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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.03%3A_Rate_Determining_Step |
The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate determining step can be compared to the neck of a funnel. The rate at which water flows through a funnel is limited/ determined by the width of the neck of the funnel and not by the rate at which the water is poured into the funnel. Like the neck of the funnel, the slow step of a reaction determines Not all reactions have rate determining steps and has one only if one step is significantly slower than the other steps in the reaction. Rate determining step is the slowest step within a chemical reaction. The slowest step determines the rate of chemical reaction.The slowest step of a chemical reaction can be determined by setting up a reaction mechanisms. Many reactions do not occur in a single reaction but they happen in multiple elementary steps. Consider this reaction: \[\ce{2NO2 +F2 -> 2NO2 F}, \nonumber \] which occurs via this mechanism elementary step 1: \[\ce{NO2 + F2 ->NO2F +F} \tag{slow} \] elementary step 2: \[\ce{NO2+F -> NO2F} \tag{fast} \] For elementary step 1 has a rate constant of k and for elementary step 2 it has a rate constant of k The slowest step in this mechanism is elementary step 1 which is our rate determining step. Looking at this mechanism I see Intermediates. Intermediates are molecules or elements that are found on the product of one step but are also located in the reactant of another step. In this case we have two intermediates \(\ce{NO2}\) and \(\ce{F}\). The rate equation is derived by the slowest step in the reaction. When writing a rate equation you set up the equation by writing rate is equal to the rate constant of the slowest step times the concentrations of the reactant or reactants raised to there reaction order. Lets look at elementary step one. elementary step one: \[\ce{NO2 +F2 -> NO2F + F}\nonumber \] Here in this example rate=k [NO ,F ]. For this reaction \[\ce{2NO + O2 -> 2NO2} \nonumber \] elementary step one: \[\ce{NO + NO <=> N2O2 } \tag{fast equilibrium} \] elementary step two: \[\ce{N2O2 + O2 -> 2 NO2 }\tag{slow} \] 1: N O is found on the product side and the reactant side. 2: rate= k [N O ,O ]; N O gets canceled out leaving the overall reaction rate. elementary step one: \[\ce{Br2 + M <=> Br + Br +M}\tag{fast equilibrium} \] elementary step two: \[\ce{Br +H2 -> HBr +H }\tag{slow} \] elementary step three: \[\ce{H + Br2 -> HBR + Br }\tag{fast} \] | 2,517 | 782 |
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On average, a million tons of earth contain just ten pounds of gold. This scarcity, as well as its beauty and chemical properties account for its high value from ancient times. The name for the element is of Anglo-Saxon origin and the symbol comes from the Latin aurum, meaning "shining dawn". Gold metal has a distinctive yellow color and is incredibly malleable and ductile. A single ounce of pure gold can be beaten out to a sheet that is about 300 feet square!! Pure gold is easily cut with a knife. Few elements react with gold under normal conditions and so most gold is recovered as small flakes of the pure element. Gold is a very good conductor and is often used to plate electrical contacts since it resists corrosion so well. It also is a good reflector of heat-carrying infra-red radiation. The world's oceans contain billions of tons of gold but it is too widely dispersed to be recovered (two-tenths of an ounce per million tons of water). with either \(H_2SO_{4(aq)}\) or \(HNO_{3(aq)}\), rather it art \(HNO_3\) and three parts \(HCl\). For \[Au_{(s)} + 4H^+_{(aq)} + NO^-_{3(aq)} + 4Cl^-_{(aq)} \rightarrow [AuCl_4]^-_{(aq)} + 2H_2O_{(l)} + NO_{(g)} \] | 1,190 | 784 |
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The elements of Group 1 consist of: Lithium, Sodium, Potassium, Rubidium, Cesium, and Francium. These elements are called the alkali metals because they react strongly with water and create hydroxide ions and hydrogen gas, leaving a basic solution. Group 1 metals are very reactive with oxygen and must be kept away from oxygen in order to not get oxidized. These alkali metals rapidly react with oxygen to produce several different ionic oxides. Oxides: O , peroxides: O , super oxide: O . The usual oxide, M O, can be formed with alkali metals generally by limiting the supply of oxygen. With excess oxygen, the alkali metals can form peroxides, M O , or superoxides, MO . Lithium: Reacts with oxygen to give 2Li O, lithium oxide. Reactions are shown below. 4Li(s) + O (g)→2Li O(s) Sodium: Reacts with oxygen to form mostly sodium peroxide, Na O . Na O along with Li O is used in emergency breathing devices in submarines and spacecrafts. Reactions are shown below. 2Na(s) + O (g) → Na O (s) The rest of the group, K, Rb, Cs, and Fr, forms the superoxides. M(s) + O (g) → MO (s) M= K, Rb, Cs, Fr Metal oxides, peroxides, and superoxides that dissolve in water react with water to form basic solutions. Oxide ion with water: O (aq) + H O(l) → 2OH (aq) Peroxide ion with water: O (aq) + 2H O(l) → H O (aq) + 2OH (aq) Superoxide ion with water: 2O (aq) + 2H O(l) → H O (aq) + 2OH (aq) + O (g) Tl The elements in Group 14 consist of carbon, silicon, germanium, tin, and lead. Carbon is the only nonmetal element of the group 14. Silicon is mostly nonmetallic. Germanium is a metalloid or semi-metal. Tin and lead have mainly metallic properties. Carbon: Reacts with oxygen to form oxides. The main form of oxides of carbon are carbon monoxide, CO, and carbon dioxide, CO . Carbon dioxide is the primary product of burning organic materials and also a byproduct of respiration. During photosynthesis carbon is combined with water to form carbohydrates. 6CO + 6H O C H O + 6O Carbon is the building block to many organic compounds. Carbon dioxide is the only oxide formed when carbon is burned in an excess of air. The reactions are shown below. C(s) + O (g) → CO 2C(s) + O (g) → 2CO Silicon: Forms only one stable oxide with the empirical formula SiO , silica. In silica, each Si atom is bonded to four O atoms and each O atom to two Si atoms forming a network covalent solid with a network of –Si–O–Si– bonds. Germanium: Forms germanium dioxide which is covalent network solid similar to silicon dioxide. Tin: Forms two primary oxides, SnO and SnO . By heating SnO in air, it can be converted to SnO . SnO is used as a jewerly abarasive. The reactions are shown below. Sn(s) + O (g) → SnO (s) 2Sn(s) + O (g) → 2SnO(s) Lead: Forms a several forms of oxides. The best known oxides of lead are yellow lead oxide, PbO, red-brown lead dioxide, PbO ,and red lead, Pb O . The reactions are shown below. 2Pb(s) + O (g) → 2PbO(s) Pb(s) + O (g) → PbO (s) 3Pb(s) + 2O (g) → Pb O (s) The elements in Group 15 consist of : nitrogen, phosphorus, arsenic, antimony, and bismuth. Nitrogen and phosphorus are nonmetallic, arsenic and antimony are metalloids, and bismuth is metallic. Nitrogen: Forms a sires of oxides in which the oxidation state of N can have every value ranging from +1 to +5. All these oxides are gases at room temperature except for N O , which is solid. Preparation of Oxides of Nitrogen: Reactions are shown below. N O NH NO (s) —∆→ N O(g) + 2H O(g) NO 3Cu(s) + 8H (aq) + 2NO (aq) → 3Cu (aq) + 2NO(g) + 4H O(l) N O 2NO(g) + N O (g) —-20°C→ 2N O (l) NO 2Pb(NO ) (s) —∆→ 2PbO(s) +4NO (g) + O (g) 2NO(g) + O (g) <=> 2NO (g) N O 2NO (g) <=> N O (g) N O 4HNO (l) + P O ) —-10°C→ 4HPO (s) + 2N O (s) | 3,997 | 786 |
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is the energy, or work, required to increase the surface area of a liquid due to . Since these intermolecular forces vary depending on the nature of the liquid (e.g. water vs. gasoline) or solutes in the liquid (e.g. surfactants like detergent), each solution exhibits differing surface tension properties. Whether you know it or not, you already have seen surface tension at work. Whenever you fill a glass of water too far, you may notice afterward that the level of the water in the glass is actually higher than the height of the glass. You may have also noticed that the water that you spilled has formed into pools that rise up off the counter. Both of these phenomena are due to surface tension. In a sample of water, there are two types of molecules. Those that are on the outside, exterior, and those that are on the inside, interior. The interior molecules are attracted to all the molecules around them, while the exterior molecules are attracted to only the other surface molecules and to those below the surface. This makes it so that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior. Because of this, the molecules try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. This is what creates what is referred to as surface tension. An illustration of this can be seen in Figure \(\Page {1}\). The water molecules attract one another due to the water's polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to "stick" together. This is why there is surface tension and takes a certain amount of energy to break these intermolecular bonds. Same goes for other liquids, even hydrophobic liquids such as oil. There are forces between the liquid such as Van der Waals forces that are responsible for the intermolecular forces found within the liquid. It will then take a certain amount of energy to break these forces, and the surface tension. Water is one liquid known to have a very high surface tension value and is difficult to overcome. Surface tension of water can cause things to float which are denser than water, allowing organisms to literally walk on water (Figure \(\Page {2}\)). An example of such an organism is the water strider, which can run across the surface of water, due to the intermolecular forces of the molecules, and the force of the strider which is distributed to its legs. Surface tension also allows for the formation of droplets that we see in nature. There are several other important concepts that are related to surface tension. The first of these is the idea of . Cohesive forces are those that hold the body of a liquid together with minimum surface area and adhesive forces are those that try to make a body of a liquid spread out. So if the cohesive forces are stronger than the adhesive forces, the body of water will maintain its shape, but if the opposite is true than the liquid will be spread out, its surface area. Any substance that you can add to a liquid that allows a liquid to increase its surface area is called a . In the lab there are also several important points to remember about surface tension. The first you've probably noticed before. This is the idea of a meniscus (Figure \(\Page {3a}\)). This is the concave (curved in) or convex (curved out) look that water or other liquids have when they are in test tubes. This is caused by the attraction between the glass and the liquid. With water, this causes it to climb up the sides of a test tube. This attraction is amplified as the diameter of the tubes increases; this is called . This can be seen if you take a tube with a very small diameter (a ) and lower it into a body of water. The liquid will climb up into the tube, even though there is no outside force. You may have also seen this when you put a straw into a drink and notice that the liquid level inside the straw is higher than it is in your drink. All of this however, requires that the adhesive forces (between the liquid and the capillary surface) be higher than the cohesive forces (between the liquid and itself), otherwise there will be no capillary action or the opposite can even happen. Mercury has higher cohesive forces than adhesive forces, so the level of the liquid will actually be lower in the capillary tubes than compared to the rest of the mercury (Figure \(\Page {3b}\)). | 4,462 | 787 |
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The refers to the tendency of atoms to prefer to have eight electrons in the . When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it useful for the (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with \(s^2p^6\). In 1904, Richard Abegg formulated what is now known as , which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light. A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made. The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell. The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. The formula for table salt is NaCl. It is the result of Na ions and Cl ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron. The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na and Cl , which is much less stable than Na and Cl . Atoms are more stable when they have no charge, or a small charge. | 3,356 | 788 |
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Make sure you thoroughly understand the following essential ideas: The study of gases allows us to understand the behavior of matter at its simplest: individual particles, acting independently, almost completely uncomplicated by interactions and interferences between each other. This knowledge of gases will serve as the pathway to our understanding of the far more complicated (liquids and solids) in which the theory of gases will no longer give us correct answers, but it will still provide us with a useful that will at least help us to rationalize the behavior of these more complicated states of matter. First, we know that a gas has ; a gas will fill whatever volume is available to it. Contrast this to the behavior of a liquid, which always has a distinct upper surface when its volume is less than that of the space it occupies. The other outstanding characteristic of gases is their , compared with those of liquids and solids. One mole of liquid water at 298 K and 1 atm pressure occupies a volume of 18.8 cm , whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm , more than 1000 times greater. The most remarkable property of gases, however, is that to a very good approximation, in response to changes in temperature and pressure, expanding or contracting by predictable amounts. This is very different from the behavior of liquids or solids, in which the properties of each particular substance must be determined individually. We will see later that each of these three characteristics of gases follows directly from the view— that is, from the nature of matter. The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy, but the reversal of direction ( ) imparts a to the container walls. This force, divided by the total surface area on which it acts, is the of the gas. The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. To visualize this, imagine some gas trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the gas in the container, a certain amount of weight (more precisely, a ) must be placed on the piston so as to exactly balance the force exerted by the gas on the bottom of the piston, and tending to push it up. The pressure of the gas is simply the quotient , where is the cross-section area of the piston. The unit of pressure in the SI system is the (Pa), defined as a force of one newton per square meter (1 Nm = 1 kg m s .) At the Earth's surface, the force of gravity acting on a 1 kg mass is 9.81 N. Thus if the weight is 1 kg and the surface area of the piston is 1 M , the pressure of the gas would be 9.81 Pa. A 1-gram weight acting on a piston of 1 cm cross-section would exert a pressure of 98.1 pA. (If you wonder why the pressure is higher in the second example, consider the number of cm contained in 1 m .) In chemistry, it is often common to express pressures in units of or : 1 atm = 101325 Pa = 760 torr. The older unit (mm Hg) is almost the same as the torr and is defined as one mm of level difference in a mercury barometer at 0°C. In meteorology, the pressure unit most commonly used is the : 1 bar = 10 N m = 750.06 torr = 0.987 atm. The pressures of gases encountered in nature span an exceptionally wide range, only part of which is ordinarily encountered in chemistry. Note that in the chart below, the pressure scales are logarithmic; thus 0 on the atm scale means 10 = 1 atm. The column of air above us exerts a force on each 1-cm of surface equivalent to a weight of about 1034 g. The higher into the air you go, the lower the mass of air above you, hence the lower the pressure (right). This figure is obtained by solving Newton's law \(\textbf{F} = m\textbf{a}\) for \(m\), using the acceleration of gravity for \(\textbf{a}\): \[ m = \dfrac{101375\; kg\, m^{-1} \, s^{-2}}{9.8 \, m \, s^{-2}} = 10340 \, kg\, m^{-1} =1034\; g \; cm^{-2}\] If several kilos of air are constantly pressing down on your body, why do you not feel it? Because every other part of your body (including within your lungs and insides) also experiences the same pressure, so there is no net force (other than gravity) acting on you. This was the crucial first step that led eventually to the concept of gases and their essential role in the early development of Chemistry. In the early 17th century the Italian Evangelista Torricelli invented a device — the — to measure the pressure of the atmosphere. A few years later, the German scientist and some-time mayor of Magdeburg Otto von Guericke devised a method of pumping the air out of a container, thus creating which might be considered the opposite of air: the . As with so many advances in science, idea of a vacuum — a region of nothingness — was not immediately accepted. Torricelli's invention overturned the then-common belief that air (and by extension, all gases) are weightless. The fact that we live at the bottom of a sea of air was most spectacularly demonstrated in 1654, when two teams of eight horses were unable to pull apart two 14-inch copper hemispheres (the "Magdeburg hemispheres") which had been joined together and then evacuated with Guericke's newly-invented vacuum pump. The classical barometer, still used for the most accurate work, measures the height of a column of liquid that can be supported by the atmosphere. As indicated below, this pressure is exerted directly on the liquid in the reservoir, and is transmitted hydrostatically to the liquid in the column. Metallic , being a liquid of exceptionally high density and low vapor pressure, is the ideal barometric fluid. Its widespread use gave rise to the "millimeter of mercury" (now usually referred to as the "torr") as a measure of pressure. How is the air pressure of 1034 g cm related to the 760-mm height of the mercury column in the barometer? What if water were used in place of mercury? The density of Hg is 13.6 g cm , so in a column of 1-cm cross-section, the height needed to counter the atmospheric pressure would be (1034 g × 1 cm ) / (13.6 g cm ) = 76 cm. The density of water is only 1/13.6 that of mercury, so standard atmospheric pressure would support a water column whose height is 13.6 x 76 cm = 1034 cm, or 10.3 m. You would have to read a water barometer from a fourth-story window! Water barometers were once employed to measure the height of the ground and the heights of buildings before more modern methods were adopted. A modification of the barometer, the , provides a simple device for measuring the pressure of any gas in a container. The U-tube is partially filled with mercury, one end is connected to container, while the other end can either be opened to the atmosphere. The pressure inside the container is found from the difference in height between the mercury in the two sides of the U-tube. The illustration below shows how the two kinds of manometer work. The manometers ordinarily seen in the laboratory come in two flavors: closed-tube and open-tube. In the closed-tube unit shown at the left, the longer limb of the J-tube is evacuated by filling it with mercury and then inverting it. If the sample container is also evacuated, the mercury level will be the same in both limbs. When gas is let into the container, its pressure pushes the mercury down on one side and up on the other; the difference in levels is the pressure in torr. For practical applications in engineering and industry, especially where higher pressures must be monitored, many types of mechanical and electrical are available. If two bodies are at different temperatures, heat will flow from the warmer to the cooler one until their temperatures are the same. This is the principle on which is based; the temperature of an object is measured indirectly by placing a calibrated device known as a in contact with it. When thermal equilibrium is obtained, the temperature of the thermometer is the same as the temperature of the object. A makes use of some temperature-dependent quantity, such as the density of a liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the length of a mercury column. The resulting scale of temperature is entirely arbitrary; it is defined by locating its zero point, and the size of the unit. At one point in the 18 century, 35 different temperature scales were in use! The Celsius temperature scale (formally called centigrade) locates the zero point at the freezing temperature of water; the Celsius degree is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure. The older scale placed the zero point at the coldest temperature it was possible to obtain at the time (by mixing salt and ice.) The 100° point was set with body temperature (later found to be 98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the same temperature interval that contains 100 Celsius degrees, so 1 F° = 5/9 C°. Since the zero points are also different by 32F°, conversion between temperatures expressed on the two scales requires the addition or subtraction of this offset, as well as multiplication by the ratio of the degree size. In 1787 the French mathematician and physicist Jacques Charles discovered that for each Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish by 1/273 of its volume at 0°C. The obvious implication of this is that if the temperature could be reduced to –273°C, the volume of the gas would contract to zero. Of course, all real gases condense to liquids before this happens, but at sufficiently low pressures their volumes are linear functions of the temperature ( ), and extrapolation of a plot of volume as a function of temperature predicts zero volume at -273°C. This temperature, known as , corresponds to the total absence of thermal energy. The temperature scale on which the zero point is –273.15°C was suggested by Lord Kelvin, and is usually known as the scale. Since the size of the Kelvin and Celsius degrees are the same, conversion between the two scales is a simple matter of adding or subtracting 273.15; thus room temperature, 20°, is about 293 K. Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical relations involving temperature are expressed mathematically in terms of absolute temperature. In engineering work, an absolute scale based on the Fahrenheit degree is sometimes used; this is known as the . The of a gas is simply the space in which the molecules of the gas are free to move. If we have a mixture of gases, such as air, the various gases will coexist within the same volume. In these respects, gases are very different from liquids and solids, the two condensed states of matter. The volume of a gas can be measured by trapping it above mercury in a calibrated tube known as a . The SI unit of volume is the cubic meter, but in chemistry we more commonly use the and the (mL). The (cc) is also frequently used; it is very close to 1 milliliter (mL). It's important to bear in mind, however, that the volume of a gas varies with both the temperature and the pressure, so reporting the volume alone is not very useful. A common practice is to measure the volume of the gas under the ambient temperature and atmospheric pressure, and then to correct the observed volume to what it would be at standard atmospheric pressure and some fixed temperature, usually 0° C or 25°C. ) | 11,917 | 789 |
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With the interstitial mechanism, one atom is usually much smaller than the other, so cannot successfully replace an atom in the crystals of the base metal. The smaller atoms become trapped in the spaces between the atoms in the crystal matrix, called the . This is referred to as an . Steel is an example of an interstitial alloy, because the very small carbon atoms fit into interstices of the iron matrix. Stainless steel is an example of a combination of interstitial and substitutional alloys, because the carbon atoms fit into the interstices, but some of the iron atoms are replaced with nickel and chromium atoms. | 640 | 790 |
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The enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the and allows the accurate determination of equilibrium constants (including solubility constants). Recall that the actual free-energy change for a reaction under nonstandard conditions, \(\Delta{G}\), is given as follows: \[\Delta{G} = \Delta{G°} + \ln Q \label{Eq1} \] We also know that \(ΔG = −nFE_{cell}\) (under non-standard confitions) and \(ΔG^o = −nFE^o_{cell}\) (under standard conditions). Substituting these expressions into Equation \(\ref{Eq1}\), we obtain \[−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2} \] Dividing both sides of this equation by \(−nF\), \[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3} \] Equation \(\ref{Eq3}\) is called the , after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium (\(ΔG = 0\)), then Equation \(\ref{Eq3}\) reduces to Equation \(\ref{Eq31}\) and \(\ref{Eq32}\) because \(Q = K\), and there is no net transfer of electrons (i.e., E = 0). \[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31} \] since \[E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32} \] \(\ref{Eq3}\) \[E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4} \] The Nernst Equation (\(\ref{Eq3}\)) can be used to determine the value of E , and thus the direction of spontaneous reaction, for any redox reaction under any conditions. Equation \(\ref{Eq4}\) allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation \(\ref{Eq4}\) that the cell potential changes by 0.0591/n V for each 10-fold change in the value of \(Q\) because log 10 = 1. The following reaction proceeds spontaneously under standard conditions because E° > 0 (which means that ΔG° < 0): \[\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber \] Calculate \(E_{cell}\) for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce ] = 0.013 M, [Ce ] = 0.60 M, [Cl ] = 0.0030 M, \(P_\mathrm{Cl_2}\) = 1.0 atm, and T = 25°C. balanced redox reaction, standard cell potential, and nonstandard conditions cell potential Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions. We can use the information given and the Nernst equation to calculate E . Moreover, because the temperature is 25°C (298 K), we can use Equation \(\ref{Eq4}\) instead of Equation \(\ref{Eq3}\). The overall reaction involves the net transfer of two electrons: \[2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber \] \[2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber \] so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl , and the value of E° into Equation \(\ref{Eq4}\), \[\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\
& =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \\
& =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber \] Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture Molecular oxygen will not oxidize \(MnO_2\) to permanganate via the reaction \[\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber \] Calculate \(E_{cell}\) for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, \(P_\mathrm{O_2}\)= 0.20 atm, [MNO ] = 1.0 × 10 M, and T = 25°C. E = −0.22 V; the reaction will not occur spontaneously. Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows: \[Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5} \] The reaction quotient is therefore \(Q = [Zn^{2+}]/[Cu^{2+}]\). Suppose that the cell initially contains 1.0 M Cu and 1.0 × 10 M Zn . The initial voltage measured when the cell is connected can then be calculated from Equation \(\ref{Eq4}\): \[\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\\
& =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6} \] Thus the initial voltage is greater than E° because \(Q<1\). As the reaction proceeds, [Zn ] in the anode compartment increases as the zinc electrode dissolves, while [Cu ] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn ]/[Cu ] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn ] = [Cu ], so Q = 1 and E = E° . Beyond this point, [Zn ] will continue to increase in the anode compartment, and [Cu ] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in E . When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn and 1.0 × 10 M Cu ), Q = 1.0 × 10 , and the cell potential will be reduced to 0.92 V. The variation of E with \(\log{Q}\) over this range is linear with a slope of −0.0591/n, as illustrated in Figure \(\Page {1}\). As the reaction proceeds still further, \(Q\) continues to increase, and E continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of \(Q\) when E = 0 is calculated as follows: Recall that at equilibrium, \(Q = K\). Thus the equilibrium constant for the reaction of Zn metal with Cu to give Cu metal and Zn is 1.7 × 10 at 25°C. The Nernst Equation: A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO in one compartment and 1.0 M AgNO in the other. The cell diagram and corresponding half-reactions are as follows: \[\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8} \] cathode: \[\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9} \] anode: \[\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10} \] Overall \[\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11} \] As the reaction progresses, the concentration of \(Ag^+\) will increase in the left (oxidation) compartment as the silver electrode dissolves, while the \(Ag^+\) concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of \(Ag(s)\) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E° because E° = −E° : \[\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \\[4pt] &=\textrm{0.12 V} \end{align*} \nonumber \] An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a . As the reaction proceeds, the difference between the concentrations of Ag in the two compartments will decrease, as will E . Finally, when the concentration of Ag is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (E = 0). Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl as the cathode, and a manganese electrode immersed in a 5.2 × 10 M solution of MnSO as the anode (T = 25°C). galvanic cell, identities of the electrodes, and solution concentrations voltage This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO ) do not participate in the reaction, so their identity is not important. The overall reaction is as follows: \[\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber \] For the reduction of Mn (aq) to Mn(s), n = 2. We substitute this value and the given Mn concentrations into Equation \(\ref{Eq4}\): \[ \begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \\[4pt] &=\textrm{0.047 V}\end{align*} \nonumber \] Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution. Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na SO at pH 7.00. Both cells are in contact with the atmosphere, with \(P_\mathrm{O_2}\) = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water: \[\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber \] What will be the potential when the circuit is closed? 0.41 V Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products (\(K_{sp}\)) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10 . Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods. To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure \(\Page {1}\), which is designed to measure the solubility product of silver chloride: \[K_{sp} = [\ce{Ag^{+}},\ce{Cl^{−}}]. \nonumber \] In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag ; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag ion concentration in the first compartment equals K . We can see this by dividing both sides of the equation for K by [Cl ] and substituting: \[\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \\[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber \] The overall cell reaction is as follows: Ag (aq, concentrated) → Ag (aq, dilute) Thus the voltage of the concentration cell due to the difference in [Ag ] between the two cells is as follows: \[\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \\[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \\[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align} \] By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation \(\ref{Eq122}\) for \(K_{sp}\), \[\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \\[4pt]
K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber \] Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt. To measure the solubility product of lead(II) sulfate (PbSO ) at 25°C, you construct a galvanic cell like the one shown in Figure \(\Page {1}\), which contains a 1.0 M solution of a very soluble Pb salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na SO saturated with PbSO in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is K for PbSO ? Report your answer to two significant figures. galvanic cell, solution concentrations, electrodes, and voltage K You have constructed a concentration cell, with one compartment containing a 1.0 M solution of \(\ce{Pb^{2+}}\) and the other containing a dilute solution of Pb in 1.0 M Na SO . As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb in the dilute solution to K : \[\begin{align*}[\mathrm{Pb^{2+}},\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \\
[\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber \] The reduction of Pb to Pb is a two-electron process and proceeds according to the following reaction: Pb (aq, concentrated) → Pb (aq, dilute) so \[\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \\
\textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \\
-7.77 & =\log K_\textrm{sp} \\
1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber \] A concentration cell similar to the one described in Example \(\Page {3}\) contains a 1.0 M solution of lanthanum nitrate [La(NO ) ] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the K for LaF ? Report your answer to two significant figures. 5.7 × 10 Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example \(\Page {3}\), in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to K . Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H ] using a pH meter, as illustrated below. Suppose a galvanic cell is constructed with a standard Zn/Zn couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H ] in the second compartment is unknown. The cell diagram is as follows: \[\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber \] What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C? galvanic cell, cell diagram, and cell potential pH of the solution Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H (note that Zn lies below H in ): By substituting the given values into the simplified Nernst equation (Equation \(\ref{Eq4}\)), we can calculate [H ] under nonstandard conditions: \[\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \\
\textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \\
16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \\
8.46 &=-\log[\mathrm{H^+}] \\
8.5 &=\mathrm{pH}\end{align*} \nonumber \] Thus the potential of a galvanic cell can be used to measure the pH of a solution. Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E° = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows: \[Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber \] When the circuit is closed, the cell has a measured potential of 1.62 V. Use to determine the concentration of Pb in the groundwater. \(1.2 \times 10^{−9}\; M\) The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | 18,827 | 792 |
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Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book. The chlorination of saturated hydrocarbons can be induced by light, but also can be carried out at temperatures of about \(300^\text{o}\) in the dark. Under such circumstances the mechanism is similar to that of light-induced chlorination, except that the chlorine atoms are formed by thermal dissociation of chlorine molecules. Solid carbon surfaces catalyze thermal chlorination, possibly by aiding in the cleavage of the chlorine molecules. Direct monohalogenation of saturated hydrocarbons works satisfactorily only with chlorine and bromine. For the general reaction the calculated \(\Delta H^\text{0}\) value is negative and very large for fluorine, negative and moderate for chlorine and bromine, and positive for iodine (see Table 4-7). With fluorine, the reaction evolves so much heat that it may be difficult to control, and products from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorination procedure for hydrocarbons involves diffusion of minute amounts of fluorine mixed with helium into liquid or solid hydrocarbons at low temperatures, typically \(-78^\text{o}\) (Dry Ice temperature). As fluorination proceeds, the concentration of fluorine can be increased. The process is best suited for preparation of completely fluorinated compounds, and it has been possible to obtain in this way amounts of \(\left( CF_3 \right)_4C\) and \(\left( CF_3 \right)_3 C-C \left( CF_3 \right)_3\) from 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane corresponding to \(10\)-\(15\%\) yields based on the fluorine used. Bromine generally is much less reactive toward hydrocarbons than chlorine is, both at high temperatures and with activation by light. Nonetheless, it usually is possible to brominate saturated hydrocarbons successfully. Iodine is unreactive. The chlorination of methane does not have to stop with the formation of chloromethane (methyl chloride). It is usual when chlorinating methane to obtain some of the higher chlorination products: dichloromethane (methylene chloride), trichloromethane (chloroform), and tetrachloromethane (carbon tetrachloride): In practice, one can control the degree of substitution to a considerable extent by controlling the methane-chlorine ratio. For example, for monochlorination to predominate, a high methane-chlorine ratio is necessary such that the chlorine atoms react with \(CH_4\) and not with \(CH_3Cl\). For propane and higher hydrocarbons for which more than one monosubstitution product is generally possible, difficult separation problems bay arise when a particular product is desired. For example, the chlorination of 2-methylbutane \(3\) at \(300^\text{o}\) gives all four possible monosubstitution products, \(4\), \(5\), \(6\), and \(7\): On a purely statistical basis, we may expect the ratio of products from \(3\) to correlate with the number of available hydrogens at the various positions of substitution. That is, \(4\), \(5\), \(6\), and \(7\) would be formed in the ratio 6:3:2:1 (\(50\%\):\(25\%\):\(17\%\):\(8\%\)). However, as can be seen from Table 4-6, the strengths of hydrogen bonds to primary, secondary, and tertiary carbons are not the same and, from the argument given in we would expect the weaker \(C-H\) bonds to be preferentially attacked by \(Cl \cdot\). The proportion of \(7\) formed is about three times that expected on a statistical basis which is in accord with our expectation that the tertiary \(C-H\) bond of 2-methylbutane should be the weakest of the \(C-H\) bonds. (See Table 4-6.) The factors governing selectivity in halogenation of alkanes follow: 1. The rates at which the various \(C-H\) bonds of 2-methylbutane are broken by attack of chlorine atoms approach 1:1:1 as the temperature is raised above \(300^\text{o}\). At higher temperatures both chlorine atoms and hydrocarbons become more reactive because of increases in their thermal energies. Ultimately, temperatures are attained where a chlorine atom essentially removes the first hydrogen with which it collides regardless of position on the hydrocarbon chain. In such circumstances, the composition of monochlorination products will correspond to that expected from simple statistics. 2. Bromine atoms are far more selective than chlorine atoms. This is not unexpected because is endothermic, whereas corresponding reactions with a chlorine atoms usually are exothermic (data from Table 4-6). Bromine removes only those hydrogens that are relatively weakly bonded to a carbon atom. As predicted, attack of \(Br \cdot\) on 2-methylbutane leads mostly to 2-bromo-2-methylbutane, some secondary bromide, and essentially no primary bromides: 3. The selectivity of chlorination reactions carried on in is increased markedly in the presence of benzene or alkyl-substituted benzenes because benzene and other arenes form loose complexes with chlorine atoms. This substantially cuts down chlorine-atom reactivity, thereby making the chlorine atoms behave more like bromine atoms. It is possible to achieve chlorination of alkanes using sulfuryl chloride (\(SO_2Cl_2\), bp \(69^\text{o}\)) in place of chlorine: The reaction has a radical-chain mechanism and the chains can be initiated by light or by chemicals, usually peroxides, \(ROOR\). Chemical initiation requires an with a weak bond that dissociates at temperatures between \(40\)-\(80^\text{o}\). Peroxides are good examples. The \(O-O\) bond is very weak (\(30\)-\(50 \: \text{kcal}\)) and on heating dissociates to alkoxyl radicals, \(RO \cdot\), which are reactive enough to generate the chain-propagating radicals from the reactants. The exact sequence of chemical initiation is not always known, but a plausible route in the present case would have \(RO \cdot\) abstract hydrogen from the alkane: The propagation steps that would follow are Chlorination with sulfuryl chloride of alkanes with more than one kind of hydrogen gives a mixture of alkyl chlorides resembling that obtained with chlorine itself. However, in some circumstances the mixture of chlorides is not the same mixture obtained with chlorine itself and when this is true, the hydrogen-abstraction step probably involves \(\cdot SO_2Cl\) rather than \(Cl \cdot\). The alternative propagation steps then are Different product ratios are expected from \(Cl \cdot\) and \(ClSO_2 \cdot\) for the same reason that \(Cl \cdot\) and \(Br \cdot\) lead to different product ratios ( ). Other reagents that sometimes are useful halogenating agents in radical-chain reactions include and (1977) | 7,065 | 793 |
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Since D = - ln ,
ln = - D / Differentiate both sides with respect to (1/ ) in the above equation gives, d(ln ) / (d ) = (- 1 / (d D / ) / (d )
= - D /
If and are the equilibrium constant at and respectively, show further that ln ( / ) = - (D / ) (1/ - 1/ ). This is achieved by definite integral. This relationship indicates that the plot of ln ( versus 1/( is a straight line, and the slope is - ( ). Thus, can be determined by measuring the equilibrium constant at different temperatures. = 8.312 J * 298 ln(3166)
= 20.0 kJ / hr
This example illustrates the evaluation of Gibb's energy when the equilibrium constant is known. | 690 | 794 |
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An elementary reaction is a single step reaction with a single transition state and no intermediates. Elementary reactions add up to complex reactions; non-elementary reactions can be described by multiple elementary reaction steps. A set of elementary reactions comprises a reaction mechanism, which predicts the elementary steps involved in a complex reaction. Below are two reaction coordinates of two reactions. One describes an elementary reaction, and the other describes a non-elementary reaction. \[ \text{Reactants} \rightarrow \text{Products} \nonumber \] This is a sample reaction coordinate of an elementary reaction. Note that there is one transition state and no intermediate. Elementary steps cannot be broken down into simpler reactions. \[ \text{Reactants} \rightarrow \text{Intermediates} \rightarrow \text{Products} \nonumber \] This is a sample reaction coordinate of a complex reaction. Note that it involves an intermediate and multiple transition A complex reaction can be explained in terms of elementary reactions. The of a reaction refers to the number of reactant particles involved in the reaction. Because there can only be discrete numbers of particles, the molecularity must take an integer value. Molecularity can be described as unimolecular, bimolecular, or termolecular. There are no known elementary reactions involving four or more molecules. A unimolecular reaction occurs when a molecule rearranges itself to produce one or more products. An example of this is radioactive decay, in which particles are emitted from an atom. Other examples include cis-trans isomerization, thermal decomposition, ring opening, and racemization. The rate at which a substance decomposes is dependent on its concentration. Unimolecular reactions are often first-order reactions as explained by Frederick Alexander Lindemann, which is referred to as the Lindemann mechanism. A reaction involves the collision of two particles. Bimolecular reactions are common in organic reactions such as nucleophilic substitution. The rate of reaction depends on the product of the concentrations of both species involved, which makes bimolecular reactions second-order reactions. A reaction requires the collision of three particles at the same place and time. This type of reaction is very uncommon because all three reactants must simultaneously collide with each other, with sufficient energy and correct orientation, to produce a reaction. There are three ways termolecular reactions can react, and all are third order. 1. How are non-elementary steps and elementary steps related? 2. Choose the correct statements. 3. Which of the following elementrary reactions is a termolecular reaction? 4. Which rate law corresponds to a bimolecular reaction? 5. Give an example of a reaction with a molecularity of 1/2. 6. True or False: Given species A and B inside a container, instruments detect that three (3) collisions occured before product was formed. That is, we know a reaction occured after detecting three collisions in a box. We can conclude that the reaction is a termolecular reaction (as the reaction could have been produced from A+A+B or A+B+B). | 3,186 | 795 |
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have their own classes of , different and , and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal ligands] , while names are written Prefix Ligands Metal (Oxidation State). According to the , ligands are Lewis bases since they can donate electrons to the central metal atom. The metals, in turn, are Lewis acids since they accept electrons. Coordination complexes consist of a ligand and a metal center cation. The overall charge can be positive, negative, or neutral. Coordination compounds are complex or contain complex ions, for example: A ligand can be an anion or a neutral molecule that to the complex (NH , H O, Cl ). The number of ligands that attach to a metal depends on whether the ligand is monodentate or polydentate. To begin naming coordination complexes, here are some things to keep in mind. Ligands that act as anions which end in "-ide" are replaced with an ending "-o" (e.g., Chloride → Chloro). Anions ending with "-ite" and "-ate" are replaced with endings "-ito" and "-ato" respectively (e.g., Nitrite → Nitrito, Nitrate → Nitrato). Most neutral molecules that are ligands carry their normal name. The few exceptions are the first four on the chart: ammine, aqua, carbonyl, and nitrosyl. The number of ligands present in the complex is indicated with the prefixes di, tri, etc. The exceptions are polydentates that have a prefix already in their name are the most common). When indicating how many of these are present in a coordination complex, put the ligand's name in parentheses and use (for two ligands), (for three ligands), and (for four ligands). Prefixes always go before the ligand name; they are taken into account when putting ligands in alphabetical order. Note that "mono" often is not used. For example, \(\ce{[FeCl(CO)2(NH3)3]^{2+}}\) would be called triamminedicarbonylchloroiron(III) ion. Remember that ligands are always , before the metal is. What is the name of this complex ion: \(\ce{[CrCl2(H2O)4]^{+}}\)? Let's start by identifying the ligands. The ligands here are Cl and H O. Therefore, we will use the monodentate ligand names of "chloro" and "aqua". Alphabetically, aqua comes before chloro, so this will be their order in the complex's name. There are 4 aqua's and 2 chloro's, so we will add the number prefixes before the names. Since both are monodentate ligands, we will say "tetra[aqua]di[chloro]". Now that the ligands are named, we will name the metal itself. The metal is Cr, which is chromium. Therefore, this coordination complex is called tetraaquadichlorochromium(III) ion. See the next section for an explanation of the (III). What is the name of this complex ion: \(\ce{[CoCl_2(en)_2]^{+}}\)? We take the same approach. There are two chloro and ethylenediamine ligands. The metal is Co, cobalt. We follow the same steps, except that \(en\) is a polydentate ligand with a prefix in its name (ethylene amine), so "bis" is used instead of "di", and parentheses are added. Therefore, this coordination complex is called dichlorobis(ethylenediamine)cobalt(III) ion. When naming the metal center, you must know the formal metal name and the oxidation state. To show the oxidation state, we use Roman numerals inside parenthesis. For example, in the problems above, chromium and cobalt have the oxidation state of +3, so that is why they have (III) after them. Copper, with an oxidation state of +2, is denoted as copper(II). If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. Copper +2 will change into cuprate(II). The following change to their Latin names when part of an anion complex: The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present). Finally, when a complex has an overall charge, "ion" is written after it. This is not necessary if it is neutral or part of a coordination compound (Example \(\Page {3}\)). Here are some examples with determining oxidation states, naming a metal in an anion complex, and naming coordination compounds. What is the name of [Cr(OH) ] ? Immediately we know that this complex is an anion. There is only one monodentate ligand, hydroxide. There are four of them, so we will use the name "tetrahydroxo". The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". The oxidation state of the metal is 3 (x+(-1)4=-1). Write this with Roman numerals and parentheses (III) and place it after the metal to get tetrahydroxochromate(III) ion. What is the name of \(\ce{[CuCl4]^{2-}}\)? tetrachlorocuprate(II) ion A last little side note: when naming a coordination compound, it is important that you name the cation first, then the anion. You base this on the charge of the ligand. Think of NaCl. Na, the positive cation, comes first and Cl, the negative anion, follows. What is the name of \([\ce{Pt(NH3)4},\ce{Pt(Cl)4}]\)? NH is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH first, followed by the one with Cl (the same order as the formula). This coordination compound is called tetraammineplatinum(II) tetrachloroplatinate(II). What is the name of \(\ce{[CoCl(NO2)(NH3)4]^{+}}\) ? This coordination complex is called tetraamminechloronitrito-N-cobalt(III). N comes before the O in the symbol for the nitrite ligand, so it is called nitrito-N. If an O came first, as in [CoCl(ONO)(NH ) ] , the ligand would be called nitrito-O, yielding the name tetraamminechloronitrito-O-cobalt(III). Nitro (for NO ) and nitrito (for ONO) can also be used to describe the nitrite ligand, yielding the names tetraamminechloronitrocobalt(III) and tetraamminechloronitritocobalt(III). While chemistry typically follow the nomenclature rules for naming complexes and compounds, there is disagreement with the rules for constructing formulas of inorganic complex. The order of ligand names in their formula has been ambiguous with different conventions being used (charged vs neutral, number of each ligand, etc.). In 2005, adopted the recommendation that all ligand names in formulas be listed (in the same way as in the naming convention) irrespective of the charge or number of each ligand type. However, this rule is not adhered to in many chemistry laboratories. For practice, the order of the ligands in chemical formulas does not matter as long as you write the transition metal first, which is the stance taken here. Write the chemical formulas for: Write the chemical formulas for | 6,882 | 796 |
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Formal logic deals with relationships among propositions, where a is any statement of (alleged) fact. Any proposition can be expressed as an ordinary English sentence, although it may be more convenient to use mathematical symbols or some other notation. The following are all propositions: A proposition need not be true. The last of these examples is a false proposition. We represent an arbitrary proposition by any convenient symbol, usually a letter of the alphabet. Thus, we could stipulate that “\(p\)” represents any of the propositions above. Once we have associated a symbol with a particular proposition, the symbol itself is taken to represent an assertion that the proposition is true. It is an axiom of ordinary logic that any proposition must be either true or false. If we associate the symbol “\(p\)” with a particular proposition, we write “\(\sim p\)” to represent the statement: “The proposition represented by the symbol ‘\(p\)’ is false.” \(\sim p\) is called the . We can use the negation of \(p\), \(\sim p\), to state the axiom that a proposition must be either true or false. To do so, we write: Either \(p\) or \(\sim p\) is true. We can write this as the proposition “\(p\) or \(\sim p\)”. The negation of the negation of \(p\) is an assertion that \(p\) is true; that is, \(\sim \ \sim \ p\ =\ p\). Logic is concerned with relationships among propositions. One important relationship is that of . If a proposition, \(q\), follows logically from another proposition, \(p\), we say that \(q\) is implied by \(p\). Equivalently, we say that proposition \(p\) implies proposition \(q\). The double-shafted arrow, \(\mathrm{\Rightarrow }\), is used to symbolize this relationship. We write “\(p\Rightarrow q\)” to mean, “That proposition \(p\) is true implies that proposition \(q\) is true.” We usually read this more tersely, saying, “\(p\) implies \(q\).” Of course, “\(p\Rightarrow q\)” is itself a proposition; it asserts the truth of a particular logical relationship between propositions \(p\) and \(q\). For example, let \(p\) be the proposition, “Figure A is a square.” Let \(q\) be the proposition, “Figure A is a rectangle.” Then, writing out the proposition, \(p\Rightarrow q\), we have: Figure A is a square implies figure A is a rectangle. This is, of course, a valid implication; for this example, the proposition \(p\Rightarrow q\) is true. For reasons that will become clear shortly, \(p\Rightarrow q\) is called the of \(p\) and \(q\). Proposition \(p\) is often called a , while proposition \(q\) is called a . That is, the truth of \(p\) is sufficient to establish the truth of \(q\). \[\text{sufficient condition}\, \mathrm{\Rightarrow} \, \text{necessary condition}\] Now, if proposition \(p\Rightarrow q\) is true, and proposition \(q\) is also true, can we infer that proposition \(p\) is true? We most certainly cannot! In the example we just considered, the fact that figure A is a rectangle does not prove that figure A is a square. We call \(q\Rightarrow p\) the converse of \(p\Rightarrow q\). The conditional of \(p\) and \(q\) can be true while the converse is false. Of course, it can happen that both \(p\Rightarrow q\) and \(q\Rightarrow p\) are true. We often write “\(p\Leftrightarrow q\)” to express this relationship of mutual implication. We say that, “\(p\) implies \(q\) and conversely.” What if \(p\Rightarrow q\), and \(q\) is false? That is, \(\sim q\) is true. In this case, \(p\) must be false! If \(\sim q\) is true, it must also be that \(\sim p\) is true. Using our notation, we can express this fact as \[(p\Rightarrow q\, \text{and} \sim q) \Rightarrow \sim p\] Equivalently, we can write \[(p\Rightarrow q) \mathrm{\Leftrightarrow } (\sim q\Rightarrow \sim p)\] That is, \(p\Rightarrow q\) and \(\sim q\Rightarrow \sim p\) are equivalent propositions; if one is true, the other must be true. \(\sim q\Rightarrow \sim p\) is called the of \(p\Rightarrow q\). The equivalence of the conditional and its contrapositive is a theorem that can be proved rigorously in an axiomatic formulation of logic. In our later reasoning about thermodynamic principles, we use the equivalence of the conditional and the contrapositive of \(p\) and \(q\). The equivalence of the conditional, \(p\Rightarrow q\), and the contrapositive, \(\sim q\Rightarrow \sim p\), is the reason that \(q\) is called a necessary condition. If \(p\Rightarrow q\), it is necessary that \(q\) be true for \(p\) to be true. (If figure A is to be a square, it must be a rectangle.) It is also intimately related to proof by contradiction. Suppose that we know \(p\) to be true. If, by assuming that \(q\) is false (\(\sim q\) is true), we can validly demonstrate that \(p\) must also be false \((\sim q\mathrm{\Rightarrow }\sim p\), so that \(\sim p\) is true), we have the contradiction that \(p\) is both true and false (\(p\) and \(\sim p\)). Since \(p\) cannot be both true and false, it must be false that is false (\(\sim \sim q=q\)). Otherwise stated, the equivalence of the conditional and the contrapositive leads not only to (\(p\) and \(\sim p\)) but also to (\(q\) and \(\sim q\)). \[\sim q \mathrm{\Rightarrow } \sim p\] implies \[p \mathrm{\Rightarrow } q.\] In summary, since we know p to be true, our assumption that \(q\) is false, together with the valid implication \(\sim q\mathrm{\Rightarrow }\sim p\), leads to the conclusion that \(q\) is true, which contradicts our original assumption, so that the assumption is false, and \(q\) is true. | 5,524 | 797 |
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Since a discrete distribution is completely specified by the probabilities of each of its events, we can represent it by a bar graph. The probability of each event is represented by the height of one bar. We can generalize this graphical representation to represent continuous distributions. To see what we have in mind, let us consider a particular example. Let us suppose that we have a radar gun and that we decide to interest ourselves in the typical speeds of cars on a highway just outside of town. As we think about this project, we recognize that speeds might vary with the time of day and the day of the week. Random variations in many other factors might also be important; these include weather conditions and accidents in the vicinity. To eliminate as many atypical factors as possible, we might decide that typical speeds are those of cars going north between 1:00 pm and 4:00 pm on weekdays when the road surface is dry and there are no disabled vehicles in view. If we have a lot of time and the road is busy, we could collect a lot of data. Let us suppose that we record the speeds of \(10,000\) cars. Each datum would be the speed of a car on the road at a time when the selected conditions are satisfied. To use this data, we want to summarize it in a form that is easy to visualize. One way to do this is to aggregate the data to give the number of cars in each \(20\) mph range; the results might look something like the data in Table 2. Figure 3 is a five-channel bar graph that displays the number of cars in each \(20\) mph range. A great deal of information is lost in the aggregating process. In particular, nothing on the graph represents the number of automobiles in narrower speed intervals. Now, suppose that we repeat this task, but that we do not have enough time to collect data on as many as \(10,000\) more cars. We will be curious about the extent to which our two samples agree with one another. Since the total number of vehicles will be different, the appropriate way to go about this is obviously to compare the fraction of cars in each speed range. In fact, using fractions enables us to compare any number of such studies. To the extent that these studies measure the same thing—typical speeds under the specified conditions—the fraction of automobiles in any particular speed interval should be approximately constant. Dividing the number of automobiles in each speed interval by the total number of automobiles gives a representation that focuses attention on the proportion of automobiles with various speeds. The shape of the bar graph remains the same; all that changes is the scale we use to label the ordinate. (See Figure 4.) Insofar as any repetition of this experiment gives nearly the same results, this is a useful change. However, the fundamental limitations of the graph remain. For example, if we want to use the graph to estimate how speeds are distributed in any other set of intervals, we have to read values off the ordinate and manipulate them in ways that may not be very satisfactory. To estimate the fraction with speeds between \(20\) mph and \(40\) mph, we might assign half of the automobiles in the \(10-30\) mph interval and half of those in the \(30-50\) mph interval to the new interval. This enables us to estimate that the fraction in the \(20-40\) mph interval is \(0.165\). This estimate is much less reliable than one that could be made by going back to the raw data for all \(10,000\) automobiles. The data can also be represented as a . In a histogram, the information is represented by the area rather than the height of the bar. In the present case, the only visible change to the graph is another change in the numerical values on the ordinate. In Figure 5, the area of a bar represents the fraction of automobiles with speeds in the given interval. As the speed interval is made smaller, any of these bar graphs looks increasingly like a continuous curve. (See Figure 6.) The histogram has the advantage that, as the curve becomes continuous, the interpretation remains constant: the area under the curve between any two speeds always represents the fraction of automobiles with speeds in this interval. It turns out that we are adept at visually estimating the relative areas of different parts of a histogram. That is, from a quick glance at a histogram, we are able to obtain a good semi-quantitative appreciation of the significance of the underlying data. If the histogram captures our experience, and we expect future events to have the same characteristics, the histogram becomes an expression of probability. All that is necessary is that we construct the histogram so that the total area under the graph is unity. If we let \(f\left(u\right)\) be the area under the graph from \(u=-\infty \) to \(u=u\), then \(f\left(u\right)\) represents the probability that the speed of a randomly selected automobile will lie between \(-\infty\) and \(u\). For any \(a\) and , the probability that \(u\) lies in the interval \(a<b\)> is \(f\left(b\right)-f\left(a\right)\). The function \(f\left(u\right)\) is called the , because its value for any \(u\) is the fraction of automobiles that have a speed less than \(u\). \(f\left(a\right)\) is the frequency with which we observe values of the random variable, \(u\), that are less than \(a\). Equivalently, we can say that \(f\left(u\right)\) is the probability that any randomly selected automobile will have a speed less than \(u\). If we let the width of every interval go to zero, the bar graph representation of the histogram becomes a curve, and the histogram becomes a continuous function of the random variable, \(u\). (See Figure 7.) \(\boldsymbol{f}\left(\boldsymbol{u}\right)\) \(\boldsymbol{\ \ }\boldsymbol{f}\left(\boldsymbol{u}\right)\) | 5,799 | 798 |
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Recrystallization, also known as fractional crystallization, is a procedure for purifying an impure compound in a solvent. The method of purification is based on the principle that the solubility of most solids increases with increased temperature. This means that as temperature increases, the amount of solute that can be dissolved in a solvent increases. An impure compound is dissolved (the impurities must also be soluble in the solvent), to prepare a highly concentrated solution at a high temperature. The solution is cooled. Decreasing the temperature causes the solubility of the impurities in the solution and the substance being purified to decrease. The impure substance then crystallizes before the impurities- assuming that there was more impure substance than there were impurities. The impure substance will crystallize in a purer form because the impurities won't crystallize yet, therefore leaving the impurities behind in the solution. A filtration process must be used to separate the more pure crystals at this point. The procedure can be repeated. Solubility curves can be used to predict the outcome of a recrystallization procedure. Recrystallization works best when The slower the rate of cooling, the larger the crystals are that form. The disadvantage of recrystallization is that it takes a long time. Also, it is very important that the proper solvent is used. This can only be determined by trial and error, based on predictions and observations. The solution must be soluble at high tempratures and insoluble at low temperatures. The advantage or recrystallization is that, when carried out correctly, it is a very effective way of obtaining a pure sample of some product, or precipitate. These are the important steps to the recrsytallization process. | 1,801 | 800 |
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Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid). Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber\] giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\Page {1}\)). The remaining weak acid is present in the nonionized form. For acetic acid, at equilibrium: \[K_\ce{a}=\ce{\dfrac{[H3O+,CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber\] Table \(\Page {1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1. At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH ) N, in water reacts according to the equation: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber\] This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure \(\Page {2}\)). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, \(K_b\), is called the of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+,OH- ]}{[(CH3)3N]}} \nonumber\] The ionization constants of several weak bases are given in Table \(\Page {2}\) and Table E2. Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH CO H] = 0.0787 and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). What is the value of \(K_a\) for acetic acid? We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \] \[\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+,CH3CO2- ]}{[CH3CO2H]}} \\[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \\[4pt] &=1.77×10^{−5} \end{align*}\] What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers: \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber\] In one mixture of NaHSO and Na SO at equilibrium, \(\ce{[H3O+]}\) = 0.027 ; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\). \(K_a\) for \(\ce{HSO_4^-}= 1.2 ×\times 10^{−2}\) Caffeine, C H N O is a weak base. What is the value of for caffeine if a solution at equilibrium has [C H N O ] = 0.050 , \(\ce{[C8H10N4O2H+]}\) = 5.0 × 10 , and [OH ] = 2.5 × 10 ? At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber \] so \[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+,OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber\] What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base: \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber \] In a solution containing a mixture of \(\ce{NaH2PO4}\) and \(\ce{Na2HPO4}\) at equilibrium with: for \(\ce{HPO4^2-}=1.6×10^{−7} \) The pH of a 0.0516- solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its \(K_a\)? \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber\] We determine an equilibrium constant starting with the initial concentrations of HNO , \(\ce{H3O+}\), and \(\ce{NO2-}\) as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.) We can solve this problem with the following steps in which is a change in concentration of a species in the reaction: We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration): To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH: \[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber \] The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H O , which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). The initial concentration of \(\ce{H3O+}\) is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0). The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). For each 1 mol of \(\ce{H3O+}\) that forms, 1 mol of \(\ce{NO2-}\) forms. The equilibrium concentration of HNO is equal to its initial concentration plus the change in its concentration. Now we can fill in the ICE table with the concentrations at equilibrium, as shown here: Finally, we calculate the value of the equilibrium constant using the data in the table: \[K_\ce{a}=\ce{\dfrac{[H3O+,NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber\] The pH of a solution of household ammonia, a 0.950- solution of NH is 11.612. What is for NH . \(K_b = 1.8 × 10^{−5}\) Formic acid, HCO H, is the irritant that causes the body’s reaction to ant stings. What is the concentration of hydronium ion and the pH in a 0.534- solution of formic acid? \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber \] . The equilibrium expression is: \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \] The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color): At equilibrium: \[\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+,HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}\] Now solve for \(x\). Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x << 0.534\), which us to simplify the denominator term as \((0.534 − x) = 0.534\). This gives: \[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber\] Solve for \(x\) as follows: \[\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \\[4pt] &=9.6×10^{−5} \\[4pt] x &=\sqrt{9.6×10^{−5}} \\[4pt] &=9.8×10^{−3} \end{align*}\] To check the assumption that \(x\) is small compared to 0.534, we calculate: \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}\] \(x\) is less than 5% of the initial concentration; the assumption is valid. We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &=9.8×10^{−3}\:M \end{align*}\] The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so: \[pH = −\log(9.8×10^{−3})=2.01 \nonumber\] Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100- solution of acetic acid, CH CO H? \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber\] Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\). percent ionization = 1.3% The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid. Find the concentration of hydroxide ion in a 0.25- solution of trimethylamine, a weak base: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber\] Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example \(\Page {6}\). The reactants and products will be different and the numbers will be different, but the logic will be the same: . The table shows the changes and concentrations: . At equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+,OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber\] If we assume that is small relative to 0.25, then we can replace (0.25 − ) in the preceding equation with 0.25. Solving the simplified equation gives: \[x=4.0×10^{−3} \nonumber \] This change is less than 5% of the initial concentration (0.25), so the assumption is justified. Recall that, for this computation, \(x\) is equal to the equilibrium concentration of in the solution (see earlier tabulation): \[\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \\[4pt] &=4.0×10^{−3}\:M \end{align*}\] Then calculate pOH as follows: \[\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber\] Using the relation introduced in the previous section of this chapter: \[\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber \] permits the computation of pH: \[\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber\] Check the work. A check of our arithmetic shows that \(K_b = 6.3 \times 10^{−5}\). \(7.56 × 10^{−4}\, M\), 2.33% 2.33% Some weak acids and weak bases ionize to such an extent that the simplifying assumption that is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation. Sodium bisulfate, NaHSO , is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50- solution of \(\ce{HSO4-}\)? \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \] We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps: . This table shows the changes and concentrations: . As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\). At equilibrium: \[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+,SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber\] If we assume that is small and approximate (0.50 − ) as 0.50, we find: \[x=7.7×10^{−2} \nonumber \] When we check the assumption, we confirm: \[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber\] which for this system is \[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber\] The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\). The equation: \[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \] gives \[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber\] or \[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \] This equation can be solved using the quadratic formula. For an equation of the form \[ax^{2+} + bx + c=0, \nonumber\] \(x\) is given by the quadratic equation: \[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber\] In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\). Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root: \[x=7.2×10^{−2} \nonumber\] Now determine the hydronium ion concentration and the pH: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*}\] The pH of this solution is: \[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber\] \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber\] It will be necessary to convert [OH ] to \(\ce{[H3O+]}\) or pOH to pH toward the end of the calculation. pH 11.16 Strong acids, such as \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\), all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl , Br , and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\) differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order \(\ce{HCl < HBr < HI}\), and so \(\ce{HI}\) is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the . Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O , and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water: \[\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq)\] \[\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq)\] Thus, O and \(\ce{NH2-}\) appear to have the same base strength in water; they both give a 100% yield of hydroxide ion. In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is \(\ce{HF < HCl < HBr < HI}\). Likewise, for group 16, the order of increasing acid strength is H O < H S < H Se < H Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is \(\ce{CH4 < NH3 < H2O < HF}\); across the third row, it is \(\ce{SiH4 < PH3 < H2S < HCl}\) (see Figure \(\Page {3}\)). Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O E(OH) , and include sulfuric acid, \(\ce{O2S(OH)2}\), sulfurous acid, \(\ce{OS(OH)2}\), nitric acid, \(\ce{O2NOH}\), perchloric acid, \(\ce{O3ClOH}\), aluminum hydroxide, \(\ce{Al(OH)3}\), calcium hydroxide, \(\ce{Ca(OH)2}\), and potassium hydroxide, \(\ce{KOH}\): If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond between the element and oxygen is more readily broken than bond between oxygen and hydrogen. Hence bond is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH) and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds. If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond relatively strongly covalent. The oxygen-hydrogen bond, bond , is thereby weakened because electrons are displaced toward E. Bond is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called . Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H SO , or O S(OH) (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H SO , or OS(OH) (with a sulfur oxidation number of +4). Likewise nitric acid, HNO , or O NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO , or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\Page {4}\)). Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate \(\ce{Al(H2O)3(OH)3}\), is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, \(\ce{Al(H2O)3(OH)3}\), is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion: \[[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber\] In this reaction, a proton is transferred from one of the aluminum-bound H O molecules to a hydroxide ion in solution. The \(\ce{Al(H2O)3(OH)3}\) compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion: \[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber\] In this case, protons are transferred from hydronium ions in solution to \(\ce{Al(H2O)3(OH)3}\), and the compound functions as a base. The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH < NH < H O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H SO < H SO ). The strengths of oxyacids also increase as the electronegativity of the central element increases [H SeO < H SO ]. | 20,666 | 801 |
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As we have seen in the previous section, it may be easy to construct the carbon skeleton of the target compound of a synthesis, but with a reactive functional group at the wrong carbon. Therefore it is important also to have practice at shifting reactive entry points around to achieve the final desired product. We shall illustrate this form of molecular chess with reactions from previous post. A typical problem may be to devise syntheses for achieving the following conversions: If you proceed as in the previous section, you will see that the starting material and products have the same number of carbons and the same general bonding arrangement of those carbons. Getting the functionality to the right carbons is now the problem. If you review the reactions discussed up to now, you will find that the only good way of getting a reactive group at the end of a chain starting with a reactive group in the middle of the chain is borane isomerization ( ). The borane, \(6\), can be obtained from the starting material, \(5\), by hydroboration ( ) and, on heating, \(6\) will be converted to \(7\): Production of 4-methyl-1-pentanol from the substituted alkylborane, \(7\), can be achieved by oxidation ( ): The three steps - hydroboration, isomerization, and oxidation - thus constitute a reasonable synthesis of the first desired compound. The second desired product is a little more tricky because the isomerization of \(6\) to \(7\) cannot be stopped at the alkylborane, \(8\): The best procedure to get the desired product is to generate the 1-alkene from the borane with 1-decene ( ) and then add hydrogen bromide by a mechanism ( ). Incursion of radical-chain addition must be avoided because it would give 1-bromo-4-methylpentane ( ): A very brief summary of the transformations that we have studied so far, which do not change the carbon skeleton, is given in Table 13-5 along with appropriate section references. In using this table, it is necessary to check the specific sections to be sure the reaction is applicable to the conversion that you wish to achieve and to determine the proper conditions for the reaction. and (1977) | 2,165 | 802 |
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or are molecules that posess a metal center that is bound to ligands (atoms, ions, or molecules that donate electrons to the metal). These complexes can be neutral or charged. When the complex is charged, it is stabilized by neighboring counter-ions. Coordination chemistry emerged from the work of Alfred Werner, a Swiss chemist who examined different compounds composed of cobalt (III) chloride and ammonia. Upon the addition of hydrochloric acid, Werner observed that ammonia could not be completely removed. He then proposed that the ammonia must be bound more tightly to the central cobalt ion. However, when aqueous silver nitrate was added, one of the products formed was solid silver chloride. The amount of silver chloride formed was related to the number of ammonia molecules bound to the cobalt (III) chloride. For example, when silver nitrate was added to CoCl ·6NH all three chlorides were converted to silver chloride. However, when silver nitrate was added to CoCl ·5NH , only 2 of the 3 chlorides formed silver chloride. When CoCl ·4NH was treated with silver nitrate, one of the three chlorides precipitated as silver chloride. The resulting observations suggested the formation of or . In the , which is also referred to in some texts as the , ligands are directly bound to the central metal. In the , sometimes referred to as the , other ions are attached to the complex ion. Werner was awarded the Nobel Prize in 1913 for his coordination theory. As the table above shows, the complex ion [Co(NH ) ] is countered by the three chloride ions. The multi-level binding of coordination complexes play an important role in determining the dissociation of these complexes in aqueous solution. For example, \(\ce{[Co(NH3)5Cl]Cl2}\) dissociates into three ions while \(\ce{[Co(NH3)4Cl2 ]Cl}\) dissociates into two ions. By applying a current through the aqueous solutions of the resulting complex compounds, Werner measured the electrical conductivity and thus the dissociation properties of the complex compounds. The results confirmed his hypothesis of the formation of complex compounds. It is important to note that the above compounds have a coordination number of 6, which is a common coordination number for many inorganic complexes. Coordination numbers for complex compounds typically range from 1 to 16. Some methods of verifying the presence of complex ions include studying its chemical behavior. This can be achieved by observing the compounds' color, solubility, absorption spectrum, magnetic properties, etc. The properties of complex compounds are separate from the properties of the individual atoms. By forming coordination compounds, the properties of both the metal and the ligand are altered. Metal-ligand bonds are typically thought of Lewis acid-base interactions. The metal atom acts as an electron pair acceptor (Lewis acid), while the ligands act as electron pair donors (Lewis base). The nature of the bond between metal and ligand is stronger than intermolecular forces because they form directional bonds between the metal ion and the ligand, but are weaker than covalent bonds and ionic bonds. Monodentate ligands donate one pair of electrons to the central metal atoms. An example of these ligands are the haldide ions (F , Cl , Br , I ). Polydentate ligands, also called chelates or chelating agents, donate more than one pair of electrons to the metal atom forming a stronger bond and a more stable complex. A common chelating agent is ethylenediamine (en), which, as the name suggests, contains two ammines or :NH sites which can bind to two sites on the central metal. An example of a tridentate ligand is bis-diethylenetriammine. An example of such a coordination complex is bis-diethylenetriamine cobalt III. Complex ions can form many compounds by binding with other complex ions in multiple ratios. This leads to many combinations of coordination compounds. The structures of certain coordination compounds can also have isomers, which can change their interactions with other chemical agents. The binding between metal and ligands is studied in metals, tetrahedral, and octahedral structures. There are many pharmaceutical and biological applications of coordination complexes and their isomers. | 4,277 | 803 |
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Pressure is the average force that material (gas, liquid or solid) exert upon the surface, e.g. walls of a container or other confining boundary. Vapor pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate or sublime into a gaseous form and all gases have a tendency to condense back to their liquid or solid form. When a liquid is in a confined, closed, container, an equilibrium exists between the liquid and its gaseous phase. This equilibrium exists regardless of the temperature inside the container and the temperature of the liquid. The equilibrium exists due to the fact that some of the particles in the liquid, essentially at any temperature, will always have enough energy to escape the intrinsic cohesive forces and enter the gaseous phase (Figure \(\Page {1}\)). The fact that the vapor pressure is equal to the external pressure can become important when talking about boiling temperatures at various altitudes. At higher altitudes the pressure is lower than at sea level and therefore liquids such as water boil at lower temperatures. This would translate it into a longer time to cook something in the water as compared to cooking the same object at sea level. The opposite is also true if one boils water at an altitude lower than sea level.
