head/tail configurations of the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss resulting in a particular macrostate is proportional to the number of microstates corresponding to the macrostate, and is equal to this number, divided by the total number of possible microstates (in this example, 2 4 =16). An important assumption here is that all microstates are equally probable; that is, the toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an entirely random way.
The greater the number of microstates that correspond to a given macrostate , the greater the probability of that macrostate .
Disorder is more probable than order because there are so many more ways of achieving it. Thus coins and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to become more scattered about a teenager’s room during the course of daily living. But there are some important differences between these large-scale mechanical, or macro systems, and the collections of sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically as molecules . Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move around. Thus, in chemical systems ,
- We are dealing with huge numbers of particles . This is important because statistical predictions are always more accurate for larger samples. Thus although for the four tosses there is a good chance (62%) that the H/T ratio will fall outside the range of 0.45 - 0.55, this probability becomes almost zero for 1000 tosses. To express this in a different way, the chances that 1000 gas molecules moving about randomly in a container would at any instant be distributed in a sufficiently non-uniform manner to produce a detectable pressure difference between the two halves of a container will be extremely small. If we increase the number of molecules to a chemically significant number (around 10 20 , say), then the same probability becomes indistinguishable from zero.
- Once the change begins, it proceeds spontaneously. That is, no external agent (a tosser, shuffler, or teenager) is needed to keep the process going. Gases will spontaneously expand if they are allowed to, and reactions, one started, will proceed toward equilibrium.
- Thermal energy is continually being exchanged between the particles of the system, and between the system and the surroundings. Collisions between molecules result in exchanges of momentum (and thus of kinetic energy) amongst the particles of the system, and (through collisions with the walls of a container, for example) with the surroundings.
- Thermal energy spreads rapidly and randomly throughout the various energetically accessible microstates of the system. The direction of spontaneous change is that which results in the maximum possible spreading and sharing of thermal energy.
The importance of these last two points is far greater than you might at first think, but to fully appreciate this, you must recall the various ways in which thermal energy is stored in molecules— hence the following brief review.
The Spreading of Energy
Thermal energy is the portion of a molecule's energy that is proportional to its temperature , and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation . Since there are three directions in space, all molecules possess three modes of translational motion.
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation ; a linear molecule such as CO 2 in which the atoms are all laid out along the x -axis can rotate along the y - and z -axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes.
Finally, molecules consisting of two or more atoms can undergo internal vibrations . For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at a given temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. It is readily apparent that virtually all the thermal energy resides in translational states.
Notice the greatly different spacing of the three kinds of energy levels. This is extremely important because it determines the number of energy quanta that a molecule can accept, and, as the following illustration shows, the number of different ways this energy can be distributed amongst the molecules.
Accessible energy states . The more closely spaced the quantized energy states of a molecule, the greater will be the number of ways in which a given quantity of thermal energy can be shared amongst a collection of these molecules.
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states. This is a feature of the particle-in-a-box model, which predicts that the separation of the energy states of a gas confined within a box depends on the inverse square of the box length (and on the inverse of the particle mass as well.)
Quantum states, microstates, and energy spreading
At the atomic and molecular level, all energy is quantized ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute (roughly 10 –30 J) that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 600 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
- Addition of energy quanta (higher temperature),
- Increase in the number of molecules (resulting from dissociation, for example).
- the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Increasing temperature increases the number of microstates in a state and hence the entropy of the state.
Energy-spreading changes the world
Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery.
The profundity of this conclusion was recognized around 1900, when it was first described as the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever change. Not a happy thought!
Gases Spontaneously Expand and never contract
Everybody knows that a gas, if left to itself, will tend to expand so as to fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of energy states (and thus microstates) its thermal energy can occupy. Since all such microstates within the thermally accessible range of energies are equally probable, the expansion of the gas can viewed as a consequence of the tendency of energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
The same can in fact be said for even other highly probable distributions, such as having 49.999% of the molecules in the left half of the container and 50.001% in the right half. Even though the number of possible configurations that would yield this distribution of molecules is inconceivably great, it is essentially negligible compared to the number that would correspond to an exact 50-percent distribution.
The illustration represents the allowed thermal energy states of an ideal gas. The larger the volume in which the gas is confined, the more closely-spaced are these states, resulting in a huge increase in the number of microstates into which the available thermal energy can reside; this can be considered the origin of the thermodynamic "driving force" for the spontaneous expansion of a gas.
Heat Spontaneously flows from Hot to Cold
Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler always operates in the direction “warmer-to-cooler” because this allows thermal energy to occupy a larger number of energy states as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”.
When two bodies at different temperatures are placed in thermal contact, thermal energy flow from the warmer into the cooler one until they reach the same temperature.
- Schematic depiction of the thermal energy states in two separated identical bodies at different temperatures (indicated by shading.)
- When the bodies are brought into thermal contact, thermal energy flows from the higher occupied levels in the warmer object into the unoccupied levels of the cooler one until equal numbers are occupied .
As you might expect, the increase in the amount of energy spreading and sharing is proportional to amount of heat transferred q , but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T , the degree of dilution of the thermal energy is given by q /T
To understand why we have to divide by the temperature, consider the effect of very large and very small values of T in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are occupied, so the amount of energy spreading can be very great. Conversely, if the temperature is initially large, the number of new thermal energy states that become occupied will be negligible compared to the number already active.
Energy spreading and sharing in chemical reactions
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
- The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H 2 → 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H 2 . Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and H 2 would not exist. must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in H, as indicated by the vertical displacement of the right half in each of the four panels below.
Energy levels in H 2 and 2 H. The number of thermally accessible energy states (indicated by the shading in the diagram below) increases with temperature, but because 2 moles of H possess twice as many translational states as one mole of H 2 , dissociation becomes increasingly favored at higher temperatures as more of these H states become thermally accessible.
This argument can be generalized to all molecules, illustrating an important principle:
All molecules spontaneously absorb heat and dissociate at high temperatures.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
- As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T 1 , the number of populated states of H 2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
- At some temperature T 2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H 2 and “2H” in equal amounts; that is, the mole ratio of H 2 /H will be 1:2.
- As the temperature rises to T 3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant.
The result is exactly what the Le Chatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures. This is all very well for helping you understand the direct connection between energy spreading when a chemical reaction occurs, but it is of little help in achieving our goal of predicting the direction and extent of chemical change. For this, we need to incorporate the concept of energy spreading into thermodynamics.
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libretexts
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2025-03-17T19:53:17.740108
| 2013-10-03T01:38:12 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.01%3A_Energy_Spreading_Drives_Spontaneous_Change",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.1: Energy Spreading Drives Spontaneous Change",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.02%3A_Entropy_Rules
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15.2: Entropy Rules
You are expected to be able to define and explain the significance of terms identified in bold .
- A reversible process is one carried out in infinitessimal steps after which, when undone, both the system and surroundings (that is, the world) remain unchanged (see the example of gas expansion-compression below). Although true reversible change cannot be realized in practice, it can always be approximated.
- ((in which a process is carried out.
- As a process is carried out in a more reversible manner, the value of w approaches its maximum possible value, and q approaches its minimum possible value.
- Although q is not a state function, the quotient q rev /T is, and is known as the entropy .
- energy within a system.
- The entropy of a substance increases with its molecular weight and complexity and with temperature. The entropy also increases as the pressure or concentration becomes smaller. Entropies of gases are much larger than those of condensed phases.
- The absolute entropy of a pure substance at a given temperature is the sum of all the entropy it would acquire on warming from absolute zero (where S =0) to the particular temperature.
Entropy is one of the most fundamental concepts of physical science, with far-reaching consequences ranging from cosmology to chemistry. It is also widely mis-represented as a measure of "disorder", as we discuss below. The German physicist Rudolf Clausius originated the concept as "energy gone to waste" in the early 1850s, and its definition went through a number of more precise definitions over the next 15 years.
Previously, we explained how the tendency of thermal energy to disperse as widely as possible is what drives all spontaneous processes, including, of course chemical reactions . We now need to understand how the direction and extent of the spreading and sharing of energy can be related to measurable thermodynamic properties of substances— that is, of reactants and products .
You will recall that when a quantity of heat q flows from a warmer body to a cooler one, permitting the available thermal energy to spread into and populate more microstates, that the ratio q / T measures the extent of this energy spreading. It turns out that we can generalize this to other processes as well, but there is a difficulty with using q because it is not a state function; that is, its value is dependent on the pathway or manner in which a process is carried out. This means, of course, that the quotient q / T cannot be a state function either, so we are unable to use it to get differences between reactants and products as we do with the other state functions. The way around this is to restrict our consideration to a special class of pathways that are described as reversible .
Reversible and irreversible changes
A change is said to occur reversibly when it can be carried out in a series of infinitesimal steps, each one of which can be undone by making a similarly minute change to the conditions that bring the change about. For example, the reversible expansion of a gas can be achieved by reducing the external pressure in a series of infinitesimal steps; reversing any step will restore the system and the surroundings to their previous state. Similarly, heat can be transferred reversibly between two bodies by changing the temperature difference between them in infinitesimal steps each of which can be undone by reversing the temperature difference.
The most widely cited example of an irreversible change is the free expansion of a gas into a vacuum. Although the system can always be restored to its original state by recompressing the gas, this would require that the surroundings perform work on the gas. Since the gas does no work on the surrounding in a free expansion (the external pressure is zero, so P Δ V = 0,) there will be a permanent change in the surroundings. Another example of irreversible change is the conversion of mechanical work into frictional heat; there is no way, by reversing the motion of a weight along a surface, that the heat released due to friction can be restored to the system.
These diagrams show the same expansion and compression ±ΔV carried out in different numbers of steps ranging from a single step at the top to an "infinite" number of steps at the bottom. As the number of steps increases, the processes become less irreversible; that is, the difference between the work done in expansion and that required to re-compress the gas diminishes. In the limit of an ”infinite” number of steps (bottom), these work terms are identical, and both the system and surroundings (the “world”) are unchanged by the expansion-compression cycle. In all other cases the system (the gas) is restored to its initial state, but the surroundings are forever changed.
A reversible change is one carried out in such as way that, when undone, both the system and surroundings (that is, the world) remain unchanged.
It should go without saying, of course, that any process that proceeds in infinitesimal steps would take infinitely long to occur, so thermodynamic reversibility is an idealization that is never achieved in real processes, except when the system is already at equilibrium, in which case no change will occur anyway! So why is the concept of a reversible process so important?
The answer can be seen by recalling that the change in the internal energy that characterizes any process can be distributed in an infinity of ways between heat flow across the boundaries of the system and work done on or by the system, as expressed by the First Law Δ U = q + w . Each combination of q and w represents a different pathway between the initial and final states. It can be shown that as a process such as the expansion of a gas is carried out in successively longer series of smaller steps, the absolute value of q approaches a minimum, and that of w approaches a maximum that is characteristic of the particular process.
Thus when a process is carried out reversibly, the w -term in the First Law expression has its greatest possible value, and the q -term is at its smallest. These special quantities w max and q min (which we denote as q rev and pronounce “q-reversible”) have unique values for any given process and are therefore state functions.
Work and reversibility
For a process that reversibly exchanges a quantity of heat q rev with the surroundings, the entropy change is defined as
\[ \Delta S = \dfrac{q_{rev}}{T} \label{23.2.1}\]
This is the basic way of evaluating Δ S for constant-temperature processes such as phase changes, or the isothermal expansion of a gas. For processes in which the temperature is not constant such as heating or cooling of a substance, the equation must be integrated over the required temperature range, as discussed below.
If no real process can take place reversibly, what use is an expression involving q rev ? This is a rather fine point that you should understand: although transfer of heat between the system and surroundings is impossible to achieve in a truly reversible manner, this idealized pathway is only crucial for the definition of Δ S ; by virtue of its being a state function, the same value of Δ S will apply when the system undergoes the same net change via any pathway. For example, the entropy change a gas undergoes when its volume is doubled at constant temperature will be the same regardless of whether the expansion is carried out in 1000 tiny steps (as reversible as patience is likely to allow) or by a single-step (as irreversible a pathway as you can get!) expansion into a vacuum.
The physical meaning of entropy
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system. This “spreading and sharing” can be spreading of the thermal energy into a larger volume of space or its sharing amongst previously inaccessible microstates of the system. The following table shows how this concept applies to a number of common processes.
| system and process | source of entropy increase of system |
|---|---|
| A deck of cards is shuffled, or 100 coins, initially heads up, are randomly tossed. | This has nothing to do with entropy because macro objects are unable to exchange thermal energy with the surroundings within the time scale of the process |
| Two identical blocks of copper, one at 20°C and the other at 40°C, are placed in contact. | The cooler block contains more unoccupied microstates, so heat flows from the warmer block until equal numbers of microstates are populated in the two blocks. |
| A gas expands isothermally to twice its initial volume. | A constant amount of thermal energy spreads over a larger volume of space |
| 1 mole of water is heated by 1C°. |
The increased thermal energy makes additional microstates accessible. (The increase is by a factor of about
10 20,000,000,000,000, 000,000,000 .) |
| Equal volumes of two gases are allowed to mix. | The effect is the same as allowing each gas to expand to twice its volume; the thermal energy in each is now spread over a larger volume. |
| One mole of dihydrogen, H 2 , is placed in a container and heated to 3000K. | Some of the H 2 dissociates to H because at this temperature there are more thermally accessible microstates in the 2 moles of H. |
| The above reaction mixture is cooled to 300K. | The composition shifts back to virtually all H 2 because this molecule contains more thermally accessible microstates at low temperatures. |
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Entropy and "disorder"
Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical sense this is correct, since the spreading and sharing of thermal energy does have the effect of randomizing the disposition of thermal energy within a system. But to simply equate entropy with “disorder” without further qualification is extremely misleading because it is far too easy to forget that entropy (and thermodynamics in general) applies only to molecular-level systems capable of exchanging thermal energy with the surroundings. Carrying these concepts over to macro systems may yield compelling analogies, but it is no longer science. it is far better to avoid the term “disorder” altogether in discussing entropy.
Entropy and Probability
The distribution of thermal energy in a system is characterized by the number of quantized microstates that are accessible (i.e., among which energy can be shared); the more of these there are, the greater the entropy of the system. This is the basis of an alternative (and more fundamental) definition of entropy
\[\color{red} S = k \ln Ω \label{23.2.2}\]
in which k is the Boltzmann constant (the gas constant per molecule, 1.38 10 –23 J K –1 ) and Ω (omega) is the number of microstates that correspond to a given macrostate of the system. The more such microstates, the greater is the probability of the system being in the corresponding macrostate. For any physically realizable macrostate, the quantity Ω is an unimaginably large number, typically around \(10^{10^{25}}\) for one mole. By comparison, the number of atoms that make up the earth is about \(10^{50}\). But even though it is beyond human comprehension to compare numbers that seem to verge on infinity, the thermal energy contained in actual physical systems manages to discover the largest of these quantities with no difficulty at all, quickly settling in to the most probable macrostate for a given set of conditions.
The reason S depends on the logarithm of Ω is easy to understand. Suppose we have two systems (containers of gas, say) with S 1 , Ω 1 and S 2 , Ω 2 . If we now redefine this as a single system (without actually mixing the two gases), then the entropy of the new system will be
\[S = S_1 + S_2\]
but the number of microstates will be the product Ω 1 Ω 2 because for each state of system 1, system 2 can be in any of Ω 2 states. Because
\[\ln(Ω_1Ω_2) = \ln Ω_1 + \ln Ω_2\]
Hence, the additivity of the entropy is preserved.
If someone could make a movie showing the motions of individual atoms of a gas or of a chemical reaction system in its equilibrium state, there is no way you could determine, on watching it, whether the film is playing in the forward or reverse direction. Physicists describe this by saying that such systems possess time-reversal symmetry ; neither classical nor quantum mechanics offers any clue to the direction of time.
However, when a movie showing changes at the macro scopic level is being played backward, the weirdness is starkly apparent to anyone; if you see books flying off of a table top or tea being sucked back up into a tea bag (or a chemical reaction running in reverse), you will immediately know that something is wrong. At this level, time clearly has a direction, and it is often noted that because the entropy of the world as a whole always increases and never decreases, it is entropy that gives time its direction. It is for this reason that entropy is sometimes referred to as "time's arrow".
But there is a problem here: conventional thermodynamics is able to define entropy change only for reversible processes which, as we know, take infinitely long to perform. So we are faced with the apparent paradox that thermodynamics, which deals only with differences between states and not the journeys between them, is unable to describe the very process of change by which we are aware of the flow of time.
The direction of time is revealed to the chemist by the progress of a reaction toward its state of equilibrium; once equilibrium is reached, the net change that leads to it ceases, and from the standpoint of that particular system, the flow of time stops. If we extend the same idea to the much larger system of the world as a whole, this leads to the concept of the "heat death of the universe" that was mentioned briefly in the previous lesson.
Absolute Entropies
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298K and 1 atm pressure as zero. The same is not true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the Third law of thermodynamics , which states that the entropy of a perfectly-ordered solid at 0 K is zero.
The entropy of a perfectly-ordered solid at 0 K is zero.
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat q required to bring the substance from 0 K to the temperature of interest, and then summing the ratios q / T . Two kinds of experimental measurements are needed:
- The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with melting, for example, is jus t ΔH fusion /T m .
- The heat capacity C of a phase expresses the quantity of heat required to change the temperature by a small amount Δ T , or more precisely, by an infinitesimal amount dT . Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities C dT/T for each increment of temperature dT . This is of course just the integral
\[ S_{0^o \rightarrow T^o} = \int _{o^o}^{T^o} \dfrac{C_p}{T} dt \]
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of C on T be used in the above integral in place of a constant C .
\[ S_{0^o \rightarrow T^o} = \int _{o^o}^{T^o} \dfrac{C_p(T)}{T} dt \]
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments Δ T and measure the area under each section of the curve in Figure \(\PageIndex{3}\).
The area under each section of the plot represents the entropy change associated with heating the substance through an interval Δ T . To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of C p for temperatures near zero are not measured directly, but can be estimated from quantum theory.
/ T b are added to obtain the absolute entropy at temperature T . As shown in Figure \(\PageIndex{4}\) above, the entropy of a substance increases with temperature, and it does so for two reasons:
- As the temperature rises, more microstates become accessible, allowing thermal energy to be more widely dispersed. This is reflected in the gradual increase of entropy with temperature.
- The molecules of solids, liquids, and gases have increasingly greater freedom to move around, facilitating the spreading and sharing of thermal energy. Phase changes are therefore accompanied by massive and discontinuous increase in the entropy.
Standard Entropies of substances
The standard entropy of a substance is its entropy at 1 atm pressure. The values found in tables are normally those for 298K, and are expressed in units of J K –1 mol –1 . The table below shows some typical values for gaseous substances.
| He | 126 | H 2 | 131 | CH 4 | 186 |
|---|---|---|---|---|---|
| Ne | 146 | N 2 | 192 | H 2 O(g) | 187 |
| Ar | 155 | CO | 197 | CO 2 | 213 |
| Kr | 164 | F 2 | 203 | C 2 H 6 | 229 |
| Xe | 170 | O 2 | 205 | n -C 3 H 8 | 270 |
| Cl 2 | 223 | n -C 4 H 10 | 310 |
Note especially how the values given in this Table \(\PageIndex{2}\): illustrate these important points:
- Although the standard internal energies and enthalpies of these substances would be zero, the entropies are not. This is because there is no absolute scale of energy, so we conventionally set the “energies of formation” of elements in their standard states to zero. Entropy, however, measures not energy itself, but its dispersal amongst the various quantum states available to accept it, and these exist even in pure elements.
- It is apparent that entropies generally increase with molecular weight. For the noble gases, this is of course a direct reflection of the principle that translational quantum states are more closely packed in heavier molecules, allowing of them to be occupied.
- The entropies of the diatomic and polyatomic molecules show the additional effects of rotational quantum levels.
| C(diamond) | C(graphite) | Fe | Pb | Na | S(rhombic) | Si | W |
|---|---|---|---|---|---|---|---|
| 2.5 | 5.7 | 27.1 | 51.0 | 64.9 | 32.0 | 18.9 | 33.5 |
The entropies of the solid elements are strongly influenced by the manner in which the atoms are bound to one another. The contrast between diamond and graphite is particularly striking; graphite, which is built up of loosely-bound stacks of hexagonal sheets, appears to be more than twice as good at soaking up thermal energy as diamond, in which the carbon atoms are tightly locked into a three-dimensional lattice, thus affording them less opportunity to vibrate around their equilibrium positions. Looking at all the examples in the above table, you will note a general inverse correlation between the hardness of a solid and its entropy. Thus sodium, which can be cut with a knife, has almost twice the entropy of iron; the much greater entropy of lead reflects both its high atomic weight and the relative softness of this metal. These trends are consistent with the oft-expressed principle that the more “disordered” a substance, the greater its entropy.
| solid | liquid | gas |
|---|---|---|
| 41 | 70 | 186 |
Gases , which serve as efficient vehicles for spreading thermal energy over a large volume of space, have much higher entropies than condensed phases. Similarly, liquids have higher entropies than solids owing to the multiplicity of ways in which the molecules can interact (that is, store energy.)
How Entropy depends on Concentration
As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy per mole of solute (the molar entropy) will of course always increase as the solution becomes more dilute.
For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by
\[ \Delta S = R \ln \left( \dfrac{V_2}{V_1} \right) \label{23.2.4}\]
Note: If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.
Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation to express the entropy change associated with a change in the pressure of a perfect gas:
\[ \Delta S = R \ln \left( \dfrac{P_1}{P_2} \right) \label{23.2.5}\]
Expressing the entropy change directly in concentrations, we have the similar relation
\[ \Delta S = R \ln \left( \dfrac{c_1}{c_2} \right) \label{23.2.6}\]
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
How thermal energy is stored in molecules
Thermal energy is the portion of a molecule's energy that is proportional to its temperature , and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation . Since there are three directions in space, all molecules possess three modes of translational motion .
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation ; a linear molecule such as CO 2 in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal vibrations . For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the quantized translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at a given temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. It is readily apparent that virtually all the thermal energy resides in translational states.
Notice the greatly different spacing of the three kinds of energy levels. This is extremely important because it determines the number of energy quanta that a molecule can accept, and, as the following illustration shows, the number of different ways this energy can be distributed amongst the molecules.
The more closely spaced the quantized energy states of a molecule, the greater will be the number of ways in which a given quantity of thermal energy can be shared amongst a collection of these molecules.
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states.
Quantum states, microstates, and energy spreading
At the atomic and molecular level, all energy is quantized ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 600 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
- Addition of energy quanta (higher temperature),
- Increase in the number of molecules (resulting from dissociation, for example).
- the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Heat Death: Energy-spreading changes the world
Energy is conserved ; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery. The profundity of this conclusion was recognized around 1900, when it was first described at the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever happen.
Why do gases tend to expand, but never contract?
Everybody knows that a gas, if left to itself, will tend to expand and fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of microstates its thermal energy can occupy. Since all such states within the thermally accessible range of energies are equally probable, the expansion of the gas can be viewed as a consequence of the tendency of thermal energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
Imagine a gas initially confined to one half of a box (Figure \(\PageIndex{7}\)). The barrier is then removed so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space. In terms of the spreading of thermal energy, Figure 23.2.X may be helpful. The tendency of a gas to expand is due to the more closely-spaced thermal energy states in the larger volume .
Entropy of mixing and dilution
Mixing and dilution really amount to the same thing, especially for idea gases. Replace the pair of containers shown above with one containing two kinds of molecules in the separate sections (Figure \(\PageIndex{9}\)). When we remove the barrier, the "red" and "blue" molecules will each expand into the space of the other. (Recall Dalton's Law that "each gas is a vacuum to the other gas".) However, notice that although each gas underwent an expansion, the overall process amounts to what we call "mixing".
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. But bear in mind that whereas the enthalpy associated with the expansion of a perfect gas is by definition zero, Δ H 's of mixing of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute. But what's really dramatic is that when just one molecule of a second gas is introduced into the container ( in Figure \(\PageIndex{8}\)), an unimaginably huge number of new configurations become possible, greatly increasing the number of microstates that are thermally accessible (as indicated by the pink shading above).
Why heat flows from hot to cold
Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler one always operates in the direction “warmer-to-cooler” because this allows thermal energy to populate a larger number of energy microstates as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”.
When the bodies are brought into thermal contact ( b ), thermal energy flows from the higher occupied levels in the warmer object into the unoccupied levels of the cooler one until equal numbers are occupied in both bodies, bringing them to the same temperature. As you might expect, the increase in the amount of energy spreading and sharing is proportional to the amount of heat transferred q , but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T , the degree of dilution of the thermal energy is given by
\[\dfrac{q}{T}\]
To understand why we have to divide by the temperature, consider the effect of very large and very small values of T in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are initially occupied, so the amount of energy spreading into vacant states can be very great. Conversely, if the temperature is initially large, more thermal energy is already spread around within it, and absorption of the additional energy will have a relatively small effect on the degree of thermal disorder within the body.
Chemical reactions: why the equilibrium constant depends on the temperature
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
- The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H 2 → 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H 2 . Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and this molecule would not exist.
- In order for this dissociation to occur, however, a quantity of thermal energy (heat) q =Δ U must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in H, as indicated by the vertical displacement of the right half in each of the four panels below.
In Figure \(\PageIndex{11}\) a re schematic representations of the translational energy levels of the two components H and H 2 of the hydrogen dissociation reaction. The shading shows how the relative populations of occupied microstates vary with the temperature, causing the equilibrium composition to change in favor of the dissociation product.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
- As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T 1 , the number of populated states of H 2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
- At some temperature T 2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H 2 and “2H” in equal amounts; that is, the mole ratio of H 2 /H will be 1:2.
- As the temperature rises to T 3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant, thus favoring dissociation.
The result is exactly what the LeChatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures.
The following table generalizes these relations for the four sign-combinations of Δ H and Δ S . (Note that use of the standard Δ H ° and Δ S ° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.)
This combustion reaction , like most such reactions, is spontaneous at all temperatures . The positive entropy change is due mainly to the greater mass of CO 2 molecules compared to those of O 2 .
< 0
- Δ H ° = –46.2 kJ
- Δ S ° = –389 J K –1
- Δ G ° = –16.4 kJ at 298 K
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures . Thus higher T , which speeds up the reaction, also reduces its extent.
> 0
- Δ H ° = 55.3 kJ
- Δ S ° = +176 J K –1
- Δ G ° = +2.8 kJ at 298 K
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures . Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
< 0
- Δ H ° = 33.2 kJ
- ΔS° = –249 J K – 1
- Δ G ° = +51.3 kJ at 298 K
This reaction is not spontaneous at any temperature , meaning that its reverse is always spontaneous . But because the reverse reaction is kinetically inhibited, NO 2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Phase changes
Everybody knows that the solid is the stable form of a substance at low temperatures, while the gaseous state prevails at high temperatures. Why should this be? The diagram in Figure \(\PageIndex{12}\) shows that
- the density of energy states is smallest in the solid and greatest (much, much greater) in the gas, and
- the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.
Changes of phase involve exchange of energy with the surroundings (whose energy content relative to the system is indicated (with much exaggeration!) by the height of the yellow vertical bars in Figure \(\PageIndex{13}\). When solid and liquid are in equilibrium (middle section of diagram below), there is sufficient thermal energy (indicated by pink shading) to populate the energy states of both phases. If heat is allowed to flow into the surroundings, it is withdrawn selectively from the more abundantly populated levels of the liquid phase, causing the quantity of this phase to decrease in favor of the solid. The temperature remains constant as the heat of fusion is returned to the system in exact compensation for the heat lost to the surroundings. Finally, after the last trace of liquid has disappeared, the only states remaining are those of the solid. Any further withdrawal of heat results in a temperature drop as the states of the solid become depopulated.
Colligative Properties of Solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare , connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of why these phenomena occur.
Basically, these all result from the effect of dilution of the solvent on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities . The temperatures at which this occurs are depicted by the shading.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases.
Effects of pressure on the entropy: Osmotic pressure
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised (Figure \(\PageIndex{16}\)). The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase.
Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
This phenomenon can explain osmotic pressure . Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this pressure \(\Pi\) is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium.
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libretexts
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2025-03-17T19:53:17.891616
| 2013-10-03T01:38:13 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.02%3A_Entropy_Rules",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.2: Entropy Rules",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.03%3A_The_Second_Law_of_Thermodynamics
|
15.3: The Second Law of Thermodynamics
You are expected to be able to define and explain the significance of terms identified in green type .
- In any macroscopic change , the entropy of the world (that is, system + surroundings) always increases; it never decreases.
- Processes that do not exchange heat with the surroundings (such as the free expansion of a gas into a vacuum) involve entropy change of the system alone, and are always spontaneous.
- A heat engine is a device that converts heat into work. The fraction of heat that can be converted into work is limited by the fall in temperature between the input to the engine and the exhaust.
- According to the Second Law of Thermodynamics , complete conversion of heat into work by a spontaneous cyclic process is impossible.
The First Law of thermodynamics , expressed as Δ U = q + w , is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process.
Why is the First Law not enough?
For simple mechanical operations on macroscopic objects, the First Law, conservation of energy, is all we usually need to determine such things as how many joules of energy is required to lift a weight or to boil some water, how many grams of glucose you must metabolize in order to climb a hill, or how much fuel your car needs to drive a given distance.
But if you think about it, there are a number of "simple mechanical operations" that never occur, even though they would not violate energy conservation.
- Suppose you drop a book onto a table top. The kinetic energy contained in the falling book is dispersed as thermal energy, slightly warming the book and the table top. According to the First Law, there is no reason why placing pre-warmed book on a warmed table top should not be able to propel the book back into the air. Similarly, why can't the energy imparted to the nail (and to the wood) by a hammer not pop the nail back out?
- One might propose a scheme to propel a ship by means of a machine that takes in seawater, extracts part of its thermal energy which is used to rotate the propeller, and then tosses the resulting ice cubes overboard. As long as the work done to turn the propeller is no greater than the heat required to melt the ice, the First Law is satisfied.
- Because motion of the air molecules is completely random, there is no reason why all of the molecules in one half of a room cannot suddenly "decide" to move into the other half, asphyxiating the unfortunate occupants of that side. (To the extent that air behaves as a perfect gas, this doesn't involve the First Law at all.)
What do all these scenarios that conform to the First Law but are nevertheless never seen to occur have in common? In every case, energy becomes less spread out, less "diluted". In the first two examples, thermal energy (dispersed) gets concentrated into organized kinetic energy of a macroscopic object— a book, a propeller. In the third case, the thermal energy gets concentrated into a smaller volume as the gas contracts.
The second law of thermodynamics says in effect, that the extent to which any natural process can occur is limited by the dilution of thermal energy (increase in entropy) that accompanies it, and once the change has occurred, it can never be un-done without spreading even more energy around. This is one of the most profound laws of nature, and should be a part of every educated person's world view. It is unfortunate that this law is so widely misrepresented as simply ordaining the increase in "disorder". A more brief statement of the Second Law (for those who know the meaning of "entropy") is
Second Law of Thermodynamics: The entropy of the world only increases and never decreases.
The more formal and historical ways of stating the Second Law will be presented farther below after we introduce the topic of heat engines. It is also worth knowing this important consequence of the Second Law: Just because the energy is “there” does not mean it will be available to do anything useful.
Entropy and Spontaneous Change
We explained how processes that take place spontaneously always proceed in a direction that leads to the spreading and sharing of thermal energy.
- A book falls to the tabletop (rather than absorbing heat and jumping up from it) because its kinetic energy changes into thermal energy which is widely dispersed into the molecules of the book and the table.
- A gas expands and solutions mix because the thermal energy their molecules possess get spread over a larger volume of space.
- Hydrogen gas dissociates into H atoms which share thermal energy amongst more particles and a greater volume of space. (But only if the temperature is high enough to make the huge number of new microstates energetically accessible.)
Because all natural processes lead to the spreading and sharing of thermal energy, and because entropy is a measure of the extent to which energy is dispersed in the world, it follows that:
In any spontaneous macroscopic change, the entropy of the world increases.
All natural processes that allow the free exchange of thermal energy amongst chemically-significant numbers of particles are accompanied by a spreading or “dilution” of energy that leaves the world forever changed. In other words, all spontaneous change leads to an increase in the entropy of the world . At first sight, this might seem to be inconsistent with our observations of very common instances in which there is a clear decrease in entropy, such as the freezing of a liquid, the formation of a precipitate, or the growth of an organism.
System + surroundings = the world!
... but it’s the entropy of the system plus surroundings that counts! It is important to understand that the criterion for spontaneous change is the entropy change of the system and the surroundings— that is, of the “world”, which we denote by Δ S total :
\[ΔS_{total} = ΔS_{system} + ΔS_{surroundings} \label{23.1}\]
The only way the entropy of the surroundings can be affected is by exchange of heat with the system :
\[ΔS_{surroundings} = \dfrac{q_{surr}}{ T} \label{23.2}\]
Thus the freezing of water is accompanied by a flow of heat (the heat of fusion) into the surroundings, causing ΔS surr to increase. At temperatures below the freezing point, this increase more than offsets the decrease in the entropy of the water itself, so ΔSworld exceeds zero and the process is spontaneous. The problem example below works this out in detail for a specific example.
Note that it does not matter whether the change in the system occurs reversibly or irreversibly; as mentioned previously, it is always possible to define an alternative (irreversible) pathway in which the amount of heat exchanged with the surroundings is the same as q rev ; because Δ S is a state function, the entropy change of the surroundings will have the same value as for the unrealizable reversible pathway.
If there is no flow of heat into or out of the surroundings, the entropy change of the system and that of the world are identical. Examples of such processes, which are always spontaneous , are the free expansion of an ideal gas into a vacuum, and the mixing of two ideal gases. In practice, almost all processes involving mixing and diffusion can be regarded as driven exclusively by the entropy increase of the system.
Most processes involving chemical and phase changes involve the exchange of heat with the surroundings, so their tendency to occur cannot always be predicted by focusing attention on the system alone. Further, owing to the – q/T term in Δ S surroundings , the spontaneity of all such processes will depend on the temperature, as we illustrated for the dissociation of H 2 previously.
As a quantitative example, let us consider the freezing of water . We know that liquid water will spontaneously change into ice when the temperature drops below 0°C at 1 atm pressure. Since the entropy of the solid is less than that of the liquid, we know the entropy of the water (the system here) will decrease on freezing. The amount of decrease is found by dividing the heat of fusion of ice by the temperature for the reversible pathway, which occurs at the normal freezing point:
\[ΔS_{system} = \dfrac{-6000 \; J/mol}{273 \;K} = -21.978 \; J/mol\]
If the process is actually carried at 0°C, then the heat of fusion is transferred to the surroundings at the same temperature, and the entropy of the surroundings increases by
\[ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 21.979\; J/mol\]
so that Δ S total = 0. Under these conditions the process can proceed in either direction (freezing or melting) without affecting the entropy of the world; this means that both ice and liquid water can be present simultaneously without any change occurring; the system is said to be in equilibrium .
Suppose now that the water is supercooled to –1°C before it freezes. The entropy change of the water still corresponds to the reversible value q rev /T = (–6000J)/(273K). The entropy change of the surroundings, however, is now given by
\[ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 22.059\; J/mol\]
The total entropy change is now
\[ΔS_{total} = (–21.978 + 22.059) J;\ K^{–1} mol^{–1} = +0.081\; J \;K^{–1} mol^{–1}\]
indicating that the process can now occur (“is spontaneous”) only in the one direction.
Why did we use 273 K when evaluating Δ S system and 272 K for calculating Δ S surroundings ? In the latter case it is possible to formulate a reversible pathway by which heat can be transferred to the surroundings at any temperature. Δ S system , however, is a state function of water, and will vary with temperature only slightly.
Note that in order to actually freeze water, it must be cooled to very slightly below its normal freezing point, a condition known as supercooling. Freezing of supercooled water is of course an irreversible process (once it starts, it cannot be stopped except by raising the temperature by a finite amount), and the positive value of \(ΔS_{total}\) tells us that this process will occur spontaneously at temperatures below 273 K. Under these conditions, the process is driven by the entropy increase of the surroundings resulting from flow of the heat of fusion of water into the surroundings.
Does the entropy of the world ever decrease?
The principle that thermal energy (and the molecules carrying it) tends to spread out is based on simple statistics. It must be remembered, however, that the laws of probability have meaningful application only to systems made up of large numbers of independent actors. If you trap a hundred flies in a bottle, they will generally distribute themselves more or less uniformly throughout the container; if there are only four flies, however, it is quite likely that all of them will occasionally be located in one particular half of the bottle.
Why the sky is blue
Similarly, you can trust with complete certainty that the spontaneous movement of half the molecules of the air to one side of the room you now occupy will not occur, even though the molecules are moving randomly and independently. On the other hand, if we consider a box whose dimensions are only a few molecular diameters, then we would expect that the random and short-term displacement of the small number of particles it contains to one side of the box would occur quite frequently. This is, in fact, the cause of the blueness of the sky: random fluctuations in the air density over tiny volumes of space whose dimensions are comparable with the wavelength of light results in selective scattering of the shorter wavelengths, so that blue light is scattered out, leaving the red light for the enjoyment of sunset-watchers to the east.
Brownian motion
This refers to the irregular zig-zag-like movement of extremely small particles such as plant pollen when they are suspended in a drop of liquid. Any such particle is continually being buffeted by the thermal motions of the surrounding liquid molecules. If size of the particle is very large compared to that the the liquid molecules, the forces that result from collisions of these molecules with the particle will cancel out and the particle remains undisturbed. If the particle is very small, however (perhaps only a thousand times larger than a molecule of the liquid), then the chances that it will undergo sufficiently more hits from one direction than from another during a brief interval of time become significant.
In these two examples, the entropy of the system decreases without any compensating flow of heat into the surroundings, leading to a net (but only temporary) decrease in the entropy of the world. This does not represent a failure of the Second Law, however, because no one has ever devised a way to extract useful work from these processes.
Heat Engines
The Industrial Revolution of the 19th century was largely driven by the invention of the steam engine. The first major use of such engines was to pump water out of mines, whose flooding from natural seepage seriously limited the depths to which they could be driven, and thus the availability of the metal ores that were essential to the expansion of industrial activities. The steam engine is a type of heat engine, a device that converts heat, provided by burning a fuel, into mechanical work, typically delivered through the motion of a piston in opposition to an opposing force. An engine is therefore an energy conversion device in which, ideally, every joule of heat released by combustion of the fuel could be extracted as work at the output shaft; such an engine would operate at 100 percent efficiency.
However, engineers of the time were perplexed to find that the efficiencies of steam engines were rather low (usually around 20%), with most of the heat being exhausted uselessly to the environment. Everyone understood that an efficiency exceeding 100% would be impossible (that would violate conservation of energy, and thus the First Law), but it was not clear why efficiencies could not rise significantly beyond the small values observed even as mechanical designs improved
The answer was found by a young French engineer, Sadi Carnot, who in 1824 published an analysis of an idealized heat engine that is generally considered to be the foundation of the science of thermodynamics— notwithstanding the fact that Carnot still accepted the belief that heat is a fluid-like substance called “caloric”. We will not replicate his analysis here (this is normally done in more advanced courses in physical chemistry), but will simply state his conclusion in his own [translated] words:
"The production of motive power is then due in steam-engines not to an actual consumption of caloric, but to its transportation from a warm body to a cold body...the production of heat alone is not sufficient to give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be useless. The ultimate attainable efficiency of any heat engine will depend on the temperatures at which heat is supplied to and removed from it."
The left side of the figure represents a generalized heat engine into which a quantity of heat q H , extracted from a source or “reservoir” at temperature T H is partly converted into work w . The remainder of the heat q L is exhausted to a reservoir at a lower temperature T L . In practice, T H would be the temperature of the steam in a steam engine, or the temperature of the combustion mixture in an internal combustion or turbine engine. The low temperature reservoir is ordinarily that of the local environment. The efficiency ε ( epsilon ) of a heat engine is the fraction of the heat abstracted from the high temperature reservoir that can be converted into work:
\[ ε = \dfrac{w}{q_H} \label{3.3}\]
Carnot’s crucial finding (for which he would certainly have deserved a Nobel prize if these had existed at the time) is that the efficiency is proportional to the "distance'' in temperature that the heat can “fall” as it passes through the engine:
\[ ε = 1 - \dfrac{T_L}{T_H} \label{3.4}\]
This is illustrated graphically in the right half of the figure just above, in which the efficiency is simply the fraction of the “complete” fall (in temperature) to absolute zero (arrow b ) that the heat undergoes in the engine (arrow a .) Clearly, the only way to attain 100% efficiency would be to set the temperature of the exhaust reservoir to 0°K, which would be impossible. For most terrestrial heat engines, T L is just the temperature of the environment, normally around 300 K, so the only practical way to improve the efficiency is to make T H as high as possible. This is the reason that high pressure (superheated) steam is favored in commercial thermal power plants. The highest temperatures (and the greatest operating efficiencies) are obtained in gas turbine engines. However, as operating temperatures rise, the costs of dealing with higher steam pressures and the ability of materials such as turbine blades to withstand high temperatures become significant factors, placing an upper limit of around 600K on T H , thus imposing a maximum of around 50 percent efficiency on thermal power generation.
For nuclear plants , in which safety considerations require lower steam pressures, the efficiency is lower. One consequence of this is that a larger fraction of the heat is exhausted to the environment, which may result in greater harm to aquatic organisms when the cooling water is returned to a stream or estuary.
Several proposals have been made for building a heat engine that makes use of the temperature differential between the surface waters of the ocean and cooler waters that, being more dense, reside at greater depth. If the exhaust temperature is 5°C, what is the maximum amount of work that could be extracted from 1000 L of surface water at 10°C? (The specific heat capacity of water is 4.184 J g –1 K –1 .)
Solution
The amount of heat (q H ) that must be extracted to cool the water by 5 K is (4.184 J g –1 K –1 )(10 6 g)(5 K) = 2.09 × 10 7 J. The ideal thermodynamic efficiency is given by
\[ 1 -\dfrac{278 \;K}{283\; K} = 0.018\]
The amount of work that could be done would be
\[(0.018)(2.09 \times 10^7 \;J) = 3.7 \times 10^6 \;J\]
Comment : It may be only 1.8% efficient, but it’s free!
The drinking bird as a heat engine
Few toys illustrate as many principles of physical science as this popular device that has been around for many years. At first glance it might appear to be a perpetual motion machine, but it's really just a simple heat engine. Modern "dippy birds" (as they are sometimes called) utilize dichloromethane as the working fluid.
This liquid boils at 39° C, and therefore has a rather high vapor pressure at room temperature. The liquid (to which a dye is often added for dramatic effect) is stored in a reservoir at the bottom of the bird. The bird's beak is covered with felt which, when momentarily dipped in water, creates a cooling effect as the water evaporates. This causes some of the CH 2 Cl 2 vapor to condense in the head, reducing the pressure inside the device, causing more liquid to boil off and re-condense in the head. The redistribution of fluid upsets the balance, causing the bird to dip its beak back into the water. Once the head fills with liquid, it drains back into the bottom, tipping the bird upright to repeat the cycle.
We will leave it to you to relate this to the heat engine diagram above by identifying the heat source and sink, and estimate the thermodynamic efficiency of the engine.
Heat Pumps
If a heat engine is run “in reverse” by performing work on it (that is, changing “work out” to “work in” in Fig 8), it becomes a device for transporting heat against a thermal gradient. Refrigerators and air conditioners are the most commonly-encountered heat pumps. A heat pump can also be used to heat the interior of a building. In this application, the low temperature reservoir can be a heat exchanger buried in the earth or immersed in a well. In this application heat pumps are more efficient than furnaces or electric heating, but the capital cost is rather high.
The Second Law: what it means
It was the above observation by Carnot that eventually led to the formulation of the Second Law of Thermodynamics near the end of the 19th Century. One statement of this law (by Kelvin and Planck) is as follows:
It is impossible for a cyclic process connected to a reservoir at one temperature to produce a positive amount of work in the surroundings.
To help you understand this statement and how it applies to heat engines, consider the schematic heat engine in the figure in which a working fluid (combustion gases or steam) expands against the restraining force of a weight that is mechanically linked to the piston. From a thermodynamic perspective, the working fluid is the system and everything else is surroundings. Expansion of the fluid occurs when it absorbs heat from the surroundings; return of the system to its initial state requires that the surrounding do work on the system. Now re-read the above statement of the Second Law, paying special attention to the italicized phrases which are explained below:
- A cyclic process is one in which the system returns to its initial state. A simple steam engine undergoes an expansion step (the power stroke), followed by a compression (exhaust stroke) in which the piston, and thus the engine, returns to its initial state before the process repeats.
- “ At one temperature ” means that the expansion and compression steps operate isothermally. This means that ΔU = 0; just enough heat is absorbed by the system to perform the work required to raise the weight, so for this step q = –w.
- “ A positive amount of work in the surroundings ” means that the engine does more work on the surroundings than the surroundings do on the engine. Without this condition the engine would be useless.
Note carefully that the Second Law applies only to a cyclic process — isothermal expansion of a gas against a non-zero pressure always does work on the surroundings, but an engine must repeat this process continually; to do so it must be returned to its initial state at the end of every cycle. When operating isothermally, the work – w it does on the surroundings in the expansion step (power stroke) is nullified by the work + w the surroundings must do on the system in order to complete the cycle. The Second Law can also be stated in an alternative way:
It is impossible to construct a machine operating in cycles that will convert heat into work without producing any other changes.
Thus the Second Law does allow an engine to convert heat into work, but only if “other changes” (transfer of a portion of the heat directly to the surroundings) are allowed. And since heat can only flow spontaneously from a source at a higher temperature to a sink at a lower temperature, the impossibility of isothermal conversion of heat into work is implied.
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libretexts
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2025-03-17T19:53:18.067792
| 2013-10-03T01:38:14 |
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"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.03%3A_The_Second_Law_of_Thermodynamics",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.3: The Second Law of Thermodynamics",
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.04%3A_Free_Energy_and_the_Gibbs_Function
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15.4: Free Energy and the Gibbs Function
- Gibbs Energy is a state function defined as \(G = H – TS\).
- The practical utility of the Gibbs function is that \(ΔG\) for any process is negative if it leads to an increase in the entropy of the world. Thus spontaneous change at a given temperature and pressure can only occur when it would lead to a decrease in \(G\).
- The sign of the standard free energy change \(ΔG^o\) of a chemical reaction determines whether the reaction will tend to proceed in the forward or reverse direction.
- Similarly, the relative signs of \(ΔH^o\) and \(ΔH^o\) determine whether the spontaneity of a chemical reaction will be affected by the temperature, and if so, in what way.
- The existence of sharp melting and boiling points reflects the differing temperature dependancies of the free energies of the solid, liquid, and vapor phases of a pure substance, which are in turn reflect their differing entropies.
Previously, we saw that it is the sum of the entropy changes of the system and surroundings that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to focus our attention on the system rather than the surroundings, and would like to avoid having to calculate the entropy change of the surroundings explicitly.
In this unit we introduce a new thermodynamic function, the free energy , which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. However, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as "Gibbs energy." The free energy enables us to do this for changes that occur at a constant temperature and pressure (the Gibbs energy ) or constant temperature and volume (the Helmholtz energy .)
Free energy: the Gibbs function
The Gibbs energy (also known as the Gibbs function or Gibbs Potential ) is defined as
\[G = H – T S \label{23.4.1}\]
in which \(S\) refers to the entropy of the system . Since \(H\), \(T\) and \(S\) are all state functions, so is \(G\). Thus for any change in state (under constant temperature), we can write the extremely important relation
\[ΔG = ΔH – T ΔS \label{23.4.2}\]
How does this simple equation encompass the entropy change of the world \(ΔS_{total}\), which we already know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition
\[ΔS_{total} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}\]
we would first like to get rid of \(ΔS_{surr}\). How can a chemical reaction (a change in the system ) affect the entropy of the surroundings ? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat q p across the system boundary. The enthalpy change of the reaction \(ΔH\) is defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be \(–q_p\) which will cause the entropy of the surroundings to change by \(–q_p / T = –ΔH/T\). We can therefore rewrite Equation \(\ref{23.4.3}\) as
\[ΔS_{total} = \dfrac{- ΔH}{T} + ΔS_{sys} \label{23.4.4}\]
Multiplying each side by \(-T\), we obtain
\[-TΔS_{total} = ΔH - TΔS_{sys} \label{23.4.5}\]
which expresses the entropy change of the world in terms of thermodynamic properties of the system exclusively. If \(-TΔS_{total}\) is denoted by \(ΔG\), then we have Equation \(\ref{23.4.2}\) which defines the Gibbs energy change for the process.
From the foregoing, you should convince yourself that \(G\) will decrease in any process occurring at constant temperature and pressure which is accompanied by an overall increase in the entropy. The constant temperature is a consequence of the temperature and the enthalpy appearing in the preceding Equation \(\ref{23.4.5}\). Since most chemical and phase changes of interest to chemists take place under such conditions, the Gibbs energy is the most useful of all the thermodynamic properties of a substance, and (as we shall see in the lesson that follows this one) it is closely linked to the equilibrium constant.
Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether \(ΔH\) or the \(TΔS\) term dominates. This is technically correct, but misleading because it disguises the important fact that \(ΔS_{total}\), which this equation expresses in an indirect way, is the only criterion of spontaneous change.
We will deal only with the Gibbs energy in this course. The Helmholtz free energy is of interest mainly to chemical engineers (whose industrial-scale processes are often confined to tanks and reactors of fixed volume) and some geochemists whose interest is centered on the chemistry that occurs deep within the earth's surface.
Gibbs Energy and Chemical Change
Remember that \(ΔG\) is meaningful only for changes in which the temperature and pressure remain constant . These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction.
\(ΔG\) serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Moreover, it determines the direction and extent of chemical change.
In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the \(TΔS\) term).
\[\ce{H_2O(l) \rightarrow H2O(s)} \label{23.5.6}\]
water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H 2 O diminishes, and the process proceeds spontaneously.
In a spontaneous change, Gibbs energy always decreases and never increases.
An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows:
- \(ΔG < 0\): reaction can spontaneously proceed to the right: \[A \rightarrow B \nonumber\]
- \(ΔG > 0\): reaction can spontaneously proceed to the left: \[A \leftarrow B \nonumber\]
- \(ΔG = 0\): the reaction is at equilibrium and both \([A]\) and \([B]\) will not change: \[A \rightleftharpoons B. \nonumber\]
No need to find the value of ΔG for a Specific Reaction!
This might seem strange, given the key importance \(ΔG\) in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate \(ΔG\). As we will show in the lesson that follows this one, it is far more convenient to work with the equilibrium constant of a reaction, within which \(ΔG\) is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of \(ΔG\) depends on the proportions of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with \(ΔH\).
Recalling the condition for spontaneous change
\[ΔG = ΔH – TΔS < 0\]
it is apparent that the temperature dependence of Δ G depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of \(ΔH\) and \(ΔS\) are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation the sign of the entropy change determines whether the reaction becomes more or less spontaneous as the temperature is raised. For any given reaction, the sign of \(ΔH\) can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process:
Case 1: Δ H < 0 and Δ S > 0
Both enthalpic \(\Delta H\) and entropic \(-T\Delta S\) terms will be negative, so \(ΔG\) will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures.
Case 2: Δ H < 0 and Δ S < 0
If the reaction is sufficiently exothermic it can force \(ΔG\) negative only at temperatures below which \(|TΔS| < |ΔH|\). This means that there is a temperature \(T = ΔH / ΔS\) at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition.
Case 3: Δ H > 0 and Δ S > 0
This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that \(TΔS > ΔH\). Since the effect of the temperature is to "magnify" the influence of a positive \(ΔS\), the process will be spontaneous at temperatures above \(T = ΔH / ΔS\). (Think of melting and boiling.)
Case 4: Δ H > 0 and Δ S < 0
With both \(ΔH\) and \(ΔS\) working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form.
The plots above are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of Δ H and Δ S .
- Their most important differentiating features are the position of the Δ H line (above or below the is T Δ S line), and the slope of the latter, which of course depends on the sign of Δ S .
- The reaction A → B will occur spontaneously only when Δ G is negative (blue arrows pointing down.)
- T Δ S plots are not quite straight lines as shown here. Similarly, the lines representing Δ H are even more curved.
The other two plots on each diagram are only for the chemistry-committed.
- Each pair of energy-level diagrams depicts the relative spacing of the microscopic energy levels in the reactants and products as reflected by the value of Δ S °. (The greater the entropy, the more closely-spaced are the quantized microstates.)
- The red shading indicates the range of energy levels that are accessible to the system at each temperature. The spontaneous direction of the reaction will always be in the direction in which the red shading overlaps the greater number of energy levels, resulting in the maximum dispersal of thermal energy.
- Note that the vertical offsets correspond to Δ H ° for the reaction.
- Never forget that it is the ability of thermal energy to spread into as many of these states as possible that determines the tendency of the process to take place. None of this is to scale, of course!
The Standard Gibbs Energy
You have already been introduced to the terms such as \(ΔU^o\) and \(ΔH^o\) in which the \(^o\) sign indicates that all components (reactants and products) are in their standard states . This concept of standard states is especially important in the case of the free energy, so let's take a few moments to review it. More exact definitions of the conventional standard states can be found in most physical chemistry textbooks. In specialized fields such as biochemistry and oceanography, alternative definitions may apply. For example, the "standard pH" of zero (corresponding to \([H^{+}] = 1\,M\)) is impractical in biochemistry, so pH = 7 is commonly employed. For most practical purposes, the following definitions are good enough:
- gases: 1 atmosphere partial pressure
- pure liquids: the liquid under a total (hydrostatic) pressure of 1 atm.
- solutes: an effective concentration of 1 mol L –1 (1 mol dm –3 ). ("Effective" concentrations approach real concentrations as the latter approach zero; for practical purposes, these can be considered identical at real concentrations smaller than about 10 –4 molar.)
- solids: the pure solid under 1 atm pressure
- There is actually no "standard temperature", but because most thermodynamics tables list values for 298.15 K (25° C), this temperature is usually implied.
- These same definitions apply to standard enthalpies and internal energies.
- Do not confuse these thermodynamic standard states with the "standard temperature and pressure" (STP) widely employed in gas law calculations.
To make use of Gibbs energies to predict chemical changes, we need to know the free energies of the individual components of the reaction. For this purpose we can combine the standard enthalpy of formation and the standard entropy of a substance to get its standard free energy of formation
\[ΔG_f^o = ΔH_f^o – TΔS_f^o \label{23.4.7}\]
Recall that the symbol ° refers to the standard state of a substance measured under the conditions of 1 atm pressure or an effective concentration of 1 mol L –1 and a temperature of 298 K. Then determine the standard Gibbs energy of the reaction according to
\[ ΔG^o = \sum ΔG_f^o \;(\text{products})– \sum ΔG_f^o \;(\text{reactants}) \label{24.4.8}\]
As with standard heats of formation, the standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 K. Standard Gibbs free energies of formation are normally found directly from tables. Once the values for all the reactants and products are known, the standard Gibbs energy change for the reaction is found by Equation \(\ref{23.4.7}\). Most tables of thermodynamic values list \(ΔG_f^o\) values for common substances (e.g., Table T2 ), which can, of course, always be found from values of Δ H f ° and Δ S f ° .
Find the standard Gibbs energy change for the reaction
\[\ce{CaCO3(s) \rightarrow CaO (s) + CO2(g)} \nonumber\]
The \(ΔG_f^o°\) values for the three components of this reaction system are \(\ce{CaCO3(s)}\): –1128 kJ mol –1 , CaO (s) : –603.5 kJ mol –1 , CO 2 (g) : –137.2 kJ mol –1 .
Solution
Substituting into Equation \(\ref{23.4.7}\), we have
\[ΔG^o = (–603.5 –137.2) – (–1128) kJ\, mol^{–1} = +130.9\, kJ\, mol^{–1} \nonumber \]
This indicates that the process is not spontaneous under standard conditions (i.e., solid calcium carbone will not form solid calcium oxide and CO 2 at 1 atm partial pressure at 25° C).
Comment: This reaction is carried out on a huge scale to manufacture cement, so it is obvious that the process can be spontaneous under different conditions.
The practical importance of the Gibbs energy is that it allows us to make predictions based on the properties (Δ G° values) of the reactants and products themselves, eliminating the need to experiment. But bear in mind that while thermodynamics always correctly predicts whether a given process can take place (is spontaneous in the thermodynamic sense), it is unable to tell us if it will take place at an observable rate.
When thermodynamics says "no", it means exactly that. When it says "yes", it means "maybe".
The reaction
\[\ce{ 1/2 O2(g) + H2(g) → H2O(l)} \nonumber\]
is used in fuel cells to produce an electrical current. The reaction can also be carried out by direct combustion.
Thermodynamic data : molar entropies in J mol –1 K –1 : O 2 (g) 205.0; H 2 (g)130.6; H 2 O(l) 70.0; H 2 O(l) Δ H ° f = –285.9 kJ mol –1 .
Use this information to find
- The amount of heat released when the reaction takes place by direct combustion;
- The amount of electrical work the same reaction can perform when carried out in a fuel cell at 298 K under reversible conditions;
- The amount of heat released under the same conditions.
Solution
First, we need to find \(ΔH^o\) and \(ΔS^o\) for the process. Recalling that the standard enthalpy of formation of the elements is zero,
\[\begin{align*} ΔH^o &= ΔH^p_f(\text{products}) – ΔH^°_f(\text{reactants}) \\[4pt] &= –285.9\, kJ\, mol^{–1} – 0 \\[4pt] &= –285.9 \,kJ \,mol^{–1} \end{align*}.\]
Similarly,
\[\begin{align*} ΔS^o &= S^o_f(\text{products}) – S^o_f(\text{reactants}) \\[4pt] &= (70.0) – (½ \times 205.0 + 130.6) \\[4pt] &= –163\, J\, K^{–1}mol^{–1} \end{align*}\]
- When the hydrogen and oxygen are combined directly, the heat released will be \(ΔH^o = –285.9\, kJ\, mol^{–1}\).
- The maximum electrical work the fuel cell can perform is given by \[\begin{align*}ΔG^o &= ΔH^o – TΔS^o \\[4pt] &= –285.9 \,kJ\, mol^{–1} – (298\, K)(–163\, JK^{–1}mol^{–1}) \\[4pt] &= –237.2 \,kJ\, mol^{–1}.\end{align*}.\]
- The heat released in the fuel cell reaction is the difference between the enthalpy change (the total energy available) and the reversible work that was expended: \[\begin{align*} ΔH^o – ΔG^o &= TΔS^o \\[4pt] &= (298\, K)(–163\, JK^{–1}mol^{–1}) \\[4pt] &= –48,800\, J\, mol^{–1} \\[4pt] &=–48.8 \,kJ\, mol^{–1}.\end{align*}.\]
The foregoing example illustrates an important advantage of fuel cells. Although direct combustion of a mole of hydrogen gas yields more energy than is produced by the same net reaction within the fuel cell, the latter, in the form of electrical energy, can be utilized at nearly 100-percent energy efficiency by a motor or some other electrical device. If the thermal energy released by direct combustion were supplied to a heat engine, second-law considerations would require that at least half of this energy be "wasted" to the surroundings.
Δ G° refer to single, specific chemical changes in which all components (reactants and products) are in their standard states .
The \(ΔG_f^o\) of a substance, like \(ΔH_f^o\), refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. Both of these terms are by definition zero for the elements in their standard states. There are only a few common cases in which this might create some ambiguity:
| Stable Form | \(ΔG_f^o\) (kJ mol –1 ) | Unstable Form | \(ΔG_f^o\) (kJ mol –1 ) |
|---|---|---|---|
| \(\ce{O2(g)}\) | 0 | \(\ce{O3(g)}\) | 163.2 |
| \(\ce{C(graphite)}\) | 0 | \(\ce{C(diamond)}\) | 2.9 |
| \(\ce{S(rhombic)}\) | 0 | \(\ce{S(monoclinic)}\) | 0.1 |
| \(\ce{P(white)}\) | 0 | \(\ce{P4(g)}\) | 24.4 |
Ions in aqueous solution are a special case; their standard free energies are relative to the hydrated hydrogen ion \(\ce{H^{+}(aq)}\) which is assigned \(ΔG_f^o = 0\).
\(ΔG\) is very different from ΔG°. The distinction is nicely illustrated in Figure \(\PageIndex{5}\) in which Δ G is plotted on a vertical axis for two hypothetical reactions having opposite signs of Δ G° . The horizontal axis schematically expresses the relative concentrations of reactants and products at any point of the process. Note that the origin corresponds to the composition at which half of the reactants have been converted into products.
Take careful note of the following:
- for the Δ G° > 0 reaction. Notice that there are an infinite number of these values, depending on the progress of the reaction. In contrast there is only a single value of Δ G° , corresponding to the composition at which Δ G = 0 ( ).
- At this point, some products have been formed, but the composition is still dominated by reactants.
- If we begin at a composition to the left of , Δ G will be negative and the composition will move to the right. Similarly if we begin with a composition to the right of , Δ G will be positive and the composition will move to the left.
- The plot on the right is for the Δ G ° < 0 reaction, for which Δ G ° is shown at . At its equilibrium point , there are more products than reactants. If we start at a composition to the right of , the composition will tend to move to the left. If the initial composition is to the left of , the reaction will tend to proceed to the right.
- What would happen if Δ G° were 0? The equilibrium point of such a reaction would be at the origin, corresponding to half the reactants being converted to products.
The important principle you should understand from this is that a negative Δ G° does not mean that the reactants will be completely transformed into products. By the same token, a positive Δ G° does not mean that no products are formed at all.
It should now be clear from the discussion above that a given reaction carried out under standard conditions is characterized by a single value of Δ G° .
The reason for the Gibbs energy minimum at equilibrium relates to the increase in entropy when products and reactants coexist in the same phase. As seen in the plot, even a minute amount of "contamination" of products by reactants reduces the free energy below that of the pure products. In contrast, composition of a chemical reaction system undergoes continual change until the equilibrium state is reached. So the a single reaction can have an infinite number of Δ G values, reflecting the infinite possible compositions between the extremes of pure reactants (zero extent of reaction) and pure products (unity extent of reaction).
In the example of a reaction A → B, depicted in the above diagram, the standard free energy of the products is smaller than that of the reactants , so the reaction will take place spontaneously. T his does not mean that each mole of pure A will be converted into one mole of pure B . For reactions in which products and reactants occupy a single phase (gas or solution), the meaning of "spontaneous" is that the equilibrium composition will correspond to an extent of reaction greater than 0.5 but smaller than unity.Note, however, that for Δ G° values in excess of about ±50 kJ mol –1 , the equilibrium composition will be negligibly different from zero or unity extent-of-reaction. The physical meaning of Δ G is that it tells us how far the free energy of the system has changed from G ° of the pure reactants . As the reaction proceeds to the right, the composition changes, and Δ G begins to fall. When the composition reaches , Δ G reaches its minimum value and further reaction would cause it to rise. But because free energy can only decrease but never increase, this does not happen. The composition of the system remains permanently at its equilibrium value.
A G vs . extent-of-reaction diagram for a non-spontaneous reaction can be interpreted in a similar way; the equilibrium composition will correspond to an extent of reaction greater than zero but less than 0.5. In this case, the minimum at reflects the increase in entropy when the reactants are "contaminated" by a small quantity of products.
If all this detail about Δ G seems a bit overwhelming, do not worry: it all gets hidden in the equilibrium constant and reaction quotient that we discuss in the next lesson!
Interpretation of Standard Gibbs energy chan ges
Although it is \(ΔG\) rather than \(ΔG^o\) that serves as a criterion for spontaneous change at constant temperature and pressure, \(ΔG^o\) values are so readily available that they are often used to get a rough idea of whether a given chemical change is possible. This is practical to do in some cases, but not in others:
It generally works for reactions such as
\[\ce{4 NH_3(g) + 5 O_2(g) → 4 NO(g) + 6 H_2O(g)} \nonumber\]
with \(ΔG^o = –1,010\, kJ\).
(industrially important for the manufacture of nitric acid) because \(ΔG^o\) is so negative that the reaction will be spontaneous and virtually complete under just about any reasonable set of conditions.
The following reaction expresses the fact that the water molecule is thermodynamically stable:
\[\ce{2 H_2(g) + 1/2 O_2(g)→ H_2O(l)} \nonumber\]
with \(ΔG^o = –237.2 \,kJ\).
Note that this refers to liquid water (the standard state of H 2 O at 25°). If you think about it, a negative standard Gibbs energy of formation (of which this is an example) can in fact be considered a definition of molecular stability.
Similarly, dissociation of dihydrogen into its atoms is highly unlikely under standard conditions:
\[\ce{H_2O(g) → 2 H(g) + O(g)} \nonumber\]
with \(ΔG^o = +406.6\, kJ\).
Again, an analogous situation would apply to any stable molecule.
Now consider the dissociation of dinitrogen tetroxide
\[\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\]
with \(ΔG^o = +2.8 kJ\).
in which the positive value of Δ G° tells us that N 2 O 4 at 1 atm pressure will not change into two moles of NO 2 at the same pressure, but owing to the small absolute value of Δ G° , we can expect the spontaneity of the process to be quite sensitive to both the temperature (as shown in the table below) and to the pressure in exactly the way the Le Chatelier principle predicts.
For reactions involving dissolved ions, one has to be quite careful. Thus for the dissociation of the weak hydrofluoric acid
\[\ce{HF(aq) → H^+(aq) + F^–(aq)} \nonumber\]
with \(ΔG^o = –317 \,kJ\).
it is clear that a 1 mol/L solution of HF will not dissociate into 1M ions, but this fact is not very useful because if the HF is added to water, the initial concentration of the fluoride ion will be zero (and that of H + very close to zero), and the Le Chatelier principle again predicts that some dissociation will be spontaneous.
It is common knowledge that dissociation of water into hydrogen- and hydroxyl ions occurs only very sparingly:
\[\ce{H_2O(l) → H^+(aq) + OH^–(aq) } \nonumber\]
with \(ΔG^o = 79.9 \,kJ\).
which correctly predicts that the water will not form 1M (effective concentration) of the ions, but this is hardly news if you already know that the product of these ion concentrations can never exceed 10 –14 at 298K.
Finally, consider this most familiar of all phase change processes, the vaporization of liquid water:
\[\ce{H_2O(l) → H_2O(g)} \nonumber\]
with \(ΔG^o = 8.58 \,kJ \).
Conversion of liquid water to its vapor at 1 atm partial pressure does not take place at 25° C, at which temperature the equilibrium partial pressure of the vapor (the "vapor pressure") is only 0.031 atm (23.8 torr.) Gaseous H 2 O at a pressure of 1 atm can only exist at 100° C. Of course, water left in an open container at room temperature will spontaneously evaporate if the partial pressure of water vapor in the air is less than 0.031 atm, corresponding to a relative humidity of under 100%
Finding the Equilibrium Temperature
A reaction is in its equilibrium state when
\[ΔG = ΔH – TΔS = 0 \label{23.4.1a}\]
The temperature at which this occurs is given by
\[T = \dfrac{ΔH}{ΔS} \label{23.4.1b}\]
If we approximate \(ΔH\) by \(ΔH^o\) and \(ΔS\) by \(ΔS^o\), so Equation \ref{23.4.1a} would be
\[ΔG \approx ΔH^o – TΔS^o = 0 \label{23.4.1aa}\]
We can then estimate the normal boiling point of a liquid. From the following thermodynamic data for water:
Because ΔH° values are normally expressed in kilojoules while ΔS° is given in joules, a very common student error is to overlook the need to express both in the same units.
We find that liquid water is in equilibrium with water vapor at a partial pressure of 1 atm when the temperature is
\[T = \dfrac{44,100\, J}{118.7\, J\, K^{–1}} = 371.5\, K\]
But " the normal boiling point of water is 373 K ", you say? Very true. The reason we are off here is that both Δ H ° and Δ S ° have their own temperature dependencies; we are using the "standard" 25° values without correcting them to 100° C. Nevertheless, if you think about it, the fact that we can estimate the boiling point of a liquid from a table of thermodynamic data should be rather impressive! Of course, the farther one gets from 298 K, the more unreliable will be the result. Thus for the dissociation of dihydrogen into its atoms,
All one can say here is that H 2 will break down at something over 3000 K or so. (You may already know that all molecules will dissociate into their atoms at high temperatures.) We tend to think of high temperatures as somehow "forcing" molecules to dissociate into their atoms, but this is wrong. In order to get the H–H bond to vibrate so violently through purely thermal excitation that the atoms would fly apart, a temperature more like 30,000 K would be required. The proper interpretation is at the temperature corresponding to Δ H /TΔ S , the molecule spontaneously absorbs energy from the surroundings sufficient to overcome the H-H bond strength.
Predicting the Effects of Temperature
The \(T\Delta S\) term interacts with the \(ΔH\) term in \(\Delta G\) to determine whether the reaction can take place at a given temperature. This can be more clearly understood by examining plots of \(TΔS^o\) and \(ΔH^o\) as functions of the temperature for some actual reactions. Of course these parameters refer to standard states that generally do not correspond to the temperatures, pressures, or concentrations that might be of interest in an actual case. Nevertheless, these quantities are easily found and they can usefully predict the way that temperature affects these systems.
Case 1: Exothermic reaction, Δ S ° > 0
\[\ce{C(graphite) + O_2(g) → CO_2(g)} \nonumber \]
- \(ΔH^o = –393\, kJ\)
- \(ΔG^o = –394 \,kJ\) at \(298\, K\)
This combustion reaction , like most such reactions, is spontaneous at all temperatures . The positive entropy change is due mainly to the greater mass of \(\ce{CO2}\) molecules compared to those of \(\ce{O2}\).
Case 2: Exothermic reaction, Δ S ° < 0
\[\ce{3 H_2 + N_2 → 2 NH_3(g) } \nonumber\]
- \(ΔH^o = –46.2\, kJ\)
- \(ΔG^o = –16.4\, kJ\) at \(298\, K\)
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures . Thus higher T, which speeds up the reaction, also reduces its extent.
Case 3: Endothermic reaction, Δ S ° > 0
\[ \ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\]
- \(ΔH^o = 55.3\, kJ\)
- \(ΔG^o = +2.8\, kJ\) at \(298\, K\)
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures . Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
Case 4: Endothermic Reaction, ΔS° < 0
\[\ce{ 1/2 N_2 (g) + O_2 (g)→ NO_2(g)} \nonumber\]
- \(ΔH^° = 33.2\, kJ\)
- \(ΔS^o = –249\, J\, K^{–1}\)
- \(ΔG^o = +51.3\, kJ\) at \(298\, K\)
This reaction is not spontaneous at any temperature , meaning that its reverse is always spontaneous . But because the reverse reaction is kinetically inhibited, NO 2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Concluding remarks on Gibbs Energy
The appellation “free energy” for \(G\) has led to so much confusion that many scientists now refer to it simply as the Gibbs energy . The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: \(ΔG\) is the maximum amount of energy, which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible.
A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although \(G\) has the units of energy (joules, or in its intensive form, J mol –1 ), it lacks one of the most important attributes of energy in that it is not conserved . Thus, although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy anywhere else. Referring to \(G\) as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy. The quantity \(–Δ G \) associated with a process represents the quantity of energy that is “shared and spread”, which as we have already explained is the meaning of t he increase in the entropy. The quotient \(–ΔG/T\) is in fact identical with \(ΔS_{total}\), the entropy change of the world, whose increase is the primary criterion for an y kind of change.
\(G\) differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas \(H\) and \(S\) can be related to the quantity and distribution of energy in a collection of molecules. The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.
- Gibbs Energy is not free energy
- Gibbs Energy is not energy
- Gibbs Energy is not even "real"
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libretexts
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2025-03-17T19:53:18.217843
| 2013-10-03T01:38:13 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.04%3A_Free_Energy_and_the_Gibbs_Function",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.4: Free Energy and the Gibbs Function",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.05%3A_Thermodynamics_of_Mixing_and_Dilution
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15.5: Thermodynamics of Mixing and Dilution
The free energy of a pure liquid or solid at 1 atm pressure is just its molar free energy of formation Δ G multiplied by the number of moles present. For gases and substances in solution, we have to take into account the concentration (which, in the case of gases, is normally expressed in terms of the pressure). We know that the lower the concentration, the greater the entropy, and thus the smaller the free energy. The following excerpt from this lesson serves as the starting point for the rest of the present lesson.
Entropy Depends on Concentration
As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy per mole of solute (the molar entropy) will of course always increase as the solution becomes more dilute.
For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by
\[\Delta S = R\ln \left( \dfrac{V_2}{V_1} \right) \label{2-4}\]
(If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.)
Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation to express the entropy change associated with a change in the pressure of a perfect gas:
\[\Delta S = R\ln \left( \dfrac{P_1}{P_2} \right) \label{2-5}\]
Expressing the entropy change directly in concentrations \(c\), we have the similar relation
\[\Delta S = R\ln \left( \dfrac{C_1}{C_2} \right) \label{2-6}\]
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
The Gibbs energy of a Gas: standard states
The pressure of a perfect gas does not affect its enthalpy, but it does affect the entropy (box at left), and thus, through the –T Δ S term, the free energy. When the pressure of such a gas changes from \(P_1\) to \(P_2\), the Gibbs energy change is
\[ \Delta G = \Delta H - T \Delta S = 0 - RT \ln \left( \dfrac{P_1}{P_2} \right) \label{4.8}\]
How can we evaluate the free energy of a specific sample of a gas at some arbitrary pressure? First, recall that the standard molar free energy G ° that you would look up in a table refers to a pressure of 1 atm. The free energy per mole of our sample is just the sum of this value and any change in free energy that would occur if the pressure were changed from 1 atm to the pressure of interest
\[ G = G^o + RT \ln \left( \dfrac{P_1}{1\; atm} \right) \label{4.9}\]
which we normally write in abbreviated form
\[G = G^o + RT \ln P \label{4-10}\]
The term escaping tendency is not commonly used in traditional thermodynamics because it is essentially synonymous with the free energy, but it is worth knowing because it helps us appreciate the physical significance of free energy in certain contexts. The higher the pressure of a gas, the greater will be the tendency of its molecules to leave the confines of the container; we will call this the escaping tendency . The above equation tells us that the pressure of a gas is a directly observable measure of its free energy ( G , not G °!). Combining these two ideas, we can say that the free energy of a gas is also a measure of its escaping tendency .
From Gases to Solutions: Mixing and Dilution
All substances, given the opportunity to form a homogeneous mixture with other substances, will tend to become more dilute. This can be rationalized simply from elementary statistics; there are more equally probable ways of arranging one hundred black marbles and one hundred white marbles, than two hundred marbles of a single color. For massive objects like marbles this has nothing to do with entropy, of course. However, when we are dealing with huge numbers of molecules capable of storing, exchanging and spreading thermal energy, mixing and expansion are definitely entropy-driven processes.
It can be argued, in fact, that mixing and expansion are really very similar ; after all, when we mix two gases, each is expanding into the space formerly occupied exclusively by the other.
Suppose, for example, that we have a gas initially confined to one half of a box, and we then remove the barrier so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space; the actual increase, according to Equations \(\ref{2-4}\) and \(\ref{2-5}\) above, is \(R \ln 2\). And from Equation \(\ref{4-10}\), the change in \(G\) will be \(–RT \ln 2\).
Now let us repeat the experiment, but starting this time with "red" molecules in one half of the container and "blue" ones in the right half. Because we now have two gases expanding into twice their initial volumes, the changes in S and G will be twice as great:
\[ΔS = 2 R \ln 2\]
\[ΔG = –2 RT \ln 2\]
However, notice that although each gas underwent an expansion, the overall process in this case is equivalent to mixing .
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. An important qualification here is that the solution must be an ideal one, meaning that the strength of interactions between all type of molecules (solutes A and B, and the solvent) must be the same. Remember that the enthalpy associated with the expansion of a perfect gas is by definition zero. In contrast, the \(ΔH_{mixing}\) of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute.
Given this proviso, we can define the Gibbs energy of dilution or mixing by substituting this equation into the definition of Δ G :
\[ \Delta G_{dilution} = \Delta H_{dilution} - RT\ln \left( \dfrac{C_1}{C_2} \right) \label{4-11}\]
If the substance in question forms an ideal solution with the other components, then \(ΔH_{diution}\) is zero, and we can write
\[ \Delta G_{dilution} = RT\ln \left( \dfrac{C_2}{C_1} \right) \label{4-12}\]
These relations tell us that the dilution of a substance from an initial concentration \(C_1\) to a more dilute concentration \(C_2\) is accompanied by a decrease in the free energy , and thus will occur spontaneously. By the same token, the spontaneous “un-dilution” of a solution will not occur (we do not expect the tea to diffuse back into the tea bag!) However, un-dilution can be forced to occur if some means can be found to supply to the system an amount of energy (in the form of work) equal to \(\Delta G_{dilution} \).
An important practical example of this is the metabolic work performed by the kidneys in concentrating substances from the blood for excretion in the urine.
To find the Gibbs energy of a solute at some arbitrary concentration, we proceed in very much the same way as we did for a gas: we take the sum of the standard free energy plus any change in the free energy that would accompany a change in concentration from the standard state to the actual state of the solution. From Equation \(\ref{4-12}\) it is easy to derive an expression analogous to Equation \(\ref{4-10}\):
\[G = G^o + RT \ln C \label{4-13}\]
which gives the free energy of a solute at some arbitrary concentration \(C\) in terms of its value \(G^o\) in its standard state.
Although Equation \(\ref{4-13}\) has the same simple form as Equation \(\ref{4-10}\), its practical application is fraught with difficulties, the major one being that it does not usually give values of \(G\) that are consistent with experiment, especially for solutes that are ionic or are slightly soluble. This is because most solutions (especially those containing dissolved ions) are far from ideal ; intermolecular interactions between solute molecules and between solute and solvent bring back the enthalpy term that we left out in deriving Equation \(\ref{4-12}\). In addition, the structural organization of the solution becomes concentration dependent, so that the entropy depends on concentration in a more complicated way than is implied by the concentration analog of Equation \(\ref{4-12}\).
Chemical Reactions and Mixing
We characterize the tendency for a chemical reaction A → B to occur at constant temperature and pressure by the value of its standard Gibbs energy change Δ G °. If this quantity is negative, we know that the reaction will take place spontaneously. However, have you ever wondered why it is that substance A is not completely transformed into B if the latter is thermodynamically more stable? The answer is that if the reaction takes place in a single phase (gas or liquid), something else is going on: A and B are mixing together, and this process creates its own Gibbs energy change \(ΔG_{mixing}\). For a simple binary mixture of A and B (without any reaction), the changes in \(S\) and \(G\) can be represented by these simple plots:
We will not try to prove it here, but it turns out that no matter how much lower the Gibbs energy of the products compared to that of the reactants, the free energy of the system can always be reduced even more if some of the reactants remain in the solution to contribute a Δ G mixing term. This is the reason that a plot of G as a function of the composition of such a system has a minimum at some point short of complete conversion.
D iffusion refers to the transport of a substance across a concentration gradient. The direction is always toward the region of lower concentration. You should now see that from a thermodynamic standpoint, these processes are identical in that they both represent the spontaneous "escape" of molecules from a region of higher concentration (lower entropy, higher Gibbs energy) to a region of lower concentration.
Instead of complicating G° by trying to correct for all of these effects, chemists have chosen to retain its simple form by making a single small change in the form of \(\ref{4-13}\):
\[G = G^o + RT \ln a \label{4-14}\]
This equation is guaranteed to work, because a , the activity of the solute, is its thermodynamically effective concentration. The relation between the activity and the concentration is given by
\[a = \gamma c \label{4-15}\]
where \(\gamma\) is the activity coefficient . As the solution becomes more dilute, the activity coefficient approaches unity:
\[ \lim _{c \rightarrow 0} \gamma =1 \label{4.16}\]
The price we pay for this simplicity is that the relation between the concentration and the activity at higher concentrations can be quite complicated, and must be determined experimentally for every different solution.
The question of what standard state we choose for the solute (that is, at what concentration is G ° defined, and in what units is it expressed?) is one that you will wish you had never asked. We might be tempted to use a concentration of 1 molar, but a solution this concentrated would be subject to all kinds of intermolecular interaction effects, and would not make a very practical standard state. These effects could be eliminated by going to the opposite extreme of an “infinitely dilute” solution, but by Equation \(\ref{4-12}\) this would imply a free energy of minus infinity for the solute, which would be awkward. Chemists have therefore agreed to define the standard state of a solute as one in which the concentration is 1 molar, but all solute-solute interactions are magically switched off, so that \(\gamma\) is effectively unity. Since this is impossible, no solution corresponding to this standard state can actually exist, but this turns out to be only a small drawback, and seems to be the best compromise between convenience, utility, and reality.
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libretexts
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2025-03-17T19:53:18.296807
| 2013-10-03T01:38:14 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.05%3A_Thermodynamics_of_Mixing_and_Dilution",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.5: Thermodynamics of Mixing and Dilution",
"author": "Stephen Lower"
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.06%3A_Free_energy_and_Equilibrium
|
15.6: Free Energy and Equilibrium
You are expected to be able to define and explain the significance of terms identified in italic type .
- As a homogeneous chemical reaction proceeds, the Gibbs energies of the reactants become more negative and those of the products more positive as the composition of the system changes.
- The total Gibbs energy of the system (reactants + products) always becomes more negative as the reaction proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium composition of the system, after which time no further net change will occur.
-
The
equilibrium constant
for the reaction is determined the standard Gibbs energy change:
ΔG° = - RT ln K p - The sign of the temperature dependence of the equilibrium constant is governed by the sign of Δ H °. This is the basis of the Le Chatelier Principle .
- The Gibbs energies of solid and liquid components are constants that do not change with composition. Thus in heterogeneous reactions such as phase changes, the total Gibbs energy does not pass through a minimum and when the system is not at equilibrium only all-products or all-reactants will be stable.
- Two reactions are coupled when the product of one reaction is consumed in the other. If Δ G ° for the first reaction is positive, the overall process can still be spontaneous if Δ G ° for the second reaction is sufficiently negative— in which case the second reaction is said to "drive" the first reaction.
Under conditions of constant temperature and pressure, chemical change will tend to occur in whatever direction leads to a decrease in the value of the Gibbs energy . In this lesson we will see how G varies with the composition of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The equilibrium composition of the mixture is determined by ΔG° which also defines the equilibrium constant K.
The Road to Equilibrium is Down the Gibbs Energy Hill
This means, of course, that if the total Gibbs energy \(G\) of a mixture of reactants and products goes through a minimum value as the composition changes, then all net change will cease— the reaction system will be in a state of chemical equilibrium . You will recall that the relative concentrations of reactants and products in the equilibrium state is expressed by the equilibrium constant . In this lesson we will examine the relation between the Gibbs energy change for a reaction and the equilibrium constant.
To keep things as simple as possible, we will consider a homogeneous chemical reaction of the form
\[A + B \rightleftharpoons C + D\]
in which all components are gases at the temperature of interest. If the sum of the standard Gibbs energies of the products is less than that of the reactants, Δ G ° for the reaction will be negative and the reaction will proceed to the right. But how far? If the reactants are completely transformed into products, the equilibrium constant would be infinity. The equilibrium constants we actually observe all have finite values, implying that even if the products have a lower Gibbs energy than the reactants, some of the latter will always remain when the process comes to equilibrium.
A homogeneous reaction is one in which everything takes place in a single gas or liquid phase.
To understand how equilibrium constants relate to Δ G ° values, assume that all of the reactants are gases, so that the Gibbs energy of gas A, for example, is given at all times by
\[G_A = G_A^° + RT \ln P_A \label{5-1}\]
The Gibbs energy change for the reaction is sum of the Gibbs energies of the products, minus the sum of Gibbs energies of the reactants:
\[\Delta G = \underbrace{G_C + G_D}_{\text{products}} \underbrace{– G_A – G_B}_{\text{reactants}} \label{5-2}\]
Using Equation \(\ref{5-1}\) to expand each term on the right of Equation \ref{5-2}, we have
\[\Delta G = (G^°_C + RT \ln P_C) + (G^°_D + RT \ln P_D) – (G^°_B + RT \ln P_B) – (G^°_A + RT \ln P+A) \label{5-3}\]
We can now express the \(G^°\) terms collectively as \(\Delta G^°\), and combine the logarithmic pressure terms into a single fraction
\[ \Delta G = \Delta G° + RT \ln \left( \dfrac{P_CP_D}{P_AP_B} \right) \label{5-4}\]
which is more conveniently expressed in terms of the reaction quotient \(Q\).
\[\Delta{G} = \Delta G^° + RT \ln Q \label{5-5}\]
The Gibbs energy \(G\) is a quantity that becomes more negative during the course of any natural process. Thus as a chemical reaction takes plac e, \(G\) only falls and will never become more positive. Eventually a point is reached where any further transformation of reactants into products would cause \(G\) to increase. At this point \(G\) is at a minimum (see below), and no further net change can take place; the reaction is then at equilibrium.
Although Equations \ref{5-1}-\ref{5-5} are strictly correct only for perfect gases, we will see later that equations of similar form can be applied to many liquid solutions by substituting concentrations for pressures.
Consider the gas-phase dissociation reaction
\[\ce{N_2O_4 \rightarrow 2 NO_2 } \nonumber\]
which is a simple example of the Gibbs energy relationships in a homogeneous reaction.
The Gibbs energy of 1 mole of N 2 O 4 (1) is smaller than that of 2 moles of NO 2 (2) by 5.3 kJ; thu s \( \Delta G^o = +5.3\, \text{kJ}\) fo r the complete transformation of reactants into products. The straight diagonal line shows the Gibbs energy of all possible compositions if the two gases were prevented from mixing. The red curved line show the Gibbs energy of the actual reaction mixture. This passes through a minimum at (3) where 0.814 mol of \(N_2O_4\) are in equilibrium with 0.372 mol of \(NO_2\). The difference (4) corresponds to the Gibbs energy of mixing of reactants and products which always results in an equilibrium mixture whose Gibbs energy is lower than that of either pure reactants or pure products. Thus some amount of reaction will occur even if Δ G ° for the process is positive.
