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https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Catabolism/Calvin-Benson-Bassham_Cycle |
Photosynthesis is responsible for creating NADPH and ATP and the Calvin-Benson-Bassham cycle (CBB) uses those high energy molecules to drive the production of glyceraldehyde-3-phosphate (G-3-P). G-3-P can then be used to synthesize hexose sugars which are the primary source of nutrients for heterotrophs. Because ATP and NADPH are required for the CBB to proceed it is necessary for photosynthesis to occur prior. Photosynthesis (a light dependent reaction) uses light energy to produce ATP and NADPH which can then be used to drive synthesis of of carbohydrate molecules in the CBB, namely glyceraldehyde-3-phosphate. Although the CBB cycle has been given the nick name the "dark reaction" the enzymes involved are activated by light. Light stimulates changes in pH in the different regions of the plant cell which then in turn create a better environment for the CBB enzymes. The enzymes in the CBB cycle are very similar to other enzymes found in other metabolic path ways with the exception that they are found in the stoma instead of in the cytoplasm like in glycolysis. The diagram directly below is an extremely abbreviated version of the CBB cycle. This stage is very similar to the isomerization phase of . This phase of CBB very closely resembles part of . This phase of CBB very closely resembles the rearrangement phase of PPP. | 1,364 | 3,824 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Rechargeable_Batteries |
Rechargeable batteries (also known as secondary cells) are batteries that potentially consist of reversible cell reactions that allow them to recharge, or regain their cell potential, through the work done by passing currents of electricity. As opposed to primary cells (not reversible), rechargeable batteries can charge and discharge numerous times. Secondary cells encompass the same mechanism as the primary cells with the only difference being that the Redox reaction of the secondary cell could be reversed with sufficient amount of energy placed into the equation. The figure below illustrates the mechanism of a charging secondary cell. The Charger shown on the top of the diagram is pulling the negative charges toward the right side of the separator. This makes it seem like the positive charges are compiling on the other side of the cell which is not allowed to pass the separator. This disequilibrium is the representation of the cell potential that, when allowed, could once again approach equilibrium through the transferring of the electrons. Different secondary batteries provide various functions. For long-term use (followed by discharging and charging), long storage time when not in use, remote activation, and use under harsh weather conditions are just a few obstacles of creating such secondary cells. Unfortunately, there are no batteries that are capable of encompassing all functions mentioned above. Therefore, the user must decide which application is the most important for a specific task in order to determine the most compatible version of rechargeable batteries. Lead-acid batteries are one of the most common secondary batteries, used primarily for storing large cell potential. These are commonly found in automobile engines. Its advantages include low cost, high voltage and large storage of cell potential; and disadvantages include heavy mass, incompetence under low-temperatures, and inability to maintain its potential for long periods of time through disuse. The reactions of a lead-acid battery are shown below: \[PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{19.90}\] \[Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{19.92}\] Discharging occurs when the engine is started and where the cell potential equals 2.02V. Charging occurs when the car is in motion and where the electrode potential equals -2.02V, a non- spontaneous reaction which requires an external electrical source. The reverse reaction takes place during charging. The nickel-cadmium (NiCd) battery is another common secondary battery that is suited for low-temperature conditions with a long shelf life. However, the nickel-cadmium batteries are more expensive and their capacity in terms of watt-hours per kilogram is less than that of the nickel-zinc rechargeable batteries. \[2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{19.86}\] \[Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{19.87}\] \[Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{19.88}\] Advantages of the nickel-zinc battery are its long life span, high voltage, and the sufficient energy to mass to volume ratio. These characteristics make the nickel-zinc battery more attractive than some. However, it is not yet made in sealed form. | 3,425 | 3,825 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.07%3A_The_Hybrid_Orbital_Model_II |
Make sure you thoroughly understand the following essential ideas: This is a continuation of the previous page which introduced the model and illustrated its use in explaining how valence electrons from atomic orbitals of s and p types can combine into equivalent shared-electron pairs known as , , and hybrid orbitals. In this lesson, we extend this idea to compounds containing double and triple bonds, and to those in which atomic d electrons are involved (and which do not follow the octet rule.) We have already seen how hybridization in carbon leads to its combining power of four in the methane molecule. Two such tetrahedrally coordinated carbons can link up together to form the molecule . In this molecule, each carbon is bonded in the same way as the other; each is linked to four other atoms, three hydrogens and one carbon. The ability of carbon-to-carbon linkages to extend themselves indefinitely and through all coordination positions accounts for the millions of organic molecules that are known. Carbon and hydrogen can also form a compound in which each carbon atom is linked to only three other atoms. Here, we can regard carbon as being trivalent. We can explain this trivalence by supposing that the orbital hybridization in carbon is in this case not , but is instead; in other words, only two of the three orbitals of carbon mix with the 2 orbital to form hybrids; the remaining p-orbital, which we will call the i orbital, remains unhybridized. Each carbon is bonded to three other atoms in the same kind of plane trigonal configuration that we saw in the case of boron trifluoride, where the same kind of hybridization occurs. Notice that the bond angles around each carbon are all 120°. This alternative hybridization scheme explains how carbon can combine with four atoms in some of its compounds and with three other atoms in other compounds. You may be aware of the conventional way of depicting carbon as being tetravalent in all its compounds; it is often stated that carbon always forms four bonds, but that sometimes, as in the case of ethylene, one of these may be a double bond. This concept of the multiple bond preserves the idea of tetravalent carbon while admitting the existence of molecules in which carbon is clearly combined with fewer than four other atoms. These three views of the ethylene molecule emphasize different aspects of the disposition of shared electron pairs in the various bonding orbitals of ethene (ethylene). The "backbone" structure consisting of σ ( ) bonds formed from the three -hybridized orbitals on each carbon. The π ( ) bonding system formed by overlap of the unhybridized p orbital on each carbon. The π orbital has two regions of electron density extending above and below the plane of the molecule. A cutaway view of the combined σ and π system. The σ ( ) bond has its maximum electron density along the line-of-centers joining the two atoms (below left). Viewed end-on, the σ bond is cylindrically symmetrical about the line-of-centers. It is this symmetry, rather than its parentage, that defines the σ bond, which can be formed from the overlap of two -orbitals, from two -orbitals arranged end-to-end, or from an - and a -orbital. They can also form when hybrid orbitals on two atoms overlap end-to-end. orbitals, on the other hand, require the presence of two atomic orbitals on adjacent atoms. Most important, the charge density in the orbital is concentrated above and below the molecular plane; it is almost zero along the line-of-centers between the two atoms. It is this perpendicular orientation with respect to the molecular plane (and the consequent lack of cylindrical symmetry) that defines the π orbital. The combination of a σ bond and a π bond extending between the same pair of atoms constitutes the double bond in molecules such as ethylene. We have not yet completed our overview of multiple bonding, however. Carbon and hydrogen can form yet another compound, acetylene (ethyne), in which each carbon is connected to only two other atoms: a carbon and a hydrogen. This can be regarded as an example of divalent carbon, but is usually rationalized by writing a triple bond between the two carbon atoms. We assume here that since two geometrically equivalent bonds are formed by each carbon, this atom must be -hybridized in acetylene. On each carbon, one hybrid bonds to a hydrogen and the other bonds to the other carbon atom, forming the σ bond skeleton of the molecule. In addition to the hybrids, each carbon atom has two half-occupied orbitals oriented at right angles to each other and to the interatomic axis. These two sets of parallel and adjacent orbitals can thus merge into two sets of π orbitals. The triple bond in acetylene is seen to consist of one σ bond joining the line-of-centers between the two carbon atoms, and two π bonds whose lobes of electron density are in mutually-perpendicular planes. The acetylene molecule is of course linear, since the angle between the two hybrid orbitals that produce the skeleton of the molecule is 180°. Multiple bonds can also occur between dissimilar atoms. For example, in each carbon atom has two unhybridized atomic orbitals, and each oxygen atom still has one orbital available. When the two O-atoms are brought up to opposite sides of the carbon atom, one of the orbitals on each oxygen forms a π bond with one of the carbon -orbitals. In this case, -hybridization is seen to lead to two double bonds. Notice that the two C–O π bonds are mutually perpendicular. Similarly, in , HCN, we assume that the carbon is -hybridized, since it is joined to only two other atoms, and is hence in a divalent state. One of the -hybrid orbitals overlaps with the hydrogen 1 orbital, while the other overlaps end-to-end with one of the three unhybridized orbitals of the nitrogen atom. This leaves us with two nitrogen -orbitals which form two mutually perpendicular π bonds to the two atomic orbitals on the carbon. Hydrogen cyanide thus contains one single and one triple bond. The latter consists of a σ bond from the overlap of a carbon hybrid orbital with a nitrogen orbital, plus two mutually perpendicular π bonds deriving from parallel atomic orbitals on the carbon and nitrogen atoms. bond delocalization furnishes a means of expressing the structures of other molecules that require more than one electron-dot or structural formula for their accurate representation. A good example is the nitrate ion, which contains 24 electrons: The electron-dot formula shown above is only one of three equivalent resonance structures that are needed to describe trigonal symmetry of this ion. Nitrogen has three half-occupied orbitals available for bonding, all perpendicular to one another. Since the nitrate ion is known to be planar, we are forced to assume that the nitrogen outer electrons are -hybridized. The addition of an extra electron fills all three hybrid orbitals completely. Each of these filled orbitals forms a σ bond by overlap with an empty oxygen 2 orbital; this, you will recall, is an example of , in which one of the atoms contributes both of the bonding electrons. The empty oxygen 2 orbital is made available when the oxygen electrons themselves become hybridized; we get three filled hybrid orbitals, and an empty 2 atomic orbital, just as in the case of nitrogen. The π bonding system arises from the interaction of one of the occupied oxygen orbitals with the unoccupied 2 orbital of the nitrogen. Notice that this, again, is a coordinate covalent sharing, except that in this instance it is the oxygen atom that donates both electrons. bonds can form in this way between the nitrogen atom and any of the three oxygens; there are thus three equivalent π bonds possible, but since nitrogen can only form one complete π bond at a time, the π bonding is divided up three ways, so that each N–O bond has a bond order of 4/3. We have seen that the π bonding orbital is distinctly different in shape and symmetry from the σ bond. There is another important feature of the π bond that is of far-reaching consequence, particularly in organic and coordination chemistry. Consider, for example, an extended hydrocarbon molecule in which alternate pairs of carbon atoms are connected by double and single bonds. Each non-terminal carbon atom forms two σ bonds to two other carbons and to a hydrogen (not shown.) This molecule can be viewed as a series of ethylene units joined together end-to-end. Each carbon, being hybridized, still has a half-filled atomic orbital. Since these orbitals on adjacent carbons are all parallel, we can expect them to interact with each other to form π bonds between alternate pairs of carbon atoms as shown below. But since each carbon atom possesses a half-filled orbital, there is nothing unique about the π bond arrangement; an equally likely arrangement might be one in which the π bonding orbitals are shifted to neighboring pairs of carbons (middle illustration above). You will recall that when there are two equivalent choices for the arrangements single and double bonds in a molecule, we generally consider the structure to be a . In keeping with this idea, we would expect the electron density in a π system of this kind to be extended or shared out evenly along the entire molecular framework, as shown in the bottom figure. A system of alternating single and double bonds, as we have here, is called a . Chemists say that the π bonds in a conjugated system are ; they are, in effect, “smeared out” over the entire length of the conjugated part of the molecule. Each pair of adjacent carbon atoms is joined by a σ bond and "half" of a π bond, resulting in an a C-C of 1.5. An even higher degree of conjugation exists in compounds containing extended (C=C) chains. These compounds, known as , exhibit interesting electrical properties, and whose derivatives can act as "organic wires". The classic example of π bond delocalization is found in the cyclic molecule (C H ) which consists of six carbon atoms bound together in a hexagonal ring. Each carbon has a single hydrogen atom attached to it. The lines in this figure represent the σ bonds in benzene. The basic ring structure is composed of σ bonds formed from overlap of hybrid orbitals on adjacent carbon atoms. The unhybridized carbon orbitals project above and below the plane of the ring. They are shown here as they might appear if they did not interact with one another. But what happens, of course, is that the lobes of these atomic orbitals meld together to form circular rings of electron density above and below the plane of the molecule. The two of these together constitute the "second half" of the carbon-carbon double bonds in benzene. This computer-generated plot of electron density in the benzene molecule is derived from a more rigorous theory that does not involve hybrid orbitals; the highest electron density (blue) appears around the periphery of the ring, while the lowest (red) is in the "doughnut hole" in the center. In atoms that are below those in the first complete row of the periodic table, the simple octet rule begins to break down. For example, we have seen that PCl does conform to the octet rule but PCl does not. We can describe the bonding in PCl very much as we do NH : four -hybridized orbitals, three of which are shared with electrons from other atoms and the fourth containing a nonbonding pair. The molecule exemplifies one of the most common types of -orbital hybridization. The six bonds in this octahedrally-coordinated molecule are derived from mixing six atomic orbitals into a hybrid set. The easiest way to understand how these come about is to imagine that the molecule is made by combining an imaginary S ion (which we refer to as the ) with six F ions to form the neutral molecule. These now-empty 3 and 3 orbitals then mix with two 3 orbitals to form the hybrids. Some of the most important and commonly encountered compounds which involve the orbitals in bonding are the . The term “complex” in this context means that the molecule is composed of two or more kinds of species, each of which can have an independent existence. For example, the ions Pt and Cl can form the ion [PtCl ] . To understand the hybridization scheme, it helps to start with the neutral Pt atom, then imagine it losing two electrons to become an ion, followed by grouping of the two unpaired 5 electrons into a single orbital, leaving one vacant.This vacant orbital, along with the 6 and two of the 6 orbitals, can then accept an electron pair from four chlorines. All of the four-coordinated molecules we have discussed so far have tetrahedral geometry around the central atom. Methane, CH , is the most well known example. It may come as something as a surprise, then, to discover that the has an essentially two-dimensional configuration. This type of bonding pattern is quite common when the parent central ion (Pt in this case) contains only eight electrons in its outmost -subshell. Many of the most commonly encountered transition metal ions accept electron pairs from donors such as CN and NH (or lacking these, even from H O) to form octahedral coordination complexes. The cation depicted below is typical. In hybridization the bonding orbitals are derived by mixing atomic orbitals having the same principal quantum number ( = 4 in the preceding example). A slightly different arrangement, known as hybridization, involves orbitals of lower principal quantum number. This is possible because of the rather small energy differences between the orbitals in one “shell” with the and orbitals of the next higher one — hence the term “inner orbital” complex which is sometimes used to describe ions such as , shown below.. Both arrangements produce octahedral coordination geometries. In some cases, the same central atom can form either inner or outer complexes depending on the particular ligand and the manner in which its electrostatic field affects the relative energies of the different orbitals.Thus the ion utilizes the iron 3 orbitals, whereas achieves a lower energy by accepting two H O molecules in its 4 orbitals. As is the case with any scientific model, the hybridization model of bonding is useful only to the degree to which it can predict phenomena that are actually observed. Most models contain weaknesses that place limits on their general applicability. The need for caution in accepting this particular model is made more apparent when we examine the shapes of the molecules below the first full row of the periodic table. For example, we would expect the bonding in hydrogen sulfide to be similar to that in water, with tetrahedral geometry around the sulfur atom. Experiments, however, reveal that the H–S–H bond angle is only 92°. Hydrogen sulfide thus deviates much more from tetrahedral geometry than does water, and there is no apparent and clear reason why it should. It is certainly difficult to argue that electron-repulsion between the two nonbonding orbitals is pushing the H–S bonds closer together (as is supposed to happen to the H–O bonds in water); many would argue that this repulsion would be less in hydrogen sulfide than in water, since sulfur is a larger atom and is hence less electronegative. The justification we gave for invoking hybridization in molecules such as BeH , BF and CH was that the bonds in each are geometrically and chemically equivalent, whereas the atomic - and -orbitals on the central atoms are not. By combining these into new orbitals of , and types we obtain the required number of completely equivalent orbitals. This seemed easy enough to do on paper; we just drew little boxes and wrote “ ” or whatever below them. But what is really going on here? The full answer is beyond the scope of this course, so we can only offer the following very general explanation. First, recall what we mean by “orbital”: a mathematical function ψ having the character of a standing wave whose square ψ is proportional to the probability of finding the electron at any particular location in space. The latter, the , can be observed (by X-ray scattering, for example), and in this sense is the only thing that is “real”. A given standing wave (ψ-function) can be synthesized by combining all kinds of fundamental wave patterns (that is, atomic orbitals) in much the same way that a color we observe can be reproduced by combining different sets of primary colors in various proportions. In neither case does it follow that these original orbitals (or colors) are actually present in the final product. So one could well argue that hybrid orbitals are not “real”; they simply turn out to be convenient for understanding the bonding of simple molecules at the elementary level, and this is why we use them. It turns out, in fact, that the electron distribution and bonding in ethylene can be equally well described by assuming no hybridization at all. The "bent bond" model requires only that the directions of some of the atomic- orbitals be distorted sufficiently to provide the overlap needed for bonding; these are sometimes referred to as " ". The smallest of the closed-ring hydrocarbons is cyclopropane, a planar molecule in which the C–C bond angles are 120°— quite a departure from the tetrahedral angle of 109.5° associated with hybridization! Theoretical studies suggest that the bent-bond model does quite well in predicting its properties. | 17,603 | 3,827 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/01%3A_The_Basics/1.03%3A_Temperature |
Another important variable that describes the state of a system it the system’s . Like pressure, temperature scales experienced an important process of development over time. Three of the most important temperature scales in US culture are the Fahrenheit, Celsius, and Kelvin scales. G. Daniel Fahrenheit wanted to develop a temperature scale that would be convenient to use in his laboratory. He wanted it to be of convenient magnitude and wanted to avoid having to use any negative values for temperature. So he define the zero of his temperature scale to be the lowest temperature he could create in his laboratory, which was in a saturated brine/water/ice slurry. He then defined 100 °F as his own body temperature. As a result, using his temperature scale, water has a normal melting point (the temperature at 1.00 atm pressure at which water ice melts) of 32 °F. Similarly, water boils (again at 1 atm pressure) at a temperature of 212 °F. The difference between these values is 180 °F. Anders Celsius also thought a 100 degree temperature scale made sense, and was given the name “the centigrade scale”. He defined 0 °C on his scale as the normal boiling point of water, and 100 °C as the normal freezing point. By today’s standards, this inverted temperature scale makes little sense. The modern Celsius temperature scale defines 0 °C as the normal freezing point of water and 100 °C as the normal boiling point. The difference is 100 °C. Comparing this to the Fahrenheit scale, one can easily construct a simple equation to convert between the two scales. \[ 212\, °F = 100\, °C(m) + b \nonumber \] \[32 °F = 0 \,°C (m) + b \nonumber \] Solving these equations for \(m\) and \(b\) yields \[ m= \dfrac{9 \, °F}{5\, °C} \nonumber \] \[b= 32\,°F \nonumber \] And so conversion between the two scales is fairly simple. \[ y\, °F = x\, \cancel{°C} \left( \dfrac{9 \, °F}{5\, \cancel{°C}} \right) + 32\,°F \nonumber \] \[x\,°C = (y \, \cancel{°F} - 32 \cancel{°F}) \left( \dfrac{5 \, °C}{9\,\cancel{ °F}} \right) \nonumber \] Many physical properties of matter suggest that there is an absolute minimum temperature that can be attained by any sample. This minimum temperature can be shown by several types or experiments to be -273.15 °C. An absolute temperature scale is one that assigns the minimum temperature a value of 0. One particularly useful scale is named after William Lord Kelvin (Kelvin, Lord William Thomson (1824-1907) , 2007). The Kelvin scale fixes the normal melting temperature of water at 273.15 K and the boiling point at 373.15 K. As such, temperatures can be converted using the following expression: \[ z\, K = x\, \cancel{°C} \left( \dfrac{1 \, K}{1\, \cancel{°C}} \right) + 273.15 \,K \nonumber \] | 2,740 | 3,828 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Rearrangement_Reactions/Keto-Enol_Tautomerism |
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing S 2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pK > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pK > 45), NaNH (sodium amide, pK = 34), and LiN[CH(CH ) ] (lithium diisopropylamide, LDA, pK 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane. Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced. Examples If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide. NaOCH CH = Na OCH CH = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. However under acidic and basic conditions the equilibrium can be shifted to the right Acid conditions 1) Protonation of the Carbonyl 2) Enol formation Basic conditions 1) Enolate formation 2) Protonation | 3,016 | 3,829 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Slater's_Rules_for_Effective_Nuclear_Charge |
Effective nuclear charge is really important, because it determines the size and energy of orbitals, which determine most properties of atoms. So it's useful to be able to predict effective nuclear charge! Slater's rules give a simple approximation of effective nuclear charge that works pretty well. Based on the last section, we can expect that effective nuclear charge will depend on the number of electrons that might get between, so it depends on the electron we are looking at. For any electron, to find the effective nuclear charge it feels, we need to know how many other electrons might get in the way, and how much time it spends near the nucleus. Based on these, we will calculate a , S. Then, \[Z_{eff}=Z-S\] where Z is the actual nuclear charge (which is the same as the atomic number) and Z is the effective nuclear charge. To calculate S, we will write out all the electrons in atom until we get to the group of the electron we want, like this: (1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p) etc. The outcome of this is that Z changes suddenly when going from one period to another. As you go from Li to Be, Z (for the new electron) increases, because you add one proton (Z + 1) and it is only shielded 35% (S + 0.35). When you get to B, you added one proton, and it still shields 35%. So Z increases until you go from Ne to Na. Now, suddenly, the (1s) electrons shield 100% instead of 85%, and the (2s,2p) shield 85% instead of 35%! So Z goes down suddenly. From Na to Ar, Z increases slowly again. From Ar to K, it drops again. For an example, let's calcalate Z for a d electron in Zn, atomic number 30. Notice that although 4s is filled, we don't include it because it comes to the right of the d electrons we are looking at. (1s)(2s,2p)(3s,3p)(3d) \[S=18(1)+9(0.35)=21.15\] \[Z_{eff}=31-26=5\] You can see that just like changing periods (going to a new shell), going from the d-block to the p-block also gives a drop in Z (partly because you actually are going to a new shell, as well as subshell). | 2,046 | 3,830 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.03%3A_Entropy |
In addition to learning that the efficiency of a Carnot engine depends only on the high and low temperatures, more interesting things can be derived through the exploration of this system. For example, consider the total heat transferred in the cycle: \[ q_{tot} = nRT_h \ln \left( \dfrac{V_2}{V_1} \right) - nRT_l \ln \left( \dfrac{V_4}{V_3} \right) \nonumber \] Making the substitution \[ \dfrac{V_2}{V_1} = \dfrac{V_3}{V_4} \nonumber \] the total heat flow can be seen to be given by \[ q_{tot} = nRT_h \ln \left( \dfrac{V_4}{V_3} \right) - nRT_l \ln \left( \dfrac{V_4}{V_3} \right) \nonumber \] It is clear that the two terms do not have the same magnitude, unless \(T_h = T_l\). This is sufficient to show that \(q\) is , since it’s net change around a closed cycle is not zero (as any value of a state function must be.) However, consider what happens when the sum of \(q/T\) is considered: \[ \begin{align*} \sum \dfrac{q}{T} &= \dfrac{nR \cancel{T_h} \ln \left( \dfrac{V_4}{V_3} \right)}{\cancel{T_h}} - \dfrac{nR \cancel{T_l} \ln \left( \dfrac{V_4}{V_3} \right)}{ \cancel{T_l}} \\[4pt] &= nR \ln \left( \dfrac{V_4}{V_3} \right) - nR \ln \left( \dfrac{V_4}{V_3} \right) \\[4pt] & = 0 \end{align*} \] This is the behavior expected for a state function! It leads to the definition of entropy in differential form, \[ dS \equiv \dfrac{dq_{rev}}{T} \nonumber \] In general, \(dq_{rev}\) will be larger than \(dq\) (since the reversible pathway defines the maximum heat flow.) So, it is easy to calculate entropy changes, as one needs only to define a reversible pathway that connects the initial and final states, and then integrate \(dq/T\) over that pathway. And since \(\ S\) is defined using \(q\) for a reversible pathway, \(\Delta S\) is of the actual path a system follows to undergo a change. | 1,821 | 3,831 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.24%3A_Addition_Polymers |
Addition polymers are usually made from a monomer containing a double bond. We can think of the double bond as "opening out" in order to participate in two new single bonds in the following way: Thus, if ethene is heated at moderate temperature and pressure in the presence of an appropriate catalyst, it polymerizes: The result is the familiar waxy plastic called polyethylene, which at a molecular level consists of a collection of long-chain alkane molecules, most of which contain tens of thousands of carbon atoms. There is only an occasional short branch chain. Polyethylene is currently manufactured on a very large scale, larger than any other polymer, and is used for making plastic bags, cheap bottles, toys, etc. Many of its properties are what we would expect from its molecular composition. The fact that it is a mixture of molecules each of slightly different chain length (and hence slightly different melting point) explains why it softens over a range of temperatures rather than having a single melting point. Because the molecules are only held together by London forces, this melting and softening occurs at a rather low temperature. (Some of the cheaper varieties of polyethylene with shorter chains and more branch chains will even soften in boiling water.) The same weak London forces explain why polyethylene is soft and easy to scratch and why it is not very ‘strong mechanically.' The table above lists some other well-known addition polymers and also some of their uses. You can probably find at least one example of each of them in your home. Except for Teflon, all these polymers derive from a monomer of the form The resulting polymer thus has the general form
By varying the nature of the R group, the physical properties of the polymer can be controlled rather precisely. | 1,828 | 3,832 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Oxacyclopropane_Synthesis |
Oxacyclopropane rings, also called rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with . Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH. Peroxycarboxylic acids are generally unstable. An exception is , shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions. Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as , , , or is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal , a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions. Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal , a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one S and one R . But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of S, S and R, R . 1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid) a) in acetone solvent. b) in an aqueous medium with acid or base catalyst present. 2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent. 3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent. 4. Predict the reaction of propene with MCPBA. a) in acetone solvent b) after aqueous work-up. 5. Predict the reaction of cis-2-butene in chloroform solvent. 1. a) Cis-2-methyl-3-propyloxacyclopropane b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol 2. Trans-3-ethyl-2-methyloxacyclopropane. 3. Trans-3,4-diethyloxacyclopropane. 4. a) 1-ethyl-oxacyclopropane b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol 5. Cis-2,3-dimethyloxacyclopropane | 3,645 | 3,833 |
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We have already stated several times that solubility in water is a characteristic property of many ionic compounds, and provides further confirmation of this fact. We have also presented experimental evidence that ions in solution are nearly independent of one another. This raises an important question, though, because we have also stated that attractive forces between oppositely charged ions in a crystal lattice are large. The high melting and boiling points of ionic compounds provide confirmation of the expected difficulty of separating oppositely charged ions. How, then, can ionic compounds dissolve at room temperature? Surely far more energy would be required for an ion to escape from the crystal lattice into solution than even the most energetic ions would possess. The resolution of this apparent paradox lies in the interactions between ions and the molecules of water or other polar solvents. The negative (oxygen) side of a dipolar water molecule attracts and is attracted by any positive ion in solution. Because of this , water molecules cluster around positive ions, as shown in Figure \(\Page {1}\) . Similarly, the positive (hydrogen) ends of water molecules are attracted to negative ions. This process, in which either a positive or a negative ion attracts water molecules to its immediate vicinity, is called . When water molecules move closer to ions under the influence of their mutual attraction, there is a net lowering of the potential energy of the microscopic particles. This counteracts the increase in potential energy which occurs when ions are separated from a crystal lattice against their attractions for other ions. Thus the process of dissolving an ionic solid may be divided into the two hypothetical steps shown in Figure \(\Page {2}\). First, the crystalline salt is separated into gaseous ions. The heat energy absorbed when the ions are separated this way is called the (or sometimes the lattice energy). Next, the separate ions are placed in solution; that is, water molecules are permitted to surround the ions. The enthalpy change for this process is called the . Since there is a lowering of the potential energy of the ions and water molecules, heat energy is given off and hydration enthalpies are invariably negative. The heat energy absorbed when a solute dissolves (at a pressure of 1.00 atm) is called the . It can be calculated using Hess' law, provided the lattice enthalpy and hydration enthalpy are known. Using data given in Figure \(\Page {2}\), calculate the enthalpy of solution for NaCl( ). According to the figure, the lattice enthalpy is 773 kJ mol . The hydration enthalpy is – 769 kJ mol . Thus we can write the thermo-chemical equations \(\text{NaCl}(s) \rightarrow \text{Na}^{+}(g) + \text{Cl}^{-}(g) \) \(\triangle H_f = 773 \frac{kJ}{mol}\) \(\underline{\text{Na}^{+}(g) + \text{Cl}^{-}(g) \rightarrow \text{Na}^{+}(aq) + \text{Cl}^-(aq)}\) \(\triangle H_h = -769 \frac{kJ}{mol}\) \(\text{NaCl}(s) \rightarrow \text{Na}^{+}(aq) + \text{Cl}^{-}(aq)\) \(\triangle H_s=\triangle H_f + \triangle H_h\) \(\triangle H_s = (773-769)\frac{kJ}{mol} = +4\frac{kJ}{mol} \) When NaCl( ) dissolves, 773 kJ is required to pull apart a mole of Na ions from a mole of Cl ions, but almost all of this requirement is provided by the 769 kJ released when the mole of Na and the mole of Cl becomes surrounded by water dipoles. Only 4 kJ of heat energy is absorbed from the surroundings when a mole of NaCl( ) dissolves. You can verify the small size of this enthalpy change by putting a few grains of salt on your moist tongue. The quantity of heat energy absorbed as the salt dissolves is so small that you will feel no cooling, even though your tongue is quite a sensitive indicator of temperature changes. Few molecules are both small enough and polar to cluster around positive and negative ions in solution as water does. Consequently water is one of the few liquids which readily dissolves many ionic solids. Hydration of Na , Cl and other ions in aqueous solution prevents them from attracting each other into a crystal lattice and precipitating. | 4,133 | 3,834 |
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The SI base units are not always of convenient size for a particular measurement. For example, the meter would be too big for reporting the thickness of this page, but rather small for the distance from Chicago to Detroit. To overcome this obstacle the SI includes a series of prefixes, each of which represents a power of 10. These allow us to reduce or enlarge the SI base units to convenient sizes. The figures below show how these prefixes can be applied to the meter to cover almost the entire range of lengths we might wish to measure. One non-SI unit of length, the angstrom (Å), is convenient for chemists and will continue to be used for a limited time. Since 1Å = 10 m, the angstrom corresponds roughly to the diameters of atoms and small molecules. Such dimensions are also conveniently expressed in picometers, 1 pm = 10 m = 0.01Å, but the angstrom is widely used and very familiar. Therefore we will usually write atomic and molecular dimensions in both angstroms and picometers. The SI base unit of mass, the kilogram, is unusual because it already contains a prefix. The standard kilogram is a cylinder of corrosion-resistant platinum-iridium alloy which is kept at the International Bureau of Weights and Measures near Paris. The kilogram was chosen instead of a gram because the latter would have made an inconveniently small piece of platinum-iridium and would have been difficult to handle. Also, units of force, pressure, energy, and power have been derived using the kilogram instead of the gram. Despite the fact that the kilogram is the SI unit of mass, the standard prefixes are applied to the when larger or smaller mass units are needed. For example, the quantity 10 kg (1 million kilograms) can be written as 1 Gg (gigagram) but as 1 Mkg (megakilogram). The operative rule here is that one and only one prefix should be attached to the name for a unit. Figure 1.6 illustrates the use of this rule in expressing the wide range of masses available in the universe. Note that the masses of atoms and molecules are usually so small that scientific notation must be used instead of prefixes. | 2,131 | 3,835 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.E%3A_Gases_(Exercises) |
Assuming the form of the Maxwell distribution allowing for motion in three directions to be \[ f(v) = N v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right) \label{MBFullN}\] derive the correct expression for \(N\) such that the distribution is normalized. Hint: a table of definite integrals indicates \[ \int_0^{\infty} x^{2n} e^{-ax^2} dx = \dfrac{1}{4} \dfrac{\sqrt{\pi}}{a^{3/2}}\] Dry ice (solid CO ) has a density of 1.6 g/cm . Assuming spherical molecules, estimate the collisional cross section for \(CO_2\). How does it compare to the value listed in the text? Calculate the pressure exerted by 1.00 mol of Ar, N , and CO as an ideal gas, a van der Waals gas, and a Redlich-Kwong gas, at 25 °C and 24.4 L. The compression factor \(Z\) for CO at 0 °C and 100 atm is 0.2007. Calculate the volume of a 2.50 mole sample of CO at 0 °C and 100 atm. What is the maximum pressure that will afford a N molecule a mean-free-path of at least 1.00 m at 25 °C? In a Knudsen cell, the effusion orifice is measured to be 0.50 mm . If a sample of naphthalene is allowed to effuse for 1.0 hr at a temperature of 40.3 °C, the cell loses 0.0236 g. From this data, calculate the vapor pressure of naphthalene at this temperature. The vapor pressure of scandium was determined using a Knudsen cell [Kirkorian, , , 1586 (1963)]. The data from the experiment are given below. From this data, find the vapor pressure of scandium at 1555.4 K. A thermalized sample of gas is one that has a distribution of molecular speeds given by the Maxwell-Boltzmann distribution. Considering a sample of N at 25 C what fraction of the molecules have a speed less than Assume that a person has a body surface area of 2.0 m . Calculate the number of collisions per second with the total surface area of this person at 25 °C and 1.00 atm. (For convenience, assume air is 100% N ) Two identical balloons are inflated to a volume of 1.00 L with a particular gas. After 12 hours, the volume of one balloon has decreased by 0.200 L. In the same time, the volume of the other balloon has decreased by 0.0603 L. If the lighter of the two gases was helium, what is the molar mass of the heavier gas? Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for \(CO_2\). Find an expression in terms of van der Waals coefficients for the Boyle temperature. ( : use the viral expansion of the van der Waals equation to find an expression for the second viral coefficient!) Consider a gas that follows the equation of state \[p =\dfrac{RT}{V_m - b}\] Using a virial expansion, find an expression for the second virial coefficient. Consider a gas that obeys the equation of state \[ p =\dfrac{nRT}{V_m - b}\] where a and b are non-zero constants. Does this gas exhibit critical behavior? If so, find expressions for \(p_c\), \(V_c\), and \(T_c\) in terms of the constants \(a\), \(b\), and \(R\). Consider a gas that obeys the equation of state \[ p = \dfrac{nRT}{V- nB}-\dfrac{an}{V}\] | 3,001 | 3,836 |
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The macroscopic, descriptive approach to chemical knowledge has led to a great deal of factual information. Right now more than 100 million chemical compounds and their properties are on file at the Chemical Abstracts Service of the American Chemical Society . Anyone who wants information about these substances can look it up, although in practice it helps to have a computer do the looking! Even with a computer’s memory it is hard to keep track of so many facts–no single person can remember more than a fraction of the total. Fortunately these millions of facts are interrelated in numerous ways, and the relationships are helpful in remembering the facts. To illustrate this point, we shall present part of the descriptive chemistry of about 20 elements. Although each element has unique physical and chemical properties, it will be obvious that certain groups of elements are closely related. Members of each group are more like each other than they are like any member of another group. Because of this close relationship a special name has been assigned to these collections of elements. It is also possible to write general equations which apply to members of a family of elements. Practical laboratory experience with one member gives a fairly accurate indication of how each of the others will behave. Because of their similarities, lithium, sodium, potassium, rubidium, and cesium are grouped together and called the . What do we mean when we say these elements are similar? The following video of lithium, sodium, and potassium reacting in water demonstrates the similarities quite well. In the video, each metal is added to a dish containing water. All three metals react with the water, flying over the surface as H is produced (albeit at different rates). LiOH, NaOH, and KOH, are each produced respectively. These hydroxides are evidenced through the use of phenolphthalein indicator, which turns pink in the presence of OH ions. While the reactivity of each metal differs (potassium is clearly more reactive with water than sodium, which is more reactive than lithium) all three are undergoing the same reaction. The three reactions are: \[ 2 \text{Li} (s) + 2 \text{H}_{2} \text{O} (l) \rightarrow 2 \text{LiOH} (aq) + \text{H}_{2} (g) \nonumber \] \[ 2 \text{Na} (s) + 2 \text{H}_{2} \text{O} (l) \rightarrow 2 \text{NaOH} (aq) + \text{H}_{2} (g) \nonumber \] \[ 2 \text{K} (s) + 2\text{H}_{2} \text{O} (l) \rightarrow 2 \text{KOH} (aq) + \text{H}_{2} (g) \nonumber \] Indeed, all alkali metals react with water in this exact way, according to a general equation: \[ 2 \text{M} (s) + 2\text{H}_{2} \text{O} (l) \rightarrow 2 \text{MOH} (aq) + \text{H}_{2} (g)~~~~~~~~~ \text{M = Li, Na, K, Rb, or Cs} \nonumber \] Physical similarities are also apparent in the video. All three are metallic, silver-gray in color, and all three metals are less dense than water. All float on the surface while they react. Other properties not obvious in the video exist. Alkali metals are soft and easily cut. All are solid at room temperature, but melt below 200°C, low for a metal. Beyond similar reactions with water, all alkali metals undergo analogous reactions with oxygen from the atmosphere, forming oxides, M O. Alkali metals react with hydrogen to form hydrides, MH, and sulfur to form sulfides, M S. In all of these compounds, the alkali metals are positive ions, Li , Na , K , Rb or Cs . Each member of the chemical family of alkali metals has physical and chemical properties very similar to all the others. In most cases The peroxides and superoxides are exceptions to this rule, but formulas for oxides and each of the other types of compounds we have described are identical except for the chemical symbol of each alkali metal. For a demonstration of alkali metals on a larger (and more exciting scale), check out what Mythbusters did with Alkali metals. \[ 2 \text{M} + \text{X}_{2} \rightarrow 2 \text{MX}~~~~~~~~~~~ \text{M = Li, Na, K, Rb, or Cs and X = F, Cl, Br, I} \nonumber \] All of the resulting compounds have similar properties as well, such as tasting "salty". Another example of similarity in chemical properties is with mercury, whose reaction with bromine was discussed in the section covering . Mercury reacts with other halogens in the same way: The halogens also react directly with hydrogen, yielding the hydrogen halides: In all these examples, the halogens are anions, F , Cl , Br , and I . From these examples, it should be clear that the halogens, like the alkali metals, represent a grouping of elements which share a number of chemical and physical properties. There are several other examples of related groups of elements. Beryllium, magnesium, calcium, strontium, barium, and radium all show similarities, and are called . All alkaline earths are silvery-gray metals which are ductile and relatively soft. However, they are much denser than the alkali metals, and their melting points are significantly higher. They are also harder than the alkali metals. Further, they all form form hydrides, MH ; oxides, MO; halides, MX ; with M = Mg, Ca, Sr, Ba, or Ra and X = F, Cl, Br, I. In all these compounds the alkaline-earth elements occur as dipositive ions, Mg , Ca , Sr , or Ba . One group of elements, the (helium, neon, argon, krypton, xenon, and radon), forms almost no chemical compounds. Although small concentrations of the noble gases are present in the earth’s atmosphere, they were not discovered until 1894, largely because they underwent no reactions. Fluorine is sufficiently reactive to combine with pure samples of xenon, radon, and (under special conditions) krypton. The only other element that has been shown conclusively to occur in compounds with the noble gases is oxygen, and no more than a couple of dozen noble-gas compounds of all types are known. This group of elements is far less reactive chemically than any other. | 5,917 | 3,837 |
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Hardly a year goes by without the appearance in the media of the old cliché that the average human body is worth about $3.57 (perhaps a little more, allowing for inflation). Such figures are based on the elemental composition of the human body reported in the table. Considering just the market value of the elements, one obtains a ridiculously low price. What has been ignored, of course, is how the atoms of those elements are put together. The raw materials for a fine watch are not worth much either-what we pay for is mainly the skill and intelligence with which they are combined. Before considering how the elements from which we all are made have been combined, however, it is worth thinking a little about why those particular elements in the table are involved. This is especially important because the same elements in very similar ratios are found in nearly all living systems. As shown in the biological periodic table (Figure \(\Page {1}\) ) a great many other elements are simply not involved in the chemistry of life-at least not to the extent that the necessity of their presence can be demonstrated experimentally. More than 99 percent of the atoms in the human body, or any other organism for that matter, are H, O, N, and C. This does not appear to be mere happenstance. The indications are that life as we know it could not be based on any other four elements. These elements have the lightest and smallest atoms which can form one, two, three, and four covalent bonds, respectively. Because of their small radii these atoms can approach very closely and bond very strongly to each other. The most important of these four elements is undoubtedly carbon. Carbon atoms have a special capacity for binding with each other to form long chains and rings. This capacity allows carbon to form a very large number of stable compounds whose molecular structures are different but nevertheless closely related. No other element has this capacity. Perhaps the closest contender is silicon. Although a number of compounds containing Si—Si bonds such as are known, none of these compounds are very stable. All are readily converted to compounds containing Si—O bonds. The bond enthalpy of the bond (368 kJ mol ) is much larger than that of the Si—Si bond (176 kJ mol ) and also larger than that of the Si—H bond (318 kJ mol ). The production of Si—O bonds is thus exothermic and thermodynamically favorable. Another factor is the rapid rate at which these chemical reactions can occur. Silicon is capable of expanding its valence shell through the use of d orbitals to allow more than four bonds. This enables it to form an activated complex in which both the bond being made and the bond being broken feature, but which requires very little activation energy. Equivalent reactions involving carbon require very much higher activation energies and usually proceed so slowly as to be imperceptible at room temperature, even when thermodynamically permitted. Carbon is not the only element with unique properties in biological molecules. Hydrogen is also special and plays two important roles. You will recall that alkanes are chemically unreactive. We can attribute this to the large C—H bond enthalpy of 413 kJ mol . Only fluorine forms a stronger bond than this with carbon. It is thus quite difficult to replace a hydrogen atom attached to a carbon atom with a more stable alternative. When an organic compound reacts chemically, it is almost always a which undergoes a change. The presence of C—H bonds renders a large proportion of the carbon chain unreactive and restricts reaction almost entirely to those sites which include an atom other than carbon or hydrogen. In biological molecules these other atoms are almost invariably oxygen and nitrogen, and this is no accident either. Oxygen and nitrogen are the two most electronegative elements which have a valence greater than 1. They are able to form bonds both to carbon on the one hand and to hydrogen on the other. In groups like and hydrogen atoms fulfill their second important role in biological molecules— between different molecules or between different parts of the same molecule. This ability of different functional groups on a carbon chain to hydrogen bond with each other is a particularly important aspect of biological molecules. You will recall from figure 2 in the section on molecular equilibrium that a molecule containing a chain of carbon atoms is capable of a very large number of conformations due to free rotation around the single bonds. Since many biological molecules contain very long chains, they are capable of adopting an almost infinite number of shapes. In practice only a few of these shapes are useful, and the molecules can be "frozen" into such a useful conformation through hydrogen-bonding between various segments of the chain. A very good example of this is the enzyme trypsin. If the amino acid chain in this molecule were not held in the particular conformation shown in the figure by means of hydrogen bonds between adjacent parts of the chain, the various segments making up the active site would no longer be grouped together and the molecule would be unable to function as a catalyst. | 5,216 | 3,840 |
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Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower , but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst ( e \(\Page {1}\)). Nevertheless, because of its lower , the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes. A catalyst affects , not Δ . In , the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency. An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in \(\Page {2}\) the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H is substantially lower on the catalyst surface. \(\Page {2}\) shows a process called , in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter. Several important examples of industrial heterogeneous catalytic reactions are in \(\Page {1}\). Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface. In , the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds ( \(\Page {2}\)), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis. Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a . Because enzymes can increase reaction rates by enormous factors (up to 10 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ( \(\Page {3}\)). cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research. Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. | 7,265 | 3,841 |
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After completing this section, you should be able to This course uses IUPAC nomenclature; therefore, you need not usually memorize a large number of trivial names. However, you will encounter some trivial names so frequently in books and articles that they soon become familiar. An alkene that can exhibit geometric isomerism has not been properly named unless its name specifies whether the double bond (or bonds) is (or are) cis or trans. The most effective way of giving this information is discussed, and more details of cis and trans follow in . Alkenes contain carbon-carbon double bonds and are hydrocarbons with the molecular formula is C H ; this is also the same molecular formula as cycloalkanes. The parent chain of an alkene is the longest chain containing both carbon atoms of the double bond. Alkenes are named by dropping the ending of the parent and adding . Also, the position of double bond in the parent chain of the alkene is indicated with a number. For straight chain alkenes, it is the same basic rules as nomenclature of alkanes apply except the suffix is changed to . 1) Find the longest carbon chain that contains both carbons of the double bond. 2) Start numbering from the end of the parent chain which gives the lowest possible number to the double bond. If the double bond is equidistant from both ends of the parent chain, number from the end which gives the substituents the lowest possible number. The double bond in cycloalkenes do not need to number because it is understood that they are in the one position. 3) Place the location number of the double bond directly before the parent name. The location number indicates the position of the first carbon of the double bond. Add substituents and their position to the alkene as prefixes. Remember substituents are written in alphabetical order. The presence of multiple double bonds is indicated by using the appropriate suffix such as , , ect. Each of the multiple bonds receives a location number. Also, only is removed from the parent alkane chain name leaving an "a" in the name. Overall, the name of an alkene should look like: (Location number of substituent)-(Name of substituent)-(Location number of double bond)-(Name of parent chain) + ene In 1993 IUPAC updated their naming recommendation to place the location number of the double bond before the -ene suffix of alkene names. The provides names such as hex-2-ene rather than 2-hexene. The newer system is slowly being accepted so it may occasionally be encountered. Because there are no chain ends in cycloalkenes, the double bond is assumed to numbered C1 and C2 and its location number is not required in the name. The direction of the numbering is determined by which will give the substituent closest to the double bond the lowest number. If multiple double bonds are present, it may be necessary to include their location numbers in the name. One of the double bonds will be number C1 and C2 and the numbering direction is determined by which gives the remaining double bonds the lowest possible number. Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring. Cyclopentene is an example of an endocyclic double bond. Methylenecylopentane is an example of an exocyclic double bond. Name the following compounds... 1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene. 1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene. Try to draw structures for the following compounds... Some alkene containing fragments have common names which should be recognized. These common names can be used to simplify naming much the alkyl fragments discussed in . Some of these fragments are the methylene group (H C=), the vinyl group (H C=CH-), and the allyl group (H C=CH-CH -). In addition, the common name some small alkene compounds are still accepted by IUPAC. It is important to be able to identify them. Name the following compounds using common fragment names. a) b) C) a) 2-Vinyl-1,3-cyclohexadiene b) Methylenecylopentane c) 3-Allylcyclohexene Both these compounds have double bonds, making them alkenes. In example (1) the longest chain consists of six carbons, so the root name of this compound will be . Three methyl substituents (colored red) are present. Numbering the six-carbon chain begins at the end nearest the double bond (the left end), so the methyl groups are located on carbons 2 & 5. The IUPAC name is therefore: . In example (2) the longest chain incorporating both carbon atoms of the double bond has a length of five. There is a seven-carbon chain, but it contains only one of the double bond carbon atoms. Consequently, the root name of this compound will be . There is a propyl substituent on the inside double bond carbon atom (#2), so the IUPAC name is: . The double bond in example (3) is located in the center of a six-carbon chain. The double bond would therefore have a locator number of 3 regardless of the end chosen to begin numbering. The right hand end is selected because it gives the lowest first-substituent number (2 for the methyl as compared with 3 for the ethyl if numbering were started from the left). The IUPAC name is assigned as shown. Example (4) is a diene (two double bonds). Both double bonds must be contained in the longest chain, which is therefore five- rather than six-carbons in length. The second and fourth carbons of this 1,4-pentadiene are both substituted, so the numbering begins at the end nearest the alphabetically first-cited substituent (the ethyl group). These examples include rings of carbon atoms as well as some carbon-carbon triple bonds. Example (6) is best named as an alkyne bearing a cyclobutyl substituent. Example (7) is simply a ten-membered ring containing both a double and a triple bond. The double bond is cited first in the IUPAC name, so numbering begins with those two carbons in the direction that gives the triple bond carbons the lowest locator numbers. Because of the linear geometry of a triple bond, a-ten membered ring is the smallest ring in which this functional group is easily accommodated. Example (8) is a cyclooctatriene (three double bonds in an eight-membered ring). The numbering must begin with one of the end carbons of the conjugated diene moiety (adjacent double bonds), because in this way the double bond carbon atoms are assigned the smallest possible locator numbers (1, 2, 3, 4, 6 & 7). Of the two ways in which this can be done, we choose the one that gives the vinyl substituent the lower number. Name the following alkenes. a) 2-ethylhept-1-ene or 2-ethyl-1-heptene b) 1,2-dimethylcycloheptene c) 2,5-dimethyloct-2-ene or 2,5-dimethyl-1-octene Draw structures for the following compounds from the given names. | 6,970 | 3,844 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.7%3A_Ions_as_Acids_and_Bases |
We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as \(Na^+\), replaces the proton on the acid. An example is the reaction of \(\ce{CH3CO2H}\), a weak acid, with \(\ce{NaOH}\), a strong base: \[\underset{acid}{\ce{CH3CO2H(l)}} +\underset{base}{\ce{NaOH(s)}} \overset{\ce{H2O}}{\longrightarrow} \underset{salt}{\ce{H2OCH3CO2Na(aq)} }+\underset{water}{\ce{H2O(l)}} \label{16.35}\] Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution. When a salt such as \(NaCl\) dissolves in water, it produces \(Na^+_{(aq)}\) and \(Cl^−_{(aq)}\) ions. Using a Lewis approach, the \(Na^+\) ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The \(Cl^−\) ion is the conjugate base of the strong acid \(HCl\), so it has essentially no basic character. Consequently, dissolving \(NaCl\) in water has no effect on the \(pH\) of a solution, and the solution remains neutral. Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations (\(K^+\) and \(Na^+\)) have essentially no acidic character, but the anions (\(CN^−\) and \(CH_3CO_2^−\)) are weak bases that can react with water because they are the conjugate bases of the weak acids \(HCN\) and acetic acid, respectively. \[ \ce{CN^{-}(aq) + H2O(l) <<=> HCN(aq) + OH^{-}(aq)}\] \[ \ce{CH3CO^{-}2(aq) + H2O(l) <<=> CH3CO2H(aq) + OH^{-}(aq)}\] Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both \(HCN\) and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the \(pH\) of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table E1 and E2, we can see that \(CN^−\) is a stronger base (\(pK_b = 4.79\)) than acetate (\(pK_b = 9.24\)), which is consistent with \(KCN\) producing a more basic solution than sodium acetate at the same concentration. In contrast, the conjugate acid of a weak base should be a weak acid. For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with \(HCl\). As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows: \[ \ce{NH^{+}4(aq) + H2O(l) <<=> NH3(aq) + H3O^{+}(aq)}\] \[ \ce{C5H5NH^{+}(aq) + H2O(l) <<=> C5H5NH(aq) + H3O^{+}(aq)}\] Figure \(\Page {2}\) shows that \(H_3O^+\) is a stronger acid than either \(NH_4^+\) or \(C_5H_5NH^+\), and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The \(H_3O^+\) concentration produced by the reactions is great enough, however, to decrease the \(pH\) of the solution significantly: the \(pH\) of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the \(pH\), while according to Figure 16.9, the acetate ion will raise the \(pH\). This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a \(pH\) < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a \(pH\) > 7.00. Ions as Acids and Bases: Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce \(H_3O^+\). Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (Figure \(\Page {1a}\)). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton. Second, the positive charge on the \(Al^{3+}\) ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the \(\ce{O–H}\) bonds, as shown in Figure \(\Page {1b}\). With less electron density between the \(O\) atoms and the H atoms, the \(\ce{O–H}\) bonds are weaker than in a free \(H_2O\) molecule, making it easier to lose a \(H^+\) ion. The magnitude of this effect depends on the following two factors (Figure \(\Page {3}\)): Thus aqueous solutions of small, highly charged metal ions, such as \(Al^{3+}\) and \(Fe^{3+}\), are acidic: \[[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36}\] The \([Al(H_2O)_6]^{3+}\) ion has a \(pK_a\) of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as \(Li^+\) and \(Mg^{2+}\) or \(Ca^{2+}\) and \(Y^{3+}\), have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well. Solutions of small, highly charged metal ions in water are acidic. Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions. A hydrolysis reaction is an acid–base reaction. Predict whether aqueous solutions of these compounds are acidic, basic, or neutral. : compound : acidity or basicity of aqueous solution : : a b. c. Predict whether aqueous solutions of the following are acidic, basic, or neutral. neutral acidic basic (due to the reaction of \(HS^−\) with water to form \(H_2S\) and \(OH^−\)) A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (\(A^−\)), the conjugate acid of a weak base as the cation (\(BH^+\)), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction. | 7,778 | 3,846 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Free_Radical_Reactions/Bromination_of_Methane |
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light. \[CH_4 + Br_2 \rightarrow CH_3Br + HBr\] The organic product is bromomethane. One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms. The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going.The over-all process is known as free radical substitution, or as a free radical chain reaction. Br 2Br CH + Br CH + HBr CH + Br CH Br + Br 2Br Br CH + Br CH Br CH + CH CH CH Jim Clark ( ) | 1,035 | 3,848 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.09%3A_Indicators |
We mentioned that in most it is necessary to add an indicator which produces a sudden color change at the equivalence point. A typical indicator for acid-base titrations is phenolphthalein, HC H O . Phenolphthalein, whose structure is shown below, is a colorless weak acid ( = 3 × 10 mol/L). Its conjugate base, C H O has a strong pinkish-red color. In order to simplify, we will write the phenolphthalein molecule as HIn (protonated indicator) and its pink conjugate base as . In aqueous solution, phenolphthalein will present the following equilibrium Clearly there must be some intermediate situation where half the phenolphthalein is in the acid form and half in the colored conjugate-base form. That is, at some pH \[\text{pH}=\text{p}K_{a}\text{ + log}\frac{[\text{ In}^{-}]}{[\text{ HIn }]} \nonumber \] \[\text{pH}=\text{p}K_{a}\text{+log1}=\text{p}K_{a} \nonumber \] For phenolphthalein, we have \[\text{pH}=\text{p}K_{a}=-\text{log(3 }\times \text{ 10}^{-10}\text{)}=\text{9.5} \nonumber \] The way in which both the color of phenolphthalein and the fraction present as the conjugate base varies with the pH is shown in detail in Figure \(\Page {1}\). The change of color occurs over quite a limited range of pH―roughly p ± 1. In other words the color of phenolphthalein changes perceptibly between about pH 8.3 and 10.5. Observe the actual color change for this indicator in Figure \(\Page {2}\). Other indicators behave in essentially the same way, but for many of them both the acid and the conjugate base are colored. Their p ’s also differ from phenolphthalein, as shown in the following table. The indicators listed have been selected so that their p values are approximately two units apart. Consequently, they offer a series of color changes spanning the whole pH range.