The change in vapor pressure of a pure substance as temperature changes can be described using the equation known as the : \[ ln \dfrac{P_2}{P_1} = \dfrac{\Delta H_{vap}}{R} \left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right) \label{CC}\] 293 K 300 K Step 1: Use the Clausius Clapeyron equation (Equation \ref{CC}). Assume 293 K to be T and 17.5 mmHg to be P and 300 K to be T . We know the enthalpy of vaporization of water is 44000 J mol . Therefore we plug in everything we are given into the equation. \[ ln\dfrac{P_2}{17.5mmHg} = \dfrac{44000 J mol^{-1}}{8.3145 Jmol^{-1}K^{-1}}\ \left(\dfrac{1}{293K}-\dfrac{1}{300K} \right) \] Step 2: Calculate everything possible, which is everything on the right side of the equation. \[ ln \dfrac{P_2}{17.5 mmHg} = (5291.96 K)(0.0000796 K^{-1}) \] Step 3: To isolate the variable, we need to get rid of the natural log function on the left side. To do so, we must exponentiate both sides of the equation after calculating the numerical value of the right side of the equation: \[ \dfrac{P_2}{17.5 mmHg} = e^{0.421} \] Step 4: To solve for P , we multiply both sides of the equation by 17.5mmHg. \[ P_2 = 26.7 mmHg \] 298K Step 1: Once again this can be solved using the Clausius-Clapeyron equation. If we set P1 = 23.8 mmHg, T1 = 298K, and P2 = 1075 mmHg, all we need to do is to solve for T . So plug everything we know into the Clausius-Clapeyron Equation and we get: \[ \ln\dfrac{1075 mmHg}{23.8 mmHg} = \dfrac{44000 Jmol^{-1}}{8.3145 Jmol^{-1}K^{-1}}\ \left (\dfrac{1}{298 K}-\dfrac{1}{T_2} \right) \] Step 2: We can solve the right side of the equation for a numerical answer and we can simplify the right side of the equation to: \[ 3.81 = (5291.96 K)(\dfrac{1}{298K}-\dfrac{1}{T_2}) \] Step 3: Further simplifying the equation by distributing the 5291.96 K, we get: \[ 3.81 = (17.76) - \dfrac{5291.96K}{T_{2}} \] Step 4: After subtracting 17.76 from both sides and multiplying both sides by T to isolate the T term, we get: \[ -13.95T_2 = -5291.96K \] Step 5: Solving for T we get: \[ T_2 = 383K \] *Note: There may be some slight variations in this answer if you try working this problem out. The rounding in this problem will make a relatively large difference in mmHg (10-20 mmHg). While the Clausius-Clapeyron equation is useful for describing the vapor pressure behavior of a pure substance, it does not quite help us when we need to describe the vapor pressure of a solution comprised of two ore more different liquids with different vapor pressures, that is where comes in. Raoult's Law is: \[ \displaystyle P_{total} = \sum_{i} P_{i}X_{i} \label{Rlaw} \] Where P is the vapor pressure of that particular substance and X is the corresponding mole fraction of that substance. "i" is an indexing component that keeps track of each substance in the solution.
Essentially what Raoult's Law states is that the vapor pressure of a solution with two or more components is directly proportional to the vapor pressures of each component and their respective amounts in the solution (Figure 3). However, Raoult's Law is used to describe solutions that are essentially ideal solutions, meaning it is assumed that there are no interactions between the components of the solution. If the solution is non-ideal, it will deviate from the relationship described by Raoult's Law. H O 293 K 293 K Calculate the mole fractions (moles of each substance divided by total moles) of each substance in the solution. \[ X_{water} = 0.5000/3.500 = 0.143 \] \[ X_{ethanol} = 1.000/3.500 = 0.286 \] \[ X_{acetaldehyde} = 2.000/3.500 = 0.571 \] Plug in all the values and solve for P \[ P_{total} = (18mmHg)(0.143) + (67.5 mmHg)(0.286) + (740mmHg)(0.571mmHg) \] \[ P_{total} = 444.4 mmHg \] A solution is comprised of only water and ethylene glycol and is at 293 K. Water's mole fraction in this solution is 0.379 and water's vapor pressure at this temperature is 18 mmHg (assume the water and ethylene glycol mix homogeneously). The total vapor pressure of this solution is 9.15 mmHg. Calculate the vapor pressure of pure ethylene glycol at this temperature. Since this solution is only comprised of water and ethylene glycol, we can easily calculate the mole fraction of ethylene glycol in this solution by subtracting water's mole fraction from 1. \[ X_{ethylene glycol} = 1 - 0.379 = 0.621 \] Once we have the mole fraction of ethylene glycol, we have everything we need to solve for the partial pressure of pure ethylene glycol at 293 K using Raoult's Law: \[ 9.15 = (18mmHg)(0.379) + (P_{ethylene glycol})(0.621) \] Solving for \(P_{ethylene glycol}\) we get: \[ P_{ethylene glycol} = 3.75mmHg \] Both the Clausius-Clapeyron Equation (Equation \ref{CC}) and Raoult's Law (Equation \ref{Rlaw}) describe liquids without any significant solutes in them. So what happens if the liquid contains a nonvolatile solute, in other words, a solute that does not evaporate and merely stays in the solution? Will the vapor pressure of the liquid increase or decrease? It turns out that having a nonvolatile solute in the liquid will decrease the vapor pressure of the liquid because the solutes will interfere with the high-energy liquid molecules' path to the surface to break free into the gaseous phase. An example of a nonvolatile solute in a liquid would be glucose in water (Figure \(\Page {4}\): Henry's Law). Henry's Law can also be used the describe the partial pressure of a volatile solute in a liquid as a function of its concentration in the liquid. An example of this would be oxygen or carbon dioxide in a soda. Henry's Law is as follows: \[ C = (K_{H})P_{gas} \] Where P is the partial pressure of either the volatile solute or of the solvent with a nonvolatile solute in it and C is the solubility of the volatile solute or the concentration of the solvent with a nonvolatile solute in it. K is Henry's Constant and is different for various substances and differs with temperature and comes in many different units. Assume at 273 K and a CO pressure of 3.6 atm, the aqueous solubility of CO is 24.8 ml CO per liter. What is the molarity of a saturated water solution when the CO is under its normal partial pressure in air of 0.000395 atm. We need to calculate Henry's Constant for CO when the partial pressure of CO is 3.6 atm. To do this, we must first calculate the molarity of CO under these conditions: \[ Molarity = \dfrac{0.0248L CO_{2} * \dfrac{1 mol CO_{2}}{22.4L CO_{2}}}{1 L solution} = 0.00111 M CO_{2} \] Next we must solve for Henry's Constant under these conditions: \[ K_H = \dfrac{C}{P_{CO_2}} = 0.000308 \dfrac{M CO_2}{atm} \] Solve for the concentration (molarity) using Henry's Law: \[ C = K_H * P_{CO_2} = 0.000308M CO_{2}atm^{-1} * 0.000395atm = 1.22 M CO_2 \] One great use of Henry's Law is to calculate the various solubilities of gases in solutions under different pressures. One example of this is what is colloquially referred to as (scientifically called decompression sickness). When divers go underwater, they are subjected to much higher pressures than they would be at sea level therefore their blood is much more soluble to gases and as such much larger quantities of gases such as nitrogen dissolve in the blood. If the divers then rise up back to the surface too quickly, these gases will be forced out of their dissolved form and form bubbles in the blood vessels. These gas bubbles can then cause the blood vessels to pop, causing much physical pain and serious medical issues. To avoid this, divers rise from deep depths slowly so their bodies can expel the extraneous gas through the lungs and allow the amounts of gas in the blood to equilibrate with the external pressure. This way the divers can avoid gas bubbles forming in blood vessels and therefore any personal injury. The vapor pressure of water at 283 K is 9.2 mmHg, at what temperature is the vapor pressure of water 546 mmHg? | 9,308 | 804 |
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Electrospray ionization is a soft ionization technique that is typically used to determine the molecular weights of proteins, peptides, and other biological macromolecules. Soft ionization is a useful technique when considering biological molecules of large molecular mass, such as the aformetioned, because this process does not fragment the macromolecules into smaller charged particles, rather it turns the macromolecule being ionized into small droplets. These droplets will then be further desolvated into even smaller droplets, which creates molecules with attached protons. These protonated and desolvated molecular ions will then be passed through the mass analyzer to the detector, and the mass of the sample can be determined. Electrospray ionization mass spectrometry is a desorption ionization method. Desorption ionization methods can be performed on solid or liquid samples, and allows for the sample to be nonvolatile or thermally unstable. This means that ionization of samples such as proteins, peptides, olgiopeptides, and some inorganic molecules can be performed. Electrospray ionization mass spectrometry requires that a molecule be of a fairly large mass. The instrument has a small mass range that it is able to detect, so therefore the mass of the unknown injected sample can easily be determined; as it must be in the range of the instrument. This quantitative analysis is done by considering the mass to charge ratios of the various peaks in the spectrum (Figure 1). The spectrum is shown with the mass-to-charge (m/z) ratio on the x-axis, and the relative intensity (%) of each peak shown on the y-axis. Calculations to determine the unknown mass, M , from the spectral data can then be performed using \[ p = \dfrac{m}{z}\] \[p_1 = \dfrac{M_r + z_1}{ z_1} \] \[p_2 = \dfrac{M_r + (z_1 - 1)}{z_1 - 1}\] where p and p are adjacent peaks. Peak p comes before peak p in the spectrum, and has a lower m/z value. The z value represents the charge of peak one. It should be noted that as the m/z value increases, the number of protons attached to the molecular ion decreases. Figure 1 below illustrates these concepts. Electrospray ionization mass spectrometry research was pioneered by the analytical chemistry professor John Bennet Fenn, who shared the Nobel Prize in Chemistry with Koichi Tanaka in 2002 for his work on the subject. Calculations for m/z in spectrum [M + 6H] [M + 5H] [M + 4H] m/z = 15006/6 = 2501 m/z = 15005/5 = 3001 m/z = 15004/4 = 3751 [M + 3H] [M + 2H] [M + H] m/z = 15003/3 = 5001 m/z = 15002/2 = 7501 m/z = 15001/1= 15001 Samples for injection into the electrospray ionization mass spectrometer work the best if they are first purified. The reason purity in a sample is important is because this technique does not work well when mixtures are used as the analyte. For this reason a means of purification is often employed to inject a homogeneous sample into the capillary needle. , , and are methods of choice for this purpose. The chosen purification method is then attached to the capillary needle, and the sample can be introduced directly. There are some clear advantages to using electrospray ionization mass spectrometry as an analytical method. One advantage is its ability to handle samples that have large masses. Another advantage is that this ionization method is one of the softest ionization methods available, therefore it has the ability to analyze biological samples that are defined by non-covalent interactions. A quadrupole mass analyzer can also be used for this method, which means that a sample's structure can be determined fairly easily. The m/z ratio range of the quadrupole instrument is fairly small, which means that the mass of the sample can be determined to with a high amount of accuracy. Finally, the sensitivity for this instrument is impressive and therefore can be useful in accurate quantitative and qualitative measurements. Some disadvantages to electrospray ionization mass spectrometry are present as well. A major disadvantage is that this technique cannot analyze mixtures very well, and when forced to do so, the results are unreliable. The apparatus is also very difficult to clean and has a tendency to become overly contaminated with residues from previous experiments. Finally, the multiple charges that are attached to the molecular ions can make for confusing spectral data. This confusion is further fueled by use of a mixed sample, which is yet another reason why mixtures should be avoided when using an electrospray ionization mass spectrometer. The capillary needle is the inlet into the apparatus for the liquid sample. Once in the capillary needle, the liquid sample is nebulized and charged. There is a large amount of pressure being applied to the capillary needle, which in effect nebulizes the liquid sample forming a mist. The stainless steel capillary needle is also surrounded by an electrode that retains a steady voltage of around 4000 volts. This applied voltage will place a charge on the droplets. Therefore, the mist that is ejected from the needle will be comprised of charged molecular ions. The molecular ions are oxidized upon entering the desolvating capillary, and a continual voltage is applied to the gas chamber in which this capillary is located. Here the desolvation process begins, through the use of a dry gas or heat, and the desolvation process continues through various pumping stages as the molecular ion travels towards the mass analyzer. An example of a dry gas would be an N gas that has been dehydrated. The gas or heat then provides means of evaporation, or desolvation, for the ionized droplets. As the droplets become smaller in size, their electric field densities become more concentrated. The increase in electric field density causes the like charges to repel one another, which induces an increase in surface tension. The point where the droplet can no longer support this increase in surface tension is known as the Rayleigh limit. At this point, the droplet divides into smaller droplets of either positive or negative chrage. This process is refered to as either a coulombic explosion or the ions are described as exiting the droplet through the "Taylor cone". Once the molecular ions have reached the entrance to the mass analyzer, they have been effectively reduced through protonation. are used to determine the mass-to-charge ratio (m/z), this ratio is used to differentiate between molecular ions that were formed in the desolvating capillary. In order for a mass-to-charge ratio to be determined, the mass analyzer must be able to separate even the smallest masses. The ability of the analyzer to resolve the mass peaks can be defined with the following equation; \[R = \dfrac{m}{\Delta m} \] This equation represents the mass of the first peak (m), divided by the difference between the neighboring peaks \(\Delta m\). The better the resolution, the more useful the data. The mass analyzer must also be able to measure the ion currents produced by the multiply charged particles that are created in this process. Mass analyzers use electrostatic lenses to direct the beam of molecular ions to the analyzer. A vacuum system is used to maintain a low pressure environment in order to prevent unwanted interactions between the molecular ions and any components that may be present in the atmosphere. These atmospheric components can effect the determined mass-to-charge ratio, so it is best to keep them to a minimum. The mass-to-charge ratio is then used to determine quantitative and qualitative properties of the liquid sample. The molecular ions pass through the mass analyzer to the detector. The detector most commonly used in conjunction with the quadrupole mass analyzer is a high energy dynode (HED), which is a electron multiplier with some slight variations. In an HED detector, the electrons are passed through the system at a high voltage and the electrons are measured at the end of the funnel shaped apparatus; otherwise known as the anode. A HED detector differs from the electron multiplier in that it operates at a much higher sensitivity for samples with a large mass than does the electron multiplier detector. Once the analog signal of the mass-to-charge ratio is recorded, it is then converted to a digital signal and a spectrum representing the data run can be analyzed. Using the above spectrum, calculate the mass of the protein. For p shown in spectrum, the m/z is 7501 and for p the m/z is 15001. (Hint: Use the above equations, and the charge of p for z ) | 8,566 | 805 |
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There are two general types of bulk intermolecular forces: Wetting agents are substances that reduce the of water to allow it to spread drops onto a surface, increasing the spreading abilities of a liquid. Lowering the surface tension lowers the energy required to spread drops onto a film, thus weakening the cohesive properties of the liquid and strengthening its adhesive properties. One example of how wetting agents work is in the formation of micelles. Micelles consist of hydrophilic heads forming an outer layer around lipophilic tails. When in water, the micelles' tails can surround an oil droplet while the heads are attracted to the water. There are four main types of wetting agents: anionic, cationic, amphoteric, and nonionic. One method of knowing whether or not a liquid has a wetting agent in it is to spread the liquid on a surface that is coated in grease. If the liquid does not contain a wetting agent, the its cohesive forces would overpower adhesive forces, causing the liquid to for droplets on the surface (Figure \(\Page {2}\), left). If the liquid does contain a wetting agent, the grease would be dissolved and the surface tension of the liquid would be lowered, causing the adhesive forces to overpower the cohesive forces. This would result in the liquid spreading evenly along the surface (Figure \(\Page {2}\), right). Another method is to place the liquid in a test tube and observe the liquid's meniscus (Figure \(\Page {3}\)). If the liquid contains a wetting agent, its adhesive forces are stronger than cohesive forces, which means the liquid molecules are more inclined to stick to the surface than other liquid molecules. This results in a concave meniscus. If the liquid does not contain a wetting agent and is naturally very cohesive, like mercury, it forms a convex meniscus. This is caused by the fact the the molecules of the liquid have a stronger attraction to each other than to the surface of the test tube. Intermolecular forces also cause a phenomenon called , which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a ), as shown in Figure \(\Page {3}\). When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. | 2,464 | 807 |
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To consider theories for fluctuations in the number of particles we require an ensemble that keeps V, T, and the chemical potential, constant, a grand canonical ensemble. To construct the grand canonical ensemble, the system is enclosed in a container that is permeable both to heat and to the passage of particles. The number of particles in the system can range over all possible values. As in the canonical ensemble we have occupation numbers a describing the number of systems that have energy E and N particles. There are two indices for summation since neither the energy of the system nor the number of particles is the same in all of the systems. We can specify the state of the ensemble by specifying that a , a , a , of the systems are in states 1, 2, 3, …, respectively, with energies E , E , E ,…. depending on N, the number of particles in each system. In the grand canonical ensemble the occupation numbers obey three conditions. Following the principle of a priori probabilities, we assume that every distribution of occupation number is equally probable. As we have seen for the canonical ensemble we can use the method of the most probable distribution to derive the form of the distribution function and the partition function for the grand ensemble. The number of ways W(a) or W(a , a , a ,….) that any particular distribution of a ’s can be achieved is given by As before we assume that the systems are macroscopic and are distinguishable objects that can be distributed among available states. In any particular distribution a /A is the fraction of systems of the canonical ensemble in the jth energy state containing N particles. The overall probability P that a system is in the jth quantum state with N particles is obtained by averaging a /A over all the allowed distributions. The notation of summing over a means that the value of a depends on the distribution and that the summations are over all distributions that satisfy the constraints. The most probable distribution is the distribution that maximizes W. The maximum W will be found by setting the derivative ( lnW/ a ) = 0 subject to the constraints above that the a must sum to A, the total energy E is equal to the sum of a E , and the total number of particles is a N. This implies that In other words, there is no change in the total number of systems A, total energy E, and the total number of particles, with respect to changes in the occupation numbers. The procedure followed here is analogous to that used for the canonical ensemble; we maximize W subject to the constraints. The difference is that there is one addition constraint on the number of particles, that was not present in the canonical ensemble. To maximize subject to constraints we use the method of LaGrange undetermined multipliers. where we have moved the summation symbol in front of the three terms. The constants , – , and are the undetermined multipliers. We first carry out the derivative and then find the value of the multipliers. We can evaluate ( lnW/ a ) = ( lnA!/ a ) – ( ln a !/ a ) using Stirling's approximation: lnx! xlnx – x. To simplify ( lnW/ a ) the first step is to note that lnW = lnA! - ln a ! = AlnA – A - a ln a - a Since A = a the last two terms cancel to give lnW = AlnA - a ln a Note that exactly the same procedure and algebra are used to show that the derivative of ln W is equal to the most probable distribution in the canonical ensemble, and so we have: The most probable distribution is: Now we only need to find the undetermined multipliers , and . By summing both sides the indices N and j we can obtain . The left-hand side is equal to one and e = 1/ where is the grand canonical partition function or the grand partition function (for short). The Boltzmann distribution in this ensemble can be written The star indicates that this is the most probable distribution as shown above by maximizing W with respect to the occupation numbers. The averages of mechanical properties E, P, and N are We have shown that = 1/k T where k is Boltzmann’s constant. This is true for all of the ensembles. A similar approach can be used to show that = - /k T. The differential of the grand partition function is Using the definitions of mechanical properties from above we have The last term is the ensemble-averaged work done by the system. We add d( E ) + d( N) to both sides The thermodynamic equation dE = TdS - PdV + dN can be rearranged to TdS = dE - dN + PdV Comparing these two equations by dividing through by we find that = - = - /kT from the second term in each equation. Since we can replace / by - we have which gives the entropy as Since G = N = E + PV - TS We can use the above equation to determine that There are several ways to express the grand partition function. Starting with the definition, we can define the canonical partition function for N particles as and then insert this expression into the grand partition function We know also that e = e . The quantity e is often denoted . For indistinguishable particles Q = q /N! where q is the molecular partition function. Therefore, ( ) | 5,194 | 808 |
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Contact angle is one of the common ways to measure the wettability of a surface or material. Wetting refers to the study of how a liquid deposited on a solid (or liquid) substrate spreads out or the ability of liquids to form boundary surfaces with solid states. The wetting, as mentioned before is determined by measuring the contact angle, which the liquid forms in contact with the solids or liquids. The wetting tendency is larger, the smaller the contact angle or the is. A wetting liquid is a liquid that forms a contact angle with the solid which is smaller then 90º. A non-wetting liquid creates a contact angle between 90º and 180º with the solid. The contact angle is an angle that a liquid creates with a solid surface or capillary walls of a porous material when both materials come in contact together. This angle is determined by both properties of the solid and the liquid and the interaction and repulsion forces between liquid and solid and by the three phase interface properties (gas, liquid and solid). Those interactions are described by forces which are intermolecular forces. The balance between the cohesive forces of similar molecules such as between the liquid molecules (i.e. hydrogen bonds and Van der Waals forces) and the adhesive forces between dissimilar molecules such as between the liquid and solid molecules (i.e. mechanical and electrostatic forces) will determine the contact angle created in the solid and liquid interface. The traditional definition of a contact angle is the angle a liquid creates with the solid or liquid when it is deposited on it. A less traditional definition is the angle a liquid creates with the sides of a capillary when it rises in it to create a meniscus. Thomas Young (13 June 1773 – 10 May 1829) proposed treating the contact angle of a liquid with a surface as the mechanical equilibrium of a drop resting on a plane solid surface under the restrains of three surface tensions: \[ \gamma_{sv} - \gamma_{sl} = \gamma_{lv} \cos \theta \label{1}\] This equation is deceptively simple, but there are a few difficulties with it because of the definitions of the surface tension of the solid-vapor and solid liquid phases. Another approach avoids specifying the field of intermolecular forces between solid and liquid and instead offers a thermodynamic solution. This lead to Young and Dupre’ equation which introduces the reversible work of adhesion of liquid and solid and it’s relation to the surface tension between liquid and vapor phases and the contact angle: \[ W_{sl}^{adh} = \gamma_{lv} (1 + \cos \theta) \label{2}\] Where \(WA\) is the reversible work of adhesion of the liquid to the solid when coated with an adsorbed film of the saturated vapor. Capillarity is the ability of a substance to draw another substance into it and It occurs when the adhesive intermolecular forces between the liquid and a substance are stronger than the cohesive intermolecular forces inside the liquid. The effect causes a concave meniscus to form where the substance is touching a vertical surface (Figure \(\Page {3}\)). The same effect is what causes porous materials to soak up liquids. Capillary forces pull a wetting liquid has a low contact angle with the surface, it wets the surface. Assuming that there are no other factors involved (e.g. roughness), a low contact angle with water means that the surface is hydrophilic. A completely wetting liquid forms a zero contact angle into a capillary by creating a curved meniscus at the rising liquid front. This phenomenon can be described with the and the Laplace pressure inside a capillary. Many measuring methods have been described in the literature but only a few of them have been found to be widely applicable. The most frequently used is the goniometer-telescope measurement of sessile-drop contact angles. Commercial contact angle goniometers employ a microscope objective to view the angle directly. In the static method a drop is deposited on a surface and the contact angle can be measured by looking at the drop through a goniometer (an instrument that measures contact angles). The dynamic method is similar to the static one but the drop of liquid which is deposited on a surface is modified. The droplet is being deposited via a syringe and the droplet’s volume is changed dynamically without increasing its solid-liquid interface area and this maximum angle is the advancing angle. Volume is then removed to produce the smallest possible angle, which is called the receding angle. The difference between those two measured angles is called contact . The capillary rise on a Wilhelmy plate is a good way to obtain the contact angle by measuring the height of the meniscus created on a partially immersed plate. The contact angle is obtained directly form the height trough: \[ \sin ( \theta) = 1 - \rho g h^2 / 2 \gamma \label{3} \] where: The common direct measurement of contact angles of a liquid drop on a flat and smooth solid is not applicable to powders and dry porous foods. The common method of measuring the contact angle in capillary rise in porous media is by using the Lucas-Washburn’s equation which is derived from of liquid flow in capillary rise: \[ h^2 = r c(\gamma \cos \theta / 2 \eta) t \label{4}\] \[ w^2 = c (\rho^2 \gamma \cos \theta / \eta) t \label{5}\] Where: The common method to measure the contact angles in this case is to conduct capillary rise experiments where your sample is hang bellow a balance and you dip it in the tested liquid. The main deficiency of this approach (using Washburn’s equation) is its inability to separate between those two variables (r and cos θ or c and cos θ). The common method to overcome the problem of the unknown term r•cos(θ) is by using a reference liquid that completely wets the sample (θ=0, cos(θ)=1). It was discovered however that dynamic advancing contact angle is generally larger than the static one, even for a total wetting liquid and as a result this method can be erroneous. Seibold et al, (2000) suggested a way to overcome this problem. It was observed that this constant term varies as a function of the liquid used, in contradiction with the Washburn approach which uses Hexane as a completely wetting liquid in order to find the radius of the pores. In this study, it was proposed to get the actual constant term r in Washburn equation by plotting the measured value r cos θ versus the rate of rise of the alkanes (the slope in Equations \ref{4} and \ref{5}). The intercept at zero velocity gives the value of r. Siebold el al. (2000) carried out capillary rise experiments with different n-alkanes, which are all considered completely wetting, due to their low surface tension. In each case, a linear relationship was obtained between squared height of the rising liquid and time, their results of the term r•cos(θ) were calculated from the slopes of those curves. This finding was transformed into a method to calculate or measure the contact angle at zero velocity. Siebold et al. (2000) suggested that for porous media the term r•cos(θ) for each alkane (the term was calculated from Washburn equation) can be ploted against the initial front rate of each liquid. The curve that was created can be extrapolated to velocity zero (and therefore, cos(θ)=1) and this enables the determination of the representative radius r. After finding r, which is a property of the solid sample and doesn’t change as a function of the liquids used, we can conduct capillary rise experiment with our test liquid and calculate the contact angles this liquid creates with the solid porous sample. The interest in contact angles is because it plays a significant role in a number of technological, environmental and biological phenomena. Water imbibition into porous media theory has been shown to have a multidisciplinary validity in food, soil physics, geology, printings and more. Imbibition of a liquid by a porous solid is a phenomenon highly dependent on wetting. Capillary imbibition is a mechanism that plays a significant role during rehydration of dry food particles that are considered as porous media. Imbibition is highly dependent on the wettability of the porous media, which is usually determined by measuring contact angles which liquids form with the solid. | 8,298 | 809 |
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The and introduced in Chemical Energy are very useful chemical properties. We have already mentioned some basic rules regarding the quantities and and their preceding equations. If both sides of the equations are multiplied by a factor to alter the number of moles, or for the equation should be multiplied by the same factor, since they are quantities per equation as written. Thus, for the equation \(\mathrm{C_{\large{(graphite)}} + 0.5\, O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -110\: kJ/mol}\). We can write it in any of the following forms: \(\mathrm{
2 C_{\large{(graphite)}} + O_2 \rightarrow 2 CO, \hspace{20px} \mathit{\Delta H}^\circ = -220\: kJ/mol\: (multiplied\: by\: 2) \\
6 C_{\large{(graphite)}} + 3 O_2 \rightarrow 6 CO, \hspace{20px} \mathit{\Delta H}^\circ = -660\: kJ/mol\: (multiplied\: by\: 6)}\) For the reverse reaction, the signs of these quantities are changed (multiply by -1). The equation implies the following: \(\mathrm{
CO \rightarrow C_{\large{(graphite)}} + 0.5 O_2, \hspace{20px} \mathit{\Delta H}^\circ = 110\: kJ/mol\\
2 CO \rightarrow 2 C_{\large{(graphite)}} + O_2, \hspace{20px} \mathit{\Delta H}^\circ = 220\: kJ/mol}\) states that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless of whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction. For example, in the diagram below, we look at the oxidation of carbon into \(\ce{CO}\) and \(\ce{CO2}\). The direct oxidation of carbon (graphite) into \(\ce{CO2}\) yields an enthalpy of -393 kJ/mol. When carbon is oxidized into \(\ce{CO}\) and then \(\ce{CO}\) is oxidized to \(\ce{CO2}\), the enthalpies are -110 and -283 kJ/mol respectively. The sum of enthalpy in the two steps is exactly -393 kJ/mol, same as the one-step reaction. The two-step reactions are: \(\mathrm{
C + \dfrac{1}{2} O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -110\: kJ/mol\\
CO + \dfrac{1}{2} O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -283\: kJ/mol}\) Adding the two equations together and canceling out the intermediate, \(\ce{CO}\), on both sides leads to \(\mathrm{C + O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = (-110)+(-283) = -393\: kJ/mol}\) The real merit is actually to evaluate the enthalpy of formation of \(\ce{CO}\) as we shall see soon. Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure. In the above example, it is very difficult to control the oxidation of graphite to give pure \(\ce{CO}\). However, enthalpy for the oxidation of graphite to \(\ce{CO2}\) can easily be measured. So can the enthalpy of oxidation of \(\ce{CO}\) to \(\ce{CO2}\). The application of Hess's law enables us to estimate the enthalpy of formation of \(\ce{CO}\). Since, \(\mathrm{
C + O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -393\: kJ/mol\\
CO + \dfrac{1}{2} O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -283\: kJ/mol}\) Subtracting the second equation from the first gives \(\mathrm{C + \dfrac{1}{2} O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -393 -(-283) = -110\: kJ/mol}\) The equation shows the standard enthalpy of formation of \(\ce{CO}\) to be -110 kJ/mol. Application of Hess's law enables us to calculate °, and for chemical reactions that impossible to measure, providing that we have all the data of related reactions. Some more examples are given below to illustrate the applications of Hess Law. | 3,790 | 810 |
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A particle in a 1-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it escape. The particle in a box problem is a common application of a quantum mechanical model to a simplified system consisting of a particle moving horizontally within an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of E and \(\psi\) that the particle can possess. E represents allowed energy values and \(\psi(x)\) is a wavefunction, which when squared gives us the probability of locating the particle at a certain position within the box at a given energy level. To solve the problem for a particle in a 1-dimensional box, we must follow our : The potential energy is (V=0 for 0<x<L) and (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box. Doing so significantly simplifies our later mathematical calculations as we employ these when solving the Schrödinger Equation. The time-independent Schrödinger equation for a particle of mass \(m\) moving in one direction with energy \(E\) is \[-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \label{5.5.1}\] with This equation can be modified for a particle of mass \(m\) free to move parallel to the x-axis with zero potential energy (V = 0 everywhere) resulting in the quantum mechanical description of free motion in one dimension: \[ -\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x) \label{5.5.2}\] This equation has been well studied and gives a general solution of: \[\psi(x) = A\sin(kx) + B\cos(kx) \label{5.5.3}\] where A, B, and k are constants. The solution to the Schrödinger equation we found above is the general solution for a 1-dimensional system. We now need to apply our to find the solution to our particular system. According to our boundary conditions, the probability of finding the particle at \(x=0\) or \(x=L\) is zero. When \(x=0\), then \(\sin(0)=0\) and \(\cos(0)=1\); therefore, \(B\) to fulfill this boundary condition giving: \[\psi(x) = A\sin(kx) \label{5.5.4}\] We can now solve for our constants (\(A\) and \(k\)) systematically to define the wavefunction.
Differentiate the wavefunction with respect to \(x\): \[\dfrac{d\psi}{dx} = kA\cos(kx) \label{5.5.5}\] Differentiate the wavefunction algain with respect to \(x\): \[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}A\sin(kx) \label{5.5.6}\] Since \(\psi(x) = A\sin(kx)\), then \[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}\psi \label{5.5.7}\] If we then solve for k by comparing with the Schrödinger equation above, we find: \[k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{1/2} \label{5.5.8}\] Now we plug \(k\) into our wavefunction (Equation \ref{5.5.4}): \[\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.9}\]
To determine A, we have to apply the boundary conditions again. Recall that the When x = L: \[0 = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L \label{5.5.10}\] This is only true when \[\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L = n\pi \label{5.5.11}\] where \(n = 1,2,3, …\) Plugging this back in gives us: \[\psi = A\sin{\dfrac{n\pi}{L}}x \label{5.5.12}\] To determine \(A\), recall that the total probability of finding the particle inside the box is 1, meaning there is no probability of it being outside the box. When we find the probability and set it equal to 1, we are the wavefunction. \[\int^{L}_{0}\psi^{2}dx = 1 \label{5.5.13}\] For our system, the normalization looks like: \[A^2 \int^{L}_{0}\sin^2\left(\dfrac{n\pi}{L}\right)x\,dx = 1 \label{5.5.14}\] Using the solution for this integral from an integral table, we find our normalization constant, \(A\): \[A = \sqrt{\dfrac{2}{L}} \label{5.5.15}\] Which results in the normalized wavefunctions for a particle in a 1-dimensional box: \[\psi_n = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x \label{5.5.16}\] where \(n = 1,2,3, …\) Solving for the energy of each \(\psi\) requires substituting Equation \(\ref{5.5.16}\) into Equation \(\ref{5.5.2}\) to get the allowed energies for a particle in a box: \[E_n = \dfrac{n^{2}h^{2}}{8mL^{2}} \label{5.5.17}\] Equation \(\ref{5.5.17}\) is a very important result and tells us that: This is also consistent with the : if the particle had zero energy, we would know where it was in both space and time. The wavefunction for a particle in a box at the \(n=1\) and \(n=2\) energy levels look like this: The probability of finding a particle a certain spot in the box is determined by squaring \(\psi\). The probability distribution for a particle in a box at the \(n=1\) and \(n=2\) energy levels looks like this: Notice that the number of (places where the particle has zero probability of being located) increases with increasing energy n. Also note that as the energy of the particle becomes greater, the quantum mechanical model breaks down as the energy levels get closer together and overlap, forming a continuum. This continuum means the particle is free and can have any energy value. At such high energies, the classical mechanical model is applied as the particle behaves more like a continuous wave. Therefore, the particle in a box problem is an example of . What is the \(\Delta E\) between the \(n = 4\) and \(n = 5\) states for an \(F_2\) molecule trapped within in a one-dimension well of length 3.0 cm? At what value of \(n\) does the energy of the molecule reach \(¼k_BT\) at 450 K, and what is the separation between this energy level and the one immediately above it? Since this is a one-dimensional particle in a box problem, the particle has only kinetic energy (V = 0), so the permitted energies are: \[E_n = \dfrac{n^2 h^2}{8 m L^2} \nonumber\] with \(n=1,2,...\) The energy difference between \(n = 4\) and \(n=5\) is then \[\Delta E = E_5 - E_4=\dfrac{5^2 h^2}{8 m L^2} - \dfrac{4^2 h^2}{8 m L^2}\nonumber\] Using Equation \(\ref{5.5.17}\) with the mass of \(F_2\) (37.93 amu = \(6.3 \times 10^{-26}\;kg \)) and the length of the box (\(L= 3 \times 3.0\times10^{-2}\;\text{m}^2\): \[\Delta E=\dfrac{9 h^2}{8 m L^2} = \dfrac{9 (6.626\times10^{-34}\;\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1})^2 }{8(6.30938414\times10^{-26}\;\text{kg})(3.0\times10^{-2}\;\text{m}^2)^2 }\nonumber\] \[\Delta E=8.70\times10^{-39}\; J \nonumber\] The \(n\) value for which the energy reaches \(\frac{1}{4} k_B T\): \[\dfrac{n^2 h^2}{8 m L^2} = \dfrac{1}{4} k_B T \nonumber\] \[n=1.79 \times 10^9 \nonumber\] The separation between \(n+1\) and \(n\): \[\Delta E = E_{n+1} - E_{n} = \dfrac{(n+1)^2h^2 - (n)^2h^2}{8mL^2} \nonumber\] \[\Delta E = \dfrac{(2n+1)h^2}{8mL^2} \nonumber\] \[\Delta E = 3.47\times10^{-30}\;J \nonumber\] | 6,826 | 811 |
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The ability to predict the outcomes of acid-base reactions, which are very common in chemistry, is extremely beneficial. Many different things can change the outcome of an acid-base reaction including heat and pressure. Knowing the K (acid dissociation constant) and the K (base association constant) is the best way to predict the direction of an Acid-Base Reaction. Given the reaction: Using the K and K values, one can predict which molecule will act as the acid in this reaction and which molecule will act as the base. For OH the K is extremely small in relation to K , so it will act as a base in this reaction. For CH CH SH, the K is extremely small in relation to K so CH CH SH will act like an acid. For problems 1 through 3: In which direction will an acid-base reaction move, given the following factor? K is Big K is Small K=Q (For extra review, check out the ChemWiki page on K and Q !) For problems 4-6: Given the following information, finish the equation and determine the acid and the base NaOH + HCl ⇔ ? + ? H O + HCl ⇔ ? + ? H O + HC H O ⇔ ? + ? | 1,095 | 812 |
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In determining the structural formula of an alkene, it is often necessary to find the location of the double bond within a given carbon framework. One way of accomplishing this would be to selectively break the double bond and mark the carbon atoms that originally formed that bond. For example, there are three isomeric alkenes that all give 2-methylbutane on catalytic hydrogenation. These are 2-methyl-2-butene (compound A), 3-methyl-1-butene (compound B) and 2-methyl-1-butene (compound C), shown in the following diagram. If the double bond is cleaved and the fragments marked at the cleavage sites, the location of the double bond is clearly determined for each case. A reaction that accomplishes this useful transformation is known. It is called Ozone, O , is an allotrope of that adds rapidly to carbon-carbon double bonds. Since the overall change in ozonolysis is more complex than a simple addition reaction, its mechanism has been extensively studied. Reactive intermediates called ozonides have been isolated from the interaction of ozone with alkenes, and these unstable compounds may be converted to stable products by either a reductive workup (Zn dust in water or alcohol) or an oxidative workup (hydrogen peroxide). The results of an oxidative workup may be seen by clicking the " " button a second time. Continued clicking of this button repeats the cycle. The chief difference in these conditions is that reductive workup gives an aldehyde product when hydrogen is present on a double bond carbon atom, whereas oxidative workup gives a carboxylic acid or carbon dioxide in such cases. The following equations illustrate ozonide formation, a process that is believed to involve initial syn-addition of ozone, followed by rearrangement of the extremely unstable molozonide addition product. They also show the decomposition of the final ozonide to carbonyl products by either a reductive or oxidative workup. From this analysis and the examples given here, you should be able to deduce structural formulas for the alkenes that give the following ozonolysis products: The vicinal glycols prepared by alkene hydroxylation (reaction with osmium tetroxide or permanganate) are cleaved to aldehydes and ketones in high yield by the action of (Pb(OAc) ) or (HIO ). This oxidative cleavage of a carbon-carbon single bond provides a two-step, high-yield alternative to ozonolysis, that is often preferred for small scale work involving precious compounds. A general equation for these oxidations is shown below. As a rule, cis-glycols react more rapidly than trans-glycols, and there is evidence for the intermediacy of heterocyclic intermediates (as shown), although their formation is not necessary for reaction to occur. | 2,757 | 813 |
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The motion of the electron in the hydrogen atom is not free. The electron is bound to the atom by the attractive force of the nucleus and consequently quantum mechanics predicts that the total energy of the electron is quantized. The expression for the energy is: \[E_n =\dfrac{-2 \pi^2 m e^4Z^2}{n^2h^2} \label{1}\] with \(n = 1,2,3,4...\) where \(m\) is the mass of the electron, \(e\) is the magnitude of the electronic charge, \(n\) is a quantum number, \(h\) is Planck's constant and \(Z\) is the atomic number (the number of positive charges in the nucleus). Equation \(\ref{1}\) applies to any one-electron atom or ion. For example, He is a one-electron system for which = 2. We can again construct an energy level diagram listing the allowed energy values (Figure \(\Page {1}\)). These are obtained by substituting all possible values of into Equation \(\ref{1}\). As in our previous example, we shall represent all the constants which appear in the expression for \(E_n\) by a constant and we shall set \(Z = 1\), i.e., consider only the hydrogen atom. \[ E_n = \dfrac{-K}{n^2} \label{2}\] with \(n = 1,2,3,4...\) Since the motion of the electron occurs in three dimensions we might correctly anticipate three quantum numbers for the hydrogen atom. But the energy depends only on the quantum number \(n\) and for this reason it is called the . In this case, the energy is inversely dependent upon , and as is increased the energy becomes less negative with the spacings between the energy levels decreasing in size. When \(n = 0\), then \(E = 0\) and the electron is free of the attractive force of the nucleus. The average distance between the nucleus and the electron (the average value of ) increases as the energy or the value of increases. Thus energy must be supplied to pull the electron away from the nucleus. The parallelism between increasing energy and increasing average value of \(r\) is a useful one. In fact, when an electron loses energy, we refer to it as "falling" from one energy level to a lower one on the energy level diagram. Since the average distance between the nucleus and the electron also decreases with a decrease in , then the electron literally does fall in closer to the nucleus when it "falls" from level to level on the energy level diagram. The energy difference between \(E_{\infty}\) and \(E_1\): \[E_{n= \infty} - E_{n=1} = 0 -(-K) = K = \dfrac{-2 \pi^2 m e^4}{n^2h^2} = \dfrac{e^2}{2a_o}\label{3}\] is called the ionization energy and is the energy required to pull the electron completely away from the nucleus and is, therefore, the energy of the reaction: \[ H \rightarrow H^+ + e^-\] with \(\Delta E = 13.60 \,eV = 0.5 \,a.u.\) As mentioned earlier, hydrogen gas emits colored light when a high voltage is applied across a sample of the gas contained in a glass tube fitted with electrodes. The electrical energy transmitted to the gas causes many of the hydrogen molecules to dissociate into atoms: \[ H_2 \rightarrow H + H \] The electrons in the molecules and in the atoms absorb energy and are excited to high energy levels. lonization of the gas also occurs. When the electron is in a quantum level other than the lowest level (with = 1) the electron is said to be excited, or to be in an excited level. The lifetime of such an excited level is very brief, being of the order of magnitude of only 10 sec. The electron loses the energy of excitation by falling to a lower energy level and at the same time emitting a photon to carry off the excess energy. We can easily calculate the frequencies which should appear in the emitted light by calculating the difference in energy between the two levels and making use of Bohr's frequency condition (E =h\nu\): \[ \nu = \dfrac{E_n' - E_n}{h} \] with \(n' > n\) Suppose we consider all those frequencies which appear when the electron falls to the lowest level, \(n = 1\): \[ \nu = \dfrac{E_1 - E_n}{h} = \dfrac{K}{h} \left( \dfrac{1}{1} - \dfrac{1}{n^2} \right) \label{4}\] with \(n= 2, 3, 4 ... \) very value of \(n\) substituted into Equation \(\ref{3}\) gives a distinct value for . In Figure \(\Page {1}\) we illustrate the changes in energy which result when the electron emits a photon by an arrow connecting the excited level (of energy ) with the ground level (of energy ). The frequency resulting from each drop in energy will be directly proportional to the length of the arrow. Just as the arrows increase in length as is increased, so increases. However, the spacings between the lines decrease as is increased, and the spectrum will appear as shown directly below the energy level diagram in Figure \(\Page {1}\).