What’s the difference between Δ G and Δ G °?
It’s very important to be aware of this distinction; that little ° symbol makes a world of difference! First, the standard Gibbs energy change Δ G ° has a single value for a particular reaction at a given temperature and pressure; this is the difference
\[ \sum G^°_{f} (\text{products}) – \sum G^°_{f}(\text{reactants}) \]
that are tabulated in thermodynamic tables . It corresponds to the Gibbs energy change for a process that never really happens: the complete transformation of pure N 2 O 4 into pure NO 2 at a constant pressure of 1 atm.
The other quantity \(\Delta G\), defined by Equation \(\ref{5-5}\), represents the total Gibbs energies of all substances in the reaction mixture at any particular system composition. In contrast to \(\Delta G^°\) which is a constant for a given reaction, \(\Delta G\) varies continuously as the composition changes, finally reaching zero at equilibrium. \(\Delta G\) is the “distance” (in Gibbs energy) from the equilibrium state of a given reaction. Thus for the limiting cases of pure \(\ce{N_2O_4}\) or \(\ce{NO_2}\) (a s far from the equilibrium state as the system can be!),
\[Q = \dfrac{[NO_2]^2}{[N_2O_4]} = \pm\infty\]
which makes the logarithm in Equation \(\ref{5-5}\), and thus the value o f \(\Delta G\), approach the same asymptotic limits (1) or (2) . As the reaction proceeds in the appropriate direction \(\Delta G\) approaches zero; once there (3) , the system is at its equilibrium composition and no further net change will occur.
The standard molar Gibbs energy change for this very simple reaction is –2.26 kJ, but mixing of the unreacted butane with the product brings the Gibbs energy of the equilibrium mixture down to about –3.1 kJ mol –1 at the equilibrium composition corresponding to 77 percent conversion.
Notice particularly that
- The sum of the Gibbs energies of the two gases ( n -butane and iso -butane) separately varies linearly with the composition of the mixture (red line ).
- The green curve adds the Gibbs energy of mixing to the above sum; its minimum defines the equilibrium composition.
- As the composition approaches the equilibrium value , \(ΔG\) (which denotes how much farther the Gibbs energy of the system can fall) approaches zero.
The detailed calculations that lead to the values shown above can be found here .
Why reactions lead to mixtures of reactants and products
We are now in a position to answer the question posed earlier: if Δ G ° for a reaction is negative, meaning that the Gibbs energies of the products are more negative than those of the reactants, why will some of the latter remain after equilibrium is reached? The answer is that no matter how low the Gibbs energy of the products, the Gibbs energy of the system can be reduced even more by allowing some of the products to be "contaminated" (i.e., diluted) by some reactants. Owing to the entropy associated with mixing of reactants and products, no homogeneous reaction will be 100% complete. An interesting corollary of this is that any reaction for which a balanced chemical equation can be written can in principle take place to some extent, however minute that might be.
Gibbs energies of mixing of products with reactants tend to be rather small, so for reactions having Δ G ° values that are highly negative or positive (±20 kJ mol –1 , say), the equilibrium mixture will, for all practical purposes, be either [almost] "pure" reactants or products.
The Equilibrium Constant
Now let us return to Equation \(\ref{5-5}\) which we reproduce here:
\[\Delta{G} = \Delta{G^°} + RT \ln Q \]
As the reaction approaches equilibrium, \(\Delta G\) becomes less negative and finally reaches zero. At equilibrium \(\Delta{G} = 0\) and \(Q = K\), so we can write ( must know this!)
\[\Delta{G^°} = –RT \ln K_p \label{5-6}\]
in which \(K_p\), the equilibrium constant expressed in pressure units, is the special value of \(Q\) that corresponds to the equilibrium composition.
This equation is one of the most important in chemistry because it relates the equilibrium composition of a chemical reaction system to measurable physical properties of the reactants and products. If you know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the mechanism of the reaction.
Instead of writing Equation \(\ref{5-6}\) in terms of K p , we can use any of the other forms of the equilibrium constant such as K c (concentrations), K x (mole fractions), K n (numbers of moles), etc. Remember, however, that for ionic solutions especially, only the K a , in which activities are used, will be strictly valid.
It is often useful to solve Equation \(\ref{5-6}\) for the equilibrium constant, yielding
\[ K = \exp {\left ( {-\Delta G \over RT} \right )} \label{5-7}\]
This relation is most conveniently plotted against the logarithm of \(K\) as shown in Figure \(\PageIndex{3}\), where it can be represented as a straight line that passes through the point (0,0).
Calculate the equilibrium constant for the reaction from the following thermodynamic data :
\[\ce{H^{+}(aq) + OH^{–}(aq) <=> H_2O(l)} \nonumber\]
|
\(H^+(aq)\)
|
|---|
* Note that the standard entropy of the hydrogen ion is zero by definition. This reflects the fact that it is impossible to carry out thermodynamic studies on a single charged species. All ionic entropies are relative to that of \(\ce{H^{+}(aq)}\), which explains why some values (as for aqueous hydroxide ion) are negative.
Solution
From the above data, we can evaluate the following quantities:
\[\begin{align*} \Delta{H}^o &= \sum \Delta H^o_{f}(\text{products}) - \sum \Delta H^o_{f}(\text{reactants}) \\[4pt] &= (–285.8) - (-230) \\[4pt] &= –55.8\, kJ \; mol^{-1} \end{align*}\]
\[\begin{align*}\Delta{S}^o &= \sum \Delta S^o (\text{products}) - \sum \Delta S° (\text{reactants}) \\[4pt] &= (70.0) – (–10.9) \\[4pt] &= +80.8\, J \; K^{-1}\; mol^{-1} \end{align*}\]
The value of \(\Delta{G}°\) at 298 K is
\[\begin{align*} \Delta H^o – T\Delta S^o &= (–55800) – (298)(80.8) \\[4pt] &= –79900\, J\, mol^{–1} \end{align*}\]
From Equation \(\ref{5-7}\) we have
\[\begin{align*} K &= \exp\left(\dfrac{–79900}{8.314 \times 298}\right) \\[4pt] &= e^{32.2} = 1.01 \times 10^{–14} \end{align*}\]
Equilibrium and Temperature
We have already discussed how changing the temperature will increase or decrease the tendency for a process to take place, depending on the sign of Δ S °. This relation can be developed formally by differentiating the relation
\[ \Delta G^° = \Delta H^° – T\Delta S^° \label{5-8}\]
with respect to the temperature:
\[ \dfrac{d(-\Delta G^°)}{dT} = -\Delta S^° \label{5-9}\]
Hence, the sign of the entropy change determines whether the reaction becomes more or less allowed as the temperature increases.
We often want to know how a change in the temperature will affect the value of an equilibrium constant whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value \(K_1\) at temperature \(T_1\) and we wish to estimate \(K_2\) at temperature \(T_2\). Expanding Equation \(\ref{5-7}\) in terms of \(\Delta H^°\) and \(\Delta S^°\), we obtain
\[–RT_1 \ln K_1 = \Delta H^ ° – T_1 \Delta S^° \]
and
\[–RT_2 \ln K_2 = \Delta H ^° – T_2 \Delta S^°\]
Dividing both sides by RT and subtracting, we obtain
\[ \ln K_1 - \ln K_2 = - \left( \dfrac{\Delta H^°}{RT_1} -\dfrac{\Delta H^°}{RT_2} \right) \label{5-10}\]
Which is most conveniently expressed as the ratio
\[ \ln \dfrac{K_1}{K_2} = - \dfrac{\Delta H^°}{R} \left( \dfrac{1}{T_1} -\dfrac{1}{T_2} \right) \label{5-11}\]
This is its theoretical foundation of Le Ch a telier's Principle with respect to the effect of the temperature on equilibrium:
- if the reaction is exothermic \(\Delta H^° < 0\), then increasing temperature will make the second exponential term smaller and \(K\) will decrease. The equilibrium will then “shift to the left”.
- If \(\Delta H^° > 0\), then increasing T will make the exponent less negative and \(K\) will increase and the equilibrium will “shift to the right”.
This is an extremely important relationship, but not just because of its use in calculating the temperature dependence of an equilibrium constant. Even more important is its application in the “reverse” direction to experimentally determine Δ H ° from two values of the equilibrium constant measured at different temperatures. Direct calorimetric determinations of heats of reaction are not easy to make; relatively few chemists have the equipment and experience required for this rather exacting task. Measurement of an equilibrium constant is generally much easier, and often well within the capabilities of anyone who has had an introductory Chemistry course. Once the value of ΔH° is determined it can be combined with the Gibbs energy change (from a single observation of K , through Equation \(\ref{5-7}\)) to allow Δ S ° to be calculated through Equation \(\ref{5-9}\).
Equilibrium Without Mixing: it's all or nothing
You should now understand that for homogeneous reactions (those that take place entirely in the gas phase or in solution) the equilibrium composition will never be 100% products, no matter how much lower their Gibbs energy relative to the reactants. As was summarized in the N 2 O 4 -dissociation example discussed previously. This is due to "dilution" of the products by the reactants. In heterogeneous reactions (those which involve more than one phase) this dilution, and the effects that flow from it, may not be possible.
A particularly simple but important type of a heterogeneous process is phase change . Consider, for example, an equilibrium mixture of ice and liquid water. The concentration of H 2 O in each phase is dependent only on the density of the phase; there is no way that ice can be “diluted” with water, or vice versa. This means that at all temperatures other than the freezing point, the lowest Gibbs energy state will be that corresponding to pure ice or pure liquid. Only at the freezing point, where the Gibbs energies of water and ice are identical, can both phases coexist, and they may do so in any proportion.
Gibbs energy of the ice-water system
Only at 0°C can ice and liquid water coexist in any proportion. Note that in contrast to the homogeneous N 2 O 4 example, there is no Gibbs energy minimum at intermediate compositions.
Coupled Reactions
Two reactions are said to be coupled when the product of one of them is the reactant in the other:
\[A \rightarrow B \nonumber\]
and
\[B \rightarrow C \nonumber\]
If the standard Gibbs energy of the first reaction is positive but that of the second reaction is sufficiently negative, then for the overall process will be negative and we say that the first reaction is “driven” by the second one. This, of course, is just another way of describing an effect that you already know as the Le Chatelier principle: the removal of substance B by the second reaction causes the equilibrium of the first to “shift to the right”. Similarly, the equilibrium constant of the overall reaction is the product of the equilibrium constants of the two steps.
| 1 Cu 2 S(s) → 2 Cu(s) + S(s) | Δ G ° = + 86.2 kJ | Δ H ° = + 76.3 kJ |
| 2 S(s) + O 2 (g)→ SO 2 (g) | Δ G ° = –300.1 kJ | Δ H ° = + 296.8 kJ |
| 3 Cu 2 S(s)→ 2 Cu(s) + SO 2 (g) | Δ G ° = –213.9 kJ | Δ H ° = – 217.3 kJ |
In the above example, reaction 1 is the first step in obtaining metallic copper from one of its principal ores. This reaction is endothermic and it has a positive Gibbs energy change, so it will not proceed spontaneously at any temperature. If Cu 2 S is heated in the air, however, the sulfur is removed as rapidly as it is formed by oxidation in the highly spontaneous reaction 2 , which supplies the Gibbs energy required to drive 1 . The combined process, known as roasting , is of considerable industrial importance and is one of a large class of processes employed for winning metals from their ores.
|
libretexts
|
2025-03-17T19:53:18.403194
| 2017-02-24T04:39:16 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.06%3A_Free_energy_and_Equilibrium",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.6: Free Energy and Equilibrium",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.07%3A_Some_Applications_of_Entropy_and_Free_Energy
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15.7: Some Applications of Entropy and Free Energy
You are expected to be able to define and explain the significance of terms identified in bold
- What do all of the colligative properties of solutions have in common in terms of entropy?
- At the melting and boiling points, the two phases in equilibrium have equal numbers of energetically accessible microstates. How does the addition of a nonvolatile solute upset this balance in the two cases?
- Describe the effect of increasing the hydrostatic pressure on a liquid has on its vapor pressure, and suggest a reason for this.
- Under what conditions does osmotic flow occur? What is the fundamental driving force? What is the definition of osmotic pressure ?
- Explain the critical role of temperature in the extraction of metals from their oxide ores.
- Describe the bioenergetic cycle of free energy in terms of the roles of glucose, electron-acceptors, ATP, and photosynthesis.
- Define aerobic and non-aerobic oxidation, and explain why they differ in efficiency.
- Sketch a simple proton-free energy diagram showing why an acid reacts with a base.
- Sketch a simple electron-free energy diagram showing why a metal reacts with an oxidizing agent that is below it in the electromotive series.
- What thermodynamic factors are involved in the spontaneous contraction of an elastic material such as a rubber band?
Thermodynamics may appear at first to be a rather esoteric subject, but when you think about it, almost every chemical (and biological) process is governed by changes in entropy and free energy. Examples such as those given below should help you connect these concepts with the real world.
Colligative properties of solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute , and because of this commonality they are referred to as colligative properties (Lat. co ligare , connected to.) The key role of the solvent concentration is obscured by the expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered in the unit on solutions. Our purpose here is to offer a more complete explanation of why these phenomena occur.
The conventional explanation is that dilution of a liquid by a non-volatile solute reduces the vapor pressure or "escaping tendency" of the liquid in that phase, leading to a net transport of material into that phase. Equilibrium can then be re-established by adjusting the temperature of applying a hydrostatic pressure to the solution. The stable phase at any temperature will be the one having the lowest free energy, and from which molecules have the smallest escaping tendency.
A more fundamental approach is to recall that dilution of a liquid creates uncountable numbers of new microstates, increasing the density of quantum states in the solution compared to that in the pure liquid. To the extent that these new states are thermally accessible, they will become populated at the expense of some of the microstates of the other phase. Equilibrium between two phases can be restored by adjusting the temperature or pressure so that equal numbers of microstates are occupied in each phase.
Boiling point elevation and freezing point depression
These effects are readily understood in terms of the schematic diagrams shown below. The red shading indicates the temperatures required to make equal numbers of microstates thermally accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases. The shifts of the boiling- and and freezing points in opposite directions reflects the higher energies of vapor microstates and lower energies of solid microstates in relation to those of the liquid. Note that these diagrams are purely schematic and nowhere near to scale!
As mentioned previously, the more conventional explanation of bp elevation and fp depression is given in terms of vapor pressures. The latter are measures of the free energies of molecules in a phase, and the relationships are best understood by showing plots of – T Δ S ° for each phase in terms of the temperature:
Melting and boiling points of a pure substance.
The stable phase at any temperature will be the one having the lowest free energy, and from which molecules have the smallest escaping tendency.
Melting- and boiling points of a solvent containing a non-volatile solute
.
The solute "dilutes" the solvent, reducing its free energy (purple line) and shifting the transition temperatures in opposite directions.
The pressure on a liquid affects its volatility
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25° C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of the quantized energy states in the liquid phase.
Why pressure affects the volatility of a liquid . Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
Osmotic flow and osmotic pressure
Molecules in a fluid are always subject to random thermal motions. If a given kind of molecule is present in greater concentration in one region of a fluid, the effect of these random motions will be to produce a net migration into the region of lower concentration; this migration is known as thermal diffusion .
Osmotic flow (commonly known simply as osmosis ) occurs when molecules of a solvent diffuse through a membrane that is permeable only to those molecules, and into a solution in which the solvent is "diluted" by the presence of solute molecules. In the simplest example, there is pure solvent on the left side of the membrane; on the other side the same solvent contains a solute whose molecules are too large to pass through the membrane. Because the solvent concentration on the right side will always be smaller than that on the left, osmotic flow will continue indefinitely if the right side can accommodate the increased volume.
If, however, the liquid on the right side of the membrane is placed under hydrostatic pressure, the driving force for osmotic flow will diminish. If the pressure is raised sufficiently high, osmotic flow will come to a halt; the system is then said to be in osmotic equilibrium. The pressure required to bring about osmotic equilibrium (and thus to stop osmotic flow) is known as the osmotic pressure , commonly denoted by Π (Greek capital pi ).
Osmotic equilibrium, like any kind of equilibrium, occurs when the free energies (that is, the escaping tendencies) of the diffusible molecules are the same on the two sides of the membrane. The free energy on the right side, initially lower due to the lower solvent concentration, is raised by the hydrostatic pressure, making it identical with that of the pure liquid on the left.
From the standpoint of microscopic energy states, the effect of the applied pressure is to slightly increase the spacing of solvent energy states on right side of the membrane so as to equalize the number of accessible states on the two sides, as shown here in a very schematic way.
Extraction of Metals from their Oxides
Since ancient times, the recovery of metals from their ores has been one of the most important applications of chemistry to civilization and culture. The oldest, and still the most common smelting process for oxide ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal. The basic reactions are:
\[MO + C \rightleftharpoons M + CO \label{2.1}\]
\[MO + ½ O_2 \rightleftharpoons M + ½ CO_2 \label{2.2}\]
\[MO + CO \rightleftharpoons M + CO_2 \label{2.3}\]
Each of these can be regarded as a pair of coupled reactions in which the metal M and the carbon are effectively competing for the oxygen atom. Using reaction \(\ref{2.1}\) as an example, it can be broken down into the following two parts:
\[MO \rightleftharpoons M + ½ O_2 \;\;\; ΔG^o > 0 \label{2.4}\]
\[C + ½ O_2 \rightleftharpoons CO \;\;\; ΔG^o < 0 \label{2.5}\]
At ordinary environmental temperatures, reaction \(\ref{2.4}\) is always spontaneous in the reverse direction (that is why ores form in the first place!), so Δ G ° of Reaction \(\ref{2.4}\) will be positive. Δ G ° for reaction \(\ref{2.5}\) is always negative, but at low temperatures it will not be sufficiently negative to drive \(\ref{2.4}\).
The smelting process depends on the different ways in which the free energies of reactions like \(\ref{2.4}\) and \(\ref{2.4}\) vary with the temperature. This temperature dependence is almost entirely dominated by the T Δ S ° term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δ n g , the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always accompanied by a large increase in the system volume so Δ S for this step is always positive and the reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that take up oxygen vary, however (Table 23.6.X).
|
Reaction
|
Δ
n
g
|
d
(Δ
G
°)/
dT
|
|---|---|---|
| C + ½ O 2 → CO |
0.5
|
<0>
|
| C + O 2 → CO 2 |
0
|
0
|
| CO + ½ O 2 →CO 2 |
–0.5
|
>0
|
A plot of the temperature dependences of the free energies of these reactions, superimposed on similar plots for the oxygen removal reactions \(\ref{2.4}\) is called an Ellingham diagram . For a given oxide MO to be smeltable, the temperature must be high enough that reaction \(\ref{2.4}\) falls below that of at least one of the oxygen-consuming reactions. The slopes of the lines on this diagram are determined by the sign of the entropy change.
Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were mainly those such as copper and lead, which can be obtained by smelting at the relatively low temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the development of technology capable of producing higher temperatures, such as the blast furnace. Smelting of aluminum oxide by carbon requires temperatures too high to be practical; commercial production of aluminum is accomplished by electrolysis of the molten ore.
Bioenergetics
Many of the reactions that take place in living organisms require a source of free energy to drive them. The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most bacteria, is the sugar glucose . Oxidation of glucose to carbon dioxide and water is accompanied by a large negative free energy change
\[\ce{C6H12O6 + 6 O2 \rightarrow 6 CO2 + 6 H2O} \quad ΔG^o = – 2,880 \,kJ\, mol^{–1} \label{6-1}\]
Of course it would not do to simply “burn” the glucose in the normal way; the energy change would be wasted as heat, and rather too quickly for the well-being of the organism! Effective utilization of this free energy requires a means of capturing it from the glucose and then releasing it in small amounts when and where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more steps in which the energy liberated in each stage is captured by an “energy carrier” molecule, of which the most important is adenosine diphosphate , known as ADP . At each step in the breakdown of glucose, an ADP molecule reacts with inorganic phosphate (denoted by P i ) and changes into ATP :
\[ADP + P_i \rightarrow ATP\]
\[ΔG^° = +30 \;kJ\; mol^{–1} \label{(6-2}\]
The 30 kJ mol –1 of free energy stored in each ATP molecule is released when the molecule travels to a site where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP, which eventually finds its way back the site of glucose metabolism for recycling back into ATP. The complete breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP according to the overall reaction
\[\ce{C6H12O6 + 6 O2 + 38 Pi + 38 ADP 38 ATP + 6CO2 + 44 H2O} \label{6-3}\]
For each mole of glucose metabolized, 38 × (30 kJ) = 1140 kJ of free energy is captured as ATP, representing an energy efficiency of 1140/2880 = 0.4. That is, 40% of the free energy obtainable from the oxidation of glucose is made available to drive other metabolic processes. The rest is liberated as heat.
Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose and starches that, like glucose, have the empirical formula {CH 2 O}. Animals obtain this food by eating plants or other animals. Ultimately, all food comes from plants, most of which are able to make their own glucose from CO 2 and H 2 O through the process of photosynthesis. This is just the reverse of Eq 40 in which the free energy is supplied by the quanta of light absorbed by chlorophyll and other photosynthetic pigments.
This describes aerobic respiration , which evolved after the development of photosynthetic life on Earth began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at the cellular level, and it is believed that aerobic respiration developed as a means to protect organisms from this peril. Those that did not adapt have literally “gone underground” and constitute the more primitive anaerobic bacteria.
The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available, but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead of CO 2 :
\[\ce{C6H12O6 + 2 ADP → 2 CH3CH(OH)COOH} \quad ΔG^o = –218\, kJ\, mol^{–1} \label{6-4}\]
In this process, only (2 × 30 kJ) = 60 kJ of free energy is captured, so the efficiency is only 28% on the basis of this reaction, and it is even lower in relation to glucose. In “aerobic” exercising, one tries to maintain sufficient lung capacity and cardiac output to supply oxygen to muscle cells at a rate that promotes the aerobic pathway.
Respiration and many other metabolic processes involve electron-transfer reactions.
Acid-base reactions: the fall of the proton
According to the widely useful Brønsted-Lowry concept , an acid is a proton donor and a base is a proton acceptor. In 1953, Gurney showed how this idea could be made even more useful by placing acid-base conjugate pairs on a proton-free energy scale .
In this view, acids are proton sources and bases are proton sinks . Protons fall spontaneously from acids to fill sinks in which the proton free energy levels are lower. The pH is a measure of the average proton free energy in the solution; when this quantity is the same as the proton free energy level of a conjugate pair, the two species are present in equal concentrations (this corresponds, of course to the equality of pH and pK a in the conventional theory.)
The proton-free energy concept is commonly employed in aquatic environmental chemistry in which multiple acid-base systems must be dealt with on a semi-quantitative bases. It is, however, admirably adapted to any presentation of acid-base chemistry, even at the first-year college level, and it seems a shame that it never seems to have made its way into the ordinary curriculum.
Oxidation-reduction: the fall of the electron
Oxidation-reduction reactions proceed in a direction that allows the electron to “fall” (in free energy) from a “source” to a “sink”. Later on when you study electrochemistry you will see how this free energy can manifest itself as an electrical voltage and be extracted from the system as electrical work.
Electron-free energy levels and the electromotive series
You may already have seen an electromotive force table that shows the relative tendencies of different reducing agents to donate electrons. The same information can be presented in another way that relates electron-donating (reducing) power to the "fall" in free energy that electrons undergo when they are transferred to an oxidizing agent.
Any available sink on the right side will tend to drain electrons from a source above it. For example, immersion of metallic zinc in a solution of CuSO 4 allow electrons to "fall" from the high free energy they possess in Zn down to the lower free energy level in the newly-introduced Cu 2 + ions. This will result in the reduction of Cu 2 + to metallic copper and the oxidation of the zinc (red arrows.)
Similarly, addition of chlorine to water will introduce a new electron sink (Cl 2 ) that lies below the free energy of the accessible electrons in H 2 O, draining them away and producing O 2 and Cl – (blue arrows.) Note especially the positions of the H 2 / H + and H 2 O/O 2 ,H + couples on this chart, as they define the range of E °s for substances that will not decompose water. The zero for G ° is arbitrarily set at the electron activity at which the H 2 /H + couple is at equilibrium; this corresponds to E ° = 0 volts on the ordinary electromotive scale.
Coupled redox reactions are central to bioenergetics
\[\{CH_2O\} → CO_2 + H_2O \label{6-5}\]
In the above chart, the zero for G ° is arbitrarily set at the electron free energy required to maintain the H 2 /H + couple is at equilibrium; this corresponds to E° = 0 volts on the ordinary electromotive scale.
Organisms that live in environments where oxygen is lacking, such as marshes, muddy soils, and the intestinal tracts of animals, must utilize other electron acceptors to extract free energy from carbohydrate. A wide variety of inorganic ions such as sulfate and nitrate, as well as other carbon compounds can serve as electron acceptors, yielding the gaseous products like H 2 S, NH 3 and CH 4 which are commonly noticed in such locations. From the location of these acceptors on the scale, it is apparent that the amount of energy they can extract from a given quantity of carbohydrate is much less than for O 2 . One reason that aerobic organisms have dominated the earth is believed to be the much greater energy-efficiency of oxygen as an electron acceptor.
What did organisms use for food before there was a widespread supply of carbohydrate in the world? Any of the electron sources near the upper left of the table can in theory serve this function, although at reduced energy efficiency. As a matter of fact, there are still a number of these autotrophic bacteria around whose “food” is CH 4 , CH 3 OH, FeCO 3 , and even H 2 !
Thermodynamics of Rubber Bands
Rubber is composed of random-length chains of polymerized isoprene molecules. The poly(isoprene) chains are held together partly by weak intermolecular forces, but are joined at irregular intervals by covalent disulfide bonds so as to form a network. The intermolecular forces between the chain fragments tend to curl them up, but application of a tensile force can cause them to elongate. The disulfide cross-links prevent the chains from slipping apart from one another, thus maintaining the physical integrity of the material. Without this cross-linking, the polymer chains would behave more like a pile of spaghetti.
The ability of rubber bands and other elastic substances to undergo a change in physical dimensions in response to a change in the applied stretching force is subject to the same laws of thermodynamics as any other physical process. You can investigate this for yourself.
Hold a rubber band (the thicker the better) against your upper lip, and notice how the temperature changes when the band is stretched, and then again when it is allowed to contract.
- Use the results of this observation to determine the signs of Δ H , Δ G and Δ S for the process \[rubber_{stretched} \rightarrow rubber_{unstretched}\]
- How will the tendency of the stretched rubber to contract be changed if the temperature is raised?
Solution
- Contraction is obviously a spontaneous process, so Δ G is negative. You will have observed that heat is given off when the band is stretched, meaning that contraction is endothermic, so Δ H > 0. Thus according to Δ G = Δ H – TΔ S , Δ S for the contraction process must be positive.
- Because Δ S > 0, contraction of the rubber becomes more spontaneous as the temperature is raised.
Why is Rubber Elastic?
When an ordinary material is placed under tension, the strain energy is taken up by bond distortions and is stored as electrostatic potential energy which rises very rapidly so as to greatly inhibit further elongation. In rubber-like polymers, this does not happen; the strain energy is instead stored as thermal (kinetic) energy.
Free polymer chains naturally tend to curl up in complex and ever-changing ways as thermal energy allows random bond rotations to take place. In a rubber-like material in its relaxed state, the portions of the polymer chains between cross-links are continually jumping between different randomly-coiled configurations.
When the rubber is stretched, the polymer segments straighten out as the applied force overcomes the weak dispersion force interactions that caused the strands to curl. Each chain segment is pulled into an almost-straight conformation, thus greatly reducing the quantity of thermal energy it can store. The excess thermal energy spreads into the material and is lost in the form of heat. When the rubber relaxes, the polymer strands curl up again and soak up thermal energy.
The spontaneous contraction of rubber is largely an entropy-driven process. The number of energetically-equivalent ways of distributing thermal energy amongst the nearly-linear polymer chains of the stretched state of rubber is insignificant compared to those available when the chains are curled up in random ways, so the un-stretched form of rubber is statistically the most likely one by overwhelming odds. As noted in part (b) of the above problem example, the gain in entropy when the rubber contracts drives Δ G more negative at higher temperatures. This means that a rubber band, held at constant tension in stretched state, will contract when it is heated.
This fact can be put to use in an interesting way. Replace the spokes of a bicycle wheel with rubber bands, and shine a heat lamp on one side of the wheel. The contraction of the heated bands will shift the wheel off-center, causing it to rotate. This rotation will continue indefinitely as long as the heat source is present. The device has become a heat engine whose working "fluid" is rubber!
This recalls the classic perpetual motion machine design in which a wheel is caused to rotate by continually-shifting unbalanced weights (the one depicted here uses hinged vials of mercury). That, as we saw, would violate the Second Law by producing work in a cyclic process without degrading heat to a lower temperature. The rubber-band heat engine avoids this pitfall by absorbing heat from an external source on one side of the wheel, and releasing it at a lower temperature on the unheated side.
|
libretexts
|
2025-03-17T19:53:18.512975
| 2013-10-03T01:38:14 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.07%3A_Some_Applications_of_Entropy_and_Free_Energy",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.7: Some Applications of Entropy and Free Energy",
"author": "Stephen Lower"
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.08%3A_Quantum_states_Microstates_and_Energy_spreading_in_Reactions
|
15.8: Quantum states, Microstates, and Energy spreading in Reactions
You are expected to be able to define and explain the significance of terms identified in bold .
-
energy
within a system.
- The entropy of a substance increases with its molecular weight and complexity and with temperature. The entropy also increases as the pressure or concentration becomes smaller. Entropies of gases are much larger than those of condensed phases.
The Physical Meaning of Entropy
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system. This “spreading and sharing” can be spreading of the thermal energy into a larger volume of space or its sharing amongst previously inaccessible microstates of the system. The following table shows how this concept applies to a number of common processes.
| system and process | source of entropy increase of system |
|---|---|
| A deck of cards is shuffled, or 100 coins, initially heads up, are randomly tossed. | This has nothing to do with entropy because macro objects are unable to exchange thermal energy with the surroundings within the time scale of the process |
| Two identical blocks of copper, one at 20°C and the other at 40°C, are placed in contact. | The cooler block contains more unoccupied microstates, so heat flows from the warmer block until equal numbers of microstates are populated in the two blocks. |
| A gas expands isothermally to twice its initial volume. | A constant amount of thermal energy spreads over a larger volume of space |
| 1 mole of water is heated by 1C°. | The increased thermal energy makes additional microstates accessible. (The increase is by a factor of about 10 20,000,000,000,000, 000,000,000 .) |
| Equal volumes of two gases are allowed to mix. | The effect is the same as allowing each gas to expand to twice its volume; the thermal energy in each is now spread over a larger volume. |
| One mole of dihydrogen, H 2 , is placed in a container and heated to 3000K. | Some of the H 2 dissociates to H because at this temperature there are more thermally accessible microstates in the 2 moles of H. |
| The above reaction mixture is cooled to 300K. | The composition shifts back to virtually all H 2 because this molecule contains more thermally accessible microstates at low temperatures. |
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Thermal energy is the portion of a molecule's energy that is proportional to its temperature , and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation . Since there are three directions in space, all molecules possess three modes of translational motion .
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation ; a linear molecule such as CO 2 in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal vibrations . For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the quantized translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at an arbitrary temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels.
Notice that the spacing between the quantized translational levels is so minute that they can be considered nearly continuous. This means that at all temperatures, the thermal energy of a collection of molecules resides almost exclusively in translational microstates. At ordinary temperatures (around 25° C), most of the molecules are in their zero-level vibrational and rotational states (corresponding to the bottom-most bars in the diagram.) The prevalence of vibrational states is so overwhelming that we can effectively equate the thermal energy of molecules with translational motions alone.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics. A very important point to bear in mind is that the number of discrete microstates that can be populated by an arbitrary quantity of energy depends on the spacing of the states. As a very simple example, suppose that we have two molecules (depicted by the orange dots) in a system total available thermal energy is indicated by the yellow shading.
In the system with the more closely-spaced energy levels, there are three possible microstates, while in the one with the more widely-spaced levels, only two possibilities are available.
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states.
At the atomic and molecular level, all energy is quantized ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 600 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
- Addition of energy quanta (higher temperature),
- Increase in the number of molecules (resulting from dissociation, for example).
- the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Using Entropy to Understand Spontaneous Processes
Heat Death
Energy is conserved ; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery. The profundity of this conclusion was recognized around 1900, when it was first described at the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever happen.
Gases Expansions
Everybody knows that a gas, if left to itself, will tend to expand and fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of microstates its thermal energy can occupy. Since all such states within the thermally accessible range of energies are equally probable, the expansion of the gas can be viewed as a consequence of the tendency of thermal energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
Imagine a gas initially confined to one half of a box (Figure \(\PageIndex{4}\)). The barrier is then removed so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space. In terms of the spreading of thermal energy, Figure \(\PageIndex{5}\) may be helpful. The tendency of a gas to expand is due to the more closely-spaced thermal energy states in the larger volume .
Mixing and dilution
Mixing and dilution really amount to the same thing, especially for idea gases. Replace the pair of containers shown above with one containing two kinds of molecules in the separate sections (Figure \(\PageIndex{6}\)). When we remove the barrier, the "red" and "blue" molecules will each expand into the space of the other. (Recall Dalton's Law that "each gas is a vacuum to the other gas".) However, notice that although each gas underwent an expansion, the overall process amounts to what we call "mixing".
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. But bear in mind that whereas the enthalpy associated with the expansion of a perfect gas is by definition zero, Δ H 's of mixing of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute.
It's unfortunate the the simplified diagrams we are using to illustrate the greater numbers of energetically accessible microstates in an expanded gas or a mixture of gases fail to convey the immensity of this increase. Only by working through the statistical mathematics of these processes (beyond the scope of first-year Chemistry!) can one gain an appreciation of the magnitude of the probabilities of these spontaneous processes.
It turns out that when just one molecule of a second gas is inroduced into the container of another gas, an unimaginably huge number of new configurations becom available. This happens because the added molecule (indicated by the blue arrow in the diagram) can in principle replace any one of the old (red) ones, each case giving rise to a new microstate.
Heat Flow
ust as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler one always operates in the direction “warmer-to-cooler” because this allows thermal energy to populate a larger number of energy microstates as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”. In this simplified schematic diagram, the "cold" and "hot" bodies differ in the numbers of translational microstates that are occupied, as indicated by the shading. When they are brought into thermal contact, a hugely greater number of microstates are created, as is indicated by their closer spacing in the rightmost section of the diagram, which represents the combined bodies in thermal equilibrium. The thermal energy in the initial two bodies fills these new microstates to a level (and thus, temperature) that is somewhere between those of the two original bodies.
Note that this explanation applies equally well to the case of two solids brought into thermal contact, or two the mixing of two fluids having different temperatures.
As you might expect, the increase in the amount of energy spreading and sharing is proportional to the amount of heat transferred q , but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T , the degree of dilution of the thermal energy is given by
\[\dfrac{q}{T}\]
To understand why we have to divide by the temperature, consider the effect of very large and very small values of \(T\) in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are initially occupied, so the amount of energy spreading into vacant states can be very great. Conversely, if the temperature is initially large, more thermal energy is already spread around within it, and absorption of the additional energy will have a relatively small effect on the degree of thermal disorder within the body.
Chemical Reactions and Equilibrium
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
- The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H 2 → 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H 2 . Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and this molecule would not exist.
- In order for this dissociation to occur, however, a quantity of thermal energy (heat) \( q =Δ U\) must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in \(H\), as indicated by the vertical displacement of the right half in each of the four panels below.
In Figure \(\PageIndex{8}\) a re schematic representations of the translational energy levels of the two components H and H 2 of the hydrogen dissociation reaction. The shading shows how the relative populations of occupied microstates vary with the temperature, causing the equilibrium composition to change in favor of the dissociation product.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
- As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T 1 , the number of populated states of H 2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
- At some temperature T 2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H 2 and “2H” in equal amounts; that is, the mole ratio of H 2 /H will be 1:2.
- As the temperature rises to T 3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant, thus favoring dissociation.
The result is exactly what the Le Chatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures.
The following table generalizes these relations for the four sign-combinations of ΔH and ΔS. (Note that use of the standard ΔH° and ΔS° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.)
> 0
This
combustion reaction
, like most such reactions, is
spontaneous at all temperatures
. The positive entropy change is due mainly to the greater mass of CO
2
molecules compared to those of O
2
.
< 0
- Δ H ° = –46.2 kJ
- Δ S ° = –389 J K –1
- Δ G ° = –16.4 kJ at 298 K
The decrease in moles of gas in the
Haber ammonia synthesis
drives the entropy change negative, making the reaction
spontaneous only at low temperatures
. Thus higher
T
, which speeds up the reaction, also reduces its extent.
> 0
- Δ H ° = 55.3 kJ
- Δ S ° = +176 J K –1
- Δ G ° = +2.8 kJ at 298 K
Dissociation reactions
are typically endothermic with positive entropy change, and are therefore
spontaneous at high temperatures
.
Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
< 0
- Δ H ° = 33.2 kJ
- ΔS° = –249 J K – 1
- Δ G ° = +51.3 kJ at 298 K
This reaction is not spontaneous at any temperature , meaning that its reverse is always spontaneous . But because the reverse reaction is kinetically inhibited, NO 2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Phase Changes
Everybody knows that the solid is the stable form of a substance at low temperatures, while the gaseous state prevails at high temperatures. Why should this be? The diagram in Figure \(\PageIndex{9}\) shows that
- the density of energy states is smallest in the solid and greatest (much, much greater) in the gas, and
- the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.
Changes of phase involve exchange of energy with the surroundings (whose energy content relative to the system is indicated (with much exaggeration!) by the height of the yellow vertical bars in Figure \(\PageIndex{13}\). When solid and liquid are in equilibrium (middle section of diagram below), there is sufficient thermal energy (indicated by pink shading) to populate the energy states of both phases. If heat is allowed to flow into the surroundings, it is withdrawn selectively from the more abundantly populated levels of the liquid phase, causing the quantity of this phase to decrease in favor of the solid. The temperature remains constant as the heat of fusion is returned to the system in exact compensation for the heat lost to the surroundings. Finally, after the last trace of liquid has disappeared, the only states remaining are those of the solid. Any further withdrawal of heat results in a temperature drop as the states of the solid become depopulated.
Colligative Properties of Solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare , connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of why these phenomena occur.
Basically, these all result from the effect of dilution of the solvent on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities . The temperatures at which this occurs are depicted by the shading.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases.
Osmotic Pressure: Effects of Pressure on the Entropy
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised (Figure \(\PageIndex{13}\)). The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase.
Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
This phenomenon can explain osmotic pressure . Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this pressure \(\Pi\) is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium.
|
libretexts
|
2025-03-17T19:53:18.611712
| 2018-01-19T04:58:35 |
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"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.08%3A_Quantum_states_Microstates_and_Energy_spreading_in_Reactions",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "15.8: Quantum states, Microstates, and Energy spreading in Reactions",
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry
|
16: Electrochemistry
Electrochemistry is the study of electricity and how it relates to chemical reactions. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as redox reaction, or oxidation-reduction reaction .
-
- 16.2: Galvanic cells and Electrodes
- We can measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials. This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, each immersed each in a solution containing a dissolved salt of the corresponding metal.
-
- 16.3: Cell Potentials and Thermodynamics
- It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. For Example, zinc is more active because it can displace (precipitate) copper from solution. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
-
- 16.4: The Nernst Equation
- The standard cell potentials we discussed in a previous section refer to cells in which all dissolved substances are at unit activity, which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure (known as the fugacity) of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know.
-
- 16.5: Applications of the Nernst Equation
- We ordinarily think of the oxidation potential being controlled by the concentrations of the oxidized and reduced forms of a redox couple, as given by the Nernst equation. Under certain circumstances it becomes more useful to think of E as an independent variable that can be used to control the value of Q in the Nernst equation. This usually occurs when two redox systems are present, one being much more concentrated or kinetically active than the other.
-
- 16.6: Batteries and Fuel Cells
- One of the oldest and most important applications of electrochemistry is to the storage and conversion of energy. You already know that a galvanic cell converts chemical energy to work; similarly, an electrolytic cell converts electrical work into chemical free energy. Devices that carry out these conversions are called batteries. In ordinary batteries the chemical components are contained within the device itself. If the reactants are supplied from an external source, the device is a fuel cell.
-
- 16.7: Timeline of Battery Development
- Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive.
-
- 16.8: Electrochemical Corrosion
- Corrosion can be defined as the deterioration of materials by chemical processes. Of these, the most important by far is electrochemical corrosion of metals, in which the oxidation process M → M+ + e– is facilitated by the presence of a suitable electron acceptor, sometimes referred to in corrosion science as a depolarizer. In a sense, corrosion can be viewed as the spontaneous return of metals to their ores.
-
- 16.10: Electrolytic Cells and Electrolysis
- Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds.
|
libretexts
|
2025-03-17T19:53:18.680145
| 2016-02-06T18:42:36 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16: Electrochemistry",
"author": "Stephen Lower"
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.01%3A_Chemistry_and_Electricity
|
16.1: Chemistry and Electricity
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- Electroneutrality principle - Bulk matter cannot have a chemically-significant unbalance of positive and negative ions.
- Dissolution of a metal in water can proceed to a measurable extent only if some means is provided for removing the excess negative charge that remains. This can be by electron-acceptor ions in solution, or by drawing electrons out of the metal through an external circuit.
- Interfacial potentials - these exist at all phase boundaries. In the case of a metal in contact with an electrolyte solution, the interfacial region consists of an electric double layer.
- The potential difference between a metal and the solution is almost entirely located across the very thin double layer, leading to extremely large potential gradients in this region.
The connection between chemistry and electricity is a very old one, going back to ALESSANDRO VOLTA'S discovery, in 1793, that electricity could be produced by placing two dissimilar metals on opposite sides of a moistened paper.
Electroneutrality
Nature seems to strongly discourage any process that would lead to an excess of positive or negative charge in matter. Suppose, for example, that we immerse a piece of zinc metal in pure water. A small number of zinc atoms go into solution as \(Zn^{2+}\) ions, leaving their electrons behind in the metal:
\[Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.1}\]
As this process goes on, the electrons which remain in the zinc cause a negative charge to build up within the metal which makes it increasingly difficult for additional positive ions to leave the metallic phase. A similar buildup of positive charge in the liquid phase adds to this inhibition. Very soon, therefore, the process comes to a halt, resulting in a solution in which the concentration of \(Zn^{2+}\) is still too low (around 10 –10 M ) to be detected by ordinary chemical means.
There would be no build-up of this opposing charge in the two phases if the excess electrons could be removed from the metal or the positive ions consumed as the electrode reaction proceeds. For example, we could drain off the electrons left behind in the zinc through an external circuit that forms part of a complete electrochemical cell ; this we will describe later. Another way to remove these same electrons is to bring a good electron acceptor (that is, an oxidizing agent ) into contact with the electrode. A suitable acceptor would be hydrogen ions; this is why acids attack many metals. For the very active metals such as sodium, water itself is a sufficiently good electron acceptor.
The degree of charge unbalance that is allowed produces differences in electric potential of no more than a few volts, and corresponds to unbalances in the concentrations of oppositely charged particles that are not chemically significant. There is nothing mysterious about this prohibition, known as the electroneutrality principle; it is a simple consequence of the thermodynamic work required to separate opposite charges, or to bring like charges into closer contact. The additional work raises the free energy change of the process, making it less spontaneous.
The only way we can get the oxidation of the metal to continue is to couple it with some other process that restores electroneutrality to the two phases. A simple way to accomplish this would be to immerse the zinc in a solution of copper sulfate instead of pure water. The zinc metal quickly becomes covered with a black coating of finely-divided metallic copper. The reaction is a simple reduction-oxidation reaction involving a transfer of two electrons from the zinc to the copper:
\[ Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.2} \]
and
\[ Cu^{2+} + 2e^– \rightarrow Cu(s) \label{1.1.3}\]
The dissolution of the zinc is no longer inhibited by a buildup of negative charge in the metal, because the excess electrons are removed from the zinc by copper ions that come into contact with it. At the same time, the solution remains electrically neutral, since for each \(Zn\) ion introduced to the solution, one \(Cu\) ion is removed. The net reaction
\[Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s) \label{1.1.4}\]
quickly goes to completion.
Potential differences at interfaces
The transition region between two phases consists of a region of charge unbalance known as the electric double layer . As its name implies, this consists of an inner monomolecular layer of adsorbed water molecules and ions, and an outer diffuse region that compensates for any local charge unbalance that gradually merges into the completely random arrangement of the bulk solution. In the case of a metal immersed in pure water, the electron fluid within the metal causes the polar water molecules to adsorb to the surface and orient themselves so as to create two thin planes of positive and negative charge. If the water contains dissolved ions, some of the larger (and more polarizable) anions will loosely bond ( chemisorb ) to the metal, creating a negative inner layer which is compensated by an excess of cations in the outer layer.
Electrochemistry is the study of reactions in which charged particles (ions or electrons) cross the interface between two phases of matter, typically a metallic phase (the electrode ) and a conductive solution, or electrolyte . A process of this kind can always be represented as a chemical reaction and is known generally as an electrode process . Electrode processes (also called electrode reactions ) take place within the double layer and produce a slight unbalance in the electric charges of the electrode and the solution. Much of the importance of electrochemistry lies in the ways that these potential differences can be related to the thermodynamics and kinetics of electrode reactions. In particular, manipulation of the interfacial potential difference affords an important way of exerting external control on an electrode reaction.
The interfacial potential differences which develop in electrode-solution systems are limited to only a few volts at most. This may not seem like very much until you consider that this potential difference spans a very small distance. In the case of an electrode immersed in a solution, this distance corresponds to the thin layer of water molecules and ions that attach themselves to the electrode surface, normally only a few atomic diameters. Thus a very small voltage can produce a very large potential gradient. For example, a potential difference of one volt across a typical 10 –8 cm interfacial boundary amounts to a potential gradient of 100 million volts per centimeter— a very significant value indeed!
Interfacial potentials are not confined to metallic electrodes immersed in solutions; they can in fact exist between any two phases in contact, even in the absence of chemical reactions. In many forms of matter, they are the result of adsorption or ordered alignment of molecules caused by non-uniform forces in the interfacial region. Thus colloidal particles in aqueous suspensions selectively adsorb a given kind of ion, positive for some colloids, and negative for others. The resulting net electric charge prevents the particles from coming together and coalescing, which they would otherwise tend to do under the influence of ordinary van der Waals attractions.
Interfacial potential differences are not directly observable. The usual way of measuring a potential difference between two points is to bring the two leads of a voltmeter into contact with them. It's simple enough to touch one lead of the meter to a metallic electrode, but there is no way you can connect the other lead to the solution side of the interfacial region without introducing a second electrode with its own interfacial potential, so you would be measuring the sum of two potential differences. Thus single electrode potentials , as they are commonly known, are not directly observable.
What we can observe, and make much use of, are potential differences between pairs of electrodes in electrochemical cells . This is the topic of the next page in this series.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- Electroneutrality principle - Bulk matter cannot have a chemically-significant unbalance of positive and negative ions.
- Dissolution of a metal in water can proceed to a measurable extent only if some means is provided for removing the excess negative charge that remains. This can be by electron-acceptor ions in solution, or by drawing electrons out of the metal through an external circuit.
- Interfacial potentials - these exist at all phase boundaries. In the case of a metal in contact with an electrolyte solution, the interfacial region consists of an electric double layer.
- The potential difference between a metal and the solution is almost entirely located across the very thin double layer, leading to extremely large potential gradients in this region.
|
libretexts
|
2025-03-17T19:53:18.747226
| 2013-10-02T00:42:26 |
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"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.01%3A_Chemistry_and_Electricity",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.1: Chemistry and Electricity",
"author": "Stephen Lower"
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.02%3A_Galvanic_cells_and_Electrodes
|
16.2: Galvanic cells and Electrodes
It is physically impossible to measure the potential difference between a piece of metal and the solution in which it is immersed. We can, however, measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell , and we are measuring the sum of the two half-cell potentials .
This arrangement is called a galvanic cell . A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through.
If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn 2 + ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu 2 + ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:
\[Zn_{(s)} + Cu^{2+} \rightarrow Zn^{2+} + Cu_{(s)}\]
but this time, the oxidation and reduction steps (half reactions) take place in separate locations:
|
left electrode:
|
Zn (s) → Zn 2 + + 2 e – | oxidation |
|
right electrode:
|
Cu 2 + + 2 e – → Cu (s) | reduction |
Electrochemical cells allow measurement and control of a redox reaction
The reaction can be started and stopped by connecting or disconnecting the two electrodes. If we place a variable resistance in the circuit, we can even control the rate of the net cell reaction by simply turning a knob. By connecting a battery or other source of current to the two electrodes, we can force the reaction to proceed in its non-spontaneous, or reverse direction. By placing an ammeter in the external circuit, we can measure the amount of electric charge that passes through the electrodes, and thus the number of moles of reactants that get transformed into products in the cell reaction.
Electric charge q is measured in coulombs. The amount of charge carried by one mole of electrons is known as the Faraday , which we denote by F . Careful experiments have determined that 1 F = 96467 C. For most purposes, you can simply use 96,500 Coulombs as the value of the faraday. When we measure electric current, we are measuring the rate at which electric charge is transported through the circuit. A current of one ampere corresponds to the flow of one coulomb per second.
Charge Transport within the Cell
For the cell to operate, not only must there be an external electrical circuit between the two electrodes, but the two electrolytes (the solutions) must be in contact. The need for this can be understood by considering what would happen if the two solutions were physically separated. Positive charge (in the form of Zn 2 + ) is added to the electrolyte in the left compartment, and removed (as Cu 2 + ) from the right side, causing the solution in contact with the zinc to acquire a net positive charge, while a net negative charge would build up in the solution on the copper side of the cell. These violations of electroneutrality would make it more difficult (require more work) to introduce additional Zn 2 + ions into the positively-charged electrolyte or for electrons to flow into right compartment where they are needed to reduce the Cu 2 + ions, thus effectively stopping the reaction after only a chemically insignificant amount has taken place.
In order to sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. This means that we must provide a path for ions to move directly from one cell to the other. This ionic transport involves not only the electroactive species Cu 2 + and Zn 2 + , but also the counterions , which in this example are nitrate, NO 3 - . Thus an excess of Cu 2 + in the left compartment could be alleviated by the drift of these ions into the right side, or equally well by diffusion of nitrate ions to the left. More detailed studies reveal that both processes occur, and that the relative amounts of charge carried through the solution by positive and negative ions depends on their relative mobilities , which express the velocity with which the ions are able to make their way through the solution. Since negative ions tend to be larger than positive ions, the latter tend to have higher mobilities and carry the larger fraction of charge.
In the simplest cells, the barrier between the two solutions can be a porous membrane, but for precise measurements, a more complicated arrangement, known as a salt bridge , is used. The salt bridge consists of an intermediate compartment filled with a concentrated solution of KCl and fitted with porous barriers at each end. The purpose of the salt bridge is to minimize the natural potential difference, known as the junction potential , that develops (as mentioned in the previous section) when any two phases (such as the two solutions) are in contact. This potential difference would combine with the two half-cell potentials so as introduce a degree of uncertainty into any measurement of the cell potential. With the salt bridge, we have two liquid junction potentials instead of one, but they tend to cancel each other out.
Cell description conventions
In order to make it easier to describe a given electrochemical cell, a special symbolic notation has been adopted. In this notation the cell we described above would be
Zn (s) | Zn 2 + (aq) || Cu 2 + (aq) | Cu (s)
There are several other conventions relating to cell notation and nomenclature that you are expected to know:
- The anode is where oxidation occurs, and the cathode is the site of reduction. In an actual cell, the identity of the electrodes depends on the direction in which the net cell reaction is occurring.
- If electrons flow from the left electrode to the right electrode (as depicted in the above cell notation) when the cell operates in its spontaneous direction, the potential of the right electrode will be higher than that of the left, and the cell potential will be positive.
- "Conventional current flow" is from positive to negative, which is opposite to the direction of the electron flow. This means that if the electrons are flowing from the left electrode to the right, a galvanometer placed in the external circuit would indicate a current flow from right to left.
Electrodes and Electrode Reactions
An electrode reaction refers to the net oxidation or reduction process that takes place at an electrode. This reaction may take place in a single electron-transfer step, or as a succession of two or more steps. The substances that receive and lose electrons are called the electroactive species.
This process takes place within the very thin interfacial region at the electrode surface, and involves quantum-mechanical tunneling of electrons between the electrode and the electroactive species. The work required to displace the H 2 O molecules in the hydration spheres of the ions constitutes part of the activation energy of the process.
In the example of the Zn/Cu cell we have been using, the electrode reaction involves a metal and its hydrated cation; we call such electrodes metal-metal ion electrodes. There are a number of other kinds of electrodes which are widely encountered in electrochemistry and analytical chemistry.
Ion-ion Electrodes
Many electrode reactions involve only ionic species, such as \(Fe^{2+}\) and \(Fe^{3+}\). If neither of the electroactive species is a metal, some other metal must serve as a conduit for the supply or removal of electrons from the system. In order to avoid complications that would arise from electrode reactions involving this metal, a relatively inert substance such as platinum is commonly used. Such a half cell would be represented as
Pt(s) | Fe 3 + (aq) , Fe 2 + (aq) || ...
and the half-cell reaction would be
\[Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e^-\]
The reaction occurs at the surface of the electrode (Fig 4 above). The electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it by van der Waals and Coulombic forces . In doing so, the waters of hydration that are normally attached to any ionic species must be displaced. This process is always endothermic, sometimes to such an extent that only a small fraction of the ions be able to contact the surface closely enough to undergo electron transfer, and the reaction will be slow. The actual electron-transfer occurs by quantum-mechanical tunnelling .
Gas Electrodes
Some electrode reactions involve a gaseous species such as \(H_2\), \(O_2\), or \(Cl_2\). Such reactions must also be carried out on the surface of an electrochemically inert conductor such as platinum. A typical reaction of considerable commercial importance is
\[Cl^-(aq) \rightarrow ½ Cl_2(g) + e^- \]
Similar reactions involving the oxidation of \(Br_2\) or \(I_2\) also take place at platinum surfaces.
Insoluble–salt Electrodes
A typical electrode of this kind consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water. The electrode reaction consists in the oxidation and reduction of the silver:
\[AgCl(s) + e^– → Ag(s) + Cl^–(aq)\]
The half cell would be represented as
\[ ... || Cl^– (aq) | AgCl (s) | Ag (s)\]
Although the usefulness of such an electrode may not be immediately apparent, this kind of electrode finds very wide application in electrochemical measurements, as we shall see later.
Reference Electrodes
In most electrochemical experiments our interest is concentrated on only one of the electrode reactions. Since all measurements must be on a complete cell involving two electrode systems, it is common practice to employ a reference electrode as the other half of the cell. The major requirements of a reference electrode are that it be easy to prepare and maintain, and that its potential be stable. The last requirement essentially means that the concentration of any ionic species involved in the electrode reaction must be held at a fixed value. The most common way of accomplishing this is to use an electrode reaction involving a saturated solution of an insoluble salt of the ion. One such system, the silver-silver chloride electrode has already been mentioned:
\[Ag | AgCl(s) | Cl^–(aq) || ...\]
\[Ag(s) + Cl^–(aq) →AgCl(s) + e^–\]
This electrode usually takes the form of a piece of silver wire coated with AgCl. The coating is done by making the silver the anode in an electrolytic cell containing HCl; the Ag + ions combine with Cl – ions as fast as they are formed at the silver surface.
The other common reference electrode is the calomel electrode ; calomel is the common name for mercury(I) chloride. Such a half cell would be represented as
\[Hg | Hg^{2+}(aq) | KCl || ...\]
and the half-cell reaction would be
\[Hg(l) + Cl^– → ½ HgCl2(s) + e^–\]
The potentials of both of these electrodes have been very accurately determined against the hydrogen electrode. The latter is seldom used in routine electrochemical measurements because it is more difficult to prepare; the platinum surface has to be specially treated by preliminary electrolysis. Also, there is need for a supply of hydrogen gas which makes it somewhat cumbersome and hazardous.
Summary and additional notes
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- A galvanic cell (sometimes more appropriately called a voltaic cell ) consists of two half-cells joined by a salt bridge or some other path that allows ions to pass between the two sides in order to maintain electroneutrality .
- The conventional way of representing an electrochemical cell of any kind is to write the oxidation half reaction on the left and the reduction on the right. Thus for the reaction
Zn (s) + Cu 2 + → Zn 2 + + Cu (s)
we write
Zn (s) | Zn 2 + (aq) || Cu 2 + (aq) | Cu (s)
in which the single vertical bars represent phase boundaries . The double bar denotes a liquid-liquid boundary which in laboratory cells consists of a salt bridge or in ion-permeable barrier. If the net cell reaction were written in reverse, the cell notation would become
Cu (s) | Cu 2 + (aq) || Zn 2+ (aq) | Zn (s)
Remember: the R eduction process is always shown on the R ight .
-
at the electrode surface. The energy required to displace water molecules from the hydration shell of an ion as it approaches the electrode surface constitutes an
activation energy
which can slow down the process. Even larger activation energies (and slower reactions) occur when a molecule such as O
2
is formed or consumed.
|
libretexts
|
2025-03-17T19:53:18.831984
| 2013-10-02T00:42:29 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.02%3A_Galvanic_cells_and_Electrodes",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.2: Galvanic cells and Electrodes",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.03%3A_Cell_potentials_and_Thermodynamics
|
16.3: Cell Potentials and Thermodynamics
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.
- When we refer to the "standard potential of a half-cell" or "couple" M 2 + /M, we mean the potential difference E right – E left of the cell
- If the potential difference of this cell is positive ( E right – E left >0), electrons will flow through an external circuit from the Pt/H 2 electrode to the M electrode and the cell reaction will spontaneously proceed in the direction written. The more positive the cell potential, the greater the tendency of this reaction to occur and the stronger the oxidizing agent M 2 + .
- Through the relation E° = – Δ G° /nF it is apparent that a standard half-cell reduction potential is simply the decrease in the free energy per mole of electrons transferred to H + ions under the conditions that define the SHE. Strong reducing agents (good electron donors) have more negative E° s, while strong oxidizing agents (good acceptors) have more positive E° s.
-
For a more general cell
X
(s)
| X
+
|| M
2
+
| M
(s)
,
E°
is similarly the fall in free energy per electron-mole when M
2
+
is reduced by X. This reaction can proceed spontaneously only if the cell potential is positive (Δ
G° negative.) - An electron free energy diagram that displays various redox couples on a vertical scale of free energies relative to H + serves as a convenient means of visualizing the possible reactions when two or more redox-active pairs are present in a solution. The position of a redox couple in relation to those of the H 2 / H + and H 2 O/O 2 ,H + couples is especially significant because it indicates whether a given species will be thermodynamically stable in water.
- Latimer diagrams provide a convenient means of correlating the various oxidation states of a particular element.
It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. The classic example is the one we have already mentioned on the preceding page:
\[\ce{ Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)} \nonumber\]
Here zinc is more active because it can displace (precipitate) copper from solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a black coating of finely-divided copper, and the blue color of the hydrated copper(II) ion diminishes. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
|
The activity series has long been used to predict the direction of oxidation-reduction reactions. Consider, for example, the oxidation of Cu by metallic zinc that we have mentioned previously. The fact that zinc is near the top of the activity series means that this metal has a strong tendency to lose electrons. By the same token, the tendency of Zn to accept electrons is relatively small. Copper, on the other hand, is a poorer electron donor, and thus its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the reaction
\[Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)\]
to proceed in the direction indicated, rather than in the reverse direction. An old-fashioned way of expressing this is to say that "zinc will displace copper from solution". The above table is of limited practical use because it does not take into account the concentrations of the dissolved species. In order to treat these reactions quantitatively, it is convenient to consider the oxidation and reduction steps separately.
Standard half-cell potentials
When a net reaction proceeds in an electrochemical cell, oxidation occurs at one electrode (the anode) and reduction takes place at the other electrode (the cathode.) We can think of the cell as consisting of two half-cells joined together by an external circuit through which electrons flow and an internal pathway that allows ions to migrate between them so as to preserve electroneutrality.
Reduction potentials
Each half-cell has associated with it an electrode-solution potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved electroactive species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. To express them in a uniform way, we adopt the convention that half-cell potentials are always defined for the reduction direction. Thus the half-cell potential for the Zn/Zn 2 + electrode (or couple as it is sometimes called) always refers to the reduction reaction
\[Zn^{2+} + 2e^– \rightarrow Zn (s)\]
In the cell Zn (s) | Zn 2 + (aq) || Cu 2 + (aq) | Cu (s) the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional half-cell potential. More generally, we can define the cell potential (or cell EMF) as
\[E_{cell} = \Delta V = E_{right} – E_{left} \label{1}\]
in which "right" and "left" refer to the cell notation convention (" r eduction on the r ight") and not, of course, to the physical orientation of a real cell in the laboratory. If we expand the above expression we see that the cell potential
\[E_{cell} = V_{Cu} – V_{solution} + V_{solution} – V_{Zn}\]
is just the difference between the two half-cell potentials \(E_{right}\) and \(E_{left}\).
Reference half-cells
The fact that individual half-cell potentials are not directly measurable does not prevent us from defining and working with them. Although we cannot determine the absolute value of a half-cell potential, we can still measure its value in relation to the potentials of other half cells. In particular, if we adopt a reference half-cell whose potential is arbitrarily defined as zero, and measure the potentials of various other electrode systems against this reference cell, we are in effect measuring the half-cell potentials on a scale that is relative to the potential of the reference cell.
The reference cell that has universally been adopted for this purpose is the hydrogen half-cell
\[Pt | ½ H_{2(g)} | H^+_{(aq)} || \;... \]
in which hydrogen gas is allowed to bubble over a platinum electrode having a specially treated surface which catalyzes the reaction
\[½ H_{2(g)} → H^+ + e^–\]
When this electrode is operated under standard conditions of 1 atm H 2 pressure, 25°C, and pH = 0, it becomes the standard hydrogen electrode, sometimes abbreviated SHE.
To measure the relative potential of some other electrode couple M 2 + /M, we can set up a cell
\[Pt | H_{2(g)} | H^+ || M^{2+}_{ (aq)} | M_{(s)}\]
whose net reaction is
H 2 (g) + M 2 + (aq) → 2H + + M (s)
the potential difference between the platinum and M electrodes will be
E cell = V M – V solution + V solution – V Pt
but since the difference V solution – V Pt is by definition zero for the hydrogen half-cell, the cell potential we measure corresponds to
E cell = V M – V solution
which is just the potential (relative to that of the SHE) of the half-cell whose reaction is
\[M^{2+} + 2e^– → M_{(s)}\]
|
Measurement of a standard reduction potential. The M 2 + /M half-cell is on the left, and the standard hydrogen cell is on the right. The two half-cells are joined through the salt bridge in the middle. The more "active" the metal M (the greater its tendency to donate electrons to H + ), the more negative will be E cell = Δ V = E right – E left |
Standard [reduction] potentials are commonly denoted by the symbol E° . E° values for hundreds of electrodes have been determined (mostly in the period 1925-45, during which time they were referred to as "oxidation potentials") and are usually tabulated in order of increasing tendency to accept electrons (increasing oxidizing power.)
|
oxidant
(electron acceptor) |
reductant
(elecron donor) |
E°, volts |
|---|---|---|
|
Na
+
|
Na
(s)
|
–2.71 |
|
Zn
2
+
|
Zn
(s)
|
–.76 |
|
Fe
2
+
|
Fe
(s)
|
–.44 |
|
Cd
2
+
|
Cd
(s)
|
–.40 |
|
Pb
2
+
|
Pb
(s)
|
–.126 |
0.000
+.222
+.268
+.337
+.535
+.771
+.799
+1.23
+1.36
Note particularly that
- Unlike the activity series mentioned above, this table includes non-metallic substances , and it is quantitative .
- The more positive the half-cell EMF, the greater the tendency of the reductant to donate electrons, and the smaller the tendency of the oxidant to accept electrons.
- A species in the leftmost column can act as an oxidizing agent to any species above it in the reductant column.
- Oxidants such as Cl 2 that are below H 2 O will tend to decompose water.
Given the E° values for two half reactions, you can easily predict the potential difference of the corresponding cell: simply add the reduction potential of the reduction half-cell to the negative of the reduction potential (that is, to the oxidation potential) of the oxidation reaction.
Find the standard potential of the cell
Cu (s) | Cu 2 + || Cl – | AgCl (s) | Ag (s)
and predict the direction of electron flow when the two electrodes are connected.
Solution
The net reaction corresponding to this cell will be:
Cu (s) + 2AgCl (s) → 2 Ag (s) + 2 Cl – (aq) + Cu 2 + (aq)
Where Cu(s)/Cu 2 + is being Oxidized and Ag(s)/Ag + is being Reduced
E cell = E reduction + E oxidation Or Written another way E cell = E right - E left =(.222 + (– .337)) v = -0 .115 v
Since E cell is negative, the reaction will run in the opposite direction. The correct net reaction will be:
2 Ag (s) + 2 Cl – (aq) + Cu 2 + (aq) → 2AgCl (s) + Cu (s)
Where Ag(s)/Ag + is being Oxidized and Cu(s)/Cu 2 + is being Reduced
E cell = E reduction + E oxidation Or Written another way E cell = E right - E left =(.337 + (– .222)) v = +0 .115 v
Since this potential is positive, the reaction will proceed to the right; electrons will be withdrawn from the silver electrode and flow through the external circuit into the copper electrode. Note carefully that in combining these half-cell potentials, we did not multiply E° for the Cu 2 + /Cu couple by two. The reason for this will be explained later.
Cell potentials and Free (Gibbs) energy
From the above, it should be apparent that the potential difference between the electrodes of a cell is a measure of the tendency for the cell reaction to take place: the more positive the cell potential, the greater the tendency for the reaction to proceed to the right. But we already know that the standard free energy change expresses the tendency for any kind of process to occur under the conditions of constant temperature and pressure. Thus Δ G ° and E° measure the same thing, and are related in a simple way:
\[\Delta G° = –nFE° \label{2}\]
... or in more detail (see below for explanations of the units given for voltage)
A few remarks are in order about this very fundamental and important relation:
- The negative sign on the right indicates that a positive cell potential (according to the sign convention discussed previously) implies a negative free energy change , and thus that the cell reaction will spontaneously proceed to the right .
- Electrical work is done when an electric charge q moves through a potential difference Δ V . The right side of Eq. 2 refers to the movement of n moles of charge across the cell potential E° , and thus has the dimensions of work .
- The value of Δ G° expresses the maximum useful work that a system can do on the surroundings. "Useful" work is that which can be extracted from the cell by electrical means to operate a lamp or some other external device. This excludes any P-V work that is simply a consequence of volume change (which could conceivably be put to some use!) and which would be performed in any case, even if the reactants were combined directly. This quantity of work –ΔG° can only be extracted from the system under the limiting conditions of a thermodynamically reversible change, which for an electrochemical cell implies zero current. The more rapidly the cell operates, the less electrical work it can supply.
- If F is expressed in coulombs per mole of electrons, the electrical work is in joules per mole. To relate these units to electrical units, recall that the coulomb is one amp-sec, and that power , which is the rate at which work is done, is measured in watts, which is the product of amps and volts
1 J = 1 watt-sec = 1 (amp-sec) × volts
Thus the volt has the dimensions of joules/coulomb– the energy produced per quantity of charge passing through the cell. Because voltage is the quotient of two extensive quantities, it is itself intensive . When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged. This explains why we do not have to multiply the E° s of the anode and cathode reactions by stoichiometric factors when we are finding the potential of a complete cell.
If Eq. 2 is solved for E° , we have
\[ E° =\dfrac{\Delta G°}{nF} \label{3}\]
This states explicitly that the cell potential is a measure of the free energy change per mole of electrons transferred , which is a brief re-statement of the principle explained immediately above. To see this more clearly, consider the cell
Cu (s) | Cu 2 + || Cl – | AgCl (s) | Ag (s)
for which we list the standard reduction potentials and ΔG °s of the half-reactions:
|
reaction |
E° |
- nFE° = ΔG° |
|
cathode: 2 × [AgCl(s) + e – → Ag(s) + Cl – ] anode: Cu(s) → Cu 2 + + 2 e – |
+.222 v –(+.337) V |
–42800 J +65000 J |
|
net: 2 Ag (s) + 2 Cl – (aq) + Cu 2 + (aq) → AgCl (s) + Cu (s) cell: Cu (s) | Cu 2 + (aq) || AgCl (s) | Cl – (aq) | Ag (s) |
–.115 v |
+22200 J |
Here we multiply the cathodic reaction by two in order to balance the charge. Because the anodic reaction is written as an oxidation, we reverse the sign of its E° and obtain E cell = E right – E left = –.115 volt for the cell potential. The negative cell potential tells us that this reaction will not proceed spontaneously.
When the electrons don't cancel out
Note, however, that if we are combining two half reactions to obtain a third half reaction, the E° values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use Δ G° = –nFE° to find the cell potential.
Calculate E° for the electrode Fe 3 + /Fe (s) from the standard potential of the couples Fe 3 + /Fe 2 + and Fe 2 + /Fe (s)
Solution
Tabulate the values and calculate the Δ G° s as follows:
| (i) Fe 3 + + e – → Fe 2 + | E° 1 = .771 v , Δ G° 1 = –.771 F |
| (ii) Fe 2 + + 2 e – → Fe(s) | E° 2 = –.440 v , Δ G° 2 = +.880 F |
| (iii) Fe 3 + + 3 e – → Fe(s) | E° 3 = ? , Δ G° 3 = +.109 F |
The Gibbs energy for half-reaction (iii) is 0.109 n F , so E° 3 = –.109/ 3 = –.036 v
The fall of the electron
A table of standard half-cell potentials summarizes a large amount of chemistry, for it expresses the relative powers of various substances to donate and accept electrons by listing reduction half-reactions in order of increasing E° values, and thus of increasing spontaneity. The greater the value of E°, the greater the tendency of the substance on the left to acquire electrons, and thus the stronger this substance is as an oxidizing agent.
If you have studied elementary chemical thermodynamics, you will have learned about the role that a quantity called the Gibbs free energy , usually referred to as simply the " free energy ", plays in determining the direction of any chemical change. The rule is that all spontaneous change (that is, all reactions that proceed to the "right") is associated with a fall in the free energy, and the greater the degree of that fall (Δ G° ), the greater will be the tendency for the reaction to take place.
If you are not familiar with the concept of free energy, just think of it as something like potential energy, which similarly decreases when spontaneous mechanical events occur, such as the dropping of a weight.
Since oxidation-reduction processes involve the transfer of an electron from a donor to an acceptor, it makes sense to focus on the electron and to consider that it falls from a higher-free energy environment (the reductant, or "source") to a lower-free energy one (the oxidant, or "sink".)
As can be seen from the diagram below, this model makes it far easier to predict what will happen when two or more oxidants and reducants are combined; the electron "falls" as far as it can, filling up oxidizing agents (sinks) from the bottom up, very much in the same way as electrons fill atomic orbitals as we build up larger atoms.
|
Electron-free energy diagram of redox couples
This chart is essentially an abbreviated form of a table of standard potentials in which the various couples are displayed on a vertical scale corresponding to
|
A more detailed table with a more complete explanation can be seen on the " Fall of the electron" tutorial page ; it is strongly recommended that you take the time to acquire a thorough understanding of this concept.
At this point, it might be worth calling your attention to the similar way of depicting acid-base reactions as representing the "fall of the proton" as shown below and described much more thoroughly here .
|
Proton-free energy diagram of acid-base systems Acids are proton sources (donors), bases are proton sinks . Protons "fall" (in free energy) whenever a base is present that presents proton-empty free energy levels. The red arrows show what happens when acetic acid is titrated with a strong base; the results are acetate ion and water. Note here again the crucial role of water, both as a proton acceptor (forming hydronium ion) and as a proton donor (forming hydroxide ion.) Note also that the pH of a solution is a direct measure of the average free energy of protons in the solution (relative to H 3 O + .) |
An important difference between proton transfer and electron transfer reactions is that the latter can vary greatly in speed, from almost instantaneous to so slow as to be unobservable. Acid-base reactions are among the fastest known.
Considerable insight into the chemistry of a single element can be had by comparing the standard electrode potentials (and thus the relative free energies) of the various oxidation states of the element. The most convenient means of doing this is the Latimer diagram . As examples, diagrams for iron and chlorine are shown below.
The formulas of the species that represent each oxidation state of the element are written from left to right in order of decreasing oxidation number, and the standard potential for the reduction of each species to the next on the right is written in between the formulas. Potentials for reactions involving hydrogen ions will be pH dependent, so separate diagrams are usually provided for acidic and alkaline solutions (effective hydrogen ion concentrations of 1M and 10 –14 M, respectively).
The more positive the reduction potential, the greater will be the tendency of the species on the left to be reduced to the one on the right. To see how Latimer diagrams are used, look first at the one for iron in acid solution. The line connecting Fe 3 + and Fe 2 + represents the reaction
Fe 3 + + e – → Fe 2 +
whose positive E° (.440 v) indicates that metallic iron will dissolve in acidic solution to form Fe 2 + . Because the oxidation of this species to the +3 state has a negative potential (-0.771v; moving to the left on the diagram reverses the sign), the +2 state will be the stable oxidation state of iron under these conditions.
Disproportionation
An important condition to recognize in a Latimer diagram is when the potential on the left of a species is less positive than that on the right. This indicates that the species can oxidize and reduce itself, a process known as disproportionation . As an example, consider Cl 2 in alkaline solution. The potential for its reduction to Cl – is sufficiently positive (+1.35 v) to supply the free energy necessary for the oxidation of one atom of chlorine to hypochlorite. Thus elemental chlorine is thermodynamically unstable with respect to disproportionation in alkaline solution, and the same it true of the oxidation product, ClO – (hypochlorite ion).
Behavior of chlorine in water
Cl 2 can oxidize water (green arrows, top) and also undergo disproportionation (purple arrows, bottom). In the latter process, one Cl 2 molecule donates electrons to another. Bear in mind that many oxidation-reduction reactions, unlike most acid-base reactions, tend to be very slow, so the fact that a species is thermodynamically unstable does not always mean that it will quickly decompose. Thus the two reactions shown in the figure are normally very slow.
Thermodnamics of Galvanic cells
The free energy change for a process represents the maximum amount of non- PV work that can be extracted from it. In the case of an electrochemical cell, this work is due to the flow of electrons through the potential difference between the two electrodes. Note, however, that as the rate of electron flow (i.e., the current) increases, the potential difference must decrease; if we short-circuit the cell by connecting the two electrodes with a conductor having negligible resistance, the potential difference is zero and no work will be done. The full amount of work can be realized only if the cell operates at an infinitessimal rate; that is, reversibly.
You should recall that this is exactly analogous to the expansion of an ideal gas. The full amount of work w = PdV is extracted only under the special condition that the external pressure P opposing expansion is only infinitessimally smaller than the pressure of the gas itself. If the gas is allowed to expand into a vacuum ( P = 0), no work will be done.
The total amount of energy a reaction can supply under standard conditions at constant pressure and temperature is given by ΔH °. If the reaction takes place by combining the reactants directly (no cell) or in a short-circuited cell, no work is done and the heat released is Δ H . If the reaction takes place in a cell that performs electrical work, then the heat released is diminished by the amount of electrical work done. In the limit of reversible operation, the heat released becomes
Δ H = Δ G° + T Δ S
Note
Pt | H 2 (g) | H + || M 2+ (aq) | M (s)
whose left half consists of a standard hydrogen electrode (SHE) and whose net reaction is
H 2 (g) + M 2+ (aq) → 2H + + M (s)
|
libretexts
|
2025-03-17T19:53:18.986222
| 2013-10-02T00:42:43 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.03%3A_Cell_potentials_and_Thermodynamics",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.3: Cell Potentials and Thermodynamics",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.04%3A_The_Nernst_Equation
|
16.4: The Nernst Equation
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- The Nernst equation relates the effective concentrations ( activities ) of the components of a cell reaction to the standard cell potential. For a simple reduction of the form M n + + ne – → M, it tells us that a half-cell potential will change by 59/ n mV per 10-fold change in the activity of the ion.
- Ionic concentrations can usually be used in place of activities when the total concentration of ions in the solution does not exceed about about 0.001 M .
- In those reactions in which H + or OH – ions take part, the cell potential will also depend on the pH. Plots of E vs. pH showing the stability regions of related species are very useful means of summarizing the redox chemistry of an element.
The standard cell potentials we discussed in a previous section refer to cells in which all dissolved substances are at unit activity , which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure (known as the fugacity ) of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know.
Cell Potentials Depend on Concentrations
Suppose, for example, that we reduce the concentration of \(Zn^{2+}\) in the \(Zn/Cu\) cell from its standard effective value of 1 M to an to a much smaller value:
\[Zn(s) | Zn^{2+}(aq, 0.001\,M) || Cu^{2+}(aq) | Cu(s)\]
This will reduce the value of \(Q\) for the cell reaction
\[Zn(s) + Cu^{2+} → Zn^{2+} + Cu(s)\]
thus making it more spontaneous, or "driving it to the right" as the Le Chatelier principle would predict, and making its free energy change \(\Delta G\) more negative than \(\Delta G°\), so that \(E\) would be more positive than \(E^°\). The relation between the actual cell potential \(E\) and the standard potential \(E^°\) is developed in the following way. We begin with the equation derived previously which relates the standard free energy change (for the complete conversion of products into reactants) to the standard potential
\[\Delta G° = –nFE° \]
By analogy we can write the more general equation
\[\Delta G = –nFE\]
which expresses the change in free energy for any extent of reaction— that is, for any value of the reaction quotient \(Q\). We now substitute these into the expression that relates \(\Delta G\) and \(\Delta G°\) which you will recall from the chapter on chemical equilibrium:
\[\Delta G = \Delta G° + RT \ln Q\]
which gives
\[–nFE = –nFE° + RT \ln Q \]
which can be rearranged to
\[ \color{red} {\underbrace{E=E° -\dfrac{RT}{nF} \ln Q}_{\text{applicable at all temperatures}}} \label{Nernst Long}\]
This is the Nernst equation that relates the cell potential to the standard potential and to the activities of the electroactive species. Notice that the cell potential will be the same as \(E°\) only if \(Q\) is unity. The Nernst equation is more commonly written in base-10 log form and for 25 °C:
\[ \color{red} {\underbrace{E=E° -\dfrac{0.059}{n} \log_{10} Q}_{\text{Applicable at only 298K}}} \label{Nernst Short}\]
Significance of the Nernst Equation
The Nernst equation tells us that a half-cell potential will change by 59 millivolts per 10-fold change in the concentration of a substance involved in a one-electron oxidation or reduction; for two-electron processes, the variation will be 28 millivolts per decade concentration change. Thus for the dissolution of metallic copper
\[Cu_{(s)} \rightarrow Cu^{2+} + 2e^–\]
the potential
\[E = (– 0.337) – 0.0295 \log_{10} [Cu^{2+}]\]
becomes more positive (the reaction has a greater tendency to take place) as the cupric ion concentration decreases. This, of course, is exactly what the Le Chatelier Principle predicts; the more dilute the product, the greater the extent of the reaction.
Electrodes with poise
The equation just above for the Cu/Cu 2 + half-cell raises an interesting question: suppose you immerse a piece of copper in a solution of pure water. With Q = [Cu 2 + ] = 0, the potential difference between the electrode and the solution should be infinite! Are you in danger of being electrocuted? You need not worry; without any electron transfer, there is no charge to zap you with. Of course it won't be very long before some Cu 2 + ions appear in the solution, and if there are only a few such ions per liter, the potential reduces to only about 20 volts. More to the point, however, the system is so far from equilibrium (for example, there are not enough ions to populate the electric double layer) that the Nernst equation doesn't really give meaningful results. Such an electrode is said to be un poised . What ionic concentration is needed to poise an electrode? I don't really know, but I would be suspicious of anything much below 10 –6 M.
Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10 –3 M.
How the cell potential really depends on concentration! The Nernst equation accurately predicts cell potentials only when the equilibrium quotient term Q is expressed in activities . Ionic activities depart increasingly from concentrations when the latter exceed 10 –4 to 10 –5 M, depending on the sizes and charges of the ions.
If the Nernst equation is applied to more concentrated solutions, the terms in the reaction quotient Q must be expressed in "effective concentrations" or activities of the electroactive ionic species. The activity coefficient \(\gamma\)) relates the concentration of an ion to its activity a in a given solution through the relation
\[a = \gamma c\]
Since electrode potentials measure activities directly, activity coefficients can be determined by carrying out appropriate EMF measurements on cells in which the concentration of the ion of interest is known. The resulting Es can then be used to convert concentrations into activities for use in other calculations involving equilibrium constants.
Cell potentials and pH: stability diagrams
As most of us recall from our struggles with balancing redox equations in beginning chemistry courses, many electron-transfer reactions involve hydrogen ions and hydroxide ions. The standard potentials for these reactions therefore refer to the pH, either 0 or 14, at which the appropriate ion has unit activity. Because multiple numbers of H + or OH – ions are often involved, the potentials given by the Nernst equation can vary greatly with the pH. It is frequently useful to look at the situation in another way by considering what combinations of potential and pH allow the stable existence of a particular species. This information is most usefully expressed by means of a E -vs.-pH diagram, also known as a Pourbaix diagram .
As was noted in connection with the shaded region in the figure below, water is subject to decomposition by strong oxidizing agents such as Cl 2 and by reducing agents stronger than H 2 . The reduction reaction can be written either as
\[2H^+ + 2e^– \rightarrow H_{2}(g) \]
or, in neutral or alkaline solutions as
\[2H_2O + 2 e^– \rightarrow H_{2}(g) + 2 OH^–\]
These two reactions are equivalent and follow the same Nernst equation
\[E_{H^+/H_2} = E_{H^+/H_2}^o + \dfrac{RT}{nF} \ln \left( \dfrac{[H^+]^2} {P_{H_2}} \right)\]
which, at 25°C and unit H 2 partial pressure reduces to
\[E = E° - \dfrac{0.059}{2} × 2 pH = –0.059\; pH\]
Similarly, the oxidation of water
\[2H_2O \rightarrow O_{2}(g) + 4 H^+ + 2 e^–\]
is governed by the Nernst equation
\[ E_{O_2/H_2O} = E_{O_2/H_2O}^o + \dfrac{RT}{nF} \ln \left( P_{O_2}[H^+]^4 \right)\]
which similarly becomes E = 1.23 – 0.059 pH, so the E-vs -pH plots for both processes have identical slopes and yield the stability diagram for water in Figure \(\PageIndex{2}\). This Pourbaix diagram has special relevance to electrochemical corrosion of metals. Thus metals above hydrogen in the activity series will tend to undergo oxidation (corrosion) by reducing H + ions or water.
Because chlorine is widely used as a disinfectant for drinking water, swimming pools, and sewage treatment, it is worth looking at its stability diagram. Note that the effective bactericidal agent is not Cl 2 itself, but its oxidation product hypochlorous acid HOCl which predominates at pH values below its pK a of 7.3. Note also that
- Cl 2 is unstable in water except at very low pH; it decomposes into HOCl and Cl – .
- Hypochlorous acid and its anion are stronger oxidants than O 2 and thus subject to decomposition in water. The only stable chlorine species in water is Cl – .
- Decomposition of HOCl occurs very slowly in the dark, but is catalyzed by sunlight. For this reason the chlorine in outside swimming pools must be frequently renewed.
- Decomposition of Cl 2 and HOCl by reaction with organic material in municipal water supply systems sometimes makes it necessary to inject additional chlorine at outlying locations.
Each solid line represents a combination of E and pH at which the two species on either side of it can coexist; at all other points, only a single species is stable. Note that equilibria between species separated by diagonal lines are dependent on both E and pH, while those separated by horizontal or vertical lines are affected by pH only or E only, respectively.
Stability diagrams are able to condense a great amount of information into a compact representation, and are widely employed in geochemistry and corrosion engineering. The Pourbaix diagram for iron is one of the more commonly seen examples.
Pourbaix diagram for iron. Three oxidation states of iron (0, +2 and +3) are represented on this diagram. The stability regions for the oxidized iron states are shown only within the stability region of H 2 O. Equilibria between species separated by vertical lines are dependent on pH only.
The +3 oxidation state is the only stable one in environments in which the oxidation level is controlled by atmospheric O 2 . This is the reason the Earth’s crust contains iron oxides, which developed only after the appearance of green plants which are the source of O 2 . Iron is attacked by H + to form H 2 and Fe(II); the latter then reacts with O 2 to form the various colored Fe(III) oxides that constitute “rust”. Numerous other species such as oxides and hydrous oxides are not shown. A really “complete” diagram for iron would need to have at least two additional dimensions showing the partial pressures of O 2 and CO 2 .
Concentration Cells
From your study of thermodynamics you may recall that the process
solute (concentrated) → solute (dilute)
is accompanied by a fall in free energy, and therefore is capable of doing work on the surroundings; all that is required is some practical way of capturing this work. One way of doing this is by means of a concentration cell such as
Cu (s) | CuNO 3 (.1 M) || CuNO 3 (.01 M) | Cu (s)
cathode: Cu 2 + (.1 M) + 2 e – → Cu (s)
anode: Cu (s) → Cu 2 + (.01 M) + 2 e –
net: Cu 2 + (.1 M) → Cu 2 + (.01 M)
which represents the transport of cupric ion from a region of higher concentration to one of lower concentration.
The driving force for this process is the free energy change ΔG associated with the concentration gradient (C 2 – C 1 ), sometimes known as the free energy of dilution:
\[ΔG_{dilution} = RT \ln(C_2 – C_1)\]
Note, however, that Cu 2 + ions need not physically move between the two compartments; electron flow through the external circuit creates a "virtual" flow as copper ions are created in the low-concentration side and discharged at the opposite electrode. Nitrate ions must also pass between the cells to maintain electroneutrality . The Nernst equation for this cell is
\[E = E^° - \left(\dfrac{0.059}{n}\right) \log_{10} Q = 0 - 0.29 \log_{10} 0.1 = +0.285 \,V\]
Note that \(E^°\) for a concentration cell is always zero, since this would be the potential of a cell in which the electroactive species are at unit activity in both compartments.