If a careful choice of both colors and p is made, it is possible to mix several indicators and obtain a which changes color continuously over a very wide pH range. With such a mixture it is possible to find the approximate pH of any solution within this range. So-called pH paper, as seen below, is impregnated with one or several indicators. When a strip of this paper is immersed in a solution, its pH can be judged from the resulting color. What indicator, from those listed in the table, would you use to determine the approximate pH of the following solutions: | 2,381 | 3,850 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/16%3A_Thermodynamics/16.3%3A_The_Second_and_Third_Laws_of_Thermodynamics |
In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy (Δ > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include , we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true: \[ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{1} \] To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process: These results lead to a profound statement regarding the relation between entropy and spontaneity known as the : A summary of these three relations is provided in Table \(\Page {1}\). For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, \(q_{surr}\) is a good approximation of \(q_{rev}\), and the second law may be stated as the following: \[ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4} \] We may use this equation to predict the spontaneity of a process as illustrated in Example \(\Page {1}\). The entropy change for the process \[\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber \] is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C? We can assess the spontaneity of the process by calculating the entropy change of the universe. If Δ is positive, then the process is spontaneous. At both temperatures, Δ = 22.1 J/K and = −6.00 kJ. At −10.00 °C (263.15 K), the following is true: \[\begin{align*}
ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\
&=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K}
\end{align*} \nonumber \] \(S_{univ} < 0\), so melting is nonspontaneous ( spontaneous) at . At 10.00 °C (283.15 K), the following is true: \[ \begin{align*} ΔS_\ce{univ} &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\[4pt] &=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K \end{align*} \nonumber \] \(S_{univ} > 0\), so melting spontaneous at Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of ? Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K. The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal ( = 1). According to the Boltzmann equation, the entropy of this system is zero. \[S=k\ln W=k\ln(1)=0 \label{5} \] This limiting condition for a system’s entropy represents the : We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. are given the label \(S^\circ_{298}\) for values determined for one mole of substance, isolated in its pure form in its own container, at a pressure of 1 bar and a temperature of 298 K. The thermodynamic standard state of a substance refers to an isolated sample of that substance, in its own container, at 1.000 bar (0.9869 atm) pressure. If the substance is a solute, the most common standard state is one in which the concentration of the solute is 1.000 molal (sometimes approximated with 1.000 M). There is no defined temperature for the standard state, but most discussions about standard state assume that the temperature is 298.15 K (25ºC) unless otherwise noted. This may seem like a strange definition, because it requires that each of the reactants and each of the products of a reaction are kept separate from one another, unmixed. The entropy of mixing must be determined separately. The for any process may be computed from the standard entropies of its reactant and product species like the following: \[ΔS°=\sum νS^\circ_{298}(\ce{products})−\sum νS^\circ_{298}(\ce{reactants}) \label{\(\Page {6}\)} \] Here, ν represents stoichiometric coefficients in the balanced equation representing the process. For example, Δ ° for the following reaction at room temperature is computed as the following: \[=[xS^\circ_{298}(\ce{C})+yS^\circ_{298}(\ce{D})]−[mS^\circ_{298}(\ce{A})+nS^\circ_{298}(\ce{B})] \label{\(\Page {8}\)} \] Table \(\Page {2}\) lists some standard entropies at 298.15 K. You can find additional standard entropies in Tables T1 or T2. Calculate the standard entropy change for the following process: \[\ce{H2O}(g)⟶\ce{H2O}(l) \nonumber \] The value of the standard entropy change at room temperature, \(ΔS^\circ_{298}\), is the difference between the standard entropy of the product, H O( ), and the standard entropy of the reactant, H O( ). \[\begin{align*}
ΔS^\circ_{298}&=S^\circ_{298}(\ce{H2O}(l))−S^\circ_{298}(\ce{H2O}(g))\\[4pt]
&= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})=−118.8\:J\:mol^{−1}K^{−1} \end{align*} \nonumber \] The value for \(ΔS^o_{298}\) is negative, as expected for this phase transition (condensation), which the previous section discussed. Calculate the standard entropy change for the following process: \[\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g) \nonumber \] −120.6 J mol K Calculate the standard entropy change for the combustion of methanol, CH OH at room temperature: \[\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l) \nonumber \] The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients. \[ \begin{align*} ΔS^\circ &=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants}) \\[4pt] &=[2S^\circ_{298}(\ce{CO2}(g))+4S^\circ_{298}(\ce{H2O}(l))]−[2S^\circ_{298}(\ce{CH3OH}(l))+3S^\circ_{298}(\ce{O2}(g))] \\[4pt] &=\{[2(213.8)+4×70.0]−[2(126.8)+3(205.03)]\}=−161.1\:J/mol⋅K \end{align*} \nonumber \] Calculate the standard entropy change for the following reaction: \[\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l) \nonumber \] 24.7 J/mol•K The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, > 0. If Δ < 0, the process is nonspontaneous, and if Δ = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process. | 7,654 | 3,851 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.04%3A_Zero-Order_Reactions |
A is one whose rate is independent of concentration; its differential rate law is rate = . We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1}\] Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of − . The value of is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of , a positive value. The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form \[[A] = [A]_0 − kt \label{14.4.2}\] where [A] is the initial concentration of reactant A. has the form of the algebraic equation for a straight line, = + , with = [A], = − , and = [A] .) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second. Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N O on a platinum (Pt) surface to produce N and O , which occurs at temperatures ranging from 200°C to 400°C: \[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3}\] Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N O to react with the entire Pt surface, doubling or quadrupling the N O concentration will have no effect on the reaction rate. At very low concentrations of N O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N O concentration. The reaction rate is as follows: \[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4}\] Thus the rate at which N O is consumed and the rates at which N and O are produced are independent of concentration. As shown in , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally. A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is where NAD (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in \(\Page {3}\)). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in \(\Page {3}\)). These examples illustrate two important points: Zero-Order Reactions: | 4,579 | 3,852 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/11%3A_Electrochemical_Methods/11.06%3A_Additional_Resources |
The following set of experiments introduce students to the applications of electrochemistry. Experiments are grouped into four categories: general electrochemistry, preparation of electrodes, potentiometry, coulometry, and voltammetry and amperometry. The following general references providing a broad introduction to electrochemistry. These short articles provide a good introduction to important principles of electrochemistry. Additional information on potentiometry and ion-selective electrodes can be found in the following sources. The following sources provide additional information on electrochemical biosensors. A good source covering the clinical application of electrochemistry is listed below. Coulometry is covered in the following texts. For a description of electrogravimetry, see the following resource. The following sources provide additional information on polarography and pulse polarography. Additional Information on stripping voltammetry is available in the following text. The following papers discuss the numerical simulation of voltammetry. Gathered together here are many useful resources for cyclic voltammetry, including experiments. | 1,176 | 3,854 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Nomenclature_of_Amino_acids |
There are 20 common amino acids. They are composed of C, H, O, N and S atoms. They are structurally and chemically different, and also differ in size and volume. Some are branched structures, some are linear, some have ring structures. One of the 20 common amino acids is actually an imino acid. A typical grouping of their chemical nature is as follows: The amino acids have a name, as well as a three letter or single letter mnemonic code: Leu, L Ile, I Val, V Ala, A Met, M Phe, F Trp, W Pro, P Gly, G (note: sometimes included in polar group) Ser, S Asn, N Gln, Q Thr, T Cys, C Tyr, Y Asp, D Glu, E Lys, K Arg, R His, H In addition to the 20 common amino acids, there are several uncommon ones found: Some amino acids are made that are not intended for incorporation into proteins, rather they have important functionalities on their own Thumbnail: 3D model of L-tryptophan. (Public Domain; ). | 917 | 3,855 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Making_Esters_From_Alcohols |
This page looks at esterification - mainly the reaction between alcohols and carboxylic acids to make esters. It also looks briefly at making esters from the reactions between acyl chlorides (acid chlorides) and alcohols, and between acid anhydrides and alcohols. Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester the hydrogen in this group is replaced by a hydrocarbon group of some kind. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring). The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is: Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group on the end. In each case, be sure that you can see how the names and formulae relate to each other. Notice that the acid is named by counting up the total number of carbon atoms in the chain - including the one in the -COOH group. So, for example, CH CH COOH is propanoic acid, and CH CH COO is the propanoate group. Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring). If you are a UK A level student you won't have to worry about these. The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is: So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be: Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulfuric acid in order to observe the smell of the esters formed. You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes. Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker. Apart from the very small ones, esters are fairly insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer. Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues). As the esters get bigger, the smells tend towards artificial fruit flavoring - "pear drops", for example. If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones. To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulfuric acid, and distil off the ester as soon as it is formed. This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces. Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulfuric acid in the mixture by fractional distillation. Esters can also be made from the reactions between alcohols and either acyl chlorides (acid chlorides) or acid anhydrides.If you add an acyl chloride to an alcohol, you get a vigorous (even violent) reaction at room temperature producing an ester and clouds of steamy acidic fumes of hydrogen chloride. For example, if you add the liquid ethanoyl chloride to ethanol, you get a burst of hydrogen chloride produced together with the liquid ester ethyl ethanoate. \[ CH_3COCl + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + HCl\] The reactions of acid anhydrides are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture. Taking ethanol reacting with ethanoic anhydride as a typical reaction involving an alcohol. There is a slow reaction at room temperature (or faster on warming). There is no visible change in the colorless liquids, but a mixture of ethyl ethanoate and ethanoic acid is formed. \[(CH_3CO)_2O + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + CH_3COOH\] Jim Clark ( ) | 4,873 | 3,857 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Aldol_Reaction |
A useful carbon-carbon bond-forming reaction known as the is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as . For this reaction to occur at least one of the reactants must have α hydrogens. A three step mechanism: : Enolate formation : Nucleophilic attack by the enolate : Protonation The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this \(\beta\) reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed . Hence, the following examples are properly referred to as . Overall the general reaction involves a dehydration of an aldol product to form an alkene: Going from reactants to products simply example 1) Form enolate 2) Form enone When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl. Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of \(\alpha\) hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred. As with other aldol reaction the addition of heat causes an aldol condensation to occur. The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called or reactions, and under certain conditions such crossed aldol condensations can be effective. The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens. The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult. CH CHO + CH CHO + NaOH → – + – + – + – The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the reaction. | 3,557 | 3,858 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Acetoacetic_Ester_Synthesis |
Acetoacetic ester (ethyl acetoacetate) is an extremely useful molecule that can be used to make ketones and other molecules. You’ll even use this later on in amino acid synthesis, so let’s break down the way it reacts. How do we accomplish this transformation See those two carbonyls there? Each carbonyl has something called an alpha-carbon, and each alpha-carbon has hydrogens that are easily abstracted. The pKa of the green alpha-hydrogen is about 20, and the pKa of the blue alpha-hydrogen is actually about 10. Why? Because of the resonance structures the anions can form! Whenever you have a beta-dicarbonyl like this one, the enolate will preferentially form on the shared alpha-carbon. The anion on the blue alpha-carbon above can form more resonance structures than the anion on the green alpha-carbon can, so the blue hydrogen’s pKa will be lower (more acidic). That all sounds cool, but can we just use any ol’ base to form our enolate? Definitely not! Let’s say we were to try using NaOH. Instead of forming the enolate, we’d actually wind up with a competing reaction: , a type of nucleophilic acyl substitution. Notice that the hydroxide replaces the ethoxy group. So, how can we specifically avoid that type of acyl substitution? We can use a bulky base like LDA or the anionic version of our alkoxy group! See how we’ve got an ethoxy group (—OEt) in our starting material? In order to prevent any substitutions of that group, we can actually use NaOEt. Those ethoxy groups are totally exchanging, but the same molecule is produced. Okay, cool! These enolates are pretty good at SN2 reactions. They can act as nucleophiles on alkyl halides, acyl (acid) chlorides, and more! Let’s try adding a propyl group. Once we’ve got our alkyl group on there, we can actually get rid of the ester entirely through a mechanism called decarboxylation if we want to. All it takes is some heat and a little bit of aqueous acid. It could be written a ton of different ways—H SO (aq), HCl (aq), or even generically as H O . First we hydrolyze the ester to make a beta ketoacid, and then we heat things up to lose CO . After acidic hydrolysis, the enol (vinyl alcohol) that results will tautomerize back into a substituted ketone. Boom! There’s our product, a substituted ketone, in the green box! Not so bad, right? Of course, there are tons of different ways to use this molecule. We’ve just walked through the steps for a single alkylation, but there’s nothing stopping us from adding different groups. We’ve added one alkyl group, but what if we want to add another one? Well, we just have to follow the same steps! So, let’s start from the beginning. Let’s first add a propyl group and then an ethyl group. Once we’ve added the propyl group, all we need to do is add another equivalent of base and then the ethyl group. Here’s what the order of reagents looks like: And here’s what the mechanism would look like: Okay, but what if we want to add a cyclic group to our molecule? Well, luckily that’s not so bad either. We just need a molecule that has two leaving groups at terminal positions. Basically, it’s going to be very similar to the double alkylation but with just one equivalent of our alkyl molecule being added. Here’s what the reagents look like: And here’s what the mechanism looks like: Let’s take a step back and use the same enolate we used in the alkylation, but let’s use an acyl chloride instead of an alkyl halide this time. This follows basically the same pattern as the alkylation, but I’m going to rotate the acetoacetic ester a little bit and highlight the acid chloride so that it’s easier to follow. See how we just followed the same pattern? Form the enolate, provide an electrophile, and cleave off the ester by adding acid and heating it up! | 3,787 | 3,859 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.03%3A_pH_and_pOH |
The calculations we have just done show that the concentrations of hydronium and hydroxide ions in aqueous solution can vary from about 1 mol L down to about 1 × 10 mol L , and perhaps over an even wider range. The numbers used to express [H O ] and [OH ] in the units mole per liter will often include large negative powers of 10. Consequently it is convenient to define the following: \[ \text{pH}=-\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{1 mol L}^{-\text{1}}}\\\text{ } \\ \text{pOH}=-\text{log}\frac{\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{1 mol L}^{-\text{1}}}\text{ } \\ \nonumber \] Note carefully what these equations tell us to do. To obtain pH, for example, we divide [H O ] by the units mole per liter. This gives a pure number, and so we can take its logarithm. (It does not make sense to take the logarithm of a unit, such as mole per cubic decimeter.) The minus sign insures that we will obtain a positive result most of the time. The logarithm of a number is the power to which 10 must be raised to give the number itself. Therefore the definitions of pH and pOH mean that we can deal with powers of 10 rather than numerical values. Since the numbers needed to express [H O ] and [OH ] are usually between 1 and 10 pH and pOH values are usually between 0 and 14. Calculate the pH and the pOH of each of the following aqueous solutions: (a) 1.00 HNO ; 0.306 Ba(OH) Our previous discussion showed that for this solution [H O ] = 1.00 mol/L and [OH ] = 1.00 x 10 . Applying the definitions of pH and pOH, we have \[\begin{align}\text{pH}=-\text{log}\frac{\text{1}\text{.00 mol L}^{-\text{1}}}{\text{1 mol L}^{-\text{1}}}=-\text{log(10 }^{\text{0}}\text{)}\\\text{ }=-\text{(0)}=\text{0}\text{.00}\\\text{ }\\\text{pOH}=-\text{log}\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol L}^{-\text{1}}}{\text{1 mol L}^{-\text{1}}}=-\text{log(10}^{-\text{14}}\text{)}\\\text{ }=-\text{(}-\text{14)}=14.\text{00}\end{align} \nonumber \] In the example in the section on , we found for this solution [H O ] = 1.63 × 10 mol/L and [OH ] = 6.12 × 10 mol/L. Thus \[\begin{align}\text{pH} = -\text{log}{ 1.63}\times{10^{-14}}=-({-13.788})=13.788\\\text{ }\\\text{pOH} = -\text{log}{ 6.12}\times{10^{-1}}=-({-0.213})=0.213\end{align} \nonumber \] In the laboratory it is convenient to measure the pH of a solution using a pH meter. Such a device works on a different principle from the conductivity measurements we have already mentioned, and an accurate explanation of how it works is beyond the scope of the present discussion. While greater accuracy can be obtained when great care and special instruments are used, pH is usually measured to an accuracy of ± 0.01. Therefore pH values are usually rounded to the second decimal place; the results of Example 1 would commonly be rounded to pH = 13.79 and pOH = 0.21. Because pH measurements are so easily made, it is essential that you be able to convert from pH to [H O ]. This is the reverse of finding pH from [H O ]. Consequently it involves antilogs instead of logs. From the definition \[\text{pH}=-\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber \] we have \[\text{pH}=\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber \] Taking the antilog of both sides, we have \[\text{antilog}\left( -\text{pH} \right)\text{ = antilog}-\text{pH}=\left\{ \text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \right\} \nonumber \] so that \[\text{antilog}\left( -\text{pH} \right)=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber \] remembering that antilog = 10 , we can write this expression as \[\text{10}^{-\text{pH}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber \] or \[\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^\text{-pH}\text{ mol}\text{ L}^{-1}\label{15} \] An alternative method of writing this equation is \[\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }=\frac{\text{1}}{\text{10}^{\text{pH}}}\text{ mol L}^{-\text{1}} \nonumber \] The pH of a solution is found to be 3.40. Find the hydronium-ion concentration of the solution. If you have a calculator which has an antilog or 10 button, the problem is very simple. You enter – 3.40 and hit the button. The number thus obtained, 3.9822 x 10 is the number of moles of hydronium ion per liter. This follows from Eq. \(\ref{15}\): \(\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^{-\text{pH}}\text{ mol}\text{ L}^{-1}=\text{10}^{-3.4}\text{ mol}\text{ L}^{-1}=\text{3.98}\times\text{10}^{-4}\text{ mol}\text{ L}^{-1}\) The same result is almost as easy to find using Eq. (1 ). \(\text{10}^{-\text{pH}}=\text{antilog}\left(\text{pH}\right)=\text{antilog 3.40}=\text{antilog 3}\times \text{antilog 0.40}=\text{10}^{3}\times \text{2.51}\) Thus \(\text{10}^{-\text{pH}}=\frac{\text{1}}{\text{10}^{\text{pH}}}=\frac{\text{1}}{\text{2}\text{.51 }\times \text{ 10}^{\text{3}}}=\text{3}\text{.98 }\times \text{ 10}^{-\text{4}}\) in other words, \(\left[\text{H}_{3}\text{O}^{+}\right]=\text{3.98}\times\text{10}^{-4}\text{ mol}\text{ L}^{-1}\) There is a very simple relationship between the pH and the pOH of an aqueous solution at 25°C. We know that at this temperature \[{K}_{w}={K}_{c}\left(\text{55.5 mol}\text{ L}^{-1}\right)^{2}\left[\text{H}_{3}\text{O}^{+}\right] \left[\text{OH}^{-}\right]={K}_{w}=\text{ 10}^{-14} \text{mol}^{2}\text{L}^{-2} \nonumber \] Dividing both sides by mol L , we obtain \[\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}}\text{ }\times \text{ }\frac{[\text{OH}^{-}]}{\text{mol L}^{-\text{1}}}=\text{10}^{-\text{14}} \nonumber \] Taking logs and multiplying both by – 1, we then have \[\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}}\text{ }-\text{ }\log \text{ }\frac{[\text{OH}^{-}]}{\text{mol L}^{-\text{1}}}=-\text{log}(\text{10}^{-\text{14}}\text{)} \nonumber \] or \[\text{pH} + \text{pOH}= \text{14.00} \nonumber \] This simple relationship is often useful in finding the pH of solutions containing bases, as the following example shows. If 3.53 g of pure NaOH is dissolved in 10 L of H O find the pH of the resulting solution. We first calculate the concentration of the NaOH. \(n_{\text{NaOH}}=\text{3}\text{.53 }\times \text{ }\frac{\text{1 mol}^{-\text{1}}}{\text{40}\text{.0 g}}=\text{0}\text{.088 25 mol}\) so that \(c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V}=\frac{\text{0}\text{.088 25 mol}}{\text{10 L}^{\text{1}}}=\text{8}\text{.82 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}}\text{ }\) Since NaOH is a strong base, each mole of NaOH dissolved produces 1 mol OH ions, so that \(\left[\text{OH}^{-}\right]=\text{8.82 }\times \text{10}^{-3}\text{mol L}^{-1}\) Thus \(\text{pOH}=-\text{log }\left(\text{8.82}\times \text{10}^{-3}\right)=-\left(\text{0.95}-\text{3.00}\right)=+\text{2.05}\) From which \(\text{pH} =\text{14.00 - pOH} =\text{11.95}\) While the ability to calculate the pH of a solution from the hydronium-ion concentration and vice versa is useful, it is not the only thing we need to understand about pH. If someone gives you a solution whose pH is 14.74, it is true that the hydronium-ion concentration must be 1.82 × 10 mol L but it is perhaps more important to know that the solution is corrosively basic. In general, then, we need not only to be able to calculate a pH but also to have some realization of what kind of solutions have what kind of pH, as displayed in the following table. This table is part of our collection of acid and base constants. Battery acid In pure water at 25°C the hydronium-ion concentration is close to 1.00 × 10 mol/L, so that the pH is 7. In consequence any solution, not only pure water, which has a is described as being . An solution, as we know, is one in which the hydronium-ion concentration is greater than that of pure water, i.e., 10 mol/L. In pH terms this translates into a pH which is (because the pH is a negative logarithm). Small pH values are thus characteristic of acidic solutions; the smaller the pH, the more acidic the solution. By contrast, a solution is one in which the hydroxide-ion concentration is greater than 10 mol/L. In such a solution the hydronium-ion concentration is 10 mol/L, so that the pH of a basic solution is . Large pH values are thus characteristic of basic solutions. The larger the pH, the more basic the solution. | 8,783 | 3,860 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.08%3A_Temperature_and_Molecular__Velocities |
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? . This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure 10.7.1. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure 10.7.1 were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases. | 1,186 | 3,861 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Monosaccharides/Glucose_(Dextrose) |
Glucose is by far the most common carbohydrate and classified as a monosaccharide, an aldose, a hexose, and is a reducing sugar. It is also known as dextrose, because it is dextrorotatory (meaning that as an optical isomer is rotates plane polarized light to the right and also an origin for the D designation. Glucose is also called blood sugar as it circulates in the blood at a concentration of 65-110 mg/dL of blood. Glucose is initially synthesized by chlorophyll in plants using carbon dioxide from the air and sunlight as an energy source. Glucose is further converted to starch for storage. Up until now we have been presenting the structure of glucose as a chain. In reality, an aqueous sugar solution contains only 0.02% of the glucose in the chain form, the majority of the structure is in the cyclic chair form. Since carbohydrates contain both alcohol and functional groups, the straight-chain form is easily converted into the chair form - ring structure. Due to the tetrahedral geometry of carbons that ultimately make a 6 membered stable ring , the -OH on carbon #5 is converted into the ether linkage to close the ring with carbon #1. This makes a 6 member ring - five carbons and one oxygen. The chair structures are always written with the orientation depicted below to avoid confusion. The position of the -OH group on the anomeric carbon (#1) is an important distinction for carbohydrate chemistry. The Beta position is defined as the -OH being on the same side of the ring as the C #6. In the chair structure this results in a horizontal projection. The Alpha position is defined as the -OH being on the opposite side of the ring as the C #6. In the chair structure this results in a downward projection. The alpha and beta label is not applied to any other carbon - only the anomeric carbon, in this case #1. | 1,858 | 3,862 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/02%3A_Atoms_Molecules_and_Ions_(Exercises) |
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton’s atomic theory. Which one? The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed. Which postulate of Dalton’s theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced. Identify the postulate of Dalton’s theory that is violated by the following observations: 59.95% of one sample of titanium dioxide is titanium; 60.10% of a different sample of titanium dioxide is titanium. This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio. Samples of compound X, Y, and Z are analyzed, with results shown here. Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z? 1 Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties. 2 Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged. 3 Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged. 4. (a) The plum pudding model indicates that the positive charge is spread uniformly throughout the atom, so we expect the α particles to (perhaps) be slowed somewhat by the positive-positive repulsion, but to follow straight-line paths (i.e., not to be deflected) as they pass through the atoms. (b) Higher-energy α particles will be traveling faster (and perhaps slowed less) and will also follow straight-line paths through the atoms. (c) The α particles followed straight-line paths through the plum pudding atom. There was no apparent slowing of the α particles as they passed through the atoms. 5. (a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. (b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. (c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the α particles match the predictions from (a), (b), and (c). In what way are isotopes of a given element always different? In what way(s) are they always the same? Write the symbol for each of the following ions: (a) Cs ; (b) I ; (c) P ; (d) Co Write the symbol for each of the following ions: Open the and click on the Atom icon. (a) Carbon-12, C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen. Open the (a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom. (b) Now add two more electrons to make an ion and give the symbol for the ion you have created. Open the (a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Lithium-6 and give the isotope symbol for this atom. (b) Now remove one electron to make an ion and give the symbol for the ion you have created. (a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is Li or \(\ce{^6_3Li}\). (b) Li or \(\ce{^6_3Li+}\) Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses: (a) atomic number 9, mass number 18, charge of 1− (b) atomic number 43, mass number 99, charge of 7+ (c) atomic number 53, atomic mass number 131, charge of 1− (d) atomic number 81, atomic mass number 201, charge of 1+ (e) Name the elements in parts (a), (b), (c), and (d). The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them. (a) atomic number 26, mass number 58, charge of 2+ (b) atomic number 53, mass number 127, charge of 1− (a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes: (a) \(\ce{^{10}_5B}\) (b) \(\ce{^{199}_{80}Hg}\) (c) \(\ce{^{63}_{29}Cu}\) (d) \(\ce{^{13}_6C}\) (e) \(\ce{^{77}_{34}Se}\) Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes: (a) \(\ce{^7_3Li}\) (b) \(\ce{^{125}_{52}Te}\) (c) \(\ce{^{109}_{47}Ag}\) (d) \(\ce{^{15}_7N}\) (e) \(\ce{^{31}_{15}P}\) (a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons on the and select the “Mix Isotopes” tab, hide the “Percent Composition” and “Average Atomic Mass” boxes, and then select the element boron. (a) Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts. (b) Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice. (c) Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on “More” and then move the sliders to the appropriate amounts. (d) Reveal the “Percent Composition” and “Average Atomic Mass” boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction. (e) Select “Nature’s” mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match “Nature’s” amounts as closely as possible. Repeat using an element that has three naturally occurring isotopes. Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are 9% Ne-22, 91% Ne-20, and only a trace of Ne-21. The average mass would be 20.18 amu. Checking the nature’s mix of isotopes shows that the abundances are 90.48% Ne-20, 9.25% Ne-22, and 0.27% Ne-21, so our guessed amounts have to be slightly adjusted. An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element. Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes Br and Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments. 79.904 amu Variations in average atomic mass may be observed for elements obtained from different sources. Lithium provides an example of this. The isotopic composition of lithium from naturally occurring minerals is 7.5% Li and 92.5% Li, which have masses of 6.01512 amu and 7.01600 amu, respectively. A commercial source of lithium, recycled from a military source, was 3.75% Li (and the rest Li). Calculate the average atomic mass values for each of these two sources. The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consists of two isotopes with accurately known masses ( B, 10.0129 amu and B, 11.0931 amu). The actual atomic mass of boron can vary from 10.807 to 10.819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries. Turkey source: 0.2649 (of 10.0129 amu isotope); US source: 0.2537 (of 10.0129 amu isotope) The O: O abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. Is the average mass of an oxygen atom in these meteorites greater than, less than, or equal to that of a terrestrial oxygen atom? Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ. The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O , contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms. Explain why the symbol for the element sulfur and the formula for a molecule of sulfur differ. Write the molecular and empirical formulas of the following compounds: (a) (b) (c) (d) (a) molecular CO , empirical CO ; (b) molecular C H , empirical CH; (c) molecular C H , empirical CH ; (d) molecular H SO , empirical H SO Write the molecular and empirical formulas of the following compounds: (a) (b) (c) (d) Determine the empirical formulas for the following compounds: (a) C H N O; (b) C H O ; (c) HO; (d) CH O; (e) C H O Determine the empirical formulas for the following compounds: Write the empirical formulas for the following compounds: (a) (b) (a) CH O; (b) C H O Open the and select the “Larger Molecules” tab. Select an appropriate atoms “Kit” to build a molecule with two carbon and six hydrogen atoms. Drag atoms into the space above the “Kit” to make a molecule. A name will appear when you have made an actual molecule that exists (even if it is not the one you want). You can use the scissors tool to separate atoms if you would like to change the connections. on “3D” to see the molecule, and look at both the space-filling and ball-and-stick possibilities. Use the to repeat , but build a molecule with two carbons, six hydrogens, and one oxygen. (a) ethanol (b) methoxymethane, more commonly known as dimethyl ether (c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers. Use the to repeat , but build a molecule with three carbons, seven hydrogens, and one chlorine. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal: (a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element; (d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal: Using the periodic table, identify the lightest member of each of the following groups: (a) He; (b) Be; (c) Li; (d) O Using the periodic table, identify the heaviest member of each of the following groups: (a) krypton, Kr; (b) calcium, Ca; (c) fluorine, F; (d) tellurium, Te Use the periodic table to give the name and symbol for each of the following elements: Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each. (a) \(\ce{^{23}_{11}Na}\); (b) \(\ce{^{129}_{54}Xe}\); (c) \(\ce{^{73}_{33}As}\); (d) \(\ce{^{226}_{88}Ra}\) Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each. Using the periodic table, predict whether the following chlorides are ionic or covalent: KCl, NCl , ICl, MgCl , PCl , and CCl . Ionic: KCl, MgCl ; Covalent: NCl , ICl, PCl , CCl Using the periodic table, predict whether the following chlorides are ionic or covalent: SiCl , PCl , CaCl , CsCl, CuCl , and CrCl . For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved: (a) covalent; (b) ionic, Ba , O ; (c) ionic, \(\ce{NH4+}\), \(\ce{CO3^2-}\); (d) ionic, Sr , \(\ce{H2PO4-}\); (e) covalent; (f) ionic, Na , O For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions involved: For each of the following pairs of ions, write the symbol for the formula of the compound they will form: (a) CaS; (b) (NH ) CO ; (c) AlBr ; (d) Na HPO ; (e) Mg (PO ) For each of the following pairs of ions, write the symbol for the formula of the compound they will form: Name the following compounds: (a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide; (f) aluminum fluoride Name the following compounds: Write the formulas of the following compounds: (a) RbBr; (b) MgSe; (c) Na O; (d) CaCl ; (e) HF; (f) GaP; (g) AlBr ; (h) (NH ) SO Write the formulas of the following compounds: Write the formulas of the following compounds: (a) ClO ; (b) N O ; (c) K P; (d) Ag S; (e) AlN; (f) SiO Write the formulas of the following compounds: Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds: (a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) oxide; (f) molybdenum(IV) sulfide Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds: The following ionic compounds are found in common household products. Write the formulas for each compound: (a) K PO ; (b) CuSO ; (c) CaCl ; (d) TiO ; (e) NH NO ; (f) NaHSO The following ionic compounds are found in common household products. Name each of the compounds: What are the IUPAC names of the following compounds? (a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide | 15,775 | 3,864 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Barbiturates_and_Benzodiazepines |
Hypnotic and sedative drugs are non-selective, general depressants of the central nervous system. If the dose is relatively low, a sedative action results in a reduction in restlessness and emotional tension. A larger dose of the same drug produces a hypnotic sleep inducing effect. As the dosage is increased further, the result is anesthesia or death if the dosage is sufficiently high. The barbiturates once enjoyed a long period of extensive use as sedative-hypnotic drugs; however, except for a few specialized uses, they have been largely replaced by the much safer . Barbiturates are central nervous system depressants and are similar, in many ways, to the depressant effects of alcohol. To date, there are about 2,500 derivatives of barbituric acid of which only 15 are used medically. The first barbiturate was synthesized from barbituric acid in 1864. The original use of barbiturates was to replace drugs such as opiates, bromides, and alcohol to induce sleep. The hyponotic and sedative effects produced by barbiturates are usually ascribed to their interference of nerve transmission to the cortex. Various theories for the action of barbiturates include: changes in ion movements across the cell membrane; interactions with cholinergic and non cholinergic receptor sites; impairment of biochemical reactions which provide energy; and depression of selected areas of the brain. The structures of the barbiturates can be related to the duration of effective action. Although over 2000 derivatives of barbituric acid have been synthesized only about a dozen are currently used. All of the barbiturates are related to the structure of barbituric acid shown below. The duration of effect depends mainly on the alkyl groups attached to carbon # 5 which confer lipid solubility to the drug. The duration of effective action decreases as the total number of carbons at C # 5 increases. To be more specific, a long effect is achieved by a short chain and/or phenyl group. A short duration effect occurs when there are the most carbons and branches in the alkyl chains The term benzodiazepine refers to the portion of the structure composed of a benzene ring (A) fused to a seven-membered diazepine ring (B). However, since all of the important benzodiazepines contain a aryl substituent ring C) and a 1, 4-diazepine ring, the term has come to mean the aryl-1,4-benzodiazepines. There are several useful benzodiazepines available: chlordiazepoxide (Librium) and diazapam (Valium). The actions of benzodiazepines are a result of increased activation of receptors by gamma-aminobutyric acid (GABA). Benzodiazepine receptors are located on the alpha subunit of the GABA receptor located almost exclusively on postsynaptic nerve endings in the CNS (especially cerebral cortex). Benzodiazepines enhance the GABA transmitter in the opening of postsynaptic chloride channels which leads to hyperpolarization of cell membranes. That is, they "bend" the receptor slightly so that GABA molecules attach to and activate their receptors more effectively and more often. | 3,085 | 3,865 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/11.E%3A_Solutions_and_Colloids_(Exercises) |
How do solutions differ from compounds? From other mixtures? A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. Which of the principal characteristics of solutions can we see in the solutions of \(\ce{K2Cr2O7}\) shown in The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K Cr O in water can vary. When KNO is dissolved in water, the resulting solution is significantly colder than the water was originally. (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K and \(\ce{NO3-}\) ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. Give an example of each of the following types of solutions: (a) CO in water; (b) O in N (air); (c) bronze (solution of tin or other metals in copper) Indicate the most important types of intermolecular attractions in each of the following solutions: (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C H , nonpolar solvent): (a) heptane; (b) water; (c) water Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes. Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat. Solutions of hydrogen in palladium may be formed by exposing Pd metal to H gas. The concentration of hydrogen in the palladium depends on the pressure of H gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal. Explain why the ions Na and Cl are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules. Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive. HBr is an acid and so its molecules react with water molecules to form H O and Br ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive. Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO ) ( )? (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO ) . (a) Fe(NO ) is a strong electrolyte, thus it should completely dissociate into Fe and \(\ce{(NO3- )}\) ions. Therefore, (z) best represents the solution. (b) \(\ce{Fe(NO3)3}(s)⟶\ce{Fe^3+}(aq)+\ce{3NO3- }(aq)\) Compare the processes that occur when methanol (CH OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution. Methanol, \(CH_3OH\), dissolves in water in all proportions, interacting via hydrogen bonding. \[CH_3OH_{(l)}+H_2O_{(l)}⟶CH_3OH_{(aq)}\] Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions. \[HCl{(g)}+H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)}\] Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively. \[NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^−_{(aq)}\] What is the expected electrical conductivity of the following solutions? (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved) Why are most ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a (molten) ionic compound to be electrically conductive or nonconductive? Explain. A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces Suppose you are presented with a clear solution of sodium thiosulfate, Na S O . How could you determine whether the solution is unsaturated, saturated, or supersaturated? Add a small crystal of \(Na_2S_2O_3\). It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution. Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures. The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating. Suggest an explanation for the observations that ethanol, C H OH, is completely miscible with water and that ethanethiol, C H SH, is soluble only to the extent of 1.5 g per 100 mL of water. The hydrogen bonds between water and C H OH are much stronger than the intermolecular attractions between water and C H SH. Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C using the following figure for useful data, and report the computed percentage to one significant digit.
At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water. \[\%\; KBr =\dfrac{60\; g\; KBr}{(60+100)\;g\; solution} = 40\%\] Which of the following gases is expected to be most soluble in water? Explain your reasoning. (c) CHCl is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules. At 0 °C and 1.00 atm, as much as 0.70 g of O can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O dissolve in 1 L of water? This problem requires the application of Henry’s law. The governing equation is \(C_g = kP_g\). \[k=\dfrac{C_g}{P_g}=\dfrac{0.70\;g}{1.00\; atm} =0.70\;g\; atm^{−1}\] Under the new conditions, \(C_g=0.70\;g\;atm^{−1} \times 4.00\; atm = 2.80\; g\). Refer to following figure for the following three questions: (a) It decreased as some of the CO gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO gas above the beverage. The reduced CO gas pressure, per Henry’s law, lowers the solubility for CO . (c) The dissolved CO concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated. The Henry’s law constant for CO is \(3.4 \times 10^{−2}\; M/atm\) at 25 °C. What pressure of carbon dioxide is needed to maintain a CO concentration of 0.10 in a can of lemon-lime soda? \[P_g=\dfrac{C_g}{k}=\dfrac{0.10\; M}{3.4 \times 10^{−2}\;M/atm} =2.9\; atm\] The Henry’s law constant for O is \(1.3\times 10^{−3}\; M/atm\) at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O is 0.21 atm? Start with Henry's law \[C_g=kP_g\] and apply it to \(O_2\) \[C(O_2)=(1.3 \times 10^{−3}\; M/atm) (0.21\;atm)=2.7 \times 10^{−4}\;mol/L\] The total amount is \((2.7 \times 10^{−4}\; mol/L)(40\;L=1.08 \times 10^{−2} \;mol\] The mass of oxygen is \((1.08 \times 10^{−2}\; mol)(32.0\; g/mol)=0.346\;g\) or, using two significant figures, \(0.35\; g\). How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20- solution of hydrochloric acid? First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required. Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion? What is the microscopic explanation for the macroscopic behavior illustrated in ? The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region. Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume. A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C H (OH) ), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical? Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions. What are the mole fractions of H PO and water in a solution of 14.5 g of H PO in 125 g of water? What are the mole fractions of HNO and water in a concentrated solution of nitric acid (68.0% HNO by mass)? Calculate the mole fraction of each solute and solvent: a. \(583\:g\:\ce{H2SO4}\times\dfrac{1\:mole\:\ce{H2SO4}}{98.08\:g\:\ce{H2SO4}}=5.94\:mole\:\ce{H2SO4}\) \(\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\) \(1.50\:kg\:\ce{H2O}\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mole\:\ce{H2O}}{18.02\:g\:\ce{H2O}}=83.2\:moles\:\ce{H2O}\) Calculate the mole fraction of each solute and solvent: Calculate the mole fractions of methanol, CH OH; ethanol, C H OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.) What is the difference between a 1 solution and a 1 solution? In a 1 solution, the mole is contained in exactly 1 L of solution. In a 1 solution, the mole is contained in exactly 1 kg of solvent. What is the molality of phosphoric acid, H PO , in a solution of 14.5 g of H PO in 125 g of water? What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO by mass)? (a) Determine the molar mass of HNO . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 Calculate the molality of each of the following solutions: Calculate the molality of each of the following solutions: (a) 6.70 × 10 ; (b) 5.67 ; (c) 2.8 ; (d) 0.0358 The concentration of glucose, C H O , in normal spinal fluid is \(\mathrm{\dfrac{75\:mg}{100\:g}}\). What is the molality of the solution? A 13.0% solution of K CO by mass has a density of 1.09 g/cm . Calculate the molality of the solution. 1.08 What is the boiling point of a solution of 9.04 g of I in 75.5 g of benzene, assuming the I is nonvolatile? What is the freezing temperature of a solution of 115.0 g of sucrose, C H O , in 350.0 g of water, which freezes at 0.0 °C when pure? (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C What is the freezing point of a solution of 9.04 g of I in 75.5 g of benzene? What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO ) in water at 25 °C? The volume of the solution is 275 mL. (a) Determine the molar mass of Ca(NO ) ; determine the number of moles of Ca(NO ) in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol ) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin? What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; = 5.02 °C/ ) that boils at 81.5 °C at 1 atm? (a) Determine the molal concentration from the change in boiling point and ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × 10 g mol A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound. A 1.0 solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain. No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ = (1.0 m)(5.14 °C/ ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. A solution contains 5.00 g of urea, CO(NH ) , a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. 144 g mol Arrange the following solutions in order by their decreasing freezing points: 0.1 Na PO , 0.1 C H OH, 0.01 CO , 0.15 NaCl, and 0.2 CaCl . Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na SO , and 0.030 mol of MgCl , assuming complete dissociation of these electrolytes. 0.870 °C How could you prepare a 3.08 aqueous solution of glycerin, C H O ? What is the freezing point of this solution? A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS ( = 2.43 °C/ ). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide? S In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI ? Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10 atm at 25 °C. What is the molar mass of lysozyme? 1.39 × 10 g mol The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin? The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C H O , is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C? 54 g What is the freezing point of a solution of dibromobenzene, C H Br , in 0.250 kg of benzene, if the solution boils at 83.5 °C? What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C? 100.26 °C The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and for ethanol is 1.20 °C/ . What is the molecular formula of fructose? The vapor pressure of methanol, CH OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C H OH, is 44 torr at the same temperature. (a) \(X_\mathrm{CH_3OH}=0.590\); \(X_\mathrm{C_2H_5OH}=0.410\); (b) Vapor pressures are: CH OH: 55 torr; C H OH: 18 torr; (c) CH OH: 0.75; C H OH: 0.25 The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air? Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature? The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. for camphor is 37.7 °C/ . What is the molecular formula of the solute? Show your calculations. A sample of HgCl weighing 9.41 g is dissolved in 32.75 g of ethanol, C H OH ( = 1.20 °C/ ). The boiling point elevation of the solution is 1.27 °C. Is HgCl an electrolyte in ethanol? Show your calculations. \(\mathrm{Δbp}=K_\ce{b}m=(1.20\:°\ce C/m)\mathrm{\left(\dfrac{9.41\:g×\dfrac{1\:mol\: HgCl_2}{271.496\:g}}{0.03275\:kg}\right)=1.27\:°\ce C}\) The observed change equals the theoretical change; therefore, no dissociation occurs. A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations. Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby. Distinguish between dispersion methods and condensation methods for preparing colloidal systems. Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds. How do colloids differ from solutions with regard to dispersed particle size and homogeneity? Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale. Explain the cleansing action of soap. Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules. How can it be demonstrated that colloidal particles are electrically charged? If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. ). | 20,901 | 3,866 |
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Make sure you thoroughly understand the following essential ideas which have been presented. The answer to this question would ideally be a simple, easily understood theory that would not only explain why atoms bind together to form molecules, but would also predict the three-dimensional structures of the resulting compounds as well as the energies and other properties of the bonds themselves. Unfortunately, no one theory exists that accomplishes these goals in a satisfactory way for all of the many categories of compounds that are known. Moreover, it seems likely that if such a theory does ever come into being, it will be far from simple. When we are faced the need to find a scientific explanation for a complex phenomenon such as bonding, experience has shown that it is often best to begin by developing a . A is something like a theory in that it should be able to explain observations and to make useful predictions. But whereas a theory can be discredited by a single contradictory case, a model can be useful even if it does not encompass all instances of the effects it attempts to explain. We do not even require that a model be a credible representation of reality; all we ask is that it be able to explain the behavior of those cases to which it is applicable in terms that are consistent with the model itself. An example of a model that you may already know about is the kinetic molecular theory of gases. Despite its name, this is really a model (at least at the level that beginning students use it) because it does not even try to explain the observed behavior of real gases. Nevertheless, it serves as a tool for developing our understanding of gases, and as an essential starting point for more elaborate treatments. One thing is clear: chemical bonding is basically in nature, the result of attraction between bodies of opposite charge; bonding occurs when outer-shell electrons are simultaneously attracted to the positively-charged nuclei of two or more nearby atoms. The need for models arises when we try to understand why Given the extraordinary variety of ways in which atoms combine into aggregates, it should come as no surprise that a number of useful bonding models have been developed. Most of them apply only to certain classes of compounds or attempt to explain only a restricted range of phenomena. In this section we will provide brief descriptions of some of the bonding models; the more important of these will be treated in much more detail in later lessons in this unit. Napoleon's definition of history as a set of lies agreed on by historians seems to have a parallel with chemical bonding and chemists. At least in Chemistry, we can call the various explanations "models" and get away with it even if they are demonstrably wrong, as long as we find them useful. In a provocative article ( 1990 67(4) 280-298), J. F. Ogilvie tells us that there are no such things as orbitals, or, for that matter, non-bonding electrons, bonds, or even uniquely identifiable atoms within molecules. This idea disturbed a lot of people (teachers and textbook authors preferred to ignore it) and prompted a spirited rejoinder ( 1992 69(6) 519-521) from Linus Pauling, father of the modern quantum-mechanical view of the chemical bond. But the idea has never quite gone away. Richard Bader of McMaster University has developed a quantitative "atoms in molecules" model that depicts molecules as a collection of point-like nuclei embedded in a diffuse cloud of electrons. There are no "bonds" in this model, but only "bond paths" that correspond to higher values of electron density along certain directions that are governed by the manner in which the positive nuclei generate localized distortions of the electron cloud. By , we mean models that do not take into account the quantum behavior of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics. Although this completely ignores what has been learned about the nature of the electron since the development of quantum theory in the 1920’s, these classical models have not only proven extremely useful, but the major ones also serve as the basis for the chemist’s general classification of compounds into “covalent” and “ionic” categories. Ever since the discovery early in the 19th century that solutions of salts and other electrolytes conduct electric current, there has been general agreement that the forces that hold atoms together must be electrical in nature. Electrolytic solutions contain having opposite electrical charges; opposite charges attract, so perhaps the substances from which these ions come consist of positive and negatively charged atoms held together by electrostatic attraction. It turns out that this is not true generally, but a model built on this assumption does a fairly good job of explaining a rather small but important class of compounds that are called . The most well known example of such a compound is sodium chloride, which consists of two interpenetrating lattices of Na and Cl ions arranged in such as way that every ion of one type is surrounded (in three dimensional space) by six ions of opposite charge. One can envision the formation of a solid NaCl unit by a sequence of events in which one mole of gaseous Na atoms lose electrons to one mole of Cl atoms, followed by condensation of the resulting ions into a crystal lattice: Note: positive energy values denote endothermic processes, while negative ones are exothermic. Since the first two energies are known experimentally, as is the energy of the sum of the three processes, the lattice energy can be found by difference. It can also be calculated by averaging the electrostatic forces exerted on each ion over the various directions in the solid, and this calculation is generally in good agreement with observation, thus lending credence to the model. The sum of the three energy terms is clearly negative, and corresponds to the liberation of heat in the net reaction (bottom row of the table), which defines the Na–Cl “bond” energy. The ionic solid is more stable than the equivalent number of gaseous atoms simply because the three-dimensional NaCl structure allows more electrons to be closer to more nuclei. This is the criterion for the stability of any kind of molecule; all that is special about the “ionic” bond is that we can employ a conceptually simple electrostatic model to predict the bond strength. The main limitation of this model is that it applies really well only to the small class of solids composed of Group 1 and 2 elements with highly electronegative elements such as the halogens. Although compounds such as CuCl dissociate into ions when they dissolve in water, the fundamental units making up the solid are more like polymeric chains of covalently-bound CuCl molecules that have little ionic character. This model originated with the theory developed by G.N. Lewis in 1916, and it remains the most widely-used model of chemical bonding. It is founded on the idea that a pair of electrons shared between two atoms can create a mutual attraction, and thus a chemical bond. Usually each atom contributes one electron (one of its ) to the pair, but in some cases both electrons come from one of the atoms. For example, the bond between hydrogen and chlorine in the hydrogen chloride molecule is made up of the single 1 electron of hydrogen paired up with one of chlorine's seven valence (3 ) electrons. The stability afforded by this sharing is thought to derive from the noble gas configurations (helium for hydrogen, argon for chlorine) that surround the bound atoms. The origin of the electrostatic binding forces in this model can best be understood by examining the simplest possible molecule. This is the H , which consists of two nuclei and one electron. First, however, think what would happen if we tried to make the even simpler molecule H . Since this would consist only of two protons whose electrostatic charges would repel each other at all distances, it is clear that such a molecule cannot exist; something more than two nuclei are required for bonding to occur. A purely covalent bond can only be guaranteed when the (electron-attracting powers) of the two atoms are identical. When atoms having different electronegativities are joined, the electrons shared between them will be displaced toward the more electronegative atom, conferring a polarity on the bond which can be described in terms of percent ionic character. This is an extension of the ionic model to compounds that are ordinarily considered to be non-ionic. Combined hydrogen is always considered to exist as the hydride ion H , so that methane can be treated as if it were . This is not as bizarre as it might seem at first if you recall that the proton has almost no significant size, so that it is essentially embedded in an electron pair when it is joined to another atom in a covalent bond. This model, which is not as well known as it deserves to be, has surprisingly good predictive power, both as to bond energies and structures. The “valence shell electron repulsion” model is not so much a model of chemical bonding as a scheme for explaining the shapes of molecules. It is based on the quantum mechanical view that bonds represent electron clouds— physical regions of negative electric charge that repel each other and thus try to stay as far apart as possible. We will explore this concept in much greater detail in a later unit. These models of bonding take into account the fact that a particle as light as the electron cannot really be said to be in any single location. The best we can do is define a region of space in which the probability of finding the electron has some arbitrary value which will always be less than unity. The shape of this volume of space is called an and is defined by a mathematical function that relates the probability to the ( coordinates of the molecule. Like other models of bonding, the quantum models attempt to show how more electrons can be simultaneously close to more nuclei. Instead of doing so through purely geometrical arguments, they attempt this by predicting the nature of the orbitals which the valence electrons occupy in joined atoms. This was developed by in 1931 and was the first quantum-based model of bonding. It is based on the premise that if the atomic , , and orbitals occupied by the valence electrons of adjacent atoms are combined in a suitable way, the hybrid orbitals that result will have the character and directional properties that are consistent with the bonding pattern in the molecule. The rules for bringing about these combinations turn out to be remarkably simple, so once they were worked out it became possible to use this model to predict the bonding behavior in a wide variety of molecules. The hybrid orbital model is most usefully applied to the -block elements in the first few rows of the periodic table, and is especially important in organic chemistry. This model takes a more fundamental approach by regarding a molecule as a collection of valence electrons and positive cores. Just as the nature of atomic orbitals derives from the spherical symmetry of the atom, so will the properties of these new molecular orbitals be controlled by the interaction of the valence electrons with the multiple positive centers of these atomic cores. These new orbitals, unlike those of the hybrid model, are delocalized; that is, they do not “belong” to any one atom but extend over the entire region of space that encompasses the bonded atoms. The available (valence) electrons then fill these orbitals from the lowest to the highest, very much as in the Aufbau principle that you learned for working out atomic electron configurations. For small molecules (which are the only ones we will consider here), there are simple rules that govern the way that atomic orbitals transform themselves into molecular orbitals as the separate atoms are brought together. The real power of molecular orbital theory, however, comes from its mathematical formulation which lends itself to detailed predictions of bond energies and other properties. A common theme uniting all of the models we have discussed is that bonding depends on the fall in potential energy that occurs when opposite charges are brought together. In the case of covalent bonds, the shared electron pair acts as a kind of “electron glue” between the joined nuclei. In 1962, however, it was shown that this assumption is not strictly correct, and that instead of being concentrated in the space between the nuclei, the electron orbitals become even more concentrated around the bonded nuclei. At the same time however, they are free to “move” between the two nuclei by a process known as . This refers to a well-known quantum mechanical effect that allows electrons (or other particles small enough to exhibit wavelike properties) to pass (“tunnel”) through a barrier separating two closely adjacent regions of low potential energy. One result of this is that the effective volume of space available to the electron is increased, and according to the this will reduce the kinetic energy of the electron. According to this model, the bonding electrons act as a kind of fluid that concentrates in the region of each nucleus (lowering the potential energy) and at the same time is able to freely flow between them (reducing the kinetic energy). Despite its conceptual simplicity and full acknowledgment of the laws of quantum mechanics, this model is less known than it deserves to be and is unfortunately absent from most textbooks. | 13,743 | 3,867 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.05%3A_Wave_Mechanics |