Each line in the spectrum is placed beneath the arrow which represents the change in energy giving rise to that particular line. Free electrons with varying amounts of kinetic energy (½ ) can also fall to the \( = 1\) level. The energy released in the reversed ionization reaction ( ): \[H^+ + e^- \rightarrow H\] will equal , the difference between and , plus ½ , the kinetic energy originally possessed by the electron. Since this latter energy is not quantized, every energy value greater than should be possible and every frequency greater than that corresponding to \[\nu = \dfrac{K}{h}\] The hydrogen atom does possess a spectrum identical to that predicted by Equation \(\ref{3}\), and the observed value for agrees with the theoretical value. This particular series of lines, called the Lyman series, falls in the ultraviolet region of the spectrum because of the large energy changes involved in the transitions from the excited levels to the lowest level. The first few members of a second series of lines, a second line spectrum, falls in the visible portion of the spectrum. It is called the Balmer series and arises from electrons in excited levels falling to the second quantum level. Since E equals only one quarter of , the energy jumps are smaller and the frequencies are correspondingly lower than those observed in the Lyman series. Four lines can be readily seen in this series: red, green, blue, and violet. Each color results from the electrons falling from a specific level, to the = 2 level: red ; green, ; blue, and violet . Other series, arising from electrons falling to the = 3 and = 4 levels, can be found in the infrared (frequencies preceding the red end or long wavelength end of the visible spectrum). The fact that the hydrogen atom exhibits a line spectrum is visible proof of the quantization of energy on the atomic level. ( / ) | 6,582 | 814 |
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Infrared spectroscopy, an analytical technique that takes advantage of the vibrational transitions of a molecule, has been of great significance to scientific researchers in many fields such as protein characterization, nanoscale semiconductor analysis and space exploration. Infrared spectroscopy is the study of interaction of infrared light with matter, which can be used to identify unknown materials, examine the quality of a sample or determine the amount of components in a mixture. Infrared light refers to electromagnetic radiation with wavenumber ranging from 13000 – 10 cm (corresponding wavelength from 0.78 – 1000 μm). Infrared region is further divided into three subregions: near-infrared (13000 – 4000 cm or 0.78 – 2.5 μm), mid-infrared (4000 – 400 cm or 2.5 – 25 μm) and far-infrared (400 – 10 cm or 25 – 1000 μm). The most commonly used is the middle infrared region, since molecules can absorb radiations in this region to induce the vibrational excitation of functional groups. Recently, applications of near infrared spectroscopy have also been developed. By passing infrared light through a sample and measuring the absorption or transmittance of light at each frequency, an infrared spectrum is obtained, with peaks corresponding to the frequency of absorbed radiation. Since all groups have their characteristic vibrational frequencies, information regarding molecular structure can be gained from the spectrum. Infrared spectroscopy is capable of analyzing samples in almost any phase (liquid, solid, or gas), and can be used alone or in combination with other instruments following different sampling procedures. Besides fundamental vibrational modes, other factors such as overtone and combination bands, Fermi resonance, coupling and vibration-rotational bands also appear in the spectrum. Due to the high information content of its spectrum, infrared spectroscopy has been a very common and useful tool for structure elucidation and substance identification. Most commonly used instruments in infrared spectroscopy are dispersive infrared spectrometer and Fourier transform infrared spectrometer. Dispersive infrared spectrometer is mainly composed of radiation source, monochromator and detector. For mid-infrared region, Globar (silicon carbide), Nernst glower (oxides of zirconium, yttrium and erbium) and metallic helices (chromium-nickel alloy or tungsten) are frequently used as radiation sources. Tungsten-halogen lamps and metallic conductors coated with ceramic are utilized as sources for near-infrared region. A mercury high-pressure lamp is suitable for far-infrared region. Monochromator in conjunction with slits, mirrors and filters separates the wavelengths of light emitted. The dispersive elements within monochromator are prisms or gratings. Gratings have gradually replaced prisms due to their comparatively low cost and good quality. As shown in , radiation passes through both a sample and a reference path. Then the beams are directed to a diffraction grating (splitter), which disperses the light into component frequencies and directs each wavelength through a slit to the detector. The detector produces an electrical signal and results in a recorder response. Figure 1. Schematic illustration of dispersive infrared spectrometer. Figure from . Two types of detectors are employed in dispersive infrared spectrometer, namely, thermal detectors and photon detectors. Thermal detectors include thermocouples, thermistors, and pneumatic devices, which measure the heating effect generated by infrared radiation. Photon detectors are semiconductor-based. Radiation is able to promote electrons in photon detectors from valence band to conduction band, generating a small current. Photon detectors have faster response and higher sensitivity than do thermal detectors but are more susceptible to thermal noise. Dispersive infrared spectrometer has many limitations because it examines component frequencies individually, resulting in slow speed and low sensitivity. Fourier transform infrared (FTIR) spectrometer is preferred over dispersive spectrometer, since it is capable of handling all frequencies simultaneously with high throughput, reducing the time required for analysis. The radiation sources used in dispersive infrared spectrometer can also be used in FTIR spectrometer. In contrast with the monochromator in dispersive spectrometer, FTIR spectrometer as shown in employs an interferometer. The beamsplitter within the interferometer splits the incoming infrared beam into two beams, one of which is reflected by a fixed mirror, while the other one reflected by a moving mirror perpendicular to the fixed one. The length of path one beam travels is fixed and that of the other one is changing as the mirror moves, generating an optical path difference between the two beams. After meeting back at the beamsplitter, the two beams recombine, interfere with each other, and yield an interferogram. The interferogram produces inference signal as a function of optical path difference. It is converted to a spectrum of absorbance or transmittance versus wavenumber or frequency by Fourier transform. Figure 2. Schematic representation of Fourier transform infrared spectrometer. Figure from . Detectors used in FTIR spectrometers are mainly pyroelectric and photoconductive detectors. The former are constructed of crystalline materials (such as deuterated triglycine sulfate) whose electric polarization rely on temperature. The change in temperature leads to change in charge distribution of the detector and electric signal is produced. The latter (such as mercury cadmium telluride) provide better sensitivity and faster speed than do pyroelectric detectors over a broad spectral range. However, liquid nitrogen is needed for cooling of photoconductive detectors. Since different molecules with different combination of atoms produce their unique spectra, infrared spectroscopy can be used to qualitatively identify substances. In addition, the intensity of the peaks in the spectrum is proportional to the amount of substance present, enabling its application for quantitative analysis. For qualitative identification purposes, the spectrum is commonly presented as transmittance versus wavenumber. Functional groups have their characteristic fundamental vibrations which give rise to absorption at certain frequency range in the spectrum ( ). Figure 3.Infrared spectrum of 1-hexanol. Each band in a spectrum can be attributed to stretching or bending mode of a bond. Almost all the fundamental vibrations appear in the mid-infrared region. For instance, 4000 – 2500 cm region usually can be assigned to stretching modes of O-H, N-H or C-H. Triple-bond stretching modes appear in the region of 2500 – 2000 cm . C=C and C=O stretching bands fall in the 2000 – 1500 cm region. Hence, characterization of functional groups in substances according to the frequencies and intensities of absorption peaks is feasible, and also structures of molecules can be proposed. This method is applicable to organic molecules, inorganic molecules, polymers, etc. A detailed frequency list of functional groups is shown in . Note that several functional groups may absorb at the same frequency range, and a functional group may have multiple-characteristic absorption peaks, especially for 1500 – 650 cm , which is called the fingerprint region. Figure 4. Bands in the near-infrared region are overtones or combination bands. They are weak in intensity and overlapped, making them not as useful for qualitative analysis as those in mid-infrared region. However, absorptions in this region are helpful in exploiting information related to vibrations of molecules containing heavy atoms, molecular skeleton vibrations, molecular torsions and crystal lattice vibrations. Besides structural elucidation, another qualitative application of infrared spectroscopy is the identification of a compound with a reference infrared spectrum. If all the peaks of the unknown match those of the reference, the compound can be identified. Additional reference spectra are available online at databases such as . Absorbance is used for quantitative analysis due to its linear dependence on concentration. Given by , absorbance is directly proportional to the concentration and pathlength of sample: \[A=\epsilon c l\] where is absorbance, the molar extinction coefficient or molar absorptivity which is characteristic for a specific substance, the concentration and the pathlength (or the thickness) of sample. The conversion from transmittance to absorbance is given by \[ A=-\log T \] where is transmittance. For quantitative analysis of liquid samples, usually an isolated peak with high molar absorptivity that appears in the spectrum of the compound is chosen. A calibration curve of absorbance at the chosen frequency against concentration of the compound is acquired by measuring the absorbance of a series of standard compound solution with known concentrations. These data are then graphed to get a linear plot, from which the concentration of the unknown can be calculated after measuring its absorbance at the same frequency. The number of functional groups can also be calculated in this way, since the molar absorptivity of the band is proportional to the number of functional groups that are present in the compound. For solid samples, an internal standard with a constant known amount is added to the unknown sample and the standards. Then similar procedures as those with liquid samples are carried out except that the calibration curve is a graph of the ratio of absorbance of analyte to that of the internal standard versus concentration of the analyte. A multi-component analysis of the mixture is also feasible since different components have different values of molar absorptivity at the same frequency. However, infrared spectroscopy may be more susceptible to deviation from Beer's law than is UV-Vis spectroscopy because of its narrow bands, complex spectra, weak incident beam, low transducer sensitivity and solvent absorption. | 10,123 | 817 |
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Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus. Each element can be represented by the notation \(^A_Z \textrm X\), where , the mass number, is the sum of the number of protons and the number of neutrons, and , the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called , and an atom with a particular number of protons and neutrons is called a . Nuclides with the same number of protons but different numbers of neutrons are called . Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways: Because the number of neutrons is equal to − , we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s . For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be , emitting subatomic particles and electromagnetic radiation. The emissions are collectively called and can be measured. Isotopes that emit radiation are called . The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the , an extremely powerful but very short-range attractive force between nucleons ( ). All stable nuclei except the hydrogen-1 nucleus ( H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability. The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in . The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, H and He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive. As shown in , more than half of the stable nuclei (166 out of 279) have numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called 2, 8, 20, 50, 82, and 126. For example, tin ( = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium ( = 49) and antimony ( = 51), have only 2 stable isotopes each. Nuclei with magic numbers of protons neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element. Most stable nuclei contain numbers of both neutrons and protons The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in , the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei. Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the and the . Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text. Classify each nuclide as stable or radioactive. : mass number and atomic number : predicted nuclear stability : Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide. : a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in , its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is. b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is. c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable. d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive. Classify each nuclide as stable or radioactive. radioactive stable radioactive stable In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the , with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with = 114 and = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with = 292 that was found in Th. With an estimated half-life greater than 10 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements. Subatomic particles of the nucleus (protons and neutrons) are called . A is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is , and its emissions are collectively called . Isotopes that emit radiation are called . Each nucleon is attracted to other nucleons by the . Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain of protons and neutrons are often especially stable. , with atomic numbers near 126, may even be stable enough to exist in nature. | 9,058 | 819 |
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Cytochrome c occupies a prominent place in the mitochondrial electron-transport chain. Its water solubility, low molecular weight (12.4 kDa), stability, and ease of purification have allowed many experiments, which, when taken together, present a detailed picture of the structure and biological function of this electron carrier. X-ray structures of oxidized and reduced tuna cytochrome c are very similar; most of the differences are confined to changes in the orientations of the side chains of some surface-exposed amino acids and sub-Ångström adjustments of some groups in the protein interior. Upon reduction, the heme active site becomes slightly more ordered (Figure 6.33). Two-dimensional NMR studies confirm this interpretation of the x-ray data, and further establish that the crystal and solution structures of cytochrome c differ in only minor respects. Cytochrome c exhibits several pH-dependent conformational states. In particular, an alkaline transition with a pK ~ 9.1 has been observed for ferricytochrome c. This transition is believed to be associated with the dissociation of Met-80; the reduction potential decreases dramatically, and the 695-nm absorption band, associated with a sulfur → iron charge-transfer transition, disappears. The H NMR resonance due to ( H C-) Met-80 in deuterium-enriched ferricytochrome c disappears from its hyperfine-shifted upfield position without line broadening, and reappears coincident with the ( H C-)Met-65 resonance. In contrast, ferrocytochrome c maintains an ordered structure over the pH range 4 to 11. The heme iron in ferricytochrome c remains low-spin throughout this transition, and a new strong-field ligand must therefore replace Met-80. It has been suggested that an e-amino nitrogen of a nearby Lys provides the new donor atom, but this has not been confirmed. However, it is clear that reduction of ferricytochrome c at alkaline pH values below 11 causes a drastic conformational change at the heme site. The unknown sixth ligand must be displaced by Met-80 in order for the reduced protein to assume a structure similar to the one at neutral pH. This structural change is accompanied by a decrease in the rate of reduction of ferricytochrome c by hydrated electrons, as expected. How does the protein control the reduction potential of the iron center in cytochrome c? Factors that appear to play a role include the nature of the axial ligands, the stability and solvent accessibility of the heme crevice, and the hydrophobicities of the amino acids that line the heme crevice. These issues have been addressed theoretically and experimentally using cytochrome c variants engineered by protein semisynthesis or site-directed mutagenesis. Results for horse heart cytochrome c are set out in Table 6.6. Point mutations at either of positions 78 or 83 do not significantly alter E°'; however, the double mutant (Thr-78 → Asn-78; Tyr-83 → Phe-83) exhibits a substantially lower redox potential. Evidently, the results of such changes are not necessarily additive; great care must be taken in drawing conclusions about structure-function relations in engineered proteins. Finally, the ~310 mV difference between the values for the heme octapeptide and the native protein (the axial ligands are the same in both) provides a dramatic illustration of protein environmental effects on the redox potential: shielding the heme from the solvent is expected to stabilize Fe and therefore result in an increase in E°'. During the last fifteen years, much has been learned about the interaction of cytochrome c with its redox partners. Cytochrome c is a highly basic protein (pI = 10.05); lysine residues constitute most of the cationic amino acids. Despite the indication from the x-ray structures that only ~1 percent of the heme surface is solvent-exposed, the asymmetric distribution of surface charges, particularly a highly conserved ring of Lys residues surrounding the exposed edge of the heme crevice, led to the suggestion that electron-transfer reactions of cytochrome c (and other Class I cytochromes as well) occur via the exposed heme edge. Chemical modification of the surface Lys residues of cytochrome c has afforded opportunities to alter the properties of the surface \(\varepsilon\)-amino groups suspected to be involved in precursor complex formation. Margoliash and coworkers used a 4-carboxy-2,6-dinitrophenol (CDNP) modification of the Lys residues to map out the cytochrome c interaction domains with various transition-metal redox reagents and proteins. These experiments have shown that cytochrome c interacts with inorganic redox partners near the exposed heme edge. Numerous studies of cytochrome c with physiological reaction partners are in accord with electrostatic interactions featured in the model cytochrome c/cytochrome b complex discussed earlier. Similar types of interactions have been proposed for cytochrome c/flavodoxin and cytochrome c/cytochrome c peroxidase complexes. (Recent x-ray crystal structure work has shed new light on this problem.) Theoretical work additionally suggests that electrostatic forces exert torques on diffusing protein reactants that "steer" the proteins into a favorable docking geometry. However, the domains on cytochrome c for interaction with physiological redox partners are not identical, as Figure 6.34 illustrates. Reactions between cytochrome c and its physiological redox partners at low ionic strength generally are very fast, ~10 M s , even though the thermodynamic driving force may be as low as 20 mV, as it is for the reduction of cytochrome a in cytochrome c oxidase. Such rates are probably at the diffusioncontrolled limit for such protein-protein reactions. A more detailed understanding of these reactions will require studies that focus on the dynamical (rather than static) features of complexes of cytochrome c with other proteins. For example, there is evidence that a cytochrome c conformational change in the vicinity of the heme edge accompanies the formation of the complex with cytochrome c oxidase. Studies of the influence of geometry changes on activation energies are of particular importance in elucidating the mechanisms of protein-protein reactions. Intramolecular electron transfer in cytochrome c has been investigated by attaching photoactive Ru complexes to the protein surface. Ru(bpy) (CO ) (bpy = 2,2'-bipyridine) has been shown to react with surface His residues to yield, after addition of excess imidazole (im), Ru(bpy) (im)(His) . The protein-bound Ru complexes are luminescent, but the excited states (*Ru ) are rather short lived (\(\tau \leq\) 100 ns). When direct electron transfer from *Ru to the heme cannot compete with excited-state decay, electron-transfer quenchers (e.g., Ru(NH ) ) are added to the solution to intercept a small fraction (1-10%) of the excited molecules, yielding (with oxidative quenchers) Ru . If, before laser excitation of the Ru site, the heme is reduced, then the Fe to Ru reaction (k ) can be monitored by transient absorption spectroscopy. The k values for five different modified cytochromes have been reported: (Ru(His- 33), 2.6(3) x 10 ; Ru(His-39), 3.2(4) x 10 ; Ru(His-62), 1.0(2) x 10 ; Ru(His- 72), 9.0(3) x 10 ; and Ru(His-79), > 10 s ). According to Equation (6.27), rates become activationless when the reaction driving force (- \(\Delta\)G°) equals the reorganization energy (A), The driving force (0.74 eV) is approximately equal to the reorganization energy (0.8 eV) estimated for the Ru(bpy) (im)(His)-cyt c reactions. The activationless (maximum) rates (k ) are limited by H , where H is the electronic matrix element that couples the reactants and products at the transition state. Values of k and H for the Fe to Ru reactions are given in Table 6.7. 14.0 (11C) (1H) 13.9 (11C) (1H) 17.6 (7C) (1S) 20.6 (16C) (2H) Calculations that explicitly include the structure of the intervening medium have been particularly helpful in developing an understanding of distant electronic couplings. As discussed in Section IV.A, the couplings in proteins can be interpreted in terms of pathways comprised of covalent, H-bond, and through-space contacts. An algorithm has been developed that searches a protein structure for the best pathways coupling two redox sites (the pathways between the histidines (33, 39, 62, 72, 79) and the heme are shown in Figure 6.35). A given coupling pathway consisting of covalent bonds, H-bonds, and through-space jumps can be described in terms of an equivalent covalent pathway with an effective number of covalent bonds (n ). Multiplying the effective number of bonds by 1.4 Å/bond gives a-tunneling lengths (\(\sigma\)1) for the five pathways (Table 6.7) that correlate well with the maximum rates (one-bond limit set at 3 x 10 s ; slope of 0.71 Å ) (Figure 6.35). The 0.71 Å decay accords closely with related distance dependences for covalently coupled donor-acceptor molecules. Photosynthetic bacteria produce only one type of reaction center, unlike green plants (which produce two different kinds linked together in series), and are therefore the organisms of choice in photosynthetic electon-transfer research. As indicated in Section I.B, the original reaction center structure (Figure 6.15) lacked a quinone (Q ). Subsequent structures for reaction centers from other photosynthetic bacteria contain this quinone (Figure 6.36 See color plate section, page C-13.). The reaction center contains ten cofactors and three protein subunits. (Note that the structure contains a cytochrome subunit as well.) The cofactors are arrayed so that they nearly span the 40-Å-thick membrane (Figure 6.37 See color plate section, page C-13.). The iron atom is indicated by the red dot near the cytoplasmic side of the membrane (bottom). In spite of the near two-fold axis of symmetry, electron transfer proceeds along a pathway that is determined by the A branch. In particular, BChl and BPhe do not appear to play an important role in the electron transfers. It was demonstrated long ago that (BChl) is the primary electron donor and that ubiquinone (or metaquinone) is the ultimate electron acceptor. Transient flash photolysis experiments indicate that several electron-transfer steps occur in order to translocate the charge across the membrane (Figure 6.38). Curiously, the high-spin ferrous iron appears to play no functional role in the Q to Q electron transfer. In addition, the part played by BChl is not understood—it may act to promote reduction of BPhe via a superexchange mechanism. Cytochromes supply the reducing equivalents to reduce the special pair (BChl) . Estimated rate constants for the various electron-transfer steps, together with approximate reduction potentials, are displayed in Figure 6.39. For each step, the forward rate is orders of magnitude faster than the reverse reaction. The rapid rates suggest that attempts to obtain x-ray structures of intermediates (especially the early ones!) will not be successful. However, molecular dynamics methods are being explored in computer simulations of the structures of various intermediates. Within a few years we may begin to understand why the initial steps are so fast. | 11,277 | 821 |
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A common interpretation of entropy is that it is somehow a measure of chaos or randomness. There is some utility in that concept. Given that entropy is a measure of the dispersal of energy in a system, the more chaotic a system is, the greater the dispersal of energy will be, and thus the greater the entropy will be. Ludwig Boltzmann (1844 – 1906) (O'Connor & Robertson, 1998) understood this concept well, and used it to derive a statistical approach to calculating entropy. Boltzmann proposed a method for calculating the entropy of a system based on the number of energetically equivalent ways a system can be constructed. Boltzmann proposed an expression, which in its modern form is: \[S = k_b \ln(W) \label{Boltz} \] \(W\) is the number of available microstates in a macrostate (ensemble of systems) and can be taken as the quantitative measure of energy dispersal in a macrostate: \[W=\frac{A!}{\prod_j{a_j}!} \nonumber \] Where \(a_j\) is the number of systems in the ensemble that are in state \(j\) and \(A\) represents the total number of systems in the ensemble: \[A=\sum_j{a_j} \nonumber \] Equation 20.5.1 is a rather famous equation etched on Boltzmann’s grave marker in commemoration of his profound contributions to the science of thermodynamics (Figure \(\Page {1}\)). Calculate the entropy of a carbon monoxide crystal, containing 1.00 mol of \(\ce{CO}\), and assuming that the molecules are randomly oriented in one of two equivalent orientations. : Using the Boltzmann formula (Equation \ref{Boltz}): \[S = nK \ln (W) \nonumber \] And using \(W = 2\), the calculation is straightforward. \[ \begin{align*} S &= \left(1.00 \, mol \cot \dfrac{6.022\times 10^{23}}{1\,mol} \right) (1.38 \times 10^{-23} J/K) \ln 2 \\ &= 5.76\, J/K \end{align*} \nonumber \] | 1,803 | 822 |
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The fundamental thermodynamic equations follow from five primary thermodynamic definitions and describe internal energy, enthalpy, Helmholtz energy, and Gibbs energy in terms of their natural variables. Here they will be presented in their differential forms. The fundamental thermodynamic equations describe the thermodynamic quantities U, H, G, and A in terms of their natural variables. The term "natural variable" simply denotes a variable that is one of the convenient variables to describe U, H, G, or A. When considered as a whole, the four fundamental equations demonstrate how four important thermodynamic quantities depend on variables that can be controlled and measured experimentally. Thus, they are essentially equations of state, and using the fundamental equations, experimental data can be used to determine sought-after quantities like \(G\) or \(H\). The first law of thermodynamics is represented below in its differential form \[ dU = đq+đw \] where The "đ" symbol represent and indicates that both \(q\) and \(w\) are path functions. Recall that \(U\) is a state function. The first law states that internal energy changes occur only as a result of heat flow and work done. It is assumed that w refers only to PV work, where \[ w = -\int{pdV}\] The fundamental thermodynamic equation for internal energy follows directly from the first law and the principle of Clausius: \[ dU = đq + đw\] \[ dS = \dfrac{\delta q_{rev}}{T} \] we have \[ dU = TdS + \delta w\] Since only \(PV\) work is performed, \[ dU = TdS - pdV \label{DefU}\] The above equation is the fundamental equation for \(U\) with natural variables of entropy \(S\) and volume\(V\). The states that the entropy change of a system is equal to the ratio of heat flow in a reversible process to the temperature at which the process occurs. Mathematically this is written as \[ dS = \dfrac{\delta q_{rev}}{T}\] where Mathematically, enthalpy is defined as \[ H = U + pV \label{DefEnth}\] where \(H\) is enthalpy of the system, p is pressure, and V is volume. The fundamental thermodynamic equation for enthalpy follows directly from it deffinition (Equation \(\ref{DefEnth}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)) : \[ dH = dU + d(pV)\] \[ = dU + pdV + VdP\] \[ dU = TdS - pdV\] \[ dH = TdS - pdV + pdV + Vdp\] \[ dH = TdS + Vdp\] The above equation is the fundamental equation for H. The natural variables of enthalpy are S and p, entropy and pressure. The mathematical description of Gibbs energy is as follows \[ G = U + pV - TS = H - TS \label{Defgibbs}\] where \(G\) is the Gibbs energy of the system. The fundamental thermodynamic equation for Gibbs Energy follows directly from its definition \(\ref{Defgibbs}\) and the fundamental equation for enthalpy \(\ref{DefEnth}\): \[ dG = dH - d(TS)\] \[ = dH - TdS - SdT\] Since \[ dH = TdS + Vdp\] \[ dG = TdS + Vdp - TdS - SdT\] \[ dG = Vdp - SdT\] The above equation is the fundamental equation for G. The natural variables of Gibbs energy are p and T, pressure and temperature. Mathematically, Helmholtz energy is defined as \[ A = U - TS \label{DefHelm}\] where \(A\) is the Helmholtz energy of the system, which is often written as the symbol \(F\). The fundamental thermodynamic equation for Helmholtz energy follows directly from its definition (Equation \(\ref{DefHelm}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)): \[ dA = dU - d(TS)\] \[ = dU - TdS - SdT\] Since \[ dU = TdS - pdV\] \[ dA = TdS - pdV -TdS - SdT\] \[ dA = -pdV - SdT\] The above equation is the fundamental equation for A with natural variables of \(V\) and \(T\). For the definitions to hold, it is assumed that PV work is done and that processes are used. These assumptions are required for the first law and the principle of Clausius to remain valid. Also, these equations do not account include n, the number of moles, as a variable. When \(n\) is included, the equations appear different, but the essence of their meaning is captured without including the n-dependence. The fundamental equations derived above were not dependent on changes in the amounts of species in the system. Below the n-dependent forms are presented . \[ dU = TdS - PdV + \sum_{i=1}^{N}\mu_idn_i \] \[ dH = TdS + VdP + \sum_{i=1}^{N}\mu_idn_i \] \[ dG = -SdT + Vdp + \sum_{i=1}^{N}\mu_idn_i \] \[ dA = -SdT - PdV + \sum_{i=1}^{N}\mu_idn_i\] where μ is the chemical potential of species i and dn is the change in number of moles of substance i. The differential fundamental equations describe U, H, G, and A in terms of their natural variables. The natural variables become useful in understanding not only how thermodynamic quantities are related to each other, but also in analyzing relationships between measurable quantities (i.e. P, V, T) in order to learn about the thermodynamics of a system. Below is a table summarizing the natural variables for U, H, G, and A: The fundamental thermodynamic equations are the means by which the Maxwell relations are derived . The Maxwell Relations can, in turn, be used to group thermodynamic functions and relations into more general "families" . See the sample problems and the Maxwell Relation section for details. \[ \left (\dfrac{\partial H}{\partial P} \right)_{T,n} = -T \left(\dfrac{\partial V}{\partial T} \right)_{P,n} +V \] Then apply this equation to an ideal gas. Does the result seem reasonable?
5. Using the definition of Gibbs energy and the conditions observed at phase equilibria, derive the Clapeyron equation. | 5,551 | 823 |
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This module refers to a finite amount of particles placed in a closed container (i.e. no volume change) in which boiling cannot occur. The inability for boiling to occur- because the particles in the container are not exposed to the atmosphere, results in the incessant increase of temperature and pressure. The critical point is the temperature and pressure at which the distinction between liquid and gas can no longer be made. At the critical point, the particles in a closed container are thought to be vaporizing at such a rapid rate that the density of liquid and vapor are equal, and thus form a . As a result of the high rates of change, the surface tension of the liquid eventually disappears. You will have noticed that this liquid-vapor equilibrium curve has a top limit (labeled as C in the phase diagram in Figure 1), which is known as the critical point. The temperature and pressure corresponding to this are known as the critical temperature and critical pressure. If you increase the pressure on a gas (vapor) at a temperature lower than the critical temperature, you will eventually cross the liquid-vapor equilibrium line and the vapor will condense to give a liquid. The condensation of a gas will never occur above the critical point. A massive amount of pressure can be applied to a gas in a closed container, and it may become highly dense, but will not exhibit a meniscus. Molecules at critical temperatures possess high kinetic energy, and as a result the intermolecular forces in the molecules are weakened. A novel discovery made by the University of Manchester, identified that lower critical temperatures are existent in polymer solutions. It has been manifested that hydrocarbon polymers integrated with a hydrocarbon solution portrays what the university terms a "L.C.S.T." or a lower critical solution temperature. This lower critical solution temperature of polymers has been proclaimed to be in a range near the gas- liquid critical point of the polymer's solvent, and can reach up to 170 degrees Celsius. Such a lower critical solution temperature can be contributed to the assimilation of the heat and volume of the substance n-pentane with most hydrocarbon polymers at room temperature (Freeman, P.I., Rowlinson, J.S.). When a fluid is present in two phases, in a container, and a critical point is near establishment, contact with the imminently forming third phase does not occur. This phenomena can be accounted for by examining the other two existing phases; the third phase does not immediately form because one of the other two phases wets the third phase, causing it to be eliminated. This wetting phase will continually occur when a phase is not entirely stable as a whole. Temperature and vapor pressure are essential to the stimulation of a critical point; the following problems interconnect the two concepts. In correlation with , the gold vapor can be interpreted as a solitary gas occurring at the given volume of Neon. The ideal-gas law will be utilized to determine the vapor pressure. \[PV=nRT\] Convert to mol the molar mass of Au \[n= 0.213\; \cancel{g} \left( \dfrac{1 \;mol\; Au}{196.9\; \cancel{g}\; Au}\right) = 0.00108\; mol\; Au\] \(PV=nRT\) can then be algebraically converted into \(P=\dfrac{nRT}{V}\). Substitute you attained values. \[P= \dfrac{ (0.00108 \;mol) \times (0.08206\; L \; atm / mol K) \times (2209 \; K)}{ 200 \; L}\] \[P= 9.78\times 10^{-4}\; atm\] Taking into account the density of water, 1 g/ml, a 0.423 g sample of \(H_2O\) would amount to 0.42 ml. Considering this, it can be concluded that the water in the flask could not solely exist as a liquid at equilibrium since a 0.42 ml sample of water is incapable of accounting for the overall volume present in the flask (0.726L). The ideal gas law can be applied to determine whether or not the water in the flask could solely exist in the vapor form. Rearrange the Ideal Gas Law: \[P=\dfrac{nRT}{V}\] \[P= \dfrac{(0.0235 \;mol) \times (0.08206 \;L\; atm/mol\; K) \times (335\; K)}{0.726\; L}\] \[P=0.889 atm\] The actual Vapor pressure of water at 60 °C is \(149.4\; mmHg\). To make a comparison to the vapor pressure attained at 62°C (which is within a close range of the temperature of 60 degrees Celsius), convert the vapor pressure calculated in atm to mmHg. \[(0.889\;\cancel{atm})* \left(\dfrac{760\;mmHg}{1\; \cancel{atm}}\right)=676.27 \;mmHg\] Comparing this vapor pressure with the actual vapor pressure of water at 60 °C, it can be concluded that it is improbable that water solely exists in the flask as vapor since this attained vapor pressure exceeds that of the actual vapor pressure that occurs naturally. As the prior explications and calculations have proven, the sample of water is incapable of existing in the flask as either a liquid or vapor alone. Therefore vapor and liquid must exist together at 60 °C and 149.4 mmHg. To evaluate the pressure of water at 28 °C, the must be used. \[\ln \dfrac{P_1}{P_2}= \dfrac{H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)\] Begin by designating the variables in the equation to the given values in the problem. If \(T_2 =30 °C\) and \(P_2= 46.2\; mmHg\), what is \(P_1\) when \(T_1=28 °C\) The units for temperature must be in Kelvin. The enthalpy of vaporization (44 kJ/ mol) should be converted to J/ mol to allow for unit cancelation. \[44 \;\cancel{kJ}/mol \left(\dfrac{1000\; J}{1\; \cancel{kJ}}\right)= 44.0 \times 10^{3} J/mol.\] R= 8.3145 J/mol K Substituting the values in the equation one you obtain: \[\ln \left(\dfrac{46.2\; mmHg}{P_1}\right) = \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)\] To eliminate the natural logarithm, take the exponentl of both sides: \[ e^{\ln \left(\dfrac{46.2\; mmHg}{P_1}\right) }= e^{ \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)}\] \[ \dfrac{46.2\; mmHg}{P_1} = e^{ \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)}\] \[\dfrac{46.2\; mmHg}{P_1} = e^{0.11593} \] \[\dfrac{46.2\; mmHg}{P_1} = 1.1229\] \[\dfrac{1}{ P_1} =\dfrac{1.1229 }{46.2\; mmHg}\] \[P_1 = \dfrac{46.2\; mmHg}{1.1229}\] \(P_1 = 41.143\; mmHg\), which gets rounded to two significant digits (from the two significant digits in 44 kJ/mol). \(P_1\) vapor pressure is 41 mmHg. | 6,406 | 824 |
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The Aufbau section discussed how electrons fill the lowest energy orbitals first, and then move up to higher energy orbitals only after the lower energy orbitals are full. However, there is a problem with this rule. Certainly, 1s orbitals should be filled before 2s orbitals, because the 1s orbitals have a lower value of \(n\), and thus a lower energy. What about filling the three different 2p orbitals? In what order should they be filled? The answer to this question involves Hund's rule. Hund's rule states that: When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. In visualizing this process, consider how electrons exhibit the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals, they first try to get as far as possible from each other before having to pair up. Consider the correct electron configuration of the nitrogen (Z = 7) atom: 1s 2s 2p The p orbitals are half-filled; there are three electrons and three p orbitals. This is because the three electrons in the 2p subshell will fill all the empty orbitals first before pairing with electrons in them. Keep in mind that elemental nitrogen is found in nature typically as molecular nitrogen, \(\ce{N2}\), which requires molecular orbitals instead of atomic orbitals as demonstrated above. Next, consider oxygen (Z = 8) atom, the element after nitrogen in the same period; its electron configuration is: 1s 2s 2p Oxygen has one more electron than nitrogen; as the orbitals are all half-filled, the new electron must pair up. Keep in mind that elemental oxygen is found in nature typically as molecular oxygen, \(\ce{O_2}\), which has molecular orbitals instead of atomic orbitals as demonstrated above. According to the first rule, electrons always enter an empty orbital before they pair up. Electrons are negatively charged and, as a result, they repel each other. Electrons tend to minimize repulsion by occupying their own orbitals, rather than sharing an orbital with another electron. Furthermore, quantum-mechanical calculations have shown that the electrons in singly occupied orbitals are less effectively screened or shielded from the nucleus. Electron shielding is further discussed in the next section. For the second rule, unpaired electrons in singly occupied orbitals have the same spins. Technically speaking, the first electron in a sublevel could be either "spin-up" or "spin-down." Once the spin of the first electron in a sublevel is chosen, however, the spins of all of the other electrons in that sublevel depend on that first spin. To avoid confusion, scientists typically draw the first electron, and any other unpaired electron, in an orbital as "spin-up." Consider the electron configuration for carbon atoms: 1s 2s 2p : The two 2s electrons will occupy the same orbital, whereas the two 2p electrons will be in different orbital (and aligned the same direction) in accordance with Hund's rule. Consider also the electron configuration of oxygen. Oxygen has 8 electrons. The electron configuration can be written as 1s 2s 2p . To draw the orbital diagram, begin with the following observations: the first two electrons will pair up in the 1s orbital; the next two electrons will pair up in the 2s orbital. That leaves 4 electrons, which must be placed in the 2p orbitals. According to Hund’s rule, all orbitals will be singly occupied before any is doubly occupied. Therefore, two p orbital get one electron and one will have two electrons. Hund's rule also stipulates that all of the unpaired electrons must have the same spin. In keeping with convention, the unpaired electrons are drawn as "spin-up", which gives (Figure 1). When atoms come into contact with one another, it is the outermost electrons of these atoms, or valence shell, that will interact first. An atom is least stable (and therefore most reactive) when its valence shell is not full. The valence electrons are largely responsible for an element's chemical behavior. Elements that have the same number of valence electrons often have similar chemical properties. Electron configurations can also predict stability. An atom is most stable (and therefore unreactive) when all its orbitals are full. The most stable configurations are the ones that have full energy levels. These configurations occur in the noble gases. The noble gases are very stable elements that do not react easily with any other elements. Electron configurations can assist in making predictions about the ways in which certain elements will react, and the chemical compounds or molecules that different elements will form. | 4,884 | 825 |
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There is wide diversity in the nature of organohalogen compounds but, of necessity, we have restricted this chapter to alkyl, cycloalkyl, alkenyl, alkynyl, and aryl halides. Some of the chemistry of the carbon-halogen bonds already will be familiar to you because it involves the addition, substitution, and elimination reactions discussed in previous chapters. To some extent, we will amplify these reactions and consider nucleophilic substitution by what are called the and mechanisms. Subsequently, we will discuss the formation of carbon-metal bonds from carbon-halogen bonds. The latter type of reaction is of special value because compounds that have carbon-metal bonds are potent reagents for the formation of carbon-carbon bonds, as we will show later in this chapter. Although large numbers of organohalogens are known, very few of them occur naturally. Thyroid hormones (e.g., thyroxine) that contain iodine are exceptions; other organohalogens are found as mold metabolites (such as griseofulvin) and in marine organisms: Almost all of the organohalogen compounds in use today are synthetic in origin. You may wonder why, if nature doesn’t choose to make them, man elects to do so. The main interest to us here is that they are very useful intermediates for the synthesis of a wide range of other compounds. However, vast quantities of synthetic halogen compounds, particularly polyhalogen compounds, are used as pesticides, cleaning solvents, anaesthetics, aerosol propellants, refrigerants, polymers, and so on. The wisdom of this massive use of materials that are foreign to our natural environment gradually is being reevaluated as the long-term detrimental effects of many of these chemicals become known. For example, many of the chlorinated hydrocarbons such as DDT, Chlordane, and Lindane, which have been used very widely as insecticides, now are at least partially banned because of concern for their long-term effects on nontarget species, including man. Sometimes the long-term effects are quite unexpected and difficult to predict. For example, millions of kilograms of \(\ce{CF_2Cl_2}\), which is used as a propellant, have been released into the atmosphere from aerosol cans. This compound appears to be wholly free of direct adverse physiological effects. However, as the substance diffuses into the upper atmosphere, it is slowly decomposed by sunlight to produce chlorine atoms. Serious danger then is possible because chlorine atoms are known to catalyze the decomposition of ozone, and it is the ozone layer in the upper atmosphere that absorbs most of the sun’s ultraviolet radiation that is strongly harmful to life. and (1977) | 2,683 | 826 |
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Nuclear reactions are associated with changes in both mass and energy. In this module, the relationship between these two concepts are examined on a nuclear level. Albert Einstein’s \[ E=mc^2 \label{Einstein} \] Each time an energy change occurs, there is also a mass change that is related by the constant c (the speed of light squared). nergy changes in chemical reactions are small, making the mass change insignificant for chemical reactions. However, on a nuclear level, there is a significant amount of energy change in comparison and therefore a discernible mass change. In Albert Einstein’s mass-energy equivalence (Equation \ref{Einstein}), “m” is the net change in mass in kilograms and “c” is a constant (the speed of light) in meters per second. Two common units to express nuclear energy are joules (\(J\)) and megaelectronvolts (\(MeV\)). \[1.6022 \times 10^{-13}\, J = 1 \,MeV\] The energy equivalent of 1 atomic mass unit (u), via Equation \ref{Einstein}, is: \[\text{1 atomic mass unit (u)} = 1.4924 \times 10^{-10} J = 931.5\, MeV\] By knowing the mass change in amu, the energy released can be directly calculated using these conversion factors, which have already taken into account mass conversions and the value of \(c^2\). Keep in mind that the sum of mass-energy equivalents of reactants and products must equal each other. Figure \(\Page {1}\) provides an example of a . That is, the sum of the mass of the individual protons and neutrons differs from the overall mass of the nucleus. The nucleus weighs less than the masses of the individual subatomic particles. The sum of the mass and energy of the reactants are equivalent to the sum of the mass and energy of the products. Where does this mass go when a nucleus is formed? Recall Einstein's mass-energy equivalence and how matter and energy are essentially different configurations of one another. Mass is lost and as a result, energy is released as the nucleons come together to form the nucleus. This energy is known as the . Einstein's mass-energy equivalence can be rewritten in the following terms: \[\text{Nuclear Binding Energy} = \text{Mass Defect} \times c^2\] or \[E = Δm \times c^2 \label{Einstein2}\] The mass is converted into the energy required to bind the protons and neutrons together to make a nuclei. Nuclear binding energy is also the defined as the energy required to break apart a nucleus. Note that Δm can be either positive or negative (depends on where you set your zero point), but for our purposes, it will not really matter, and we should simply always use a positive mass deficit and say that is the energy released. One additional thing to keep in mind is what a is and what a is. A (plural ) is the center of the atom, and is composed of (which are both the protons and neutrons: a nucleon is what makes up the nuclei). This means the of an atom of Carbon-14 ( C)would contain 14 a of Uranium-238 ( U) would contain 238 etc. The mass number 60 is the maximum binding energy for each nucleon. (In other words, nuclei of mass number of approximately 60 require the most energy to dismantle). This means that the binding energy increases when small nuclei join together to form larger nuclei in a process known as . For nuclei with mass numbers greater than 60, the heavier nuclei will break down into smaller nuclei in a process known as . For fusion processes, the binding energy per nucleon will increase and some of the mass will be converted and released as energy (Figure \(\Page {1}\)). Fission processes also release energy when heavy nuclei decompose into lighter nuclei. The driving force behind fission and fusion is for an atomic nuclei to become more stable. So nuclei with a mass number of approximately 60 will be the most stable, which explains why iron is the most stable element in the universe. Elements with mass numbers around 60 will also be stable elements, while elements with extremely large atomic masses will be unstable. Nuclear fusion can release more energy than nuclear fission, especially when fusing small nuclei like hydrogen and helium into bigger nuclei. A certain nuclear reaction gives off 22.1 MeV. Calculate the energy released in Joules. This is a simple conversion problem. Use 1.6022 x 10 J = 1MeV. 22.1 Mev x 1.6022 x 10 J/1 MeV = Suppose that the nuclear mass of N is reported as 13.998947 amu. Calculate the binding energy per nucleon. Nitrogen-14 has 7 protons and 7 neutrons. The combined mass of the subatomic particles is 7 * (proton mass) + 7 * (neutron mass)
7 * (1.00728 amu) + 7 * (1.00866 amu ) = 14.11158 amu The reported mass of Nitrogen is 13.998947 amu, so the mass defect (difference) is 14.11158 amu - 13.998947 amu = 0.112633 amu Using Equation \ref{Einstein2}, E = ((0.112633g/1mol)(1kg/1000g)(1mol/6.022*10 nuclei)(1nuclei/14nucleon)) x c = You have a pebble with a mass of 1.0 gram. If of the mass was suddenly turned into energy, how much energy would be released? The mass-energy equivalent of 1.0 gram is E = .001 kg x ( 2.9979x10 m/s ) = . This is roughly equal to the energy released by the Fat Man atomic bomb! Note that this problem tells us that it doesn't matter what's turned into the energy, if it happens, it will release the same amount of energy. | 5,281 | 827 |
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One of the more important reactions of organohalogen compounds is the formation of compounds by replacement of the halogen by a metal atom. Carbon is positive in carbon-halogen bonds and becomes negative in carbon-metal bonds, and therefore carbon is considered to be in formation of an organometallic compound ( ): This transformation is of value because it makes an electrophilic carbon into a nucleophilic carbon. Organometallic compounds are a convenient source of nucleophilic carbon. A typical example of their utility is the way the achieve addition of nucleophilic carbon to carbonyl groups with formation of carbon-carbon bonds: In this chapter we will restrict our discussion of organometallic compounds to the alkyl and aryl compounds of magnesium and lithium, and the sodium and potassium salts of 1-alkynes. These substances normally are derived directly or indirectly from organohalogen compounds and are used very widely in organic synthesis. Organometallic compounds of transition metals and of boron are discussed in Chapters 11 and 31. and (1977) | 1,087 | 828 |
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Azeotropes are a mixture of at least two different liquids. Their mixture can either have a higher boiling point than either of the components or they can have a lower boiling point. Azeotropes occur when fraction of the liquids cannot be altered by distillation. Typically when dealing with mixtures, components can be extracted out of solutions by means of Fractional Distillation, or essentially repeated distillation in stages (hence the idea of 'fractional'). The more volatile component tends to vaporize and is collected separately while the least volatile component remains in the distillation container and ultimately, the result is two pure, separate solutions. Ideal solutions are uniform mixtures of components that have physical properties connected to their pure components. These solutions are supported by Raoult’s law stating that interactions between molecules of solute and molecules of solvent are the same as those molecules each are by themselves. An example of ideal solutions would be benzene and toluene. Azeotropes fail to conform to this idea because, when boiling, the component ratio of unvaporized solution is equal to that of the vaporized solution. So an azeotrope can be defined as a solution whose vapor has the same composition its liquid. As you can imagine, it is extremely difficult to distil this type of substance. In fact, the most concentrated form of ethanol, an azeotrope, is around 95.6% ethanol by weight because pure ethanol is basically nonexistent. Azeotropes exist in solution at a . This is best represented graphically and the phase diagram of a maximum-boiling point azeotrope can be seen in the following figure. The point Z represents where the azeotrope exists at a certain boiling point. Imagine that at point Z, the A-B solution is 64% B by mass while component A is water. If that same solution contained any less than 64%, the solution would then be water + the azeotrope. Conversely, if it were to be greater than 64% then the solution would be component B + the azeotrope. This demonstrates that an azeotrope can only exist at one temperature because any higher or lower temperature would result in a different concentration of component A or B. Also, a maximum-boiling point azeotrope is said to be a negative azeotrope because the boiling point of the azeotrope itself is higher than the boiling point of its components. As you can imagine, a positive azeotrope would have a lower boiling point than any of its components. If pure ethanol has a boiling point of 78.3 °C and its azeotrope has a boiling point of 78.174 °C, what would its graph look like? Since the azeotrope BP < pure ethanol BP, the azeotrope is a positive azeotrope and would have a graph that looks like the above figure upside-down (U shaped). | 2,796 | 830 |
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We have not given you much evidence to decide why it is that some [2 + 2] cycloadditions occur but not others. What is special about fluoroalkenes, allenes, and ketenes in these reactions? One possibility is that Mobius rather than the Huckel transition states are involved, but the Mobius transition states are expected to suffer from steric hindrance ( ). It is also possible that [2 + 2] cycloadditions, unlike the Diels-Alder additions, proceed by stepwise mechanisms. This possibility is strongly supported by the fact that these reactions generally are not stereospecific. Thus with tetrafluoroethene and -2,5-hexadiene products are formed, which differ in that the 1-propenyl group is to the methyl group in one adduct, \(45\), and in the other, \(46\): A stepwise reaction involving a accounts for the formation of \(45\) and\(46\). In the biradical mechanism the first step is formation of just one \(\ce{C-C}\) bond between the reactants, and this could occur in two different ways to give \(47\) or \(48ab\): Of these, \(48ab\) is predicted to have substantial electron delocalization because of the nearly equivalent VB structures \(48a\) and \(48b\). By the simple MO theory \(48ab\) should have a delocalization energy of \(16 \: \text{kcal mol}^{-1}\) ( ). The biradical \(47\) has no comparable electron delocalization and would be expected to be formed much less readily. Collapse of \(48\) through formation of the second \(\ce{C-C}\) bond would give \(45\) and an overall stereospecific addition. However, rotation around the \(\ce{C-C}\) single bond of \(48\) forms a different radical conformation, \(49\), which would collapse to the other stereoisomer, \(46\): If the reaction is stepwise, why is it stepwise? In the first place, as we have seen ( ), there are theoretical reasons why [2 + 2] cycloadditions may not occur in a concerted manner. Second, there are thermodynamic reasons why some alkenes undergo stepwise [2 + 2] additions and others do not. Regarding the second point, we can estimate that \(2 \ce{CH_2=CH_2} \rightarrow \cdot \ce{CH_2-CH_2-CH_2-CH_2} \cdot\) has \(\Delta H^0 \sim 37 \: \text{kcal}\), which is too high to achieve at a useful rate at those temperatures where the equilibrium constant is favorable for cyclobutane formation. In other words, when \(K_\text{eq}\) is favorable, the rate is too slow, and when the rate is fast enough, \(K_\text{eq}\) is unfavorable. In contrast, \(2 \ce{CF_2=CF_2} \rightarrow \cdot \ce{CF_2-CF_2-CF_2-CF_2} \cdot\) is estimated to have \(\Delta H^0 = -7 \: \text{kcal}\)! This tells us that \(\ce{CF_2=CF_2}\) has an abnormally low \(\ce{C=C}\) \(\pi\)-bond energy and, in fact, \(\Delta H^0\) for addition of hydrogen to one mole of tetrafluoroethene \(\left( -55 \: \text{kcal} \right)\) is \(22 \: \text{kcal}\) more negative than \(\Delta H^0\) for ethene \(\left( -33 \: \text{kcal} \right)\). If formation of \(\cdot \ce{CF_2-CF_2-CF_2-CF_2} \cdot\) from \(2 \ce{CF_2=CF_2}\) actually is exothermic, then it may seem surprising that \(\ce{CF_2=CF_2}\) can be kept in a container without immediately reacting with itself. That it can is because fairly high-energy collisions are required to overcome the nonbonded repulsions that resist bringing the carbons close enough together to permit the formation of the biradical. Nonetheless, \(\ce{CF_2=CF_2}\) generally is regarded as a hazardous and unpredictable chemical by virtue of its unusually low \(\ce{C=C}\) \(\pi\)-bond strength. 1,2-Propadiene also appears to have the potential for much easier formation of a biradical than does ethene. Not all [2 + 2] cycloadditions proceed by biradical mechanisms, some clearly occur by stepwise reactions involving ionic intermediates. and (1977) | 3,764 | 831 |
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At an elementary level, we can think about atoms being held together by simple electrostatic attraction. It is a fundamental principal that opposite charges attract. A positively charged ion and a negatively charged ion are held together by this force of attraction. Maybe the carbons are negatively charged and the hydrogens are positively charged, but what is holding the two carbons together if they have like charges? A similar problem is encountered in diatomic molecules such as \(H_2\). Is one of these hydrogen atoms negative while the other hydrogen atom is positive? In the early twentieth century, G. N. Lewis noticed a trend in the characteristics of compounds that he called "the rule of two." If you were to count up the total number of electrons in any stable compound, you would always come up with an even number -- that is, some number that is divisible by two. Perhaps, Lewis reasoned, this predominance of even numbers arises because electrons need to be in pairs. What does an element do if it has an odd number of electrons? One solution is to steal an electron from another element, or to allow one to be stolen away; these arrangements lead to ionic bonds. However, those elements not adept at stealing electrons may have a problem; they may need to share them instead. In order to share electrons, elements will have to form close associations with each other. They will become bonded together. Lewis took this idea further. If you count up the valence shell electrons around each of the atoms in stable compounds, not only are there even numbers, but there are almost always the same number of electrons as there are in one of the noble gases: He, Ne, Ar, Kr (2, 8, 8, or 18). This observation is sometimes called the Lewis because so many common atoms that form stable compounds obtain 8 electrons in their outermost shell as a result. Neon is the nearest noble gas to carbon, oxygen and nitrogen, and all of these atoms adopt 8-electron configurations in stable compounds. Lewis structures illustrate how atoms can maintain these numbers of electrons by sharing with other atoms. These simple structural drawings are used to convey most of our ideas about molecular chemistry. However, additional information can often be found through quantum mechanics and a molecular orbital approach to bonding. , | 2,342 | 832 |
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For main group molecules, chemists (like Pauling) thought a long time ago that hypervalence is due to expanded s p octets. The consensus is now clear that d orbitals are NOT involved in bonding in molecules like SF any more than they are in SF and SF . In all three cases, there is a small and roughly identical participation of d-orbitals in the wavefunctions. This has been established in both MO and VB theory. However using Hybrid orbitals with d-orbital contributions equips us with a language which can pragmatically describe the geometries of highly coordinated substances. While hybrid orbitals are a powerful tool to describe the geometries and shape of molecules and metal complexes. However, in "real" molecules, their significance may be debated. Often with a more realistically molecular orbitals approach is needed. However, f | 861 | 833 |
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Many problems in organic chemistry require consideration of structures in three dimensions, and it is very helpful to use molecular models to visualize the relative positions of the atoms in space. Unfortunately, we are forced to communicate three-dimensional concepts by means of drawings in two dimensions, and not all of us are equally gifted in making or visualizing such drawings. Obviously, communication by means of drawings, such as those in Figure 5-5 and 5-7, would be impractically difficult and time consuming, thus some form of abbreviation is necessary. Two styles of abbreviating the eclipsed and staggered conformations of ethane are shown in Figure 5-9; in each, the junction of lines representing bonds is assumed to be a carbon atom. Using the "sawhorse" convention, we always consider that we are viewing the molecule slightly from above and from the right, and it is understood that the central \(C-C\) bond is perpendicular to the plane of the paper. With the "Newman" convention, we view the molecule directly down the \(C-C\) bond axis so the carbon in front hides the carbon behind. The circle is only a visual aid to help distinguish the bonds of the back carbon from those of the front carbon. The rear atoms in the eclipsed conformation are drawn slightly offset from a truly eclipsed view so the bonds to them can be seen. The staggered conformations of butane are shown in Figure 5-10 in both the sawhorse and Newman conventions. There is little to choose between the two conventions for simple ethane derivatives, but the sawhorse convention is strongly favored for representing the conformations of ring compounds such as cyclohexane. The resemblance between the gauche forms of butane and the most stable conformation of cyclohexane is strikingly apparent in the sawhorse representations of both, as shown in Figure 5-10. Notice that the ring carbons of cyclohexane do not lie in one plane and that all the bond angles are tetrahedral. The conformations of this interesting and important molecule are discussed in detail in . Despite the usefulness of the sawhorse-type drawing, cyclic molecules often are drawn with planar rings and distorted bond angles even though the rings actually may not be planar. The reason for this is partly that planar rings are easier to draw and partly to emphasize the configuration of attached groups, irrespective of the conformation. Typical examples follow: Generally we shall avoid such drawings and suggest that it is much better to learn to draw molecules in as nearly correct perspective as possible. Once the sawhorse representation of cyclohexane is mastered, it is almost as easy to drawn \(14\) as \(13\), and \(14\) is much more informative about the shape of the molecule: We have indicated how the enantiomers of 2-butanol differ by drawing their strutures \(5\) and \(6\) (Section 5-1D) in perspective to show the tetrahedral configuration of substituents at the chiral carbon. This configuration also can be represented by the sawhorse or Newman formulas using any one of the several possible staggered conformations such as \(5a\) and \(6a\) or \(5b\) and \(6b\): These drawings are clear but can be cumbersome, particularly for more complex molecules, and we shortly shall describe other means of representing the configurations of chiral molecules. Planar molecules such as benzene, ethene, and methanal are best drawn in the plane of the paper with bond angles of about \(120^\text{o}\). When it is desired to draw them as viewed on edge (out of plane) care must be taken to provide proper perspective. The forward bonds can be drawn with slightly heavier lines; a tapered bond indicates direction, the wide end pointing toward the viewer and the narrow end away from the viewer (Figure 5-11). Barred lines are used here to indicate a rear or receding bond (many writers use dashed lines, but these may be confused with other uses of dashed lines, as for partial bonds). However, you will find other representations of planar carbons with rather grossly distorted bond angles. For example, methanoic acid is planar with nearly \(120^\text{o}\) bond angles, but often is drawn with \(H-C-O\) angles of \(90^\text{o}\) and \(180^\text{o}\): The distorted structures commonly are used to save space and, regretfully, we have to use them very frequently for this reason. The sawhorse or Newman representations of 2-butanol, \(5a\) and \(5b\) and \(6a\) and \(6b\), are excellent for showing the arrangements of the atoms in conformations, but are needlessly complex for representing the stereochemical . are widely used to show configurations and are quite straightforward, once one gets the idea of what they represent.. The projection formulas of 2-butanol are \(5c\) and \(6c\): and (1977) | 4,797 | 835 |
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Enzymes are biological catalysts and functional proteins. Enzymes contain specificity in its protein structure in order to have its specialized function. It usually contains more than one subunit and they are critical to sustain life. Enzymes can increase the chemical reactions in living cells. However, enzymes are not consumed in the reaction and their main function is to assist in bringing the substrates together so they can undergo normal reaction faster. The first enzyme was found in the process of fermentation in milk and alcohol during the nineteenth century. Later in the early 1830s, the term enzyme was used to replace the term ferment. Some scientists believe that ferments must contain living cells and some think ferments could be non-living cells. Finally, in the 1920s, Sumner purified the structure of enzyme and then properties of enzyme then was more clearly understood. Until today, enzymes are still the popular research field that many people are subject to study. x, \(\text{ES}\), is \[\dfrac{d \left[ \text{ES} \right]}{dt} = k_1 \left[ \text{E} \right] \left[ \text{S} \right] - k_{-1} \left[ \text{ES} \right]_{ss} - k_2 \left[ \text{ES} \right]_{ss} \approx 0 \label{Eq22} \] \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{ES} \right]_{ss} = \dfrac{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]}{\left[ \text{S} \right] + \dfrac{k_{-1} + k_2}{k_1}} = \dfrac{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]}{\left[ \text{S} \right] + K_M} \label{Eq26} \] \[\dfrac{d \left[ \text{P} \right]}{dt} = r_\text{max} = k_2 \left[ \text{E} \right]_0 \label{Eq27} \] \[\dfrac{1}{r} = \dfrac{K_M + \left[ \text{S} \right]}{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]} = \dfrac{K_M}{k_2 \left[ \text{E} \right]_0} \dfrac{1}{\left[ \text{S} \right]} + \dfrac{1}{k_2 \left[ \text{E} \right]_0} \label{Eq28} \] ( ) | 1,906 | 836 |
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The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called . The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events: \[\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}\] A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 10 kJ of energy per mole of \(\ce{^4_2He}\) produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 10 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ). It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, \(^2_1H\) and a triton, \(^3_1H\), undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron: \[\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}\] This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 10 kilojoules per mole of \(\ce{^4_2He}\) formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide. The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure \(\Page {2}\)). Calculate the energy released in each of the following hypothetical processes. Fusion of \(\ce{He}\) to give \(\ce{C}\) releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy. Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur. Another much more beneficial way to create fusion reactions is in a , a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure \(\Page {3}\)). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors ). | 4,387 | 839 |
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Discovered by von Reichenstein in 1782, tellurium is a brittle metalloid that is relatively rare. It is named from the Latin tellus for "earth". Tellurium can be alloyed with some metals to increase their machinability and is a basic ingredient in the manufacture of blasting caps. Elemental tellurium is occasionally found in nature but is more often recovered from various gold ores, all containing \(AuTe_2\). The ore was known as "Faczebajer weißes blättriges Golderz" (white leafy gold ore from Faczebaja) or antimonalischer Goldkies (antimonic gold pyrite). In 1782, while serving as the Austrian chief inspector of mines in Transylvania Franz-Joseph Müller von Reichenstein concluded that ore did not contain antimony, but that it contained bismuth sulfide. However the following year, he reported that this was erroneous and that the ore contained mostly gold and an unknown metal very similar to antimony. After 3 years of testing Müller determined the specific gravity of the mineral and noted the radish-like odor of the white smoke, which passed off, when the new metal was heated. In 1789, another Hungarian scientist, Pál Kitaibel, also discovered the element independently in an ore from Deutsch-Pilsen which had been regarded as argentiferous molybdenite, but later he gave the credit to Müller. In 1798, it was named by Martin Heinrich Klaproth who earlier isolated it from the mineral calaverite. Tellurium is a semimetallic, lustrous, crystalline, brittle, silver-white element. It is usually available as a dark grey powder and has metal and non-metal properties. Te forms many compounds corresponding to those of sulfur and selenium. When burned in the air tellurium has a greenish-blue flame and forms tellurium dioxide as a result. Tellurium is unaffected by water or hydrochloric acid, but dissolves in nitric acid. It as an atomic mass of 127.6 g/mol and a density of 6.24 g-cm . It's boiling point is at 450 degrees Celsius and its melting point is at 1390 °C. There are eight naturally occurring isotopes of Tellurium, of which three are radioactive. Tellurium is among the rarest stable solid elements in the Earth's crust. At 0.005 ppm, it is comparable to in abundance. However, tellurium is far more abundant in the wider universe. Tellurium was originally and is most commonly found in gold tellurides. However, the largest sources for modern production of tellurium is as a byproduct of blister copper refinement. The treatment of 500 tons of copper ore results in 0.45 kg of tellurium. Tellurium can also be found in lead deposits. Other tellurium sources, known as subeconomic deposits because the cost of abstraction outweighs the yield in tellurium, are lower-grade copper and some coal. Originally, the copper tellurium ore is treated with sodium bicarbonate and elemental oxygen to produce a tellurium oxide salt, copper oxide, and carbon dioxide: \[Cu_2Te + Na_2CO_3 + 2O_2 \rightarrow 2CuO + Na_2TeO_3 + CO_2\] Then, the sodium tellurium oxide is treated with sulfuric acid to precipitate out tellurium dioxide which can be treated with aqueous sodium hydroxide to reduce to pure tellurium and oxygen gas: \[TeO_2 + 2NaOH \rightarrow Na_2TeO_3 + H_2O \rightarrow Te + 2NaOH + O_2\] Tellurium has many unique industrial and commercial uses that improve product quality and quality-of-life. Many of these technologies that utilize tellurium have important uses for the energy industry, the military, and health industries. Tellurium is used to color glass and ceramics and can improve the machining quality of metal products. When added to copper alloys, tellurium makes the alloy more ductile, whereas it can prevent corrosion in lead products. Tellurium is an important component of infrared detectors used by the military as well as x-ray detectors used by a variety of fields including medicine, science, and security. In addition, tellurium-based catalysts are used to produce higher-quality rubber. CdTe films are one of the highest efficiency photovoltaics, metals that convert sunlight directly into electrical power, at 11-13% efficiency and are, therefore, widely used in solar panels. CdTe is a thin-film semiconductor that absorbs sunlight. Tellurium can be replaced by other elements in some of its uses. For many metallurgical uses, selenium , bismuth , or lead are effective substitutes. Both selenium and sulfur can replace tellurium in rubber production. Technologies based on tellurium have global impacts. As a photovoltaic, CdTe is the second most utilized solar cell in the world, soon said to surpass crystalline silicon and become the first. According to the US military, the tellurium-based infrared detectors are the reason that the military has such an advantage at night, an advantage which, in turn, has an effect on global and domestic politics. Tellurium extraction, as a byproduct of copper refinement, shares environmental impacts associated with copper mining and extraction. While a generally safe process, the removal of copper from other impurities in the ore is can lead to leaching of various hazardous sediments. In addition, the mining of copper tends to lead to reduced water flow and quality, disruption of soils and erosion of riverbanks, and reduction of air quality. About 215-220 tons of tellurium are mined across the globe every year. In 2006, the US produced 40% of production, Peru produced 30%, Japan produced 20%, and Canada produced 10% of the world's tellurium supply (since the chart can't be any bigger). The leading countries in production are the United States with 50 tons per year, Japan with 40 tons per year, Canada with 16 tons per year, and Peru with 7 tons per year (year 2009). When pure, tellurium costs $24 per 100 grams. Because tellurium is about as rare as platinum on earth, the United States Department of Energy expects a supply shortfall by the year 2025, despite the always improving extraction methods. As demand increases to provide the tellurium needed for solar panels and other such things, supply will continue to decrease and thus the price will skyrocket. This will cause waves in the sustainable energy movement as well as military practices and modern medicine. | 6,211 | 841 |
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which has become so useful in identification, estimation, and structure determination of compounds draws its strength from being able to identify the various vibrational modes of a molecule. A complete description of these vibrational normal modes, their properties and their relationship with the molecular structure is the subject of this article. We are familiar with resolving a translational vector into its three components along the x-, y-, and z- axes. Similarly a rotational motion can also be resolved into its components. Likewise the same is true for vibrational motion. The complex vibration that a molecule is making is really a superposition of a number of much simpler basic vibrations called “normal modes”. Before we take up any further description of “normal modes” it is necessary to discuss the degrees of freedom. Degree of freedom is the number of variables required to describe the motion of a particle completely. For an atom moving in 3-dimensional space, three coordinates are adequate so its degree of freedom is three. Its motion is purely translational. If we have a molecule made of N atoms (or ions), the degree of freedom becomes 3N, because each atom has 3 degrees of freedom. Furthermore, since these atoms are bonded together, all motions are not translational; some become rotational, some others vibration. For non-linear molecules, all rotational motions can be described in terms of rotations around 3 axes, the rotational degree of freedom is 3 and the remaining 3N-6 degrees of freedom constitute vibrational motion. For a linear molecule however, rotation around its own axis is no rotation because it leave the molecule unchanged. So there are only 2 rotational degrees of freedom for any linear molecule leaving 3N-5 degrees of freedom for vibration. \(\int\psi_A\psi_B \;dR= 0\) (integration is done over the entire space) The number of vibrational normal modes can be determined for any molecule from the formula given above. For a diatomic molecule, N = 2 so the number of modes is \(3\times 2-5 = 1\). For a triatomic linear molecule (CO ), it is \(3 \times 3-5 = 4\) and triatomic nonlinear molecule (H O), it is \(3 \times 3-6 = 3\) and so on. A linear molecule will have another bend in a different plane that is degenerate or has the same energy. This accounts for the extra vibrational mode. It is important to note that there are many different kinds of bends, but due to the limits of a 2-dimensional surface it is not possible to show the other ones. The frequency of these vibrations depend on the inter atomic binding energy which determines the force needed to stretch or compress a bond. We discuss this problem in the next section. The determination of the nature of the relative displacement of each atom with respect to each other is more complicated and beyond the scope of this article. However, such motion can be seen in some common molecules as shown below. For studying the energetics of molecular vibration we take the simplest example, a diatomic heteronuclear molecule AB. Homonuclear molecules are not IR active so they are not a good example to select. Let the respective masses of atoms A and B be \(m_A\) and \(m_B\). So the reduced mass \(\mu_{AB}\) is given by: \[ \mu_{AB}=\dfrac{m_A\, m_B}{m_A+m_B} \] The equilibrium internuclear distance is denoted by \(r_{eq}\). However as a result of molecular vibrations, the internuclear distance is continuously changing; let this distance be called \(r(t)\). Let \(x(t)=r(t)-r_{eq}\). When \(x\) is non-zero, a restoring force \(F\) exists which tries to bring the molecule back to \(x=0\), that is equilibrium. For small displacements this force can be taken to be proportional to \(x\). \[ F=-kx \] where \(k\) is the force constant. The negative sign arises from the fact that the force acts in the direction opposite to \(x\). This is indeed a case of Simple where the following well known relations hold. \[x(t)= A \sin \left( 2\pi \nu t \right) \] where \[ \nu=\dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu_{AB}}}\] The potential energy is given by \( V=\frac{1}{2}kx^2\). The total energy \(E\) (Kinetic+Potential) is obtained by solving the Schrödinger equation: \[-\dfrac{h^2}{8\pi^2\mu_{AB}} \dfrac{d^2\psi}{dx^2}+\dfrac{1}{2} kx^2\psi = E\psi\] \(\psi_n\)) s \(E_n\) are o \(E_n=(n+(1/2))hv\) where \(n\) As show above, the energy difference between adjacent vibrational energy levels is hv . On the other hand, the photon energy is hv . Energy conservation requires that the first condition for photon absorption be, Hv = hv or v = v . Such photons are in IR region of the electromagnetic spectrum. In addition, two more conditions must be met. Spectroscopy in the IR region can determine the frequency and intensity of absorption. These frequencies are generally specific for a specific bonds such as c-c, c(double bond)c, c(triple bond)c, c-o, c(double bond)o, etc. So the IR absorption data is very useful in structure determination. The intensity depends on the concentration of the resposble spec. So it is useful for quantitative estimation and for identification. 1.) NH – 6 C H –30 C H –48 CH –9 C H – 7 2.) N – IR inactive C0 – active C0 (stretching) – inactive HCl – active 3.) m = m xm /(m +m ) = 11.395x10 v = (1/2pi)(k/m ) = 2143.3 cm 4.) Energy of the mode for n = 0 E = (1/2)hv = 2.13x10 J Energy per mole = 2.13x10 x6.022x10 = 12.8KJ/mole 5.) v for D O will be lower because v is inversely proportional to 1/(m ), where m is the reduced mass. | 5,519 | 845 |
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This page explains what the fingerprint region of an infra-red spectrum is, and how it can be used to identify an organic molecule. This is a typical infra-red spectrum: Each trough is caused because energy is being absorbed from that particular frequency of infra-red radiation to excite bonds in the molecule to a higher state of vibration - either stretching or bending. Some of the troughs are easily used to identify particular bonds in a molecule. For example, the big trough at the left-hand side of the spectrum is used to identify the presence of an oxygen-hydrogen bond in an -OH group. The region to the right-hand side of the diagram (from about 1500 to 500 cm-1) usually contains a very complicated series of absorptions. These are mainly due to all manner of bending vibrations within the molecule. This is called the . It is much more difficult to pick out individual bonds in this region than it is in the "cleaner" region at higher wavenumbers. The importance of the fingerprint region is that each different compound produces a different pattern of troughs in this part of the spectrum. Compare the infra-red spectra of propan-1-ol and propan-2-ol. Both compounds contain exactly the same bonds. Both compounds have very similar troughs in the area around 3000 cm-1 - but compare them in the fingerprint region between 1500 and 500 cm-1. The pattern in the fingerprint region is completely different and could therefore be used to identify the compound. To positively identify an unknown compound, use its infra-red spectrum to identify what sort of compound it is by looking for specific bond absorptions. That might tell you, for example, that you had an alcohol because it contained an -OH group. You would then compare the fingerprint region of its infra-red spectrum with known spectra measured under exactly the same conditions to find out which alcohol (or whatever) you had. | 1,912 | 846 |
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Make sure your volume is turned up!