\(E^°\) for a concentration cell is always zero,
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libretexts
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2025-03-17T19:53:19.077115
| 2013-10-02T00:42:44 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.04%3A_The_Nernst_Equation",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.4: The Nernst Equation",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.05%3A_Applications_of_the_Nernst_Equation
|
16.5: Applications of the Nernst Equation
Make sure you thoroughly understand the following essential ideas:
- Atmospheric oxygen is a strong oxidizing agent, but in the absence of a suitable catalyst this reaction is ordinarily too slow to be observable.
- The pE is a useful index of electron availability, especially in natural waters and biological systems in which multiple redox systems are usually present.
- Solubility products can often be measured by constructing a cell in which one of the electrodes involves the sparingly soluble salt, and the net cell equation corresponds to the solubility reaction.
- Potentiometric titrations are widely used to measure concentrations of species that are readily oxidized or reduced.
- Measurement of pH is accomplished with an electrode consisting of a thin membrane of glass in which Na + ions are exchanged with H + .
- Membrane (Donnan) potentials arise when passage of an ion across a semipermeable membrane is selectively facilitated or inhibited.
- Nerve conduction occurs not by a flow of charge through the nerve, but by a wave of depolarization resulting from the concerted action of active channels that govern the passage of ions (mainly K + and Na + ) across the membrane enclosing the nerve.
We ordinarily think of the oxidation potential being controlled by the concentrations of the oxidized and reduced forms of a redox couple, as given by the Nernst equation. Under certain circumstances it becomes more useful to think of E as an independent variable that can be used to control the value of Q in the Nernst equation. This usually occurs when two redox systems are present, one being much more concentrated or kinetically active than the other. By far the most important example of this is the way atmospheric oxygen governs the composition of the many redox systems connected with biological activity.
Oxygen and the Aquatic Environment
The presence of oxygen in the atmosphere has a profound effect on the redox properties of the aquatic environment— that is, on natural waters exposed directly or indirectly to the atmosphere, and by extension, on organisms that live in an aerobic environment.This is due, of course, to its being an exceptionally strong oxidizing agent and thus a low-lying sink for electrons from most of the elements and all organic compounds. Those parts of the environment that are protected from atmospheric oxygen are equally important because it is only here that electrons are sufficiently available to produce the "reducing" conditions that are essential for processes varying from photosynthesis to nitrogen fixation.
Estimate the redox potential of a natural water that is in equilibrium with the atmosphere at pH 7 and 298 K. What fraction of a dilute solution Fe 2 + will be in its oxidized form Fe 3 + in such a water?
Solution
The relevant E °s are
- 1.23 v for O 2 (g) + 4H + + 4 e – → 2H 2 O and
- 0.77 V for the Fe 3 + /Fe 2 + couple.
(a) The potential (with respect to the SHE, of course) is given by the Nernst equation
which works out to E = 0.82 volt. As the Le Chatelier principle predicts, the higher pH (lower [H + ] compared to that at the "standard" pH of zero) reduces the electron-accepting tendency of oxygen.
(b) The Nernst equation for the reduction of Fe 3 + is E = .77–.059 log Q, in which Q is the ratio [Fe 2 + ]/[Fe 3+ ]. With E set by the O 2 /H 2 O couple, this becomes
0.82 = 0.77 – 0.059 log Q
which gives Q = 10 –0.85 or [Fe 2 + ]/[Fe 3 + ] = 0.14/1, so the fraction of the iron in its oxidized form is 1/1.14 = 0.88.
If we can have pH, why not pE?
As you will recall from your study of acid-base chemistry, the pH of a solution (defined as –log {H + }) is a measure of availablity (technically, the activity ) of protons in the solution. As is explained in more detail here , protons tend to "fall" (in free energy) from filled donor levels (acids) to lower acceptor levels (bases.) Through the relation
\[[H^+] \approx K_a \dfrac{C_a}{C_b}\]
which can be rewritten as
\[\dfrac{C_a}{C_b} \approx \dfrac{[H^+]}{K_a}\]
in which the pH is treated as an independent variable that controls the ratio of the conjugate forms of any acid-base pairs in the solution:
\[\log \left(\dfrac{C_a}{C_b}\right) \approx pH – pK_a\]
In the same way, we can define the pE as the negative log of the electron activity in the solution:
\[pE = –\log{e^–}\]
of these particles (but not to their "concentrations") when we are considering their availability to donors and acceptors. We will not get into the details of how pE is actually calculated (it is of course related to the ordinary standard electrode potential). To get an idea of its significance, consider the following chart that shows the pE° values of some redox systems that are of immense importance in the aquatic environment.Environmentally-important redox systems
- These pE values refer to typical environmental conditions with pH=7 and oxygen partial pressure of 0.21 atm. The scale below shows the free energy of a mole of electrons relative to their level in H 2 O.
- The two conjugate forms of any redox pair are present in equal concentrations when the pE is at the level at which the pair is shown. At pE's above or below this level, the reduced or oxidized form will predominate.
- The sugar glucose , denoted by the general formula for carbohydrates {CH 2 O}, is the source of chemical energy for most organisms. Note that it is thermodynamically stable (and thus capable of being formed by photosynthesis) only under highly reducing conditions.
- Organisms derive their metabolic free energy when electrons fall from glucose to a lower-lying acceptor on the right.
- Delivery of electrons from glucose to O 2 ( 8 ) is the source of metabolic free energy for all aerobic organisms, yielding 125 kJ per mole of electrons transferred.
A few other points about this plot are worth noting:
- Anaerobic organisms must make do with electron-acceptors above oxygen. The poor bacteria that depend on reducing hydrogen ions ( 2 ) have it worst of all, gaining only a tiny amount of metabolic energy to produce a tiny puff of hydrogen gas.
- Reaction ( 3 ) is not much more efficient, but it is the vital first link in the process of natural nitrogen fixation that gets carried out in the protected electron-rich environment of the organisms that live in the root nodules of legumes.
- The consequences of ( 4 ) are readily apparent if you have ever noticed the rotten-egg odor of some poorly-aerated muddy soils.
- Reaction ( 6 ) is known as fermentation ; it takes place in the anarobic soils of marshes and bogs (hence the nickname "marsh gas" for methane) and in the insides of animals from termites to cows, and sometimes to our embarrassment, in ourselves. Notice that glucose plays the double role of electron donor and acceptor here; this is a disproportionation reaction.
- Finally, if you have ever had much to do with babys' diapers, you have likely noticed the smell of ammonia produced as the anaerobic bacteria from within deliver electrons to nitrate ions ( 7 ). It's important to bear in mind that the reactions discussed above are mediated by living organisms; without the necessary enzymes to catalyze them, their rates are essentially zero.
For a more detailed chart, see Falling through the respiratory chain .
Analytical chemistry applications
A very large part of Chemistry is concerned, either directly or indirectly, with determining the concentrations of ions in solution. Any method that can accomplish such measurements using relatively simple physical techniques is bound to be widely exploited. Cell potentials are fairly easy to measure, and although the Nernst equation relates them to ionic activities rather than to concentrations, the difference between them becomes negligible in solutions where the total ionic concentration is less than about 10 –3 M .
The concentrations of ions in equilibrium with a sparingly soluble salt are sufficiently low that their direct determination can be quite difficult. A far simpler and common procedure is to set up a cell in which one of the electrode reactions involves the insoluble salt, and whose net cell reaction corresponds to the dissolution of the salt. For example, to determine the K sp for silver chloride, we could use the cell
\[Ag_{(s)} | Ag^+(?\; M) || Ag^+,Cl^– | AgCl_{(s)} | Ag_{(s)}\]
whose net equation corresponds to the dissolution of silver chloride:
|
cathode:
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AgCl (s) + e – → Ag (s) + Cl – (aq) | E° = +0.222 v |
|
anode:
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Ag (s) → Ag + (aq) + e – | E° = –(+0.799) v |
|
net:
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AgCl (s) → Ag + + Cl – | E° = -0.577 v |
The standard potential for the net reaction refers to a hypothetical solution in which the activities of the two ions are unity. The cell potential we actually observe corresponds to E in the Nernst equation, which is then solved for Q which gives K sp directly.
Potentiometric titrations
In many situations, accurate determination of an ion concentration by direct measurement of a cell potential is impossible due to the presence of other ions and a lack of information about activity coefficients. In such cases it is often possible to determine the ion indirectly by titration with some other ion. For example, the initial concentration of an ion such as Fe 2 + can be found by titrating with a strong oxidizing agent such as Ce 4 + . The titration is carried out in one side of a cell whose other half is a reference electrode:
Pt (s) | Fe 2 + , Fe 3 + || reference electrode
Initially the left cell contains only Fe 2 + . As the titrant is added, the ferrous ion is oxidized to Fe 3 + in a reaction that is virtually complete:
\[Fe^{2+} + Ce^{4+} → Fe^{3+} + Ce^{3+}\]
The cell potential is followed as the Fe 2 + is added in small increments. Once the first drop of ceric ion titrant has been added, the potential of the left cell is controlled by the ratio of oxidized and reduced iron according to the Nernst equation
\[E = 0.68 - 0.059 \; \log \dfrac{]Fe^{3+}]}{[Fe^{2+}]}\]
which causes the potential to rise as more iron becomes oxidized.
When the equivalence point is reached, the Fe 2 + will have been totally consumed (the large equilibrium constant ensures that this will be so), and the potential will then be controlled by the concentration ratio of Ce 3 + /Ce 4 + . The idea is that both species of a redox couple must be present in reasonable concentrations poise an electrode (that is, to control its potential according to the Nernst equation.) If one works out the actual cell potentials for various concentrations of all these species, the resulting titration curve looks much like the familiar acid-base titration curve. The end point is found not by measuring a particular cell voltage, but by finding what volume of titrant gives the steepest part of the curve.
Measurement of pH
Since pH is actually defined in terms of hydrogen ion activity and not its concentration, a hydrogen electrode allows a direct measure of {H + } and thus of –log {H + }, which is the pH. All you need is to measure the voltage of a cell
H 2 (g, 1 atm) | Pt | H + (? M ) || reference electrode
In theory this is quite simple, but when it was first employed in the pre-electronics era, it required some rather formidable-looking apparatus (such as the L&N vibrating-reed electrometer setup from the 1920's shown here) and the use of explosive hydrogen gas. Although this arrangement (in which the reference electrode could be a standard hydrogen electrode) has been used for high-precision determinations since that time, it would be impractical for routine pH measurements of the kinds that are widely done, especially outside the research laboratory.
The glass electrode for pH measurements
In 1914 it was discovered that a thin glass membrane enclosing a solution of HCl can produce a potential that varies with the hydrogen ion activity {H + } in about the same way as that of the hydrogen electrode. Glass electrodes are manufactured in huge numbers for both laboratory and field measurements. They contain a built-in Ag-AgCl reference electrode in contact with the HCl solution enclosed by the membrane.
The potential of a glass electrode is given by a form of the Nernst equation very similar to that of an ordinary hydrogen electrode, but of course without the H 2 :
E membrane = A + (RT/F) ln ( {H + } + B )
in which A and B are constants that depend on the particular glass membrane.
The reason a glass membrane would behave in this way was not understood until around 1970. It now appears that hydrogen ions in the external solution diffuse through the glass and push out a corresponding number of the Na + ions which are normally present in most glasses. These sodium ions diffuse to whichever side of the membrane has the lower concentration, where they remain mostly confined to the surface of the glass, which has a porous, gelatinous nature. It is the excess charge produced by these positive ions that gives rise to the pH-dependent potential. The first commercial pH meter was developed by Arnold Beckman (1900-2004) while he was a Chemistry professor at CalTech. He was unable to interest any of the instrumentation companies in marketing it, so he founded his own company and eventually became a multi-millionaire philanthropist.
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Arnold Beckman
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An early Beckman pH meter
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A typical modern pH meter
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Ion-selective electrodes
The function of the membrane in the glass electrode is to allow hydrogen ions to pass through and thus change its potential, while preventing other cations from doing the same thing (this selectivity is never perfect; most glass electrodes will respond to moderate concentrations of sodium ions, and to high concentrations of some others.) A glass electrode is thus one form of ion-selective electrode. Since about 1970, various other membranes have been developed which show similar selectivities to certain other ions. These are widely used in industrial, biochemical, and environmental applications.
Membrane potentials and nerve conduction
You may recall the phenomena of osmosis and osmotic pressure that are observed when two solutions having different solute concentrations are separated by a thin film or membrane whose porosity allows small ions and molecules to diffuse through, but which holds back larger particles. If one solution contains a pair of oppositely-charged ionic species whose sizes are very different, the smaller ions may pass through the semipermeable membrane while the larger ones are retained. This will produce a charge imbalance between the two solutions, with the original solution having the charge sign of the larger ion. Eventually the electrical work required to bring about further separation of charges becomes too large to allow any further net diffusion to take place, and the system settles into an equilibrium state in which a constant potential difference (usually around a volt or less) is maintained. This potential difference is usually called a membrane potential or Donnan potential after the English chemist who first described this phenomenon around 1930.
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Origin of a membrane potential If the smaller ions are able to diffuse through the membrane but the larger ions cannot, a potential difference will develop between the two solutions. This membrane potential can be observed by introducing a pair of platinum electrodes. |
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The figure shows a simple system containing the potassium salt of a protein on one side of a membrane, and potassium chloride on the other. The proteinate anion, being too large to diffuse through the membrane, gives rise to the potential difference. The value of this potential difference can be expressed by a relation that is essentially the same as the Nernst equation, although its derivation is different. The membrane potential can be expressed in terms of the ratio of either the K + or Cl – ion activities:
The membrane surrounding most living cells contains sites or "channels" through which K + ions are selectively transported so that the concentration of K + inside the cell is 10-30 times that of the intracellular fluid. Taking the activity ratio as about 20, the above equation predicts that the potential difference θ inside - θ outside will be
which is consistent with observed values. Transport of an ion such as K + from a region of low concentration into the more concentrated intercellular fluid requires a source of free energy, which is supplied by ATP under enzymatic control. The metabolic processes governing this action are often referred to as "ion pumps".
Nerve conduction
Transmission of signals through the nervous system occurs not by the movement of a charge carrier through the nerve, but by waves of differential ion concentrations that travel along the length of the nerve. These concentration gradients are reduced by protein-based ion channels and ATP-activated (and energy-consuming) ion pumps specific to K + and Ca 2 + ions.
We sometimes think of our nerves as the body's wiring, but the "electricity" that they transmit is not a flow of electrons, but a rapidly-traveling wave of depolarization involving the transport of ions through the nerve membrane. The normal potential difference between the inner and outer parts of nerve cells is about –70 mv as estimated above. Transmissin of a nerve impulse is initiated by a reduction of this potential difference to about –20 mv. This has the effect of temporarily opening the Na + channel; the influx of these ions causes the membrane potential of the adjacent portion of the nerve to collapse, leading to an effect that is transmitted along the length of the nerve. As this pulse passes, K + and Na + pumps restore the nerve to its resting condition.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- Atmospheric oxygen is a strong oxidizing agent, but in the absence of a suitable catalyst this reaction is ordinarily too slow to be observable. In organisms that possess the appropriate enzymes, the reaction is sufficiently fast to control the availability of electrons to other redox systems.
- The pE is a useful index of electron availability, especially in natural waters and biological systems in which multiple redox systems are usually present. Although electrons tend to "fall" (in free energy) from sources to sinks, the outcome depends very strongly on the rates of the various reactions. This is quite different from proton-exchange (acid-base) reactions which are uniformly fast.
- Solubility products can often be measured by constructing a cell in which one of the electrodes involves the sparingly soluble salt, and the net cell equation corresponds to the solubility reaction.
- Potentiometric titrations are widely used to measure concentrations of species that are readily oxidized or reduced. The reaction must be fast and have a very large equilibrium constant. The equivalance point is detected by the rapid change in potential that occurs when control of the cell potential passes from the redox system of the analyte to that of the titrant.
- Measurement of pH is accomplished with an electrode consisting of a thin membrane of glass in which Na + ions are exchanged with H + .
- Membrane (Donnan) potentials arise when passage of an ion across a semipermeable membrane is selectively facilitated or inhibited. The classic example is a sodium proteinate solution in which the protein anion is too large to pass through the membrane. In organisms, ion-specific channels or "pumps" have a similar effect.
- Nerve conduction occurs not by a flow of charge through the nerve, but by a wave of depolarization resulting from the concerted action of active channels that govern the passage of ions (mainly K + and Na + ) across the membrane enclosing the nerve.
|
libretexts
|
2025-03-17T19:53:19.257149
| 2013-10-02T00:42:44 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.05%3A_Applications_of_the_Nernst_Equation",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.5: Applications of the Nernst Equation",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.06%3A_Batteries_and_Fuel_Cells
|
16.6: Batteries and Fuel Cells
Make sure you thoroughly understand the following essential ideas which have been presented below.
- A battery is a galvanic cell in which some of the free energy change associated with a spontaneous electron-transfer reaction is captured in the form of electrical energy.
- A secondary or storage battery is one in which the electron-transfer reaction can be reversed by applying a charging current from an external source.
- A fuel cell is a special type of battery in which the reactants are supplied from an external source as power is produced. In most practical fuel cells, H + ions are produced at the anode (either from H 2 or a hydrocarbon) and oxygen from the air is reduced to H2O at the cathode.
- The cathodic reduction of O 2 is kinetically limited, necessitating the use of electrode surfaces having high catalytic activity.
- The electrodes in batteries must have very high effective surface areas, and thus be highly porous. This requirement may conflict with the other important one of efficient diffusion of reactants and products in the narrow channels within the pores.
- Batteries and fuel cells designed to power vehicles and portable devices need to have high charge-to-weight and charge-tovolume ratios.
One of the oldest and most important applications of electrochemistry is to the storage and conversion of energy. You already know that a galvanic cell converts chemical energy to work; similarly, an electrolytic cell converts electrical work into chemical free energy . Devices that carry out these conversions are called batteries . In ordinary batteries the chemical components are contained within the device itself. If the reactants are supplied from an external source as they are consumed, the device is called a fuel cell .
Introduction
The term battery derives from the older use of this word to describe physical attack or "beating"; Benjamin Franklin first applied the term to the electrical shocks that could be produced by an array of charged glass plates. In common usage, the term "call" is often used in place of battery. For portable and transportation applications especially, a battery or fuel cell should store (and be able to deliver) the maximum amount of energy at the desired rate (power level) from a device that has the smallest possible weight and volume. The following parameters are commonly used to express these attributes:
- Storage capacity or charge density , coulombs/liter or coulombs/kg;
- Energy density , J/kg or watt-hour/lb
- Power density , watts/kg
- Voltage efficiency , ratio of output voltage to E°
- Lifetime: shelf-life (resistance to self-discharge) or charge/recharge cycles
Primary and Secondary Batteries
A secondary or storage battery is capable of being recharged; its electrode reactions can proceed in either direction. During charging, electrical work is done on the cell to provide the free energy needed to force the reaction in the non-spontaneous direction. A primary cell , as expemplified by an ordinary flashlight battery, cannot be recharged with any efficiency, so the amount of energy it can deliver is limited to that obtainable from the reactants that were placed in it at the time of manufacture.
The lead-acid storage cell
The most well-known storage cell is the lead-acid cell, which was invented by Gaston Planté in 1859 and is still the most widely used device of its type. The cell is represented by
\[Pb(s) | PbSO_4(s) | H_2SO_4(aq) || PbSO_4(s), PbO_2(s) | Pb(s)\]
and the net cell reaction is
\[Pb(s) + PbO_2(s) + 2 H_2SO_4(aq) → 2 PbSO_4(s) + 2 H_2O\]
The reaction proceeds to the right during discharge and to the left during charging. The state of charge can be estimated by measuring the density of the electrolyte; sulfuric acid is about twice as dense as water, so as the cell is discharged, the density of the electrolyte decreases.
The technology of lead-acid storage batteries has undergone remarkably little change since the late 19th century. Their main drawback as power sources for electric vehicles is the weight of the lead; the maximum energy density is only about 35 Ah/kg, and actual values may be only half as much. There are also a few other problems:
- The sulfuric acid electrolyte becomes quite viscous when the temperature is low, inhibiting the flow of ions between the plates and reducing the current that can be delivered. This effect is well-known to anyone who has had difficulty starting a car in cold weather.
- These batteries tend to slowly self-discharge, so a car left idle for several weeks might be unable to start.
- Over time, PbSO 4 that does not get converted to PbO 2 due to lack of complete discharge gradually changes to an inert form which limits the battery capacity. Also, "fast" charging causes rapid evolution of hydrogen from the water in the electrolyte; the bubbles form on the lead surface and can tear PbO 2 off the positive plate. Eventually enough solid material accumulates at the bottom of the electrolyte to short-circuit the battery, leading to its permanent demise.
The LeClanché "dry cell"
The most well-known primary battery has long been the common "dry cell" that is widely used to power flashlights and similar devices. The modern dry cell is based on the one invented by Georges Leclanché in 1866. The electrode reactions are
\[Zn → Zn^{2+} + 2e^–\]
\[2 MnO_2 + 2H^+ + 2e^– → Mn_2O_3 + H_2O\]
Despite its name, this cell is not really "dry"; the electrolyte is a wet paste containing NH 4 Cl to supply the hydrogen ions. The chemistry of this cell is more complicated than it would appear from these equations, and there are many side reactions and these cells have limited shelf-lifes due to self discharge. (In some of the older ones, attack by the acidic ammonium ion on the zinc would release hydrogen gas, causing the battery to swell and rupture, often ruining an unused flashlight or other device.) A more modern version, introduced in 1949, is the alkaline cell which employs a KOH electrolyte and a zinc-powder anode which permits the cell to deliver higher currents and avoids the corrosive effects of the acidic ammonium ion on the zinc.
What is the cell notation for the LeClanché dry cell?
Physical limitations of Battery Performance
The most important of these are:
- Effective surface area of the electrode. A 1-cm 2 sheet of polished metal presents far less active surface than does one that contains numerous surface projections or pores. All useful batteries and fuel cells employ highly porous electrodes. Recent advances in nanotechnology are likely to greatly improve this parameter.
- Current density of electrode surface . Expressed in amperes m –2 , this is essentially a measure of the catalytic ability of the electrode, that is, its ability to reduce the activation energy of the electron transfer process.
- Rate at which electroactive components can be delivered to or depart from the active electrode surface. These processes are controlled by thermal diffusion and are inhibited by the very narrow pores that are needed to produce the large active surface area.
- Side reactions and irreversible processes. The products of the discharge reaction may tend to react with the charge-storing components. Thermal diffusion can also cause self-discharge, limiting the shelf life of the battery. Recharging of some storage batteries may lead to formation of less active modifications of solid phases, thus reducing the number of charge/discharge cycles possible.
Clearly, these are all primarily kinetic and mechanistic factors which require a great deal of experimentation to understand and optimize.
The Fuel Cell
Conventional batteries supply electrical energy from the chemical reactants stored within them; when these reactants are consumed, the battery is "dead". An alternative approach would be to feed the reactants into the cell as they are required, so as to permit the cell to operate continuously. In this case the reactants can be thought of as "fuel" to drive the cell, hence the term fuel cell.
Although fuel cells were not employed for practical purposes until space exploration began in the 1960's, the principle was first demonstrated in 1839 by Sir William Grove, a Welsh lawyer and amateur chemist. At the time, it was already known that water could be decomposed into hydrogen and oxygen by electrolysis; Grove tried recombining the two gases in a simple apparatus, and discovered what he called "reverse electrolysis"— that is, the recombination of H 2 and O 2 into water— causing a potential difference to be generated between the two electrodes:
| H 2 (g) → 2 H + + 2 e – | E° = 0 v | |
| ½ O 2 + 2 H + + 2 e – → H 2 O (l) | E° = +1.23 v | |
| H 2 (g) + ½ O 2 (g) → H 2 O (l) | E° = +1.23 v |
It was not until 1959 that the first working hydrogen-oxygen fuel cell was developed by Francis Thomas Bacon in England. Modern cells employ an alkaline electrolyte, so the electrode reactions differ from the one shown above by the addition of OH – to both sides of the equations (note that the net reaction is the same):
| H 2 (g) + 2 OH – → 2 H 2 O + 2 e – | E° = 0 v | |
| ½ O 2 (g) + 2 H 2 O + 2 e – → 2 OH – | E° = +1.23 v | |
| H 2 (g) + ½ O 2 (g) → H 2 O | E° = +1.23 v |
Although hydrogen has the largest energy-to-mass ratio of any fuel, it cannot be compressed to a liquid at ordinary temperatures. If it is stored as a gas, the very high pressures require heavy storage containers, greatly reducing its effective energy density. Some solid materials capable of absorbing large amount of H 2 can reduce the required pressure. Other fuels such as alcohols, hydrocarbon liquids, and even coal slurries have been used; methanol appears to be an especially promising fuel.
One reason for the interest in fuel cells is that they offer a far more efficient way of utilizing chemical energy than does conventional thermal conversion. The work obtainable in the limit of reversible operation of a fuel cell is 229 kJ per mole of H 2 O formed. If the hydrogen were simply burned in oxygen, the heat obtainable would be ΔH = 242 kJ mol –1 , but no more than about half of this heat can be converted into work so the output would not exceed 121 kJ mol –1 . This limit is a consequence of the Second Law of Thermodynamics . The fraction of heat that can be converted into work (\(\eta\)) is a function of how far (in temperature) the heat falls as it flows through the engine and into the surroundings; this fraction is given by
\[\eta=\dfrac{1 - T_{high}}{T_{low}}\]
At normal environmental temperatures of around 300 K, this would have to be at least 600 K for 50% thermal efficiency.
The major limitation of present fuel cells is that the rates of the electrode reactions, especially the one in which oxygen is reduced, tend to be very small, and thus so is the output current per unit of electrode surface. Coating the electrode with a suitable catalytic material is almost always necessary to obtain usable output currents, but good catalysts are mostly very expensive substances such as platinum, so that the resulting cells are too costly for most practical uses. There is no doubt that if an efficient, low-cost catalytic electrode surface is ever developed, the fuel cell would become a mainstay of the energy economy.
Microbobial Fuel Cells
Certain types of bacteria are able to oxidize organic compounds to carbon dioxide while directly transferring electrons to electrodes. These so-called electricigen organisms may make it possible to convert renewable biomass and organic waste directly into electricity without the wasted energy and pollution produced by direct combustion. In one experiment, a graphite electrode immersed in ordinary mud (containing humic materials) was able to produce measurable amounts of electricity.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.
- A battery is a galvanic cell in which some of the free energy change associated with a spontaneous electron-transfer reaction is captured in the form of electrical energy.
- A secondary or storage battery is one in which the electron-transfer reaction can be reversed by applying a charging current from an external source.
- A fuel cell is a special type of battery in which the reactants are supplied from an external source as power is produced. In most practical fuel cells, H + ions are produced at the anode (either from H 2 or a hydrocarbon) and oxygen from the air is reduced to H2O at the cathode.
- The cathodic reduction of O 2 is kinetically limited, necessitating the use of electrode surfaces having high catalytic activity.
- The electrodes in batteries must have very high effective surface areas, and thus be highly porous. This requirement may conflict with the other important one of efficient diffusion of reactants and products in the narrow channels within the pores.
- Batteries and fuel cells designed to power vehicles and portable devices need to have high charge-to-weight and charge-tovolume ratios.
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libretexts
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2025-03-17T19:53:19.342599
| 2013-10-02T00:42:45 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.06%3A_Batteries_and_Fuel_Cells",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.6: Batteries and Fuel Cells",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.07%3A_Timeline_of_Battery_Development
|
16.7: Timeline of Battery Development
Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive.
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"Baghdad Battery" - 1000 BCE?
Drawing of the three pieces. (CC-BY-SA 2.5; Ironie) |
Earthenware jars containing an iron rod surrounded by a copper cylinder were discovered near Baghdad in 1938. They are believed to have been used by the Parthian civilization that occupied the region about 2000 years ago as a source of electricity to plate gold onto silver. |
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Allesandro Volta 1782
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His "Voltaic pile", a stack of zinc and silver disks separated by a wet cloth containing a salt or a weak acid solution, was the first battery known to Western civilization. |
| Sir Humphry Davy 1813 | Davy builds a 2000-plate battery that occupies 889 square feet in the basement of Britain's Royal Society. His earlier batteries provided power for the first public demonstration of electric lighting (carbon arc). |
| Michael Faraday, 1830's | Faraday discovered the fundamentals of galvanic cells and electrolysis that put electrochemistry on a firm scientific basis. |
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1836 - Daniell cell (also known as a Crow's Foot or Gravity cell.)
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John Daniell (English chemist and meterologist) developed the first modern storage cell based on Faraday's principles. This consists of a large glass jar with a copper star-shaped electrode in the bottom and a zinc "crow's foot" shaped electrode suspended near the top. The bottom of the jar was filled with a concentrated copper sulfate solution. On top of this was poured dilute sulfuric acid, whose lower density kept it on top. This was the first practical battery to find wide use to power telegraphs and railway signaling systems and home doorbells. |
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1839 - William Grove (Welsh)
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Grove was best known in the 19th century for his "nitric acid battery" which came into wide use in early telegraphy.
Now, however, he is most famous for his "gas voltaic battery" in which discovered "reverse electrolysis": the recombination of H 2 and O 2 following electrolysis of water at platinum electrodes. This was the first demonstration of what we now know as the hydrogen-oxygen fuel cell (see below.) |
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1859 - Gaston Planté (French)
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Invents the first lead-acid storage cell which consisted of two sheets of lead separated by a rubber sheet, rolled into a spiral and immersed in dilute sulfuric acid. |
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1866 - Georges Leclanché (French)
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By 1868 twenty thousand Leclanché cells were being used in telegraph systems. The original Leclanché cells were built in porous pots which were heavy and subject to breakage. Within twenty years other inventors had modified the design into what we now know as "dry cells" which became widely used in the first flashlights (1909) and in battery-powered radios of the 1920s. |
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1881 - Faure and others
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Development of the first practical lead-acid storage cell . The major improvement over Planté's design was the addition of a paste of PbSO 4 to the positive plate. |
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1905 Nickel-iron cell
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Edison, who was as much a chemist as an all-around inventor, thought that the lead in Planté-type cells made them too heavy, and that having acid in contact with any metal was an inherently bad idea. After much experimentation, he developed a successful alkaline battery. The Edison cell uses an iron anode, nickel oxide cathode, and KOH electrolyte. This cell is extremely rugged and is still used in certain industrial applications, but it was never able to displace the lead-acid cell as Edison had hoped. |
| 1950s | A similar cell, employing a nickel anode instead of iron, was the first rechargeable cell that was small enough to be used in portable consumer devices. Its main disadvantage is that it is ruined by complete discharge. |
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1949 - Alkaline dry cell - Lew Urry
(Eveready Battery Co.) |
First commercial alkaline dry cell. These substitute KOH for the corrosive NH 4 Cl used in the older dry cells and last 5-8 times longer. |
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1947 - Mercury cell
(Ruben and Mallory, 1950's) |
This was one of the first "button"-type cells which were widely used in cameras and hearing aids. The constancy of the 1.34 v output made them popular for use in sensitive instruments and cardiac pacemakers. The net cell reaction is Zn (s) + HgO (s) → ZnO (s) + Hg (l) Most countries have outlawed sales of these cells in order to reduce mercury contamination of the environment. |
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Nickel-Cadmium (NiCad) cells
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The NiCad cell quickly become one of the most popular rechargeable batteries for small consumer devices. They can deliver high current and undergo hundreds of charge/discharge cycles. Because cadmium is an environmental toxin, their use is being discouraged. |
| 1959 - Fuel cell - Francis Bacon (UK) | The first practical fuel cell was developed by British engineer Francis Bacon (1904-1992). This hydrogen-oxygen cell used an alkaline electrolyte and inexpensive nickel electrodes. |
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Late 1960's - Nickel-metal hydride cells
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The hydride ion H – would be an ideal cathode material except for the fact that its oxidation product H 2 is a gas. The discovery that certain compounds such as LiNi 5 and ZrNi 2 can act as "hydrogen sponges" made it practical to employ metal hydrides as a cathode material. One peculiarity of Ni-MH cells is that recharging them is an exothermic process, so that proper dissipation of heat must be allowed for. These batteries are widely used in cell phones, computers, and portable power tools. The electrode reactions take place in a concentrated KOH electrolyte: Cathode (+): NiOOH + H 2 O + e – → Ni(OH) 2 + OH – Anode (-): (1/x) MH x + OH – → (1/x) M + H 2 O + e – |
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1990s - Lithium cells (Sony Corp.)
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Lithium is an ideal anode material owing to its low density and high reduction potential, making Li-based cells the most compact ways of storing electrical energy. Lithium cells are used in wristwatches, cardiac pacemakers and digital cameras. Both primary (non-rechargeble) and rechargeable types have been available for some time. More recent applications are in portable power tools and— perhaps most importantly, in electric-powered or hybrid automobiles. Modern lithium cells operate by transporting Li + ions between electrodes into which the ions can be inserted or intercalated. Cathodes are lithium transition-metal oxides such as LiCoO 3 , while anodes are lithium-containing carbon, LiC 6 . The species that undergoes oxidation-reduction is not lithium, but the transition metal, e.g. Co(III)-Co(IV). |
Lithium batteries as incendiary devices
There have been numerous reports of fires and explosions associated with lithium batteries. In 2006, the Dell Corporation had to recall 4.1 million Sony batteries that had been shipped with Dell's laptop computers and were judged to be at risk owing to a manufacturing defect. This illustrates the difficulty of concentrating a large amount of chemical energy into a small package, which is of course the goal of all battery developers eager to meet commercial demands ranging from consumer personal electronics to electrically-powered cars. The fully-charged Li + -deficient lithium cobalt oxide cathodes are inherently unstable, held in check only by a thin insulating membrane which, if accidentally breached, can lead to thermal runaway involving gaseous oxygen, carbon, organic solvents, and (in some cases) lithium chlorate— all the components necessary for a fierce fire.
Much research has gone into the development of fail-safe membranes. In one type, made by ExxonMobil and targeted at the automotive market, the pores are designed to close up and thus inhibit the passage of lithium ions when the temperature rises above a safe level.
Biological Batteries
Finally, we should mention the biological batteries that are found in a number of electric fish . The "electric organs" of these fish are modified muscle cells known as electrocytes which are arranged in long stacks. A neural signal from the brain causes all the electrocytes in a stack to become polarized simultaneously, in effect creating a battery made of series-connected cells. Most electric fish produce only a small voltage which they use for navigation, much in the way that bats use sound for echo-location of prey. The famous electric eel, however, is able to produce a 600-volt jolt that it employs to stun nearby prey.
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libretexts
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2025-03-17T19:53:19.423101
| 2017-07-05T16:57:08 |
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"title": "16.7: Timeline of Battery Development",
"author": "Stephen Lower"
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.08%3A_Electrochemical_Corrosion
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16.8: Electrochemical Corrosion
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
- Electrochemical corrosion of metals occurs when electrons from atoms at the surface of the metal are transferred to a suitable electron acceptor or depolarizer . Water must be present to serve as a medium for the transport of ions.
- The most common depolarizers are oxygen, acids, and the cations of less active metals.
- Because the electrons flow through the metallic object itself, the anodic and cathodic regions (the two halves of the electrochemical cell) can be at widely separated locations.
- Anodic regions tend to develop at locations where the metal is stressed or is protected from oxygen.
- Contact with a different kind of metal, either direct or indirect, can lead to corrosion of the more active one.
- Corrosion of steel can be inhibited by galvanizing , that is, by coating it with zinc, a more active metal whose dissolution leaves a negative charge on the metal which inhibits the further dissolution of Fe 2+ .
- Cathodic protection using an external voltage source is widely used to protect underground structures such as tanks, pipelines and piers. The source can be a sacrificial anode of zinc or aluminum, or a line-operated or photovoltaic power supply.
Corrosion can be defined as the deterioration of materials by chemical processes. Of these, the most important by far is electrochemical corrosion of metals, in which the oxidation process M → M + + e – is facilitated by the presence of a suitable electron acceptor, sometimes referred to in corrosion science as a depolarizer.
In a sense, corrosion can be viewed as the spontaneous return of metals to their ores; the huge quantities of energy that were consumed in mining, refining, and manufacturing metals into useful objects is dissipated by a variety of different routes. The economic aspects of corrosion are far greater than most people realize; the estimated cost of corrosion in the U.S. alone was $276 billion per year. Of this, about $121 billion was spent to control corrosion, leaving the difference of $155 billion as the net loss to the economy. Utilities, especially drinking water and sewer systems, suffer the largest economic impact, with motor vehicles and transportation being a close second.
Corrosion Cells and Reactions
The special characteristic of most corrosion processes is that the oxidation and reduction steps occur at separate locations on the metal. This is possible because metals are conductive, so the electrons can flow through the metal from the anodic to the cathodic regions (Figure \(\PageIndex{1}\)). The presence of water is necessary in order to transport ions to and from the metal, but a thin film of adsorbed moisture can be sufficient.
A corrosion system can be regarded as a short-circuited electrochemical cell in which the anodic process is something like
\[\ce{Fe(s) \rightarrow Fe^{2+}(aq) + 2 e^{-}} \label{1.7.1}\]
and the cathodic steps may invove the reduction of oxygen gas
\[ \ce{O_2} + \ce{2 H_2O} + \ce{4e^{-}} \rightarrow \ce{4 OH^{-}} \label{1.7.2}\]
or the reduction of protons
\[ \ce{H^{+} + e^{-}} \rightarrow \ce{1/2 H2(g)} \label{1.7.2b} \]
or the reduction of a metal ion
\[\ce{M^{2+} + 2 e^{–}} \rightarrow \ce{M(s)} \label{1.7.2c}\]
where \(\ce{M}\) is a metal.
Which parts of the metal serve as anodes and cathodes can depend on many factors, as can be seen from the irregular corrosion patterns that are commonly observed. Atoms in regions that have undergone stress, as might be produced by forming or machining, often tend to have higher free energies, and thus tend to become anodic.
If one part of a metallic object is protected from the atmosphere so that there is insufficient \(\ce{O2}\) to build or maintain the oxide film, this "protected" region will often be the site at which corrosion is most active. The fact that such sites are usually hidden from view accounts for much of the difficulty in detecting and controlling corrosion.
In contrast to anodic sites, which tend to be localized to specific regions of the surface, the cathodic part of the process can occur almost anywhere. Because metallic oxides are usually semiconductors, most oxide coatings do not inhibit the flow of electrons to the surface, so almost any region that is exposed to \(\ce{O2}\) or to some other electron acceptor can act as a cathode. The tendency of oxygen-deprived locations to become anodic is the cause of many commonly-observed patterns of corrosion.
Rusted-out Cars and Bathroom Stains
Anyone who has owned an older car has seen corrosion occur at joints between body parts and under paint films. You will also have noticed that once corrosion starts, it tends to feed on itself. One reason for this is that one of the products of the O 2 reduction reaction is hydroxide ion. The high pH produced in these cathodic regions tends to destroy the protective oxide film, and may even soften or weaken paint films, so that these sites can become anodic. The greater supply of electrons promotes more intense cathodic action, which spawns even more anodic sites, and so on.
A very common cause of corrosion is having two dissimilar metals in contact, as might occur near a fastener or at a weld joint. Moisture collects at the junction point, acting as an electrolyte and forming a cell in which the two metals serve as electrodes. Moisture and conductive salts on the outside surfaces provide an external conductive path, effectively short-circuiting the cell and producing very rapid corrosion; this is why cars rust out so quickly in places where salt is placed on roads to melt ice.
Dissimilar-metal corrosion can occur even if the two metals are not initially in direct contact. For example, in homes where copper tubing is used for plumbing, there is always a small amount of dissolved \(\ce{Cu^{2+}}\) in the water. When this water encounters steel piping or a chrome-plated bathroom sink drain, the more-noble copper will plate out on the other metal, producing a new metals-in-contact corrosion cell. In the case of chrome bathroom sink fittings, this leads to the formation of \(\ce{Cr^{3+}}\) salts which precipitate as greenish stains.
Control of Corrosion
Since both the cathodic and anodic steps must take place for corrosion to occur, prevention of either one will stop corrosion. The most obvious strategy is to stop both processes by coating the object with a paint or other protective coating. Even if this is done, there are likely to be places where the coating is broken or does not penetrate, particularly if there are holes or screw threads. A more sophisticated approach is to apply a slight negative charge to the metal, thus making it more difficult for the reaction to take place:
\[\ce{M -> M^{2+} + 2 e^{-}}.\]
Protection Method 1: Sacrificial Coatings
One way of supplying this negative charge is to apply a coating of a more active metal. Thus a very common way of protecting steel from corrosion is to coat it with a thin layer of zinc; this process is known as galvanizing.The zinc coating, being less noble than iron, tends to corrode selectively. Dissolution of this sacrificial coating leaves behind electrons which concentrate in the iron, making it cathodic and thus inhibiting its dissolution.
The effect of plating iron with a less active metal provides an interesting contrast. The common tin-plated can (on the right) is a good example. As long as the tin coating remains intact, all is well, but exposure of even a tiny part of the underlying iron to the moist atmosphere initiates corrosion. The electrons released from the iron flow into the tin, making the iron more anodic so now the tin is actively promoting corrosion of the iron! You have probably observed how tin cans disintegrate very rapidly when left outdoors.
Protection Method 2: Cathodic Protection
A more sophisticated strategy is to maintain a continual negative electrical charge on a metal, so that its dissolution as positive ions is inhibited. Since the entire surface is forced into the cathodic condition, this method is known as cathodic protection . The source of electrons can be an external direct current power supply (commonly used to protect oil pipelines and other buried structures), or it can be the corrosion of another, more active metal such as a piece of zinc or aluminum buried in the ground nearby, as is shown in the illustration of the buried propane storage tank below.
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libretexts
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2025-03-17T19:53:19.493373
| 2013-10-02T00:42:45 |
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"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.08%3A_Electrochemical_Corrosion",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.8: Electrochemical Corrosion",
"author": "Stephen Lower"
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.09%3A_Corrosion_Gallery
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16.9: Corrosion Gallery
Corrosion gallery
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Corrosion of a nail The nails are immersed in agar which forms a moist solid gel. The agar also contains phenolphthalein and hexacyanoiron(III) Fe(CN 6 ) – which forms a deep blue color ("prussian blue") in the presence of Fe 2 + . The blue colors are clearly associated with those parts of the nail that have been stressed, thus faciliting the anodic release of Fe 2 + from the metal. The the pink color shows the cathodic regions that have been made alkaline by the reaction O 2 + 2 H 2 O + 4 e – → 4 OH – This clearly shows the separation between the anodic and cathodic processes in corrosion. [Illustration from U of West Indies: link] |
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Water distribution main If you live in the older part of a city where the mains are 50-100 years old, the water you drink may well have passed through a pipe in this condition! Severe corrosion like this is more common in areas where the water is acidic. Such water comes from mountain snowmelt and runoff, and usually acquires its acidity from dissolved atmospheric carbon dioxide. Waters from rivers, lakes, and especially groundwaters from wells have usually been in sufficiently long contact with carbonate-containing sediments to have been neutralized. Water-works administrators like to make the water slightly alkaline and slightly supersaturated in calcium carbonate in order to maintain a thin coating of solid carbonate on the interior of the pipe which acts to protect it from corrosion. |
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Corrosion of reinforcing bars in concrete All large concrete structures contain steel reinforcing bars ("rebars") that help ensure structural integrity under varying load conditions and especially during earthquakes. Intrusion of water, even in the form of fog or mists, can lead to serious corrosion damage, as seen in this picture of this column which supports a highway overpass. |
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Corrosion at metallic joints The picture shows two steel structural members joined by cast iron flanges which have been bolted together. For some reason, one of the pieces has become more anodic than the other, leading to extensive corrosion of the upper part. |
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Bacterial-assisted corrosion This gas pipe was buried in a red soil that contained iron pyrites (FeS.) The bacterium thiobacillus ferrooxidans derives its energy by oxidizing Fe 2 + to the more soluble Fe 3 + , transferring the electrons to O 2 . It also oxidizes the sulfur, producing sulfuric acid. The resulting chemical cocktail has eaten a hole into the pipe. |
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| These galvanized bolts were used to join wooden beams together. Subsequent movement of the beams due to varying load conditions abraded the zinc coating. A lack of oxygen near the centers of the bolts also likely contributed to the corrosion by preventing the formation of a protective oxide film. |
Pitting corrosion
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libretexts
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2025-03-17T19:53:19.557134
| 2017-07-05T17:08:49 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.09%3A_Corrosion_Gallery",
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"title": "16.9: Corrosion Gallery",
"author": "Stephen Lower"
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.10%3A_Electrolytic_Cells_and_Electrolysis
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16.10: Electrolytic Cells and Electrolysis
Make sure you thoroughly understand the following essential ideas.
- In electrolysis, an external power source supplies the free energy required to drive a cell reaction in its non-spontaneous direction. An electrolytic cell is in this sense the opposite of a galvanic cell. In practice, the products of electrolysis are usually simpler than the reactants, hence the term electro- lysis .
- Transport of ions in the electrolyte in response to the potential difference between the electrodes ("drift") is largely restricted to the regions very close to the electrodes. Ionic transport in the greater part of the electrolyte is by ordinary thermal diffusion— the statistical tendency of concentrations to become uniform.
- When an aqueous solution is subjected to electrolysis, the oxidation or reduction of water can be a competing process and may dominate if the applied voltage is sufficiently great. Thus an attempt to electrolyze a solution of NaNO 3 will produce only H 2 and O 2 .
- A large number of electrolysis processes are employed by industry to refine metals and to produce both inorganic and organic products. The largest of these are the chloralkali industry (chlorine and "caustic"), and the refining of aluminum; the latter consumes approximately 5% of the electrical power generated in North America.
Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds.
Electrolysis of molten alkali halides is the usual industrial method of preparing the alkali metals:
| Na + + e – → Na (l) | E° = –2.71 V |
| Cl – → ½ Cl 2 (g) + e – | E° = –1.36 V |
| Na + + Cl – → Na(l) + ½ Cl 2 (g) | E° = –4.1 V |
Ions in aqueous solutions can undergo similar reactions. Thus if a solution of nickel chloride undergoes electrolysis at platinum electrodes, the reactions are
| Ni 2 + + 2 e – → Ni(s) | E° = –0.24 V |
| 2 Cl – → Cl 2 (g) + 2 e – | E° = –1.36 V |
| Ni 2 + + 2 Cl – → Ni(s) + Cl 2 (g) | E° = –1.60 v |
Both of these processes are carried out in electrochemical cells which are forced to operate in the "reverse", or non-spontaneous direction, as indicated by the negative for the above cell reaction. The free energy is supplied in the form of electrical work done on the system by the outside world (the surroundings). This is the only fundamental difference between an electrolytic cell and the galvanic cell in which the free energy supplied by the cell reaction is extracted as work done on the surroundings.
A common misconception about electrolysis is that "ions are attracted to the oppositely-charged electrode." This is true only in the very thin interfacial region near the electrode surface. Ionic motion throughout the bulk of the solution occurs mostly by diffusion, which is the transport of molecules in response to a concentration gradient. Migration— the motion of a charged particle due to an applied electric field, is only a minor player, producing only about one non-random jump out of around 100,000 random ones for a 1 volt cm –1 electric field. Only those ions that are near the interfacial region are likely to undergo migration.
Electrolysis in aqueous solutions
Water is capable of undergoing both oxidation
\[H_2O \rightarrow O_{2(g)} + 4 H^+ + 2 e^– \;\;\; E^o = -1.23 V\]
and reduction
\[2 H_2O + 2 e^– \rightarrow H_{2(g)} + 2 OH^– \;\;\; E^o = -0.83 \;V\]
Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute. For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium:
| H 2 O + 2 e – → H 2 (g) + 2 OH – | E =+0.41 V ([OH – ] = 10-7 M) | |
| Cl – → ½ Cl 2 (g) + e – | E° = –1.36 V | |
| Cl – + H 2 O → 2 H 2 (g) + ½ Cl 2 (g) + 2 OH – | E = –0.95 V |
[This illustration is taken from the excellent Purdue University Chemistry site ]Reduction of Na+ (E° = –2.7 v) is energetically more difficult than the reduction of water (–1.23 V), so in aqueous solution the latter will prevail.
Electrolysis of salt ("brine") is carried out on a huge scale and is the basis of the chloralkali industry.
Electrolysis of water
Pure water is an insulator and cannot undergo significant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO 3 results in the decomposition of water at both electrodes:
| H 2 O + 2 e – → H 2 (g) + 2 OH – | E =+0.41 V ([OH – ] = 10 -7 M) | |
| 2 H 2 O → O 2 (g) + 4 H+ + 2 e – | E° = -0.82 V | |
| 2 H 2 O(l) → 2 H 2 (g) + O 2 (g) | E = -1.23 V |
Electrolytic production of hydrogen is usually carried out with a dilute solution of sulfuric acid. This process is generally too expensive for industrial production unless highly pure hydrogen is required. However, it becomes more efficient at higher temperatures, where thermal energy reduces the amount of electrical energy required, so there is now some interest in developing high-temperature electrolytic processes. Most hydrogen gas is manufactured by the steam reforming of natural gas.
Faraday's laws of electrolysis
One mole of electric charge (96,500 coulombs), when passed through a cell, will discharge half a mole of a divalent metal ion such as Cu 2 + . This relation was first formulated by Faraday in 1832 in the form of two laws of electrolysis :
- The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte.
- The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance.
The equivalent weight of a substance is defined as the molar mass, divided by the number of electrons required to oxidize or reduce each unit of the substance. Thus one mole of V 3+ corresponds to three equivalents of this species, and will require three faradays of charge to deposit it as metallic vanadium.
Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units:
- current (amperes) is the rate of charge transport ; 1 amp = 1 C/sec.
- power (watts) is the rate of energy production or consumption; \[1 W = 1 J/sec = 1 volt-amp; 1 watt-sec = 1 J, 1 kW-h = 3600 J.\]
A metallic object to be plated with copper is placed in a solution of CuSO 4 .
- To which electrode of a direct current power supply should the object be connected?
- What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?
Solution
- Since Cu 2 + ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!)
- The amount of charge passing through the cell is
(0.22 amp) × (5400 sec) = 1200 C
or
(1200 C) ÷ (96500 c F –1 ) = 0.012 F
Since the reduction of one mole of Cu 2 + ion requires the addition of two moles of electrons, the mass of Cu deposited will be
(63.54 g mol –1 ) (0.5 mol Cu/F) (.012 F) = 0.39 g of copper
How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency?
Solution
- moles of Cl 2 produced: (10 6 g) ÷ 70 g mol –1 = 14300 mol Cl 2
- faradays of charge: (2 F/mol) × (14300 mol) = 28600 F
- charge in coulombs: (96500 C/F) × (28600 F) = 2.76 × 10 9 C
- duration of electrolysis: (3600 s/h) x (24 h) = 86400 s
- current (rate of charge delivery): (2.76 × 10 9 amp-sec) ÷ (86400 sec) = 32300 amps
- power (volt-amps): (2.0 V) × (32300 amp) = 64.6 kW
- energy in kW-h: (64.6 kW) × (24 h) = 1550 kW-h
- energy in joules: (1550 kW-h) × (3.6 MJ/kW-h) = 5580 MJ (megajoules)
(In the last step, recall that 1 W = 1 J/s, so 1 kW-h = 3.6 MJ)
Industrial Electrolytic Processes
For many industrial-scale operations involving the oxidation or reduction of both inorganic and organic substances, and especially for the production of the more active metals such as sodium, calcium, magnesium, and aluminum, the most cost-effective reducing agent is electrons supplied by an external power source. The two most economically important of these processes are described below.
The chloralkali industry
The electrolysis of brine is carried out on a huge scale for the industrial production of chlorine and caustic soda (sodium hydroxide). Because the reduction potential of Na + is much higher than that of water, the latter substance undergoes decomposition at the cathode, yielding hydrogen gas and OH – .
|
anode
reactions |
2 Cl – → Cl 2 (g) + 2 e – | -1.36 V | i |
|
4 OH – → O 2 (g) + 2 H 2 O + 4 e – |
-0.40 V | ii | |
|
cathode
|
Na+ + e – → Na(s) | -2.7 V | iii |
|
H 2 O + 2 e – → H 2 (g) + 2 OH – |
+.41 V | iv |
A comparison of the E° s would lead us to predict that the reduction ( ii ) would be favored over that of ( i ). This is certainly the case from a purely energetic standpoint, but as was mentioned in the section on fuel cells, electrode reactions involving O 2 are notoriously slow (that is, they are kinetically hindered), so the anodic process here is under kinetic rather than thermodynamic control. The reduction of water ( iv ) is energetically favored over that of Na + ( iii ), so the net result of the electrolysis of brine is the production of Cl 2 and NaOH ("caustic"), both of which are of immense industrial importance:
\[\ce{2 NaCl + 2 H2O -> 2 NaOH + Cl2(g) + H2(g)} \nonumber\]
Since chlorine reacts with both OH – and H 2 , it is necessary to physically separate the anode and cathode compartments. In modern plants this is accomplished by means of an ion-selective polymer membrane, but prior to 1970 a more complicated cell was used that employed a pool of mercury as the cathode. A small amount of this mercury would normally find its way into the plant's waste stream, and this has resulted in serious pollution of many major river systems and estuaries and devastation of their fisheries.
Electrolytic refining of aluminum
Aluminum is present in most rocks and is the most abundant metallic element in the earth's crust (eight percent by weight.) However, its isolation is very difficult and expensive to accomplish by purely chemical means, as evidenced by the high E° (–1.66 V) of the Al 3 + /Al couple. For the same reason, aluminum cannot be isolated by electrolysis of aqueous solutions of its compounds, since the water would be electrolyzed preferentially. And if you have ever tried to melt a rock, you will appreciate the difficulty of electrolyzing a molten aluminum ore! Aluminum was in fact considered an exotic and costly metal until 1886, when Charles Hall (U.S.A) and Paul Hérault (France) independently developed a practical electrolytic reduction process.
The Hall-Hérault process takes advantage of the principle that the melting point of a substance is reduced by admixture with another substance with which it forms a homogeneous phase. Instead of using the pure alumina ore Al 2 O 3 which melts at 2050°C, it is mixed with cryolite, which is a natural mixture of NaF and AlF 3 , thus reducing the temperature required to a more manageable 1000°C. The anodes of the cell are made of carbon (actually a mixture of pitch and coal), and this plays a direct role in the process; the carbon gets oxidized (by the oxide ions left over from the reduction of Al 3 + to CO, and the free energy of this reaction helps drive the aluminum reduction, lowering the voltage that must be applied and thus reducing the power consumption. This is important, because aluminum refining is the largest consumer of industrial electricity, accounting for about 5% of all electricity generated in North America. Since aluminum cells commonly operated at about 100,000 amperes, even a slight reduction in voltage can result in a large saving of power.
The net reaction is
\[\ce{2 Al_2O_3 + 3 C \rightarrow 4 Al + 3 CO_2} \nonumber\]
However, large quantities of CO and of HF (from the cryolite), and hydrocarbons (from the electrodes) are formed in various side reactions, and these can be serious sources of environmental pollution.
|
libretexts
|
2025-03-17T19:53:19.656553
| 2013-10-02T00:42:46 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.10%3A_Electrolytic_Cells_and_Electrolysis",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "16.10: Electrolytic Cells and Electrolysis",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics
|
17: Chemical Kinetics and Dynamics
The chapters in this section deal with the rates and mechanisms of chemical change. These topics stand in contrast to the subjects of equilibrium and thermodynamics that control the direction of chemical change. Chemical change is guided and driven by energetics (thermodynamics), but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products.
-
- 17.1: Rates of reactions and rate laws
- Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products. The energetic aspects of change are governed by the laws of thermodynamics.
-
- 17.3: Collision and activation- the Arrhenius law
- Why are some reactions so much faster than others, and why are reaction rates independent of the thermodynamic tendency of the reaction to take place? These are the central questions we address in this unit. In doing so, we open the door to the important topic of reaction mechanisms: what happens at the microscopic level when chemical reactions take place? To keep things as simple as possible, we will restrict ourselves to reactions that take place in the gas phase.
-
- 17.4: Reaction Mechanisms
- The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collision") or as the breaking-up of a molecule ("dissociation") into simpler units.
-
- 17.5: Kinetics of Reactions in Solution
- The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions.
-
- 17.6: Catalysts and Catalysis
- Catalysts play an essential role in our modern industrial economy, in our stewardship of the environment, and in all biological processes. This lesson will give you a glimpse into the wonderful world of catalysts, helping you to understand what they are and how they work.
-
- 17.7: Experimental methods of chemical kinetics
- Studies of the dynamics of chemical processes impinge on almost every area of chemistry and biochemistry. It is useful for students even at the general chemistry level to have some understanding of the experimental techniques that have informed what you have already learned about kinetics.
|
libretexts
|
2025-03-17T19:53:19.725783
| 2016-02-26T07:44:02 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17: Chemical Kinetics and Dynamics",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.01%3A_Rates_of_reactions_and_rate_laws
|
17.1: Rates of reactions and rate laws
Make sure you thoroughly understand the following essential ideas:
- Describe the contrasting roles of thermodynamics and kinetics in understanding chemical change.
- Given a balanced net equation, write an expression for the rate of a reaction .
- Sketch a curve showing how the instantaneous rate of a reaction might change with time.
- Determine the order of a reaction of the form A → B + C from experimental data for the concentrations of its products at successive times.
- Describe the initial rate and isolation methods of determining the orders of the individual reactants in a reaction involving multiple reactants.
- Explain the difference between differential and integral rate laws .
- Sketch out a plot showing how the concentration of a component ([A] or ln [A]) that follows first-order kinetics will change with time. Indicate how the magnitude of the rate constant affects this plot
- Define the half-life of a reaction.
- Given the half-life for a first-order reaction A → products along with the initial value of [A] o , find [A] t at a subsequent time an integral number of half-lifes later.
- Describe the conditions under which a reaction can appear to have an order of zero.
Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products.
The energetic aspects of change are governed by the laws of thermodynamics (the "dynamics" part of this word is related to the historical origins of the field and is not a part of dynamics in the sense of these lessons.)
Energetics + dynamics = chemical change
Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the free energy of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium .
The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need.
Half the Story
Thermodynamics points the way and makes it possible...but it says nothing about how long it will take to get there!
The stoichiometric equation for the reaction says nothing about its mechanism. By mechanism, we mean, basically, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products:
The inner workings of the black box are ordinarily hidden from us, are highly unpredictable and can only be inferred by indirect means. Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements.
The Balanced equation says nothing about a Reaction's Mechanism
The thermodynamics of these reactions are all similar (they are all highly exothermic), but their dynamics (their kinetics and mechanisms) could not be more different.
|
|
|
|
|
Careful experiments, carried out over many years, are consistent with the simplest mechanism: a single collision between the two reactant molecules results in a rearrangement of the bonds: |
One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br. |
The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively. |
What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation.
The rates of chemical reactions
Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed:
\[\color{red} \text{rate} = \dfrac{\Delta \text{concentration}}{\Delta \text{time}} \label{2-1}\]
For a reaction of the form A + B → C, the rate can be expressed in terms of the change in concentration of any of its components:
\[\text{rate} = \dfrac{-\Delta[A]}{\Delta t} \label{Eq1a}\]
\[\text{rate} = \dfrac{-\Delta[B]}{\Delta t} \label{Eq1b}\]
\[\text{rate} = \dfrac{\Delta[C]}{\Delta t} \label{Eq1c}\]
in which Δ[A] is the difference between the concentration of \(A\) over the time interv al \(\Delta t = t_2 – t_1\):
\[Δ[A] = [A]_2 – [A]_1 \label{2-2}\]
Notice the minus signs in Equations \(\ref{Eq1a}\) and \(\ref{Eq1a}\); the concentration of a reactant always decreases with time, so Δ[A] and Δ[B] are both negative. Since negative rates does not make much sense, rates expressed in terms of a reactant concentration are always preceded by a minus sign in order to make the rate come out positive.
Consider now a reaction in which the coefficients are different:
\[A + 3B → 2D \]
It is clear that [B] decreases three times as rapidly as [A], so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient:
\[\text{rate} = \dfrac{-\Delta[A]}{\Delta t}= \dfrac{-\Delta[B]}{3 \Delta t} = \dfrac{+\Delta[D]}{2\Delta t}\]
Each of the above quotients is a legitimate expression of the rate of this particular reaction; they all yield the same number. Which one you employ when doing a calculation is largely a matter of convenience.
For the oxidation of ammonia
\[4 NH_3 + 3O_2 \rightarrow 2 N_2 + 6 H_2O \nonumber\]
it was found that the rate of formation of N 2 was 0.27 mol L –1 s –1 .
- At what rate was water being formed?
- At what rate was ammonia being consumed?
Solution:
- From the equation stoichiometry, Δ[H 2 O] = 6/2 Δ[N 2 ], so the rate of formation of H 2 O is 3 × (0.27 mol L –1 s –1 ) = 0.81 mol L –1 s –1 .
- 4 moles of NH 3 are consumed for every 2 moles of N 2 formed, so the rate of disappearance of ammonia is 2 × (0.27 mol L –1 s –1 ) = 0.54 mol L –1 s –1 .
Comment : Because of the way this question is formulated, it would be acceptable to express this last value as a negative number.
Instantaneous rates
Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δ t . Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The instantaneous rate of a reaction is given by the slope of a tangent to the concentration-vs.-time curve. Three such rates have been identified in this plot.
An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an initial rate (label (1) here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are derivatives whose values can very at each point on the curve, so that these instantaneous rates are really limiting rates defined as
\[rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta[A]}{\Delta t}\]
However, if you does not know calculus, just bear in mind that the larger the time interval Δ t , the smaller will be the precision of the instantaneous rate.
Instantaneous rates are also known as differential rates.
Rate Laws and Reaction Order
The relation between the rate of a reaction and the concentrations of reactants is expressed by its rate law . For example, the rate of the gas-phase decomposition of dinitrogen pentoxide
\[2N_2O_5 → 4NO_2 + O_2\]
has been found to be directly proportional to the concentration of \(N_2O_5\):
\[\text{rate} = k [N_2O_5]\]
Be very careful about confusing equilibrium constant expressions with those for rate laws. The expression for \(K_{eq}\) can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by
\[ K_{eq} = \dfrac{[NO_2]^2 [O_2]}{[N_2O_5]^2}\]
In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined experimentally . More generally, for a reaction of the form
\[n_A A + n_B B + ... → products\]
the rate law will be
\[rate = [A]^a[B]^b ... \]
, n B .Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let \(p\) be the sum of the exponents of the concentration terms in the rate law
\[p = a + b + ...\]
then k will have the dimensions (concentration 1–p /time).
Reaction order
The order of a rate law is the sum of the exponents in its concentration terms. For the N 2 O 5 decomposition with the rate law k[ N 2 O 5 ], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. We can also say that the reaction is "first order in N 2 O 5 ". For more complicated rate laws, we can speak of the overall reaction order and also the orders with respect to each component. As an example, consider a reaction
\[A + 3B + 2C → \text{products}\]
whose experimental rate law is
\[rate = k[A] [B]^2\]
We would describe this reaction as
- third-order overall,
- first-order in A,
- second-order in B, and
- zero-order in C.
Zero-order means that the rate is independent of the concentration of a particular reactant. However, enough C must be present to allow the equilibrium mixture to form.
The rate of oxidation of bromide ions by bromate in an acidic aqueous solution
\[6H^+ + BrO_3^– + 5Br^– → 3 Br_2 + 3 H_2O \nonumber\]
is found to follow the rate law
\[rate = k[Br^–][BrO_3^–][H^+]^2\]
What happens to the rate if, in separate experiments,
- [ BrO 3 – ] is doubled;
- the pH is increased by one unit;
- the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?
Solution
- Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.
- Increasing the pH by one unit will decrease the [H + ] by a factor of 10. Since the reaction is second-order in [H + ], this will decrease the rate by a factor of 100.
- Dilution reduces the concentrations of both Br 2 and BrO 3 – to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentration will reduce the rate by a factor of 4, to (½)×(½) = ¼ of its initial value.
How reaction orders are observed
Observing rate-vs.-concentration proportionality
In order to determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form
A → products
in which the rate is – d [A]/ dt , we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A].
- If doubling the concentration of A doubles the rate, then the reaction is first-order in A.
- If doubling the concentration results in a four-fold rate increase, the reaction is second-order in A.
Use the tabulated experimental data to determine the order of the reaction
\[2 N_2O_5 → 4 NO_2 + O_2 \nonumber\]
| Time (min) | p(N 2 O 5 ) | [N 2 O 5 ] mol L -1 |
Rate (mol L -1 min -1 ) |
|---|---|---|---|
| 0 | 301.6 | 0.0152 | |
| 10 | 224.8 | 0.0113 | 3.4 × 10 –4 |
| 20 | 166.7 | 0.0084 | 2.5 |
| 30 | 123.2 | 0.0062 | 1.8 |
| 40 | 92.2 | 0.0046 | 1.3 |
| 69.1 | 69.1 | 0.0035 | 1.0 |
Solution
The ideal gas law can be used to convert the partial pressures of \(N_2O_5\) to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of \([N_2O_5]\) and \([N)2O)5]^2\). It is apparent that the rates are directly proportional to \([N)2O)5]^1\), indicating that this is a first-order reaction.
Initial rate method
When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found.
We then plot the five initial rates of consumption of \(N_2O_5\) as a function of its molar concentration. As before, we see that these rates are directly proportional to \([N_2O_5]\). The slope of this plot gives the value of the rate constant.
rate = (5.2 × 10 –3 ) [N2O5] mol L –1 s –1
A study of the gas-phase reduction of nitric oxide by hydrogen
\[2 NO + 2 H_2 → N_2 + 2 H_2O \nonumber\]
yielded the following initial-rate data (all pressures in torr):
| experiment | P(NO) | P(H 2 ) |
initial rate (torr s–1) |
|---|---|---|---|
|
|
359 | 300 | 1.50 |
|
|
300 | 300 | 1.03 |
|
|
152 | 300 | 0.25 |
|
|
300 | 289 | 1.00 |
|
|
300 | 205 | 0.71 |
|
|
200 | 147 | 0.51 |
Find the order of the reaction with respect to each component. In looking over this data, take note of the following:
- The six runs recorded here fall into two groups, in which the initial pressures of H 2 and of NO, respectively, are held constant.
- All the data are expressed in pressures, rather than in concentrations. We can do this because the reactants are gases, whose concentrations are directly proportional to their partial pressures when T and V are held constant. And since we are only interested in comparing the ratios of pressures and rates, the units cancel out and does notmatter. It is far easier experimentally to adjust and measure pressures than concentrations.
Solution
Experiments 2 and 3 : Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.
Experiments 4 and 6 : Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.
The rate law is thus
\[rate = k[NO]^2[H_2]\]
Dealing with multiple reactants: the isolation method
It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is
\[A + B + C → \text{products}\]
and we need to find the order with respect to [B] in the rate law. If we set [B] o to 0.020 M and let [A] o = [C] o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested.
|
libretexts
|
2025-03-17T19:53:19.843298
| 2016-02-26T07:45:19 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.01%3A_Rates_of_reactions_and_rate_laws",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.1: Rates of reactions and rate laws",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.02%3A_Reaction_Rates_Typically_Change_with_Time
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17.2: Reaction Rates Typically Change with Time
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
- Describe the contrasting roles of thermodynamics and kinetics in understanding chemical change.
- Given a balanced net equation, write an expression for the rate of a reaction .
- Sketch a curve showing how the instantaneous rate of a reaction might change with time.
- Determine the order of a reaction of the form A → B + C from experimental data for the concentrations of its products at successive times.
- Describe the initial rate and isolation methods of determining the orders of the individual reactants in a reaction involving multiple reactants.
- Explain the difference between differential and integral rate laws .
- Sketch out a plot showing how the concentration of a component ([A] or ln [A]) that follows first-order kinetics will change with time. Indicate how the magnitude of the rate constant affects this plot
- Define the half-life of a reaction.
- Given the half-life for a first-order reaction A → products along with the initial value of [A] o , find [A] t at a subsequent time an integral number of half-lifes later.
- Describe the conditions under which a reaction can appear to have an order of zero.
On this page we extend the concept of differential rate laws introduced on the previous page to integral rate laws and reaction half-life that are of great importance in most practical applications of kinetics.
Differential and integral rate laws
Measuring instantaneous rates as we have described on the previous page of this unit is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision.
- If the reaction is very fast, its rate may change more rapidly than the time required to measure it; the reaction may be finished before even an initial rate can be observed.
- In the case of very slow reactions, observable changes in concentratons occur so slowly that the observation of a truly "instantaneous" rate becomes impractical.
The ordinary rate law (more precisely known as the instantaneous or differential rate law) tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants ( and of products ) change with time. For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products.
This is the purpose of an integrated rate law .
Integrating the rate law
This is easy to do, but only some courses expect you to know how to do it. For a quick run-through, click here . If you have had even a bit of calculus, here is an opportunity to put it to use!
Expressing the "speed" of a reaction: the half-life
How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case,
How do we define the point at which the reaction is "completed"?
A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing. If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion.
"How long?" may be too long
If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%.
The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially imporant in industrial processes in which the balances of these costs affect the profitability of the operation.
The half-life of a reaction
Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the half-life (or half-time) of the reaction.
First-order reactions
The law of exponential change
The rate at which a reactant is consumed in a first-order process is proportional to its concentration at that time. This general relationship, in which a quantity changes at a rate that depends on its instantaneous value , is said to follow an exponential law .
Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms.
The reason that the exponential function y=e x so efficiently describes such changes stems from the remarkable property that dy/dx = e x ; that is, e x is its own derivative, so the rate of change of y is identical to its value at any point. A nice discussion of this property can be found here .
The first-order integrated rate law
The integrated rate law for a first-order reaction
A → products
is a common example of the law of exponential change. For a reactant A, its concentration [A] t at time t is given by
[A] t = [A] o × e – kt
in which [A] o is its initial concentration and k is the first-order rate constant.
The "e" in the exponential term is of course the base of the natural logarithms, and the negative sign in its exponent means that the value of this term diminishes as t increases, as we would expect for any kind of a decay process.
A more convenient form of the integrated rate law is obtained by taking the natural logarithm of both sides:
ln [A] = – kt + ln [A] o (4-1)
This has the form of an equation for a straight line
y = mx + b
in which the slope m corresponds to the rate constant k . This means that, for a first-order reaction, a plot of ln [A] as a function of time gives a straight line with a slope of – k .
Half-life
After a period of one half-life, t = t ½ and we can write
\[\dfrac{[\mathrm{A}]_{1 / 2}}{[\mathrm{~A}]_0}=\dfrac{1}{2}=\mathrm{e}^{-k t_{1 / 2}} \nonumber \]
(in which we express the exponential as a function in order to make it stand out more prominently.) Taking logarithms of both sides (remember that
ln e
x
=
x
) yields
\[\ln 0.5=-k t \nonumber \]
Solving for the half-life, we obtain the simple relation
\[t_{1 / 2}=\dfrac{0.693}{k} \nonumber \]
which tells us that the half-life of a first-order reaction is a constant . This means that 100,000 molecules of a reactant will be reduced to 50,000 in the same time interval needed for ten molcules to be reduced to five.
It should be clear that the rate constant and the half life of a first-order process are inversely related.
The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?
Solution:
From the above equation, k = –0.693/(600 s) = 0.00115 s –1
The decay of radioactive nuclei is always a first-order process.
The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.
a) What fraction of the Am 241 in a smoke detector will have decayed after 50 years?
b) How long will it take for the activity to decline to 80% of its initial value?
c) What would be the "seventh-life" of Am 241 ?
Second-order reactions
Integration of the second-order rate law
\[-\dfrac{d[\mathrm{~A}]}{d t}=-k|\mathrm{~A}|^2 \nonumber \]
yields
\[-\dfrac{1}{\mid \mathrm{A}]}=\dfrac{1}{\left|\mathrm{~A}_{\mathrm{o}}\right|}+k t \nonumber \]
which is easily rearranged into a form of the equation for a straight line (try showing this yourself!) and yields plots similar to the one shown on the left below.
The half-life is given by
\[t_{1/2}=\dfrac{1}{k\left|\mathrm{~A}_0\right|} \nonumber \]
(see here for details)
Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to its constancy for a first-order reaction. For this reason, the concept of half-life for a second-order reaction is far less useful.
Zero-order processes
In some reactions, the rate is apparently independent of the reactant concentraton, in which case
Integration of the differential rate law yields [A] t = [A] 0 – k
\[rate =k[\mathrm{~A}]^0=k \nonumber \]
Note the word "apparently" in the preceding sentence; zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions.
Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.
There are two general conditions that can give rise to zero-order rates:
This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface ( heterogeneous catalysis ) or to an enzyme.
For example, the decomposition of nitrous oxide
N 2 O (g) → N 2 (g) + ½ O 2 (g)
in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional kinetics when carried out entirely in the gas phase.
In this case, the N 2 O molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.
Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex . If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.
This is most often seen when two or more reactants are involved. Thus if the reaction
A + B → products
is first-order in both reactants so that
rate = k [A][B]
then if B is present in great excess, the reaction will appear to be zero order in B (and first order overall). This commonly happens when B is H 2 O and the reaction is carried out in aqueous solution.
- Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
- When two or more reactants are involved, the concentrations of some are much greater than those of others
Summary
The following table compares the rate parameters of zero-, first-, and second-order reactions of the form a A → products . See also this excellent page from Purdue University that shows plots of [A], ln [A], and 1/[A] for each reaction order.
| zero order | first order | second order | |
|---|---|---|---|
| differential rate law | rate = k | rate = k [A] | rate = k [A] 2 |
| integrated rate law | \([A]=[A]_0=-a k t\) | \(\ln \dfrac{[\mathrm{A}]_0}{[\mathrm{~A}]}=a k t\) | \(\dfrac{1}{[\mathrm{~A}]}-\dfrac{1}{[\mathrm{~A}]_0}=a k t\) |
| half-life | \(\dfrac{[\mathrm{A}]_0}{2 a k}\) | \(\frac{\ln 2}{a k}=\frac{.693}{a k}\) | \(\frac{1}{a k[\mathrm{~A}]_0}\) |
| plot that gives straight line | [A] vs. t | ln [A] vs. t | 1/[A] vs. t |
| plot, showing interpretation of slope and intercept |
|
libretexts
|
2025-03-17T19:53:19.945910
| 2016-02-26T07:45:31 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.02%3A_Reaction_Rates_Typically_Change_with_Time",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.2: Reaction Rates Typically Change with Time",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.03%3A_Collision_and_activation-_the_Arrhenius_law
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17.3: Collision and activation- the Arrhenius law
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
- Explain the meaning of a reaction mechanism and define elementary step and intermediate.
- Describe the role of collisions in reaction mechanisms, and explain why not all collisions lead to the formation of products.
- Sketch out activation energy diagrams for simple reactions that are endothermic or exothermic,
- Explain how an activated complex differs from an intermediate.
- Define catalyst, and sketch out an activation energy diagram that illustrates how catalysts work.
- Explain the significance of the various terms that appear in the Arrhenius Law.
- Sketch out a typical Arrhenius Law plot for a hypothetical reaction at higher and lower temperatures.
- Explain how the activation energy of a reaction can be determined experimentally.
- Explain the significance of the various terms that appear in the pre-exponential factor of the Arrhenius equation.
Why are some reactions so much faster than others, and why are reaction rates independent of the thermodynamic tendency of the reaction to take place? These are the central questions we address in this unit. In doing so, we open the door to the important topic of reaction mechanisms : what happens at the microscopic level when chemical reactions take place? We can thank Prof. Svante Arrhenius for unlocking this door! To keep things as simple as possible, we will restrict ourselves to reactions that take place in the gas phase. The same principles will apply to reactions in liquids and solids, but with added complications that we will discuss in a later unit.
Reaction mechanisms
The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collison") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an intermediate — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation.
A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called elementary reactions ) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as
| unimolecular | A → | by far the most common |
| bimolecular | A + B → | |
| termolecular | A + B + C → | very rare |
Collision theory of chemical change: Molecules must collide before they can react
This fundamental rule must guide any analysis of an ordinary chemical reaction mechanism.
This explains why termolecular processes are so uncommon. The kinetic theory of gases tells us that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction. Consider a simple bimolecular step
\[\ce{A + B -> products} \nonumber\]
Clearly, if two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. We call such an encounter a collision .
The frequency of collisions between A and B in a gas will be proportional to the concentration of each; if we double [A], the frequency of A-B collisions will double, and doubling [B] will have the same effect. So if all collisions lead to products, than the rate of a bimolecular process will be first-order in A and B, or second-order overall:
\[\text{rate} = k[\ce{A}][\ce{B}] \nonumber\]
However, not all collisions are equal. In a gas at room temperature and normal atmospheric pressure, there will be about 10 33 collisions in each cubic centimetre every second. If every collision between two reactant molecules yielded products, all reactions would be complete in a fraction of a second.
When two billiard balls collide, they simply bounce off of each other. This is also the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, collisions must be sufficiently energetic (kinetic energy) to bring about this bond disruption. More about this further on.
And there is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N 2 O must hit the nitrogen end of NO; reversing the orientation of either molecule prevents the reaction.
Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. But of course, the more critical this orientational requirement is, the fewer collisions will be effective.
Anatomy of a collision
Energetic collisions between molecules cause interatomic bonds to stretch and bend farther, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds. Chemical bonds have some of the properties of mechanical springs, whose potential energy depends on the extent to which they are stretched or compressed. Each atom-to-atom bond can be described by a potential energy diagram that shows how its energy changes with its length. When the bond absorbs energy (either from heating or through a collision), it is elevated to a higher quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length oscillates between the extended limits corresponding to the curve.
A particular collision will typically excite a number of bonds in this way. Within about 10 –13 second this excitation gets distributed among the other bonds in the molecule in rather complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction.
Consider, for example, the isomerization of cyclopropane to propene which takes place at fairly high temperatures in the gas phase.
We can imagine the collision-to-product sequence in the following [grossly oversimplified] way:
Note that
- To keep things simple, we do not show the hydrogen atoms here. This is reasonable because C–C bonds are weaker then C–H bonds and thus less likely to be affected.
- The collision at will usually be with another cyclopropane molecule, but because no part of the colliding molecule gets incorporated into the product, it can in principle be a noble gas or some other non-reacting species;
- Although the C–C bonds in cyclopropane are all identicial, the instantaneous localization of the collisional energy can distort the molecule in various ways ( ), leading to a configuration sufficiently unstable to initiate the rearrangement to the product.
The cyclopropane isomerization described above is typical of many decomposition reactions that are found to follow first-order kinetics, implying that the process is unimolecular. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are really rather complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as pseudounimolecular .
Activation energy
Higher temperatures, faster reactions
The chemical reactions associated with most food spoilage are catalyzed by enzymes produced by the bacteria which mediate these processes. It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.
It is not hard to understand why this should be. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelyhood of bond cleavages and rearrangements as described above.
Activation energy diagrams
Most reactions involving neutral molecules cannot take place at all until they have acquired the energy needed to stretch, bend, or otherwise distort one or more bonds. This critical energy is known as the activation energy of the reaction. Activation energy diagrams of the kind shown below plot the total energy input to a reaction system as it proceeds from reactants to products.
In examining such diagrams, take special note of the following:
- The " reaction coordinate " plotted along the abscissa represents the changes in atomic coordinates as the system progresses from reactants to products. In the very simplest elementary reactions it might correspond to the stretching or twisting of a particular bond, and be shown to a scale. In general, however, the reaction coordinate is a rather abstract concept that cannot be tied to any single measurable and scaleable quantity.
- The activated complex (also known as the transition state ) represents the structure of the system as it exists at the peak of the activation energy curve. It does not correpond to an identifiable intermediate structure (which would more properly be considered the product of a separate elementary process), but rather to whatever configuration of atoms exists during the collision, which lasts for only about 0.1 picosecond.
- Activation energy diagrams always incorporate the energetics (Δ U or Δ H ) of the net reaction, but it is important to understand that the latter quantities depend solely on the thermodynamics of the process which are always independent of the reaction pathway. This means that the same reaction can exhibit different activation energies if it can follow alternative pathways.
- With a few exceptions for very simple processes, activation energy diagrams are largely conceptual constructs based on our standard collision model for chemical reactions. It would be unwise to read too much into them.
Gallery of activation energy plots
Activation energy diagrams can describe both exothermic and endothermic reactions:
... and the activation energies of the forward reaction can be large, small, or zero (independently, of course, of the value of Δ H ):
Processes with zero activation energy most commonly involve the combination of oppositely-charged ions or the pairing up of electrons in free radicals, as in the dimerization of nitric oxide (which is an odd-electron molecule).
In this plot for the dissociation of bromine, the E a is just the enthalpy of atomization
Br 2 (g) → 2 Br· (g)
and the reaction coordinate corresponds roughly to the stretching of the vibrationally-excited bond. The "activated complex", if it is considered to exist, is just the last, longest "stretch". The reverse reaction, being the recombination of two radicals, occurs immediately on contact.
Where does the activation energy come from?
In most cases, the activation energy is supplied by thermal energy, either through intermolecular collisions or (in the case of thermal dissocation) by thermal excitation of a bond-stretching vibration to a sufficiently high quantum level.
As products are formed, the activation energy is returned in the form of vibrational energy which is quickly degraded to heat. It's worth noting, however, that other sources of activation energy are sometimes applicable:
- Absorption of light by a molecule (photoexcitation) can be a very clean and efficient, but it doesn't always work. It's not enough that the wavelength of the light correspond to the activation energy; it must also fall within the absorption spectrum of the molecule, and (in a complex molecule) enough of it must end up in the right part of the molecule, such as in a particular bond.
- Electrochemical activation . Molecules capable of losing or gaining electrons at the surface of an electrode can undergo activation from an extra potential (known as the overvoltage ) between the electrode and the solution. The electrode surface often plays an active role, so the process is also known as electrocatalysis .
Catalysts can reduce activation energy
A catalyst is usually defined as a substance that speeds up a reaction without being consumed by it. More specifically, a catalyst provides an alternative, lower activation energy pathway between reactants and products. As such, they are vitally important to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various kind. In addition, most biochemical processes that occur in living organisms are mediated by enzymes , which are catalysts made of proteins.
It is important to understand that a catalyst affects only the kinetics of a reaction; it does not alter the thermodynamic tendency for the reaction to occur. Thus there is a single value of Δ H for the two pathways depicted in the plot on the right
Temperature and kinetic energy
A review of the principles of gas molecular velocities and the Boltzmann distribution can be found on the "KMT-classic" page .
In the vast majority of cases, we depend on thermal actvation, so the major factor we need to consider is what fraction of the molecules possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a opulation of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.
The two distribution plots shown here are for a lower temperature T 1 and a higher temperature T 2 . The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of E a that are shown.
It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.
2 The Arrhenius law
By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann disribution law into one of the most important relationships in physical chemistry:
Take a moment to focus on the meaning of this equation, neglecting the A factor for the time being.
First, note that this is another form of the exponential decay law we discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –E a / RT . And what is the significance of this quantity? If you recall that RT is the average kinetic energy , it will be apparent that the exponent is just the ratio of the activation energy E a to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign.) This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. And because these terms occur in an exponent, their effects on the rate are quite substantial.
The two plots below show the effects of the activation energy (denoted here by E ‡ ) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10 8 .
The logarithmic scale in the right-hand plot leads to nice straight lines, as described under the next heading below.
Looking at the role of temperature, we see a similar effect. (If the x -axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.)
Determining the activation energy
The Arrhenius equation
\[k=A \mathrm{e}^{-E_{a} / R T}\]
can be written in a non-exponential form which is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields
\[\ln k=\ln \left(A \mathrm{e}^{-E_{\mathrm{a}} / R T}\right)=\ln A+\ln \left(\mathrm{e}^{-E_{\mathrm{a}} / R T}\right)\]
\[\ln k=\ln A-\dfrac{E_{a}}{R T}\]
which is the equation of a straight line whose slope is \(–E_a /R\). This affords a simple way of determining the activation energy from values of \(k\) observed at different temperatures; we just plot \(\ln k\) as a function of \(1/T\).
Thus for the isomerization of cyclopropane to propene
the following data were obtained (calculated values shaded in pink):
| T, °C | 477 | 523 | 577 | 623 |
|---|---|---|---|---|
| 1/ T , K –1 × 10 3 | 1.33 | 1.25 | 1.18 | 1.11 |
| k , s –1 | 0.00018 | 0.0027 | 0.030 | 0.26 |
| ln k | –8.62 | –5.92 | –3.51 | –1.35 |
From the calculated slope, we have
– ( E a / R ) = –3.27 × 10 4 K
E a =– (8.314 J mol –1 K –1 ) (–3.27 × 10 4 K) = 273 kJ mol –1
Comment: This activation energy is rather high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature.
You don't always need a plot
(... if you are willing to live a bit dangerously!) Since the ln k -vs.-1/ T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that
\[\ln k_{2}-\ln k_{1}=\left(\ln A-\frac{E_{a}}{R T_{2}}\right)-\left(\ln A-\frac{E_{a}}{R T_{1}}\right)=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \nonumber\]
(... in which we have made the ln- A term disappear by subtracting the expressions for the two ln- k terms.) Solving the expression on the right for the activation energy yields
\[E_{a}=\dfrac{R \ln \dfrac{k_{2}}{k_{1}}}{\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}} \nonumber\]
A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten-C° rise in the temperature approximately doubles the rate. (This is obviously not generally true, especially when a strong covalent bond must be broken.) But for a reaction that does show this behavior, what would the activation energy be?
Solution
We will center our ten-degree interval at 300 K. Substituting into the above expression yields
= (8.314)(0.693) / (.00339 - 0.00328)
= (5.76 J mol –1 K –1 ) / (0.00011 K –1 ) = 52400 J mol –1 = 52.4 kJ mol –1
It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein.