During the middle of the nineteen-twenties some scientists began to realize that electrons must move around the nucleus in a very different way from that in which planets move around the sun. They abandoned the idea that an electron traces out a definite orbit or trajectory. Instead they adopted the point of view that it was impossible to describe the exact path of a particle whose mass was as small as that of an electron. Rather than think of the motion in planetary terms, they suggested it was much more useful to think of this motion in terms of a which could fold itself around the nucleus only in certain specific three-dimensional patterns. This new way of approaching the behavior of electrons (and other particles too) became known as or . In order to familiarize you with some of the concepts and terminology of wave mechanics, we shall consider the simple, though somewhat artificial, example illustrated in Figure 1. This is usually referred to as a . We consider the particle (which could be an electron) to have a mass and to be restricted in its movement by a narrow but absolutely straight tube of length d into which it can just fit. This container, or box, is closed at both ends and insures that the particle can move in only one dimension within its length. An everyday object like a marble or a billiard ball could move back and forth in this container, bouncing off either end. If there were no friction to slow it down, the particle would oscillate indefinitely, maintaining a constant speed and a constant of value ½ . The actual magnitude of (and hence ) would depend on how large or how small a “push” the particle was given initially, to start it moving. In order to look at this particle from a wave-mechanical point of view, we apply an idea originally suggested in 1924 by Louis de Broglie (1892 to 1987). He proposed that a wave of wavelength λ is associated with every particle. The larger the mass of the particle and the faster it is moving, the smaller this wavelength becomes. The exact relationship is given by the formula \[\lambda =\dfrac{h}{\mu} \label{1} \] where μ is the momentum of the electron, the product of its mass and velocity (μ = m * ) and the constant of proportionality h is known as Planck’s constant (h = 6.626 × 10 J s). The wave-mechanical view no longer pictures the particle as the oscillating billiard ball of Figure \(\Page {1}\) . Instead we must begin to think of it as having a behavior similar to that of the guitar string illustrated in Figure \(\Page {1}\) , or . Elsewhere you can see a series of videos demonstrate and describe the aspects of wave mechanics. They show the different wavelengths that can be formed by a string anchored at both ends. If the string is attached to both ends of the box, only those waves or vibrations in which the ends of the string do not move are possible. The length of the box can thus correspond to a single half wavelength (Figure \(\Page {1}\) ), to two half wavelengths (Figure \(\Page {1}\) ), to three half wavelengths (Figure \(\Page {1}\)d, etc., but to the intermediate situation shown in Figure \(\Page {1}\) . In other words, the length of the tube must correspond to an integral number of half wavelengths, or \[d=n\dfrac{\lambda}{\text{2}} \label{2} \] where = 1, 2, 3, 4, or some larger whole number. If = 1, d = λ/2; if = 2, d = λ; and so on. Rearranging Equation \(\ref{2}\), we then obtain \[\lambda =\dfrac{\text{2}d}{n} \label{3} \] Since the right-hand sides of Equations \(\ref{1}\) and \(\ref{3}\) are both equal to λ, they may be set equal to each other, giving \[\dfrac{\text{2}}{n} d = \dfrac{h}{\mu} = \dfrac{h}{mv} \nonumber \] which rearranges to give \[v=\dfrac{nh}{\text{2}md} \text{ } n = \text{1, 2, 3, 4,} \ldots \nonumber \] We can now calculate the kinetic energy of our wave-particle. It is given by the formula \[E_{k}=\tfrac{\text{1}}{\text{2}}mv^{\text{2}}=\tfrac{\text{1}}{\text{2}}m\left( \dfrac{nh}{\text{2}md} \right)^{\text{2}} \nonumber \] or \[E_{k}=n^{\text{2}}\left( \dfrac{h^{\text{2}}}{\text{8}md^{\text{2}}} \right) \label{4} \] Since the value of is restricted to positive whole numbers, we arrive at the interesting result that the kinetic energy of the electron can have only certain values and not others. Thus, if our particle is an electron ( = 9.1 × 10 kg) and the one-dimensional box is about the size of an atom ( = 1 × 10 m), the allowed values of the energy are given by \[E_{k}=n^{\text{2}}\left( \dfrac{h^{\text{2}}}{\text{8}md^{\text{2}}} \right)=n^{\text{2}}\text{ }\times \text{ }\dfrac{\text{(6}\text{.624 }\times \text{ 10}^{-\text{34}}\text{ J s)}^{\text{2}}}{\text{8 }\times \text{ 9}\text{.1 }\times \text{ 10}^{-\text{31}}\text{ kg }\times \text{ 1 }\times \text{ 10}^{-\text{20}}\text{ m}^{\text{2}}} \nonumber \] \[=n^{\text{2}}\text{ }\times \text{ 6}\text{.0 }\times \text{ 10}^{-\text{18}}\text{ }\dfrac{\text{J}^{\text{2}}\text{ s}^{\text{2}}}{\text{kg m}^{\text{2}}}=n^{\text{2}}\text{ }\times \text{ 6}\text{.0 }\times \text{ 10}^{-\text{18}}\text{ J} \nonumber \] \[\text{ (recall: 1 J}=\text{1 kg m}^{\text{2}}\text{ s}^{-\text{2}}\text{)} \nonumber \] Thus if , = × 6.0 × 10 J = 6.0 aJ If , = × 6.0 × 10 J = 24.0 aJ If , = × 6.0 × 10 J = 54.0 aJ and so on. This result means that by treating the electron as a wave, its energy is automatically restricted to certain specific values and not those in between. Although the electron can have an energy of 6.0 or 24.0 attojoules (aJ), it cannot have an intermediate energy such as 7.3 or 11.6 aJ. We describe this situation by saying that the energy of the electron is . We can now begin to glimpse some of the advantages of looking at the electron in wave-mechanical terms. If an electron is in some sense a wave, then only certain kinds of motion, i.e., only certain kinds of wave patterns, governed by whole numbers, are possible. As we shall shortly see, when an electron is allowed to move in three dimensions around a nucleus, this kind of behavior easily translates into a shell structure. An electron can be in shell 1 or shell 2 but not in an intermediate shell like 1.386. One more property of the electron arises out of the wave analogy. If the electron in a box behaves like a guitar string, we can no longer state that the electron is located at a specific position within the box or is moving in one direction or the other. Indeed the electron seems to be all over the box at once! All we can say is that the wave (or the vibration of the string) has a certain at any point along the box. This intensity is larger in some places than in others and is always zero at both ends of the box. | 6,714 | 3,868 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Lasers/Laser_Theory |
For reasons that we cannot explain, it appears that all things in nature prefer to go to the lowest energy state available to them. This seems to be how nature behaves. Apples fall all the way to the ground, once they are let go by the tree branches they grew on. Of course unless they fall into a hole, in which case they go even lower than the ground level - to the bottom of the hole. To raise the energy of the fallen apple someone, or something, has to intervene. A child must pick it up off the ground, for example. Otherwise, the apple will remain at its lowest point. Similarly, atoms tend to prefer to always stay in their ground state, unless some intervening force causes them to reach an excited state. In the language used in the study of lasers, any process that feeds energy into a collection of atoms or molecules and causes them to vacate their ground state is referred to as an energy pump, or simply as a pump. In most lamps electricity is used, by one mechanism or other, to pump atoms out of their ground-state into some excited state. We have already seen that incandescent bulbs produce light from Radiative transitions after excitation via collisions of the electrons in an electric current with the filament atoms. As we've also seen, other light sources, like the fluorescent lamps or neon tubes, use gas or vapor for their medium of excitation instead of a solid filament. In most lamps, the pumping mechanism is electricity. This is also the case for most lasers, but in some lasers optical pumping is the mechanism that is used to cause excitation. In optical pumping a light source generates photons with enough energy that they are able to get absorbed by the atoms in the lasing medium and cause them to go into an excited state. Evidently, all excitations occur "instantaneously", but de-excitations can lag by a measurable time interval that depends on the properties of the excited state. That is to say, an atom in an excited state does not instantaneously de-excite. The time that it spends, on average, in that excited level is called the lifetime of that state. Lifetimes can vary in duration depending on the atom and on the energy level. Excited state lifetimes are typically a few nanoseconds (10-9 s, or a billionth of a second), but they could be as short as a picosecond (10-12 s, or a thousand times shorter still) or as long as a few milliseconds (10-3 s). The ground state, of course, has an infinitely long lifetime since an atom in its ground state can no longer decrease its energy. So, the most stable of states is the ground state. Long-lived states are referred to as meta-stable states. In the case of Radiative emission, atoms happen to take two very distinctly different approaches: spontaneous emission, or stimulated emission. Spontaneous emission refers to the case when the excited atom de-excites, rather randomly whenever it "feels like it", and emits a photon. This photon has an energy equal to the difference between the two energy levels of the transition, but its direction of travel and its other properties, such as polarization are random. Stimulated emission was first theorized by Albert Einstein in 1917. For stimulated emission to occur a second, non-participating yet stimulating, photon must be present. The energy of this second photon must exactly match the allowed energy of the transition. Then the emitted photon will not only have the same energy as the stimulating one, but it will also travel in the same direction, and will be essentially identical to it. So, independent of what type of medium is used in a laser, in the absence of a pump the atoms or molecules are almost all in their ground state. Let us imagine that we could count the number of atoms in our laser medium that are in each energy state and denote the ones which are in their ground state by Ngs , those in the first excite state by N1, those in the second excited state by N2 , and so on. Then in the absence of a pump we are mostly certain that N1= N2 = ... = 0, and Ngs = total number of all the atoms in the medium. Once the pump is turned on it will deplete the number of atoms that were originally in the ground state and increases the number of atoms in the excited states that it is pumping to. Because excited atoms de-excite quickly and return to their ground states by spontaneous emission, in almost all lasers even when the pump is feeding energy into the medium the number of atoms in their ground state remains many times more than atoms in any other energy state. That is to say, almost always we could safely state that: N much bigger than N . To make a laser we need to not only excite the atoms in the laser medium, but somehow encourage them to undergo a decay through stimulated emission. In stimulated emission a passer-by photon which has an energy exactly equal to the transition energy stimulates the atom to emit a photon, identical to the passer-by photon, instantly. The problem with this is that the same passer-by photon could instead get absorbed by a de-excited atom. So, aside from pumping the atoms to excited states, we need to use clever procedures to insure that there are more excited atoms that could use the passer-by photon for stimulated emission than there are de-excited atoms which could absorb it; i.e. we need to generate a population inversion. Although Einstein predicted stimulation emission in 1917, it was not found experimentally for over 10 years and took over another 30 years just to predict the possibility of a laser. This is basically because it was not considered possible to produce a population inversion because of the above inequality. For the sake of our studies, let's first consider a laser medium whose atoms have only two energy states: a ground state and one excited state. In such an idealized atom the only possible transitions are excitation from the ground state to the excited state, and de-excitation from the excited state back into ground state. Could such an atom be used to make a laser? There are several important conditions that our laser must satisfy. First of all, the light that it produces must be coherent. That is to say, it must emit photons that are in-phase with one another. Secondly, it should emit monochromaticlight, i.e. photons of the same frequency (or wavelength).Thirdly, it would be desirable if our laser's output were collimated, producing a sharply defined "pencil-like" beam of light (this is not crucial, but clearly a desirable condition). Lastly, it would also be desirable for our laser to be efficient, i.e. the higher the ratio of output energy - to - input energy, the better. Let us begin by examining the requirements for our first condition for lasing, coherence. This condition is satisfied only when the lasing transition occurs through stimulated emission. As we have already seen, stimulated emission produces identical photons that are of equal energy and phase and travel in the same direction. But for stimulated emission to take place a "passer-by" photon whose energy is just equal to the de-excitation energy must approach the excited atom before it de-excites via spontaneous emission. Typically, a photon emitted by the spontaneous emission serves as the seed to trigger a collection of stimulated emissions. Still, if the lifetime of the excited state is too short, then there will not be enough excited atoms around to undergo stimulated emission. So, the first criteria that we need to satisfy is that the upper lasing state must have a relatively long lifetime, otherwise known as a meta-stablestate, with typical lifetimes in the milliseconds range. In addition to the requirement of a long lifetime, we need to ensure that the likelihood of absorption of the "passer-by" photons is minimized. This likelihood is directly related to the ratio of the atoms in their ground state versus those in the excited state. The smaller this ratio, the more likely that the "passer-by" photon will cause a stimulated emission rather than get absorbed. So, to satisfy this requirement, we need to produce a population inversion: create more atoms in the excited state than those in the ground state. Achieving population inversion in a two-level atom is not very practical. Such a task would require a very strong pumping transition that would send any decaying atom back into its excited state. This would be similar to reversing the flow of water in a water fall. It can be done, but is very energy costly and inefficient. In a sense, the pumping transition would have to work against the lasing transition. It is clear, from the above diagram, that in the two-level atom the pump is, in a way, the laser itself! Such a two-level laser would work only in jolts. That is to say, once the population inversion is achieved the laser would lase. But immediately it would end up with more atoms in the lower level. Such two-level lasers involve a more complicated process. We will see, in later material, examples of these in the context of excimer lasers, which are pulsed lasers. For a continuous laser action we need to consider other possibilities, such as a three-level atom. In fact, the first laser that was demonstrated to operate was a three-level laser, Maiman's ruby laser. In the above diagram of a three level laser the pump causes an excitation from the ground state to the second excited state. This state is a rather short-lived state, so that the atom quickly decays into the first excited level. [Decays back to the ground state also occur, but these atoms can be pumped back to the second excited state again.] The first excited state is a long-lived (i.e. metastable) state which allows the atom to "wait" for the "passer-by" photon while building up a large population of atoms in this state. The lasing transition, in this laser, is due to the decay of the atom from this first excited metastable state to the ground state. If the number of atoms in the ground state exceeds the number of atoms that are pumped into the excited state, then there is a high likelihood that the "lasing photon" will be absorbed and we will not get sustained laser light. The fact that the lower lasing transition is the ground state makes it rather difficult to achieve efficient population inversion. In a ruby laser this task is accomplished by providing the ruby crystal with a very strong pulsating light source, called a flash lamp. The flash lamp produces a very strong pulse of light that is designed to excite the atoms from their ground state into any short-lived upper level.t In this way the ground state is depopulated and population inversion is achieved until a pulse of laser light is emitted. In the ruby laser the flash lamp light lasts for about 1/1000 of a second (1 ms) and can be repeated about every second. The duration of the laser pulse is shorter than this, typically 0.1 ms. In some pulsed lasers the pulse duration can be tailored using special methods to be much shorter than this, down to about 10 fs (where 1 fs = 10-15 s or one thousandth of a millionth of a millionth of a second). So, the output of a three-level laser is not continuous, but consists of pulses of laser light. To achieve a continuous beam of laser light a four-level laser is required. Here, the lower laser level is not the ground state. As a result, even a pump that may not be very efficient could produce population inversion, so long as the upper level of the laser transition is longer lived than the lower level. Of course, all attempts are made to design a pump that maximizes the number of excited atoms. A typical four-level laser is the helium-neon (He-Ne) gas laser. In these lasers electric pumping excites helium atoms to an excited state whose energy is roughly the same as the upper short-lived state in the neon atom. The sole purpose of the helium atoms is to exchange energy with neon atoms via collisional excitation. As it turns out, this is a very efficient way of getting neon atoms to lase. All lasers have three primary components: The laser medium can be gaseous, liquid, or a solid. These could include atoms, molecules, or collections of atoms that would be involved in a laser transition. Typically, a laser is distinguished by its medium, even though two lasers using different media may have more in common than two which have similar media. There are three different laser pumps: electromagnetic, optical, and chemical. Most lasers are pumped electro-magnetically, meaning via collisions with either electrons or ions. Dye lasers and many solid state lasers are pumped optically; however, solid state lasers are typically pumped with a broad band (range of wavelengths) flash-lamp type light source, or with a narrow band semiconductor laser. Chemically pumped lasers, using chemical reactions as an energy source, are not very efficient. So far, these lasers have been made to work not so much for their usefulness as for their curious operation. Up to now in our discussion of laser theory we have not really seen how the beam is generated. We know that photons emitted by stimulated emission travel coherently in the same direction, but what is it that defines the beam direction and what allows the intensity of the laser light to get large? The answer to these two questions is coupled together in the resonant cavity. Laser resonant cavities usually have two flat or concave mirrors, one on either end, that reflect lasing photons back and forth so that stimulated emission continues to build up more and more laser light. The "back" mirror is made as close to 100% reflective as possible, while the "front" mirror typically is made only 95 - 99% reflective so that the rest of the light is transmitted by this mirror and leaks out to make up the actual laser beam outside the laser device. The resonant cavity thus accounts for the directionality of the beam since only those photons that bounce back and forth between the mirrors lead to amplification of the stimulated emission. Once the beam escapes through the front mirror it continues as a well-directed laser beam. However, as the beam exits the laser it undergoes diffraction and does have some degree of spreading. Typically this beam divergence is as small as 0.05o but even this small amount will be apparent if the beam travels long distances. Even more, the resonant cavity also accounts for the amplification of the light since the path through the laser medium is elongated by repeated passes back and forth. Typically this amplification grows exponentially, similar to the way compound interest works in a bank. The more money in your bank account, with compound interest, the faster you earn more interest dollars. Similarly, the more photons there are to produce stimulated emission, the larger the rate at which new coherent photons are produced. The term used for laser light is gain, or the number of additional photons produced per unit path length. The last question to address in this section is: why is the resonant cavity called by that name? What does resonance have to do with having mirrors on either end of a region containing the laser medium? Recall that when we discussed resonance on a string, we spoke about the wave traveling one way along the string (say to the right) interfering with the wave reflected at the end traveling back to the left. At a resonant frequency, there are points at which the two waves exactly add or cancel all the time, leading to a standing wave. At other frequencies the waves will randomly add or cancel and the wave will not have a large amplitude. The case of a light wave traveling back and forth in the resonant cavity is exactly analogous in that only at certain resonant frequencies will the light wave be amplified. The required condition is easy to see. The mirror separation distance, L, must be equal to a multiple of half a wavelength of light, just as we saw in the case of a string. In symbols, we have that L = nl/2, where l is the wavelength of the light and n is some integer. In the case of light, because of the small wavelength n is a very large number, implying that there are a huge number of resonant frequencies. On the other hand, only those resonant frequencies that are amplified by the laser medium will have large amplitudes and so usually there are only a few so-called laser modes or laser resonant frequencies present in the light from a laser, as shown in the figure. Jay Newman and Seyffie Maleki ( ) | 16,494 | 3,870 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Chemical_Compounds |
Chemical compounds can generally be classified into two broad groups: molecular compounds and ionic compounds. Molecular compounds involve atoms joined by covalent bonds and can be represented by a variety of formulas. Ionic compounds are composed of ions joined by ionic bonding, and their formulas are generally written using oxidation states. Molecular compounds are composed of atoms that are held together by covalent bonds. These bonds are formed when electrons are shared between two atoms. The concept of chemical formulas was created to describe many characteristics of molecular compounds in a simple manner. A normal chemical formula encompasses factors such as which elements are in the molecule and how many atoms of each element there are. The number of atoms of each element is denoted by a subscript, a small number that is written to the left of the element. \[ CH_3COOH\] In the preceding formula, the subscript “3” denotes the fact that there are three hydrogen atoms present in the molecule. An empirical formula represents the proportions of atoms in a molecule. It gives important information about a molecule, because it displays the ratios of atoms that are present within the molecule. However, its limitations exist in the sense that it does not represent the exact number of atoms present in the molecule as the molecular formula does. In certain situations, the molecular and the empirical formula can be the same, but in other situations, the molecular formula is a multiple of the ratios of atoms indicated in the empirical formula. Since empirical formulas can be derived from molecular formulas, molecular formulas are generally more useful than empirical formulas. C H O is a possible empirical formula, because a ratio of 5:7:1 cannot be simplified any further. In this particular case, the empirical formula could also be the molecular formula, if there are exactly 5 carbon atoms, 7 hydrogen atoms, and 1 oxygen atom per molecule. However, another possible molecular formula for this same molecule is C H O , because while there are 10 carbon atoms, 14 hydrogen atoms, and 2 oxygen atoms present, the ratio 10:14:2 can be simplified to 5:7:1, giving way to the same empirical formula. Additionally, C H O is not the only possibility of a molecular formula for this molecule; any formula with the same relative proportions of these atoms that can be simplified to a 5:7:1 ratios is a possible molecular formula for this molecule. When given adequate information, the empirical formula and molecular formula can be quantitatively ascertained. A is written to denote the details of individual atoms’ bonding. More specifically, it clarifies what types of bonds exist, between which atoms these bonds exist, and the order of the atoms’ bonding within the molecule. Covalent bonds are denoted by lines. A single line represents a single bond, two lines represent a double bond, three lines represent a triple bond, and onwards. A single covalent bond occurs when two electrons are shared between atoms, a double occurs when four electrons are shared between two atoms, etc. In this sense, the higher the number of bonds, the stronger the bond between the two atoms. A is a less graphical way of representing the same characteristics displayed by a structural formula. In this type of formula, the molecule is written as a molecular formula with the exception that it indicates where the bonding occurs. All the representations discussed thus far have not addressed how to show a molecule’s three-dimensional structure. The two ways to illustrate a spatial structure are through the use of the ball-and-stick model as well as the space-filling model. The ball-and-stick model uses balls to spatially represent a molecule. The balls are the atoms in a molecule and sticks are the bonds between specific atoms. The space-filling model is also a method of spatially displaying a molecule and its characteristics. A space-filling model shows atoms’ sizes relative sizes to one another. Ionic compounds are composed of positive and negative ions that are joined by ionic bonds. Ionic bonds are generally formed when electrons are transferred from one atom to another, causing individual atoms to become charged particles, or ions. Ions can be referred to as either monatomic or polyatomic. Monatomic ions such as Cl are composed of only one ion, while polyatomic ions such as NO are defined as polyatomic ions. A combination of these ions that forms a compound whose charge is equal to zero is known as a formula unit of an ionic compound. Ionic compounds generally tend to form crystallized salts. They generally have high boiling/melting points, and are good conductors of electricity. The formulas of ionic compounds are always written with the cation first, followed by the anion. The formula can then be completed with reference to the of the elements present. | 4,928 | 3,871 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Haloalkanes |
The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a atom ( , , , or ). There is a fairly large distinction between the structural and physical properties of haloalkanes and the . As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size. With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge. The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases. The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases. When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The incease in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point. Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons. The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain. | 2,879 | 3,872 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.08%3A__The_Mole |
Because atoms and molecules are extremely small, there are a great many of them in any macroscopic sample. The 1 cm of mercury referred to in would contain 4.080 x 10 mercury atoms, for example, and the 3.47 cm of bromine would contain twice as many (8.160 x 10 ) bromine atoms. The very large numbers involved in counting microscopic particles are inconvenient to think about or to write down. Therefore chemists have chosen to count atoms and molecules using a unit called the mole. One (abbreviated mol) is 6.022 x 10 of the microscopic particles which make up the substance in question. Thus 6.022 x 10 Br atoms is referred to as 1 mol Br. The 8.160 x 10 atoms in the sample we have been discussing would be \[\dfrac {8.160\cdot10^{22}} {6.022\cdot10^{23}\text{ mol Br}} = \text {0.1355 mol Br} \nonumber \] The idea of using a large number as a unit with which to measure how many objects we have is not unique to chemists. Eggs, doughnuts, and many other things are sold by the dozen—a unit of twelve items. Smaller objects, such as pencils, may be ordered in units of 144, that is, by the gross, and paper is packaged in reams, each of which contains 500 sheets. A chemist who refers to 0.1355 mol Br is very much like a bookstore manager who orders 2½ dozen sweat shirts, 20 gross of pencils, or 62 reams of paper. There is a difference in degree, however, because the chemist’s unit, 6.022 x 10 , is so large. A stack of paper containing a mole of sheets would extend more than a million times the distance from the earth to the sun, and 6.022 x 10 grains of sand would cover all the land in the world to a depth of nearly 2 ft. Obviously there are a great many particles in a mole of anything. Why have chemists chosen such an unusual number as 6.022 x 10 as the unit with which to count the number of atoms or molecules? Surely some nice round number would be easier to remember. The answer is that . For example, 1 mol of mercury atoms not only contains 6.022 x 10 atoms, but its mass of 200.59 g is conveniently obtained by adding the unit gram to the . Some other examples are \[\begin{align} &\text{1 mol H contains 6.022} \times 10^{23} \text{H atoms;} & \text{its mass is 1.008 g.} \\&\text{1 mol C contains 6.022} \times 10^{23} \text{C atoms;} &\text{its mass is 12.01 g.} \\&\text{1 mol O contains 6.022} \times 10^{23} \text{O atoms;} &\text{its mass is 16.00 g.} \\&\text{1 mol Br contains 6.022} \times 10^{23} \text{Br atoms;} &\text{its mass is 79.90 g.} \end{align} \nonumber \] Here and in subsequent calculations atomic weights are rounded to two decimal places, unless, as in the case of H, fewer than four significant figures would remain. The mass of a mole of can also be obtained from atomic weights. Just as a dozen eggs will have a dozen whites and a dozen yolks, a mole of CO molecules will contain a mole of C atoms and a mole of O atoms. The mass of a mole of CO is thus \[ \text{Mass of 1 mol C + mass of 1 mol O = mass of 1 mol CO} \nonumber \] \[ \text{12.01 g + 16.00 g = 28.01 g} \nonumber \] The molecular weight of CO (28.01) expressed in grams is the mass of a mole of CO. Some other examples are in Table \(\Page {1}\). It is important to specify to what kind of particle a mole refers. A mole of Br atoms, for example, has only half as many atoms (and half as great a mass) as a mole of Br molecules. It is best not to talk about a mole of bromine without specifying whether you mean 1 mol Br or 1 mol Br . | 3,486 | 3,873 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.15%3A_Hess's_Law_of_Heat_Summation |
Calculating the energy involved in the operation of an acetylene torch is no simple matter. Since there is a complex series of reactions taking place, simple methods for determining the heat of reaction will not work. We need to develop new approaches to these calculations. It is sometimes very difficult or even impossible to measure the enthalpy change for a reaction directly in the laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because multiple side reactions may occur at the same time. Fortunately, it is possible to measure the enthalpy change for a reaction with an indirect method. states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a heat of reaction for the final equation. An example will illustrate how Hess's law can be used. Acetylene \(\left( \ce{C_2H_2} \right)\) is a gas that burns at an extremely high temperature \(\left( 3300^\text{o} \text{C} \right)\) and is used in welding. On paper, acetylene gas can be produced by the reaction of solid carbon (graphite) with hydrogen gas. \[2 \ce{C} \left( s, graphite \right) + \ce{H_2} \left( g \right) \rightarrow \ce{C_2H_2} \left( g \right) \: \: \: \: \: \Delta H = ?\nonumber \] Unfortunately, this reaction would be virtually impossible to perform in the laboratory because carbon would react with hydrogen to form many different hydrocarbon products simultaneously. There is no way to create conditions under which only acetylene would be produced. However, enthalpy changes for combustion reactions are relatively easy to measure. The heats of combustion for carbon, hydrogen, and acetylene are shown below, along with each balanced equation. \[\begin{array}{ll} \ce{C} \left( s, graphite \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) & \Delta H = -393.5 \: \text{kJ} \\ \ce{H_2} \left( g \right) + \frac{1}{2} \ce{O_2} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H = -285.8 \: \text{kJ} \\ \ce{C_2H_2} \left( g \right) + \frac{5}{2} \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) & \Delta H = -1301.1 \: \text{kJ} \end{array}\nonumber \] To use Hess's law, we need to determine how the three equations above can be manipulated so that they can be added together to result in the desired equation (the formation of acetylene from carbon and hydrogen). In order to do this, we will go through the desired equation, one substance at a time—choosing the combustion reaction from the equations above that contains that substance. It may be necessary to either reverse a combustion reaction, or multiply it by some factor in order to make it "fit" to the desired equation. The first reactant is carbon and in the equation for the desired reaction, the coefficient of the carbon is a 2. So, we will write the first combustion reaction, doubling all of the coefficients and the \(\Delta H\): \[2 \ce{C} \left( s, graphite \right) + 2 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = 2 \left( -393.5 \right) = -787.0 \: \text{kJ}\nonumber \] The second reactant is hydrogen and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that reaction will be used as written. \[\ce{H_2} \left( g \right) + \frac{1}{2} \ce{O_2} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -285.8 \: \text{kJ}\nonumber \] The product of the reaction is \(\ce{C_2H_2}\) and its coefficient is also a 1. In combustion reaction #3, the acetylene is a reactant. Therefore, we will reverse reaction 3, changing the sign of the \(\Delta H\): \[2 \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{C_2H_2} \left( g \right) + \frac{5}{2} \ce{O_2} \left( g \right) \: \: \: \: \: \Delta H = 1301.1 \: \text{kJ}\nonumber \] Now, these three reactions can be summed together. Any substance that appears in equal quantities as a reactant in one equation and a product in another equation cancels out algebraically. The values for the enthalpy changes are likewise added. \[\begin{array}{rccclclrcl} 2 \ce{C} \left( s, graphite \right) & + & \cancel{2 \ce{O_2} \left( g \right)} & \rightarrow & \cancel{2 \ce{CO_2} \left( g \right)} & & & \Delta H & = & -787.0 \: \text{kJ} \\ \ce{H_2} \left( g \right) & + & \cancel{\frac{1}{2} \ce{O_2} \left( g \right)} & \rightarrow & \cancel{\ce{H_2O} \left( l \right)} & & & \Delta H & = & -285.8 \: \text{kJ} \\ \cancel{2 \ce{CO_2} \left( g \right)} & + & \cancel{\ce{H_2O} \left( l \right)} & \rightarrow & \ce{C_2H_2} \left( g \right) & + & \cancel{\frac{5}{2} \ce{O_2} \left( g \right)} & \Delta H & = & 1301.1 \: \text{kJ} \\ \hline 2 \ce{C} \left( s, graphite \right) & + & \ce{H_2} \left( g \right) & \rightarrow & \ce{C_2H_2} \left( g \right) & & & \Delta H & = & 228.3 \: \text{kJ} \end{array}\nonumber \] So, the heat of reaction for the combination of carbon with hydrogen to produce acetylene is \(228.3 \: \text{kJ}\). When one mole of acetylene is produced, \(228.3 \: \text{kJ}\) of heat is absorbed, making the reaction endothermic. | 5,348 | 3,875 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Psychoactive_Drugs |
A sedative drug decreases activity, moderates excitement, and calms the recipient. A hypnotic drug produces drowsiness and facilitates the onset and maintenance of a state of sleep that resembles natural sleep in its electrocephalographic characteristics and from which the recipient may be easily aroused; the effect is sometimes called hypnosis. Sedation, pharmacological hypnosis, and general anesthesia are usually regarded as only increasing depths of a continuum of central nervous system (CNS) depression. Indeed, most sedative or hypnotic drugs, when used in high doses, can induce general anesthesia. One important exception is the benzodiazepines The term benzodiazepine refers to the portion of the structure composed of a benzene ring (A) fused to a seven-membered diazepine ring (B). However, since all of the important benzodiazepines contain a aryl substituent (ring C) and a 1, 4-diazepine ring, the term has come to mean the aryl-1,4-benzodiazepines. There are several useful benzodiazepines available. The skeletal structure and two examples are shown below. . The effects of the benzodiazepines virtually all result from action of these drugs on the CNS, even when lethal doses are used. The most prominent of these effects are sedation, hypnosis, decreased anxiety, muscle relaxation, and anticonvulsant activity. As the dose of a benzodiazepine is increased, sedation progresses to hypnosis and hypnosis to stupor. They are used as sedatives, hypnotics, antianxiety agents (in panic disorder), anticonvulsants, muscle relaxants, in anesthesia and in alcoholism. The actions of benzodiazepines are a result of potentiation of neural inhibition that is mediated by gamma-aminobutyric acid ( ). This view is supported by behavioral and electrophysiological evidence that the effects of benzodiazepines are reduced or prevented by prior treatment with antagonists of GABA or inhibitors of the synthesis of the transmitter. Benzodiazepine receptors are located on the alpha subunit of the GABA receptor (see figure below) located almost exclusively on postsynaptic nerve endings in the CNS (especially cerebral cortex). Benzodiazepines enhance the GABA transmitter in the opening of postsynaptic chloride channels which leads to hyperpolarization of cell membranes. That is, they "bend" the receptor slightly so that GABA molecules attach to and activate their receptors more effectively and more often. The remarkable safety of the benodiazepines can be accounted for by the self-limited nature of neuronal depression that requires the release of an endogenous inhibitory neurotransmitter to be expressed. That is, they do not directly act to open chloride ion channels, and therefore, are not lethal in over dose as are barbiturates. Benzodiazepines are highly lipid soluble. There is rapid diffusion into the CNS followed by redistribution to inactive tissue sites. Benzodiazepines have a high volume of distribution and rapidly cross the placenta. The duration of action is determined by rate of metabolism and elimination. Diazepam is not water soluble and is dissolved in propylene glycol; therefore it may cause venous irritation and thrombophlebitis. Diazepam also has unpredictable absorption after IM injection. Benzodiazepines are extensively bound to albumin. The barbiturates once enjoyed a long period of extensive use as sedative-hypnotic drugs; however, except for a few specialized uses, they have been largely replaced by the much safer benzodiazepines. are CNS depressants and are similar, in many ways, to the depressant effects of alcohol. To date, there are about 2,500 derivatives of barbituric acid of which only 15 are used medically. The first barbiturate was synthesized from barbituric acid in 1864. The original use of barbiturates was to replace drugs such as opiates, bromides, and alcohol to induce sleep. Barbiturates are broken down chemically within the liver and eliminated via the kidneys at different rates according to their type. With regular use, the body develops a tolerance to barbiturates that translates into a need for larger and more frequent doses to attain the desired affect. However, while the tolerance increases in terms of realizing a desired effect, tolerance to the lethal level does not. In general, structural changes that increase lipid solubility decrease duration of action, decrease latency to onset of activity, accelerate metabolic degradation, and increase hypnotic potency. Thus, large aliphatic groups at R2 confer greater activity than do methyl groups, but the compounds have a shorter duration of action; however, groups longer than seven carbons tend to have convulsive activity. Introduction of polar groups, such as ether, keto, hydroxyl, amino, or carboxyl groups, into alkyl side chains decreases lipid solubility and abolishes hypnotic activity. Barbiturates facilitate GABA-ergic inhibition in a way that resembles some of the actions of the benzodiazepines, discussed above. However, barbiturates do not displace benzodiazepines from their binding sites. Instead, they enhance such binding by increasing the affinity for benzodiazepines; they also enhance the binding of GABA and its agonist analogs to specific sites in neural membranes. These effects are almost completely dependent upon the presence of chloride or other anions that are known to permeate through chloride channels and they are completely inhibited by picritoxin. (The picrotoxin group of toxins are naturally-occurring GABA antagonists which can cause death due to convulsions.) While both barbiturates and benzodiazepines are capable of potentiating GABA-induced increases in chloride conductance, significant differences in their modes of action can be detected. Pentobarbital appears to increase the lifetime of the open state of the chloride channels that are regulated by GABA-ergic receptors; the magnitude of this effect more than offsets a barbiturate-induced decrease in the frequency of channel openings. By contrast, high concentrations of diazepam increase the frequency of channel openings with little effect on the lifetime of the open state. It is thought that barbiturates prolong the activation of the channel by decreasing the rate of dissociation of GABA from its receptor. Schizophrenia comes in many varieties. One of the most common types is seen in the person who hears voices and has delusions of grandeur, intense fear, or other types of feelings that are unreal. Many schizophrenics are highly paranoid, with a sense of persecution from outside sources. Schizophrenia appears to result from excessive excitement of a group of neurons that secrete dopamine in the behavioral centers of the brain, including in the frontal lobes. An alternative possibility is either hypersensitive or excess D2 dopamine receptors. Therefore, drugs used to treat this disorder decrease the level of dopamine excreted from these neurons or antagonize dopamine. Dopamine has been implicated as a cause of schizophrenia because many patients with Parkinson's disease develop schizophrenic-like symptoms when they are treated with the drug L-DOPA. Also, drugs known to enhance central dopamine activity can worsen symptoms and even produce psychotic-like signs in normal individuals. It has been suggested that in schizophrenia, excess dopamine is secreted by a group of dopamine-secreting neurons whose cell bodies lie in the mesencephalon, medial to the substantia nigra. These neurons give rise to the so-called mesolinic dopaminergic system that projects nerve fibers to the medial and anterior portions of the limbic system, especially to the hippocampus, amygdala, anterior caudate nucleus, and portions of the prefrontal lobes. All of these are powerful behavioral control centers. An even more compelling reason for believing that schizophrenia is caused by excess production of dopamine is that many drugs that are effective in treating schizophrenia-such as chlorpromazine, haloperidol, and thiothizene-all decrease the secretion of dopamine at the dopaminergic nerve endings or decrease the effect of dopamine on subsequent neurons. The phenothiazines as a class, and especially chlorpromazine, the prototype, are among the most widely used drugs in medical practice and are primarily employed in the management of patients with serious psychiatric illnesses. In addition, many members of the group have other clinically useful properties, including antiemetic, antinausea, and antihistaminic effects and the ability to potentiate analgesics, sedatives and general anesthetics. Phenothiazine compounds were synthesized in Europe in the late nineteenth century as part of the development of aniline dyes such as methylene blue. In the late 1930s a derivative of phenothiazine was found to have antihistamine and a strong sedative effect and so the drug was introduced as into clinical anesthesia. It was noted that chlorpromazine by itself did not cause a loss of consciousness but produced only a tendency to sleep and a lack of interest in what was going on. These central actions became known as neuroleptic soon after. Phenothiazine has a tricyclic structure in which two benzene rings are linked by a sulfur and a nitrogen atom (see figures below). Substitution of an electron-withdrawing group at R2 (but not at position 3 or 4) increases the efficacy of phenothiazines and other tricyclic congeners. Neuroleptic drugs reduce initiative and interest in the environment, and they reduce displays of emotion or affect. Initially there may be some slowness in response to external stimuli and drowsiness. However subject are easily aroused, capable of giving appropriate answers to direct questions, and seem to have intact intellectual functions; there is no ataxia, incoordination, or dysathria at ordinary doses. Psychotic patients become less agitated and restless, and withdrawn or autistic patients sometimes become more responsive and communicative. Aggressive and impulsive behavior diminishes. Gradually (over a period of days). psychotic symptoms of hallucinations, delusions, and disorganized or incoherent thinking tend to disappear. The most prominent observable effects of typical neuroleptic agents are strikingly similar. In low doses, operant behavior is reduced but spinal reflexes are unchanged. Exploratory behavior is diminished, and responses to a a variety of stimuli are fewer, slower, and smaller, although the ability to discriminate stimuli is retained. Conditioned avoidance behaviors are selectively inhibited, while unconditioned excape or avoidance responses are not. Behavioral activation, stimulated environmentally of pharmacologically, is blocked. Feeding is inhibited. Most neuroleptics block the emesis and aggression induced by apomorphine--a dopaminergic agonist. In high doses, most neuroleptic agents induce characteristic cataleptic immobility that allows the animal to be placed in abnormal postures that persist. Muscle tone is altered, and ptosis (drooping of the eyelids) is typical. Even very high doses of most neuroleptics do not induce comma, and the lethal dose is extraordinarily high. . It is well established that benzodiazepines block dopaminergic receptors in the brain. There are three major central dopamine pathways: the , which is affiliated with motor effects produced by antipsychotic drugs; the , which is associated with the endocrine effects of neuroleptics; and the , which is the most likely to relate to the symptoms of schizophrenia. Of the three central dopamine receptor subtypes D1, D2 and D3, the D2 receptor, is believed to be most relevant to antipsychotic drug action. Most interesting, however, is that both D1 and D2 are altered in drug-naive schizophrenics. Though neuroleptic drugs are D1 and D2 antagonists, in vitro D2 effects are achieved at 103 lower concentrations. D2 receptors are also located outside the blood brain barrier. One area is in the the chemoreceptor trigger zone of the medulla which is the reason that many of the phenothiazine drugs work as antiemetics and stop nausea. Side Effects. There are several resulting syndromes which can occur from using antipsychotic drugs. A parkinsonian syndrome that may be indistinguishable from idiopathic parkinsonism my develop during administration of antipsychotic drugs. The most noticeable signs are rigidity and tremor at rest, especially involving the upper extremities. Tardive dyskenesia is a late-appearing neurological syndrome also associated with antipsychotic drug use. It is characterized by stereotypical involuntary movements consisting in sucking and smacking of the lips, lateral jaw movements, and fly-catching dartings of the tongue. There may be purposeless, quick movements of the extremities and slower, more dystonic movements and postures of the extremities, trunk, and neck may also be seen. All of these movements disappear during sleep as they do in parkinsonism. Symptoms of these conditions my persists indefinitely after discontinuation of the medication, although sometimes they disappear with time. In 1994 an addition tot he antipsychotic drugs is risperidone (Risperdal). This drug antagonises D2 and serotonin type 2 receptors. The drug also antagonizes for other receptors such as a adrenergic and histaminergic H1 receptors. Major depression is the most common of the major mental illnesses, and it must be distinguished from normal grief, sadness, and disappointment. Major depression is characterized by feelings of intense sadness and despair, mental slowing and loss of concentration, pessimistic worry, agitation, and self-depreciation. Physical changes also occur, such as weight loss, decreased libido, and disruption of hormonal circadian rhythms. Before the advent of psychotherapy in the 1950s, treatment of depression consisted of stimulants such as caffeine and amphetamines to ameliorate the depressive phases and barbiturates to allay agitation, anxiety, and insomnia. At best, such attempts at therapy may have offered transient relief to some patients. Suffering usually decreased little. Monoamine oxidase inhibitors were the first effective antidepressants used. These were discussed in detail in the section on and therefore will not be further discussed here. Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. The cell bodies for serotonergic neurons are found in the raphe region in the brainstem/pons region. Lesions of this area can be made using 5,6 or 5,7-dihydroxytryptamine in a similar manner to 6-hydroxydopamine and have helped define the CNS pathways for 5-HT. The monoamine serotonin is itself a precursor for melatonin production in the pineal gland. The biosynthesis of serotonin from the amino acid tryptophan is similar to that found for the catecholamines, and 5-hydroxytryptophan can cross the BBB to increase central levels of 5-HT. Although some of the serotonin is metabolized by monoamine oxidase to 5-hydroxyindole acetic acid, most of the serotonin released into the post-synaptic space is removed by the neuron through a reuptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressants such as fluoxetine and sertraline. Serotonin receptors are diverse and numerous. Over the past several years, over fourteen different serotonin receptors have been cloned and sequenced through molecular biological techniques. Overall, there are seven distinct families of 5-HT receptors, with as many as five within a particular family. Only one of the 5-HT receptors is a ligand-gated ion channel (the 5-HT3 receptor), and the other six families are all G protein-coupled receptors. Imipramine, amitriptylin, and other closely related drugs are among the drugs currently most widely used for the treatment of major depression. Because of there structure ( see below). They are often referred to as the tricyclic antidepressants. Although these compounds seem to be similar to the phenothiazines chemically, the ethylene group of imiprimine's middle ring imparts dissimilar stereochemical properties and prevents conjegation of the rings, as occurs with the phenothiazines. One might expect an effective antidepressant drug to have a stimulating or mood-elevating effect when given to a normal subject. Although this may occur with the MAOIs, it is not true of the tricyclic antidepressants. If a dose of imiprimine given to a normal subject, he feels sleepy and tends to be quieter, his blood pressure falls slightly, and he feels light headed. These drug effects are usually perceived to be unpleasant, and cause a feeling of unhappiness and increased anxiety. Repeated administration for several days may lead to accentuation of these symptoms and, in addition, to difficulty in concentrating and thinking. In contrast, if the drug is given over a period of time ( two to three weeks) to depressed patients an elevated mood occurs. For this reason, the tricyclics are not prescribed on an "as-needed" basis. . All tricyclic antidepressants in current use in the U.S. potentiate the actions of biogenic amines in the CNS by blocking its re-uptake at nerve terminals. However, the potency and selectivity for the inhibition of the uptake of norepinephrine, serotonin, and dopamine vary greatly among the agents. The tertiary amine tricyclics seem to inhibit the serotonin uptake pump, whereas the secondary amine ones seem better in switching off the NE pump (see table below). For instance, imipramine and amitriptyline are potent and selective blockers of serotonin transport with small effects on NE uptake, while desipramine and nortriptyline inhibit the uptake of norepinephrine and exert smaller effects on serotonin inhibition. None of these agents is very effective as an inhibitor of dopamine transport; this contrasts with the rather nonselective inhibitory actions of cocaine and amphetamine on the uptake of both norepinephrine and dopamine. These are poor antidepressants, despite the fact that it has a stimulant and even euphoriant effect in some people. In recent years, selective serotonin reuptake inhibitors (SSRIs) have been introduced for the treatment of depression. Prozac is the most famous drug in this class. Lilly's sales of Prozac in 1993 exceeded 1 billion US dollars. Clomiprimine, fluoxetine (Prozac), sertraline and paroxetine selectively block the reuptake of serotonin, thereby increasing the levels of serotonin in the central nervous system. Note the similarities and differences between the tricyclic antidepressants and the selective serotonin reuptake inhibitors. The SSRIs generally have fewer anticholinergic side effects, but caution is still necessary when co-administering any drugs that affect serotonergic systems (e.g., monoamine oxidase inhibitors). Some of the newer, SSRIs (e.g., clomipramine) have been useful in the treatment of obsessive-compulsive disorders. Here are some data to give you an idea of what transport systems are likely to be altered by the different antidepressants: From: Hyttel J. Pharmacological characterization of selective serotonin reuptake inhibitors (SSRIs). International Clinical Psychopharmacology. 9 Suppl 1: 19-26, 1994 Mar. Lithium is widely used as \(\ce{Li2CO3}\) to control manic behavior in manic-depressive patients. No totally acceptable mechanism for its action exists. Postulations involve action that would likely adjust overactive catecholaminergic activity, which is the accepted occurrence in mania. Recent research has shown that \(\ce{Li^{+}}\) is a glutamate reuptake inhibitor. Other research has shown that \(\ce{Li^{+}}\) may inhibit glutamate stimulation of nerve cells. Several studies showing that \(\ce{Li^{+}}\) has some antidepressant effects are known. Some weak biphasic alterations of norepinephrine and serotonin turnover in the brain were established. Amphetamines were discussed under the topic of and therefore will not be further discussed here. Caffeine, theophylline and theobromine share in common several pharmacological actions of therapeutic interest. They stimulate the central nervous system, act on the kidney to produce diuresis, stimulate cardiac muscle, and relax smooth muscle, notably bronchial muscle. Because the various xanthines differ markedly in the intensity of their action on various structures, one particular xanthine has been used more than another for a particular therapeutic effect. Since theobromine displays low potency in these pharmacological actions, it has all but disappeared from the therapeutic scene. Caffeine, theophyline, and theobromine occur naturally in plants widely distributed geographically. Caffeine is found in the coffee bean, tea leaves, guarana, and other plants. It is probably the most-used of all psychoactive drugs. From the figure below, we can see that the methylxanthines have a structure which is very similar to adenine. We have already discussed the role of the second messenger, cAMP, in the response to norepinephrine and epinephrine in the section on Adrenergic Mechanisms. The mechanism of action of caffeine and other methylxanthines is inhibiting the degradation of cAMP. | 21,384 | 3,876 |
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Place the sample to be sublimed in the bottom of the sublimation apparatus. Lightly grease all joints. Use thick-walled tubing to attach to the vacuum arm, and apply the vacuum. The setup should not hiss or there is a leak. Fill the cold finger, or run water through the condenser. Be sure to apply the vacuum first, then coolant. If cooled before the vacuum, condensation may occur on the cold finger. Wave a heat gun or Bunsen burner on the apparatus to heat the sample. Sublimation should begin within a few minutes. Coax solid deposited on the side of the glassware toward the cold finger by waving the heat gun/burner on the sides of the glass. When the sublimation is complete: Remove the coolant. Allow the apparatus to come to room temperature. Delicately reinstate the air pressure, noting that an abrupt opening of the vessel will cause air to knock crystals off the cold finger. Remove the cold finger. | 925 | 3,877 |
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How can DNA and RNA molecules act as blueprints for the manufacture of proteins? The exact details were unraveled in the early 1960s mainly by Marshall Nirenberg (born 1927) at the National Institutes of Health and H. G. Khorana (born 1922) at the University of Wisconsin, work which earned them the Nobel prize in 1968. They showed that each amino acid in a protein is determined by a specific of three nitrogenous bases in the DNA or RNA chain. The details of this are given in the table below. As an example of how this code works, let us take the section of RNA shown in Figure 3 on Nucleic Acid Structure. This has the sequence UCAUGG. This is part of the instructions for building a polypeptide chain containing the amino acid serine (UCA) followed by the amino acid tryptophan (UGG). \(\Page {1}\) The Genetic Code for RNA ( ) A termination codon is indicated by TERM. ( ) AUG, the codon for methionine is also the initiation codon. All protein synthesis begins at this codon, though this initial methionine is often removed during post-transcriptional processing. There are three additional features of the genetic code. First, AUG, the codon for Methionine also serves as an , and, with help from other signals, is where protein systhesis begins. A second feature is that reading RNA for protein synthesis goes from the 5' carbon end of the nucleic acid to the 3' carbon end. A final important feature of the genetic code is the existence of three . These correspond to an instruction for ending a polypeptide chain. How these features work is best illustrated by an example. Decode the RNA fragment 5' A C C U U A U G A C G C C U G U C C A U U A A C G A U 3' First, we must decide which direction to read the RNA code. Synthesis goes from the 5' end to the 3' end, so this segment is read left to right. Had it been displayed 3' to 5', we would have needed to read it from right to left. Second, we need to look for an initiation codon, AUG. This codon appears starting at the sixth letter in. Thus, we can divide the sequence up like this, with the start codon bold: Third, let us see if there is a stop codon in this sequence. Sure enough, the fifth codon after the start codon, UAA is a stop codon. Thus, the entire sequence to be translated, in bold: which translates to the amino acid sequence: Notice in the example, that if we had not started with the initiation codon, an entirely different protein would have been formed. Look at what would have happened if we had simply started at the beginning of the sequence: a stop codon appears in a new place, and the translated protien is: This highlights the importance of the , the place where codons start being read. Notice that, since codons are 3 bases long, any sequence has three different reading frames. Without the initiation codon, there would be no way to identify the correct reading frame. In addition to the AUG initiation codon, other element regulate initiation. In bacteria, a sequence of bases before the initiation codon, called the precedes the AUG codon, specifying where to begin translation. A different set up occurs in eukaryotes. An initiation complex forms, but instead of having a specific sequence connected to the initiation codon, the complex slides along the mRNA strand, until it finds the AUG initiation codon. | 3,331 | 3,878 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.E%3A_Solutions_and_Colloids_(Exercises) |
How do solutions differ from compounds? From other mixtures? A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. Which of the principal characteristics of solutions can we see in the solutions of \(\ce{K2Cr2O7}\) shown in The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K Cr O in water can vary. When KNO is dissolved in water, the resulting solution is significantly colder than the water was originally. (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K and \(\ce{NO3-}\) ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. Give an example of each of the following types of solutions: (a) CO in water; (b) O in N (air); (c) bronze (solution of tin or other metals in copper) Indicate the most important types of intermolecular attractions in each of the following solutions: (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C H , nonpolar solvent): (a) heptane; (b) water; (c) water Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes. Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat. Solutions of hydrogen in palladium may be formed by exposing Pd metal to H gas. The concentration of hydrogen in the palladium depends on the pressure of H gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal. Explain why the ions Na and Cl are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules. Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive. HBr is an acid and so its molecules react with water molecules to form H O and Br ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive. Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO ) ( )? (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO ) . (a) Fe(NO ) is a strong electrolyte, thus it should completely dissociate into Fe and \(\ce{(NO3- )}\) ions. Therefore, (z) best represents the solution. (b) \(\ce{Fe(NO3)3}(s)⟶\ce{Fe^3+}(aq)+\ce{3NO3- }(aq)\) Compare the processes that occur when methanol (CH OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution. Methanol, \(CH_3OH\), dissolves in water in all proportions, interacting via hydrogen bonding. \[CH_3OH_{(l)}+H_2O_{(l)}⟶CH_3OH_{(aq)}\] Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions. \[HCl{(g)}+H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)}\] Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively. \[NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^−_{(aq)}\] What is the expected electrical conductivity of the following solutions? (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved) Why are most ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a (molten) ionic compound to be electrically conductive or nonconductive? Explain. A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces Suppose you are presented with a clear solution of sodium thiosulfate, Na S O . How could you determine whether the solution is unsaturated, saturated, or supersaturated? Add a small crystal of \(Na_2S_2O_3\). It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution. Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures. The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating. Suggest an explanation for the observations that ethanol, C H OH, is completely miscible with water and that ethanethiol, C H SH, is soluble only to the extent of 1.5 g per 100 mL of water. The hydrogen bonds between water and C H OH are much stronger than the intermolecular attractions between water and C H SH. Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C using the following figure for useful data, and report the computed percentage to one significant digit.