If you liked the one on the bottom, you have just experienced the excitement of how different amounts of chemicals can affect things. We often use the concept of limiting reagents in everyday life without even realizing it, especially while cooking or preparing a grocery list! What do and have to do with chemistry? When you cook, often times you can encounter a limiting reagent calculation that you have been doing in your head before you can remember. This page will use the familiarity of an everyday example to teach the steps of a limiting reagent problem. Here are two things to look for when trying to identify a limiting reagent problem: 1) it should be a task that you start with at least two starting materials and 2) it should form at least one new product. Another main concept found in limiting reagent questions is there will be starting material that when the task you are doing is complete. There are three methods (A, B, and C) that can be used to . This is the most important step of this type of question. As you progress through this page, choose the method that is most intuitive to you.
I want to have friends over for lunch on Saturday and make grilled cheese sandwiches that require two slices of bread and one slice of cheese. I open the refrigerator to find that I have 40 slices of cheese. I look in the bread box to find that I have 16 slices of bread. Question 1: Which of my ingredients is the limiting the number of sandwiches I can make? Question 2: How many sandwiches can I make? Question 3: How much of my starting material is left over once I am done making sandwiches? These three steps are the first steps to solving this type of question and will be used in conjunction with ALL three methods A, B, and C. + → 2 slices of bread + 1 slice of cheese → 1 grilled cheese sandwich What we know:
1) We know we have 40 slices of cheese
2) We know we have 16 slices of bread. According to the equation written in step 1, these ratios can be written by using the number in front of the ingredient, also known as a .
2 slices of bread is required for one slice of cheese.
Remember, the powerful thing about ratios is that we can also write them "upside-down".
The following ratios or unit conversions that can be written from this equation are: \(\begin{align}
\frac{\text{1 slice of cheese}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 slice of cheese}} \\
& \\
\frac{\text{1 sandwich}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 sandwich}} \\
& \\
\frac{\text{1 sandwich}}{\text{1 slice of cheese}}&=\frac{\text{1 slice of cheese}}{\text{1 sandwich}} \\
\end{align}\) Useful Math Review: Units cancel out just like numbers. If there is a unit in the numerator (on top of the ratio) and the same unit in the denominator (on the bottom of the ratio), they will cancel each other out.
Question 1: Which of my ingredients is the limiting the number of sandwiches I can make? You maybe just did this calculation in your head without even realizing the math that you did. Use , , and/or to answer question 1.
Let's use and our unit conversion to lead us to what we . Never start a math problem with a unit conversion if you don't have to. What we know:
1) We know we have 40 slices of cheese
2) We know we have 16 slices of bread. Let's set up equations using the to figure out how much of the other starting material we would need to make our sandwiches. \(\begin{align}
& n_{\text{bread needed}}=\text{40 slices of cheese}\times \frac{\text{2 slices of bread}}{\text{1 slice of cheese}}=\text{80 slices of bread} \\
& \\
& n_{\text{cheese needed}}=\text{16 slices of bread}\times \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}=\text{8 slices of cheese} \\
\end{align}\) These calculations show us that we need 80 slices of bread . . . do we have that many? No.
These calculations show us that we need 16 slices of cheese . . . do we have that many? Yes.
Because we don't have enough bread, bread is considered to be our limiting factor, also known as the .
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to , to answer .
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting before moving on to the final step of calculating the amount of product you can make.
We must compare the stoichiometric ratio X/Y with the ratio of amounts of X and Y which were initially mixed together. Whichever ratio is LESS than the theoretical stoichiometric ratio given by the co-efficients in the equation is considered to be the limiting reagent.
In Example 1 this ratio of initial amounts (using the materials we actually have)
\(\frac{n_{\text{slices of bread (actual)}}}{n_{\text{slices of cheese (actual)}}}=\frac{\text{16 slices of cheese}}{\text{40 slices of bread}}=\frac{\text{0.4 slices of bread}}{\text{1 slice of cheese}}\) was LESS than the theoretical stoichiometric ratio (in an ideal world we would have all the materials necessary)
\(\text{Theoretical}\left( \frac{\text{1 slices of bread}}{\text{2 slices of cheese}} \right)=\frac{\text{0.5 slice of bread}}{\text{1 slices of cheese}}\) \(\frac{\text{0.4 slices of bread}}{\text{1 slice of cheese}}<\frac{\text{0.5 slice of bread}}{\text{1 slices of cheese}}\)
This indicates that there is not enough bread to react with all the cheese and the bread is the . The corresponding general rule, for any reagents X and Y, is \(\begin{align}
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \\
& \\
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \\
\end{align}\) Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to , to answer .
These calculations can also be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Option 1: Use all of the cheese n/a There can't be a negative amount of something
Option 2: Use all of the bread Because Option 2 leaves all of the products with positive numbers, the bread is the . Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We can now move on to , to answer .
Now we can answer using the Limiting Reagent that we found using Method A, B or C.
Calculate how many sandwiches we can make (must start with limiting reagent)
\(\begin{align}
& n_{\text{sandwiches made}}=\text{16 slices of bread}\times \frac{\text{1 sandwich}}{\text{2 slice of bread}}=\text{8 sandwiches} \\
\end{align}\) When we are finished making sandwiches, how much cheese will be left over? (This is a two-step calculation)
\(\begin{align}
& n_{\text{cheese USED}}=\text{16 slices of bread}\times \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}=\text{8 slices of cheese USED} \\
\end{align}\) Next, simply subtract how much cheese you used from how much you started with. \(\begin{align}
& \text{40 slices of cheese to START}-\text{8 slices of cheese USED}=\text{32 slices of cheese left over} \\
\end{align}\)
A follow up question could be:
If I wanted to make sure that I used up all of the left over cheese and make a special trip to the store. How much bread should I get at the store.
(This calculation would be similar to Question/Answer 2.)
\(\begin{align}
& n_{\text{bread needed}}=\text{32 slices of cheese}\times \frac{\text{2 slices of bread}}{\text{1 slice of cheese}}=\text{64 more slices of bread need to be purchased} \\
\end{align}\)
(Of course, when the actual and theoretical molar ratio of X and Y are equivalent, both reagents will be completely consumed at the same time, and neither is in excess.). One of my favorite breakfast foods are ham and cheese omelets in the morning. I am on a diet so I always measure the amount of each ingredient I use. I always make two in case someone else wants one. Below is the recipe for my 'perfect' omelet. Recipe:
6 Large eggs - 200. g per one egg
1 cup of ham - 125. g per one cup
2 cups of shredded cheese - 50. g per one cup
I open the refrigerator this morning to find an excess of large eggs (50 eggs), 400. g of ham and 250. g of cheese.
Question 1: How many omelets can I make for breakfast with all of the ingredients I pulled out of the refrigerator?
Question 2: What remains when I am done cooking all of the omelets?
Question 3: What is the total mass of our cooked omelets? 6 Large eggs + 2 cups of cheese + 1 cup of ham → 2 omelets According to the equation written in step 1, these molar ratios can be written by using the number in front of the ingredient, also known as a .
Once again, when you are first starting out, it is helpful to write the ratios or the stoichiometric ratios connecting all of the components of the recipe.
It makes it easier to pick out which one ratio you will use as a unit conversion to help you calculate what the question is asking.
As you can see, there are more molar ratios this time because there are more starting materials. Additionally, the question gave us information about the mass per unit food item. Therefore there are 8 unit conversions in this question. Then, we need to convert all of the starting materials from into the associated with recipe or equation to be able to compare the starting materials.
The actual quantity of ham and cheese can be calculated using the mass per one unit of material. For instance, in this case, 1 cup of cheese is 50. grams.
In chemistry terms, this is called the . Molar mass can be used as a . Refer above for useful unit conversions
To relate the idea of molar mass to everyday life is something we do all of the time without even thinking about it.
For instance, you have an assortment of candy in a bucket. Three kids come up to you and ask you for some candy.
Do you weigh the candy and give the three kids equivalent masses? No. That would be unrealistic and probably a little messy.
Instead, we just decide that one candy bar should be given to each kid. Kid #1 receives one Snickers bar, kid #2 receives one Milky Way and kid #3 receives one Kit Kat bar. They are all happy because they received their OWN candy bar; they don't care that one might weigh a little bit more than the other candy bars.
To summarize, Snickers, Milky Way, and Kit Kat all represent ONE candy bar, however they have different masses...but that's okay. To relate it even more to chemistry, one mole of any compound is equivalent to one mole of any other compound, they just have a different mass depending on what substance you are talking about; just substitute these words and you are back to our everyday example; 'mole' = 'candy bar' and 'compound' = 'type of candy bar'.
You can use this idea in reverse, if you know the mass, then the amount of candy bars or 'moles' can be calculated. \(\begin{align}
& \text{ }n_{\text{cheese (actual)}}=\text{250. g of cheese}\times \frac{\text{1 cup of cheese}}{\text{50. g of cheese}}=\text{5.00 cups of cheese} \\
\\
& \text{ }n_{\text{ham (actual)}}=\text{400. g of ham}\times \frac{\text{1 cup of ham}}{\text{125 g of ham}}=\text{3.2 cups of ham} \\
\end{align}\)
Let's use and our unit conversion to lead us to what we . Never start a math problem with a unit conversion if you don't have to. What we know:
1) We know we have 5.0 cups of cheese
2) We know we have 3.2 cups of ham. Let's set up equations using the to figure out how much of the other starting material we would need to make our sandwiches. \(\begin{align}
& n_{\text{cheese needed}}=\text{3.2 cups of ham}\times \frac{\text{2 cups of cheese}}{\text{1 cup of ham}}=\text{6.4 cups of cheese} \\
& \\
& n_{\text{ham needed}}=\text{5.0 cups of cheese}\times \frac{\text{1 cup of ham}}{\text{2 cups of cheese}}=\text{2.5 cups of ham} \\
\end{align}\) These calculations show us that we need 6.4 cups of cheese . . . do we have that much? No.
These calculations show us that we need 2.5 cups of ham . . . do we have that many? Yes.
Because we don't have enough cheese, cheese is considered to be our limiting factor, also known as the .
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to , to answer .
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting before moving on to the final step of calculating the amount of product you can make.
Since the eggs, or one starting material is in it doesn't need to be involved in finding the limiting reagent.
We must compare the stoichiometric ratio X/Y with the ratio of amounts of X and Y which were initially mixed together. Whichever ratio is LESS than the theoretical stoichiometric ratio given by the co-efficients in the equation is considered to be the limiting reagent.
In Example 1 this ratio of initial amounts (using the materials we actually have)
\(\frac{n_{\text{cups of cheese (actual)}}}{n_{\text{cups of ham (actual)}}}=\frac{\text{5.0 cups of cheese}}{\text{3.2 cups of ham}}=\frac{\text{1.5625 cups of cheese}}{\text{1 cup of ham}}\) was LESS than the theoretical stoichiometric ratio (in an ideal world we would have all the materials necessary)
\(\text{Theoretical}\left( \frac{\text{2 cups of cheese}}{\text{1 cup of ham}} \right)=\frac{\text{2.0 cups of cheese}}{\text{1 cup of ham}}\) \(\frac{\text{1.5625 cups of cheese}}{\text{1 cup of ham}}<\frac{\text{2.0 cups of cheese}}{\text{1 cup of ham}}\)
This indicates that there is not enough cups of cheese to react with all the ham and the cheese is the . The corresponding general rule, for any reagents X and Y, is \(\begin{align}
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \\
& \\
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \\
\end{align}\) Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to , to answer .
These calculations can also be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Note: Don't forget about the ratios when using the table method.
Option 1: Use all of the ham n/a There can't be a negative amount of something
Option 2: Use all of the cheese
Option 3: Use all of the bread n/a There can't be a negative amount of something Because Option 2 leaves us with all positive numbers, the cheese is the . Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We can now move on to , to answer . Now we can answer using the Limiting Reagent that we found using Method A, B or C.
How many omelets can I make for breakfast with all of the ingredients I pulled out of the refrigerator? (Start with limiting reagent)
\(\begin{align}
& n_{\text{omelets made}}=\text{5.0 cups of cheese}\times \frac{\text{2 omelets}}{\text{2 cups of cheese}}=\text{5 omelets} \\
\end{align}\)
What remains when I am done cooking the omelets? (This is a two-step calculation)
\(\begin{align}
& n_{\text{ham USED}}=\text{5.0 cups of cheese}\times \frac{\text{1 cup of ham}}{\text{2 cups of cheese}}=\text{2.5 cups of ham USED} \\
\end{align}\)
\(\begin{align}
& n_{\text{eggs USED}}=\text{5.0 cups of cheese}\times \frac{\text{6 Large eggs}}{\text{2 cups of cheese}}=\text{15 Large eggs USED} \\
\end{align}\) Next, simply subtract how much ham and eggs you used from how much you started with. \(\begin{align}
& \text{3.2 cups of ham to START}-\text{2.5 cups of ham USED}=\text{0.7 cups of ham left over} \\
\end{align}\) \(\begin{align}
& \text{50 Large eggs to START}-\text{15 Large eggs USED}=\text{35 Large eggs left over} \\
\end{align}\)
What is the total weight of one omelet
Use the answer to Question 1 in order to start this problem. \(\begin{align}
& {\text{Mass of all omelets}}=\text{5.0 omelets}\times \frac{\text{(1200 g of egg + 125 g. of ham + 100. g of cheese)}}{\text{2 omelets}}=\text{3562.5 g of omelets} \\
\end{align}\)
A follow up question could be:
How much does one omelet weigh?
(This calculation would be similar to the explained above.)
\(\begin{align}
& {\text{Mass of one omelet}}=\frac{\text{3562.5 g of omelet}}{\text{5 omelets}}=\text{712.5 g per one omelet} \\
\end{align}\) Ore hematite, Fe O , is the chief iron ore used in production of iron metal. Reacting the ore hematite (Fe O ) with coke (C) produces iron metal (Fe) and CO . As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe O and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have? If you get stuck, go back to and review each step using terms you are familiar with. You will use the exact same process and calculation to solve Example 3 as you used in Example 2. Write a balanced equation 2 Fe O + 3 C → 3 CO + 4 Fe Find Moles and Identify useful Unit Conversions
The initial amounts of C and Fe O are calculated using appropriate molar masses found on the periodic table \(\begin{align}
& \text{ }n_{\text{Carbon (actual)}}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g C}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g C}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \\
& \\
& n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{ (actual)}}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g Fe}_{\text{2}}\text{O}_{\text{3}}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \\
\end{align}\) The stoichiometric ratio connecting C and Fe O is \(\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}= \frac{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{3 mol C}}\) Compare to Find the
Let's use and our unit conversion to lead us to what we . Never start a math problem with a unit conversion if you don't have to. What we know:
1) We know we have 1.28 x 10 mol of Fe O
2) We know we have 2.36 x 10 mol C. Let's set up equations using the to figure out how much of the other starting material we would need to make our sandwiches. \(\begin{align}
& \text{ }n_{\text{Carbon needed}}=\text{1.28}\times \text{10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\times \frac{\text{3 mol C}}{\text{2}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{1.92}\times \text{10}^{\text{5}}\text{ mol C needed} \\
& \\
& n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{ needed}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{ mol C}\times \frac{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{3 mol C}}=\text{1.57}\times \text{10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ needed} \\
\end{align}\)
These calculations show us that we need 1.92 x 10 mol C . . . do we have that ? Yes. We have 2.36 x 10 mol C
These calculations show us that we need 1.57 x 10 mol Fe O . . . do we have that? No. We only have 1.28 x 10 mol Fe O
Because we don't have enough Fe O , Fe O is considered to be our limiting factor, also known as the .
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to , to answer . Their ratio is \(\frac{n_{\text{C}}\text{(actual)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(actual)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) was MORE than the stoichiometric ratio
\(\text{Theoretical}\left( \frac{\text{3 C}}{\text{2 Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{1.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) \(\frac{\text{1.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}>\frac{\text{1.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\)
This indicates that there is not enough Fe O to react with all of the carbon (C). Therefore Fe O is the . In other words, you have more than enough C to react with all the Fe O . Fe O is the limiting reagent, and you will want to order more of it since it will be consumed first. The corresponding general rule, for any reagents X and Y, is \(\begin{align}
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \\
& \\
& \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \\
\end{align}\) We could now move on to , to answer . 2 Fe O + 3 C → 3 CO + 4 Fe Option 1: Use all of the iron (III) oxide
Option 2: Use all of the carbon Amounts cannot be negative Because Option 1 leaves all of the amounts with positive numbers, the Fe O is the . Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was . Some of the excess reactant C will be left over, but all the initial amount of Fe O will be consumed. Therefore we use the actual \(n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\) to calculate how much Fe can be obtained Four general steps:
Step 1) Write and equation
Step 2) Find moles
Step 3) Use molar ratio to identify limiting reagent
\(m_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\xrightarrow{divide M_{\text{Fe}}}n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{times \text{ (Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{times M_{\text{Fe}}}\text{ }m_{\text{Fe}}\) \(n_{\text{mol Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{2.56}\times \text{ 10}^{\text{5}}\text{ mol Fe}\)
\(m_{\text{ g Fe}}=\text{2}\text{.56 }\times \text{ 10}^{\text{5}}\text{ mol Fe}\times \frac{\text{55}\text{.85 g}}{\text{1 mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}\)
We will make 1.43 × 10 g Fe, or 14.3 Mg, Fe with this amount of reagents.
Check out these simulations to help you with this concept. Follow the four steps below.
1) Write a BALANCED chemical equation. The equation in the simulation is not balanced.
2) Choose two starting masses for your starting materials.
3) Work through the steps to calculate how much of each product will be left.
4) Work through the steps to calculate how much of a reactant will be left once the reaction is complete.
5) "Run Trial" to check if you are right!
As you can see from all of these examples, in a case where there is a limiting reagent, . Using the initial amount of a reagent present in excess would be , because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams.
Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the laboratory as well as the environment, inexpensive reagents like atmospheric O are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the . | 26,225 | 848 |
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Goal: Density measurements are presented at two temperatures (22 °C and 40 °C) for mixtures of water and ethanol. The data are analyzed in order to calculate the partial molar volumes of each component and the excess volume. Prerequisites: An introductory knowledge of the thermodynamics of mixtures and using density data to determine the volume of a solution. Resources you will need: This exercise should be carried out within a software environment that is capable of simple data manipulation and graphing. Every bartender knows well that mixing 50 ml of water with 50 ml of ethanol does not give an alcohol drink of 100 ml; the difference is called the excess volume. This excess volume can be expressed in terms of the mole fraction of either ethanol or water, and the magnitude of the excess volume provides a measure of non-ideal mixing of water and ethanol molecules. The existence of the excess volume indicates that molar volumes are non-additive. Instead, thermodynamics dictates that only the partial molar properties are additive. Partial molar (or molal) quantities relate changes in extensive properties of the solution (such as V, G, H, S, and A) to changes in concentration. Of all the extensive thermodynamic properties, the volume is easiest to visualize.
For a two component system composed of \(A\) and \(B\), the partial molar volumes are defined as \[ \bar{V}_A = \left( \dfrac{\partial V}{\partial n_A} \right)_{n_B} \label{1}\] and \[ \bar{V}_B = \left( \dfrac{\partial V}{\partial n_B} \right)_{n_A} \label{2}\] The total differential statement for volume changes in a two component mixture is therefore \[dV = \bar{V}_{A}\,dn_A + \bar{V}_{B}\,dn_B \label{3}\] Integration of Equation \ref{3} gives the Euler relation \[V = n_A \bar{V}_{A} + n_B \bar{V}_{B} \label{4}\] indicating that partial molar quantities are indeed additive.
The use of Equation \ref{4} requires an accurate knowledge of the partial molar volumes of each component at a given concentration. One of the simplest methods for experimentally determining partial molar volumes involves the careful measurement of solution density at two nearby concentrations. To see how this works, consider two hypothetical liquid mixtures of \(A\) and \(B\): Let us further assume that the molar masses of \(A\) and \(B\) are 35.00 g/mol and 24.00 g/mol, respectively. How do we use this information to determine the partial molar volume of component \(A\)? We first need to think carefully about the meaning of partial molar volume that is implied by Equations \ref{1} and \ref{2}; the value of the partial derivative can be estimated by calculating how much the volume changes when the moles of only one component is changed by a small amount (holding the moles of the other component constant). Stated symbolically, the partial molar volume of \(A\) can be estimated as \[ \bar{V}_A = \left( \dfrac{\partial V}{\partial n_A} \right)_{n_B} \approx \left( \dfrac{\Delta V}{\Delta n_A} \right)_{n_B} = \left( \dfrac{V_2-V_1}{n_{A,1}-n_{A,2}} \right)_{per\,1\,mol\, B} \label{5}\] In Equation \ref{5}, \(V_2\) and \(V_1\) are the volumes occupied by the two mixtures per 1 mol of component \(B\) and \(n_{A,2}\) and \(n_{A,1}\) are the moles of component A per 1 mol of component B; these quantities, in turn, are obtained by dividing the actual volumes and the actual moles of \(A\) by the actual moles of \(B\). Using the molar masses and densities from above, we have the following WORKED EXAMPLE. For the two solutions above estimate the partial molar volume of \(A\). For a 40% \(A\) Solution (assume 100 grams of solution): \[n_{A,1}= \dfrac{n_A}{n_B} = 0.457\nonumber \] \[ \text{mass(per 1 mol of B)} = \dfrac{(0.457\, mol \times 35.0 \,g/mol)}{1\, mol} + \dfrac{1.000\, mol \times 24.0\, g/mol}{1\, mol} = 39.95 \,g/mol \nonumber\] \[V_1 = \dfrac{mass}{density} = \dfrac{39.95\, g/mol}{0.8800 \,g/mL} = 45.40\, mL/mol \nonumber \] For a 50% \(A\) Solution (assume 100 grams of solution): \[n_{A,2}= \dfrac{n_{A}}{n_B} = 0.686 \nonumber \] \[ \text{mass(per 1 mol of B)} = \dfrac{0.686\, mol \times 35.0\, g/mol}{1\, mol} + \dfrac{1.000\, mol \times 24.0 \,g/mol}{1\, mol} = 48.01\, g/mol \nonumber\] \[V_2 = \dfrac{mass}{density}= \dfrac{48.01\, g/mol}{0.8751\, g/mL}= 54.86\, mL/mol \nonumber\] Plugging into Equation \ref{5} yields the partial molar volume of \(A\); \[ \bar{V}_A = \dfrac{45\,mL/mol - 54.86\, mL/mol}{0.457 - 0.686} = 41.3\,mL/mol \nonumber\] It is important to realize that this value of partial molar volume of \(A\) represents a numerical estimate (valid between the 40 – 50 weight percent A range). In the exercise below, you will also calculate the excess molar volume, \(V^E\), which is also related to solution density according to the following equation: \[ V^E = \dfrac{\chi_AM_A + \chi_B M_B}{\rho_L} - \chi_A V_{m,A} - \chi_B V_{m,B} \label{6}\] where \(ρ_L\) is the density of the mixture and \(\chi_A\), \(V_{m,A}\), \(M_A\), \(\chi_B\), \(V_{m,B}\) and \(M_B\) are the mole fraction, molar volume, and molecular weight of pure compounds \(A\) and \(B\), respectively. The density of water-ethanol mixtures were measured by students at the UW-Green Bay campus using a precision density meter (Anton Paar DMA 4500, Ashland, VA). The exact theoretical details about how this instrument works will not be dealt with here. In brief, the instrument has a glass U-tube that is filled with the sample. The U-tube is made to resonate and the resultant vibration frequency is measured. The vibration frequency provides a sensitive measure of the density of the solution (i.e. the denser the solution, the smaller the vibration frequency). The apparatus is computer controlled; and the computer converts the vibration frequency into the measured density using a polynomial calibration equation. Density is a strong function of temperature, so the instrument also measures the temperature and applies a correction to the calculated density. The density meter is calibrated using reagent grade water. The temperature of the sample compartment can be varied, so the instrument is capable of measuring densities over a range of temperatures. The following data was collected for ethanol-water mixtures at 22 °C and 40 °C. | 6,254 | 850 |
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Enzymes are large, three-dimensional proteins that catalyze a wide range of biochemical reactions. Enzymes normally show a remarkable specificity for catalyzing the reaction of only certain substrates. This specificity exists because the mechanism for catalysis involves the binding of the enzyme and a particular substrate in a “lock and key” fashion (i.e. evolution has favored the development of certain enzymes that have a three-dimensional surface that contours to the shape of a specific substrate). The rate of the reaction increases because intermolecular interactions that accompany binding lower the activation energy.
A simple model for enzyme action is represented by
where E is the enzyme, S is the substrate, ES is called the enzyme-substrate complex, and P is the product. The rate equation associated with this mechanism is worked out in most physical chemistry textbooks, and is given by
(1)
where R is the rate of reaction and [E]t represents the concentration of both bound and unbound enzyme in solution (i.e. [E]t = [E] + [ES]). The ratio of rate constants that appears in the denominator of equation (1) is simply equal to another constant. Consequently, equation (1) can be written more compactly as
(2)
where Km is called the Michaelis constant, and equation (2) is called the Michaelis-Menten equation.
In real biochemical systems, the enzyme is normally present at extraordinarily small concentrations in comparison to the substrate (i.e. [E]<<[S]). As the substrate concentration is increased, very little unbound enzyme will remain. In other words, the enzyme in the system will eventually become saturated with substrate (i.e. [E]t = [ES]) and the maximum possible rate for the reaction will be observed in this limit. Referring to equation (2), the maximum rate is achieved when [S] >> Km , meaning Km can be neglected in the denominator and equation (2) becomes
(3)
With the definition of Rmax, we can rewrite the Michaelis-Menten rate equation in the form
(4)
The constants Km and Rmax each quantify a characteristic of our enzyme-substrate system. Km can be interpreted as the concentration of substrate that is required to achieve one-half the maximum rate. This can be verified by substituting Km = [S] in equation (4) which yields
(5)
If an enzyme-substrate system has a relative small value of Km, then we know that that enzyme has a very high affinity for binding that substrate. Alternatively, the constant Rmax is simply a measure of the inherent ability of the enzyme to act as a catalyst (an enzyme-substrate system with a large Rmax value is going to occur at a relatively fast rate). Equation (4) is plotted in figure (1) for arbitrary values of Km and Rmax; initially, the rate increases at a rapid rate with increasing [S], but levels out as the enzyme becomes saturated (i.e. Rconverges on Rmax as [S] becomes large).
How does one evaluate the constants Km and Rmax for a particular enzyme-substrate system? The most common method involves measuring the rate of reaction at several different concentrations of substrate and subsequently fitting the data to equation (4). This process is made easier by taking the reciprocal of equation (4), which yields
(6)
In other words, a plot of 1/R versus 1/[S] should be linear (if the enzyme-substrate system obeys the Michaelis-Menten mechanism), and the slope and intercept can be used to evaluate the parameters Kmand Rmax. Such a plot is called a .
In the exercise below, you will be presented with some kinetic data for the oxidation of ethanol to acetaldehyde. This reaction is catalyzed in the human body by an enzyme called alcohol dehydrogenase. The data will be analyzed within the context of the Michaelis-Menten mechanism to determine the constants Km and Rmax. The following data where obtained from K. Bendinskas, C. DiJiacomo, A. Krill, and E. Vitz, Journal of Chemical Education, 82(7), 1068 (2005). In this study, the enzyme alcohol dehydrogenase (ADH) is used to catalyze the conversion of ethanol (the substrate) to acetaldehyde (the product). Eight kinetic trials were carried out in a pH 9.0 buffer; only the concentration of ethanol was varied from one trial to the next. The reaction was followed spectrophotometrically, although in an indirect fashion. Specifically, the direct conversion of ethanol to acetaldehyde only gives rise to absorbance changes in the UV portion of the spectrum, which is difficult to follow with standard lab spectrometers and cuvettes. The authors were able to coupled the ethanol reaction to another reaction that gives rise to a color change in the visible region (through a coupling scheme that can be read about in the above reference, dichlorophenolindophenol undergoes a synchronous redox reaction which causes the solution to go from blue to colorless; the authors monitored this change at 635 nm).
The initial rate of reaction was defined as the slope of absorbance at 635 nm versus time plot that was recorded for each trial (these dA/dt values are given in the table below, and represent the rate of reaction, R). 1. Enter the raw data into an appropriate quantitative analysis software environment and plot R versus [S]. Estimate values for Km and Rmax for this system by simply inspecting the graph.
2. Manipulate the data into a form that will allow you to generate a Lineweaver-Burk plot. Generate this plot, include a best-fit line, and use the slope and intercept to calculate values of Km and Rmax. Do these numbers agree with the estimates that you obtained in step (1)?
3. Lineweaver-Burk plots, which are double-reciprocal plots, are sometimes problematic because they tend to cluster all the large substrate concentration data to one side of the plot. Also, the few points corresponding to smaller substrate concentrations tend to be noisy because they are obtained by taking the reciprocal of a small number that contains a relatively high degree of random error. These difficulties can sometimes be avoided by plotting the data in an alternate fashion. One such method is known as a Hanes-Woolf plot, and is obtained by multiplying both sides of equation (6) by [S], which yields
(7)
Generate a Hanes-Woolf plot, include a best-fit line, and calculate values of Km and Rmax. Comment about the similarities (dissimilarities) of the plots and the values of Km and Rmax obtained in steps (2) and (3).
4. MORE CHALLENGING: Using a software environment that is capable of determining the standard deviation in the slope and intercept of a best-fit line, and by propagating that error into Km and Rmax, determine whether the Hanes-Woolf method yields Km and Rmax values with more or less uncertainty in comparison to the Lineweaver-Burk method. | 6,786 | 851 |
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The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic behavior including paramagnetism, diamagnetism, and ferromagnetism. An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number \(m_s\) as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic. Ferromagnetism is the basic mechanism by which certain materials (such as iron) form . This means the compound shows permanent magnetic properties rather than exhibiting them only in the presence of an external magnetic field (Figure \(\Page {1}\)). In a ferromagnetic element, electrons of atoms are grouped into domains in which each domain has the same charge. In the presence of a magnetic field, these domains line up so that charges are parallel throughout the entire compound. Whether a compound can be ferromagnetic or not depends on its number of unpaired electrons and on its atomic size. Ferromagnetism, the permanent magnetism associated with nickel, cobalt, and iron, is a common occurrence in everyday life. Examples of the knowledge and application of ferromagnetism include Aristotle's discussion in 625 BC, the use of the compass in 1187, and the modern-day refrigerator. Einstein demonstrated that electricity and magnetism are inextricably linked in his . refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, \(O_2\) is a good example of (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet: As shown in the video, molecular oxygen (\(\ce{O2}\)) is paramagnetic and is attracted to the magnet. , molecular nitrogen (\(\ce{N_2}\)) has no unpaired electrons and is diamagnetic; it is unaffected by the magnet. Diamagnetic substances are characterized by paired electrons, e.g., no unpaired electrons. According to the which states that no two electrons may occupy the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly by a magnetic field as demonstrated with the carbon sheet in Figure \(\Page {2}\). The magnetic properties of a substance can be determined by examining its electron configuration: If it has unpaired electrons, then the substance is paramagnetic and if all electrons are paired, the substance is then diamagnetic. This process can be broken into three steps: For Cl atoms, the electron configuration is 3s 3p Ignore the core electrons and focus on the valence electrons only. There is one unpaired electron. Since there is an unpaired electron, \(\ce{Cl}\) atoms are paramagnetic (albeit, weakly). For Zn atoms, the electron configuration is 4s 3d There are no unpaired electrons. Because there are no unpaired electrons, \(\ce{Zn}\) atoms are diamagnetic. Jim Clark ( ) | 4,537 | 852 |
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The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Notice that enthalpy change of neutralization is always measured per mole of water formed. Enthalpy changes of neutralization are negative - heat is released when an acid and and alkali react. For reactions involving strong acids and alkalis, the values are always very closely similar, with values between -57 and -58 kJ mol . That varies slightly depending on the acid-alkali combination (and also on what source you look it up in!). We make the assumption that strong acids and strong alkalis are fully ionized in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralized by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is: \[ NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O (l)\] but what is actually happening is: \[ OH^-(aq) + H^+(aq) \rightarrow H_2O (l)\] If the reaction is the same in each case of a strong acid and a strong alkali, it is not surprising that the enthalpy change is similar. In a weak acid, such as acetic acid, at ordinary concentrations, something like 99% of the acid is not actually ionized. That means that the enthalpy change of neutralization will include other enthalpy terms involved in ionizing the acid as well as the reaction between the hydrogen ions and hydroxide ions. And in a weak alkali like ammonia solution, the ammonia is also present mainly as ammonia molecules in solution. Again, there will be other enthalpy changes involved apart from the simple formation of water from hydrogen ions and hydroxide ions. For reactions involving acetic acid or ammonia, the measured enthalpy change of neutralization is a few kJ less exothermic than with strong acids and bases. For example, one source which gives the enthalpy change of neutralization of sodium hydroxide solution with HCl as -57.9 kJ mol : \[ NaOH_{(aq)} + HCl_{(aq)} \rightarrow Na^+_{(aq)} + Cl^-_{(aq)} + H_2O\] the enthalpy change of neutralization for sodium hydroxide solution being neutralized by acetic acid -56.1 kJ mol : \[ NaOH_{(aq)} + CH_3COOH_{(aq)} \rightarrow Na^+_{(aq)} + CH_3COO^-_{(aq)} + H_2O\] For very weak acids, like hydrogen cyanide solution, the enthalpy change of neutralization may be much less. A different source gives the value for hydrogen cyanide solution being neutralized by potassium hydroxide solution as -11.7 kJ mol , for example. \[ NaOH_{(aq)} + HCN_{(aq)} \rightarrow Na^+_{(aq)} + CN^-_{(aq)} + H_2O\] The experimentally measured enthalpy change of neutralization is a few kJ less exothermic than with strong acids and bases. Jim Clark ( ) | 3,175 | 853 |
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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{\(\Page {1}\)}\] 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: \[ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{\(\Page {2}\)}\] \[ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{\(\Page {3}\)}\] 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, is a good approximation of , 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{\(\Page {4}\)}\] 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: < 0, so melting is nonspontaneous ( spontaneous) at −10.0 °C. At 10.00 °C (283.15 K), the following is true: \[ΔS_\ce{univ}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \nonumber\] \[\mathrm{=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K} \nonumber\] > 0, so melting spontaneous at 10.00 °C. 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. One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine the entropy change for the system the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard . This new property is called the (or simply the ), and it is defined in terms of a system’s enthalpy and entropy as the following: \[G=H−TS \label{2}\] Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following: \[ΔG=ΔH−TΔS \label{3}\] (For simplicity’s sake, the subscript “sys” will be omitted henceforth.) We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression: \[ΔS_\ce{univ}=ΔS+\dfrac{q_\ce{surr}}{T} \label{4}\] The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following: \[ΔS_\ce{univ}=ΔS−\dfrac{ΔH}{T} \label{\(\Page {5}\)}\] ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following: \[−TΔS_\ce{univ}=ΔH−TΔS \label{\(\Page {6}\)}\] Comparing this equation to the previous one for free energy change shows the following relation: \[ΔG=−TΔS_\ce{univ} \label{\(\Page {7}\)}\] The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, \(ΔS_{univ}\). Table \(\Page {2}\) expands on Table \(\Page {2}\) and summarizes the relation between the spontaneity of a process and the arithmetic signs of \(\Delta G\) and \(\Delta S\) indicators. Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes. Gibbs energy is a state function, so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in Example \(\Page {1}\). \[ ΔG°=ΔH°−TΔS° \label{\(\Page {7}\)}\] The Definition of Gibbs Free Energy: Change from ΔH° and ΔS° Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process? The process of interest is the following: \[\ce{H2O}(l)⟶\ce{H2O}(g) \label{\(\Page {8}\)} \nonumber\] The standard change in free energy may be calculated using the following equation: \[ΔG^\circ_{298}=ΔH°−TΔS° \label{\(\Page {9}\)} \nonumber\] From Appendix G, here is the data: Combining at 298 K: \[ΔH°=ΔH^\circ_{298}=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\\
=\mathrm{[−241.82\: kJ−(−285.83)]\:kJ/mol=44.01\: kJ/mol} \nonumber\]
\[ΔS°=ΔS^\circ_{298}=S^\circ_{298}(\ce{H2O}(g))−S^\circ_{298}(\ce{H2O}(l)) \nonumber\\
=\mathrm{188.8\:J/mol⋅K−70.0\:J/K=118.8\:J/mol⋅K} \nonumber\]
\[ΔG°=ΔH°−TΔS° \nonumber\] Converting everything into kJ and combining at 298 K: \[ΔG^\circ_{298}=ΔH°−TΔS° \nonumber\] \[\mathrm{=44.01\: kJ/mol−(298\: K×118.8\:J/mol⋅K)×\dfrac{1\: kJ}{1000\: J}} \nonumber\] \[\mathrm{44.01\: kJ/mol−35.4\: kJ/mol=8.6\: kJ/mol} \nonumber\] At 298 K (25 °C) \(ΔG^\circ_{298}>0\), and so boiling is nonspontaneous (not spontaneous). Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process? \[\ce{C2H6}(g)⟶\ce{H2}(g)+\ce{C2H4}(g) \nonumber\] Answer: \(ΔG^\circ_{298}=\mathrm{102.0\: kJ/mol}\); the reaction is nonspontaneous (not spontaneous) at 25 °C. Free energy changes may also use the standard free energy of formation \( (ΔG^\circ_\ce{f})\), for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, \( (ΔG^\circ_\ce{f})\) is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction \[m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D},\] the standard free energy change at room temperature may be calculated as \[ΔG^\circ_{298}=ΔG°=∑νΔG^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants})\] \[=[xΔG^\circ_\ce{f}(\ce{C})+yΔG^\circ_\ce{f}(\ce{D})]−[mΔG^\circ_\ce{f}(\ce{A})+nΔG^\circ_\ce{f}(\ce{B})].\] Consider the decomposition of yellow mercury(II) oxide. \[\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+\dfrac{1}{2}\ce{O2}(g) \nonumber\] Calculate the standard free energy change at room temperature, \(ΔG^\circ_{298}\), using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions? The required data are available in and are shown here. (a) Using free energies of formation: \[ΔG^\circ_{298}=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \nonumber\] \[=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\] \[\mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \nonumber\] (b) Using enthalpies and entropies of formation: \[ΔH^\circ_{298}=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \nonumber\] \[=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\] \[\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \nonumber\] \[ΔS^\circ_{298}=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \nonumber\] \[=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\] \[\mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \nonumber\] \[ΔG°=ΔH°−TΔS°=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \nonumber\] \[ΔG°=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \nonumber\] Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature. Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C? \[\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) \nonumber\] −141.5 kJ/mol, nonspontaneous 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. Gibbs free energy ( ) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for Δ indicates a spontaneous process; a positive Δ indicates a nonspontaneous process; and a Δ of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible. ). | 11,887 | 854 |
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This page looks at Le Chatelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium. It is important in understanding everything on this page to realize that Le Chatelier's Principle is no more than a useful guide to help you work out what happens when you change the conditions in a reaction in dynamic equilibrium. It does not explain why the system moved to equilibrium. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Suppose you have an equilibrium established between four substances A, B, C and D. \[ \ce{ A + 2B <=> C +D} \nonumber\] According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of A decreases again - by reacting it with B and turning it into C + D. The position of equilibrium moves to the right. This is a useful way of converting the maximum possible amount of B into C and D. You might use it if, for example, B was a relatively expensive material whereas A was cheap and plentiful. According to Le Chatelier, the position of equilibrium will move so that the concentration of A increases again. That means that more C and D will react to replace the A that has been removed. The position of equilibrium moves to the left. This is essentially what happens if you remove one of the products of the reaction as soon as it is formed. If, for example, you removed C as soon as it was formed, the position of equilibrium would move to the right to replace it. If you kept on removing it, the equilibrium position would keep on moving rightwards - turning this into a one-way reaction. This is not in any way an explanation of why the position of equilibrium moves in the ways described. All Le Chatelier's Principle gives you is a quick way of working out what happens. This only applies to reactions involving gases: \[ \ce{A(g) + 2B(g) <=> C(g) + D(g)} \nonumber\] According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the pressure is reduced again. Pressure is caused by gas molecules hitting the sides of their container. The more molecules you have in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules. In this case, there are 3 molecules on the left-hand side of the equation, but only 2 on the right. By forming more C and D, the system causes the pressure to reduce. Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules. The equilibrium will move in such a way that the pressure increases again. It can do that by producing more molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction. In this case, increasing the pressure has no effect whatsoever on the position of the equilibrium. Because you have the same numbers of molecules on both sides, the equilibrium can't move in any way that will reduce the pressure again. Again, this is not an explanation of why the position of equilibrium moves in the ways described. For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our forward reaction is exothermic (heat is evolved): This shows that 250 kJ is evolved (hence the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic by exactly the same amount. According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the temperature is reduced again. Suppose the system is in equilibrium at 300°C, and you increase the temperature to 500°C. How can the reaction counteract the change you have made? How can it cool itself down again? To cool down, it needs to absorb the extra heat that you have just put in. In the case we are looking at, the back reaction absorbs heat. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D. If you were aiming to make as much C and D as possible, increasing the temperature on a reversible reaction where the forward reaction is exothermic is not a good idea! The equilibrium will move in such a way that the temperature increases again. Suppose the system is in equilibrium at 500°C and you reduce the temperature to 400°C. The reaction will tend to heat itself up again to return to the original temperature. It can do that by favoring the exothermic reaction. The position of equilibrium will move to the right. More A and B are converted into C and D at the lower temperature. Catalysts have sneaked onto this page under false pretenses, because adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Chatelier's Principle does not apply to them. This is because a catalyst speeds up the forward and back reaction to the same extent. Because adding a catalyst does not affect the relative rates of the two reactions, it can't affect the position of equilibrium. So why use a catalyst? For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction have to become equal. This does not happen instantly. For a very slow reaction, it could take years! A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium. Increasing the temperature of a system in dynamic equilibrium favors the endothermic reaction. The system counteracts the change you have made by absorbing the extra heat. Decreasing the temperature of a system in dynamic equilibrium favors the exothermic reaction. The system counteracts the change you have made by producing more heat. Again, this is not in any way an explanation of why the position of equilibrium moves in the ways described. It is only a way of helping you to work out what happens. Jim Clark ( ) | 6,475 | 856 |
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Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, zeroth o know how to determine the reaction order from experimental data. 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: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \tag{14.15} \] 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 \[[A] = [A]_0 − kt \tag{14.16} \] 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: \[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \tag{14.17} \] 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: \[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \tag{14.18} \] 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 ). 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 ). These examples illustrate two important points: In a , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows: If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows: \[[A] = [A]_0e^{−kt} \tag{14.20} \] where [A] is the initial concentration of reactant A at = 0; is the rate constant; and is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and : \[\ln[A] = \ln[A]_0 − kt \tag{14.21} \] Because has the form of the algebraic equation for a straight line, = + , with = ln[A] and = ln[A] , a plot of ln[A] versus for a first-order reaction should give a straight line with a slope of − and an intercept of ln[A] . Either the differential rate law ( ) or the integrated rate law ( ) can be used to determine whether a particular reaction is first order. First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin (Figure 14.6) and the reaction of -butyl bromide with water to give -butanol (Equation 14.10). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows: Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in have been studied extensively to find ways of maximizing the concentration of the active species. If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in . The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in shows that the reaction rate doubles [(1.8 × 10 M/min) ÷ (9.0 × 10 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 M/min) ÷ (9.0 × 10 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = [cisplatin] . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of . For example, substituting the values for Experiment 3 into , 3.6 × 10 M/min = (0.024 M) 1.5 × 10 min = Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug. Example 4 At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction: Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction. balanced chemical equation, initial concentrations of reactant, and initial rates of reaction reaction order and rate constant Use measured concentrations and rate data from any of the experiments to find the rate constant. The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate. Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH CH Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH CH Cl]. This behavior is characteristic of a first-order reaction, for which the rate law is rate = [CH CH Cl]. We can calculate the rate constant ( ) using any row in the table. Selecting Experiment 1 gives the following: 1.60 × 10 M/s = (0.010 M) 1.6 × 10 s = Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: SO Cl (g) → SO (g) + Cl (g) Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: SO Cl (g) → SO (g) + Cl (g) Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. first order; = 2.2 × 10 s We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for = 100 min ([cisplatin] = 0.0086 M) and = 1000 min ([cisplatin] = 0.0022 M), The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions. Example 5 Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.) initial concentration, rate constant, and time interval concentration at specified time and time required to obtain particular concentration The exponential form of the integrated rate law for a first-order reaction ( ) is [A] = [A] . Having been given the initial concentration of ethyl chloride ([A] ) and having calculated the rate constant in Example 4 ( = 1.6 × 10 s ), we can use the rate law to calculate the concentration of the reactant at a given time . Substituting the known values into the integrated rate law, We could also have used the logarithmic form of the integrated rate law ( ): To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for . gives the following: In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO Cl ) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO Cl that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO Cl to decompose? 0.0252 M; 29 h The simplest kind of is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer). The differential rate law for the simplest second-order reaction in which 2A → products is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \tag{14.22} \] Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M ·s ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s). For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time: \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \tag{14.23} \] Because has the form of an algebraic equation for a straight line, = + , with = 1/[A] and = 1/[A] , a plot of 1/[A] versus for a simple second-order reaction is a straight line with a slope of and an intercept of 1/[A] . Second-order reactions generally have the form 2A → products or A + B → products. Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO to NO and O and the decomposition of HI to I and H . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture. Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows: For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( ) or the integrated rate law ( ). To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7: Because (1.7) = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration. rate ∝ [monomer] This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following: We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in . The measurements show that the concentration of the monomer (initially 5.4 × 10 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus should be a straight line, as shown in part (b) in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, = 4.1 M ·min , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally. Dimerization of a Monomeric Compound, a Second-Order Reaction. For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in . Example 6 At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen. Experimental data for the reaction at 300°C and four initial concentrations of NO are listed in the following table: Determine the reaction order and the rate constant. balanced chemical equation, initial concentrations, and initial rates reaction order and rate constant We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 ) ÷ (1.35 × 10 ) = 4.0], which means that the reaction rate is proportional to [NO ] . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO ] . This behavior is characteristic of a second-order reaction. We have rate = [NO ] . We can calculate the rate constant ( ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following: When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: \[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \] The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: \[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \] The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: as Determine the reaction order and the rate constant. second order in HO ; = 1.4 × 10 M ·s If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order. Example 7 If a flask that initially contains 0.056 M NO is heated at 300°C, what will be the concentration of NO after 1.0 h? How long will it take for the concentration of NO to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in Example 6. balanced chemical equation, rate constant, time interval, and initial concentration final concentration and time required to reach specified concentration We know and [NO ] , and we are asked to determine [NO ] at = 1 h (3600 s). Substituting the appropriate values into , Thus [NO ] = 5.1 × 10 M. In this case, we know and [NO ] , and we are asked to calculate at what time [NO ] = 0.1[NO ] = 0.1(0.056 M) = 0.0056 M. To do this, we solve for , using the concentrations given. NO decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min. In the exercise in Example 6, you calculated the rate constant for the decomposition of HO as = 1.4 × 10 M ·s . This high rate constant means that HO decomposes rapidly under the reaction conditions given in the problem. In fact, the HO molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO , calculate the concentration of HO that remains after 1.0 h at 25°C. How long will it take for 90% of the HO to decompose? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in the exercise in Example 6. 2.0 × 10 M; 6.4 × 10 s In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A,\textrm B] \tag{14.24} \] Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction \(A + B \rightarrow products\), for example, we need to determine and the exponents and in the following equation: \[\text{rate} = k[A]^m[B]^n \tag{14.25} \] To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type \(A + B \rightarrow products\) are given in . Rate Data for a Hypothetical Reaction of the Form \(A + B \rightarrow products\) The general rate law for the reaction is given in . We can obtain or directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in . \[\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber \] Inserting the appropriate values from , Because 1.00 to any power is 1, [1.00 M] = 1.00 M. We can cancel like terms to give 0.25 = [0.50] , which can also be written as 1/4 = [1/2] . Thus we can conclude that = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for . Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for . \(\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}\) Substituting the appropriate values from , Canceling leaves 1.0 = [0.50] , which gives \(n = 0\); that is, the reaction is zeroth order in \(B\). The experimentally determined rate law is therefore rate = [A] [B] = [A] We can now calculate the rate constant by inserting the data from any row of into the experimentally determined rate law and solving for \(k\). Using Experiment 2, we obtain 19 × 10 M/min = (0.75 M) 3.4 × 10 M ·min = k You should verify that using data from any other row of gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same. Example 8 Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with \(O_2\) to give \(NO_2\), which then reacts rapidly with water to give nitrous acid and nitric acid: These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O at 25°C: \[2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber \] Determine the rate law for the reaction and calculate the rate constant. balanced chemical equation, initial concentrations, and initial rates rate law and rate constant Comparing Experiments 1 and 2 shows that as [O ] is doubled at a constant value of [NO ], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O ] , so the reaction is first order in O . Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O ] is held constant. That is, the reaction rate is proportional to [NO] , which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction: rate = [NO] [O ] The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives Alternatively, using Experiment 2 gives The difference is minor and associated with significant digits and likely experimental error in making the table. The overall reaction order \((m + n) = 3\), so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases. The peroxydisulfate ion (S O ) is a potent oxidizing agent that reacts rapidly with iodide ion in water: S O (aq) + 3I (aq) → 2SO (aq) + I (aq) The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant. rate = [S O ,I ]; = 20 M ·s The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism. If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant. : \(\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k\) : [A] = [A] − : \(\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]\) : [A] = [A] e : ln[A] = ln[A] − : \(\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2\) : \(\frac{1}{[\textrm A]}=\frac{1}{[\textrm A]_0}+kt\) 2Fe (soln) + 2I (soln) → 2Fe (soln) + I (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? What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction? C H Br + S O → C H S O + 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? | 30,969 | 857 |
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The is very important. The central limit theorem says that if we average enough values from any distribution, the distribution of the averages we calculate will be the normal distribution. The probability density function for the normal distribution is \[\frac{df}{du} =\frac{1}{\sigma \sqrt{2\pi }}\mathrm{exp}\left[\frac{{-\left(u-\mu \right)}^2}{2{\sigma }^2}\right]\] The integral of the normal distribution from \(u=-\infty\) to \(u=\infty\) is unity. However, the definite integral between arbitrary limits cannot be obtained as an analytical function. This turns out to be true for some other important distributions also; this is one reason for working with probability density functions rather than the corresponding cumulative probability functions. Of course, the definite integral can be calculated to any desired accuracy by numerical methods, and readily available tables give values for definite integrals from \(u=-\infty\) to \(u=u\). (We mention normal curve of error tables in , where we introduce a method for testing whether a given set of data conforms to the normal distribution equation.) | 1,127 | 859 |
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We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibriums in terms of the Brønsted–Lowry model and then proceed to a quantitative description in . In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pair. For example, in the reaction of HCl with water ( ), HCl, the , donates a proton to a water molecule, the , thereby forming Cl . Thus HCl and Cl constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl ion in solution acts as a base to accept a proton from H O , forming H O and HCl. Thus H O and H O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain conjugate acid–base pairs, which in this case are HCl/Cl and H O /H O. All acid–base reactions contain conjugate acid–base pairs. Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, H O is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH CO H/CH CO ) and the parent base and its conjugate acid (H O /H O). In the reaction of ammonia with water to give ammonium ions and hydroxide ions ( ), ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH /NH and H O/OH . Some common conjugate acid–base pairs are shown in . The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A is its conjugate base, is as follows:
\[HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \tag{16.2.1}\] The equilibrium constant for this dissociation is as follows:
\[K=\dfrac{[H_3O^+,A^−]}{[H_2O,HA]} \tag{16.2.2}\] As we noted earlier, the concentration of water is essentially constant for all reactions in aqueous solution, so [H O] in can be incorporated into a new quantity, the acid ionization constant ( ) A (aq) , also called the :
\[K_a=K[H_2O]=\dfrac{[H_3O^+,A^−]}{[HA]} \tag{16.2.3}\] Thus the numerical values of and differ by the concentration of water (55.3 M). Again, for simplicity, H O can be written as H in . Keep in mind, though, that free H does not exist in aqueous solutions and that a proton is transferred to H O in all acid ionization reactions to form H O . The larger the , the stronger the acid and the higher the H concentration at equilibrium. The values of for a number of common acids are given in . Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH is its conjugate acid:
\[B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \tag{16.2.4}\] The equilibrium constant for this reaction is the base ionization constant ( ) , also called the :
\[K_b=K[H_2O]=[BH^+,OH^−,B] \tag{16.2.5}\] Once again, the concentration of water is constant, so it does not appear in the equilibrium constant expression; instead, it is included in the . The larger the , the stronger the base and the higher the OH concentration at equilibrium. The values of for a number of common weak bases are given in There is a simple relationship between the magnitude of for an acid and for its conjugate base. Consider, for example, the ionization of hydrocyanic acid (HCN) in water to produce an acidic solution, and the reaction of CN with water to produce a basic solution:
\[HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \tag{16.2.6}\] \[CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \tag{16.2.7}\] The equilibrium constant expression for the ionization of HCN is as follows:
\[K_a=\dfrac{[H^+,CN^−]}{[HCN]} \tag{16.2.8}\] The corresponding expression for the reaction of cyanide with water is as follows:
\[K_b=\dfrac{[OH^−,HCN]}{[CN^−]} \tag{16.2.9}\] If we add and , we obtain the following (recall from that the equilibrium constant for the of two reactions is the of the equilibrium constants for the individual reactions): \[\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]} \notag \] \[\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]} \notag \] \[H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+,OH^−] \notag \] In this case, the sum of the reactions described by and is the equation for the autoionization of water, and the product of the two equilibrium constants is :
\[K_aK_b = K_w \tag{16.2.10}\] Thus if we know either for an acid or for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair. Just as with pH, pOH, and p , we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining p as follows:
\[pKa = −\log_{10}K_a \tag{16.2.11}\] \[K_a=10^{−pK_a} \tag{16.2.12}\] and \(pK_b\) as
\[pK_b = −\log_{10}K_b \tag{16.2.13}\] \[K_b=10^{−pK_b} \tag{16.2.14}\] Similarly, , which expresses the relationship between and , can be written in logarithmic form as follows:
\[pK_a + pK_b = pK_w \tag{16.2.15}\] At 25°C, this becomes \[pK_a + pK_b = 14.00 \tag{16.2.16}\] The values of p and p are given for several common acids and bases in and , respectively, and a more extensive set of data is provided in Tables and . Because of the use of negative logarithms, smaller values of p correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (HNO ), with a p of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a p of 9.21. Conversely, smaller values of p correspond to larger base ionization constants and hence stronger bases. The relative strengths of some common acids and their conjugate bases are shown graphically in . The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of p . This order corresponds to decreasing strength of the conjugate base or increasing values of p . At the bottom left of are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base. The conjugate base of a strong acid is a weak base and vice versa. We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: : \[\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \notag \] In an acid–base reaction, the proton always reacts with the stronger base. For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce H O and Cl ; only negligible amounts of HCl molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
\[HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \tag{16.2.17}\] In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of H O and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows: \[ \ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \notag \] Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left: \[H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \notag \] All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base. p and corresponding and p , and p The constants and are related as shown in . The p and p for an acid and its conjugate base are related as shown in and . Use the relationships p = −log and = 10 ( and ) to convert between and p or and p . \[4.83+pK_b=14.00 \notag \] \[pK_b=14.00−4.83=9.17 \notag \] Because p = −log , is 10 = 6.8 × 10 . \[K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14} \notag \] \[K_a=1.9 \times 10^{−11} \notag \] Because p = −log , we have p = −log(1.9 × 10 ) = 10.72. We could also have converted to p to obtain the same answer: \[pK_b=−\log(5.4 \times 10^{−4})=3.27 \notag \] \[pKa+pK_b=14.00 \notag \] \[pK_a=10.73 \notag \] \[K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11} \notag \] If we are given any one of these four quantities for an acid or a base ( , p , , or p ), we can calculate the other three. Exercise Lactic acid [CH CH(OH)CO H] is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its p is 3.86 at 25°C. Calculate for lactic acid and p and for the lactate ion. = 1.4 × 10 for lactic acid; p = 10.14 and = 7.2 × 10 for the lactate ion You will notice in that acids like H SO and HNO lie the hydronium ion, meaning that they have p values less than zero and are stronger acids than the H O ion. In fact, all six of the common strong acids that we first encountered in have p values less than zero, which means that they have a greater tendency to lose a proton than does the H O ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibriums for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the H O ion and the conjugate base of the acid. Although for HI is about 10 greater than for HNO , the reaction of either HI or HNO with water gives an essentially stoichiometric solution of H O and I or NO . In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M H O , . This phenomenon is called the leveling effect : any species that is a stronger acid than the conjugate acid of water (H O ) is leveled to the strength of H O in aqueous solution because H O is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO in aqueous solution, and an alternative approach must be used to determine their relative acid strengths. One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than H O. Measurements of the conductivity of 0.1 M solutions of both HI and HNO in acetic acid show that HI is completely dissociated, but HNO is only partially dissociated and behaves like a weak acid . This result clearly tells us that HI is a stronger acid than HNO . The relative order of acid strengths and approximate and p values for the strong acids at the top of were determined using measurements like this and different nonaqueous solvents. In aqueous solutions, [H O ] is the strongest acid and OH is the strongest base that can exist in equilibrium with H O. The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH is leveled to the strength of OH because OH is the strongest base that can exist in equilibrium with water. Salts such as K O, NaOCH (sodium methoxide), and NaNH (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in , are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of OH and the corresponding cation:
\[K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \tag{16.2.18}\] \[NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \tag{16.2.19}\] \[NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \tag{16.2.20}\] Other examples that you may encounter are potassium hydride (KH) and organometallic compounds such as methyl lithium (CH Li). As you learned in , such as H SO , H PO , and H CO contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the p increases. Consider H SO , for example: \[HSO^−_{4 (aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \notag \] The equilibrium in the first reaction lies far to the right, consistent with H SO being a strong acid. In contrast, in the second reaction, appreciable quantities of both HSO and SO are present at equilibrium. For a polyprotic acid, acid strength decreases and the p increases with the sequential loss of each proton. The hydrogen sulfate ion (HSO ) is both the conjugate base of H SO and the conjugate acid of SO . Just like water, HSO can therefore act as an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion (SO ) is a that is capable of accepting two protons in a stepwise manner: \[SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^- \notag \] \[HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \notag \] Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by p + p = p . Consider, for example, the HSO / SO conjugate acid–base pair. From , we see that the p of HSO is 1.99. Hence the p of SO is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas OH is a strong base, so the equilibrium shown above lies to the left. The HSO ion is also a very weak base [p of H SO = 2.0, p of HSO = 14 − (−2.0) = 16], which is consistent with what we expect for the conjugate base of a strong acid. Thus the equilibrium also lies almost completely to the left. Once again, equilibrium favors the formation of the weaker acid–base pair. Predict whether the equilibrium for each reaction lies to the left or the right as written. balanced chemical equation equilibrium position Identify the conjugate acid–base pairs in each reaction. Then refer to , and to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair. \[ \underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \notag \] Exercise Predict whether the equilibrium for each reaction lies to the left or the right as written. We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. In , you learned that a 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 CH CO H, a weak acid, with NaOH, a strong base:
\[\underset{acid}{CH_3CO_2H_{(l)}} +\underset{base}{NaOH_{(s)}} \overset{H_2O}{\longrightarrow} \underset{salt}{H_2OCH_3CO_2Na_{(aq)} }+\underset{water}{H_2O_{(l)}} \tag{16.2.22}\] 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 CO ) are weak bases that can react with water because they are the conjugate bases of the weak acids HCN and acetic acid, respectively. \[ CN^-_{(aq)} + H_2O_{(l)} \ce{ <<=>} HCN_{(aq)} + OH^-_{(aq)} \tag{16.2.23}\] \[ CH_3CO^2_{2(aq)} + H_2O_{(l)} \ce{<<=>} CH_3CO_2H_{(aq)} + OH^-_{(aq)} \tag{16.2.24}\] 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 and , we can see that CN is a stronger base (p = 4.79) than acetate (p = 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: \[ NH^+_{4(aq)} + H_2O_{(l)} \ce{ <<=>} HH_{3(aq)} + H_3O^+_{(aq)} \tag{16.2.25}\] \[ C_5H_5NH^+_{(aq)} + H_2O_{(l)} \ce{<<=>} C_5H_5NH_{(aq)} + H_3O^+_{(aq)} \tag{16.2.26}\] .1 shows that H O is a stronger acid than either NH or C H NH , 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 O 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. This is consistent with the information shown in , indicating that the pyridinium ion is more acidic than the ammonium ion. What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation the anion can react separately with water to produce an acid and a base, respectively? According to Eq 16.2.25, the ammonium ion will lower the pH, while according to Eq 16.2.24, 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 (p ≈ p ). 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. 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 O . 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 (part (a) in ), as discussed in . 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 ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the O–H bonds, as shown in part (b) in . With less electron density between the O atoms and the H atoms, the O–H bonds are weaker than in a free H O molecule, making it easier to lose a H ion. The magnitude of this effect depends on the following two factors ( ): Thus aqueous solutions of small, highly charged metal ions, such as Al and Fe , are acidic:
\[[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \tag{16.2.27}\] The [Al(H O) ] ion has a p 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 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 or Ca and Y , 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 Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the pH of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution. If the anion is the conjugate base of a strong acid, it will not affect the pH of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic. The NO anion is the conjugate base of a strong acid, so it has essentially no basic character ( ). Hence neither the cation nor the anion will react with water to produce H or OH , and the solution will be neutral. The Br anion is a very weak base (it is the conjugate base of the strong acid HBr), so it does not affect the pH of the solution. Hence the solution will be acidic. In contrast, SO is the conjugate base of HSO , which is a weak acid. Hence the SO ion will react with water as shown in to give a slightly basic solution. Exercise Predict whether aqueous solutions of the following are acidic, basic, or neutral. Two species that differ by only a proton constitute a . The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the . Similarly, the equilibrium constant for the reaction of a weak base with water is the . For any conjugate acid–base pair, = . Smaller values of p correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of p correspond to larger base ionization constants and hence stronger bases. At 25°C, p + p = 14.00. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than H O and no base stronger than OH can exist in aqueous solution, leading to the phenomenon known as the . Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base. 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 . : \(K_a=K[H_2O]=\dfrac{[H_3O^+,A^−]}{[HA]} \) : \(K_b=K[H_2O]=[BH^+,OH^−,B] \) : \(K_aK_b = K_w \) : \( pKa = −\log_{10}K_a \) : \(K_a=10^{−pK_a} \) : \( pK_b = −\log_{10}K_b \) : \( K_b=10^{−pK_b} \) : \(pK_a + pK_b = pK_w \) : \( pK_a + pK_b = 14.00 \; \text{at 25°C} \) Identify the conjugate acid–base pairs in each equilibrium. Identify the conjugate acid–base pairs in each equilibrium. Salts such as NaH contain the hydride ion (H ). When sodium hydride is added to water, it produces hydrogen gas in a highly vigorous reaction. Write a balanced chemical equation for this reaction and identify the conjugate acid–base pairs. Write the expression for for each reaction. Write an expression for the ionization constant for each reaction. Predict whether each equilibrium lies primarily to the left or to the right. Species that are strong bases in water, such as CH , NH , and S , are leveled to the strength of OH , the conjugate base of H O. Because their relative base strengths are indistinguishable in water, suggest a method for identifying which is the strongest base. How would you distinguish between the strength of the acids HIO , H SO , and HClO ? Is it accurate to say that a 2.0 M solution of H SO , which contains two acidic protons per molecule, is 4.0 M in H ? Explain your answer. The alkalinity of soil is defined by the following equation: alkalinity = [HCO ] + 2[CO ] + [OH ] − [H ]. The source of both HCO and CO is H CO . Explain why the basicity of soil is defined in this way. Why are aqueous solutions of salts such as CaCl neutral? Why is an aqueous solution of NaNH basic? Predict whether aqueous solutions of the following are acidic, basic, or neutral. When each compound is added to water, would you expect the pH of the solution to increase, decrease, or remain the same? Which complex ion would you expect to be more acidic—Pb(H O) or Sn(H O) ? Why? Would you expect Sn(H O) or Sn(H O) to be more acidic? Why? Is it possible to arrange the hydrides LiH, RbH, KH, CsH, and NaH in order of increasing base strength in aqueous solution? Why or why not? Arrange these acids in order of increasing strength. Given solutions with the same initial concentration of each acid, which would have the highest percent ionization? Arrange these bases in order of increasing strength: Given solutions with the same initial concentration of each base, which would have the highest percent ionization? Calculate the and the p of the conjugate acid of a base with each p value. Benzoic acid is a food preservative with a p of 4.20. Determine the and the p for the benzoate ion. Determine and p of boric acid [B(OH) ], solutions of which are occasionally used as an eyewash; the p of its conjugate base is 4.80. acid B < acid C < acid A (strongest) = 6.3 × 10 p = 9.20 | 28,455 | 860 |
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The Chernobyl disaster was a nuclear accident that occurred at the Chernobyl Nuclear Power Plant in on April 26, 1986. It is considered the worst nuclear power plant disaster in history. A nuclear meltdown in one of the reactors caused a fire that sent a plume of radioactive fallout that eventually spread all over Europe. On April 26, 1986, a test was scheduled at the Chernobyl Nuclear Power Plant to test a method of keeping the reactors properly cooled in the event of a power grid failure. If the test had gone as planned, the risk to the plant was very small. When things did go wrong, though, the potential for disaster was miscalculated and the test was continued even as serious problems arose. Meltdown occurred at 1:23 AM, starting a fire that dispersed large quantities of radioactive materials into the atmosphere. The amount of radioactive material released was 400 times more than the amount the atomic bombing of Hiroshima released. The fallout would be detected in almost all parts of Europe. Nuclear reactors require active cooling in order to remove the heat generated by radioactive decay. Even when not generating power, reactors still generate some heat, which must be removed in order to prevent damage to the reactor core. Cooling is usually accomplished through fluid flow, water in Chernobyl's case. The problem at the Chernobyl plant was that following an emergency shutdown of all power, diesel generators were needed to run the cooling pumps. These generators took about a minute to attain full speed, which was deemed an unacceptably long time for the reactor to be without cooling. It was suggested that the rotational momentum of the winding down steam turbine be used to power the pumps in the time between shutdown and the generators being ready. A test was devised to test this method in 1982, but the turbine did not prove to be successful in providing the required voltage as it spooled down. Two more tests would be conducted in the following years, but would also be unsuccessful. The fourth test was scheduled to be run on April 25, 1986. The experiment was devised in such a way that if it had gone as planned, the disruption and danger to the plant would be very minimal. First, the reactors would be brought down to low power, between 700 and 800 megawatts. Then the steam turbine would be run up to full speed and then turned off. The power generated by the winding down generators would then be measured to determine if it was sufficient to power the cooling pumps in the time before the diesel generators got up to full speed. By 1986, the plant had been running for two years without the implementation of a method to keep the cooling pumps running continuously following an emergency shutdown. This was an important safety measure that the plant was lacking, which presumably gave the plant managers a considerable amount of urgency in completing another test. The experiment was scheduled to run during the day shift of 1985, while the night shift would only have to maintain cooling of the radioactive decay in the shut-down plant. However, another power generator nearby unexpectedly shut down, necessitating the need for the Chernobyl plant to delay the test and continue producing power. The experiment would be resumed at 11:04 PM, by which time the day shift had departed and the evening shift was about to leave. This meant that the experiment would be conducted in the middle of two shifts, leaving very little time for the night shift employees to be briefed about the experiment and told what to do. The power reduction of reactor 4 to 700 MW was accomplished at 00:05 AM on the 26th of April. However, the natural production of a neutrino absorber, Xenon-135, led to a further decrease in power. When the power dropped to about 500 MW, the night shift operator committed an error and inserted the reactor control rods too far. This caused the reactor to go into a near-shutdown state, dropping power output to around 30 MW. Since this was too low for the test, it was decided to restore power by extracting the control rods. Power would eventually rise and stabilize at around 200 MW. The operation of the reactor at such a low power level would lead to unstable temperature and flow. Numerous alarms and warnings were recorded regarding emergency measures taken to keep the reactor stable. In the time between 0:35 and 0:45 AM, alarm signals regarding thermal-hydraulic parameters were ignored in order to preserve the reactor's power level. The test continued, and at 1:05 AM extra water pumps were activated in order to increase the water flow. The increased coolant flow rate led to an increase of the coolant temperature in the core, reducing the safety margin. The extra water flow also led to a decrease in the core's temperature and increased the neutron absorption rate, decreasing the reactor's power output. Operators removed the manual control rods in order to maintain power. All these actions led to the reactor being in an unstable state that was clearly outside safe operation protocol. Almost all the control rods had been removed, which reduced the effectiveness of inserting safety rods in an emergency shutdown. The water was very close to boiling, which meant that any power increase would cause it to boil. If it started boiling, it would be less effective at absorbing neutrons, further increasing the reactor's power output. The experiment was started at 1:23:04 AM. The steam to the turbines was shut off, causing the turbines to start spooling down. Four of the eight cooling pumps were also shut down. The diesel generator was started and began powering the cooling pumps after at 1:23:43. Between this time, the four pumps were powered by the slowing steam turbines. As the turbines slowed down, their power output decreased, slowing the cooling pumps. This lead to increased formation of steam voids in the core, reducing the ability of the cooling water to absorb neutrons. This increased the power output of the reactor, which caused more water to boil into steam, further increasing the reactor's power. However, during this time the automatic control system was successful in limiting power increase through the insertion of control rods. At 1:23:40, a button was pressed that initiated the emergency shutdown of the reactor and the insertion of all control rods. It is believed that this was done as a routine method to shut down the reactor to conclude the experiment and not as an emergency measure. The process of inserting the control rods was initiated, but it took about 20 seconds for the rods to be completely inserted. A flawed design in the graphite-tip control rod meant that coolant was displaced before the neutron absorbing material could be fully inserted and slow down the reaction. This meant that the process of inserting the control rods actually increased the reaction rate in the lower half of the core. A massive power spike occurred, causing the core to overheat. Some of the fuel rods fractured, causing the control rods to become stuck before they were fully inserted. Within three seconds the core's power output rose to above 500 MW. According to simulation, it is estimated that power output then rose to 30 GW, ten times the normal power output. This was caused by the rising power output causing massive steam buildup, which destroyed fuel elements and ruptured their channels. It is not possible to know precisely what sequence of events led to the destruction of the reactor. It is believed that the steam buildup entered the reactor's inner structure and lifted the 2000 ton upper plate. This steam explosion further ruptured fuel channels, resulting in more coolant turning into steam and leaving the reactor core. This loss of coolant further increased the reactor's power. A nuclear excursion (an increasing nuclear chain reaction) caused a second, even more powerful explosion. The explosion destroyed the core and scattered its contents in the surrounding area, igniting the red-hot graphite blocks. Against safety regulations, a flammable material, bitumen, had been used in roof of the reactor. When this was ignited and scattered into the surrounding area, it started several fires on reactor 3. Those working there were not aware of the damage that had been done and continued running the reactor until it was shut down at 5:00 AM. In the worst-hit parts of the reactor building, radiation levels were high enough to cause fatal doses in a matter of minutes. However, all dosimeters available to the workers did not have the ability to read radiation levels so high and thus read "off scale." Thus, the crew did not know exactly how much radiation they were being exposed to. It was assumed that radiation levels were much lower than they actually were, leading to the reactor crew chief to believe that the reactor was still intact. He and his crew would try to pump water into the reactor for several hours, causing most of them to receive fatal doses of radiation. Fire crew were called in to protect the remaining buildings from catching fire and to extinguish the still burning reactor 4. While some firefighters were not aware of the harmful doses of radiation they were receiving and had assumed it to be a simple electrical fire, others knew that they would probably receive fatal doses of radiation. However, their heroic efforts were necessary in order to try to contain the large amounts of radiation being released into the atmosphere. The fires in the surrounding buildings were extinguished by 5:00 AM, but it would take firefighters until May 10 before they could fully extinguish reactor 4. In order to prevent a steam explosion from occurring, volunteers were needed to swim through radioactive water and drain a pool of water under the reactor core. While they were successful, they would later succumb to the high doses of radiation that they had received. The worst of the radioactive debris was shoveled back into the reactor by crew wearing heavy protective gear. In total, 600,000 people worked in the cleanup, about 250,000 of which received their lifetimes' limit of radiation. It is estimated that over 10,000 eventually died from the radiation. By December, a concrete sarcophagus had been completed that sealed off the reactor. This was never meant to be a completely permanent solution, however, and is now in danger of collapsing. A collapse could cause a large amount of radioactive material to once again be released and spread around the world. This is why it is necessary that a new structure be constructed to contain reactor 4. Pripyat, a city nearby the power plant, was not immediately evacuated. At first, the government denied that the reactor had exploded and insisted that it was only a small accident. By April 27, though, investigators were forced to acknowledge that the reactor had exploded and ordered Pripyat to be immediately evacuated. 400 times more radiation was released by the disaster than had been by the atomic bombing of Hiroshima. The radiation would later be detected in almost all parts of Europe. Over one million people could have been adversely affected by the radiation. The radiation would cause numerous problems, including Down's Syndrome, chromosomal aberrations, mutations, leukemia, thyroid cancer, and birth defects. The radiation would affect all parts of the environment surrounding the plant, killing plants and animals and infecting the soil and groundwater. Life has returned to the area and seems to be flourishing, possibly due to the lack of human intrusion. Remarkably, numerous species have been reported to have adapted to their environment and have developed increased tolerance of radiation, making it possible for them to live with the radiation that is still prevalent in the soil and plants around the plant. It has even been reported that radiotrophic fungi have been growing on the walls of reactor 4. Today, radiation levels are still higher than normal in the areas surrounding the plant, but have dropped considerably from the levels that they were at twenty years ago. It is now considered safe to visit the areas immediately surrounding the plant for short periods of time. However, it is estimated that it will take 20,000 years for reactor 4's core to be completely safe. | 12,323 | 861 |
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When a conjugated diene is attacked by an electrophile, the resulting products are a mixture of 1,2 and 1,4 isomers. Kinetics and Thermodynamics control a reaction when there are two products under different reaction conditions. The Kinetic product (Product A) will be formed fast, and the Thermodynamic product (Product B) will be formed more slowly. Usually the first product formed is the more stable favored product, but in this case, the slower product formed is the more stable product; Product B. Like nonconjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated electrophiles experience relatively greater kinetic reactivity when reacted with electrophiles than nonconjugated dienes do. Upon electrophilic addition, the conjugated diene forms a mixture of two products—the kinetic product and the thermodynamic product—whose ratio is determined by the conditions of reaction. A reaction yielding more thermodynamic product is under thermodynamic control, and likewise, a reaction that yields more kinetic product is under kinetic control. Kinetic products form the fastest. They usually occur at or below 0°C. This is also known as the 1,2-adduct because the substituents are added to the first and second carbons. Kinetic products contain a terminal double bond and the reaction is irreversible. Thermodynamic products form at higher temperatures, generally greater than 40 °C. These are known are the 1,4-adducts because they add to the first and fourth carbons. Thermodynamic products contain an internal double bond and the reaction is reversible. Also, when reactions are carried out, thermodynamic products are more stable than kinetic products because they are more substituted. : Conditions at 0 °C: : The reason for the two products is the difference in their activation energies. The reaction will go through to completion on the easiest path and in this case that is the Kinetic path. This path has a much lower Activation Energy which means that less is required to have the product formed. This product- Product A, is not the most stable form however. Over time, the amount of Product B- the more stable one, will increase and the Product A will decrease. : Potential Energy profiles : Conjugated Dienese: Kinetics vs. Thermodynamic Conditions To ensure the greatest possible yield of thermodynamic products, the reaction should be carried out at a temperature of 40°C or greater. This is known as thermodynamic control. At higher temperatures and longer reaction times, thermodynamic products are favored. On the contrary, at lower temperatures, one would tend to see a greater yield of kinetic products. These products are generally formed at or around 0°C. Carrying out reactions around these temperatures is known as kinetic control and kinetic products form before thermodynamic products. Since the thermodynamic product contains an internal double bond, it is more stable than the kinetic product, and this is due to with neighboring atoms. Additionally, a higher activation energy results in the thermodynamic product forming slower than the kinetic product. Therefore, a thermodynamically controlled reaction gives a more stable product and kinetically controlled reaction gives a less stable product. The conjugated diene has 2 double bonds with one single C-C bond between them. This structure offers stability because the two pi bonds can transfer electrons through the two carbons that are sp hybridized with a single bond which results in electron delocalization. Extended P orbital sharing makes this diene more stable than the isolated dienes. The more stable molecule also has lower energy and a shorter bond length. Figure 4: The p electrons reach out to the electrophile and form a bond that in turn forms a Carbocation. The Markovnikov Addition states that the most stable carbocation is most likely to be formed with the charge going on the more substituted carbon. The difference between a conjugated diene and an alkene is that there is still a double bond left after the reaction has completed. The reactivity of conjugated dienes (hydrocarbons that contain two double bonds) varies depending on the location of double bonds and temperature of the reaction.These reactions can produce both thermodynamic and kinetic products. Isolated double bonds provide dienes with less stability thermodynamically than conjugated dienes. However, they are more reactive kinetically in the presence of electrophiles and other reagents. This is a result of Markovnikov addition to one of the double bonds. A carbocation is formed after a double bond is opened. This carbocation has two resonance structures and addition can occur at either of the positive carbons. 5. Addition of 1 equivalent of Bromine to 2,4-hexadiene at 0 degrees C gives 4,5-dibromo-2-hexene plus an isomer. Which of the following is that isomer: 6. Which of the following will be the kinetically favored product from the depicted reaction? 7. Addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene may occur in the absence or presence of peroxides. In each case two isomeric C H Br products are obtained. Which of the following is a common product from both reactions? 8. and 9. 8. The kinetically controlled product in the above reaction is: 9. For the reaction in question 8, which one is the result of 1,4-addition? 1. A) Same product for both modes of addition. B) Both cis and trans isomers will form. Addition of the HX to unsubstituted cycloalka-1,3-dienes in either 1,2- or 1,4- manner gives the same product becasuse of symmetry. 2. Yes. the Kinetic Product will still form faster but in this case there will be enough energy to form the thermodynamic product because the thermodynamic product is still more stable. 3. The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond. This form offers stability to the overall structure. 4. All of these isomers are viable. 5. D 6. C 7. D 8. A 9. B | 6,015 | 862 |
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The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure \(\Page {1}\)). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. The entropy of any perfectly ordered, crystalline substance at absolute zero is . The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (C ) as a function of temperature and then plotting the quantity C /T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously. One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol•K)] and 298 K. As shown in Table \(\Page {1}\), for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol•K), whereas S° for water vapor is 188.8 J/(mol•K). Likewise, S° is 260.7 J/(mol•K) for gaseous I and 116.1 J/(mol•K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure \(\Page {2}\), which is a generalized plot of the entropy of a substance versus temperature. Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures. A closer examination of Table \(\Page {1}\) also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol•K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol•K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH OH(l) and CH CH OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure. ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°. To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example \(\Page {1}\) illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C H ; 2,2,4-trimethylpentane). Use the data in Table \(\Page {1}\) to calculate ΔS° for the reaction of liquid isooctane with O (g) to give CO (g) and H O(g) at 298 K. : standard molar entropies, reactants, and products : ΔS° : Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table \(\Page {1}\). Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction. : The balanced chemical equation for the complete combustion of isooctane (C H ) is as follows: We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant: \begin{align}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\\ &=515.3\;\mathrm{J/K}\end{align} ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products. Use the data in Table \(\Page {1}\) to calculate ΔS° for the reaction of H (g) with liquid benzene (C H ) to give cyclohexane (C H ). : −361.1 J/K We can also calculate a change in entropy using a thermodynamic cycle. As you learned previously, the (C ) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, C is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in Figure \(\Page {2}\) is approximately proportional to the heat capacity of the substance. Recall that the entropy change (ΔS) is related to heat flow (q ) by ΔS = q /T. Because q = nC ΔT at constant pressure or nC ΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T to T is as follows: \[\Delta S=\dfrac{q_{\textrm{rev}}}{T}=nC_\textrm p\dfrac{\Delta T}{T}\hspace{4mm}(\textrm{constant pressure})\] As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in Chapter 11 "Liquids". \[\Delta S=nC_\textrm p\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant pressure}) \label{18.20}\] Similarly, \[\Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \label{18.21}\] Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample. We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in Figure \(\Page {3}\)): an orthorhombic form with a highly ordered structure (S ) and a less-ordered monoclinic form (S ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from S to S can be described by the thermodynamic cycle shown in part (b) in Figure \(\Page {3}\), in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to S (−ΔS = ΔS in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS can be calculated from the overall entropy change (ΔS ) for the S –S transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of S and S , ΔH , and the melting point of S .) If we know the melting point of S (T = 115.2°C = 388.4 K) and ΔS for the overall phase transition [calculated to be 1.09 J/(mol•K) in the exercise in Example 6], we can calculate ΔS from the values given in part (b) in Figure \(\Page {3}\) where C = 22.70 J/mol•K and C = 24.77 J/mol•K (subscripts on ΔS refer to steps in the cycle): \(\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4
\\ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right)
\\ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right)
\\ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right)
\\ &=[1.194\;\mathrm{J/(mol\cdot K)}]+[4.434\;\mathrm{J/(mol\cdot K)}]+\Delta S_3+[-1.303\;\mathrm{J/(mol\cdot K)}]\end{align}\) Solving for ΔS gives a value of −3.24 J/(mol•K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS is negative. The of a substance at any temperature above 0 K must be determined by calculating the increments of heat required to bring the substance from 0 K to the temperature of interest, and then summing the ratios / . Two kinds of experimental measurements are needed: \[ S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p}{T} dt \] Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of on be used in the above integral in place of a constant . \[ S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p(T)}{T} dt \] When this is not known, one can take a series of heat capacity measurements over narrow temperature increments Δ and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval Δ . To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (taken from the experimental plot on the left) are then plotted as a function of , and any phase-change entropies such as S = H / T are added to obtain the absolute entropy at temperature . The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products. | 12,758 | 863 |
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Before beginning a more detailed examination of the spectroscopy and magnetism of transition meal complexes, it is worth while reviewing how far a simple CFT approach will take us.
When electromagnetic radiation is absorbed by atoms or molecules it promotes them to an excited state. Microwave and infrared radiation correspond to lower energy quanta and so initiate rotational and vibrational excitation. Visible and UV light have much higher frequencies and can cause excitations characterstic of electronic excitation: the promotion of an electron from one orbital to another. We expect therefore that molecules will absorb light when the energy corresponds to the energy differences between occupied and unoccupied orbitals. For transition metal ions, the simplest case is Ti(III), solutions of which appear violet. A = ε c l The most common (and cheapest) sample cells have a 1 cm path length and since A is unitless then we can see that the units of ε are mol l cm . To move this to an acceptable SI set of units requires converting ε to units of m mol and this involves a factor of 1/10.
Thus an ε of 5 mol l cm is equivalent to ε of 0.5 m mol .
Given that the separation between the t and e levels is Δ then whether there is 1 d electron or several d electrons the simple Crystal Field Theory model would suggest that there is only 1 energy gap hence all spectra should consist of 1 peak. That this is not found in practise means that the theory is not sophisticated enough. What is required is an extension of the theory that allows for multi-electron systems where the energy levels are modified to include electron-electron interactions. This can be achieved by looking at the various quantum numbers for each of the electrons involved and using a system called the Russell-Saunders coupling scheme to describe an electronic state that can adequately describe the energy levels available to a group of electrons that includes these interactions. | 1,993 | 864 |
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In this section we shall be mainly interested in the reactions of arenes that involve attack on the carbon atoms of the aromatic ring. We shall not elaborate now on the reactions of substituent groups around the ring. The principal types of reactions involving aromatic rings are substitution, addition, and oxidation. Of these, the most common type is electrophilic substitution. A summary of the more important substitution reactions of benzene is given in Figure 22-7. Many of the reagents used to achieve these substitutions will be familiar to you in connection with electrophilic addition reactions to alkenes (e.g., \(\ce{Cl_2}\), \(\ce{Br_2}\), \(\ce{H_2SO_4}\), and \(\ce{HOCl}\); ). Electrophilic addition to alkenes and electrophilic aromatic substitution are both polar, stepwise processes, and the key step for each is attack of an electrophile at carbon to form a cationic intermediate. We may represent this type of reaction by the following general equations, in which the attacking reagent is represented either formally as a cation, \(\ce{X}^\ominus\), or as a neutral but polarized molecule, \(\overset{\delta \oplus}{\ce{X}}\)---\(\overset{\delta \ominus}{\ce{Y}}\): (first step) (first step) The intermediate shown for aromatic substitution no longer has an aromatic structure; rather, it is a cation with four \(\pi\) electrons delocalized over five carbon nuclei, the sixth carbon being saturated with \(sp^3\)-hybrid bonds. It may be formulated in terms of the following contributing structures, which are assumed to contribute essentially equally: The importance of writing the hybrid structure with the partial charges at these three positions will become evident later. This kind of ion is referred to as a or a . The aromatic ring is regenerated from this cationic intermediate by loss of a proton from the \(sp^3\)-hybridized carbon. The electron pair of this \(\ce{C-H}\) bond then becomes part of the aromatic \(\pi\)-electron system and a substitution product of benzene, \(\ce{C_6H_5X}\), is formed. (second step) The gain in stabilization attendant on regeneration of the aromatic ring is sufficiently advantageous that this, rather than combination of the cation with \(\ce{Y}^\ominus\), normally is the favored course of reaction. Herein lies the difference between aromatic substitution and alkene addition. In the case of alkenes there usually is no substantial resonance energy to be gained by loss of a proton from the intermediate, which tends therefore to react by combination with a nucleophilic reagent. (second step) \[\overset{\oplus}{\ce{C}} \ce{H_2-CH_2X} + \ce{Y}^\ominus \rightarrow \ce{YCH_2-CH_2X}\] It is important to realize that in aromatic substitution the actual electrophilic substituting agent, \(\overset{\oplus}{\ce{X}}\) or \(\overset{\delta \oplus}{\ce{X}}-\overset{\delta \ominus}{\ce{Y}}\), is not necessarily the reagent that is added to the reaction mixture. For example, nitration in mixtures of nitric and sulfuric acids is not brought about by attack of the nitric acid molecule on the aromatic compound, but by attack of a more electrophilic species, the nitronium ion, \(\ce{NO_2^+}\). This ion is formed from nitric acid and sulfuric acid according to the following equation: \[\ce{HNO_3} + 2 \ce{H_2SO_4} \rightleftharpoons \ce{NO_2^+} + \ce{H_3O^+} + 2 \ce{HSO_4^-}\] The nitronium ion attacks the aromatic ring to give first a nitrobenzenium ion and then an aromatic nitro compound: In general, the function of a catalyst (which is so often necessary to promote aromatic substitution) is to generate an electrophilic substituting agent from the given reagents. Thus it is necessary to consider carefully for each substitution reaction what the actual substituting agent may be. This problem does not arise to the same degree in electrophilic additions to alkenes, because alkenes are so much more reactive than arenes that the reagents employed (e.g., \(\ce{Br_2}\), \(\ce{Cl_2}\), \(\ce{HCl}\), \(\ce{HOCl}\), \(\ce{HOBr}\), \(\ce{H_3O}^\oplus\)) themselves are sufficiently electrophilic to react with alkenes without the aid of a catalyst. In fact, conditions that lead to substitution of arenes, such as nitration in mixtures of nitric and sulfuric acid, often will degrade the carbon skeleton of alkenes. Now we shall consider the individual substitution reactions listed in Figure 22-1 with regard to the nature of the substituting agent and the utility for synthesis of various classes of aromatic compounds. The nitronium ion, \(\ce{NO_2^+}\), is the active nitrating agent in nitric acid-sulfuric acid mixtures. The nitration of methylbenzene (toluene) is a typical example of a nitration that proceeds well using nitric acid in a 1:2 mixture with sulfuric acid. The nitration product is a mixture of 2-, 3-, and 4-nitromethylbenzenes: The presence of appreciable amounts of water in the reaction mixture is deleterious because water tends to reverse the reaction by which nitronium ion is formed: \[\ce{NO_2^+} + \ce{H_2O} \overset{\ce{HSO_4^-}}{\rightleftharpoons} \ce{HNO_3} + \ce{H_2SO_4}\] For this reason the potency of a nitric-sulfuric acid mixture can be considerably increased by using fuming nitric and fuming sulfuric acids. With such mixtures nitration of relatively unreactive compounds can be achieved. For example, 4-nitromethylbenzene is far less reactive than methylbenzene, but when heated with an excess of nitric acid in fuming sulfuric acid, it can be converted successively to 2,4-dinitromethylbenzene and to 2,4,6-trinitromethylbenzene (TNT): There are several interesting features about the nitration reactions thus far discussed. For instance, the conditions required for nitration of 4-nitromethylbenzene would rapidly oxidize an alkene by cleavage of the double bond: Also the mononitration of methylbenzene does not lead to equal amounts of the three possible products. The methyl substituent apparently orients the entering substituent preferentially to the 2 and 4 positions. This aspect of aromatic substitution will be discussed in in conjunction with the effect of substituents on the reactivity of aromatic compounds. Some compounds are sufficiently reactive that they can be nitrated with nitric acid in ethanoic acid. Pertinent examples are 1,3,5-trimethylbenzene and naphthalene: Other convenient nitrating reagents are benzoyl nitrate, \(\ce{C_6H_5COONO_2}\), and ethanoyl nitrate, \(\ce{CH_3COONO_2}\). These reagents provide a source of \(\ce{NO_2^+}\) and have some advantage over \(\ce{HNO_3} \cdot \ce{H_2SO_4}\) mixtures in that they are soluble in organic solvents such as ethanenitrile or nitromethane. Having homogeneous solutions is especially important for kinetic studies of nitration. The reagents usually are prepared in solution as required from the corresponding acyl chloride and silver nitrate or from the acid anhydride and nitric acid. Such reagents are hazardous materials and must be handles with care. Nitronium salts of the type \(\ce{NO_2^+} \ce{X^-}\) are very powerful nitrating agents. The counterion, \(\ce{X^-}\), must be non-nucleophilic and usually is fluoroborate, \(\ce{BF_4^-}\) or \(\ce{SbF_4^-}\): To some degree we have oversimplified electrophilic substitution by neglecting the possible role of the 1:1 charge-transfer complexes that most electrophiles form with arenes (see for discussion of analogous complexes of alkenes): With halogens, especially iodine, complex formation is visually evident from the color of solutions of the halogen in arenes. Although complex formation may assist substitution by bringing the halogen and arene in close proximity, substitution does not necessarily occur. A catalyst usually is required, and the catalysts most frequently used are metal halides that are capable of accepting electrons (i.e., Lewis acids such as \(\ce{FeBr_3}\), \(\ce{AlCl_3}\), and \(\ce{ZnCl_2}\)). Their catalytic activity may be attributed to their ability to polarize the halogen-halogen bond in the following way: \[\overset{\delta \oplus}{\ce{Br}} \cdots \overset{\delta \ominus}{\ce{Br}} \cdots \ce{FeBr_3}\] The positive end of the dipole attacks the aromatic compound while the negative end is complexed with the catalyst. We can represent the reaction sequence as follows, with the slow step being formation of a \(\sigma\) bond between \(\ce{Br}^\oplus\) and the aromatic ring: The order of reactivity of the halogens is \(\ce{F_2} > \ce{Cl_2} > \ce{Br_2} > \ce{I_2}\). Fluorine is too reactive to be of practical use for the preparation of aromatic fluorine compounds and indirect methods are necessary (see ). Iodine usually is unreactive. It has been stated that iodination fails because the reaction is reversed as the result of the reducing properties of the hydrogen iodide that is formed: \[\ce{C_6H_6} + \ce{I_2} \overset{\rightarrow}{\longleftarrow} \ce{C_6H_5I} + \ce{HI}\] This view is not correct because, as Kekule himself showed, iodobenzene is not reduced by hydroiodic acid except at rather high temperatures. The way to achieve direct iodination in the absence of powerful activating substituent groups is to convert molecular iodine to some more active species (perhaps \(\ce{H_2OI}^\oplus\) or \(\ce{I}^\oplus\)) with an oxidizing agent such as nitric acid or hydrogen peroxide: \[\begin{align} \ce{I_2} + 4 \ce{HNO_3} &\rightarrow 2 \ce{H_2O-I^+} + 2 \ce{NO_2} + 2 \ce{NO_3^-} \\ \ce{I_2} + \ce{H_2O_2} + 2 \ce{H^+} &\rightarrow 2 \ce{H_2OI^+} \end{align}\] With combinations of this kind good yields of iodination products are obtained: Halogen substitution reactions with chlorine or bromine must be carried out with adequate protection from strong light. If such precautions are not taken, an benzene will react rapidly with halogen by a photochemical process to substitute a hydrogen of the alkyl group rather than of the aromatic ring. The reaction has a light-induced, radical-chain mechanism of the kind discussed for the chlorination of propene ( ). Thus methylbenzene reacts with bromine when illuminated to give phenylmethyl bromide; but when light is excluded and a Lewis acid catalyst is present, substitution occurs to give principally the 2- and 4-bromomethylbenzenes. Much less of the 3-bromomethylbenzene is formed: Benzene itself can be induced to halogens on strong irradiation to give polyhalocyclohexanes (see and ): An important method of synthesis of alkylbenzenes utilizes an alkyl halide as the alkylating agent and a metal halide, usually aluminum chloride, as catalyst: This class of reaction is called in honor of its discoverers, C. Friedel (a French chemist) and J. M. Crafts (an American chemist). The metal-halide catalyst functions much as it does in halogenation reactions to provide a source of a positive substituting agent, which in this case is a carbocation: Alkylation is not restricted to alkyl halides; alcohols and alkenes may be used to advantage in the presence of acidic catalysts such as \(\ce{H_3PO_4}\), \(\ce{H_2SO_4}\), \(\ce{HF}\), \(\ce{BF_3}\), or \(\ce{HF-BF_3}\). Ethylbenzene is made commercially from benzene and ethene using phosphoric acid as the catalyst. Isopropylbenzene is made similarly from benzene and propene: Under these conditions the carbocation, which is the active substituting agent, is generated by protonation of the alkene: \[\begin{align} \ce{CH_2=CH_2} + \ce{H^+} &\rightleftharpoons \ce{CH_3CH_2^+} \\ \ce{CH_3CH=CH_2} + \ce{H^+} &\rightleftharpoons \ce{CH_3} \overset{+}{\ce{C}} \ce{HCH_3} \end{align}\] With alcohols the electrophile can be formed by initial protonation by the acid catalyst and subsequent cleavage to a carbocation: There are several factors that limit the usefulness of alkylation reactions. First, it may be difficult to limit reaction to monosubstitution because introduction of one alkyl substituent tends to activate the ring towards a second substitution (see ). Therefore, to obtain reasonable yields of a monoalkylbenzene, it usually is necessary to use a large excess relative to the alkylating agent: A second limitation is the penchant for the alkylating reagent to give rearrangement products. As an example, the alkylation of benzene with 1-chloropropane leads to a mixture of propylbenzene and isopropylbenzene. We may write the reaction as first involving formation of a propyl cation, which is a carbocation: \[\ce{CH_3CH_2CH_2Cl} + \ce{AlCl_3} \rightarrow \ce{CH_3CH_2CH_2^+} + \overset{-}{\ce{Al}} \ce{Cl_4}\] This ion either can alkylate benzene to give propylbenzene, \[\ce{C_6H_6} + \ce{CH_3CH_2CH_2^+} \rightarrow \ce{C_6H_5CH_2CH_2CH_3} + \ce{H^+}\] or it can rearrange to a more stable secondary ion by the transfer of a hydrogen from a neighboring carbon together with its bonding electron pair (i.e., 1,2-hydride shift). The positive charge is thereby transferred from \(\ce{C_1}\) to \(\ce{C_2}\): Alkylation of benzene with the isopropyl cation then produces isopropylbenzene: \[\ce{C_6H_6} + \ce{CH_3} \overset{\oplus}{\ce{C}} \ce{HCH_3} \rightarrow \ce{C_6H_5CH(CH_3)_2} + \ce{H}^\oplus\] Rearrangements of this type involving carbocation intermediates often occur in Friedel-Crafts alkylations with primary and secondary alkyl groups larger than \(\ce{C_2}\) and \(\ce{C_3}\). Related carbocation rearrangements are discussed in and . Further complications arise from the fact that the alkylation reactions sometimes are under equilibrium control rather than kinetic control. Products often isomerize and disproportionate, particularly in the presence of large amounts of catalyst. Thus 1,2- and 1,4-dimethylbenzenes ( - and -xylenes) are converted by large amounts of Friedel-Crafts catalysts into 1,3-dimethylbenzene ( -xylene): Ethylbenzene disproportionates under the influence of excess \(\ce{HF-BF_3}\) to benzene and 1,3-diethylbenzene: Acylation and alkylation of arenes are closely related. Both reactions were developed as the result of the collaboration between Friedel and Crafts, in 1877. The acylation reaction introduces an acyl group, \(\ce{RCO}\), into an aromatic ring and the product is an aryl ketone: The acylating reagents commonly used are carboxylic acid halides, \(\ce{RCOCl}\), anhydrides, \(\ce{(RCO)_2O}\), or the acid itself, \(\ce{RCO_2H}\). A strong proton or other Lewis-acid catalyst is essential. The catalyst functions to generate the acyl cation: The catalyst most commonly used with acyl halides and anhydrides is aluminum chloride: Acylation differs from alkylation in that the reaction usually is carried out in a solvent, commonly carbon disulfide, \(\ce{CS_2}\), or nitrobenzene. Furthermore, acylation requires more catalyst than alkylation, because much of the catalyst is tied up and inactivated by complex formation with the product ketone: Unlike alkylation, acylation is controlled easily to give monosubstitution, because once an acyl group is attached to a benzene ring, it is not possible to introduce a second acyl group into the same ring. Because of this, a convenient synthesis of alkylbenzenes starts with acylation, followed by reduction of the carbonyl group with zinc and hydrochloric acid ( ). For example, propylbenzene is prepared best by this two-step route because, as we have noted, the direct alkylation of benzene with propyl chloride produces considerable amounts of isopropylbenzene and polysubstitution products: In the acylation of alkylbenzene the product almost always is the isomer. The synthesis of (4- -butylphenyl)ethanone illustrates this as well as the sequential use of alkylation and acylation reactions: Chemists are inclined to give names to reactions that associate them either with their discoverers or with the products they give. This practice can be confusing because many named reactions (or "name reactions") which once were thought to be quite unrelated, have turned out to have very similar mechanisms. Thus we have two closely related acylation reactions: one is the Friedel-Crafts ketone synthesis, in which the electrophile is \(\ce{R} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O}\); and the other is the , in which the electrophile is \(\ce{H} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O}\): The latter reaction utilizes carbon monoxide and \(\ce{HCl}\) under pressure in the presence of aluminum chloride. The electrophile may be considered to be formed as follows: \[\ce{C=O} + \ce{HCl} + \ce{AlCl_3} \rightleftharpoons \ce{H} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O} + \overset{\ominus}{\ce{Al}} \ce{Cl_4}\] Substitution of the sulfonic acid \(\left( \ce{-SO_3H} \right)\) group for a hydrogen of an aromatic hydrocarbon can be carried out by heating the hydrocarbon with a slight excess of concentrated or fuming sulfuric acid: The actual sulfonating agent normally is the \(\ce{SO_3}\) molecule, which, although neutral, has a powerfully electrophilic sulfur atom: Sulfonation is reversible and the \(\ce{-SO_3H}\) group can be removed by hydrolysis at \(180^\text{o}\): A useful alternative preparation of sulfonyl derivatives is possible with chlorosulfonic acid: This procedure has an advantage over direct sulfonation in that sulfonyl chlorides usually are soluble in organic solvents and may be easily separated from the reaction mixture. Also, the sulfonyl chloride is a more useful intermediate than the sulfonic acid, but can be converted to the acid by hydrolysis if desired: Sulfonation is important in the commercial production of an important class of detergents - the sodium alkylbenzenesulfonates: The synthesis illustrates several important types of reactions that we have discussed in this and previous chapters. First, the alkyl group \(\ce{R}\) usually is a \(\ce{C_{12}}\) group derived from the straight-chain hydrocarbon, dodecane, which on photochlorination gives a mixture of chlorododecanes: This mixture of chlorododecanes is used to alkylate benzene, thereby giving a mixture of isomeric dodecylbenzenes, called : Sulfonation of the detergent alkylate gives exclusively the 4-dodecylbenzenesulfonic acids, which with sodium hydroxide form water-soluble dodecylbenzenesulfonates: In many countries it is prohibited by law to market detergents of this type, which have highly branched alkyl groups. The reason is that quaternary carbons and, to a lesser extent, tertiary carbons are not degraded readily by bacteria in sewage treatment plants: It is possible to replace the ring hydrogens of many aromatic compounds by exchange with strong acids. When an isotopically labeled acid such as \(\ce{D_2SO_4}\) is used, this reaction is an easy way to introduce deuterium. The mechanism is analogous to other electrophilic substitutions: Perdeuteriobenzene\(^3\) can be made from benzene in good yield if a sufficiently large excess of deuteriosulfuric acid is used. Deuteration might appear to be competitive with sulfonation, but deuteration actually occurs under much milder conditions. Because metals are electropositive elements they can be considered potential electrophiles. Their reactions with arenes have been investigated most thoroughly for mercury. Benzene can be substituted with \(\ce{HgX}^\oplus\) derived from a mercuric salt, \(\ce{HgX_2}\), in the presence of an acid catalyst. The salt most commonly used is mercuric ethanoate, \(\ce{Hg(OOCCH_3)_2}\). The catalyst is considered to function by assisting the generation of the active electrophile, \(\ce{HgX}^\oplus\). Other metals that may be introduced directly into an aromatic ring in this manner include thallium and lead. \(^3\)The prefix , as in perdeuterio- or perfluoro-, means that the hydrogens have been replaced with the named substituent, \(\ce{D}\) or \(\ce{F}\). Perhydro means saturated or fully hydrogenated. and (1977) | 19,700 | 865 |
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The nomenclature and physical properties of cycloalkanes covers the the IUPAC systems of naming cycloalkanes. The physical properties of cycloalkanes can explain each cycloalkane molecular structure and the relative size from simple propane to multiple carbon containing cycloakane like cyclononane. For a review, contain rings of of carbon atoms linked together by a single bonds. The simple cycloalkanes of formula \(\ce{(CH_2)}_n\) make up a particularly homologous series in which the chemical properties change in a much more dramatic way with increasing \(n\) than do those of the acyclic hydrocarbons \(\ce{CH_3(CH_2)}_{n-1}\ce{H}\). The cycloalkanes with small rings (\(n=3\)-\(6\)) are of special interest in exhibiting chemical properties intermediate between those of alkanes and alkenes. In this chapter we will show how this behavior can be explained in terms of angle strain and steric hindrance, concepts that have been introduced previously and will be used with increasing frequency as we proceed further. The IUPAC system for naming cycloalkanes and cycloalkenes was presented in some detail in Sections 3-2 and 3-3, and you may wish to review that material before proceeding further. Additional procedures are required for naming polycyclic compounds, which have rings with common carbons, and these will be discussed later in this chapter. Furthermore, given below are the table of physical properties of cycloalkanes from propane to cyclononane. The melting and boiling points of cycloalkanes (Table 12-1) are somewhat higher than those of the corresponding alkanes. In contrast to the more rigid cyclic compounds, the general “floppiness” of open-chain hydrocarbons makes them harder to fit into a crystal lattice (hence their lower melting points) and less hospitable toward neighboring molecules of the same type (hence their lower boiling points). The nomenclature and physical properties of cycloalkanes also includes conformation of cycloalkanes, especially cyclohexane in later discussion because of their importance to the chemistry of many kinds of naturally occurring organic compounds. The nomenclature and physical properties of cycloalkanes will also cover polycyclic compounds, substances with more than one ring, and to the cycloalkenes and cycloalkynes. and (1977) | 2,324 | 866 |