Solution:
The ratio of the rate constants at the elevations of LA and Denver is 4.5/3.0 = 1.5, and the respective temperatures are 373K and 365K . With the subscripts 2 and 1 referring to LA and Denver respectively, we have
E a = (8.314)(ln 1.5) / (1/365 – 1/273) = (8.314)(.405) / (0.00274 – 0.00366)
= (3.37 J mol –1 K –1 ) / (0.000923 K –1 ) = 3650 J mol –1 = 3.65 kJ mol –1
Comment : This rather low value seems reasonable because protein denaturation involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken
Crickets and popcorn
- Many biological processes exhibit a temperature dependence that follows the Arrhenius law, and can thus be characterized by an activation energy. See this interesting Dartmouth U. page that looks at the kinetics of cricket chirps .
The pre-exponential factor
It is now time to focus in on the pre-exponential term A in the Arrhenius equation. We have been neglecting it because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. But since A multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate.
and of the temperature.
If this fraction were unity, the Arrhenius law would reduce to
\[k = A\]
In other words, A is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded E a — admittedly, an uncommon scenario.
It's all about collisions
So what would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor Z . In some reactions, the relative orientation of the molecules at the point of collision is important, so we can also define a geometrical or steric factor (commonly denoted by ρ (Greek lower case rho ).
In general, we can express A as the product of these two factors:
\[A = Zρ\]
Values of \(ρ\) are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which A is assumed to be the same as Z .
Direction makes a difference
The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other. We showed one example of this near the top of the page, but for another, consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane. This kind of electrophilic addition reaction is well-known to all students of organic chemistry.
Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below.
The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to the σ ( sigma ) and π ( pi ) molecular orbitals. The latter, which extends above and below the plane of the C 2 H 4 molecule, interacts with and attracts the HCl molecule.
If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to bounce away from each other before any reaction can take place. The same thing happens in ; the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom.
The lesson you should take from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called "chemical intuition" which you can apply to a wide variety of problems. This is far more important than memorizing specific examples. Now that you know what it takes to get a reaction started, you are ready for the next lesson that describes their actual mechanisms .
|
libretexts
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2025-03-17T19:53:20.069878
| 2016-02-26T07:45:44 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.03%3A_Collision_and_activation-_the_Arrhenius_law",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.3: Collision and activation- the Arrhenius law",
"author": "Stephen Lower"
}
|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.04%3A_Reaction_Mechanisms
|
17.4: Reaction Mechanisms
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the green-highlighted terms in the context of this topic.
- Explain what is meant by the mechanism of a reaction.
- Define an elementary reaction , and state how it differs from an ordinary net chemical reaction .
- Sketch out an activation energy diagram for a multistep mechanism involving a rate-determining step , and relate this to the activation energy of the overall reaction.
- Write the rate law expression for a two-step mechanism in which the rate constants have significantly different magnitudes.
- Write the rate law expression for a three-step reaction in which one step is a rapid equilibrium , and the other two steps have significantly different magnitudes.
- Define a chain reaction , and list some of the different kinds of steps such a reaction will involve.
- Define a branching chain reaction , and explain how such reactions can lead to explosions.
We are now ready to open up the "black box" that lies between the reactants and products of a net chemical reaction. What we find inside may not be very pretty, but it is always interesting because it provides us with a blow-by-blow description of how chemical reactions take place.
The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collision") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an intermediate — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation.
For an example of a mechanism, consider the decomposition of nitrogen dioxide into nitric oxide and oxygen. The net balanced equation is
\[\ce{2 NO2(g) → 2 NO(g) + O2(g)} \nonumber\]
The mechanism of this reaction is believed to involve the following two elementary steps:
\[ \begin{align*} \ce{2 NO2} &→ \ce{NO3 + NO} \label{step 1} \\[4pt] \ce{NO3} &→ \ce{NO + O2} \label{step 2} \end{align*}\]
Note that the intermediate species \(\ce{NO3}\) has only a transient existence and does not appear in the net equation.
A useful reaction mechanism
- consists of a series of elementary steps (defined below) that can be written as chemical equations, and whose sum gives the net balanced reaction equation;
- must agree with the experimental rate law;
- can rarely if ever be proved absolutely.
It is important to understand that the mechanism of a given net reaction may be different under different conditions. For example, the dissociation of hydrogen bromide
\[\ce{2 HBr(g) → H2(g) + Br2(g)} \nonumber\]
proceeds by different mechanisms (and follows different rate laws) when carried out in the dark ( thermal decomposition ) and in the light ( photochemical decomposition ). Similarly, the presence of a catalyst can enable an alternative mechanism that greatly speeds up the rate of a reaction.
Elementary Steps
A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called elementary reactions ) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as
| unimolecular | \(A →\) | by far the most common |
| bimolecular | \(A + B →\) | common |
| termolecular | \(A + B + C →\) | very rare |
Elementary reactions differ from ordinary net chemical reactions in two important ways:
- The rate law of an elementary reaction can be written by inspection. For example, a bimolecular process always follows the second-order rate law \(rate=k[A][B]\).
- Elementary steps often involve unstable or reactive species that do not appear in the net reaction equation.
Some net reactions do proceed in a single elementary step, at least under certain conditions. However, without careful experimentation, one can never be sure. The gas-phase formation of \(\ce{HI}\) from its elements was long thought to be a simple bimolecular combination of \(\ce{H2}\) and \(\ce{I2}\), but it was later found that under certain conditions, it follows a more complicated rate law.
Multi-step (Consecutive) Reactions
Mechanisms in which one elementary step is followed by another are very common.
\[\ce{ A + B → \cancel{Q} } \tag{step 1}\]
\[\ce{B + \cancel{Q} → C} \tag{step 2}\]
\[\ce{A + 2B → C} \tag{net reaction}\]
(As must always be the case, the net reaction is just the sum of its elementary steps.)
In this example, the species \(Q\) is an intermediate , usually an unstable or highly reactive species. If both steps proceed at similar rates, rate law experiments on the net reaction would not reveal that two separate steps are involved here. The rate law for the reaction would be
\[rate = k[A][B]^2 \nonumber\]
(Bear in mind that intermediates such as \(Q\) cannot appear in the rate law of a net reaction.)
When the rates are quite different, things can get interesting and lead to quite varied kinetics as well as some simplifying approximations. When the rate constants of a series of consecutive reactions are quite different, a number of relationships can come into play that greatly simplify our understanding of the observed reaction kinetics.
Approximation 1: The Rate-Determining Step Approximation
The rate-determining step is also known as the rate-limiting step . We can generally expect that one of the elementary reactions in a sequence of consecutive steps will have a rate constant that is smaller than the others. The effect is to slow the rates of all the reactions — very much in the way that a line of cars creeps slowly up a hill behind a slow truck.
The three-step reaction depicted here involves two intermediate species I 1 and I 2 , and three activated complexes numbered X 1-3 . Although the step I 2 → products has the smallest individual activation energy E a 3 , the energy of X 3 with respect to the reactants determines the activation energy of the overall reaction, denoted by the leftmost vertical arrow . Thus the rate-determining step is
\[X_1 → X_2. \nonumber\]
Chemists often refer to elementary reactions whose forward rate constants have large magnitudes as "fast", and those with forward small rate constants as "slow". Always bear in mind, however, that as long as the steps proceed in single file (no short-cuts!), all of them will proceed at the same rate . So even the "fastest" members of a consecutive series of reactions will proceed as slowly as the "slowest" one.
Approximation 2: The Rapid Equilibrium Approximation
In many multi-step processes, the forward and reverse rate constants for the formation of an intermediate \(Q\) are of similar magnitudes and sufficiently large to make the reaction in each direction quite rapid. Decomposition of the intermediate to product is a slower process:
\[\ce{A <=>[k_1][k_{-1}] Q ->[k_2] B} \nonumber\]
This is often described as a rapid equilibrium in which the concentration of \(Q\) can be related to the equilibrium constant
\[K = \dfrac{k_1}{k_{–1}} \nonumber\]
This is just the Law of Mass Action . It should be understood, however, that true equilibrium is never achieved because \(Q\) is continually being consumed; that is, the rate of formation of \(Q\) always exceeds its rate of decomposition. For this reason, the steady-state approximation described below is generally preferred to treat processes of this kind.
Approximation 3: The Steady-State Approximation
Consider a mechanism consisting of two sequential reactions
\[\ce{A ->[k_1] Q ->[k_2] B} \nonumber\]
in which \(Q\) is an intermediate. The time-vs-concentration profiles of these three substances will depend on the relative magnitudes of \(k_1\) and \(k_2\), as shown in the following diagrams. Construction of these diagrams requires the solution of sets of simultaneous differential equations , which is [fortunately!] beyond the scope of this course.
The steady-state approximation is usually not covered in introductory courses, although it is not particularly complicated mathematically.
In the left-hand diagram, the rate-determining step is clearly the conversion of the rapidly-formed intermediate into the product, so there is no need to formulate a rate law that involves \(Q\). But on the right side, the formation of \(Q\) is rate-determining, but its conversion into \(B\) is so rapid that \([Q]\) never builds up to a substantial value. (Notice how the plots for \([A]\) and \([B]\) are almost mutually inverse.) The effect is to maintain the concentration of \(Q\) at an approximately constant value. This steady-state approximation can greatly simplify the analysis of many reaction mechanisms, especially those that are mediated by enzymes in organisms.
Converting Mechanisms into Rate Laws
We are unable to look directly at the elementary steps hidden within the "black box" of the reaction mechanism, we are limited to proposing a sequence that would be consistent with the reaction order which we can observe. Chemical intuition can guide us in this, for example by guessing the magnitudes of some of the activation energies. In the end, however, the best we can do is to work out a mechanism that is plausible ; we can never "prove" that what we come up with is the actual mechanism.
Consider the following reaction:
\[A + B → C\nonumber\]
One possible mechanism might involve two intermediates \(Q\) and \(R\):
| Step 1 | \(\ce{A + B ->[k_1] Q}\) | (slow, rate-determining) |
| Step 2 | \(\ce{Q + A ->[k_2] R }\) | (fast) |
| Step 3 | \(\ce{ R + B ->[k_3] C }\) | (fast) |
The rate law corresponding to this mechanism would be that of the rate-determining step:
\[\text{rate} = k_1[A][B]. \nonumber\]
If the first step in a mechanism is rate-determining, it is easy to find the rate law for the overall expression from the mechanism. If the second or a later step is rate-determining, determining the rate law is slightly more complicated and often requires either of the two approximation above to identify.
An alternative mechanism for the following reaction:
\[A + B → C \nonumber\]
in which the rate-determining step involves one of the intermediates would display third-order kinetics:
| Step 1 | \(\ce{A + B <=>[k_1][k_{-1}] Q}\) | (rapid equilibrium) |
| Step 2 | \(\ce{Q + A ->[k_2] R }\) | (slow, rate-determining) |
| Step 3 | \(\ce{ R + B ->[k_3] C }\) | (fast) |
Since intermediates cannot appear in rate law expressions, we must express \([Q]\) in the rate-determining step in terms of the other reactants. To do this, we make use of the fact that Step 1 involves an equilibrium constant \(K_1\):
\[K_1 = \dfrac{k_1}{k_{-1}} = \dfrac{[Q]}{[A][B]} \nonumber\]
Solving this for \([Q]\), we obtain
\[[Q] = K_1[A][B]. \nonumber \]
We can now express the rate law for 2 as
\[\begin{align*} \text{rate} &= k_2 K_1[A][B][A] \\[4pt] &= k[A]^2[B] \end{align*}\]
in which the constants \(k_2\) and \(K_1\) have been combined into a single constant \(k\).
Consider the following reaction:
\[\ce{ F2 + 2 NO2 → 2 NO2F }\nonumber\]
Application of the "chemical intuition" mentioned in the above box would lead us to suspect that any process that involves breaking of the strong F–F bond would likely be slow enough to be rate limiting, and that the resulting atomic fluorines would be very fast to react with another odd-electron species:
| Step 1 | \(\ce{F2 + 2 NO2 ->[k_1] NO2F + F + NO2}\) | (slow, rate-determining) |
| Step 2 | \(\ce{F + NO2 ->[k_2] NO2F}\) | (very fast) |
If this mechanism is correct, then the rate law of the net reaction would be that of the rate-determining step:
\[\text{rate} = k_1[F_2][NO_2] \nonumber\]
Ozone is an unstable allotrope of oxygen that decomposes back into ordinary dioxygen according to the net reaction
\[\ce{2 O3 → 3 O2} \nonumber\]
A possible mechanism would be the simple one-step bimolecular collision suggested by the reaction equation, but this would lead to a second-order rate law which is not observed. Instead, experiment reveals a more complicated rate law:
\[\text{rate} = [O_3]^2[O_2]^{–1} \nonumber\]
What's this? It looks as if O 2 actually inhibits the reaction in some way. The generally-accepted mechanism for this reaction is:
| Step 1 | \(\ce{O3 <=>[k_1][k_{-1}] O2+ O}\) | (rapid equilibrium) |
| Step 2 | \(\ce{O + O3 ->[k_2] 2 O2}\) | (rate-determining) |
Does this seem reasonable? Note that
- The equilibrium in Step 1 should be quite rapid because ozone's instability predicts a low activation energy for the forward process. The same can be said for the reverse process which involves the highly reactive oxygen atom which we would expect to rapidly gobble up one of the nonbonding electron pairs on the O 2 molecule.
- The inhibitory effect of O 2 can be explained by the Le Chatelier effect — being a product of the equilbrium in 1 , buildup of oxygen forces it back to the left.
- In Step 2 , one might expect the O atom to react with ozone as rapidly as it does with O 2 in the reverse of Step 1, but other studies show that this is not the case. The rate constant k 2 is probably fairly large. Given the strong covalent bond in O 2 , we not expect any significant reverse reaction.
To translate this mechanism into a rate law, we first write the equilibrium constant for Step 1 and solve it for the concentration of the intermediate:
\[ K =\dfrac{[O_2][O]}{[O_3]} \nonumber\]
\[[O] = \dfrac{k_1[O_3]}{k_{-1}[O_2]} \nonumber\]
We substitute this value of \(\ce{[O]}\) into the rate expression \(\ce{[O][O3]}\) for Step 2, which yields the experimentally-obtained rate law
\[ rate = k_1K \dfrac{[O_3]^2}{[O_2]} \nonumber\]
Consider the gas-phase oxidation of nitric oxide:
\[\ce{2 NO + O_2 → 2 NO_2} \nonumber\]
This reaction, like most third-order reactions, is not termolecular but rather a combination of an equilibrium followed by a subsequent bimolecular step:
| Step 1 | \(\ce{ NO + NO <=>[k_1][k_{-1}] N2O2}\) | (equilibrium, eq. const. K ) |
| Step 2 | \(\ce{ N2O2 + O2 ->[k_2] 2 NO2}\) | (rate-limiting) |
Since the intermediate \(\ce{N2O2}\) may not appear in the rate equation, we need to express its concentration in terms of the reactant \(NO\). As in the previous example, we do this through the equilibrium constant of Step 1 :
\[K = \dfrac{[N_2O_2]}{[NO]^2} \nonumber\]
\[[N_2O_2] = K [NO]^2 \nonumber\]
\[ \begin{align*} \text{rate} &= k_2 [N_2O_2 ][O_2] \\[4pt] &= k_2 K [NO]^2 [O_2] \end{align*} \]
The unusual feature of this net reaction is that its rate diminishes as the temperature increases, suggesting that the activation energy is negative. Reaction 1 involves bond formation and is exothermic, so as the temperature rises, \(K\) decreases (Le Chatelier effect). At the same time, k 2 increases, but not sufficiently to overcome the decrease in \(K\). So the apparently negative activation energy of the overall process is simply an artifact of the magnitudes of the opposing temperature coefficients of k 2 and \(K\).
Chain Reactions (Positive Feedback Mechanisms)
Many important reaction mechanisms, particularly in the gas phase, involve intermediates having unpaired electrons, commonly known as free radicals . Free radicals are often fairly stable thermodynamically and may be quite long-lived by themselves, but they are highly reactive, and hence kinetically labile. The dot ·, representing the unpaired electron, is not really a part of the formula, and is usually shown only when we want to emphasize the radical character of a species. The "atomic" forms of many elements that normally form diatomic molecules are free radicals; H·, O·, and Br· are common examples. The simplest and most stable (Δ G = +87 kJ/mol) molecular free radical or " odd-electron molecule " is nitric oxide, NO·.
The most important chemical property of a free radical is its ability to pass the odd electron along to another species with which it reacts. This chain-propagation process creates a new radical which becomes capable of initiating another reaction. Radicals can, of course, also react with each other, destroying both (" chain termination ") while creating a new covalent-bonded species.
Much of the pioneering work in this field, of which the \(\ce{HBr}\) synthesis is a classic example, was done by the German chemist Max Bodenstein (1871-1942)
The synthesis of hydrogen bromide from its elements illustrates the major features of a chain reaction. The figures in the right-hand column are the activation energies per mole.
| Step 1 | Br 2 + M → 2 Br· | chain initiation (Δ H ° = +188 kJ) |
| Step 2 | Br· + H 2 → H· + HBr | chain propagaton (+75 kJ) |
| Step 3 | H· + Br 2 → HBr + Br· | chain propagation (–176 kJ) |
| Step 4 | H· + HBr → H 2 + Br· | chain inhibition (–75 kJ) |
| Step 5 | 2 Br· + M → M* + Br 2 | chain termination (–188 kJ) |
Note the following points:
- To start the reaction, a free radical must be formed ( 1 ). If the temperature of the gas is fairly high, then Br· can be formed from the more energetic collisions of Br 2 molecules with some other molecule M (most likely a second Br 2 ). This is known as thermal activation ; Another way of creating free radicals is photochemical activation .
- Reactions 2 and 3 consume a free radical but form another, thus "propagating" the chain.
- In reaction 4 , a molecule of the product is destroyed, thus partially un-doing the net process.
- If the only the first four reactions were active, then the cycle would continue indefinitely. But reaction 5 consumes the Br· chain carrier without producing new radicals, thus terminating the chain. The function of the molecule M is to absorb some of the kinetic energy of the collision so that the two bromine atoms do not simply bounce apart. The product M* represents a thermally-excited M which quickly dissipates its energy to other molecules.
- Other reactions, such as the recombination of hydrogen atoms, also take place, but their contribution to the overall kinetics is usually very small.
The rate laws for chain reactions tend to be very complex, and often have non-integral orders.
Branching Chain Reactions: Explosions
The gas-phase oxidation of hydrogen has been extensively studied over a wide range of temperatures and pressures.
\[\ce{H2(g) + 1/2 O2(g) → H2O(g)}\quad ΔH^o = –242\, kJ/mol \nonumber\]
This reaction does not take place at all when the two gases are simply mixed at room temperature. At temperatures around 500-600°C it proceeds quite smoothly, but when heated above 700° or ignited with a spark, the mixture explodes. As with all combustion reactions, the mechanism of this reaction is extremely complex (you do not want to see the rate law!) and varies somewhat with the conditions. Some of the major radical formation steps are
| Step 1 | H 2 + O 2 → HO 2 · + H· | chain initiation |
| Step 2 | H 2 + HO 2 · → HO· + H 2 O | chain propagaton |
| Step 3 | H· + O 2 → HO· + O· | chain propagaton + branching |
| Step 4 | O· + H 2 → HO· + H· | chain propagation + branching |
| Step 5 | HO· + H 2 → H 2 O + H· | chain propagation |
| The species HO 2 · is known as perhydroxyl ; HO· is hydroxyl (not to be confused with the hydroxyl ion HO: – ) |
Reactions 3 and 4 give birth to more radicals than they consume, so when these are active, each one effectively initiates a new chain process causing the rate overall rate to increase exponentially, producing an explosion.
An explosive reaction is a highly exothermic process that once initiated, goes to completion very rapidly and cannot be stopped. The destructive force of an explosion arises from the rapid expansion of the gaseous products as they absorb the heat of the reaction. There are two basic kinds of chemical explosions:
- Thermal explosions occur when heat is released by a reaction more rapidly than it can escape from the reaction space. This causes the reaction to proceed more rapidly, releasing even more heat, resulting in an ever-accelerating runaway rate.
- Chain-branching explosions occur when the number of chain carriers increases exponentially, effectively seeding new reaction centers in the mixture. The process then transforms into a thermal explosion.
Whether or not a reaction proceeds explosively depends on the balance between formation and destruction of the chain-carrying species. This balance depends on the temperature and pressure, as illustrated here for the hydrogen-oxygen reaction.
- Direct recombination of chain carriers generally requires a three-body collision with another molecule to absorb some of the kinetic energy; such ternary processes are unlikely at very low pressures. Thus below the lower explosion limit , the radicals (including those produced by a spark) are usually able to reach the walls of the container and combine there — or in the case of an open-air explosion, simply combine with other molecules as they exit the active volume of the reaction.
- Within the explosion zone between and (which, for most gases, are known as the lower - and upper explosion limits ), the propagation and branching processes operate efficiently and explosively, even when the mixture is heated homogeneously.
- Above , the concentration of gas molecules is sufficient to enable ternary collisions which allow chain-termination processes to operate efficiently, thus suppressing branching. Above this upper limit, reactions involving most gases proceed smoothly.
-
Hydrogen is unusual in that it exhibits a third explosion limit
. The cause of this was something of a mystery for some time, but it is now believed that explosions in this region do not involve branching, but are thermally induced by the reaction
HO 2 · + H 2 → H 2 O 2 + H· .
The lower explosion limit of gas mixtures varies with the size, shape, and composition of the enclosing container. Needless to say, experimental determination of explosion limits requires some care and creativity. Upper and lower explosion limits for several common fuel gases are shown below.
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libretexts
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2025-03-17T19:53:20.199703
| 2016-02-26T07:45:58 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.04%3A_Reaction_Mechanisms",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.4: Reaction Mechanisms",
"author": "Stephen Lower"
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.05%3A_Kinetics_of_Reactions_in_Solution
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17.5: Kinetics of Reactions in Solution
Make sure you thoroughly understand the following essential ideas:
- Describe some of the major differences between the kinetics of reactions in the gas phase, compared with those in liquid solutions.
- What role do solvent cages play in solution kinetics?
- Explain the distinction between diffusion-control and activation-control of reaction rates in solutions.
- How can the polarity of a solvent affect the energetics of a reaction mechanism?
The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions.
What's different about kinetics in liquid solutions?
Most of the added complications of kinetics and rate processes in liquid solutions arise from the much higher density of the liquid phase. In a typical gas at atmospheric pressure, the molecules occupy only about 0.2 per cent of the volume; the other 99.8 percent is empty space. In a liquid, molecules may take up more than half the volume, and the "empty" spaces are irregular and ever-changing as the solvent molecules undergo thermal motions of their own.
In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10 –11 sec) confined to a "hole" within the liquid. This trapping becomes especially important when the solvent is strongly hydrogen-bonded as is the case with water or alcohol.
When thermal motions occasionally release a solute molecule from this trap, it will jump to a new location. The jumps are very fast (10 –12 - 10 –13 sec) and short (usually a few solvent-molecule diameters), and follow an entirely random pattern, very much as in Brownian motion . Consider a simple bimolecular process A + B → products. The reactant molecules will generally be jumping from hole to hole in the solvent matrix, only occasionally finding themselves in the same solvent cage where thermal motions are likely to bring them into contact.
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| A pair of reactants end up in the same solvent cage, where they bounce around randomly and exchange kinetic energy with the solvent molecules. | Eventually the two reactants form an encounter pair . If they fail to react the first time, they have many more opportunities during the lifetime of the cage. | The products form and begin to move away from each other. | Finally, after about 10 –11 sec, the solvent cage breaks up and the products diffuse away. |
The process can be represented as
\[A + B \rightarrow \{AB\} → \text{products}\]
in which the \(\{AB\}\) term represents the caged reactants including the encounter pair and the activated complex.
Contrast this scenario with a similar reaction taking place in the gas phase; the molecules involved in the reaction will often be the only ones present, so a significant proportion of the collisions will be \(A\)-\(B\) encounters. However, if the collision should fail to be energetically or geometrically viable, the reactant molecules fly apart and are unlikely to meet again anytime soon. In a liquid, however, the solute molecules are effectively in a constant state of collision — if not with other reactants, then with solvent molecules which can exchange kinetic energy with the reactants. So once an A-B encounter pair forms, the two reactants get multiple whacks at each other, greatly increasing the probability that they will obtain the kinetic energy needed to kick them over the activation hump before the encounter pair disintegrates.
Limiting Cases: Diffusion-Controlled and Activation-Controlled Reactions
The encounter pair model introduces some new rate parameters:
\[\ce{A + B <=>[k1][k_{-1}] {AB} -> products}\]
The first step is an equilibrium between the reactants outside and inside the solvent cage. The rate constants \(k_1\) and \(k_2\) reflect those relating to diffusion of molecules through the solvent; their values are strongly dependent on the viscosity (and thus the temperature) of the solvent. (Note that \(k_1\) is a second-order rate constant, while \(k_2\) is first-order.)
Diffusion is the transport of a substance through a concentration gradient; that is, from a region of higher concentration to one of lower concentration. Think of the way the color of tea spreads out when a tea bag is immersed in hot water. Diffusion occurs because random thermal motions are statistically more likely to move molecules out of a region of higher concentration than in the reverse direction, simply because in the latter case fewer molecules are available to make the reverse trip. Eventually the concentrations become uniform and equilibrium is attained.
As molecules diffuse through a liquid, they must nudge neighboring molecules out of the way. The work required to do this amounts to an activation energy, so diffusion can be thought of as a kinetic process with its own rate constant k d and activation energy. These parameters depend on the sizes of the solute and solvent molecules and on how strongly the latter interact with each other. This suggests two important limiting cases for reactions in solution.
For water at room temperature, k 1 is typically 10 9 -10 10 dm –3 mol –1 s –1 and k 2 is around 10 –9 -10 –10 dm –3 mol –1 s –1 . Given these values, k 3 > 10 12 s –1 implies diffusion control, while values < 10 9 s –1 are indicative of activation control.
- Diffusion Controlled (\(k_3 \gg k_2\)): If the activation energy of the A+B reaction is very small or if escape of molecules from the {AB} cage is difficult, the kinetics will be dominated by k 1 , and thus by the activation energy of diffusion. Such a process is said to be diffusion controlled . Reactions in aqueous solution in which E a > 20 kJ/mol are likely to fall into this category.
- Activation Controlled ( \(k_3 \ll k_2\)): Alternatively, if the activation energy of the A+B reaction dominates the kinetics, and the reaction is activation-controlled .
Several general kinds of reactions are consistently very "fast" and thus are commonly found to be diffusion-controlled in most solvents:
Units Matter
Gas-phase rate constants are normally expressed in units of mol s –1 , but rate constants of reactions in solution are conventionally given in mol/L units, or dm 3 mol –1 s –1 . Conversion between them depends on a number of assumptions and is non-trivial.
- Recombination of atoms and radicals
For example the formation of I 2 from I atoms in hexane at 298 K has k 3 = 1.3×10 12 dm 3 mol –1 s –1 .
- Acid-base reactions which involve the transport of H + and OH – ions tend to be very fast.
The most famous of these is one of the fastest reactions known:\[H^+ + OH^– → H_2O \nonumber \] for which k 3 = 1.4×10 11 dm 3 mol –1 s –1 at 298 K.
Polar solvents such as water and alcohols interact with ions and polar molecules through attractive dipole-dipole and ion-dipole interactions, leading to lower-energy solvated forms which stabilize these species. In this way, a polar solvent can alter both the thermodynamics and kinetics (rate) of a reaction.
Solvent Thermodynamic Effect
If the products of the reaction are markedly more or less polar than the reactants, solvent polarity can change the overal thermodynamics (equilibrium constant) of the reaction. Nowhere is this more apparent than when an ionic solid such as salt dissolves in water. The Na + and Cl – ions are bound together in the solid through strong coulombic forces; pulling the solid apart in a vacuum or in a nonpolar solvent is a highly endothermic process. In contrast, dissolution of NaCl in water is slightly exothermic and proceeds spontaneously.
The water facilitates this process in two important ways. First, its high dielectric constant of 80 reduces the force between the separated ions to 1/80 of its normal value. Secondly, the water molecules form a solvation shell around the ions (lower left), rendering them energetically (thermodynamically) more stable than they were in the NaCl solid.
Solvent Kinetic Effect
In the same way, a reaction whose mechanism involves the formation of an intermediate or activated complex having a polar or ionic character will have its activation energy, and thus its rate, subject to change as the solvent polarity is altered. As an example we will consider an important class of reactions that you will hear much about if you take a course in organic chemistry. When an aqueous solution of a strong base such as KOH is added to a solution of tertiary -butyl chloride in ethanol, the chlorine is replaced by a hydroxyl group, leaving t -butyl alcohol as a product:
This reaction is one of a large and important class known as S N 1 nucleophilic substitution processes that are discussed in most organic chemistry courses. In these reactions, a species that possesses a pair of non-bonding electrons (also called a nucleophile or Lewis base ) uses them to form a new bond with an electrophile — a compound in which a carbon atom has a partial positive charge owing to its bonds to electron-withdrawing groups. In the example here, other nucleophiles such as NH 3 or even H 2 O would serve as well.
In order to reflect the generality of this process and to focus on the major changes that take place, we will represent this reaction as
Extensive studies of this class of reactions in the 1930's revealed that it proceeds in two activation energy-controlled steps, followed by a simple dissociation into the products:
In step , which is rate-determining, the chlorine leaves the alkyl chloride which becomes an intermediate known as a carbocation ("cat-ion"). These ions, in which the central carbon atom lacks a complete octet, are highly reactive, and in step the carbocation is attacked by the hydroxide ion which supplies the missing electron. The immediate product is another cation in which the positive charge is on the oxygen atom. This oxonium ion is unstable and rapidly dissociates ( )into the alcohol and a hydrogen ion.
The reaction coordinate diagram helps us understand the effect of solvent polarity on this reaction. Polar solvent molecules interact most strongly with species in which the electric charge is concentrated in one spot. Thus the carbocation is stabilized to a greater degree than are the activated complexes in which the charge is spread out between the positive and negative ends. As the heavy green arrows indicate, a more polar solvent will stabilize the carbocation more than it will either of the activated complexes; the effect is to materially reduce the activation energy of the rate-determining step, and thus speed up the reaction. Because neither the alkyl chloride nor the alcohol is charged, the change in solvent polarity has no effect on the equilibrium constant of the reaction. This is dramatically illustrated by observing the rate of the reaction in solvents composed of ethanol and water in varying amounts:
| % water | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|
|
|
1.7 | 9.1 | 40.3 | 126 | 367 | 1294 |
|
libretexts
|
2025-03-17T19:53:20.285055
| 2016-02-26T07:46:12 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.05%3A_Kinetics_of_Reactions_in_Solution",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.5: Kinetics of Reactions in Solution",
"author": "Stephen Lower"
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.06%3A_Catalysts_and_Catalysis
|
17.6: Catalysts and Catalysis
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
- What are catalysts , and how do they work in terms altering the parameters of a reaction?
- Describe the similarities and differences between the three principal classes of catalysts.
- Define physisorption and chemisorption , and explain the role of the latter in initiating a catalytic event.
- What is the meaning and significance of the Sabatier Principle ?
- What are the principal differences between the Langmuir-Hinshelwood and Eley-Rideal mechanisms of heterogeneous catalysis?
- Describe the role of the enzyme-substrate complex .
- How might the particular properties of the amino-acids surrounding the active site of an enzyme contribute to its catalytic effect?
- Describe the lock-and-key and induced-fit models of enzyme action.
- Explain the function and significance of allosteric sites on an enzyme.
It almost seems like magic! A mixture of gaseous H 2 and O 2 can coexist indefinitely without any detectable reaction, but if a platinum wire is heated and then immersed in the gaseous mixture, the reaction proceeds gently as the heat liberated by the reaction makes the wire glow red-hot. Catalysts play an essential role in our modern industrial economy, in our stewardship of the environment, and in all biological processes. This lesson will give you a glimpse into the wonderful world of catalysts, helping you to understand what they are and how they work.
What are Catalysts?
Catalysts have no effect on the equilibrium constant and thus on the equilibrium composition. Catalysts are substances that speed up a reaction but which are not consumed by it and do not appear in the net reaction equation. Also — and this is very important — catalysts affect the forward and reverse rates equally ; this means that catalysts have no effect on the equilibrium constant and thus on the composition of the equilibrium state . Thus a catalyst (in this case, sulfuric acid) can be used to speed up a reversible reaction such as ester formation or its reverse, ester hydrolysis:
The catalyst has no effect on the equilibrium constant or the direction of the reaction. The direction can be controlled by adding or removing water (Le Chatelier principle).
Catalysts function by allowing the reaction to take place through an alternative mechanism that requires a smaller activation energy. This change is brought about by a specific interaction between the catalyst and the reaction components. You will recall that the rate constant of a reaction is an exponential function of the activation energy, so even a modest reduction of \(E_a\) can yield an impressive increase in the rate.
Catalysts provide alternative reaction pathways
Catalysts are conventionally divided into two categories: homogeneous and heterogeneous . Enzymes , natural biological catalysts, are often included in the former group, but because they share some properties of both but exhibit some very special properties of their own, we will treat them here as a third category.
Some common examples of catalysis
How to burn a Sugar Cube
When heated by itself, a sugar cube (sucrose) melts at 185°C but does not burn. But if the cube is rubbed in cigarette ashes, it burns before melting owing to the catalytic action of trace metal compounds in the ashes.
Platinum as an Oxidation Catalyst
The surface of metallic platinum is an efficient catalyst for the oxidation of many fuel vapors. This property is exploited in flameless camping stoves (left). The image at the right shows a glowing platinum wire heated by the slow combustion of ammonia on its surface. However, if you dip a heated Pt wire into liquid ammonia, you get a miniature explosion: see video below.
This is oxidation of ammonia by oxygen catalysed by warmed platinum wire. Oxygen is bubbled through ammonia solution, in which it mixes with ammonia gas present. The reaction causes the platinum wire to glow, and the hot wire ignites a mixture of ammonia and oxygen.
Decomposition of Hydrogen Peroxide
Hydrogen peroxide is thermodynamically unstable according to the reaction
\[\ce{2 H2O2 → 2 H2O + O2 } \quad \quad ΔG^o = –210\, kJ\, mol^{–1}\]
In the absence of contaminants this reaction is very slow, but a variety of substances, ranging from iodine, metal oxides, trace amount of metals, greatly accelerate the reaction, in some cases almost explosively owing to the rapid release of heat. The most effective catalyst of all is the enzyme catalase , present in blood and intracellular fluids; adding a drop of blood to a solution of 30% hydrogen peroxide induces a vigorous reaction.
Rapid liberation of O 2 can result in a spectacular bubble bath if some soap is added. [ This same reaction has been used to power a racing car!
Potassium iodide efficiently catalyzes H 2 O 2 deposition. This short video shows what happens when some colored soap, H 2 O 2 , and KI are combined. But "don't try this at [your] home"!
Each kind of catalyst facilitates a different pathway with its own activation energy. Because the rate is an exponential function of E a (Arrhenius equation), even relatively small differences in E a 's can have dramatic effects on reaction rates. Note especially the values for catalase ; the chemist is still a rank amateur compared to what Nature can accomplish through natural selection!
| catalyst |
Ea
|
relative rate |
|---|---|---|
| no catalyst | 75 | 1 |
| iodide ion | 56 | 2140 |
| colloidal platinum | 50 | 24,000 |
| catalase (enzyme) | 21 | 2,900,000,000 |
How catalytic activity is expressed
Changes in the rate constant or of the activation energy are obvious ways of measuring the efficacy of a catalyst. But two other terms have come into use that have special relevance in industrial applications.
Turnover number
The turnover number (TON) is an average number of cycles a catalyst can undergo before its performance deteriorates (see below ). Reported TONs for common industrial catalysts span a very wide range from perhaps 10 to well over 10 5 , which approaches the limits of diffusion transport.
Turnover Frequency
This term, which was originally applied to enzyme-catalyzed reactions, has come into more general use. It is simply the number of times the overall catalyzed reaction takes place per catalyst (or per active site on an enzyme or heterogeneous catalyst) per unit time:is defined as
The number of active sites S on a heterogeneous catalyst is often difficult to estimate, so it is often replaced by the total area of the exposed catalyst, which is usually experimentally measurable. TOFs for heterogeneous reactions generally fall between 10 –2 to 10 2 s –1 .
Homogeneous Catalysis
As the name implies, homogeneous catalysts are present in the same phase (gas or liquid solution) as the reactants. Homogeneous catalysts generally enter directly into the chemical reaction (by forming a new compound or complex with a reactant), but are released in their initial form after the reaction is complete, so that they do not appear in the net reaction equation.
Iodine-catalyzed cis-trans isomerization
Unless you are taking an organic chemistry course in which your instructor indicates otherwise, don't try to memorize these mechanisms . They are presented here for the purpose of convincing you that catalysis is not black magic, and to familiarize you with some of the features of catalyzed mechanisms. It should be sufficient for you to merely convince yourself that the individual steps make chemical sense.
You will recall that cis-trans isomerism is possible when atoms connected to each of two doubly-bonded carbons can be on the same ( cis ) or opposite ( trans ) sides of the bond. This reflects the fact that rotation about a double bond is not possible.
Conversion of an alkene between its cis - and trans forms can only occur if the double bond is temporarily broken, thus freeing the two ends to rotate. Processes that cleave covalent bonds have high activation energies, so cis-trans isomerization reactions tend to be slow even at high temperatures. Iodine is one of several catalysts that greatly accelerate this process, so the isomerization of butene serves as a good introductory example of homogeneous catalysis.
The mechanism of the iodine-catalyzed reaction is believed to involve the attack of iodine atoms (formed by the dissociation equilibrium on one of the carbons in Step :
During its brief existence, the free-redical activated complex can undergo rotation about the C—C bond, so that when it decomposes by releasing the iodine ( ), a portion of the reconstituted butene will be in its trans form. Finally, the iodine atom recombine into diiodine. Since processes and cancel out, iodine does not appear in the net reaction equation — a requirement for a true catalyst.
Acid-base catalysis
Many reactions are catalyzed by the presence of an acid or a base; in many cases, both acids and bases will catalyze the same reaction. As one might expect, the mechanism involves the addition or removal of a proton, changing the reactant into a more kinetically labile form. A simple example is the addition of iodine to propanone
I 2 + (CH 3 ) 2 –C=O → (CH 2 I)(CH 3 )–C=O
The mechanism for the acid-catalyzed process involves several steps. The role of the acid is to provide a proton that attaches to the carbonyl oxygen, forming an unstable oxonium ion . The latter rapidly rearranges into an enol (i.e., a carbon connected to both a double bond ( ene ) and a hydroxyl ( ol ) group.) This completes the catalytic part of the process, which is basically an acid-base (proton-transfer) reaction in which the role of the proton is to withdraw an electron from the ketone oxygen.
In the second stage, the enol reacts with the iodine. The curved arrows indicate shifts in electron locations. In below, an electron is withdrawn from the π orbital of the double bond by one of the atoms of the I 2 molecule. This induces a shift of electrons in the latter, causing half of this molecule to be expelled as an iodide ion. The other half of the iodine is now an iodonium ion I + which displaces a proton from one of the methyl groups. The resultant carbonium ion then expels the -OH proton to yield the final neutral product.
Perhaps the most well-known acid-catalyzed reaction is the hydrolysis (or formation) of an ester — a reaction that most students encounter in an organic chemistry laboratory course. This is a more complicated process involving five steps; its mechanism is discussed here . See also this U. Calgary site, which describes both the acid- and base-catalyzed reaction.
Oxidation-reduction catalysis
Many oxidation-reduction (electron-transfer) reactions, including direct oxidation by molecular oxygen, are quite slow. Ions of transition metals capable of existing in two oxidation states can often materially increase the rate. An example would be the reduction of Fe 3+ by the vanadium ion V 3+ :
\[\ce{V^{3+} + Fe^{3+} → V^{4+} + Fe^{2+}}\]
This reaction is catalyzed by either Cu + or Cu 3+ , and the rate is proportional to the concentration of V 3+ and of the copper ion, but independent of the Fe 3+ concentration. The mechanism is believed to involve two steps:
| 1 V 3+ + Cu 2+ V 4+ + Cu + | (rate-determining) |
| 2 Fe 3+ + Cu + Fe 2+ + Cu 2+ | (very fast) |
(If Cu + is used as the catalyst, it is first oxidized to Cu 2+ by step 2.)
Hydrogen Peroxide, again
Ions capable of being oxidized by an oxidizing agent such as H 2 O 2 can serve as catalysts for its decomposition. Thus H 2 O 2 oxidizes iodide ion to iodate, when then reduces another H 2 O 2 molecule, returning an I – ion to start the cycle over again:
\[H_2O_2 + I^– → H_2O + IO^–\]
\[H_2O_2 + IO^– → H_2O + O_2 + I^–\]
Iron(II) can do the same thing. Even traces of metallic iron can yield enough Fe 2+ to decompose solutions of hydrogen peroxide.
\[H_2O_2 + Fe^{2+} → H_2O + Fe^{3+}\]
\[H_2O_2 + Fe^{3+} + 2H^+ → H_2O + O_2 + Fe^{2+} + 2H^+\]
Heterogeneous catalysts
As its name implies, a heterogeneous catalyst exists as a separate phase (almost always a solid) from the one (most commonly a gas) in which the reaction takes place. The catalytic affect arises from disruption (often leading to dissociation) of the reactant molecules brought about by their interaction with the surface of the catalyst.
← Model of a catalyst consisting of clusters of 8-10 platinum atoms (blue) deposited on an aluminum oxide surface. This catalyst efficiently removes hydrogen atoms from propane, converting it into the industrially-important propylene. [source]
Unbalanced forces at surfaces
You will recall that one universal property of matter is the weak attractive forces that arise when two particles closely approach each other. ( See here for a quick review.) When the particles have opposite electric charges or enter into covalent bonding, these far stronger attraction dominate and define the "chemistry" of the interacting species.
The molecular units within the bulk of a solid are bound to their neighbors through these forces which act in opposing directions to keep each under a kind of "tension" that restricts its movement and contributes to the cohesiveness and rigidity of the solid.
At the surface of any kind of condensed matter, things are quite different. The atoms or molecules that reside on the surface experience unbalanced forces which prevents them from assuming the same low potential energies that characterize the interior units. (The same thing happens in liquids, and gives rise to a variety of interfacial effects such as surface tension.)
But in the case of a solid, in which the attractive forces tend to be stronger, something much more significant happens. The molecular units that reside on the surface can be thought of as partially buried in it, with their protruding parts (and the intermolecular attractions that emerge from them) exposed to the outer world. The strength of the attractive force field which emanates from a solid surface varies in strength depending on the nature of the atoms or molecules that make up the solid.
Don't confuse a d sorption with a b sorption; the latter refers to the bulk uptake of a substance into the interior of a porous material. At the microscopic level, of course, absorption also involves adsorption. The process in which molecules in a gas or a liquid come into contact with and attach themselves to a solid surface is known as adsorption . Adsorption is almost always an exothermic process and its strength is conventionally expressed by the enthalpy or "heat" of adsorption Δ H ads .
Chemisorption and Physisorption
Two general categories of adsorption are commonly recognized, depending on the extent to which the electronic- or bonding structure of the attached molecule is affected. When the attractive forces arise from relatively weak van der Waals interactions, there is little such effect and Δ H ads tends to be small. This condition is described as physical adsorption ( physisorption ). Physisorption of a gas to a surface is energetically similar to the condensation of the gas to a liquid, it usually builds up multiple layers of adsorbed molecules, and it proceeds with zero activation energy.
Of more relevance to catalytic phenomena is chemisorption , in which the adsorbate is bound to the surface by what amounts to a chemical bond. The resulting disruption of the electron structure of the adsorbed species "activates" it and makes it amenable to a chemical reaction (often dissociation) that could not be readily achieved through thermal activation in the gas or liquid phase. In contrast to physisorption, chemisorption generally involves an activation energy (supplied by Δ H ads ) and the adorbed species is always a monolayer.
Mechanisms of reactions on surfaces
Dissociative adsorption
The simplest heterogeneous process is chemisorption followed by bond-breaking as described above. The most common and thoroughly-studied of these is the dissociation of hydrogen which takes place on the surface of most transition metals. The single 1 s electron of each hydrogen atom coordinates with the d orbitals of the metal, forming a pair of chemisorption bonds (indicated by the red dashed lines). Although these new bonds are more stable than the single covalent bond they replace, the resulting hydrogen atoms are able to migrate along the surface owing to the continuous extent of the d -orbital conduction band.
The Langmuir-Hinshelwood mechanism
Although the adsorbed atoms (" adatoms ") are not free radicals, they are nevertheless highly reactive, so if a second, different molecular species adsorbs onto the same surface, an interchange of atoms may be possible. Thus carbon monoxide can be oxidized to CO 2 by the process illustrated below:
In this example, only the O 2 molecule undergoes dissociation . The CO molecule adsorbs without dissociation , configured perpendicular to the surface with the chemisorption bond centered over a hollow space between the metal atoms. After the two adsorbed species have migrated near each other , the oxygen atom switches its attachment from the metal surface to form a more stable C=O bond with the carbon , followed by release of the product molecule.
The Eley-Rideal mechanism
An alternative mechanism eliminates the second chemisorption step; the oxygen adatoms react directly with the gaseous CO molecules by replacing the chemisorption bond with a new C–O bond as they swoop over the surface:
Examples of both mechanisms are known, but the Langmuir-Hinshelwood mechanism is more importantin that it exploits the activation of the adsorbed reactant. In the case of carbon monoxide oxdation, studies involving molecular beam experiments support this scheme. A key piece of evidence is the observation of a short time lag between contact of a CO molecule with the surface and release of the CO 2 , suggesting that CO remains chemisorbed during the interval.
The Sabatier Principle
To be effective, these processes of adsorption, reaction, and desorption must be orchestrated in a way that depends critically on the properties of the catalyst in relation to the chemisorption properties (Δ H ads ) of the reactants and products.
- Adsorption of the reactant onto the catalytic surface ( 2 ) must be strong enough to perturb the bonding within the species to dissociate or activate it;
- If the resulting fragments must migrate to other locations on the surface ( 3-4 ), their chemisorption must be weak enough to allow this movement but not so small that they escape before they have a chance to react;
- The product species must have sufficiently small Δ H ads values to ensure their rapid desorption from the catalyst ( 5 ) so that surface is freed up to repeat the cycle
The importance of choosing a catalyst that achieves the proper balance of the heats of adsorption of the various reaction components is known as the Sabatier Principle , but is sometimes referred to as the "just-right" or " Goldilocks principle ". Remember the story of Goldilocks and the Three Bears ? ... or see this UTube video.
In its application to catalysis, this principle is frequently illustrated by a "volcano diagram" in which the rate of the catalyzed reaction is plotted as a function of Δ H ads of a substrate such as H 2 on a transition metal surface.
The plot at the left shows the relative effectiveness of various metals in catalyzing the decomposition of formic acid HCOOH. The vertical axis is plotted as temperature, the idea being that the better the catalyst, the lower the temperature required to maintain a given rate.
The catalytic cycle
This term refers to the idealized sequence of steps between the adsorption of a reactant onto the catalyst and the desorption of the product, culminating in restoration of the catalyst to its original condition. A typical catalytic cycle for the hydrogenation of propene is illustrated below.
This particular reaction
H 3 C–CH=CH 2 + H 2 → H 3 C–CH 2 –CH 3
takes place spontaneously only in the reverse direction, but it is representative of the process used to hydrogenate the carbon-carbon double bonds in vegetable oils to produce solid saturated fats such as margarine .
Catalyst poisoning and breakdown
Catalyst poisoning, brought about by irreversible binding of a substance to its surface, can be permanent or temporary. In the latter case the catalyst can be regenerated, usually by heating to a high temperature. In organisms, many of the substances we know as "poisons" act as catalytic poisons on enzymes. If catalysts truly remain unchanged, they should last forever, but in actual practice, various events can occur that limit the useful lifetime of many catalysts.
- Impurities in the feedstock or the products of side reactions can bind permanently to a sufficient number of active sites to reduce catalytic efficiency over time, "poisoning" the catalyst.
- Physical deterioration of the catalyst or of its support, often brought about by the high temperatures sometimes used in industrial processes, can reduce the effective surface area or the accessibility of reactants to the active sites.
Catalysts tend to be rather expensive, so it is advantageous if they can be reprocessed or regenerated to restore their activity. It is a common industrial practice to periodically shut down process units to replace spent catalysts.
How heterogeneous catalysts work
The actual mechanisms by which adsorption of a molecule onto a catalytic surface facilitates the cleavage of a bond vary greatly from case to case.
We give here only one example, that of the dissociation of dixoygen O 2 on the surface of a catalyst capable of temporarily donating an electron which enters an oxygen antibonding molecular orbital that will clearly destabilize the O–O bond. (Once the bond has been broken, the electron is given back to the catalyst.)
Types of catalytically active surfaces
Heterogeneous catalysts mostly depend on one or more of the followng kinds of surfaces:
| Active surface type | Remarks |
|---|---|
| Atoms at surfaces or crystal edges in macromolecular 2- and 3-D networks such as graphite or quartz may have free valences | |
| Surfaces and edges are sites of intense electric fields able to interact with ions and polar molecules | |
| Oxides can acquire H + and/or OH – groups able to act as acid- or base catalyts | |
| "Electron gas" at metal surfaces can perturb bonding in substrate molecules | |
| Vacant d orbitals can provide a variety of coordination sites for activation. | |
| Semiconductors (including many oxides) can supply electrons, thermally excited through reasonably small (<50 kJ) band gaps. |
Some factors affecting catalyst efficacy
Since heterogeneous catalysis requires direct contact between the reactants and the catalytic surface, the area of active surface goes at the top of the list. In the case of a metallic film, this is not the same as the nominal area of the film as measured by a ruler; at the microscopic level, even apparently smooth surfaces are highly irregular, and some cavities may be too small to accommodate reactant molecules.
Consider, for example, that a 1-cm cube of platinum (costing roughly $1000) has a nominal surface area of only 6 cm 2 . If this is broken up into 10 12 smaller cubes whose sides are 10 –6 m, the total surface area would be 60,000 cm 2 , capable in principle of increasing the rate of a Pt-catalyzed reaction by a factor of 10 4 . These very finely-divided (and often very expensive) metals are typically attached to an inert supporting surface to maximize their exposure to the reactants.
Surface topography. At the microscopic level, even an apparently smooth surface is pitted and uneven, and some sites will be more active than others. Penetration of molecules into and out of some of the smaller channels of a porous surface may become rate-limiting.
An otherwise smooth surface will always possess a variety of defects such as steps and corners which offer greater exposure and may be either the only active sites on the surface, or overly active so as to permanently bind to a reactant, reducing the active area of the surface. In one study, it was determined that kink defects constituting just 5 percent of platinum surface were responsible for over 90% of the catalytic activity in a certain reaction.
Steric factors
When chemisorbtion occurs at two or more locations on the reactant, efficient catalysis requires that the spacing of the active centers on the catalytic surface be such that surface bonds can be formed without significant angular distortion.
Thus activation of the ethylene double bond on a nickel surface proceeds efficiently because the angle between the C—Ni bonds and the C—C is close to the tetrahedral value of 109.5° required for carbon sp 3 hybrid orbital formation. Similarly, we can expect that the hydrogenation of benzene should proceed efficiently on a surface in which the active sites are spaced in the range of 150 and 250 pm.
This is one reason why many metallic catalysts exhibit different catalytic activity on different crystal faces.
Some special types of heterogeneous catalysts
Metal-cluster catalysts
As the particle size of a catalyst is reduced, the fraction of more highly exposed step, edge, and corner atoms increases. An extreme case occurs with nano-sized (1-2 nm) metal cluster structures composed typically of 10-50 atoms.
[link] →
Metallic gold, well known for its chemical inertness, exhibits very high catalytic activity when it is deposited as metallic clusters on an oxide support. For example, O 2 dissociates readily on Au 55 clusters which have been found to efficiently catalyze the oxidation of hydrocarbons [article].
Zeolite catalysts
Zeolites are clay-like aluminosilicate solids that form open-framework microporous structures that may contain linked cages, cavities or channels whose dimensions can be tailored to the sizes of the reactants and products. To those molecules able to diffuse through these spaces, zeolites are in effect "all surface", making them highly efficient. This size-selectivity makes them important for adsorption, separation, ion-exchange, and catalytic applications. Many zeolites occur as minerals, but others are made synthetically in order to optimize their properties.
As catalyts, zeolites offer a number of advantages that has made them especially important in "green chemistry" operations in which the number of processing steps, unwanted byproducts, and waste stream volumes are minimized.
Enzymes and biocatalysis
"A Jug of Wine, a Loaf of Bread, and . . ." Enzymes!
This distortion of Robert FitzGerald's already-distorted translation of the famous quatrain from the wonderful Rubaiyat of Omar Khayyam underlines the central role that enzymes and their technology have played in civilization since ancient times.
← Illustration by Edmund Dulac (1882-1953)
Fermentation and wine-making have been a part of human history and culture for at least 8000 years, but recognition of the role of catalysis in these processes had to wait until the late nineteenth century. By the 1830's, numerous similar agents, such as those that facilitate protein digestion in the stomach, had been discovered. The term "enzyme", meaning "from yeast", was coined by the German physiologist Wilhelm Kühne in 1876. In 1900, Eduard Buchner (1860-1917, 1907 Nobel Prize in Chemistry) showed that fermentation, previously believed to depend on a mysterious "life force" contained in living organisms such as yeast, could be achieved by a cell-free "press juice" that he squeezed out of yeast.
By this time it was recognized that enzymes are a form of catalyst (a term introduced by Berzelius in 1835), but their exact chemical nature remained in question. They appeared to be associated with proteins, but the general realization that enzymes are proteins began only in the 1930s when the first pure enzyme was crystallized, and did not become generally accepted until the 1950s. It is now clear that nearly all enzymes are proteins, the major exception being a small but important class of RNA-based enzymes known as ribozymes.
Proteins are composed of long sequences of amino acids strung together by amide bonds; this sequence defines the primary structure of the protein.. Their huge size (typically 200-2000 amino acid units, with total molecular weights 20,000 - 200,000) allows them to fold in complicated ways (known as secondary and tertiary structures) whose configurations are essential to their catalytic function.
Because enzymes are generally very much larger than the reactant molecules they act upon (known in biochemistry as substrates ), enzymatic catalysis is in some ways similar to heterogeneous catalysis. The main difference is that the binding of a subtrate to the enzyme is much more selective.
Precursors and cofactors
Most enzymes come into being as inactive precursors ( zymogens ) which are converted to their active forms at the time and place they are needed.
- For example, the enzymes that lead to the clotting of blood are supposed to remain inactive until bleeding actually begins; a major activating factor is exposure of the blood to proteins in the damaged vessel wall.
- The enzyme that catalyzes the formation of lactose (milk sugar) in the mammary gland is formed during pregnancy, but it remains inactive until the time of birth, when hormonal changes cause a modifier unit to bind to and activate the enzyme.
Conversion to the active form may involve a simple breaking up of the protein by hydrolysis of an appropriate peptide bond or the addition of a phosphate or similar group to one of the amino acid residues.
Many enzyme proteins also require "helper" molecules, known as cofactors , to make them catalytically active. These may be simple metal ions (many of the trace nutrient ions of Cu, Mn, Mo, V, etc.) or they may be more complex organic molecules which are called coenzymes . Many of the latter are what we commonly refer to as vitamins . Other molecules, known as inhibitors , decrease enzyme activity; many drugs and poisons act in this way.
The enzyme-substrate complex
The standard model of enzyme kinetics consists of a two-step process in which an enzyme binds reversibly to its substrate S (the reactant) to form an enzyme-substrate complex ES:
The enzyme-substrate complex plays a role similar to that of the activated complex in conventional kinetics, but the main function of the enzyme is to stabilize the transition state.
In the second, essentially irreversible step, the product and the enzyme are released:
The basic kinetic treatment of this process involves the assumption that the concentrations [E] and [ES] reach steady-state values which do not change with time. (The detailed treatment, which is beyond the scope of this course, can be found here.)
The overall process is described by the Michaelis-Menten equation which is plotted here. The Michaelis constant k M is defined as shown, but can be simplified to the ES dissociation constant k –1 / k 1 in cases when dissociation of the complex is the rate-limiting step. The quantity v max is not observed directly, but can be determined from k M as shown here.
Enzymes are proteins
In order to understand enzymes and how they catalyze reactions, it is first necessary to review a few basic concepts relating to proteins and the amino acids of which they are composed.
Amino acids: polar, nonpolar, positive, negative.
The 21 amino acids that make up proteins all possess the basic structure shown here, where R represents either hydrogen or a side chain which may itself contain additional –NH 2 or –COOH groups. Both kinds of groups can hydrogen-bond with water and with the polar parts of substrates, and therefore contribute to the amino acid's polarity and hydophilic nature. Side chains that contain longer carbon chains and especially benzene rings have the opposite effect, and tend to render the amino acid non-polar and hydrophobic.
Both the –NH 2 and –COOH groups are ionizable (i.e., they can act as proton donors or acceptors) and when they are in their ionic forms, they will have an electric charge. The –COOH groups have pKa's in the range 1.8-2.8, and will therefore be in their ionized forms –COO – at ordinary cellular pH values of around 7.4. The amino group pKa's are around 8.8-10.6, so these will also normally be in their ionized forms NH 3 + .
This means that at ordinary cellular pH, both the carboxyl and amino groups will be ionized. But because the charges have opposite signs, an amino acid that has no extra ionizable groups in its side chain will have a net charge of zero. But if the side chain contains an extra amino or carboxyl group, the amino acid can carry a net electric charge. The following diagram illustrates typical amino acids that fall into each of the groups described above.
Proteins
Proteins are made up of one or more chains of amino acids linked to each other through peptide bonds by elimination of a water molecule.
The product shown above is called a peptide , specifically it is a dipeptide because it contains two amino acid residues (what's left after the water has been removed.) Proteins are simply very long polypeptide chains, or combinations of them. (The distinction between a long polypeptide and a small protein is rather fuzzy!)
Globular proteins
Most enzymes fall into the category of globular proteins . In contrast to the fibrous proteins that form the structural components of tissues, globular proteins are soluble in water and rarely have any systematic tertiary structures. They are made up of one or more amino-acid (" peptide ") chains which fold into various shapes that can roughly be described as spherical — hence the term "globular", and the suffix "globin" that is frequently appended to their names, as in "hemoglobin".
Protein folding is a spontaneous process that is influenced by a number of factors. One of these is obviously their primary amino-acid sequence that facilitates formation of intramolecular bonds between amino acids in different parts of the chain. These consist mostly of hydrogen bonds, although disulfide bonds S—S between sulfur-containing amino acids are not uncommon.
In addition to these intramolecular forces, interactions with the surroundings play an important role. The most import of these is the hydrophobic effect , which favors folding conformations in which polar amino acids (which form hydrogen bonds with water) are on the outside, while the so-called hydrophobic amino acids remain in protected locations within the folds.
How enzymes work: the active site
The catalytic process mediated by an enzyme takes place in a depression or cleft that exposes the substrate to only a few of the hundreds-to-thousands of amino acid residues in the protein chain. The high specificity and activity of enzyme catalysis is sensitively dependent on the shape of this cavity and on the properties of the surrounding amino acids
In 1894, long before it was clear that enzymes are proteins, the German chemist Emil Fischer suggested the so-called lock-and-key model as a way of understanding how a given enzyme can act specificilly on only one kind of substrate molecule. This model is essentially an elaboration of the one we still use for explaining heterogeneous catalysis.
Although the basic lock-and-key model continues to be useful, it has been modified into what is now called the induced-fit model . This assumes that when the substrate enters the active site and interacts with the surrounding parts of the amino acid chain, it reshapes the active site (and perhaps other parts of the enzyme) so that it can engage more fully with the substrate.
One important step in this process is to squeeze out any water molecules that are bound to the substrate and which would interfere with its optimal positioning.
Within the active site, specific interactions between the substrate and appropriately charged, hydrophlic and hydrophobic amino acids of the active site then stabilize the transition state by distorting the substrate molecule in such a way as to lead to a transition state having a substantially lower activation energy than can be achieved by ordinay non-enzymatic catalysis. Beyond this point, the basic catalytic steps are fairly conventional, with acid/base and nucleophilic catalysis being the most common.
For a very clear and instructive illustration of a typical multi-step sequence of a typical enzyme-catalyzed reaction, see this page from Mark Bishop's online textbook, from which this illustration is taken.
Enzyme regulation and inhibition
If all the enzymes in an organism were active all the time, the result would be runaway chaos. Most cellular processes such as the production and utilization of energy, cell division, and the breakdown of metabolic products must operate in an exquisitely choreographed, finely-tuned manner, much like a large symphony orchestra; no place for jazz-improv here!
Nature has devised various ways of achieving this; we described the action of precursors and coenzymes above . Here we focus on one of the most important (and chemically-interesting) regulatory mechanisms.
Allosteric regulation: tweaking the active site
There is an important class of enzymes that possess special sites (distinct from the catalytically active sites) to which certain external molecules can reversibly bind. Although these allosteric sites , as they are called, may be quite far removed from the catalytic sites, the effect of binding or release of these molecules is to trigger a rapid change in the folding pattern of the enzyme that alters the shape of the active site. The effect is to enable a signalling or regulatory molecule (often a very small one such as NO) to modulate the catalytic activity of the active site, effectively turning the enzyme on or off.
In some instances, the product of an enzyme-catalyzed reaction can itself bind to an allosteric site, decreasing the activity of the enzyme and thus providing negative feedback that helps keep the product at the desired concentration. It is believed that concentrations of plasma ions such as calcium, and of energy-supplying ATP are, are regulated in this way.
Allosteric enzymes are more than catalysts: they act as control points in metabolic and cellular signalling networks.Allosteric enzymes frequently stand at the beginning of a sequence of enzymes in a metabolic chain, or at branch points where two such chains diverge, acting very much like traffic signals at congested intersections.
Enzyme inhibition
As is the case with heterogeneous catalysts, certain molecules other than the normal substrate may be able to enter and be retained in the active site so as to competitively inhibit an enzyme's activity. This is how penicillin and related antibiotics work; these molecules covalently bond to amino acid residues in the active site of an enzyme that catalyzes the formation of an essential component of bacterial cell walls. When the cell divides, the newly-formed progeny essentially fall apart.
Enzymes outside the cell
Enzymes have been widely employed in the food, pulp-and-paper, and detergent industries for a very long time, but mostly as impure whole-cell extracts.
In recent years, developments in biotechnology and the gradual move of industry from reliance on petroleum-based feedstocks and solvents to so-called "green" chemistry have made enzymes more attractive as industrial catalysts. Compared to the latter, purified enzymes tend to be expensive, difficult to recycle, and unstable outside of rather narrow ranges of temperature, pH, and solvent composition.
Immobilized enzymes
Many of the problems connected with the use of free enzymes can be overcome by immobilizing the enzyme. This can be accomplished in several ways:
- Binding the enzyme to an inert solid supporting material
- This was first done in 1916, using charcoal or aluminum hydroxide. since the 1960's when this method became more popular, synthetic polymers, often engineered for a specific application, have come into wide use. Some natural biopolymers such as cellulose or starch, and inorganic solids such as silica and alumina have also been employed.
- Trapping the enzyme in a porous material
- Polymerization of acrylamide in the presence of an enzyme yields a porous gel which alllows free diffusion of substrate and products, Other materils used for this purpose include natural gels such as alginate (obtained from seaweed), porous silica or ceramic solids, and zeolites.
- Cross-linked enzymes
- If the enzyme is chemically linked to another substance (usually a very small molecule), the resulting solid, known as a cross-linked enzyme aggregate, can retain high catalytic efficiency over a very wide range of conditions. A more recent related development involves precipitating a less-purified form of the enzyme (by addition of ethylene glycol or ammonium sulfate, for example) and then cross-linking the resulting aggregates.
Catalysis
- Gadi Rothenberg- Catalysis: Concepts and Green Applications (Wiley-VCH, 2008)
- R.A. van Santan, M. Neurock - Molecular Heterogeneous Catalysis: A conceptual and computational approach (Wiley-VCH, 2006)
- History of Catalysis: List of Web pages, including Fifty Years of Catalysis - A list of major advances in the field 1949-1999. More detailed History pages from UK
- Catalysis and its applications - a very readable and interesting account of the applications of catalysis to various industrially-important fields.
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libretexts
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2025-03-17T19:53:20.538096
| 2016-02-26T07:46:29 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/3.0/",
"url": "https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.06%3A_Catalysts_and_Catalysis",
"book_url": "https://commons.libretexts.org/book/chem-3482",
"title": "17.6: Catalysts and Catalysis",
"author": "Stephen Lower"
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.07%3A_Experimental_methods_of_chemical_kinetics
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17.7: Experimental methods of chemical kinetics
Studies of the dynamics of chemical processes impinge on almost every area of chemistry and biochemistry. It is useful for students even at the general chemistry level to have some understanding of the experimental techniques that have informed what you have already learned about kinetics.
The time domain of chemical kinetics
Kinetics deals with rates of change, so as a preface to this section, it is useful to consider the range of times within which the study of chemical processes takes place. Prior to about 1945, the practical range for kinetic investigations began at about 10 3 sec and extended out to around 10 8 sec, this upper limit being governed by the length of time that a graduate student (who usually does the actual work) can expect to get a Ph.D. degree. The development of flash photolysis and flow methods in the mid-1940s moved the lower time limit down to the millisecond range. The most dramatic changes began to occur in the 1970s as the availablity of lasers capable of very short-lived pulses and fast electronics gradually opened up the study of reactions that are completed in nanoseconds, picoseconds, and even femtoseconds (10 –15 sec).
The basic way of obtaining the information needed to determine rate constants and reaction orders is to bring the reactants together and then measure successive changes in concentration of one of the components as a function of time. Two important requirements are
- The time required to take a measurement must be very short compared to the time the reaction takes to run to completion;
- The temperature must be held constant — something than can pose a problem if the reaction is highly exothermic.
Measuring the concentration of a reactant or product directly — that is by chemical analysis, is awkward and seldom necessary. When it cannot be avoided, the reaction sample must usually be quenched in some way in order to stop any further change until its composition can be analyzed. This may be accomplished in various ways, depending on the particular reaction. For reactions carried out in solution, especially enzyme-catalyzed ones, it is sometimes practical to add a known quantity of acid or base to change the pH, or to add some other inhibitory agent. More commonly, however, the preferred approach is to observe some physical property whose magnitude is proportional to the extent of the reaction.
Classical and laboratory-class methods
These methods are applicable to reactions that are not excessively fast, typically requiring a few minutes or hours to run to completion. They were about the only methods available before 1950. Most first-year laboratory courses students will include at least one experiment based on one of these methods.
Optical methods
Light absorption is perhaps the most widely-used technique. If either a reactant or a product is colored, the reaction is easily followed by recording the change in transmission of an appropriate wavelength after the reaction is started.
When a beam of light passes through a solution containing a colored substance, the fraction that is absorbed is directly proportional to the concentration of that substance and to the length of the light's path through the solution. The latter can be controlled by employing a cell or cuvette having a fixed path length.
If I 0 is the intensity of the light incident on the cell and I is the intensity that emerges on the other side, then the percent absorption is just 100 × I / I 0 . Because a limited 1-100 scale of light absorption is often inadequate to express the many orders of magnitude frequently encountered, a logarithmic term optical density is often employed. The relation between this, the cell path length, concentration, and innate absorption ability of the colored substance is expressed by Beer's law .
The simplest absorbance measuring device is a colorimeter in which a beam of white light from an incandescent lamp is passed through a cell (often just a test tube) and onto a photodetector whose electrical output is directly proportional to the light intensity. Before beginning the experiment, a zeroing control sets the meter to zero when the light path is blocked off, and a sensitivity control sets it to 100 when a cell containing an uncolored solution (a "blank") is inserted.
The sensitivity and selectivity of such an arrangement is greatly enhanced by adjusting the wavelength of the light to match the absorption spectrum of the substance being measured. Thus if the substance has a yellow color, it is because blue light is being absorbed, so a blue color filter is placed in the light path.
More sophisticated absorption spectrophotometers employ two cells (one for the sample and another reference cell for the blank. They also allow one to select the particular wavelength range that is most strongly absorbed by the substance under investigation, which can often extend into the near-ultraviolet region.
Light scattering measurements (made with a nephthelometer ) can be useful for reactions that lead to the formation a fine precipitate/ A very simple student laboratory experiment of this kind can be carried out by placing a conical flask containing the reaction mixture on top of a marked piece of paper. The effects of changing the temperature or reactant concentrations can be made by observing how long it takes for precipitate formation to obscure the mark on the paper.
Other optical methods such a fluorescence and polarimetry (measurement of the degree to which a solution rotates the plane of polarized light) are also employed when applicable.
Other methods
- Measurement of gases : The classical method of following reactions that produce changes in the number of moles of gases is to observe changes in pressure or volume. An alternative method for following the loss of a gas is to place the reaction container on an electronic balance and monitor the loss in weight. This is generally less accurate than pressure measurements, but is sometimes used in student experiments.
- pH measurements : Many reactions yield or consume hydrogen ions and are conveniently followed by means of a pH meter.
- Electrical Conductance: Reactions that yield or consume ionic substances are often studied by measuring the electrical conductance of the solution. Conductimetry is usually carried out by balancing the solution conductance with a known resistance in a bridge arrangement. An audio-frequency alternating current is used in order to avoid electrolysis
The study of rapid reactions
Rapid mixing
The traditional experimental methods described above all assume that we can follow the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can we do if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity.
If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined.
Although this method is very simple in principle, it can be rather complicated in practice, as the illustration shows. Owing to the rather large volumes required, his method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred.
Stopped-flow and Quenched-flow methods
These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates.
The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed.
Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to quench (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the actvity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner.
The quenched-flow technique works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution.
Too fast to mix? No problem!
In order to investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated in situ . Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium.
Flash photolysis
Many reactions are known which do not take place in the absence of light whose wavelength is sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl 2 with H 2 , which procedes explosively when the system is illuminated with visible light. In flash photolysis , a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means.
Photolysis refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to thermolysis (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner", meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly specific ; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present.
All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species producd by photolysis. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH 2 radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy.
Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work.Nanosecond flash photolysis setup.
In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Flash durations of around 1 millisecond permitted one to follow processes having lifetimes in the microsecond range, but the advent of fast lasers gradually extended this to picoseconds and femtoseconds.
Flash photolysis revolutionized the study of organic photochemistry, especially that relating to the chemistry of free radicals and other reactive species that cannot be isolated or stored, but which can easily be produced by photolysis of a suitable precursor. It has proven invaluable for understanding the complicated kinetics relating to atmospheric chemistry and smog formation. More recently, flash photolysis has become an important tool in biochemistry and cellular physiology.
Perturbation-relaxation methods
Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water
\[2 H_2O → H_3O^+ + OH^–\]
and the formation of the triiodide ion in aqueous solution
\[I^– + I_2 → I_3^–\]
Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system.
For example, if the reaction A B is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process.
For the general case illustrated here, the quantity " x " being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a first-order process , x will vary with time according to
x t = x o e – kt
After the abrupt perturbation at time t o , the relaxation time t* is defined as the half-time for the return to equilibrium — that is, as the time required for xo to decrease by Δ x /e = Δ x /2.718. The derivation of t* and the relations highlighted in yellow can be found in most standard kinetics textbooks. Temperature jumps are probably most commonly used. They can be brought about in various ways:
This is the method that Manfred Eigen pioneered when, in the early 1960's, he measured the rate constant of what was then the fastest reaction ever observed:
H + + OH – → H 2 O k = 1.3 × 10 11 M –1 sec –1
A detailed description of the kinetics of this process can be seen here. Eigen shared the 1967 Nobel Prize in Chemistry with Porter and Norrish, who developed flash photolysis. Eigen's Nobel lecture, in keeping with his legendary sense of humor, was titled "immeasurably fast reactions". His later work has focused on self-organizing systems and the origin of life, molecular genetics, and neurology.
- high-voltage electric discharge : A capacitor, charged to 5-10 kV, is discharged through a solution to which an electrolyte has been added to provide a conductive path.
- laser irradiation : The sample is irradiated with a laser whose wavelength corresponds to an absorption peak in the sample. Infrared lasers are often used for this purpose.
- mixing of two pre-equilibrated solutions : Two solutions, otherwise identical but at different temperatures, are rapidly mixed in a stopped-flow type of apparatus. Although this method is not as fast, it has the advantage of allowing both negative and positive T-jumps. The device shown here uses 0.1-mL samples and provides jumps of up to ±40 C° over a few microseconds. Observation times, however, are limited to 1-2 milliseconds owing to thermal dissipation.
Pressure jumps
According to the Le Chatelier principle, a change in the applied pressure will shift the equilibrium state of any reaction which involves a change in the volume of a system. Aside from the obvious examples associated with changes in the number of moles of gases, there are many more subtle cases involving formation of complexes, hydration shells and surface adsorption, and phase changes. One area of considerable interest is the study of protein folding, which has implications in diseases such as Parkinson's and Alzheimer's.
The pressure-jump is applied to the cell through a flexible membrane that is activated by a high-pressure gas supply, or through an electrically-actuated piezoelectric crystal. The latter method is employed in the device shown here, which can produce P-jumps of around 1GPa over sub-millisecond time intervals.
More on P-jumps: this 1999 review covers many non-biochemical applications. See also this article on protein folding and aggregation.
Shock tubes: extreme jumping
When a change in pressure propagates through a gas at a rate greater than the ordinary compressions and rarefactions associated with the travel of sound, a moving front (a shock wave ) of very high pressure forms. This in turn generates an almost instantaneous rise in the temperature that can approach several thousand degrees in magnitude.
A shock tube is an apparatus in which shock waves can be generated and used to study the kinetics of gas-phase reactions that are otherwise inaccessible to kinetic measurements. Since all molecules tend to dissociate at high temperatures, shock tubes are widely used to study dissociation processes and the chemistry of the resulting fragments. For example, the shock-induced decomposition of carbon suboxide provides an efficient means of investigating carbon atom reactions:
\[C_3O_2 → C + CO\]
Shock tube techniques are also useful for studying combustion reactions, including those that proceed explosively.
The shock tube itself consists of two sections separated by a breakable diaphragm of metal or plastic. One section is filled with a "driver" gas at a very high pressure, commonly helium, but often mixed with other inert gases to adjust the properties of the shock. The other, longer section of the tube contains the "driven" gas — the reactants — at a low pressure, usually less than 1 atmosphere.
The reaction is initiated by causing the diaphragm to rupture, either by means of a mechanical plunger or by raising the pressure beyond its bursting point. The kinetics of the reaction are monitored by means of an absorption or other optical monitoring device that is positioned at a location along the reaction tube that is appropriate to the time course of the reaction.
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libretexts
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2025-03-17T19:53:20.628700
| 2016-02-26T07:46:43 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics
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1: The Nature of Science and Physics
In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single mathematical language.
-
- 1.0: Prelude to Science and the Realm of Physics, Physical Quantities, and Units
- In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single mathematical language.
-
- 1.1: Physics- An Introduction
- Science seeks to discover and describe the underlying order and simplicity in nature. Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions. Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws of nature are rules that all natural processes appear to follow.
-
- 1.2: Physical Quantities and Units
- Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units. The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for electric current). These units are part of the metric system, which
-
- 1.3: Accuracy, Precision, and Significant Figures
- Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to the correct value for that measurement. The precision of a measurement system refers to how close the agreement is between repeated measurements (which are repeated under the same conditions).
-
- 1.4: Approximation
- As you develop problem-solving skills, you will also develop skills at approximating. You will develop these skills through thinking more quantitatively, and by being willing to take risks. As with any endeavor, experience helps, as well as familiarity with units. These approximations allow us to rule out certain scenarios or unrealistic numbers. Approximations also allow us to challenge others and guide us in our approaches to our scientific world.
Thumbnail: This parabola-shaped lava flow illustrates the application of mathematics in physics—in this case, Galileo's law of falling bodies. (Jim D. Griggs, HVO (USGS) staff photographer @ http://pubs.usgs.gov/dds/dds-80 ).
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libretexts
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2025-03-17T19:53:20.692419
| 2015-10-27T19:11:52 |
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"title": "1: The Nature of Science and Physics",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.00%3A_Prelude_to_Science_and_the_Realm_of_Physics_Physical_Quantities_and_Units
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1.0: Prelude to Science and the Realm of Physics, Physical Quantities, and Units
What is your first reaction when you hear the word “physics”? Did you imagine working through difficult equations or memorizing formulas that seem to have no real use in life outside the physics classroom? Many people come to the subject of physics with a bit of fear. But as you begin your exploration of this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice.
For example, take a look at the image above. This image is of the Andromeda Galaxy, which contains billions of individual stars, huge clouds of gas, and dust. Two smaller galaxies are also visible as bright blue spots in the background. At a staggering 2.5 million light years from the Earth, this galaxy is the nearest one to our own galaxy (which is called the Milky Way). The stars and planets that make up Andromeda might seem to be the furthest thing from most people’s regular, everyday lives. But Andromeda is a great starting point to think about the forces that hold together the universe. The forces that cause Andromeda to act as it does are the same forces we contend with here on Earth, whether we are planning to send a rocket into space or simply raise the walls for a new home. The same gravity that causes the stars of Andromeda to rotate and revolve also causes water to flow over hydroelectric dams here on Earth. Tonight, take a moment to look up at the stars. The forces out there are the same as the ones here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything we can see and know in this universe.
Think now about all of the technological devices that you use on a regular basis. Computers, smart phones, GPS systems, MP3 players, and satellite radio might come to mind. Next, think about the most exciting modern technologies that you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light around them, and microscopic robots that fight cancer cells in our bodies. All of these groundbreaking advancements, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must understand how wind forces affect a flight path and a physical therapist must understand how the muscles in the body experience forces as they move and bend. As you will learn in this text, physics principles are propelling new, exciting technologies, and these principles are applied in a wide range of careers.
In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single mathematical language. Finally, you will study the limits of our ability to be accurate and precise, and the reasons scientists go to painstaking lengths to be as clear as possible regarding their own limitations.
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libretexts
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2025-03-17T19:53:20.759640
| 2015-11-01T04:29:52 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.01%3A_Physics-_An_Introduction
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1.1: Physics- An Introduction
Learning Objectives
By the end of this section, you will be able to:
- Explain the difference between a principle and a law.
- Explain the difference between a model and a theory.
The physical universe is enormously complex in its detail. Every day, each of us observes a great variety of objects and phenomena. Over the centuries, the curiosity of the human race has led us collectively to explore and catalog a tremendous wealth of information. From the flight of birds to the colors of flowers, from lightning to gravity, from quarks to clusters of galaxies, from the flow of time to the mystery of the creation of the universe, we have asked questions and assembled huge arrays of facts. In the face of all these details, we have discovered that a surprisingly small and unified set of physical laws can explain what we observe. As humans, we make generalizations and seek order. We have found that nature is remarkably cooperative—it exhibits the underlying order and simplicity we so value.
It is the underlying order of nature that makes science in general, and physics in particular, so enjoyable to study. For example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws, permitting an understanding beyond just the memorization of lists of facts.
The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of this text. In learning to apply these laws, you will, of course, study the most important topics in physics. More importantly, you will gain analytical abilities that will enable you to apply these laws far beyond the scope of what can be included in a single book. These analytical skills will help you to excel academically, and they will also help you to think critically in any professional career you choose to pursue. This module discusses the realm of physics (to define what physics is), some applications of physics (to illustrate its relevance to other disciplines), and more precisely what constitutes a physical law (to illuminate the importance of experimentation to theory).
Science and the Realm of Physics
Science consists of the theories and laws that are the general truths of nature as well as the body of knowledge they encompass. Scientists are continually trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics is concerned with describing the interactions of energy, matter, space, and time, and it is especially interested in what fundamental mechanisms underlie every phenomenon. The concern for describing the basic phenomena in nature essentially defines the realm of physics .
Physics aims to describe the function of everything around us, from the movement of tiny charged particles to the motion of people, cars, and spaceships. In fact, almost everything around you can be described quite accurately by the laws of physics. Consider a smart phone (Figure \(\PageIndex{2}\)). Physics describes how electricity interacts with the various circuits inside the device. This knowledge helps engineers select the appropriate materials and circuit layout when building the smart phone. Next, consider a GPS system. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it takes to travel that distance. When you use a GPS device in a vehicle, it utilizes these physics equations to determine the travel time from one location to another.
Applications of Physics
You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers (Figures \(\PageIndex{3}\) and \(\PageIndex{4}\)). Physics allows you to understand the hazards of radiation and rationally evaluate these hazards more easily. Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals through our body’s nervous system are much easier to understand when you think about them in terms of basic physics.
Physics is the foundation of many important disciplines and contributes directly to others. Chemistry, for example—since it deals with the interactions of atoms and molecules—is rooted in atomic and molecular physics. Most branches of engineering are applied physics. In architecture, physics is at the heart of structural stability, and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer in the Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cell walls and cell membranes (Figures \(\PageIndex{5}\) and \(\PageIndex{6}\)). On the macroscopic level, it can explain the heat, work, and power associated with the human body. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; for example, cancer radiotherapy uses ionizing radiation. Physics can also explain sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers can transmit information.
It is not necessary to formally study all applications of physics. What is most useful is knowledge of the basic laws of physics and a skill in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics has retained the most basic aspects of science, so it is used by all of the sciences, and the study of physics makes other sciences easier to understand.
Models, Theories, and Laws; The Role of Experimentation
The laws of nature are concise descriptions of the universe around us; they are human statements of the underlying laws or rules that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and so cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort (Figures \(\PageIndex{7}\) and \(\PageIndex{8}\)). The cornerstone of discovering natural laws is observation; science must describe the universe as it is, not as we may imagine it to be.
We all are curious to some extent. We look around, make generalizations, and try to understand what we see—for example, we look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how the data may be organized and unified. We then formulate models, theories, and laws based on the data we have collected and analyzed to generalize and communicate the results of these experiments.
A model is a representation of something that is often too difficult (or impossible) to display directly. While a model is justified with experimental proof, it is only accurate under limited situations. An example is the planetary model of the atom in which electrons are pictured as orbiting the nucleus, analogous to the way planets orbit the Sun. (See Figure 9.) We cannot observe electron orbits directly, but the mental image helps explain the observations we can make, such as the emission of light from hot gases (atomic spectra). Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario and perform a calculation, or they can be used to represent a situation in the form of a computer simulation. A theory is an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers. Some theories include models to help visualize phenomena, whereas others do not. Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, we picture them mentally to understand what our instruments tell us about the behavior of gases.
A law uses concise language to describe a generalized pattern in nature that is supported by scientific evidence and repeated experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence. However, the designation law is reserved for a concise and very general statement that describes phenomena in nature, such as the law that energy is conserved during any process, or Newton’s second law of motion, which relates force, mass, and acceleration by the simple equation
\[F =ma.\]
A theory, in contrast, is a less concise statement of observed phenomena. For example, the Theory of Evolution and the Theory of Relativity cannot be expressed concisely enough to be considered a law. The biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law describes a single action, whereas a theory explains an entire group of related phenomena. And, whereas a law is a postulate that forms the foundation of the scientific method, a theory is the end result of that process.
Less broadly applicable statements are usually called principles (such as Pascal’s principle, which is applicable only in fluids), but the distinction between laws and principles often is not carefully made.
MODELS, THEORIES, AND LAWS
Models, theories, and laws are used to help scientists analyze the data they have already collected. However, often after a model, theory, or law has been developed, it points scientists toward new discoveries they would not otherwise have made.
The models, theories, and laws we devise sometimes imply the existence of objects or phenomena as yet unobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables scientists to make such spectacular predictions. However, if experiment does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to perform every imaginable experiment in order to confirm a law in every possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable experiment contradicts a well-established law, then the law must be modified or overthrown completely.
The study of science in general and physics in particular is an adventure much like the exploration of uncharted ocean. Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained.
THE SCIENTIFIC METHOD
As scientists inquire and gather information about the world, they follow a process called the scientific method. This process typically begins with an observation and question that the scientist will research. Next, the scientist typically performs some research about the topic and then devises a hypothesis. Then, the scientist will test the hypothesis by performing an experiment. Finally, the scientist analyzes the results of the experiment and draws a conclusion. Note that the scientific method can be applied to many situations that are not limited to science, and this method can be modified to suit the situation.
Consider an example. Let us say that you try to turn on your car, but it will not start. You undoubtedly wonder: Why will the car not start? You can follow a scientific method to answer this question. First off, you may perform some research to determine a variety of reasons why the car will not start. Next, you will state a hypothesis. For example, you may believe that the car is not starting because it has no engine oil. To test this, you open the hood of the car and examine the oil level. You observe that the oil is at an acceptable level, and you thus conclude that the oil level is not contributing to your car issue. To troubleshoot the issue further, you may devise a new hypothesis to test and then repeat the process again.
The Evolution of Natural Philosophy into Modern Physics
Physics was not always a separate and distinct discipline. It remains connected to other sciences to this day. The word physics comes from Greek, meaning nature. The study of nature came to be called “natural philosophy.” From ancient times through the Renaissance, natural philosophy encompassed many fields, including astronomy, biology, chemistry, physics, mathematics, and medicine. Over the last few centuries, the growth of knowledge has resulted in ever-increasing specialization and branching of natural philosophy into separate fields, with physics retaining the most basic facets (Figures \(\PageIndex{10}\)-Figure \(\PageIndex{12}\)). Physics as it developed from the Renaissance to the end of the 19th century is called classical physics . It was transformed into modern physics by revolutionary discoveries made starting at the beginning of the 20th century.
Classical physics is not an exact description of the universe, but it is an excellent approximation under the following conditions: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields, such as the field generated by the Earth, can be involved. Because humans live under such circumstances, classical physics seems intuitively reasonable, while many aspects of modern physics seem bizarre. This is why models are so useful in modern physics—they let us conceptualize phenomena we do not ordinarily experience. We can relate to models in human terms and visualize what happens when objects move at high speeds or imagine what objects too small to observe with our senses might be like. For example, we can understand an atom’s properties because we can picture it in our minds, although we have never seen an atom with our eyes. New tools, of course, allow us to better picture phenomena we cannot see. In fact, new instrumentation has allowed us in recent years to actually “picture” the atom.