At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water. \[\%\; KBr =\dfrac{60\; g\; KBr}{(60+100)\;g\; solution} = 40\%\] Which of the following gases is expected to be most soluble in water? Explain your reasoning. (c) CHCl is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules. At 0 °C and 1.00 atm, as much as 0.70 g of O can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O dissolve in 1 L of water? This problem requires the application of Henry’s law. The governing equation is \(C_g = kP_g\). \[k=\dfrac{C_g}{P_g}=\dfrac{0.70\;g}{1.00\; atm} =0.70\;g\; atm^{−1}\] Under the new conditions, \(C_g=0.70\;g\;atm^{−1} \times 4.00\; atm = 2.80\; g\). Refer to following figure for the following three questions: (a) It decreased as some of the CO gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO gas above the beverage. The reduced CO gas pressure, per Henry’s law, lowers the solubility for CO . (c) The dissolved CO concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated. The Henry’s law constant for CO is \(3.4 \times 10^{−2}\; M/atm\) at 25 °C. What pressure of carbon dioxide is needed to maintain a CO concentration of 0.10 in a can of lemon-lime soda? \[P_g=\dfrac{C_g}{k}=\dfrac{0.10\; M}{3.4 \times 10^{−2}\;M/atm} =2.9\; atm\] The Henry’s law constant for O is \(1.3\times 10^{−3}\; M/atm\) at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O is 0.21 atm? Start with Henry's law \[C_g=kP_g\] and apply it to \(O_2\) \[C(O_2)=(1.3 \times 10^{−3}\; M/atm) (0.21\;atm)=2.7 \times 10^{−4}\;mol/L\] The total amount is \((2.7 \times 10^{−4}\; mol/L)(40\;L=1.08 \times 10^{−2} \;mol\] The mass of oxygen is \((1.08 \times 10^{−2}\; mol)(32.0\; g/mol)=0.346\;g\) or, using two significant figures, \(0.35\; g\). How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20- solution of hydrochloric acid? First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required. Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion? What is the microscopic explanation for the macroscopic behavior illustrated in ? The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region. Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume. A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C H (OH) ), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical? Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions. What are the mole fractions of H PO and water in a solution of 14.5 g of H PO in 125 g of water? What are the mole fractions of HNO and water in a concentrated solution of nitric acid (68.0% HNO by mass)? Calculate the mole fraction of each solute and solvent: a. \(583\:g\:\ce{H2SO4}\times\dfrac{1\:mole\:\ce{H2SO4}}{98.08\:g\:\ce{H2SO4}}=5.94\:mole\:\ce{H2SO4}\) \(\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\) \(1.50\:kg\:\ce{H2O}\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mole\:\ce{H2O}}{18.02\:g\:\ce{H2O}}=83.2\:moles\:\ce{H2O}\) Calculate the mole fraction of each solute and solvent: Calculate the mole fractions of methanol, CH OH; ethanol, C H OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.) What is the difference between a 1 solution and a 1 solution? In a 1 solution, the mole is contained in exactly 1 L of solution. In a 1 solution, the mole is contained in exactly 1 kg of solvent. What is the molality of phosphoric acid, H PO , in a solution of 14.5 g of H PO in 125 g of water? What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO by mass)? (a) Determine the molar mass of HNO . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 Calculate the molality of each of the following solutions: Calculate the molality of each of the following solutions: (a) 6.70 × 10 ; (b) 5.67 ; (c) 2.8 ; (d) 0.0358 The concentration of glucose, C H O , in normal spinal fluid is \(\mathrm{\dfrac{75\:mg}{100\:g}}\). What is the molality of the solution? A 13.0% solution of K CO by mass has a density of 1.09 g/cm . Calculate the molality of the solution. 1.08 What is the boiling point of a solution of 9.04 g of I in 75.5 g of benzene, assuming the I is nonvolatile? What is the freezing temperature of a solution of 115.0 g of sucrose, C H O , in 350.0 g of water, which freezes at 0.0 °C when pure? (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C What is the freezing point of a solution of 9.04 g of I in 75.5 g of benzene? What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO ) in water at 25 °C? The volume of the solution is 275 mL. (a) Determine the molar mass of Ca(NO ) ; determine the number of moles of Ca(NO ) in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol ) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin? What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; = 5.02 °C/ ) that boils at 81.5 °C at 1 atm? (a) Determine the molal concentration from the change in boiling point and ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × 10 g mol A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound. A 1.0 solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain. No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ = (1.0 m)(5.14 °C/ ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. A solution contains 5.00 g of urea, CO(NH ) , a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. 144 g mol Arrange the following solutions in order by their decreasing freezing points: 0.1 Na PO , 0.1 C H OH, 0.01 CO , 0.15 NaCl, and 0.2 CaCl . Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na SO , and 0.030 mol of MgCl , assuming complete dissociation of these electrolytes. 0.870 °C How could you prepare a 3.08 aqueous solution of glycerin, C H O ? What is the freezing point of this solution? A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS ( = 2.43 °C/ ). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide? S In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI ? Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10 atm at 25 °C. What is the molar mass of lysozyme? 1.39 × 10 g mol The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin? The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C H O , is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C? 54 g What is the freezing point of a solution of dibromobenzene, C H Br , in 0.250 kg of benzene, if the solution boils at 83.5 °C? What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C? 100.26 °C The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and for ethanol is 1.20 °C/ . What is the molecular formula of fructose? The vapor pressure of methanol, CH OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C H OH, is 44 torr at the same temperature. (a) \(X_\mathrm{CH_3OH}=0.590\); \(X_\mathrm{C_2H_5OH}=0.410\); (b) Vapor pressures are: CH OH: 55 torr; C H OH: 18 torr; (c) CH OH: 0.75; C H OH: 0.25 The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air? Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature? The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. for camphor is 37.7 °C/ . What is the molecular formula of the solute? Show your calculations. A sample of HgCl weighing 9.41 g is dissolved in 32.75 g of ethanol, C H OH ( = 1.20 °C/ ). The boiling point elevation of the solution is 1.27 °C. Is HgCl an electrolyte in ethanol? Show your calculations. \(\mathrm{Δbp}=K_\ce{b}m=(1.20\:°\ce C/m)\mathrm{\left(\dfrac{9.41\:g×\dfrac{1\:mol\: HgCl_2}{271.496\:g}}{0.03275\:kg}\right)=1.27\:°\ce C}\) The observed change equals the theoretical change; therefore, no dissociation occurs. A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations. Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby. Distinguish between dispersion methods and condensation methods for preparing colloidal systems. Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds. How do colloids differ from solutions with regard to dispersed particle size and homogeneity? Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale. Explain the cleansing action of soap. Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules. How can it be demonstrated that colloidal particles are electrically charged? If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. ). | 20,901 | 3,879 |
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In a potentiometric method of analysis we determine an analyte’s concentration by measuring the potential of an electrochemical cell under static conditions in which no current flows and the concentrations of species in the electrochemical cell remain fixed. Dynamic techniques, in which current passes through the electrochemical cell and concentrations change, also are important electrochemical methods of analysis. In this section we consider coulometry. Voltammetry and amperometry are covered in . is based on an exhaustive electrolysis of the analyte. By exhaustive we mean that the analyte is oxidized or reduced completely at the working electrode, or that it reacts completely with a reagent generated at the working electrode. There are two forms of coulometry: , in which we apply a constant potential to the electrochemical cell, and , in which we pass a constant current through the electrochemical cell. During an electrolysis, the total charge, , in coulombs, that passes through the electrochemical cell is proportional to the absolute amount of analyte by \[Q=n F N_{A} \label{11.1}\] where is the number of electrons per mole of analyte, is Faraday’s constant (96 487 C mol ), and is the moles of analyte. A coulomb is equivalent to an A•sec; thus, for a constant current, , the total charge is \[Q=i t_{e} \label{11.2}\] where is the electrolysis time. If the current varies with time, as it does in controlled-potential coulometry, then the total charge is \[Q=\int_{0}^{t_e} i(t) d t \label{11.3}\] In coulometry, we monitor current as a function of time and use either Equation \ref{11.2} or Equation \ref{11.3} to calculate . Knowing the total charge, we then use Equation \ref{11.1} to determine the moles of analyte. To obtain an accurate value for , all the current must oxidize or reduce the analyte; that is, coulometry requires 100% or an accurate measurement of the current efficiency using a standard. Current efficiency is the percentage of current that actually leads to the analyte’s oxidation or reduction. The easiest way to ensure 100% current efficiency is to hold the working electrode at a constant potential where the analyte is oxidized or reduced completely and where no potential interfering species are oxidized or reduced. As electrolysis progresses, the analyte’s concentration and the current decrease. The resulting current-versus-time profile for controlled-potential coulometry is shown in Figure 11.3.1
. Integrating the area under the curve (Equation \ref{11.3}) from = 0 to = gives the total charge. In this section we consider the experimental parameters and instrumentation needed to develop a controlled-potential coulometric method of analysis. To understand how an appropriate potential for the working electrode is selected, let’s develop a constant-potential coulometric method for Cu based on its reduction to copper metal at a Pt working electrode. \[\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightleftharpoons \mathrm{Cu}(s) \label{11.4}\] Figure 11.3.2
shows a ladder diagram for an aqueous solution of Cu . From the ladder diagram we know that reaction \ref{11.4} is favored when the working electrode’s potential is more negative than +0.342 V versus the standard hydrogen electrode. To ensure a 100% current efficiency, however, the potential must be sufficiently more positive than +0.000 V so that the reduction of H O to H does not contribute significantly to the total current flowing through the electrochemical cell. We can use the Nernst equation for reaction \ref{11.4} to estimate the minimum potential for quantitatively reducing Cu . \[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\frac{0.05916}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \label{11.5}\] So why are we using the concentration of Cu in Equation \ref{11.5} instead of its activity? In potentiometry we use activity because we use to determine the analyte’s concentration. Here we use the Nernst equation to help us select an appropriate potential. Once we identify a potential, we can adjust its value as needed to ensure a quantitative reduction of Cu . In addition, in coulometry the analyte’s concentration is given by the total charge, not the applied potential. If we define a quantitative electrolysis as one in which we reduce 99.99% of to Cu, then the concentration of Cu at is \[\left[\mathrm{Cu}^{2+}\right]_{t_{e}}=0.0001 \times\left[\mathrm{Cu}^{2+}\right]_{0} \label{11.6}\] where [Cu ] is the initial concentration of Cu in the sample. Substituting Equation \ref{11.6} into Equation \ref{11.5} allows us to calculate the desired potential. \[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-\frac{0.05916}{2} \log \frac{1}{0.0001 \times\left[\mathrm{Cu}^{2+}\right]} \nonumber\] If the initial concentration of Cu is \(1.00 \times 10^{-4}\) M, for example, then the working electrode’s potential must be more negative than +0.105 V to quantitatively reduce Cu to Cu. Note that at this potential H O is not reduced to H , maintaining 100% current efficiency. Many controlled-potential coulometric methods for Cu use a potential that is negative relative to the standard hydrogen electrode—see, for example, Rechnitz, G. A. , Macmillan: New York, 1963, p.49. Based on the ladder diagram in you might expect that applying a potential <0.000 V will partially reduce H O to H , resulting in a current efficiency that is less than 100%. The reason we can use such a negative potential is that the reaction rate for the reduction of H O to H is very slow at a Pt electrode. This results in a significant —the need to apply a potential more positive or a more negative than that predicted by thermodynamics—which shifts for the H O /H redox couple to a more negative value. In controlled-potential coulometry, as shown in , the current decreases over time. As a result, the rate of electrolysis—recall from that current is a measure of rate—becomes slower and an exhaustive electrolysis of the analyte may require a long time. Because time is an important consideration when designing an analytical method, we need to consider the factors that affect the analysis time. We can approximate the current’s change as a function of time in as an exponential decay; thus, the current at time is \[i_{t}=i_{0} e^{-k t} \label{11.7}\] where is the current at = 0 and is a rate constant that is directly proportional to the area of the working electrode and the rate of stirring, and that is inversely proportional to the volume of solution. For an exhaustive electrolysis in which we oxidize or reduce 99.99% of the analyte, the current at the end of the analysis, , is \[i_{t_{e}} \leq 0.0001 \times i_{0} \label{11.8}\] Substituting Equation \ref{11.8} into Equation \ref{11.7} and solving for gives the minimum time for an exhaustive electrolysis as \[t_{e}=-\frac{1}{k} \times \ln (0.0001)=\frac{9.21}{k} \nonumber\] From this equation we see that a larger value for reduces the analysis time. For this reason we usually carry out a controlled-potential coulometric analysis in a small volume electrochemical cell, using an electrode with a large surface area, and with a high stirring rate. A quantitative electrolysis typically requires approximately 30–60 min, although shorter or longer times are possible. A three-electrode potentiostat is used to set the potential in controlled-potential coulometry (see ). The working electrodes is usually one of two types: a cylindrical Pt electrode manufactured from platinum-gauze (Figure 11.3.3
), or a Hg pool electrode. The large overpotential for the reduction of H O at Hg makes it the electrode of choice for an analyte that requires a negative potential. For example, a potential more negative than –1 V versus the SHE is feasible at a Hg electrode—but not at a Pt electrode—even in a very acidic solution. Because mercury is easy to oxidize, it is less useful if we need to maintain a potential that is positive with respect to the SHE. Platinum is the working electrode of choice when we need to apply a positive potential. The auxiliary electrode, which often is a Pt wire, is separated by a salt bridge from the analytical solution. This is necessary to prevent the electrolysis products generated at the auxiliary electrode from reacting with the analyte and interfering in the analysis. A saturated calomel or Ag/AgCl electrode serves as the reference electrode. The other essential need for controlled-potential coulometry is a means for determining the total charge. One method is to monitor the current as a function of time and determine the area under the curve, as shown in . Modern instruments use electronic integration to monitor charge as a function of time. The total charge at the end of the electrolysis is read directly from a digital readout. If the product of controlled-potential coulometry forms a deposit on the working electrode, then we can use the change in the electrode’s mass as the analytical signal. For example, if we apply a potential that reduces Cu to Cu at a Pt working electrode, the difference in the electrode’s mass before and after electrolysis is a direct measurement of the amount of copper in the sample. As we learned in , we call an analytical technique that uses mass as a signal a gravimetric technique; thus, we call this . A second approach to coulometry is to use a constant current in place of a constant potential, which results in the current-versus-time profile shown in Figure 11.3.4
. Controlled-current coulometry has two advantages over controlled-potential coulometry. First, the analysis time is shorter because the current does not decrease over time. A typical analysis time for controlled-current coulometry is less than 10 min, compared to approximately 30–60 min for controlled-potential coulometry. Second, because the total charge simply is the product of current and time (Equation \ref{11.2}), there is no need to integrate the current-time curve in Figure 11.3.4
. Using a constant current presents us with two important experimental problems. First, during electrolysis the analyte’s concentration—and, therefore, the current that results from its oxidation or reduction—decreases continuously. To maintain a constant current we must allow the potential to change until another oxidation reaction or reduction reaction occurs at the working electrode. Unless we design the system carefully, this secondary reaction results in a current efficiency that is less than 100%. The second problem is that we need a method to determine when the analyte's electrolysis is complete. As shown in , in a controlled-potential coulometric analysis we know that electrolysis is complete when the current reaches zero, or when it reaches a constant background or residual current. In a controlled-current coulometric analysis, however, current continues to flow even when the analyte’s electrolysis is complete. A suitable method for determining the reaction’s endpoint, , is needed. To illustrate why a change in the working electrode’s potential may result in a current efficiency of less than 100%, let’s consider the coulometric analysis for Fe based on its oxidation to Fe at a Pt working electrode in 1 M H SO . \[\mathrm{Fe}^{2+}(a q) \rightleftharpoons \text{ Fe}^{3+}(a q)+e^{-} \nonumber\] Figure 11.3.5
shows the ladder diagram for this system. At the beginning of the analysis, the potential of the working electrode remains nearly constant at a level near its initial value. As the concentration of Fe decreases and the concentration of Fe increases, the working electrode’s potential shifts toward more positive values until the oxidation of H O begins. \[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \text{ O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} \nonumber\] Because a portion of the total current comes from the oxidation of H O, the current efficiency for the analysis is less than 100% and we cannot use Equation \ref{11.1} to determine the amount of Fe in the sample. Although we cannot prevent the potential from drifting until another species undergoes oxidation, we can maintain a 100% current efficiency if the product of that secondary oxidation reaction both rapidly and quantitatively reacts with the remaining Fe . To accomplish this we add an excess of Ce to the analytical solution. As shown in Figure 11.3.6
, when the potential of the working electrode shifts to a more positive potential, Ce begins to oxidize to Ce \[\mathrm{Ce}^{3+}(a q) \rightleftharpoons \text{ Ce}^{4+}(a q)+e^{-} \label{11.9}\] The Ce that forms at the working electrode rapidly mixes with the solution where it reacts with any available Fe . \[\mathrm{Ce}^{4+}(a q)+\text{ Fe}^{2+}(a q) \rightleftharpoons \text{ Ce}^{3+}(a q)+\text{ Fe}^{3+}(a q) \label{11.10}\] Combining reaction \ref{11.9} and reaction \ref{11.10} shows that the net reaction is the oxidation of Fe to Fe \[\mathrm{Fe}^{2+}(a q) \rightleftharpoons \text{ Fe}^{3+}(a q)+e^{-} \nonumber\] which maintains a current efficiency of 100%. A species used to maintain 100% current efficiency is called a . Adding a mediator solves the problem of maintaining 100% current efficiency, but it does not solve the problem of determining when the analyte's electrolysis is complete. Using the analysis for Fe in Figure 11.3.6
, when the oxidation of Fe is complete current continues to flow from the oxidation of Ce , and, eventually, the oxidation of H O. What we need is a signal that tells us when no more Fe is present in the solution. For our purposes, it is convenient to treat a controlled-current coulometric analysis as a reaction between the analyte, Fe , and the mediator, Ce , as shown by reaction \ref{11.10}. This reaction is identical to a redox titration; thus, we can use the end points for a redox titration—visual indicators and potentiometric or conductometric measurements—to signal the end of a controlled-current coulometric analysis. For example, ferroin provides a useful visual endpoint for the Ce mediated coulometric analysis for Fe , changing color from red to blue when the electrolysis of Fe is complete. Reaction \ref{11.10} is the same reaction we used in to develop our understanding of redox titrimetry. Controlled-current coulometry normally is carried out using a two-electrode galvanostat, which consists of a working electrode and a counter electrode. The working electrode—often a simple Pt electrode—also is called the generator electrode since it is where the mediator reacts to generate the species that reacts with the analyte. If necessary, the counter electrode is isolated from the analytical solution by a salt bridge or a porous frit to prevent its electrolysis products from reacting with the analyte. Alternatively, we can generate the oxidizing agent or the reducing agent externally, and allow it to flow into the analytical solution. Figure 11.3.7
shows one simple method for accomplishing this. A solution that contains the mediator flows into a small-volume electrochemical cell with the products exiting through separate tubes. Depending upon the analyte, the oxidizing agent or the reducing reagent is delivered to the analytical solution. For example, we can generate Ce using an aqueous solution of Ce , directing the Ce that forms at the anode to our sample. shows an example of a manual galvanostat. Although a modern galvanostat uses very different circuitry, you can use and the accompanying discussion to understand how we can use the working electrode and the counter electrode to control the current. includes an optional reference electrode, but its presence or absence is not important if we are not interested in monitoring the working electrode’s potential. There are two other crucial needs for controlled-current coulometry: an accurate clock for measuring the electrolysis time, , and a switch for starting and stopping the electrolysis. An analog clock can record time to the nearest ±0.01 s, but the need to stop and start the electrolysis as we approach the endpoint may result in an overall uncertainty of ±0.1 s. A digital clock allows for a more accurate measurement of time, with an overall uncertainty of ±1 ms. The switch must control both the current and the clock so that we can make an accurate determination of the electrolysis time. A controlled-current coulometric method sometimes is called a coulometric titration because of its similarity to a conventional titration. For example, in the controlled-current coulometric analysis for Fe using a Ce mediator, the oxidation of Fe by Ce (reaction \ref{11.10}) is identical to the reaction in a redox titration. There are other similarities between controlled-current coulometry and titrimetry. If we combine Equation \ref{11.1} and Equation \ref{11.2} and solve for the moles of analyte, , we obtain the following equation. \[N_{A}=\frac{i}{n F} \times t_{e} \label{11.11}\] Compare Equation \ref{11.11} to the relationship between the moles of analyte, , and the moles of titrant, , in a titration \[N_{A}=N_{T}=M_{T} \times V_{T} \nonumber\] where and are the titrant’s molarity and the volume of titrant at the end point. In constant-current coulometry, the current source is equivalent to the titrant and the value of that current is analogous to the titrant’s molarity. Electrolysis time is analogous to the volume of titrant, and is equivalent to the a titration’s end point. Finally, the switch for starting and stopping the electrolysis serves the same function as a buret’s stopcock. For simplicity, we assumed above that the stoichiometry between the analyte and titrant is 1:1. The assumption, however, is not important and does not effect our observation of the similarity between controlled-current coulometry and a titration. Coulometry is used for the quantitative analysis of both inorganic and organic analytes. Examples of controlled-potential and controlled-current coulometric methods are discussed in the following two sections. The majority of controlled-potential coulometric analyses involve the determination of inorganic cations and anions, including trace metals and halides ions. Table 11.3.1
summarizes several of these methods. Rechnitz, G. A. , Macmillan: New York, 1963. Electrolytic reactions are written in terms of the change in the analyte’s oxidation state. The actual species in solution depends on the analyte. The ability to control selectivity by adjusting the working electrode’s potential makes controlled-potential coulometry particularly useful for the analysis of alloys. For example, we can determine the composition of an alloy that contains Ag, Bi, Cd, and Sb by dissolving the sample and placing it in a matrix of 0.2 M H SO along with a Pt working electrode and a Pt counter electrode. If we apply a constant potential of +0.40 V versus the SCE, Ag(I) deposits on the electrode as Ag and the other metal ions remain in solution. When electrolysis is complete, we use the total charge to determine the amount of silver in the alloy. Next, we shift the working electrode’s potential to –0.08 V versus the SCE, depositing Bi on the working electrode. When the coulometric analysis for bismuth is complete, we determine antimony by shifting the working electrode’s potential to –0.33 V versus the SCE, depositing Sb. Finally, we determine cadmium following its electrodeposition on the working electrode at a potential of –0.80 V versus the SCE. We also can use controlled-potential coulometry for the quantitative analysis of organic compounds, although the number of applications is significantly less than that for inorganic analytes. One example is the six-electron reduction of a nitro group, –NO , to a primary amine, –NH , at a mercury electrode. Solutions of picric acid—also known as 2,4,6-trinitrophenol, or TNP, a close relative of TNT—is analyzed by reducing it to triaminophenol. Another example is the successive reduction of trichloroacetate to dichloroacetate, and of dichloroacetate to monochloroacetate \[\text{Cl}_3\text{CCOO}^-(aq) + \text{H}_3\text{O}^+(aq) + 2 e^- \rightleftharpoons \text{Cl}_2\text{HCCOO}^-(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber\] \[\text{Cl}_2\text{HCCOO}^-(aq) + \text{ H}_3\text{O}^+(aq) + 2 e^- \rightleftharpoons \text{ ClH}_2\text{CCOO}^-(aq) + \text{ Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber\] We can analyze a mixture of trichloroacetate and dichloroacetate by selecting an initial potential where only the more easily reduced trichloroacetate reacts. When its electrolysis is complete, we can reduce dichloroacetate by adjusting the potential to a more negative potential. The total charge for the first electrolysis gives the amount of trichloroacetate, and the difference in total charge between the first electrolysis and the second electrolysis gives the amount of dichloroacetate. The use of a mediator makes a coulometric titration a more versatile analytical technique than controlled-potential coulometry. For example, the direct oxidation or reduction of a protein at a working electrode is difficult if the protein’s active redox site lies deep within its structure. A coulometric titration of the protein is possible, however, if we use the oxidation or reduction of a mediator to produce a solution species that reacts with the protein. Table 11.3.2
summarizes several controlled-current coulometric methods based on a redox reaction using a mediator. \(\textbf{Fe}(\mathbf{C N})_\textbf{6}^\textbf{4–}(a q)+\text{ Ce}^{4+}(a q) \rightleftharpoons \\ \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)+\text{ Ce}^{3+}(a q)\) Note: The electrochemically generated reagent and the analyte are shown in . For an analyte that is not easy to oxidize or reduce, we can complete a coulometric titration by coupling a mediator’s oxidation or reduction to an acid–base, precipitation, or complexation reaction that involves the analyte. For example, if we use H O as a mediator, we can generate H O at the anode \[6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 4 \mathrm{H}_{3} \text{O}^{+}(a q)+\text{ O}_{2}(g)+4 e^{-} \nonumber\] and generate OH at the cathode. \[2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \rightleftharpoons 2 \mathrm{OH}^{-}(a q)+\text{ H}_{2}(g) \nonumber\] If we carry out the oxidation or reduction of H O using the generator cell in , then we can selectively dispense H O or OH into a solution that contains the analyte. The resulting reaction is identical to that in an acid–base titration. Coulometric acid–base titrations have been used for the analysis of strong and weak acids and bases, in both aqueous and non-aqueous matrices. Table 11.3.3
summarizes several examples of coulometric titrations that involve acid–base, complexation, and precipitation reactions. In comparison to a conventional titration, a coulometric titration has two important advantages. The first advantage is that electrochemically generating a titrant allows us to use a reagent that is unstable. Although we cannot prepare and store a solution of a highly reactive reagent, such as Ag or Mn , we can generate them electrochemically and use them in a coulometric titration. Second, because it is relatively easy to measure a small quantity of charge, we can use a coulometric titration to determine an analyte whose concentration is too small for a conventional titration. The absolute amount of analyte in a coulometric analysis is determined using Faraday’s law (Equation \ref{11.1}) and the total charge given by Equation \ref{11.2} or by Equation \ref{11.3}. The following example shows the calculations for a typical coulometric analysis. To determine the purity of a sample of Na S O , a sample is titrated coulometrically using I as a mediator and \(\text{I}_3^-\) as the titrant. A sample weighing 0.1342 g is transferred to a 100-mL volumetric flask and diluted to volume with distilled water. A 10.00-mL portion is transferred to an electrochemical cell along with 25 mL of 1 M KI, 75 mL of a pH 7.0 phosphate buffer, and several drops of a starch indicator solution. Electrolysis at a constant current of 36.45 mA requires 221.8 s to reach the starch indicator endpoint. Determine the sample’s purity. As shown in , the coulometric titration of \(\text{S}_2 \text{O}_3^{2-}\) with \(\text{I}_3^-\) is \[2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\text{ I}_{3}^{-}(a q)\rightleftharpoons \text{ S}_{4} \mathrm{O}_{6}^{2-}(a q)+3 \mathrm{I}^{-}(a q) \nonumber\] The oxidation of \(\text{S}_2 \text{O}_3^{2-}\) to \(\text{S}_4 \text{O}_6^{2-}\) requires one electron per \(\text{S}_2 \text{O}_3^{2-}\) ( = 1). Combining Equation \ref{11.1} and Equation \ref{11.2}, and solving for the moles and grams of Na S O gives \[N_{A} =\frac{i t_{e}}{n F}=\frac{(0.03645 \text{ A})(221.8 \text{ s})}{\left(\frac{1 \text{ mol } e^{-}}{\text{mol Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}\right)\left(\frac{96487 \text{ C}}{\text{mol } e^{-}}\right)} =8.379 \times 10^{-5} \text{ mol Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \nonumber\] This is the amount of Na S O in a 10.00-mL portion of a 100-mL sample; thus, there are 0.1325 grams of Na S O in the original sample. The sample’s purity, therefore, is \[\frac{0.1325 \text{ g} \text{ Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}{0.1342 \text{ g} \text { sample }} \times 100=98.73 \% \text{ w} / \text{w } \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \nonumber\] Note that for Equation \ref{11.1} and Equation \ref{11.2} it does not matter whether \(\text{S}_2 \text{O}_3^{2-}\) is oxidized at the working electrode or is oxidized by \(\text{I}_3^-\). To analyze a brass alloy, a 0.442-g sample is dissolved in acid and diluted to volume in a 500-mL volumetric flask. Electrolysis of a 10.00-mL sample at –0.3 V versus a SCE reduces Cu to Cu, requiring a total charge of 16.11 C. Adjusting the potential to –0.6 V versus a SCE and completing the electrolysis requires 0.442 C to reduce Pb to Pb. Report the %w/w Cu and Pb in the alloy. The reduction of Cu to Cu requires two electrons per mole of Cu ( = 2). Using Equation \ref{11.1}, we calculate the moles and the grams of Cu in the portion of sample being analyzed. \[N_{C u}=\frac{Q}{n F}=\frac{16.11 \text{ C}}{\frac{2 \text{ mol } e^{-}}{\mathrm{mol} \text{ Cu}} \times \frac{96487 \text{ C}}{\text{ mol } e^{-}}}=8.348 \times 10^{-5} \text{ mol Cu} \nonumber\] \[8.348 \times 10^{-5} \text{ mol Cu} \times \frac{63.55 \text{ g Cu} }{\text{mol Cu}}=5.301 \times 10^{-3} \text{ g Cu} \nonumber\] This is the Cu from a 10.00 mL portion of a 500.0 mL sample; thus, the %/w/w copper in the original sample of brass is \[\frac{5.301 \times 10^{-3} \text{ g Cu} \times \frac{500.0 \text{ mL}}{10.00 \text{ mL}}}{0.442 \text{ g sample} } \times 100=60.0 \% \text{ w/w Cu} \nonumber\] For lead, we follow the same process; thus \[N_{\mathrm{Pb}}=\frac{Q}{n F}=\frac{0.422 \text{ C}}{\frac{2 \text{ mol } e^-}{\text{mol Pb}} \times \frac{96487 \text{ C}}{\text{mol } e^{-}}}=2.19 \times 10^{-6} \text{ mol Pb} \nonumber\] \[2.19 \times 10^{-6} \text{ mol Pb}\times \frac{207.2 \text{ g Pb} }{\text{mol Cu} }=4.53 \times 10^{-4} \text{ g Pb} \nonumber\] \[\frac{4.53 \times 10^{-4} \text{ g Pb} \times \frac{500.0 \text{ mL}}{10.00 \text{ mL}}}{0.442 \text{ g sample}} \times 100=5.12 \% \text{ w/w Pb} \nonumber\] The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of \(\text{Cr}_2 \text{O}_7^{2-}\) provides an instructive example of a typical procedure. The description here is based on Bassett, J.; Denney, R. C.; Jeffery, G. H.; Mendham, J. , Longman: London, 1978, p. 559–560. Thee concentration of \(\text{Cr}_2 \text{O}_7^{2-}\) in a sample is determined by a coulometric redox titration using Fe as a mediator and electrogenerated Fe as the titrant. The endpoint of the titration is determined potentiometrically. The electrochemical cell consists of a Pt working electrode and a Pt counter electrode placed in separate cells connected by a porous glass disk. Fill the counter electrode’s cell with 0.2 M Na SO , keeping the level above that of the solution in the working electrode’s cell. Connect a platinum electrode and a tungsten electrode to a potentiometer so that you can measure the working electrode’s potential during the analysis. Prepare a mediator solution of approximately 0.3 M NH Fe(SO ) . Add 5.00 mL of sample, 2 mL of 9 M H SO , and 10–25 mL of the mediator solution to the working electrode’s cell, and add distilled water as needed to cover the electrodes. Bubble pure N through the solution for 15 min to remove any O that is present. Maintain the flow of N during the electrolysis, turning if off momentarily when measuring the potential. Stir the solution using a magnetic stir bar. Adjust the current to 15–50 mA and begin the titration. Periodically stop the titration and measure the potential. Construct a titration curve of potential versus time and determine the time needed to reach the equivalence point. 1. Is the platinum working electrode the cathode or the anode? Reduction of Fe to Fe occurs at the working electrode, making it the cathode in this electrochemical cell. 2. Why is it necessary to remove dissolved oxygen by bubbling N through the solution? Any dissolved O will oxidize Fe back to Fe , as shown by the following reaction. \[4\text{Fe}^{2+}(aq) + \text{ O}_2 + \text{ 4H}_3\text{O}^+(aq) \rightleftharpoons 4\text{Fe}^{3+}(aq) + 6\text{H}_2\text{O}(l) \nonumber\] To maintain current efficiency, all the Fe must react with \(\text{Cr}_2 \text{O}_7^{2-}\). The reaction of Fe with O means that more of the Fe mediator is needed, increasing the time to reach the titration’s endpoint. As a result, we report the presence of too much \(\text{Cr}_2 \text{O}_7^{2-}\). 3. What is the effect on the analysis if the NH Fe(SO ) is contaminated with trace amounts of Fe ? How can you compensate for this source of Fe ? There are two sources of Fe : that generated from the mediator and that present as an impurity. Because the total amount of Fe that reacts with \(\text{Cr}_2 \text{O}_7^{2-}\) remains unchanged, less Fe is needed from the mediator. This decreases the time needed to reach the titration’s end point. Because the apparent current efficiency is greater than 100%, the reported concentration of \(\text{Cr}_2 \text{O}_7^{2-}\) is too small. We can remove trace amount of Fe from the mediator’s solution by adding H O and heating at 50–70 C until the evolution of O ceases, converting the Fe to Fe . Alternatively, we can complete a blank titration to correct for any impurities of Fe in the mediator. 4. Why is the level of solution in the counter electrode’s cell maintained above the solution level in the working electrode’s cell? This prevents the solution that contains the analyte from entering the counter electrode’s cell. The oxidation of H O at the counter electrode produces O , which can react with the Fe generated at the working electrode or the Cr resulting from the reaction of Fe and \(\text{Cr}_2 \text{O}_7^{2-}\). In either case, the result is a positive determinate error. One useful application of coulometry is determining the number of electrons involved in a redox reaction. To make the determination, we complete a controlled-potential coulometric analysis using a known amount of a pure compound. The total charge at the end of the electrolysis is used to determine the value of using Faraday’s law (Equation \ref{11.1}). A 0.3619-g sample of tetrachloropicolinic acid, C HNO Cl , is dissolved in distilled water, transferred to a 1000-mL volumetric flask, and diluted to volume. An exhaustive controlled-potential electrolysis of a 10.00-mL portion of this solution at a spongy silver cathode requires 5.374 C of charge. What is the value of for this reduction reaction? The 10.00-mL portion of sample contains 3.619 mg, or \(1.39 \times 10^{-5}\) mol of tetrachloropicolinic acid. Solving Equation \ref{11.1} for and making appropriate substitutions gives \[n=\frac{Q}{F N_{A}}=\frac{5.374 \text{ C}}{\left(96478 \text{ C/mol } e^{-}\right)\left(1.39 \times 10^{-5} \text{ mol } \mathrm{C}_{6} \mathrm{HNO}_{2} \mathrm{Cl}_{4}\right)} = 4.01 \text{ mol e}^-/\text{mol } \mathrm{C}_{6} \mathrm{HNO}_{2} \mathrm{Cl}_{4} \nonumber\] Thus, reducing a molecule of tetrachloropicolinic acid requires four electrons. The overall reaction, which results in the selective formation of 3,6-dichloropicolinic acid, is A coulometric method of analysis can analyze a small absolute amount of an analyte. In controlled-current coulometry, for example, the moles of analyte consumed during an exhaustive electrolysis is given by Equation \ref{11.11}. An electrolysis using a constant current of 100 μA for 100 s, for example, consumes only \(1 \times 10^{-7}\) mol of analyte if = 1. For an analyte with a molecular weight of 100 g/mol, \(1 \times 10^{-7}\) mol of analyte corresponds to only 10 μg. The concentration of analyte in the electrochemical cell, however, must be sufficient to allow an accurate determination of the endpoint. When using a visual end point, the smallest concentration of analyte that can be determined by a coulometric titration is approximately 10 M. As is the case for a conventional titration, a coulometric titration using a visual end point is limited to major and minor analytes. A coulometric titration to a preset potentiometric endpoint is feasible even if the analyte’s concentration is as small as 10 M, extending the analysis to trace analytes [Curran, D. J. “Constant-Current Coulometry,” in Kissinger, P. T.; Heineman, W. R., eds., , Marcel Dekker Inc.: New York, 1984, pp. 539–568]. In controlled-current coulometry, accuracy is determined by the accuracy with which we can measure current and time, and by the accuracy with which we can identify the end point. The maximum measurement errors for current and time are about ±0.01% and ±0.1%, respectively. The maximum end point error for a coulometric titration is at least as good as that for a conventional titration, and is often better when using small quantities of reagents. Together, these measurement errors suggest that an accuracy of 0.1%–0.3% is feasible. The limiting factor in many analyses, therefore, is current efficiency. A current efficiency of more than 99.5% is fairly routine, and it often exceeds 99.9%. In controlled-potential coulometry, accuracy is determined by current efficiency and by the determination of charge. If the sample is free of interferents that are easier to oxidize or reduce than the analyte, a current efficiency of greater than 99.9% is routine. When an interferent is present, it can often be eliminated by applying a potential where the exhaustive electrolysis of the interferents is possible without the simultaneous electrolysis of the analyte. Once the interferent is removed the potential is switched to a level where electrolysis of the analyte is feasible. The limiting factor in the accuracy of many controlled-potential coulometric methods of analysis is the determination of charge. With electronic integrators the total charge is determined with an accuracy of better than 0.5%. If we cannot obtain an acceptable current efficiency, an electrogravimetric analysis is possible if the analyte—and only the analyte—forms a solid deposit on the working electrode. In this case the working electrode is weighed before beginning the electrolysis and reweighed when the electrolysis is complete. The difference in the electrode’s weight gives the analyte’s mass. Precision is determined by the uncertainties in measuring current, time, and the endpoint in controlled-current coulometry or the charge in controlled-potential coulometry. Precisions of ±0.1–0.3% are obtained routinely in coulometric titrations, and precisions of ±0.5% are typical for controlled-potential coulometry. For a coulometric method of analysis, the calibration sensitivity is equivalent to in Equation \ref{11.1}. In general, a coulometric method is more sensitive if the analyte’s oxidation or reduction involves a larger value of . Selectivity in controlled-potential and controlled-current coulometry is improved by adjusting solution conditions and by selecting the electrolysis potential. In controlled-potential coulometry, the potential is fixed by the potentiostat, and in controlled-current coulometry the potential is determined by the redox reaction with the mediator. In either case, the ability to control the electrolysis potential affords some measure of selectivity. By adjusting pH or by adding a complexing agent, it is possible to shift the potential at which an analyte or interferent undergoes oxidation or reduction. For example, the standard-state reduction potential for Zn is –0.762 V versus the SHE. If we add a solution of NH , forming \(\text{Zn(NH}_3\text{)}_4^{2+}\), the standard state potential shifts to –1.04 V. This provides an additional means for controlling selectivity when an analyte and an interferent undergo electrolysis at similar potentials. Controlled-potential coulometry is a relatively time consuming analysis, with a typical analysis requiring 30–60 min. Coulometric titrations, on the other hand, require only a few minutes, and are easy to adapt to an automated analysis. Commercial instrumentation for both controlled-potential and controlled-current coulometry is available, and is relatively inexpensive. Low cost potentiostats and constant-current sources are available for approximately $1000. | 37,648 | 3,880 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/13%3A_Air/13.10%3A__Who_Pollutes_Who_Pays |
The Clean Air Act, which was last amended in 1990, requires EPA to set National Ambient Air Quality Standards (40 CFR part 50) for pollutants considered harmful to public health and the environment. The Clean Air Act identifies two types of national ambient air quality standards. provide public health protection, including protecting the health of "sensitive" populations such as asthmatics, children, and the elderly. provide public welfare protection, including protection against decreased visibility and damage to animals, crops, vegetation, and buildings. The EPA has set National Ambient Air Quality Standards for six principal pollutants, which are called criteria" air pollutants. Periodically, the standards are reviewed and may be revised. The current standards are listed in Table \(\Page {1}\). Units of measure for the standards are parts per million (ppm) by volume, parts per billion (ppb) by volume, and micrograms per cubic meter of air (µg/m ). Table \(\Page {2}\) list the sources and harmful effects of criteria pollutants. Visibility impairment Fine particles (PM ) are the main cause of reduced visibility (haze) in parts of the United States, including many of our treasured national parks and wilderness areas. Environmental damage Particles can be carried over long distances by wind and then settle on ground or water. Depending on their chemical composition, the effects of this settling may include: · making lakes and streams acidic · changing the nutrient balance in coastal waters and large river basins · depleting the nutrients in soil · damaging sensitive forests and farm crops · affecting the diversity of ecosystems · contributing to . Materials damage PM can stain and damage stone and other materials, including culturally important objects such as statues and monuments. Some of these effects are related to . NO and other NO interact with water, oxygen and other chemicals in the atmosphere to form acid rain. Acid rain harms sensitive ecosystems such as lakes and forests. NO in the atmosphere contributes to nutrient pollution in coastal waters. At high concentrations, gaseous SOx can harm trees and plants by damaging foliage and decreasing growth. SO and other sulfur oxides can contribute to acid rain which can harm sensitive ecosystems. www.epa.gov The likelihood of immediate reactions to air pollutants depends on several factors. Age and preexisting medical conditions are two important influences. Some sensitive individuals appear to be at greater risk for air pollution-related health effects, for example, those with pre-existing heart and lung diseases (e.g., heart failure/ischemic heart disease, asthma, emphysema, and chronic bronchitis), diabetics, older adults, and children. In other cases, whether a person reacts to a pollutant depends on individual sensitivity, which varies tremendously from person to person. Some people can become sensitized to biological pollutants after repeated exposures, and it appears that some people can become sensitized to chemical pollutants as well. Pollution has a cost. Manufacturing activities that cause air pollution impose health and clean-up costs on the whole of society, whereas the neighbors of an individual who chooses to fire-proof his home may benefit from a reduced risk of a fire spreading to their own homes. A manufacturing activity that causes air pollution is an example of a negative externality in production. A negative externality in production occurs “when a firm’s production reduces the well-being of others who are not compensated by the firm." For example, if a laundry firm exists near a polluting steel manufacturing firm, there will be increased costs for the laundry firm because of the dirt and smoke produced by the steel manufacturing firm. If external costs exist, such as those created by pollution, the manufacturer will choose to produce more of the product than would be produced if the manufacturer were required to pay all associated environmental costs. Because responsibility or consequence for self-directed action lies partly outside the self, an element of externalization is involved. If there are external benefits, such as in public safety, less of the good may be produced than would be the case if the producer were to receive payment for the external benefits to others. However, goods and services that involve negative externalities in production, such as those that produce pollution, tend to be over-produced and underpriced since the externality is not being priced into the market. Pollution can also create costs for the firms producing the pollution. Sometimes firms choose, or are forced by regulation, to reduce the amount of pollution that they are producing. The associated costs of doing this are called abatement costs, or marginal abatement costs if measured by each additional unit. In 2005 pollution abatement capital expenditures and operating costs in the US amounted to nearly $27 billion. ( ) | 4,998 | 3,881 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.25%3A_Rotary_Evaporation |
Be sure there is ice in the water circulator (if used). Fill a round bottomed flask no greater than half-full. Connect to the bump trap with a plastic clip. Lower the flask into the water bath to submerge the liquid (don't submerge the plastic clip). Turn on the vacuum source (vacuum will hiss). Rotate the flask at a moderate rate (one-third the maximum value). Close the stopcock in the apparatus (hissing will stop). Evaporate until solid forms or liquid level doesn't appear to change anymore, then evaporate an extra few minutes for good measure. To stop evaporation, reverse all steps: | 604 | 3,883 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.05%3A_Calorimetry |
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called . To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object. We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The of the temperature change depends on both the of thermal energy transferred ( ) and the of the object. Its heat capacity ( ) is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of are joules per degree Celsius (J/°C). Note that a degree Celsius is exactly the same as a Kelvin, so the heat capacities can be expresses equally well, and perhaps a bit more correctly in , as joules per Kelvin, J/K The change in temperature (\(ΔT\)) is \[ \Delta T = \dfrac{q}{C} \label{5.5.1} \] where is the amount of heat (in joules), is the heat capacity (in joules per degree Celsius), and \(ΔT\) is \(T_{final} − T_{initial}\) (in degrees Celsius). Note that \(ΔT\) is written as the final temperature minus the initial temperature. Since the scaling for Kelvin (K) and degrees Celsius (°C) are exactly the same, the \[ΔT = T_{final} − T_{initial} \nonumber \] is the same is either is used for temperature calculations, but make sure not to mix these two temperature units for \(T_{final}\) and \(T_{initial}\). The value of \(C\) is intrinsically a positive number, but \(ΔT\) and \(q\) can be either positive or negative, and they both must have the sign. If \(ΔT\) and \(q\) are positive, then . If \(ΔT\) and \(q\) are negative, then . The heat capacity of an object depends on both its and its . For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (\(C_p\)) is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of \(C_p\) are thus J/(mol•°C). The specific heat ( ) is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change either via : \[q = nC_pΔT \label{5.5.2} \] e \(n\) is the n \(C_p\) is the molar heat capacity or via : \[q = mC_sΔT \label{5.5.3} \] The specific heats of some common substances are given in Table \(\Page {1}\). Note that the specific heat values of most solids are less than 1 J/(g•°C), whereas those of most liquids are about 2 J/(g•°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g•°C), is one of the highest known. The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure \(\Page {1}\)). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1 °C. Consequently, the mechanism for maintaining our body temperature at about 37 °C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat. A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0 °C. During the course of the day, the temperature of the water rises to 38.0 °C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0 °C is 0.998 g/mL.) volume and density of water and initial and final temperatures amount of energy stored The mass of water is \[ \begin{align*} \text{mass of } \ce{H2O} &= 400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) \\[4pt] &= 3.99\times 10^{5}g\; \ce{H2O} \end{align*} \nonumber \] The temperature change (\(ΔT\)) is \[38.0 ^oC − 22.0 ^oC = +16.0^oC. \nonumber \] From Table \(\Page {1}\), the specific heat of water is 4.184 J/(g•°C). From Equation \(\ref{5.5.3}\), the heat absorbed by the water is thus \[ \begin{align*} q &=mC_{s}\Delta T \\[4pt] &= \left ( 3.99 \times 10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right ) \left ( 16.0 \; \cancel{^{o}C} \right ) \\[4pt] &= 2.67 \times 10^{7}\,J = 2.67 \times 10^{4}\,kJ \end{align*} \nonumber \] Both \(q\) and \(ΔT\) are positive, consistent with the fact that the water has absorbed energy. Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0 °C to 34.5 °C during the day? Assume that the specific heat of sandstone is the same as that of quartz (\(\ce{SiO2}\)) in Table \(\Page {1}\). \(2.7 \times 10^4 \,kJ\) Even though the mass of sandstone is more than six times the mass of the water in Example \(\Page {1}\), the amount of thermal energy stored is the same to two significant figures. When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process: \[q_{cold} + q_{hot} = 0 \label{5.5.4} \] The equation implies that the amount of heat that flows a warmer object is the same as the amount of heat that flows a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite: \[q_{cold} = −q_{hot} \label{5.5.5} \] Thus heat is conserved in any such process, consistent with the law of conservation of energy. Equation \(\ref{5.5.5}\) argues that the amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. Substituting for from Equation \(\ref{5.5.2}\) into Equation \(\ref{5.5.5}\) gives \[ \left [ mC_{s} \Delta T \right ] _{cold} + \left [ mC_{s} \Delta T \right ] _{hot}=0 \label{5.5.6} \] which can be rearranged to give \[ \left [ mC_{s} \Delta T \right ] _{cold} = - \left [ mC_{s} \Delta T \right ] _{hot} \label{5.5.7} \] When two objects initially at different temperatures are placed in contact, we can use Equation \(\ref{5.5.7}\) to calculate the final temperature if we know the chemical composition and mass of the objects. If a 30.0 g piece of copper pipe at 80.0 °C is placed in 100.0 g of water at 27.0 °C, what is the final temperature? Assume that no heat is transferred to the surroundings. mass and initial temperature of two objects final temperature Using Equation \(\ref{5.5.6}\) and writing \(ΔT\) as \(T_{final} − T_{initial}\) for both the copper and the water, substitute the appropriate values of \(m\), \(C_s\), and \(T_{initial}\) into the equation and solve for \(T_{final}\). \[ \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \nonumber \] Substituting the data provided in the problem and Table \(\Page {1}\) gives \[ \newcommand{\celsius}{\,^\circ\text{C}} \begin{align*} \left [ \left (30 \; g \right ) \left (0.385 \; J \right ) \left (T_{final} - 80.0 \celsius \right ) \right ] _{Cu} + \left [ (100\,g) (4.18\, J/\celsius) \left (T_{final} - 27.0 \celsius \right ) \right ] _{\ce{H2O}} &=0 \\[4pt] \left[T_{final}\left ( 11.6 \; J/ \celsius \right ) -924 \; J \right] + \left[ T_{final}\left ( 418.4 \; J/\celsius \right ) -11,300 \; J \right]&= 0 \\[4pt] T_{final}\left ( 430 \; J/\left ( g \cdot \celsius \right ) \right ) &= 12,224 \; J \\[4pt] T_{final} &= 28.4 \; \celsius \end{align*} \nonumber \] It is expected that the water will increase in temperature since the added copper pipe was hotter. However, the amount of the increase is not great and that is because there is more mass of water (three-fold), but mostly because the greater specific heat of water (order of magnitude) vs. copper. If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0 °C, what is the final temperature if no heat is transferred to the surroundings? 80.0°C A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0 °C. If the final temperature of the water is 24.0 °C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.) 90.6°C In Example \(\Page {1}\), radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that the thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (\(q_{rxn} < 0\)), then heat is absorbed by the calorimeter (\(q_{calorimeter} > 0\)) and its temperature increases. Conversely, if the reaction absorbs heat (\(q_{rxn} > 0\)), then heat is transferred from the calorimeter to the system (\(q_{calorimeter} < 0\)) and the temperature of the calorimeter decreases. In both cases, . The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants. Because \(ΔH\) is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter give \(ΔH\) values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a (Figure \(\Page {2}\)), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 °C). Because the heat released or absorbed at constant pressure is equal to \(ΔH\), the relationship between heat and \(ΔH_{rxn}\) is \[ \Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{5.5.8} \] The use of a constant-pressure calorimeter is illustrated in Example \(\Page {3}\). When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0 °C to 34.7 °C. The density of water in this temperature range averages 0.9969 g/cm . What is \(ΔH_{soln}\) (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. mass of substance, volume of solvent, and initial and final temperatures Δ To calculate Δ , we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is \[ \left (100.0 \; \cancel{mL}\; \ce{H2O} \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; \ce{KOH}=104.72 \; g \nonumber \] The temperature change is therefor \[(34.7\, ^oC − 23.0 \,^oC) = +11.7\, ^oC. \nonumber \] Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus \[ \begin{align*} q_{calorimater}&=mC_{s} \Delta T \\[4pt] &=\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right ) \\[4pt] &=5130 \; J = 5.13 \; kJ \end{align*} \nonumber \] The temperature of the solution increased because heat was absorbed by the solution (\(q > 0\)). Where did this heat come from? It was released by \(\ce{KOH}\) dissolving in water. From Equation \(\ref{5.5.1}\), we see that \[ΔH_{rxn} = −q_{calorimeter} = −5.13\, kJ \nonumber \] This experiment tells us that dissolving 5.03 g of \(\ce{KOH}\) in water is accompanied by the of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of \(\ce{KOH}\) in water must be exothermic. The last step is to use the molar mass of \(\ce{KOH}\) to calculate \(ΔH_{soln}\), i.e., the heat released when dissolving 1 mol of \(\ce{KOH}\): \[ \begin{align*} \Delta H_{soln} &= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right ) \\[4pt] &=-57.2 \; kJ/mol \end{align*} \nonumber \] A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7 °C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example \(\Page {3}\), find \(ΔH_{soln}\) for \(\ce{NH4Br }\) (in kilojoules per mole). 16.6 kJ/mol Conservation of Energy: Coffee Cup Calorimetry: Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter shown schematically in Figure \(\Page {3}\)). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in (\(ΔU\)) rather than the enthalpy change (\(ΔH\)); \(ΔU\) is related to \(ΔH\) by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that \(ΔU < ΔH\), the relationship between the measured temperature change and \(ΔH_{comb}\) is given in Equation \(\ref{5.5.9}\), where is the total heat capacity of the steel bomb and the water surrounding it: \[ \Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{5.5.9} \] To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (\(\ce{C6H5CO2H}\)) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its \(ΔH_{comb} = −26.38\, kJ/g\)). This value and the measured increase in temperature of the calorimeter can be used in Equation \(\ref{5.5.9}\) to determine \(C_{bomb}\). The use of a bomb calorimeter to measure the \(ΔH_{comb}\) of a substance is illustrated in Example \(\Page {4}\). The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08 °C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the \(ΔH_{comb}\) of glucose? mass and \(ΔT\) for combustion of standard and sample \(ΔH_{comb}\) of glucose The first step is to use Equation \(\ref{5.5.9}\) and the information obtained from the combustion of benzoic acid to calculate . We are given \(ΔT\), and we can calculate from the mass of benzoic acid: \[ \begin{align*} q_{comb} &= \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) \\[4pt] &= - 15.3 \; kJ \end{align*} \nonumber \] From Equation \(\ref{5.5.9}\), \[ \begin{align*} -C_{bomb} &= \dfrac{q_{comb}}{\Delta T} \\[4pt] &= \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} \\[4pt] &=- 7.34 \; kJ/^{o}C \end{align*} \nonumber \] According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose: \[ \begin{align*} q_{comb} &=-C_{bomb}\Delta T \\[4pt] &= \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right ) \\[4pt] &=- 26.7 \; kJ \end{align*} \nonumber \] Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the Δ of glucose is \[ \begin{align*} \Delta H_{comb} &=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right ) \\[4pt] &= -2780 \; kJ/mol \\[4pt] &=2.78 \times 10^{3} \; J/mol \end{align*} \nonumber \] This result is in good agreement (< 1% error) with the value of \(ΔH_{comb} = −2803\, kJ/mol\) that calculated using enthalpies of formation. When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75 °C is observed. When 1.932 g of methylhydrazine (CH NHNH ) is ignited in the same calorimeter, the temperature increase is 4.64 °C. Calculate the Δ of methylhydrazine, the fuel used in the maneuvering jets of the space shuttle. −1.30 × 10 kJ/mol Conservation of Energy: Bomb Calorimetry: is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the can be measured. A negative Δ means that heat flows from a system to its surroundings; a positive Δ means that heat flows into a system from its surroundings. Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called , which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a , which gives Δ values directly, or a , which operates at constant volume and is particularly useful for measuring enthalpies of combustion. | 20,233 | 3,884 |
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Isomers are different compounds that have the same molecular formula. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. For example, in the case of the C H hydrocarbons, most of the isomers are constitutional. Shorthand structures for four of these isomers are shown below with their IUPAC names. Note that the twelve atoms that make up these isomers are bonded in very different ways. As is true for all constitutional isomers, each different compound has a different IUPAC name. Furthermore, the molecular formula provides information about some of the structural features that must be present in the isomers. Since the formula C H has two fewer hydrogens than the four-carbon alkane butane (C H ), all the isomers having this composition must incorporate either a ring or a double bond. A fifth possible isomer of formula C H is CH CH=CHCH . This would be named 2-butene according to the IUPAC rules; however, a close inspection of this molecule indicates it has two possible structures. These isomers may be isolated as distinct compounds, having characteristic and different properties. They are shown here with the designations and . The bonding patterns of the atoms in these two isomers are essentially equivalent, the only difference being the relative configuration of the two methyl groups and the two associated hydrogen atoms about the double bond. In the isomer the methyl groups are on the same side; whereas they are on opposite sides in the isomer. Isomers that differ only in the spatial orientation of their component atoms are called stereoisomers. Stereoisomers always require that an additional nomenclature prefix be added to the IUPAC name in order to indicate their spatial orientation. Stereoisomers are also observed in certain disubstituted (and higher substituted) cyclic compounds. Unlike the relatively flat molecules of alkenes, substituted cycloalkanes must be viewed as three-dimensional configurations in order to appreciate the spatial orientations of the substituents. By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring, and a hatched line for bonds to atoms or groups located below the ring. As in the case of the 2-butene stereoisomers, disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as and . The stereoisomeric 1,2-dibromocyclopentanes below are an example. In general, if any two carbons in a ring have two different substituent groups (not counting other ring atoms) stereoisomerism is possible. This is similar to the substitution pattern that gives rise to stereoisomers in alkenes; indeed, one might view a double bond as a two-membered ring. Four other examples of this kind of stereoisomerism in cyclic compounds are shown below. If more than two ring carbons have different substituents (not counting other ring atoms) the stereochemical notation distinguishing the various isomers becomes more complex. However, we can always state the relationship of any two substituents using cis or trans. For example, in the trisubstitutued cyclohexane below, we can say that the methyl group is to the ethyl group, and to the chlorine. We can also say that the ethyl group is to the chlorine. We cannot, however, designate the entire molecule as a or trans . ), | 3,488 | 3,885 |