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A solution with a strong electrolyte, such as NaCl in water, is perhaps one of the most obvious systems to consider but, unfortunately, is also one of the more difficult ones. The reason is that the electrolyte produces two solutes, Na and Cl (both in hydrated form), in solution. We need to consider the dissociation process and stoichiometry as we are bringing more than one solute species into solution. We also need to consider electrostatic interactions between solutes. The charges introduce a strong interaction that falls off with r , as opposed to ~r if only neutral species are present. This causes a very serious divergence from ideality even at very low concentrations. Consider a salt going into solution: \[C_{ν_+}A_{ν_-} \rightarrow ν_+C^{z+} + ν_-A^{z-} \label{Eq1} \] where \(ν_+\) and \(ν_-\) are the and \(z_+\) and \(z_-\) are the formal charges of the cation and anion, respectively. As we shall see, the stoichiometric coefficients involved in the dissociation process are important for a proper description of the thermodynamics of strong electrolytes. Charge neutrality demands: \[ν_+z_+ + ν_- z_- = 0 \label{Eq2} \] For the salt, we can write: \[μ_2 =μ_2^o + RT \ln a_2 \label{Eq3} \] However, we need to take into account the dissociation of the salt. To do so, we write: \[μ_2 = ν_+μ_+ + ν_-μ_- \label{Eq4} \] This implies: \[μ_2^o = ν_+μ_+^o + ν_-μ_-^o \label{Eq5} \] where \[μ_+ =μ_-^o + RT\ln a_+ \label{Eq6} \] \[μ_- =μ_-^o + RT\ln a_- \label{Eq7} \] Usually Henry's law is taken as standard state for both type of ions. However, we cannot measure the activities of the ions separately as it is impossible to add one without adding the other. Nevertheless, we can derive a useful formalism that takes into account the dissociation process. If we substitute the last two equations in the ones above we get: \[ν_+\ln a_+ + ν_- \ln a_-=\ln a_2 \label{Eq8} \] Taking the exponent of either side of Equation \(\ref{Eq8}\), we get: \[a_2 =a_+^{ν_+}a_-^{ ν-} \label{Eq9} \] Notice that the stoichiometric coefficients (Equation \(\ref{Eq1}\)) are in Equation \(\ref{Eq9}\). We now introduce the sum of the stoichiometric coefficients: \[ν_+ + ν_- = ν \label{Eq10} \] and the \(a_±\) as: \[a_±^ ν ≡ a_2 =a_+^{ν+}a_-^{ ν-} \nonumber \] The mean ionic activity \(a_\pm\) and the activity of the salt are closely related but the relationship involves due to stoichiometric coefficients involved in the dissociation process. For example: All this remains a formality unless we find a way to relate it back to the concentration of the salt. Usually is used as a convenient concentration measure rather than molarity because we are dealing with pretty strong deviations from ideal behavior and that implies that volume may not be an additive quantity. Molality does not involve volume in contrast to molarity. Working with molalities, we can define activity coefficients for both ions, even though we have no hope to determine them separately: \[a_+ =γ_+ m_+ \label{Eq11} \] \[a_- =γ_- m_- \label{Eq12} \] Stoichiometry dictates the molalities of the individual ions must be related to the \(m\) by: \[m_-=ν_-m \label{Eq13} \] \[m_+=ν_+m \label{Eq14} \] We cannot measure the activities of the ions separately because it is impossible to add one without adding the other Analogous to the mean ionic activity, we can define a as: \[m_{\pm}^ν ≡ m_+^{ν+}m_-^{ ν-} \label{Eq15} \] We can do the same for the : \[γ_{\pm}^ν = γ_+^{ν+}γ_-^{ν-} \label{Eq16} \] Using this definitions we can rewrite: \[a_2=a_{\pm}^ν=a_+^{ν+}a_-^{ν-} \label{Eq17} \] as: \[a_2=a_{\pm}^ ν =γ_{\pm}^ ν m_{\pm}^ ν \label{Eq18} \] Note that when preparing a salt solution of molality \(m\), we should \[m_-=ν_-m \nonumber \] \[m_+=ν_+m \nonumber \] into: \[m_±^ν ≡ m_+^{ν+}m_-^{ ν-} \nonumber \] For Al (SO ) we get: So: As you can see the stoichiometry enter into the exponents into the calculation of the molality. Notice that the activity of the salt now goes as the fifth power of its overall molality (on top of the dependency of γ of exp(√m) as shown below). In contrast to the individual coefficients, the \(γ_{\pm}\) is a quantity that . In fact we can use the same Gibbs-Duhem trick we did for the sucrose problem to do so. We simply measure the water vapor pressure above the salt solution and use: \[\ln γ_{\pm} = φ -1 + \int_{m'=0}^m [ φ -1 ]m' \,dm' \nonumber \] The fact that the salt itself has a negligible vapor pressure does not matter. Particularly for ions with high charges, the deviations from ideality are very strong even at tiny concentrations. Admittedly doing these vapor pressure measurements in pretty tedious, there are some other procedures involving electrochemical potentials. However, they too are tedious. | 4,774 | 867 |
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This page is going to discuss the structure of the molecule cesium chloride (\(\ce{CsCl}\)), which is a white hydroscopic solid with a mass of 168.36 g/mol. Cesium Chloride is a type of unit cell that is commonly mistaken as Body-Centered Cubic. This misconception is easy to make, since there is a center atom in the unit cell, but CsCl is really a type. CsCl has a boiling point of 1303 degrees Celsius, a melting point of 646 degrees Celsius, and is very soluble in water. For the most part this molecule is stable, but is not compatible with strong oxidizing agents and strong acids. Although it is not hazardous, one should not prolong their exposure to CsCl. CsCl is an ionic compound that can be prepared by the reaction: \[\ce{Cs2CO3 + 2HCl -> 2 CsCl + H2O + CO2}\] There is one atom in CsCl. To determine this, the following equation is given: Cesium chloride is used in centrifugation, a process that uses the centrifugal force to separate mixtures based on their molecular density. It is also used in the preparation of electrically conducting glasses. Radioactive CsCl is used in some types of radiation therapy for cancer patients, although it is blamed for some deaths. | 1,204 | 868 |
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Generally magnetic properties of diatomic molecules or ions whose total number electrons lie in the range (1-20) can be evaluated with the help of Molecular orbital theory (MO theory) . The present study involves three (03) new formulae by just manipulating the number of unpaired electrons (n) for determination of magnetic properties without MO theory using mod function (based on Applied Mathematics) and by means of these n values one can easily stumble the magnetic moment values in Bohr-Magneton using spin only formula \[\mu_s = \sqrt{n(n+2)} \mu_B\] where First of all we classify the molecules or ions depending on the total number of electrons present in them in the following three (03) sets. Then for different set we have to use three different formulae to calculate the number of unpaired electrons which have been presented in and thus magnetic moment (\(\mu_s\)) can be evaluated in the following way: Here, ND = next digit i.e. digit next to minimum digit and ‘I I’ indicates mod function. Eg:Molecules or ions having (1-3)electrons, in this case ND = 2 because here minimum digit is 1. He (3electrons), the total number of electrons will be 3, ND = 2, Hence, unpaired electron n = I (ND - total electrons) I = I (2-3) I = 1. Hence, Magnetic Moment μ = √n(n+2) \(\mu_B\) = √ 1(1+2) BM = √3 BM = 1.73BM. For the molecules or ions containing (3-5)electrons, (5-7)electrons, (7-10)electrons, and (13-16)electrons the ND value will be 4, 6, 8 and 14 respectively. Hence, the value of n = [ I (4-total electrons) I ]; [ I (6- total electrons) I ] [ I (8- total electrons) I ] and [ I (14- total electrons) I ] respectively. Here, PD = Penultimate electron digit (i.e. before last electron). The \(C_2^-\) diatomic ion has 13 electrons, so PD = 12. Hence, unpaired electron n = I (12 - total electrons) I = I (12-13) I = 1 Hence, Magnetic Moment μ = √n(n+2) \(\mu_B\) = √ 1(1+2) BM = √3 BM = 1.73BM The \(F_2\) diatomic molecules has 18 electrons, the total number of electrons will be 18, PD = 18. Hence, unpaired electron n = I (18 - total electrons) I = I (18-18) I = 0 Hence, Magnetic Moment μ = √n(n+2) \(\mu_B\) = √ 0(0+2) BM = 0 BM = Diamagnetic in nature. The \(Ne_2\) diatomic molecules has 20 electrons, the total number of electrons will be 20. Hence, unpaired electron n = (20 - total electrons) = (20-20) = 0 Hence, Magnetic Moment μ = √n(n+2) \(\mu_B\) = √ 0(0+2) BM = 0 BM = Diamagnetic in nature. Species (Molecules or ions) Total Number of electrons Number of unpaired electrons (n) Magnetic moment (μs) in Bohr Magneton (\(\mu_B\)) Magnetic Behavior H 1 1 1.73 Paramagnetic H , He 2 0 0 Diamagnetic H ,He 3 1 1.73 Paramagnetic He , 4 0 0 Diamagnetic Li ,He 5 1 1.73 Paramagnetic Li , He , Be 6 0 0 Diamagnetic Be ,Li 7 1 1.73 Paramagnetic Be ,Li 8 0 0 Diamagnetic Be ,B 9 1 1.73 Paramagnetic B , Be , HF 10 2 2.82 Paramagnetic B ,C 11 1 1.73 Paramagnetic C ,B ,N , CN 12 0 0 Diamagnetic C ,N 13 1 1.73 Paramagnetic N ,CO,NO ,C ,CN ,O 14 0 0 Diamagnetic N ,NO,O 15 1 1.73 Paramagnetic NO ,O 16 2 2.82 Paramagnetic O 17 1 1.73 Paramagnetic F ,O ,HCl 18 0 0 Diamagnetic F 19 1 1.73 Paramagnetic Ne 20 0 0 Diamagnetic | 3,164 | 869 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/06._One_Dimensional_Harmonic_Oscillator/Kinetic_Isotope_Effects |
Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects. Research was first introduced on this topic over 50 years ago and has grown into an enormous field. The scientists behind much of the understanding and development of kinetic isotope effects were Jacob Bigeleisen and Maria Goeppert Mayer who published the first paper on isotope effects [J. Chem. Phys., 15, 261 (1947)]. Kinetic isotope effects specifically explore the change in rate of a reaction due to isotopic substitution. An element is identified by its symbol, mass number, and atomic number. The atomic number is the number of protons in the nucleus while the mass number is the total number of protons and neutrons in the nucleus. Isotopes are two atoms of the same element that have the same number of protons but different numbers of neutrons. Isotopes are specified by the mass number. As an example consider the two isotopes of chlorine, you can see that their mass numbers vary, with Cl being the most abundant isotope, while their atomic numbers remain the same at 17. \[ ^{35}Cl \;\text{and}\; ^{37}Cl\] The most common isotope used in light atom isotope effects is hydrogen (\(^{1}H\)) commonly replaced by its isotope deuterium (\(^{2}H\)). Note: Hydrogen also has a third isotope, tritium (\(^{2}H\)). Isotopes commonly used in heavy atom isotope effects include carbon (\(^{12}C\), \(^{13}C\), nitrogen (\(^{14}N\), \(^{15}N\)), oxygen, sulfur, and bromine. Not all elements exhibit reasonably stable isotopes (i.e. Fluorine, \(^{19}F\)), but those that due serve as powerful tools in isotope effects. Understanding potential energy surfaces is important in order to be able to understand why and how isotope effects occur as they do. The harmonic oscillator approximation is used to explain the vibrations of a diatomic molecule. The energies resulting from the quantum mechanic solution for the help to define the internuclear potential energy of a diatomic molecule and are \[ E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{1}\] where The Morse potential is an analytic expression that is used as an approximation to the intermolecular potential energy curves: \[ V(l) = D_e{\left(1-e^{-\beta(l-l_o)}\right)}^2 \label{2}\] where The \(D_e\), \( \beta \), and \(l_o\) variables can be looked up in a textbook or CRC handbook. Below is an example of a Morse potential curve with the zero point vibrational energies of two isotopic molecules (for example R-H and R-D where R is a group/atom that is much heavier than H or D). The y-axis is potential energy and the x axis is internuclear distance. In this figure respond to the zero point energies of deuterium and hydrogen. The zero point energy is the lowest possible energy of a system and equates to the ground state energy. Zero point energy is dependent upon the reduced mass of the molecule as will be shown in the next section. The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero point energy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy. We see this is the figure below where deuterium is heavier than hydrogen and therefore has the lower zero point energy. This results in different bond dissociation energies for R-D and R-H. The bond dissociation energy for R-D (E ) is greater than the bond dissociation energy of R-H (E ). This difference in energy due to isotopic replacement results in differing rates of reaction, the effect that is measured in kinetic isotope effects. The reaction rate for the conversion of R-D is slower than the reaction rate for the conversion of R-H. p> It is important to note that isotope replacement does not change the electronic structure of the molecule or the potential energy surfaces of the reactions the molecule may undergo. Only the rate of the reaction is affected. The energy of the vibrational levels of a vibration (i.e., a bond) in a molecule is given by \[ E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{3} \] where we assume that the molecule is in its ground state and we can compare zero-point vibrational energies, \[ E_o = \left(\dfrac{1}{2}\right)hv \label{4}\] Using the harmonic oscillator approximation the fundamental vibrational frequency is \[ \nu = \dfrac{1}{2 \pi} \sqrt{ \dfrac {k}{\mu} } \label{5}\] where \[ \mu = \dfrac{m_1m_2}{m_1+m_2} \label{6}\] The is used to determine reaction rates and activation energies and since we are interested in the change in rate of reactions with different isotopes, this equation is very important, \[ k = Ae^{-\frac{E_a}{kT}} \label{7} \] where The Arrhenius equation can be used to compare the rates of a reaction with R-H and R-D, \[ k_H = A_He^{-\frac{E_a^H}{kT}} \label{8}\] \[ k_D = A_De^{-\frac{E_a^D}{kT}} \label{9} \] where k and k are the rates of reaction associated with R-H and the isotope substituted R-D. We will then assume the Arrhenius constants are equal (\(A_H=A_D\)). The ratio of the rates of reaction gives an approximation for the isotope effect resulting in: \[ \dfrac{k_H}{k_D} = e^{-\frac{E_a^H - E_a^D}{kT}} \label{10}\] By using the relationship that for both R-H and R-D \[ E_o = \left(\dfrac{1}{2}\right)h\nu \label{11} \] a substitution can be made resulting in \[\dfrac{k_H}{k_D} = e^{\frac{h(\nu_H - \nu_D)}{2kT}} \label{12}\] The vibrational frequency (Equation 5) can then be substituted for R-H and R-D and the value of the expected isotope effect can be calculated. \[\dfrac{k_H}{k_D} = e^{\dfrac {h \left( \dfrac{k_{RH}}{\mu_{RH}} - \dfrac{k_{RD}}{\mu_{RD}} \right)}{4\pi kT}} \label{13}\] The same general procedure can be followed for any isotope substitution. In summary, the greater the mass the more energy is needed to break bonds. A heavier isotope forms a stronger bond. The resulting molecule has less of a tendency to dissociate. The increase in energy needed to break the bond results in a slower reaction rate and the observed isotope effect. Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio k /k (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units). Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction. Consider the : kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made. When hydrogen was replaced with deuterium in this reaction a \(k_H \over k_D \) of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected. A rule of thumb for heavy atom isotope effects is that the maximum isotopic rate ratio is proportional to the square root of the inverse ratio of isotopic masses. Secondary kinetic isotope effects are rate changes due to isotopic substitutions at a site other than the bond breaking site in the rate determining step of the reaction. These come in three forms: \(\alpha\), \(\beta\), and \(\gamma\) effects. \(\beta\) secondary isotope effects occur when the isotope is substituted at a position next to the bond being broken. \[\ce{(CH3)2CHBr + H2O ->[k_H] (CH3)2CHOH}\] \[\ce{(CD3)2CHBr + H2O ->[k_D] (CD3)2CHOH}\] This is thought to be due to hyperconjugation in the transition state. Hyperconjugation involves a transfer of electron density from a sigma bond to an empty p orbital (for more on hyperconjugation see outside links). Reactions may be affected by the type of solvent used (for example H O to D O or ROH to ROD). There are three main ways solvents effect reactions: | 8,974 | 870 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.04%3A_Conventions_for_Indicating_Ring_Size_and_Anomer_Configurations_of_Monosaccharides |
The oxide ring is six-membered in some sugars and five-membered in others, and it is helpful to use names that indicate the ring size. The five- and six-membered oxide rings bear a formal relationship to oxa-2,5-cyclohexadiene and oxa-2,4-cyclopentadiene that commonly are known as pyran and furan, respectively: There is an important question as to which one of the two anomeric forms of a sugar should be designated as \(\alpha\) and which one as \(\beta\). The convention is simple; the \(\alpha\) anomer is the one that has the configuration of the \(\ce{OH}\) at the anomeric carbon as the carbon which determines the configuration of the sugar itself: and (1977) | 688 | 871 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.10%3A_Evaluation_of_Standard_Molar_Quantities |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Some of the most useful experimentally-derived data for thermodynamic calculations are values of standard molar reaction enthalpies, standard molar reaction Gibbs energies, and standard molar reaction entropies. The values of these quantities for a given reaction are related, as we know (Eq. 11.8.21), by \begin{equation} \Delsub{r}G\st=\Delsub{r}H\st-T\Delsub{r}S\st \tag{12.10.1} \end{equation} and \(\Delsub{r}S\st\) can be calculated from the standard molar entropies of the reactants and products using Eq. 11.8.22: \begin{equation} \Delsub{r}S\st=\sum_i\nu_i S_i\st \tag{12.10.2} \end{equation} The standard molar quantities appearing in Eqs. 12.10.1 and 12.10.2 can be evaluated through a variety of experimental techniques. Reaction calorimetry can be used to evaluate \(\Delsub{r}H\st\) for a reaction (Sec. 11.5). Calorimetric measurements of heat capacity and phase-transition enthalpies can be used to obtain the value of \(S_i\st\) for a solid or liquid (Sec. 6.2.1). For a gas, spectroscopic measurements can be used to evaluate \(S_i\st\) (Sec. 6.2.2). Evaluation of a thermodynamic equilibrium constant and its temperature derivative, for any of the kinds of equilibria discussed in this chapter (vapor pressure, solubility, chemical reaction, etc.), can provide values of \(\Delsub{r}G\st\) and \(\Delsub{r}H\st\) through the relations \(\Delsub{r}G\st=-RT\ln K\) and \(\Delsub{r}H\st=-R\dif\ln K/\dif(1/T)\). In addition to these methods, measurements of cell potentials are useful for a reaction that can be carried out reversibly in a galvanic cell. Section 14.3.3 will describe how the standard cell potential and its temperature derivative allow \(\Delsub{r}H\st\), \(\Delsub{r}G\st\), and \(\Delsub{r}S\st\) to be evaluated for such a reaction. An efficient way of tabulating the results of experimental measurements is in the form of standard molar enthalpies and Gibbs energies of . These values can be used to generate the values of standard molar reaction quantities for reactions not investigated directly. The relations between standard molar reaction and formation quantities (Sec. 11.3.2) are \begin{equation} \Delsub{r}H\st= \sum_i\nu_i\Delsub{f}H\st(i) \qquad \Delsub{r}G\st= \sum_i\nu_i\Delsub{f}G\st(i) \tag{12.10.3} \end{equation} and for ions the conventions used are \begin{equation} \Delsub{f}H\st\tx{(H\(^+\), aq)}=0 \qquad \Delsub{f}G\st\tx{(H\(^+\), aq)}=0 \qquad S\m\st\tx{(H\(^+\), aq)}=0 \tag{12.10.4} \end{equation} Appendix H gives an abbreviated set of values of \(\Delsub{f}H\st\), \(S\m\st\), and \(\Delsub{f}G\st\) at \(298.15\K\). For examples of the evaluation of standard molar reaction quantities and standard molar formation quantities from measurements made by various experimental techniques, see Probs. 12.18–12.20, 14.3, and 14.4. | 10,150 | 872 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/Hesss_Law |
(or just ) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a . Hess's Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess's law is due to enthalpy being a , which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in and The heat of any reaction \(\Delta{H^°_f}\) for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction: (Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.) Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: \[C_{(s)} + 2 H_2O_{(g)} \rightarrow CO_{2\, (g)} + 2 H_{2\, (g)} \tag{2}\] Calorimetry reveals that this reaction requires the input of 90.1 kJ of heat for every mole of \(C_{(s)}\) consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, \(q > 0\) for an endothermic reaction. When heat is evolved, the reaction is exothermic and \(q < 0\) by convention. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, \(C_{(s)}\), into another, \(H_{2(g)}\). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)} \tag{3}\] produces \(393.5\, kJ\) for one mole of carbon burned; hence \(q=-393.5\, kJ\). The reaction \[2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \tag{4}\] produces 483.6 kJ for two moles of hydrogen gas burned, so q=-483.6 kJ. It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in equation [2], 90.1 kJ, is exactly equal to the difference between the heat evolved, -393.5 kJ, in the combustion of carbon and the heat evolved, -483.6 kJ, in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen, we get \[C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}\] \[2 H_2O_{(g)} \rightarrow 2 H_{2(g)} + O_{2(g)}\] \[C_{(s)} + O_{2(g)} + 2 H_2O_{(g)} \rightarrow CO_{2(g)} + 2 H_{2(g)} + O_{2(g)} \tag{5}\] Canceling the \(O_{2(g)}\) from both sides, since it is net neither a reactant nor product, equation [5] is equivalent to equation [2]. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields equation [2]. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of equation [2]. By studying many chemical reactions in this way, we discover that this result, known as Hess's Law, is general. A pictorial view of Hess's Law as applied to the heat of equation [2] is illustrative. In figure 1, the reactants C(s) + 2 H O(g) are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products CO (g) + 2 H (g) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing C(s), O (g), and 2 H (g). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in equation [3] and equation [4]. This picture of Hess's Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing \(C_{(s)}\), \(O_{2(g)}\), and 2 \(H_{2(g)}\). A consequence of our observation of Hess's Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product (this statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions). A slightly different view of figure 1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants. By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, H, so that the change in the function, ΔH = H - H , is equal to the heat of reaction q under constant pressure conditions. H, which we call the enthalpy, is a state function, since its value depends only on the state of the materials under consideration, that is, the temperature, pressure and composition of these materials. The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus a state function, the elevation gain is independent of the path. Now, the existence of an energy state function H is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in subfigure 2.1, with reactants R being converted to products P. We wish to calculate the heat absorbed or released in this reaction, which is ΔH. Since H is a state function, we can follow any path from R to P and calculate ΔH along that path. In subfigure 2.2, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in figure 1, the atoms involved in the reaction are C, H, and O, each of which are represented in the intermediate state in elemental form. We can see in subfigure 2.2 that the ΔH for the overall reaction is now the difference between the ΔH in the formation of the products P from the elements and the ΔH in the formation of the reactants R from the elements. The ΔH values for formation of each material from the elements are thus of general utility in calculating ΔH for any reaction of interest. We therefore define the standard formation reaction for reactant R, as elements in standard state R and the heat involved in this reaction is the standard enthalpy of formation, designated by ΔH °. The subscript f, standing for "formation," indicates that the ΔH is for the reaction creating the material from the elements in standard state. The superscript ° indicates that the reactions occur under constant standard pressure conditions of 1 atm. From subfigure 2.2, we see that the heat of any reaction can be calculated from \[\Delta{H^°_f} = \Delta{H^°_{f,products}} -\Delta{H^°_{f,reactants}} \tag{6}\] Extensive tables of ΔH° values ( ) have been compiled that allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react. The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants. Enthalpy is an extensive property and hence changes when the size of the sample changes. This means that the enthalpy of the reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ 2H (g) + O (g) → 2H O (g) ΔH° = -1144kJ The sign of the reaction enthalpy changes when a process is reversed. H (g) + 1/2O (g) → H O (g) ΔH° = -572 kJ When switched: H O (g) → H (g) + 1/2O (g) ΔH° = +572 kJ Since enthalpy is a state function, it is path independent. Therefore, it does not matter what reactions one uses to obtain the final reaction. | 10,512 | 874 |
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Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques employed mainly by inorganic and organic chemists due to its usefulness in determining structures of compounds and identifying them. Chemical compounds have different chemical properties due to the presence of different functional groups. Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques. Absorbing groups in the infrared region absorb within a certain wavelength region. The absorption peaks within this region are usually sharper when compared with absorption peaks from the ultraviolet and visible regions. In this way, IR spectroscopy can be very sensitive to determination of functional groups within a sample since different functional group absorbs different particular frequency of IR radiation. Also, each molecule has a characteristic spectrum often referred to as the fingerprint. A molecule can be identified by comparing its absorption peak to a data bank of spectra. IR spectroscopy is very useful in the identification and structure analysis of a variety of substances, including both organic and inorganic compounds. It can also be used for both qualitative and quantitative analysis of complex mixtures of similar compounds. The use of infrared spectroscopy began in the 1950's by Wilbur Kaye. He had designed a machine that tested the near-infrared spectrum and provided the theory to describe the results. Karl Norris started using IR Spectroscopy in the analytical world in the 1960's and as a result IR Spectroscopy became an accepted technique. There have been many advances in the field of IR Spec, the most notable was the application of Fourier Transformations to this technique thus creating an IR method that had higher resolution and a decrease in noise. The year this method became accepted in the field was in the late 1960's. There are three main processes by which a molecule can absorb radiation. and each of these routes involves an increase of energy that is proportional to the light absorbed. The first route occurs when absorption of radiation leads to a higher rotational energy level in a rotational transition. The second route is a vibrational transition which occurs on absorption of quantized energy. This leads to The energy levels can be rated in the following order: electronic > vibrational > rotational. Each of these transitions differs by an order of magnitude. Rotational transitions occur at lower energies (longer wavelengths) and this energy is insufficient and cannot cause vibrational and electronic transitions but vibrational (near infra-red) and electronic transitions (ultraviolet region of the electromagnetic spectrum) require higher energies. The energy of IR radiation is weaker than that of visible and ultraviolet radiation, and so the type of radiation produced is different. Absorption of IR radiation is typical of molecular species that have a small energy difference between the rotational and vibrational states. A criterion for IR absorption is a net change in dipole moment in a molecule as it vibrates or rotates. Using the molecule HBr as an example, the charge distribution between hydrogen and bromine is not evenly distributed since bromine is more electronegative than hydrogen and has a higher electron density. \(HBr\) thus has a large dipole moment and is thus polar. The dipole moment is determined by the magnitude of the charge difference and the distance between the two centers of charge. As the molecule vibrates, there is a fluctuation in its dipole moment; this causes a field that interacts with the electric field associated with radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule, absorption occurs and this alters the amplitude of the molecular vibration. This also occurs when the rotation of asymmetric molecules around their centers results in a dipole moment change, which permits interaction with the radiation field. Molecules such as O , N , Br , do not have a changing dipole moment (amplitude nor orientation) when they undergo rotational and vibrational motions, as a result, they cannot The absorption of IR radiation by a molecule can be likened to two atoms attached to each other by a massless spring. Considering simple diatomic molecules, only one vibration is possible. The Hook's law potential on the other hand is based on an ideal spring \[\begin{align} F &= -kx \label{1} \\[4pt] &= -\dfrac{dV(x)}{dx} \label{2} \end{align}\] this results in one dimensional space \[ V(r) = \dfrac{1}{2} k(r-r_{eq})^2 \label{3}\] One thing that the Morse and Harmonic oscillator have in common is the small displacements (\(x=r-r_{eq}\)) from the equilibrium. Solving the Schrödinger equation for the harmonic oscillator potential results in the energy levels results in \[ E_v = \left(v+\dfrac{1}{2}\right)hv_e \label{4}\] with \(v=0,1,2,3,...,\,infinity\) \[ v_e = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5}\] When calculating the energy of a diatomic molecule, factors such as anharmonicity (has a similar curve with the harmonic oscillator at low potential energies but deviates at higher energies) are considered. The energy spacing in the harmonic oscillator is equal but not so with the anharmonic oscillator. The anharmonic oscillator is a deviation from the harmonic oscillator. Other considered terms include; centrifugal stretching, vibrational and rotational interactions have to be taken into account. The energy can be expressed mathematically as \[ E_v = \underset{\text{Harmonic Oscillator}}{\left(v+\dfrac{1}{2}\right)hv_e} - \underset{\text{anharmonicity}}{\left(v+\dfrac{1}{2}\right)^2 X_e hv_e} + \underset{\text{Rigid Rotor}}{B_e J (J+1)} - \underset{\text{centrifugal stretching}}{D_e J^2 (J+1)^2} -\alpha_e \underset{\text{rovibrational coupling}}{\left(v+\dfrac{1}{2}\right) J(J+1)} \label{6}\] The first and third terms represent the and rigid rotor behavior of a diatomic molecule such as HCl. The second term represents anharmonicity and the fourth term represents centrifugal stretching. The fifth term represents the interaction between the vibration and rotational interaction of the molecule. The bond of a molecule experiences various types of vibrations and rotations. This causes the atom not to be stationary and to fluctuate continuously. Vibrational motions are defined by stretching and bending modes. These movements are easily defined for diatomic or triatomic molecules. This is not the case for large molecules due to several vibrational motions and interactions that will be experienced. When there is a continuous change in the interatomic distance along the axis of the bond between two atoms, this process is known as a stretching vibration. A change in the angle occurring between two bonds is known as a bending vibration. Four bending vibrations exist namely, wagging, twisting, rocking and scissoring. A CH group is used as an example to illustrate stretching and bending vibrations below. Types of Vibrational Modes. To ensure that no center of mass motion occurs, the center atom (yellow ball) will also move. Figure from Diagram of Stretching and Bending Modes for H O. H O molecule is a non-linear molecule due to the uneven distribution of the electron density. O is more electronegative than H and carries a negative charge, while H has a partial positive charge. The total degrees of freedom for H O will be 3(3)-6 = 9-6 = 3 degrees of freedom which correspond to the following stretching and bending vibrations. The vibrational modes are illustrated below: . CO is a linear molecule and thus has the formula (3N-5). It has 4 modes of vibration (3(3)-5). CO has 2 stretching modes, symmetric and asymmetric. The CO symmetric stretch is not IR active because there is no change in dipole moment because the net dipole moments are in opposite directions and as a result, they cancel each other. In the asymmetric stretch, O atom moves away from the C atom and generates a net change in dipole moments and hence absorbs IR radiation at 2350 cm . The other IR absorption occurs at 666 cm . CO symmetry with \(D_{\infty h}\) CO has a total of four of stretching and bending modes but only two are seen. Two of its bands are degenerate and one of the vibration modes is symmetric hence it does not cause a dipole moment change because the polar directions cancel each other. The vibrational modes are illustrated below: The second law of Newton states that \[F = ma\label{7}\] where m is the mass and a is the acceleration, acceleration is a 2nd order differential equation of distance with respect to time. Thus "a" can be written as \[a = \dfrac{d^2 y}{d t} \label{8}\] Substituting this into Equation \ref{1} gives \[\dfrac{m d^2 y}{d t^2}= - k y \label{9}\] the 2nd order differential equation of this equation is equal to \(\dfrac{-k}{m}\) displacement of mass and time can be stated as \[y = A\cos 2 \pi \nu_m t \label{10}\] where v is the natural vibrational frequency and A is the maximum amplitude of the motion. On differentiating a second time the equation becomes \(\dfrac{d^2 y}{d t^2} = - 4 \pi^2 \nu_m^2 A \cos 2 \pi \nu_m t \label{11}\) substituting the two equations above into Newton's second law for a harmonic oscillator, \[m*\left (-4\pi^{2}\nu_{m}^{2} A \textrm{cos }2\pi\nu_{m}t \right ) = -k * \left ( A\textrm{cos }2\pi\nu_{m}t \right ) \label{12}\] If we cancel out the two functions \(y\), \[4m\pi^{2}\nu_{m}^{2} = k \] from above, we obtain the natural frequency of the oscillation. \[\nu_m = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} \label{13}\] \(\nu_m\) which is the natural frequency of the mechanical oscillator which depends on the force constant of the spring and the mass of the attached body and independent of energy imparted on the system. when there are two masses involved in the system then the mass used in the above equation becomes \[\mu = \dfrac{m_1 m_2}{m_1+m_2} \label{14}\] The vibrational frequency can be rewritten as \[\nu_m = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{15}\] Using the harmonic oscillator and wave equations of quantum mechanics, the energy can be written as \[E = \left(v+\dfrac{1}{2}\right) \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{16}\] where h is Planck's constant and v is the vibrational quantum number and ranges from 0,1,2,3.... infinity. \[E = \left(v+\dfrac{1}{2}\right)hv_m \label{17}\] where \(\nu_m\) is the vibrational frequency. Transitions in vibrational energy levels can be brought about by absorption of radiation, provided the energy of the radiation exactly matches the difference in energy levels between the vibrational quantum states and provided the vibration causes a change in dipole moment. This can be expressed as \[{\triangle E} = hv_m = \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{18}\] At room temperature, the majority of molecules are in the ground state v = 0, from the equation above \[E_o = \dfrac{1}{2}hv_m \label{19}\] following the selection rule, when a molecule absorbs energy, there is a promotion to the first excited state \[E_1 = \dfrac{3}{2} hv_m \label{20}\] \[\left(\dfrac{3}{2} hv_m - \dfrac{1}{2} hv_m \right) = hv_m \label{21}\] The frequency of radiation v that will bring about this change is identical to the classical vibrational frequency of the bond v and it can be expressed as \[E_{radiation} = hv = {\triangle E} = hv_m = \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{22}\] The above equation can be modified so that the radiation can be expressed in wave numbers \[\widetilde{\nu} = \dfrac{h}{2\pi c} \sqrt{\dfrac{k}{\mu}} \label{23}\] where Molecular vibrational frequencies lie in the IR region of the electromagnetic spectrum, and they can be measured using the IR technique. In IR, polychromatic light (light having different frequencies) is passed through a sample and the intensity of the transmitted light is measured at each frequency. When molecules absorb IR radiation, transitions occur from a ground vibrational state to an excited vibrational state (Figure 1). For a molecule to be IR active there must be a change in dipole moment as a result of the vibration that occurs when IR radiation is absorbed. The dipole moment can be expressed mathematically as \(\mu = er \label{24}\) The relationship between IR intensity and dipole moment is given as \(I_{IR} \propto \left(\dfrac{d\mu}{dQ}\right)^2 \label{25}\) relating this to intensity of the IR radiation, we have have the following equation below. where \(\mu\) is the dipole moment and \(Q\) is the vibrational coordinate. The that gives information about the probability of a transition occurring, for IR can also be written as \(\langle \psi_ | \hat{M}| \psi_f \rangle \label{26}\) \(i\) and \(f\) represent are initial and final states. \(\psi_i\) is the wave function. Relating this to IR intensity we have \(I_{IR} \propto \langle \psi_ | \hat{M}| \psi_f \rangle \label{27}\) where \(\hat{M}\) is the dipole moment and has the Cartesian coordinates, \(\hat {M_x}\),\(\hat {M_y}\), \(\hat{M_z}\). In order for a transition to occur by dipole selection rules , at least one of the integrals must be non zero. The IR region of the electromagnetic spectrum ranges in wavelength from . Conventionally the IR region is subdivided into three regions, near IR, mid IR and far IR. Most of the IR used originates from the mid IR region. The table below indicates the IR spectral regions IR deals with the interaction between a molecule and radiation from the electromagnetic region ranging (4000- 40 cm ). The cm is the wave number scale and it can also be defined as 1/wavelength in cm. A linear wavenumber is often used due to its direct relationship with both frequency and energy. The frequency of the absorbed radiation causes the molecular vibrational frequency for the absorption process. The relationship is given below \[\bar{v}(cm^{-1}) = \dfrac{1}{\lambda(\mu m)} \times 10^4 (\dfrac{\mu m}{cm}) = \dfrac{v(Hz)}{c(cm/s)} \label{28}\] IR spectroscopy is a great method for identification of compounds, especially for identification of functional groups. Therefore, we can use group frequencies for structural analysis. Group frequencies are vibrations that are associated with certain functional groups. It is possible to identify a functional group of a molecule by comparing its vibrational frequency on an IR spectrum to an IR stored data bank. Here, we take the IR spectrum of Formaldehyde for an example. Infrared spectroscopy can also be applied in the field of quantitative analysis, although sometimes it's not as accurate as other analytical methods, like gas chromatography and liquid chromatography. The main theory of IR quantification is Beer's law or , which is written as \[ A= \log \left ( \dfrac{I_0}{I} \right ) =\epsilon lc \label{29}\] Where A is the absorbance of the sample, I is the intensity of transmitted light, I is the intensity of incident light, l is the path length, a is the molar absorptivity of the substance, and c is the concentration of the substance. From the Beer's Law, we could figure out the relation between the absorbance and the concentration of the sample since the analytes have a particular molar absorptivity at a particular wavelength. Therefore, we could use IR spectroscopy and Beer's Law to find the concentration of substance or the components of mixture. This is how the IR quantification operated. \(\left(\dfrac{d\mu}{dr}\right)_{r_{eq}} \not= 0 \label{30}\) \(\triangle v = +1\) and \(\triangle J = +1 \label{31}\) \(k = \left(\dfrac{d^2 V(r)}{dr^2}\right)_{r_{eq}} \label{32}\) where k is the force constant and indicates the strength of a bond. One of the most importance applications of IR spectroscopy is structural assignment of the molecule depending on the relationship between the molecule and observed IR absorption bands. Every molecule is corresponding to one particular symmetry point group. Then we can predict which point group the molecule is belonging to if we know its IR vibrational bands. Vice versa, we can also find out the IR active bands from the spectrum of the molecule if we know its symmetry. These are two main applications of group theory. How do you distinguish whether the structure of transition metal complex molecule M(CO) L is or by inspection of the CO stretching region of the IR spectra? For cis- Since A1 has a basis on z axis and B1 has a basis on x axis, there are two IR vibrational bands observed in the spectrum. Since Therefore, from what have been discussed above, we can distinguish these two structures based on the number of IR bands. The frequency of C=O stretching is higher than that of C=C stretching. The Intensity of C=O stretching is stronger than that of C=C stretching. Explain it. | 16,797 | 875 |
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We noted that light, color, and photochemistry owe their existence to the electronic structure of atoms. It may not be surprising, therefore, that the same wave/particle model that is used for light can be applied to electronic structure, and that we can get some insight into the energy involved in photosynthesis by studying the light that is absorbed. Let's see how this wave model developed, and review the wave model. Not all light is effective in allowing plants to carry out photosynthesis. Plants generally absorb light in roughly the same region that is visible to the Human eye (350 - 700 nm). Shorter wavelengths are absorbed by the ozone layer, or if they make it through, have enough energy to cause cell damage. Longer wavelengths do not provide enough energy for photosynthesis. The absorbance spectrum of some biomolecules involved in photosynthesis is shown here. Note that "chlorophyll a" absorbs strongly in the red at ~680 nm and in the blue at ~440 nm. Photosynthesis cannot occur in plants irradiated with light in the yellow region (570-590 nm), because it is not absorbed, no matter how bright it is. What is the energy of photons with a wavelength of 440 nm? Solution: \(\text{E} = \text {h} \times \nu =\frac{\text{h}\times\text{c}}{\lambda}\) \( \text{E} = [(6.626 \times 10^{-34} \text{ J s} )(3 \times 10^{8})] / 440 \times 10^{-9} \times m = 4.52 \times 10^{-19} \text{J}\) The energy supplied by each photon of 680 nm light is 2.92 x 10 J, and that of 580 nm light is 3.43 x 10 J. Even though it lies between the energies of the two photons that cause photosynthesis, it is completely ineffective, no matter how bright (how many photons per second). In the early 1900s Frederick Frost Blackman and Gabrielle Matthaei<rf>en. .org/wiki/Photosynthesis</ref> found that at constant temperature, the rate of carbon assimilation varies with light intensity, initially increasing as the intensity increases. How can this be consisten with the energy of light being related to just its frequency or wavelength? The paradox is solved by noting that intensity is the number of photons per second. If one photon has enough energy to initiate photosynthesis in one chlorophyll molecule, then many photons will cause more photosynthesis. But if the photons were 580 nm light, they would have no effect, no matter what their intensity. Some botanists claim that blue light at 440 nm is most effective in promoting leaf growth through photosynthesis, while red light around 680 nm is most effective in causing flowering . | 2,553 | 876 |
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Selenium was discovered by Berzelius in 1818. It is named for the Greek for "moon", selene. Selenium can exist in multiple allotropes that are essentially different molecular forms of an element with varying physical properties. For example, one allotrope of selenium can be seen as an amphorous (“without crystalline shape”) red powder. Selenium also takes a crystalline hexagonal structure, forming a metallic gray allotrope which is known to be stable. The most thermodynamically stable allotrope of selenium is trigonal selenium and also appears as a gray. Most selenium is recovered from the electrolytic copper refining process. This is usually in the form of the red allotrope. Selenium is mostly noted for its important chemical properties, especially those dealing with electricity. Unlike sulfur, selenium is a semiconductor, meaning that it conducts some electricity, but not as well as conductors. Selenium is a photoconductor, which means it has the ability to change light energy into electrical energy. Not only is selenium able to convert light energy into electrical energy, but it also displays the property of photoconductivity. Photoconductivity is the idea that the electrical conductivity of selenium increases due to the presence of light or in other words, it becomes a better photoconductor as light intensity increases. Isotopes of an element are atoms that have the same atomic numbers but a different number of neutrons (different mass numbers) in their nuclei. Selenium is known to have over 20 different isotopes; however, only 5 of them are stable. The five stable isotopes of selenium are Se, Se, Se, Se, Se. Due to selenium’s property of photoconductivity, it is known to be used in photocells, exposure meters in photography, and also in solar cells. Selenium can also be seen in its production in plain-paper photocopiers, laser printers and photographic toners. Besides its uses in the electronic industry, selenium is also popular in the glass-making industry. When selenium is added to glass, it is able to negate the color of other elements found in the glass and essentially decolorizes it. Selenium is also able to create a ruby-red colored glass when added. The element can also be used in the production of alloys and is an additive to stainless steel. Selenium, a trace element, is important in the diet and health of both plants and animals, but can be only taken in very small amounts. Exposure to an excess amount of selenium is known to be toxic and cause health problems. With a tolerable upper intake level of 400 micrograms per day, too much selenium can lead to selenosis and may result in health problems and even death. Compounds of selenium are also known to be carcinogenic. Selenium forms hydrogen selenide, H Se, a colorless flammable gas when reacted with hydrogen. Selenium burns in air displaying a blue flame and forms solid selenium dioxide. \[Se_{8(s)} + 8O_{2(g)} \rightarrow 8SeO_{2(s)}\] Selenium is also known to form selenium trioxide, SeO . Selenium reacts with fluorine, F , and burns to form the selenium hexafluoride. \[Se_{8(s)} + 24F_{2(g)} \rightarrow 8SeF_{6(l)}\] Selenium also reacts with chlorine and bromine to form diselenium dichloride, \(Se_2Cl_2\) and diselenium dibromide, \(Se_2Br_2\). \[Se_8 + 4Cl_2 \rightarrow 4Se_2Cl_{2(l)}\] \[Se_8 + 4Br_2 \rightarrow 4Se_2Br_{2(l)}\] Selenium also forms \(SeF_4\), \(SeCl_2\) and \(SeCl_4\), Selenium reacts with metals to form selenides. Example: Aluminum selenide \[3 Se_8 + 16 Al \rightarrow 8 Al_2Se_3\] Selenium reacts to form salts called selenites, e.g., silver selenite (Ag SeO ) and sodium selenite (Na SeO ) | 3,677 | 877 |
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Polonium was discovered in 1898 by Marie Curie and named for her native country of Poland. The discovery was made by extraction of the remaining radioactive components of pitchblende following the removal of uranium. There is only about 10 g per ton of ore! Current production for research purposes involves the synthesis of the element in the lab rather than its recovery from minerals. This is accomplished by producing Bi-210 from the abundant Bi-209. The new isotope of bismuth is then allowed to decay naturally into Po-210. The sample pictured above is actually a thin film of polonium on stainless steel. Although radioactive, polonium has a few commercial uses. You can buy your own sample of polonium at a photography store. It is part of the special anti-static brushes for dusting off negatives and prints. | 838 | 878 |
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Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He ) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation. The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling. In contrast, is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions: \[\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}} \] Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials ( ). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle: \[\text{1 MeV/particle} = \text{96 billion J/mol}. \nonumber \] The effects of ionizing radiation depend on four factors: The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in . Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal. Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source the body. Alpha particles, however, are the most damaging if their source is the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in . There are many different ways to measure radiation exposure, or the dose. The , which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the ; the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram: \[\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}} \] Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle. Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10 rem). Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics. We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr ( ). One component of background radiation is , high-energy particles and \(\gamma\) rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure. A second component of background radiation is , produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as \(\ce{^{14}C}\): \[\ce{^{14}_7 N + ^1_0 n \rightarrow ^{14}_6 C + ^1_1p }\label{Eq3} \] The carbon atoms react with oxygen atoms to form CO , which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 10 of the carbon atoms in our bodies is radioactive C, which decays by beta emission. About 5000 C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium ( H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to C is estimated to be 1 mrem/yr, while that due to H is about 1000 times less. The third major component of background radiation is , which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as \(\ce{^{232}Th}\) and \(\ce{^{238}U}\) as well as radioactive daughter isotopes, such as\(\ce{^{226}Ra}\). The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the \(K^+\) ion. Naturally occurring potassium contains 0.0117% \(\ce{^{40}K}\), which decays by emitting both a β particle and a (\gamma\) ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 \(\ce{^{40}K}\) nuclei disintegrated in your body. By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, Rn undergoes two successive alpha-decay events to give Pb: \[\ce{^{222}_{86} Rn \rightarrow ^4_2\alpha + ^{218}_{84} Po + ^4_2\alpha + ^{214}_{82} Pb } \label{Eq4} \] Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of \(\ce{^{218}Po}\) releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The \(\ce{^{218}Po}\) isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States. In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium ( H) and promethium ( Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight). Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring K in his or her body, which contains about 140 g of potassium (as the K ion). The natural abundance of K is 0.0117%. Each 1.00 mol of K undergoes 1.05 × 10 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle. mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle annual radiation dose in rads The number of moles of K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as K divided by the atomic mass of K: \[\textrm{moles }^{40}\textrm K= 140\textrm{ g K} \times \dfrac{0.0117\textrm{ mol }^{40}\textrm K}{100\textrm{ mol K}}\times\dfrac{1\textrm{ mol K}}{40.0\textrm{ g K}}=4.10\times10^{-4}\mathrm{\,mol\,^{40}K} \nonumber \] We are given the number of atoms of K that decay per second in 1.00 mol of K, so the number of decays per year is as follows: The total energy the body receives per year from the decay of K is equal to the total number of decays per year multiplied by the energy associated with each decay event: We use the definition of the rad (1 rad = 10 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows: This corresponds to almost half of the normal background radiation most people experience. Because strontium is chemically similar to calcium, small amounts of the Sr ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca . Consequently, the radioactive strontium ( Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of Sr undergoes 4.55 × 10 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were Sr? 5.7 × 10 rad/yr (which is 10 times the fatal dose) One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in . Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid. Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess. The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in , but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure. Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. | 19,117 | 879 |
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Activity is a measure of the concentration of a species under non-ideal (e.g., concentrated) conditions. This determines the real chemical potential for a real solution rather than an ideal one. Activities and concentrations can both be used to calculate equilibrium constants and reaction rates. However, most of the time we use concentration even though activity is also a measure of composition, similar to concentration. It is satisfactory to use concentration for diluted solutions, but when you are dealing with more concentrated solutions, the difference in the observed concentration and the calculated concentration in equilibrium increases. This is the reason that the activity was initially created. \[ \large a=e^{\frac {\mu -\mu_o}{RT}} \tag{1}\] where Fugacity is the effective pressure for a non-ideal gas. The pressures of an ideal gas and a real gas are equivalent when the chemical potential is the same. The equation that relates the non-ideal to the ideal gas pressure is: \[ f = \phi P \tag{2}\] with For an ideal gas, the fugacity coefficient is 1. We have become accustomed to using the equation \( pH= -\log[H^+] \), but this equation not accurate at all concentrations. A better expression for pH is \( pH= -\log[a_H^+] \) which accounts for the , a. The only reason that other indicators may correctly seem to measure the acidity, which was equivalent to \(–\log[H^+]\), is because of the use of , which uses concentration rather than activity. | 1,491 | 880 |
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These are homework exercises to accompany of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap. In Section 17-3, we derived an expression for the expectation value of energy, \( \langle E \rangle \), by applying Equation 17.20 to \( Q(N, V, T) \) given by Equation 17.22. Now, apply Equation 17.21 to \( Q(N, V, T) \) to give the same result, except with \( \beta \) replaced by \( \dfrac{1}{k_bT} \). Beginning with Equation 17.21 \( \langle E \rangle = k_bT^2 (\dfrac{\partial{ln Q}}{\partial T})_{N,V} \) We can use the partition function in the problem to find \( (\dfrac{\partial{ ln Q}}{\partial T})_{N,V} \): \( Q(N, V, T) = \dfrac{V^N}{N!} ( \dfrac{2\pi mk_b}{h^2} )^{\dfrac{3N}{2}} T^{\dfrac{3N}{2}} \) \( ln Q = \dfrac{3N}{2} ln T \) \( ( \dfrac{\partial_{ln Q}}{\partial T} )_{N,V} = \dfrac{3N}{2T} \) Substituting this last result into Equation 17.21 yields \( \langle E \rangle = k_bT^2 \dfrac{3N}{2T} = \dfrac{3}{2}(Nk_bT) \) A gas absorbed on a surface can sometimes be modeled as a two-dimensional ideal gas. We will learn in that the partition function of a two-dimensional ideal gas is: \[Q(N,A,T) = \dfrac{1}{N!}\big(\dfrac{2\pi mk_{B}T}{h^{2}}\big)^{N}A^{N}\] where \(A\) is the area of the surface. Derive an expression for \(\bar{C}_{V}\). \(\bar{C}_{V} = \big(\dfrac{\partial U}{\partial T} \big)_{N,A} \) and \(U= K_{b}T^{2} \big(\dfrac{\partial ln (Q)}{\partial T} \big)_{N,A}\) Given the partition function of a monatomic can der Waals gas: \[ Q(N,V,T)=\frac{1}{N!}(\frac{2\pi m k_B T}{h^2})^{\frac{3N}{2}} (V-Nb)^N e^{\frac{aN^2}{V k_b T}} \] what is the average energy of this gas? The average energy of a gas is given by \[<E> = k_b T^2 (\frac{\partial ln Q}{\partial T})_{N,V}\] taking the natural log of the partition function gives us \[lnQ = \frac{3N}{2} ln({\frac{2\pi m k_B T}{h^2}}) + \frac{aN^2}{V k_b T} +\frac{1}{N!}N ln(V-Nb)\] Now we take the partial derivative with respect to T while holding N and V constant. This yields \[(\dfrac{\partial ln Q}{\partial T})_{N,V} = \frac{3Nh^2 2 \pi m k_b}{4\pi m k_b T} - \frac{aN^2}{V k_b T^2} = \frac{3nh^2}{2T} -\frac{aN^2}{V k_b T^2} \] Substituting this value into the Average Energy equation, we get \[<E> = \frac{3N k_b T}{2} - \frac{aN^2}{V} \]. Given the following equation: \[Q(N,V,T) = \dfrac{(V)^{N}}{N!}\Big(\dfrac{2\pi mk_{B}}{h^{2}}\Big)^{3N/2}T^{3N/2} \] An approximate partition function for a gas of hard spheres can be obtained from the partition function of a monatomic gas by replacing \(V\) in the given equation with \(V-b\), where \(b\) is related to the volume of the \(N\) hard spheres. Derive expressions for the energy and the pressure of this system. We can use the partition function specified in the problem to find \[ Q(N,V,T) = \dfrac{(V-b)^{N}}{N!}\Big(\dfrac{2\pi mk_{B}}{h^{2}}\Big)^{3N/2}T^{3N/2} \] \[ ln Q = \dfrac{3N}{2} ln T + terms \: not \: involving \: T \] After substitution into the given equation, we find that the energy (E) is the same as that for a monatomic ideal gas: \(3Nk_{B}T/2\). We can use the partition function specified in the problem to find \[ Q(N,V,T) = \dfrac{1}{N!}\Big(\dfrac{2\pi mk_{B}T}{h^{2}}\Big)^{3N/2}(V-b)^{N} \] \[ ln Q = N ln(V-b) + terms \: not \: involving \: V \] Similar substitution allows us to find \[ \langle P \rangle = k_{B}T\Big(\dfrac{\partial ln Q}{\partial V}\Big)_{N,B} = \dfrac{Nk_{B}T}{V-b} \] Using the partition function \[Q(N,\:,A\:,\:T) = \dfrac{1}{N!}\Big(\dfrac{2\pi mk_{B}T}{h^{2}}\Big)^{N}A^{N}\], calculate the heat capacity of a two-dimensional ideal gas. First, the energy must be found. This can be done by finding \(\Big(\dfrac{\partial ln Q}{\partial T}\Big)_{N,\:V}\). \[ln[Q] = N ln[T] + ...\] where the ... refers to terms that do not depend on T. Because they do not depend on T, the partial with respect to those terms are 0. The partial derivative is \[\Big(\dfrac{\partial ln Q}{\partial T}\Big)_{N,\:V} = \dfrac{N}{T}\]. The energy can be expressed as \[E = \Big(\dfrac{\partial ln Q}{\partial T}\Big)_{N,\:V}*k_{B}T^{2}\] \[E = \dfrac{Nk_{B}T^{2}}{T}\] \[E = Nk_{B}T\] In the case of our two-dimensional ideal gas, \[E = U\], so we can use the relationship \[C_{v} = \Big(\dfrac{\partial U}{\partial T}\Big)_{N,V}\] \[C_{v} = Nk_{B}\] The partition function of the rigid rotator-harmonic oscillator model of an ideal diatomic gas is given by $$Q(N, V, \beta) = \frac{[q(V, \beta)]^N}{N!}$$ where $$q(V, \beta) = (\frac{2\pi m}{h^2\beta})^(\frac{3}{2})*V*\frac{8\pi^2I}{h^2\beta}*\frac{e^(-\beta h\nu/2)}{1-e^(-\beta h\nu/2)}$$ Find the expression for the pressure of an ideal atomic gas. $$Q(N, V, \beta)= \frac{1}{N!}(\frac{2\pi m}{h^2\beta})^N*V^N*\frac{8\pi^2I}{h^2\beta}^N*\frac{e^(-\beta h\nu/2)}{1-e^(-\beta h\nu/2)}^N$$ $$lnQ = NlnV$$ $$\frac{dlnQ}{dV} = \frac{N}{V}$$ $$\langle P\rangle = \frac{Nk_{B}T}{V}$$ which is the ideal gas law. a.) Prove that a simplified form of the molar heat capacity of system of independent particles can be written as: \[\overline{C_{V}} = R(\beta\varepsilon)^{2} \dfrac{e^{-\beta\varepsilon}}{(1+e^{-\beta\varepsilon})^{2}} \] b.) Verify by plotting \(\overline{C_{V}} \) versus \(\beta\varepsilon \) that the maximum value occurs when \(\beta\varepsilon = 2.40 \). a.) We can write the partition function of the system as: \[q = e^{-\dfrac{\varepsilon_{0}}{k_{B}T}} + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \] If we assume that the ground quantum state, \(\varepsilon_{0} \) is equal to zero, we get: \[ q = e^{-\dfrac{0}{k_{B}T}} + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} = 1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \] We can then write the average energy of the system in terms of q: \[ \langle E \rangle = RT^{2} \left( \dfrac{\partial ln(q)}{\partial T} \right)_{V} = RT^{2} \left( \dfrac{\partial ln \left( 1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \right)}{\partial T} \right)_{V} = RT^{2}\dfrac{1}{1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}} \left( -\dfrac{\varepsilon_{1}}{k_{B}} \right) \left( -\dfrac{1}{T^{2}} \right) e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} = R \left( \dfrac{\varepsilon_{1}}{k_{B}} \right) \dfrac{e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}}{1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}} \] Using the definition of molar heat capacity, we can then plug in our expression for \( \langle E \rangle \): \[\overline{C_{V}} = \left( \dfrac{\partial \langle E \rangle}{\partial T} \right)_{V} = \left( \dfrac{\partial \left( R \left( \dfrac{\varepsilon_{1}}{k_{B}} \right) \dfrac{e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}}{1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}} \right)}{\partial T} \right)_{V} \] \[ = R \left( \dfrac{\varepsilon_{1}}{k_{B}} \right) \left( -\dfrac{\varepsilon_{1}}{k_{B}} \right) \left( -\dfrac{1}{T^{2}} \right) e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \left( 1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \right)^{-2} = R \left( \dfrac{\varepsilon_{1}}{k_{B}T} \right)^{2} \dfrac{e^{-\dfrac{\varepsilon_{1}}{k_{B}T}}}{\left( 1 + e^{-\dfrac{\varepsilon_{1}}{k_{B}T}} \right)^{2}} \] If we define \(\varepsilon_{1} = \varepsilon \) and \(\dfrac{1}{k_{B}T} = \beta \), we can simplify the above expression for the molar heat capacity as: \[\overline{C_{V}} = R(\beta\varepsilon)^{2} \dfrac{e^{-\beta\varepsilon}}{(1 + e^{-\beta\varepsilon})^{2}} \] which is what the original problem asked us to show. b.) Plotting \(\overline{C_{V}} \) versus \(\beta\varepsilon \) results in the following graph, with the maximum at \(\beta\varepsilon = 2.40 \) verified. (All calculations and graphs were done in MATLAB and produced by me) Each of the N atoms of the crystal is assumed to vibrate independently about its lattice position, so that the crstal is pictures as N independent harmonic oscillators, each vibrating in three directions. The partition function of a harmonic oscillator is \[q_{ho}(T) = \sum_{\nu=0}^\infty e^{-\beta (\nu+\dfrac{1}{2}) hv} \] \[e^{-\beta hv/2}\sum_{\nu=0}^\infty e^{\beta \nu hv}\] This summation is easy to evaluate if you recognize it as the so-called geometric series \[ \sum_{\nu=0}^\infty x^\nu = \dfrac{1}{1-x}\] Show that \[q_{ho}(T) = \dfrac{e^{-\beta h \nu/2}}{1-e^{-\beta h \nu}} \] and that \[ Q = e^{\beta U_0} \bigg( \dfrac{e^{-\beta h \nu/2}}{1-e^{-\beta h \nu}} \bigg)^3N \] where U simply represents the zero-of-energy, where all N atoms are infinitely separated. let \(x = e^{-\beta h \nu} then \sum_{\nu=0}^\infty e^{-\beta h \nu} = \dfrac{1}{1- e^{-\beta h \nu}}\) \(Q = q^N = (q_{vib}q_{elec})^N \) \[q_{elec}=e^{-\beta U_0 /N}+q_{vib} = (q_{ho}^3)] \[Q = e^{-\beta U_0}\bigg( \dfrac{e^{-\beta h \nu/2}}{1-e^{-\beta h \nu}} \bigg)^{3N}\] Evaluate \[S = \sum_{i=0}^{2} \,\sum_{j=0}^{1}\,x^{i+j}= x(1+y)+x^2(1+y) = (x+x^2)(1+y)\] by summing over j first then over i. Now obtain the same result by writing S as a product of two separate summations. By evaluating the formula given in the problem, S can be written as a product of two separate summations, given by \[ S = \sum_{i=0}^{2} \,\sum_{j=0}^{1}\ = (x+x^2)(y+1)\] Evaluate: \[S = \sum_{i=0}^{4} \,\sum_{j=1}^{i}\,5+j\] \[(1+5) +[(1+5)+(2+5)]+[(1+5)+(2+5)+(3+5)]+[(1+5)+(2+5)+(3+5)+(4+5)]=6+13+21+30=70\] Consider a system of two noninteracting identical femions, each has energy states of \(\varepsilon_1\), \(\varepsilon_2\) and \(\varepsilon_3\). How many terms are there in the unrestriced evaluation of Q(2,V,T)? Q(2,V,T) = \(\Sigma e^{-\beta(\varepsilon_i+\varepsilon_j)}\) Total terms = 9. From these 9 terms that will appear in the unrestricted evaluation of Q, only 3 of them will be allowed for 2 identical femions: \(\varepsilon_1\) + \(\varepsilon_2\) \(\varepsilon_1\) + \(\varepsilon_3\) \(\varepsilon_2\) + \(\varepsilon_3\) Looking at problem 17.24, how many allowed terms are there in the case of bosons instead of fermions? There are six allowable terms, three that were found in Q17.24 and an additional three, they are as follows: \[\epsilon_1 + \epsilon_3 \] \[\epsilon_1 + \epsilon_2 \] \[\epsilon_2 + \epsilon_3 \] \[\epsilon_1 + \epsilon_1 \] \[\epsilon_2 + \epsilon_2 \] \[\epsilon_3 + \epsilon_3 \] Consider a system of three noninteracting identical bosons, each of which has states with energies \(\varepsilon_{1}\), \(\varepsilon_{2}\), and \(\varepsilon_{3}\). How many terms are there in the unrestricted evaluation of \(Q(3,V,T)?\) How many terms occur in \(Q(3,V,T)\) when the boson restriction is taken into account? Enumerate the allowed total energies in the summation of Equation 17.37: \[Q(N,V,T) = \sum_{i,j,k..} e^{-\beta(\varepsilon_{i}+\varepsilon_{j}+\varepsilon_{k}...)} \] There are ten allowed terms, given by the total energies \[1. \varepsilon_{1} + \varepsilon_{2} + \varepsilon_{3} \] \[2. \varepsilon_{1} + \varepsilon_{1} + \varepsilon_{3} \] \[3. \varepsilon_{2} + \varepsilon_{2} + \varepsilon_{3} \] \[4. \varepsilon_{2} + \varepsilon_{2} + \varepsilon_{1} \] \[5. \varepsilon_{2} + \varepsilon_{2} + \varepsilon_{2} \] \[6. \varepsilon_{2} + \varepsilon_{2} + \varepsilon_{3} \] \[7. \varepsilon_{1} + \varepsilon_{3} + \varepsilon_{3} \] \[8. \varepsilon_{3} + \varepsilon_{2} + \varepsilon_{3} \] \[9. \varepsilon_{1} + \varepsilon_{1} + \varepsilon_{1} \] \[10. \varepsilon_{3} + \varepsilon_{3} + \varepsilon_{3} \] Consider a system of 2 noninteracting bosons, each having a state with energy \(\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4}\). How many terms are in the unrestricted evaluation \(Q\big(2,V,T\big)\) Write the allowed energies in the summation form?. There are 10 allowed terms, given by the total energies \(\epsilon_{1} +\epsilon_{2}\) \(\epsilon_{2} +\epsilon_{2}\) \(\epsilon_{1} +\epsilon_{3}\) \(\epsilon_{2} +\epsilon_{3}\) \(\epsilon_{1} +\epsilon_{4}\) \(\epsilon_{2} +\epsilon_{4}\) \(\epsilon_{1} +\epsilon_{1}\) \(\epsilon_{3} +\epsilon_{4}\) \(\epsilon_{4} +\epsilon_{4}\) \(\epsilon_{3} +\epsilon_{3}\) | 11,849 | 881 |