LIMITS ON THE LAWS OF CLASSICAL PHYSICS
For the laws of classical physics to apply, the following criteria must be met: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields (such as the field generated by the Earth) can be involved.
Some of the most spectacular advances in science have been made in modern physics. Many of the laws of classical physics have been modified or rejected, and revolutionary changes in technology, society, and our view of the universe have resulted. Like science fiction, modern physics is filled with fascinating objects beyond our normal experiences, but it has the advantage over science fiction of being very real. Why, then, is the majority of this text devoted to topics of classical physics? There are two main reasons: Classical physics gives an extremely accurate description of the universe under a wide range of everyday circumstances, and knowledge of classical physics is necessary to understand modern physics.
Modern physics itself consists of the two revolutionary theories, relativity and quantum mechanics. These theories deal with the very fast and the very small, respectively. Relativity must be used whenever an object is traveling at greater than about 1% of the speed of light or experiences a strong gravitational field such as that near the Sun. Quantum mechanics must be used for objects smaller than can be seen with a microscope. The combination of these two theories is relativistic quantum mechanics, and it describes the behavior of small objects traveling at high speeds or experiencing a strong gravitational field. Relativistic quantum mechanics is the best universally applicable theory we have. Because of its mathematical complexity, it is used only when necessary, and the other theories are used whenever they will produce sufficiently accurate results. We will find, however, that we can do a great deal of modern physics with the algebra and trigonometry used in this text.
Exercise \(\PageIndex{1}\)
A friend tells you he has learned about a new law of nature. What can you know about the information even before your friend describes the law? How would the information be different if your friend told you he had learned about a scientific theory rather than a law?
- Answer
-
Without knowing the details of the law, you can still infer that the information your friend has learned conforms to the requirements of all laws of nature: it will be a concise description of the universe around us; a statement of the underlying rules that all natural processes follow. If the information had been a theory, you would be able to infer that the information will be a large-scale, broadly applicable generalization.
PHET EXPLORATIONS: EQUATION GRAPHER
Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. \(y=bx\)) to see how they add to generate the polynomial curve.
Summary
- Science seeks to discover and describe the underlying order and simplicity in nature.
- Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions.
- Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws of nature are rules that all natural processes appear to follow.
Glossary
- classical physics
- physics that was developed from the Renaissance to the end of the 19th century
- physics
- the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon
- model
- representation of something that is often too difficult (or impossible) to display directly
- theory
- an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers
- law
- a description, using concise language or a mathematical formula, a generalized pattern in nature that is supported by scientific evidence and repeated experiments
- scientific method
- a method that typically begins with an observation and question that the scientist will research; next, the scientist typically performs some research about the topic and then devises a hypothesis; then, the scientist will test the hypothesis by performing an experiment; finally, the scientist analyzes the results of the experiment and draws a conclusion
- modern physics
- the study of relativity, quantum mechanics, or both
- relativity
- the study of objects moving at speeds greater than about 1% of the speed of light, or of objects being affected by a strong gravitational field
- quantum mechanics
- the study of objects smaller than can be seen with a microscope
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libretexts
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2025-03-17T19:53:20.846060
| 2015-11-01T04:30:17 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.01%3A_Physics-_An_Introduction",
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"title": "1.1: Physics- An Introduction",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.02%3A_Physical_Quantities_and_Units
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1.2: Physical Quantities and Units
Learning Objectives
By the end of this section, you will be able to:
- Perform unit conversions both in the SI and English units.
- Explain the most common prefixes in the SI units and be able to write them in scientific notation.
The range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of the Earth, from the tiny sizes of sub-nuclear particles to the vast distance to the edges of the known universe, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of \(10\) to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities and equations for physical principles allows us to understand nature much more deeply than does qualitative description alone. To comprehend these vast ranges, we must also have accepted units in which to express them. And we shall find that (even in the potentially mundane discussion of meters, kilograms, and seconds) a profound simplicity of nature appears—all physical quantities can be expressed as combinations of only four fundamental physical quantities: length, mass, time, and electric current.
We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we define distance and time by specifying methods for measuring them, whereas we define average speed by stating that it is calculated as distance traveled divided by time of travel.
Measurements of physical quantities are expressed in terms of units , which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way (Figure \(\PageIndex{2}\)).
There are two major systems of units used in the world: SI units (also known as the metric system) and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. Virtually every other country in the world now uses SI units as the standard; the metric system is also the standard system agreed upon by scientists and mathematicians. The acronym “SI” is derived from the French Système International .
SI Units: Fundamental and Derived Units
Table \(\PageIndex{1}\) gives the fundamental SI units that are used throughout this textbook. This text uses non-SI units in a few applications where they are in very common use, such as the measurement of blood pressure in millimeters of mercury (mm Hg). Whenever non-SI units are discussed, they will be tied to SI units through conversions.
| Length | Mass | Time | Electric Current |
|---|---|---|---|
| meter (m) | kilogram (kg) | second (s) | ampere (A) |
It is an intriguing fact that some physical quantities are more fundamental than others and that the most fundamental physical quantities can be defined only in terms of the procedure used to measure them. The units in which they are measured are thus called fundamental units . In this textbook, the fundamental physical quantities are taken to be length, mass, time, and electric current. (Note that electric current will not be introduced until much later in this text.) All other physical quantities, such as force and electric charge, can be expressed as algebraic combinations of length, mass, time, and current (for example, speed is length divided by time); these units are called derived units .
Units of Time, Length, and Mass: The Second, Meter, and Kilogram
The Second
The SI unit for time, the second (abbreviated s), has a long history. For many years it was defined as 1/86,400 of a mean solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or constant, physical phenomenon (because the solar day is getting longer due to very gradual slowing of the Earth’s rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed and counted. In 1967, the second was redefined as the time required for 9,192,631,770 of these vibrations (Figure \(\PageIndex{3}\)). Accuracy in the fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units and can be no more accurate than are the fundamental units themselves.
The Meter
The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more accurate and precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum-iridium bar now kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its present definition (partly for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a second (Figure \(\PageIndex{4}\)). This change defines the speed of light to be exactly 299,792,458 meters per second. The length of the meter will change if the speed of light is someday measured with greater accuracy.
The Kilogram
The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum-iridium cylinder kept with the old meter standard at the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram are also kept at the United States’ National Institute of Standards and Technology, or NIST, located in Gaithersburg, Maryland outside of Washington D.C., and at other locations around the world. The determination of all other masses can be ultimately traced to a comparison with the standard mass.
Electric current and its accompanying unit, the ampere, will be introduced in Introduction to Electric Current, Resistance, and Ohm's Law when electricity and magnetism are covered. The initial modules in this textbook are concerned with mechanics, fluids, heat, and waves. In these subjects all pertinent physical quantities can be expressed in terms of the fundamental units of length, mass, and time.
Metric Prefixes
SI units are part of the metric system . The metric system is convenient for scientific and engineering calculations because the units are categorized by factors of 10. Table 2 gives metric prefixes and symbols used to denote various factors of 10.
Metric systems have the advantage that conversions of units involve only powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple—there are 12 inches in a foot, 5280 feet in a mile, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by using an appropriate metric prefix. For example, distances in meters are suitable in construction, while distances in kilometers are appropriate for air travel, and the tiny measure of nanometers are convenient in optical design. With the metric system there is no need to invent new units for particular applications.
The term order of magnitude refers to the scale of a value expressed in the metric system. Each power of 1 0 in the metric system represents a different order of magnitude. For example, 10 1 , 10 2 , 10 3 , and so forth are all different orders of magnitude. All quantities that can be expressed as a product of a specific power of 10 are said to be of the same order of magnitude. For example, the number 800 can be written as 8×10 2 , and the number 450 can be written as 4.5×10 2 . Thus, the numbers 800 and 450 are of the same order of magnitude: 10 2 . Order of magnitude can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10 -9 m, while the diameter of the Sun is on the order of 10 9 m.
THE QUEST FOR MICROSCOPIC STANDARDS FOR BASIC UNITS
The fundamental units described in this chapter are those that produce the greatest accuracy and precision in measurement. There is a sense among physicists that, because there is an underlying microscopic substructure to matter, it would be most satisfying to base our standards of measurement on microscopic objects and fundamental physical phenomena such as the speed of light. A microscopic standard has been accomplished for the standard of time, which is based on the oscillations of the cesium atom.
The standard for length was once based on the wavelength of light (a small-scale length) emitted by a certain type of atom, but it has been supplanted by the more precise measurement of the speed of light. If it becomes possible to measure the mass of atoms or a particular arrangement of atoms such as a silicon sphere to greater precision than the kilogram standard, it may become possible to base mass measurements on the small scale. There are also possibilities that electrical phenomena on the small scale may someday allow us to base a unit of charge on the charge of electrons and protons, but at present current and charge are related to large-scale currents and forces between wires.
| Prefix | Symbol | Value | Examples (some are Approximate) | |
|---|---|---|---|---|
| exa | E | \(10^18\) | exameter Em \(10^{18}m\) | distance light travels in a century |
| peta | P | \(10^15\) | petasecond Ps \(10^{15} s\) | 30 million years |
| tera | T | \(10^12\) | terawatt TW \(10^{12} W\) | powerful laser output |
| giga | G | \(10^9\) | gigahertz GHz \(10^9 Hz\) | a microwave frequency |
| mega | M | \(10^6\) | megacurie MCi \(10^{6 }Ci\) | high radioactivity |
| kilo | k | \(10^3\) | kilometer km \(10^3 m\) | about 6/10 mile |
| hecto | h | \(10^2\) | hectoliter hL \(10^2 L\) | 26 gallons |
| deka | da | \(10^1\) | dekagram dag \(10^g\) | teaspoon of butter |
| — | — | \(10^0 (=1)\) | ||
| deci | d | \(10^{−1}\) | deciliter dL \(10^{−1} L\) | less than half a soda |
| centi | c | \(10^{−2}\) | centimeter cm \(10^{−2} m\) | fingertip thickness |
| milli | m | \(10^{−3}\) | millimeter mm \(10^{−3} m\) | flea at its shoulders |
| micro | µ | \(10^{−6}\) | micrometer µm \(10^{−6} m\) | detail in microscope |
| nano | n | \(10^{−9}\) | nanogram ng \(10^{−9} g\) | small speck of dust |
| pico | p | \(10^{−12}\) | picofarad pF \(10^{−12} F\) | small capacitor in radio |
| femto | f | \(10^{−15}\) | femtometer fm \(10^{−15} m\) | size of a proton |
| atto | a | \(10^{−18}\) | attosecond as \(10^{−18} s\) | time light crosses an atom |
Known Ranges of Length, Mass, and Time
The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, and times in Table \(\PageIndex{2}\). Examination of this table will give you some feeling for the range of possible topics and numerical values (Figures \(\PageIndex{5}\) and \(\PageIndex{6}\)).
Unit Conversion and Dimensional Analysis
It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or, perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you will need to convert units of feet to miles. Let us consider a simple example of how to convert units.
Let us say that we want to convert 80 meters (\(m\)) to kilometers (\(km\)).
- The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in meters and we want to convert to kilometers .
- Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in 1 minute, and so on. In this case, we know that there are 1,000 meters in 1 kilometer.
- Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel out, as shown: \[80\,\cancel{m} \times \dfrac{1\,km}{1000\,\cancel{m}} =0.08\, km\] Note that the unwanted \(m\) unit cancels, leaving only the desired km unit. You can use this method to convert between any types of unit.
Click Appendix C for a more complete list of conversion factors.
| lengths in meters | Masses in kilograms (more precise values in parentheses) | Times in seconds (more precise values in parentheses) | |||
|---|---|---|---|---|---|
| 10 −18 | Present experimental limit to smallest observable detail |
10
−30
|
Mass of an electron (9.11×10 −31 kg) |
10
−23
|
Time for light to cross a proton |
|
10
−15
|
Diameter of a proton |
10
−27
|
Mass of a hydrogen atom (1.67×10 −27 kg) |
10
−22
|
Mean life of an extremely unstable nucleus |
|
10
−14
|
Diameter of a uranium nucleus |
10
−15
|
Mass of a bacterium |
10
−15
|
Time for one oscillation of visible light |
|
10
−10
|
Diameter of a hydrogen atom |
10
−5
|
Mass of a mosquito |
10 −13 |
Time for one vibration of an atom in a solid |
|
10
−8
|
Thickness of membranes in cells of living organisms |
10
−2
|
Mass of a hummingbird |
10
−8
|
Time for one oscillation of an FM radio wave |
|
10
−6
|
Wavelength of visible light |
1
|
Mass of a liter of water (about a quart) |
10
−3
|
Duration of a nerve impulse |
|
10
−3
|
Size of a grain of sand |
10
2
|
Mass of a person |
1
|
Time for one heartbeat |
|
1
|
Height of a 4-year-old child |
10
3
|
Mass of a car |
10
5
|
One day (8.64×10 4 s) |
|
10
2
|
Length of a football field |
10
8
|
Mass of a large ship |
10
7
|
One year (y) (3.16×10 7 s) |
|
10
4
|
Greatest ocean depth |
10
12
|
Mass of a large iceberg |
10
9
|
About half the life expectancy of a human |
|
10
7
|
Diameter of the Earth |
10
15
|
Mass of the nucleus of a comet |
10
11
|
Recorded history |
|
10
11
|
Distance from the Earth to the Sun |
10
23
|
Mass of the Moon (7.35×10 22 kg) |
10
17
|
Age of the Earth |
|
10
16
|
Distance traveled by light in 1 year (a light year) |
10
25
|
Mass of the Earth (5.97×10 24 kg) |
10
18
|
Age of the universe |
|
10 21 |
Diameter of the Milky Way galaxy |
10
30
|
Mass of the Sun (1.99×10 30 kg) | ||
|
10
22
|
Distance from the Earth to the nearest large galaxy (Andromeda) |
10
42
|
Mass of the Milky Way galaxy (current upper limit) | ||
|
10
26
|
Distance from the Earth to the edges of the known universe |
10
53
|
Mass of the known universe (current upper limit) |
Example \(\PageIndex{1}\): Unit Conversions: A Short Drive Home
Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel.)
Strategy
First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.
Solution for (a)
(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now—average speed and other motion concepts will be covered in a later module.) In equation form,
\[\text{average speed} =\dfrac{distance}{time}. \nonumber\]
(2) Substitute the given values for distance and time.
\[ \begin{align*} \text{average speed} &=\dfrac{10.0\, km}{20.0\, min} \\[5pt] &=0.500 \dfrac{km}{ min}.\end{align*} \]
(3) Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min/hr . Thus,
\[\begin{align*} \text{average speed} &=0.500 \dfrac{km}{ min}×\dfrac{60\, min}{1 \,h} \\[5pt] &=30.0 \dfrac{km}{ h} \end{align*} \]
Discussion for (a)
To check your answer, consider the following:
(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:
\[\dfrac{km}{min}×\dfrac{1\, hr}{60\, min}=\dfrac{1}{60} \dfrac{km⋅hr}{ min^2}, \nonumber\]
which are obviously not the desired units of km/h.
(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.
(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect.
(4) Next, check whether the answer is reasonable. Let us consider some information from the problem—if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.
Solution for (b)
There are several ways to convert the average speed into meters per second.
(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed—one to convert hours to seconds, and another to convert kilometers to meters.
(2) Multiplying by these yields
\[\begin{align*} \text{Average speed} &=30.0\dfrac{\bcancel{km}}{\cancel{h}}×\dfrac{1\,\cancel{h}}{3,600 \,s}×\dfrac{1,000\,m}{1\, \bcancel{km}} \\[5pt] &=8.33 \,m/s \end{align*}\]
Discussion for (b)
If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s.
You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions.
NONSTANDARD UNITS
While there are numerous types of units that we are all familiar with, there are others that are much more obscure. For example, a firkin is a unit of volume that was once used to measure beer. One firkin equals about 34 liters. To learn more about nonstandard units, use a dictionary or encyclopedia to research different “weights and measures.” Take note of any unusual units, such as a barleycorn, that are not listed in the text. Think about how the unit is defined and state its relationship to SI units.
Exercise \(\PageIndex{1}\)
Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird to beat its wings once. Which fundamental unit should the scientist use to describe the measurement? Which factor of 10 is the scientist likely to use to describe the motion precisely? Identify the metric prefix that corresponds to this factor of 10.
- Answer
-
The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat so fast, the scientist will probably need to measure in milliseconds, or 10 −3 seconds. (50 beats per second corresponds to 20 milliseconds per beat.)
Exercise \(\PageIndex{2}\)
One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system?
- Answer
-
The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter is dependent on the measure of a centimeter.
Summary
- Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements.
- Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units.
- The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for electric current). These units are part of the metric system, which uses powers of 10 to relate quantities over the vast ranges encountered in nature.
- The four fundamental units are abbreviated as follows: meter, m; kilogram, kg; second, s; and ampere, A. The metric system also uses a standard set of prefixes to denote each order of magnitude greater than or lesser than the fundamental unit itself.
- Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using conversion factors, which are ratios relating equal quantities of different units.
Glossary
- physical quantity
- a characteristic or property of an object that can be measured or calculated from other measurements
- units
- a standard used for expressing and comparing measurements
- SI units
- the international system of units that scientists in most countries have agreed to use; includes units such as meters, liters, and grams
- English units
- system of measurement used in the United States; includes units of measurement such as feet, gallons, and pounds
- fundamental units
- units that can only be expressed relative to the procedure used to measure them
- derived units
- units that can be calculated using algebraic combinations of the fundamental units
- second
- the SI unit for time, abbreviated (s)
- meter
- the SI unit for length, abbreviated (m)
- kilogram
- the SI unit for mass, abbreviated (kg)
- metric system
- a system in which values can be calculated in factors of 10
- order of magnitude
- refers to the size of a quantity as it relates to a power of 10
- conversion factor
- a ratio expressing how many of one unit are equal to another unit
|
libretexts
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2025-03-17T19:53:20.986008
| 2015-11-01T04:30:42 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.03%3A_Accuracy_Precision_and_Significant_Figures
|
1.3: Accuracy, Precision, and Significant Figures
Learning Objectives
By the end of this section, you will be able to:
- Determine the appropriate number of significant figures in both addition and subtraction, as well as multiplication and division calculations.
- Calculate the percent uncertainty of a measurement.
Accuracy and Precision of a Measurement
Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard computer paper. The packaging in which you purchased the paper states that it is 11.0 inches long. You measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate.
The precision of a measurement system refers to how close the agreement is between repeated measurements (which are repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured values. One way to analyze the precision of the measurements would be to determine the range, or difference, between the lowest and the highest measured values. In that case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by at most 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had been 10.9, 11.1, and 11.9, then the measurements would not be very precise because there would be significant variation from one measurement to another.
The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider an example of a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target, and think of each GPS attempt to locate the restaurant as a black dot. In Figure \(\PageIndex{3}\), you can see that the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision, high accuracy measuring system. However, in Figure 4, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system.
Accuracy, Precision, and Uncertainty
The degree of accuracy and precision of a measuring system are related to the uncertainty in the measurements. Uncertainty is a quantitative measure of how much your measured values deviate from a standard or expected value. If your measurements are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11 in., plus or minus 0.2 in. The uncertainty in a measurement, A, is often denoted as δA (“delta A”), so the measurement result would be recorded as A ± δA. In our paper example, the length of the paper could be expressed as 11 in.± 0.2.
The factors contributing to uncertainty in a measurement include:
- Limitations of the measuring device,
- The skill of the person making the measurement,
- Irregularities in the object being measured,
- Any other factors that affect the outcome (highly dependent on the situation).
In our example, such factors contributing to the uncertainty could be the following: the smallest division on the ruler is 0.1 in., the person using the ruler has bad eyesight, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a measurement must be based on a careful consideration of all the factors that might contribute and their possible effects.
MAKING CONNECTIONS: REAL-WORLD CONNECTIONS – FEVERS OR CHILLS?
Uncertainty is a critical piece of information, both in physics and in many other real-world applications. Imagine you are caring for a sick child. You suspect the child has a fever, so you check his or her temperature with a thermometer. What if the uncertainty of the thermometer were 3.0ºC? If the child’s temperature reading was 37.0ºC (which is normal body temperature), the “true” temperature could be anywhere from a hypothermic 34.0ºC to a dangerously high 40.0ºC. A thermometer with an uncertainty of 3.0ºC would be useless.
Percent Uncertainty
One method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with uncertainty, \(δA\), the percent uncertainty (%uncertainty) is defined to be
\[\% \,\text{unc} =\dfrac {δA}{A} \times 100\%\]
Example \(\PageIndex{1}\): Calculating Percent Uncertainty: A Bag of Apples
A grocery store sells 5-lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the following measurements:
Week 1 weight:
4.8 lb
Week 2 weight:
5.3 lb
Week 3 weight:
4.9 lb
Week 4 weight:
5.4 lb
You determine that the weight of the 5-lb bag has an uncertainty of ±0.4lb. What is the percent uncertainty of the bag’s weight?
Strategy
First, observe that the expected value of the bag’s weight, \(A\), is 5 lb. The uncertainty in this value, \(δA\), is 0.4 lb. We can use the following equation to determine the percent uncertainty of the weight:
\(\text{% unc} =\frac{δA}{A}×100%\).
Solution
Plug the known values into the equation:
\(\text{% unc} =\frac{0.4 lb}{5 lb}×100%=8%\).
Discussion
We can conclude that the weight of the apple bag is \(5lb±8%\). Consider how this percent uncertainty would change if the bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when calculating percent uncertainty, always remember that you must multiply the fraction by 100%. If you do not do this, you will have a decimal quantity, not a percent value.
Uncertainties in Calculations
There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements going into the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used for multiplication or division. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation . For example, if a floor has a length of 4.00m and a width of 3.00m, with uncertainties of 2% and 1%, respectively, then the area of the floor is 12.0m2 and has an uncertainty of 3%. (Expressed as an area this is 0.36m2, which we round to \(0.4\,m^2\) since the area of the floor is given to a tenth of a square meter.)
Exercise \(\PageIndex{1}\)
A high school track coach has just purchased a new stopwatch. The stopwatch manual states that the stopwatch has an uncertainty of ±0.05s. Runners on the track coach’s team regularly clock 100-m sprints of 11.49 s to 15.01 s. At the school’s last track meet, the first-place sprinter came in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach’s new stopwatch be helpful in timing the sprint team? Why or why not?
- Answer
-
No, the uncertainty in the stopwatch is too great to effectively differentiate between the sprint times.
Precision of Measuring Tools and Significant Figures
An important factor in the accuracy and precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter, while a caliper can measure length to the nearest 0.01 millimeter. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise and accurate the measurements can be.
When we express measured values, we can only list as many digits as we initially measured with our measuring tool. For example, if you use a standard ruler to measure the length of a stick, you may measure it to be 36.7cm. You could not express this value as 36.71cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in between 36.6cm and 36.7cm, and he or she must estimate the value of the last digit. Using the method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. In order to determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value 36.7cm has three digits, or significant figures. Significant figures indicate the precision of a measuring tool that was used to measure a value.
Zeros
Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant, because they are only placekeepers that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placekeepers but are significant—this number has five significant figures. The zeros in 1300 may or may not be significant depending on the style of writing numbers. They could mean the number is known to the last digit, or they could be placekeepers. So 1300 could have two, three, or four significant figures. (To avoid this ambiguity, write 1300 in scientific notation.) Zeros are significant except when they serve only as placekeepers .
Exercise \(\PageIndex{2}\)
Determine the number of significant figures in the following measurements:
- 0.0009
- 15,450.0
- × 10 3
- 87.990
- 30.42
Solutions
- 1; the zeros in this number are placekeepers that indicate the decimal point
- 6; here, the zeros indicate that a measurement was made to the 0.1 decimal point, so the zeros are significant
- 1; the value 3 signifies the decimal place, not the number of measured values
- 5; the final zero indicates that a measurement was made to the 0.001 decimal point, so it is significant
- 4; any zeros located in between significant figures in a number are also significant
Significant Figures in Calculations
When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value . There are two different rules, one for multiplication and division and the other for addition and subtraction, as discussed below.
1. For multiplication and division: The result should have the same number of significant figures as the quantity having the least significant figures entering into the calculation. For example, the area of a circle can be calculated from its radius using A=πr2. Let us see how many significant figures the area has if the radius has only two—say, r=1.2m. Then,
\[A=πr2=(3.1415927...)×(1.2m)^2=4.5238934\,m^2\]
is what you would get using a calculator that has an eight-digit output. But because the radius has only two significant figures, it limits the calculated quantity to two significant figures or
\[A=4.5\,m^2,\]
even though \( is good to at least eight digits.
2. For addition and subtraction: The answer can contain no more decimal places than the least precise measurement. Suppose that you buy 7.56-kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg. Then you drop off 6.052-kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:
kg − 6.052\, kg + 13.7\, kg \, 15.208\, kg = 15.2\, kg .\]
Next, we identify the least precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg.
Significant Figures in this Text
In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. You will note that an answer given to three digits is based on input good to at least three digits, for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more accurate numbers are needed and more than three significant figures will be used. Finally, if a number is exact , such as the two in the formula for the circumference of a circle, \( = 2 π r ,\) it does not affect the number of significant figures in a calculation.
Exercise \(\PageIndex{3}\)
Perform the following calculations and express your answer using the correct number of significant digits.
- A woman has two bags weighing 13.5 pounds and one bag with a weight of 10.2 pounds. What is the total weight of the bags?
- The force \(F\) on an object is equal to its mass m multiplied by its acceleration \(a\). If a wagon with mass 55 kg accelerates at a rate of \(0.0255 m/s^2\), what is the force on the wagon? (The unit of force is called the newton, and it is expressed with the symbol N.)
- Answer
-
(a) 37.2 pounds; Because the number of bags is an exact value, it is not considered in the significant figures.
(b) 1.4 N; Because the value 55 kg has only two significant figures, the final value must also contain two significant figures.
PHET EXPLORATIONS: ESTIMATION
Explore size estimation in one, two, and three dimensions! Multiple levels of difficulty allow for progressive skill improvement.
Summary
- Accuracy of a measured value refers to how close a measurement is to the correct value. The uncertainty in a measurement is an estimate of the amount by which the measurement result may differ from this value.
- Precision of measured values refers to how close the agreement is between repeated measurements.
- The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement increment, the more precise the tool.
- Significant figures express the precision of a measuring tool.
- When multiplying or dividing measured values, the final answer can contain only as many significant figures as the least precise value.
- When adding or subtracting measured values, the final answer cannot contain more decimal places than the least precise value.
Glossary
- accuracy
- the degree to which a measured value agrees with correct value for that measurement
- method of adding percents
- the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation
- percent uncertainty
- the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage
- precision
- the degree to which repeated measurements agree with each other
- significant figures
- express the precision of a measuring tool used to measure a value
- uncertainty
- a quantitative measure of how much your measured values deviate from a standard or expected value
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:21.084788
| 2015-11-01T04:31:06 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.04%3A_Approximation
|
1.4: Approximation
Learning Objectives
By the end of this section, you will be able to:
- Make reasonable approximations based on given data.
On many occasions, physicists, other scientists, and engineers need to make approximations or “guesstimates” for a particular quantity. What is the distance to a certain destination? What is the approximate density of a given item? About how large a current will there be in a circuit? Many approximate numbers are based on formulae in which the input quantities are known only to a limited accuracy. As you develop problem-solving skills (that can be applied to a variety of fields through a study of physics), you will also develop skills at approximating. You will develop these skills through thinking more quantitatively, and by being willing to take risks. As with any endeavor, experience helps, as well as familiarity with units. These approximations allow us to rule out certain scenarios or unrealistic numbers. Approximations also allow us to challenge others and guide us in our approaches to our scientific world. Let us do two examples to illustrate this concept.
Example \(\PageIndex{1}\): Approximate the Height of a Building
Can you approximate the height of one of the buildings on your campus, or in your neighborhood? Let us make an approximation based upon the height of a person. In this example, we will calculate the height of a 39-story building.
Strategy
Think about the average height of an adult male. We can approximate the height of the building by scaling up from the height of a person.
Solution
Based on information in the example, we know there are 39 stories in the building. If we use the fact that the height of one story is approximately equal to about the length of two adult humans (each human is about 2-m tall), then we can estimate the total height of the building to be
Discussion
You can use known quantities to determine an approximate measurement of unknown quantities. If your hand measures 10 cm across, how many hand lengths equal the width of your desk? What other measurements can you approximate besides length?
Example \(\PageIndex{2}\): Approximating Vast Numbers - a Trillion Dollar
The U.S. federal debt in the 2008 fiscal year was a little less than $10 trillion. Most of us do not have any concept of how much even one trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100-bill stacks and used them to evenly cover a football field (between the end zones), make an approximation of how high the money pile would become. (We will use feet/inches rather than meters here because football fields are measured in yards.) One of your friends says 3 in., while another says 10 ft. What do you think?
Strategy
When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you might see in movies or at a bank. Since this is an easy-to-approximate quantity, let us start there. We can find the volume of a stack of 100 bills, find out how many stacks make up one trillion dollars, and then set this volume equal to the area of the football field multiplied by the unknown height.
Solution
(1) Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100 of these is about 0.5 in. thick. So the total volume of a stack of 100 bills is:
\[ \begin{align*} \text{volume of stack} &=length×width×height \\[5pt] \text{volume of stack}&=6 \,in.×3 \,in.×0.5 \,in., \\[5pt] \text{volume of stack}&=9 \,in.^3 \end{align*}\]
(2) Calculate the number of stacks. Note that a trillion dollars is equal to \(\$1×10^{12}\), and a stack of one-hundred $100 bills is equal to $10,000, or \(\$1×10^4\). The number of stacks you will have is:
(3) Calculate the area of a football field in square inches. The area of a football field is \(100\, yd×50\, yd\), which gives \(5,000 \,yd^2\). Because we are working in inches, we need to convert square yards to square inches:
\[ \begin{align*} Area &=5,000\, yd^2 × \frac{3\,ft}{1\,yd}×\frac{3\,ft}{1\, yd}×\frac{12\,in.}{1\, ft}×\frac{12\,in}{1 \,ft} \\[5pt] &=6,480,000\, in.^2, \\[5pt] &≈6×10^6\, in.^2. \end{align*}\]
This conversion gives us \(\displaystyle 6×10^6\,in^2\) for the area of the field. (Note that we are using only one significant figure in these calculations.)
(4) Calculate the total volume of the bills. The volume of all the $100 -bill stacks is
\[9\, in.^3 /stack \times 10^8\, \text{stacks} =9 \times 10^8 \,in.^3. \nonumber\]
(5) Calculate the height. To determine the height of the bills, use the equation:
\[ \begin{align*} \text{volume of bills} &=\text{area of field}×\text{height of money} \\[5pt] \text{Height of money} &=\frac{\text{volume of bills}}{\text{area of field}} \\[5pt] \text{Height of money} &=\frac{9×10^8in.^3}{6×10^6in.^2}=1.33×10^2in \\[5pt] \text{Height of money} &≈1 \times 10^2\,in.=100\, in. \end{align*}\]
The height of the money will be about 100 in. high. Converting this value to feet gives
Discussion
The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough “guesstimates” versus carefully calculated approximations?
Exercise \(\PageIndex{1}\)
Using mental math and your understanding of fundamental units, approximate the area of a regulation basketball court. Describe the process you used to arrive at your final approximation
- Answer
-
An average male is about two meters tall. It would take approximately 15 men laid out end to end to cover the length, and about 7 to cover the width. That gives an approximate area of \(\displaystyle 420 m^2\).
Summary
- Scientists often approximate the values of quantities to perform calculations and analyze systems.
Glossary
- approximation
- an estimated value based on prior experience and reasoning
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2025-03-17T19:53:21.153123
| 2015-11-01T04:31:27 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/01%3A_The_Nature_of_Science_and_Physics/1.E%3A_The_Nature_of_Science_and_Physics_(Exercises)
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1.E: The Nature of Science and Physics (Exercises)
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Conceptual Questions
1.1: Physics: An Introduction
1. Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by humans. What is a model?
2. How does a model differ from a theory?
3. If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)?
4. What determines the validity of a theory?
5. Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result?
6. Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a theory or a law?
7. Classical physics is a good approximation to modern physics under certain circumstances. What are they?
8. When is it necessary to use relativistic quantum mechanics?
9. Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not.
1.3: Accuracy, Precision, and Significant Figures
11. What is the relationship between the accuracy and uncertainty of a measurement?
12. Prescriptions for vision correction are given in units called diopters (D). Determine the meaning of that unit. Obtain information (perhaps by calling an optometrist or performing an internet search) on the minimum uncertainty with which corrections in diopters are determined and the accuracy with which corrective lenses can be produced. Discuss the sources of uncertainties in both the prescription and accuracy in the manufacture of lenses.
Problems & Exercises
1.2: Physical Quantities and Units
13.. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?
Solution:
a. \(27.8 m/s\)
b. \(62.1 mph\)
14. A car is traveling at a speed of \(33 m/s\)
(a) What is its speed in kilometers per hour?
(b) Is it exceeding the \(90 km/h\)
15. Show that \(1.0 m/s=3.6 km/h\)
[Hint: Show the explicit steps involved in converting \(1.0 m/s=3.6 km/h\)].
Solution
\(\frac{1.0 m}{s}=\frac{1.0 m}{s}×\frac{3600 s}{1 hr}×\frac{1 km}{1000 m}=3.6 km/h\)
16. American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
17. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
Solution
length: \(377 ft; 4.53×10^3 in: width: 280 ft; 3.3×10^3 in\)
18. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)
19. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
Solution
8.847 km
20. The speed of sound is measured to be 342 m/s on a certain day. What is this in km/h?
21. Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year.
(a) What distance does it move in 1 s at this speed?
(b) What is its speed in kilometers per million years?
Solution
(a) \(1.3×10^{−9} m\)
(b) \(40 km/My\)
22. (a) Refer to Table to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second.
(b) What is this in meters per second?
1.3: Accuracy, Precision, and Significant Figures
23. Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?
Solution
2 kg
24. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?
25. (a) A car speedometer has a \(5.0%\) uncertainty. What is the range of possible speeds when it reads \(90km/h\)?
(b) Convert this range to miles per hour. (\(1 km=0.6214 mi\))
Solution
a. 85.5to 94.5 km/h
b. 53.1to 58.7 mi/h
26. An infant’s pulse rate is measured to be \(130±5\) beats/min. What is the percent uncertainty in this measurement?
27. (a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?
Solution
(a) \(7.6×10^7\) beats
(b) \(7.57×10^7\) beats
(c) \(7.57×10^7\) beats
28. A can contains 375 mL of soda. How much is left after 308 mL is removed?
29. State how many significant figures are proper in the results of the following calculations:
(a) \((106.7)(98.2)/(46.210)(1.01)\)
(b) \((18.7)^2\)
(c) \((1.60×10^{−19})(3712)\)
Solution
a. 3
b. 3
c.3
30. (a) How many significant figures are in the numbers 99 and 100?
(b) If the uncertainty in each number is 1, what is the percent uncertainty in each?
(c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?
31. (a) If your speedometer has an uncertainty of \(2.0km/h\) at a speed of \(90km/h\). what is the percent uncertainty?
(b) If it has the same percent uncertainty when it reads \(60km/h\), what is the range of speeds you could be going?
Solution
(a) \(2.2%\)
(b) 59 to 61 km/h
32. (a) A person’s blood pressure is measured to be \(120±2\) mm Hg. What is its percent uncertainty?
(b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of \(80\) mm Hg?
33. A person measures his or her heart rate by counting the number of beats in \(30s\). If \(40±1\) beats are counted in \(30.0±0.5s\) , what is the heart rate and its uncertainty in beats per minute?
Solution
\(80±3\) beats/min
34. What is the area of a circle \(3.102cm\)in diameter?
35. If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?
Solution
\(2.8h\)
36. A marathon runner completes a \(42.188-km\) course in \(2h\), 30 min, and \(12s\). There is an uncertainty of \(25m\) in the distance traveled and an uncertainty of 1 s in the elapsed time.
(a) Calculate the percent uncertainty in the distance.
(b) Calculate the uncertainty in the elapsed time.
(c) What is the average speed in meters per second?
(d) What is the uncertainty in the average speed?
37. The sides of a small rectangular box are measured to be \(1.80±0.01cm, 2.05±0.02cm,\) and \(3.1±0.1 cm\) long. Calculate its volume and uncertainty in cubic centimeters.
Solution
\(11±1cm^3\)
38. When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where \(1lbm=0.4539kg\)
(a) If there is an uncertainty of \(0.0001kg\) in the pound-mass unit, what is its percent uncertainty?
(b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?
39. The length and width of a rectangular room are measured to be \(3.955±0.005m\) and \(3.050±0.005m\). Calculate the area of the room and its uncertainty in square meters.
Solution
\(12.06±0.04m^2\)
40. A car engine moves a piston with a circular cross section of \(7.500±0.002cm\) diameter a distance of \(3.250±0.001cm\) to compress the gas in the cylinder.
(a) By what amount is the gas decreased in volume in cubic centimeters?
(b) Find the uncertainty in this volume.
1.4: Approximation
41. How many heartbeats are there in a lifetime?
Solution
Sample answer: \(2×10^9\) heartbeats
42. A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?
43. How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of \(10^{−22}s\).)
Solution
Sample answer: \(2×10^{31}\) if an average human lifetime is taken to be about 70 years.
44. Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of \(10^{−27} kg\) and the mass of a bacterium is on the order of \(10^{−15} kg\).)
This color-enhanced photo shows Salmonella typhimurium (red) attacking human cells. These bacteria are commonly known for causing foodborne illness. Can you estimate the number of atoms in each bacterium? (credit: Rocky Mountain Laboratories, NIAID, NIH)
45. Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?
Solution
Sample answer: 50 atoms
46. (a) What fraction of Earth’s diameter is the greatest ocean depth?
(b) The greatest mountain height?
47. (a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium.
(b) Making the same assumption, how many cells are there in a human?
Solution
Sample answers:
(a) \(10^{12}\) cells/hummingbird
(b) \(10^{16}\) cells/human
48. Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?
Contributors and Attributions
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Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:21.242852
| 2018-05-04T02:45:16 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics
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2: Kinematics
Objects are in motion everywhere we look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves motion. When you are resting, your heart moves blood through your veins. And even in inanimate objects, there is continuous motion in the vibrations of atoms and molecules. Questions about motion are interesting in and of themselves: How long will it take for a space probe to get to Mars? Where will a football land if it is thrown at a certain angle? But an understanding of motion is also key to understanding other concepts in physics. An understanding of acceleration, for example, is crucial to the study of force. Kinematics is the branch of classical mechanics which describes the motion of points, bodies, and systems of bodies without consideration of the masses of those objects, nor the forces that may have caused the motion.
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- 2.0: Prelude to One-Dimensional Kinematics
- Our formal study of physics begins with kinematics which is defined as the study of motion without considering its causes. In one-dimensional kinematics and Two-Dimensional Kinematics we will study only the motion of a football, for example, without worrying about what forces cause or change its motion. Such considerations come in other chapters. In this chapter, we examine the simplest type of motion—namely, motion along a straight line, or one-dimensional motion.
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- 2.2: Vectors, Scalars, and Coordinate Systems
- A vector is any quantity that has magnitude and direction. A scalar is any quantity that has magnitude but no direction. Displacement and velocity are vectors, whereas distance and speed are scalars. In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like.
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- 2.3: Time, Velocity, and Speed
- There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we add definitions of time, velocity, and speed to expand our description of motion.
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- 2.4: Acceleration
- Acceleration is the rate at which velocity changes. In symbols, average acceleration is a= Δv/Δt. The SI unit for acceleration is m/s². Acceleration is a vector, and thus has a both a magnitude and direction. Acceleration can be caused by either a change in the magnitude or the direction of the velocity. Instantaneous acceleration a is the acceleration at a specific instant in time. Deceleration is an acceleration with a direction opposite to that of the velocity.
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- 2.5: Motion Equations for Constant Acceleration in One Dimension
- We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
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- 2.6: Problem-Solving Basics for One-Dimensional Kinematics
- The ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics.
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- 2.7: Falling Objects
- An object in free-fall experiences constant acceleration if air resistance is negligible. On Earth, all free-falling objects have an acceleration due to gravity g, which averages g=9.80 m/s2. Whether the acceleration a should be taken as +g or −g is determined by your choice of coordinate system. Since acceleration is constant, the kinematic equations above can be applied with the appropriate +g or −g substituted for a. For objects in free-fall, up is normally taken as positive.
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- 2.8: Graphical Analysis of One-Dimensional Motion
- Graphs of motion can be used to analyze motion. Graphical solutions yield identical solutions to mathematical methods for deriving motion equations. The slope of a graph of displacement x vs. time t is velocity v. The slope of a graph of velocity v vs. time t graph is acceleration a. Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
Thumbnail: Kinematics of a classical particle of mass \(m\), position \(r\), velocity \(v\), acceleration \(a\). (Public domain; Maschen ).
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libretexts
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2025-03-17T19:53:21.310229
| 2015-10-27T19:12:25 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.00%3A_Prelude_to_One-Dimensional_Kinematics
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2.0: Prelude to One-Dimensional Kinematics
Objects are in motion everywhere we look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves motion. When you are resting, your heart moves blood through your veins. And even in inanimate objects, there is continuous motion in the vibrations of atoms and molecules. Questions about motion are interesting in and of themselves: How long will it take for a space probe to get to Mars? Where will a football land if it is thrown at a certain angle? But an understanding of motion is also key to understanding other concepts in physics. An understanding of acceleration, for example, is crucial to the study of force.
Our formal study of physics begins with kinematics which is defined as the study of motion without considering its causes . The word “kinematics” comes from a Greek term meaning motion and is related to other English words such as “cinema” (movies) and “kinesiology” (the study of human motion). In one-dimensional kinematics and Two-Dimensional Kinematics we will study only the motion of a football, for example, without worrying about what forces cause or change its motion. Such considerations come in other chapters. In this chapter, we examine the simplest type of motion—namely, motion along a straight line, or one-dimensional motion. In Two-Dimensional Kinematics , we apply concepts developed here to study motion along curved paths (two- and three-dimensional motion); for example, that of a car rounding a curve.
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libretexts
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2025-03-17T19:53:21.367437
| 2015-11-01T04:32:06 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.01%3A_Displacement
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2.1: Displacement
Learning Objectives
- Define position, displacement, distance, and distance traveled.
- Explain the relationship between position and displacement.
- Distinguish between displacement and distance traveled.
- Calculate displacement and distance given initial position, final position, and the path between the two.
These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change in position, or displacement, in the frame of reference.
Position
In order to describe the motion of an object, you must first be able to describe its position—where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor’s position could be described in terms of where she is in relation to the nearby white board. In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame.
Displacement
If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object’s position changes. This change in position is known as displacement. The word “displacement” implies that an object has moved, or has been displaced.
Definition: Displacement
Displacement is the change in position of an object:
\(Δx=x_f−x_0,\)
where \(Δx\) is displacement, \(x_f\) is the final position, and \(x_0\) is the initial position.
In this text the upper case Greek letter size \(Δ\)(delta) always means “change in” whatever quantity follows it; thus, size \(Δx\) means change in position. Always solve for displacement by subtracting initial position size \(x_0\) from final position \(x_f\).
Note that the SI unit for displacement is the meter (\(m\)) (see Section on Physical Quantities and Units ), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.
Note that displacement has a direction as well as a magnitude. The professor’s displacement in Figure \( \PageIndex{2}\) is 2.0 m to the right, and the airline passenger’s displacement is 4.0 m toward the rear in Figure \( \PageIndex{3}\). In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor’s initial position is size x 0 = 1.5 m and her final position is x f = 3.5 m. Thus her displacement is
\[\begin{align*} Δx &=x_f−x_0 \\[5pt] &=3.5\, m−1.5\, m \\[5pt] &=+2.0\, m. \end{align*}\]
In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger’s initial position is \(\displaystyle x_0=6.0 m\) and his final position is \(\displaystyle x_f=2.0 m\), so his displacement is
\[\begin{align*} Δx&=x_f−x_0 \\[5pt] &= 2.0\, m−6.0\, m \\[5pt] &=−4.0\, m.\end{align*}\]
His displacement is negative because his motion is toward the rear of the plane, or in the negative size 12{x} {} direction in our coordinate system.
Distance
Although displacement is described in terms of direction, distance is not. Distance is defined to be the magnitude or size of displacement between two positions . Note that the distance between two positions is not the same as the distance traveled between them. Distance traveled is the total length of the path traveled between two positions. Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.
Misconception Alert: Distance Traveled vs. Magnitude of Displacement
It is important to note that the distance traveled, however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks.
Exercise \(\PageIndex{1}\)
A cyclist rides 3 km west and then turns around and rides 2 km east.
- What is her displacement?
- What distance does she ride?
- What is the magnitude of her displacement?
- Answer
- Answer a
-
The rider’s displacement is \(\displaystyle Δ_x=x_f−x_0=−1 km\). (The displacement is negative because we take east to be positive and west to be negative.)
- Answer b
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The distance traveled is \(\displaystyle 3 km+2 km=5 km\).
- Answer c
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The magnitude of the displacement is \(\displaystyle 1 km\).
Summary
- Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.
- Displacement is the change in position of an object.
- In symbols, displacement \(\displaystyle Δx\) is defined to be
\(\displaystyle Δx=x_f−x_0\),
where \(\displaystyle x_0\) is the initial position and xf is the final position. In this text, the Greek letter \(\displaystyle Δ\) (delta) always means “change in” whatever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude.
- When you start a problem, assign which direction will be positive.
- Distance is the magnitude of displacement between two positions.
- Distance traveled is the total length of the path traveled between two positions.
Glossary
- kinematics
- the study of motion without considering its causes
- position
- the location of an object at a particular time
- displacement
- the change in position of an object
- distance
- the magnitude of displacement between two positions
- distance traveled
- the total length of the path traveled between two positions
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libretexts
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2025-03-17T19:53:21.444097
| 2015-11-01T04:32:34 |
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"title": "2.1: Displacement",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.02%3A_Vectors_Scalars_and_Coordinate_Systems
|
2.2: Vectors, Scalars, and Coordinate Systems
Learning Objectives
By the end of this section, you will be able to:
- Define and distinguish between scalar and vector quantities.
- Assign a coordinate system for a scenario involving one-dimensional motion.
What is the difference between distance and displacement? Whereas displacement is defined by both direction and magnitude, distance is defined only by magnitude. Displacement is an example of a vector quantity. Distance is an example of a scalar quantity. A vector is any quantity with both magnitude and direction . Other examples of vectors include a velocity of 90 km/h east and a force of 500 newtons straight down.
The direction of a vector in one-dimensional motion is given simply by a plus (+) or minus (−) sign. Vectors are represented graphically by arrows. An arrow used to represent a vector has a length proportional to the vector’s magnitude (e.g., the larger the magnitude, the longer the length of the vector) and points in the same direction as the vector.
Some physical quantities, like distance, either have no direction or none is specified. A scalar is any quantity that has a magnitude, but no direction. For example, a 20ºC temperature, the 250 kilocalories (250 Calories) of energy in a candy bar, a 90 km/h speed limit, a person’s 1.8 m height, and a distance of 2.0 m are all scalars—quantities with no specified direction. Note, however, that a scalar can be negative, such as a −20ºC temperature. In this case, the minus sign indicates a point on a scale rather than a direction. Scalars are never represented by arrows.
Coordinate Systems for One-Dimensional Motion
In order to describe the direction of a vector quantity, you must designate a coordinate system within the reference frame. For one-dimensional motion, this is a simple coordinate system consisting of a one-dimensional coordinate line. In general, when describing horizontal motion, motion to the right is usually considered positive, and motion to the left is considered negative. With vertical motion, motion up is usually positive and motion down is negative. In some cases, however, as with the jet in Figure \PageIndex{2}\), it can be more convenient to switch the positive and negative directions. For example, if you are analyzing the motion of falling objects, it can be useful to define downwards as the positive direction. If people in a race are running to the left, it is useful to define left as the positive direction. It does not matter as long as the system is clear and consistent. Once you assign a positive direction and start solving a problem, you cannot change it.
Exercise \(\PageIndex{1}\)
A person’s speed can stay the same as he or she rounds a corner and changes direction. Given this information, is speed a scalar or a vector quantity? Explain.
- Answer
-
Speed is a scalar quantity. It does not change at all with direction changes; therefore, it has magnitude only. If it were a vector quantity, it would change as direction changes (even if its magnitude remained constant).
Summary
- A vector is any quantity that has magnitude and direction.
- A scalar is any quantity that has magnitude but no direction.
- Displacement and velocity are vectors, whereas distance and speed are scalars.
- In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like.
Glossary
- scalar
- a quantity that is described by magnitude, but not direction
- vector
- a quantity that is described by both magnitude and direction
|
libretexts
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2025-03-17T19:53:21.508948
| 2015-11-01T04:35:03 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.02%3A_Vectors_Scalars_and_Coordinate_Systems",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.2: Vectors, Scalars, and Coordinate Systems",
"author": "OpenStax"
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.03%3A_Time_Velocity_and_Speed
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2.3: Time, Velocity, and Speed
Learning Objectives
By the end of this section, you will be able to:
- Explain the relationships between instantaneous velocity, average velocity, instantaneous speed, average speed, displacement, and time.
- Calculate velocity and speed given initial position, initial time, final position, and final time.
- Derive a graph of velocity vs. time given a graph of position vs. time.
- Interpret a graph of velocity vs. time.
There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we add definitions of time, velocity, and speed to expand our description of motion.
Time
As discussed in Physical Quantities and Units , the most fundamental physical quantities are defined by how they are measured. This is the case with time. Every measurement of time involves measuring a change in some physical quantity. It may be a number on a digital clock, a heartbeat, or the position of the Sun in the sky. In physics, the definition of time is simple— time is change , or the interval over which change occurs. It is impossible to know that time has passed unless something changes.
The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We might, for example, observe that a certain pendulum makes one full swing every 0.75 s. We could then use the pendulum to measure time by counting its swings or, of course, by connecting the pendulum to a clock mechanism that registers time on a dial. This allows us to not only measure the amount of time, but also to determine a sequence of events.
How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an airplane passenger to get from his seat to the back of the plane. To find elapsed time, we note the time at the beginning and end of the motion and subtract the two. For example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed time would be 50 min. Elapsed time \(Δt\) is the difference between the ending time and beginning time,
\(Δt=t_f−t_0\),
where \(Δt\) is the change in time or elapsed time, \(t_f\) is the time at the end of the motion, and \(t_0\) is the time at the beginning of the motion. (As usual, the delta symbol, \(Δ\), means the change in the quantity that follows it.)
Life is simpler if the beginning time \(t_0\) is taken to be zero, as when we use a stopwatch. If we were using a stopwatch, it would simply read zero at the start of the lecture and 50 min at the end. If \(t_0=0\), then
\[Δt=t_f≡t.\]
In this text, for simplicity’s sake,
- motion starts at time equal to zero (\(t_0=0\))
- the symbol \(t\) is used for elapsed time unless otherwise specified (\(Δt=t_f≡t\))
Velocity
Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small amount of time you have a large velocity, and that velocity has units of distance divided by time, such as miles per hour or kilometers per hour.
Definition: AVERAGE VELOCITY
Average velocity is displacement (change in position) divided by the time of travel,
\[\bar{v}=\frac{Δx}{Δt}=\frac{x_f−x_0}{t_f−t_0}.\]
where \(\bar{v}\) is the average (indicated by the bar over the \(v\)) velocity, \(Δx\) is the change in position (or displacement), and \(x_f\) and \(x_0\) are the final and beginning positions at times \(t_f\) and \(t_0\), respectively. If the starting time \(t_0\) is taken to be zero, then the average velocity is simply
\[\bar{v}=\frac{Δx}{t}.\]
Notice that this definition indicates that velocity is a vector because displacement is a vector . It has both magnitude and direction. The SI unit for velocity is meters per second or m/s, but many other units, such as km/h, mi/h (also written as mph), and cm/s, are in common use. Suppose, for example, an airplane passenger took 5 seconds to move −4 m (the negative sign indicates that displacement is toward the back of the plane). His average velocity would be
The minus sign indicates the average velocity is also toward the rear of the plane.
The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals.
The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical conclusion, we are left with an infinitesimally small interval. Over such an interval, the average velocity becomes the instantaneous velocity or the velocity at a specific instant . A car’s speedometer, for example, shows the magnitude (but not the direction) of the instantaneous velocity of the car. (Police give tickets based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to use average velocity.) Instantaneous velocity \(v\) is the average velocity at a specific instant in time (or over an infinitesimally small time interval).
Mathematically, finding instantaneous velocity, \(v\), at a precise instant \(t\) can involve taking a limit, a calculus operation beyond the scope of this text. However, under many circumstances, we can find precise values for instantaneous velocity without calculus.
Speed
In everyday language, most people use the terms “speed” and “velocity” interchangeably. In physics, however, they do not have the same meaning and they are distinct concepts. One major difference is that speed has no direction. Thus speed is a scalar . Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and average speed.
Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant had an instantaneous velocity of −3.0 m/s (the minus meaning toward the rear of the plane). At that same time his instantaneous speed was 3.0 m/s. Or suppose that at one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your instantaneous speed at that instant would be 40 km/h—the same magnitude but without a direction. Average speed, however, is very different from average velocity. Average speed is the distance traveled divided by elapsed time.
We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity, which is displacement divided by time. For example, if you drive to a store and return home in half an hour, and your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero, because your displacement for the round trip is zero. (Displacement is change in position and, thus, is zero for a round trip.) Thus average speed is not simply the magnitude of average velocity.
Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be very useful. For example, for this trip to the store, the position, velocity, and speed-vs.-time graphs are displayed in Figure \(\PageIndex{4}\). (Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip, which is unrealistic given that we’ll probably stop at the store. But for simplicity’s sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store and the house is a perfectly straight line.)
MAKING CONNECTIONS: TAKE-HOME INVESTIGATION - GETTING A SENSE OF SPEED
If you have spent much time driving, you probably have a good sense of speeds between about 10 and 70 miles per hour. But what are these in meters per second? What do we mean when we say that something is moving at 10 m/s? To get a better sense of what these values really mean, do some observations and calculations on your own:
- calculate typical car speeds in meters per second
- estimate jogging and walking speed by timing yourself; convert the measurements into both m/s and mi/h
- determine the speed of an ant, snail, or falling leaf
Exercise \(\PageIndex{1}\)
A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is approximately 40 miles. What is
- the average velocity of the train, and
- the average speed of the train in m/s?
- Answer
-
(a) The average velocity of the train is zero because \(x_f=x_0\); the train ends up at the same place it starts.
(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 miles.
\(\frac{distance}{time}=\frac{80 miles}{105 minutes}\)
Summary
- Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is \[Δt=t_f−t_0 \nonumber,\] where \(t_f\) is the final time and \(t_0\) is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the elapsed time is then just \(t\).
- Average velocity \(\bar{v}\) is defined as displacement divided by the travel time. In symbols, average velocity is \[\bar{v}=\frac{Δx}{Δt}=\frac{x_f−x_0}{t_f−t_0} \nonumber.\]
- The SI unit for velocity is m/s.
- Velocity is a vector and thus has a direction.
- Instantaneous velocity \(v\) is the velocity at a specific instant or the average velocity for an infinitesimal interval.
- Instantaneous speed is the magnitude of the instantaneous velocity.
- Instantaneous speed is a scalar quantity, as it has no direction specified.
- Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is a scalar quantity; it has no direction associated with it.
Glossary
- average speed
- distance traveled divided by time during which motion occurs
- average velocity
- displacement divided by time over which displacement occurs
- instantaneous velocity
- velocity at a specific instant, or the average velocity over an infinitesimal time interval
- instantaneous speed
- magnitude of the instantaneous velocity
- time
- change, or the interval over which change occurs
- model
- simplified description that contains only those elements necessary to describe the physics of a physical situation
- elapsed time
- the difference between the ending time and beginning time
|
libretexts
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2025-03-17T19:53:21.590376
| 2015-11-01T04:36:46 |
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"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.3: Time, Velocity, and Speed",
"author": "OpenStax"
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.04%3A_Acceleration
|
2.4: Acceleration
Learning Objectives
By the end of this section, you will be able to:
- Define and distinguish between instantaneous acceleration, average acceleration, and deceleration.
- Calculate acceleration given initial time, initial velocity, final time, and final velocity.
In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration , the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.
Definition: Average Acceleration
The average acceleration is the rate at which velocity changes,
\[\bar{a}=\frac{Δv}{Δt}=\frac{v_f−v_0}{t_f−t_0}\]
where \(\bar{a}\) is average acceleration, \(v\) is velocity, and \( t\) is time. (The bar over the \(a\) means average .)
Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are \( m/s^2\), meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second.
Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.
ACCELERATION AS A VECTOR
Acceleration is a vector in the same direction as the change in velocity, \(Δv\). Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed or direction, or both.
Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion . When an object slows down, its acceleration is opposite to the direction of its motion. This is known as deceleration .
MISCONCEPTION ALERT: DECELERATION VS. NEGATIVE ACCELERATION
Deceleration always refers to acceleration in the direction opposite to the direction of the velocity. Deceleration always reduces speed. Negative acceleration, however, is acceleration in the negative direction in the chosen coordinate system. Negative acceleration may or may not be deceleration, and deceleration may or may not be considered negative acceleration. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down. For example, consider Figure \(\PageIndex{2}\).
Example \(\PageIndex{1}\): Calculating Acceleration: A Racehorse Leaves the Gate
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?
Strategy
First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.
We can solve this problem by identifying \(Δv\) and \(Δt\) from the given information and then calculating the average acceleration directly from the Equation \ref{averagea}:
\[\bar{a}=\dfrac{Δv}{Δt}=\dfrac{v_f−v_0}{t_f−t_0}. \nonumber\]
Solution
- Identify the knowns. \(v_0=0 , v_f=−15.0 m/s\) (the negative sign indicates direction toward the west), \( Δt=1.80 s\).
- Find the change in velocity. Since the horse is going from zero to \( −15.0 m/s\), its change in velocity equals its final velocity: \[Δv=v_f=−15.0\, m/s .\nonumber\]
- Plug in the known values ( \(Δv\) and \(Δt\) ) and solve for the unknown \(\bar{a}\).
\(\bar{a}=\frac{Δv}{Δt}=\frac{−15.0 m/s}{1.80 s}=−8.33 m/s^2\).
Discussion
The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of \(8.33\, m/s^2\) due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as \( 8.33\, m/s^2\). This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.
Instantaneous Acceleration
Instantaneous acceleration \( a\), or the acceleration at a specific instant in time, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed —that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure \(\PageIndex{6}\) shows graphs of instantaneous acceleration versus time for two very different motions. In Figure \(\PageIndex{6a}\), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about \(1.8 m/s^2\)). In Figure \(\PageIndex{6b}\), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of \( +3.0\, m/s^2\) and \( –2.0 \,m/s^2\), respectively.
The next several examples consider the motion of the subway train shown in Figure \(\PageIndex{7}\). In Figure \(\PageIndex{7a}\) the shuttle moves to the right, and in Figure \(\PageIndex{7b}\) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.
Example \(\PageIndex{2}\): Calculating Displacement - A Subway Train
What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure \(\PageIndex{7}\)?
Strategy
A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation \( Δx=x_f−x_0\). This is straightforward since the initial and final positions are given.
Solution
- Identify the knowns. In the figure we see that \(x_f=6.70\, km \) and \(x_0=4.70\, km\) for part (a), and \(x'_f=3.75\, km \) and \(x'_0=5.25\, km \) for part (b).
- Solve for displacement in part (a).\[\begin{align*} Δx &=x_f−x_0 \\[5pt] &=6.70\, km−4.70\, km \\[5pt] &=+2.00 \,km \end{align*}\]
- Solve for displacement in part (b).\[\begin{align*} Δx' &=x'_f−x'_0 \\[5pt] &=3.75\, km−5.25\, km \\[5pt] &=−1.50\, km \end{align*}\]
Discussion
The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.
Example \(\PageIndex{3}\): Comparing Distance Traveled with Displacement - A Subway Train
What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure \(\PageIndex{7}\) ?Strategy
To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example \(\PageIndex{2}\). Distance traveled is the total length of the path traveled between the two positions (see Section on Displacement ). In the case of the subway train shown in Figure \(\PageIndex{7}\), the distance traveled is the same as the distance between the initial and final positions of the train.
Solution
1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.
2. The displacement for part (b) was \( −1.5 km\). Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.
Discussion
Distance is a scalar. It has magnitude but no sign to indicate direction.
Example \(\PageIndex{4}\): Calculating Acceleration: A Subway Train Speeding Up
Suppose the train in Figure \(\PageIndex{7a}\) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?
Strategy
It is worth it at this point to make a simple sketch:
Solution
- Identify the knowns. \( v_0=0\) (the trains starts at rest), \( v_f=30.0 km/h \), and \(Δt=20.0 s \).
- Calculate \(Δv\). Since the train starts from rest, its change in velocity is \(Δv=+30.0 km/h \), where the plus sign means velocity to the right.
- Plug in known values and solve for the unknown, \(\bar{ a}\). \[\bar{a}=\dfrac{Δv}{Δt}=\dfrac{+30.0 km/h}{20.0 s}\nonumber\]
- Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.) \[\bar{a}=\left(\dfrac{+30 km/h}{20.0 s}\right)\left(\dfrac{10^3m}{1 km}\right)\left(\dfrac{1h}{3600 s}\right)=0.417 m/s^2 \nonumber\]
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.
Example \(\PageIndex{5}\): Calculate Acceleration
A Subway Train Slowing Down: Now suppose that at the end of its trip, the train in Figure \(\PageIndex{7a}\) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?
Strategy
In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.
Solution
- Identify the knowns. \(v_0=30.0 km/h , v_f=0 km/h\) (the train is stopped, so its velocity is 0), and \(Δt=8.00 s \).
- Solve for the change in velocity, \(Δv\). \[Δv=v_f−v_0=0−30.0 km/h=−30.0 km/h \nonumber\]
- Plug in the knowns, \( Δv\) and \(Δt\), and solve for \(\bar{a}\). \[\bar{a}=\dfrac{Δv}{Δt}=\dfrac{−30.0 km/h}{8.00 s} \nonumber\]
- Convert the units to meters and seconds. \[\bar{a}=\dfrac{Δv}{Δt}=\left(\dfrac{−30.0 km/h}{8.00 s}\right)\left(\dfrac{10^3m}{1 km}\right)\left(\dfrac{1 h}{3600 s}\right)=−1.04 m/s^2. \nonumber\]
Discussion
The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.
The graphs of position, velocity, and acceleration vs. time for the trains in Example \(\PageIndex{4}\) and \(\PageIndex{5}\) are displayed in Figure \(\PageIndex{10}\). (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)
Example \(\PageIndex{6}\): Calculating Average Velocity: The Subway Train
What is the average velocity of the train in part b of Example \(\PageIndex{2}\), and shown again below, if it takes 5.00 min to make its trip?
Strategy
Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.
Solution
- Identify the knowns. \[ x'_f=3.75 km,\, x'_0=5.25 km,\, Δt=5.00 min. \nonumber\]
- Determine displacement, \(Δx'\). We found \( Δx'\) to be \( −1.5 km\) in Example \(\PageIndex{7}\).
- Solve for average velocity. \[\bar{v}=\dfrac{Δx'}{Δt}=\dfrac{−1.50 km}{5.00 min} \nonumber\]
- Convert units. \[\bar{v}=\dfrac{Δx'}{Δt}=(\dfrac{−1.50 km}{5.00 min})(\dfrac{60 min}{1 }h)=−18.0 km/h \nonumber\]
Discussion
The negative velocity indicates motion to the left.
Example \(\PageIndex{7}\): Calculating Deceleration: The Subway Train
Finally, suppose the train in Figure \(\PageIndex{7}\) slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?
Strategy
Once again, let’s draw a sketch:
Solution
- Identify the knowns. \(v_0=−20\, km/h\), \(v_f=0\, km/h\), \(Δt=10.0\, s\).
- Calculate \(Δv\). The change in velocity here is actually positive, since \[Δv=v_f−v_0=0−(−20\, km/h)=+20\, km/h. \nonumber\]
- Solve for \(\bar{ a}\). \[\bar{a}=\dfrac{Δv}{Δt}=\dfrac{+20.0\, km/h}{10.0\, s} \nonumber\]
- Convert units. \[\bar{a}=\left(\dfrac{+20.0 \,km/h}{10.0\, s}\right)\left(\dfrac{10^3\,m}{1\, km}\right)\left(\dfrac{1\, h}{3600 \,s}\right)=+0.556\, m/^s2 \nonumber\]
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the change in velocity, which is positive here. As in Example \(\PageIndex{5}\), this acceleration can be called a deceleration since it is in the direction opposite to the velocity.
Sign and Direction
Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in Example \(\PageIndex{5}\), where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure Figure \(\PageIndex{11}\) is sped up by an acceleration to the left. In that case, both v and a are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down.
Exercise \(\PageIndex{1}\)
An airplane lands on a runway traveling east. Describe its acceleration.
- Answer
-
If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity.
PHET EXPLORATIONS: MOVING MAN SIMULATION
Learn about position, velocity, and acceleration graphs with the PhET Moving Man simulation . Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you.
Summary
- Acceleration is the rate at which velocity changes. In symbols, average acceleration \(\bar{ a}\) is \(\bar{a}=\dfrac{Δv}{Δt}=\dfrac{v_f−v_0}{t_f−t_0}\).
- The SI unit for acceleration is 2 .
- Acceleration is a vector, and thus has a both a magnitude and direction.
- Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
- Instantaneous acceleration is the acceleration at a specific instant in time.
- Deceleration is an acceleration with a direction opposite to that of the velocity.
Glossary
- acceleration
- the rate of change in velocity; the change in velocity over time
- average acceleration
- the change in velocity divided by the time over which it changes
- instantaneous acceleration
- acceleration at a specific point in time
- deceleration
- acceleration in the direction opposite to velocity; acceleration that results in a decrease in velocity
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libretexts
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2025-03-17T19:53:21.687474
| 2015-11-01T04:37:07 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.04%3A_Acceleration",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.4: Acceleration",
"author": "OpenStax"
}
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.05%3A_Motion_Equations_for_Constant_Acceleration_in_One_Dimension
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2.5: Motion Equations for Constant Acceleration in One Dimension
Learning Objectives
By the end of this section, you will be able to:
- Calculate displacement of an object that is not accelerating, given initial position and velocity.
- Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
- Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.
We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
Notation: t , x , v , a
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is \(\displaystyle Δt=t_f−t_0\), taking \(\displaystyle t_0=0\) means that \(\displaystyle Δt=t_f\), the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, \(\displaystyle x_0\) is the initial position and \(\displaystyle v_0\) is the initial velocity . We put no subscripts on the final values. That is, \(\displaystyle t\) is the final time , \(\displaystyle x\) is the final position , and \(\displaystyle v\) is the final velocity . This gives a simpler expression for elapsed time—now, \(\displaystyle Δt=t\). It also simplifies the expression for displacement, which is now \(\displaystyle Δx=x−x_0\). Also, it simplifies the expression for change in velocity, which is now \(\displaystyle Δv=v−v_0\). To summarize, using the simplified notation, with the initial time taken to be zero,
\(\displaystyle Δt=t\)
\(\displaystyle Δx=x−x_0\)
\(\displaystyle Δv=v−v_0\)
where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.
We now make the important assumption that acceleration is constant . This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,
\[\bar{a}=a=constant,\]
so we use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.
SOLVING FOR DISPLACEMENT (Δx) AND FINAL POSITION (x) FROM AVERAGE VELOCITY WHEN ACCELERATION (a) IS CONSTANT
To get our first two new equations, we start with the definition of average velocity:
\[\displaystyle \bar{v}=\frac{Δx}{Δt}\]
Substituting the simplified notation for \(\displaystyle Δx\) and \(\displaystyle Δt\) yields
Solving for \(\displaystyle x\) yields
where the average velocity is
\[\displaystyle \bar{v}=\frac{v_0+v}{2} \label{eq5}\]
with constant \(a\).
Equation \ref{eq5} reflects the fact that, when acceleration is constant, \(v\) is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) to check this, we see that
\(\displaystyle \bar{v}=\frac{v_0+v}{2}=\frac{30 km/h+60 km/h}{2}=45 km/h,\)
which seems logical.
Example \(\PageIndex{1}\): Calculating Displacement - How Far does the Jogger Run?
A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?
Strategy
Draw a sketch.
The final position is given by the equation
\(\displaystyle x=x_0+\bar{v}t\).
To find \(\displaystyle x\), we identify the values of \(\displaystyle x_0, \bar{v}\), and \(\displaystyle t\) from the statement of the problem and substitute them into the equation.
Solution
- Identify the knowns. \(\displaystyle \bar{v}=4.00 m/s, Δt=2.00 min\), and \(\displaystyle x_0=0 m\).
- Enter the known values into the equation.
Discussion
Velocity and final displacement are both positive, which means they are in the same direction.
The equation \(\displaystyle x=x_0+\bar{v}t\) gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on \(\displaystyle \bar{v}\) rather than on \(\displaystyle \bar{v}\) raised to some other power, such as \(\displaystyle \bar{v}^2\). When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.
SOLVING FOR FINAL VELOCITY
We can derive another useful equation by manipulating the definition of acceleration.
\(\displaystyle a=\frac{Δv}{Δt}\)
Substituting the simplified notation for \(\displaystyle Δv\) and \(\displaystyle Δt\) gives us
\[\displaystyle a=\frac{v−v_0}{t}\] (constant a).
Solving for \(\displaystyle v\) yields
\[\displaystyle v=v_0+at\](constanta).
Example \(\PageIndex{2}\):Calculating Final Velocity: An Airplane Slowing Down after Landing
An airplane lands with an initial velocity of 70.0 m/s and then decelerates at \(\displaystyle 1.50 m/s^2\) for 40.0 s. What is its final velocity?
Strategy
Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.
Solution
1. Identify the knowns. \(\displaystyle v_0=70.0 m/s, a=−1.50 m/s^2, t=40.0s.\)
2. Identify the unknown. In this case, it is final velocity, \(\displaystyle v_f\).
3. Determine which equation to use. We can calculate the final velocity using the equation \(\displaystyle v=v_0+at\).
4. Plug in the known values and solve.
\(\displaystyle v=v_0+at=70.0 m/s+(−1.50 m/s^2)(40.0 s)=10.0 m/s\)
Discussion
The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.
In addition to being useful in problem solving, the equation \(\displaystyle v=v_0+at\) gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that
- final velocity depends on how large the acceleration is and how long it lasts
- if the acceleration is zero, then the final velocity equals the initial velocity (\(\displaystyle v=v_0\)), as expected (i.e., velocity is constant)
- if a is negative, then the final velocity is less than the initial velocity
(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)
MAKING CONNECTIONS: REAL-WORLD CONNECTION
An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.
SOLVING FOR FINAL POSITION WHEN VELOCITY IS NOT CONSTANT (a≠0)
We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with
\(\displaystyle v=v_0+at.\)
Adding \(\displaystyle v_0\) to each side of this equation and dividing by 2 gives
\(\displaystyle \frac{v_0+v}{2}=v_0+\frac{1}{2}at\).
Since \\frac{(v_0+v}{2}=\bar{v}\) for constant acceleration, then
\(\displaystyle \bar{v}=v_0+\frac{1}{2}at\).
Now we substitute this expression for \(\displaystyle \bar{v}\) into the equation for displacement, \(\displaystyle x=x_0+\bar{v}t\), yielding
\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)(constant a).
Example \(\PageIndex{3}\): Calculating Displacement of an Accelerating Object - Dragsters
Dragsters can achieve average accelerations of \(\displaystyle 26.0 m/s^2\). Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?
Strategy
Draw a sketch.
We are asked to find displacement, which is \(\displaystyle x\) if we take \(\displaystyle x_0\) to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) once we identify \(\displaystyle v_0, a,\) and \(\displaystyle t\) from the statement of the problem.
Solution
1. Identify the knowns. Starting from rest means that \(\displaystyle v_0=0, a\) is given as \(\displaystyle 26.0m/s^2\) and t is given as 5.56 s.
2. Plug the known values into the equation to solve for the unknown x:
\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\).
Since the initial position and velocity are both zero, this simplifies to
\(\displaystyle x=\frac{1}{2}at^2\).
Substituting the identified values of a and t gives
\(\displaystyle x=\frac{1}{2}(26.0 m/s^2)(5.56 s)^2\),
yielding
\(\displaystyle x=402 m.\)
Discussion
If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.
What else can we learn by examining the equation \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)? We see that:
- displacement depends on the square of the elapsed time when acceleration is not zero. In Example, the dragster covers only one fourth of the total distance in the first half of the elapsed time
- if acceleration is zero, then the initial velocity equals average velocity (\(\displaystyle v_0=\bar{v}\)) and \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) becomes \(\displaystyle x=x_0+v_0t\)
SOLVING FOR FINAL VELOCITY WHEN VELOCITY IS NOT CONSTANT (a≠0)
A fourth useful equation can be obtained from another algebraic manipulation of previous equations.
If we solve \(\displaystyle v=v_0+at\) for \(\displaystyle t\), we get
\(\displaystyle t=\frac{v−v_0}{a}\).
Substituting this and \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) into \(\displaystyle x=x_0+\bar{v}t\), we get
\(\displaystyle v^2=v^2_0+2a(x−x_0)\)(constant a).
Example \(\PageIndex{4}\): Calculating Final Velocity: Dragsters
Calculate the final velocity of the dragster in Example \(\PageIndex{3}\) without using information about time.
Strategy
Draw a sketch.
The equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.
Solution
1. Identify the known values. We know that \(\displaystyle v_0=0\), since the dragster starts from rest. Then we note that \(\displaystyle x−x_0=402 m\) (this was the answer in Example). Finally, the average acceleration was given to be \(\displaystyle a=26.0 m/s^2\).
2. Plug the knowns into the equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) and solve for \(\displaystyle v\).
\(\displaystyle v^2=0+2(26.0 m/s^2)(402 m).\)
Thus
\(\displaystyle v^2=2.09×10^4m^2/s^2.\)
To get \(\displaystyle v\), we take the square root:
\(\displaystyle v=\sqrt{2.09×10^4m^2/s^2}=145 m/s\).
Discussion
145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.
An examination of the equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) can produce further insights into the general relationships among physical quantities:
- The final velocity depends on how large the acceleration is and the distance over which it acts
- For a fixed deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)
Putting Equations Together
In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.
SUMMARY OF KINEMATIC EQUATIONS (CONSTANT a)
\(\displaystyle x=x_0+\bar{v}t\)
\(\displaystyle \bar{v}=\frac{v_0+v}{2}\)
\(\displaystyle v=v_0+at\)
\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)
\(\displaystyle v^2=v^2_0+2a(x−x_0)\)
Example \(\PageIndex{5}\):Calculating Displacement: How Far Does a Car Go When Coming to a Halt?
On dry concrete, a car can decelerate at a rate of \(7.00 m/s^2\), whereas on wet concrete it can decelerate at only \(5.00 m/s^2\). Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h)
- on dry concrete and
- on wet concrete.
- Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
Strategy
Draw a sketch.
In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.
Solution for (a)
1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=30.0 m/s; v=0; a=−7.00m/s^2\) (\(\displaystyle a\) is negative because it is in a direction opposite to velocity). We take \(\displaystyle x_0\) to be \(\displaystyle 0\). We are looking for displacement \(\displaystyle Δx\), or \(\displaystyle x−x_0\).
2. Identify the equation that will help up solve the problem. The best equation to use is
\(\displaystyle v^2=v^2_0+2a(x−x_0)\).
This equation is best because it includes only one unknown, \(\displaystyle x\). We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for \(\displaystyle x\), but they require us to know the stopping time, \(\displaystyle t\), which we do not know. We could use them but it would entail additional calculations.)
3. Rearrange the equation to solve for \(\displaystyle x\).
\(\displaystyle x−x_0=\frac{v^2−v^2_0}{2a}\)
4. Enter known values.
\(\displaystyle x−0=\frac{0^2−(30.0 m/s)^2}{2(−7.00 m/s^2)}\)
Thus,
\(\displaystyle x=64.3 m\) on dry concrete.
Solution for (b)
This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is \(\displaystyle –5.00 m/s^2\). The result is
\(\displaystyle x_{wet}=90.0 m\) on wet concrete.
Solution for (c)
Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver’s reaction time.
1. Identify the knowns and what we want to solve for. We know that \(\displaystyle \bar{v}=30.0 m/s; t_{reaction}=0.500s; a_{reaction}=0\). We take \(\displaystyle x_{0−reaction}\) to be 0. We are looking for \(\displaystyle x_{reaction}\).
2. Identify the best equation to use.
\(\displaystyle x=x_0+\bar{v}t\) works well because the only unknown value is \(\displaystyle x\), which is what we want to solve for.
3. Plug in the knowns to solve the equation.
\(\displaystyle x=0+(30.0 m/s)(0.500 s)=15.0 m.\)
This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.
4. Add the displacement during the reaction time to the displacement when braking.
\(\displaystyle x_{braking}+x_{reaction}=x_{total}\)
a. 64.3 m + 15.0 m = 79.3 m when dry
b. 90.0 m + 15.0 m = 105 m when wet
Discussion
The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.
Example \(\PageIndex{5}\): Calculating Time - A Car Merges into Traffic
Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at \(\displaystyle . 00 m/s^ 2\) , how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)
Strategy
Draw a sketch.
Solution
1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=10 m/s; a=2.00 m/s^2\); and \(\displaystyle x=200 m.\)
2. We need to solve for t. Choose the best equation. \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) works best because the only unknown in the equation is the variable t for which we need to solve.
3. We will need to rearrange the equation to solve for \(t\). In this case, it will be easier to plug in the knowns first.
\(\displaystyle 200 m=0 m+(10.0 m/s)t+\frac{1}{2}(2.00 m/s^2)t^2\)
4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking \(\displaystyle t=ts\), where \(\displaystyle t\) is the magnitude of time and s is the unit. Doing so leaves
\(\displaystyle 200=10t+t^2.\)
5. Use the quadratic formula to solve for \(\displaystyle t\).
(a) Rearrange the equation to get 0 on one side of the equation.
\(\displaystyle t^2+10t−200=0\)
This is a quadratic equation of the form
\(\displaystyle at^2+bt+c=0,\)
where the constants are \(\displaystyle a=1.00,b=10.0\),and \(\displaystyle c=−200.\)
(b) Its solutions are given by the quadratic formula:
\(\displaystyle t=\frac{−b±\sqrt{b^2−4ac}}{2a}\).
This yields two solutions for \(\displaystyle t\), which are
\(\displaystyle t=10.0\) and \(\displaystyle −20.0\).
In this case, then, the time is \(\displaystyle t=t\) in seconds, or
\(\displaystyle t=10.0s\) and \(\displaystyle −20.0s\).
A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,
\(\displaystyle t=10.0s.\)
Discussion
Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.
With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task.
MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—BREAKING NEWS
We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, \(\displaystyle \bar{a}=Δv/Δt\). While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.
Exercise \(\PageIndex{1}\)
A manned rocket accelerates at a rate of \(\displaystyle 20 m/s^2\) during launch. How long does it take the rocket to reach a velocity of 400 m/s?
- Answer
-
To answer this, choose an equation that allows you to solve for time \(\displaystyle t\), given only \(\displaystyle a, v_0\), and \(\displaystyle v\).
\(\displaystyle v=v_0+at\)
Rearrange to solve for \(\displaystyle t\).
\(\displaystyle t=\frac{v−v_0}{a}=\frac{400 m/s−0 m/s}{20 m/s^2}=20 s\)
Summary
- To simplify calculations we take acceleration to be constant, so that \(\displaystyle \bar{a}=a\) at all times.
- We also take initial time to be zero.
- Initial position and velocity are given a subscript 0; final values have no subscript. Thus,
\(\displaystyle Δt=t\)
\(\displaystyle Δx=x−x_0\)
\(\displaystyle Δv=v−v_0\)
- The following kinematic equations for motion with constant a are useful:
\(\displaystyle x=x_0+\bar{v}t\)
\(\displaystyle \bar{v}=\frac{v_0+v}{2}\)
\(\displaystyle v=v_0+at\)
\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)
\(\displaystyle v^2=v^2_0+2a(x−x_0)\)
- In vertical motion, \(\displaystyle y\) is substituted for \(\displaystyle x\).
|
libretexts
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2025-03-17T19:53:21.882312
| 2015-11-01T04:37:33 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.05%3A_Motion_Equations_for_Constant_Acceleration_in_One_Dimension",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.5: Motion Equations for Constant Acceleration in One Dimension",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.06%3A_Problem-Solving_Basics_for_One-Dimensional_Kinematics
|
2.6: Problem-Solving Basics for One-Dimensional Kinematics
Learning Objectives
By the end of this section, you will be able to:
- Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics.
- Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause.
Problem-Solving Basics for One-Dimensional Kinematics
Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and professional life.
Problem-Solving Steps
While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.
Step 1
Examine the situation to determine which physical principles are involved . It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.
Step 2
Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.
Step 3
Identify exactly what needs to be determined in the problem (identify the unknowns) . In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.
Step 4
Find an equation or set of equations that can help you solve the problem . Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.
Step 5
Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units . This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.
Step 6
Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.
When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.
Unreasonable Results
Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a person starting a foot race accelerates at \(0.40 m/s^2\) for 100 s, his final speed will be \(40 m/s\) (about 150 km/h)—clearly unreasonable because the time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving—it also builds intuition in judging whether nature is being accurately described.
Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause.
Step 1
Solve the problem using strategies as outlined and in the format followed in the worked examples in the text . In the example given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is,
\[v=v0+at=0+(0.40m/s2)(100s)=40m/s.\]
Step 2
Check to see if the answer is reasonable . Is it too large or too small, or does it have the wrong sign, improper units, …? In this case, you may need to convert meters per second into a more familiar unit, such as miles per hour.
\[(40 ms)(3.28 ftm)(1 mi5280 ft)(60 smin)(60 min1 h)=89 mph\]This velocity is about four times greater than a person can run—so it is too large.
Step 3
If the answer is unreasonable, look for what specifically could cause the identified difficulty . In the example of the runner, there are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number means. If someone accelerates at 0.40 m/s2, their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant rate of 0.40 m/s2 for 100 s (almost two minutes).
Summary
- The six basic problem solving steps for physics are:
Step 1. Examine the situation to determine which physical principles are involved.
Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
Step 4. Find an equation or set of equations that can help you solve the problem.
Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.
Step 6. Check the answer to see if it is reasonable: Does it make sense?
|
libretexts
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2025-03-17T19:53:21.957739
| 2015-11-01T04:37:53 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.06%3A_Problem-Solving_Basics_for_One-Dimensional_Kinematics",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.6: Problem-Solving Basics for One-Dimensional Kinematics",
"author": "OpenStax"
}
|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.07%3A_Falling_Objects
|
2.7: Falling Objects
Learning Objectives
By the end of this section, you will be able to:
- Describe the effects of gravity on objects in motion.
- Describe the motion of objects that are in free fall.
- Calculate the position and velocity of objects in free fall.
Falling Objects
Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Gravity
The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones.
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall .
The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity . The acceleration due to gravity is constant , which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, . It is constant at any given location on Earth and has the average value
\[g=9.80 m/s^2.\]
Although varies from . 78 m/s 2 to 9.83 m/s2, depending on latitude, altitude, underlying geological formations, and local topography, the average value of 9.80 m/s2 will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate system. If we define the upward direction as negative, then a=−g=−9.80 m/s2, and if we define the downward direction as positive, then = g = 9 . 80 m/s 2 .
One-Dimensional Motion Involving Gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g. We will also represent vertical displacement with the symbol y and use x for horizontal displacement.
KINEMATIC EQUATIONS FOR OBJECTS IN FREE-FALL WHERE ACCELERATION = -G
\(v=v_0−gt\)
\(y=y_0+v_0t−\frac{1}{2}gt^2\)
\(v^2=v^2_0−2g(y−y_0)\)
Example \(\PageIndex{1}\): Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward
A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s . The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.
Strategy
Draw a sketch.
We are asked to determine the position \(y\) at various times. It is reasonable to take the initial position \(y_0\) to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so a is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these as \(y_1\) and \(v_1; y_2\) and \(v_2\); and \(y_3\) and \(v_3\).
Solution for Position \(y_1\)
1. Identify the knowns. We know that \(y_0=0; v_0=13.0 m/s; a=−g=−9.80 m/s^2\); and \(t=1.00 s.\)
2. Identify the best equation to use. We will use \(y=y_0+v_0t+\frac{1}{2}at^2\) because it includes only one unknown, \(y\) (or \(y_1\), here), which is the value we want to find.
3. Plug in the known values and solve for \(y_1\).
\(y_1=0+(13.0 m/s)(1.00 s)+\frac{1}{2}(−9.80m/s^2)(1.00 s)^2=8.10m\)
Discussion
The rock is 8.10 m above its starting point at \(t=1.00 s\), since \(y_1>y_0\). It could be moving up or down; the only way to tell is to calculate \(v_1\) and find out if it is positive or negative.
Solution for Velocity \(v_1\)
1. Identify the knowns. We know that \(y_0=0; v_0=13.0 m/s; a=−g=−9.80 m/s^2\); and \(t=1.00 s\). We also know from the solution above that \(y_1=8.10 m.\)
2. Identify the best equation to use. The most straightforward is \(v=v_0−gt\) (from \(v=v_0+at\), where a=gravitational acceleration=−g).
3. Plug in the knowns and solve.
\(v_1=v_0−gt=13.0 m/s−(9.80 m/s^2)(1.00 s)=3.20 m/s\)
Discussion
The positive value for \(v_1\) means that the rock is still heading upward at \(t=1.00s\). However, it has slowed from its original 13.0 m/s, as expected.