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The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry. We construct the periodic table by following the aufbau principle (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available . We use the orbital energy diagram of Figure 6.29, recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be . Here is a schematic orbital diagram for a hydrogen atom in its ground state: A neutral helium atom, with an atomic number of 2 ( = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1 orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore written as 1 , where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle. The next element is lithium, with = 3 and three electrons in the neutral atom. We know that the 1 orbital can hold two of the electrons with their spins paired. Figure 6.29 tells us that the next lowest energy orbital is 2 , so the orbital diagram for lithium is This electron configuration is written as 1 2 . The next element is beryllium, with = 4 and four electrons. We fill both the 1 and 2 orbitals to achieve a 1 2 electron configuration: When we reach boron, with = 5 and five electrons, we must place the fifth electron in one of the 2 orbitals. Because all three 2 orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1 2 2 : At carbon, with = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2 orbital that already has an electron, or should it go in one of the empty 2 orbitals? If it goes in an empty 2 orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2 orbitals are degenerate? Because of electron-electron repulsions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s rule (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1 2 2 , is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons. When we get to nitrogen ( = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is with three unpaired electrons. The electron configuration of nitrogen is thus 1 2 2 . At oxygen, with = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2 orbitals, which gives us two unpaired electrons and a 1 2 2 electron configuration. Because all the 2 orbitals are degenerate, it doesn’t matter which one has the pair of electrons. Similarly, fluorine has the electron configuration 1 2 2 : When we reach neon, with = 10, we have filled the 2 subshell, giving a 1 2 2 electron configuration: Notice that for neon, as for helium, all the orbitals through the 2 level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see. Electron Configuration of Atoms: As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1 2 2 electron configuration of neon ( = 10), so the electron configuration of sodium, with = 11, which is 1 2 2 3 , is written as [Ne]3 : Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electrons. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have: It is readily apparent that both sodium and lithium have one electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case. As we continue to build the eight elements of period 3, the 3 and 3 orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3 3 , corresponding to a filled valence shell. Draw an orbital diagram and use it to derive the electron configuration of phosphorus, = 15. What is its valence electron configuration? atomic number orbital diagram and valence electron configuration for phosphorus Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus. The additional five electrons are placed in the next available orbitals, which Figure 6.29 tells us are the 3 and 3 orbitals: Because the 3 orbital is lower in energy than the 3 orbitals, we fill it first: Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3 orbitals separately but with their spins aligned: The electron configuration is [Ne]3 3 . We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3 3 . Draw an orbital diagram and use it to derive the electron configuration of chlorine, = 17. What is its valence electron configuration? [Ne]3 3 ; 3 3 Definition of Valence Electrons: The general order in which orbitals are filled is depicted in Figure \(\Page {1}\). Subshells corresponding to each value of are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4 orbital is filled prior to the 3 orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4 , and the configuration of calcium is [Ar]4 . Five 3 orbitals are filled by the next 10 elements, the transition metals, followed by three 4 orbitals. Notice that the last member of this row is the noble gas krypton ( = 36), [Ar]4 3 4 = [Kr], which has filled 4 , 3 , and 4 orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5 , 4 , and 5 orbitals are filled sequentially. The sixth row of the periodic table will be different from the preceding two because the 4 orbitals, which can hold 14 electrons, are filled between the 6 and the 5 orbitals. The elements that contain 4 orbitals in their valence shell are the lanthanides. When the 6 orbitals are finally filled, we have reached the next (and last known) noble gas, radon ( = 86), [Xe]6 4 5 6 = [Rn]. In the last row, the 5 orbitals are filled between the 7 and the 6 orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive. Write the electron configuration of mercury ( = 80), showing all the inner orbitals. atomic number complete electron configuration Using the orbital diagram in Figure \(\Page {1}\) and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed. By placing the electrons in orbitals following the order shown in Figure \(\Page {1}\) and using the periodic table as a guide, we obtain After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure \(\Page {2}\), we need to fill the 6 (2 electrons), 4 (14 electrons), and 5 (10 electrons) orbitals. The result is mercury’s electron configuration: with a filled 5 subshell, a 6 4 5 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.) Although element 114 is not stable enough to occur in nature, two isotopes of element 114 were created for the first time in a nuclear reactor in 1999 by a team of Russian and American scientists. Write the complete electron configuration for element 114. 1 2 2 3 3 4 3 4 5 4 5 6 4 5 6 7 5 6 7 The electron configurations of the elements are presented in Figure \(\Page {3}\), which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by Figure \(\Page {1}\). Some of these anomalies occur as the 3 orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4 3 rather than the predicted [Ar]4 3 . Similarly, the observed electron configuration of copper is [Ar]4 3 instead of [Ar] 3 . The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled ( , , , ) or filled ( , , , ) subshell. Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to some of the additional anomalies indicated in Figure \(\Page {3}\). For example, cerium has an electron configuration of [Xe]6 4 5 , which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences. Additional stability is associated with half-filled or filled subshells. Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the ), which gives rise to a particular arrangement of electrons for each element (its ). says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the . | 11,734 | 3,886 |
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The atomic molecular theory provides us a particulate understanding of matter. Each element is characterized as consisting of identical, indestructible atoms with atomic weights which have been determined. Compounds consists of identical molecules, each made up from a specific number of atoms of each of the component elements. We also know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. Finally, we know that the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus than the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell. The shell model of the atom is a good start in understanding the differences in the chemical properties of the atoms of different elements. For example, we can understand the periodicity of chemical and physical properties from our model, since elements in the same group have the same number of electrons in the valence shell. However, there are many details missing from our description. Other than a very crude calculation of "distance" of the shells from the nucleus, we have no description of what the differences are between the electrons in different shells. What precisely is a "shell"? Most importantly, the arrangement of elements into groups and the periodicity of chemical properties both depend on the concept that a shell is "filled" by a certain number of electrons. Looking at the number of elements in each period, the number of electrons which fills a shell depends on which shell is being filled. In some cases, a shell is filled by eight electrons, in others, it appears to be 18 electrons. What determines how many electrons can "fit" in a shell? Why is there a limit at all? Finally, a closer look at the ionization energies in Figure 4.3 reveals that our shell model must be incomplete. Our model implies that the elements of the second period from Lithium to Neon have their valence electrons in the second shell. With increasing nuclear charge, the ionization energy of these atoms should increase from Lithium to Neon. As a general trend, this is true, but there are variations. Note that the ionization energy of Oxygen atoms is less than that of Nitrogen atoms. We need to pursue addition detail in our model of the structure of the atom. To begin, we need to know a little about light. All forms of electromagnetic radiation travel as an oscillating wave, with an electric field component perpendicular to a magnetic field component. As a wave, the radiation can be characterized by its "wavelength", symbolized as \(\lambda\), which is the distance between adjacent peaks in the wave. Different wavelengths correspond to different forms of electromagnetic radiation. For example, microwave radiation has wavelength in the range of \(10^{-2}\) to \(10^{-3}\) meters, whereas x-ray radiation has wavelength in the range \(10^{-9}\) to \(10^{-10}\) meters. Radiation which is visible to the human eye has wavelength in the very narrow range from \(3.8 \times 10^{7}\) to \(7.8 \times 10^{-7}\) meters. Radiation can also be characterized by the frequency of the electromagnetic wave, which is the number of peaks in the wave which pass a point in space per second. Frequency is symbolized by \(\nu\). The speed which light travels in a vacuum is the same for all forms of electromagnetic radiation, \(c = 2.997 \times 10^8 \: \frac{\text{m}}{\text{s}}\). As such, we can relate the frequency of light to the wavelength of light by the equation \[\lambda \left( \text{m} \right) \times \nu \left( \text{s}^{-1} \right) = c \left( \frac{\text{m}}{\text{s}} \right)\] The longer the wavelength \(\lambda\), the lower the frequency \(\nu\). This makes sense when we remember that light travels at a fixed speed. When the wavelength is longer, fewer peaks will pass a point in space in a second. From this equation, there is a specific relationship between frequency and wavelength, and either or both can be used to characterize the properties of radiation. With this background in hand, we can use our understanding of light to pursue more data about the energies of electrons in atoms. Ionization energies tell us how much energy is required to remove an electron from an atom, but do not tell us what happens if an electron changes its energy in an atom. To analyze this, we need a means to measure the energies gained or lost by an atom. One way to do so is to analyze the "spectrum" of an atom, which is the set of frequencies of light emitted by the atom. Since hydrogen is the simplest atom, we analyze the hydrogen spectrum first. We find that, if we pass a current of electricity through a sample of hydrogen gas, light is emitted. Careful analysis shows that, although some of this light is emitted by \(\ce{H_2}\) molecules, some of the light is also emitted by \(\ce{H}\) atoms. Since the light is a form of energy, then these \(\ce{H}\) atoms must release energy supplied to them by the electrons in the current. Most importantly, if we pass the light emitted by the hydrogen gas sample through a prism, we can separate the colors as in a rainbow, each with a characteristic frequency. The resultant image of separated colors is called the of hydrogen. We find in this experiment that there are only four frequencies (four colors) of light in the emission that are visible. The most intense of the lines in the spectrum is bright red, but there are blue and violet lines. It turns out that there are also many other frequencies of light emitted which are invisible to the human eye. Careful observation and analysis reveals that every frequency in the hydrogen atom spectrum can be predicted by a very simple formula, called the Rydberg equation: \[\nu = R \times \left( \frac{1}{n^2} - \frac{1}{m^2} \right)\] where \(R\) is the Rydberg constant \(\left( 3.29 \times 10^{15} \: \text{s}^{-1} \right)\). \(n\) and \(m\) are integers (1,2,3,...). Each choice of \(n\) and \(m\) predicts a single observed frequency in the hydrogen atom spectrum. The atoms of all elements emit radiation when energized in an electric current, and as do all molecules of all compounds. However, we find that the specific frequencies of light emitted are characteristic of each atom or molecule. In other words, the spectrum of each element is unique to each element or compound. As a result, the spectrum of each substance can be used to identify that substance. (Note that the Rydberg equation tells us only the spectrum of hydrogen.) Our interest is in the fact that the radiation emitted by an atom tells us about the amounts of energy which can be released by an atom. For a hydrogen atom, for example, these changes in energy must correspond to the amounts of energy which the electrons inside the atom can gain or lose. At this point, we need to relate the frequency of radiation emitted by an atom to the amount of energy lost by the electron in the atom. We thus examine some observations about the energy of radiation. When a light source is directed at a metal surface, it is found under many circumstances that electrons are ejected from the surface. This phenomenon is called the "photoelectric effect". These electrons can be collected to produce a usable electric current. (This effect has a variety of common practical applications, for example, in "electric eye" devices.) It is reasonable to expect that a certain amount of energy is required to liberate an electron from a metal surface, since the electron is attracted to the positively charged nuclei in the metal. Thus, in order for the electron to escape, the light must supply sufficient energy to the electron to overcome this attraction. The following experimental observations are found when studying the photoelectric effect. First, in order for the effect to be observed, the light must be of at least a minimum frequency which we call the , \(\nu_0\). This frequency is characteristic for a given metal. That is, it is the same value for each sample of that metal, but it varies from one metal to the next. For low frequency light, photoelectrons are not observed in any number, no matter how intense the light source is. For light with frequency above \(\nu_0\), the number of photoelectrons emitted by the metal (measured by the photoelectric current, \(\Phi\)) increases directly with the intensity of the light. These results are shown in Figure 5.1. a. b. Second, we can measure the energies of the electrons emitted by the metal. For a given metal, all photoelectrons have the same kinetic energy for a fixed frequency of light above \(\nu_0\). This fixed kinetic energy is independent of the intensity of the light source. As the frequency of the light is increased, the kinetic energy of the emitted electrons increases proportionally. These results are shown in Figure 5.2. a. b. Are these results surprising? To the physicists at the end of the nineteenth century, the answer was yes, very surprising indeed. They expected that the energy of the light source should be determined by its intensity. Hence, the energy required to eject a photoelectron should be supplied by light of high intensity, no matter how low the frequency of the radiation. Thus, there should be no threshold frequency, below which no electrons are emitted. Moreover, the kinetic energy of the electrons should increase with intensity, not with light frequency. These predictions are not observed, so the results are counter to physical intuition. We can account for these results in a straightforward but perhaps non-obvious manner. (Einstein provided the explanation in 1905.) Since the kinetic energy of the emitted photoelectrons increases proportionally with increases in the frequency of the light above the threshold frequency, we can conclude from conservation of total energy that the energy supplied by the light to the ejected electron must be proportional to its frequency: \(E \propto \nu\). This does not immediately account for the existence of the threshold frequency, though, since it would still seem to be the case that even low frequency light would possess high energy if the intensity were sufficient. By this reasoning, high intensity, low frequency light should therefore produce as many photoelectrons as are produced by low intensity, high frequency light. But this is not observed. This is a very challenging puzzle, and an analogy helps to reveal the subtle answer. Imagine trying to knock pieces out of a wall by throwing objects at it. We discover that, no matter how many ping pong balls we throw, we cannot knock out a piece of the wall. On the other hand, only a single bowling ball is required to accomplish the task. The results of this "experiment" are similar to the observations of the photoelectric effect: very little high frequency light can accomplish what an enormous amount of low frequency light cannot. The key to understanding our imaginary experiment is knowing that, although there are many more ping pong balls than bowling balls, it is only the impact of each individual particle with the wall which determines what happens. Reasoning from this analogy, we must conclude that the energy of the light is supplied in "bundles" or "packets" of constant energy, which we will call . We have already concluded that the light supplies energy to the electron which is proportional to the light frequency. Now we can say that the energy of each photon is proportional to the frequency of the light. The intensity of the light is proportional to the number of these packets. This now accounts for the threshold frequency in a straightforward way. For a photon to dislodge a photoelectron, it must have sufficient energy, by itself, to supply to the electron to overcome its attraction to the metal. Although increasing the intensity of the light does increase the total energy of the light, it does not increase the energy of an individual photon. Therefore, if the frequency of the light is too low, the photon energy is too low to eject an electron. Referring back to the analogy, we can say that a single bowling ball can accomplish what many ping pong balls cannot, and a single high frequency photon can accomplish what many low frequency photons cannot. The important conclusion for our purposes is that . The amount of energy in each photon is given by Einstein's equation, \[E = h \nu\] where \(h\) is a constant called Planck's constant. We can combine the observation of the hydrogen atom spectrum with our deduction that light energy is quantized into packets to reach an important conclusion. Each frequency of light in the spectrum corresponds to a particular energy of light and, therefore, to a particular energy by a hydrogen atom, since this light energy is quantized into packets. Furthermore, since only certain frequencies are observed, then only certain energy losses are possible. This is only reasonable if the energy of each hydrogen atom is restricted to certain specific values. If the hydrogen atom could possess any energy, then it could lose any amount of energy and emit a photon of any energy and frequency. But this is not observed. Therefore, the energy of the electron in a hydrogen atom must be restricted to certain . The Hydrogen atom spectrum also tells us what these energy levels are. Recall that the frequencies of radiation emitted by Hydrogen atoms are given by the Rydberg equation. Each choice of the positive integers \(n\) and \(m\) predicts a single observed frequency in the hydrogen atom spectrum. Each emitted frequency must correspond to an energy \(h \nu\) by Einstein's equation. This photon energy must be the between two energy levels for a hydrogen electron, since that is the amount of energy released by the electron moving from one level to the other. If the energies of the two levels are \(E_m\) and \(E_n\), then we can write that \[h \nu = E_m - E_n\] By comparing this to the Rydberg equation, each energy level must be given by the formula \[E_n = \left( -h \right) R \frac{1}{n^2}\] We can draw two conclusions. First, the electron in a hydrogen atom can exist only with certain energies, corresponding to motion in what we now call a state or an . Second, the energy of a state can be characterized by an integer , \(n = 1, 2, 3, ...\) which determines its energy. These conclusions are reinforced by similar observations of spectra produced by passing a current through other elements. Only specific frequencies are observed for each atom, although only the hydrogen frequencies obey the Rydberg formula. We conclude that the energies of electrons in atoms are "quantized", that is, restricted to certain values. We now need to relate this quantization of energy to the existence of shells, as developed in Module 4. The ionization energy of an atom tells us the energy of the electron or electrons which are at highest energy in the atom and are thus easiest to remove from the atom. To further analyze the energies of the electrons more tightly bound to the nucleus, we introduce a new experiment. The photoelectric effect can be applied to ionize atoms in a gas, in a process often called . We shine light on an atom and measure the minimum frequency of light, corresponding to a minimum energy, which will ionize an electron from an atom. When the frequency of light is too low, the photons in that light do not have enough energy to ionize electrons from an atom. As we increase the frequency of the light, we find a threshold at which electrons begin to ionize. Above this threshold, the energy \(h \nu\) of the light of frequency \(\nu\) is greater than the energy required to ionize the atom, and the excess energy is retained by the ionized electron as kinetic energy. In photoelectron spectroscopy, we measure the kinetic energy of the electrons which are ionized by light. This provides a means of measuring the ionization energy of the electrons. By conservation of energy, the energy of the light is equal to the ionization energy \(IE\) plus the kinetic energy \(KE\) of the ionized electron: \[h \nu = IE + KE\] Thus, if we use a known frequency \(n\) and measure \(KE\), we can determine \(IE\). The more tightly bound an electron is to the atom, the higher the ionization energy and the smaller the kinetic energy of the ionized electron. If an atom has more than one electron and these electrons have different energies, then for a given frequency of light, we can expect electrons to be ejected with different kinetic energies. The higher kinetic energies correspond to the weakly bound outer electrons, and the lower kinetic energies correspond to the tightly bound inner electrons. The ionization energies for the first twenty elements are given in Table 5.1. We note that there is a single ionization energy for hydrogen and helium. This is consistent with the shell model of these atoms since, in both of these atoms, the electron or electrons are in the innermost shell. The energies of these electrons correspond to the \(n = 1\) energy level of the hydrogen atom. In lithium and beryllium, there are two ionization energies. Again, this is consistent with the shell model, since now there are electrons in both of the first two shells. Note also that the ionization energy of the inner shell electrons increases as we go from hydrogen to lithium to beryllium, because of the increase in nuclear charge. The lower energy electrons correspond to the \(n = 1\) energy level of hydrogen and the higher energy electrons correspond to the \(n = 2\) energy level. Surprisingly, though, boron has three ionization energies, which does not seem consistent with the shell model. From the hydrogen atom energy levels, we would have expected that all \(n = 2\) electrons would have the same energy. We can note that the two smaller ionization energies in boron are comparable in magnitude and smaller by more than a factor of ten than the ionization energy of the electrons in the inner shell. Thus, the electrons in the outer \(n = 2\) shell apparently have comparable energies, but they are not identical. The separation of the second shell into two groups of electrons with two comparable but different energies is apparent for elements boron to neon. As such, we conclude from the experimental data that the second shell of electrons should be described as two with slightly different energies. For historical reasons, these subshells are referred to as the "\(2s\)" and "\(2p\)" subshells, with \(2s\) electrons slightly lower in energy than \(2p\) electrons. The energies of the \(2s\) and \(2p\) electrons decrease from boron to neon, consistent with the increase in the nuclear charge. Beginning with sodium, we observe four distinct ionization energies, and beginning with aluminum there are five. Note for these elements that the fourth and fifth ionization energies are again roughly a factor of ten smaller than the second and third ionization energies, which are in turn at least a factor of ten less than the first ionization energy. Thus, it appears that there are three shells of electrons for these atoms, consistent with our previous shell model. As with \(n = 2\), the \(n = 3\) shell is again divided into two subshells, now called the \(3s\) and \(3p\) subshells. These data also reveal how many electrons can reside in each subshell. In each \(n\) level, there are two elements which have only the ionization energy for the \(s\) subshell. Hence, \(s\) subshells can hold two electrons. By contrast, there are 6 elements which have both the \(s\) and \(p\) subshell ionization energies, so the \(p\) subshell can hold 6 electrons. The shell and subshell organization of electron energies can also be observed by measuring the "electron affinity" of the atoms. Electron affinity is the energy released when an electron is added to an atom: \[\ce{A} \left( g \right) + \ce{e^-} \left( g \right) \rightarrow \ce{A^-} \left( g \right)\] If there is a strong attraction between the atom \(\ce{A}\) and the added electron, then a large amount of energy is released during this reaction, and the electron affinity is a large positive number. (As a note, this convention is the opposite of the one usually applied for energy changes in reactions: exothermic reactions, which give off energy, conventionally have negative energy changes.) The electron affinities of the halogens are large positive values: the electron affinities of \(\ce{F}\), \(\ce{Cl}\), and \(\ce{Br}\) are \(328.0 \: \text{kJ/mol}\), \(348.8 \: \text{kJ/mol}\), and \(324.6 \: \text{kJ/mol}\). Thus, the attached electrons are strongly attracted to the nucleus in each of these atoms. This is because there is room in the current subshell to add an additional electron, since each atom has 5 \(p\) electrons, and the core charge felt by the electron in that subshell is large. By contrast, the electron affinities of the inert gases are : the addition of an electron to an inert gas atom actually requires the of energy, in effect, to force the electron into place. This is because the added electron cannot fit in the current subshell and must be added to a new shell, farther from the nucleus. As such, the core charge felt by the added electron is very close to zero. Similarly, the electron affinities of the elements \(\ce{Be}\), \(\ce{Mg}\), and \(\ce{Ca}\) are all negative. This is again because the \(s\) subshell in these atoms already has two electrons, so the added electron must go into a higher energy subshell with a much smaller core charge. We now have a fairly detailed description of the energies of the electrons in atoms. What we do not have is a model which tells us what factors determine the energy of an electron in a shell or subshell. Nor do we have a model to explain why these energies are similar but different for electrons in different subshells. A complete answer to these questions requires a development of the quantum theory of electron motion in atoms. Because the postulates of this quantum theory cannot be readily developed from experimental observations, we will concern ourselves with a few important conclusions only. The first important conclusion is that the motion of an electron in an atom is described by a wave function. Interpretation of the wave motion of electrons is a very complicated proposition, and we will only deal at present with a single important consequence, namely the . A characteristic of wave motion is that, unlike a particle, the wave does not have a definite position at a single point in space. By contrast, the location of a particle is precise. Therefore, since an electron travels as a wave, we must conclude that we cannot determine the precise location of the electron in an atom. This is, for our purposes, the uncertainty principle of quantum mechanics. We make measurements of the location of the electron, but we find that each measurement results in a different value. We are then forced to accept that we cannot determine the precise location. We are allowed, however, to determine a for where the electron is observed. This probability distribution is determined by quantum mechanics. The motion of the electron in a hydrogen atom is described by a function, often called the or the and typically designated by the symbol \(\Psi\). \(\Psi\) is a function of the position of the electron \(r\), and quantum mechanics tells us that \(\left( \left| \Psi \right| \right)^2\) is the of observing the electron at the location \(r\). Each electron orbital has an associated constant value of the electronic energy, \(\ce{E_n}\), in agreement with our earlier conclusions. In fact, quantum mechanics exactly predicts the energy shells and the hydrogen atom spectrum we observe. The energy of an electron in an orbital is determined primarily by two characteristics of the orbital. The first, rather intuitive, property determines the average potential energy of the electron: an orbital which has substantial probability in regions of low potential energy will have a low total energy. By Coulomb's law, the potential energy arising from nucleus-electron attraction is lower when the electron is nearer the nucleus. In atoms with more than one electron, electron-electron repulsion also contributes to the potential energy, as Coulomb's law predicts an increase in potential energy arising from the repulsion of like charges. A second orbital characteristic determines the contribution of kinetic energy, via a more subtle effect arising out of quantum mechanics. As a consequence of the uncertainty principle, quantum mechanics predicts that, the more confined an electron is to a smaller region of space, the higher must be its average kinetic energy. Since we cannot measure the position of the electron precisely, we define the uncertainty in the measurement as \(\Delta \left( x \right)\). Quantum mechanics also tells us that we cannot measure the momentum of an electron precisely either, so there is an uncertainty \(\Delta \left( p \right)\) in the momentum. In mathematical detail, the uncertainty principle states that these uncertainties are related by an inequality: \[\Delta \left( x \right) \Delta \left( p \right) \ge \frac{h}{4 \pi}\] where \(h\) is Planck's constant, \(6.62 \times 10^{-34} \: \text{J} \cdot \text{s}\) (previously seen in Einstein's equation for the energy of a photon). This inequality reveals that, when an electron moves in a small area with a correspondingly small uncertainty \(\Delta \left( x \right)\), the uncertainty in the momentum \(\Delta \left( p \right)\) must be large. For \(\Delta \left( p \right)\) to be large, the momentum must also be large, and so must be the kinetic energy. Therefore, the more compact an orbital is, the higher will be the average kinetic energy of an electron in that orbital. This extra kinetic energy, which can be regarded as the , is comparable in magnitude to the average potential energy of electron-nuclear attraction. Therefore, in general, an electron orbital provides a compromise, somewhat localizing the electron in regions of low potential energy but somewhat delocalizing it to lower its confinement energy. We need to account for the differences in energies of the electrons in different subshells, since we know that, in a Hydrogen atom, the orbital energy depends only on the \(n\) quantum number. We recall that, in the Hydrogen atom, there is a electron. The energy of that electron is thus entirely due to its kinetic energy and its attraction to the nucleus. The situation is different in all atoms containing more than one electron, because the energy of the electrons is affected by their mutual repulsion. This repulsion is very difficult to quantify, but our model must take it into account. A simple way to deal with the effect of electron-electron repulsion is to examine the shell structure of the atom. The two \(n = 1\) electrons in beryllium are in a shell with a comparatively short average distance from the nucleus. Therefore, the two \(n = 2\) electrons are in a shell which is, on average, "outside" of the \(n = 1\) shell. The \(n = 1\) electrons are thus the "core" and the \(n = 2\) electrons are in the valence shell. This structure allows us to see in a simple way the effect of electron-electron repulsion on the energies of the \(n = 2\) electrons. Each \(n = 2\) electron is attracted by the +4 charge on the tiny beryllium nucleus, but is repelled by the two -1 charges from the inner shell formed by the two \(n = 1\) electrons. Net, then, an \(n = 2\) electron effectively "sees" roughly a +2 nuclear charge. We refer to this +2 as the "core charge" since it is the net charge on the core resulting from the balance of attraction to the nucleus and repulsion from the core electrons. The nucleus is partially "shielded" from the valence electrons by the core electrons. This shielding effect does not seem to account for the difference in ionization energies between the \(2s\) and \(2p\) or for the lower ionization energy of boron compared to beryllium, since, in each atom, the valence electrons are in the \(n = 2\) shell. However, the shielding effect is not perfect. Recall that we only know the for observing the positions of the electrons. Therefore, we cannot definitely state that the \(n = 2\) electrons are outside of the \(n = 1\) core. In fact, there is some probability that an \(n = 2\) electron might be found inside the \(n = 1\) core, an effect called "core penetration". When an \(n = 2\) electron is very strongly attracted to the nucleus and its energy is thus lowered. What is the extent of this penetration? We must consult quantum theory. The answer is in Figure 5.3, which shows the probability of finding an electron a distance \(r\) away from the nucleus for each of the \(1s\), \(2s\), and \(2p\) orbitals. We can see that there is a greater probability (though small) for the \(2s\) electron to penetrate the core than for the \(2p\) electron to do so. As a result of the core penetration, an electron in a \(2s\) orbital feels a greater "effective nuclear charge" than just the core charge, which was approximated by assuming perfect shielding. Thus the effective nuclear charge for a \(2s\) electron is greater than the effective nuclear charge for a \(2p\) electron. Therefore, the energy of an electron in the \(2s\) orbital in beryllium is lower than it would be in the \(2p\) orbital. A detailed analysis from quantum mechanics gives the following ordering of orbitals in order of increasing energy: \[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < \ldots\] This ordering can be rationalized on the basis of effective nuclear charge, shielding, and core penetration. The photoelectric effect demonstrates that radiation energy is quantized into "packets" or photons. Explain how and why this observation is of significance in understanding the structure of atoms. Explain how we can know that higher frequency light contains higher energy photons. Electron affinity is the energy released when an electron is attached to an atom. If an atom has a positive electron affinity, the added electron is attracted to the nucleus to form a stable negative ion. Why doesn't a Beryllium atom have a positive electron affinity? Explain how this demonstrates that the energy of a \(2s\) orbital is less than the energy of a \(2p\) orbital. Why does an inert gas atom have a high ionization energy but a low electron affinity? Why do these properties combine to make the atoms of inert gases unreactive? Consider electrons from two different subshells in the same atom. In photoelectron spectroscopy, the lower energy electron has a higher ionization energy but is observed to have lower kinetic energy after ionization. Reconcile the lower kinetic energy with the higher ionization energy. Chlorine atoms have 5 distinct ionization energies. Explain why. Predict the number of ionization energies for Bromine atoms, and explain you answer. (Hint: examine the structure of the periodic table.) Why does a Bromine atom have a much smaller radius than a Potassium atom, even though a \(\ce{Br}\) atom has 16 more electrons than does a \(\ce{K}\) atom? Explain why electrons confined to smaller orbitals are expected to have higher kinetic energies. Define "shielding" in the context of electron-electron repulsion. What is the significance of shielding in determining the energy of an electron? How is this affected by core penetration? ; Chemistry) | 31,686 | 3,887 |
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The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases. Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a (impulse) from molecules to the walls (Figure \(\Page {2}\)). The to the wall perpendicular to \(x\) axis as a molecule with an initial velocity \(u_x\) in \(x\) direction hits is expressed as: \[\Delta p_x=2mu_x \label{6.7.1}\] The , a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume. \[f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{6.7.2}\] The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency. \[P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{6.7.3}\] At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace \(u_x^2\) in the expression above with the average value of \(u_x^2\), which is denoted by \(\overline{u_x^2}\). The overbar designates the average value over all molecules. The exact expression for pressure is given as : \[P=\dfrac{N}{V}m\overline{u_x^2} \label{6.7.4}\] Finally, we must consider that there is nothing special about \(x\) direction. We should expect that \(\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}\). Here the quantity \(\overline{u^2}\) is called the defined as the average value of square-speed (\(u^2\)) over all molecules. Since \(u^2=u_x^2+u_y^2+u_z^2\) for each molecule, \(\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}\). By substituting \(\dfrac{1}{3}\overline{u^2}\) for \(\overline{u_x^2}\) in the expression above, we can get the final expression for the pressure: \[P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{6.7.5}\] Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules. Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the kinetic energy of the molecules of a gas \((\overline{e_K})\), which can be represented as \((T)\) \[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{6.7.6}\] where \(N_A\) is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by \(N_A\): \[N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{6.7.7}\] where \(M\) is the molar mass of the gas molecules and is related to the molecular mass by \(M=N_Am\). By rearranging the equation, we can get the relationship between the root-mean square speed (\(u_{\rm rms}\)) and the temperature. The rms speed (\(u_{\rm rms}\)) is the square root of the sum of the squared speeds divided by the number of particles: \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{6.7.8}\] where \(N\) is the number of particles and \(u_i\) is the speed of particle \(i\). The relationship between \(u_{\rm rms}\) and the temperature is given by: \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{6.7.9}\] In this equation, \(u_{\rm rms}\) has units of meters per second; consequently, the units of molar mass \(M\) are kilograms per mole, temperature \(T\) is expressed in kelvins, and the ideal gas constant \(R\) has the value 8.3145 J/(K•mol). The equation shows that \(u_{\rm rms}\) of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones. At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed. The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)). particle speeds average speed (\(v_{\rm av}\)), root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)) Use Equation 6.7.6 to calculate the average speed and Equation 6.7.8 to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving. The average speed is the sum of the speeds divided by the number of particles: \[v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s\] The rms speed is the square root of the sum of the squared speeds divided by the number of particles: \[v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s\] The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence \(v_{\rm m}=6.0\) m/s. The \(v_{\rm rms}\) of the particles, which is related to the average kinetic energy, is greater than their average speed. At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in Chapter 14 "Chemical Kinetics". This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure \(\Page {1}\). Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure \(\Page {1}\) were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. The temperature of a 4.75 L container of N gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the temperatures and volume effect of increase in temperature Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas. A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases Kinetic-Molecular Theory of Gases: The behavior of ideal gases is explained by the . Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same . The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a , in which some molecules have higher or lower speeds (and kinetic energies) than average. | 9,836 | 3,888 |
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Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula \(C_nH_{2n+2}\) and can be subdivided into the following three groups: the , , and . Alkanes are also . When we smell something, the information travels to us via molecules, and almost always these are organic molecules. These molecules can be detected by a variety of organs, including noses in dogs and antennae in crickets, but no matter what organ is sensing the smell, one of the crucial factors in determining how an organism reacts to a compound is the shape of the molecule. The sense of smell depends on thousands of different receptors in the organ, working conceptually on a lock-and-key basis: a molecule with a given shape can fit into a given receptor, and when it does, a signal is sent telling the nervous system that the organism has encountered that particular type of molecule, and the organism reacts appropriately. What gives a molecule its shape? Given a structural formula, could you determine the shape of the corresponding molecule in three dimensions? Could you predict its biological activity, including not only its smell, but also a host of other behaviors linked to the shape of molecular messengers, such as anti-cancer activity or narcotic properties? These questions are at the cutting edge of biological chemistry. Although they are best answered through computer modeling, we can develop some of the qualitative ideas used in these models. In the eclipsed conformation of ethane the smallest dihedral angle is 0º, whereas in the staggered conformation it is 60º, which can be seen more clearly in the respective Newman projection. In the case of ethane, conformational changes are very subtle, but in others they are more obvious. Butane (CH CH CH CH ) has four tetrahedral carbons and three carbon-carbon bonds connecting them together. Let's number the carbons along the chain C1, C2, C3 and C4. Rotating around the C-C bonds connected to the terminal carbons -- C1-C2 and C3-C4 -- only subtle changes in shape would be apparent. However, rotating about C2-C3 produces some pretty obvious shape changes. Pay attention to where the two methyl groups are with respect to each other. If we call C1-C2-C3-C4 the dihedral angle, then at 0 degrees the molecule is in an eclipsed conformation, apparent when looking at the Newman projection. Looked at from other vantage points, the molecule is curled up into the shape of a letter C (Figure \(\Page {2}\)). At 60 degrees, the molecule is no longer eclipsed, and just as in ethane the energy is a little bit lower, but the overall shape when viewed from the side is still a sort of twisted C (Figure \(\Page {3}\)). At 120 degrees the molecule is eclipsed again, but from the side it has now twisted almost into a shape like the letter Z ( Figure \(\Page {4}\)). At 180 degrees, the molecule is staggered again and has settled into a regular, zig-zag, letter Z shape. These conformations of butane are really pretty different. Which shape would the molecule prefer? From what we learned about ethane, we could probably rule out the eclipsed conformations. Each of those would have 3 kcal/mol of torsional strain. However, there is another factor that destabilizes the initial conformation at 0 degrees, in favor of the Z-shaped one with the methyl groups 180 degrees apart from each other. This factor is called "sterics" and it refers to the idea that molecules, or parts of molecules, take up space, and so two parts of the butane can't occupy the same place at the same time. Put more simply, sterics refers to crowdedness. When the two methyl groups in butane are too close together, they are too crowded, and they are at higher energy. When they get farther apart, crowding subsides and the energy in the molecule goes down. For butane, that means getting those two methyl groups away from each other and keeping the bonds staggered. There is some additional jargon that is used to describe these butane conformations: An important way to generate alkanes is the hydrogenation of an alkene (discussed later) The double bond of an consists of a ( ) bond and a pi ( ) bond. Because the carbon-carbon bond is relatively weak, it is quite reactive and can be easily broken and reagents can be added to carbon. Reagents are added through the formation of single bonds to carbon in an addition reaction. An example of an alkene addition reaction is a process called hydrogenation.In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated . Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability. Although the hydrogenation of an alkene is a thermodynamically favorable reaction, it will not proceed without the addition of a . Common catalysts used are insoluble metals such as palladium and platinum. Hydrogenation reactions are extensively used to create commercial goods.Hydrogenation is used in the food industry to make a large variety of manufactured goods, like spreads and shortenings, from liquid oils. This process also increases the chemical stability of products and yields semi-solid products like margarine. Hydrogenation is also used in coal processing. Solid coal is converted to a liquid through the addition of hydrogen. Liquefying coal makes it available to be used as fuel. , | 5,747 | 3,889 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.06%3A_Critical_Values_for_Dixon's_Q-Test |
The following table provides critical values for \(Q(\alpha, n)\), where \(\alpha\) is the probability of incorrectly rejecting the suspected outlier and \(n\) is the number of samples in the data set. There are several versions of Dixon’s -Test, each of which calculates a value for where is the number of suspected outliers on one end of the data set and is the number of suspected outliers on the opposite end of the data set. The critical values for here are for a single outlier, , where \[Q_\text{exp} = Q_{10} = \frac {|\text{outlier's value} - \text{nearest value}|} {\text{largest value} - \text{smallest value}} \nonumber\] The suspected outlier is rejected if is greater than \(Q(\alpha, n)\). For additional information consult Rorabacher, D. B. “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s ‘ ’ Parameter and Related Subrange Ratios at the 95% confidence Level,” , , 139–146. | 947 | 3,890 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/10%3A_Spectroscopic_Methods/10.05%3A_Emission_Spectroscopy |
An analyte in an excited state possesses an energy, , that is greater than its energy when it is in a lower energy state, . When the analyte returns to its lower energy state—a process we call relaxation—the excess energy, \(\Delta E\) \[\Delta E=E_{2}-E_{1} \nonumber\] is released. shows a simplified picture of this process. The amount of time an analyte, , spends in its excited state—what we call the excited state's —is short, typically 10 –10 s for an electronic excited state and 10–15 s for a vibrational excited state. Relaxation of the analyte's excited state, *, occurs through several mechanisms, including collisions with other species in the sample, photochemical reactions, and the emission of photons. In the first process, which we call vibrational relaxation or nonradiative relaxation, the excess energy is released as heat. \[A^{*} \longrightarrow A+\text { heat } \nonumber\] Relaxation by a photochemical reaction may involve simple decomposition \[A^{*} \longrightarrow X+Y \nonumber\] or a reaction between * and another species \[A^{*}+Z \longrightarrow X+Y \nonumber\] In both cases the excess energy is used up in the chemical reaction or released as heat. In the third mechanism, the excess energy is released as a photon of electromagnetic radiation. \[A^{*} \longrightarrow A+h \nu \nonumber\] The release of a photon following thermal excitation is called emission and that following the absorption of a photon is called photoluminescence. In chemiluminescence and bioluminescence, excitation results from a chemical or a biochemical reaction, respectively. Spectroscopic methods based on photoluminescence are the subject of the next section and atomic emission is covered in . | 1,732 | 3,891 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.05%3A_Precipitation_Titrations |
Thus far we have examined titrimetric methods based on acid–base, complexation, and oxidation–reduction reactions. A reaction in which the analyte and titrant form an insoluble precipitate also can serve as the basis for a titration. We call this type of titration a . One of the earliest precipitation titrations—developed at the end of the eighteenth century—was the analysis of K CO and K SO in potash. Calcium nitrate, Ca(NO ) , was used as the titrant, which forms a precipitate of CaCO and CaSO . The titration’s end point was signaled by noting when the addition of titrant ceased to generate additional precipitate. The importance of precipitation titrimetry as an analytical method reached its zenith in the nineteenth century when several methods were developed for determining Ag and halide ions. A precipitation titration curve follows the change in either the titrand’s or the titrant’s concentration as a function of the titrant’s volume. As we did for other titrations, we first show how to calculate the titration curve and then demonstrate how we can sketch a reasonable approximation of the titration curve. Let’s calculate the titration curve for the titration of 50.0 mL of 0.0500 M NaCl with 0.100 M AgNO . The reaction in this case is \[\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightleftharpoons \text{AgCl}(s) \nonumber\] Because the reaction’s equilibrium constant is so large \[K = (K_\text{sp})^{-1} = (1.8 \times 10^{-10})^{-1} = 5.6 \times 10^9 \nonumber\] we may assume that Ag and Cl react completely. By now you are familiar with our approach to calculating a titration curve. The first task is to calculate the volume of Ag needed to reach the equivalence point. The stoichiometry of the reaction requires that \[\text{mol Ag}^+ = M_\text{Ag}V_\text{Ag} = M_\text{Cl}V_\text{Cl} = \text{mol Cl}^- \nonumber\] Solving for the volume of Ag \[V_{eq} = V_\text{Ag} = \frac{M_\text{Cl}V_\text{Cl}}{M_\text{Ag}} = \frac{(0.0500 \text{ M})(50.0 \text{ mL})}{0.100 \text{ M}} = 25.0 \text{ mL} \nonumber\] shows that we need 25.0 mL of Ag to reach the equivalence point. Before the equivalence point the titrand, Cl , is in excess. The concentration of unreacted Cl after we add 10.0 mL of Ag , for example, is \[[\text{Cl}^-] = \frac{(\text{mol Cl}^-)_\text{initial} - (\text{mol Ag}^+)_\text{added}}{\text{total volume}} = \frac{M_\text{Cl}V_\text{Cl} - M_\text{Ag}V_\text{Ag}}{V_\text{Cl} + V_\text{Ag}} \nonumber\] \[[\text{Cl}^-] = \frac{(0.0500 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 2.50 \times 10^{-2} \text{ M} \nonumber\] which corresponds to a pCl of 1.60. At the titration’s equivalence point, we know that the concentrations of Ag and Cl are equal. To calculate the concentration of Cl we use the for AgCl; thus \[K_\text{sp} = [\text{Ag}^+,\text{Cl}^-] = (x)(x) = 1.8 \times 10^{-10} \nonumber\] Solving for gives [Cl ] as \(1.3 \times 10^{-5}\) M, or a pCl of 4.89. After the equivalence point, the titrant is in excess. We first calculate the concentration of excess Ag and then use the expression to calculate the concentration of Cl . For example, after adding 35.0 mL of titrant \[[\text{Ag}^+] = \frac{(\text{mol Ag}^+)_\text{added} - (\text{mol Cl}^-)_\text{initial}}{\text{total volume}} = \frac{M_\text{Ag}V_\text{Ag} - M_\text{Cl}V_\text{Cl}}{V_\text{Ag} + V_\text{Cl}} \nonumber\] \[[\text{Ag}^+] = \frac{(0.100 \text{ M})(35.0 \text{ mL}) - (0.0500 \text{ M})(50.0 \text{ mL})}{35.0 \text{ mL} + 50.0 \text{ mL}} = 1.18 \times 10^{-2} \text{ M} \nonumber\] \[[\text{Cl}^-] = \frac{K_\text{sp}}{[\text{Ag}^+]} = \frac{1.8 \times 10^{-10}}{1.18 \times 10^{-2}} = 1.5 \times 10^{-8} \text{ M} \nonumber\] or a pCl of 7.81. Additional results for the titration curve are shown in Table 9.5.1
and Figure 9.5.1
. When calculating a precipitation titration curve, you can choose to follow the change in the titrant’s concentration or the change in the titrand’s concentration. Calculate the titration curve for the titration of 50.0 mL of 0.0500 M AgNO with 0.100 M NaCl as pAg versus , and as pCl versus . The first task is to calculate the volume of NaCl needed to reach the equivalence point; thus \[V_{eq} = V_\text{NaCl} = \frac{M_\text{Ag}V_\text{Ag}}{M_\text{NaCl}} = \frac{(0.0500 \text{ M})(50.0 \text{ mL})}{0.100 \text{ M}} = 25.0 \text{ mL} \nonumber\] Before the equivalence point the titrand, Ag , is in excess. The concentration of unreacted Ag after adding 10.0 mL of NaCl, for example, is \[[\text{Ag}^+] = \frac{(0.0500 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 2.50 \times 10^{-2} \text{ M} \nonumber\] which corresponds to a pAg of 1.60. To find the concentration of Cl we use the for AgCl; thus \[[\text{Cl}^-] = \frac{K_\text{sp}}{[\text{Ag}^+]} = \frac{1.8 \times 10^{-10}}{2.50 \times 10^{-2}} = 7.2 \times 10^{-9} \text{ M} \nonumber\] or a pCl of 8.14. At the titration’s equivalence point, we know that the concentrations of Ag and Cl are equal. To calculate their concentrations we use the sp expression for AgCl; thus \[K_\text{sp} = [\text{Ag}^+,\text{Cl}^-] = (x)(x) = 1.8 \times 10^{-10} \nonumber\] Solving for gives the concentration of Ag and the concentration of Cl as \(1.3 \times 10^{-5}\) M, or a pAg and a pCl of 4.89. After the equivalence point, the titrant is in excess. For example, after adding 35.0 mL of titrant \[[\text{Cl}^-] = \frac{(0.100 \text{ M})(35.0 \text{ mL}) - (0.0500 \text{ M})(50.0 \text{ mL})}{35.0 \text{ mL} + 50.0 \text{ mL}} = 1.18 \times 10^{-2} \text{ M} \nonumber\] or a pCl of 1.93. To find the concentration of Ag we use the for AgCl; thus \[[\text{Ag}^+] = \frac{K_\text{sp}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{1.18 \times 10^{-2}} = 1.5 \times 10^{-8} \text{ M} \nonumber\] or a pAg of 7.82. The following table summarizes additional results for this titration. To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a precipitation titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.0500 M NaCl with 0.100 M AgNO . This is the same example that we used in developing the calculations for a precipitation titration curve. You can review the results of that calculation in and . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 25.0 mL. Next we draw our axes, placing pCl on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point’s volume, we draw a vertical line that intersects the -axis at 25.0 mL of AgNO . Figure 9.5.2
a shows the result of this first step in our sketch. Before the equivalence point, Cl is present in excess and pCl is determined by the concentration of unreacted Cl . As we learned earlier, the calculations are straightforward. Figure 9.5.2
b shows pCl after adding 10.0 mL and 20.0 mL of AgNO . After the equivalence point, Ag is in excess and the concentration of Cl is determined by the solubility of AgCl. Again, the calculations are straightforward. Figure 9.5.2
c shows pCl after adding 30.0 mL and 40.0 mL of AgNO . Next, we draw a straight line through each pair of points, extending them through the vertical line that represents the equivalence point’s volume (Figure 9.5.2
d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.5.2
e). A comparison of our sketch to the exact titration curve (Figure 9.5.2
f) shows that they are in close agreement. At the beginning of this section we noted that the first precipitation titration used the cessation of precipitation to signal the end point. At best, this is a cumbersome method for detecting a titration’s end point. Before precipitation titrimetry became practical, better methods for identifying the end point were necessary. There are three general types of indicators for a precipitation titration, each of which changes color at or near the titration’s equivalence point. The first type of indicator is a species that forms a precipitate with the titrant. In the for Cl using Ag as a titrant, for example, a small amount of K CrO is added to the titrand’s solution. The titration’s end point is the formation of a reddish-brown precipitate of Ag CrO . The Mohr method was first published in 1855 by Karl Friedrich Mohr. Because \(\text{CrO}_4^{2-}\) imparts a yellow color to the solution, which might obscure the end point, only a small amount of K CrO is added. As a result, the end point is always later than the equivalence point. To compensate for this positive determinate error, an analyte-free reagent blank is analyzed to determine the volume of titrant needed to affect a change in the indicator’s color. Subtracting the end point for the reagent blank from the titrand’s end point gives the titration’s end point. Because \(\text{CrO}_4^{2-}\) is a weak base, the titrand’s solution is made slightly alkaline. If the pH is too acidic, chromate is present as \(\text{HCrO}_4^{-}\) instead of \(\text{CrO}_4^{2-}\), and the Ag CrO end point is delayed. The pH also must be less than 10 to avoid the precipitation of silver hydroxide. A second type of indicator uses a species that forms a colored complex with the titrant or the titrand. In the for Ag using KSCN as the titrant, for example, a small amount of Fe is added to the titrand’s solution. The titration’s end point is the formation of the reddish-colored Fe(SCN) complex. The titration is carried out in an acidic solution to prevent the precipitation of Fe as Fe(OH) . The Volhard method was first published in 1874 by Jacob Volhard. The third type of end point uses a species that changes color when it adsorbs to the precipitate. In the for Cl using Ag as a titrant, for example, the anionic dye dichlorofluoroscein is added to the titrand’s solution. Before the end point, the precipitate of AgCl has a negative surface charge due to the adsorption of excess Cl . Because dichlorofluoroscein also carries a negative charge, it is repelled by the precipitate and remains in solution where it has a greenish-yellow color. After the end point, the surface of the precipitate carries a positive surface charge due to the adsorption of excess Ag . Dichlorofluoroscein now adsorbs to the precipitate’s surface where its color is pink. This change in the indicator’s color signals the end point. The Fajans method was first published in the 1920s by Kasimir Fajans. Another method for locating the end point is a potentiometric titration in which we monitor the change in the titrant’s or the titrand’s concentration using an ion-selective electrode. The end point is found by visually examining the titration curve. For a discussion of potentiometry and ion-selective electrodes, see . Although precipitation titrimetry rarely is listed as a standard method of analysis, it is useful as a secondary analytical method to verify other analytical methods. Most precipitation titrations use Ag as either the titrand or the titrant. A titration in which Ag is the titrant is called an . Table 9.5.2
provides a list of several typical precipitation titrations. AgNO AgNO and KSCN Mohr or Fajans Volhard AgNO AgNO and KSCN Mohr or Fajarns Volhard AgNO AgNO and KSCN Fajans Volhard When two titrants are listed (AgNO and KSCN), the analysis is by a back titration; the first titrant, AgNO , is added in excess and the excess is titrated using the second titrant, KSCN. For those Volhard methods identified with an asterisk (*), the precipitated silver salt is removed before carrying out the back titration. The quantitative relationship between the titrand and the titrant is determined by the stoichiometry of the titration reaction. If you are unsure of the balanced reaction, you can deduce the stoichiometry from the precipitate’s formula. For example, in forming a precipitate of Ag CrO , each mole of \(\text{CrO}_4^{2-}\) reacts with two moles of Ag . A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.3172-g sample is dissolved in 50 mL of water and titrated to the Ag CrO end point, requiring 36.85 mL of 0.1120 M AgNO . A blank titration requires 0.71 mL of titrant to reach the same end point. Report the %w/w KCl in the sample. To find the moles of titrant reacting with the sample, we first need to correct for the reagent blank; thus \[V_\text{Ag} = 36.85 \text{ mL} - 0.71 \text{ mL} = 36.14 \text{ mL} \nonumber\] \[(0.1120 \text{ M})(0.03614 \text{ L}) = 4.048 \times 10^{-3} \text{ mol AgNO}_3 \nonumber\] Titrating with AgNO produces a precipitate of AgCl and AgBr. In forming the precipitates, each mole of KCl consumes one mole of AgNO and each mole of NaBr consumes one mole of AgNO ; thus \[\text{mol KCl + mol NaBr} = 4.048 \times 10^{-3} \text{ mol AgNO}_3 \nonumber\] We are interested in finding the mass of KCl, so let’s rewrite this equation in terms of mass. We know that \[\text{mol KCl} = \frac{\text{g KCl}}{74.551 \text{g KCl/mol KCl}} \nonumber\] \[\text{mol NaBr} = \frac{\text{g NaBr}}{102.89 \text{g NaBr/mol NaBr}} \nonumber\] which we substitute back into the previous equation \[\frac{\text{g KCl}}{74.551 \text{g KCl/mol KCl}} + \frac{\text{g NaBr}}{102.89 \text{g NaBr/mol NaBr}} = 4.048 \times 10^{-3} \nonumber\] Because this equation has two unknowns—g KCl and g NaBr—we need another equation that includes both unknowns. A simple equation takes advantage of the fact that the sample contains only KCl and NaBr; thus, \[\text{g NaBr} = 0.3172 \text{ g} - \text{ g KCl} \nonumber\] \[\frac{\text{g KCl}}{74.551 \text{g KCl/mol KCl}} + \frac{0.3172 \text{ g} - \text{ g KCl}}{102.89 \text{g NaBr/mol NaBr}} = 4.048 \times 10^{-3} \nonumber\] \[1.341 \times 10^{-2}(\text{g KCl}) + 3.083 \times 10^{-3} - 9.719 \times 10^{-3} (\text{g KCl}) = 4.048 \times 10^{-3} \nonumber\] \[3.69 \times 10^{-3}(\text{g KCl}) = 9.65 \times 10^{-4} \nonumber\] The sample contains 0.262 g of KCl and the %w/w KCl in the sample is \[\frac{0.262 \text{ g KCl}}{0.3172 \text{ g sample}} \times 100 = 82.6 \text{% w/w KCl} \nonumber\] The analysis for I using the Volhard method requires a back titration. A typical calculation is shown in the following example. The %w/w I in a 0.6712-g sample is determined by a Volhard titration. After adding 50.00 mL of 0.05619 M AgNO and allowing the precipitate to form, the remaining silver is back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach the end point. Report the %w/w I in the sample. There are two precipitates in this analysis: AgNO and I form a precipitate of AgI, and AgNO and KSCN form a precipitate of AgSCN. Each mole of I consumes one mole of AgNO and each mole of KSCN consumes one mole of AgNO ; thus \[\text{mol AgNO}_3 = \text{mol I}^- + \text{mol KSCN} \nonumber\] Solving for the moles of I we find \[\text{mol I}^- = \text{mol AgNO}_3 - \text{mol KSCN} = M_\text{Ag} V_\text{Ag} - M_\text{KSCN} V_\text{KSCN} \nonumber\] \[\text{mol I}^- = (0.05619 \text{ M})(0.0500 \text{ L}) - (0.05322 \text{ M})(0.03514 \text{ L}) = 9.393 \times 10^{-4} \nonumber\] The %w/w I in the sample is \[\frac{(9.393 \times 10^{-4} \text{ mol I}^-) \times \frac{126.9 \text{ g I}^-}{\text{mol I}^-}}{0.6712 \text{ g sample}} \times 100 = 17.76 \text{% w/w I}^- \nonumber\] A 1.963-g sample of an alloy is dissolved in HNO and diluted to volume in a 100-mL volumetric flask. Titrating a 25.00-mL portion with 0.1078 M KSCN requires 27.19 mL to reach the end point. Calculate the %w/w Ag in the alloy. The titration uses \[(0.1078 \text{ M KSCN})(0.02719 \text{ L}) = 2.931 \times 10^{-3} \text{ mol KSCN} \nonumber\] The stoichiometry between SCN and Ag is 1:1; thus, there are \[2.931 \times 10^{-3} \text{ mol Ag}^+ \times \frac{107.87 \text{ g Ag}}{\text{mol Ag}} = 0.3162 \text{ g Ag} \nonumber\] in the 25.00 mL sample. Because this represents 1⁄4 of the total solution, there are \(0.3162 \times 4\) or 1.265 g Ag in the alloy. The %w/w Ag in the alloy is \[\frac{1.265 \text{ g Ag}}{1.963 \text{ g sample}} \times 100 = 64.44 \text{% w/w Ag} \nonumber\] The scale of operations, accuracy, precision, sensitivity, time, and cost of a precipitation titration is similar to those described elsewhere in this chapter for acid–base, complexation, and redox titrations. Precipitation titrations also can be extended to the analysis of mixtures provided there is a significant difference in the solubilities of the precipitates. Figure 9.5.3
shows an example of a titration curve for a mixture of I and Cl using Ag as a titrant. | 16,844 | 3,893 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics/14.10%3A_Reaction_Mechanisms |
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism. In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water: \[\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1} \] For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction. The overall sequence of elementary reactions is the mechanism of the reaction. To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide. \[\ce{NO2(g) + (g) -> NO(g) + CO2 (g)} \label{14.6.2} \] From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of \(\ce{NO2}\) with a molecule of \(\ce{CO}\) that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows: \[rate = k[\ce{NO2}]^2 \label{14.6.3} \] The fact that the reaction is second order in \([\ce{NO2}]\) and independent of \([\ce{CO}]\) tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be \[rate = k[\ce{NO2},\ce{CO}]. \nonumber \] The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The \(\ce{NO3}\) molecule is an in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. The sum of the elementary reactions in a reaction mechanism give the overall balanced chemical equation of the reaction. The of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as ; if there are two reactant molecules, it is ; and if there are three reactant molecules (a relatively rare situation), it is . Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity ( \(\Page {1}\)). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is \[rate = k[A]. \nonumber \] For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in e \(\Page {1}\). For a bimolecular elementary reaction of the form A + B → products, the general rate law is \[rate = k[A,B]. \nonumber \] For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step). Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the , that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. The experimentally determined rate law for the reaction of \(NO_2\) with \(CO\) is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so \(k\) for the overall reaction must equal \(k_1\). That is, NO is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. In an alternative mechanism for the reaction of \(\ce{NO2}\) with \(\ce{CO}\) with \(\ce{N2O4}\) appearing as an intermediate. Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = [NO ] )? elementary reactions rate law for each elementary reaction and overall rate law The rate law for step 1 is rate = [NO ] ; for step 2, it is rate = [N O ,CO]. If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = [NO ] . This is the same as the experimentally determined rate law. Hence this mechanism, with N O as an intermediate, and the one described previously, with NO as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO and N O , directly. Iodine monochloride (\(\ce{ICl}\)) reacts with \(\ce{H2}\) as follows: \[\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber \] The experimentally determined rate law is \(rate = k[\ce{ICl},\ce{H2}]\). Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: \(\ce{HI}\) is an intermediate.) This mechanism is consistent with the experimental rate law if the first step is the rate-determining step. Assume the reaction between \(\ce{NO}\) and \(\ce{H_2}\) occurs via a three-step process: Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: \[\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed} \] The overall reaction is then \[\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber \] Reaction Mechanism (Slow step followed by fast step): A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its , the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. | 9,520 | 3,894 |