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Boltzmann was brilliant, undoubtedly a genius, far ahead of his time in theory. Of course he was not infallible. Most important for us moderns to realize, he was still very limited by the science of his era; dominant physical chemist and later Nobelist Ostwald named his estate “Energie” but did not believe in the physical reality of molecules nor in Boltzmann’s treatment of them. Some interesting but minor details that are not too widely known: Even though Boltzmann died in 1906, there is no evidence that he ever saw, and thus certainly never calculated entropy values via the equation Planck published in a 1900 article, S= R/N ln W. It was first printed in 1906 in a book by Planck as \( S=k_{B}lnW \), and subsequently carved on Boltzmann’s tombstone. Planck’s nobility in allowing R/N to be called ‘Boltzmann’s constant’, k , was uncharacteristic of most scientists of that day, as well as now. The important question is “what are the bases for Boltzmann’s introduction of order to disorder as a key to understanding spontaneous entropy change?” That 1898 idea came from two to three pages of a conceptual description, a common language summary, that follow over 400 pages of detailed theory in Brush’s translation of Boltzmann’s 1896-1898 “Lectures on Gas Theory” (University of California Press, 1964). The key paragraph should be quoted in full. (The preceding and following phrases and sentences, disappointingly, only expand on it or support it without additional meaningful technical details or indications of Boltzmann’s thought processes. I have inserted an explanatory clause from the preceding paragraph in brackets, and put in italics Boltzmann’s surprisingly naïve assumptions about all or most initial states as “ordered”.) “In order to explain the fact that the calculations based on this assumption [“…that by far the largest number of possible states have the characteristic properties of the Maxwell distribution…”] correspond to actually observable processes, one must assume that an enormously complicated mechanical system represents a good picture of the world, . When this is the case, and when left to itself it rapidly proceeds to the disordered most probable state.” (Final paragraph of #87, p. 443.) [Pitzer’s calculation of a mole of any substance at near 0 K shows that can be more ordered than having the possibility of 10 different accessible microstates! (Pitzer, , 3rd edition, 1995, p. 67.)] Thus, today we know that system above 0 K has "order" in correct thermodynamic descriptions of systems of energetic molecules. The common older textbook comparison of orderly crystalline ice to disorderly liquid water is totally deceptive, It is a visual "Boltzmann error" not a proper thermodynamic evaluation. If liquid water at 273 K, with its 10 accessible microstates (quantized molelcular arrangements) is considered "disorderly", how can ice at 273 K that has 10 accessible microstates be considered "orderly"? Obviously, using such common words is inappropriate in measuring energetic microstates and thus in discussing entropy change conceptually. That slight, innocent paragraph of a sincere man — but before modern understanding of q /T via knowledge of molecular behavior (Boltzmann believed that molecules perhaps could occupy only an infinitesimal volume of space), or , or the — that paragraph and its similar nearby words are the foundation of all dependence on “entropy is a measure of disorder”. Because of it, uncountable thousands of scientists and non-scientists have spent endless hours in thought and argument involving ‘disorder’and entropy in the past century. Apparently never having read its astonishingly overly-simplistic basis, they believed that somewhere there was some profound base. Somewhere. There isn’t. Boltzmann was the source and no one bothered to challenge him. Why should they? Boltzmann’s concept of entropy change was accepted for a century primarily because skilled physicists and thermodynamicists focused on the fascinating relationships and powerful theoretical and practical conclusions arising from entropy’s relation to the behavior of matter. They were not concerned with conceptual, non-mathematical answers to the question, “What is entropy, really?” that their students occasionally had the courage to ask. Their response, because it was what had been taught to them, was “Learn how to calculate changes in entropy. Then you will understand what entropy ‘really is’.” There is no basis in physical science for interpreting entropy change as involving order and disorder. The original definition of entropy (change) involves a transfer of heat from a thermal reservoir to a system via a virtually reversible energy flow process. Although Clausius described it and his equation of dq /T or q /T as a “Verwandlung” or “transformation”, he limited it and “disgregation” to discussions of fusion or vaporization where the “disgregation values” changed. Thus, Clausius was observing phase change, but he made no statements about “orderly crystalline substances” being transformed into “disorderly” liquids, an obvious claim for him to make from his observation. Unfortunately, Clausius did not see that his dq, an amount of ‘heat’ energy, initially relatively localized in a thermal reservoir, was transformed in any process that allowed heat to become more . That is what happens when a warm metal bar is placed in contact with a similar barely cooler metal bar — or when any system is warmed by its slightly warmer surroundings. The final state of the “universe” in both of these examples is at equilibrium and at a uniform temperature. The internal energy of the atoms or molecules in the reservoir has become less localized, more dispersed in the greater final three-dimensional space than it was in the initial state. (More profoundly, of course, that energy has become more dispersed in phase-space, and spread over more energy levels in the once-cooler object than was its dispersal decreased in the once-hotter surroundings.) That is also what happens when ideal gases A and B with their individually different internal energy content (S values) but comparably energetic, constantly moving molecules in adjacent chambers are allowed access to one another’s chambers at 298 K. With no change in temperature, they will mix spontaneously because, on the lowest level of interpretation, the translational energy of the A and B molecules can thereby become more spread out in the larger volume. On a more sophisticated level, their energy is more widely distributed in phase-space. From the quantum-mechanical view of the occupancy of energy levels by individual molecules, each type of molecule has additional energy levels in the greater volume, because the energy levels become closer together. But the same causal description of energy spontaneously spreading out can be used as in the naïve view of seeing mobile molecules always moving to occupy newly available 3-D volume: the energy of the molecules is more dispersed, more spread out, now in terms of dispersal over more energy levels. (Of course, this energy dispersal can best be described in terms of additional accessible microstates. The greater the number of possible arrangements of molecular energies over energy levels, the greater the entropy increase — because the system in any one arrangement at one instant has more probability of being in a different arrangement the next instant. The total energy of the system is unchanged over time, but there is a continuing ‘temporal dance’ of the system over a minute fraction of the hyper-astronomic number of accessible arrangements.) The increase of entropy in either A or B can readily be shown to be equal to R ln V /V , or more fundamentally, —nR(x ln x ). This result is not specific to gases, of course. What is shown to be significant by the basic equation is that separation of molecules of one type from its own kind is an entropy increase due to spreading out of its intrinsic internal energy . Further, this increased dispersal of energy is interpretable in terms of the increased number of accessible arrangements of the system’s energy at any instant and thus, a greater number of chances for change in the next instant — a greater ‘temporal dance’ by the system over greater possibilities and a consequent entropy increase. Boltzmann’s sense of “increased randomness” as a criterion of the final equilibrium state for a system compared to initial conditions was not wrong. What failed him (and succeeding generations) was his surprisingly simplistic conclusion: “Disorder” was the consequence, to Boltzmann, of an initial “order” not — as is obvious today — of what can only be called a “prior, lesser but still humanly-unimaginable, large number of accessible microstates” Clearly, a great advantage of introducing chemistry students to entropy increase as due to molecular energy spreading out in space, if it is not constrained, begins with the ready parallels to spontaneous behavior of kinds of energy that are well-known to beginners: the light from a light bulb, the sound from a stereo, the waves from a rock dropped in a swimming pool, the air from a punctured tire. However, its profound “added value” is its continued pertinence at the next level in the theoretical interpretation of energy dispersal in thermal or non-thermal events, i.e., when the quantization of molecular energies on energy levels, their distributions, and accessible microstates become the focus. When a system is heated, and its molecules move more rapidly, their probable distributions on energy levels change so that previous higher levels are more occupied and additional high levels become accessible. The molecular energy of the heated system therefore has become more widely spread out . The dispersal of energy on energy levels is comparable in adiabatic processes that some authors characterize as involving “positional” or “configurational” entropy. When a larger volume is made available to ideal components in a system — by expansion of a gas, by fluids mixing (or even by a solute dissolving) — the energy levels of the final state of each constituent are closer together, denser than in the initial state. This means that more energy levels are occupied in the final state despite no change in the total energy of any constituent.. Thus, the initial energy of the system has become more spread out, more widely dispersed in the final state.. The ‘Boltzmann’ equation for entropy is S = k ln W, where W is the number of different ways or microstates in which the energy of the molecules in a system can be arranged on energy levels. Then, ΔS would equal k ln W / W for the thermal or expansion or mixing processes just mentioned. A most important ΔS value in chemistry is the standard state entropy for a mole of any substance at 298 K, S , that can be determined by calorimetric measurement of increments of heat/T added reversibly to the substance from 0 K to 298 K. Any transition state or phase change/T is also added. Obviously, therefore, considerable energy is ‘stored’ in any substance on many different energy levels when it is in its standard state. (A. Jungermann, , 83, 1686-1694.) Further, for example, if energy from 398 K surroundings is spread out to a mole of nitrogen at 298 K, the nitrogen’s molecules become differently arranged on previous energy levels and spread to some higher levels. If, while at any fixed temperature, the nitrogen is allowed to expand into a vacuum or to mix with another gas, its energy levels in the new larger volume will be closer together. Even in fixed volume or steady temperature situations, the constantly colliding molecules in a mole of any gas are clearly not just in one unique arrangement on energy levels for more than an instant. They are continually changing from one arrangement to another due to those collisions — within the unchanged total energy content at a given temperature and a distribution on energy levels consistent with a Boltzmann distribution. Thus, because WInitial at 0 K is arbitrarily agreed to be 1, nitrogen’s S of 191.6 J/K mole = 1.380 x 10 J/K ln W . Then, W = 10 to the exponent of microstates, a number of arrangements for the nitrogen molecules at 298.15 K that is humanly beyond comprehension — except in the terms of manipulating or comparing that number mathematically. It should be emphasized that these gigantic numbers are significant guides mathematically and physically and conceptually — i.e., that a greater or smaller such number indeed indicates a difference in real physical systems of molecules and we should sense that However, conceptually, we should also realize that in real time, it is impossible for a system to be more than a few quadrillion different states in a few seconds, perhaps spending most time in a few billion or million, It is impossible even in almost-infinite time for a system to explore all possible microstates or that even a tiny fraction of the possible microstates could be explored in a century. (It is even less probable that a gigantic number would be visited because the most probably and frequently occupied microstates constitute an extremely narrow peak in a ‘probability spectrum’.) However, the concepts are clear. At one instant, of the molecules are only in energetic arrangement — an instantaneous ‘freeze-frame’ photo of molecular energies on energy levels (that is an abstraction derived from an equally impossible photo of the actual molecules with their speeds and locations in space at one instant.) Then, in the next instant, a collision between even two molecules will change the arrangement into a different microstate. In the next instant to another. Then to another. (Leff has called this sequence of instantaneous changes “a temporal dance” by the system over some of its possible accessible microstates.) Even though the calculated number of possible microstates is so great that there is no chance that more than a small fraction of that number could be explored or “danced in” over finite time, that calculated number influences how many chances there are for the system’s energy arrangement to be in at the next moment. The greater the number of microstates, the more chances a system has for its energy next to be in a different microstate. In this sense, the greater the number of microstates possible for a system, the less probable it is that it could return to a previously visited microstate and thus, the more dispersed is its energy over time. (In no way is this evaluation and conclusion a novel/radical introduction of time into thermodynamic considerations! It is simply that in normal thermo measurements lasting seconds, minutes or hours, the detailed behavior of molecules in maintaining a macro state is of no interest.) For example, heating a mole of nitrogen only one degree, say from the standard state of 298.15 K to 299.15 K, in view of its heat capacity of 29 J/K, results in an entropy increase of 0.097 J/K, increasing the microstates from 6,027 x 10 to 6,034 x 10 . Thus, even a slight macro change in a system is signified by an equal change in the number of microstates — in the number of chances for the system to be in a different microstate in the next instant than in the previous moment.. T The most frequently used example to show entropy increase as greater “disorder” in elementary chemistry texts for many years was that of ice melting to water. Sparkling orderly crystalline ice to disorderly mobile liquid water — that is indeed a striking visual and crystallographic impression, but appearance is not the criterion of entropy change. Entropy increase is dependent on the dispersal of energy — in three-dimensional space (an easily understandable generality for all beginning students.) Then, more capable students can be led to see how entropy increase due to heat transfer involves molecular energies occupying more and higher energy levels while entropy increase in gas expansion and all mixing is characterized by occupancy of denser energy levels within the original energy span of the system. Finally, advanced students can be shown that any increase in entropy in a final system or universe that has as the ultimate correlation of entropy increase with theory, quantitatively derivable from molecular thermodynamics. Crystalline ice at 273 K has an S of 41.34 J/K mol, and thus, via S = k ln W, there are 10 to an exponent of 1,299,000,000,000,000,000,000,000 possible accessible microstates for ice. Because the S for liquid water at 273 K = 63.34 J/K mole, there are 10 to an even larger exponent of 1,991,000,000,000,000,000,000,000 accessible microstates for water. Does this clearly show that water is “disorderly” compared to crystalline ice? Of course not. For ice to have fewer accessible microstates than water at the same temperature means primarily — so far as entropy considerations are concerned — that any pathway to change ice to water will result in an increase in entropy in the system and therefore is favored thermodynamically. Gibbs’ use of the phrase “mixed-upness” is totally irrelevant to ‘order-disorder’ in thermodynamics or any other discussion. It comes from a posthumous fragment of writing, unconnected with any detailed argument or logical support for the many fundamental procedures and concepts developed by Gibbs Finally, the idea that there is any ‘order’ or simplicity in the distribution of energy in an initial state of real substance under real conditions is destroyed by Pitzer’s calculation of numbers for microstates in his “Thermodynamics” (Third edition, 1995, p. 67). As near to 0 K and thus, to as ‘practical’ a zero entropy as can be achieved in a laboratory, Pitzer shows that there must be 10 of possible accessible microstates for any substance. | 17,882 | 886 |
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The entropy of a substance at any temperature T is not complex or mysterious. It is simply a measure of the total amount of energy that had to be dispersed within the substance (from the surroundings) from 0 K to T, incrementally and reversibly and divided by T for each increment, so the substance could exist as a solid or (with additional reversible energy input for breaking intermolecular bonds in phase changes) as a liquid or as a gas at the designated temperature. Because the heat capacity at a given temperature is the energy dispersed in a substance per unit temperature, integration from 0 K to T of C /T dT (+ q/T for any phase change) yields the total entropy. This result, of course, is equivalent to the area under the curve to T in Figure 5. To change a solid to a liquid at its melting point requires large amounts of energy to be dispersed from the warmer surroundings to the solid for breaking the intermolecular bonds to the degree required for existence of the liquid at the fusion temperature. (“To the degree required” has special significance in the melting of ice. Many, but not all of the hydrogen bonds in crystalline ice are broken. The rigid tetrahedral structure is no longer present in liquid water but the presence of a large number of hydrogen bonds is shown by the greater density of water than ice due to the even more compact hydrogen-bonded clusters of H O.) Quantitatively, the entropy increase in this isothermal dispersal of energy from the surroundings is ΔH /T. Because melting involves bond-breaking, it is an entropy increase in the energy of the substance involved. (This potential energy remains unchanged in a substance throughout heating, expansion, mixing, subsequent phase change to a vapor, or mixing. Of course, it is released when the temperature of the system drops below the freezing/melting point).The process is isothermal, and therefore there is no energy transferred to the system to increase motional energy. However, a change in the motional energy — not an increase in the of energy — occurs from the transfer of vibrational energy in the ice crystal to the liquid. When the liquid forms, there is rapid breaking of hydrogen bonds (trillionths of a second) and forming new ones with adjacent molecules. This might be compared to a fantastically huge dance in which the individual participants don't move very far (takes a water molecule >12 hours to move a cm at 298K) but they are holding hands and then releasing to grab new partners far more frequently than billions of times a second. Thus, that previous motional energy of intermolecular vibration that was in the crystal is now distributed among a far greater number of new translational energy levels, and that means that there are many more accessible microstates than in the solid. Similarly, a liquid at its vaporization temperature has the same energy as its gas molecules. (All of the enthalpy of vaporization is needed to break intermolecular bonds in the liquid.) However in the case of liquid to vapor, there is a huge expansion (a thousand times increase)in volume. Therefore, this means closer energy levels, far more than were available for the motional energy in the liquid — and a greatly increased number of microstates for the vapor. The entropy effects in gas expansion into a vacuum, as described previously, are qualitatively similar to gases mixing.. From a macro viewpoint, the initial energy of each constituent becomes more dispersed in the new larger volume provided by the combined volumes of the components. Then, on a molecular basis, because the density (closeness) of energy levels increases in the larger volume, and therefore there is greater dispersal of the molecular energies on those additional levels, there are more possible arrangements (more microstates) for the mixture than for the individual constituents. Thus, the mixing process for gases is actually a spontaneous process due to an increase in volume. The entropy increases ( ). Meyer vigorously pointed toward this same statement in "the spontaneity of processes generally described as "mixing", that is, combination of two different gases initially at pressure, , and finally at pressure, , has to do with the mixing itself of either" ( ). The same cause — of increase of the system resulting in a greater density of energy levels — obviously cannot apply to liquids in which there is little or no volume increase when they are mixed. However, as Craig has well said, "The "entropy of mixing" might better be called the "entropy of dilution" ( ). By this one could mean that the molecules of either constitutent in the mixture become to some degree separated from one another and thus their energy levels become closer together by an amount determined by the amount of the other constituent that is added. Whether or not this is true, ‘configurational entropy' is the designation in statistical mechanics for considering entropy change when two or more substances are mixed to form a solution. The model uses combinatorial methods to determine the number of possible “cells” (that are identified as microstates, and thus each must correspond to one accessible arrangement of the energies of the substances involved). This number is shown to depend on the mole fractions of each component in the final solution and thus the entropy is found to be: ΔS = - R (n ln X + n ln X ) with n and n the moles of pure solute and solvent, and X and X the mole fractions of solute and solvent ( , ). In general chemistry texts, configurational entropy is called ‘positional entropy' and is contrasted to the classic entropy of Clausius that is then called ‘thermal entropy'. The definition of Clausius is fundamental; positional entropy is derivative in that its conclusions can be derived from thermal entropy concepts/procedures, but the reverse is not possible. Most important is the fact that positional entropy in texts often is treated as just that: the positions of molecules in space determine the entropy of the system, as though their locations — totally divorced from any motion or any energy considerations — were causal in entropy change. This is misleading. Any count of ‘positions in space' or of ‘cells' implicitly includes the fact that molecules being counted are particles with energy. Although the initial energy of a system is unchanged when it increases in volume or when constituents are mixed to form it, "Escaping tendency" or chemical potentials or graphs that are complex to a beginner are often used to explain the freezing point depression and boiling point elevation of solutions. These topics can be far more clearly explained by first describing that an entropy increase occurs when a non-volatile solute is added to a solvent — the solvent's motional energy becomes more dispersed compared to the pure solvent, just as it does when any non-identical liquids are mixed. (This is the fundamental basis for a solvent's decreased "escaping tendency" when it is in a solution. If the motional energy of the solvent in a solution is less localized, more spread out, the solvent less tends to “escape” from the liquid state to become a solid when cooled or a vapor when heated.) Considering the most common example of aqueous solutions of salts: Because of its greater entropy in a solution (i.e., its energy more ‘spread out' at 273.15 K and less tending to have its molecules ‘line up' and give out that energy in forming bonds of solid ice), liquid water containing a solute that is insoluble in ice is not ready for equilibrium with solid ice at 273.15 K. Somehow, the more-dispersed energy in the water of the solution must be decreased for the water to change to ice. But that is easy, conceptually — all that has to be done is to cool the solution below 273 K because, contrary to making molecules move move rapidly and spread their energy when heated, cooling a pure liquid or a solution obviously will make them move more slowly and their motional energy become less spread out, more like the energy in crystalline ice. Interestingly, the heat capacity of water of an aqueous solution (75 J/mol) is about twice that of ice (38 J/mol). That means that when the temperature of the surroundings decrease by one degree, the solution disperses much more energy to the surroundings than the ice — and the motional energy of water and any possible ice becomes closer. Therefore, when the temperature decreases a degree or so (and finally to –1.86 C/kg.mol!), the solution's higher entropy has rapidly decreased to be in thermodynamic equilibium with ice and the surroundings amd freezing can begin. As energy (the enthalpy of fusion) continues to be dispersed to the cold surroundings, the liquid water freezes. Of course, the preceding explanation could have been framed in terms of entropy change or numbers of microstates, but keeping the focus on what is happening to molecular motion, on energy and its dispersal is primary rather than derivative. The elevation of boiling points of solutions having a non-volatile solute is as readily rationalized as is freezing point depression. The more dispersed energy (and greater entropy) of a solvent in the solution means that the expected thermodynamic equilibrium (the equivalence of the solvent's vapor pressure at the normal boiling point with atmospheric pressure) cannot occur at that boiling point. For example, the energy in water in a solution at 373 K is more widely dispersed due to the increased number of microstates for a solution than for pure water at 373 K. Therefore, water molecules in a solution less tend to leave the solution for the vapor phase than from pure water. Energy must be transferred from the surroundings to an aqueous solution to increase its energy content, thereby compensating for the greater dispersion of the water's energy due to its being in a solution, and to allow water molecules to escape readily to the surroundings. (More academically, “to raise the vapor pressure of the water in the solution”.) As energy is dispersed to the solution from the surroundings and the temperature rises above 373 K, at some point a new equilibrium temperature for phase transition is reached. The water then boils because the greater vapor pressure of the water in the solution with its increased motional energy now equals the atmospheric pressure. Although the hardware system of osmosis in the chemistry laboratory and in industry is unique, the process that occurs in it is merely a special case of the mixing of a solvent with a solution — ‘special' because of the existence of such marvels as semi-permeable membranes through which a solvent can pass but a solute cannot. As would be deduced from the discussion about mixing two liquids or mixing a solid solute and a liquid, the solvent in a solution has a greater entropy than a sample of pure solvent. Its energy is more dispersed in the solution. Thus, if there is a semi-permeable membrane between a solution made with a particular solvent and some pure solvent, the solvent will spontaneously move through the membrane to the solution because its energy becomes more dispersed in the solution and thus, its entropy becomes increased if it mixes with the solution. | 11,270 | 888 |
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Polarity is a physical property of compounds which relates other physical properties such as melting and boiling points, solubility, and intermolecular interactions between molecules. For the most part, there is a direct correlation between the polarity of a molecule and number and types of polar or non-polar covalent bonds which are present. In a few cases, a molecule may have polar bonds, but in a symmetrical arrangement which then gives rise to a non-polar molecule such as carbon dioxide. Polarity results from the uneven partial charge distribution between various atoms in a compound. Atoms, such as nitrogen, oxygen, and halogens, that are more electronegative have a tendency to have partial negative charges. Atoms, such as carbon and hydrogen, have a tendency to be more neutral or have partial positive charges. Electrons in a polar covalent bond are unequally shared between the two bonded atoms, which results in partial positive and negative charges. The separation of the partial charges creates a dipole. The word dipole means two poles: the separated partial positive and negative charges. A polar molecule results when a molecule contains polar bonds in an unsymmetrical arrangement. Nonpolar molecules are of two types. Molecules whose atoms have equal or nearly equal electronegativities have zero or very small dipole moments. A second type of nonpolar molecule has polar bonds, but the molecular geometry is symmetrical allowing the bond dipoles to cancel each other out. | 1,518 | 890 |
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Thus, \(2H_{\large\textrm f} - (2H_{\large\textrm{sub}} + H_{\large\textrm{vap}} + 2IP + D) = 2EA + 2E_{\large\textrm{cryst}}\) Note that we have used two moles of \(\ce{NaBr}\) in the above diagram. This scheme shows that we can calculate the lattice energy of \(\ce{NaBr}\) from some known thermodynamic data. The same can be calculated from reaction equations and their associated energies. This is illustrated below Thus, \(2 H_{\textrm f} - 2H_{\textrm{sub}} - 2 IP - H_{\textrm{vap}} - D - 2 EA = 2 E_{\textrm{cryst}}\)
\(E_{\textrm{cryst}} = H_{\textrm f} - H_{\textrm{sub}} - IP - \dfrac{(H_{\textrm{vap}} + D)}{2} - EA\) This is the same result as shown in the diagram. \(\begin{align}
&E_{\large\textrm{cryst}}
= \mathrm{-411-\left(108+496+\dfrac{244}{2}\right)-(-349)\: kJ/mol}\\
&\hspace{44px} = \mathrm{-788\: kJ/mol}\\
&U = - E_{\large\textrm{cryst}}\\
&\hspace{12px} = \textrm{788 kJ/mol (lattice energy)}
\end{align}\)
The value calculated for depends on the data used. Data from various sources differ slightly, and so does the result. The lattice energies for \(\ce{NaCl}\) most often quoted in other texts is about 765 kJ/mol. Compare with the method shown below There is a another method based on principles of physics to evaluate the lattice energy, and some examples are given in the discussion enthalpy of hydration and lattice energy. The Born-Haber cycle enables us to calculate lattice energies of various compounds. For salts containing polyatomic ions, the Born-Haber cycle is not as useful. Some other means have to be used to evaluate the lattice energy or energy of crystallization. The lattice energies of some salts are given in the table on the right here. Among the mono-valent salts, the lattice energies decrease when the sizes of the ions increase. Comparing the lattice energy for salts with one divalent ion leads to the same conclusion. The lattice energies decrease when the sizes of the ions increase. The lattice energy of salts involving a divalent ion are much higher than those of monovalent salts, because much more energy is required to separate these ions. | 2,137 | 891 |
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In this section we shall be mainly interested in the reactions of arenes that involve attack on the carbon atoms of the aromatic ring. We shall not elaborate now on the reactions of substituent groups around the ring. The principal types of reactions involving aromatic rings are substitution, addition, and oxidation. Of these, the most common type is electrophilic substitution. A summary of the more important substitution reactions of benzene is given in Figure 22-7. Many of the reagents used to achieve these substitutions will be familiar to you in connection with electrophilic addition reactions to alkenes (e.g., \(\ce{Cl_2}\), \(\ce{Br_2}\), \(\ce{H_2SO_4}\), and \(\ce{HOCl}\); ). Electrophilic addition to alkenes and electrophilic aromatic substitution are both polar, stepwise processes, and the key step for each is attack of an electrophile at carbon to form a cationic intermediate. We may represent this type of reaction by the following general equations, in which the attacking reagent is represented either formally as a cation, \(\ce{X}^\ominus\), or as a neutral but polarized molecule, \(\overset{\delta \oplus}{\ce{X}}\)---\(\overset{\delta \ominus}{\ce{Y}}\): (first step) (first step) The intermediate shown for aromatic substitution no longer has an aromatic structure; rather, it is a cation with four \(\pi\) electrons delocalized over five carbon nuclei, the sixth carbon being saturated with \(sp^3\)-hybrid bonds. It may be formulated in terms of the following contributing structures, which are assumed to contribute essentially equally: The importance of writing the hybrid structure with the partial charges at these three positions will become evident later. This kind of ion is referred to as a or a . The aromatic ring is regenerated from this cationic intermediate by loss of a proton from the \(sp^3\)-hybridized carbon. The electron pair of this \(\ce{C-H}\) bond then becomes part of the aromatic \(\pi\)-electron system and a substitution product of benzene, \(\ce{C_6H_5X}\), is formed. (second step) The gain in stabilization attendant on regeneration of the aromatic ring is sufficiently advantageous that this, rather than combination of the cation with \(\ce{Y}^\ominus\), normally is the favored course of reaction. Herein lies the difference between aromatic substitution and alkene addition. In the case of alkenes there usually is no substantial resonance energy to be gained by loss of a proton from the intermediate, which tends therefore to react by combination with a nucleophilic reagent. (second step) \[\overset{\oplus}{\ce{C}} \ce{H_2-CH_2X} + \ce{Y}^\ominus \rightarrow \ce{YCH_2-CH_2X}\] It is important to realize that in aromatic substitution the actual electrophilic substituting agent, \(\overset{\oplus}{\ce{X}}\) or \(\overset{\delta \oplus}{\ce{X}}-\overset{\delta \ominus}{\ce{Y}}\), is not necessarily the reagent that is added to the reaction mixture. For example, nitration in mixtures of nitric and sulfuric acids is not brought about by attack of the nitric acid molecule on the aromatic compound, but by attack of a more electrophilic species, the nitronium ion, \(\ce{NO_2^+}\). This ion is formed from nitric acid and sulfuric acid according to the following equation: \[\ce{HNO_3} + 2 \ce{H_2SO_4} \rightleftharpoons \ce{NO_2^+} + \ce{H_3O^+} + 2 \ce{HSO_4^-}\] The nitronium ion attacks the aromatic ring to give first a nitrobenzenium ion and then an aromatic nitro compound: In general, the function of a catalyst (which is so often necessary to promote aromatic substitution) is to generate an electrophilic substituting agent from the given reagents. Thus it is necessary to consider carefully for each substitution reaction what the actual substituting agent may be. This problem does not arise to the same degree in electrophilic additions to alkenes, because alkenes are so much more reactive than arenes that the reagents employed (e.g., \(\ce{Br_2}\), \(\ce{Cl_2}\), \(\ce{HCl}\), \(\ce{HOCl}\), \(\ce{HOBr}\), \(\ce{H_3O}^\oplus\)) themselves are sufficiently electrophilic to react with alkenes without the aid of a catalyst. In fact, conditions that lead to substitution of arenes, such as nitration in mixtures of nitric and sulfuric acid, often will degrade the carbon skeleton of alkenes. Now we shall consider the individual substitution reactions listed in Figure 22-1 with regard to the nature of the substituting agent and the utility for synthesis of various classes of aromatic compounds. The nitronium ion, \(\ce{NO_2^+}\), is the active nitrating agent in nitric acid-sulfuric acid mixtures. The nitration of methylbenzene (toluene) is a typical example of a nitration that proceeds well using nitric acid in a 1:2 mixture with sulfuric acid. The nitration product is a mixture of 2-, 3-, and 4-nitromethylbenzenes: The presence of appreciable amounts of water in the reaction mixture is deleterious because water tends to reverse the reaction by which nitronium ion is formed: \[\ce{NO_2^+} + \ce{H_2O} \overset{\ce{HSO_4^-}}{\rightleftharpoons} \ce{HNO_3} + \ce{H_2SO_4}\] For this reason the potency of a nitric-sulfuric acid mixture can be considerably increased by using fuming nitric and fuming sulfuric acids. With such mixtures nitration of relatively unreactive compounds can be achieved. For example, 4-nitromethylbenzene is far less reactive than methylbenzene, but when heated with an excess of nitric acid in fuming sulfuric acid, it can be converted successively to 2,4-dinitromethylbenzene and to 2,4,6-trinitromethylbenzene (TNT): There are several interesting features about the nitration reactions thus far discussed. For instance, the conditions required for nitration of 4-nitromethylbenzene would rapidly oxidize an alkene by cleavage of the double bond: Also the mononitration of methylbenzene does not lead to equal amounts of the three possible products. The methyl substituent apparently orients the entering substituent preferentially to the 2 and 4 positions. This aspect of aromatic substitution will be discussed in in conjunction with the effect of substituents on the reactivity of aromatic compounds. Some compounds are sufficiently reactive that they can be nitrated with nitric acid in ethanoic acid. Pertinent examples are 1,3,5-trimethylbenzene and naphthalene: Other convenient nitrating reagents are benzoyl nitrate, \(\ce{C_6H_5COONO_2}\), and ethanoyl nitrate, \(\ce{CH_3COONO_2}\). These reagents provide a source of \(\ce{NO_2^+}\) and have some advantage over \(\ce{HNO_3} \cdot \ce{H_2SO_4}\) mixtures in that they are soluble in organic solvents such as ethanenitrile or nitromethane. Having homogeneous solutions is especially important for kinetic studies of nitration. The reagents usually are prepared in solution as required from the corresponding acyl chloride and silver nitrate or from the acid anhydride and nitric acid. Such reagents are hazardous materials and must be handles with care. Nitronium salts of the type \(\ce{NO_2^+} \ce{X^-}\) are very powerful nitrating agents. The counterion, \(\ce{X^-}\), must be non-nucleophilic and usually is fluoroborate, \(\ce{BF_4^-}\) or \(\ce{SbF_4^-}\): To some degree we have oversimplified electrophilic substitution by neglecting the possible role of the 1:1 charge-transfer complexes that most electrophiles form with arenes (see for discussion of analogous complexes of alkenes): With halogens, especially iodine, complex formation is visually evident from the color of solutions of the halogen in arenes. Although complex formation may assist substitution by bringing the halogen and arene in close proximity, substitution does not necessarily occur. A catalyst usually is required, and the catalysts most frequently used are metal halides that are capable of accepting electrons (i.e., Lewis acids such as \(\ce{FeBr_3}\), \(\ce{AlCl_3}\), and \(\ce{ZnCl_2}\)). Their catalytic activity may be attributed to their ability to polarize the halogen-halogen bond in the following way: \[\overset{\delta \oplus}{\ce{Br}} \cdots \overset{\delta \ominus}{\ce{Br}} \cdots \ce{FeBr_3}\] The positive end of the dipole attacks the aromatic compound while the negative end is complexed with the catalyst. We can represent the reaction sequence as follows, with the slow step being formation of a \(\sigma\) bond between \(\ce{Br}^\oplus\) and the aromatic ring: The order of reactivity of the halogens is \(\ce{F_2} > \ce{Cl_2} > \ce{Br_2} > \ce{I_2}\). Fluorine is too reactive to be of practical use for the preparation of aromatic fluorine compounds and indirect methods are necessary (see ). Iodine usually is unreactive. It has been stated that iodination fails because the reaction is reversed as the result of the reducing properties of the hydrogen iodide that is formed: \[\ce{C_6H_6} + \ce{I_2} \overset{\rightarrow}{\longleftarrow} \ce{C_6H_5I} + \ce{HI}\] This view is not correct because, as Kekule himself showed, iodobenzene is not reduced by hydroiodic acid except at rather high temperatures. The way to achieve direct iodination in the absence of powerful activating substituent groups is to convert molecular iodine to some more active species (perhaps \(\ce{H_2OI}^\oplus\) or \(\ce{I}^\oplus\)) with an oxidizing agent such as nitric acid or hydrogen peroxide: \[\begin{align} \ce{I_2} + 4 \ce{HNO_3} &\rightarrow 2 \ce{H_2O-I^+} + 2 \ce{NO_2} + 2 \ce{NO_3^-} \\ \ce{I_2} + \ce{H_2O_2} + 2 \ce{H^+} &\rightarrow 2 \ce{H_2OI^+} \end{align}\] With combinations of this kind good yields of iodination products are obtained: Halogen substitution reactions with chlorine or bromine must be carried out with adequate protection from strong light. If such precautions are not taken, an benzene will react rapidly with halogen by a photochemical process to substitute a hydrogen of the alkyl group rather than of the aromatic ring. The reaction has a light-induced, radical-chain mechanism of the kind discussed for the chlorination of propene ( ). Thus methylbenzene reacts with bromine when illuminated to give phenylmethyl bromide; but when light is excluded and a Lewis acid catalyst is present, substitution occurs to give principally the 2- and 4-bromomethylbenzenes. Much less of the 3-bromomethylbenzene is formed: Benzene itself can be induced to halogens on strong irradiation to give polyhalocyclohexanes (see and ): An important method of synthesis of alkylbenzenes utilizes an alkyl halide as the alkylating agent and a metal halide, usually aluminum chloride, as catalyst: This class of reaction is called in honor of its discoverers, C. Friedel (a French chemist) and J. M. Crafts (an American chemist). The metal-halide catalyst functions much as it does in halogenation reactions to provide a source of a positive substituting agent, which in this case is a carbocation: Alkylation is not restricted to alkyl halides; alcohols and alkenes may be used to advantage in the presence of acidic catalysts such as \(\ce{H_3PO_4}\), \(\ce{H_2SO_4}\), \(\ce{HF}\), \(\ce{BF_3}\), or \(\ce{HF-BF_3}\). Ethylbenzene is made commercially from benzene and ethene using phosphoric acid as the catalyst. Isopropylbenzene is made similarly from benzene and propene: Under these conditions the carbocation, which is the active substituting agent, is generated by protonation of the alkene: \[\begin{align} \ce{CH_2=CH_2} + \ce{H^+} &\rightleftharpoons \ce{CH_3CH_2^+} \\ \ce{CH_3CH=CH_2} + \ce{H^+} &\rightleftharpoons \ce{CH_3} \overset{+}{\ce{C}} \ce{HCH_3} \end{align}\] With alcohols the electrophile can be formed by initial protonation by the acid catalyst and subsequent cleavage to a carbocation: There are several factors that limit the usefulness of alkylation reactions. First, it may be difficult to limit reaction to monosubstitution because introduction of one alkyl substituent tends to activate the ring towards a second substitution (see ). Therefore, to obtain reasonable yields of a monoalkylbenzene, it usually is necessary to use a large excess relative to the alkylating agent: A second limitation is the penchant for the alkylating reagent to give rearrangement products. As an example, the alkylation of benzene with 1-chloropropane leads to a mixture of propylbenzene and isopropylbenzene. We may write the reaction as first involving formation of a propyl cation, which is a carbocation: \[\ce{CH_3CH_2CH_2Cl} + \ce{AlCl_3} \rightarrow \ce{CH_3CH_2CH_2^+} + \overset{-}{\ce{Al}} \ce{Cl_4}\] This ion either can alkylate benzene to give propylbenzene, \[\ce{C_6H_6} + \ce{CH_3CH_2CH_2^+} \rightarrow \ce{C_6H_5CH_2CH_2CH_3} + \ce{H^+}\] or it can rearrange to a more stable secondary ion by the transfer of a hydrogen from a neighboring carbon together with its bonding electron pair (i.e., 1,2-hydride shift). The positive charge is thereby transferred from \(\ce{C_1}\) to \(\ce{C_2}\): Alkylation of benzene with the isopropyl cation then produces isopropylbenzene: \[\ce{C_6H_6} + \ce{CH_3} \overset{\oplus}{\ce{C}} \ce{HCH_3} \rightarrow \ce{C_6H_5CH(CH_3)_2} + \ce{H}^\oplus\] Rearrangements of this type involving carbocation intermediates often occur in Friedel-Crafts alkylations with primary and secondary alkyl groups larger than \(\ce{C_2}\) and \(\ce{C_3}\). Related carbocation rearrangements are discussed in and . Further complications arise from the fact that the alkylation reactions sometimes are under equilibrium control rather than kinetic control. Products often isomerize and disproportionate, particularly in the presence of large amounts of catalyst. Thus 1,2- and 1,4-dimethylbenzenes ( - and -xylenes) are converted by large amounts of Friedel-Crafts catalysts into 1,3-dimethylbenzene ( -xylene): Ethylbenzene disproportionates under the influence of excess \(\ce{HF-BF_3}\) to benzene and 1,3-diethylbenzene: Acylation and alkylation of arenes are closely related. Both reactions were developed as the result of the collaboration between Friedel and Crafts, in 1877. The acylation reaction introduces an acyl group, \(\ce{RCO}\), into an aromatic ring and the product is an aryl ketone: The acylating reagents commonly used are carboxylic acid halides, \(\ce{RCOCl}\), anhydrides, \(\ce{(RCO)_2O}\), or the acid itself, \(\ce{RCO_2H}\). A strong proton or other Lewis-acid catalyst is essential. The catalyst functions to generate the acyl cation: The catalyst most commonly used with acyl halides and anhydrides is aluminum chloride: Acylation differs from alkylation in that the reaction usually is carried out in a solvent, commonly carbon disulfide, \(\ce{CS_2}\), or nitrobenzene. Furthermore, acylation requires more catalyst than alkylation, because much of the catalyst is tied up and inactivated by complex formation with the product ketone: Unlike alkylation, acylation is controlled easily to give monosubstitution, because once an acyl group is attached to a benzene ring, it is not possible to introduce a second acyl group into the same ring. Because of this, a convenient synthesis of alkylbenzenes starts with acylation, followed by reduction of the carbonyl group with zinc and hydrochloric acid ( ). For example, propylbenzene is prepared best by this two-step route because, as we have noted, the direct alkylation of benzene with propyl chloride produces considerable amounts of isopropylbenzene and polysubstitution products: In the acylation of alkylbenzene the product almost always is the isomer. The synthesis of (4- -butylphenyl)ethanone illustrates this as well as the sequential use of alkylation and acylation reactions: Chemists are inclined to give names to reactions that associate them either with their discoverers or with the products they give. This practice can be confusing because many named reactions (or "name reactions") which once were thought to be quite unrelated, have turned out to have very similar mechanisms. Thus we have two closely related acylation reactions: one is the Friedel-Crafts ketone synthesis, in which the electrophile is \(\ce{R} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O}\); and the other is the , in which the electrophile is \(\ce{H} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O}\): The latter reaction utilizes carbon monoxide and \(\ce{HCl}\) under pressure in the presence of aluminum chloride. The electrophile may be considered to be formed as follows: \[\ce{C=O} + \ce{HCl} + \ce{AlCl_3} \rightleftharpoons \ce{H} \ce{-} \overset{\oplus}{\ce{C}} \ce{=O} + \overset{\ominus}{\ce{Al}} \ce{Cl_4}\] Substitution of the sulfonic acid \(\left( \ce{-SO_3H} \right)\) group for a hydrogen of an aromatic hydrocarbon can be carried out by heating the hydrocarbon with a slight excess of concentrated or fuming sulfuric acid: The actual sulfonating agent normally is the \(\ce{SO_3}\) molecule, which, although neutral, has a powerfully electrophilic sulfur atom: Sulfonation is reversible and the \(\ce{-SO_3H}\) group can be removed by hydrolysis at \(180^\text{o}\): A useful alternative preparation of sulfonyl derivatives is possible with chlorosulfonic acid: This procedure has an advantage over direct sulfonation in that sulfonyl chlorides usually are soluble in organic solvents and may be easily separated from the reaction mixture. Also, the sulfonyl chloride is a more useful intermediate than the sulfonic acid, but can be converted to the acid by hydrolysis if desired: Sulfonation is important in the commercial production of an important class of detergents - the sodium alkylbenzenesulfonates: The synthesis illustrates several important types of reactions that we have discussed in this and previous chapters. First, the alkyl group \(\ce{R}\) usually is a \(\ce{C_{12}}\) group derived from the straight-chain hydrocarbon, dodecane, which on photochlorination gives a mixture of chlorododecanes: This mixture of chlorododecanes is used to alkylate benzene, thereby giving a mixture of isomeric dodecylbenzenes, called : Sulfonation of the detergent alkylate gives exclusively the 4-dodecylbenzenesulfonic acids, which with sodium hydroxide form water-soluble dodecylbenzenesulfonates: In many countries it is prohibited by law to market detergents of this type, which have highly branched alkyl groups. The reason is that quaternary carbons and, to a lesser extent, tertiary carbons are not degraded readily by bacteria in sewage treatment plants: It is possible to replace the ring hydrogens of many aromatic compounds by exchange with strong acids. When an isotopically labeled acid such as \(\ce{D_2SO_4}\) is used, this reaction is an easy way to introduce deuterium. The mechanism is analogous to other electrophilic substitutions: Perdeuteriobenzene\(^3\) can be made from benzene in good yield if a sufficiently large excess of deuteriosulfuric acid is used. Deuteration might appear to be competitive with sulfonation, but deuteration actually occurs under much milder conditions. Because metals are electropositive elements they can be considered potential electrophiles. Their reactions with arenes have been investigated most thoroughly for mercury. Benzene can be substituted with \(\ce{HgX}^\oplus\) derived from a mercuric salt, \(\ce{HgX_2}\), in the presence of an acid catalyst. The salt most commonly used is mercuric ethanoate, \(\ce{Hg(OOCCH_3)_2}\). The catalyst is considered to function by assisting the generation of the active electrophile, \(\ce{HgX}^\oplus\). Other metals that may be introduced directly into an aromatic ring in this manner include thallium and lead. \(^3\)The prefix , as in perdeuterio- or perfluoro-, means that the hydrogens have been replaced with the named substituent, \(\ce{D}\) or \(\ce{F}\). Perhydro means saturated or fully hydrogenated. and (1977) | 19,700 | 892 |
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When the wavefunctions that solve Schrödinger's equation \(\psi(x,y,z)\) are expressed in spherical coordinates, then the wavefunctions can be separated into a radial and angular components \[ \psi(r,\theta,\phi) =R(r) Y(\theta, \phi) \label{1.1}\] The \(Y (\theta , \varphi )\) functions provide information about where the electron is around the proton and the radial function \(R(r)\) describes how far the electron is away from the proton. Both \(R(r)\) and \(Y (\theta , \varphi )\) depend on all three quantum numbers \(n, l, m_l\), although energy of each wavefunction (\(E_n\)) depends on only \(n\). In general, the radial wavefunction (\(R(r)\)) can be expressed as \[ R(r) = Q(r) e^{-r/a} \label{1.2}\] where \(Q(r)\) is a polynomial. \(R(r)\) is typically normalized so that it can used to calculate electron probability just like proper wavefunction (but only with respect to the radial, \(r\) dimension only). A more complete 3D description requires addressing the complete wavefunction in Equation \(\ref{1.1}\). Since the square of a particle's wave function at some location yields the probability that the particle is at that location, one might think that we can determine the probability that the electron is within some range of radii by integrating; however this is incorrect. \[ P ( r_1 < r < r_2) \neq \int_{r_1}^{r_2} R(r)^2 dr = \int_{r_1}^{r_2} Q(r)^2 e^{-2r/a} dr \label{1.3}\] The trouble is that once we leave the familiar territory of Cartesian coordinates and venture into the Land of Spherical Coordinates, the volume element changes. In one dimension, things are dead simple: a single coordinate describes everything. If we want to integrate, we must move a little extra distance in all dimensions ... which means a little extra distance . That's all there is. The "differential volume element" is simply \[ dV = dx\] Actually this is the differential , but the connection to volume is better described below. In two dimensions, thing are more interesting because there are several ways to identify a location in space. The Cartesian system uses variables and . To integrate, we move a little extra distance in each dimension: This sweeps out a little square box. The differential volume element is the area of this box: \[ dV = dx\, dy\] Actually, this is the , but the connection to volume is described below. Now, we can also use polar coordinates and to describe the location of a point in a plane. If we increase each of these variables by a teeny bit to integrate ... ... we sweep out an area which is not square. Instead, it has a curve along the inside and outside edges. The area of this curvy box, and thus the differential volume element in polar coordinates on a plane, is \[ dV = dr\, r d\theta\] Actually, this is the . I hope you'll believe me when I say that the differential volume element in three-dimensional Cartesian coordinates is \[dV = dx\,dx\,dz\] On the other hand, in three-dimensional spherical coordinates, we describe the position of an object with radius \(r\) and the angles \(θ\) and \(φ\). If we extend each of these three variables just a bit, we sweep out a funnel-shaped box. In one direction, the box has straight lines of length . In the other directions, the edges are curved. We can compute the length of these tiny arcs using the arclength formulae in spherical coordinates, and end up with a differential volume element \[ dV = (dr) (rd\theta)(r \sin \theta \,d \phi)\] \[ = r^2 \sin \theta \, dr\, d \theta \, d \phi\] So, back to the electron within the hydrogen atom. The probability the electron is at location \((r, θ, φ)\) within a tolerance of \((dr, dθ, dφ)\) is \[ P ( r; \theta; \phi) = \int (\psi( r; \theta; \phi))^2 dV\] \[ P ( r; \theta; \phi) = \int (\psi( r; \theta; \phi))^2 r^2 \sin \theta \, dr\, d \theta \, d \phi \label{Int1}\] If we can separate the wavefunction \(\ref{1.1}\) further into the product of functions, each one dealing with only a single variable, \[ \psi(r,\theta,\phi) =R(r) \Theta(\theta) \Phi(\phi) \] then the integral in Equation \(\ref{Int1}\) becomes \[ P ( r; \theta; \phi) = \int ( R(r) \Theta(\theta) \Phi(\phi) )^2 r^2 \sin \theta \, dr\, d \theta \, d \phi \label{Int2}\] \[ = \int (R(r)^2 r^2 dr) (\Theta(\theta)^2 \sin \theta d \theta ) ( \Phi(\phi)^2 d \phi) \label{Int3}\] \[ = \int R(r)^2 r^2 dr \int \Theta(\theta)^2 \sin \theta d \theta \int \Phi(\phi)^2 d \phi \label{Int4}\] In the simplest case, corresponding to the ground state \(1s\), the wave function depend on either angle. That means the angular functions in Equation \(\ref{Int4}\) are simply equal to 1; and that makes doing those integrals trivial. \[ \int 1 \sin \theta d\theta \label{Int5}\] \[ \int 1 d\phi \label{Int6}\] What are the limits and values of the integrals in Equations \(\ref{Int5}\) and \(\ref{Int6}\)? So the probability of finding the electron within some range of radii (Equation \(\ref{1.3}\)) becomes \[ P ( r_1 < r < r_2) = 4\pi \int_{r_1}^{r_2} R(r)^2 r^2 dr \label{Int7}\] Recall that the radial function (Equation \(\ref{1.2}\)) can be separated into two compontents. For the electron in the 1s wavefunction, \(Q\) is a and then \[ R(r) = Q e^{-r/a}\] We can simplify this just a little bit by making a substitution: let \[u = \dfrac{2r}{a}.\] then \[ du =\dfrac{2}{a} dr\] and Equation \(\ref{Int7}\) can be rewritten as \[ P ( r_1 < r < r_2) = 4\pi \left(Q^2 \dfrac{a^3}{8} \right) \int_{2r_1/a}^{2r_2/a} u^2 e ^{-u} du \label{Int8a}\] You can look up that integral in a table; the result isn't too ugly. \[ P ( r_1 < r < r_2) = 4\pi \left(Q^2 \dfrac{a^3}{8} \right) \left[ -e^{-u}(u^2 + 2u +2) \right]_{2r_1/a}^{2r_2/a} \label{Int9a}\] So, where is the electron most likely to be? While the \(R^2(r)\) plot (Figure 1; red curve) tracks the probability along a vector away from the nucleus, the \(r^2r^2(r)\) plot (blue curve) tracks the probability of a finding the electron at a radius \(r\). The peak of this curve is the MOST LIKELY radius for the electron. You can find more examples of wave functions and probability distributions in textbooks and on web sites. Below are some examples made using Paul Fastad's hydrogen atom applet. In each example, the radial function is shown in the top-right panel of the figure below, labelled "Rnl(r)". The bottom-left panel provides an illustration of this function, looking down at the atom from above. | 6,410 | 893 |
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With \(U\), \(A\), \(H\) and \(G\) in hand we have potentials as a functions of whichever variable pair we want: \(S\) and \(V\), to \(T\) and \(P\). Additional Legendre transforms will provide us with further potentials in case we have other variables (such as surface area \(A\), length \(L\), magnetic moment \(M\), etc.). Thermodynamic problems always involve computing a variable of interest. It may be a derivative if it is an intensive variable, or even a second derivative (higher derivatives are rarely of interest). 1 order ones like \[\left( \dfrac{\partial G}{\partial P} \right)_T=V\] or 2 order ones like \[ \left( \dfrac{\partial^2 G}{\partial P^2} \right)_T= \left( \dfrac{\partial V}{\partial P} \right)_T = \kappa V\] The solution procedure is thus: Easy as 1-2-3! We now turn to two methods to manipulate the thermodynamic derivations: | 874 | 894 |