Solution for Remaining Times
The procedures for calculating the position and velocity at \(t=2.00s\) and \(3.00 s\) are the same as those above. The results are summarized in Table and illustrated in Figure.
| Time, t | Position, y | Velocity, v | Acceleration, a |
|---|---|---|---|
| . 00 s | . 10 m | . 20 m/s\) | 9 . 80 m/s^ 2\) |
| . 00 s | . 40 m | 6 . 60 m/s\) | 9 . 80 m/s^ 2\) |
| . 00 s | 5 . 10 m | 16 . 4 m/s\) | 9 . 80 m/s^ 2\) |
Graphing the data helps us understand it more clearly.
\(PageIndex{3}\): Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down.
Discussion
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since \(y_1\) and \(v_1\) are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both \(y_3\) and \(v_3\) are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still \(−9.80 m/s^2\). Its acceleration is \(−9.80 m/s^2\) for the whole trip—while it is moving up and while it is moving down. Note that the values for y are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since y1 and v1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both y3 and v3 are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still −9.80 m/s2. Its acceleration is −9.80 m/s2 for the whole trip—while it is moving up and while it is moving down. Note that the values for y are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.
MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—REACTION TIME
A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?
Example \(\PageIndex{2}\):Calculating Velocity of a Falling Object: A Rock Thrown Down
What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.
Strategy
Draw a sketch.
Since up is positive, the final position of the rock will be negative because it finishes below the starting point at \(y_0=0\). Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.
Solution
1. Identify the knowns. \(y_0=0; y_1=−5.10 m; v_0=−13.0 m/s; a=−g=−9.80 m/s^2\).
2. Choose the kinematic equation that makes it easiest to solve the problem. The equation \(v^2=v^2_0+2a(y−y_0)\) works well because the only unknown in it is \(v\). (We will plug \(y_1\) in for \(y\).)
3. Enter the known values
\(v^2=(−13.0 m/s)^2+2(−9.80 m/s^2)(−5.10 m−0 m)=268.96 m^2/s^2,\)
where we have retained extra significant figures because this is an intermediate result.
Taking the square root, and noting that a square root can be positive or negative, gives
\(v=±16.4 m/s.\)
The negative root is chosen to indicate that the rock is still heading down. Thus,
\(v=−16.4 m/s.\)
Discussion
Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example and Figure(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example) when the initial velocity is 13.0 m/s straight up, a result of \(±3.20 m/s\) is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.
Another way to look at it is this: In Example, the rock is thrown up with an initial velocity of \(13.0 m/s\). It rises and then falls back down. When its position is \(y=0\) on its way back down, its velocity is \(−13.0 m/s\). That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of \(y=−5.10\) m to be the same whether we have thrown it upwards at \(+13.0 m/s\) or thrown it downwards at \(−13.0 m/s\). The velocity of the rock on its way down from \(y=0\) is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.
Example \(\PageIndex{3}\):Find g from Data on a Falling Object
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.
Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?
Strategy
Draw a sketch.
We need to solve for acceleration \(a.\) Note that in this case, displacement is downward and therefore negative, as is acceleration.
Solution
1. Identify the knowns. \(y_0=0; y=–1.0000 m; t=0.45173; v_0=0.\)
2. Choose the equation that allows you to solve for \(a\) using the known values.
\(y=y_0+v_0t+\frac{1}{2}at^2\)
3. Substitute 0 for \(v_0\) and rearrange the equation to solve for \(a\). Substituting 0 for \(v_0\)yields
\(y=y_0+\frac{1}{2}at^2\).
Solving for \(a\) gives
\(a=\frac{2(y−y_0)}{t^2}\).
4. Substitute known values yields
\(a=\frac{2(−1.0000 m – 0)}{(0.45173 s)^2}=−9.8010 m/s^2,\)
so, because \(a=−g\) with the directions we have chosen,
\(g=9.8010 m/s^2.\)
Discussion
The negative value for a indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of \(9.80 m/s^2\), so \(9.8010 m/s^2\) makes sense. Since the data going into the calculation are relatively precise, this value for g is more precise than the average value of \(9.80 m/s%2\); it represents the local value for the acceleration due to gravity.
Exercise \(\PageIndex{1}\)
A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?
- Answer
-
We know that initial position \(y_0=0\), final position \(y=−30.0 m\), and \(a=−g=−9.80 m/s^2\). We can then use the equation \(y=y_0+v_0t+\frac{1}{2}at^2\) to solve for \(t\). Inserting \(a=−g\), we obtain
\(y=0+0−\frac{1}{2}gt^2\)
\(t^2=\frac{2y}{−g}\)
\(t=±\sqrt{\frac{2y}{−g}}=±\sqrt{\frac{2(−30.0 m)}{−9.80 m/s^2}}=±\sqrt{6.12s^2}=2.47 s≈2.5 s\)
where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.
PHET EXPLORATIONS: EQUATION GRAPHER
Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. \(y=bx\)) to see how they add to generate the polynomial curve.
Summary
- An object in free-fall experiences constant acceleration if air resistance is negligible.
- On Earth, all free-falling objects have an acceleration due to gravity \(g\), which averages
\(g=9.80 m/s^2\).
- Whether the acceleration a should be taken as \(+g\) or \(−g\) is determined by your choice of coordinate system. If you choose the upward direction as positive, \(a=−g=−9.80 m/s^2\) is negative. In the opposite case, \(a=+g=9.80 m/s^2\) is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate \(+g\) or \(−g\) substituted for a.
- For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.
Glossary
- free-fall
- the state of movement that results from gravitational force only
- acceleration due to gravity
- acceleration of an object as a result of gravity
|
libretexts
|
2025-03-17T19:53:22.063137
| 2015-11-01T04:38:11 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.07%3A_Falling_Objects",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.7: Falling Objects",
"author": "OpenStax"
}
|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.08%3A_Graphical_Analysis_of_One-Dimensional_Motion
|
2.8: Graphical Analysis of One-Dimensional Motion
Learning Objectives
By the end of this section, you will be able to:
- Describe a straight-line graph in terms of its slope and y -intercept.
- Determine average velocity or instantaneous velocity from a graph of position vs. time.
- Determine average or instantaneous acceleration from a graph of velocity vs. time.
- Derive a graph of velocity vs. time from a graph of position vs. time.
- Derive a graph of acceleration vs. time from a graph of velocity vs. time.
A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.
Slopes and General Relationships
First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axis the x-axis and the vertical axis the y-axis, as in Figure \(\PageIndex{1}\), a straight-line graph has the general form
\[y=mx+b.\]
Here \(m\) is the slope, defined to be the rise divided by the run of the straight line (Figure \(\PageIndex{1}\)). The letter \(b\) is used for they-intercept, which is the point at which the line crosses the vertical axis.
Graph of Displacement vs. Time ( a = 0, so v is constant)
Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have on the vertical axis and on the horizontal axis. Figure \(\PageIndex{2}\) is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.
Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity \(\bar{v}\) and the intercept is displacement at time zero—that is, \(x_0\). Substituting these symbols into \(y=mx+b\) gives
\[x=\bar{v}t+x_0\]or
\[x=x_0+ \bar{v}t.\]
Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.
THE SLOPE OF \(X\) VS. \(T\)
The slope of the graph of displacement \(x\) vs. time \(t\) is velocity \(v\).
Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension .
From the figure we can see that the car has a displacement of 25 m at 0.50 s and 2000 m at 6.40 s. Its displacement at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.
Example \(\PageIndex{1}\):Determining Average Velocity from a Graph of Displacement versus Time: Jet Car
Find the average velocity of the car whose position is graphed in Figure \(\PageIndex{2}\).
Strategy
The slope of a graph of \(x\) vs. \(t\) is average velocity, since slope equals rise over run. In this case, rise = change in position and run = change in time, so that
Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)
Solution
- Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)
- Substitute the x and t values of the chosen points into the equation. Remember in calculating change (Δ) we always use final value minus initial value. \[\displaystyle \bar{v}=\frac{Δx}{Δt}=\frac{2000 m−525 m}{6.4 s−0.50 s}, \nonumber\] yielding \[\displaystyle v−=250 m/s. \nonumber\]
Discussion
This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.
Graphs of Motion when is constant but ≠ 0
The graphs in Figure \(\PageIndex{3}\) below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.
Figure \(\displaystyle \PageIndex{3}\): Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an \(\displaystyle x\) vs. \(\displaystyle t\) graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the \(\displaystyle v\) vs. \(\displaystyle t\) graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of \(\displaystyle 5.0 m/s^2\) over the time interval plotted.
The graph of displacement versus time in Figure \(\PageIndex{3a}\) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure \(\PageIndex{3a}\). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure \(\PageIndex{3b}\) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure \(\PageIndex{3c}\).
Example \(\PageIndex{2}\):
Calculate the velocity of the jet car at a time of 25 s by finding the slope of the \(\displaystyle vs. \(\displaystyle graph in the graph below
Strategy
The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure, where Q is the point at \(\displaystyle t=25 s\).
Solution
- Find the tangent line to the curve at \(\displaystyle t=25 s\).
- Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.
- Plug these endpoints into the equation to solve for the slope, .
\(\displaystyle slope=v_Q=\frac{Δx_Q}{Δt_Q}=\frac{(3120 m−1300 m)}{(32 s−19 s)}\)
Thus,
Discussion
This is the value given in this figure’s table for v at \(\displaystyle t=25 s\). The value of 140 m/s for \(\displaystyle v_Q\) is plotted in Figure. The entire graph of \(\displaystyle v\) vs. \(\displaystyle t\) can be obtained in this fashion.
Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a \(\displaystyle v\) vs. \(\displaystyle t\) graph, rise = change in velocity \(\displaystyle Δv\) and run = change in time \(\displaystyle Δt\).
THE SLOPE OF V VS. T
The slope of a graph of velocity \(\displaystyle v\) vs. time \(\displaystyle t\) is acceleration \(\displaystyle a\).
\(\displaystyle slope=\frac{Δv}{Δt}=a\)
Since the velocity versus time graph in Figure \(\PageIndex{3b}\) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure(c).
Additional general information can be obtained from Figure and the expression for a straight line, \(\displaystyle y=mx+b.\)
In this case, the vertical axis \(\displaystyle y\) is \(\displaystyle V\), the intercept \(\displaystyle b\) is \(\displaystyle v_0\), the slope \(\displaystyle m\) is \(\displaystyle a\), and the horizontal axis \(\displaystyle x\) is \(\displaystyle t\). Substituting these symbols yields
\[v=v_0+at. \nonumber\]
A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension .
It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.
Graphs of Motion Where Acceleration is Not Constant
Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure \(\PageIndex{6}\). Time again starts at zero, and the initial position and velocity are 2900 m and 165 m/s, respectively. (These were the final position and velocity of the car in the motion graphed in Figure \(\PageIndex{4}\)) Acceleration gradually decreases from \(\displaystyle 5.0 m/s^2\) to zero when the car hits 250 m/s. The slope of the \(\displaystyle x\) vs. \(\displaystyle t\) graph increases until \(\displaystyle t=55 s\), after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.
Example \(\PageIndex{3}\):Calculating Acceleration from a Graph of Velocity versus Time
Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the \(\displaystyle v\) vs. \(\displaystyle t\) graph in Figure \(\PageIndex{6b}\).
Strategy
The slope of the curve at \(\displaystyle t=25 s\) is equal to the slope of the line tangent at that point, as illustrated in Figure \(\PageIndex{6b}\).
Solution
Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, .
Discussion
Note that this value for a is consistent with the value plotted in Figure(c) at \(\displaystyle t=25 s\).
A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.
Exercise \(\displaystyle \PageIndex{1}\):Check Your Understanding
A graph of velocity vs. time of a ship coming into a harbor is shown below.
- Describe the motion of the ship based on the graph.
- What would a graph of the ship’s acceleration look like?
- Answer a
-
(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.
Solution b
A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.
Summary
- Graphs of motion can be used to analyze motion.
- Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
- The slope of a graph of displacement \(\displaystyle x\) vs. time \(\displaystyle t\) is velocity \(\displaystyle v\).
- The slope of a graph of velocity \(\displaystyle v\) vs. time \(\displaystyle t\) graph is acceleration \(\displaystyle a\).
- Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
Glossary
- independent variable
- the variable that the dependent variable is measured with respect to; usually plotted along the \(x\)-axis
- dependent variable
- the variable that is being measured; usually plotted along the \(y\)-axis
- slope
- the difference in \(y\)-value (the rise) divided by the difference in \(x\)-value (the run) of two points on a straight line
- y-intercept
- the \(y\)-value when \(x\)=0, or when the graph crosses the \(y\)-axis
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libretexts
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2025-03-17T19:53:22.152205
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.08%3A_Graphical_Analysis_of_One-Dimensional_Motion",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "2.8: Graphical Analysis of One-Dimensional Motion",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/02%3A_Kinematics/2.E%3A_Kinematics_(Exercises)
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2.E: Kinematics (Exercises)
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Conceptual Questions
2.1: Displacement
1. Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify each quantity in your example.
2. Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?
3. Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to \(\displaystyle 50 μm/s(50×10^{−6}m/s)\) have been observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this?
2.2: Vectors, Scalars, and Coordinate Systems
4. A student writes, “ A bird that is diving for prey has a speed of \(\displaystyle −10m/s\)” What is wrong with the student’s statement? What has the student actually described? Explain.
5. What is the speed of the bird in Exercise?
6. Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain.
7. A weather forecast states that the temperature is predicted to be \(\displaystyle −5ºC\) the following day. Is this temperature a vector or a scalar quantity? Explain.
2.3: Time, Velocity, and Speed
8. Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
9. There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
10. Does a car’s odometer measure position or displacement? Does its speedometer measure speed or velocity?
11. If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity? Under what circumstances are these two quantities the same?
12. How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
2.4: Acceleration
13. Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation.
14. Is it possible for velocity to be constant while acceleration is not zero? Explain.
15. Give an example in which velocity is zero yet acceleration is not.
16. If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?
17. Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?
2.6: Problem-Solving Basics for One-Dimensional Kinematics
18. What information do you need in order to choose which equation or equations to use to solve a problem? Explain.
19. What is the last thing you should do when solving a problem? Explain.
2.7: Falling Objects
20. What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?
21. An object that is thrown straight up falls back to Earth. This is one-dimensional motion.
(a) When is its velocity zero?
(b) Does its velocity change direction?
(c) Does the acceleration due to gravity have the same sign on the way up as on the way down?
22. Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.
23. If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not negligible, how would its speed upon return compare with its initial speed? How would the maximum height to which it rises be affected?
24. The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)?
25. How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational acceleration on the Moon is about 1/6 of \(\displaystyle g\) on Earth)?
2.8: Graphical Analysis of One-Dimensional Motion
23. (a) Explain how you can use the graph of position versus time in Figure to describe the change in velocity over time. Identify
(b) the time (\(\displaystyle t_a, t_b, t_c, t_d,\) or \(\displaystyle t_e\)) at which the instantaneous velocity is greatest,
(c) the time at which it is zero, and
(d) the time at which it is negative.
24. (a) Sketch a graph of velocity versus time corresponding to the graph of position versus time given in Figure.
(b) Identify the time or times (\(\displaystyle t_a, t_b, t_c\), etc.) at which the instantaneous velocity is greatest.
(c) At which times is it zero?
(d) At which times is it negative?
25. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure.
(b) Based on the graph, how does acceleration change over time?
26. (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure.
(b) Identify the time or times (\(\displaystyle t_a, t_b, t_c,\) etc.) at which the acceleration is greatest.
(c) At which times is it zero?
(d) At which times is it negative?
27. Consider the velocity vs. time graph of a person in an elevator shown in Figure. Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of
(a) position vs. time and
(b) acceleration vs. time for this trip.
28. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
Problems & Exercises
2.1: Displacement
29. Find the following for path A:
(a) The distance traveled.
(b) The magnitude of the displacement from start to finish.
(c) The displacement from start to finish.
Solution
(a) 7 m
(b) 7 m
(c) \(\displaystyle +7m\)
30. Find the following for path B:
(a) The distance traveled.
(b) The magnitude of the displacement from start to finish.
(c) The displacement from start to finish.
31. Find the following for path C:
(a) The distance traveled.
(b) The magnitude of the displacement from start to finish.
(c) The displacement from start to finish.
Solution
(a) 13 m
(b) 9 m
(c) \(\displaystyle +9m\)
32. Find the following for path D:
(a) The distance traveled.
(b) The magnitude of the displacement from start to finish
(c) The displacement from start to finish
2.3: Time, Velocity, and Speed
33. (a) Calculate Earth’s average speed relative to the Sun.
(b) What is its average velocity over a period of one year?
Solution
(a) \(\displaystyle 3.0×10^4m/s\)
(b) 0 m/s
34. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.
(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference
(b) What is its average velocity over one revolution?
35. The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?
Solution
\(\displaystyle 2×10^7years\)
36. Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?
37. On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?
Solution
\(\displaystyle 34.689 m/s=124.88 km/h\)
38. Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by \(\displaystyle 3.84×10^6m\)(1%)?
39. A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min.
(a) What was her average speed?
(b) If the straight-line distance from her home to the university is 10.3 km in a direction \(\displaystyle 25.0º\) south of east, what was her average velocity?
(c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?
Solution
(a) \(\displaystyle 40.0 km/h\)
(b) 34.3 km/h, \(\displaystyle 25º\) \(\displaystyle S\) of \(\displaystyle E\).
(c) average speed = \(\displaystyle 3.20 km/h,\bar{v}=0\).
40. The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?
41.
Conversations
ith astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (\(\displaystyle 3.00×10^8m/s\))
Solution
384,000 km
42. A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity
(a) for each of the three intervals and
(b) for the entire motion.
43. The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron in a circular orbit \(\displaystyle 1.06×10^{−10}m\) in diameter.
(a) If the average speed of the electron in this orbit is known to be \(\displaystyle 2.20×10^6m/s\), calculate the number of revolutions per second it makes about the nucleus.
(b) What is the electron’s average velocity?
Solution
(a) \(\displaystyle 6.61×10^{15}rev/s\)
(b) 0 m/s
2.4: Acceleration
44. A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?
Solution
\(\displaystyle 4.29m/s^2\)
45. Professional Application
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of \(\displaystyle g (9.80 m/s^2)\) by taking its ratio to the acceleration of gravity.
46. A commuter backs her car out of her garage with an acceleration of \(\displaystyle 1.40 m/s^2\)
(a) How long does it take her to reach a speed of 2.00 m/s?
(b) If she then brakes to a stop in 0.800 s, what is her deceleration?
Solution
(a) \(\displaystyle 1.43 s\)
(b) \(\displaystyle −2.50m/s^2\)
47. Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in \(\displaystyle m/s^2\)and in multiples of \(\displaystyle g (9.80 m/s^2)\)?
2.5: Motion Equations for Constant Acceleration in One Dimension
48. An Olympic-class sprinter starts a race with an acceleration of \(\displaystyle 4.50 m/s^2\)
(a) What is her speed 2.40 s later?
(b) Sketch a graph of her position vs. time for this period.
Solution
(a) \(\displaystyle 10.8m/s\)
(b)
49. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is \(\displaystyle 2.10×10^4m/s^2\), and 1.85 ms \(\displaystyle (1 ms=10^{−3}s)\) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?
Solution
38.9 m/s (about 87 miles per hour)
50. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of \(\displaystyle 6.20×10^5m/s^2\) for \(\displaystyle 8.10×10^{−4}s\). What is its muzzle velocity (that is, its final velocity)?
51. (a) A light-rail commuter train accelerates at a rate of \(\displaystyle 1.35 m/s^2\). How long does it take to reach its top speed of 80.0 km/h, starting from rest?
(b) The same train ordinarily decelerates at a rate of \(\displaystyle 1.65 m/s^2\). How long does it take to come to a stop from its top speed?
(c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in \(\displaystyle m/s^2\)?
Solution
(a) \(\displaystyle 16.5 s\)
(b) \(\displaystyle 13.5 s\)
(c) \(\displaystyle −2.68 m/s^2\)
52. While entering a freeway, a car accelerates from rest at a rate of \(\displaystyle 2.40 m/s^2\) for 12.0 s.
(a) Draw a sketch of the situation.
(b) List the knowns in this problem.
(c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable.
(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.
53. At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of \(\displaystyle 2.00 m/s^2\).
(a) How far does she travel in the next 5.00 s?
(b) What is her final velocity?
(c) Evaluate the result. Does it make sense?
Solution
(a) \(\displaystyle 20.0 m\)
(b) \(\displaystyle −1.00 m/s\)
(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at \(\displaystyle 2.00 m/s^2\) then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.
54. Professional Application:
Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.
(a) Make a sketch of the situation.
(b) List the knowns in this problem.
(c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.
(d) Is the answer reasonable when compared with the time for a heartbeat?
55. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes \(\displaystyle 3.33×10^{−2}s\), calculate the distance over which the puck accelerates.
Solution
0.799 m
56. A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.
(a) What is its average acceleration?
(b) How far does it travel in that time?
57. Freight trains can produce only relatively small accelerations and decelerations.
(a) What is the final velocity of a freight train that accelerates at a rate of \(\displaystyle 0.0500 m/s^2\) for 8.00 min, starting with an initial velocity of 4.00 m/s?
(b) If the train can slow down at a rate of \(\displaystyle 0.550 m/s^2\), how long will it take to come to a stop from this velocity?
(c) How far will it travel in each case?
Solution
(a) \(\displaystyle 28.0 m/s\)
(b) \(\displaystyle 50.9 s\)
(c) 7.68 km to accelerate and 713 m to decelerate
58. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.
(a) How long did the acceleration last?
(b) Calculate the acceleration.
59. A swan on a lake gets airborne by flapping its wings and running on top of the water.
(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of \(\displaystyle 0.350 m/s^2\), how far will it travel before becoming airborne?
(b) How long does this take?
Solution
(a) \(\displaystyle 51.4m\)
(b) \(\displaystyle 17.1 s\)
60. Professional Application:
A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.
(a) Find the acceleration in \(\displaystyle m/s^2\) and in multiples of \(\displaystyle g(g=9.80m/s^2)\).
(b) Calculate the stopping time.
(c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of \(\displaystyle g\)?
61. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.
(a) What is his deceleration?
(b) How long does the collision last?
Solution
(a) \(\displaystyle −80.4m/s^2\)
(b) \(\displaystyle 9.33×10^{−2}s\)
62. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.
63. Consider a grey squirrel falling out of a tree to the ground.
(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.
(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.
Solution
(a) \(\displaystyle 7.7 m/s\)
(b) \(\displaystyle −15×10^2m/s^2\). This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!
64. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of \(\displaystyle 0.150 m/s^2\) as it goes through. The station is 210 m long.
(a) How long is the nose of the train in the station?
(b) How fast is it going when the nose leaves the station?
(c) If the train is 130 m long, when does the end of the train leave the station?
(d) What is the velocity of the end of the train as it leaves?
65. Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example and Example.
(a) Calculate the average acceleration for such a dragster.
(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.
(c) Why is the final velocity greater than that used to find the average acceleration?
Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.
Solution
(a) \(\displaystyle 32.6 m/s^2\)
(b) \(\displaystyle 162 m/s\)
(c) \(\displaystyle v>v_{max}\), because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at \(\displaystyle 32.6 m/s^2\) during the last few meters, but substantially less, and the final velocity would be less than 162 m/s.
66. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of \(\displaystyle 0.500 m/s^2\) for 7.00 s.
(a) What is his final velocity?
(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?
(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?
67. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?
Solution
104 s
68. (a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.
(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
Solution
(a) \(\displaystyle v=12.2 m/s; a=4.07 m/s^2\)
(b) \(\displaystyle v=11.2 m/s\)
2.7: Falling Objects
Assume air resistance is negligible unless otherwise stated.
69. Calculate the displacement and velocity at times of
(a) 0.500,
(b) 1.00,
(c) 1.50, and
(d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be \(\displaystyle y_0=0\)
Solution
(a) \(\displaystyle y_1=6.28 m; v_1=10.1 m/s\)
(b) \(\displaystyle y_2=10.1 m; v_2=5.20 m/s\)
(c) \(\displaystyle y_3=11.5 m; v_3=0.300 m/s\)
(d) \(\displaystyle y_4=10.4 m; v_4=−4.60 m/s\)
70. Calculate the displacement and velocity at times of
(a) 0.500,
(b) 1.00,
(c) 1.50,
(d) 2.00, and
(e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
71. A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?
Solution
\(\displaystyle v_0=4.95 m/s\)
72. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water.
(a) List the knowns in this problem.
(b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
73. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.
(a) List the knowns in this problem.
(b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.
(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.
Solution
(a) \(\displaystyle a=−9.80 m/s^2; v_0=13.0 m/s; y_0=0 m\)
(b) \(\displaystyle v=0m/s\). Unknown is distance \(\displaystyle y\) to top of trajectory, where velocity is zero. Use equation \(\displaystyle v^2=v^2_0+2a(y−y_0)\) because it contains all known values except for \(\displaystyle y\), so we can solve for \(\displaystyle y\). Solving for \(\displaystyle y\) gives
\(\displaystyle v^2−v^2_0=2ay−y_0)\)
\(\displaystyle \frac{v^2−v^2_0}{2a}=y−y_0\)
\(\displaystyle y=y0+\frac{v^2−v^2_0}{2a}=0m+\frac{(0 m/s)^2−(13.0 m/s)^2}{2(−9.80 m/s^2)}=8.62 m\)
Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.
(c) \(\displaystyle 2.65 s\)
74. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.
(a) How long are her feet in the air?
(b) What is her highest point above the board?
(c) What is her velocity when her feet hit the water?
75. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.
(b) How long would it take to reach the ground if it is thrown straight down with the same speed?
Solution
(a) 8.26 m
(b) 0.717 s
76 . A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?
77 . You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?
Solution
1.91 s
78. A kangaroo can jump over an object 2.50 m high.
(a) Calculate its vertical speed when it leaves the ground.
(b) How long is it in the air?
79. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later.
(a) How far above the hiker is the rock when he can see it?
(b) How much time does he have to move before the rock hits his head?
Solution
(a) 94.0 m
(b) 3.13 s
80. An object is dropped from a height of 75.0 m above ground level.
(a) Determine the distance traveled during the first second.
(b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.
81. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff.
(a) How fast will it be going when it strikes the ground?
(b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.
Solution
(a) -70.0 m/s (downward)
(b) 6.10 s
82. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball's velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.
83. Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.
(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.
(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.
Solution
(a) \(\displaystyle 19.6 m\)
(b) \(\displaystyle 18.5 m\)
84. A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m.
(a) Calculate its velocity just before it strikes the floor.
(b) Calculate its velocity just after it leaves the floor on its way back up.
(c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms \(\displaystyle (8.00×10^{−5}s)\)
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
85 . A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find
(a) the maximum height reached,
(b) its position and velocity 4.00 s after being released, and
(c) the time before it hits the ground.
Solution
(a) 305 m
(b) 262 m, -29.2 m/s
(c) 8.91 s
86. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m.
(a) Calculate its velocity just before it strikes the floor.
(b) Calculate its velocity just after it leaves the floor on its way back up.
(c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms \(\displaystyle (3.50×10^{−3}s)\)
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
2.8: Graphical Analysis of One-Dimensional Motion
Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.
87. (a) By taking the slope of the curve in Figure, verify that the velocity of the jet car is 115 m/s at \(\displaystyle t=20\) s.
(b) By taking the slope of the curve at any point in Figure, verify that the jet car’s acceleration is \(\displaystyle 5.0 m/s^2\)
Solution
(a) \(\displaystyle 115 m/s\)
(b) \(\displaystyle 5.0 m/s^2\)
88. Using approximate values, calculate the slope of the curve in Figure to verify that the velocity at \(\displaystyle t=10.0 s\) is 0.208 m/s. Assume all values are known to 3 significant figures.
89. Using approximate values, calculate the slope of the curve in above Figure to verify that the velocity at \(\displaystyle t=30.0 s\) is approximately 0.24 m/s.
Solution
\(\displaystyle v=\frac{(11.7−6.95)×10^3m}{(40.0 – 20.0)s}=238 m/s\)
90. By taking the slope of the curve in Figure, verify that the acceleration is \(\displaystyle 3.2 m/s^2\) at \(\displaystyle t=10 s\)
91. Construct the position graph for the subway shuttle train as shown in [link] (a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
92. (a) Take the slope of the curve in Figure to find the jogger’s velocity at \(\displaystyle t=2.5 s\).
(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure.
93. A graph of \(\displaystyle v(t)\) is shown for a world-class track sprinter in a 100-m race. (See Figure).
(a) What is his average velocity for the first 4 s?
(b) What is his instantaneous velocity at \(\displaystyle t=5 s\)?
(c) What is his average acceleration between 0 and 4 s?
(d) What is his time for the race?
Solution
(a) \(\displaystyle 6 m/s\)
(b) \(\displaystyle 12 m/s\)
(c) \(\displaystyle 3 m/s^2\)
(d) \(\displaystyle 10 s\)
94. Figure shows the position graph for a particle for 6 s.
(a) Draw the corresponding Velocity vs. Time graph.
(b) What is the acceleration between 0 s and 2 s?
(c) What happens to the acceleration at exactly 2 s?
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:22.298434
| 2017-02-01T07:41:47 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics
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3: Two-Dimensional Kinematics
The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics.
-
- 3.0: Prelude to Two-Dimensional Kinematics
- Motion not confined to a plane, such as a car following a winding mountain road, is described by three-dimensional kinematics. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected insights about nature.
-
- 3.2: Vector Addition and Subtraction- Graphical Methods
- A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.
-
- 3.3: Vector Addition and Subtraction- Analytical Methods
- Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made.
-
- 3.4: Projectile Motion
- Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.
-
- 3.5: Addition of Velocities
- Velocities in two dimensions are added using the same analytical vector techniques. Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).
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libretexts
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2025-03-17T19:53:22.363148
| 2015-10-27T19:12:55 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.00%3A_Prelude_to_Two-Dimensional_Kinematics
|
3.0: Prelude to Two-Dimensional Kinematics
The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics. Motion not confined to a plane, such as a car following a winding mountain road, is described by three-dimensional kinematics. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected insights about nature.
|
libretexts
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2025-03-17T19:53:22.419844
| 2015-11-01T04:40:31 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.01%3A_Kinematics_in_Two_Dimensions_-_An_Introduction
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3.1: Kinematics in Two Dimensions - An Introduction
Learning Objectives
By the end of this section, you will be able to:
- Observe that motion in two dimensions consists of horizontal and vertical components.
- Understand the independence of horizontal and vertical vectors in two-dimensional motion.
Two-Dimensional Motion: Walking in a City
Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure \(\PageIndex{2}\).
The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a two-dimensional path, such as the one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line distance?
An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem,
\[a^2 + b^2 = c^2\]
can be used to find the straight-line distance.
The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is \(\sqrt{(9 blocks)^2+ (5 blocks)^2}= 10.3 blocks\), considerably shorter than the 14 blocks you walked. (Note that we are using three significant figures in the answer. Although it appears that “9” and “5” have only one significant digit, they are discrete numbers. In this case “9 blocks” is the same as “9.0 or 9.00 blocks.” We have decided to use three significant figures in the answer in order to show the result more precisely.)
The fact that the straight-line distance (10.3 blocks) in Figure is less than the total distance walked (14 blocks) is one example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.)
As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector’s magnitude. The arrow’s length is indicated by hash marks in Figure and Figure. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure. The first represents a 9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.)
The Independence of Perpendicular Motions
The person taking the path shown in Figure walks east and then north (two perpendicular directions). How far he or she walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward.
INDEPENDENCE OF MOTION
The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.
This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving movement in two directions at once. For example, let’s compare the motions of two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A stroboscope has captured the positions of the balls at fixed time intervals as they fall.
It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, and not by any horizontal forces.) Careful examination of the ball thrown horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is thrown. This result means that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal conditions. In the real world, air resistance will affect the speed of the balls in both directions.
The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile motion , is to resolve (break) it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. We shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas of physics.
PHET EXPLORATIONS: LADYBUG MOTION 2D
Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior.
Summary
- The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and vertical components.
- The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.
Glossary
- vector
- a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction
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libretexts
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2025-03-17T19:53:22.486258
| 2015-11-01T04:41:09 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.02%3A_Vector_Addition_and_Subtraction-_Graphical_Methods
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3.2: Vector Addition and Subtraction- Graphical Methods
Learning Objectives
By the end of this section, you will be able to:
- Understand the rules of vector addition, subtraction, and multiplication.
- Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.
Vector Addition and Subtraction: Graphical Methods
Vectors in Two Dimensions
A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.
Figure shows such a graphical representation of a vector , using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as \(D\), stands for a vector. Its magnitude is represented by the symbol in italics, \(D\), and its direction by \(θ\).
VECTORS IN THIS TEXT
In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector \(F\), which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as \(F\), and the direction of the variable will be given by an angle \(θ\).
Figure \(\PageIndex{2}\): A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle .1 º north of east.
Vector Addition: Head-to-Tail Method
The head-to-tail method is a graphical way to add vectors, described in Figure below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.
Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor .
Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector.
Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail .
Step 4. Draw an arrow from the tail of the first vector to the head of the last vector . This is the resultant , or the sum, of the other vectors.
Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)
Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)
The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.
Example \(\PageIndex{1}\):Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk
Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction north of east. Then, she walks 23.0 m heading north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.
Strategy
Represent each displacement vector graphically with an arrow, labeling the first , the second , and the third , making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted .
Solution
(1) Draw the three displacement vectors.
(2) Place the vectors head to tail retaining both their initial magnitude and direction.
(3) Draw the resultant vector, .
(4) Use a ruler to measure the magnitude of , and a protractor to measure the direction of . While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.
In this case, the total displacement is seen to have a magnitude of 50.0 m and to lie in a direction south of east. By using its magnitude and direction, this vector can be expressed as = 50.0 m and = 7 . 0 º south of east.
Discussion
The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure and we will still get the same solution.
Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative . Vectors can be added in any order.
\(A+B=B+A.\)
(This is true for the addition of ordinary numbers as well—you get the same result whether you add + 3 or + 2 , for example).
Vector Subtraction
Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract from , written – B , we must first define what we mean by subtraction. The negative of a vector is defined to be ; that is, graphically the negative of any vector has the same magnitude but the opposite direction , as shown in Figure. In other words, has the same length as , but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.
The subtraction of vector from vector is then simply defined to be the addition of to . Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.
This is analogous to the subtraction of scalars (where, for example, ( –2 ) ) . Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.
Example \(\PageIndex{1}\):Subtracting Vectors Graphically: A Woman Sailing a Boat
A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction north of east from her current location, and then travel 30.0 m in a direction north of east (or west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.
Strategy
We can represent the first leg of the trip with a vector , and the second leg of the trip with a vector . The dock is located at a location + B . If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance (30.0 m) in the direction – 112 º = 68 º south of east. We represent this as , as shown below. The vector has the same magnitude as but is in the opposite direction. Thus, she will end up at a location + ( – B ) , or – B .
We will perform vector addition to compare the location of the dock, + B , with the location at which the woman mistakenly arrives, + ( – B ) .
Solution
(1) To determine the location at which the woman arrives by accident, draw vectors and .
(2) Place the vectors head to tail.
(3) Draw the resultant vector .
(4) Use a ruler and protractor to measure the magnitude and direction of .
In this case, = 23 . 0 m and = 7 . 5 º south of east.
(5) To determine the location of the dock, we repeat this method to add vectors and . We obtain the resultant vector ' :
In this case = 52.9 m and = 90.1 º north of east.
We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.
Discussion
Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.
Multiplication of Vectors and Scalars
If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk × 27 . 5 m , or 82.5 m, in a direction . 0 º north of east. This is an example of multiplying a vector by a positive scalar . Notice that the magnitude changes, but the direction stays the same.
If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector is multiplied by a scalar ,
- the magnitude of the vector becomes the absolute value of ,
- if is positive, the direction of the vector does not change,
- if is negative, the direction is reversed.
In our case, = 3 and = 27.5 m . Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.
Resolving a Vector into Components
In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x - and y -components, or the north-south and east-west components.
For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction .0 º north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton’s Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.
PHET EXPLORATIONS: MAZE GAME
Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.
- The graphical method of adding vectors and involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector is defined such that + B = R . The magnitude and direction of are then determined with a ruler and protractor, respectively.
- The graphical method of subtracting vector from involves adding the opposite of vector , which is defined as B . In this case, – B = A + ( – B ) = R . Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector .
- Addition of vectors is commutative such that + B = B + A .
- The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
- If a vector is multiplied by a scalar quantity , the magnitude of the product is given by . If is positive, the direction of the product points in the same direction as ; if is negative, the direction of the product points in the opposite direction as .
Glossary
- component (of a 2-d vector)
- a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components
- commutative
- refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum
- direction (of a vector)
- the orientation of a vector in space
- head (of a vector)
- the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”
- head-to-tail method
- a method of adding vectors in which the tail of each vector is placed at the head of the previous vector
- magnitude (of a vector)
- the length or size of a vector; magnitude is a scalar quantity
- resultant
- the sum of two or more vectors
- resultant vector
- the vector sum of two or more vectors
- scalar
- a quantity with magnitude but no direction
- tail
- the start point of a vector; opposite to the head or tip of the arrow
|
libretexts
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2025-03-17T19:53:22.596132
| 2015-11-01T04:41:28 |
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"title": "3.2: Vector Addition and Subtraction- Graphical Methods",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.03%3A__Vector_Addition_and_Subtraction-_Analytical_Methods
|
3.3: Vector Addition and Subtraction- Analytical Methods
Learning Objectives
By the end of this section, you will be able to:
- Understand the rules of vector addition and subtraction using analytical methods.
- Apply analytical methods to determine vertical and horizontal component vectors.
- Apply analytical methods to determine the magnitude and direction of a resultant vector.
Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.
Resolving a Vector into Perpendicular Components
Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like \(\displaystyle A\) in Figure \(\PageIndex{1}\), we may wish to find which two perpendicular vectors, \(\displaystyle A_x\) and \(\displaystyle A_y\), add to produce it.
\(\displaystyle A_x\) and \(\displaystyle A_y\) are defined to be the components of \(\displaystyle A\) along the x- and y-axes. The three vectors \(\displaystyle A, A_x\), and \(\displaystyle A_y\) form a right triangle:
\[\displaystyle A_x + A_y = A.\]
Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if \(\displaystyle A_x=3 m\) east, \(\displaystyle A_y=4 m\) north, and \(\displaystyle A=5 m \)north-east, then it is true that the vectors \(\displaystyle A_x + A_y = A\). However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,
\[\displaystyle 3 m+4 m ≠ 5 m\]
Thus,
\[\displaystyle A_x+A_y≠A\]
If the vector \(\displaystyle A\) is known, then its magnitude \(\displaystyle A\) (its length) and its angle \(\displaystyle θ\) (its direction) are known. To find \(\displaystyle A_x\) and \(\displaystyle A_y\), its x- and y-components, we use the following relationships for a right triangle.
\[\displaystyle A_x=Acosθ\]
and
\[\displaystyle A_y=Asinθ.\]
Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.
Then \(\displaystyle A=10.3\) blocks and \(\displaystyle θ=29.1º\), so that
\[\displaystyle A_x=A\cos θ=(10.3 blocks)(\cos 29.1º)=9.0 \quad blocks \]
\[\displaystyle A_y=A\sin θ=(10.3 blocks)(\sin 29.1º)=5.0 \quad blocks \]
Calculating a Resultant Vector
If the perpendicular components \(\displaystyle A_x\) and \(\displaystyle A_y\) of a vector \(\displaystyle A\) are known, then A can also be found analytically. To find the magnitude \(\displaystyle A\) and direction \(\displaystyle θ\) of a vector from its perpendicular components \(\displaystyle A_x\) and \(\displaystyle A_y\), we use the following relationships:
\[\displaystyle A=\sqrt{A_{x^2}+A_{y^2}}\]
\[\displaystyle θ=tan^{−1}(A_y/A_x)\]
Note that the equation \(\displaystyle A=\sqrt{A^2_x+A^2_y}\) is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if \(\displaystyle A_x\) and \(\displaystyle A_y\) are 9 and 5 blocks, respectively, then \(\displaystyle A=\sqrt{9^2+5^2}=10.3\) blocks, again consistent with the example of the person walking in a city. Finally, the direction is \(\displaystyle θ=tan^{–1}(5/9)=29.1º\), as before.
DETERMINING VECTORS AND VECTOR COMPONENTS WITH ANALYTICAL METHODS
Equations \(\displaystyle A_x=Acosθ\) and \(\displaystyle A_y=Asinθ\) are used to find the perpendicular components of a vector—that is, to go from \(\displaystyle A\) and \(\displaystyle θ\) to \(\displaystyle A_x\) and \(\displaystyle A_y\). Equations \(\displaystyle A=\sqrt{A^2_x+A^2_y}\) and \(\displaystyle θ=tan^{–1}(A_y/A_x)\) are used to find a vector from its perpendicular components—that is, to go from \(\displaystyle A_x\) and \(\displaystyle A_y\) to \(\displaystyle A\) and \(\displaystyle θ\). Both processes are crucial to analytical methods of vector addition and subtraction.
Adding Vectors Using Analytical Methods
To see how to add vectors using perpendicular components, consider Figure \(\PageIndex{5}\), in which the vectors \(\displaystyle A\) and \(\displaystyle B\) are added to produce the resultant \(\displaystyle R\).
If \(\displaystyle A\) and \(\displaystyle B\) represent two legs of a walk (two displacements), then \(\displaystyle R\) is the total displacement. The person taking the walk ends up at the tip of R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, \(\displaystyle R_x\) and \(\displaystyle R_y\). If we know \(\displaystyle R_x\) and \(\displaystyle R_y\), we can find \(\displaystyle R\) and \(\displaystyle θ\) using the equations \(\displaystyle A=\sqrt{A_x^2+A_y^2}\) and \(\displaystyle θ=tan^{–1}(A_y/A_x)\). When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.
Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations \(\displaystyle A_x=Acosθ\) and \(\displaystyle A_y=Asinθ\) to find the components. In Figure, these components are \(\displaystyle A_x, A_y, B_x\), and \(\displaystyle B_y\). The angles that vectors \(\displaystyle A\) and \(\displaystyle B\) make with the x-axis are \(\displaystyle θ_A\) and \(\displaystyle θ_B\), respectively.
Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure \(\PageIndex{7}\ ,
\[\displaystyle R_x=A_x+B_x\]
and
\[\displaystyle R_y=A_y+B_y.\]
Components along the same axis, say the x -axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y -axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R are known, its magnitude and direction can be found.
Step 3. To get the magnitude \(\displaystyle R\) of the resultant, use the Pythagorean theorem:
\[\displaystyle R=\sqrt{R^2_x+R^2_y}\]
Step 4. To get the direction of the resultant:
\[\displaystyle θ=tan^{−1}(R_y/R_x)\]
The following example illustrates this technique for adding vectors using perpendicular components.
Example \(\displaystyle \PageIndex{1}\): Adding Vectors Using Analytical Methods
Add the vector \(\displaystyle A\) to the vector \(\displaystyle B\) shown in Figure, using perpendicular components along the x - and y- axes. The x - and y -axes are along the east–west and north–south directions, respectively. Vector \(\displaystyle A\) represents the first leg of a walk in which a person walks \(\displaystyle 53.0 m\) in a direction \(\displaystyle 20.0º\) north of east. Vector \(\displaystyle B\) represents the second leg, a displacement of \(\displaystyle 34.0 m\) in a direction \(\displaystyle 63.0º\) north of east.
Figure \(\PageIndex{8}\)
:
Vector \(\displaystyle A\) has magnitude \(\displaystyle 53.0 m\) and direction \(\displaystyle 20.0º\) north of the
x
-axis. Vector B has magnitude \(\displaystyle 34.0 m\) and direction \(\displaystyle 63.0º\) north of the x-axis. You can use analytical methods to determine the magnitude and direction of \(\displaystyle R\).
Strategy
The components of \(\displaystyle A\) and \(\displaystyle B\) along the x- and y -axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.
Solution
Following the method outlined above, we first find the components of \(\displaystyle A\) and \(\displaystyle B\) along the x- and y- axes. Note that \(\displaystyle A=53.0 m, θ_A=20.0º, B=34.0 m,\) and \(\displaystyle θ_B=63.0º\). We find the x- components by using \(\displaystyle A_x=Acosθ\), which gives
\[\displaystyle A_x=Acosθ_A=(53.0 m)(cos 20.0º)(53.0 m)(0.940)=49.8 m\]
and
\[\displaystyle B_x=Bcosθ_B=(34.0 m)(cos 63.0º)(34.0 m)(0.454)=15.4 m.\]
Similarly, the y- components are found using \(\displaystyle A_y=Asinθ_A\):
\[\displaystyle A_y=Asinθ_A=(53.0 m)(sin 20.0º)(53.0 m)(0.342)=18.1 m\]
and
\[\displaystyle B_y=Bsinθ_B=(34.0 m)(sin 63.0º)(34.0 m)(0.891)=30.3 m.\]
The x- and y- components of the resultant are thus
\[\displaystyle R_x=A_x+B_x=49.8 m+15.4 m=65.2 m\]
and
\[\displaystyle R_y=A_y+B_y=18.1 m+30.3 m=48.4 m.\]
Now we can find the magnitude of the resultant by using the Pythagorean theorem:
\[\displaystyle R=\sqrt{R^2_x+R^2_y}=\sqrt{(65.2)^2+(48.4)^2m}\]
so that
\[\displaystyle R=81.2 m.\]
Finally, we find the direction of the resultant:
\[\displaystyle θ=tan^{−1}(R_y/R_x)=+tan^{−1}(48.4/65.2).\]
Thus,
\[\displaystyle θ=tan^{−1}(0.742)=36.6º.\]
Discussion
This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.
Subtraction of vectors is accomplished by the addition of a negative vector. That is, \(\displaystyle A−B≡A+(–B)\). Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of \(\displaystyle –B\) are the negatives of the components of \(\displaystyle B\). The x- and y- components of the resultant \(\displaystyle A−B = R\) are thus
\[\displaystyle R_x=A_x+(–B_x)\]
and
\[\displaystyle R_y=A_y+(–B_y)\]
and the rest of the method outlined above is identical to that for addition. (See Figure \(\PageIndex{10}\).)
Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.
PHET EXPLORATIONS: VECTOR ADDITION
Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.
- The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
- The steps to add vectors \(\displaystyle A\) and \(\displaystyle B\) using the analytical method are as follows:
Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations
\(\displaystyle A_x=Acosθ\)
\(\displaystyle B_x=Bcosθ\)
and
\(\displaystyle A_y=Asinθ\)
\(\displaystyle B_y=Bsinθ.\)
Step 2: Add the horizontal and vertical components of each vector to determine the components Rx and Ry of the resultant vector, R:
\(\displaystyle R_x=A_x+B_x\)
and
\(\displaystyle R_y=A_y+B_y\).
Step 3: Use the Pythagorean theorem to determine the magnitude, R, of the resultant vector R:
\(\displaystyle R=\sqrt{R^2_x+R^2_y}\).
Step 4: Use a trigonometric identity to determine the direction, \(\displaystyle θ\), of R :
\(\displaystyle θ=tan^{−1}(R_y/R_x)\).
Glossary
- analytical method
- the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities
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libretexts
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2025-03-17T19:53:22.686328
| 2015-11-01T04:41:51 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.03%3A__Vector_Addition_and_Subtraction-_Analytical_Methods",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "3.3: Vector Addition and Subtraction- Analytical Methods",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.04%3A_Projectile_Motion
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3.4: Projectile Motion
Learning Objectives
By the end of this section, you will be able to:
- Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
- Determine the location and velocity of a projectile at different points in its trajectory.
- Apply the principle of independence of motion to solve projectile motion problems.
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .
The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure illustrates the notation for displacement, where \(\displaystyle s\) is defined to be the total displacement and \(\displaystyle x\) and \(\displaystyle y\) are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation \(\displaystyle A\) to represent a vector with components \(\displaystyle A_x\) and \(\displaystyle A_y\). If we continued this format, we would call displacement \(\displaystyle s\) with components \(\displaystyle s_x\) and \(\displaystyle s_y\). However, to simplify the notation, we will simply represent the component vectors as \(\displaystyle x\) and \(\displaystyle y\).)
Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: \(\displaystyle a_y=–g=–9.80 m/s^2\)(. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, \(\displaystyle a_x=0\). Both accelerations are constant, so the kinematic equations can be used.
REVIEW OF KINEMATIC EQUATIONS (CONSTANT )
\(\displaystyle x=x_0+\bar{v}t\)
\(\displaystyle \bar{v}=\frac{v_0+v}{2}\)
\(\displaystyle v=v_0+at\)
\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)
\(\displaystyle v^2=v^2_0+2a(x−x_0)\).
Given these assumptions, the following steps are then used to analyze projectile motion:
Step 1 . Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so \(\displaystyle A_x=Acosθ\) and \(\displaystyle A_y=Asinθ\) are used. The magnitude of the components of displacement \(\displaystyle s\) along these axes are \(\displaystyle x\) and \(\displaystyle y\). The magnitudes of the components of the velocity \(\displaystyle v\) are \(\displaystyle v_x=vcosθ\) and \(\displaystyle v_y=vsinθ\), where \(\displaystyle v\) is the magnitude of the velocity and θ is its direction, as shown in Figure. Initial values are denoted with a subscript 0, as usual.
Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:
\(\displaystyle \text{Horizontal Motion}(a_x=0)\)
\(\displaystyle x=x_0+v_xt\)
\(\displaystyle v_x=v_{0x}=v_x=\text{velocity is a constant}\).
\(\displaystyle \text{Vertical Motion(assuming positive is up}a_y=−g=−9.80m/s^2)\)
\(\displaystyle y=y_0+\frac{1}{2}(v_{0y}+v_y)t\)
\(\displaystyle v_y=v_{0y}−gt\)
\(\displaystyle y=y_0+v_{0y}t−\frac{1}{2}gt^2\)
\(\displaystyle v^2_y=v^2_{0y}−2g(y−y_0)\).
Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.
Step 4. Recombine the two motions to find the total displacement \(\displaystyle s\) and velocity \(\displaystyle v\) . Because the x - and y - motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing \(\displaystyle A=\sqrt{A^2_x+A^2_y}\) and \(\displaystyle θ=tan^{−1}(A_y/A_x)\) in the following form, where \(\displaystyle θ\) is the direction of the displacement \(\displaystyle s\) and \(\displaystyle θ_v\) is the direction of the velocity \(\displaystyle v\):
Total displacement and velocity
\(\displaystyle s=\sqrt{x^2+y^2}\)
\(\displaystyle θ=tan^{−1}(y/x)\)
\(\displaystyle v=\sqrt{v^2_x+v^2_y}\)
\(\displaystyle θ_v=tan^{−1}(v_y/v_x)\).
Example \(\displaystyle \PageIndex{1}\): A Fireworks Projectile Explodes High and Away
During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in Figure. The fuse is timed to ignite the shell just as it reaches its highest point above the ground.
- Calculate the height at which the shell explodes.
- How much time passed between the launch of the shell and the explosion?
- What is the horizontal displacement of the shell when it explodes?
Strategy
Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which \(\displaystyle a_x=0\) and \(\displaystyle a_y=–g\). We can then define \(\displaystyle x_0\) and \(\displaystyle y_0\) to be zero and solve for the desired quantities.
Solution for (a)
By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when \(\displaystyle v_y=0\). Since we know the initial and final velocities as well as the initial position, we use the following equation to find \(\displaystyle y\):
\(\displaystyle v^2_y=v^2_{0y}−2g(y−y_0)\).
Because \(\displaystyle y_0\) and \(\displaystyle v_y\) are both zero, the equation simplifies to
\(\displaystyle 0=v^2_{0y}−2gy.\)
Solving for \(\displaystyle y\) gives
\(\displaystyle y=\frac{v^2_{0y}}{2g}\).
Now we must find \(\displaystyle v_{0y}\), the component of the initial velocity in the y- direction. It is given by \(\displaystyle v_{0y}=v_0sinθ\), where \(\displaystyle v_{0y}\) is the initial velocity of 70.0 m/s, and \(\displaystyle θ_0=75.0º\) is the initial angle. Thus,
\(\displaystyle v_{0y}=v_0sinθ_0=(70.0 m/s)(sin 75º)=67.6 m/s.\)
and \(\displaystyle y\) is
\(\displaystyle y=\frac{(67.6 m/s)^2}{2(9.80 m/s^2)}\),
so that
\(\displaystyle y=233m.\)
Discussion for (a)
Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.
Solution for (b)
As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use \(\displaystyle y=y_0+\frac{1}{2}(v_{0y}+v_y)t\). Because y0 is zero, this equation reduces to simply
\(\displaystyle y=\frac{1}{2}(v_{0y}+v_y)t\).
Note that the final vertical velocity, \(\displaystyle v_y\), at the highest point is zero. Thus,
\(\displaystyle t=\frac{2y}{(v_{0y}+v_y)}=\frac{2(233 m)}{(67.6 m/s)}=6.90 s.\)
Discussion for (b)
This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using \(\displaystyle y=y_0+v_{0y}t−\frac{1}{2}gt^2\), and solving the quadratic equation for \(\displaystyle t\).)
Solution for (c)
Because air resistance is negligible, \(\displaystyle a_x=0\) and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by \(\displaystyle x=x_0+v_xt\), where \(\displaystyle x_0\) is equal to zero:
\(\displaystyle x=v_xt,\)
where \(\displaystyle v_x\) is the x- component of the velocity, which is given by \(\displaystyle v_x=v_0cosθ_0\). Now,
\(\displaystyle v_x=v_0cosθ_0=(70.0 m/s)(cos 75.0º)=18.1 m/s.\)
The time \(\displaystyle t\) for both motions is the same, and so \(\displaystyle x\) is
\(\displaystyle x=(18.1 m/s)(6.90 s)=125 m.\)
Discussion for (c)
The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.
In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height \(\displaystyle y=h\); then,
\(\displaystyle h=\frac{v^2_{0y}}{2g}\).
This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.
DEFINING A COORDINATE SYSTEM
It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the \(\displaystyle x\) and \(\displaystyle y\) positions. Often, it is convenient to choose the initial position of the object as the origin such that \(\displaystyle x_0=0\) and \(\displaystyle y_0=0\). It is also important to define the positive and negative directions in the \(\displaystyle x\) and \(\displaystyle y\) directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, \(\displaystyle g\), takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, g takes a positive value.
Example \(\displaystyle \PageIndex{2}\): Calculating Projectile Motion: Hot Rock Projectile
Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle above the horizontal, as shown in Figure. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?
Strategy
Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for \(\displaystyle t\) first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain \(\displaystyle v\) and \(\displaystyle θ_v\) at the final time t determined in the first part of the example.
Solution for (a)
While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using
\(\displaystyle y=y_0+v_{0y}t−\frac{1}{2}gt^2\).
If we take the initial position \(\displaystyle y_0\) to be zero, then the final position is \(\displaystyle y=−20.0 m\). Now the initial vertical velocity is the vertical component of the initial velocity, found from \(\displaystyle v_{0y}=v_0sinθ_0 = (25.0 m/s)(sin 35.0º) = 14.3 m/s.\) Substituting known values yields
\(\displaystyle −20.0 m=(14.3 m/s)t−(4.90 m/s^2)t^2\).
Rearranging terms gives a quadratic equation in \(\displaystyle t\):
\(\displaystyle (4.90 m/s^2)t^2−(14.3 m/s)t−(20.0 m)=0.\)
This expression is a quadratic equation of the form \(\displaystyle at^2+bt+c=0\), where the constants are \(\displaystyle a=4.90, b=–14.3\), and \(\displaystyle c=–20.0.\) Its solutions are given by the quadratic formula:
\(\displaystyle t=\frac{−b±\sqrt{b^2−4ac}}{2a}\).
This equation yields two solutions: \(\displaystyle t=3.96\) and \(\displaystyle t=–1.03\). (It is left as an exercise for the reader to verify these solutions.) The time is \(\displaystyle t=3.96s\) or \(\displaystyle –1.03s\). The negative value of time implies an event before the start of motion, and so we discard it. Thus,
\(\displaystyle t=3.96 s.\)
Discussion for (a)
The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.
Solution for (b)
From the information now in hand, we can find the final horizontal and vertical velocities \(\displaystyle v_x\) and \(\displaystyle v_y\) and combine them to find the total velocity v and the angle \(\displaystyle θ_0\) it makes with the horizontal. Of course, \(\displaystyle v_x\) is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:
\(\displaystyle v_x=v_0cosθ_0=(25.0 m/s)(cos 35º)=20.5 m/s.\)
The final vertical velocity is given by the following equation:
\(\displaystyle v_y=v_{0y}−gt\),
where \(\displaystyle v_{0y}\) was found in part (a) to be 14.3 m/s . Thus,
\(\displaystyle v_y=14.3 m/s−(9.80 m/s^2)(3.96 s)\)
so that
\(\displaystyle v_y=−24.5 m/s.\)
To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:
\(\displaystyle v=\sqrt{v^2_x+v^2_y}=\sqrt{(20.5 m/s)^2+(−24.5 m/s)^2}\),
which gives
\(\displaystyle v=31.9 m/s.\)
The direction θv is found from the equation:
\(\displaystyle θ_v=tan^{−1}(v_y/v_x)\)
so that
\(\displaystyle θ_v=tan^{−1}(−24.5/20.5)=tan^{−1}(−1.19).\)
Thus,
\(\displaystyle θ_v=−50.1º.\)
Discussion for (b)
The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure.)
One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.
How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed \(\displaystyle v_0\), the greater the range, as shown in Figure(a). The initial angle \(\displaystyle θ_0\) also has a dramatic effect on the range, as illustrated in Figure(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with \(\displaystyle θ_0=45º\). This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately \(\displaystyle 38º\). Interestingly, for every initial angle except \(\displaystyle 45º\), there are two angles that give the same range—the sum of those angles is \(\displaystyle 90º\). The range also depends on the value of the acceleration of gravity \(\displaystyle g\). The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range \(\displaystyle R\) of a projectile on level ground for which air resistance is negligible is given by
\(\displaystyle R=\frac{v^2_0sin2θ_0}{g}\),
where \(\displaystyle v_0\) is the initial speed and \(\displaystyle θ_0\) is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.
When we speak of the range of a projectile on level ground, we assume that \(\displaystyle R\) is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure.) If the initial speed is great enough, the projectile goes into orbit. This is called exit velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.
Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.
PHET EXPLORATIONS: PROJECTILE MOTION
Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.
Summary
- Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
- To solve projectile motion problems, perform the following steps:
- Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x and y, and the components of the velocity v are given by vx=vcosθ and vy=vsinθ, where v is the magnitude of the velocity and θ is its direction.
2. Analyze the motion of the projectile in the horizontal direction using the following equations:
\(\displaystyle \text{Horizontal motion}(a_x=0)\)
\(\displaystyle x=x_0+v_xt\)
\(\displaystyle v_x=v_{0x}=v_x=\text{velocity is a constant}\).
3. Analyze the motion of the projectile in the vertical direction using the following equations:
\(\displaystyle \text{Vertical motion(Assuming positive direction is up;}a_y=−g=−9.80 m/s^2)\)
\(\displaystyle y=y_0+\frac{1}{2}(v_{0y}+v_y)t\)
\(\displaystyle v_y=v_{0y}−gt\)
\(\displaystyle y=y_0+v_{0y}t−\frac{1}{2}gt^2\)
\(\displaystyle v^2_y=v^2_{0y}−2g(y−y_0)\).
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
\(\displaystyle s=\sqrt{x^2+y^2}\)
\(\displaystyle θ=tan^{−1}(y/x)\)
\(\displaystyle v=\sqrt{v^2_x+v^2_y}\)
\(\displaystyle θ_v=tan^{−1}(v_y/v_x)\).
- The maximum height \(\displaystyle h\) of a projectile launched with initial vertical velocity \(\displaystyle v_{0y}\) is given by
\(\displaystyle h=\frac{v^2_{0y}}{2g}\).
- The maximum horizontal distance traveled by a projectile is called the range . The range \(\displaystyle R\) of a projectile on level ground launched at an angle \(\displaystyle θ_0\) above the horizontal with initial speed \(\displaystyle v_0\) is given by
\(\displaystyle R=\frac{v^2_0sin2θ_0}{g}\).
Glossary
- air resistance
- a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero
- kinematics
- the study of motion without regard to mass or force
- motion
- displacement of an object as a function of time
- projectile
- an object that travels through the air and experiences only acceleration due to gravity
- projectile motion
- the motion of an object that is subject only to the acceleration of gravity
- range
- the maximum horizontal distance that a projectile travels
- trajectory
- the path of a projectile through the air
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libretexts
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2025-03-17T19:53:22.786619
| 2015-11-01T04:42:24 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.04%3A_Projectile_Motion",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "3.4: Projectile Motion",
"author": "OpenStax"
}
|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.05%3A_Addition_of_Velocities
|
3.5: Addition of Velocities
Learning Objectives
By the end of this section, you will be able to:
- Apply principles of vector addition to determine relative velocity.
- Explain the significance of the observer in the measurement of velocity.
Relative Velocity
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonallyrelative to the shore, as in Figure \(\displaystyle \PageIndex{1}\). The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure \(\displaystyle \PageIndex{2}\). The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figures \(\displaystyle \PageIndex{1}\) and \(\displaystyle \PageIndex{2}\). These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed inVector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the velocity of the ball is 35 m/s relative to the stationary, profusely sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (\(\displaystyle v\) and \(\displaystyle θ\) ) and its components (\(\displaystyle v_x\) and \(\displaystyle v_y\) ) along the x- and y-axes of an appropriately chosen coordinate system:
\[v_x=v\cos θ\]
\[v_y=v\sin θ\]
\[ v=\sqrt{v^2x+v^2} \]
\[ θ=\tan^{−1}\left(\dfrac{v_y}{v_x}\right).\]
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.
TAKE-HOME EXPERIMENT: RELATIVE VELOCITY OF A BOAT
Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.
Example \(\displaystyle \PageIndex{1}\): Adding Velocities - A Boat on a River
Refer to Figure \(\displaystyle \PageIndex{4}\), which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, \(\displaystyle v_{tot}\). The velocity of the boat, vboat , is 0.75 m/s in the y -direction relative to the river and the velocity of the river, \(\displaystyle v_{river}\), is 1.20 m/s to the right.
Strategy
We start by choosing a coordinate system with its \(\displaystyle x\)-axis parallel to the velocity of the river, as shown in Figure. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the y-axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations \(\displaystyle v_{tot}=\sqrt{v^2_x+v^2_y}\) and \(\displaystyle θ=tan^{−1}(v_y/v_x)\) directly.
Solution
The magnitude of the total velocity is
\(\displaystyle v_{tot}=\sqrt{v^2_x+v^2_y}\),
where
\(\displaystyle v_x=v_{river}=1.20 m/s\)
and
\(\displaystyle v_y=v_{boat}=0.750 m/s.\)
Thus,
\(\displaystyle v_{tot}=\sqrt{(1.20 m/s)^2+(0.750 m/s)^2}\)
yielding
\(\displaystyle v_{tot}=1.42 m/s.\)
The direction of the total velocity \(\displaystyle θ\) is given by:
\(\displaystyle θ=tan^{−1}(v_y/v_x)=tan^{−1}(0.750/1.20).\)
This equation gives
\(\displaystyle θ=32.0º.\)
Discussion
Both the magnitude v and the direction \(\displaystyle θ\) of the total velocity are consistent with Figure. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 32.0º ) the total velocity has relative to the riverbank.
Example \(\displaystyle \PageIndex{2}\): Calculating Velocity - Wind Velocity Causes an Airplane to Drift
Calculate the wind velocity for the situation shown in Figure \(\displaystyle \PageIndex{5}\). The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0º west of north.
Strategy
In this problem, somewhat different from the previous example, we know the total velocity \(\displaystyle v_{to}\)t and that it is the sum of two other velocities, \(\displaystyle v_w\) (the wind) and \(\displaystyle v_p\) (the plane relative to the air mass). The quantity \(\displaystyle v_p\) is known, and we are asked to find vw. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of \(\displaystyle v_w\), then we can combine them to solve for its magnitude and direction. As shown in Figure, we choose a coordinate system with its x- axis due east and its y- axis due north (parallel to \(\displaystyle v_p\)). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods.)
Solution
Because \(\displaystyle v_{tot}\) is the vector sum of the \(\displaystyle v_w\) and \(\displaystyle v_p\), its x- and y- components are the sums of the x- and y- components of the wind and plane velocities. Note that the plane only has vertical component of velocity so \(\displaystyle v_{px}=0\) and \(\displaystyle v_{py}=v_p\). That is,
\(\displaystyle v_{tot}x=v_{wx}\)
and
\(\displaystyle v_{toty}=v_{wy}+v_p\).
We can use the first of these two equations to find \(\displaystyle v_{wx}\):
\(\displaystyle v_{wx}=v_{totx}=v_{tot}cos 110º.\)
Because \(\displaystyle v_{tot}=38.0 m/s\) and \(\displaystyle cos 110º=–0.342\) we have
\(\displaystyle v_{wx}=(38.0 m/s)(–0.342)=–13 m/s.\)
The minus sign indicates motion west which is consistent with the diagram.
Now, to find \(\displaystyle v_{wy}\) we note that
\(\displaystyle v_{toty}=v_{wy}+v_p\)
Here \(\displaystyle v_{toty}=v_{tot}sin 110º\); thus,
\(\displaystyle v_{wy}=(38.0 m/s)(0.940)−45.0 m/s=−9.29 m/s.\)
This minus sign indicates motion south which is consistent with the diagram.
Now that the perpendicular components of the wind velocity \(\displaystyle v_{wx}\) and \(\displaystyle v_{wy}\) are known, we can find the magnitude and direction of \(\displaystyle v_w\). First, the magnitude is
\(\displaystyle v_w=\sqrt{v^2_{wx}+v^2_{wy}}=\sqrt{(−13.0 m/s)^2+(−9.29 m/s)^2}\)
so that
\(\displaystyle v_w=16.0 m/s.\)
The direction is:
\(\displaystyle θ=tan^{−1}(v_{wy}/v_{wx})=tan^{−1}(−9.29/−13.0)\)
giving
\(\displaystyle θ=35.6º.\)
Discussion
The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.
Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.
Relative Velocities and Classical Relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame . These velocities are called relative velocities . For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity , which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.
Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than . Most things we encounter in daily life move slower than this speed.
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (Figure \(\displaystyle \PageIndex{1}\); blue curve) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the red curved path shown in Figure \(\displaystyle \PageIndex{6}\). Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.
Example \(\displaystyle \PageIndex{3}\): Calculating Relative Velocity: An Airline Passenger Drops a Coin
An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?
Strategy
Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.
Solution for (a)
Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:
\(\displaystyle v_y^2=v_{0y}^2−2g(y−y_0).\)
Substituting known values into the equation, we get
\(\displaystyle v_y^2=0^2−2(9.80m/s^2)(−1.50m−0 m)=29.4m^2/s^2\)
yielding
\(\displaystyle v_y=−5.42 m/s.\)
We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.
Solution for (b)
Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is \(\displaystyle v_y=−5.42m/s\), the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and \(\displaystyle v_x=260 m/s\). The x- and y- components of velocity can be combined to find the magnitude of the final velocity:
\(\displaystyle v=\sqrt{v_x^2+v_y^2}\).
Thus,
\(\displaystyle v=\sqrt{(260 m/s)^2+(−5.42 m/s)^2}\)
yielding
\(\displaystyle v=260.06 m/s.\)
The direction is given by:
\(\displaystyle θ=tan^{−1}(v_y/v_x)=tan^{−1}(−5.42/260)\)
so that
\(\displaystyle θ=tan^{−1}(−0.0208)=−1.19º.\)
Discussion
In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not \(\displaystyle (260 – 5.42) m/s\); rather, it is \(\displaystyle 260.06 m/s.\) The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.
MAKING CONNECTIONS: RELATIVITY AND EINSTEIN
Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more surprises await.
PHET EXPLORATIONS: MOTION IN 2D
Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle).
Summary
- Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as
\(\displaystyle v_x=vcosθ\)
\(\displaystyle v_y=vsinθ\)
\(\displaystyle v=\sqrt{v^2_x+v^2+y}\)
\(\displaystyle θ=tan^{−1}(v_y/v_x)\).
- Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
- Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).
Glossary
- classical relativity
- the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s
- relative velocity
- the velocity of an object as observed from a particular reference frame
- relativity
- the study of how different observers moving relative to each other measure the same phenomenon
- velocity
- speed in a given direction
- vector addition
- the rules that apply to adding vectors together
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libretexts
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2025-03-17T19:53:22.878870
| 2015-11-01T04:42:51 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.05%3A_Addition_of_Velocities",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "3.5: Addition of Velocities",
"author": "OpenStax"
}
|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics/3.E%3A_Two-Dimensional_Kinematics_(Exercises)
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3.E: Two-Dimensional Kinematics (Exercises)
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Conceptual Questions
3.2: Vector Addition and Subtraction: Graphical Methods
1. Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?
2. Give a specific example of a vector, stating its magnitude, units, and direction.
3. What do vectors and scalars have in common? How do they differ?
4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?
5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure. What other information would he need to get to Sacramento?
6. Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A+B the sum of the lengths of the two steps?
7. Explain why it is not possible to add a scalar to a vector
8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more?
3.3: Vector Addition and Subtraction: Analytical Methods
9. Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?
10. Give an example of a nonzero vector that has a component of zero.
11. Explain why a vector cannot have a component greater than its own magnitude.
12. If the vectors A and B are perpendicular, what is the component of A along the direction of B ? What is the component of B along the direction of A ?
3.4: Projectile Motion
13. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º ):
(a) Is the velocity ever zero?
(b) When is the velocity a minimum? A maximum?
(c) Can the velocity ever be the same as the initial velocity at a time other than at \(t=0\)?
(d) Can the speed ever be the same as the initial speed at a time other than at \(t=0\)?
14. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither \(0º\) nor \(90º\)):
(a) Is the acceleration ever zero?
(b) Is the acceleration ever in the same direction as a component of velocity?
(c) Is the acceleration ever opposite in direction to a component of velocity?
15. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?
16. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.
3.5: Addition of Velocities
17. What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane?
18. A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn’t he need to keep his eyes on the ball?
19. If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?
20. The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger’s frame of reference. Draw its path as viewed by a stationary observer.
21. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers.
Problems & Exercises
3.2: Vector Addition and Subtraction: Graphical Methods
Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.
22. Find the following for path A in Figure:
(a) the total distance traveled, and
(b) the magnitude and direction of the displacement from start to finish.
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
solution:
(a)
480 m
(b)
379 m
,
18.4º
east of north
23. Find the following for path B in Figure:
(a) the total distance traveled, and
(b) the magnitude and direction of the displacement from start to finish.
24. Find the north and east components of the displacement for the hikers shown in Figure.
Solution
north component 3.21 km, east component 3.83 km
25. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure, then this problem asks you to find their sum R=A+B .)
The two displacements
A
and
B
add to give a total displacement
R
having magnitude \(R\) and direction \(θ\).
26.
Suppose you first walk 12.0 m in a direction
20º
west of north and then 20.0 m in a direction
40.0º
south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements
A
and
B
, as in Figure, then this problem finds their sum
R = A + B
.)
Solution
\(19.5 m, 4.65º\) south of west
27. Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A size 12{A} {}, which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A .)
28. (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R'=A−B ).
(b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R''=B−A=−R' ). Show that this is the case.
Solution
(a) \(26.6 m, 65.1º\) north of east
(b) \(26.6 m, 65.1º\) south of west
29. Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A , B and C , all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A , B , and C can be added; choose only one.)
30. Show that the sum of the vectors discussed in Example gives the result shown in Figure.
Solution
\(52.9 m, 90.1º\) with respect to the x-axis.
31. Find the magnitudes of velocities \(v_A\) and \(v_B\) in Figure
The two velocities \(v_A\) and \(v_B\) add to give a total \(v_{tot}\).
32. Find the components of \(v_{tot}\) along the x- and y-axes in Figure.
Solution
x-component 4.41 m/s
y-component 5.07 m/s
33. Find the components of \(v_{tot}\) along a set of perpendicular axes rotated \(30º\) counterclockwise relative to those in Figure.
3.3: Vector Addition and Subtraction: Analytical Methods
34. Find the following for path C in Figure:
(a) the total distance traveled and
(b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
Solution
(a) 1.56 km
(b) 120 m east
35. Find the following for path D in Figure:
(a) the total distance traveled and
(b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
36. Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure.
Solution
North-component 87.0 km, east-component 87.0 km
37. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure, then this problem asks you to find their sum R=A+B .)
The two displacements
A
and
B
add to give a total displacement
R
having magnitude \(R\) and direction \(θ\).
Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.
38. Repeat Exercise using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.
Solution
30.8 m, 35.8 west of north
You drive 7.50 km in a straight line in a direction 15º .
(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.)
(b) Show that you still arrive at the same point if the east and north legs are reversed in order.
39. a) Do Exercise again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A —that is, finding R'=A – B )
(b) Repeat again, but now you first walk 25.0 m size 12{"25" "." "0 m"} {} north and then 18.0 m size 12{"18" "." "0 m"} {} east. (This is equivalent to subtract A size 12{A} {} from B size 12{B} {} —that is, to find A=B+C size 12{A=B+C} {}. Is that consistent with your result?)
Solution
(a) \(30.8 m, 54.2º\) south of west
(b) \(30.8 m, 54.2º\) north of east
40. A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure. She then correctly calculates the length and orientation of the third side \(C\). What is her result?
41. You fly 32.0 km size 12{"32" "." "0 km"} {} in a straight line in still air in the direction 35.0º size 12{"35"°} {} south of west.
(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)
(b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º .
Solution
18.4 km south, then 26.2 km west(b) 31.5 km at
45.0º
south of west, then 5.56 km at
45.0º
west of north
42. A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A , B , and C in Figure, and then correctly calculates the length and orientation of the fourth side \(D\). What is his result?
43. In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30km 25.0º south of west; then 5.10 km straight east; then 1.70km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?
Solution
\(7.34 km, 63.5º\) south of east
44. Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure. Find her total distance \(R\) from the starting point and the direction \(θ\) of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.
3.4: Projectile Motion
45. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the \(x\) and \(y\) distances from where the projectile was launched to where it lands?
Solution
\(x=1.30 m×10%2\)
\(y=30.9 m.\)
46. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction.
(a) At what speed does the ball hit the ground?
(b) For how long does the ball remain in the air?
(c)What maximum height is attained by the ball?
47. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance.
(a) How long is the ball in the air?
(b) What must have been the initial horizontal component of the velocity?
(c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?
Solution
(a) 3.50 s
(b) 28.6 m/s
(c) 34.3 m/s
(d) 44.7 m/s,
50.2º
below horizontal
48. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s ( 144 km/h ). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long?
(b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)
49. An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow.
(a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems.
(b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?
Solution
(a)
18.4º
(b) The arrow will go over the branch.
50. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
51 . Verify the ranges for the projectiles in Figure(a) for \(θ=45º\) and the given initial velocities.
Solution
\(R=\frac{v^2_0}{sin2θ_0g}\)
For \(θ=45º,R=\frac{v^2_0}{g}\)
\(R=91.8m\) for \(v_0=30m/s; R=163m\) for \(v_0=40m/s; R=255m\) for \(v_=50m/s\).
52. Verify the ranges shown for the projectiles in Figure(b) for an initial velocity of 50 m/s at the given initial angles.
53. The cannon on a battleship can fire a shell a maximum distance of 32.0 km.
(a) Calculate the initial velocity of the shell.
(b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.)
(c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is \(6.37×10^3km\). How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?
Solution
(a) 560 m/s
(b) \(8.00×10^3m\)
(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).
54. An arrow is shot from a height of 1.5 m toward a cliff of height \(H\) size 12{H} {}. It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?
55. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, \(g\) size 12{g} {}. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Solution
1.50 m, assuming launch angle of
45º
56. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.
57. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle \(θ\) below the horizontal. The base line is 11.9 m from the net, which is 0.91 m high. What is the angle \(θ\) such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?
Solution
\(θ=6.1º\)
yes, the ball lands at 5.3 m from the net
58. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield.
(a) If the ball is thrown at an angle of 25º relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground?
(b) How long does it take to get to the receiver?
(c) What is its maximum height above its point of release?
59. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range.
(a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s.
(b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.
Solution
(a) −0.486 m
(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.
60. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.
61. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º size 12{"30º} below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.
Solution
4.23 m. No, the owl is not lucky; he misses the nest.
62. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.
63. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.
Solution
No, the maximum range (neglecting air resistance) is about 92 m.
64. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.
65. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)
Solution
15.0 m/s
66. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity.
(a) What vertical velocity does he need to rise 0.750 m above the floor?
(b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?
67. A football player punts the ball at a 45.0º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally.
(a) What is the initial speed of the ball?
(b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?
Solution
(a) 24.2 m/s
(b) The ball travels a total of 57.4 m with the brief gust of wind.
68. Prove that the trajectory of a projectile is parabolic, having the form \(y=ax+bx^2\). To obtain this expression, solve the equation \(x=v_{0x}t\) for \(t\) and substitute it into the expression for \(y=v_{0y}t–(1/2)gt^2\) (These equations describe the \(x\) and \(y\) positions of a projectile that starts at the origin.) You should obtain an equation of the form \(y=ax+bx^2\) where \(a\) and \(b\) are constants.
69. Derive \(R=\frac{v^2_0sin2θ_0}{g}\) for the range of a projectile on level ground by finding the time \(t\) at which \(y\) becomes zero and substituting this value of \(t\) into the expression for \(x−x_0\), noting that \(R=x−x_0\)
Solution
\(y−y_0=0=v_{0y}t−\frac{1}{2}gt^2=(v_0sinθ)t−\frac{1}{2}gt^2\)
so that \(t=\frac{2(v0sinθ)}{g}\)
\(x−x_0=v_0xt=(v_0cosθ)t=R,\) and substituting for \(t\) gives:
\(R=v_0cosθ(\frac{2v_0sinθ}{g})=\frac{2v^2_0sinθcosθ}{g}\)
since \(2sinθcosθ=sin2θ\), the range is:
\(R=\frac{v_0^2sin2θ}{g}\)
70. Unreasonable Results
(a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s.
(b) What is unreasonable about the range you found?
(c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer.
(d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.
71. Construct Your Own Problem
Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.
3.5: Addition of Velocities
72. Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.
(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction \(45º\) south of east. What was his total displacement?
(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?
(c) What was his total displacement relative to the air mass?
Solution
(a) \(35.8 km, 45º\) south of east
(b) \(5.53 m/s, 45º\) south of east
(c) \(56.1 km, 45º\) south of east
73. A seagull flies at a velocity of 9.00 m/s straight into the wind.
(a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind?
(b) If the bird turns around and flies with the wind, how long will he take to return 6.00 km?
(c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind.
74. Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s.
(a) What is the velocity of the second runner relative to the first?
(b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity?
(c) What distance ahead will the winner be when she crosses the finish line?
Solution
(a) 0.70 m/s faster
(b) Second runner wins
(c) 4.17 m
75. Verify that the coin dropped by the airline passenger in the Example travels 144 m horizontally while falling 1.50 m in the frame of reference of the Earth.
76. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of \(25.0º\) relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback ?
Solution
\(17.0 m/s, 22.1º\)
77. A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction \(40.0º\) north of east. What is the velocity of the ship relative to the Earth?
78. (a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction \(5.0º\) south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction \(15º\) south of east. What is the velocity of the airplane relative to the Earth?
(b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane’s path.
Solution
(a) \(230 m/s, 8.0º\) south of west
(b) The wind should make the plane travel slower and more to the south, which is what was calculated.
79. (a) In what direction would the ship in Exercise have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains \(7.00 m/s\)?
(b) What would its speed be relative to the Earth?
80. (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20º size 12{"20º"} {} south of east (as in Exercise). Its direction of motion relative to the Earth is 45.0º size 12{"45.0º"} {} south of west, while its direction of travel relative to the air is 5.00º size 12{5.00º} {} south of west. What is the airplane’s speed relative to the air mass? (b) What is the airplane’s speed relative to the Earth?
Solution
(a) 63.5 m/s
(b) 29.6 m/s
81. A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship:
(a) relative to the ship and
(b) relative to a stationary observer on shore.
(c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck.
82. The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction \(30.0º\) east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of \(50.0º\) south of west relative to the Earth. What is the velocity of the wind relative to the water?
Solution
\(6.68 m/s, 53.3º\) south of west
83. The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. Figure illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities:
(a) relative to galaxy 2 and
(b) relative to galaxy 5.
The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information.
Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 Mly across.
84. (a) Use the distance and velocity data in Figure to find the rate of expansion as a function of distance.
(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.
Solution
(a) \(H_{average}=14.9\frac{km/s}{Mly}\)
(b) 20.2 billion years
85. An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground?
86. A ship sailing in the Gulf Stream is heading \(25.0º\) west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is \(4.80 m/s 5.00º\) west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)
Solution
\(1.72 m/s, 42.3º\)north of east
87. An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a \(90.0º\) angle relative to his path as shown in Figure. What angle must the puck’s velocity make relative to the player (in his frame of reference) to hit the center of the goal?
An ice hockey player moving across the rink must shoot backward to give the puck a velocity toward the goal.
88. Unreasonable Results
Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth.
(a) At what velocity must the supplies be launched?
(b) What is unreasonable about this velocity?
(c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height?
(d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer.
89. Unreasonable Results
A commercial airplane has an air speed of \(280 m/s\) due east and flies with a strong tailwind. It travels 3000 km in a direction \(5º\) south of east in 1.50 h.
(a) What was the velocity of the plane relative to the ground?
(b) Calculate the magnitude and direction of the tailwind’s velocity.
(c) What is unreasonable about both of these velocities?
(d) Which premise is unreasonable?
90. Construct Your Own Problem
Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway.
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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2025-03-17T19:53:23.089357
| 2017-04-20T14:41:13 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion
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4: Dynamics- Force and Newton's Laws of Motion
-
- 4.4: Newton’s Third Law of Motion- Symmetry in Forces
- There is a passage in the musical Man of la Mancha that relates to Newton’s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, “Of course I hit her back, Your Grace, but she’s a lot harder than me and you know what they say, ‘Whether the stone hits the pitcher or the pitcher hits the stone, it’s going to be bad for the pitcher.’” This is exactly what happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and
-
- 4.5: Normal, Tension, and Other Examples of Forces
- Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text.
-
- 4.6: Problem-Solving Strategies
- Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples.
Thumbnail: For every action there is a reaction. (CC BY-SA 3.0; Benjamin Crowell )
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libretexts
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2025-03-17T19:53:23.158336
| 2015-10-27T19:13:29 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.00%3A_Prelude_to_Dynamics-_Newtons_Laws_of_Motion
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4.0: Prelude to Dynamics- Newton’s Laws of Motion
Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular motion, such as that of a dolphin jumping out of the water, or a pole vaulter, or the flight of a bird, or the orbit of a satellite. The study of motion is kinematics, but kinematics only describes the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of principles under which nature functions. They are also universal laws in that they apply to similar situations on Earth as well as in space.
Isaac Newton’s (1642–1727) laws of motion were just one part of the monumental work that has made him legendary. The development of Newton’s laws marks the transition from the Renaissance into the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. For many centuries natural philosophers had debated the nature of the universe based largely on certain rules of logic with great weight given to the thoughts of earlier classical philosophers such as Aristotle (384–322 BC). Among the many great thinkers who contributed to this change were Newton and Galileo.
Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than “logical” argument. Galileo’s use of the telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons orbiting Jupiter and made other observations that were inconsistent with certain ancient ideas and religious dogma. For this reason, and because of the manner in which he dealt with those in authority, Galileo was tried by the Inquisition and punished. He spent the final years of his life under a form of house arrest. Because others before Galileo had also made discoveries by observing the nature of the universe, and because repeated observations verified those of Galileo, his work could not be suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the church and scientific communities.
Galileo also contributed to the formation of what is now called Newton’s first law of motion. Newton made use of the work of his predecessors, which enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great contributions to the theories of light and color. It is amazing that many of these developments were made with Newton working alone, without the benefit of the usual interactions that take place among scientists today.
It was not until the advent of modern physics early in the 20th century that it was discovered that Newton’s laws of motion produce a good approximation to motion only when the objects are moving at speeds much, much less than the speed of light and when those objects are larger than the size of most molecules (about m in diameter). These constraints define the realm of classical mechanics, as discussed in Introduction to the Nature of Science and Physics . At the beginning of the 20 th century, Albert Einstein (1879–1955) developed the theory of relativity and, along with many other scientists, developed quantum theory. This theory does not have the constraints present in classical physics. All of the situations we consider in this chapter, and all those preceding the introduction of relativity in Special Relativity , are in the realm of classical physics.
MAKING CONNECTIONS: PAST AND PRESENT PHILOSOPHY
The importance of observation and the concept of cause and effect were not always so entrenched in human thinking. This realization was a part of the evolution of modern physics from natural philosophy. The achievements of Galileo, Newton, Einstein, and others were key milestones in the history of scientific thought. Most of the scientific theories that are described in this book descended from the work of these scientists. The importance of observation and the concept of cause and effect were not always so entrenched in human thinking. This realization was a part of the evolution of modern physics from natural philosophy. The achievements of Galileo, Newton, Einstein, and others were key milestones in the history of scientific thought. Most of the scientific theories that are described in this book descended from the work of these scientists.
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2025-03-17T19:53:23.247832
| 2015-11-01T04:45:56 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.01%3A_Development_of_Force_Concept
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4.1: Development of Force Concept
Learning Objectives
By the end of this section, you will be able to:
- Understand the definition of force.
Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definition of force. Our intuitive definition of force —that is, a push or a pull—is a good place to start. We know that a push or pull has both magnitude and direction (therefore, it is a vector quantity) and can vary considerably in each regard. For example, a cannon exerts a strong force on a cannonball that is launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea. Our everyday experiences also give us a good idea of how multiple forces add. If two people push in different directions on a third person, as illustrated in Figure \(\PageIndex{1}\), we might expect the total force to be in the direction shown. Since force is a vector, it adds just like other vectors, as illustrated in Figure \(\PageIndex{1a}\) for two ice skaters. Forces, like other vectors, are represented by arrows and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in Two-Dimensional Kinematics .