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. In addition to these individual basis; please contact What information can you obtain by studying the chemical kinetics of a reaction? Does a balanced chemical equation provide the same information? Why or why not? Kinetics gives information on the reaction rate and reaction mechanism; the balanced chemical equation gives only the stoichiometry of the reaction. If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you? Studying chemical kinetics determines whether to proceed with a reaction as it measures the rate of a reaction. Reactions conducted in an industrial facility mix compounds together, heating and stirring them for a while, before moving to the next process. When the compounds are then moved to the next phase of the process it is important to know how long to hold the reaction at one stage before continuing, to make sure a reaction has finished before starting the next one What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst? Reaction rate and the concentration of reactants have a direct relationship; as concentration goes up, so does reaction rate. This is because a higher concentration causes the reactants to have a higher probability to collide with a another reactant particle, possibly inducing the chemical reaction. This comes from the Collision theory. Reaction temperature: Reaction rate and the reaction temperature also have a direct relationship; as temperature goes up, so does reaction rate. A higher temperature makes the particles move at a faster speed so when two or more reactant particles collide, they collide with more energy and are more likely to reach the activation energy threshold, starting a chemical reaction. Physical properties of the reactants: If the reactants have the same physical properties, they are more likely to react, increasing the reaction rate, because the reactants mix and are more likely to collide another one of the reactants. Physical and chemical properties of the solvent: The properties of the solvent also affect the rates of a reaction. High viscosity means that particles more slowly than in low viscosity solvents, so reaction rates are slower in high viscosity solvents than in low viscosity solvents. This indicates an inverse relationship between viscosity and reaction rate. Presence of a catalyst: The chemical definition of a catalyst in a substance that increases the rate of a reaction without being used up in the reaction. Therefore, if a catalyst is present, reaction rate increases. A slurry is a mixture of a finely divided solid with a liquid in which it is only sparingly soluble. As you prepare a reaction, you notice that one of your reactants forms a slurry with the solvent, rather than a solution. What effect will this have on the reaction rate? What steps can you take to try to solve the problem? Why does the reaction rate of virtually all reactions increase with an increase in temperature? If you were to make a glass of sweetened iced tea the old-fashioned way, by adding sugar and ice cubes to a glass of hot tea, which would you add first? Increasing the temperature increases the average kinetic energy of molecules and ions, causing them to collide more frequently and with greater energy, which increases the reaction rate. First dissolve sugar in the hot tea, and then add the ice. In a typical laboratory setting, a reaction is carried out in a ventilated hood with air circulation provided by outside air. A student noticed that a reaction that gave a high yield of a product in the winter gave a low yield of that same product in the summer, even though his technique did not change and the reagents and concentrations used were identical. What is a plausible explanation for the different yields? A very active area of chemical research involves the development of solubilized catalysts that are not made inactive during the reaction process. Such catalysts are expected to increase reaction rates significantly relative to the same reaction run in the presence of a heterogeneous catalyst. What is the reason for anticipating that the relative rate will increase? Water has a dielectric constant more than two times greater than that of methanol (80.1 for H O and 33.0 for CH OH). Which would be your solvent of choice for a substitution reaction between an ionic compound and a polar reagent, both of which are soluble in either methanol or water? Why? Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction? Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive. Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why? Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 10 L/(mol·s). Are the reactions expected to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H O + OH → 2H O and H O + N(CH ) → H O + HN(CH ) in aqueous solution. Which would have the higher rate constant? Why? Faster in a less viscous solvent because the rate of diffusion is higher; the H O /OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants. What information can be obtained from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation? During the hydrolysis reaction A + H O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How would this effect be reflected in the overall reaction order? The reaction rate of a particular reaction in which A and B react to make C is as follows: \(\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=\dfrac{1}{2}\left ( \dfrac{\Delta[\textrm C]}{\Delta t} \right )\) Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C? While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min? Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation. Cleavage of C H to produce two CH · radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with = 5.46 × 10 s . How long will it take for the reaction to go to 15% completion? to 50% completion? 298 s; 1270 s Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere. \[\mathrm{N_2^+}+\mathrm{O_2}\xrightarrow{k_1}\mathrm{N_2}+\mathrm{O_2^+}\] \[\mathrm{O_2^+}+\mathrm{O}\xrightarrow{k_2}\mathrm{O_2}+\mathrm{O^+}\] \[(\mathrm{O^+}+\mathrm{N_2}\xrightarrow{k_3}\mathrm{NO^+}+\mathrm{N}\] Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = [N ,O ], which one of the steps is rate limiting? The oxidation of aqueous iodide by arsenic acid to give I and arsenous acid proceeds via the following reaction: Write an expression for the initial rate of decrease of [I ], Δ[I ]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: k /k = [H AsO ,I ]/[H AsO ,I ] [H ] . Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions? What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order? Predict whether the following reactions are zeroth order and explain your reasoning. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form? If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order? The reaction of NO with O is found to be second order with respect to NO and first order with respect to O . What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate? Iodide reduces Fe(III) according to the following reaction: \[2Fe^{3+}(soln) + 2I^−(soln) → 2Fe^{2+}(soln) + I_2(soln)\] Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order? First order in Fe ; second order in I ; third order overall; rate = [Fe ,I ] . Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows: What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction? The general rate law is: rate = k [Benzoyl Peroxide] . In order to find the reaction order with respect to benzoyl peroxide, divide two rate laws and solve for m: \(\dfrac{rate_2}{rate_1}=\dfrac{k_2}{k_1} \left ( \dfrac{[Benzoyl Peroxide]_2}{[Benzoyl Peroxide]_1} \right )^m\) \(\dfrac{1.64 × 10^-4 \dfrac{M}{s}}{2.22 × 10^-4 \dfrac{M}{s}}=( \dfrac{0.070 M}{1.00 M})^m\) \(0.738=(0.7)^m\) \(m=0.85\) rate law: rate = k [Benzoyl Peroxide] 1-Bromopropane is a colorless liquid that reacts with S O according to the following reaction: \[C_3H_7Br + S_2O_3^{2−} → C_3H_7S_2O_3^− + Br^−\] The reaction is first order in 1-bromopropane and first order in S O , with a rate constant of 8.05 × 10 M ·s . If you began a reaction with 40 mmol/100 mL of C H Br and an equivalent concentration of S O , what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be? 1.29 × 10 M/s; 3.22 × 10 M/s The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δ = [A] [B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally. Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy? Activation energy is required for a collision between molecules to result in a chemical reaction, as well it is related to the rate due to the Arrhenius equation Activation energy is the threshold of energy needed in order for a reaction to occur. Reactant particles must collide with enough energy to be able to break chemical bonds, that will then allow the creation of new bonds. If the particles do not have enough energy, when they collide they will simply bounce off one another. k=Ae . The increase in temperature increases the rate of reaction despite the fact the average kinetic is less than the activation energy since it increase the rate of collision between molecules. With an increase in temperature, there is a greater distribution of kinetic energy among reactant particles. This increase of temperature allows the rate in which particles collide with one another to increase. Although the average kinetic energy is still lower than the activation energy, the increase of collisions among particles increases the chance of particles that contain enough energy to overcome the energy barrier to collide. Thus, the reaction rate increases due to this increased rate of collisions. Additionally, there is a higher amount of molecules that have sufficient kinetic energy to overcome the energy barrier, despite the average energy of all these molecules still being lower than the activation energy. For any given reaction, what is the relationship between the activation energy and each of the following? 1.) Electrostatic repulsion: Electrostatic repulsion is the unfavorable interaction between two species of like charge. Activation energy is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. 2.) Bond formation in the activated complex: An activated complex is an intermediate state that is formed during the conversion of reactants into products. Bond breaking can increase activation energy as breaking bonds requires energy. 3.) Nature of the activated complex: The activation energy of a chemical reaction is the difference between the energy of the activated complex and the energy of the reactants. If the structure has a high steric hindrance the activation energy will be higher. If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction? The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility? Above , molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach ? Explain your answer. What is the relationship between , , and ? How does an increase in affect the reaction rate? Of two highly exothermic reactions with different values of , which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why? What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C? The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 2 = 2; 20°C to 70°C, the reaction rate increases by about 2 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2 . Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate? Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10 M/s at 25°C, what would the reaction rate be at the following temperatures? An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the \(E_a\) from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature? The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction? 100 kJ/mol The reaction rate at 25°C is 1.0 × 10 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10 M/s. Estimate \(E_a\) for this process. If \(E_a\) were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10 M/s, what would be the reaction rate at 75°C? How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation? What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation? When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction? If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step? Give the rate-determining step for each case. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health? Free radicals are uncharged molecules with an unpaired valence electron. The reason these are so dangerous is because they like to grab electrons from other atoms to fill their own outer shell. This allows them to impair protein function because free radicals readily oxidize proteins and cell membrane which could lead to a loss of function. It was important to purge James Bond of radicals because radicals set off chain reactions of continuously pulling electrons from molecules, which in turn, can damage cells in the body. Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows: cyclopropane cyclopropane Above approximately 500 K, the reaction between NO and CO to produce CO and NO follows the second-order rate law Δ[CO ]/Δ = [NO ,CO]. At lower temperatures, however, the rate law is Δ[CO ]/Δ = ′[NO ] , for which it is known that NO is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest? Given that NO is an intermediate in the low temperature reaction mechanism, we automatically know two things: 1) , and thus, 2) . We can also assume that since we're given an intermediate from the problem, this is the only intermediate (so we won't have to dream up any other compounds that might exist in the series of reactions.) These things being said, the will look like this: \[2NO_2(g) \rightarrow NO_3(g) + NO(g) \tag{1}\] \[CO(g) + NO_3(g) \rightarrow CO_2(g) + NO_2(g) \tag{2}\] And the overall reaction will look like this (notice how NO is not present): \[CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}\] Now that we have the reaction mechanism written out, we can go about determining which step is the slowest. It would be pretty tricky to do this if we weren't given any further information, however, we know two more things: 1) (given that the overall reaction is also equal to this rate) and 2) These two things being said, we've both confirmed that our proposed low-temperature reaction mechanism is in fact two steps, and that . By using the "guess-and-check" method we can label each step reaction one at a time as the and see if the rate matches up with the rate given to us. (see above) By using we can determine that the rate of the reaction must be in terms of its reactants, which follows: \[\textrm{k}\textrm{[CO,NO}_3\textrm{]}\] ... We can't have the overall reaction rate in terms of an intermediate. By looking at the 1st reaction, we can determine that we can sub in "[NO ] /[NO]" for "[NO ]" since by writing the full reaction rate of the first step and solving for [NO ] this is equivalent. So, we now have the overall rate of the mechanism as the following given the 2nd reaction is the "slow reaction:" \[\textrm{k}\dfrac{\textrm{[CO,NO}_2\textrm{]}^2}{\textrm{[NO}\textrm{]}}\] \[\textrm{k}\textrm{[NO}_2\textrm{]}^2\] What do you know... the rates are equal! Nitramide (O NNH ) decomposes in aqueous solution to N O and H O. What is the experimental rate law (Δ[N O]/Δ ) for the decomposition of nitramide if the mechanism for the decomposition is as follows? Assume that the rates of the forward and reverse reactions in the first equation are equal. \[rate_1 = {k_1}{[O_2NNH_2]}\] \[rate_{-1} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] Since these two rates produce and consume the same amount of \(O_2NNH^{-}\) over the same time period, we can set them equal to each other and solve for the intermediate \[rate_1 = rate_{-1}\] \[{k_1}{[O_2NNH_2]} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] \[{[O_2NNH^-]} = \frac{k_1{[O_2NNH_2]}}{k_{-1}{[H^+]}}\] Substituting this equation back into our original equation gives us \[rate=rate_2=k_2\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}\] With all of the rate constants (k), we can clean up our equation a little bit by saying \[k = \frac{k_2{k_1}}{k_{-1}}\] Leaving us with \[rate = \frac{k[O_2NNH_2]}{[H^+]}\] \(\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}\) The following reactions are given: \[\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}\] \[\mathrm{D+E}\xrightarrow{k_2}\mathrm F\] What is the relationship between the relative magnitudes of \(k_{−1}\) and \(k_2\) if these reactions have the following rate law? \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A,B,E]}{[C]}\] How does the magnitude of \(k_1\) compare to that of \(k_2\)? Under what conditions would you expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A,B]?\] Assume that the rates of the forward and reverse reactions in the first equation are equal. First, because we have broken the equations down into we can write the rate laws for each step. Step1: \[A+B\xrightarrow[]{k_{1}} C+D\] \[rate=k_{1}[A,B]\] Step 2: \[C+D \xrightarrow[]{k_{-1}} A+B\] \[rate=k_{-1}[C,D]\] Step 3: \[D+E \xrightarrow[]{k_{2}} F\] \[rate=k_{2}[D,E]\] If we add a these steps together we see that we get overall reaction \[A+B+E \rightarrow C+F\] we can see that [D] is an intermediate and \[k_{1}=k_{-1}\] Since we are not told which steps are fast or slow we need to use Steady State Approximation. If the second step is the slower step ( >> then our rate determining step would be \[rate=k_{2}[D,E]\] Since we can only write rate laws in terms of products and reactants we have to rewrite this so that we are not including an . Assume: rate of [D] formation = rate of its disappearance \[k_{1}[A,B]=k_{-1}[C,D]+k_{2}[D,E]\] \[k_{1}[A,B]=[D](k_{-1}[C]+k_{2}[E])\] Solving for [D] we find that \[[D]= \frac{k_{1}[A,B]}{(k_{-1}[C]+k_{2}[E])}\] now we can use this to substitute the intermediate [D] in the rate law to get an appropriate rate law. \[rate=\frac{k_{2} k_{1}[A,B,E]}{(k_{-1}[C]+k_{2}[E])}\] because we had already established >> \[k_{-1}[C]+k_{2}[E]\approx k_{-1}[C]\] this would give us the observed rate law \[\frac{\Delta [F]}{\Delta t}=\frac{k_{2}k_{1}[A,B,E]}{k_{-1}[C]}\] to make this clearer we can set \[k=\frac{(k_{2})(k_{1})}{(k_{-1})}\] and we can then simplify it down to \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A,B,E]}{[C]}\] we can see that all of these rate constants are related by this ratio / Since is our rate determining step >> = then we can see that >> We would expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A,B]?\] if the aka the slowest step is that corresponding to (step 1) since the rate law for this step is [A,B] and this is the exact the same as the rate law that we were given. What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor ( )? What effect does it have on the change in potential energy for the reaction? A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction. How is it possible to affect the product distribution of a reaction by using a catalyst? A heterogeneous catalyst works by interacting with a reactant in a process called . What occurs during this process? Explain how this can lower the activation energy. In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased. What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer. Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following. An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred. Catalysts are compounds that, when added to chemical reactions, reduce the activation energy and increase the reaction rate. The amount of a catalyst does not change during a reaction, as it is not consumed as part of the reaction process. Catalysts lower the energy required to reach the transition state of the reaction, allowing more molecular interactions to achieve that state. However, catalysts do not affect the degree to which a reaction progresses. In other words, though catalysts affect reaction kinetics, the equilibrium state remains unaffected. Catalysts can be classified into two types: homogenous and heterogeneous. Homogenous catalysts are those which exist in the same phase (gas or liquid) as the reactants, while heterogeneous catalysts are not in the same phase as the reactants. Typically, heterogeneous catalysis involves the use of solid catalysts placed in a liquid reaction mixture. For this question, we will be discussing homogenous catalysts. Most of the times, homogeneous catalysis involves the introduction of an aqueous phase catalyst into an aqueous solution of reactants. One reason why homogeneous catalysts are preferred over heterogeneous catalysts because homogeneous catalysts mix well in chemical reactions in comparison to heterogeneous catalysts. However, homogeneous catalyst is often irrecoverable after the reaction has run to completion. Consider the following reaction between cerium(IV) and thallium(I) ions: \[\ce{2Ce^{4+} + Tl^+ → 2Ce^{3+} + Tl^{3+}}\] This reaction is slow, but Mn catalyzes it, as shown in the following mechanism: \[\ce{Ce^{4+} + Mn^{2+} → Ce^{3+} + Mn^{3+}}\] \[\ce{Ce^{4+} + Mn^{3+} → Ce^{3+} + Mn^{4+}}\] \[\ce{Mn^{4+} + Tl^{+ }→ Tl^{3+} + Mn^{2+}}\] In what way does \(Mn^{2+}\) increase the reaction rate? The Mn ion donates two electrons to Ce , one at a time, and then accepts two electrons from Tl . Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions. The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why? Enzymes are expensive to make, denature and fail at certain temperatures, are not that stable in a solution, and are very specific to the reaction it was made for. However, scientists can use the observations from enzymes to create catalysts that are more effective in aiding the reaction and cost less to produce. Overall, catalysts still play a large part in lowering the activation energy for reactions. Creating new catalysts can help in the improvement of areas such as medical, ecological, and even commercial products. Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. ( : remember that enzymes are composed of functional amino acids.) At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (\(ES\)), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations: \[\text {enzyme (E) + substrate (S)} \rightleftharpoons \text{enzyme-substrate complex (ES)} \rightleftharpoons \text{enzyme (E) + product (P)}\] This can also be shown as follows: \[E + S \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\stackrel{k_{2}}{\rightleftharpoons}} E+P\] Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order? A particular reaction was found to proceed via the following mechanism: What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates A particular reaction has two accessible pathways (A and B), each of which favors conversion of to a different product ( and , respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst. In both cases, the product of pathway A is favored. All of the produced in the catalyzed reversible pathway B will eventually be converted to as is converted irreversibly to by pathway A. The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon. | 31,900 | 3,897 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.24%3A_Steam_Distillation |
Place large amount of plant materials in a large round bottomed flask (no more than half full), and just cover with water. Always use a Claisen adapter, as there is often turbulence in the flask. Two variations are common: Collect the distillate at a rate of . The distillate may appear cloudy, or a second layer may form on the top. If milky, the distillation can be ceased when the distillate is clear. If a steam line was used, be sure to drain the liquid from the steam trap before turning off the steam, to prevent back suction. Separated oil can be pipetted (then dried with \(\ce{Na_2SO_4}\)), while milky distillates need to be extracted. | 659 | 3,898 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.5%3A_Atomic_Mass |
The subscripts in chemical formulas, and the coefficients in chemical equations represent quantities. \(\ce{H_2O}\), for example, indicates that a water molecule comprises exactly atoms of hydrogen and atom of oxygen. The following equation: \[ \ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1}\] not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that molecule of propane reacts with molecules of oxygen to produce molecules of carbon dioxide and molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the of the compounds involved. Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen: \[\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2}\] Later, scientists discovered that water was composed of of hydrogen of oxygen. Therefore, in the above analysis, Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom: \[ \dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3}\] Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a atom has a mass of grams, and that the atom has a mass of grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the . The atomic mass unit ( ) was not standardized against hydrogen, but rather, against the C isotope of ( ). Thus, the mass of the ( H) is 1.0080 , and the mass of ( O) is 15.995 . Once the masses of atoms were determined, the could be assigned an actual value: 1 = 1.66054 x 10 conversely: 1 = 6.02214 x 10 Although the masses of the electron, the proton, and the neutron are known to a high degree of precision ( ), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of H (hydrogen) and H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions. Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses ( ). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest. The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of C. Because the masses of all other atoms are calculated relative to the C standard, C is the only atom listed in whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10 g. Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of H to the mass of C, so the of H is \[\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}\] The masses of the other elements are determined in a similar way. The lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% C (mass = 12 amu by definition) and 1.11% C (mass = 13.003355 amu). The percent abundance of C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows: \[ \rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5} \] Carbon is predominantly C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation. The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes. Finding the Averaged Atomic Weight of an Element: Naturally occurring bromine consists of the two isotopes listed in the following table: Calculate the atomic mass of bromine. : exact mass and percent abundance : atomic mass : : The atomic mass is the weighted average of the masses of the isotopes. In general, we can write atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + … Bromine has only two isotopes. Converting the percent abundances to mass fractions gives \[\ce{^{79}Br}: {50.69 \over 100} = 0.5069\] Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass: \(\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu\) \(\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu\) The sum of the weighted masses is the atomic mass of bromine is 40.00 amu + 39.90 amu = 79.90 amu This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%. Magnesium has the three isotopes listed in the following table: Use these data to calculate the atomic mass of magnesium. 24.31 amu The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. | 8,360 | 3,899 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/04%3A_The_Biosphere/4.03%3A_Gaia_-_Bioregulation_of_the_Environment |
The physical conditions under which life as we know it can exist encompass a relatively narrow range of temperature, pH, osmotic pressure, and ultraviolet radiation intensity. It seems remarkable enough that life was able to get started at all; it is even more remarkable that it has continued to thrive in the face of all the perils that have, or could have occurred, during the past 3 billion years or so. During the time that life has been evolving, the sun has also been going through the process of evolution characteristic of a typical star; one consequence of this is an increase in its energy output by about 30 percent during this time. If the sun’s output should suddenly drop to what it was 3 billion years ago, the oceans would freeze. How is it that the earth was not in a frozen state for the first 1.5 billion years of life’s existence? Alternatively, if conditions were somehow suitable 3 billion years ago, why have the oceans not long since boiled away? A rather non-traditional answer to this kind of question is that the biosphere is far from playing a passive role in which it is continually at the mercy of environmental conditions. Instead, the earth’s atmosphere, and to a lesser extent the hydrosphere, may be actively maintained and regulated by the biosphere. This view has been championed by the British geochemist J.E. Lovelock, and is known as the . Gaia is another name for the Greek earth-goddess , from which root the sciences of geography, geometry, and geology derive their names. Lovelock's book (Oxford, 1979) is a short and highly readable discussion of the hypothesis. Evidence in support of this hypothesis is entirely circumstantial, but nevertheless points to important questions that must be answered: how have the climatic and chemical conditions on the earth remained optimal for life during all this time; how can the chemical composition of the atmosphere remain in a state that is tens of orders of magnitude from equilibrium? Although the Gaia hypothesis has received considerable publicity in the popular press, it has never been very well received by the scientific community, many of whom feel that there is no justification for proposing a special hypothesis to describe a set of connections which can be quite adequately explained by conventional geochemical processes. More recently, even Lovelock has backed away from the teleological interpretation of these relations, so that the Gaia hypothesis should now be more properly described as a set of loosely connected effects, rather than as a hypothesis. Nevertheless, these effects and the mechanisms that might act to connect them are sufficiently interesting that it seems worthwhile to provide an overview of the major observations that led to the development of the hypothesis. is the doctrine that natural processes operate with a purpose. See 1988: Science 240 393-395. The increase in the oxygen content of the atmosphere as a result of the development of the eucaryotic cell was discussed above. Why has the oxygen content leveled off at 21 percent? It is interesting to note that if the oxygen concentration in the atmosphere were only four percent higher, even damp vegetation, once ignited by lightning, would continue to burn, enveloping vast areas of the earth in a firestorm. Evidence for such a worldwide firestorm that may be related to the extinction of the dinosaurs has recently been discovered. The charcoal layers found in widely distributed sediments laid down about 65 million years ago are coincident with the iridium anomaly believed to be due to the collision of a large meteor with the earth. The input of salts into the sea from streams and rivers is about 5.4 x 10 tons per year, into a total volume of about 1.2 x 10 km yr . Upwelling of juvenile water and hydrothermal action at oceanic ridges provide additional inputs of salts. With a few bizarre exceptions such as the brine shrimp and halophilic bacteria, 6 percent is about the maximum salinity level that organisms can tolerate. The internal salinities of cells must be maintained at much lower levels (around 1%) to prevent denaturation of proteins and other macromolecules whose conformations are dependent on electrostatic forces. At higher levels than this, the electrostatic interaction between the salt ions and the cell membrane destroys the integrity of the latter so that it can no longer pump out salt ions that leak in along the osmotic gradient. At the present rate of salt input, the oceans would have reached their present levels of salinity millions of years ago, and would by now have an ionic strength far to high to support life, as is presently the case in the landlocked Dead Sea. The present average salinity of seawater is 3.4 percent. The salinity of blood, and of many other intra- and intercellular fluids in animals, is about 0.8 percent. If we assume that the first organisms were approximately in osmotic equilibrium with seawater, then our body fluids might represent “fossilized” seawater as it existed at the time our predecessors moved out of the sea and onto the land. By what processes is salt removed from the oceans in order to maintain a steady-state salinity? This remains one of the major open questions of chemical oceanography. There are a number of answers, mostly based on strictly inorganic processes, but none is adequately supported by available evidence. For example, Na and Mg ions could adsorb to particulate debris as it drops to the seafloor, and become incorporated into sediments. The requirement for charge conservation might be met by the involvement of negatively charged silicate and hydroxyaluminum ions. Another possible mechanism might be the burial of salt beds formed by evaporation in shallow, isolated arms of the sea, such as the Persian Gulf. Extensive underground salt deposits are certainly found on most continents, but it is difficult to see how this very slow mechanism could have led to an unfluctuating salinity over shorter periods of highly variable climatic conditions. The possibility of biological control of oceanic salinity starts with the observation that about half of the earth’s biomass resides in the sea, and that a significant fraction of this consists of diatoms and other organisms that build skeletons of silica. When these organism die, they sink to the bottom of the sea and add about 300 million tons of silica to sedimentary rocks annually. It is for this reason that the upper levels of the sea are undersaturated in silica, and that the ratio of silica to salt in dead salt lakes is much higher than in the ocean. These facts could constitute a basis for a biological control of the silica content of seawater; any link between silica and salt could lead to the control of the latter substance as well. For example, the salt ions might adsorb onto the silica skeletons, and be carried down with them; if the growth of these silica-containing organisms is itself dependent on salinity, we would have our negative feedback mechanism. The continual buildup of biogenic sedimentary deposits on the ocean floor might possibly deform the thin oceanic crust by its weight, and cause local heating by its insulating properties. This could conceivably lead to volcanic action and the formation of new land mass, thus linking the lithosphere into Gaia. ) | 7,361 | 3,901 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./21.E%3A_Nuclear_Chemistry_(Exercises) |
. In addition to these individual basis; please contact Why are many radioactive substances warm to the touch? Why do many radioactive substances glow? Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation? Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why? Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells. Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters. Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why? List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous? Three primary naturally occurring radiations are radium, uranium and thorium, each all having long half lives. Inhalation of air, ingestion of food and water,terrestrial radation from the ground and cosmic radiation from space are all factors tat influence the does that a person receives throughout the year. Inhalation of the air is the largest contributor to exposure. Radiation can damage DNA or kill cells. When radiation is exposed to your body, it will collide with atoms and this will change and damage your DNA. Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects? Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases? Beta decay. Alfa decay can be easily stopped by paper, which means it can not be used to see inside people's body. Also, Gamma rays are really dangerous for human, that even a short period of time exploding to it will have negative effect on human body. Thus, Beta decay is the perfect choice. It can be used to see through human's body and stopped by aluminum or some other metals. Since all these radioactive decays are harmful for human body, if the half time of these reactions are short, the time exploding to these reactions will be short too. Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale. Ten exposures of 10 rem are less likely to cause major damage. A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is U, which has a half-life of 4.46 × 10 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads? There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 10 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why?
241/95 Am---> 4/2 He + 237/93Np---> 4/2He + 233/91Pa----> 1/0n+ 232/91Th---> 1/1 H + 233/92 U
By adding alpha particles to the products side of the reaction, he was able to reduce the mass number by 4 and the atomic number by 2 to get the products he wanted. Bombardment with neutrons and 1 H was required to lower to the mass number to get Th and then raise both the mass number and the atomic number to yield Uranium.
2 hours*365*2= 1460 hours of exposure.*60min/1hr*60s/1min= 5.26*10^6s of exposure
1MeV= (1.6022*10^-13 Joules) * (5.24 MeV/particle)*2 particles= 1.679*10^-12 Joules.
(1.679*10^-12 Joules) * (1 amu/ 1.4924*10^-10 Joules)= 2.51*10^-13 amu
E=mc^2
E=(2.51*10^-13 amu)(1.66*10^-22kg/amu)(2.9998*10^8m/s)^2= (3.75*10^-18 kgm^2/s)*(3.5*10^6 decays/s)= 1.31*10^-11 joules of exposure per second.
The scientist showed no ill effects from this exposure because if we multiple the energy in joules of exposure per second, 1.31*10^-11, by the total amount of seconds of exposure, 5.26*10^6s, we find that he was only exposed to 6.9*10^-5 joules of radiation throughout the span of two years. This is a very small amount of radiation for such a long span of time.
In order to plug in the values for this equation, we must convert the given MeV to Joules with the known conversion rate. Similarly, we must convert Joules to amu with another known conversion rate. Then we can plug in the values and multiply by c^2 but we must not forget to multiple the amu by the conversion rate to kg in order to yield Joules. After all of this is done, we multiple the amount of Joules of exposure per second by the total amount of exposure in seconds in order to find out the total amount of exposure over the two year span. How do chemical reactions compare with nuclear reactions with respect to mass changes? Does either type of reaction violate the law of conservation of mass? Explain your answers. Why is the amount of energy released by a nuclear reaction so much greater than the amount of energy released by a chemical reaction? Explain why the mass of an atom is less than the sum of the masses of its component particles. The stability of a nucleus can be described using two values. What are they, and how do they differ from each other? In the days before true chemistry, ancient scholars (alchemists) attempted to find the philosopher’s stone, a material that would enable them to turn lead into gold. Is the conversion of Pb → Au energetically favorable? Explain why or why not. Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome. Imagine that the universe is dying, the stars have burned out, and all the elements have undergone fusion or radioactive decay. What would be the most abundant element in this future universe? Why? Numerous elements can undergo fission, but only a few can be used as fuels in a reactor. What aspect of nuclear fission allows a nuclear chain reaction to occur? How are transmutation reactions and fusion reactions related? Describe the main impediment to fusion reactions and suggest one or two ways to surmount this difficulty. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules per mole. Using the information provided in Chapter 33, complete each reaction and then calculate the amount of energy released from each in kilojoules per mole. Using the information provided in Chapter 33, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole. Using the information provided, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole. Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for Sr if the measured mass of Sr is 86.908877 amu. The experimentally determined mass of S is 28.996610 amu. Find each of the following. Calculate the amount of energy that is released by the neutron-induced fission of U to give Ba, Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole. Calculate the amount of energy that is released by the neutron-induced fission of U to give Sr, Xe, and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole. Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction? Calculate the amount of energy released by the fusion of Li and deuterium to give two helium-4 nuclei. Express your answer in electronvolts per atom and kilojoules per mole. How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture? 1 Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements? How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed? Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers. The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer. If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation. Write a balanced nuclear reaction for the formation of each isotope. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium? What do chemists mean by the half-life of a reaction? If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare? Half-lives for the reaction A + B → C were calculated at three values of [A] , and [B] was the same in all cases. The data are listed in the following table: Does this reaction follow first-order kinetics? On what do you base your answer? Ethyl-2-nitrobenzoate (NO C H CO C H ) hydrolyzes under basic conditions. A plot of [NO C H CO C H ] versus was used to calculate ½, with the following results: Is this a first-order reaction? Explain your reasoning. Azomethane (CH N CH ) decomposes at 600 K to C H and N . The decomposition is first order in azomethane. Calculate ½ from the data in the following table: How long will it take for the decomposition to be 99.9% complete? = 1.92 × 10 s or 1920 s; 19100 s or 5.32 hrs. The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s ) for this reaction? If you started with a solution that was 7.5 × 10 M H O , what would be the initial rate of decomposition (M/s)? What would be the concentration of H O after 3.3 h? | 12,759 | 3,902 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Limiting_Reagents |
When there is not enough of one reactant in a chemical reaction, the reaction stops abruptly. To figure out the amount of product produced, it must be determined which reactant will limit the chemical reaction (the ) and which reactant is in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent. The following scenario illustrates the significance of limiting reagents. In order to assemble a car, 4 tires and 2 headlights are needed (among other things). In this example, imagine that the tires and headlights are reactants while the car is the product formed from the reaction of 4 tires and 2 headlights. \[4 \, \text{Tires} + 2 \, \text{Headlights} = 1 \, \text{Car} \nonumber\] If you have 20 tires and 14 headlights, how many cars can be made? With 20 tires, 5 cars can be produced because there are 4 tires to a car. With 14 headlights, 7 cars can be built (each car needs 2 headlights). Although more cars can be made from the headlights available, only 5 full cars are possible because of the limited number of tires available. In this case, the headlights are in excess. Because the number of cars formed by 20 tires is less than number of cars produced by 14 headlights, the tires are the limiting reagent (they limit the full completion of the reaction, in which all of the reactants are used up). This scenario is illustrated below: The initial condition is that there must be 4 tires to 2 headlights. The reactants must thus occur in that ratio; otherwise, one will limit the reaction. There are 20 tires and 14 headlights, so there are two ways of looking at this problem. For 20 tires, 10 headlights are required, whereas for 14 headlights, 28 tires are required. Because there are not enough tires (20 tires is less than the 28 required), tires are the limiting "reactant." The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction , the exact amount of reactant needed to react with another element can be calculated. If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants will be entirely consumed while another will be left over. The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant. There are two ways to determine the limiting reagent. One method is to find and compare the mole ratio of the reactants used in the reaction (approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the smallest amount of product is the limiting reagent (approach 2). Find the limiting reagent by looking at the number of moles of each reactant. Find the limiting reagent by calculating and comparing the amount of product each reactant will produce. Consider respiration, one of the most common chemical reactions on earth. \[\ce{ C6H_{12}O6 + 6 O_2 \rightarrow 6 CO2 + 6 H2O} + \rm{energy}\] What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen? When approaching this problem, observe that every 1 mole of (\(C_6H_{12}O_6\)) requires 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water. Step 1: The balanced chemical equation is already given. Step 2: \[\mathrm{25\:g \times \dfrac{1\: mol}{180.06\:g} = 0.1388\: mol\: C_6H_{12}O_6} \nonumber\] \[\mathrm{40\:g \times \dfrac{1\: mol}{32\:g} = 1.25\: mol\: O_2} \nonumber\] Step 3: a. If all of the 1.25 moles of oxygen were to be used up, there would need to be \(\mathrm{1.25 \times \dfrac{1}{6}}\) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant. \[1.25 \; \rm{mol} \; O_2 \times \dfrac{ 1 \; \rm{mol} \; C_6H_{12}O_6}{6\; \rm{mol} \; O_2}= 0.208 \; \rm{mol} \; C_6H_{12}O_6 \nonumber\] \[0.1388\; \rm{ mol}\; C_6H_{12}O_6 \times \dfrac{6 \; \rm{mol} \;O_2}{1 \; \rm{mol} \; C_6H_{12}O_6} = 0.8328 \; \rm{mol}\; O_2 \nonumber\] Therefore, the mole ratio is: (0.8328 mol O )/(0.208 mol C H O ) This gives a 4.004 ratio of \(\ce{O2}\) to \(\ce{C6H12O6}\). Step 4: For carbon dioxide produced: \(\mathrm{0.1388\: moles\: glucose \times \dfrac{6}{1} = 0.8328\: moles\: carbon\: dioxide}\). Step 5: 1.25 mol - 0.8328 mol = Calculate the mass of magnesium oxide possible if 2.40 g \(\ce{Mg}\) reacts with 10.0 g of \(\ce{O_2}\) \[\ce{ Mg +O_2 \rightarrow MgO} \nonumber\] Step 1: \[\ce{2 Mg + O_2 \rightarrow 2 MgO} \nonumber\] Step 2 and Step 3: \[\mathrm{2.40\:g\: Mg \times \dfrac{1.00\: mol\: Mg}{24.31\:g\: Mg} \times \dfrac{2.00\: mol\: MgO}{2.00\: mol\: Mg} \times \dfrac{40.31\:g\: MgO}{1.00\: mol\: MgO} = 3.98\:g\: MgO} \nonumber\] \[\mathrm{10.0\:g\: O_2\times \dfrac{1\: mol\: O_2}{32.0\:g\: O_2} \times \dfrac{2\: mol\: MgO}{1\: mol\: O_2} \times \dfrac{40.31\:g\: MgO}{1\: mol\: MgO} = 25.2\: g\: MgO} \nonumber\] Step 4: \(\ce{Mg}\) produces less \(\ce{MgO}\) than does \(\ce{O2}\) (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reagent in this reaction. Step 5: O produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O is the excess reagent in this reaction. Step 6: \[\mathrm{2.40\:g\: Mg \times \dfrac{1.00\: mol\: Mg}{24.31\:g\: Mg} \times \dfrac{1.00\: mol\: O_2}{2.00\: mol\: Mg} \times \dfrac{32.0\:g\: O_2}{1.00\: mol\: O_2} = 1.58\:g\: O_2} \nonumber\] Mass of excess reagent calculated using the mass of the product: \[\mathrm{3.98\:g\: MgO \times \dfrac{1.00\: mol\: MgO}{40.31\:g\: MgO} \times \dfrac{1.00\: mol\: O_2}{2.00\: mol\: MgO} \times \dfrac{32.0\:g\: O_2}{1.00\: mol\: O_2} = 1.58\:g\: O_2} \nonumber\] Mass of total excess reagent given – mass of excess reagent consumed in the reaction 10.0g – 1.58g = 8.42g O is in excess. What is the limiting reagent if 76.4 grams of \(C_2H_3Br_3\) were reacted with 49.1 grams of \(O_2\)? \[\ce{4 C_2H_3Br_3 + 11 O_2 \rightarrow 8 CO_2 + 6 H_2O + 6 Br_2} \nonumber\] A. \[\mathrm{76.4\:g \times \dfrac{1\: mole}{266.72\:g} = 0.286\: moles\: of\: C_2H_3Br_3} \nonumber\] \[\mathrm{49.1\: g \times \dfrac{1\: mole}{32\:g} = 1.53\: moles\: of\: O_2} \nonumber\] B. Assuming that all of the oxygen is used up, \(\mathrm{1.53 \times \dfrac{4}{11}}\) or 0.556 moles of C H Br are required. Because there are only 0.286 moles of C H Br available, C H Br is the limiting reagent. \[\mathrm{76.4\:g\: C_2H_3Br_3 \times \dfrac{1\: mol\: C_2H_3Br_3}{266.72\:g\: C_2H_3Br_3} \times \dfrac{8\: mol\: CO_2}{4\: mol\: C_2H_3Br_3} \times \dfrac{44.01\:g\: CO_2}{1\: mol\: CO_2} = 25.2\:g\: CO_2} \nonumber\] \[\mathrm{49.1\:g\: O_2 \times \dfrac{1\: mol\: O_2}{32\:g\: O_2} \times \dfrac{8\: mol\: CO_2}{11\: mol\: O_2} \times \dfrac{44.01\:g\: CO_2}{1\: mol\: CO_2} = 49.1\:g\: CO_2} \nonumber\] Therefore, by either method, . What is the limiting reagent if 78 grams of \(\ce{Na2O2}\) were reacted with 29.4 grams of \(\ce{H2O}\)? A. \[\mathrm{78\:g \times \dfrac{1\: mol}{77.96\:g} = 1.001\: moles\: of\: Na_2O_2} \nonumber\] \[\mathrm{29.4\:g \times \dfrac{1\: mol}{18\:g}= 1.633\: moles\: of\: H_2O} \nonumber\] B. Assume that all of the water is consumed, \(\mathrm{1.633 \times \dfrac{2}{2}}\) or 1.633 moles of Na O are required. Because there are only 1.001 moles of Na O it is the limiting reactant. \\[\mathrm{78\:g\: Na_2O_2 \times \dfrac{1\: mol\: Na_2O_2}{77.96\:g\: Na_2O_2} \times \dfrac{4\: mol\: NaOH}{2\: mol\: Na_2O_2} \times \dfrac{40\:g\: NaOH}{1\: mol\: NaOH} = 80.04\:g\: NaOH} \nonumber\] Using either approach gives Na O as the . How much the excess reagent remains if 24.5 grams of CoO is reacted with 2.58 grams of O ? \[\ce{4 CoO + O_2 \rightarrow 2 Co_2O_3} \nonumber\] A. \[\mathrm{24.5\:g \times \dfrac{1\: mole}{74.9\:g}= 0.327\: moles\: of\: CoO} \nonumber\] \[\mathrm{2.58\:g \times \dfrac{1\: mole}{32\:g}= 0.0806\: moles\: of\: O_2} \nonumber\] B. Assuming that all of the oxygen is used up, \(\mathrm{0.0806 \times \dfrac{4}{1}}\) or 0.3225 moles of \(CoO\) Because there are 0.327 moles of CoO, CoO is in excess and thus O is the limiting reactant. C. 0.327 mol - 0.3224 mol = 0.0046 moles left in excess. Will 28.7 grams of \(SiO_2\) react completely with 22.6 grams of \(H_2F_2\)? If not, identify the limiting reagent. \[\ce{SiO_2+ 2 H_2F_2 \rightarrow SiF_4+ 2 H_2O} \nonumber\] A. \[\mathrm{28.7\:g \times \dfrac{1\: mole}{60.08\:g} = 0.478\: moles\: of\: SiO_2} \nonumber\] \[\mathrm{22.6\:g \times \dfrac{1\: mole}{39.8\:g} = 0.568\: moles\: of\: H_2F_2} \nonumber\] B. There must be 1 mole of SiO for every 2 moles of H F consumed. Because the ratio is 0.478 to 0.568, 28.7 grams of SiO do not react with the H F . C. Assuming that all of the silicon dioxide is used up, \(\mathrm{0.478 \times \dfrac{2}{1}}\) or 0.956 moles of H F | 9,019 | 3,903 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Microscopy/Scanning_Probe_Microscopy/02_History |
Developed in 1982 by Binnig, Rohrer, Gerber, and Weibel at IBM in Zurich, Switzerland Developed in 1986 by Binnig, Quate, and Gerber as a collaboration between IBM and Stanford University. During the 20 century a world of atomic and subatomic particles opened new avenues. In order to study and manipulate material on an atomic scale there needed to be a development in new instrumentation. Physicist Richard Feynman said in his now famous in 1959: “if you want to make atomic-level manipulations, first you must be able to what’s going on.” Until the 1980s researchers lacked any method for studying the surfaces on atomic scale. It was known that the arrangement of atoms on the surface differed from the bulk, but investigators had no way to determine how they were different. The was developed in the early 1980s by Binnig, Rohrer, and co-workers. The STM provides a 3D profile of the surface on a nanoscale, by tunneling electrons between a sharp conductive probe (etched Tungsten wire) and a conductive surface. The flow of electrons is very sensitive to the probe-sample distance (1-2 nm). As the probe moves across surface features the probe position is adjusted to keep the current flow constant. From this a topographic image of the surface can be obtained on an atomic scale. Binnig and Rohrer receive the Nobel Prize in Physics (1986) for their work on the STM. They shared this award with German scientist Ernst Ruska, designer of the first electron microscope. The STM that Binnig and Rohrer had built was actually based upon the field ion microscope invented by Erwin Wilhelm Müller. A precursor instrument, the , was invented by Russell Young and colleagues between 1965 and 1971 at the National Bureau of Standards (NBS). This instrument was the fundamental tool in the development of nanotechnology. It opened the door for the ability to control, see, measure, and manipulate matter on the atomic scale. Although the STM was considered a fundamental advancement for scientific research it has limited applications, as it only works for conducting or semi-conducting samples (needed for tunneling of electrons). In 1986, Binnig, Quate, and Gerber extended the field of application to non-conducting (biological, insulators etc.) by developing an The AFM provides a 3D profile of the surface on a nanoscale, by measuring between a sharp probe (<10 nm) and surface at very short distance (0.2-10 nm probe-sample separation). The probe is supported on a flexible cantilever. The STM and AFM may be applied to samples in very different environments: These microscopes work under vacuum conditions, air, and, in liquids (with specific modifications). | 2,686 | 3,904 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Half-Reactions |
A is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Often, the concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode). Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H ions to balance the hydrogen ions in the half reaction. For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH ions to balance the H ions in the half reactions (which would give H O). Galvanic cell Consider the Galvanic cell shown in the image to the right: it is constructed with a piece of zinc (Zn) submerged in a solution of zinc sulfate (ZnSO ) and a piece of copper (Cu) submerged in a solution of copper(II) sulfate (CuSO ). The overall reaction is: At the Zn anode, oxidation takes place (the metal loses electrons). This is represented in the following oxidation half-reaction (note that the electrons are on the products side): At the Cu cathode, reduction takes place (electrons are accepted). This is represented in the following reduction half-reaction (note that the electrons are on the reactants side): Consider the example burning of magnesium ribbon (Mg). When magnesium burns, it combines with oxygen (O2) from the air to form magnesium oxide (MgO) according to the following equation: Magnesium oxide is an ionic compound containing Mg and O ions whereas Mg(s) and O (g) are elements with no charges. The Mg(s) with zero charge gains a +2 charge going from the reactant side to product side, and the O (g) with zero charge gains a -2 charge. This is because when Mg(s) becomes Mg , it loses 2 electrons. Since there are 2 Mg on left side, a total of 4 electrons are lost according to the following oxidation half reaction: On the other hand, O was reduced: its oxidation state goes from 0 to -2. Thus, a reduction half-reaction can be written for the O as it gains 4 electrons: The overall reaction is the sum of both half-reactions: When chemical reaction, especially, redox reaction takes place, we do not see the electrons as they appear and disappear during the course of the reaction. What we see is the reactants (starting material) and end products. Due to this, electrons appearing on both sides of the equation are canceled. After canceling, the equation is re-written as Two ions, positive (Mg ) and negative (O ) exist on product side and they combine immediately to form a compound magnesium oxide (MgO) due to their opposite charges (electrostatic attraction). In any given oxidation-reduction reaction, there are two half-reactions – oxidation half- reaction and reduction half-reaction. The sum of these two half-reactions is the oxidation- reduction reaction. Consider the reaction below: The two elements involved, iron and chlorine, each change oxidation state; iron from +2 to +3, chlorine from 0 to -1. There are then effectively two -reactions occurring. These changes can be represented in formulas by inserting appropriate electrons into each half-reaction: Given two half-reactions it is possible, with knowledge of appropriate electrode potentials, to arrive at the full (original) reaction the same way. The decomposition of a reaction into half-reactions is key to understanding a variety of chemical processes. For example, in the above reaction, it can be shown that this is a redox reaction in which Fe is oxidised, and Cl is reduced. Note the transfer of electrons from Fe to Cl. Decomposition is also a way to simplify the balancing of a chemical equation. A chemist can atom balance and charge balance one piece of an equation at a time. For example: It is also possible and sometimes necessary to consider a half-reaction in either basic or acidic conditions, as there may be an acidic or basic electrolyte in the redox reaction. Due to this electrolyte it may be more difficult to satisfy the balance of both the atoms and charges. This is done by adding H O, OH , e , and or H to either side of the reaction until both atoms and charges are balanced. Consider the half-reaction below: PbO → PbO OH , H O, and e can be used to balance the charges and atoms in basic conditions. 2e + H O + PbO → PbO + 2OH Again Consider the half-reaction below: PbO → PbO H , H O, and e can be used to balance the charges and atoms in acidic conditions. 2e + 2H + PbO → PbO + H O Notice that both sides are both charge balanced and atom balanced. Often there will be both H and OH present in acidic and basic conditions but that the resulting reaction of the two ions will yield water H O (shown below): H + OH → H O | 5,096 | 3,905 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/09%3A_Spices/9.04%3A_Functions_of_Salt_in_Baking |
Salt has three major functions in baking. It affects: (1) Fermentation, (2) Dough, and (3) conditioning Flavor Fermentation is salt’s major function: Salt has a binding or strengthening effect on gluten and thereby adds strength to any flour. The additional firmness imparted to the gluten by the salt enables it to hold the water and gas better, and allows the dough to expand without tearing. This influence becomes particularly important when soft water is used for dough mixing and where immature flour must be used. Under both conditions, incorporating a maximum amount of salt will help prevent soft and sticky dough. Although salt has no direct bleaching effect, its action results in a fine-grained loaf of superior texture. This combination of finer grain and thin cell walls gives the crumb of the loaf a whiter appearance. One of the important functions of salt is its ability to improve the taste and flavor of all the foods in which it is used. Salt is one ingredient that makes bread taste so good. Without salt in the dough batch, the resulting bread would be flat and insipid. The extra palatability brought about by the presence of salt is only partly due to the actual taste of the salt itself. Salt has the peculiar ability to intensify the flavor created in bread as a result of yeast action on the other ingredients in the loaf. It brings out the characteristic taste and flavor of bread and, indeed, of all foods. Improved palatability in turn promotes the digestibility of food, so it can be said that salt enhances the nutritive value of bakery products. The lack of salt or too much of it is the first thing noticed when tasting bread. In some bread 2% can produce a decidedly salty taste, while in others the same amount gives a good taste. The difference is often due to the mineralization of the water used in the dough. | 1,861 | 3,906 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./20%3A_Nuclear_Chemistry_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact 1. What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences. 2. What do chemists mean when they say a substance is ? 3. What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table? 4. In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with > 83 unstable? 5. What is the significance of a of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons? 6. Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers. 7. Potassium has three common isotopes, K, K, and K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of K. 8. Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is Sm more stable? 5. Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not. 7. Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability. 1. Write the nuclear symbol for each isotope using \(^A_Z \textrm X\) notation. a. chlorine-39 b. lithium-8 c. osmium-183 d. zinc-71 2.Write the nuclear symbol for each isotope using \(^A_Z \textrm X\) notation. a. lead-212 b. helium-5 c. oxygen-19 d. plutonium-242 3. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. a. iron-57 b. W c. potassium-39 d. Xe 4. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. a. technetium-99 b. La c. radium-227 d. Bi 5. Which of these nuclides do you expect to be radioactive? Explain your reasoning. a. Ne b. tungsten-184 c. Ti 6. Which of these nuclides do you expect to be radioactive? Explain your reasoning. a. Ag b. V c. lutetium-176 1. a. \(^{39}_{17} \textrm{Cl}\) b. \(^{8}_{3} \textrm{Li}\) c. \(^{183}_{76} \textrm{Os}\) d. \(^{71}_{30} \textrm{Zn}\) 3. a. 26 protons; 31 neutrons; 1.19 b. 74 protons; 111 neutrons; 1.50 c. 19 protons; 20 neutrons; 1.05 d. 54 protons; 77 neutrons; 1.43 Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements? How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed? Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer. If the laws of physics were different and the primary element in the universe were boron-11 ( = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation. The raw material for all elements with > 2 is helium ( = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons. Write a balanced nuclear reaction for the formation of each isotope. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium? | 4,765 | 3,908 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./04.E%3A_Aqueous_Reactions_(Exercises) |
. In addition to these individual basis; please contact What are the advantages to carrying out a reaction in solution rather than simply mixing the pure reactants? What types of compounds dissolve in polar solvents? Describe the charge distribution in liquid water. How does this distribution affect its physical properties? Must a molecule have an asymmetric charge distribution to be polar? Explain your answer. Why are many ionic substances soluble in water? Explain the phrase . What kinds of covalent compounds are soluble in water? Why do most aromatic hydrocarbons have only limited solubility in water? Would you expect their solubility to be higher, lower, or the same in ethanol compared with water? Why? Predict whether each compound will dissolve in water and explain why. Predict whether each compound will dissolve in water and explain why. Given water and toluene, predict which is the better solvent for each compound and explain your reasoning. Of water and toluene, predict which is the better solvent for each compound and explain your reasoning. Compound is divided into three equal samples. The first sample does not dissolve in water, the second sample dissolves only slightly in ethanol, and the third sample dissolves completely in toluene. What does this suggest about the polarity of ? You are given a mixture of three solid compounds— , , and —and are told that is a polar compound, is slightly polar, and is nonpolar. Suggest a method for separating these three compounds. A laboratory technician is given a sample that contains only sodium chloride, sucrose, and cyclodecanone (a ketone). You must tell the technician how to separate these three compounds from the mixture. What would you suggest? Many over-the-counter drugs are sold as ethanol/water solutions rather than as purely aqueous solutions. Give a plausible reason for this practice. What distinguishes a weak electrolyte from a strong electrolyte? Which organic groups result in aqueous solutions that conduct electricity? It is considered highly dangerous to splash barefoot in puddles during a lightning storm. Why? Which solution(s) would you expect to conduct electricity well? Explain your reasoning. Which solution(s) would you expect to conduct electricity well? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in aqueous solution? Explain your reasoning. Ionic compounds such as NaCl are held together by electrostatic interactions between oppositely charged ions in the highly ordered solid. When an ionic compound dissolves in water, the partially negatively charged oxygen atoms of the H O molecules surround the cations, and the partially positively charged hydrogen atoms in H O surround the anions. The favorable electrostatic interactions between water and the ions compensate for the loss of the electrostatic interactions between ions in the solid. An is any compound that can form ions when it dissolves in water. When a strong electrolyte dissolves in water, it dissociates completely to give the constituent ions. In contrast, when a weak electrolyte dissolves in water, it produces relatively few ions in solution. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation? Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. Which representation best corresponds to an aqueous solution originally containing each of the following? 1 M Ba(OH) + 1 M H SO Which representation in Problem 3 best corresponds to an aqueous solution originally containing each of the following? 3 What mass of precipitate would you expect to obtain by mixing 250 mL of a solution containing 4.88 g of Na CrO with 200 mL of a solution containing 3.84 g of AgNO ? What is the final nitrate ion concentration? Adding 10.0 mL of a dilute solution of zinc nitrate to 246 mL of 2.00 M sodium sulfide produced 0.279 g of a precipitate. How many grams of zinc(II) nitrate and sodium sulfide were consumed to produce this quantity of product? What was the concentration of each ion in the original solutions? What is the concentration of the sulfide ion in solution after the precipitation reaction, assuming no further reaction? 3.75 g Ag CrO ; 5.02 × 10 M nitrate Why was it necessary to expand on the Arrhenius definition of an acid and a base? What specific point does the Brønsted–Lowry definition address? State whether each compound is an acid, a base, or a salt. State whether each compound is an acid, a base, or a salt. Classify each compound as a strong acid, a weak acid, a strong base, or a weak base in aqueous solution. Decide whether each compound forms an aqueous solution that is strongly acidic, weakly acidic, strongly basic, or weakly basic. What is the relationship between the strength of an acid and the strength of the conjugate base derived from that acid? Would you expect the CH CO ion to be a strong base or a weak base? Why? Is the hydronium ion a strong acid or a weak acid? Explain your answer. What are the products of an acid–base reaction? Under what circumstances is one of the products a gas? Explain how an aqueous solution that is strongly basic can have a pH, which is a measure of the of a solution. Derive an equation to relate the hydrogen ion concentration to the molarity of a solution of a strong monoprotic acid. Derive an equation to relate the hydroxide ion concentration to the molarity of a solution of Given the following salts, identify the acid and the base in the neutralization reactions and then write the complete ionic equation: What is the pH of each solution? What is the hydrogen ion concentration of each substance in the indicated pH range? What is the hydrogen ion concentration of each substance in the indicated pH range? What is the pH of a solution prepared by diluting 25.00 mL of 0.879 M HCl to a volume of 555 mL? Vinegar is primarily an aqueous solution of acetic acid. Commercial vinegar typically contains 5.0 g of acetic acid in 95.0 g of water. What is the concentration of commercial vinegar? If only 3.1% of the acetic acid dissociates to CH CO and H , what is the pH of the solution? (Assume the density of the solution is 1.00 g/mL.) If a typical household cleanser is 0.50 M in strong base, what volume of 0.998 M strong monoprotic acid is needed to neutralize 50.0 mL of the cleanser? A 25.00 mL sample of a 0.9005 M solution of HCl is diluted to 500.0 mL. What is the molarity of the final solution? How many milliliters of 0.223 M NaOH are needed to neutralize 25.00 mL of this final solution? If 20.0 mL of 0.10 M NaOH are needed to neutralize 15.0 mL of gastric fluid, what is the molarity of HCl in the fluid? (Assume all the acidity is due to the presence of HCl.) What other base might be used instead of NaOH? Malonic acid (C H O ) is a diprotic acid used in the manufacture of barbiturates. How many grams of malonic acid are in a 25.00 mL sample that requires 32.68 mL of 1.124 M KOH for complete neutralization to occur? Malonic acid is a dicarboxylic acid; propose a structure for malonic acid. Describe how you would prepare 500 mL of a 1.00 M stock solution of HCl from an HCl solution that is 12.11 M. Using your stock solution, how would you prepare 500 mL of a solution that is 0.012 M in HCl? Given a stock solution that is 8.52 M in HBr, describe how you would prepare a 500 mL solution with each concentration. How many moles of solute are contained in each? A chemist needed a solution that was approximately 0.5 M in HCl but could measure only 10.00 mL samples into a 50.00 mL volumetric flask. Propose a method for preparing the solution. (Assume that concentrated HCl is 12.0 M.) Write the balanced chemical equation for each reaction. Write the balanced chemical equation for each reaction. A neutralization reaction gives calcium nitrate as one of the two products. Identify the acid and the base in this reaction. What is the second product? If the product had been cesium iodide, what would have been the acid and the base? What is the complete ionic equation for each reaction? [H O ] = [HA] M pH = 1.402 25 mL 0.13 M HCl; magnesium carbonate, MgCO , or aluminum hydroxide, Al(OH) 1.00 M solution: dilute 41.20 mL of the concentrated solution to a final volume of 500 mL. 0.012 M solution: dilute 12.0 mL of the 1.00 M stock solution to a final volume of 500 mL. The acid is nitric acid, and the base is calcium hydroxide. The other product is water. \(2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O\) The acid is hydroiodic acid, and the base is cesium hydroxide. The other product is water. \( HI + CsOH \rightarrow CsI + H_2O \) The complete ionic equations are \( 2H^+ + 2NO_3^- + Ca^{2+} + 2OH^- \rightarrow Ca^{2+} + 2NO_3^- + H_2O\)
\( H^+ + I^- + Cs^+ + OH^- \rightarrow Cs^+ + I^- + H_2O \) Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants? If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why? In each redox reaction, determine which species is oxidized and which is reduced: Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is a single-displacement reaction. Balance each redox reaction under the conditions indicated. Balance each redox reaction under the conditions indicated. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Dentists occasionally use metallic mixtures called for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu (aq) and nitric oxide gas. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers. Na SO Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers. Would you expect a 1.0 M solution of CaCl to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not? An alternative way to define the concentration of a solution is , abbreviated . Molality is defined as the number of moles of solute in 1 kg of . How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 solution of sucrose? Explain your answer. What are the advantages of using solutions for quantitative calculations? If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately. Calculate the number of grams of solute in 1.000 L of each solution. Calculate the number of grams of solute in 1.000 L of each solution. If all solutions contain the same solute, which solution contains the greater mass of solute? Complete the following table for 500 mL of solution. What is the concentration of each species present in the following aqueous solutions? What is the concentration of each species present in the following aqueous solutions? What is the molar concentration of each solution? What is the molar concentration of each solution? Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant. An experiment required 200.0 mL of a 0.330 M solution of Na CrO . A stock solution of Na CrO containing 20.0% solute by mass with a density of 1.19 g/cm was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na CrO using the stock solution. Calcium hypochlorite [Ca(OCl) ] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl) concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite? Phenol (C H OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol? If a tablet containing 100 mg of caffeine (C H N O ) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution? A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered? 0.48 M ClO 1.74 × 10 M caffeine The titration procedure is an application of the use of limiting reactants. Explain why this is so. Explain how to determine the concentration of a substance using a titration. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH) with 0.10 M H SO , and the other corresponds to the titration of 100 mL of 0.10 M NaOH with 0.10 M H SO . Which graph corresponds to which titration? Justify your answer. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample? What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization? A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution? | 16,134 | 3,909 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Cyclic_Voltammetry |