Figure \(\PageIndex{1b}\) is our first example of a free-body diagram , which is a technique used to illustrate all the external forces acting on a body. The body is represented by a single isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. (These forces are the only ones shown, because only external forces acting on the body affect its motion. We can ignore any internal forces within the body.) Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton’s laws of motion.
A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a standard distance. One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure \(\PageIndex{2}\), and use the force it exerts to pull itself back to its relaxed shape—called a restoring force —as a standard. The magnitude of all other forces can be stated as multiples of this standard unit of force. Many other possibilities exist for standard forces. (One that we will encounter in Magnetism is the magnetic force between two wires carrying electric current.) Some alternative definitions of force will be given later in this chapter.
TAKE HOME EXPERIMENT: FORCE STANDARDS
To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a hook. Find a small household item that could be attached to the rubber band using a paper clip, and use this item as a weight to investigate the stretch of the rubber band. Measure the amount of stretch produced in the rubber band with one, two, and four of these (identical) items suspended from the rubber band. What is the relationship between the number of items and the amount of stretch? How large a stretch would you expect for the same number of items suspended from two rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached) if the weights are also pushed to the side with a pencil?
Summary
- Dynamics is the study of how forces affect the motion of objects.
- Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and direction.
- External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body.
Glossary
- dynamics
- the study of how forces affect the motion of objects and systems
- external force
- a force acting on an object or system that originates outside of the object or system
- free-body diagram
- a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces are represented by vectors extending outward from the dot
- force
- a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force
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2025-03-17T19:53:23.319892
| 2015-11-01T04:46:22 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.02%3A_Newtons_First_Law_of_Motion_-_Inertia
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4.2: Newton’s First Law of Motion - Inertia
Learning Objectives
- Define mass and inertia.
- Understand Newton's first law of motion.
Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following:
Newton’s First Law of Motion
A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force.
Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion.
Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction) . We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down?
The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force.
Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, “That is the nature of the beast.” True perhaps, but not a useful insight.
Mass
The property of a body to remain at rest or to remain in motion with constant velocity is called inertia . Newton’s first law is often called the law of inertia . As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass . Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram.
Exercise \(\PageIndex{1}\)
Which has more mass: a kilogram of cotton balls or a kilogram of gold?
- Answer
-
They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density.
Summary
- Newton’s first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia .
- Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass.
- Mass is the quantity of matter in a substance.
Glossary
- inertia
- the tendency of an object to remain at rest or remain in motion
- law of inertia
- see Newton’s first law of motion
- mass
- the quantity of matter in a substance; measured in kilograms
- Newton’s first law of motion
- a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force; also known as the law of inertia
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2025-03-17T19:53:23.389273
| 2015-11-01T04:46:43 |
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4.3: Newton’s Second Law of Motion- Concept of a System
Learning Objectives
By the end of this section, you will be able to:
- Define net force, external force, and system.
- Understand Newton’s second law of motion.
- Apply Newton’s second law to determine the weight of an object.
Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration . Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration .
Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct — an external force acts from outside the system of interest. For example, in Figure \(\PageIndex{1a}\) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure \(\PageIndex{1a}\), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external . Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics.
Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure . In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight \(w\) and the support of the ground \(N\), and the horizontal force \(f\) represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure \(\PageIndex{1b}\) shows how vectors representing the external forces add together to produce a net force, \(F_{net}\).
To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality
\[ a \propto F_{net} \]
where the symbol \( \propto \) means “proportional to,” and \( F_{net} \) is the net external force . (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in the chapter section on Two-Dimensional Kinematics .) This proportionality states what we have said in words— acceleration is directly proportional to the net external force . Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification
Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure , the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as
\[ a \propto \dfrac{1}{m}, \]
where \(m\) is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force.
It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion.
Newton’s Second Law of Motion
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton’s second law of motion is
\[ a = \dfrac{F_{net}}{m}\]
This is often written in the more familiar form
\[ F_{net} = ma. \]
When only the magnitude of force and acceleration are considered, this equation is simply
\[ F_{net} = ma. \]
Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification.
Units of Force
\( F_{net} = ma \) is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of \( 1m/s^2\) That is, since \( F_{net} = ma, \) \[ 1N = 1 kg \cdot s^2 \]
While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.
Weight and the Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight \(w\). Weight can be denoted as a vector \(w\) because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as \(w\). Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration \(w\). Using Galileo’s result and Newton’s second law, we can derive an equation for weight.
Consider an object with mass \(m\) falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude \(w\). Newton’s second law states that the magnitude of the net external force on an object is \(F_{net} = ma\).
Since the object experiences only the downward force of gravity, \(F_{net} = w\). We know that the acceleration of an object due to gravity is \(g\), or \( a = g\). Substituting these into Newton’s second law gives
WEIGHT
This is the equation for weight - the gravitational force on mass \(m\):
\[ w = mg\]
Since weight \( g = 9.80 m/s^2 \) on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: \[ w = mg = (1.0 kg) (9.8 m/s^2) = 9.8 N. \]
Recall that \(g\) can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight.
When the net external force on an object is its weight, we say that it is in free-fall . That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object.
The acceleration due to gravity \(g\) varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only \(1.67 m/s^2\). A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body , such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness.
It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons.
COMMON MISCONCEPTIONS: MASS VS. WEIGHT
Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object \((m)\) multiplied by the acceleration due to gravity \((g)\). Like any other force, weight is measured in terms of newtons (or pounds in English units).
Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is \(1.67 m/s^2 \) (which is much less than the acceleration due to gravity on Earth, \(9.80 m/s^2\) ). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you “weigh” much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing “mass” (which in turn causes them to weigh less)
TAKE-HOME EXPERIMENT: MASS AND WEIGHT
What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon?
Example \(\PageIndex{1}\):What Acceleration Can a Person Produce when pushing a Lawn Mower?
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?
Strategy
Since \( F_{net} \) and \( m \) are given, the acceleration can be calculated directly from Newton’s second law as stated in \( F_{net} = ma \).
Solution
The magnitude of the acceleration \(a\) is \(a = \frac{F_{net}}{m}\). Entering known values gives \[ a = \dfrac{51 \, N}{24 \, kg} \]
Substituting the units \( kg \cdot m/s^2 \) for N yields \[ a = \dfrac{ 51 \, kg/s^2}{24\space kg} = 2.1\space m/s^2 \]
Discussion
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would soon be reached.
Example \(\PageIndex{2}\):What Rocket Thrust Accelerates This Sled?
Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust \(T\) for the four-rocket propulsion system shown in Figure. The sled’s initial acceleration is \(49 m/s^2\) the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.
Strategy
Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with \[ F_{net} = ma. \], where \(F_{net}\) is the net force along the horizontal direction. We can see from Figure that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is \[ F_{net} = 4T - f. \]
Substituting this into Newton’s second law gives \[ F_{net} = ma = 4T - f.\]
Using a little algebra, we solve for the total thrust 4T: \[ 4T = ma + f. \]
Substituting known values yields \[ 4T = ma + f = (2100 \, kg)(49 \, m/s^2) + 650 \, N \]
So the total thrust is \[ 1 \times 10^5 N, \]
and the individual thrusts are \[ T = \dfrac{1\times 10^5}{4} = 2.6 \times 10^4 \, N \]
Discussion
The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 \(g\)-s. (Recall that \(g\), the acceleration due gravity is \(9.80 \, m/s^2 \). When we say that an acceleration is 45 \(g\)-s, it is \(45 \times 9.80 m/s^2\), which is approximately \(440 m/s^2\) ). While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.
Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion.
Summary
- Acceleration, \(a\), is defined as a change in velocity, meaning a change in its magnitude or direction, or both.
- An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system.
- Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
- In equation form, Newton’s second law of motion is \( a = \frac{F_{net}}{m} \)
- This is often written in the more familiar form: \( F_{net} = ma. \)
- The weight \(w\) of an object is defined as the force of gravity acting on an object of mass \(m.\) The object experiences an acceleration due to gravity \(g\): \[ w = mg. \]
- If the only force acting on an object is due to gravity, the object is in free fall.
- Friction is a force that opposes the motion past each other of objects that are touching.
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libretexts
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2025-03-17T19:53:23.488866
| 2015-11-01T04:47:03 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.03%3A_Newtons_Second_Law_of_Motion-_Concept_of_a_System",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.04%3A_Newtons_Third_Law_of_Motion-_Symmetry_in_Forces
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4.4: Newton’s Third Law of Motion- Symmetry in Forces
Learning Objectives
By the end of this section, you will be able to:
- Understand Newton’s third law of motion.
- Apply Newton’s third law to define systems and solve problems of motion.
There is a passage in the musical Man of la Mancha that relates to Newton’s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, “Of course I hit her back, Your Grace, but she’s a lot harder than me and you know what they say, ‘Whether the stone hits the pitcher or the pitcher hits the stone, it’s going to be bad for the pitcher.’” This is exactly what happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in Newton’s third law of motion.
NEWTON’S THIRD LAW OF MOTION
Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts.
This law represents a certain symmetry in nature : Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the force exerted is the action and the force experienced as a consequence is the reaction. Newton’s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system.
We can readily see Newton’s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure . She pushes against the pool wall with her feet and accelerates in the direction opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not because they act on different systems . In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then \(F_{wall \space on \, feet} \) is an external force on this system and affects its motion. The swimmer moves in the direction of \(F_{wall \, on \, feet}. \) In contrast, the force \(F_{feet \, on \, wall} \) acts on the wall and not on our system of interest. Thus \(F_{feet \, on \, wall} \) does not directly affect the motion of the system and does not cancel \(F_{wall \, on \, feet}. \) Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction.
Other examples of Newton’s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust . It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho’s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent’s body.
Exercise \(\PageIndex{1}\): Getting up to speed: Choosing the Correct System
A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure. Her mass is 65.0 kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N.
Strategy
Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure. The professor pushes backward with a force \(F_{foot} \) of 150 N. According to Newton’s third law, the floor exerts a forward reaction force \(F_{floor}\) of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, \(f\) opposes the motion and is thus in the opposite direction of \(F_{floor}\). Note that we do not include the forces \(F_{prof}\) or \(F_{cart}\) because these are internal forces, and we do not include \(F_{foot}\) because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton’s second law to find the acceleration as requested. See the free-body diagram in the figure.
Solution
Newton’s second law is given by \[a = \dfrac{F_{net}}{m}. \nonumber \]
The net external force on System 1 is deduced from Figure and the discussion above to be \[F_{net} = F_{floor} - f = 150 \, N - 24.0 \, N = 126 \, N. \nonumber \]
The mass of System 1 is \[m = (65.0 + 12.0 + 7.0) = 84 \, kg.\nonumber \]
These values of \(F_{net} \) and \(m\) produce an acceleration of \[ a = \dfrac{F_{net}}{m} \nonumber \]
\[ a = \dfrac{126 \, N}{84 \, kg} = 1.5 \, m/s^2. \nonumber \]
Discussion
None of the forces between components of System 1, such as between the professor’s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.
Example \(\PageIndex{2}\): Force on the Cart: Choosing a New System
Calculate the force the professor exerts on the cart in Figure using data from the previous example if needed.
Strategy
If we now define the system of interest to be the cart plus equipment (System 2 in Figure), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, \(F_{prof}\) is an external force acting on System 2. \(F_{prof}\) was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2.
Solution
Newton’s second law can be used to find \(F_{prof}\) Starting with \[ a = \dfrac{F_{net}}{m} \nonumber \]
and noting that the magnitude of the net external force on System 2 is \[ F_{net} = F_{prof} - f, \nonumber \]
we solve for \(F_{prof}, \) the desired quantity \[ F_{net} + f.\nonumber \]
The value of \(f\) is given, so we must calculate net \(F_{net}. \) That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that \[ F_{net} = ma, \nonumber \]
where the mass of System 2 is 19.0 kg ( m = 12.0 kg + 7.0 kg) and its acceleration was found to be \(a = 1.5 \, m/s^2 \) in the previous example. Thus, \[F_{net} = ma \nonumber \]
\[ F_{net} = (19.0 \, kg)(1.5 \, m/s^2) = 29 \, N. \nonumber \]
Now we can find the desired force: \[ F_{prof} = F_{net} + f, \nonumber \]
\[F_{prof} = 29 \, N + 24.0 \, N = 53 \, N. \nonumber \]
Discussion
It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor.
The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).
PHET EXPLORATIONS: GRAVITY FORCE LAB
Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force.
Section Summary
- Newton’s third law of motion represents a basic symmetry in nature. It states: Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts.
- A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a thrust reaction force.
Glossary
- Newton’s third law of motion
- whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts
- thrust
- a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force
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libretexts
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2025-03-17T19:53:23.562984
| 2015-11-01T04:47:23 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.05%3A_Normal_Tension_and_Other_Examples_of_Forces
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4.5: Normal, Tension, and Other Examples of Forces
Learning Objectives
By the end of this section, you will be able to:
- Define normal and tension forces.
- Apply Newton’s laws of motion to solve problems involving a variety of forces.
- Use trigonometric identities to resolve weight into components.
Normal Force
Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure (a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure (b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it.
We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol \(N\). (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example.
COMMON MISCONCEPTION: NORMAL FORCE (N) VS. NEWTON (N)
In this section we have introduced the quantity normal force, which is represented by the variable \(N\) This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force \(N\) happen to be newtons (N). For example, the normal force \(N\) hat the floor exerts on a chair might be \(N = 100 \, N\) One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work \(W\) and the unit watts (W).
Example \(\PageIndex{1}\): Weight on an incline, a Two-Dimensional problem
Consider the skier on a slope shown in Figure. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?
Strategy
This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols \( \perp \) and \( \parallel \) to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled \(w,\) \(f\) and \(N\) in Figure. \(N\) is always perpendicular to the slope, and \(f\) is parallel to it. But \(w\) is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining \(w_{\parallel}\) to be the component of weight parallel to the slope and \( w_{\perp} \) the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.
Solution
The magnitude of the component of the weight parallel to the slope is \( w_{\parallel} = w\space sin \, (25^o)\) = mg \, sin \, (25^o), \) and the magnitude of the component of the weight perpendicular to the slope is \( w_{\perp} = w\space cos \, (25^o) = mg \, cos \, (25^o)\).
(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope \(w_{\parallel} \) and friction \(f\). Using Newton’s second law, with subscripts to denote quantities parallel to the slope,
\[ a_{\parallel} = \dfrac{F_{net\parallel}}{m} \] where \(F_{net\parallel} = w_{\parallel} = mg\space sin \, (25^o) \), assuming no friction for this part, so that
\[ a_{\parallel} = \dfrac{F_{net\parallel}}{m} = \dfrac{mg\space sin(25^o)}{m} = g\space sin(25^o) \]
\[ (9.80\space m/s^2)(0.4226) = 4.14 \, m/s^2 \]
is the acceleration.
(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now \[ F_{net\parallel} = w_{\parallel} - f, \] and substituting this into Newton’s second law, \(a_{\parallel} = \frac{F_{net\parallel}}{m}, \) gives
\[a_{\parallel} = \dfrac{F_{net\parallel}}{m} = \dfrac{w_{\parallel} - f}{m} = \dfrac{mg \, sin(25^o) -f}{m}. \]
We substitute known values to obtain
\[a_{\parallel} = \dfrac{(60.0 \, kg)(9.80\space m/s^2)(0.4226) - 45.0\space N}{60.0\space kg}, \]
which yields \[a_{\parallel} = 3.39 \, m/s^2, \]
which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.
Discussion
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is \(a = g\space sin\space \theta, \) regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).
RESOLVING WEIGHT INTO COMPONENTS
When an object rests on an incline that makes an angle \(\theta \) with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, \(w_{\perp} \) and a force acting parallel to the plane, \(w_{\parallel}. \) The perpendicular force of weight, \(w_{\perp}\) is typically equal in magnitude and opposite in direction to the normal force, \(N\). The force acting parallel to the plane, \(w_{\parallel}\) causes the object to accelerate down the incline. The force of friction, \(f\) opposes the motion of the object, so it acts upward along the plane.
It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle \(\theta\) to the horizontal, then the magnitudes of the weight components are
\[w_{\parallel} = w\space sin\space (\theta) = mg \, sin \, (\theta) \] and
\[ w_{\perp} = w\space sin\space (\theta) = mg \, sin \, (\theta) \]
Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle \(\theta\) of the incline is the same as the angle formed between \(w\)and \(w_{\perp} \). Knowing this property, you can use trigonometry to determine the magnitude of the weight components:
\[ cos\space (\theta) = \dfrac{w_{\perp}}{w}\]
\[ w_{\perp} = w\space cos \, (\theta) = mg \, cos \, (\theta) \]
\[ sin \, (\theta) = \dfrac{w_{\parallel}}{w} \]
\[w_{\parallel} = w\space sin \, (\theta) = mg \, sin \, (\theta) \]
TAKE-HOME EXPERIMENT: FORCE PARALLEL
To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show?
Tension
A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word “tension ” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons . Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope.
Consider a person holding a mass on a rope as shown in Figure .
Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus \( F_{net} = 0.\) The only external forces acting on the mass are its weight \(w\) and the tension \(T\) supplied by the rope. Thus, \[ F_{net} = T - w = 0, \] where \(T\) and \(w\) are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:
\[T = w = mg. \]
For a 5.00-kg mass, then (neglecting the mass of the rope) we see that
\[ T = mg = (5.00\space kg)(9.80 \, m/s^2) = 49.0 \, N \]
If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.
Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure (a) and (b).
Example \(\PageIndex{1}\): What Is the Tension in a Tightrope?
Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure.
Strategy
As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight \(w\) and the two tensions \(T_L\) (left tension) and \(T_R\) (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions \(T_L\) and \(T_R\) must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are \(T_L\) and \(T_R\). Thus, the magnitude of those forces must be equal so that they cancel each other out.
Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the \(x\)- axis and the vertical the \(y\)-axis.
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.
Consider the horizontal components of the forces (denoted with a subscript \(x\)):
\[F_{net \, x} = T_{Lx} - T_{Rx}. \]
The net external horizontal force \(F_{net \, x} = 0 \), since the person is stationary. Thus,
\[F_{net \, x} = 0 = T_{Lx} - T_{Rx}\]
\[ T_{Lx} = T_{Rx}.\]
Now, observe Figure. You can use trigonometry to determine the magnitude of \(T_L \) and \(T_R\) Notice that:
\[ cos(5.0^o) = \dfrac{T_{Lx}}{T_L} \]
\[ T_{Lx} = T_L\space cos(5.0^o) \]
\[ cos(5.0^o) = \dfrac{T_{Lx}}{T_L} \]
\[T_{Rx} = T_R \, cos(5.0^o) \]
Equating \(T_{Lx} \) and \(T_{Rx} \):
\[ T_L \, cos(5.0^o) = T_R \, cos(5.0^o) \]
\[ T_L = T_R = T \]
as predicted. Now, considering the vertical components (denoted by a subscript \(y\)), we can solve for \(T\). Again, since the person is stationary, Newton’s second law implies that net \(F_y = 0 \). Thus, as illustrated in the free-body diagram in Figure,
\[ F_{net \, y}= T_{Ly} + T_{Ry} - w = 0 \]
Observing Figure, we can use trigonometry to determine the relationship between \(T_{Ly},\space T_{Ry}\) and \( T \). As we determined from the analysis in the horizontal direction, \(T_L = T_R = T \).
\[ sin (5.0^o) = \dfrac{T_{Ly}}{T_L} \]
\[T_{Ly} = T_L \, sin (5.0^o) = T \, sin(5.0^o) \]
\[ sin (5.0^o) = \dfrac{T_{Ry}}{T_R} \]
\[ T_{Ry} = T_R \, sin(5.0^o) = T \, sin (5.0^o) \]
Now, we can substitute the values for \(T_{Ly} \) and \(T_{Ry}\), into the net force equation in the vertical direction:
\[ F_{net \, y} = T_{Ly} + T_{Ry} - w = 0 \]
\[ F_{net \, y} = T \, sin(5.0^o) + T \, sin(5.0^o) - w = 0 \]
\[ 2T \, sin(5.0^o) -w = 0 \]
\[ 2T \, sin(5.0^o) = w \]
and \[ T = \dfrac{w}{2 \, sin(5.0^o)} = \dfrac{mg}{2 \, sin(5.0^o)} \]
so that \[ T = \dfrac{(70.0 \, kg)(9.80 \, m/s^2)}{2 (0.0872)}, \]
\[ T = 3900 \, N. \]
Discussion
Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.
If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in Figure . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:
\[ T = \dfrac{w}{2 \, sin \, (\theta)} \]
We can extend this expression to describe the tension \(T\) created when a perpendicular force \((F_{\perp}) \) is exerted at the middle of a flexible connector:
\[ T = \dfrac{F_{\perp}}{2 \, sin \, (\theta)}. \]
Note that \(\theta\) is the angle between the horizontal and the bent connector. In this case, \(T\) becomes very large as \(\theta\) approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., \(\theta = 0 \) and \(sin \, \theta = 0 \)). (See Figure .)
Extended Topic: Real Forces and Inertial Frames
There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth’s frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter.
Earth’s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed.
The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.
All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text.
PHET EXPLORATIONS: FORCES IN 1 DIMENSION
Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).
Summary
- When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, \(T\)
- When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:
\[ N = mg \]
- When objects rest on an inclined plane that makes an angle \(\theta \) with the horizontal surface, the weight of the object can be resolved into components that act perpendicular \((w_{\perp})\) and parallel \(w_{\parallel}) \) to the surface of the plane. These components can be calculated using:
\[w_{\parallel} = w \, sin \, (\theta) = mg \, sin \, (\theta) \]
\[ w_{\perp} = w \, cos \, (\theta) = mg \, cos \, (\theta) \]
- The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, \(T\.) When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:
\[ T = mg. \]
- In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this chapter and all forces are real forces having a physical origin.
Glossary
- inertial frame of reference
- a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference
- normal force
- the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests
- tension
- the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension force
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libretexts
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2025-03-17T19:53:23.688856
| 2015-11-01T04:47:43 |
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"title": "4.5: Normal, Tension, and Other Examples of Forces",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.06%3A_Problem-Solving_Strategies
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4.6: Problem-Solving Strategies
Learning Objectives
By the end of this section, you will be able to:
- Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.
Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.
Problem-Solving Strategy for Newton’s Laws of Motion
Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure (a). Then, as in Figure (b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).
Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.
A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.
Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.
APPLYING NEWTON'S SECOND LAW
Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: \(F_{net} = ma\)
For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions: \[ F_{net \, x} = ma \]
\[ F_{net \, y} = 0 \]
You will need this information in order to determine unknown forces acting in a system.
Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.
Summary
- To solve problems involving Newton’s laws of motion, follow the procedure described:
- Draw a sketch of the problem.
- Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
- Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the \(x\)-direction) then \(F_{net \, x} = 0 \). If the object does accelerate in that direction, \(F_{net \, x} = ma \).
- Check your answer. Is the answer reasonable? Are the units correct?
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libretexts
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2025-03-17T19:53:23.765339
| 2015-11-01T04:48:05 |
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.07%3A_Further_Applications_of_Newtons_Laws_of_Motion
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4.7: Further Applications of Newton’s Laws of Motion
Learning Objectives
By the end of this section, you will be able to:
- Apply problem-solving techniques to solve for quantities in more complex systems of forces.
- Integrate concepts from kinematics to solve problems using Newton’s laws of motion.
There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills.
Example \(\PageIndex{1}\): Drag Force on a Barge
Suppose two tugboats push on a barge at different angles, as shown in Figure. The first tugboat exerts a force of \(2.7 \times 10^5 \, N \) in the x-direction, and the second tugboat exerts a force of \(3.6 \times 10^5 \, N \) in the y-direction.
If the mass of the barge is \(5.0 \times 10^6 \space kg \) and its acceleration is observed to be \(7.5 \times 10^{-2} \, m/s^2 \) in the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.)
Strategy
The directions and magnitudes of acceleration and the applied forces are given in Figure (a). We will define the total force of the tugboats on the barge as \(F_{app} \) so that:\[ F_{app} = F_x + F_y \nonumber \]
Since the barge is flat bottomed, the drag of the water \(F_D\) will be in the direction opposite to \(F_{app} \) as shown in the free-body diagram in Figure (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied force \(F_{app} \), and then apply Newton’s second law to solve for the drag force \(F_D\).
Solution
Since \(F_x\) and \(F_y\) are perpendicular, the magnitude and direction of \(F_{app}\) are easily found. First, the resultant magnitude is given by the Pythagorean theorem:
\[F_{app} = \sqrt{F_x^2 + F_y^2} \nonumber \]
\[F_{app} = \sqrt{(2.7 \times 10^5 \, N)^2 + (3.6 \times 10^5 \, N)^2} = 4.5 \times 10^5 \, N. \nonumber \]
The angle is given by \[ \theta = tan^{-1} \left(\dfrac{F_y}{F_x} \right) \nonumber \]
\[ \theta = tan^{-1} \left( \dfrac{3.6 \times 10^5 \, N}{2.7 \times 10^5 \, N} \right) = 53^o, \nonumber \]
which we know, because of Newton’s first law, is the same direction as the acceleration. \(F_D\) is in the opposite direction of \(F_{app} \), since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as \(F_{app} \), but its magnitude is slightly less than \(F_{app} \). The problem is now one-dimensional. From Figure (b) we can see that
\[F_{net} = F_{app} - F_D. \nonumber \] But Newton’s second law states that \[ F_{net} = ma \nonumber \]
Thus, \[ F_{app} - F_D = ma \nonumber \]
This can be solved for the magnitude of the drag force of the water \(F_D\) in terms of known quantities: \[ F_D = F_{app} - ma \nonumber \] Substituting known values gives \[ F_D = (4.5 \times 10^5 \, N) - (5.0 \times 10^6 \, kg)(7.5 \times 10^{-2} \, m/s^2) = 7.5 \times 10^4 \, N \nonumber \]
The direction of \(F_D\) has already been determined to be in the direction opposite to \(F_{app} \) or at an angle of \(53^o\) south of west.
Discussion
The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where \(F_D\) is less than 1/600th of the weight of the ship.
In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved.
Example \(\PageIndex{2}\): Different Tensions at Different Angles
Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure. Find the tension in each wire, neglecting the masses of the wires.
Strategy
The system of interest is the traffic light, and its free-body diagram is shown in Figure(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem (\(T_1\) and T_2\)), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.
Solution
First consider the horizontal or x-axis: \[F_{net \, x} = T_{2x} - T_{1x} = 0. \nonumber \]
Thus, as you might expect, \[T_{1x} = T{2x} \nonumber \]
This gives us the following relationship between \(T_1\) and \(T_2\): \[T_1 \, cos \, 30^o = T_2 \, cos\space 45^o \nonumber \]
Thus, \[ T_2 = (1.225)T_1. \nonumber \]
Note that \( T_1 \) and \(T_2\) are not equal in this case, because the angles on either side are not equal. It is reasonable that \(T_2\) ends up being greater than \(T_1\), because it is exerted more vertically than \(T_1\).
Now consider the force components along the vertical or y-axis:
\[ F_{net \, y} = T_{1y} +T_{2y} - w= 0 \nonumber \]
This implies \[ T_{1y} +T_{2y} = w \nonumber \]
Substituting the expressions for the vertical components gives \[ T_1 \, sin \, (30^o) + T_2 \, sin \, (45^o) = w. \nonumber \]
There are two unknowns in this equation, but substituting the expression for \(T_2\) in terms of \(T_1\) reduces this to one equation with one unknown:
\[T_1(0.500) + (1.225T_1)(0.707) = w = mg \nonumber \]
which yields \[ (1.366)T_1 = (15.0 \, kg)(9.80 \, m/s^2). \nonumber \]
Solving this last equation gives the magnitude of \(T_1\) to be \[ T_1 = 108 \, N. \nonumber \]
Finally, the magnitude of \(T_2\) is determined using the relationship between them, \(T_2= 1.225 T_1 \) found above. Thus we obtain
\[ T_2 = 132 \, N. \nonumber \]
Discussion
Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker).
The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example.
Example \(\PageIndex{3}\): What does the Bathroom Scale Read in an Elevator?
Figure shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of \(1.20 m/s^2 \) and (b) if the elevator moves upward at a constant speed of 1 m/s.
Strategy
If the scale is accurate, its reading will equal \(F_p\) the magnitude of the force the person exerts downward on it. Figure (a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure (b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight \(w\) and the upward force of the scale \(F_s.\) According to Newton’s third law \(F_p\) and \(F_s\) are equal in magnitude and opposite in direction, so that we need to find \(F_s\) in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,
\[F_{net} = ma \nonumber \]
From the free-body diagram we see that \(F_{net} = F_s - w, \) so that \[ F_s - w = ma. \nonumber \]
Solving for \(F_s\) gives an equation with only one unknown: \[ F_s = ma + w, \nonumber \]
or, because \(w = mg \), simply \[ F_s = ma + mg. \nonumber \]
No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in this exercise.
Solution for (a)
In this part of the problem, \(a = 1.20 m/s^2\), so that
\[ F_s = (75.0 \, kg)(1.20 \, m/s^2) + (75.0 \, kg)(9.80 \, m/s^2), \nonumber \]
yielding \[ F_s = 825 \, N. \nonumber \]
Discussion for (a)
This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:
\[ F_{net} = ma = 0 = F_s - w\nonumber \]
\[ F_s = w = mg \nonumber \]
\[F_s = (75.0 \, kg)(9.80 \, m/s^2) \nonumber \]
\[F_s = 735 \, N. \nonumber \]
So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators.
Solution for (b)
Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because \(a = \frac{\Delta v}{\Delta t} \) and \(\Delta v = 0 \).
Thus, \[F_s = ma + mg = 0 + mg. \nonumber \]
Now \[ F_s = (75.0 \, kg)(9.80 \, m/s^2), \nonumber \]
which gives \[ F_s = 735 \, N. \nonumber \]
Discussion for (b)
The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.
The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, \(a\) is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and accelerating downward at \(g\), then the scale reading will be zero and the person will appear to be weightless.
Integrating Concepts: Newton’s Laws of Motion and Kinematics
Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem:
Problem-Solving Strategy
Step 1. Identify which physical principles are involved . Listing the givens and the quantities to be calculated will allow you to identify the principles involved.
Step 2. Solve the problem using strategies outlined in the text . If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem.
Example \(\PageIndex{4}\): What Force Must a Soccer Player Exert to Reach Top Speed?
A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.
Strategy
- To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of dynamics found in this chapter.
- The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth.
Solution for (a)
We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is \(\Delta v = 8.00 \, m/s \). We are given the elapsed time, and so \(\Delta t = 2.50 \, s\). The unknown is acceleration, which can be found from its definition:
\[ a = \dfrac{\Delta v}{\Delta t}. \nonumber \]
Substituting the known values yields \[ a = \dfrac{8.00 \, m/s}{2.50 \, s} = 3.20 \, m/s^2 \nonumber \]
Discussion for (a)
This is an attainable acceleration for an athlete in good condition.
Solution for (b)
Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is,
\[ F_{net} = ma\nonumber \]
Substituting the known values of \(m\) and \(a\) gives
\[ F_{net} = (70 \, kg)(3.2 \, m/s^2) = 224 \, N. \nonumber \]
Discussion for (b)
This is about 50 pounds, a reasonable average force.
This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.
Summary
- Newton’s laws of motion can be applied in numerous situations to solve problems of motion.
- Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether \(F_{net} = ma\) or \(F_{net} = 0 \).
- The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object.
- Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion.
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2025-03-17T19:53:23.850397
| 2015-11-01T04:48:27 |
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4.8: Extended Topic- The Four Basic Forces—An Introduction
Learning Objectives
By the end of this section, you will be able to:
- Understand the four basic forces that underlie the processes in nature.
One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact, nearly all of the forces we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational force is the only force we experience directly that is not electromagnetic.) This is a tremendous simplification of the myriad of apparently different forces we can list, only a few of which were discussed in the previous section. As we will see, the basic forces are all thought to act through the exchange of microscopic carrier particles, and the characteristics of the basic forces are determined by the types of particles exchanged. Action at a distance, such as the gravitational force of Earth on the Moon, is explained by the existence of a force field rather than by “physical contact.”
The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. Their properties are summarized in Table. Since the weak and strong nuclear forces act over an extremely short range, the size of a nucleus or less, we do not experience them directly, although they are crucial to the very structure of matter. These forces determine which nuclei are stable and which decay, and they are the basis of the release of energy in certain nuclear reactions. Nuclear forces determine not only the stability of nuclei, but also the relative abundance of elements in nature. The properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly determine the chemistry of the atom. More will be said of all of these topics in later chapters.
CONCEPT CONNECTIONS: THE FOUR BASIC FORCES
The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined in Uniform Circular Motion and Gravitation , electric force in Electric Charge and Electric Field , magnetic force in Magnetism , and nuclear forces in Radioactivity and Nuclear Physics . On a macroscopic scale, electromagnetism and gravity are the basis for all forces. The nuclear forces are vital to the substructure of matter, but they are not directly experienced on the macroscopic scale.
| Force | Approximate relative strengths | Range | Attraction/Repulsion | Carrier Particle |
|---|---|---|---|---|
| Gravitational | \(10^{-38}\) | \(\infty\) | attractive only | Graviton |
| Electromagnetic | \(10^{-2}\) | \(\infty\) | attractive and repulsive | Photon |
| Weak Nuclear | \(10^{-13}\) | \(<10^{-18} \, m\) | attractive and repulsive | \(W^+, \, W^-, \, Z^0 \) |
| Strong Nuclear | \(1\) | \(<10^{-15} \, m\) | attractive and repulsive | gluons |
The gravitational force is surprisingly weak—it is only because gravity is always attractive that we notice it at all. Our weight is the gravitational force due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational force is the dominant force determining the motions of moons, planets, stars, and galaxies. The gravitational force also affects the nature of space and time. As we shall see later in the study of general relativity, space is curved in the vicinity of very massive bodies, such as the Sun, and time actually slows down near massive bodies.
Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large distances, and they nearly cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they did not cancel, electromagnetic forces would completely overwhelm the gravitational force. The electromagnetic force is a combination of electrical forces (such as those that cause static electricity) and magnetic forces (such as those that affect a compass needle). These two forces were thought to be quite distinct until early in the 19th century, when scientists began to discover that they are different manifestations of the same force. This discovery is a classical case of the unification of forces . Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to electromagnetic interactions of atoms and molecules. It is still convenient to consider these forces separately in specific applications, however, because of the ways they manifest themselves.
CONCEPT CONNECTIONS: UNIFYING FORCES
Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By “unify” we mean finding connections between the forces that show that they are different manifestations of a single force. Even if such unification is achieved, the forces will retain their separate characteristics on the macroscopic scale and may be identical only under extreme conditions such as those existing in the early universe.
Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come under the rubric of Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known that under conditions of extremely high density and temperature, such as existed in the early universe, the electromagnetic and weak nuclear forces are indistinguishable. They can now be considered to be different manifestations of one force, called the electroweak force. So the list of four has been reduced in a sense to only three. Further progress in unifying all forces is proving difficult—especially the inclusion of the gravitational force, which has the special characteristics of affecting the space and time in which the other forces exist.
While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity exists in the face of the overt complexity of the universe. There is no reason that nature must be simple—it simply is.
Action at a Distance: Concept of a Field
All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming into contact. It is also true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not actually touch. What is it that carries forces between objects? One way to answer this question is to imagine that a force field surrounds whatever object creates the force. A second object (often called a test object ) placed in this field will experience a force that is a function of location and other variables. The field itself is the “thing” that carries the force from one object to another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the test object placed in it. Earth’s gravitational field, for example, is a function of the mass of Earth and the distance from its center, independent of the presence of other masses. The concept of a field is useful because equations can be written for force fields surrounding objects (for gravity, this yields \(w = mg\) at Earth’s surface), and motions can be calculated from these equations. (See Figure.)
CONCEPT CONNECTIONS: FORCE FIELDS
The concept of a force field is also used in connection with electric charge and is presented in Electric Charge and Electric Field . It is also a useful idea for all the basic forces, as will be seen in Particle Physics . Fields help us to visualize forces and how they are transmitted, as well as to describe them with precision and to link forces with subatomic carrier particles.
The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field equations. As useful as the field concept is, however, it leaves unanswered the question of what carries the force. It has been proposed in recent decades, starting in 1935 with Hideki Yukawa’s (1907–1981) work on the strong nuclear force, that all forces are transmitted by the exchange of elementary particles. We can visualize particle exchange as analogous to macroscopic phenomena such as two people passing a basketball back and forth, thereby exerting a repulsive force without touching one another. (See Figure.)
This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of something physical actually moving between objects acting at a distance. Table lists the exchange or carrier particles , both observed and proposed, that carry the four forces. But the real fruit of the particle-exchange proposal is that searches for Yukawa’s proposed particle found it and a number of others that were completely unexpected, stimulating yet more research. All of this research eventually led to the proposal of quarks as the underlying substructure of matter, which is a basic tenet of GUTs. If successful, these theories will explain not only forces, but also the structure of matter itself. Yet physics is an experimental science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN laboratory in Switzerland are starting to test these theories using the world’s largest particle accelerator: the Large Hadron Collider. This accelerator (27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy of 14 trillion electron volts will be available. It is anticipated that some new particles, possibly force carrier particles, will be found. (See Figure .) One of the force carriers of high interest that researchers hope to detect is the Higgs boson. The observation of its properties might tell us why different particles have different masses.
Tiny particles also have wave-like behavior, something we will explore more in a later chapter. To better understand force-carrier particles from another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of years. Almost 100 years ago, Einstein predicted the existence of these waves as part of his general theory of relativity. Gravitational waves are created during the collision of massive stars, in black holes, or in supernova explosions—like shock waves. These gravitational waves will travel through space from such sites much like a pebble dropped into a pond sends out ripples—except these waves move at the speed of light. A detector apparatus has been built in the U.S., consisting of two large installations nearly 3000 km apart—one in Washington state and one in Louisiana! The facility is called the Laser Interferometer Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the relative positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these small effects to be separated from other natural phenomena, such as earthquakes. Initial operation of the detectors began in 2002, and work is proceeding on increasing their sensitivity. Similar installations have been built in Italy (VIRGO), Germany (GEO600), and Japan (TAMA300) to provide a worldwide network of gravitational wave detectors.
International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space Antenna). Earthquakes and other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO by looking at much more massive black holes through the observation of gravitational-wave sources emitting much larger wavelengths. Three satellites will be placed in space above Earth in an equilateral triangle (with 5,000,000-km sides) ( Figure ). The system will measure the relative positions of each satellite to detect passing gravitational waves. Accuracy to within 10% of the size of an atom will be needed to detect any waves. The launch of this project might be as early as 2018.
“I’m sure LIGO will tell us something about the universe that we didn’t know before. The history of science tells us that any time you go where you haven’t been before, you usually find something that really shakes the scientific paradigms of the day. Whether gravitational wave astrophysics will do that, only time will tell.” —David Reitze, LIGO Input Optics Manager, University of Florida.
The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in later chapters.
Summary
- The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces in nature.
- The properties of these forces are summarized in Table .
- Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational forces. The nuclear forces are responsible for the submicroscopic structure of matter, but they are not directly sensed because of their short ranges. Attempts are being made to show all four forces are different manifestations of a single unified force.
- A force field surrounds an object creating a force and is the carrier of that force.
Footnotes
The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational waves later in this section. The particles \(W^+, \, W^-,\) and \(Z^0\) are called vector bosons; these were predicted by theory and first observed in 1983. There are eight types of gluons proposed by scientists, and their existence is indicated by meson exchange in the nuclei of atoms.
Glossary
- carrier particle
- a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles of the electromagnetic force
- force field
- a region in which a test particle will experience a force
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2025-03-17T19:53:23.977271
| 2015-11-01T04:48:47 |
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"title": "4.8: Extended Topic- The Four Basic Forces—An Introduction",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.E%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion_(Exercises)
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4.E: Dynamics- Force and Newton's Laws of Motion (Exercises)
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Conceptual Questions
4.1: Development of Force Concept
1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
2. What properties do forces have that allow us to classify them as vectors?
4.2: Newton’s First Law of Motion: Inertia
3. How are inertia and mass related?
4. What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
4.3: Newton’s Second Law of Motion: Concept of a System
5. Which statement is correct? Explain your answer and give an example.
(a) Net force causes motion.
(b) Net force causes change in motion.
6. Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?
7. Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion.
8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.
9. system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.
10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
11. (a) Give an example of different net external forces acting on the same system to produce different accelerations.
(b) Give an example of the same net external force acting on systems of different masses, producing different accelerations.
(c) What law accurately describes both effects? State it in words and as an equation.
12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.
13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?
14. The gravitational force on the basketball in Figure is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal?
4.4: Newton’s Third Law of Motion: Symmetry in Forces
15. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)
16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?
17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton’s laws of motion apply?
18. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired?
19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.
20. Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects whether one such pair of forces cancels.
4.5: Normal, Tension, and Other Examples of Forces
21. If a leg is suspended by a traction setup as shown in Figure, what is the tension in the rope?
A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force \(\displaystyle T\) without changing its magnitude.
22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure.) (Note that the tibia is the shin bone shown in this image.)
4.7: Further Applications of Newton’s Laws of Motion
23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at \(\displaystyle g\). Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?
24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.
4.8: Extended Topic: The Four Basic Forces—An Introduction
25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force.
26. What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances?
27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.)
Problems & Exercises
4.3: Newton’s Second Law of Motion: Concept of a System
You may assume data taken from illustrations is accurate to three digits.
28. A 63.0-kg sprinter starts a race with an acceleration of \(\displaystyle 4.20 m/s^2\). What is the net external force on him?
Solution
265 N
29. If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?
30. A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of its acceleration.
Solution
\(\displaystyle 13.3 m/s^2\)
31. Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be \(\displaystyle 0.893 m/s^2\).
(a) Calculate her mass.
(b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.
32. In Figure 4.4.3, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force \(\displaystyle F\) (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force \(\displaystyle F\) is removed. How far will the mower go before stopping?
Solution
1.1 m
33. The same rocket sled drawn in Figure is decelerated at a rate of \(\displaystyle 196 m/s^2\). What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.
34. (a) If the rocket sled shown in Figure starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is \(\displaystyle 2.4×10^4\) N, and the force of friction opposing the motion is known to be 650 N.
(b) Why is the acceleration not one-fourth of what it is with all rockets burning?
Solution
(a) \(\displaystyle 11m/s^2\)
(b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.
35. What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)
36. Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.
(a) What is the system of interest if the acceleration of the child in the wagon is to be calculated?
(b) Draw a free-body diagram, including all forces acting on the system.
(c) Calculate the acceleration.
(d) What would the acceleration be if friction were 15.0 N?
Solution
(a) The system is the child in the wagon plus the wagon.
(b)
(c) \(\displaystyle a=0.130m/s^2\) in the direction of the second child’s push.
(d) \(\displaystyle a=0.00m/s^2\)
37. A powerful motorcycle can produce an acceleration of \(\displaystyle 3.50m/s^2\)while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?
38. The rocket sled shown in Figure accelerates at a rate of \(\displaystyle 49.0m/s^2\). Its passenger has a mass of 75.0 kg.
(a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio.
(b) Calculate the direction and magnitude of the total force the seat exerts against his body.
Solution
(a) \(\displaystyle 3.68×10^3N\). This force is 5.00 times greater than his weight.
(b) \(\displaystyle 3750 N; 11.3ºabove horizontal\)
39.
Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of \(\displaystyle 201 m/s^2\). In this problem, the forces are exerted by the seat and restraining belts.
40. The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth?
Solution
\(\displaystyle 1.5×10^3N,150 kg,150 kg\)
41. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.
4.4: Newton’s Third Law of Motion: Symmetry in Forces
42. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at \(\displaystyle 2.40×10^4m/s^2\)? What is the magnitude of the force exerted on the ship by the artillery shell?
Solution
Force on shell: \(\displaystyle 2.64×10^7N\)
Force exerted on ship = \(\displaystyle −2.64×10^7N\), by Newton’s third law
43. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20m/s2 size 12{1 "." "20"" m/s" rSup { size 8{2} } } {} backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.
4.5: Normal, Tension, and Other Examples of Forces
44. Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally.
(a) What is magnitude of the acceleration of the two teams?
(b) What is the tension in the section of rope between the teams?
Solution
a. \(\displaystyle 0.11 m/s^2\)
b. \(\displaystyle 1.2×10^4N\)
45. What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at \(\displaystyle 7.50 m/s^2\)? Note that the answer is independent of the velocity of the gymnast—she can be moving either up or down, or be stationary.
46. (a) Calculate the tension in a vertical strand of spider web if a spider of mass \(\displaystyle 8.00×10^{−5}kg\) hangs motionless on it.
(b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure. The strand sags at an angle of \(\displaystyle 12º\) below the horizontal. Compare this with the tension in the vertical strand (find their ratio).
Solution
(a) \(\displaystyle 7.84×10^{-4}N\)
(b) \(\displaystyle 1.89×10^{–3}N\). This is 2.41 times the tension in the vertical strand.
47. Suppose a 60.0-kg gymnast climbs a rope.
(a) What is the tension in the rope if he climbs at a constant speed?
(b) What is the tension in the rope if he accelerates upward at a rate of \(\displaystyle 1.50 m/s^2\)?
48. Show that, as stated in the text, a force \(\displaystyle F_{⊥}\) exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in Figure) gives rise to a tension of magnitude \(\displaystyle T=\frac{F_⊥}{2sin(θ)}\).
Solution
Newton’s second law applied in vertical direction gives
\(\displaystyle F_y=F−2Tsinθ=0\)
\(\displaystyle F=2Tsinθ\)
\(\displaystyle T=\frac{F}{2 sinθ}\).
49.
Consider the baby being weighed in Figure.
(a) What is the mass of the child and basket if a scale reading of 55 N is observed?
(b) What is the tension \(\displaystyle T_1\) in the cord attaching the baby to the scale?
(c) What is the tension \(\displaystyle T_2\) in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.
A baby is weighed using a spring scale.
4.6: Problem-Solving Strategies
50. A \(\displaystyle 5.00×10^5-kg\) rocket is accelerating straight up. Its engines produce \(\displaystyle 1.250×10^7N\) of thrust, and air resistance is \(\displaystyle 4.50×10^6N\). What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Solution
Using the free-body diagram:
\(\displaystyle F_{net}=T−f−mg=ma\)
so that
\(\displaystyle a=\frac{T−f−mg}{m}=\frac{1.250×10^7N−4.50×10^6N−(5.00×10^5kg)(9.80 m/s^2)}{5.00×10^5kg}=6.20m/s^2\)
51. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is \(\displaystyle 1.80 m/s^2\), what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.
52. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Solution
1.Use Newton’s laws of motion.
2. Given : \(\displaystyle a=4.00g=(4.00)(9.80 m/s^2)=39.2m/s^2; m=70.0 kg.\)
Find: \(\displaystyle F\)
3. \(\displaystyle \sum{F=+F−w=ma,}\) so that \(\displaystyle F=ma+w=ma+mg=m(a+g)\)
\(\displaystyle F=(70.0 kg)[(39.2 m/s^2)+(9.80 m/s^2)]=3.43×10^3N\) The force exerted by the high-jumper is actually down on the ground, but F size 12{F} is up from the ground and makes him jump.
4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of \(\displaystyle 10^3N\)
53. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
54. A freight train consists of two \(\displaystyle 8.00×10^4-kg\) engines and 45 cars with average masses of \(\displaystyle 5.50×10^4kg\).
(a) What force must each engine exert backward on the track to accelerate the train at a rate of \(\displaystyle 5.00×10^{–2}m/s^2\) if the force of friction is \(\displaystyle 7.50×10^5N\), assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems.
(b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?
Solution
(a) \(\displaystyle 4.41×10^5N\)
(b) \(\displaystyle 1.50×10^5N\)
55. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor.
(a) An 1800-kg tractor exerts a force of \(\displaystyle 1.75×10^4N\) backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is \(\displaystyle 0.150 m/s^2\), what is the mass of the airplane?
(b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane.
(c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.
56. A 1100-kg car pulls a boat on a trailer.
(a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of \(\displaystyle 0.550 m/s^2\)? The mass of the boat plus trailer is 700 kg.
(b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?
Solution
(a) \(\displaystyle 910 N\)
(b) \(\displaystyle 1.11×10^3N\)
57. (a) Find the magnitudes of the forces \(\displaystyle F_1\) and \(\displaystyle F_2\) that add to give the total force \(\displaystyle F_{tot}\) shown in Figure. This may be done either graphically or by using trigonometry.
(b) Show graphically that the same total force is obtained independent of the order of addition of \(\displaystyle F_1\) and \(\displaystyle F_2\)
(c) Find the direction and magnitude of some other pair of vectors that add to give \(\displaystyle F_{tot}\). Draw these to scale on the same drawing used in part (b) or a similar picture.
58. Two children pull a third child on a snow saucer sled exerting forces \(\displaystyle F_1\) and \(\displaystyle F_2\) as shown from above in Figure. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of \(\displaystyle F_1\) and \(\displaystyle F_2\).
Solution
\(\displaystyle a=0.139 m/s, θ=12.4º\) north of east
An overhead view of a child sitting on a snow saucer sled.
59. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure to pull it out.
(a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
(b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?
60. What force is exerted on the tooth in Figure if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.
Solution
1. Use Newton’s laws since we are looking for forces.
2. Draw a free-body diagram:
3. The tension is given as \(\displaystyle T=25.0 N\). Find \(\displaystyle F_{app}\). Using Newton’s laws gives: \(\displaystyle Σ F_y=0\), so that applied force is due to the
y
-components of the two tensions: \(\displaystyle F_{app}=2Tsinθ=2(25.0 N)sin(15º)=12.9 N\)
The
x
-components of the tension cancel. \(\displaystyle ∑F_x=0\).
4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.
Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, \(\displaystyle F_{app}\), points straight toward the back of the mouth.
61. Figure shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible.
(a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope.
(b) Find the tension in the rope above Superhero.
(c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.
Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?
62. A nurse pushes a cart by exerting a force on the handle at a downward angle \(\displaystyle 35.0º\) below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N.
(a) Draw a free-body diagram for the system of interest.
(b) What force must the nurse exert to move at a constant velocity?
63. Construct Your Own Problem
Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.
64. Construct Your Own Problem
Consider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.
65. Unreasonable Results
(a) Repeat Exercise, but assume an acceleration of \(\displaystyle 1.20 m/s^2\) is produced.
(b) What is unreasonable about the result?
(c) Which premise is unreasonable, and why is it unreasonable?
66. Unreasonable Results
(a) What is the initial acceleration of a rocket that has a mass of \(\displaystyle 1.50×10^6kg\) at takeoff, the engines of which produce a thrust of \(\displaystyle 2.00×10^6N\)? Do not neglect gravity.
(b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.)
(c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)
4.7: Further Applications of Newton’s Laws of Motion
67. A flea jumps by exerting a force of \(\displaystyle 1.20×10^{−5}N\) straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of \(\displaystyle 0.500×10^{−6}N\) on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is \(\displaystyle 6.00×10^{−7}kg\). Do not neglect the gravitational force.
Solution
\(\displaystyle 10.2m/s^2, 4.67º\) from vertical
68. Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?
Achilles tendon
69. A 76.0-kg person is being pulled away from a burning building as shown in Figure. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.
Solution
\(\displaystyle T_1=736 N\)
\(\displaystyle T_2=194 N\)
The force \(\displaystyle T_2\) needed to hold steady the person being rescued from the fire is less than her weight and less than the force \(\displaystyle T_1\) size 12{T rSub { size 8{1} } } {} in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force).
70. Integrated Concepts
A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)
71. Integrated Concepts
When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s.
(a) What is his final speed?
(b) How far does he travel?
Solution
(a) \(\displaystyle 7.43 m/s\)
(b) \(\displaystyle 2.97 m\)
72 . Integrated Concepts
A large rocket has a mass of \(\displaystyle 2.00×10^6kg\) at takeoff, and its engines produce a thrust of \(\displaystyle 3.50×10^7N\).
(a) Find its initial acceleration if it takes off vertically.
(b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?
(c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.
73. Integrated Concepts
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor.
(a) Calculate his velocity when he leaves the floor.
(b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m.
(c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.
Solution
(a) \(\displaystyle 4.20 m/s\)
(b) \(\displaystyle 29.4m/s^2\)
(c) \(\displaystyle 4.31×10^3N\)
74. Integrated Concepts
A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m.
(a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar.
(b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a).
(c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.
75. Integrated Concepts Repeat Exercise for a shell fired at an angle \(\displaystyle 10.0º\) from the vertical.
Solution
(a) \(\displaystyle 47.1 m/s\)
(b) \(\displaystyle 2.47×10^3m/s^2\)
(c) \(\displaystyle 6.18×10^3N\) .
76. Integrated Concepts
An elevator filled with passengers has a mass of 1700 kg.
(a) The elevator accelerates upward from rest at a rate of \(\displaystyle 1.20 m/s^2\). Calculate the tension in the cable supporting the elevator.
(b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time?
(c) The elevator decelerates at a rate of \(\displaystyle 0.600 m/s^2\) for 3.00 s. What is the tension in the cable during deceleration?
(d) How high has the elevator moved above its original starting point, and what is its final velocity?
77. Unreasonable Results
(a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of \(\displaystyle 0.400 m/s^2\) for 50.0 s?
(b) What is unreasonable about the result?
(c) Which premise is unreasonable, or which premises are inconsistent?
78. Unreasonable Results
A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s.
(a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.)
(b) What is unreasonable about the result?
(c) Which premise is unreasonable, or which premises are inconsistent?
4.8: Extended Topic: The Four Basic Forces—An Introduction
79. (a) What is the strength of the weak nuclear force relative to the strong nuclear force?
(b) What is the strength of the weak nuclear force relative to the electromagnetic force? Since the weak nuclear force acts at only very short distances, such as inside nuclei, where the strong and electromagnetic forces also act, it might seem surprising that we have any knowledge of it at all. We have such knowledge because the weak nuclear force is responsible for beta decay, a type of nuclear decay not explained by other forces.
Solution
(a) \(\displaystyle 1×10^{−13}\)
(b) \(\displaystyle 1×10^{−11}\)
80. (a) What is the ratio of the strength of the gravitational force to that of the strong nuclear force?
(b) What is the ratio of the strength of the gravitational force to that of the weak nuclear force?
(c) What is the ratio of the strength of the gravitational force to that of the electromagnetic force? What do your answers imply about the influence of the gravitational force on atomic nuclei?
81. What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? Based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes greater than the range of the strong nuclear force. At these sizes, the electromagnetic force begins to affect nuclear stability. These facts will be used to explain nuclear fusion and fission later in this text.
Solution
\(\displaystyle 10^2\)
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:24.107148
| 2017-04-20T14:38:09 |
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"title": "4.E: Dynamics- Force and Newton's Laws of Motion (Exercises)",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity
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5: Further Applications of Newton's Laws- Friction, Drag, and Elasticity
It is difficult to categorize forces into various types (aside from the four basic forces discussed in previous chapter). We know that a net force affects the motion, position, and shape of an object. It is useful at this point to look at some particularly interesting and common forces that will provide further applications of Newton’s laws of motion. We have in mind the forces of friction, air or liquid drag, and deformation.
-
- 5.1: Friction
- Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can gain.
-
- 5.2: Drag Forces
- You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object.
-
- 5.3: Elasticity - Stress and Strain
- A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the force—that is, for small deformations, Hooke’s law is obeyed.
Thumbnail: Weight (W), the frictional force (F r ), and the normal force (F n ) impacting a cube. Weight is mass (m) multiplied by gravity (g). (CC-SA-BY-3.0; Email4mobile ).
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| 2015-10-27T19:17:41 |
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"title": "5: Further Applications of Newton's Laws- Friction, Drag, and Elasticity",
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5.0: Prelude to Further Applications of Newton’s Laws
Describe the forces on the hip joint. What means are taken to ensure that this will be a good movable joint? From the photograph (for an adult) in Figure , estimate the dimensions of the artificial device.
It is difficult to categorize forces into various types (aside from the four basic forces discussed in previous chapter). We know that a net force affects the motion, position, and shape of an object. It is useful at this point to look at some particularly interesting and common forces that will provide further applications of Newton’s laws of motion. We have in mind the forces of friction, air or liquid drag, and deformation.
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libretexts
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2025-03-17T19:53:24.239561
| 2015-11-01T04:49:35 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.00%3A_Prelude_to_Further_Applications_of_Newtons_Laws",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "5.0: Prelude to Further Applications of Newton’s Laws",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.01%3A_Friction
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5.1: Friction
Learning Objectives
By the end of this section, you will be able to:
- Discuss the general characteristics of friction.
- Describe the various types of friction.
- Calculate the magnitude of static and kinetic friction.
Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which it behaves.
Friction
Friction is a force that opposes relative motion between systems in contact.
One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction . For example, friction slows a hockey puck sliding on ice. But when objects are stationary , static friction can act between them; the static friction is usually greater than the kinetic friction between the objects.
Kinetic Friction
If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.
Imagine, for example, trying to slide a heavy crate across a concrete floor—you may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction force. If you add mass to the crate, say by placing a box on top of it, you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as you might expect).
Figure is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be rough. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. A considerable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is nearly independent of speed.
\(\displaystyle \PageIndex{1}\):
Frictional forces, such as \(\displaystyle f\), always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less than those with leather soles.
The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction).
When there is no motion between the objects, the magnitude of static friction \(\displaystyle f_s\) is
\(\displaystyle f_s≤μ_sN,\)
where \(\displaystyle μ_s\) is the coefficient of static friction and \(\displaystyle N\) is the magnitude of the normal force (the force perpendicular to the surface).
MAGNITUDE OF STATIC FRICTION
Magnitude of static friction \(\displaystyle f_s\) is
\(\displaystyle f_s≤μ_sN\),
where \(\displaystyle μ_s\) is the coefficient of static friction and \(\displaystyle N\) is the magnitude of the normal force.
The symbol \(\displaystyle ≤\) means less than or equal to , implying that static friction can have a minimum and a maximum value of μsN. Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Thus
\(\displaystyle f_{s(max)}=μ_sN\).
Once an object is moving, the magnitude of kinetic friction \(\displaystyle f_k\) is given by
\(\displaystyle f_k=μ_kN\),
where \(\displaystyle μ_k\) is the coefficient of kinetic friction. A system in which \(\displaystyle f_k=μ_kN\) is described as a system in which friction behaves simply.
MAGNITUDE OF KINETIC FRICTION
The magnitude of kinetic friction \(\displaystyle f_k\) is given by
\(\displaystyle f_k=μ_kN\),
where \(\displaystyle μ_k\) is the coefficient of kinetic friction.
As seen in Table, the coefficients of kinetic friction are less than their static counterparts. That values of \(\displaystyle μ\) in Table are stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two equations.
| Static Friction | Kinetic Friction | |
| System | \(\displaystyle μ_s\) | \(\displaystyle μ_k\) |
| Rubber on dry concrete | 1.0 | 0.7 |
| Rubber on wet concrete | 0.7 | 0.5 |
| Wood on wood | 0.5 | 0.3 |
| Waxed wood on wet snow | 0.14 | 0.1 |
| Metal on wood | 0.5 | 0.3 |
| Steel on steel (dry) | 0.6 | 0.3 |
| Steel on steel (oiled) | 0.05 | 0.063 |
| Teflon on steel | 0.04 | 0.04 |
| Bone lubricated by synovial fluid | 0.016 | 0.015 |
| Shoes on wood | 0.9 | 0.7 |
| Shoes on ice | 0.1 | 0.05 |
| Ice on ice | 0.1 | 0.03 |
| Steel on ice | 0.04 | 0.02 |
Coefficients of Static and Kinetic Friction
The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, \(\displaystyle W=mg=(100 kg)(9.80m/s^2)=980 N\), perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than \(\displaystyle f_{s(max)}=μ_sN=(0.45)(980N)=440N\) to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N \(\displaystyle (f_k=μ_kN=(0.30)(980N)=290N)\) would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact.
TAKE-HOME EXPERIMENT
Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly important for drivers to note, especially after a light rain shower. Why?
Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction—often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction.
\(\displaystyle \PageIndex{2}\):
Artificial knee replacement is a procedure that has been performed for more than 20 years. In this figure, we see the post-op x rays of the right knee joint replacement. (credit: Mike Baird, Flickr)
Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor’s clinics. For example, when ultrasonic imaging is carried out, the gel that couples the transducer to the skin also serves to lubricate the surface between the transducer and the skin—thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to move freely over the skin.
Exercise \(\displaystyle \PageIndex{1}\): Skiing Exercise
A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.
Strategy
The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force \(\displaystyle N\) as \(\displaystyle f_k=μ_kN\); thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope. (See the skier and free-body diagram in Figure.)
\(\displaystyle \PageIndex{3}\): The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). \(\displaystyle N\) (the normal force) is perpendicular to the slope, and \(\displaystyle f\) (the friction) is parallel to the slope, but \(\displaystyle w\) (the skier’s weight) has components along both axes, namely \(\displaystyle w_⊥\) and \(\displaystyle W_{//}\). \(\displaystyle N\) is equal in magnitude to \(\displaystyle w_⊥\), so there is no motion perpendicular to the slope. However, \(\displaystyle f\) is less than \(\displaystyle W_{//}\) in magnitude, so there is acceleration down the slope (along the x-axis).
That is,
\(\displaystyle N=w_⊥=wcos25º=mgcos25º.\)
Substituting this into our expression for kinetic friction, we get
\(\displaystyle f_k=μ_kmgcos25º,\)
which can now be solved for the coefficient of kinetic friction \(\displaystyle μ_k\).
Solution
Solving for \(\displaystyle μ_k\) gives
\(\displaystyle μ_k=\frac{f_k}{N}=\frac{f_k}{wcos25º}=\frac{f_k}{mgcos25º}\)
Substituting known values on the right-hand side of the equation,
\(\displaystyle μ_k=\frac{45.0 N}{(62 kg)(9.80 m/s^2)(0.906)}=0.082\).
Discussion
This result is a little smaller than the coefficient listed in Table for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass m slides down a slope that makes an angle \(\displaystyle θ\) with the horizontal, friction is given by \(\displaystyle f_k=μ_kmgcosθ\). All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this chapter’s Problems and Exercises.
TAKE-HOME EXPERIMENT
An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in Example, the kinetic friction on a slope \(\displaystyle f_k=μ_kmgcosθ\). The component of the weight down the slope is equal to \(\displaystyle mgsinθ\) (see the free-body diagram in Figure). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out:
\(\displaystyle f_k=Fg_x\)
\(\displaystyle μ_kmgcosθ=mgsinθ\).
Solving for \(\displaystyle μ_k\), we find that
\(\displaystyle μ_k=\frac{mgsinθ}{mgcosθ}=tanθ\).
Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find \(\displaystyle μ_k\). Note that the coin will not start to slide at all until an angle greater than \(\displaystyle θ\) is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may affect the value for \(\displaystyle μ_k\) and its uncertainty.
We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force.
MAKING CONNECTIONS: SUBMICROSCOPIC EXPLANATIONS OF FRICTION
The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat.
Figure illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area.
\(\displaystyle \PageIndex{4}\):
Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied force, the area of actual contact increases as does friction.
But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of \(\displaystyle 10^{12}\) ) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction.
\(\displaystyle \PageIndex{5}\):
The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic nature of friction.
PHET EXPLORATIONS: FORCES AND MOTION
Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces).
\(\displaystyle \PageIndex{6}\): Forces and Motion
Summary
- Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force \(\displaystyle N\) pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction \(\displaystyle f_s\) between systems stationary relative to one another is given by
\(\displaystyle f_s≤μ_sN\),
where \(\displaystyle μ_s\) is the coefficient of static friction, which depends on both of the materials.
- The kinetic friction force \(\displaystyle f_k\) between systems moving relative to one another is given by
\(\displaystyle f_k=μ_kN\),
where \(\displaystyle μ_k\) is the coefficient of kinetic friction, which also depends on both materials.
|
libretexts
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2025-03-17T19:53:24.330041
| 2015-11-01T04:49:53 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.01%3A_Friction",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "5.1: Friction",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.02%3A_Drag_Forces
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5.2: Drag Forces
Learning Objectives
By the end of this section, you will be able to:
- Express mathematically the drag force.
- Discuss the applications of drag force.
- Define terminal velocity.
- Determine the terminal velocity given mass.
Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force \(F_D\) is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as \(F_D\propto v^2 \). When taking into account other factors, this relationship becomes
\[F_D = \dfrac{1}{2} C_{\rho} Av^2, \label{5.3.1}\]
where \(C\) is the drag coefficient, \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as \(F_D = bv^2, \) where \(b\) is a constant equivalent to \(0.5 C\rho A.\) We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.
Definition: DRAG FORCE
Drag Force \(F_D\) is found to be proportional to the square of the speed of the object. Mathematically \[F_D\propto v^2 \]
\[F_D = \dfrac{1}{2} C_{\rho} Av^2, \] where \(C\) is the drag coefficient \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid.
Athletes as well as car designers seek to reduce the drag force to lower their race times. (See Figure). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.
The value of the drag coefficient, \(C\) is determined empirically, usually with the use of a wind tunnel. (See Figure).
The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).
| Object | C |
|---|---|
| Airfoil | 0.05 |
| Toyota Camry | 0.28 |
| Ford Focus | 0.32 |
| Honda Civic | 0.36 |
| Ferrari Testarossa | 0.37 |
| Dodge Ram pickup | 0.43 |
| Sphere | 0.45 |
| Hummer H2 SUV | 0.64 |
| Skydiver (feet first) | 0.70 |
| Bicycle | 0.90 |
| Skydiver (horizontal) | 1.0 |
| Circular flat plate | 1.12 |
Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (See Figure). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.
Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as given by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity \((v_t)\). Since \(F_D\) is proportional to the speed, a heavier skydiver must go faster for \(F_D\) to equal his weight. Let’s see how this works out more quantitatively.
At the terminal velocity,
\[ F_{net} = mg - F_D = ma = 0\]
Thus,
\[ mg = F_D. \]
Using the equation for drag force, we have \[ mg = \dfrac{1}{2}\rho C Av^2. \]
Solving for the velocity, we obtain \[ v = \sqrt{\dfrac{2mg}{\rho C A}} \]
Assume the density of air is \(\rho = 1.21 \, kg/m^3. \) A 75-kg skydiver descending head first will have an area approximately \(A = 0.18 \, m^2 \) and a drag coefficient of approximately \(C = 0.70. \) We find that
\[v = \sqrt{\dfrac{2(75 \, kg)(9.80 \, m/s^2)}{(1.21 \, kg/m^3)(0.70)(0.18 \, m^2)}} \]
\[ = 98 \, m/s \]
\[ = 350 \, km/h. \]
This means a skydiver with a mass of 75 kg achieves a maximum terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.
TAKE-HOME EXPERIMENT
This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity \(v\) versus mass. Also plot \(v^2\) versus mass. Which of these relationships is more linear? What can you conclude from these graphs?
Example \(\PageIndex{1}\): The terminal Velocity
Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.
Strategy
At terminal velocity, \(F_{net} = 0 \) Thus the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find \(mg = \frac{1}{2} \rho C A v^2\).
Thus the terminal velocity \(v_t\) can be written as
\[v_t = \sqrt{\dfrac{2mg}{\rho CA}} \]
Solution
All quantities are known except the person’s projected area. This is an adult (85 kg) falling spread eagle. We can estimate the frontal area as \[ A = (2 \, m)(0.35 \, m) = 0.70 \, m^2 \]
Using our equation for \(v_t,\) we find that
\[ v_t = \sqrt{\dfrac{2(85\space kg)(9.80 \, m/s^2)}{91.21 \, kg/m^3)(1.0)(0.70 \, m^2)}} \]
\[ = 44 \, m/s\]
Discussion
This result is consistent with the value for \(v_t\) mentioned earlier. The 75-kg skydiver going feet first had a \(v = 98 \, m/s.\) He weighed less but had a smaller frontal area and so a smaller drag due to the air.
The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.
The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled “On Being the Right Size.”
To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.
The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law , which states that \[ F_S = 6 \pi r \eta v \] where \(r\) is the radius of the object, \(\eta\) is the viscosity of the fluid and \(v\) is the object's velocity.
Definition: STOKES LAW
\[ F_S = 6 \pi r \eta v \]
where \(r\) is the radius of the object, \(\eta\) is the viscosity of the fluid and \(v\) is the object's velocity.
Good examples of this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about \(1 \, \mu m\)) can be about \(2 \, m/s.\) To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell. Sediment in a lake can move at a greater terminal velocity (about \(5 \, \mu m/s,\)) so it can take days to reach the bottom of the lake after being deposited on the surface.
If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fishes, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a streamlined pattern (see Figure). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.
GALILEO’S EXPERIMENT
Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each to reach the ground. Since stopwatches weren’t readily available, how do you think he measured their fall time? If the objects were the same size, but with different masses, what do you think he should have observed? Would this result be different if done on the Moon?
Summary
- Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity \(v\) in air, the drag force is given by \[ F_D = \dfrac{1}{2}C \rho A v^2,\] where \(C\) is the drag coefficient (typical values are given in Table), \(A\) is the area of the object facing the fluid, and \(\rho\) is the fluid density.
- For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law,
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libretexts
|
2025-03-17T19:53:24.547135
| 2015-11-01T04:50:14 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.02%3A_Drag_Forces",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "5.2: Drag Forces",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.03%3A_Elasticity_-_Stress_and_Strain
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5.3: Elasticity - Stress and Strain
Learning Objectives
By the end of this section, you will be able to:
- State Hooke’s law.
- Explain Hooke’s law using graphical representation between deformation and applied force.
- Discuss the three types of deformations such as changes in length, sideways shear and changes in volume.
- Describe with examples the young’s modulus, shear modulus and bulk modulus.
- Determine the change in length given mass, length and radius.
We now move from consideration of forces that affect the motion of an object (such as friction and drag) to those that affect an object’s shape. If a bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation . Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the force—that is, for small deformations, Hooke’s law is obeyed. In equation form, Hooke’s law is given by
\[F = k \Delta L, \]
where \(\Delta L \) is the amount of deformation (the change in length, for example) produced by the force \(F\), and \(k\) is a proportionality constant that depends on the shape and composition of the object and the direction of the force. Note that this force is a function of the deformation \(\Delta L\) it is not constant as a kinetic friction force is. Rearranging this to
\[ \Delta L = \dfrac{F}{k} \]
makes it clear that the deformation is proportional to the applied force. Figure shows the Hooke’s law relationship between the extension \(\Delta L\) of a spring or of a human bone. For metals or springs, the straight line region in which Hooke’s law pertains is much larger. Bones are brittle and the elastic region is small and the fracture abrupt. Eventually a large enough stress to the material will cause it to break or fracture. Tensile strength is the breaking stress that will cause permanent deformation or fracture of a material.
HOOKE'S LAW
\[ F = k\Delta L\]
where \(\Delta L\) is the amount of deformation (the change in length, for example) produced by the force \(F\), and \(k\) is a proportionality constant that depends on the shape and composition of the object and the direction of the force. \[\Delta L = \dfrac{F}{k} \]
The proportionality constant \(k\) depends upon a number of factors for the material. For example, a guitar string made of nylon stretches when it is tightened, and the elongation \(\Delta L\) is proportional to the force applied (at least for small deformations). Thicker nylon strings and ones made of steel stretch less for the same applied force, implying they have a larger \(k\) (see Figure). Finally, all three strings return to their normal lengths when the force is removed, provided the deformation is small. Most materials will behave in this manner if the deformation is less than about 0.1% or about 1 part in \(10^3\).
STRETCH YOURSELF A LITTLE
How would you go about measuring the proportionality constant \(k\) of a rubber band? If a rubber band stretched 3 cm when a 100-g mass was attached to it, then how much would it stretch if two similar rubber bands were attached to the same mass—even if put together in parallel or alternatively if tied together in series?
We now consider three specific types of deformations: changes in length (tension and compression), sideways shear (stress), and changes in volume. All deformations are assumed to be small unless otherwise stated.
Changes in Length—Tension and Compression: Elastic Modulus
A change in length \(\Delta L\) is produced when a force is applied to a wire or rod parallel to its length \(L_0\), either stretching it (a tension) or compressing it. (See Figure.)
Experiments have shown that the change in length \((\Delta L)\) depends on only a few variables. As already noted, \(\Delta L\) is proportional to the force \(F\) and depends on the substance from which the object is made. Additionally, the change in length is proportional to the original length \(L_0\) and inversely proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these factors into one equation for \(\Delta L\):
\[\Delta L = \dfrac{1}{Y}\dfrac{F}{A} L_0,\]
where \(\Delta L\) is the change in length, \(F\) the applied force, \(Y\) is a factor, called the elastic modulus or Young’s modulus, that depends on the substance, \(A\) is the cross-sectional area, and \(L_0\) is the original length. Table lists values of \(Y\) for several materials—those with a large \(Y\) are said to have a large tensile stifness because they deform less for a given tension or compression.
| Material | Young’s modulus (tension–compression) Y \((10^9 \, N/m^2)\) | Sheer modulus S \((10^9 \, N/m^2 )\) | Bulk modulus B \((10^9 \, N/m^2)\) |
|---|---|---|---|
| Aluminum | 70 | 25 | 75 |
| Bone – tension | 16 | 80 | 8 |
| Bone – compression | 9 | ||
| Brass | 90 | 35 | 75 |
| Brick | 15 | ||
| Concrete | 20 | ||
| Glass | 70 | 20 | 30 |
| Granite | 45 | 20 | 45 |
| Hair (human) | 10 | ||
| Hardwood | 15 | 10 | |
| Iron, cast | 100 | 40 | 90 |
| Lead | 16 | 5 | 50 |
| Marble | 60 | 20 | 70 |
| Nylon | 5 | ||
| Polystyrene | 3 | ||
| Silk | 6 | ||
| Spider thread | 3 | ||
| Steel | 210 | 80 | 130 |
| Tendon | 1 | ||
| Acetone | 0.7 | ||
| Ethanol | 0.9 | ||
| Glycerin | 4.5 | ||
| Mercury | 25 | ||
| Water | 2.2 |
Young’s moduli are not listed for liquids and gases in Table because they cannot be stretched or compressed in only one direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitude \(F\) acting in opposite directions. For example, the strings in Figure are being pulled down by a force of magnitude \(w\) and held up by the ceiling, which also exerts a force of magnitude \(w\).
Example \(\PageIndex{1}\): The Stretch of a Long Cable
Suspension cables are used to carry gondolas at ski resorts. (See Figure) Consider a suspension cable that includes an unsupported span of 3020 m. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is \(3 \times 10^6 \, N\)
Strategy
The force is equal to the maximum tension, or \( F = 3 \times 10^6 \, N. \) The cross-sectional area is \(\pi r^2 = 2.46 \times 10^{-3} m^2.\) The equation \(\Delta l = \frac{1}{Y} \frac{F}{A} L_0 \) can be used to find the change in length.
Solution
All quantities are known. Thus,
\[ \Delta L = \left( \dfrac{1}{210 \times 10^9 \, N/m^2} \right) \left( \dfrac{3.0 \times 10^6 \, N}{2.46 \times 10^{-3} \, m^2} \right ) (3020 \, m)\]\[ = 18 \, m\]
Discussion
This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments.
Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behavior of bones under tension and compression is important because it determines the load the bones can carry. Bones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus the bone in the top of the femur is arranged in thin sheets separated by marrow while in other places the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in bone joints and tendons.
Another biological example of Hooke’s law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone) must stretch easily at first when a force is applied, but offer a much greater restoring force for a greater strain. Figure shows a stress-strain relationship for a human tendon. Some tendons have a high collagen content so there is relatively little strain, or length change; others, like support tendons (as in the leg) can change length up to 10%. Note that this stress-strain curve is nonlinear, since the slope of the line changes in different regions. In the first part of the stretch called the toe region, the fibers in the tendon begin to align in the direction of the stress—this is called uncrimping . In the linear region, the fibrils will be stretched, and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this chapter. Ligaments (tissue connecting bone to bone) behave in a similar way.
Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The elastic properties of the arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch when the blood is pumped out of the heart. When the aortic valve shuts, the pressure in the arteries drops and the arterial walls relax to maintain the blood flow. When you feel your pulse, you are feeling exactly this—the elastic behavior of the arteries as the blood gushes through with each pump of the heart. If the arteries were rigid, you would not feel a pulse. The heart is also an organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely and elastically when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg with no visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in elasticity starts in the early 20s.
Example \(\PageIndex{1}\): Calculating Deformation: How Much Does Your Leg Shorten
When You Stand on It?
Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius.
Strategy
The force is equal to the weight supported, or \[ F = mg = (62.0 \, kg)(9.80 \, m/s^2) = 607.6 \, N, \] and the cross-sectional area is \(\pi r^2 = 1.257 \times 10^{-3} m^2. \) The equation \(\Delta L = \frac{1}{Y} \frac{F}{A} L_0 \) can be used to find the change in length.
Solution
All quantities except \(\Delta L\) are known. Note that the compression value for Young’s modulus for bone must be used here. Thus,
\[\Delta L = \left( \dfrac{1}{9 \times 10^9 \, N/m^2} \right) \left(\dfrac{607.6 \, N}{1.257 \times 10^{-3}} \right) (0.400 \, m) \]
\[ = 2 \times 10^{-5} \, m \]
Discussion
This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table have larger values of Young’s modulus \(Y\). In other words, they are more rigid.
The equation for change in length is traditionally rearranged and written in the following form:
\[\dfrac{F}{A} = Y \dfrac{\Delta L}{L_0}.\] The ratio of force to area, \(\frac{F}{A} \) is defined as stress (measured in \(N/m^2q\), and the ratio of the change in length to length, \(\frac{\Delta L}{L_0} \) is defined as strain (a unitless quantity). In other words,
\[ stress = Y \times strain. \]
In this form, the equation is analogous to Hooke’s law, with stress analogous to force and strain analogous to deformation. If we again rearrange this equation to the form \[ F = YA \dfrac{\Delta L}{L_0}, \] we see that it is the same as Hooke’s law with a proportionality constant \[ k = \dfrac{YA}{L_0}.\]
This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in length, sideways bending, and changes in volume.
STRESS
The ratio of force to area, \(\frac{F}{A}\) is defined as stress measured in \(N/m^2\).
STRAIN
The ratio of the change in length to length, \(\frac{\Delta L}{L_0} \) is defined as strain (a unitless quantity). In other words,
\[ stress = Y \times strain \]
Sideways Stress: Shear Modulus
Figure illustrates what is meant by a sideways stress or a shearing force . Here the deformation is called \(\Delta x\) and it is perpendicular to \(L_0\), rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with similar equations. The expression for shear deformation is
\[ \Delta x = \dfrac{1}{S} \dfrac{F}{A}L_0,\]
where \(S\) is the shear modulus (see Table) and \(F\) is the force applied perpendicular to \(L_0\) and parallel to the cross-sectional area \(A\). Again, to keep the object from accelerating, there are actually two equal and opposite forces \(F\) applied across opposite faces, as illustrated in Figure. The equation is logical—for example, it is easier to bend a long thin pencil (small \(A\)) than a short thick one, and both are more easily bent than similar steel rods (large \(S\)).
SHEAR DEFORMATION
\[ \Delta x = \dfrac{1}{S} \dfrac{F}{A}L_0,\] where \(S\) is the shear modulus (see Table) and \(F\) is the force applied perpendicular to \(L_0\) and parallel to the cross-sectional area \(A\).
Examination of the shear moduli in Table reveals some telling patterns. For example, shear moduli are less than Young’s moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young’s modulus, but it is as large as that of steel. This is why bones are so rigid.
The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some of both. Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge shaped disc below the last vertebrae) is particularly at risk because of its location.
The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces.
Example \(\PageIndex{3}\): Calculating Force Required to Deform: That Nail Does Not
Bend much Under a Load.
Find the mass of the picture hanging from a steel nail as shown in Figure, given that the nail bends only \(1.80 \mu m\). (Assume the shear modulus is known to two significant figures.)
Strategy.
The force \(F\) on the nail (neglecting the nail’s own weight) is the weight of the picture \(w\). If we can find \(w\), then the mass of the picture is just \(\frac{w}{g} \). The equation \(\Delta x = \frac{1}{S} \frac{F}{A}L_0 \) can be solved for \(F\).
Solution
Solving the equation \(\Delta x = \frac{1}{S} \frac{F}{A}L_0 \) for \(F\), we see that all other quantities can be found:
\[ F = \dfrac{SA}{L_0} \Delta x. \]
\(S\) is found in Table and is \(S = 80 \times 10^9 \, N/m^2 \). The radius \(r\) is 0.750 mm (as seen in the figure), so the cross-sectional area is
\[A = \pi r^2 = 1.77 \times 10^{-6} \, m^2. \]
The value for \(L_0\) is also shown in the figure. Thus,
\[ F = \dfrac{(80 \times 10^9 \, N/m^2)(1.77 \times 10^{-6} \, m^2)}{(5.00 \times 10^{-3} \, m)}(1.80 \times 10^{-6} \, m) = 51 \]
This 51 N force is the weight \(w\) of the picture, so the picture’s mass is
\[ m = \dfrac{w}{g} = \dfrac{F}{g} = 5.2 \, kg. \]
Discussion
This is a fairly massive picture, and it is impressive that the nail flexes only \(1.80 \mu m\) —an amount undetectable to the unaided eye.
Changes in Volume: Bulk Modulus
An object will be compressed in all directions if inward forces are applied evenly on all its surfaces as in Figure. It is relatively easy to compress gases and extremely difficult to compress liquids and solids. For example, air in a wine bottle is compressed when it is corked. But if you try corking a brim-full bottle, you cannot compress the wine—some must be removed if the cork is to be inserted. The reason for these different compressibilities is that atoms and molecules are separated by large empty spaces in gases but packed close together in liquids and solids. To compress a gas, you must force its atoms and molecules closer together. To compress liquids and solids, you must actually compress their atoms and molecules, and very strong electromagnetic forces in them oppose this compression.
We can describe the compression or volume deformation of an object with an equation. First, we note that a force “applied evenly” is defined to have the same stress, or ratio of force to area \(\frac{F}{A} \) on all surfaces. The deformation produced is a change in volume \(\Delta V,\) which is found to behave very similarly to the shear, tension, and compression previously discussed. (This is not surprising, since a compression of the entire object is equivalent to compressing each of its three dimensions.) The relationship of the change in volume to other physical quantities is given by
\[ \Delta V = \dfrac{1}{B} \dfrac{F}{A} V_0, \]
where \(B\) is the bulk modulus (see Table), \(V_0\) is the original volume, and \(\frac{F}{A} \) is the force per unit area applied uniformly inward on all surfaces. Note that no bulk moduli are given for gases.
What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example illustrates.
Example \(\PageIndex{4}\): Calculating Change in Volume with Deformation: How much
is Water Compressed at Great Ocean Depths?
Calculate the fractional decrease in volume \((\frac{\Delta V}{V_0}) \) for seawater at 5.00 km depth, where the force per unit area is \(5.00 \times 10^7 \, N/m^2.\)
Strategy
Equation \(\Delta V = \frac{1}{B} \frac{F}{A} V_0 \) is the correct physical relationship. All quantities in the equation except \(\frac{\Delta V}{V_0} \) are known.
Solution
Solving for the unknown \(\frac{\Delta V}{V_0} \) gives \[ \dfrac{\Delta V}{V_0} = \dfrac{1}{B} \dfrac{F}{A}. \]
Substituting known values with the value for the bulk modulus \(B\) from Table,
\[\dfrac{\Delta V}{V_0} = \dfrac{5.00 \times 10^7 \, N/m^2}{2.2 \times 10^9 \, N/m^2} = 0.023 \]
Discussion
Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress.
Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way.
Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here.
Phet explorations: Masses & Springs
A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring.
Summary
Hooke’s law is given by \[ F = k \Delta L\] where \(L\) is the amount of deformation (the change in length), \(F\) is the applied force, and \(k\) is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as
\[ \Delta L = \dfrac{1}{Y} \dfrac{F}{A} L_0, \] where \(L\) is Young’s modulus , which depends on the substance, \(A\) is the cross-sectional area, and \(L_0\) is the original length. The ratio of force to area, \(\frac{F}{A} \) is defined as stress , measured in N/m 2 .The ratio of the change in length to length, \(\frac{\Delta L}{L_0}, \) is defined as strain (a unitless quantity). In other words, \[ stress = Y \times strain. \]
The expression for shear deformation is \[ \Delta x = \dfrac{1}{S} \dfrac{F}{A} L_0, \]
where \(S\) is the shear modulus and \(F\) is the force applied perpendicular to \(L_0\) and parallel to the cross-sectional area \(A\).
The relationship of the change in volume to other physical quantities is given by
\[\Delta V = \dfrac{1}{B} \dfrac{F}{A} V_0,\] where \(B\) is the bulk modulus, \(V_0\) is the original volume, and \(\frac{F}{A}\) is the force per unit area applied uniformly inward on all surfaces.
Footnotes
Approximate and average values. Young’s moduli \(Y\) for tension and compression sometimes differ but are averaged here. Bone has significantly different Young’s moduli for tension and compression.
Glossary
- deformation
- change in shape due to the application of force
- Hooke’s law
- proportional relationship between the force F on a material and the deformation ΔL it causes, F=kΔL
- tensile strength
- the breaking stress that will cause permanent deformation or fraction of a material
- stress
- ratio of force to area
- strain
- ratio of change in length to original length
- shear deformation
- deformation perpendicular to the original length of an objec
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2025-03-17T19:53:24.701486
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.03%3A_Elasticity_-_Stress_and_Strain",
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"title": "5.3: Elasticity - Stress and Strain",
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5.E: Further Applications of Newton's Laws (Exercises)
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Conceptual Questions
5.1: Friction
1. Define normal force. What is its relationship to friction when friction behaves simply?
2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.
3. When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction.
4. When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)
5.2: Drag Forces
5. Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits.
6. Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one?
7. As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference?
8. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?
5.3: Elasticity: Stress and Strain
9. The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood (pulsating or continuous).
10. What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference?
11. Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? What differences will dry and wet conditions make for these surfaces?