Cyclic Voltammetry (CV) is an electrochemical technique which measures the current that develops in an electrochemical cell under conditions where voltage is in excess of that predicted by the . CV is performed by cycling the potential of a working electrode, and measuring the resulting current. The potential of the working electrode is measured against a reference electrode which maintains a constant potential, and the resulting applied potential produces an excitation signal such as that of figure 1.² In the forward scan of figure 1, the potential first scans negatively, starting from a greater potential (a) and ending at a lower potential (d). The potential extrema (d) is call the switching potential, and is the point where the voltage is sufficient enough to have caused an oxidation or reduction of an analyte. The reverse scan occurs from (d) to (g), and is where the potential scans positively. Figure 1 shows a typical reduction occurring from (a) to (d) and an oxidation occurring from (d) to (g). It is important to note that some analytes undergo oxidation first, in which case the potential would first scan positively. This cycle can be repeated, and the scan rate can be varied. The slope of the excitation signal gives the scan rate used. A cyclic voltammogram is obtained by measuring the current at the working electrode during the potential scans.² Figure 2 shows a cyclic voltammogram resulting from a single electron reduction and oxidation. Consider the following reversible reaction: \[M^+ + e^- \rightleftharpoons M \] In Figure 2, the reduction process occurs from (a) the initial potential to (d) the switching potential. In this region the potential is scanned negatively to cause a reduction. The resulting current is called cathodic current (i ). The corresponding peak potential occurs at (c), and is called the cathodic peak potential (E ). The Epc is reached when all of the substrate at the surface of the electrode has been reduced. After the switching potential has been reached (d), the potential scans positively from (d) to (g). This results in anodic current (I ) and oxidation to occur. The peak potential at (f) is called the anodic peak potential (E ), and is reached when all of the substrate at the surface of the electrode has been oxidized. Electrode potential (\(E\)): \[ E = E_i + vt \tag{1}\] where When the direction of the potential sweep is switched, the equation becomes, \[ E = E_s - vt \tag{2}\] Where \(E_s\) is the potential at the switching point. Electron stoichiometry (\(n\)): \[E_p - E_{p/2} > \dfrac{0.0565}{n} \tag{3} \] where Formal Reduction Potential (E°’) is the of the \(E_{pc}\) and \(E_{pa}\) values: \[E°’ = \dfrac{E_{pa} + E_{pc}}{2}.\] In an unstirred solution, mass transport of the analyte to the electrode surface occurs by diffusion alone.¹ for mass transfer diffusion relates the distance from the electrode (x), time (t), and the reactant concentration (CA) to the diffusion coefficient (DA). \[ \dfrac{\partial c_A}{\partial t} = D_A \dfrac{\partial^2c_A}{\partial x^2} \tag{4} \] During a reduction, current increases until it reaches a peak: when all M+ exposed to the surface of an electrode has been reduced to M. At this point additional M+ to be reduced can travel by diffusion alone to the surface of the electrode, and as the concentration of M increases, the distance M+ has to travel also increases. During this process the current which has peaked, begins to decline as smaller and smaller amounts of M+ approach the electrode. It is not practical to obtain limiting currents Ipa, and Ipc in a system in which the electrode has not been stirred because the currents continually decrease with time.¹ In a stirred solution, a Nernst diffusion layer ~10 cm thick, lies adjacent to the electrode surface. Beyond this region is a laminar flow region, followed by a turbulent flow region which contains the bulk solution.¹ Because diffusion is limited to the narrow Nernst diffusion region, the reacting analytes cannot diffuse into the bulk solution, and therefore Nernstian equilibrium is maintained and diffusion-controlled currents can be obtained. In this case, Fick’s Law for mass transfer diffusion can be simplified to give the peak current \[ i_p = (2.69 \; x \; 10^5) \; n^{3/2} \; SD_A^{1/2} \; V^{1/2} \; C_A^* \tag{5} \] Here, (n) is equal to the number of electrons gained in the reduction, (S) is the surface area of the working electrode in cm², (DA) is the diffusion coefficient, (v) is the sweep rate, and (CA) is the molar concentration of A in the bulk solution. A CV system consists of an electrolysis cell, a potentiostat, a current-to-voltage converter, and a data acquisition system. The electrolysis cell consists of a working electrode, counter electrode, reference electrode, and electrolytic solution. The working electrode’s potential is varied linearly with time, while the reference electrode maintains a constant potential. The counter electrode conducts electricity from the signal source to the working electrode. The purpose of the electrolytic solution is to provide ions to the electrodes during oxidation and reduction. A potentiostat is an electronic device which uses a dc power source to produce a potential which can be maintained and accurately determined, while allowing small currents to be drawn into the system without changing the voltage. The current-to-voltage converter measures the resulting current, and the data acquisition system produces the resulting voltammogram. Cyclic Voltammetry can be used to study qualitative information about electrochemical processes under various conditions, such as the presence of intermediates in oxidation-reduction reactions, the reversibility of a reaction. CV can also be used to determine the electron stoichiometry of a system, the diffusion coefficient of an analyte, and the formal reduction potential, which can be used as an identification tool. In addition, because concentration is proportional to current in a reversible, Nernstian system, concentration of an unknown solution can be determined by generating a calibration curve of current vs. concentration. | 6,182 | 3,910 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Enzymatic_Kinetics/Ping-pong_mechanisms |
The simplest of enzymes will involve one substrate binding to the enzyme and producing a product plus the enzyme. However, the majority of enzymes are more complex and catalyze reactions involving multiple substrates. Binding of two substrates can occur through two mechanisms: sequential mechanism and non-sequential mechanism. In sequential mechanisms both substrates bind the enzyme and the reaction proceeds to form products which are then released from the enzyme. This mechanism can be further subdivided into random and ordered reactions. For random reactions the order in which the substrates bind does not matter. In ordered reactions one substrate must bind the enzyme before the second substrate is able to bind. Non-Sequential mechanism does not require both substrates to bind before releasing the first product. This page will focus on the non-sequential mechanism, which is also known as the "ping-pong" mechanism. It is called this because the enzyme bounces back and forth from an intermediate state to its standard state.The enzyme acts like a ping-pong ball, bouncing from one state to another. Ping-pong mechanism, also called a double-displacement reaction, is characterized by the change of the enzyme into an intermediate form when the first substrate to product reaction occurs. It is important to note the term intermediate indicating that this form is only temporary. At the end of the reaction the enzyme MUST be found in its original form. An enzyme is defined by the fact that it is involved in the reaction and is not consumed. Another key characteristic of the ping-pong mechanism is that one product is formed and released before the second substrate binds. The figure below explains the Ping Pong mechanism through an enzymatic reaction. This image shows that as substrate A binds to the enzyme, enzyme-substrate complex EA forms. At this point, the intermediate state, E* forms. P is released from E* , then B binds to E*. B is converted to Q, which is released as the second product. E* becomes E, and the process can be repeated. Often times, E* contains a fragment of the original substrate A.This fragment can alter the function of the enzyme, gets attached to substrate B, or both. Here is another diagram showing this same reaction: An example of the ping-pong mechanism would be the action of chymotrypsin. When reacted with p-nitrophenyl acetate (A), the reaction of chymotrypsin is seen to occur in two steps. In the first step, the substrate reacts extremely fast with the enzyme, leading to the formation of a small amount of p-nitrophenolate (P). In the second step, the substrate-enzyme interaction results in the formation of acetate ion (Q). The action of chymotrypsin is a ping-pong reaction because the binding of the two substrates causes the enzyme to switch back and forth between two states. Please refer to the section Chymotrypsin and pre-steady-state enzyme kinetics for more details on the action of chymotrypsin. Another example of an enzyme that exhibits a ping-pong mechanism is pyruvate carboxylase. This enzyme catalyzes the addition of carbon dioxide to pyruvate in order to form oxaloacetate. (leads to gluconeogenesis) This biotin-containing enzyme works by binding CO (A) to form carboxybiotin (EA). The biotin swings over towards pyruvate (E*P) and releases CO (P, due to the fact that it had been moved from its original binding site) Pyruvate (B), in close proximity to CO , attacks the partial positive of Carbon in CO (E*B). Oxaloacetate is formed within the enzyme (EQ) and gets released (Q). While this attack is occurring, biotin swings back to its initial position, (E* --> E) and is ready to bind another CO . An important factor to understand about the ping-pong mechanism is that when plotting a 1/v and 1/[A] plot at varying concentrations of B, a series of parallel lines are seen. In this case A is the first substrate and B being the second substrate. Refer to these sections on enzyme kinetics and michaelis-menten kinetics to get a better understanding of what this type of plot means. | 4,137 | 3,911 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.09%3A_Phase_Transitions |
We have now looked at the physical properties which chemists use to define the solid, liquid, and gas phases. In a solid, atoms, ions or molecules, are locked into an organized, long range lattice structure, unable to move beyond an average position due to intermolecular forces. In a liquid, this structure breaks down, molecules can slip past each other, but they are still held together by attractive forces. In a gas, these attractive forces are overcome, and the substance expands to fill space, each particle having gained mobility to break free of the others. Below, all 3 phases are shown at the submicroscopic level in animations. Notice how the movement and freedom of molecules steadily increases as attractive forces decrease from solid to liquid to gas phase. Substances can be transformed from one phase into another. Solids melt into liquids and liquids boil to form vapors at temperatures which depend on their molecular properties, so chemists are interested in these transitions between phases. We are all familiar with the changes in macroscopic properties that accompany these transitions. YouTube has time lapse movies of ice melting on a , or of the more environmentally critical arctic from 1979 to 2007. This is a familiar process. As the solid melts, the resulting liquid is able to flow and conforms to the shape of the container. Heat from a flame is needed to bring about this transition. On a microscopic level melting involves breaking the intermolecular interactions between molecules. This s, and the necessary energy is supplied by the Bunsen burner. Melting (or freezing) can, in some cases, be caused by changing just the pressure. is equally familiar. Under specific temperature and pressure conditions, liquids start to bubble, and are converted to a gaseous form. A YouTube video shows that water boils at low temperatures if the pressure is reduced. Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy. In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to boil at higher temperatures. It should be noted as well, that because there is a distribution in the kinetic energies of molecules, an is established at temperatures other than the boiling point, and this behavior is another aspect of phase transitions that chemists study. For phase transitions from solid to liquid, liquid to gas, or solid to gas, energy is required because they involve separation of particles which attract one another. Further, we can predict under which such transitions will occur. | 2,815 | 3,913 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.01%3A_Overview_of_Distillation |
Several distillation variations are used in the organic laboratory depending on the properties of the mixture to be purified. The apparatus in Figure 5.1 is used to perform a simple distillation and is used if the components have widely different boiling points (greater than a 100 °C difference in boiling points). If a simple distillation is attempted on a mixture where the components have more similar boiling points (less than a 100 °C difference in boiling points), it will fail to purify the mixture completely. Instead, fractional distillation can be used to improve the chances of purification. Vacuum distillation may be used when the boiling points of the mixture's components are very high (>150 °C), or steam distillation if the components are very water insoluble. The distillation variations are summarized in Table 5.1, and are discussed in detail in this chapter. (Butte Community College). is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International . Complete text is available . | 1,052 | 3,914 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Combination_Reactions |
describe a reaction like this: \[A + B \rightarrow C\] in which two or more reactants become one product (are combined). The problem with this term is that it doesn't give you much chemical insight because there are many different types of reactions that follow this pattern. So we'll break it into groups that reflect what's actually happening a little better. In this category, an elemental metal and an elemental non-metal react to make an ionic substance that is neutral and has each ion in its correct charge state or valence. For instance, \[2Na(s) + Cl_{2}(g) → 2NaCl(s)\] \[2Mg(s) + O_{2}(g) → 2MgO(s)\] \[2Al(s) + 3O_{2}(g) → Al_{2}O_{3}(s)\] If the metal is a transition metal, it will be much harder to predict the correct charge on the metal ion in the ionic compound. You can check the element info in the section or the links from the section. As you practice, you'll start to get a sense for what common charges are, but even then it is often good to check, because it might not be what you expect! For example, what's the charge on iron in Fe O (magnetite)? Under what circumstances do these reactions happen? Often, an elemental metal and non-metal "want" to make an ionic compound, because this is a more stable state (think about a heavy ball on a table: it can easily roll to the ground, where it has less potential energy, so the table isn't a stable state; if the heavy ball is in a small hole in the ground, it can't easily move, and if it did, it would have more potential energy, so the hole is a stable, low energy state). However, that doesn't necessarily mean the reaction will just happen on its own. That depends on how easily the reaction can happen (think about a place you want to go, but don't go because traveling there is very inconvenient). For instance, the alkali metals and the halogens react pretty easily, so they will often react even without a "push." Oxygen is very reactive, which is why things burn, but you have to light them on fire to get them started. This is good, because otherwise we would burn in air at room temperature! Many of these elemental combination reactions might need a high temperature to get started, even if they want to happen. It won't be hard to remember that alkalis, alkaline earth metals and halogens react easily, because this is why they are very hard to find in elemental form! Oxygen and nitrogen are very abundant in elemental form because it is hard for them to react even if they want to. Nitrogen in particular reacts only with lithium metal and a few complicated compounds at room temperature, although it will react with many other elements at high temperatures. Most metals aren't found in elemental form in nature (except for ones that don't want to react, like gold), but if you find them in elemental form in your house, then probably they don't react easily. These reactions involve elemental forms of elements like H, C, N, O, Cl, S, P, etc. It will often be hard to predict the product because these elements can often combine in different ratios (this is where the comes from!). You can always expect that H will have a valence of 1, and O will usually have a valence of 2. Many of these reactions will happen quickly if you get them started with a little heat, especially if oxygen or a halogen is involved. Otherwise, they might happen very slowly or not at all except under special circumstances that we will talk more about later. Some examples: \[C(s) + O_{2}(g) → CO_{2}(g)\; (fast,\; once\; lit)\] \[N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)\; (usually\; very\; slow)\] are compounds that turn into a base (a hydroxide salt) when you add water. They are metal oxides. Here's an example: \[CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(aq)\] If the metal is an alkali or alkaline earth, the reaction probably happens quickly and produces a lot of heat. If the metal is a transition metal, the reaction might not happen so easily or at all. are compounds that turn into an acid when you add water. They are non-metal oxides. These are a little more complicated than basic anhydrides, so don't worry too much about them right now. Here's an example: \[SO_{3}(g) + H_{2}O(l) \rightarrow H_{2}SO_{4}(aq)\] There are many other circumstances in which a combination reaction could happen. The types listed here are the simple ones that are good to know in the beginning. | 4,383 | 3,915 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Synthesis_of_Ethers/Williamson_Ether_Synthesis |
The Williamson Ether synthesis is the easiest, and perhaps the fastest, way to create . Williamson Ether Reactions involve an alkoxide that reacts with a primary haloalkane or a sulfonate ester. consist of the conjugate base of an alcohol and are comprised of an R group bonded to an oxygen atom. They are often written as RO , where R is the organic substituent. S 2 reactions are characterized by the of stereochemistry at the site of the leaving group. Williamson Ether synthesis is not an exception to this rule and the reaction is set in motion by the backside attack of the nucleophile. This requires that the nucleophile and the electrophile are in anti-configuration. Ethers can be synthesized in standard S 2 conditions by coupling an alkoxide with a haloalkane/sulfonate ester. The alcohol that supplies the electron rich alkoxide can be used as the solvent, as well as dimethyl sulfoxide (DMSO) or hexamethylphosphoric triamide (HMPA). For example You can also use the Williamson synthesis to produce cyclic ethers. You need a molecule that has a hydroxyl group on one carbon and a halogen atom attached to another carbon. This molecule will then undergo an S 2 reaction with itself, creating a cyclic ether and a halogen anion. Another way of deriving ethers is by converting halo alcohols into cyclic ethers. This reaction is prompted by the deprotonation of the hydrogen attached to the oxygen by an OH anion. This leads to the departure of the halogen, forming a cyclic ether and halogen radical. Another factor in determining whether a cyclic ether will be formed is ring size. Three-membered rings along with five membered rings form the fastest, followed by six, four, seven, and lastly eight membered rings. The relative speeds of ring formation are influenced by both enthalpic and entropic contributions. Ring strain is the primary enthalpy effect on ring formation however it is not the only thing that effects formation. If this were the case, rings with the most strain would be formed the slowest. The reason why this is not the trend for ring formation is because of entropy conditions. Smaller rings have less entropy making them more favorable because of less ordering of the molecule. However, the reason why ring formation does not follow this trend is because of another factor called the proximity effect. The proximity effect states that the nucleophilic part of the carbon chain is so close to the electrophilic carbon that a small amount of ring strain is evident in the ground state of the molecule. However, as the ring size increases above 4 this proximity effect is trumped by the strong reduction in . Five and six membered rings have less strain allowing them to form faster. However, as rings get larger (8,9,10 etc. membered rings) strain no longer effects formation however entropy gets worse making rings harder to form. | 2,888 | 3,916 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/03%3A_The_Atmosphere/3.02%3A_Origin_and_Evolution_of_the_Atmosphere |
The atmosphere of the Earth (and also of Venus and Mars) is generally believed to have its origin in relatively volatile compounds that were incorporated into the solids from which these planets accreted. Such compounds could include nitrides (a source of N ), water (which can be taken up by silica, for example), carbides, and hydrogen compounds of nitrogen and carbon. Many of these compounds (and also some noble gases) can form clathrate complexes with water and some minerals which are fairly stable at low temperatures. The high temperatures developed during the later stages of accretion as well as subsequent heating produced by decay of radioactive elements presumably released these gases the surface. Even at the present time, large amounts of CO , water vapor, N , HCl, SO and H S are emitted from volcanos. The more reactive of these gases would be selectively removed from the atmosphere by reaction with surface rocks or dissolution in the ocean, leaving an atmosphere enriched in its present major components with the exception of oxygen which is discussed in the next section. Any hydrogen present would tend to escape into space, causing the atmosphere to gradually became less reducing. However there is now some doubt that hydrogen and other volatiles (mainly the inert gases) were present in the newly accreted planets in anything like their cosmic abundance. The main evidence against this is the observation that gases such as helium, neon and argon, which are among the ten most abundant elements in the universe, are depleted on the earth by factors of 10 to 10 . This implies that there was a selective removal of these volatiles prior to or during the planetary accretion process. The overall oxidation state of the earth’s mantle is not consistent with what one would expect from equilibration with highly reduced volatiles, and there is no evidence to suggest that the composition of the mantle has not remained the same. If this is correct, then the primitive atmosphere may well have had about the same composition as the gases emitted by volcanos at the present time. These consist mainly of water and CO , together with small amounts of N , H , H S, SO , CO, CH , NH , HCl and HF. If water vapor was a major component of outgassing of the accreted earth, it must have condensed quite rapidly into rain. Any significant concentration of water vapor in the atmosphere would have led to a runaway greenhouse effect, resulting in temperatures as high as 400°C. Free oxygen is never more than a trace component of most planetary atmospheres. Thermodynamically, oxygen is much happier when combined with other elements as oxides; the pressure of O in equilibrium with basaltic magmas is only about 10 atm. Photochemical decomposition of gaseous oxides in the upper atmosphere is the major source of O on most planets. On Venus, for example, CO is broken down into CO and O . On the earth, the major inorganic source of O is the photolysis of water vapor; most of the resulting hydrogen escapes into space, allowing the O2 concentration to build up. An estimated 2 10 g of O per year is generated in this way. Integrated over the earth’s history, this amounts to less than 3% of the present oxygen abundance. The partial pressure of O in the prebiotic atmosphere is estimated to be no more than 10 atm, and may have been several orders of magnitude less. The major source of atmospheric oxygen on the earth is photosynthesis carried out by green plants and certain bacteria: \[\ce{H2O + CO2 -> (CH2O)_x + O2}\] A historical view of the buildup of atmospheric oxygen concentration since the beginning of the sedimentary record (3.7 10 ybp) can be worked out by making use of the fact that the carbon in the product of the above reaction has a slightly lower C content than does carbon of inorganic origin. Isotopic analysis of carbon-containing sediments thus provides a measure of the amounts of photosynthetic O produced at various times in the past. \(\Page {1}\): Carbon dioxide has probably always been present in the atmosphere in the relatively small absolute amounts now observed (around 54 x 10 mol = 54 Pmol). The reaction of CO with silicate-containing rocks to form precambrian limestones suggest a possible moderating influence on its atmospheric concentration throughout geological time. \[\ce{ CaSiO_3 + CO_2 \rightarrow CaCO_3 + SiO_2 }\] About ten percent of the atmospheric \(CO_2\) is taken up each year by . Of this, all except 0.05 percent is returned by , almost entirely due to microorganisms. The remainder leaks into the slow part of the geochemical cycle, mostly as buried carbonate sediments. Since the advent of the industrial revolution in 1860, the amount of \(CO_2\) in the atmosphere has been increasing. Isotopic analysis shows that most of this has been due to fossil-fuel combustion; in recent years, the mass of carbon released to the atmosphere by this means has been more than ten times the estimated rate of natural removal 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 \(CO_2\). The oceans have a large absorptive capacity for \(CO_2\) owing to its reaction with carbonate: \[ \ce{CO_2 + CO_3^{2–} \rightleftharpoons 2 HCO_3^–}\] There is about 60 times as much inorganic carbon in the oceans as in the atmosphere. However, efficient transfer takes place only into the topmost (100 m) wind-mixed layer of the ocean, which contains only about one atmosphere equivalent of \(CO_2\). Further uptake is limited by the very slow transport of water into the deep ocean, which takes around 1000 years. For this reason, the buffering effect of the oceans on atmospheric \(CO_2\) is not very effective; only about ten percent of the added \(CO_2\) is taken up by the oceans. ) | 5,910 | 3,917 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.11%3A_Bonding_in_Semiconductors |
The band theory of solids provides a clear set of criteria for distinguishing between conductors (metals), insulators and semiconductors. As we have seen, a conductor must posses an upper range of allowed levels that are only partially filled with valence electrons. These levels can be within a single band, or they can be the combination of two overlapping bands. A band structure of this type is known as a . An is characterized by a large band gap between the highest filled band and an even higher empty band. The band gap is sufficiently great to prevent any significant population of the upper band by thermal excitation of electrons from the lower one. The presence of a very intense electric field may be able to supply the required energy, in which case the insulator undergoes . Most molecular crystals are insulators, as are covalent crystals such as diamond. If the band gap is sufficiently small to allow electrons in the filled band below it to jump into the upper empty band by thermal excitation, the solid is known as a . In contrast to metals, whose electrical conductivity decreases with temperature (the more intense lattice vibrations interfere with the transfer of momentum by the electron fluid), the conductivity of semiconductors increases with temperature. In many cases the excitation energy can be provided by absorption of light, so most semiconductors are also . Examples of semiconducting elements are Se, Te, Bi, Ge, Si, and graphite. The presence of an impurity in a semiconductor can introduce a new band into the system. If this new band is situated within the forbidden region, it creates a new and smaller band gap that will increase the conductivity. The huge semiconductor industry is based on the ability to tailor the band gap to fit the desired application by introducing an appropriate impurity atom ( ) into the semiconductor lattice. The dopant elements are normally atoms whose valance shells contain one electron more or less than the atoms of the host crystal. At absolute zero, all of the charge carriers reside in lower of the bands below the small band gap in a semiconductor (that is, in the valence band of the illustration on the left above, or in the impurity band of the one on the right.) At higher temperatures, thermal excitation of the electrons allows an increasing fraction jump across this band gap and populate either the empty impurity band or the conduction band as shown at the right. The effect is the same in either case; the semiconductor becomes more conductive as the temperature is raised. Note that this is just the opposite to the way temperature affects the conductivity of metals. For example, a phosphorus atom introduced as an impurity into a silicon lattice possesses one more valence electron than Si. This electron is delocalized within the impurity band and serves as the charge carrier in what is known as an . In a semiconductor of the P-type, the dopant might be arsenic, which has only three valence electrons. This creates what amounts to an electron deficiency or in the electron fabric of the crystal, although the solid remains electrically neutral overall. As this vacancy is filled by the electrons from silicon atoms the vacancy hops to another location, so the charge carrier is effectively a positively charged hole, hence the designation. Substitution of just one dopant atom into 10 atoms of Si can increase the conductivity by a factor of 100,000. When P- and N-type materials are brought into contact, creating a . Holes in the P material and electrons in the N material drift toward and neutralize each other, creating a depletion region that is devoid of charge carriers. But the destruction of these carriers leaves immobile positive ions in the N material and negative ions in the P material, giving rise to an interfacial potential difference (" ") as depicted here. As this charge builds up, it acts to resist the further diffusion of electrons and holes, leaving a carrier-free , which acts as a barrier at the junction interface. | 4,062 | 3,918 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.04%3A_Kinetic_Energy |
Using expressions for \(v_{mp}\), \(v_{ave}\), or \(v_{rms}\), it is fairly simple to derive expressions for kinetic energy from the expression \[E_{kin} = \dfrac{1}{2} mv^2 \nonumber \] It is important to remember that there will be a full distribution of molecular speeds in a thermalized sample of gas. Some molecules will be traveling faster and some more slowly. It is also important to recognize that the most probable, average, and RMS kinetic energy terms that can be derived from the Kinetic Molecular Theory do not depend on the mass of the molecules (Table 2.4.1). As such, it can be concluded that the average kinetic energy of the molecules in a thermalized sample of gas depends only on the temperature. However, the average speed depends on the molecular mass. So, for a given temperature, light molecules will travel faster on average than heavier molecules. The expression for the root-mean-square molecular speed can be used to show that the Kinetic Molecular model of gases is consistent with the ideal gas law. Consider the expression for pressure \[ p =\dfrac{N_{tot}m}{3V} \langle v \rangle^2 \nonumber \] Replacing \(\langle v \rangle^2\) with the square of the RMS speed expression yields \[ p = \dfrac{N_{tot}m}{3V} \left( \dfrac{3k_BT}{m}\right) \nonumber \] which simplifies to \[ p = \dfrac{N_{tot}k_BT}{V} \nonumber \] Noting that N = n∙N , where n is the number of moles and N is Avogadro’s number \[ p = \dfrac{nN_Ak_BT}{V} \nonumber \] or \[ pV = nN_Ak_BT \nonumber \] Finally, noting that \(N_A∙k_B = R\) \[ pV = nRT \nonumber \] That’s kind of cool, no? The only assumptions (beyond the postulates of the Kinetic Molecular Theory) is that the distribution of velocities for a thermalized sample of gas is described by the Maxwell-Boltzmann distribution law. The next development will be to use the Kinetic Molecular Theory to describe molecular collisions (which are essential events in many chemical reactions.) In the derivation of an expression for the pressure of a gas, it is useful to consider the frequency with which gas molecules collide with the walls of the container. To derive this expression, consider the expression for the “collision volume”. \[V_{col} = v_x \Delta t\ \cdot A \nonumber \] All of the molecules within this volume, and with a velocity such that the x-component exceeds v (and is positive) will collide with the wall. That fraction of molecules is given by \[ N_{col} = \dfrac{N}{V} \dfrac{\langle v \rangle \Delta t \cdot A}{2} \nonumber \] and the frequency of collisions with the wall per unit area per unit time is given by \[Z_w = \dfrac{N}{V} \dfrac{\langle v \rangle}{2} \nonumber \] In order to expand this model into a more useful form, one must consider motion in all three dimensions. Considering that \[\langle v \rangle = \sqrt{\langle v_x \rangle +\langle v_y \rangle +\langle v_z \rangle} \nonumber \] and that \[\langle v_x \rangle = \langle v_y \rangle =\langle v_z \rangle \nonumber \] it can be shown that \[ \langle v \rangle = 2 \langle v_x \rangle \nonumber \] or \[ \langle v_x \rangle = \dfrac{1}{2} \langle v \rangle \nonumber \] and so \[Z_w = \dfrac{1}{4} \dfrac{N}{V} \langle v \rangle \nonumber \] The factor of N/V is often referred to as the “number density” as it gives the number of molecules per unit volume. At 1 atm pressure and 298 K, the number density for an ideal gas is approximately 2.5 x 10 molecule/cm . (This value is easily calculated using the ideal gas law.) By comparison, the average number density for the universe is approximately 1 molecule/cm . | 3,579 | 3,921 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.22%3A_Silicon_Dioxide |
Silicon dioxide, or silica, (SiO ) is another important example of a macromolecular solid. Silica can exist in six different crystalline forms. The best known of these is , whose crystal structure shown previously is shown again below. Sand consists mainly of small fragments of quartz crystals. Quartz has a very high melting point, though not so high as diamond. If you refer back to the , you can remind yourself of the reason that SiO is macromolecular. Silicon is reluctant to form multiple bonds, and so discrete molecules, analogous to , do not occur. In order to satisfy silicon’s valence of 4 and oxygen’s valence of 2, each silicon must be surrounded by four oxygens and each oxygen by two silicons. This can be represented schematically by the Lewis diagram | 786 | 3,922 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Halogenation_of_Alkanes |
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane, shown below, provides a simple example of this reaction. CH + Cl + energy → CH Cl + HCl Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following . The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride. CH + Cl + energy → CH Cl + CH Cl + CHCl + CCl + HCl The following facts must be accomodated by any reasonable mechanism for the halogenation reaction. The most plausible mechanism for halogenation is a chain reaction involving neutral intermediates such as free radicals or atoms. The weakest covalent bond in the reactants is the halogen-halogen bond (Cl-Cl = 58 kcal/mole; Br-Br = 46 kcal/mole) so the initiating step is the homolytic cleavage of this bond by heat or light, note that chlorine and bromine both absorb visible light (they are colored). A chain reaction mechanism for the chlorination of methane has been described. Bromination of alkanes occurs by a similar mechanism, but is slower and more selective because a bromine atom is a less reactive hydrogen abstraction agent than a chlorine atom, as reflected by the higher bond energy of H-Cl than H-Br. When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers? \[CH_3CH_2CH_3 + 2Cl_2 \rightarrow \text{Four} \; C_3H_6Cl_2 \; \text{isomers} + 2 HCl\] The halogenation of propane discloses an interesting feature of these reactions. . For example, propane has eight hydrogens, six of them being structurally equivalent , and the other two being . If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane. CH -CH -CH + → 45% CH -CH -CH + 55% CH -CH -CH The results of bromination ( light-induced at 25 ºC ) are even more suprising, with 2-bromopropane accounting for 97% of the mono-bromo product. CH -CH -CH + → 3% CH -CH -CH + 97% CH -CH -CH These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule. (CH ) CH + → 65% (CH ) C + 35% (CH ) CHCH It should be clear from a review of the two steps that make up the free radical chain reaction for halogenation that the first step (hydrogen abstraction) is the . Once a carbon radical is formed, subsequent bonding to a halogen atom (in the second step) can only occur at the radical site. Consequently, an understanding of the preference for substitution at 2º and 3º-carbon atoms must come from an analysis of this first step. First Step: R CH + → R C + H- Second Step: R C + → R C + Since the H-X product is common to all possible reactions, differences in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table. The difference in C-H bond dissociation energy reported for primary (1º), secondary (2º) and tertiary (3º) sites agrees with the halogenation observations reported above, in that we would expect weaker bonds to be broken more easily than are strong bonds. By this reasoning we would expect benzylic and allylic sites to be exceptionally reactive in free radical halogenation, as experiments have shown. The methyl group of toluene, C H , is readily chlorinated or brominated in the presence of free radical initiators (usually peroxides), and ethylbenzene is similarly chlorinated at the benzylic location exclusively. The hydrogens bonded to the aromatic ring (referred to as phenyl hydrogens above) have relatively high bond dissociation energies and are not substituted. Since carbon-carbon double bonds add chlorine and bromine rapidly in liquid phase solutions, free radical substitution reactions of alkenes by these halogens must be carried out in the gas phase, or by other halogenating reagents. One such reagent is N-bromosuccinimide (NBS), shown in the second equation below. By using NBS as a brominating agent, allylic brominations are readily achieved in the liquid phase. The covalent bond homolyses that define the bond dissociation energies listed above may are described by the general equation: C-H bond dissociation energies are commonly interpreted in terms of radical stability. Most importantly, in terms of the selectivity of free radical reactions, it is the energies of the bonds that matter, and not why they are what they are. | 6,266 | 3,923 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Catalytic_Hydrogenation |
The catalytic addition of hydrogen to 2-butyne not only serves as an example of such an addition reaction, but also provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown in the diagram below. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ). Since alkynes are thermodynamically less stable than alkenes, we might expect addition reactions of the former to be more exothermic and relatively faster than equivalent reactions of the latter. In the case of catalytic hydrogenation, the usual Pt and Pd hydrogenation catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, , prepared by deactivating (or poisoning) a conventional palladium catalyst by treating it with lead acetate and quinoline, permits alkynes to be converted to alkenes without further reduction to an alkane. The addition of hydrogen is stereoselectively syn (e.g. 2-butyne gives cis-2-butene). A complementary stereoselective reduction in the anti mode may be accomplished by a solution of sodium in liquid ammonia. This reaction will be discussed later in this section. Alkenes and alkynes show a curious difference in behavior toward catalytic hydrogenation. Independent studies of hydrogenation rates for each class indicate that alkenes react more rapidly than alkynes. However, careful hydrogenation of an alkyne proceeds exclusively to the alkene until the former is consumed, at which point the product alkene is very rapidly hydrogenated to an alkane. This behavior is nicely explained by differences in the stages of the hydrogenation reaction. Before hydrogen can add to a multiple bond the alkene or alkyne must be adsorbed on the catalyst surface. In this respect, the formation of stable platinum (and palladium) complexes with alkenes has been described earlier. Since alkynes adsorb more strongly to such catalytic surfaces than do alkenes, they preferentially occupy reactive sites on the catalyst. Subsequent transfer of hydrogen to the adsorbed alkyne proceeds slowly, relative to the corresponding hydrogen transfer to an adsorbed alkene molecule. Consequently, reduction of triple bonds occurs selectively at a moderate rate, followed by rapid addition of hydrogen to the alkene product. The Lindlar catalyst permits adsorption and reduction of alkynes, but does not adsorb alkenes sufficiently to allow their reduction. | 3,238 | 3,924 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.12%3A_Polypeptide_Chains |
The backbone of any protein molecule is a polypeptide chain obtained by the condensation of a large number of amino acids with the elimination of water. You will recall that the amino acids are bifunctional organic nitrogen compounds containing an acid group, —COOH, and an amine group, —NH . The amine group is attached to the carbon atom adjacent to the —COOH (the ). The three simplest amino acids are In practice, though, these acids are usually in the form of their zwitterions, and we should write them If these three amino acids are now condensed, water is eliminated and a simple polypeptide contaning three amino acids is obtained: Figure \(\Page {3}\) The two CO—NH bonds produced by this reaction are called . Notice that the peptide bond is an amide linkage. An important feature of such a peptide bond is that it is planar. This is because of the existence of two resonance structures Another important aspect of the peptide bond is the opportunity it provides for hydrogen bonding. The oxygen on the carbonyl group can bond to the hydrogen on an group further along the chain: Such a bond is somewhat stronger than a normal hydrogen bond because of the partially negative character of the oxygen atom and the partially positive character of the nitrogen atom conferred by resonance structure II. | 1,334 | 3,925 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.08_Acid-Base_Properties_of_Salts |
Salts, when placed in water, will often react with the water to produce H O or OH . This is known as a hydrolysis reaction. Based on how strong the ion acts as an acid or base, it will produce varying pH levels. When water and salts react, there are many possibilities due to the varying structures of salts. A salt can be made of either a weak acid and strong base, strong acid and weak base, a strong acid and strong base, or a weak acid and weak base. The reactants are composed of the salt and the water and the products side is composed of the conjugate base (from the acid of the reaction side) or the conjugate acid (from the base of the reaction side). In this section of chemistry, we discuss the pH values of salts based on several conditions. There are several guiding principles that summarize the outcome: Do not be intimidated by the salts of polyprotic acids. Yes, they're bigger and "badder" then most other salts. But they can be handled the exact same way as other salts, just with a bit more math. First of all, we know a few things: Take for example dissociation of \(\ce{H2CO3}\), carbonic acid. \[\ce{H2CO3(aq) + H2O(l) <=> H3O^{+}(aq) + HCO^{-}3(aq)} \nonumber\] with \(K_{a1} = 2.5 \times 10^{-4} \) \[\ce{HCO^{-}3(aq) + H2O(l) <=> H3O^{+}(aq) + CO^{2-}3(aq)} \nonumber \] with \(K_{a2} = 5.61 \times 10^{-11}\). This means that when calculating the values for K of CO , the K of the first hydrolysis reaction will be \(K_{b1} = \dfrac{K_w}{K_{a2}}\) since it will go in the reverse order. From weak bases NH , Al , Fe From strong acids: Cl , Br , I , NO , ClO From weak acids: F , NO , CN , CH COO From strong bases: Group 1 and Group 2, but not Be . From strong acids: Cl , Br , I , NO , ClO 1 \(NaOCl _{(s)} \rightarrow Na^+_{(aq)} + OCl^-_{(aq)}\) While Na will not hydrolyze, OCl will (remember that it is the conjugate base of HOCl). It acts as a base, accepting a proton from water. \(OCl^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HOCl_{(aq)} + OH^-_{(aq)}\) Na is excluded from this reaction since it is a spectator ion. Therefore, with the production of OH , it will cause a basic solution and raise the pH above 7. \(pH>7\) \[KCN_{(s)}\rightarrow K^+_{(aq)} + CN^-_{(aq)}\] K will not hydrolyze, but the CN anion will attract an H away from the water: \[CN^-_{(aq)} + H_2O_{(l)}\rightleftharpoons HCN_{(aq)} + OH^-_{(aq)}\] Because this reaction produces OH , the resulting solution will be basic and cause a pH>7. \(pH>7\) \[NH_4NO_{3(s)} \rightarrow NH^+_{4(aq)} + NO^-_{3(aq)}\] Now, NO won't attract an H because it is usually from a strong acid. This means the K will be very small. However, NH will lose an electron and act as an acid (NH is the conjugate acid of NH ) by the following reaction: \[NH^+_{4(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + H_3O^+_{(aq)}\] This reaction produces a hydronium ion, making the solution acidic, lowering the pH below 7. \(pH<7\) \(\dfrac{x^2}{0.2-x} = \dfrac{1*10^{-14}}{1.8 \times 10^{-5}}\) \(x = 1.05*10^-5 M = [H_3O^+]\) \(pH = 4.98\) The majority of the hydroxide ion will come from this first step. So only the first step will be completed here. To complete the other steps, follow the same manner of this calculation. \[\dfrac{x^2}{0.2-x}=\dfrac{1*10^-14}{3.98 \times 10{-13}}\] \[x = 0.0594 = [OH^-]\] \[pH = 12.77\] The answers to these questions can be found in the attached files section at the bottom of the page. | 3,439 | 3,926 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/London_Dispersion_Forces |
It's not too hard to see why dipole-dipole forces hold molecules like HF or H O together in the solid or liquid phase. However, let's think about the halogens. F and Cl are gases, Br is a liquid, and I is a solid at room temperature. But I has no dipole moment to make attractions between the molecules. But actually, although I has no permanent dipole moment, it can have a temporary dipole moment. We mentioned this before, when we talked about . Go back and read that section. can explain how liquids and solids form in molecules with no permanent dipole moment. "Dispersion" means the way things are distributed or spread out. Because the electrons move around a lot, sometimes they may move in a way that creates a temporary dipole moment. The more electrons an atom has, the more easily this can happen, because the electrons are held more loosely, far from the nucleus. (Basically, the energy gaps between orbitals become smaller as we move to higher shells, allowing the electrons to more easily move into excited states, occupying orbitals higher than they need to. This gives them more flexibility to move around and create temporary dipole moments.) The technical word for an element that is polarizable, or able to have temporary dipoles, is "soft". In other words, it can squish and change shape. Elements that can't polarize easily (which usually means low atomic number) are called "hard". Fluorine is really really hard. In F , both F atoms are holding all the electrons really tightly, trying to grab them and not share. In contrast, iodine is really soft. It's electrons are far away from the nucleus, and they can move around easily. If they all happen to move one direction, creating a temporary dipole, the other molecules nearby can adjust, making more dipoles to attract the first one. These are called induced dipoles, because they appear in response to the original accidental dipole. Lots of induced dipoles can create attraction between molecules, called London dispersion forces. London dispersion forces are always present, but they vary widely in strength. In light atoms, they are very small, because there aren't many electrons and they are held tightly. In large atoms, they can be very big, because the atoms are very soft and easy to polarize. Generally, London dispersion forces depend on the atomic or molecular weight of the material. Heavier atoms or molecules have more electrons, and stronger London forces. This means that they are harder to melt or boil. This explains the states of the halogen molecules at room temperature. | 2,598 | 3,927 |
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 |
The numerical examples and the images in this section have some problems ( with comments). Make sure you thoroughly understand the following essential concepts that have been presented above. 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−. 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 smaller) than 1 /L, while that of undissociated HA will be only slightly less than 1 /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 . Then, in our "1 " 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 or concentration which is commonly denoted by . 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, = 0.10 . 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 \[[HCOOH] + [HCOO^–] = C_a\] For the general case of an acid HA, we can write a \[C_a = [HA] + [A^–] \label{1-3}\] which reminds us the "A" part of the acid must always be somewhere! For a such as hydrochloric, its total dissociation means that [HCl] = 0, so the mass balance relationship in Equation \(\ref{1-3}\) reduces to the trivial expression = [Cl-]. Any acid for which [HA] > 0 is by definition a weak acid. Similarly, for a base B we can write \[C_b = [B] + [HB^+] \label{1-4}\] "Concentration of the acid" and [HA] are typical the same. 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 O). Let us represent these concentrations by . Then, in a solution containing 1 /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 . The presence of terms in both and here tells us that this is a quadratic equation. In most practical cases in which is 10 or smaller, we can assume that is much smaller than 1 , allowing us to make the simplifying approximation \[(1 – ) \approx 1 \label{1-7}\] so that \(x^2 \approx Ka\] and thus \[x = \sqrt{K_a} \label{1-8}\] This approximation will not generally be valid when the acid is or 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 , is 1 . We can easily generalize this to solutions in which 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 — either attached to the hydrogen, or in the form of the hydrated anion A . The corresponding equilibrium expression is (1-10) and the approximations (when justified) 1-3a and 1-3b become ( – ) ≈ (1-11) x ≈ ( ) (1-12) Estimate the pH of a 0.20 M solution of acetic acid, = 1.8 × 10 . For brevity, we will represent acetic acid CH 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 The rather small value of suggests that we can drop the x term in the denominator, so that The pH of the solution is –log (1.9E–3) = Even though we know that the process HA → H + A does not correctly describe the transfer of a proton to H 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). It's important to understand that whereas for a given acid is essentially a constant, \(\alpha\) of the acid. Note that these equations are also valid for weak bases if and are used in place of and . This can be shown by substituting Eq 5 into the expression for : (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 approaches zero, \(\alpha\) approaches unity and [HA] approaches . This principle is an instance of the 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 O → H O + A , thereby causing this equilibrium to shift to the right. The error here is that [H O] in most aqueous solutions is so large (55.5 M) that it can be considered constant; this is the reason the [H O] term does not appear in the expression for . Another common explanation is that dilution reduces [H 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 ) Plots of this kind are discussed in more detail in the next lesson in this set under the heading . The Le Chatelier principle predicts that the extent of the reaction \[\ce{HA → H^{+} + A^{–} }\] 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 , 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? The dissociation stoichiometry HA → H + AB tells us the concentrations [H ] and [A ] will be identical. Thus [H ] = 10 = 0.025 M = [A ]. The dissociation fraction \[α = \dfrac{[\ce{A^{–}}]}{[\ce{HA}]} = \dfrac{0.025}{0.75} = 0.033\] and thus the acid is dissociated at 0.75 M concentration. Sometimes the percent dissociation is given, and must be evaluated. A weak acid HA is 2 percent dissociated in a 1.00 M solution. Find the value of . 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 When dealing with problems involving acids or bases, bear in mind that when we speak of "the concentration", we usually mean the or concentration which is commonly denoted by . 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, = 0.10 . 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 ] = . For the general case of an acid HA, we can write 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 = [B] + [HB ](1-17) If we represent the dissociation of a solution of a weak acid by (1-18) then its dissociation constant is given by (1-19) Because the term is in the denominator here, we see that the amount of HA that dissociates varies inversely with the concentration; as approaches zero, [HA] approaches . 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 does not exceed 5% of ,
the -term in the denominator can be dropped, yielding ≈ (1-22) Note that the above equations are also valid for weak bases if Kb and are used in place of and . Compare the percent dissociation of 0.10 M and .0010 M solutions of boric acid (\(K_a = 3.8 \times 10^{-10}\)). Boric acid is sufficiently weak that we can use the approximation of Eq 1-22 to calculate a: = (5.8E–10 / .1) = 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%. 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. As we pointed out in the preceding lesson, the "effective" value of an equilibrium constant (the ) 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 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. 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 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 : (2-1) Substituting these values into the expression for \(K_a\), we obtain (2-2) If you understand the concept of mass balance on "A" expressed in (2-1), and can write the expression for , you just substitute the '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 worry! In order to predict the pH of this solution, we must first find [H ], that is, . The presence of terms in both and here tells us that this is a quadratic equation. This can be rearranged into = (1 – ) 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. 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. The reason for this is that if >> |4 |, 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 \[Q = –\dfrac{b + \pm (b) \sqrt{ b^2 – 4ac}}{2}\] from which the roots are = / and = / . (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 values of (two ) that satisfy a quadratic equation. For all acid-base equilibrium calculations that are properly set up, these roots will be , and only one will be positive; this is the one you take as the answer. We have already encountered two of these approximations in the examples of the preceding section: 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. If the acid is fairly concentrated (usually with > 10 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 reduces to [H ] / so that [H ] ( ) (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 exceeds 5 percent of the value it is being subtracted from (0.10 in the present case), then the approximation is justified. We will call this the "five percent rule". - For brevity, we will represent acetic acid CH 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 ( / 0.20) = 1.8E-5 or = (0.20 × 1.8E–5) = 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
= [H ] ≈ 1.9 × 10 M, and the pH will be –log (1.9 × 10 ) = Percent dissociation: 100% × (1.9 × 10 M) / (0.20 M) = This plot shows the combinations of and that generally yield satisfactory results with the approximation of Eq 4. Weak bases are treated in an exactly analogous way: Methylamine CH NH is a gas whose odor is noticed around decaying fish. A solution of CH NH 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 for methylamine. When methylamine "ionizes", it takes up a proton from water, forming the methylaminium ion: CH NH + H O → CH NH + OH Let = [CH NH ] = [OH ] = .064 × 0.10 = 0.0064 [CH NH ] = (0.10 – .064) = 0.094 Substitute these values into equilibrium expression for \Kb: \[K_b = \frac{0.0064^2}{0.094}\] 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
-term in the denominator. = – log \Kb = – log (4.4 × 10 ) = However, one does not always get off so easily! With a of 0.010, HClO 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. \[0.010 = \frac{x^2}{0.10 - x}\] (i) The approximation 0.10 – x ≈ 0.10 gives us ≈ ( ) = (0.010 ×0.10) = (.001) = .032 (ii) The difficulty, in this case, arises from the numerical value of differing from the nominal concentration 0.10 M by only a factor of 10. As a result,
/ = .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). In the , you start with the value of [H ] (that is, ) you calculated according to , which becomes the first approximation. You then substitute this into , 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 , = 0.010, using the method of successive approximations. The equilibrium condition is \[K_{\mathrm{a}} = \dfrac{x \cdot x}{c_{\mathrm{a}} - x}\] We solve this for (neglecting the n the denominator), resulting in the first approximation , and then successively plug each result into the previous equation, yielding approximations and : \[x_1 = \sqrt{K_{\mathrm{a}} \cdot c_\mathrm{a}} =0.032 \] \[x_2 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_1)} =0.026 \] \[x_3 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_2)} =0.027 \] The last two approximations and are within 5% of each other. Note that if we had used as the answer, the error would have been 18%. (An exact numeric solution yields the roots 0.027016 and –0.037016) 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 M solution of hypochlorous acid (HOCl, = 2.9E–8) was found by plotting the value of
y = + x + , whose roots are the two values of that correspond to = 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. 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" ( , 2.7E-11.) Of those that do, the one at the is highly recommended; others can be found 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 + x + = 0, insert values for
, and , 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. Estimate the pH of a 0.10 M aqueous solution of HClO , = 0.010. The reaction equation HClO → H + ClO – defines the equilibrium expression \[0.010 = \dfrac{x \cdot x}{0.10 - x}\] Multiplying by the denominator yields = 0.010 × (0.10 – ) = .0010 – .01 which we arrange into standard polynomial form: + 0.01 – 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 = . 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! 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 . 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. "Hydro-lysis" literally means "water splitting", as exemplified by the reaction A + H 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 — a term still used by chemists. Some examples: The HCO ion is therefore : it can both accept and donate protons, so both processes take place: However, if we compare the and of HCO , it is apparent that its basic nature wins out, so a solution of NaHCO will be slightly alkaline. (The value of is found by recalling that + = 14.) The protons can either come from the cation itself (as with the ammonium ion NH ), 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 O) → Fe(H O) OH + H (2-5) This comes about because the positive field of the metal enhances the ability of H 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. The aluminum ion exists in water as hexaaquoaluminum Al(H O) , whose = 4.9, = 10 = 1.3E–5. Setting x = [H ] = [Al(H O) OH ], the equilibrium expression is Using the above approximation, we get
≈ (1.96E–6) = 1.4E–3, corresponding to . Finally, we compute x/ = 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 : \[\ce{NH_4^{+}→ NH)3(aq) + H^{+}\lable{2-6}\] Calculate the pH of a 0.15 M solution of NH4Cl. The ammonium ion is 5.5E–10. According to Eq 6 above, we can set [NH3] = [H ] = , obtaining the same equilibrium expression as in the preceding problem. Because is quite small, we can safely use the approximation 0.15 - 1 ≈ .015, which yields pH = –log 0.90E–5 = 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 O HF + OH (2-7) Find the pH of a 0.15 M solution of NaF. (HF = 6.7E–4) The reaction is F- + H O = HF + OH ; because HF is a weak acid, the equilibrium strongly favors the right side. The usual approximation yields However, on calculating x/ = .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 + 6.7E–4 – 1.0E–4 = 0 and enter the coefficients {1 6.7E–4 –.0001} into a quadratic solver. This yields the positive root = 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! 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 and HCOO-. Formic acid, the simplest organic acid, has a of 3.7; for NH , = 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: Inspection reveals that the last equation above is the sum of the first two,plus the reverse of the dissociation of water The value of 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 is several orders of magnitude greater than or , 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. Estimate the pH of a 0.0100 M solution of ammonium formate in water. From the stoichiometry of HCOONH , [NH ] = [HCOO ] and [NH ] = [HCOOH] (i) then, from Eq 8 above, (ii) in which is the base constant of ammonia, /10 . From the formic acid dissociation equilibrium we have (iii) We now rewrite the expression for (iv) which yields (v) and thus the pH is What is interesting about this last example is that the pH of the solution is apparently independent of the concentration of the salt. If = , 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. Polyprotic acids form multiple anions; those that can themselves donate protons, and are thus amphiprotic, are called . The most widely known of these is the (hydrogen carbonate) ion, HCO , which we commonly know in the form of its sodium salt NaHCO as . The other analyte series that is widely encountered, especially in biochemistry, is those derived from phosphoric acid: 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 . The exact treatment of these systems is generally rather complicated, but for the special cases in which the successive '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 (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 's, and/or solutions that are extremely dilute, it may be necessary to consider 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. In order to keep the size of the present lesson within reasonable bounds (and to shield the sensitive eyes of beginners from the shock of confronting simultaneous equations), this material has been placed in a . A diprotic acid H A can donate its protons in two steps: H A → HA →HA and similarly, for a tripotic acid H A: In general, we can expect 2 for the "second ionization" to be smaller than 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. These acids are listed in the order of decreasing . The numbers above the arrows show the successive 's of each acid. Compare the successive values of sulfuric and oxalic acids (see their structures in the box, above right), and explain why they should be so different. The two 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 '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. With the exception of sulfuric acid (and some other seldom-encountered strong diprotic acids), most polyprotic acids have sufficiently small 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 , we must consider the three coupled equilibria \[\ce{H_2A → H^+ + HA^{–} }\,\,\, K_1\] \[\ce{HA^{–} → H^{+} + HA^{2–}} \,\,\,K_2\] \[\ce{H_2O → H+ + OH– } \,\,\,K_w\] An exact treatment of such a system of four unknowns [ ], [ ], [ ] and [H ] requires the solution of a quartic equation. If we include [OH ], it's even worse! In most practical cases, we can make some simplifying approximations: In addition to the three equilibria listed above, a solution of a polyprotic acid in pure water is subject to the following two conditions: although the distribution of species between the acid form H A and its base forms HAB and A may vary, their sum (defined as the "total acid concentration" is a constant for a particular solution: = [H A] + [HA ] + [A ] The solution may not possess a net electrical charge: [H O ] = [OH ] + [HA ] + 2 [A ] Why do we multiply [A ] by 2? moles of H O are needed in order to balance out the charge of 1 mole of A . 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. Calculate the pH of a 0.050 solution of CO in water. For H CO , = 10 = 4.5E–7, = 10 = 1.0E–14. b) Estimate the concentration of carbonate ion CO in the solution. Because and 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 ] = : (i) Can we further simplify this expression by dropping the in the denominator? Let's try: = [0.05 × (4.5E–7)] = 1.5E–4. Applying the "5-percent test", the quotient x/ must not exceed 0.05. Working this out yields (1.5E–4)/(.05) = .003, so we can avoid a quadratic. = [H ] ≈ ( )½ = [(4.5E–7) × .01] = (.001)½ = 0.032 M, and the pH = – log .032 = . We now wish to estimate [CO ] ≡ . Examining the second dissociation step, it is evident that this will consume mol/L of HCO , and produce an equivalent amount of H which adds to the quantity we calculated in . (ii) Owing to the very small value of compared to , we can assume that the concentrations of HCO 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 ]: (iii) It is of course no coincidence that this estimate of [CO ] yields a value identical with ; this is entirely a consequence of the simplifying assumptions we have made. Nevertheless, as long as << and the solution is not highly dilute, the result will be sufficiently accurate for most purposes. 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 SO . = 10 , = 0.012 Because > 1, we can assume that a solution of this acid will be completely dissociated into H O and bisulfite ions HSO . Thus the only equilibrium we need to consider is the dissociation of a 0.010 M solution of bisulfite ions. HSO + H O → SO + H O Setting [H ] = [SO ] = x, and dropping x from the denominator, yields