12. Would you expect your height to be different depending upon the time of day? Why or why not?
13. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?
14. Explain why pregnant women often suffer from back strain late in their pregnancy.
15. An old carpenter’s trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail firmly with pliers. Why does this help?
16. When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more with temperature than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a pocket of air above the vinegar would prevent the break. (This is the function of the air above liquids in glass containers.)
Problems & Exercises
5.1: Friction
17. A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?
Solution
\(\displaystyle 5.00 N\)
18. (a) When rebuilding her car’s engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force between the piston and cylinder?
(b) What is the magnitude of the force would she have to exert if the steel parts were oiled?
19. (a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee?
(b) During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.
20. Suppose you have a 120-kg wooden crate resting on a wood floor.
(a) What maximum force can you exert horizontally on the crate without moving it?
(b) If you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be?
Solution
(a) 588 N
(b) \(\displaystyle 1.96 m/s^2\)
21. (a) If half of the weight of a small \(\displaystyle 1.00×10^3kg\) utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete?
(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate?
(c) Solve both problems assuming the truck has four-wheel drive.
22. A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg.
(a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow.
(b) What is the magnitude of the acceleration once the sled starts to move?
(c) For both situations, calculate the magnitude of the force in the coupling between the dogs and the sled.
Solution
(a) \(\displaystyle 3.29 m/s^2\)
(b) \(\displaystyle 3.52 m/s^2\)
(c) 980 N; 945 N
23. Consider the 65.0-kg ice skater being pushed by two others shown in Figure.
(a) Find the direction and magnitude of \(\displaystyle F_{tot}\), the total force exerted on her by the others, given that the magnitudes \(\displaystyle F_1\) and \(\displaystyle F_2\) are 26.4 N and 18.6 N, respectively.
(b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of \(\displaystyle F_{tot}\)?
(c) What is her acceleration assuming she is already moving in the direction of \(\displaystyle F_{tot}\)? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.)
24. Show that the acceleration of any object down a frictionless incline that makes an angle \(\displaystyle θ\) with the horizontal is \(\displaystyle a=gsinθ\). (Note that this acceleration is independent of mass.)
25. Show that the acceleration of any object down an incline where friction behaves simply (that is, where \(\displaystyle f_k=μ_kN\)) is \(\displaystyle a=g(sinθ−μ_kcosθ)\). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (\(\displaystyle μ_k=0\)).
26. Calculate the deceleration of a snow boarder going up a \(\displaystyle *5.0º\). slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 25 may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies.
Solution
\(\displaystyle 1.83m/s^2\)
27. (a) Calculate the acceleration of a skier heading down a \(\displaystyle 10.0º\) slope, assuming the coefficient of friction for waxed wood on wet snow.
(b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 25 to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies.
28. If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is \(\displaystyle θ=tan^{–1}μ_s\). You may use the result of the previous problem. Assume that \(\displaystyle a=0\) and that static friction has reached its maximum value.
29. Calculate the maximum deceleration of a car that is heading down a \(\displaystyle 6º\) slope (one that makes an angle of \(\displaystyle 6º\) with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the deceleration. (Ignore rolling.) Calculate for a car:
(a) On dry concrete.
(b) On wet concrete.
(c) On ice, assuming that \(\displaystyle μ_s=0.100\), the same as for shoes on ice.
30. Calculate the maximum acceleration of a car that is heading up a \(\displaystyle 4º\) slope (one that makes an angle of \(\displaystyle 4º\) with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.)
(a) On dry concrete.
(b) On wet concrete.
(c) On ice, assuming that \(\displaystyle μ_s=0.100\), the same as for shoes on ice.
Solution
(a) \(\displaystyle 4.20 m/s^2\)
(b) \(\displaystyle 2.74 m/s^2\)
(c) \(\displaystyle –0.195 m/s^2\)
31. Repeat Exercise for a car with four-wheel drive.
32. A freight train consists of two \(\displaystyle 8.00×10^5-kg\) engines and 45 cars with average masses of \(\displaystyle 5.50×10^5kg\)
(a) What force must each engine exert backward on the track to accelerate the train at a rate of \(\displaystyle 5.00×10^{−2}m/s^2\) if the force of friction is \(\displaystyle 7.50×10^5N\), assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems.
(b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?
Solution
(a) \(\displaystyle 1.03×10^6N\)
(b) \(\displaystyle 3.48×10^5N\)
33. Consider the 52.0-kg mountain climber in Figure.
(a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms.
(b) What is the minimum coefficient of friction between her shoes and the cliff?
Part of the climber’s weight is supported by her rope and part by friction between her feet and the rock face.
34. A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure(a).
(a) Calculate the minimum force \(\displaystyle F\) he must exert to get the block moving.
(b) What is the magnitude of its acceleration once it starts to move, if that force is maintained?
Solution
(a) \(\displaystyle 51.0 N\)
(b) \(\displaystyle 0.720 m/s^2\)
35. Repeat Exercise 34 with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as shown in Figure(b).
Which method of sliding a block of ice requires less force—(a) pushing or (b) pulling at the same angle above the horizontal?
5.2: Drag Forces
36. The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of \(\displaystyle 0.140m^2\)..
Solution
\(\displaystyle 115m/s;414km/hr\)
37. A 60-kg and a 90-kg skydiver jump from an airplane at an altitude of 6000 m, both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits.
38. A 560-g squirrel with a surface area of \(\displaystyle 930cm^2\) falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance?
Solution
\(\displaystyle 25 m/s; 9.9 m/s\)
39. To maintain a constant speed, the force provided by a car’s engine must equal the drag force plus the force of friction of the road (the rolling resistance).
(a) What are the magnitudes of drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is \(\displaystyle 0.70 m^2\)
(b) What is the magnitude of drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is \(\displaystyle 2.44 m^2\)) Assume all values are accurate to three significant digits.
40. By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?
Solution
\(\displaystyle 2.9\)
41. Calculate the speed a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be \(\displaystyle 1.00×10^3kg/m^3\), and the surface area to be \(\displaystyle πr^2\)
42. Using Stokes’ law, verify that the units for viscosity are kilograms per meter per second.
Solution
\(\displaystyle [η]=\frac{[F_s]}{[r][v]}=\frac{kg⋅m/s^2}{m⋅m/s}=\frac{kg}{m⋅s}\)
43. Find the terminal velocity of a spherical bacterium (diameter \(\displaystyle 2.00 μm\) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be \(\displaystyle 1.10×10^3kg/m^3\).
44. Stokes’ law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density \(\displaystyle 7.8×10^3kg/m^3\), diameter \(\displaystyle 3.0 mm\)) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.
Solution
\(\displaystyle 0.76 kg/m⋅s\)
5.3: Elasticity: Stress and Strain
45. During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. Her mass is 60.0 kg.
Solution
\(\displaystyle 1.90×10^{−3}cm\)
46. During a wrestling match, a 150 kg wrestler briefly stands on one hand during a maneuver designed to perplex his already moribund adversary. By how much does the upper arm bone shorten in length? The bone can be represented by a uniform rod 38.0 cm in length and 2.10 cm in radius.
47. (a) The “lead” in pencils is a graphite composition with a Young’s modulus of about \(\displaystyle 1×10^9N/m^2\). Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils?
Solution
(a)1 mm
(b) This does seem reasonable, since the lead does seem to shrink a little when you push on it.
48. TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius?
49. (a) By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (b) Does the answer seem to be consistent with what you have observed for nylon ropes? Would it make sense if the rope were actually a bungee cord?
Solution
(a)9 cm
(b)This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much.
50. A 20.0-m tall hollow aluminum flagpole is equivalent in stiffness to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 900 N exerted at the top would. How far to the side does the top of the pole flex?
51. As an oil well is drilled, each new section of drill pipe supports its own weight and that of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in stiffness to a solid cylinder 5.00 cm in diameter.
Solution
8.59 mm
52. Calculate the force a piano tuner applies to stretch a steel piano wire 8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long.
53. A vertebra is subjected to a shearing force of 500 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 cm in diameter.
Solution
\(\displaystyle 1.49×10^{−7}m\)
54. A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus of \(\displaystyle 1×10^9N/m^2\). The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.
55. When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle of \(\displaystyle 20.0º\) to the horizontal.
(a) By how much does the wood flex perpendicular to its length?
(b) How much is it compressed lengthwise?
Solution
(a) \(\displaystyle 3.99×10^{−7}m\)
(b) \(\displaystyle 9.67×10^{−8}m\)
56. To consider the effect of wires hung on poles, we take data from [link], in which tensions in wires supporting a traffic light were calculated. The left wire made an angle 30.0º below the horizontal with the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow aluminum pole is equivalent in stiffness to a 4.50 cm diameter solid cylinder. (a) How far is it bent to the side? (b) By how much is it compressed?
57. A farmer making grape juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.2% (that is, \(\displaystyle ΔV/V_0=2×10^{−3}\)) relative to the space available. Calculate the magnitude of the normal force exerted by the juice per square centimeter if its bulk modulus is \(\displaystyle 1.8×10^9N/m^2\), assuming the bottle does not break. In view of your answer, do you think the bottle will survive?
Solution
\(\displaystyle 4×10^6N/m^2\). This is about 36 atm, greater than a typical jar can withstand.
58. (a) When water freezes, its volume increases by 9.05% (that is, \(\displaystyle ΔV/V_0=9.05×10^{−2}\)). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.)
(b) Is it surprising that such forces can fracture engine blocks, boulders, and the like?
59. This problem returns to the tightrope walker studied in [link], who created a tension of \(\displaystyle 3.94×10^3N\) in a wire making an angle 5.0º below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter.
Solution
1.4 cm
60. The pole in Figure is at a \(\displaystyle 90.0º\) bend in a power line and is therefore subjected to more shear force than poles in straight parts of the line. The tension in each line is \(\displaystyle 4.00×10^4N\), at the angles shown. The pole is 15.0 m tall, has an 18.0 cm diameter, and can be considered to have half the stiffness of hardwood.
(a) Calculate the compression of the pole.
(b) Find how much it bends and in what direction.
(c) Find the tension in a guy wire used to keep the pole straight if it is attached to the top of the pole at an angle of 30.0º with the vertical. (Clearly, the guy wire must be in the opposite direction of the bend.)
This telephone pole is at a \(\displaystyle 90º\) bend in a power line. A guy wire is attached to the top of the pole at an angle of \(\displaystyle 30º\) with the vertical.
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:24.797985
| 2017-04-20T14:42:20 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/05%3A_Further_Applications_of_Newton's_Laws-_Friction_Drag_and_Elasticity/5.E%3A_Further_Applications_of_Newton's_Laws_(Exercises)",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "5.E: Further Applications of Newton's Laws (Exercises)",
"author": "OpenStax"
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation
|
6: Uniform Circular Motion and Gravitation
This chapter deals with the simplest form of curved motion, uniform circular motion , motion in a circular path at constant speed. Studying this topic illustrates most concepts associated with rotational motion and leads to the study of many new topics we group under the name rotation . Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey puck moving along ice.
-
- 6.0: Prelude to Uniform Circular Motion and Gravitation
- Many motions, such as the arc of a bird’s flight or Earth’s path around the Sun, are curved. Recall that Newton’s first law tells us that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that cause it, including gravitational forces. In some ways, this chapter is a continuation of Dynamics: Newton's Laws of Motion as we study more applications of Newton’s laws of motion.
-
- 6.1: Rotation Angle and Angular Velocity
- Projectile motion is a special case of two-dimensional kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion.
-
- 6.2: Centripetal Acceleration
- We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. In this section we examine the direction and magnitude of that acceleration.
-
- 6.3: Centripetal Force
- Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.
-
- 6.4: Fictitious Forces and Non-inertial Frames - The Coriolis Force
- What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits fictitious forces—unreal forces that arise from motion and may seem real, because the observer’s frame of reference is accelerating or rotating.
-
- 6.5: Newton’s Universal Law of Gravitation
- The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
-
- 6.6: Satellites and Kepler’s Laws- An Argument for Simplicity
- there is a classical set of three laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying the two previous conditions (not just planets in our solar system). These descriptive laws are named for the German astronomer Johannes Kepler, who devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe.
Thumbnail: Two bodies of different mass orbiting a common barycenter. The relative sizes and type of orbit are similar to the Pluto–Charon system. (public domain; Zhatt).
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libretexts
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2025-03-17T19:53:24.865639
| 2015-10-27T19:21:19 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "6: Uniform Circular Motion and Gravitation",
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.00%3A_Prelude_to_Uniform_Circular_Motion_and_Gravitation
|
6.0: Prelude to Uniform Circular Motion and Gravitation
Many motions, such as the arc of a bird’s flight or Earth’s path around the Sun, are curved. Recall that Newton’s first law tells us that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that cause it, including gravitational forces. In some ways, this chapter is a continuation of Dynamics: Newton's Laws of Motion as we study more applications of Newton’s laws of motion.
This chapter deals with the simplest form of curved motion, uniform circular motion , motion in a circular path at constant speed. Studying this topic illustrates most concepts associated with rotational motion and leads to the study of many new topics we group under the name rotation . Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey puck moving along ice.
Glossary
- uniform circular motion
- the motion of an object in a circular path at constant speed
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libretexts
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2025-03-17T19:53:24.941741
| 2015-11-01T04:52:57 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.00%3A_Prelude_to_Uniform_Circular_Motion_and_Gravitation",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "6.0: Prelude to Uniform Circular Motion and Gravitation",
"author": "OpenStax"
}
|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.01%3A_Rotation_Angle_and_Angular_Velocity
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6.1: Rotation Angle and Angular Velocity
Learning Objectives
By the end of this section, you will be able to:
- Define arc length, rotation angle, radius of curvature and angular velocity.
- Calculate the angular velocity of a car wheel spin.
In Kinematics , we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion.
Rotation Angle
When objects rotate about some axis—for example, when the CD (compact disc) in Figure rotates about its center—each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle \(\Delta \theta\) to be the ratio of the arc length to the radius of curvature:
\[\Delta \theta = \dfrac{\Delta s}{r}. \]
The arc length \(\Delta s\) is the distance traveled along a circular path as shown in Figure. Note that is the radius of curvature of the circular path.We know that for one complete revolution, the arc length is the circumference of a circle of radius \(r\). The circumference of a circle is \(2 \pi r\).
Thus for one complete revolution the rotation angle is \[\Delta \theta = \dfrac{2\pi r}{r} = 2\pi. \]
This result is the basis for defining the units used to measure rotation angles, \(\Delta \theta \) to be radians (rad), defined so that
\[2 \pi \, radians = 1 \space revolution. \]
A comparison of some useful angles expressed in both degrees and radians is shown in Table \(\PageIndex{1}\).
| Degree Measure | Radian Measure |
|---|---|
| \(30^o\) | \(\dfrac{\pi}{6}\) |
| \(60^o\) | \(\dfrac{\pi}{3}\) |
| \(90^o\) | \(\dfrac{\pi}{2}\) |
| \(120^o\) | \(\dfrac{2\pi}{3}\) |
| \(135^o\) | \(\dfrac{3\pi}{4}\) |
| \(180^o\) | \(\pi\) |
If \(\Delta \theta = 2 \pi \, rad \), then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are \(360^o\) in a circle or one revolution, the relationship between radians and degrees is thus
\[ 2\pi \, rad = 360^o \]
so that
\[ 1 \, rad = \dfrac{360^o}{2\pi} \approx 57.3^o \]
Angular Velocity
How fast is an object rotating? We define angular velocity \(\omega\) as the rate of change of an angle. In symbols, this is
\[\omega = \dfrac{\Delta \theta}{\Delta t}, \]
where an angular rotation \(\Delta \theta \) takes place in a time \(\Delta t\). The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s).Angular velocity \(\omega\) is analogous to linear velocity \(v\). To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length \(\Delta s\) in a time \(\Delta t\), and so it has a linear velocity
\[v = \dfrac{\Delta s}{\Delta t}.\]
From \(\Delta \theta = \frac{\Delta s}{r} \) we see that \(\Delta s = r\Delta \theta \). Substituting this into the expression for \(v\) gives
\[v = \dfrac{r \Delta \theta}{\Delta t} = r\omega. \]
We write this relationship in two different ways and gain two different insights:
\[ v = r \omega, \, or \, \omega = \dfrac{v}{r}.\]
The first relationship in \( v = r \omega, \, or \, \omega = \dfrac{v}{r}\) states that the linear velocity \(v\) is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest \(r\)), as you might expect. We can also call this linear speed \(v\) of a point on the rim the tangential speed . The second relationship in \(v = r \omega, \, or \, \omega = \dfrac{v}{r}\) can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed \(v\) of the car. See Figure So the faster the car moves, the faster the tire spins—large \(v\) means a large \(\omega\), because \(v = r\omega\). Similarly, a larger-radius tire rotating at the same angular velocity \((\omega)\) will produce a greater linear speed \((v)\) for the car.
Example \(\PageIndex{1}\): How Fast Does the Car Tire Spin?
Calculate the angular velocity of a 0.300 m radius car tire when the car travels at \(15.0 m/s\) (about \(54 \, km/h\)). See Figure.
Strategy
Because the linear speed of the tire rim is the same as the speed of the car, we have \(v = 15.0 m/s\). The radius of the tire is given to be \(r = 0.300 \, m\). Knowing \(v\) and \(r\), we can use the second relationship in \(v = \omega r\), \(\omega = \frac{v}{r}\) to calculate the angular velocity.
Solution
To calculate the angular velocity, we will use the following relationship:
\[\omega = \dfrac{v}{r}.\]
Substituting the knowns,
\[\omega = \dfrac{15.0 \, m/s}{0.300 \, m} = 50.0 \, rad/s.\]
Discussion
When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity
\[ \omega = (15.0 \, m/s)/(1.20 \, m) = 12.5 \, rad/s.\]
Both \(\omega\) and \(v\) have directions (hence they are angular and linear velocities , respectively). Angular velocity has only two directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in Figure.
TAKE-HOME EXPERIMENT
Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities.
PHET EXPLORATIONS: LADYBUG REVOLUTION
Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs.
Section Summary
- Uniform circular motion is motion in a circle at constant speed. The rotation angle \(\delta \theta\) is defined as the ratio of the arc length to the radius of curvature:
\[\Delta \theta = \dfrac{\Delta s}{r} \]
where arc length \(\delta s\) is distance traveled along a circular path and \(r\) is the radius of curvature of the circular path. The quantity \( \Delta \theta\) is measured in units of radians (rad), for which
\[2\pi \, rad = 360^o = 1 \, revolution. \]
- The conversion between radians and degrees is \[ 1 \, rad = 57.3^o. \]
- Angular velocity \(\omega\) is the rate of change of an angle,
\[\omega = \dfrac{\Delta \theta}{\Delta t},\]
where a rotation \(\Delta \theta \) takes place in a time \(\Delta t\). The units of angular velocity are radians per second (rad/s). Linear velocity \(v\) and angular velocity \(\omega\) are related by
\[v = r\omega, \, or \, \omega = \dfrac{v}{r}. \]
Glossary
- arc length
- Δs, the distance traveled by an object along a circular path
- pit
- a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD
- rotation angle
- the ratio of the arc length to the radius of curvature on a circular path: \( Δθ=\frac{Δs}{r}\)
- radius of curvature
- radius of a circular path
- radians
- a unit of angle measurement
- angular velocity
- \(ω\), the rate of change of the angle with which an object moves on a circular path
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libretexts
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2025-03-17T19:53:25.023444
| 2015-11-01T04:53:33 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.01%3A_Rotation_Angle_and_Angular_Velocity",
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"title": "6.1: Rotation Angle and Angular Velocity",
"author": "OpenStax"
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.02%3A_Centripetal_Acceleration
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6.2: Centripetal Acceleration
Learning Objectives
By the end of this section, you will be able to:
- Establish the expression for centripetal acceleration.
- Explain the centrifuge.
We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.
Figure \(\PageIndex{1}\) shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration \(a_c\); centripetal means “toward the center” or “center seeking.”
The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii \(r\) and \(\Delta s\) are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds \( v_1 =v_2 = v \) Using the properties of two similar triangles, we obtain
\[ \dfrac{\Delta v}{v} = \dfrac{\Delta s}{r}. \]
Acceleration is \(\Delta v/\Delta t \) and so we first solve this expression for \(\delta v \):
\[\delta v = \dfrac{v}{r} \Delta s. \]
Then we divide this by \(\Delta t \), yielding
\[\dfrac{\Delta v}{\Delta t} = \dfrac{v}{r} \times \dfrac{\Delta s}{\Delta t}. \]
Finally, noting that \( \Delta v/\Delta t = a_c \) and that \(\delta s/\Delta t = v \) the linear or tangential speed, we see that the magnitude of the centripetal acceleration is
\[ a_c = \dfrac{v^2}{r}, \]
which is the acceleration of an object in a circle of radius \(r\) at a speed \(v\).
So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that \(a_c\) is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that \(a_c\) is greater for tighter turns, as you have probably noticed.It is also useful to express \(a_c\) in terms of angular velocity. Substituting \( v = r\omega \) into the above expression, we find \( a_c = (r \omega^2)/r = r \omega^2 \). We can express the magnitude of centripetal acceleration using either of two equations:
\[ a_c = \dfrac{v^2}{r}; \, a_c = r \omega^2 \]
Recall that the direction of \(a_c\) is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.
A centrifuge (Figure \(\PageIndex{2b}\)) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity \((g)\) maximum centripetal acceleration of several hundred thousand \(g\) is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.
Example \(\PageIndex{1}\): Centripetal Acceleration vs. Gravity?
What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure \(\PageIndex{2a}\).
Strategy
Because \(v\) and \(r\) are given, the first expression in \( a_c = \frac{v^2}{r}: \, a_c = r \omega^2 \) is the most convenient to use.
Solution
Entering the given values of \( v = 25.0 \, m/s and r = 500 \, m \) into the first expression for \(a_c\) gives
\[a_c = \dfrac{v^2}{r} = \dfrac{(25.0 m/s)^2}{500 \, m} = 1.25 \, m/s^2. \nonumber \]
Discussion
To compare this with the acceleration due to gravity \( (g = 9.80 \, m/s^2) \), we take the ratio of \(a_c/g = (1.25 \, m/s^2)/(9.80 \, m/s^2) = 0.128. \) Thus, \(a_c = 0.128 g \) and is noticeable especially if you were not wearing a seat belt.
Example \(\PageIndex{2}\): How Big is the Centripetal Acceleration in an Ultracentrifuge?
Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at \(7.4 \times 10^7 \, rev/min.\) Determine the ratio of this acceleration to that due to gravity. See Figure Figure \(\PageIndex{2b}\).
Strategy
The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity \(\omega\). Because \(r\) is given, we can use the second expression in the equation \(a_c = \frac{v^2}{r}; \, a_c = r\omega^2 \) to calculate the centripetal acceleration.
Solution
To convert \(7.40 \times 10^4 \, rev/min \) to radians per second, we use the facts that one revolution is \(2\pi \, rad\) and one minute is 60.0 s. Thus,
\[\omega = 7.40 \times 10^4 \dfrac{rev}{min} \times \dfrac{2 \pi \, rad}{1 \, rev} \times \dfrac{1 \, min}{60 \, sec} = 7745 \, rad/sec. \nonumber \]
Now the centripetal acceleration is given by the second expression in \( a_c = \frac{v^2}{r}; \, a_c = r\omega^2\) as
\[ a_c = r\omega^2. \nonumber\]
Converting 7.50 cm to meters and substituting known values gives
\[a_c = (0.0750 \, m)(7854 \, rad/sec)^2 = 4.50 \times 10^6 \, m/s^2. \nonumber\]
Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of \(a_c\) to \(g\) yields
\[\dfrac{a_c}{g} = \dfrac{4.63 \times 10^6}{9.80} = 4.59 \times 10^5. \nonumber\]
Discussion
This last result means that the centripetal acceleration is 472,000 times as strong as \(g\). It is no wonder that such high \(\omega\) centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.
Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In the section on Centripetal Force , we will consider the forces involved in circular motion.
PHET EXPLORATIONS: LADYBUG MOTION 2D
Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior.
Summary
- Centripetal acceleration \(a_c\) is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity \(v\) and has the magnitude \[a_c = \dfrac{v^2}{r}; \, a_c = r\omega^2. \nonumber \]
- The unit of centripetal acceleration is \(m/s^2.\)
Glossary
- centripetal acceleration
- the acceleration of an object moving in a circle, directed toward the center
- ultracentrifuge
- a centrifuge optimized for spinning a rotor at very high speeds
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libretexts
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2025-03-17T19:53:25.095373
| 2015-11-01T04:53:15 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.02%3A_Centripetal_Acceleration",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "6.2: Centripetal Acceleration",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.03%3A_Centripetal_Force
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6.3: Centripetal Force
Learning Objectives
By the end of this section, you will be able to:
- Calculate coefficient of friction on a car tire.
- Calculate ideal speed and angle of a car on a turn.
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force . The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net \(F = ma \). For uniform circular motion, the acceleration is the centripetal acceleration - \(a = a_c\). Thus, the magnitude of centripetal force \(F_c\) is \[ F_c = ma_c.\]
By using the expressions for centripetal acceleration \(a_c\) from \( a_c = \frac{v^2}{r}; \, a_c = r\omega^2\), we get two expressions for the centripetal force \(F_c\) in terms of mass, velocity, angular velocity, and radius of curvature:
\[F_c = m \dfrac{v^2}{r}; \, F_c = mr\omega^2.\]
You may use whichever expression for centripetal force is more convenient. Centripetal force \(F_c\) is always perpendicular to the path and pointing to the center of curvature, because \(a_c\) is perpendicular to the velocity and pointing to the center of curvature.Note that if you solve the first expression for \(r\), you get \[r = \dfrac{mv^2}{F_c}.\]
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
Example \(\PageIndex{1}\): What Coefficient of Friction Do Car Tires Need on a Flat Curve?
- Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
- Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure).
Strategy and Solution for (a)
We know that \(F_c = \frac{mv^2}{r}.\) Thus, \[F_c = \dfrac{mv^2}{r} = \dfrac{(900\, kg)(25.0 \, m/s)^2}{500 \, m)} = 1125 \, N. \nonumber\]
Strategy for (b)
Figure shows the forces acting on the car on an unbanked (level ground) curve.
Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is \(\mu_s N,\) where \(\mu_s\) is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground,)so that \( N = mg\). Thus the centripetal force in this situation is
\[F_c = f = \mu_sN = \mu_s mg. \nonumber\]
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for \(F_c\) from the equation
\[\begin{align*} F_c = m\dfrac{v^2}{r} \\[5pt] &= mr\omega^2 \end{align*} \]
\[m\dfrac{v^2}{r} = \mu_smg. \nonumber\]
We solve this for \(\mu_s\), noting that mass cancels, and obtain
\[\mu_s = \dfrac{v^2}{rg}. \nonumber\]
Solution for (b)
Substituting the knowns,
\[\mu_s = \dfrac{(25.0 \, m/s)^2}{(500 \, m)(9.80 \, m/s^2)} = 0.13. \nonumber\]
(Because coefficients of friction are approximate, the answer is given to only two digits.)
Discussion
We could also solve part (a) using the first expression in
\[\begin{align*} F_c &= m\dfrac{v^2}{r} \\[4pt] &= mr\omega^2 \end{align*}\]
because \(m\), \(v\) and \(r\) are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than \(\mu_sN\). A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.
Let us now consider banked curves , where the slope of the road helps you negotiate the curve. See Figure. The greater the angle \(\theta\), the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle \(\theta\) is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for \(\theta\) for an ideally banked curve and consider an example related to it.
For ideal banking , the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.
Figure shows a free body diagram for a car on a frictionless banked curve. If the angle \(\theta\) is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight \(w\) and the normal force of the road \(N\). (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude \(mv^2/r\). Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,
\[N\, \sin \, \theta = \dfrac{mv^2}{r}.\]
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is \(N\, cos \, \theta\), and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,
\[N\, \cos \, \theta = mg.\]
Now we can combine the last two equations to eliminate \(N\) and get an expression for \(\theta\), as desired. Solving the second equation for \(N = mg/(cos \, \theta) \), and substituting this into the first yields
\[mg\dfrac{\sin\, \theta}{\cos \, \theta} = \dfrac{mv^2}{r}\]
\[ mg \, tan \, \theta = \dfrac{mv^2}{r}\]
\[\tan \, \theta = \dfrac{v^2}{rg}\]
Taking the inverse tangent gives
\[\theta = \tan^{-1} \left( \dfrac{v^2}{rg} \right) \, (ideally \, banked \, curve, \, no \, friction). \]
This expression can be understood by considering how \(\theta\) depends on \(v\) and \(r\). A large \(\theta\) will be obtained for a large \(v\) and a small \(r\). That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that \(\theta\) does not depend on the mass of the vehicle.
Example \(\PageIndex{2}\): What is the Ideal Speed to Take a Steeply Banked Tight Curve?
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.
Strategy
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.
Solution
Starting with
\[\tan \, \theta = \dfrac{v^2}{rg} \nonumber\]
we get
\[v = (rg \, tan \, \theta)^{\frac{1}{2}}. \nonumber\]
Noting that tan 65.0º = 2.14, we obtain
\[ \begin{align*} v &= [(100 \, m)(9.80 \, m/s^2)(2.14)]^{\frac{1}{2}} \\[4pt] &= 45.8 \, m/s \end{align*} \]
Discussion
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.
Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.
TAKE-HOME EXPERIMENT
Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.
PHET EXPLORATIONS: GRAVITY AND ORBITS
Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!
Summary
- Centripetal force \(F_c\) is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity \(v\) and has magnitude \[F_c = ma_c \nonumber \] which can also be expressed as \[F_c = \dfrac{v^2}{r} \nonumber \] or \[F_c = mr\omega^2 \nonumber\]
Glossary
- centripetal force
- any net force causing uniform circular motion
- ideal banking
- the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction
- ideal speed
- the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road
- ideal angle
- the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed
- banked curve
- the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve
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libretexts
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2025-03-17T19:53:25.192195
| 2015-11-01T04:53:50 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.03%3A_Centripetal_Force",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "6.3: Centripetal Force",
"author": "OpenStax"
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|
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.04%3A_Fictitious_Forces_and_Non-inertial_Frames_-_The_Coriolis_Force
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6.4: Fictitious Forces and Non-inertial Frames - The Coriolis Force
Learning Objectives
By the end of this section, you will be able to:
- Discuss the inertial frame of reference.
- Discuss the non-inertial frame of reference.
- Describe the effects of the Coriolis force.
What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits fictitious forces—unreal forces that arise from motion and may seem real, because the observer’s frame of reference is accelerating or rotating.
When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you, and there is no real force backward on you. An even more common experience occurs when you make a tight curve in your car—say, to the right. You feel as if you are thrown (that is, forced ) toward the left relative to the car. Again, a physicist would say that you are going in a straight line but the car moves to the right, and there is no real force on you to the left. Recall Newton’s first law.
We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively use the car as a frame of reference, while a physicist uses Earth. The physicist chooses Earth because it is very nearly an inertial frame of reference—one in which all forces are real (that is, in which all forces have an identifiable physical origin). In such a frame of reference, Newton’s laws of motion take the form given in Dynamics: Newton's Laws of Motion. The car is a non-inertial frame of reference because it is accelerated to the side. The force to the left sensed by car passengers is a fictitious force having no physical origin. There is nothing real pushing them left—the car, as well as the driver, is actually accelerating to the right.
Let us now take a mental ride on a merry-go-round—specifically, a rapidly rotating playground merry-go-round. You take the merry-go-round to be your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force, named centrifugal force ( not to be confused with centripetal force ) , trying to throw you off. You must hang on tightly to counteract the centrifugal force. In Earth’s frame of reference, there is no force trying to throw you off. Rather you must hang on to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round.
This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, is put to good use in centrifuges (see Figure). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force.
Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball directly away from the center of the merry-go-round, as shown in Figure? The ball follows a straight path relative to Earth (assuming negligible friction) and a path curved to the right on the merry-go-round’s surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round’s frame of reference, we explain the apparent curve to the right by using a fictitious force, called the Coriolis force , that causes the ball to curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain why objects follow curved paths and allows us to apply Newton’s Laws in non-inertial frames of reference.
Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to its rotation. Yet such effects do exist—in the rotation of weather systems, for example. Most consequences of Earth’s rotation can be qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise, as does the merry-go-round in Figure. As on the merry-go-round, any motion in Earth’s northern hemisphere experiences a Coriolis force to the right. Just the opposite occurs in the southern hemisphere; there, the force is to the left. Because Earth’s angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions, such as wind patterns, it has substantial effects.
The Coriolis force causes hurricanes in the northern hemisphere to rotate in the counterclockwise direction, while the tropical cyclones (what hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms hurricane, typhoon, and tropical storm are regionally-specific names for tropical cyclones, storm systems characterized by low pressure centers, strong winds, and heavy rains. Figure helps show how these rotations take place. Air flows toward any region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the northern hemisphere, these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones of any type. Low pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure patterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern hemisphere but is less visible because high pressure is associated with sinking air, producing clear skies.
The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia and the rotation of the system underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be invented to explain the curved path. There is no identifiable physical source for these fictitious forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable source. Either view allows us to describe nature, but a view in an inertial frame is the simplest and truest, in the sense that all forces have real origins and explanations.
Section Summary
- Rotating and accelerated frames of reference are non-inertial.
- Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames.
Glossary
- fictitious force
- a force having no physical origin
- centrifugal force
- a fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of reference
- Coriolis force
- the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of reference
- non-inertial frame of reference
- an accelerated frame of reference
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libretexts
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2025-03-17T19:53:25.302819
| 2015-11-01T04:54:09 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.04%3A_Fictitious_Forces_and_Non-inertial_Frames_-_The_Coriolis_Force",
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"title": "6.4: Fictitious Forces and Non-inertial Frames - The Coriolis Force",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.05%3A_Newtons_Universal_Law_of_Gravitation
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6.5: Newton’s Universal Law of Gravitation
Learning Objectives
By the end of this section, you will be able to:
- Explain Earth’s gravitational force.
- Describe the gravitational effect of the Moon on Earth.
- Discuss weightlessness in space.
- Examine the Cavendish experiment
What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense.
Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others.
The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
MISCONCEPT ALERT
The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third law.
The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM), which will be further explored in Linear Momentum and Collisions. For two bodies having masses \(m\) and \(M\) with a distance \(r\) between their centers of mass, the equation for Newton’s universal law of gravitation is \[ F = G\dfrac{mM}{r^2},\] where \(F\) is the magnitude of the gravitational force and \(G\) is a proportionality factor called the gravitational constant . \(G\) is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be
\[G = 6.673 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2}\]
in SI units. Note that the units of \(G\) are such that a force in newtons is obtained from \(F = G\frac{mM}{r^2} \), when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of \(6.673 \times 10^{-11} \, N\).
This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of \(6 \times 10^{24} \, kg\).
Recall that the acceleration due to gravity \(g\) is about \(9.80 \, m/s^2\) on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for \(F\) in Newton’s universal law of gravitation gives
\[mg = G\dfrac{mM}{r^2}, \] where \(m\) is the mass of the object, \(M\) is the mass of Earth, and \(r\) is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure. The mass \(m\) of the object cancels, leaving an equation for \(g\):
\[g = G\dfrac{M}{r^2}. \]
Substituting known values for Earth’s mass and radius (to three significant figures),
\[g = \left(6.673 \times 10^{-11} \, \dfrac{N \cdot m^2}{kg^2} \right) \times \dfrac{5.98 \times 10^{24} \, kg}{(6.38 \times 10^6 \, m)^2},\]
and we obtain a value for the acceleration of a falling body: \[g = 9.80 \, m/s^2.\]
This is the expected value and is independent of the body’s mass . Newton’s law of gravitation takes Galileo’s observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses.
TAKE HOME EXPERIMENT
Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations.
MAKING CONNECTIONS
Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics , modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time.
In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.”
Example \(\PageIndex{1}\): Earth’s Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path
- Find the acceleration due to Earth’s gravity at the distance of the Moon.
- Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth’s gravity that you have just found.
Strategy for (a)
This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that \(r\) is the distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is \(3.84 \times 10^8 \, m\).
Solution for (a)
Substituting known values into the expression for \(g\) found above, remembering that \(M\) is the mass of Earth not the Moon, yields
\[g = G\dfrac{M}{r^2} = \left(6.67 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \right) \times \dfrac{5.98 \times 10^{24} \, kg}{3.84 \times 10^8 \, m)^2} \]
\[= 2.70 \times 10^{-3} \, m/s^2.\]
Strategy for (b)
Centripetal acceleration can be calculated using either form of
\[a_c = \dfrac{v^2}{r}\]
\[a_c = r\omega^2\]
We choose to use the second form:
\[a_c = r\omega^2,\]
where \(\omega\) is the angular velocity of the Moon about Earth.
Solution for (b)
Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using
\[1 \, d \times 24 \dfrac{hr}{d} \times 60 \dfrac{min}{hr} \times 60 \dfrac{s}{min} = 86,400 \, s\]
we see that
\[a_c = r\omega^2 = (3.84 \times 10^8 \, m)(2.66 \times 10^{-6} \, rad/s^2) \]
\[ = 2.72 \times 10^{-3} \, m/s^2. \]
The direction of the acceleration is toward the center of the Earth.
Discussion
The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.
Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see Figure). We do not sense the Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon’s gravitational force as discussed in Satellites and Kepler's Laws: An Argument for Simplicity.
Tides
Ocean tides are one very observable result of the Moon’s gravity acting on Earth. Figure is a simplified drawing of the Moon’s position relative to the tides. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well).
The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a \(90^o\) angle to the Earth-Moon alignment.
Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest and varies most rapidly, such as near black holes (see Figure). A few likely candidates for black holes have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually tear matter from a companion star.
”Weightlessness” and Microgravity
In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational force. There is no “zero gravity” in an astronaut’s orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks.
Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference does the absence of this pressure differential have upon the heart?
Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note, studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much better results.
Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in a microgravity environment.
The Cavendish Experiment: Then and Now
As previously noted, the universal gravitational constant \(G\) is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of \(G\) is very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure. Remarkably, his value for \(G\) differs by less than 1% from the best modern value.One important consequence of knowing \(G\) was that an accurate value for Earth’s mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth \(M\) from the relationship Newton’s universal law of gravitation gives
\[mg = G\dfrac{mM}{r^2},\]
where \(m\) is the mass of the object, \(M\) is the mass of Earth, and \(r\) is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure. The mass \(m\) of the object cancels, leaving an equation for \(g\):
\[g = G\dfrac{M}{r^2}. \]
Rearranging to solve for \(M\) yields
\[M = \dfrac{gr^2}{G}.\]
so \(M\) can be calculated because all quantities on the right, including the radius of Earth \(r\), are known from direct measurements. We shall see in Satellites and Kepler's Laws: An Argument for Simplicity that knowing \(G\) also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, \(G\) is by far the least well determined.
The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eötvös’ measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton’s law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed.
Summary
- Newton’s universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is
\[F = G\dfrac{mM}{r^2} \]
where F is the magnitude of the gravitational force. \(G\) is the gravitational constant, given by \(G = 6.63 \times 10^{-11} \, N \cdot m^2/kg^2\).
- Newton’s law of gravitation applies universally.
Glossary
- gravitational constant, G
- a proportionality factor used in the equation for Newton’s universal law of gravitation; it is a universal constant—that is, it is thought to be the same everywhere in the universe
- center of mass
- the point where the entire mass of an object can be thought to be concentrated
- microgravity
- an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface
- Newton’s universal law of gravitation
- every particle in the universe attracts every other particle with a force along a line joining them; the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them
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2025-03-17T19:53:25.468491
| 2015-11-01T04:54:30 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.05%3A_Newtons_Universal_Law_of_Gravitation",
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"title": "6.5: Newton’s Universal Law of Gravitation",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.06%3A_Satellites_and_Keplers_Laws-_An_Argument_for_Simplicity
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6.6: Satellites and Kepler’s Laws- An Argument for Simplicity
Learning Objectives
By the end of this section, you will be able to:
- State Kepler’s laws of planetary motion.
- Derive the third Kepler’s law for circular orbits.
- Discuss the Ptolemaic model of the universe.
Examples of gravitational orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The Moon’s orbit about Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets about the Sun are no less interesting. If we look further, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity.
All these motions are governed by gravitational force, and it is possible to describe them to various degrees of precision. Precise descriptions of complex systems must be made with large computers. However, we can describe an important class of orbits without the use of computers, and we shall find it instructive to study them. These orbits have the following characteristics:
- A small mass \(m\)orbits a much larger mass \(M\). This allows us to view the motion as if \(M\) were stationary—in fact, as if from an inertial frame of reference placed on \(M\) —without significant error. Mass \(m\) is the satellite of \(M\), if the orbit is gravitationally bound.
- The system is isolated from other masses . This allows us to neglect any small effects due to outside masses.
The conditions are satisfied, to good approximation, by Earth’s satellites (including the Moon), by objects orbiting the Sun, and by the satellites of other planets. Historically, planets were studied first, and there is a classical set of three laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying the two previous conditions (not just planets in our solar system). These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630), who devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed.
Kepler’s Laws of Planetary Motion
Kepler's First Law
The orbit of each planet about the Sun is an ellipse with the Sun at one focus.
Kepler's Second Law
Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see Figure).
Kepler's Third Law
The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun. In equation form, this is \[ \dfrac{T_1^2}{T_2^2} =\dfrac{r_1^3}{r_2^3}\]
where \(T\) is the period (time for one orbit) and \(r\) is the average radius. This equation is valid only for comparing two small masses orbiting the same large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of the equality.
Note again that while, for historical reasons, Kepler’s laws are stated for planets orbiting the Sun, they are actually valid for all bodies satisfying the two previously stated conditions.
Example \(\PageIndex{1}\): Find the Time for One Orbit of an Earth Satellite
Given that the Moon orbits Earth each 27.3 d and that it is an average distance of \(3.84 \times 10^8 \, m\) from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface.
Strategy
The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form in \(\frac{T_1^2}{T_2^2} =\frac{r_1^3}{r_2^3}\). Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find \(T_2\). The given information tells us that the orbital radius of the Moon is \(r_1 = 3.84 \times 10^8 \, m\), and that the period of the Moon is \(T_1 = 27.3 \, d\).
The height of the artificial satellite above Earth’s surface is given, and so we must add the radius of Earth (6380 km) to get \(r_2 = (1500 +6380) km = 7880 \, km\). Now all quantities are known, and so \(T_2\) can be found.
Solution
Kepler’s third law is \[ \dfrac{T_1^2}{T_2^2} =\dfrac{r_1^3}{r_2^3}.\]
To solve for \(T_2\), we cross-multiply and take the square root, yielding
\[T_2^2 = T_1^2 \left(\dfrac{r_2}{r_1} \right)^3 \]
\[T_2 = T_1 \left(\dfrac{r_2}{r_1} \right)^{\frac{3}{2}}. \]
Substituting known values yields
\[T_2 = 27.3 \, d \times \dfrac {24 \, h}{d} \times \left( \dfrac {7880 \, km}{3.84 \times 10^5 km} \right )^{ \frac{3}{2}} \]
\[ = 1.93 \, h. \]
Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will orbit in the same amount of time. This fact is related to the condition that the satellite’s mass is small compared with that of Earth.
People immediately search for deeper meaning when broadly applicable laws, like Kepler’s, are discovered. It was Newton who took the next giant step when he proposed the law of universal gravitation. While Kepler was able to discover what was happening, Newton discovered that gravitational force was the cause.
Derivation of Kepler’s Third Law for Circular Orbits
We shall derive Kepler’s third law, starting with Newton’s laws of motion and his universal law of gravitation. The point is to demonstrate that the force of gravity is the cause for Kepler’s laws (although we will only derive the third one).
Let us consider a circular orbit of a small mass \(m\) around a large mass \(M\), satisfying the two conditions stated at the beginning of this section. Gravity supplies the centripetal force to mass \(m\). Starting with Newton’s second law applied to circular motion,
\[ F_{net} = ma_c = m\dfrac{v^2}{r}. \]
The net external force on mass \(m\) is gravity, and so we substitute the force of gravity for \(F_{net}\):
\[ G\dfrac{mM}{r^2} = m\dfrac{v^2}{r}. \]
The mass \(m\) cancels, yielding
\[G\dfrac{M}{r} = v^2. \]
The fact that \(m\) cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius \(r\), all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler’s third law, we must get the period \(T\) into the equation. By definition, period \(T\) is the time for one complete orbit. Now the average speed \(v\) is the circumference divided by the period—that is,
\[v = \dfrac{2\pi r}{T}. \]
Substituting this into the previous equation gives \[G\dfrac{M}{r} = \dfrac{4\pi^2 r^2}{T^2}.\]
Solving for \(T^2\) yields \[T^2 = \dfrac{4\pi^2}{GM}r^3. \]
Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields
\[\dfrac{T_1^2}{T_2^2} = \dfrac{r_1^3}{r_2^3}. \]
This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body \(M\) cancel.
Now consider what we get if we solve \(T^2 = \frac{4\pi^2}{GM}r^3\) for the ratio \(r^3/T^2\). We obtain a relationship that can be used to determine the mass \(M\) of a parent body from the orbits of its satellites:
\[\dfrac{r^3}{T^2} = \dfrac{G}{4\pi^2}M.\]
If \(r\) and \(T\) are known for a satellite, then the mass \(M\) of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio \(r^3/T^2\) should be a constant for all satellites of the same parent body (because \(r^3/T^2 = GM/4\pi\). (See Table).
It is clear from Table that the ratio of \(r^3/T^2\) is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes—uncertainties in the \(r\) and \(T\) data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets and moons. This is another verification of Newton’s universal law of gravitation.
MAKING CONNECTIONS
Newton’s universal law of gravitation is modified by Einstein’s general theory of relativity, as we shall see in Particle Physics. Newton’s gravity is not seriously in error—it was and still is an extremely good approximation for most situations. Einstein’s modification is most noticeable in extremely large gravitational fields, such as near black holes. However, general relativity also explains such phenomena as small but long-known deviations of the orbit of the planet Mercury from classical predictions.
The Case for Simplicity
The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the scope of this text to cover that history in any detail, we note some important points. The definition of planet set in 2006 by the International Astronomical Union (IAU) states that in the solar system, a planet is a celestial body that:
- is in orbit around the Sun,
- has sufficient mass to assume hydrostatic equilibrium and
- has cleared the neighborhood around its orbit.
A non-satellite body fulfilling only the first two of the above criteria is classified as “dwarf planet.”
In 2006, Pluto was demoted to a ‘dwarf planet’ after scientists revised their definition of what constitutes a “true” planet.
| Parent | Satellite | Average orbital radius r (km) | Period T(y) | \(r^3/T^2 (km^3/y^2)\) |
|---|---|---|---|---|
| Earth | Moon |
\(3.84 \times 10^5\)
×
10
5
|
0.07481 |
\(1.01 \times 10^{19}\)
.
01
×
10
19
|
| Sun | Mercury |
\(5.79 \times 10^7\) . 79 |
0.2409 |
\( 3,34 \times 10^{24}\)
.
34
×
10
24
|
| Venus |
\(1.082 \times 10^8\)
.
082
×
10
8
|
0.6150 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Earth |
\(1.496 \times 10^8\)
.
496
×
10
8
|
1.000 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Mars |
\(2.279 \times 10^8\)
.
279
×
10
8
|
1.881 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Jupiter |
\(7.783 \times 10^8\)
.
783
×
10
8
|
11.86 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Saturn |
\(1.427 \times 10^9\)
.
427
×
10
9
|
29.46 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Neptune |
\(4.497 \times 10^9\)
.
497
×
10
9
|
164.8 |
\(3.35 \times 10^{24}\)
.
35
×
10
24
|
|
| Pluto |
\(5.90 \times 10^9\)
.
90
×
10
9
|
248.3 |
\(3.33 \times 10^{24}\)
.
33
×
10
24
|
|
| Jupiter | Io |
\(4.22 \times 10^5\)
.
22
×
10
5
|
0.00485 (1.77 d) |
\(3.19 \times 10^{21}\)
.
19
×
10
21
|
| Europa |
\(6.71 \times 10^5\)
.
71
×
10
5
|
0.00972 (3.55 d) |
\(3.20 \times 10^{21}\)
.
20
×
10
21
|
|
| Ganymede |
\(1.07 \times 10^6\)
.
07
×
10
6
|
0.0196 (7.16 d) |
\(3.19 \times 10^{21}\)
.
19
×
10
21
|
|
| Callisto |
\(1.88 \times 10^6\)
.
88
×
10
6
|
0.0457 (16.19 d) |
\(3.20 \times 10^{21}\)
.
20
×
10
21
|
The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for the gravitational force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the orbits of the planets and moons in the solar system. It epitomizes the underlying unity and simplicity of physics.
Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure (a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD. This model is characterized by a list of facts for the motions of planets with no cause and effect explanation. There tended to be a different rule for each heavenly body and a general lack of simplicity.
Figure (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident.
Summary
- Kepler’s laws are stated for a small mass \(m\) orbiting a larger mass \(M\) in near-isolation. Kepler’s laws of planetary motion are then as follows:
Kepler’s first law
The orbit of each planet about the Sun is an ellipse with the Sun at one focus.
Kepler’s second law
Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.
Kepler’s third law
The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun:
\[ \dfrac{T_1^2}{T_2^2} =\dfrac{r_1^3}{r_2^3},\]
where T is the period (time for one orbit) and \(r\) is the average radius of the orbit.
-
The period and radius of a satellite’s orbit about a larger body M are related by
\[T^2=\frac{4π^2}{GM}r^3\] or \[\dfrac{r^3}{T^2} = \dfrac{G}{4\pi^2}M.\]
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libretexts
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2025-03-17T19:53:25.629216
| 2015-11-01T04:54:48 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.06%3A_Satellites_and_Keplers_Laws-_An_Argument_for_Simplicity",
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"title": "6.6: Satellites and Kepler’s Laws- An Argument for Simplicity",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.E%3A_Uniform_Circular_Motion_and_Gravitation_(Excercise)
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6.E: Uniform Circular Motion and Gravitation (Excercise)
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Conceptual Questions
6.1: Rotation Angle and Angular Velocity
1. There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and velocity?
6.2: Centripetal Acceleration
2. Can centripetal acceleration change the speed of circular motion? Explain.
6.3: Centripetal Force
3. If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.
4. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?
5. If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.
6. Race car drivers routinely cut corners as shown in Figure. Explain how this allows the curve to be taken at the greatest speed.
Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed.
7. A number of amusement parks have rides that make vertical loops like the one shown in Figure. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:
(a) The car goes over the top at faster than this speed?
(b)The car goes over the top at slower than this speed?
Amusement rides with a vertical loop are an example of a form of curved motion.
8. What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure under the following circumstances:
(a) The car goes over the top at such a speed that the gravitational force is the only force acting?
(b) The car goes over the top faster than this speed?
(c) The car goes over the top slower than this speed?
9. As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.
10. Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.
A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will it follow path A, B, or C, as viewed from Earth’s frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round?
11. Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?
12. Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.
A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is the physical origin of the force on the string?
6.4: Fictitious Forces and Non-inertial Frames: The Coriolis Force
13. When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain?
14. Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is removed.
15. In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is spun up and the floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force. This is a fictitious force sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly one) what pins the riders to the wall, and identify all of the real forces acting on them.
16. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?
17. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 \(\displaystyle m/s^2\). Who do you agree with and why?
18. A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame?
6.5: Newton’s Universal Law of Gravitation
19. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?
20. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not \(\displaystyle 9.80 m/s^2\). Who do you agree with and why?
21. Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away.
22. Newton’s laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations?
6.6: Satellites and Kepler’s Laws: An Argument for Simplicity
23. In what frame(s) of reference are Kepler’s laws valid? Are Kepler’s laws purely descriptive, or do they contain causal information?
Problem Exercises
6.1: Rotation Angle and Angular Velocity
24. Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?
Solution
723 km
25. Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?
26. An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?
Solution
\(\displaystyle 5×10^7\) rotations
27. (a) What is the period of rotation of Earth in seconds?
(b) What is the angular velocity of Earth?
(c) Given that Earth has a radius of \(\displaystyle 6.4×10^6m\) at its equator, what is the linear velocity at Earth’s surface?
28. A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?
Solution
117 rad/s
29. In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?
30. A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?
Solution
76.2 rad/s
728 rpm
31. Integrated Concepts
When kicking a football, the kicker rotates his leg about the hip joint.
(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?
(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?
(c) Find the maximum range of the football, neglecting air resistance.
Solution
(a) 33.3 rad/s
(b) 500 N
(c) 40.8 m
32. Construct Your Own Problem
Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. Construct a problem in which you calculate the necessary angular velocity that assures the riders will not slide down the wall. Include a free body diagram of a single rider. Among the variables to consider are the radius of the cylinder and the coefficients of friction between the riders’ clothing and the wall.
6.2: Centripetal Acceleration
33. A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?
Solution
12.9 rev/min
34. A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track?
35. Taking the age of Earth to be about \(\displaystyle 4×10^9\) years and assuming its orbital radius of \(\displaystyle 1.5 ×10^{11}\) m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).
Solution
\(\displaystyle 4×10^{21}m\)
36. The propeller of a World War II fighter plane is 2.30 m in diameter.
(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?
(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?
(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of \(\displaystyle g\).
37. An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.
(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of \(\displaystyle g\).
(b) What is the linear speed of a point on its edge?
Solution
a) \(\displaystyle 3.47×10^4m/s^2, 3.55×10^3g\)
b) \(\displaystyle 51.1m/s\)
38. Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.
(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.
(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).
39. Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?
(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?
(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.
Solution
a) 31.4 rad/s
b) 118 m/s
c) 384 m/s
d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That’s quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins.
40. What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?
41. Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:
(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.
(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).
Solution
a) 0.524 km/s
b) 29.7 km/s
42. A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of \(\displaystyle 9.80m/s^2\) at the rim?
43. At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.
(a) At how many rev/min are the tires rotating?
(b) What is the centripetal acceleration at the edge of the tire?
(c) With what force must a determined \(\displaystyle 1.00×10^{−15}kg\) bacterium cling to the rim?
(d) Take the ratio of this force to the bacterium’s weight.
Solution
(a) \(\displaystyle 1.35×10^3rpm\)
(b) \(\displaystyle 8.47×10^3m/s^2\)
(c) \(\displaystyle 8.47×10^{–12}N\)
(d) 865
44. Integrated Concepts
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.
(a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.
(b) What is the centripetal acceleration at the bottom of the arc?
(c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
(d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
(e) Discuss whether the answer seems reasonable.
Solution
(a) \(\displaystyle 16.6m/s\)
(b) \(\displaystyle 19.6m/s^2\)
(c)
(d) \(\displaystyle 1.76×10^3N\) or \(\displaystyle 3.00w\), that is, the normal force (upward) is three times her weight.
(e) This answer seems reasonable, since she feels like she’s being forced into the chair MUCH stronger than just by gravity.
45. Unreasonable Results
A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass.
(a) What is the magnitude of the centripetal acceleration of the child at the low point?
(b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg?
(c) What is unreasonable about these results?
(d) Which premises are unreasonable or inconsistent?
Solution
a) \(\displaystyle 40.5m/s^2\)
b) 905 N
c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g.
d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction.
6.3: Centripetal Force
46. (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?
(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?
(c) Compare each force with her weight.
Solution
a) 483 N
b) 17.4 N
c) 2.24 times her weight, 0.0807 times her weight
47. Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.
48. What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?
Solution
\(\displaystyle 4.14º\)
48. What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?
49. (a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?
(b) Calculate the centripetal acceleration.
(c) Does this acceleration seem large to you?
Solution
a) 24.6 m
b) \(\displaystyle 36.6 m/s^2\)
c) \(\displaystyle a_c=3.73g.\) This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.
50. Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).
(a) Show that \(\displaystyle θ\) (as defined in the figure) is related to the speed v size 12{v} {} and radius of curvature r size 12{r} {} of the turn in the same way as for an ideally banked roadway—that is, \(\displaystyle θ=tan^{–1}v^2/rg\)
(b) Calculate \(\displaystyle θ\) for a 12.0 m/s turn of radius 30.0 m (as in a race).
A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.
51. A large centrifuge, like the one shown in Figure(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.
(a) At what angular velocity is the centripetal acceleration \(\displaystyle 10g\) if the rider is 15.0 m from the center of rotation?
(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure(b). At what angle \(\displaystyle θ\) below the horizontal will the cage hang when the centripetal acceleration is \(\displaystyle 10g\)?
(Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle θ size 12{θ} {} should be.)
(a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times.
Solution
a) 2.56 rad/s
b) 5.71º
bIntegrated Concepts
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).
(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.
(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
Solution
a) 16.2 m/s
b) 0.234
53. Modern roller coasters have vertical loops like the one shown in Figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?
Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than g so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place.
54. Unreasonable Results
(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s.
(b) What is unreasonable about the result?
(c) Which premises are unreasonable or inconsistent?
Solution
a) 1.84
b) A coefficient of friction this much greater than 1 is unreasonable .
c) The assumed speed is too great for the tight curve.
6.5: Newton’s Universal Law of Gravitation
55. (a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is \(\displaystyle 9.830 m/s^2\)and the radius of the Earth is 6371 km from center to pole.
(b) Compare this with the accepted value of \(\displaystyle 5.979×10^{24}kg\).
Solution
a) \(\displaystyle 5.979×10^{24}kg\)
b) This is identical to the best value to three significant figures.
56. (a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
(b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun.
(c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.
57. (a) What is the acceleration due to gravity on the surface of the Moon?
(b) On the surface of Mars? The mass of Mars is \(\displaystyle 6.418×10^{23}kg\) and its radius is \(\displaystyle 3.38×10^6m\)
Solution
a) \(\displaystyle 1.62m/s^2\)
b) \(\displaystyle 3.75m/s^2\)
58. (a) Calculate the acceleration due to gravity on the surface of the Sun.
(b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.)
59. The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.)
(a) Calculate the magnitude of the acceleration due to the Moon’s gravity at that point.
(b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.
Solution
a) \(\displaystyle 3.42×10^{–5}m/s^2\)
b) \(\displaystyle 3.34×10^{–5}m/s^2\)
The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system.
60. Solve part (b) of Example using \(\displaystyle a_c=v^2/r\).
61. Astrology, that unlikely and vague pseudo science, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.
(a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).
(b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some \(\displaystyle 6.29×10^{11}m\) away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)
Solution
a) \(\displaystyle 7.01×10^{–7}N\)
b) \(\displaystyle 1.35×10^{–6}N\)
62. The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune’s orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune’s orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune:
(a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are \(\displaystyle 4.50×10^{12}m\) apart, as they are at present. The mass of Pluto is \(\displaystyle 1.4×10^{22}kg\).
(b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about \(\displaystyle 2.50×10^{12}m\) apart, and compare it with that due to Pluto. The mass of Uranus is \(\displaystyle 8.62×10^{25}kg\).
63. (a) The Sun orbits the Milky Way galaxy once each \(\displaystyle 2.60 x 10^8y\), with a roughly circular orbit averaging \(\displaystyle 3.00 x 10^4\) light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?
(b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?
Solution
a) \(\displaystyle 1.66×10^{–10}m/s^2\)
b) \(\displaystyle 2.17×10^5m/s\)
64. Unreasonable Result
A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight.
(a) Calculate the mass of the mountain.
(b) Compare the mountain’s mass with that of Earth.
(c) What is unreasonable about these results?
(d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.)
Solution
a) \(\displaystyle 2.937×10^{17}kg\)
b) \(\displaystyle 4.91×10^{–8}\) of the Earth’s mass.
c) The mass of the mountain and its fraction of the Earth’s mass are too great.
d) The gravitational force assumed to be exerted by the mountain is too great.
6.6: Satellites and Kepler’s Laws: An Argument for Simplicity
65. A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for the moon in Table.
66. Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.
Solution
\(\displaystyle 1.98×10^{30}kg\)
67. Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass.
68. Find the ratio of the mass of Jupiter to that of Earth based on data in Table.
Solution
\(\displaystyle \frac{M_J}{M_E}=316\)
69. Astronomical observations of our Milky Way galaxy indicate that it has a mass of about \(\displaystyle 8.0×10^{11}\) solar masses. A star orbiting on the galaxy’s periphery is about \(\displaystyle 6.0×10^4\) light years from its center. (a) What should the orbital period of that star be? (b) If its period is \(\displaystyle 6.0×10^7\) instead, what is the mass of the galaxy? Such calculations are used to imply the existence of “dark matter” in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.
70. Integrated Concepts
Space debris left from old satellites and their launchers is becoming a hazard to other satellites.
(a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface.
(b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of \(\displaystyle 90º\) relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it?
(c) Given the rivet is 3.00 mm in size, how long will its collision with the satellite last?
(d) If its mass is 0.500 g, what is the average force it exerts on the satellite?
(e) How much energy in joules is generated by the collision? (The satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)
Solution
a) \(\displaystyle 7.4×10^3m/s\)
b) \(\displaystyle 1.05×10^3m/s\)
c) \(\displaystyle 2.86×10^{−7}s\)
d) \(\displaystyle 1.84×10^7N\)
e) \(\displaystyle 2.76×10^4J\)
71. Unreasonable Results
(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h.
(b) What is unreasonable about this result?
(c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?
Solution
a) \(\displaystyle 5.08×10^3km\)
b) This radius is unreasonable because it is less than the radius of earth.
c) The premise of a one-hour orbit is inconsistent with the known radius of the earth.
72. Construct Your Own Problem
On February 14, 2000, the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros.
Contributors and Attributions
-
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .
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libretexts
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2025-03-17T19:53:25.756693
| 2017-04-20T14:43:19 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/06%3A_Uniform_Circular_Motion_and_Gravitation/6.E%3A_Uniform_Circular_Motion_and_Gravitation_(Excercise)",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "6.E: Uniform Circular Motion and Gravitation (Excercise)",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources
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7: Work, Energy, and Energy Resources
There is no simple, yet accurate, scientific definition for energy. Energy is characterized by its many forms and the fact that it is conserved. We can loosely define energy as the ability to do work, admitting that in some circumstances not all energy is available to do work. Because of the association of energy with work, we begin the chapter with a discussion of work. Work is intimately related to energy and how energy moves from one system to another or changes form.
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- 7.0: Prelude to Work, Energy, and Energy Resources
- Energy plays an essential role both in everyday events and in scientific phenomena. You can no doubt name many forms of energy, from that provided by our foods, to the energy we use to run our cars, to the sunlight that warms us on the beach. You can also cite examples of what people call energy that may not be scientific, such as someone having an energetic personality. Not only does energy have many interesting forms, it is involved in almost all phenomena, and is one of the most important con
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- 7.1: Work- The Scientific Definition
- Work is the transfer of energy by a force acting on an object as it is displaced. The work \(W\) that a force \(F\) does on an object is the product of the magnitude \(F\) of the force, times the magnitude \(d\) of the displacement, times the cosine of the angle \(\theta\) between them. In symbols, \[W = Fd \space cos \space \theta. \] The SI unit for work and energy is the joule (J), where \(1 \space J = 1 \space N \cdot m = 1 \space kg \space m^2/s^2\). The work done by a force is zero if the
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- 7.2: Kinetic Energy and the Work-Energy Theorem
- The net work \(W_{net}\) is the work done by the net force acting on an object. Work done on an object transfers energy to the object. The translational kinetic energy of an object of mass \(m\) moving at speed \(v\) is \(KE = \frac{1}{2}mv^2\). The work-energy theorem states that the net work \(W_{net} \) on a system changes its kinetic energy, \(W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2\).
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- 7.3: Gravitational Potential Energy
- Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy \(\Delta PE_g\), is \(\Delta PE_g = mgh\), with \(h\) being the increase in height and \(g\) the acceleration due to gravity. The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, \(\Delta PE_g\), have physical significance. As an obje
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- 7.4: Conservative Forces and Potential Energy
- A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken. We can define potential energy \((PE\) for any conservative force, just as we defined \(PE_g\) for the gravitational force. The potential energy of a spring is \(PE_s = \frac{1}{2}kx^2\), where \(k\) is the spring’s force constant and |(x\) is the displacement from its undeformed position. Mechanical energy is defined to be \(KE = PE\) for conservative force.
-
- 7.5: Nonconservative Forces
- A nonconservative force is one for which work depends on the path. Friction is an example of a nonconservative force that changes mechanical energy into thermal energy. Work \(W_{nc}\) done by a nonconservative force changes the mechanical energy of a system. In equation form, \(W_{nc} = \Delta KE + \Delta PE \) or, equivalently, \(KE_i + PE_i + W_{nc} = KE_f + PE_f .\) When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms
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- 7.6: Conservation of Energy
- The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same. When all forms of energy are considered, conservation of energy is written in equation form as \[KE_i + PE_i + W_{nc} + OE_i = KE_f + PE_f + OE_f ,\] where \(OE\) is all other forms of energy besides mechanical energy.
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- 7.7: Power
- Power is the rate at which work is done, or in equation form, for the average power \(P\) for work \(W\) done over a time \(t\), \(P = W/t\). The SI unit for power is the watt (W), where \(1 \space W = 1 \space J/s\). The power of many devices such as electric motors is also often expressed in horsepower (hp), where \(1\space hp = 746 \space W.\)
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- 7.8: Work, Energy, and Power in Humans
- The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty tissue. The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate, and the corresponding rate when at rest is called the basal metabolic rate (BMR) The energy included in the basal metabolic rate is divided among various systems in the body, with the largest fraction going to the liver and spleen, and the brain.
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- 7.9: World Energy Use
- The relative use of different fuels to provide energy has changed over the years, but fuel use is currently dominated by oil, although natural gas and solar contributions are increasing. Although non-renewable sources dominate, some countries meet a sizeable percentage of their electricity needs from renewable resources. The United States obtains only about 10% of its energy from renewable sources, mostly hydroelectric power.
Thumbnail: One form of energy is mechanical work, the energy required to move an object of mass m a distance d when opposed by a force F, such as gravity. Image use with permission (CC-SA-BY-NC -3.0; anonymous).
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2025-03-17T19:53:25.827210
| 2015-10-27T19:22:00 |
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"title": "7: Work, Energy, and Energy Resources",
"author": "OpenStax"
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.00%3A_Prelude_to_Work_Energy_and_Energy_Resources
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7.0: Prelude to Work, Energy, and Energy Resources
Energy plays an essential role both in everyday events and in scientific phenomena. You can no doubt name many forms of energy, from that provided by our foods, to the energy we use to run our cars, to the sunlight that warms us on the beach. You can also cite examples of what people call energy that may not be scientific, such as someone having an energetic personality. Not only does energy have many interesting forms, it is involved in almost all phenomena, and is one of the most important concepts of physics. What makes it even more important is that the total amount of energy in the universe is constant. Energy can change forms, but it cannot appear from nothing or disappear without a trace. Energy is thus one of a handful of physical quantities that we say is conserved .
Conservation of energy (as physicists like to call the principle that energy can neither be created nor destroyed) is based on experiment. Even as scientists discovered new forms of energy, conservation of energy has always been found to apply. Perhaps the most dramatic example of this was supplied by Einstein when he suggested that mass is equivalent to energy (his famous equation \(E = mc^2\).
From a societal viewpoint, energy is one of the major building blocks of modern civilization. Energy resources are key limiting factors to economic growth. The world use of energy resources, especially oil, continues to grow, with ominous consequences economically, socially, politically, and environmentally. We will briefly examine the world’s energy use patterns at the end of this chapter.
There is no simple, yet accurate, scientific definition for energy. Energy is characterized by its many forms and the fact that it is conserved. We can loosely define energy as the ability to do work, admitting that in some circumstances not all energy is available to do work. Because of the association of energy with work, we begin the chapter with a discussion of work. Work is intimately related to energy and how energy moves from one system to another or changes form.
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libretexts
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2025-03-17T19:53:25.894232
| 2015-11-01T04:56:31 |
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"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.00%3A_Prelude_to_Work_Energy_and_Energy_Resources",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "7.0: Prelude to Work, Energy, and Energy Resources",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.01%3A_Work-_The_Scientific_Definition
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7.1: Work- The Scientific Definition
Learning Objectives
By the end of this section, you will be able to:
- Explain how an object must be displaced for a force on it to do work.
- Explain how relative directions of force and displacement determine whether the work done is positive, negative, or zero.
What It Means to Do Work
The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy—whenever work is done, energy is transferred. For work, in the scientific sense, to be done, a force must be exerted and there must be motion or displacement in the direction of the force.
Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as
\[W = |\vec{F}| \, \cos \, \theta |\vec{d}| \label{eq1}\]
where \(W\) is work, \(d\) is the displacement of the system, and \(\theta\) is the angle between the force vector \(\vec{F}\) and the displacement vector \(\vec{d}\), as in Figure \(\PageIndex{1}\). We can also write Equation \ref{eq1} as
\[W = F \, d \, \cos \, \theta \label{eq2} \]
To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way one-dimensional segments and add up the work done over each segment.
What is Work?
The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as
\[W = F \, d \, \cos \, \theta \]
where \(W\) is work, \(F\) is the magnitude of the force on the system, \(d\) is the magnitude of the displacement of the system, and \(\theta\) is the angle between the force vector \(F\) d the displacement vector \(d\).
To examine what the definition of work means, let us consider the other situations shown in Figure. The person holding the briefcase in Figure \(\PageIndex{1b}\)does no work, for example. Here \(d = 0\), so \(W = 0\). Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the “briefcase-Earth system” - see Gravitational Potential Energy for more details). There must be motion for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level ground in Figure \(\PageIndex{1c}\) does no work on it, because the force is perpendicular to the motion. That is, \(\cos \, 90^o = 0\), so \(W = 0\).
In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure \(\PageIndex{1d}\), work is done—energy is transferred to the briefcase. Finally, in Figure \(\PageIndex{1e}\), energy is transferred from the briefcase to a generator. There are two good ways to interpret this energy transfer. One interpretation is that the briefcase’s weight does work on the generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement downward. This makes \( \theta = 180^o\), and \(\cos \, 180^o = -1\), therefore \(W\) is negative.
Calculating Work
Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and energy are measured in newton-meters . A newton-meter is given the special name joule (J), and \(1 \, J = 1 \, N \cdot m = 1 \, kg \, m^2/s^2\). One joule is not a large amount of energy; it would lift a small 100-gram apple a distance of about 1 meter.
Example \(\PageIndex{1}\): Calculating the Work You Do to Push a Lawn Mower Across a Large Lawn
How much work is done on the lawn mower by the person in Figure (a) if he exerts a constant force of 75.0 N at an angle \(35^o\) below the horizontal and pushes the mower 25 m. on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of 10,000 kJ (about 2400 kcal ) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by \(1^o C\) and is equivalent to 4,184 J , while one food calorie ( 1 kcal ) is equivalent to 4,184 J .
Strategy
We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation \(W = Fd \, cos \, \theta\). The force, angle, and displacement are given, so that only the work \(W\) is unknown.
Solution
The equation for the work is (Equation \ref{eq2}):
\[W = Fd \, \cos \, \theta \nonumber \]
Substituting the known values gives
\[\begin{align*} W &= (75 \, N)(25.0 \, m)(cos \, 35^o) \\[5pt] &= 1536 \, J \nonumber \\[5pt] &= 1.54 \times 10^3 \, J \nonumber \end{align*} \]
Converting the work in joules to kilocalories yields \(W = (1536 \, J)(1 \, kcal/4184 \, J) = 0.367 kcal.\) The ratio of the work done to the daily consumption is
\[\dfrac{W}{2400 \, kcal} = 1.53 \times 10^{-4}. \nonumber \]
Discussion
This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat.
Summary
- Work is the transfer of energy by a force acting on an object as it is displaced.
- The work \(W\) that a force \(F\) does on an object is the product of the magnitude \(F\) of the force, times the magnitude \(d\) of the displacement, times the cosine of the angle \(\theta\) between them. In symbols, \[W = Fd \, \cos \, \theta. \]
- The SI unit for work and energy is the joule (J), where \(1 \, J = 1 \, N \cdot m = 1 \, kg \, m^2/s^2\).
- The work done by a force is zero if the displacement is either zero or perpendicular to the force.
- The work done is positive if the force and displacement have the same direction, and negative if they have opposite direction.
Glossary
- energy
- the ability to do work
- work
- the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement
- joule
- SI unit of work and energy, equal to one newton-meter
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libretexts
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2025-03-17T19:53:26.049570
| 2015-11-01T04:56:50 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.01%3A_Work-_The_Scientific_Definition",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "7.1: Work- The Scientific Definition",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.02%3A_Kinetic_Energy_and_the_Work-Energy_Theorem
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7.2: Kinetic Energy and the Work-Energy Theorem
Learning Objectives
By the end of this section, you will be able to:
- Explain work as a transfer of energy and net work as the work done by the net force.
- Explain and apply the work-energy theorem.
Work Transfers Energy
What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in [link] (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in [link] (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in [link] (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.
In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.
Net Work and the Work-Energy Theorem
We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.
Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work is the work done by the net external force \(F_{net}\). In equation form, this is \(W_{net} = F_{net}d \, cos \, \theta\), where \(\theta\) is the angle between the force vector and the displacement vector. Figure (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an \(F \, cos \, \theta\) vs. \(d\) graph. In this case, \(F \, cos \, \theta\) is constant. You can see that the area under the graph is \(F \, cos \, \theta\), or the work done. Figure (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force \((F \, cos \, \theta)_{i(ave)}\). The work done is e \((F \, cos \, \theta)_{i(ave)}d_i\) for each strip, and the total work done is the sum of the \(W_i\). Thus the total work done is the total area under the curve, a useful property to which we shall refer later.
Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure.
The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force \(F_{app}\) and the horizontal friction force \(f\). Thus, as expected, the net force is parallel to the displacement, so that \(\theta = 0\) and \(cos \, \theta = 1\), and the net work is given by
\[W_{net} = F_{net} d.\]
The effect of the net force \(F_{net}\) is to accelerate the package from \(v_0\) to \(v\) The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example.) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting \(F = ma\) from Newton’s second law gives
\[W_{net} = mad.\]
To get a relationship between net work and the speed given to a system by the net force acting on it, we take \(d = x - x_0\) and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance \(d\) if the acceleration has the constant value \(a\), namely \(v^2 = v_0^2 + 2ad\). (note that \(a\) appears in the expression for the net work). Solving for acceleration gives \(a = \frac{v^2 - v_0^2}{2d}.\) When \(a\) is substituted into the preceding expression for \(W_{net}\) we obtain
\[W_{net} = m \left(\dfrac{v^2 - v_0^2}{2d} \right)d. \]
The \(d\) cancels, and we rearrange this to obtain
\[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2. \]
This expression is called the work-energy theorem , and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity \(\frac{1}{2}mv^2\). This quantity is our first example of a form of energy.
Work-Energy Theorem
The net work on a system equals the change in the quantity \(\frac{1}{2}mv^2\).
\[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2. \]
The quantity \(\frac{1}{2}mv^2\) in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass \(m\) moving at a speed \(v\). ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,
\[KE = \dfrac{1}{2}mv^2,\]
is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.
We are aware that it takes energy to get an object, like a car or the package in Figure, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.
Example \(\PageIndex{1}\): Calculating the Kinetic Energy of a Package
Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.03.2 is moving at 0.500 m/s. What is its kinetic energy?
Strategy
Because the mass \(m\) and the speed \(v\) are given, the kinetic energy can be calculated from its definition as given in the equation \(KE = \frac{1}{2}mv^2\).
Solution
The kinetic energy is given by \[KE = \dfrac{1}{2}mv^2.\]
Entering known values gives
\[KE = 0.5(30.0 \, kg)(0.500 \, m/s)^2,\]
which yields
\[KE = 3.75 \, kg \cdot m^2/s^2 = 3.75 \, J\]
Discussion
Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.
Example \(\PageIndex{2}\): Determining the Work to Accelerate a Package
Suppose that you push on the 30.0-kg package in Figure 7.03.2. with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.
(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.
Strategy and Concept for (a)
This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.03.2.) As expected, the net work is the net force times distance.
Solution for (a)
The net force is the push force minus friction, or \(F_{net} = 120 \, N - 5.00 \, N = 115 \, N\). Thus the net work is
\[W_{net} = F_{net}d = (115 \, N)(0.800 \, m) \]
\[= 92.0 \, N \cdot m = 92.0 \, J\]
Discussion for (a)
This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.
Strategy and Concept for (b)
The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.
Solution for (b)
The applied force does work.
\[W_{app} = F_{app}d \, cos \, (0^o) = F_{app}d\]
\[= (120 \, N)(0.800 \, m) \]
\[= 96.0 \, J\]
The friction force and displacement are in opposite directions, so that \(\theta = 180^o\), and the work done by friction is
\[W_{fr} = F_{fr}d \, cos \, (180^o)\]
\[= -(5.00 \, N)(0.800 \, m)\]
\[= -4.00 \, J\]
So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,
\[W_{gr} = 0,\]
\[W_N = 0,\]
\[W_{app} = 96.0 \, J,\]
\[W_{fr} = -4 \, J.\]
The total work done as the sum of the work done by each force is then seen to be
\[W_{total} = W_{gr} + W_N + W_{app} + W_{fr} = 92.0 \, J.\]
Discussion for (b)
The calculated total work \(W_{total}\) as the sum of the work by each force agrees, as expected, with the work \(W_{net}\) done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.
Example \(\PageIndex{3}\): Determining Speed from Work and Energy
Find the speed of the package in Figure 7.03.2. at the end of the push, using work and energy concepts.
Strategy
Here the work-energy theorem can be used, because we have just calculated the net work \(W_{net}\) and the initial kinetic energy, \(\frac{1}{2}mv_0^2\) These calculations allow us to find the final kinetic energy, \(\frac{1}{2}mv^2\) and thus the final speed \(v\).
Solution
The work-energy theorem in equation form is
\[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2.\]
Solving for \(\frac{1}{2}mv^2\) gives
\[\dfrac{1}{2}mv^2 = W_{net} + \dfrac{1}{2}mv_0^2\]
Thus, \[\dfrac{1}{2}mv^2 = 92.0 \, J + 3.75 \, J = 95.75 \, J. \]
Solving for the final speed as requested and entering known values gives
\[v = \sqrt{\dfrac{2(95.75 \, J)}{m}} = \sqrt{\dfrac{191.5 \, kg \cdot m^2/s^2}{30.0 \, kg}}\]
\[= 2.53 \, m/s\]
Discussion
Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.
Example \(\PageIndex{4}\): Work and Energy Can Reveal Distance, Too
How far does the package in Figure 7.03.2. coast after the push, assuming friction remains constant? Use work and energy considerations.
Strategy
We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.
Solution
The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so \(\theta = 180^o\). To reduce the kinetic energy of the package to zero, the work \(W_{fr}\) by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus \(W_{fr} = -95.75 \, J\). Furthermore, \(W_{fr} = df' \, cos \, \theta = - Fd'\), where \(d'\) is the distance it takes to stop. Thus,
\[d' = -\dfrac{W_{fr}}{f} = \dfrac{-95.75 \, J}{5.00 \, N}, \]
and so
\[ d' = 19.2 \, m\]
Discussion
This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.
Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.
Summary
- The net work \(W_{net}\) is the work done by the net force acting on an object.
- Work done on an object transfers energy to the object.
- The translational kinetic energy of an object of mass \(m\) moving at speed \(v\) is \(KE = \frac{1}{2}mv^2\).
- The work-energy theorem states that the net work \(W_{net} \) on a system changes its kinetic energy, \(W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2\).
Glossary
- net work
- work done by the net force, or vector sum of all the forces, acting on an object
- work-energy theorem
- the result, based on Newton’s laws, that the net work done on an object is equal to its change in kinetic energy
- kinetic energy
- the energy an object has by reason of its motion, equal to \(\frac{1}{2}mv^2\) for the translational (i.e., non-rotational) motion of an object of mass \(m\) moving at speed \(v\)
|
libretexts
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2025-03-17T19:53:26.142725
| 2015-11-01T04:57:16 |
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"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.02%3A_Kinetic_Energy_and_the_Work-Energy_Theorem",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "7.2: Kinetic Energy and the Work-Energy Theorem",
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.03%3A_Gravitational_Potential_Energy
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7.3: Gravitational Potential Energy
Learning Objectives
By the end of this section, you will be able to:
- Explain gravitational potential energy in terms of work done against gravity.
- Show that the gravitational potential energy of an object of mass m at height h on Earth is given by \(PE_g = mgh\)
- Show how knowledge of the potential energy as a function of position can be used to simplify calculations and explain physical phenomena.
Work Done Against Gravity
Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.
Let us calculate the work done in lifting an object of mass \(m\) through a height \(h\) such as in Figure. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight \(mg\). The work done on the mass is then \(W = Fd = mgh\). We define this to be the gravitational potential energy \((PE_g)\) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the \(PE_g\) gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.
Converting Between Potential Energy and Kinetic Energy
Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to \(mgh\) on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of \(PE_g\) to \(KE\) without explicitly considering the intermediate step of work. (See Example \(\PageIndex{2}\).) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.
More precisely, we define the change in gravitational potential energy \(\Delta PE_g\) to be
\[\Delta PE_g = mgh, \]
where, for simplicity, we denote the change in height by \(h\) rather than the usual \(\Delta h\). Note that \(h\) is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is
\[mgh = (0.500 \, kg)(9.80 \, m/s^2)(1.00 \, m)\]
\[= 4.90 \, kg \cdot m^2/s^2 = 4.90 \, J. \]
Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the work .
Using Potential Energy to Simplify Calculations
The equation \(\Delta PE_g = mgh\) applies for any path that has a change in height of \(h\), not just when the mass is lifted straight up. (See Figure.) It is much easier to calculate \(mgh\) (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position \(h\) of a mass \(m\) is accompanied by a change in gravitational potential energy \(mgh\), and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.
Example \(\PageIndex{1}\): The Force to Stop Falling
A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.
Strategy
This person’s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial \(PE_g\) is transformed into \(KE\) as he falls. The work done by the floor reduces this kinetic energy to zero.
Solution
The work done on the person by the floor as he stops is given by
\[W = Fd \, cos \, \theta = - Fd \]
with a minus sign because the displacement while stopping and the force from floor are in opposite directions \((cos \, \theta = cos \, 180^o = -1.)\) The floor removes energy from the system, so it does negative work.
The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height \(h\):
\[KE = -\Delta PE_g = -mgh \]
The distance \(d\) that the person’s knees bend is much smaller than the height \(h\) of the fall, so the additional change in gravitational potential energy during the knee bend is ignored. The work \(W\) done by the floor on the person stops the person and brings the person’s kinetic energy to zero:
\[W = -KE = mgh\]
Combining this equation with the expression for \(W\) gives \[ -Fd = mgh.\]
Recalling that \(h\) is negative because the person fell down , the force on the knee joints is given by
\[F = -\dfrac{mgh}{d} = -\dfrac{(60.0 \, kg)(9.80 \, m/s^2)(-3.00 \, m)}{5.00 \times 10^{-3} \, m} = 3.53 \times 10^5 \, N.\]
Discussion
Such a large force (500 times more than the person’s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo's hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See Figure.)
Example \(\PageIndex{2}\): Finding the Speed of a Roller Coaster from its Height
(a) What is the final speed of the roller coaster shown in Figure, if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?
Strategy
The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance \(h\) equals the gain in kinetic energy. This can be written in equation form as \(-\Delta PE = \Delta KE \). Using the equations for \(PE_g\) and \(KE\) we can solve for the final speed \(v\), which is the desired quantity.
Solution for (a)
Here the initial kinetic energy is zero, so that \(\Delta KE = \frac{1}{2}mv^2\). The equation for change in potential energy states that \(\Delta PE_g = mgh\). Since \(h\) is negative in this case, we will rewrite this as \(\Delta PE_g = -mg|h|\) to show the minus sign clearly. Thus,\[-\Delta PE_g = \Delta KE\] becomes \[mg|h| = \dfrac{1}{2}mv^2\]
Solving for \(v\) we find that mass cancels and that \[v = \sqrt{2g|h|}. \]
Substituting known values,
\[v = \sqrt{2(9.80 \, m/s^2)(20.0 \, m)} \]
\[= 19.8 \, m/s\]
Solution for (b)
Again \(-\Delta PE_g = \Delta KE\). In this case there is initial kinetic energy, so \(\Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2\). Thus,
\[mgh = \dfrac{1}{2}mv^2 - \dfrac{1}{2}m_0^2.\]
Rearranging gives
\[\dfrac{1}{2}mv^2 = mg|h| + \dfrac{1}{2}mv_0^2.\]
This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and \[v = \sqrt{2g|h| + v_0^2}.\]
This equation is very similar to the kinematics equation \(v = \sqrt{v_0^2 + 2ad}\), but it is more general—the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives
\[v = \sqrt{2(9.80 \, m/s^2)(20.0 \, m) + (5.00)^2}\]
\[= 20.4 \, m/s.\]
Discussion and Implications
First, note that mass cancels. This is quite consistent with observations made in Falling Objects that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that speed can be found at any height along the way by simply using the appropriate value of \(h\) at the point of interest.
We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy.
Making Connections: Take-Home Investigation—Converting Potential to
Kinetic Energy
One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see Figure). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble’s kinetic energy at the bottom is proportional to its potential energy at the release point.
Summary
- Work done against gravity in lifting an object becomes potential energy of the object-Earth system .
- The change in gravitational potential energy \(\Delta PE_g\), is \(\Delta PE_g = mgh\), with \(h\) being the increase in height and \(g\) the acceleration due to gravity.
- The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, \(\Delta PE_g\) , have physical significance.
- As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that \(\Delta KE = - \Delta PE_g\).
Glossary
- gravitational potential energy
- the energy an object has due to its position in a gravitational field
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libretexts
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2025-03-17T19:53:26.257883
| 2015-11-01T04:57:38 |
{
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.03%3A_Gravitational_Potential_Energy",
"book_url": "https://commons.libretexts.org/book/phys-1419",
"title": "7.3: Gravitational Potential Energy",
"author": "OpenStax"
}
|
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