x ≈ (0.010 x .012) = (1.2E–4) = 0.0011 Applying the "five percent rule", we find that / = .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 + 0.022 – 1.2E–4 = 0 Inserting the coefficients {1 .022 .000012} into a quad-solver utility yields the roots 4.5E–3 and –0.0027. Taking the positive root, we obtain pH = – log (.0045) = Thus the second "ionization" of H SO 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, the building blocks of proteins, contain amino groups –NH 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 . The simplest of the twenty natural amino acids that occur in proteins is H N–CH –COOH, which we use as an example here. Solutions of glycine are distributed between the acidic-, zwitterion-, and basic species: (3-1) Although the zwitterionic species is amphiprotic, it differs from a typical ampholyte such as HCO in that it is electrically neutral owing to the cancellation of the opposite electrical charges on the amino and carboxyl groups. Amino acids are the most commonly-encountered kind of zwitterions, but other substances, such as quaternary ammonium compounds, also fall into this category. For more on Zwitterions, see this article or this UK . In the following development, we use the abbreviations H Gly (glycinium), HGly (zwitterion), and Gly (glycinate) to denote the dissolved forms. The two acidity constants are (3-2) 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 (3-4) If Eqs ii and iii in this Problem 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: This distribution diagram shows that the zwitterion is the predominant species between pH values corresponding to the two s given in Equation \(\ref{3-1}\). Calculate the pH and the concentrations of the various species in a 0.100 M solution of glycine. Substitution into Eq 4 above yields (i) The concentrations of the acid and base forms are found from their respective equilibrium constant expressions (Eqs 2): (ii) (iii) The small concentrations of these singly-charged species in relation to = 0.10 shows that the zwitterion is the only significant glycine species in the solution. | 37,302 | 3,928 |
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is usually defined as the capability for doing work. For example, a billiard ball can collide with a second ball, changing the direction or speed of motion of the latter. In such a process the motion of the first ball would also be altered. We would say that one billiard ball did work on (transferred energy to) the other. Energy due to motion is called kinetic energy and is represented by . For an object moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: \[ E_{k} = \frac{1}{2} mu^{2} \label{1} \] where If the two billiard balls mentioned above were studied in outer space, where friction due to their collisions with air molecules or the surface of a pool table would be negligible, careful measurements would reveal that their total kinetic energy would be the same before and after they collided. This is an example of the , which states that under the usual conditions of everyday life. Whenever there appears to be a decrease in energy somewhere, there is a corresponding increase somewhere else. Calculate the kinetic energy of a Volkswagen Beetle of mass 844 kg (1860 lb) which is moving at 13.4 m s (30 miles per hour). \(\large E_{k} = \frac{1}{2} m u^{2} = \frac{1}{2} \times 844 \text{ kg} \times ( 13.4 \text{ m} \text{ s}^{-1} )^{2} = 7.58 \times 10^{4} \text{ kg}\text{ m}^{2} \text{ s}^{-2}\) In other words the units for energy are derived from the SI base units kilogram for mass, meter for length, and second for time. A quantity of heat or any other form of energy may be expressed in kilogram meter squared per second squared. In honor of Joule’s pioneering work this derived unit 1 kg m s called the , abbreviated J. The Volkswagen in question could do nearly 76 000 J of work on anything it happened to run into. is energy that is stored by rising in height, or by other means. It frequently comes from separating things that attract, like rising birds are being separated from the Earth that attracts them, or by pulling magnets apart, or pulling an electrostatically charged balloon from an oppositely charged object to which it has clung. Potential Energy is abbreviated and gravitational potential energy is calculated as follows: \[\large E_{P} = mgh \tag{2} \] where Notice that E has the same units, kg m s or as kinetic energy. How high would the VW weighing 844 kg and moving at 30 mph need to rise (vertically) on a hill to come to a complete stop, if none of the stopping power came from friction? : The car's kinetic energy is 7.58 × 10 kg m s (from EXAMPLE \(\Page {1}\) ), so all of this would have to be converted to E . Then we could calculate the vertical height: \(\large E_{P} = mgh = 7.58 \times 10^{4} \text{ kg} \text{ m}^{2} \text{ s}^{-2} = 844 \text{ kg} \times 9.8 \text{m} \text {s}^{-2} \times h \) \( \large h = 9.2 \text{ m} \) Even when there is a great deal of friction, the law of conservation of energy still applies. If you put a milkshake on a mixer and leave it there for 10 min, you will have a warm, rather unappetizing drink. The whirling mixer blades do work on (transfer energy to) the milkshake, raising its temperature. The same effect could be produced by heating the milkshake, a fact which suggests that heating also involves a transfer of energy. The first careful experiments to determine how much work was equivalent to a given quantity of heat were done by the English physicist James Joule (1818 to 1889) in the 1840s. In an experiment very similar to our milkshake example, Joule connected falling weights through a pulley system to a paddle wheel immersed in an insulated container of water. This allowed him to compare the temperature rise which resulted from the work done by the weights with that which resulted from heating. Units with which to measure energy may be derived from the SI base units of by using Eq. \(\ref{1}\). Another unit of energy still widely used by chemists is the . The calorie used to be defined as the energy needed to raise the temperature of one gram of water from 14.5°C to 15.5°C but now it is defined as exactly 4.184 J. | 4,133 | 3,929 |
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You are expected to be able to define and explain the significance of terms identified in . 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 . In this lesson we will see how G varies with the of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The of the mixture is determined by ΔG° which also defines the K. 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 . You will recall that the relative concentrations of reactants and products in the equilibrium state is expressed by the . 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 ° for the reaction will be negative and the reaction will proceed to the right. But how far? If the reactants are 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. To understand how equilibrium constants relate to Δ ° 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 of the products, minus 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 \(Q\). \[\Delta{G} = \Delta G^° + RT \ln Q \label{5-5}\] The Gibbs energy \(G\) is a quantity that becomes more negative during the course of any natural process. Thus 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 \[\ce{N_2O_4 \rightarrow 2 NO_2 } \nonumber\] s \( \Delta G^o = +5.3\, \text{kJ}\) fo \(N_2O_4\) \(NO_2\). It’s very important to be aware of this distinction; that little ° symbol makes a world of difference! First, the standard Gibbs energy change 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 . It corresponds to the Gibbs energy change for a process that never really happens: the of pure N O into pure NO 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 for a given reaction, \(\Delta G\) varies continuously as the composition changes, finally reaching zero at equilibrium. \(\Delta G\) \(\ce{N_2O_4}\) or \(\ce{NO_2}\) (a \[Q = \dfrac{[NO_2]^2}{[N_2O_4]} = \pm\infty\] f \(\Delta G\), \(\Delta G\) 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 at the equilibrium composition corresponding to 77 percent conversion. Notice particularly that The detailed calculations that lead to the values shown above can be found . We are now in a position to answer the question posed earlier: if Δ ° 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 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 ° values that are highly negative or positive (±20 kJ mol , say), the equilibrium mixture will, for all practical purposes, be either [almost] "pure" reactants or products. 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 ( \[\Delta{G^°} = –RT \ln K_p \label{5-6}\] 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 , we can use any of the other forms of the equilibrium constant such as (concentrations), (mole fractions), (numbers of moles), etc. Remember, however, that for ionic solutions especially, only the , in which 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 \(\Page {3}\), 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 : \[\ce{H^{+}(aq) + OH^{–}(aq) <=> H_2O(l)} \nonumber\] * 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. 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 \[\begin{align*} K &= \exp\left(\dfrac{–79900}{8.314 \times 298}\right) \\[4pt] &= e^{32.2} = 1.01 \times 10^{–14} \end{align*}\] 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 °. 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, 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 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 with respect to the effect of the temperature on equilibrium: 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 ° 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 , through Equation \(\ref{5-7}\)) to allow Δ ° to be calculated through Equation \(\ref{5-9}\). You should now understand that for (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 O -dissociation example discussed previously. This is due to "dilution" of the products by the reactants. In (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 . Consider, for example, an equilibrium mixture of ice and liquid water. The concentration of H 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. Only at 0°C can ice and liquid water coexist in any proportion. Note that in contrast to the homogeneous N O example, there is no Gibbs energy minimum at intermediate compositions. Two reactions are said to be 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. In the above example, reaction 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 S is heated in the air, however, the sulfur is removed as rapidly as it is formed by oxidation in the highly spontaneous reaction , which supplies the Gibbs energy required to drive . The combined process, known as , is of considerable industrial importance and is one of a large class of processes employed for winning metals from their ores. | 12,979 | 3,930 |
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Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s law, after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps. Hess's Law argues that ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This is nothing more than arguing that ΔH is a state function. We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation \(\ref{5.6.1}\) can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.6.1, the first reaction produces 1 mol of solid aluminum oxide (Al O ) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation \(\ref{5.6.1}\)); the enthalpy change for this reaction is −732.5 kJ/mol of Fe O . The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.6.1); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation \(\ref{5.6.1}\)); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.6.1, the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation 5.6.1 gives the overall reaction, shown in part (d): \[\small \newcommand{\Celsius}{^{\circ}\text{C}} \begin{align*} \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> 2 Fe (l, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius)} & \Delta H = - 732.5\,\text{kJ} && \text{(a)} \\ \ce{2 Fe (l, 1758 \Celsius) &-> 2 Fe (s, 1758 \Celsius)} & \Delta H = -\phantom{0}27.6\,\text{kJ} && \text{(b)} \\ \ce{2 Fe (s, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius) &-> 2 Fe (s, 25 \Celsius) + Al2O3 (s, 25 \Celsius) } & \Delta H = -\phantom{0}91.0\,\text{kJ} && \text{(c)} \\[2ex] \hline \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> Al2O3 (s, 25 \Celsius) + 2 Fe (s, 25 \Celsius) } & \Delta H = -851.1\,\text{kJ} && \text{(d)} \\ \end{align*} \label{5.6.1} \tag{5.6.1}\] The net reaction in part (d) in Equation \(\ref{5.6.1}\) is identical to the equation for the thermite reaction that we saw in a previous section. By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure \(\Page {1}\). Comparing parts (a) and (d) in Equation \(\ref{5.6.1}\) also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation \(\ref{5.6.1}\)), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation \(\ref{5.6.1}\)). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation. When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure: When carbon is burned with limited amounts of oxygen gas (O ), carbon monoxide (CO) is the main product: \[ \left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ} \nonumber \] When carbon is burned in excess O , carbon dioxide (CO ) is produced: \[ \left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber \] Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O to give CO . two balanced chemical equations and their ΔH values enthalpy change for a third reaction We begin by writing the balanced chemical equation for the reaction of interest: \[ \left ( 3 \right ) \; \ce{CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H_{rxn}=? \nonumber \] There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give \[ \ce{2 CO (g) -> 2 C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ} \nonumber \] Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO , is the product we want in Equation 3: \[ \ce{C (s) + O2 (g) -> CO2 (s)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber \] Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant: \[ \ce{2 C (s) + 2 O2 (g) -> 2 CO2 (s)} \quad \Delta H=-787.0 \; \text{kJ} \nonumber \] Writing the resulting equations as a sum, along with the enthalpy change for each, gives \[ \begin{align*} \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = -\Delta H_1 = +221.0 \; \text{kJ} \\ \ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & = -2\Delta H_2 =-787.0 \; \text{kJ} \\[2ex] \hline \ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ} \end{align*} \nonumber \] Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O , and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2: \[ \ce{ CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H = -283.0 \; \text{kJ} \nonumber \] An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms: \[ \begin{alignat*}{3}
\text{(A)} \quad && \ce{ 2 C (s) + O2 (g) &-> \cancel{2 CO (g)}} \qquad & \Delta H_A &= \Delta H_1 &&= + 221.0 \; \text{kJ} \\
\text{(B)} \quad && \ce{ \cancel{2 CO (g)} + O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H_B && &= ? \\
\text{(C)} \quad && \ce{2 C (s) + 2 O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H &= 2 \Delta H_2 &= 2 \times \left ( -393.5 \; \text{kJ} \right ) &= -787.0 \; \text{kJ}
\end{alignat*} \] The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO . From Hess’s law, ΔH + ΔH = ΔH , and we are given ΔH for reactions A and C. Substituting the appropriate values gives \[ \begin{matrix}
-221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \\
\Delta H_{B} = -566.0
\end{matrix} \nonumber \] This is again the enthalpy change for the conversion of 2 mol of CO to CO . The enthalpy change for the conversion of 1 mol of CO to CO is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem. The reaction of acetylene (C H ) with hydrogen (H ) can produce either ethylene (C H ) or ethane (C H ): \[ \begin{matrix}
C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right ) & \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \\
C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2}
\end{matrix} \nonumber \] What is ΔH for the reaction of C H with H to form C H ? −136.3 kJ/mol of C H Hess’s Law: Hess's law is arguing t For a chemical reaction, the H is the difference in enthalpy between products and reactants; the units of ΔH are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔH . The magnitude of ΔH also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the H and the H , respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is . The H is the enthalpy change that occurs when a substance is burned in excess oxygen. ( ) | 9,544 | 3,931 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Huckel's_Rule |
In 1931, German chemist and physicist Erich Hückel proposed a rule to determine if a planar ring molecule would have aromatic properties. This rule states that if a cyclic, planar molecule has \(4n+2\; \pi\) electrons, it is . This rule would come to be known as Hückel's Rule. When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic. According to Hückel's , a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by \(n\)), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of \(4n+2 \; \pi\) electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 \(\pi\) electrons. Its first 2 \(\pi\) electrons fill the lowest energy orbital, and it has 4 \(\pi\) electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons. Confirm that benzene is aromatic. To apply the 4n+2 rule, first count the number of electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six electrons: 4n + 2 = 6
4n = 4
n = 1 For benzene, we find that n=1, which is a positive integer, so the rule is met. Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually electrons. Once this is figured out, the rule is quite straightforward. electrons lie in p . Sp atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the electrons. A simple way to know if an atom is sp hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is sp hybridized and has a p orbital. Let's look at our previous example, benzene: Each double bond ( bond) always contributes 2 electrons. Benzene has 3 double bonds, so it has 6 electrons. Hückel's Rule also applies to ions. As long as a compound has 4n+2 electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp hybridized is if an atom has 1 or more lone pairs and is attached to an sp hybridized atom, then that atom is sp hybridized also. This explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 electrons and fulfills the 4n+2 rule. So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp hybridized. But is the oxygen atom sp hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp hybridized atom, so it is sp hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are electrons? An sp hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp orbital. So, only 1 of oxygen's 2 lone electron pairs are electrons. Furan has 6 electrons and fulfills the 4n+2 rule. Using the criteria for aromaticity, determine if the following molecules are aromatic: | 4,622 | 3,932 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.15%3A_Redox_Reactions |
In addition to and , a third important class called is often encountered in aqueous solutions. The terms uction and idation are usually abbreviated to . Such a reaction corresponds to the transfer of electrons from one species to another. A simple redox reaction occurs when copper metal is immersed in a solution of silver nitrate. The solution gradually acquires the blue color characteristic of the hydrated Cu ion, while the copper becomes coated with glittering silver crystals. The reaction may be described by the net ionic Equation \[\ce{Cu(s) + 2Ag^+(aq) -> Cu^{2+}(aq) + Ag(s)}\label{1} \] We can regard this Equation as being made up from two hypothetical half-equations. In one, each copper atom loses 2 electrons: \[\ce{Cu -> Cu^{2+} + 2e^{-}}\label{2} \] while in the other, 2 electrons are acquired by 2 silver ions: \[\ce{2e^{-} + 2Ag^+ -> 2Ag}\label{3} \] If these two half-equations are added, the net result is Equation \(\ref{1}\). In other words, the reaction of copper with silver ions, described by Equation \(\ref{1}\), corresponds to the of electrons by the copper metal, as described by half-equation \(\ref{2}\), and the gain of electrons by silver ions, as described by Equation \(\ref{3}\). A species like copper which donates electrons in a redox reaction is called a , or . (A mnemonic for remembering this is emember, lectron onor = ucing agent.) When a reducing agent donates electrons to another species, it is said to the species to which the electrons are donated. In Equation \(\ref{1}\), for example, copper reduces the silver ion to silver. Consequently the half-equation \[\ce{2Ag^+ + 2e^{-} -> 2Ag} \nonumber \] is said to describe the of silver ions to silver. Species which accept electrons in a redox reaction are called , or . In Equation \(\ref{1}\) the silver ion, Ag , is the oxidizing agent. When an oxidizing agent accepts electrons from another species, it is said to that species, and the process of electron removal is called . The half-equation \[\ce{Cu -> Cu^{2+} + 2e^{-}} \nonumber \] thus describes the oxidation of copper to Cu ion. In summary, then, when a redox reaction occurs and electrons are transferred, there is always a reducing agent donating electrons and an oxidizing agent to receive them. The reducing agent, because it loses electrons, is said to be oxidized. The oxidizing agent, because it gains electrons, is said to be reduced. Copper is also oxidized by the oxygen present in air. The following video shows an example of this oxidation occurring. Write the following reaction in the form of half-equations. Identify each half-equation as an oxidation or a reduction. Also identify the oxidizing agent and the reducing agent in the overall reaction \[\ce{Zn + 2Fe^{3+} -> Zn^{2+} +2Fe^{2+}} \nonumber \] The half-equations are \(\ce{Zn -> Zn^{2+} + 2e^{-}}\) oxidation—loss of electrons \(\ce{2e^{-} + 2Fe^{3+} -> 2Fe^{2+}}\) reduction—gain of electrons Since zinc metal (Zn) has electrons, we can identify it as the reducing agent. Conversely, since iron(III) ion (Fe ) has accepted electrons, we identify it as the oxidizing agent. An alternative method of identification is to note that since zinc has been oxidized, the oxidizing must have been the other reactant, namely, iron(III). Also, since the iron(III) ion has been reduced, the zinc must be the reducing agent. Observe also that both the oxidizing and reducing agents are the and therefore appear on the -hand side of an Equation. A more complex redox reaction occurs when copper dissolves in nitric acid. The acid attacks the metal vigorously, and large quantities of the red-brown gas, nitrogen dioxide (NO ) are evolved. (NO is poisonous, and so this reaction should be done in a hood.) The solution acquires the blue color characteristic of the hydrated Cu ion. The reaction which occurs is \[\ce{Cu(s) + 2NO3^{-}(aq) + 4H3O^+(aq) -> Cu^{2+}(aq) + 2 NO2(g) + 6H2O(l)}\label{7} \] Merely by inspecting this net ionic Equation, it is difficult to see that a transfer of electrons has occurred. Clearly the copper metal has lost electrons and been oxidized to Cu , but where have the donated electrons gone? The matter becomes somewhat clearer if we break up Equation \(\ref{7}\) into half-equations. One must be \[\ce{Cu(s) -> Cu^{2+}(aq) +2e^{-}} \nonumber \] and the other is \[\ce{2e^{-} + 4H3O^+(aq) + 2NO3^{-}(aq) -> 2NO2(g) + 6H2O(l)}\label{9} \] You can verify that these are correct by summing them to obtain Equation \(\ref{7}\). The second half-equation shows that each NO ion has not only accepted an electron, but it has also accepted two protons. To further complicate matters, a nitrogen-oxygen bond has also been broken, producing a water molecule. With all this reshuffling of nuclei and electrons, it is difficult to say whether the two electrons donated by the copper ended up on an NO molecule or on an H O molecule. Nevertheless, it is still meaningful to call this a redox reaction. Clearly, copper atoms have lost electrons, while a combination of hydronium ions and nitrate ions have accepted them. Accordingly, we can refer to the nitrate ion (or nitric acid, HNO ) as the oxidizing agent in the overall reaction. half-equation \(\ref{9}\) is a reduction because electrons are accepted. | 5,306 | 3,933 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.15%3A_Kinetic_Theory_of_Gases-_Molecular_Speeds |
that increasing the temperature increases the speeds at which molecules move. We are now in a position to find just how large that increase is for a gaseous substance. Combining the ideal gas law with , we obtain \(\begin{align} & PV=nRT=\tfrac{\text{1}}{\text{3}}Nm\text{(}u^{\text{2}}\text{)}_{\text{ave}}\\ & \text{or } \text{3}RT=\frac{Nm}{n}\text{(}u^{\text{2}}\text{)}_{\text{ave}} \label{1}\end{align}\) Since is the number of molecules and m is the mass of each molecule, is the total mass of gas. Dividing total mass by amount of substance gives molar mass : \[M=\frac{Nm}{n} \nonumber \] Substituting in Eq. \(\ref{1}\), we have \(\begin{align} & \text{ 3}RT=M(u^{\text{2}})_{\text{ave}} \\ & \text{or }(u^{\text{2}})_{\text{ave}}=\frac{\text{3}RT}{M} \\ & \text{so that }u_{rms}=\sqrt{\text{(}u^{\text{2}}\text{)}_{\text{ave}}}=\sqrt{\frac{\text{3}RT}{M}}\text{ (2)} \end{align}\) The quantity is called the (rms) because it is the square root of the mean square velocity. The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Thus quadrupling the temperature of a given gas doubles the rms velocity of the molecules. Doubling this average velocity doubles the number of collisions between gas molecules and the walls of a container. It also doubles the impulse of each collision. Thus the pressure quadruples. This is indicated graphically in Figure \(\Page {1}\). Pressure is thus directly proportional to temperature, as required by Gay-Lussac’s law. The inverse proportionality between root-mean-square velocity and the square root of molar mass means that the heavier a molecule is, the slower it moves, which is verified by the examples below We can compare the rates of effusion or diffusion of a known gas with that of an unknown gas to determine the molar mass of the unknown gas. A convenient equation can be derived easily by considering the kinetic energy of individual molecules rather than moles of gas: Knowing that kinetic energy is proportional to temperature, if the two gases are at the same temperature, Dividing, \[\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \nonumber \] What is the molar mass of an unknown gas if the gas effuses through a pinhole into a vacuum at a rate of 2 mL/min, and H effuses at 11 mL/min. Assume that the rate of effusion is proportional to the gas molecule velocities. \[\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \\ \frac{4}{m_{2}} = \frac{2^{2}}{11^{2}} \\ m_{2} = 121 \nonumber \] Find the rms velocity for (a) H and (b) O molecules at 27°C. This problem is much easier to solve if we use SI units. Thus we choose = 8.314 J mol K = 8.314 kg m s mol K The rms velocities 1927 m s and 484 m s correspond to about 4300 miles per hour and 1080 miles per hour, respectively. The O molecules in air at room temperature move about 50 percent faster than jet planes, and H molecules are nearly 4 times speedier yet. Of course an O molecule would take a lot longer to get from New York to Chicago than a jet would. Gas molecules never go far in a straight line before colliding with other molecules. Now we can see the microscopic basis for Avogadro’s law. Most of the volume in H , O or any gas is empty space, and that empty space is the same for a given amount of any gas at the same temperature and pressure. This happens because the total kinetic energy of the molecules is the same for H or O or any other gas. The more energy they have, the more room the molecules can make for themselves by expanding against a constant pressure. This is illustrated in Figure \(\Page {2}\), where equal numbers of H and O molecules occupy separate containers at the same temperature and pressure. The volumes are seen to be the same. Because O molecules are 16 times heavier than H molecules, the average speed of H molecules is 4 times faster. H molecules therefore make 4 times as many collisions with walls. Based on mass, each collision of an H molecule with the wall has one-sixteenth the effect of an O collision, but an H collision has 4 times the effect of an O collision when molecular velocity is considered. The net result is that each H collision is only one-fourth as effective as an O collision. But since there are four times as many collisions, each one-fourth as effective, the same pressure results. Thus the same number of O molecules as H molecules is required to occupy the same volume at the same temperature and pressure. | 4,575 | 3,934 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.0%3A_Prelude_to_Solutions_and_Colloids |
Coral reefs are home to about 25% of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution we know as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans. Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions. | 1,234 | 3,937 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.1%3A_Valence_Bond_Theory |
As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts three-dimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding. There are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an orbital, a dumbbell shape for a orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization. describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap. The energy of the system depends on how much the orbitals overlap. Figure \(\Page {1}\) illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure \(\Page {1}\). The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the \(H_2\) molecule shown in Figure \(\Page {1}\), at the bond distance of 74 pm the system is \(7.24 \times 10^{−19}\, J\) lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, we know from our earlier description of thermochemistry that bond energies are often discussed on a per-mole basis. For example, it requires \(7.24 \times 10^{−19}\; J\) to break one H–H bond, but it takes \(4.36 \times 10^5\; J\) to break 1 mole of H–H bonds. A comparison of some bond lengths and energies is shown in Table \(\Page {1}\). We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH4 requires 439.3 kJ/mol, while breaking the first C–H bond in \(\ce{H–CH2C6H5}\) (a common paint thinner) requires 375.5 kJ/mol. In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure \(\Page {2}\) illustrates this for two orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle. The overlap of two orbitals (as in H ), the overlap of an orbital and a orbital (as in HCl), and the end-to-end overlap of two orbitals (as in Cl ) all produce , as illustrated in Figure \(\Page {3}\). A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory. A is a type of covalent bond that results from the side-by-side overlap of two orbitals, as illustrated in Figure \(\Page {4}\). In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a , that is, a plane with no probability of finding an electron. While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures suggest, O contains a double bond, and N contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter. As seen in Table \(\Page {1}\), an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond. Butadiene, C H , is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule. There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall. Identify each illustration as depicting a σ or π bond: (a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis. Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond. | 8,176 | 3,938 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.9%3A_Lewis_Acids_and_Bases |
Make sure you thoroughly understand the following essential ideas which have been presented. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. The Brønsted–Lowry concept of acids and bases defines a base as any species that can accept a proton, and an acid as any substance that can donate a proton. Lewis proposed an alternative definition that focuses on instead. According to Lewis: In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles. Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place. This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points: The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a from an , in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with. Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct. Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the of the ammonium ion. Because HF is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion. The bisulfite ion is and can act as an electron donor or acceptor. All Brønsted–Lowry bases (proton acceptors), such as OH , H O, and NH , are also electron-pair donors. Thus the Lewis definition of acids and bases does not contradict the Brønsted–Lowry definition. Rather, it expands the definition of acids to include substances other than the H ion. Electron-deficient molecules, such as BCl , contain less than an octet of electrons around one atom and have a strong tendency to gain an additional pair of electrons by reacting with substances that possess a lone pair of electrons. Lewis’s definition, which is less restrictive than either the Brønsted–Lowry or the Arrhenius definition, grew out of his observation of this tendency. A general Brønsted–Lowry acid–base reaction can be depicted in Lewis electron symbols as follows: The proton (H ), which has no valence electrons, is a Lewis acid because it accepts a lone pair of electrons on the base to form a bond. The proton, however, is just one of many electron-deficient species that are known to react with bases. For example, neutral compounds of boron, aluminum, and the other , which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Such compounds are therefore potent Lewis acids that react with an electron-pair donor such as ammonia to form an acid–base adduct, a new covalent bond, as shown here for boron trifluoride (BF ): The bond formed between a Lewis acid and a Lewis base is a because both electrons are provided by only one of the atoms (N, in the case of F B:NH ). After it is formed, however, a coordinate covalent bond behaves like any other covalent single bond. Species that are very weak Brønsted–Lowry bases can be relatively strong Lewis bases. For example, many of the group 13 trihalides are highly soluble in ethers (R–O–R′) because the oxygen atom in the ether contains two lone pairs of electrons, just as in H O. Hence the predominant species in solutions of electron-deficient trihalides in ether solvents is a Lewis acid–base adduct. A reaction of this type is shown in Figure \(\Page {1}\) for boron trichloride and diethyl ether: Many molecules with multiple bonds can act as Lewis acids. In these cases, the Lewis base typically donates a pair of electrons to form a bond to the central atom of the molecule, while a pair of electrons displaced from the multiple bond becomes a lone pair on a terminal atom. Identify the acid and the base in each Lewis acid–base reaction. reactants and products identity of Lewis acid and Lewis base In each equation, identify the reactant that is electron deficient and the reactant that is an electron-pair donor. The electron-deficient compound is the Lewis acid, whereas the other is the Lewis base. Identify the acid and the base in each Lewis acid–base reaction. Here are several more examples of Lewis acid-base reactions that be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction. We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base , of which the is only one (Table \(\Page {1}\)). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water. The limiting acid in a given solvent is the solvonium ion, such as H O (hydronium) ion in water. An acid which has more of a tendency to donate a hydrogen ion than the limiting acid will be a strong acid in the solvent considered, and will exist mostly or entirely in its dissociated form. Likewise, the limiting base in a given solvent is the solvate ion, such as OH (hydroxide) ion, in water. A base which has more affinity for protons than the limiting base cannot exist in solution, as it will react with the solvent. The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. Bases can exist in solution in liquid ammonia which cannot exist in aqueous solution: this is the case for any base which is than the hydroxide ion, but than the amide ion \(NH_2^-\). The limiting acid in liquid ammonia is the ammonium ion, which has a p value in water of 9.25. The limiting base is the amide ion, NH . This is a stronger base than the hydroxide ion and so cannot exist in aqueous solution. The p value of ammonia is estimated to be approximately 33. Any acid which is a stronger acid than the ammonium ion will be a strong acid in liquid ammonia. This is the case for acetic acid, which is completely dissociated in liquid ammonia solution. The addition of pure acetic acid and the addition of ammonium acetate have exactly the same effect on a liquid ammonia solution: the increase in its acidity: in practice, the latter is preferred for safety reasons. As with \(OH^-\) and \(H_3O^+\) in water, the strongest acid and base in \(NH_3\) is dictated by the corresponding autoprotolysis reaction of the solvent: \[2 NH_3 \rightleftharpoons NH_4^+ + NH_2^– \nonumber\] One use of non-aqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H O or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. Many familiar substances can serve as the basis of protonic solvent systems (Table \(\Page {1}\)). The extreme case is a , a medium in which the hydrogen ion is only very weakly solvated. The classic example is a mixture of antimony pentafluoride and liquid hydrogen fluoride: \[SbF_5 + HF \rightleftharpoons H^+ + SbF_6^−\] The limiting base, the hexfluoroantimonate anion \(SbF_6^−\), is so weakly attracted to the hydrogen ion that virtually any other base will bind more strongly: hence, this mixture can be used to protonate organic molecules which would not be considered bases in other solvents. It should noted that pH is undefined in aprotic solvents, which assumes presence of hydronium ions. In other solvents, the concentration of the respective solvonium/solvate ions should be used (e.g., \([NH_4^+]\) and \([NH_2^–]\) in \(NH_{3(l)}\). ) | 8,897 | 3,939 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.01%3A_Prelude_to_Nuclear_Chemistry |
In our discussion on we described the experimental basis for the idea that each atom has a small, very massive which contains protons and neutrons. Surrounding the nucleus are one or more which occupy most of the volume of the atom but make only a small contribution to its mass. Electrons (especially valence electrons) are the only subatomic particles which are involved in ordinary chemical changes, and we have spent considerable time describing the rearrangements they undergo when atoms and molecules combine. However, another category of reactions is possible in which the structures of atomic nuclei change. In such electronic structure is incidental—we are primarily interested in how the protons and neutrons are arranged before and after the reaction. Nuclear reactions are involved in transmutation of one element into another and in . In these sections we consider nuclear reactions in more detail, exploring their applications to , to the study of reaction mechanisms, to qualitative and quantitative analysis, and to estimation of the ages of objects as different as the Dead Sea scrolls and rocks from the moon. Nuclear reactions involve rearrangements of the protons and neutrons within atomic nuclei. During α particles, β particles, and γ rays are emitted, often in a radioactive series of successive reactions. Nuclear reactions may also be induced by bombarding nuclei with or . produced in this way may decay by positron emission or electron capture as well as by α , β or γ emission. Stability of nuclei depends on the neutron/proton ratio (usually between 1 and 1.6) and magic numbers of protons and neutrons. obeys a first-order rate law, and its rate is often reported in terms of half-life, the time necessary for half the radioactive nuclei to decompose. Known half-lives of isotopes such as \({}_{\text{6}}^{\text{14}}\text{C}\) and \({}_{\text{92}}^{\text{238}}\text{U}\) may be used to establish the ages of objects containing these elements, provided accurate measurements can be made of the quantity of radiation emitted. Geiger-Müller counters or scintillation counters are often used for such measurements. Other important applications of radioactive isotopes include tracer studies, where a particular type of atom can be labeled and followed throughout a reaction, and neutron activation analysis, which can determine extremely low concentrations of many elements. The relative stability of a nucleus is given by . This may be determined from the difference between the molar mass of the nucleus and the sum of the molar masses of its constituent protons and neutrons. Both , breaking apart of a heavy nucleus, and , combining of two light nuclei, can result in release of energy. Fission usually involves \({}_{\text{92}}^{\text{238}}\text{U}\) or , \({}_{\text{94}}^{\text{239}}\text{Pu}\) and these isotopes have been used in nuclear explosives and nuclear power plants. Fission products are highly radioactive. Because of the considerable damage done to living tissue by the ability of α, β and γ radiation to break bonds and form ions, emission of radioactive materials must be carefully controlled and fission power plants are quite expensive to construct. Although it promises much larger quantities of free energy and fewer harmful by-products than fission, nuclear fusion bas not yet been shown to be feasible for use in power plants. So far its only application has been in hydrogen bombs. | 3,471 | 3,941 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/10%3A_Spectroscopic_Methods/10.10%3A_Additional_Resources |
The following set of experiments introduce students to the applications of spectroscopy. Experiments are grouped into five categories: UV/Vis spectroscopy, IR spectroscopy, atomic absorption and atomic emission, fluorescence and phosphorescence, and signal averaging. The following sources provide additional information on spectroscopy in the following areas: general spectroscopy, Beer’s law, instrumentation, Fourier transforms, IR spectroscopy, atomic asorption and emission, luminescence, and applications. • Altermose, I. R. “Evolution of Instrumentation for UV-Visible Spectrophotometry: Part I,” , , A216–A223. . Gathered here are resources and experiments for analyzing multicomponent samples using mathematical techniques not covered in this textbook. | 776 | 3,943 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/13%3A_Air/13.04%3A_Outdoor_Air_Pollution |
refers to the introduction, into the atmosphere, of substances that have harmful effects on humans, other living organisms, and the environment either as solid particles, liquid droplets or gases. Air pollution can result from natural processes such as dust storms, forest fires, and volcanic eruptions, or from human activities such as biomass burning, vehicular emissions, mining, agriculture, and industrial processes. Improved technology and government policies have helped reduce most types of outdoor air pollution in many industrialized countries including the United States, in recent decades. However, outdoor air quality is still a problem in less industrialized nations, especially in megacities of rapidly industrializing nations such as China and India. Outdoor pollutants can come from sources or sources (Figure \(\Page {1}\)). Stationary sources have a fixed location, for example power plant smokestacks, burning, construction sites, farmlands and surface mines among others. Mobile sources of air pollutants move from place to place while emitting pollutants. Examples of mobile sources include vehicles, aircrafts, ships, and trains.
Pollutants are categorized into two major types based on how they originated namely primary and secondary pollutants. are those released directly from the source into the air in a harmful form. The primary pollutants that account for nearly all air pollution problems are carbon monoxide (58%), volatile organic compounds (VOCs, 11%), nitrogen oxides (15%), sulfur dioxides (13%), and particulate material (3%). are produced through reactions between primary pollutants and normal atmospheric compounds. For example, ground-level ozone forms over urban areas through reactions, powered by sunlight, between primary pollutants (oxides of nitrogen) and other atmospheric gases such as VOCs. Under the Clean Air Act, the Environmental Protection Agency (EPA) establishes air quality standards to protect public health and the environment. EPA has set national air quality standards for six common air pollutants namely: 1) carbon monoxide; 2) ground-level ozone; 3) nitrogen dioxide; 4) Sulfur dioxide; 5) lead; and 6) particulate matter (also known as particle pollution). Of the six pollutants, particle pollution and ground-level ozone are the most widespread health threats. EPA calls these pollutants "criteria" air pollutants because it regulates them by developing human health-based and/or environmentally-based criteria (science-based guidelines) for setting permissible levels. The set of limits based on human health is called primary standards. Another set of limits intended to prevent environmental and property damage is called secondary standards. Volatile organic compounds (VOCs) are carbon-containing chemicals emitted as gases from natural and human-made sources. Natural sources of VOCs include plants, the largest source, and bacteria in the guts of termites and ruminant animals. These compounds are generally oxidized to carbon monoxide and carbon dioxide in the atmosphere. VOCs are of great concern because they are precursors for the formation of ozone, a secondary air pollutant. A large number of synthetic organic chemicals such as benzene, toluene, formaldehyde, vinyl chloride, chloroform, and phenols are widely used as ingredients in countless household products. Paints, paint strippers, varnishes, many cleaning, disinfecting, cosmetic, degreasing, and hobby products all contain VOCs. Fuels are also made up of organic chemicals. All of these products can release organic compounds while you are using them, and, to some degree, when they are stored. The “new car smell” characteristic of new cars is from a complex mix of dozens of VOC. Also, concentrations of many VOCs are consistently higher indoors (up to ten times higher) than outdoors. They are often held responsible for sick building syndrome, an illness resulting from indoor pollution in which the specific cause is not identified. Smog is a mixture of air pollutants (sulfur dioxide, nitrogen oxides, ozone, and particulates) that often form over urban areas as a result of fossil fuel combustion. The term was coined from the terms “smoke” and “fog” referring to a brownish haze that pollutes the air, greatly reducing visibility and making it difficult for some people to breathe (Figure \(\Page {3}\) and \(\Page {4}\)). There are two main types of smog: industrial and photochemical smog. is produced primarily by the burning of fossil fuels which produces carbon dioxide (from complete combustion), carbon monoxides (from partial combustion), sulfur, and mercury. The sulfur reacts with other chemicals in the atmosphere producing several sulfur compounds including sulfur dioxide. These compounds along with particulate material make up industrial smog. is formed when sunlight drives chemical reactions between primary pollutants from automobiles and normal atmospheric compounds. The product is a mix of over 100 different chemicals with the most abundant being ground-level ozone. This type of smog will be discussed in more detail in section 13.5. Figure \(\Page {3}\): Smog over Almaty city, Kazakhstan. Photo by Igors Jefimovs. Wikicommons commons.wikimedia.org/wiki/C...dia/File:Smog_ over_Almaty.jpg Industrial smog or London-type smog is mainly a product of burning large amounts of high sulfur coal. Clean air laws passed in 1956 have greatly reduced smog formation in the United Kingdom; however, in other parts of the world London-type smog is still very prevalent. The main constituent of London-type smog is soot; however, these smogs also contain large quantities of fly ash, sulfur dioxide, sodium chloride and calcium sulfate particles. If concentrations are high enough, sulfur dioxide can react with atmospheric hydroxide to produce sulfuric acid, which will precipitate as acid rain. \[SO_2 + OH \rightarrow HOSO_2 \label{1} \] \[HOSO_2 + O_2 \rightarrow HO_2 + SO_3 \label{2} \] \[SO_3 + H_2O \rightarrow H_2SO_4 \label{3} \] Concerns about the harmful effects of acid rain have led to strong pressure on industry to minimize the release of SO and NO. For example, coal-burning power plants now use SO “scrubbers,” which trap SO by its reaction with lime (CaO) to produce calcium sulfite dihydrate (CaSO ·2H O; Figure \(\Page {5}\)). Toxic air pollutants, also known as hazardous air pollutants, are those pollutants that are known or suspected to cause cancer or other serious health effects, such as reproductive effects or birth defects, or adverse environmental effects. Examples of toxic air pollutants include , which is found in gasoline; which is emitted from some dry cleaning facilities; , which is used as a solvent and paint stripper by a number of industries; and others such as dioxin, asbestos, toluene, and metals such as cadmium, mercury, chromium, and lead compounds. Most air toxics originate from human-made sources, including mobile sources (e.g., cars, trucks, buses) and stationary sources (e.g., factories, refineries, power plants), as well as indoor sources (e.g., some building materials and cleaning solvents). Some air toxics are also released from natural sources such as volcanic eruptions and forest fires. Exposure to air toxics is mainly through breathing but some toxic air pollutants such as mercury can deposit onto soils or surface waters, where they are taken up by plants and ingested by animals and are eventually magnified up through the food chain. Like humans, animals may experience health problems if exposed to sufficiently high quantities of air toxics over time. The World Health Organization (WHO) and other international agencies recognize air pollution as a major threat to human health. Numerous scientific studies have linked air pollution to a variety of health problems (Table \(\Page {1}\)) including: aggravation of respiratory and cardiovascular diseases; decreased lung function; increased frequency and severity of respiratory symptoms such as difficulty breathing and coughing; increased susceptibility to respiratory infections; effects on the nervous system, including the brain, such as IQ loss and impacts on learning, memory, and behavior; cancer; and premature death. Immediate effects of air pollution may show up after a single exposure or repeated exposures. Other health effects may show up either years after exposure has occurred or only after long or repeated periods of exposure. Immediate effects of air pollution include irritation of the eyes, nose, and throat, headaches, dizziness, and fatigue. Such immediate effects are usually short-term and treatable. Sometimes the treatment is simply eliminating the person's exposure to the source of the pollution, if it can be identified. Symptoms of some diseases, including asthma, hypersensitivity pneumonitis, and humidifier fever , may also show up soon after exposure to some indoor air pollutants. www.epa.gov The likelihood of immediate reactions to air pollutants depends on several factors. Age and preexisting medical conditions are two important influences. Some sensitive individuals appear to be at greater risk for air pollution-related health effects, for example, those with pre-existing heart and lung diseases (e.g., heart failure/ischemic heart disease, asthma, emphysema, and chronic bronchitis), diabetics, older adults, and children. In other cases, whether a person reacts to a pollutant depends on individual sensitivity, which varies tremendously from person to person. Some people can become sensitized to biological pollutants after repeated exposures, and it appears that some people can become sensitized to chemical pollutants as well. ( ) | 9,714 | 3,944 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/04%3A_Evaluating_Analytical_Data/4.07%3A_Detection_Limits |
The International Union of Pure and Applied Chemistry (IUPAC) defines a method’s detection limit as the smallest concentration or absolute amount of analyte that has a signal significantly larger than the signal from a suitable blank [IUPAC Compendium of Chemical Technology, Electronic Version]. Although our interest is in the amount of analyte, in this section we will define the detection limit in terms of the analyte’s signal. Knowing the signal you can calculate the analyte’s concentration, , or the moles of analyte, , using the equations \[S_A = k_A C_A \text{ or } S_A = k_A n_A \nonumber\] where is the method’s sensitivity. See for a review of these equations. Let’s translate the IUPAC definition of the detection limit into a mathematical form by letting represent the average signal for a method blank, and letting \(\sigma_{mb}\) represent the method blank’s standard deviation. The null hypothesis is that the analyte is not present in the sample, and the alternative hypothesis is that the analyte is present in the sample. To detect the analyte, its signal must exceed by a suitable amount; thus, \[(S_A)_{DL} = S_{mb} \pm z \sigma_{mb} \label{4.1}\] where \((S_A)_{DL}\) is the analyte’s detection limit. If \(\sigma_{mb}\) is not known, we can replace it with ; Equation \ref{4.1} then becomes \[(S_A)_{DL} = S_{mb} \pm t s_{mb} \nonumber\] You can make similar adjustments to other equations in this section. See, for example, Kirchner, C. J. “Estimation of Detection Limits for Environme tal Analytical Procedures,” in Currie, L. A. (ed) ; American Chemical Society: Washington, D. C., 1988. The value we choose for depends on our tolerance for reporting the analyte’s concentration even if it is absent from the sample (a type 1 error). Typically, is set to three, which, from , corresponds to a probability, \(\alpha\), of 0.00135. As shown in Figure 4.7.1
a, there is only a 0.135% probability of detecting the analyte in a sample that actually is analyte-free. A detection limit also is subject to a type 2 error in which we fail to find evidence for the analyte even though it is present in the sample. Consider, for example, the situation shown in Figure 4.7.1
b where the signal for a sample that contains the analyte is exactly equal to ( ) . In this case the probability of a type 2 error is 50% because half of the sample’s possible signals are below the detection limit. We correctly detect the analyte at the IUPAC detection limit only half the time. The IUPAC definition for the detection limit is the smallest signal for which we can say, at a significance level of \(\alpha\), that an analyte is present in the sample; however, failing to detect the analyte does not mean it is not present in the sample. The detection limit often is represented, particularly when discussing public policy issues, as a distinct line that separates detectable concentrations of analytes from concentrations we cannot detect. This use of a detection limit is incorrect [Rogers, L. B. , , 3–6]. As suggested by Figure 4.7.1
, for an analyte whose concentration is near the detection limit there is a high probability that we will fail to detect the analyte. An alternative expression for the detection limit, the , minimizes both type 1 and type 2 errors [Long, G. L.; Winefordner, J. D. , , 712A–724A]. The analyte’s signal at the limit of identification, ( ) , includes an additional term, \(z \sigma_A\), to account for the distribution of the analyte’s signal. \[(S_A)_\text{LOI} = (S_A)_\text{DL} + z \sigma_A = S_{mb} + z \sigma_{mb} + z \sigma_A \nonumber\] As shown in Figure 4.7.2
, the limit of identification provides an equal probability of a type 1 and a type 2 error at the detection limit. When the analyte’s concentration is at its limit of identification, there is only a 0.135% probability that its signal is indistinguishable from that of the method blank. The ability to detect the analyte with confidence is not the same as the ability to report with confidence its concentration, or to distinguish between its concentration in two samples. For this reason the American Chemical Society’s Committee on Environmental Analytical Chemistry recommends the , ( ) [“Guidelines for Data Acquisition and Data Quality Evaluation in Environmental Chemistry,” , , 2242–2249 ]. \[(S_A)_\text{LOQ} = S_{mb} + 10 \sigma_{mb} \nonumber\] | 4,395 | 3,946 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Nonlinear_and_Two-Dimensional_Spectroscopy_(Tokmakoff) |
Spectroscopy comes from the Latin “spectron” for spirit or ghost and the Greek “σκοπιεν” for to see. These roots are very telling, because in molecular spectroscopy you use light to interrogate matter, but you actually never see the molecules, only their influence on the light. Different spectroscopies give you different perspectives. This indirect contact with the microscopic targets means that the interpretation of spectroscopy in some manner requires a model, whether it is stated or not. Linear spectroscopy commonly refers to light-matter interaction with one primary incident radiation field which is weak, and can be treated as a linear response between the incident light and the matter. From a quantum mechanical view of the light field, it is often conceived as a “one photon in/one photon out” measurement. Nonlinear spectroscopy is used to refer to cases that fall outside this view
Thumbnail: Plot of the field of an ultrashort pulse, as well as its time-averaged intensity. (CC BY- unported; and edited by LibreTexts via ) | 1,061 | 3,949 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Electrolyte_Strength |
means a solute that allows a water solution to conduct electricity. Electrolytes produce ions when they dissolve in solution. Salts are usually electrolytes, while molecular substances usually aren't, unless they are acids or bases. The situation gets a little bit more complicated, though, because of the distinction between strong electrolytes and weak electrolytes. A strong electrolyte, like NaCl, splits up completely into sodium and chloride ions in solution. Likewise, a strong acid like HCl splits up completely into hydrogen and chloride ions in solution. Salts are often strong electrolytes, and strong acids are always strong electrolytes. Weak acids are weak electrolytes, and most other molecular compounds are non-electrolytes. Many textbooks incorrectly state that all salts or ionic compounds are strong electrolytes. However, many ionic compounds or salts of transition metals or alkaline earth metals are not strong electrolytes. When they dissolve, some of the ions separate, but some stay together. Here are some examples: \[K_{2}SO_{4}(s) + water \rightarrow K^{+}(aq)\; (70\%) + KSO_{4}^{–}(aq)\; (30\%) + SO_{4}^{2–}(aq)\; (for\; a\; 0.36\; M\; solution)\] \[CdI_{2}(s) + water \rightarrow Cd^{2+}(aq)\; (2\%) + CdI^{+}(aq)\; (22\%) + CdI_{2}(aq)\; (76\%) + I^{–}(aq)\; (for\; a\; 0.5\; M\; solution)\] As you can see, these salts are electrolytes (they do produce ions) but if you do calculations assuming that they separate completely into sulfate, potassium, cadmium(II) and iodide ions, you might get very wrong answers! They are not strong electrolytes. In general, the lower the concentration and the lower the charges on the ions, the "stronger" the electrolytes will be. Alkali metals other than lithium are usually strong electrolytes especially when the anion also has a small charge, and in dilute solutions (<0.1M). Alkaline earth metal compounds are weaker electrolytes, and other metals are even weaker still. We'll talk more about why this is later. Solubility can cause some confusion here. For instance, it's possible that a compound is a strong electrolyte, but just not very soluble. For this reason, it would not be able to produce a solution with lots of ions, because it isn't soluble, not because the ions are still attached to each other in the solution. In general, it makes sense to guess that if the ions in a compound are very strongly attracted to each other, the compound will be less soluble, and also it might be a weaker electrolyte because even in solution the ions will be bonded to each other. However, electrolytes also look stronger at lower concentrations, because if the ions split up, they are less likely to find each other again. | 2,716 | 3,955 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./09.E%3A_Bonding_Theories_(Exercises) |
. In addition to these individual basis; please contact What is the main difference between the VSEPR model and Lewis electron structures? What are the differences between molecular geometry and Lewis electron structures? Can two molecules with the same Lewis electron structures have different molecular geometries? Can two molecules with the same molecular geometry have different Lewis electron structures? In each case, support your answer with an example. How does the VSEPR model deal with the presence of multiple bonds? Three molecules have the following generic formulas: AX , AX E, and AX E . Predict the molecular geometry of each, and arrange them in order of increasing X–A–X angle. Which has the smaller angles around the central atom—H S or SiH ? Why? Do the Lewis electron structures of these molecules predict which has the smaller angle? Discuss in your own words why lone pairs of electrons occupy more space than bonding pairs. How does the presence of lone pairs affect molecular geometry? When using VSEPR to predict molecular geometry, the importance of repulsions between electron pairs decreases in the following order: LP–LP, LP–BP, BP–BP. Explain this order. Draw structures of real molecules that separately show each of these interactions. How do multiple bonds affect molecular geometry? Does a multiple bond take up more or less space around an atom than a single bond? a lone pair? Straight-chain alkanes do not have linear structures but are “kinked.” Using -hexane as an example, explain why this is so. Compare the geometry of 1-hexene to that of -hexane. How is molecular geometry related to the presence or absence of a molecular dipole moment? How are molecular geometry and dipole moments related to physical properties such as melting point and boiling point? What two features of a molecule’s structure and bonding are required for a molecule to be considered polar? Is COF likely to have a significant dipole moment? Explain your answer. When a chemist says that a molecule is , what does this mean? What are the general physical properties of polar molecules? Use the VSPER model and your knowledge of bonding and dipole moments to predict which molecules will be liquids or solids at room temperature and which will be gases. Explain your rationale for each choice. Justify your answers. The idealized molecular geometry of BrF is square pyramidal, with one lone pair. What effect does the lone pair have on the actual molecular geometry of BrF ? If LP–BP repulsions were than BP–BP repulsions, what would be the effect on the molecular geometry of BrF ? Which has the smallest bond angle around the central atom—H S, H Se, or H Te? the largest? Justify your answers. Which of these molecular geometries results in a molecule with a net dipole moment: linear, bent, trigonal planar, tetrahedral, seesaw, trigonal pyramidal, square pyramidal, and octahedral? For the geometries that do not always produce a net dipole moment, what factor(s) will result in a net dipole moment? To a first approximation, the VSEPR model assumes that multiple bonds and single bonds have the same effect on electron pair geometry and molecular geometry; in other words, VSEPR treats multiple bonds like single bonds. Only when considering fine points of molecular structure does VSEPR recognize that multiple bonds occupy more space around the central atom than single bonds. Physical properties like boiling point and melting point depend upon the existence and magnitude of the dipole moment of a molecule. In general, molecules that have substantial dipole moments are likely to exhibit greater intermolecular interactions, resulting in higher melting points and boiling points. The term “polar” is generally used to mean that a molecule has an asymmetrical structure and contains polar bonds. The resulting dipole moment causes the substance to have a higher boiling or melting point than a nonpolar substance. Give the number of electron groups around the central atom and the molecular geometry for each molecule. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each species. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. What is the molecular geometry of ClF ? Draw a three-dimensional representation of its structure and explain the effect of any lone pairs on the idealized geometry. Predict the molecular geometry of each of the following. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Of the molecules Cl C=Cl , IF , and SF , which has a net dipole moment? Explain your reasoning. Of the molecules SO , XeF , and H C=Cl , which has a net dipole moment? Explain your reasoning. four electron groups, pyramidal molecular geometry The idealized geometry is T shaped, but the two lone pairs of electrons on Cl will distort the structure, making the F–Cl–F angle than 180°. Cl C=CCl : Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl C=CCl does not have a net dipole moment. IF : In this structure, the individual I–F bond dipoles cannot cancel one another, giving IF a net dipole moment. SF : The S–F bonds are quite polar, but the individual bond dipoles cancel one another in an octahedral structure. Thus, SF has no net dipole moment. Predict whether each compound is purely covalent, purely ionic, or polar covalent. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning. If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character? Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm. Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm. Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, = 228.69 pm and µ = 3.59 D; for PbO, = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state? Arrange , , and in order of increasing strength of the bond formed to a hydrogen atom. Explain your reasoning. What atomic orbitals are combined to form , , , and ? What is the maximum number of electron-pair bonds that can be formed using each set of hybrid orbitals? Why is it incorrect to say that an atom with hybridization will form only three bonds? The carbon atom in the carbonate anion is hybridized. How many bonds to carbon are present in the carbonate ion? Which orbitals on carbon are used to form each bond? If hybridization did not occur, how many bonds would N, O, C, and B form in a neutral molecule, and what would be the approximate molecular geometry? How are hybridization and molecular geometry related? Which has a stronger correlation—molecular geometry and hybridization or Lewis structures and hybridization? In the valence bond approach to bonding in BeF , which step(s) require(s) an energy input, and which release(s) energy? The energies of hybrid orbitals are intermediate between the energies of the atomic orbitals from which they are formed. Why? How are lone pairs on the central atom treated using hybrid orbitals? Because nitrogen bonds to only three hydrogen atoms in ammonia, why doesn’t the nitrogen atom use hybrid orbitals instead of hybrids? Using arguments based on orbital hybridization, explain why the CCl ion does not exist. Species such as NF and OF are unknown. If 3 atomic orbitals were much lower energy, low enough to be involved in hybrid orbital formation, what effect would this have on the stability of such species? Why? What molecular geometry, electron-pair geometry, and hybridization would be expected for each molecule? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each molecule. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each species. What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in PF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? What is the hybridization of the central atom in SF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? The promotion and hybridization process is exactly the same as shown for CH in the chapter. The only difference is that the C atom uses the four singly occupied hybrid orbitals to form electron-pair bonds with only H atoms, and an electron is added to the fourth hybrid orbital to give a charge of 1–. The electron-pair geometry is tetrahedral, but the molecular geometry is pyramidal, as in NH . , trigonal planar , pyramidal , trigonal planar The central atoms in CF , CCl , IO , and SiH are all sp hybridized. The phosphorus atom in the PF ion is hybridized, and the ion is octahedral. The PF ion is isoelectronic with SF and has essentially the same structure. It should therefore be a stable species. | 11,194 | 3,956 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Combustion_Reactions |
are common and very important. Combustion means burning, usually in oxygen but sometimes with other such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't O or F ) is called the . Combustion reactions are a type of reaction. The classic chemistry class combustion reaction involves a compound of C and H reacting with O to form CO and H O. Sometimes the reactant has some other elements, like O, S or N in it. The O will form water, the S will form usually SO and the N will often produce N , but sometimes a little bit of a nitrogen oxide. For class purposes, you can usually write equations in which carbon dioxide is produced. In real life, often some or a lot of CO is produced, depending on how much oxygen is present and other factors. In general, most elements in a compound that is combusted will form oxides, but you won't be able to say for sure how much of each oxide will be produced (CO or CO , SO or SO , etc). Here are some example equations. When you balance combustion reactions, usually start with the C, and the fuel, and do the oxygen last. \[C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)\] \[C_{6}H_{12}O_{6}(s) + 12O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(g)\] Reaction 4.2 is sugar burning, which may also represent an animal or plant using stored energy. | 1,760 | 3,958 |
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