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The central application of statistical mechanics rests on the assumption that the average of a property over a large number of systems will give the same value as the thermodynamic quantity of interest. We can distinguish between mechanical properties such as pressure, energy, volume etc. and non-mechanical properties such as entropy. Although there are a large number of particles and an extremely large number of quantum states accessible to even a small system, the state of the system can be characterized by just a few thermodynamic variables. These statements are known as the Gibbs postulate: The ensemble average of a property corresponds to the thermodynamic quantity. For example, the average energy corresponds to the internal energy, the average pressure corresponds to the thermodynamic pressure, etc. To accomplish the kind of averaging needed to calculate the pressure, for example, we must consider a large number of systems. The concept of a collection of systems, or an ensemble was first introduced by Gibbs. An ensemble consists of a very large number of systems, each constructed to be a replica on the macroscopic level. We will introduce several types of ensemble in this course depending on which variables are held fixed. Corresponding to each ensemble there is a partition function that represents the average number of states accessible at a given temperature. Symbol Ensemble Name Fixed Variables Microcanonical N, V, E Q Canonical N, V, T Isobaric-isothermal N, P, T Grand canonical , V, T The magnitude of the partition function, (E) for the microcanonical ensemble is the same as the degeneracy of the system at the given energy, E. The principle of equal a priori probabilities states that each and every quantum state of the system must be represented an equal number of times. Another way to put this is to state that in an isolated system (N, V, E, fixed) any one of the possible quantum states is equally likely. The partition function of the microcanonical ensemble is the same as the thermodynamic probability W introduced in the Boltzmann equation for the statistical entropy, S = k ln W. The same equation holds for the microcanonical ensemble, S = k ln and thus represents a direct thermodynamic connection between the partition function and the entropy. The difficulty with application of this information is that it is very difficult to obtain a set of molecules at a constant energy. In spite of this experimental difficulty, the microcanonical ensemble is useful for illustrating the number of degenerate (equal energy) states in systems of interest. The conclusion from these investigations will be that the number of quantum states accessible to a system is vast and that provides the motivation for application of statistical techniques to the calculation of average quantities, fluctuations and transport properties. We consider the three types of molecular motion From quantum mechanics we have H = E where H is the hamiltonian where h is h/2 , that is, Planck's constant divided by 2 . The hamiltonian comprises both the kinetic and potential energy of the system. For the purposes of statistical mechanics it is the energy levels and their degeneracy that are of interest. The wavefunctions, do not appear in averages and are therefore not given below. Translation is calculated using the particle-in-a-box treatment. Setting the potential U(x) = 0 over the range x = 0 to a and U(x) equal to infinity outside this range, for one dimension the hamiltonian, the energy levels are . The extension to three dimensions is easily made and is shown below. The classical Hooke's law potential is 1/2 kx and this is exactly what is used as the potential in the quantum mechanical hamiltonian. In one dimension this becomes The energy levels are where = (k/m) . The rigid rotor hamiltonian is The energy levels are I is the moment of inertia of the rotor. The essence of statistical mechanics is to connect these quantum mechanical energy levels to the macroscopically measured thermodynamic energies, pressure, and entropy. There are two important aspects of these energy levels. First, there is a ladder of increasing energy states. Second, in some cases there is a degeneracy associated with the states. For the rigid rotor solutions the degeneracy is 2J + 1. In the case of the solutions for the particle-in-a-box there is an enormous degeneracy because of the three dimensional solution. This is important for understanding ensembles and the strategy of statistical mechanics. The solution of the particle-in-a-box problem in three dimensions is The degeneracy is given by the number of ways that the integer M = 8ma /h can be written as the sum of squares of three positive integers. The degeneracy becomes a smooth function for large M. Consider a three-dimensional space spanned by n , n , and n . There is a one-to-one correspondence between energy states given by above energy equation and the points in this space. A radius in this space is given by R = n + n + n so that R = (8ma /h ) . We wish to calculate the number of lattice points that are at some fixed distance from the origin in this space. In practice, this means that we want the number of states between energy and + . To obtain the total number of states with energy less than we consider the volume of one octant (recall that the quantum numbers must be positive). This number of states is: The number of states between and + is Expand + about = 0 as follows Keeping only the first two terms we have This derivation is valid for the degeneracy of single particle. A simple calculation taking = 3kT/2 6 x 10 J, m = 10 kg, a = 1 m and = 10 gives ( ) 10 for a single particle in a of one part per billion about the energy . For an N particle system, the degeneracy is tremendously greater. To see this consider N noninteracting particles in a cube. The energy of the system is Defining the space of the quantum numbers as an N-dimensional sphere, the number of states with energy less than E is where (n) is the gamma function. The number of states between E and E + E is Even though there are a large number of levels and we will assume that the they are all equally probable, there is an energetic constraint on the system. This leads to the concept of a most probable distribution among these levels. Although the microcanonical ensemble is useful for illustrating the number of levels, the most convenient ensemble for determining the most probable distribution is the canonical ensemble (constant N, V, and T). | 6,614 | 895 |
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We have seen that the absorption of photons (especially in the ultraviolet-visible spectrum) is connected to the excitation of electrons. After excitation, a number of different relaxation pathways lead back to the ground state. Sometimes, absorption of a photon leads to a vastly different outcome. Instead of just relaxing again, the molecules may undergo bond-breaking reactions, instead. An example of this phenomenon is observed in the complex ion [Co(NH )] . Addition of UV light to this complex results in loss of ammonia. In the absence of UV light, however, the complex ion is quite stable. In many cases, loss of a ligand is followed by replacement by a new one. For example, if an aqueous solution of [Co(NH )] is photolysed, an ammonia ligand is easily replaced by water. Draw a d orbital splitting diagram for [Co(NH )] . Explain why this complex is normally inert toward substitution. Use the d orbital splitting diagram for [Co(NH )] to explain why this complex is undergoes substitution upon irradiation with UV light. Photolysis is the term used to describe the use of light to initiate bon-breaking events. Photolysis frequently involves the use of high-intensity ultraviolet lamps. The high intensity light is needed in order to provide enough photons to get higher conversion of reactant into a desired product. , | 1,348 | 898 |
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As with diatomic molecules, the energies of polyatomic molecules can be approximated by the sum of its individual degrees of freedom. Therefore, we can write the partition function as: \[ Q(N,V,T) = \frac{[q(V,T)]^2}{N!} \] We can write the polyatomic analog to diatomic molecules: \[ Q(N,V,T) = \frac{(q_\text{trans}q_\text{rot}q_\text{vib}q_\text{elec})^N}{N!} \] When derived for diatomic molecules, we assumed the rigid rotor model for rotations and the harmonic oscillator for vibrations. This allowed us to separate the rotational motion from the vibrational motion of the molecule. Polyatomic molecules are a bit more complicated, but we will still make use of these approximations. The number of translational states available to any given molecule is far greater than the number of molecules in the system. \(q_\text{trans}\) is given by: \[ q_\text{trans} = \left[\frac{2\pi MkT}{h^2}\right]^{3/2}V \] where \(M\) is the mass of the particle. Moving to the electronic partition function: \[ q_\text{elec} = \sum_i{g_{i}e^{-E_i/kT}} \] where \(E_i\) is the energy of the electronic state \(i\) and {g_i} is its degeneracy. Electronic states are typically spaced far apart from each other. The probability of the system being in any state but the ground state is extremely small. We can therefore simplify the electronic partition function to include only the ground electronic state: \[ q_\text{elec} = g_{e1}e^{D_e/kT} \] where \(-D_e\) is the energy of the ground electronic state. To complete the polyatomic partition function, we still need \(q_\text{vib}\) and \(q_\text{rot}\). We will finish this section with \(q_\text{vib}\) and talk about \(q_\text{rot}\) in the next. The vibrational motion of diatomic molecules can be expressed as a set of independent harmonic oscillators. For polyatomic molecules, the independent vibrational motions are referred to as normal modes of vibration. The vibrational energy is then the sum of the energies for each normal mode: \[ E_\text{vib} = \sum_i^\alpha{\left(v_i + \frac{1}{2}\right)h\nu} \] where \(\nu_i\) is the vibrational frequency for the \(i\)th normal mode and \(\alpha\) is the number of vibration degrees of freedom. A linear molecule has \(3n-5\) vibrational degrees of freedom and a nonlinear molecule has \(3n-6\) vibrational degrees of freedom. Because the normal modes are independent of each other, we can take out results from previous sections: \[ q_\text{vib} = \prod_i^\alpha{\frac{e^{-\theta_{\text{vib},i}/2T}}{1-e^{-\theta_{\text{vib},i}/T}}} \] \[ E_\text{vib} = Nk \sum_i^\alpha{\left( \frac{\theta_{\text{vib},i}}{2} + \theta_{\text{vib},i} \frac{e^{-\theta_{\text{vib},i}/T}}{1-e^{-\theta_{\text{vib},i}/T}} \right)} \] \[ C_{V,\text{vib}} = Nk \sum_i^\alpha{\left( \left(\frac{\theta_{\text{vib},i}}{2}\right)^2 + \frac{e^{-\theta_{\text{vib},i}/T}}{(1-e^{-\theta_{\text{vib},i}/T})^2} \right)} \] where \(\theta_{\text{vib},i}\) is the characteristic vibrational temperature defined by: \[ \theta_{\text{vib},i} =\frac{h\nu_i}{k} \] There is a great deal of utility for thermodynamic functions calculated from the vibrational normal modes of a molecule. The vibrational energy and entropy depend on the shape a multidimensional potential energy surface. If one performs a conformational search of macromolecule it is one obtains energies and structures but little direct information concerning the shape of the potential energy surface for each conformation. The vibrational entropy gives a means determining whether there are significant entropic differences in the structures and therefore whether certain conformations will be favored based on the entropy. However, it is possible to take appropriate linear combinations of the coordinates so that the cross terms are eliminated and the classical Hamiltonian as well as the operator corresponding to it contains no cross terms and in terms of the new coordinates, the Hamiltonian can be written as, \[ H = \sum_{i=1}^{f} \dfrac{h^2}{2 \mu_i} \dfrac{\partial}{\partial q_i^2}+ \sum_{i=1}^{f} \dfrac{k_i}{2} q_i^2 \label{3.81} \] Here, the degrees of freedom \(f\) is \(3N - 5\) for a linear molecule and \(3N - 6\) for a nonlinear molecule. Here, \(k_i\) is the force constant and \(μ_i\) is the reduced mass for that particular vibrational mode which is referred to as a . The Equation \(\ref{3.81}\) represents \(f\) linearly independent harmonic oscillators and the total energy for such a system is \[ \epsilon_{vib} = \sum_{i=1}^{f} \left( v_i + \dfrac{1}{2} \right) h \nu_i \nonumber \] The vibrational frequencies are given by \[ \nu_i = \dfrac{1}{2 \pi} \sqrt{\dfrac{k_i}{\mu_i}} \nonumber \] The vibrational partition function is given by the product of \(f\) vibrational functions for each frequency. \[ q_{vib} = \prod_{i=1}^f \dfrac{ e^{-\Theta_{vib,i}/2T} }{1- e^{-\Theta_{vib,i}/T}} \label{Qvib1} \] with \[ \Theta_{vib,i} = \dfrac{h\nu_i}{k_B} \nonumber \] As with the previous discussion regarding simple diatomics, \(\Theta_{vib,i}\) is called the . The molar energies and the heat capacities are given by \[ \langle E_{vib} \rangle = Nk \sum_{i=1}^f \left[ \dfrac{ \Theta_{vib,i} }{2} + \dfrac{ \Theta_{vib,i} e^{-\Theta_{vib,i}/T} }{1- e^{-\Theta_{vib,i}/T}} \right] \nonumber \] and \[ \bar{C}_V = Nk_B \sum_{i=1}^f \left( \dfrac{ \Theta_{vib,i} }{T} \right)^2 \dfrac{ e^{-\Theta_{vib,i}/T} }{\left(1- e^{-\Theta_{vib,i}/T}\right)^2} \nonumber \] The three characteristic vibrational temperatures for \(\ce{NO2}\) are 1900 K, 1980 K and 2330 K. Calculate the vibrational partition function at 300 K. The vibrational partition is (Equation \(\ref{Qvib1}\)) \[ q_{vib} = \prod_{i=1}^f \dfrac{ e^{-\Theta_{vib,i}/2T} }{1- e^{-\Theta_{vib,i}/T}} \nonumber \] If we calculate \(q_{vib}\) by taking the zero point energies as the reference points with respect to which the other energies are measured \[\begin{align*} q_{vib} &= \prod_{i=1}^f \dfrac{ 1 }{1- e^{-\Theta_{vib,i}/T}} = \left( \dfrac{ 1 }{1- e^{-1900/300}} \right) \left( \dfrac{ 1 }{1- e^{-1980/300}} \right) \left( \dfrac{ 1 }{1- e^{-2330/300}} \right) \\[4pt] & =(1.0018) ( 1.0014)(1.0004) = 1.0035 \end{align*} \nonumber \] The implication is that very few vibrational states of \(\ce{NO2}\) (other than the ground vibrational state) are accessible at 300 K. This is standard of the vibrations of most molecules. | 6,356 | 900 |
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Let's think first about the interaction of light with matter. We have all seen light shine on different objects. Some objects are shiny and some are matte or dull. Some objects are different colors. Light interacts with these objects in different ways. Sometimes, light goes straight through an object, such as a window or a piece of glass. Because chemical reactions are frequently conducted in solution, we will think about light entering a solution. Imagine sunlight shining through a glass of soda. Maybe it is orange or grape soda; it is definitely coloured. We can see that as sunlight shines through the glass, colored light comes out the other side. Also, less light comes out than goes in. Maybe some of the light just bounces off the glass, but some of it is definitely absorbed by the soda. So, what is the soda made of? Molecules. Some of these molecules are principally responsible for the colour of the soda. There are others, such as the ones responsible for the flavor or the fizziness of the drink, as well as plain old water molecules. The soda is a solution; it has lots of molecules (the solute) dissolved in a solvent (the water). Light is composed of photons. As photons shine through the solution, some of the molecules catch the photons. They absorb the light. Generally, something in the molecule changes as a result. The molecule absorbs energy from the photon and is left in an excited state. The more of these molecules there are in the solution, the more photons will be absorbed. If there are twice as many molecules in the path of the light, twice as many photons will be absorbed. If we double the concentration, we double the absorbance. Alternatively, if we kept the concentration of molecules the same, but doubled the length of the vessel through which the light traveled, it would have the same effect as doubling the concentration. Twice as much light would be absorbed. These two factors together make up part of a mathematical relationship, called Beer's Law, describing the absorption of light by a material: \[ A = \epsilon c l \] in which A = Absorbance, the percent of light absorbed; c = the concentration; l = the length of the light's path through the solution; ε = the "absorptivity" or "extinction coeficient" of the material, which is a measure of how easily it absorbs a photon that it encounters. That last factor, ε, suggests that not all photons are absorbed easily, or that not all materials are able to absorb photons equally well. There are a couple of reasons for these differences. Remember, often a particular soda will absorb light of a particular colour. That means, only certain photons corresponding to a particular colour of light are absorbed by that particular soda. How does that affect what we see? If the red light is being absorbed by the material, it isn't coming back out again. The blue and yellow light still are, though. That means the light coming out is less red, and more yellowy-blue. We see green light emerging from the glass. A "colour wheel" or "colour star" can help us keep track of the idea of complementary colours. When a colour is absorbed on one side of the star, we see mostly the colour on the opposite side of the star. What colour of photon is probably most strongly absorbed by a glass of orange soda? Why do certain materials absorb only certain colours of light? That has to do with the properties of photons. Photons have particle-wave duality, just like electrons. They have wave properties, including a wavelength. That wavelength corresponds to the energy of a photon, according to the Planck-Einstein equation: \[ E = {h c \over \lambda } \] in which E = energy of the photon, h = Planck's constant (6.625 x 10 J s ), c = speed of light (3.0 x 10 m s ), λ = wavelength of light in m. Alternatively, the Planck-Einstien equation can be thought of in terms of frequency of thr photon: as a photon passes through an object, how frequently does one of its "crests" or "troughs" encounter the object? How frequently does one full wavelength of the photon pass an object? That parameter is inversely proportional to the wavelength. The equation becomes: \[ E = h v \] in which ν = the frequency of the photon, in s . The visible spectrum, shown below, contains a very limited range of photon wavelengths, between about 400 and 700 nm. The higher the frequency, the higher the energy of the photon. The longer the wavelength, the lower the energy of the photon. As a result of this relationship, different photons have different amounts of energy, because different photons have different wavelengths. The visible spectrum ranges from photons having wavelengths from about 400 nm to 700 nm. The former is the wavelength of violet light and the latter is the wavlength of red light. Which one has higher energy: a photon of blue light or a photon of red light? Ultraviolet light -- invisible to humans and with wavelengths beyond that of violet light -- is associated with damage to skin; these are the cancer-causing rays from the sun. Explain their danger in terms of their relative energy. Different materials absorb photons of different wavelengths because absorption of a photon is an absorption of energy. Something must be done with that energy. In the case of ultraviolet and visible light, the energy is of the right general magnitude to excite an electron to a higher energy level. However, we know that energy is quantized. That means photons will be absorbed only if they have exactly the right amount of energy to promote an electron from its starting energy level to a higher one (producing an "excited state"). Just like Goldilocks, a photon with too much energy won't do the trick. Neither will a photon with too little. It has to be just right. If the absorption of a UV-visible photon is coupled to the excitation of an electron, what happens when the electron falls back down to the ground state? You might expect a photon to be released. This phenomenon was observed during the late nineteenth century, when scientists studied the "emission spectra" of metal ions. In these studies, the metal ions would be heated in a flame, producing characteristic colours. In that event, the electron would be thermally promoted to a higher energy level, and when it relaxed, a photon would be emitted corresponding to the energy of relaxation. By passing this light through a prism or grating, scientists could separate the observed colour into separate lines of different wavelengths. This evidence led directly to the idea of Niels Bohr and others that atoms had electrons in different energy levels, whci is part of our view of electronic structure today. , | 6,678 | 901 |
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If one can extract all of the vibrational information corresponds a molecule, its molecular structure can then be determined. In the field of spectroscopy, two main techniques are applied in order to detect molecular vibrational motions: (IR) and Raman spectroscopy. Raman Spectroscopy has its unique properties which have been used very commonly and widely in Inorganic, Organic, Biological systems [1] and Material Science [2], [3], etc. Generally speaking, vibrational and rotational motions are unique for every molecule. The uniqueness to molecules are in analogous to fingerprint identification of people hence the term molecular fingerprint. Study the nature of molecular vibration and rotation is particularly important in structure identification and molecular dynamics. Two of the most important techniques in studying vibration/rotation information are IR spectroscopy and Raman spectroscopy. IR is an absorption spectroscopy which measures the transmitted light. Coupling with other techniques, such as Fourier Transform, IR has been highly successful in both organic and inorganic chemistry. Unlike IR, Raman spectroscopy measures the scattered light (Figure 2). There are three types of scattered lights: Rayleigh scattering, Stokes scattering, and anti-stokes scattering. Rayleigh scattering is elastic scattering where there is no energy exchange between the incident light and the molecule. Stokes scattering happens when there is an energy absorption from the incident light, while anti-stokes scattering happens when the molecule emites energy to the incident light. Thus, Stokes scattering results in a red shift, while anti-stokes scattering results in a blue shift. (Figure 1) Stokes and Anti-Stokes scattering are called Raman scattering which can provide the vibration/rotation information The intensity of Rayleigh scattering is around 10 times that of Stokes scattering. [4]According to the Boltzmann distribution, anti-Stokes is weaker than Stokes scattering. Thus, the main difficulty of Raman spectroscopy is to detect the Raman scattering by filtering out the strong Rayleigh scattering. In order to reduce the intensity of the Rayleigh scattering, multiple monochromators are applied to selectively transmit the needed wave range. An alternative way is to use Rayleigh filters. There are many types of Rayleigh filters. One common way to filter the Rayleigh light is by interference. Because of the weakness of Raman scattering, the resolving power of a Raman spectrometer is much higher than an IR specctrometer. A resolution of 10 is needed in Raman while 10 is sufficient in IR. [5] In order to achieve high resolving power, prisms, grating spectrometers or interferometers are applied in Raman instruments. Despite the limitations above, Raman spectroscopy has some advantages over IR spectroscopy as follows: After analysis of the advantages and disadvantages of Raman Spectroscopy technique, we can begin to consider the application of Raman Spectroscopy in inorganic, organic, biological systems and Material Science, etc. X-ray diffraction (XRD) has been developed into a standard method of determining structure of solids in inorganic systems. Compared to XRD, it is usually necessary to obtain other information (NMR, electron diffraction, or UV-Visible) besides vibrational information from IR/Raman in order to elucidate the structure. Nevertheless, vibrational spectroscopy still plays an important role in inorganic systems. For example, some small reactive molecules only exist in gas phase and XRD can only be applied for solid state. Also, XRD cannot distinguish between the following bonds: –CN vs. –NC, –OCN vs. –NCO,–CNO vs. –ONC, -SCN vs. –NCS. [7] Furthermore, IR and Raman are fast and simple analytical method, and are commonly used for the first approximation analysis of an unknown compound. Raman spectroscopy has considerable advantages over IR in inorganic systems due to two reasons. First, since the laser beam used in RS and the Raman-scattered light are both in the visible region, glass (Pyrex) tubes can be used in RS. On the other hand, glass absorbs infrared radiation and cannot be used in IR. However, some glass tubes, which contain rare earth salts, will gives rises to fluorescence or spikes. Thus, using of glass tubes in RS still need to be careful. Secondly, since water is a very weak Raman scatter but has a very broad signal in IR, aqueous solution can be directly analyzed using RS. Raman Spectroscopy and IR have different selection rules. RS detects the polarizability change of a molecule, while IR detects the dipole momentum change of a molecule. Principle about the RS and IR can be found at Chemwiki and . Thus, some vibration modes that are active in Raman may not be active IR, vice versa. As a result, both of Raman and IR spectrum are provided in the stucture study. As an example, in the study of Xenon Tetrafluoride. There are 3 strong bands in IR and solid Raman shows 2 strong bands and 2 weaker bands. These information indicates that Xenon Tetrafluoride is a planar molecule and has a symmetry of D4h. [8] Another example is the application of Raman Spectroscopy in homonuclear diatomic molecules. Homonuclear diatomic molecules are all IR inactive, fortunately, the vibration modes for all the homonuclear diatomic molecules are always Raman Spectroscopy active. Unlike inorganic compounds, organic compounds have less elements mainly carbons, hydrogens and oxygens. And only a certain function groups are expected in organic specturm. Thus, Raman and IR spectroscopy are widely used in organic systems. Characteristic vibrations of many organic compounds both in Raman and IR are widely studied and summarized in many literature. [5] Qualitative analysis of organic compounds can be done base on the characteristic vibrations table. “RS is similar to IR in that they have regions that are useful for functional group detection and fingerprint regions that permit the identification of specific compounds.”[1] While from the different selection rules of Raman Spectroscopy and IR, we can get the Mutual Exclusion rule [5], which says that for a molecule with a center of symmetry, no mode can be both IR and Raman Spectroscopy active. So, if we find a strong bond which is both IR and Raman Spectroscopy active, the molecule doesn't have a center of symmetry. Although classical Raman Spectroscopy has been successfully applied in chemistry, this technique has some major limitations as follows[5]: In order to overcome the limitations, special techniques are used to modify the classical Raman Spectroscopy. These non-classical Raman Spectroscopy includes: Resonance Raman Spectroscopy, surface enhanced Raman Spectroscopy, and nonlinear coherent Raman techniques, such as hyper Raman spectroscopy The resonance effect is observed when the photon energy of the exciting laser beam is equal to the energy of the allowed electronic transition. Since only the allowed transition is affected, (in terms of group theory, these are the totally symmetric vibrational ones.), only a few Raman bands are enhanced (by a factor of 10 ). As a result, RRS can increase the resolution of the classical Raman Spectroscopy, which makes the detection of dilution solution possible (concentrations as low as 10 M). RRS is extensively used for biological molecules because of its ability to selectively study the local environment. As an example, the Resonance Raman labels are used to study the biologically active sites on the bond ligand. RRS can also be used to study the electronic excited state. For example, the excitation profile which is the Raman intensity as a function of incident laser intensity can tell the interaction between the electronic states and the vibrational modes. Also, it can be used to measure the atomic displacement between the ground state and the excited state. At 1974, Fleischmann discovered that pyridine adsorbed onto silver electrodes showed enhanced Raman signals. This phenomenon is now called surface enhanced Raman Scattering (SERS). Although the mechanism of SERS is not yet fully understood, it is believed to result from an enhancement either of transition polarizability, α,or the electric field, E, by the interaction with the rough metallic support. Unlike RRS, SERS enhances every band in the Raman spectrum and has a high sensitivity. Due to the high enhancement (by a factor of 10 ), the SERS results in a rich spectrum and is an ideal tool for trace analysis and in situ study of interfacial process. Also, it is a better tool to study highly diluted solutions. A concentration of 4x10 M was reported by Kneipp using SERS. [5] In a nonlinear process, the output is not linearly proportional to its input. This happens when the perturbation become large enough that the response to the perturbation doesn’t follows the perturbation’s magnitude. Nonlinear Raman Spectroscopy includes: Hyper Raman spectroscopy, coherent anti-Stokes Raman Spectroscopy, coherent Stokes Raman spectroscopy, stimulated Raman gain and inverse Ramen spectroscopy. Nonlinear Raman spectroscopy is more sensitive than classical Raman spectroscopy and can effectively reduce/remove the influence of fluorescence. The following paragraph will focus on the most useful nonlinear Raman spectroscopy---coherent anti-Stokes Raman Spectroscopy (CARS): | 9,377 | 903 |
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Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron. Energy of an atom is defined when the atom loses or gains energy through chemical reactions that cause the loss or gain of electrons. A chemical reaction that releases energy is called an exothermic reaction and a chemical reaction that absorbs energy is called an endothermic reaction. Energy from an exothermic reaction is negative, thus energy is given a negative sign; whereas, energy from an endothermic reaction is positive and energy is given a positive sign. An example that demonstrates both processes is when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released). When an electron is added to a neutral atom (i.e., first electron affinity) energy is released; thus, the first electron affinities are . However, more energy is required to add an electron to a negative ion (i.e., second electron affinity) which overwhelms any the release of energy from the electron attachment process and hence, second electron affinities are . \[ \ce{X (g) + e^- \rightarrow X^{-} (g)} \label{1}\] \[ \ce{X^- (g) + e^- \rightarrow X^{2-} (g)} \label{2}\] Ionization energies are always concerned with the formation of positive ions. Electron affinities are the negative ion equivalent, and their use is almost always confined to elements in groups 16 and 17 of the Periodic Table. The first electron affinity is the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous -1 ions. It is the energy released (per mole of X) when this change happens. First electron affinities have negative values. For example, the first electron affinity of chlorine is -349 kJ mol . By convention, the negative sign shows a release of energy. When an electron is added to a metal element, energy is needed to gain that electron (endothermic reaction). Metals have a less likely chance to gain electrons because it is easier to lose their valance electrons and form cations. It is easier to lose their valence electrons because metals' nuclei do not have a strong pull on their valence electrons. Thus, metals are known to have lower electron affinities. This trend of lower electron affinities for metals is described by the metals: Notice that electron affinity down the group. When nonmetals gain electrons, the energy change is usually negative because they give off energy to form an anion (exothermic process); thus, the electron affinity will be . Nonmetals have a greater electron affinity than metals because of their atomic structures: first, nonmetals have more valence electrons than metals do, thus it is easier for the nonmetals to gain electrons to fulfill a stable octet and secondly, the valence electron shell is closer to the nucleus, thus it is harder to remove an electron and it easier to attract electrons from other elements (especially metals). Thus, nonmetals have a higher electron affinity than metals, meaning they are more likely to gain electrons than atoms with a lower electron affinity. For example, nonmetals like the elements in the halogens series in have a higher electron affinity than the metals. This trend is described as below. Notice the negative sign for the electron affinity which shows that energy is released. Notice that electron affinity decreases down the group, but increases up with the period. As the name suggests, electron affinity is the ability of an atom to accept an electron. Unlike electronegativity, electron affinity is a quantitative measurement of the energy change that occurs when an electron is added to a neutral gas atom. The more negative the electron affinity value, the higher an atom's affinity for electrons. To summarize the difference between the electron affinity of metals and nonmetals (Figure \(\Page {1}\)): Electron affinity increases upward for the groups and from left to right across periods of a periodic table because the electrons added to energy levels become closer to the nucleus, thus a stronger attraction between the nucleus and its electrons. Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons. Electron affinity decreases down the groups and from right to left across the periods on the periodic table because the electrons are placed in a higher energy level far from the nucleus, thus a decrease from its pull. However, one might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table. As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately. The electron affinity is a measure of the attraction between the incoming electron and the nucleus - . The factors which affect this attraction are exactly the same as those relating to ionization energies - nuclear charge, distance and screening. The increased nuclear charge as you go down the group is offset by electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about. A fluorine atom has an electronic structure of 1s 2s 2px 2py 2pz . It has 9 protons in the nucleus.The incoming electron enters the 2-level, and is screened from the nucleus by the two 1s electrons. It therefore feels a net attraction from the nucleus of 7+ (9 protons less the 2 screening electrons). In contrast, chlorine has the electronic structure 1s 2s 2p 3s 3p 3p 3p with 17 protons in the nucleus. But again the incoming electron feels a net attraction from the nucleus of 7+ (17 protons less the 10 screening electrons in the first and second levels). There is also a small amount of screening by the 2s electrons in fluorine and by the 3s electrons in chlorine. This will be approximately the same in both these cases and so does not affect the argument in any way (apart from complicating it!). The over-riding factor is therefore the increased distance that the incoming electron finds itself from the nucleus as you go down the group. The greater the distance, the less the attraction and so the less energy is released as electron affinity. Comparing fluorine and chlorine is not ideal, because fluorine breaks the trend in the group. However, comparing chlorine and bromine, say, makes things seem more difficult because of the more complicated electronic structures involved. What we have said so far is perfectly true and applies to the fluorine-chlorine case as much as to anything else in the group, but there's another factor which operates as well which we haven't considered yet - and that over-rides the effect of distance in the case of fluorine. The incoming electron is going to be closer to the nucleus in fluorine than in any other of these elements, so you would expect a high value of electron affinity. However, because fluorine is such a small atom, you are putting the new electron into a region of space already crowded with electrons and there is a significant amount of repulsion. This repulsion lessens the attraction the incoming electron feels and so lessens the electron affinity. A similar reversal of the expected trend happens between oxygen and sulfur in Group 16. The first electron affinity of oxygen (-142 kJ mol ) is smaller than that of sulfur (-200 kJ mol-1) for exactly the same reason that fluorine's is smaller than chlorine's. As you might have noticed, the first electron affinity of oxygen (\(-142\; kJ\; mol^{-1}\)) is less than that of fluorine (\(-328\; kJ\; mol^{-1}\)). Similarly sulfur's (\(-200\; kJ\; mol^{-1}\)) is less than chlorine's (\(-349\; kJ\; mol^{-1}\)). Why? It's simply that the element has 1 less proton in the nucleus than its next door neighbor in . The amount of screening is the same in both. That means that the net pull from the nucleus is less in Group 16 than in Group 17, and so the electron affinities are less. The reactivity of the elements in group 17 falls as you go down the group - fluorine is the most reactive and iodine the least. Often in their reactions these elements form their negative ions. The first impression that is sometimes given that the fall in reactivity is because the incoming electron is held less strongly as you go down the group and so the negative ion is less likely to form. That explanation looks reasonable until you include fluorine! An overall reaction will be made up of lots of different steps all involving energy changes, and you cannot safely try to explain a trend in terms of just one of those steps. Fluorine is much more reactive than chlorine (despite the lower electron affinity) because the energy released in other steps in its reactions more than makes up for the lower amount of energy released as electron affinity. You are only ever likely to meet this with respect to the group 16 elements oxygen and sulfur which both form -2 ions. The second electron affinity is the energy required to add an electron to each ion in 1 mole of gaseous 1- ions to produce 1 mole of gaseous 2- ions. This is more easily seen in symbol terms. \[ X^- (g) + e^- \rightarrow X^{-2} (g) \label{3}\] It is the energy needed to carry out this change per mole of \(X^-\). Why is energy needed to do this? You are forcing an electron into an already negative ion. It's not going to go in willingly! \[ O_{g} + e^- \rightarrow O^- (g) \;\;\; \text{1st EA = -142 kJ mol}^{-1} \label{4}\] \[ O^-_{g} + e^- \rightarrow O^{2-} (g) \;\;\; \text{2nd EA = +844 kJ mol}^{-1} \label{5}\] The positive sign shows that you have to put in energy to perform this change. The second electron affinity of oxygen is particularly high because the electron is being forced into a small, very electron-dense space. | 10,758 | 906 |
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The modern periodic table has evolved through a long history of attempts by chemists to arrange the elements according to their properties as an aid in predicting chemical behavior. One of the first to suggest such an arrangement was the German chemist Johannes Dobereiner (1780–1849), who noticed that many of the known elements could be grouped in triads —for example, chlorine, bromine, and iodine; or copper, silver, and gold. Dobereiner proposed that all elements could be grouped in such triads, but subsequent attempts to expand his concept were unsuccessful. We now know that portions of the periodic table—the block in particular—contain triads of elements with substantial similarities. The middle three members of most of the other columns, such as sulfur, selenium, and tellurium in group 16 or aluminum, gallium, and indium in group 13, also have remarkably similar chemistry. By the mid-19th century, the atomic masses of many of the elements had been determined. The English chemist John Newlands (1838–1898), hypothesizing that the chemistry of the elements might be related to their masses, arranged the known elements in order of increasing atomic mass and discovered that every seventh element had similar properties (Figure \(\Page {1}\) ). (The noble gases were still unknown.) Newlands therefore suggested that the elements could be classified into octaves , corresponding to the rows in the main group elements. Unfortunately, Newlands’s “law of octaves” did not seem to work for elements heavier than calcium, and his idea was publicly ridiculed. At one scientific meeting, Newlands was asked why he didn’t arrange the elements in alphabetical order instead of by atomic mass, since that would make just as much sense! Actually, Newlands was on the right track—with only a few exceptions, atomic mass does increase with atomic number, and similar properties occur every time a set of subshells is filled. Despite the fact that Newlands’s table had no logical place for the -block elements, he was honored for his idea by the Royal Society of London in 1887. John Alexander Reina Newlands was an English chemist who worked on the development of the periodic table. He noticed that elemental properties repeated every seventh (or multiple of seven) element, as musical notes repeat every eighth note. The periodic table achieved its modern form through the work of the German chemist Julius Lothar Meyer (1830–1895) and the Russian chemist Dimitri Mendeleev (1834–1907), both of whom focused on the relationships between atomic mass and various physical and chemical properties. In 1869, they independently proposed essentially identical arrangements of the elements. Meyer aligned the elements in his table according to periodic variations in simple atomic properties, such as “atomic volume” (Figure \(\Page {2}\) ), which he obtained by dividing the atomic mass (molar mass) in grams per mole by the density (\(\rho\)) of the element in grams per cubic centimeter. This property is equivalent to what is today defined as molar volume, t (measured in cubic centimeters per mole): \[ \dfrac{molar\; mass\left ( \cancel{g}/mol \right )}{density\left ( \cancel{g}/cm^{3} \right )}=molar\; volume\left ( cm^{3}/mol \right ) \label{7.1.1}\] As shown in Figure \(\Page {2}\) , the alkali metals have the highest molar volumes of the solid elements. In Meyer’s plot of atomic volume versus atomic mass, the nonmetals occur on the rising portion of the graph, and metals occur at the peaks, in the valleys, and on the downslopes. When his family’s glass factory was destroyed by fire, Mendeleev moved to St. Petersburg, Russia, to study science. He became ill and was not expected to recover, but he finished his PhD with the help of his professors and fellow students. In addition to the periodic table, another of Mendeleev’s contributions to science was an outstanding textbook, , which was used for many years. Mendeleev, who first published his periodic table in 1869 (Figure \(\Page {3}\) ), is usually credited with the origin of the modern periodic table. The key difference between his arrangement of the elements and that of Meyer and others is that Mendeleev did not assume that all the elements had been discovered (actually, only about two-thirds of the naturally occurring elements were known at the time). Instead, he deliberately left blanks in his table at atomic masses 44, 68, 72, and 100, in the expectation that elements with those atomic masses would be discovered. Those blanks correspond to the elements we now know as scandium, gallium, germanium, and technetium. The groups in Mendeleev's table are determined by how many oxygen or hydrogen atoms are needed to form compounds with each element. For example, in Group I, two atoms of hydrogen, lithium, Li, sodium, Na, and potassium form compounds with one atom of oxygen. In Group VII, one atom of fluorine, F, chlorine, Cl, and bromine, Br, react with one atom of hydrogen. Notice how this approach has trouble with the transition metals. Until roughly 1960, a rectangular table developed from Mendeleev's table and based on reactivity was standard at the front of chemistry lecture halls. The most convincing evidence in support of Mendeleev’s arrangement of the elements was the discovery of two previously unknown elements whose properties closely corresponded with his predictions (Table \(\Page {1}\)). Two of the blanks Mendeleev had left in his original table were below aluminum and silicon, awaiting the discovery of two as-yet-unknown elements, -aluminum and -silicon (from the Sanskrit , meaning “one,” as in “one beyond aluminum”). The observed properties of gallium and germanium matched those of -aluminum and -silicon so well that once they were discovered, Mendeleev’s periodic table rapidly gained acceptance. When the chemical properties of an element suggested that it might have been assigned the wrong place in earlier tables, Mendeleev carefully reexamined its atomic mass. He discovered, for example, that the atomic masses previously reported for beryllium, indium, and uranium were incorrect. The atomic mass of indium had originally been reported as 75.6, based on an assumed stoichiometry of InO for its oxide. If this atomic mass were correct, then indium would have to be placed in the middle of the nonmetals, between arsenic (atomic mass 75) and selenium (atomic mass 78). Because elemental indium is a silvery-white , however, Mendeleev postulated that the stoichiometry of its oxide was really In O rather than InO. This would mean that indium’s atomic mass was actually 113, placing the element between two other metals, cadmium and tin. mp = 78°C bp* = 201°C One group of elements that absent from Mendeleev’s table is the noble gases, all of which were discovered more than 20 years later, between 1894 and 1898, by Sir William Ramsay (1852–1916; Nobel Prize in Chemistry 1904). Initially, Ramsay did not know where to place these elements in the periodic table. Argon, the first to be discovered, had an atomic mass of 40. This was greater than chlorine’s and comparable to that of potassium, so Ramsay, using the same kind of reasoning as Mendeleev, decided to place the noble gases between the halogens and the alkali metals. Despite its usefulness, Mendeleev’s periodic table was based entirely on empirical observation supported by very little understanding. It was not until 1913, when a young British physicist, H. G. J. Moseley (1887–1915), while analyzing the frequencies of x-rays emitted by the elements, discovered that the underlying foundation of the order of the elements was by the , not the atomic mass. Moseley hypothesized that the placement of each element in his series corresponded to its atomic number , which is the number of positive charges (protons) in its nucleus. Argon, for example, although having an atomic mass greater than that of potassium (39.9 amu versus 39.1 amu, respectively), was placed potassium in the periodic table. While analyzing the frequencies of the emitted x-rays, Moseley noticed that the atomic number of argon is 18, whereas that of potassium is 19, which indicated that they were indeed placed correctly. Moseley also noticed three gaps in his table of x-ray frequencies, so he predicted the existence of three unknown elements: technetium ( = 43), discovered in 1937; promethium ( = 61), discovered in 1945; and rhenium ( = 75), discovered in 1925. Moseley left his research work at the University of Oxford to join the British army as a telecommunications officer during World War I. He was killed during the Battle of Gallipoli in Turkey. Before its discovery in 1999, some theoreticians believed that an element with a of 114 existed in nature. Use Mendeleev’s reasoning to name element 114 as -______; then identify the known element whose chemistry you predict would be most similar to that of element 114. atomic number name using prefix - The = 7 row can be filled in by assuming the existence of elements with atomic numbers greater than 112, which is underneath mercury (Hg). Counting three boxes to the right gives element 114, which lies directly below lead (Pb). If Mendeleev were alive today, he would call element 114 -lead. Use Mendeleev’s reasoning to name element 112 as -______; then identify the known element whose chemistry you predict would be most similar to that of element 112. -mercury The periodic table arranges the elements according to their electron configurations, such that elements in the same column have the same valence electron configurations. Periodic variations in size and chemical properties are important factors in dictating the types of chemical reactions the elements undergo and the kinds of chemical compounds they form. The modern periodic table was based on empirical correlations of properties such as atomic mass; early models using limited data noted the existence of and of elements with similar properties. The periodic table achieved its current form through the work of Dimitri Mendeleev and Julius Lothar Meyer, who both focused on the relationship between atomic mass and chemical properties. Meyer arranged the elements by their atomic volume, which today is equivalent to the , defined as molar mass divided by molar density. The correlation with the electronic structure of atoms was made when H. G. J. Moseley showed that the periodic arrangement of the elements was determined by atomic number, not atomic mass. ( ) | 10,527 | 907 |
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There are two types of atomic bonds - and . They differ in their structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 - 200 kcal/mol). Whether two atoms can form a covalent bond depends upon their i.e. the power of an atom in a molecule to attract electrons to itself. If two atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and negatively charged ion (anion). The bond between these two ions is called an ionic bond. | 725 | 908 |
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Phase diagram is a graphical representation of the physical states of a substance under different conditions of temperature and pressure. A typical phase diagram has pressure on the y-axis and temperature on the x-axis. As we cross the lines or curves on the phase diagram, a phase change occurs. In addition, two states of the substance coexist in equilibrium on the lines or curves. A phase transition is the transition from one state of matter to another. There are three states of matter: . Phase diagrams illustrate the variations between the states of matter of elements or compounds as they relate to pressure and temperatures. The following is an example of a phase diagram for a generic single-component system: Phase diagrams plot pressure (typically in atmospheres) versus temperature (typically in degrees Celsius or Kelvin). The labels on the graph represent the stable states of a system in equilibrium. The lines represent the combinations of pressures and temperatures at which two phases can exist in equilibrium. In other words, these lines define phase change points. The red line divides the solid and gas phases, represents sublimation (solid to gas) and deposition (gas to solid). The green line divides the solid and liquid phases and represents melting (solid to liquid) and freezing (liquid to solid). The blue divides the liquid and gas phases, represents vaporization (liquid to gas) and condensation (gas to liquid). There are also two important points on the diagram, the triple point and the critical point. The triple point represents the combination of pressure and temperature that facilitates all phases of matter at equilibrium. The critical point terminates the liquid/gas phase line and relates to the critical pressure, the pressure above which a supercritical fluid forms. With most substances, the temperature and pressure related to the triple point lie below standard temperature and pressure and the pressure for the critical point lies above standard pressure. Therefore at standard pressure as temperature increases, most substances change from solid to liquid to gas, and at standard temperature as pressure increases, most substances change from gas to liquid to solid. Normally the solid/liquid phase line slopes positively to the right (as in the diagram for carbon dioxide below). However for other substances, notably water, the line slopes to the left as the diagram for water shows. This indicates that the liquid phase is more dense than the solid phase. This phenomenon is caused by the crystal structure of the solid phase. In the solid forms of water and some other substances, the molecules crystalize in a lattice with greater average space between molecules, thus resulting in a solid with a lower density than the liquid. Because of this phenomenon, one is able to melt ice simply by applying pressure and not by adding heat. Moving about the phase diagram reveals information about the phases of matter. Moving along a constant temperature line reveals relative densities of the phases. When moving from the bottom of the diagram to the top, the relative density increases. Moving along a constant pressure line reveals relative energies of the phases. When moving from the left of the diagram to the right, the relative energies increases. Imagine a substance with the following points on the phase diagram: a triple point at .5 atm and -5°C; a normal melting point at 20°C; a normal boiling point at 150°C; and a critical point at 5 atm and 1000°C. The solid liquid line is "normal" (meaning positive sloping). For this, complete the following: 1. Roughly sketch the phase diagram, using units of atmosphere and Kelvin. 1-solid, 2-liquid, 3-gas, 4-supercritical fluid, point O-triple point, C-critical point -78.5 °C (The phase of dry ice changes from solid to gas at -78.5 °C) 2. Rank the states with respect to increasing density and increasing energy. 3. Describe what one would see at pressures and temperatures above 5 atm and 1000°C. One would see a super-critical fluid, when approaching the point, one would see the meniscus between the liquid and gas disappear. 4. Describe what will happen to the substance when it begins in a vaccum at -15°C and is slowly pressurized. The substance would begin as a gas and as the pressure increases, it would compress and eventually solidify without liquefying as the temperature is below the triple point temperature. 5. Describe the phase changes from -80°C to 500°C at 2 atm. The substance would melt at somewhere around, but above 20°C and then boil at somewhere around, but above 150°C. It would not form a super-critical fluid as the neither the pressure nor temperature reach the critical pressure or temperature. 6. What exists in a system that is at 1 atm and 150°? Depending on how much energy is in the system, there will be different amounts of liquid and gas at equilibrium. If just enough energy was added to raise the temperature of the liquid to 150°C, there will just be liquid. If more was added, there will be some liquid and some gas. If just enough energy was added to change the state of all of the liquid without raising the temperature of the gas, there will just be gas. 7. Label the area 1, 2, 3, and 4 and points O and C on the diagram. 8. A sample of dry ice (solid CO ) is cooled to -100 °C, and is set on a table at room temperature (25 °C). At what temperature is the rate of sublimation and deposition the same? (Assume pressure is held constant at 1 atm). | 5,517 | 909 |
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An is one that is occurring in water. What makes water significant is that it can allow for substances to dissolve and/or be dissociated into ions within it. Water is generally the solvent found in aqueous solution, where a solvent is the substance that dissolves the solute. The solute is the substance or compound being dissolved in the solvent. A solute has fewer number of particles than a solvent, where it's particles are in random motion. Interestingly, aqueous solutions with ions conduct electricity to some degree. Pure water, having a very low concentration of ions, cannot conduct electricity. When a solute dissociates in water to form ions, it is called an , due to the solution being a good electrical conductor. When no ions are produced, or the ion content is low, the solute is a Non-electrolytes do not conduct electricity or conduct it to a very small degree. In an aqueous solution a is considered to be completely ionized, or dissociated, in water, meaning it is soluble. Strong acids and bases are usually strong electrolytes. A then is considered to be one that is not completely dissociated, therefore still containing whole compounds and ions in the solution. Weak acids and bases are generally weak electrolytes. In other words, strong electrolytes have a better tendency to supply ions to the aqueous solution than weak electrolytes, and therefore strong electrolytes create an aqueous solution that is a better conductor of electricity. Here's an example of MgCl in water: \[MgCl_2 \rightarrow Mg^{2+}_{(aq)} + 2Cl^-_{(aq)}\] The ionic compound dissociates completely to form ions in water, therefore, it is a strong electrolyte. Now let's look at a weak electrolyte: \[HC_2H_3O_2(aq) \rightleftharpoons H^+(aq) + C_2H_3O_2^-(aq)\] The ionic compound, \(HC_2H_3O_2\) in this situation, only partially dissociates, as expressed by the double arrows in the reaction. This means that the reaction is reversible and never goes to completion. The \(H^+\) cation is a proton that interacts with the \(H_2O\) molecules that it is submerged in. The interaction is called . The actual H ion does not exist in the aqueous solution. It is the , \(H_3O^+\) that interacts with water to create additional species like \(H_5O_2^+\), \(H_9O_4^+\), and \(H_7O_3^+\). \(H_3O^+\) can simply be described as the hydration of one H and one water molecule. For nonelectrolytes, all that needs to be done is write the molecular formula because no reaction or dissociation occurs. One example of a nonelectrolyte is : written as \(C_6H_{12}O_6 (aq)\). In an aqueous solution the amount of ions of a species is related to the number of moles of that species per concentration of the substance in the aqueous solution. is the number of moles of a solute (\(n\)) divided by the total volume (\(V\)) of the solution: \[M= \dfrac{n}{V}\] Molarity, or concentration, can also be represented by placing the solute within brackets (e..g, \([Cl^-]\) for the concentration of chloride ions). Determine the concentration of \(K^+\) in an aqueous solution of 0.238 M \(KNO_3\). Since there is one mole of potassium in \(KNO_3\), multiply the concentration of the species by the number of moles of the atom to obtain: \[[K^+] = (0.238\; M\; KNO_3) \times (1 \;mol\; K^+)= 0.238\; M\] Although not asked, there is also one mole of nitrite ions in one mole of \(KNO_3\), so its concentration is also 0.238 M: \[[NO_3^-] = (0.238\; M\; KNO_3) \times (1 \;mol\; NO_3^-)= 0.238\; M\] The stoichiometry always dictates the concentration, which was a simple 1:1 ratio for \(KNO_3\). However, for more complex situations, different ratios will be encountered. For instance, if consider the dissolving of \(Al_2(SO_4)_3\): \[Al_2(SO_4)_3 \rightarrow 2 Al^{3+}_{(aq)} + 3 SO^{2-}_{4(aq)}\] If the concentration of \(Al_2(SO_4)_3\) is 0.019 M, what is the concentration of \(Al^{3+}_{(aq)}\)? Simply multiply 0.019 M by the stoichiometric factor of \(Al^{3+}_{(aq)}\) in \(Al_2(SO_4)_3\), which is 2:1. The concentration of \(Al^{3+}_{(aq)}\) then becomes 0.038 M: \[[Al^{3+}] = (0.019 \; M\; Al_2(SO_4)_3) \times (2 \;mol\; Al^{3+})= 0.238\; M\] Although not asked, the concentration of \(SO_4^{2-}\) is 0.057 M via the same argument; \[[SO_4^{2-}]= (0.019 \; M) \times (3 \; mol \; SO_4^{2-}) = 0.057\; M\] Precipitation reactions occur when the resulting product of an aqueous solution is insoluble. This means that there is a solid produced, called the . The precipitate is a combination of cation and anions forming an ionic bond. Precipitates are used in manufacturing chemicals, so that certain ions can be isolated by forming precipitates with them. We can predict if a precipitate is produced using the for common ionic solids: Here's an example of a precipitation reaction: \[AgNO_{3(aq)} + NaI_{(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)}\] This is considered the "whole" reaction, because all of the species involved are recognized. However, in an aqueous solution, the particles look more like this: \[Ag^+_{(aq)} + NO^-_{3(aq)} + Na^+_{(aq)} + I^-_{(aq)} \rightarrow AgI_{(s)} + Na^+_{(aq)} + NO^-_{3(aq)}\] This is called the "ionic" form, because all the ions in the solution are shown. If the ions that are not involved in creating the solid are removed (the , a is generated: \[Ag^+_{(aq)} + I^-_{(aq)} \rightarrow AgI_{(s)}\] What are the spectator ions in this reaction? \[NaOH_{(aq)} + MgCl_{2(aq)} \rightarrow NaCl_{(aq)} + Mg(OH)_{2(s)}\] As you can see, it is to include the symbols for each species to identify what state they're in: either gaseous (g), solid (s), liquid (l) or aqueous (aq). It is also important to remember that these types of ionic equations also must have a balanced charge. You can use stoichiometric coefficients to ensure that both sides of the equation have equal net charges. An is a substance that gives off \(H^+\) ions in an aqueous solution, while a gives of OH ions. Strong acids almost completely dissociate to become \(H^+\) ions, and strong bases dissociate to become OH . There are only a few strong acids and bases. Most are weak, meaning they produce few \(H^+\) or OH ions in aqueous solution. Here is a list of common strong acids and bases: In a an acid and a base are combined to produce water and an aqueous salt. The acid and base neutralize/balance each other to get a result that is the salt. For example, determine which is the acid, base, and salt n this neutralization reaction? \[HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}\] Predict if a reaction is likely to occur, and the resulting product: Determine the concentration: | 6,646 | 911 |
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Listed below are a number of equations which give the dependence of , and on the quantum numbers , and . They refer not only to the hydrogen atom but also to any one-electron ion in general with a nuclear charge of . Thus He is a one-electron ion with = 2, Li another example with = 3. The average distance between the electron and the nucleus expressed in atomic units of length is:
Note that is proportional to for = 0 orbitals, and deviates only slightly from this for 0. The value of decreases as increases because the nuclear attractive force is greater. Thus for He would be only one half as large as for H.
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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. | 3,437 | 914 |
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In the previous section, the definition of the the partition function involves a sum of state formulism: \[Q = \sum_i e^{-\beta E_i} . \label{1} \] However, under most conditions, full knowledge of each member of an ensemble is missing and hence we have to operate with a more reduced knowledge. This is demonstrated via a simple model of two particles in a two-energy level system in Figure \(\Page {1}\). Each particle (red or blue) can occupy either the \(E_1=0\) energy level or the \(E_2=\epsilon\) energy level resulting in four possible states that describe the system. The corresponding partition function for this system is then (via Equation \ref{1}): \[Q_{\text{distinguishable}}=e^0+ e^{-\beta\epsilon} + e^{-\beta\epsilon} + e^{-2 \beta\epsilon}=q^2 \label{Q1} \] and is just the molecular partition function (\(q\)) squared. However, if the two particles are indistinguishable (e.g., both the same color as in Figure \(\Page {2}\)) then while four different combinations can be generated like in Figure \(\Page {1}\), there is no discernible way to separate the two middle states. Hence, there are effectively only three states observable for this system. The corresponding partition function for this system (again using Equation \ref{1}) can be constructed: \[ Q(N,V,\beta) =e^0+ e^{-\beta\epsilon} + e^{-2 \beta\epsilon} \neq q^2 \label{Q2} \] and this is not equal to the square of the molecular partition function. If Equation \ref{Q1} were used to describe the indistinguishable particle case, then it would overestimate the number of observable states. From combinatorics, using \(q^N\) for a large \(N\)-particle system of indistinguishable particles will overestimate the number of states by a factor of \(N!\). Therefore Equation \ref{1} requires a slight modification to account for this over counting. \[ Q(N,V,\beta) = \dfrac{\sum_i{e^{-\beta E_i}}}{N!} \label{2} \] If we have \(N\) molecules, we can perform \(N!\) permutations that should not affect the outcome. To avoid over counting (making sure we do not count each state more than once), the partition function becomes: \[Q(N,V,\beta) = \dfrac{q(V,\beta)^N}{N!} \nonumber \] As you may have noticed, using Equation \ref{2} to estimate of \(Q\) for the two-indistinguishable particle discussed case above with \(N=2\) is incorrect (i.e., Equation \ref{2} is not equal to Equation \ref{Q2}). That is because the \(N!\) factor is only applicable for large \(N\). | 2,463 | 915 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/21%3A_Binding_and_Association/21.02%3A_Statistical_Thermodynamics_of_Biomolecular_Reactions |
Statistical mechanics can be used to calculate K on the basis of the partition function. The canonical partition function Q is related to the Helmholtz free energy through \[ F= -k_bT\ln Q \] \[ Q = \sum_{\alpha}e^{-E_{\alpha}/k_BT} \] where the sum is over all microstates (a particular configuration of the molecular constituents to a macroscopic system), Boltzmann weighted by the energy of that microstate E . The chemical potential of molecular species is given by \[ \mu_i = -k_BT \left( \dfrac{\partial \ln Q}{\partial N_i} \right)_{V,T, \{ N_{j \neq i} \} } \] We will assume that we can partition Q into contributions from different molecular components of a reacting system such that \[ Q =\prod_iQ_i \] The ability to separate the partition function stems from the assumption that certain degrees of freedom are separable from each other. When two sub-systems are independent of one another, their free energies should add (F = F + F ) and therefore their partition functions are separable into products: Q = Q Q . Generally this separability is a result of being able to write the Hamiltonian as H = H + H , which results in the microstate energy being expressed as a sum of two independent parts: E = E +E . In addition to separating the different molecular species, it is also very helpful to separate the translational and internal degrees of freedom for each species, Q = Q Q . The entropy of mixing originates from the translational partition function, and therefore will be used to describe concentration dependence. For Ni non-interacting, indistinguishable molecules, we can relate the canonical and molecular partition function q for component i as \[ Q_i= \dfrac{q_i^{N_i}}{N_i!} \] and using Sterling’s approximation we obtain the chemical potential, \[ \mu_i = -RT\ln \dfrac{q_i}{N_i} \] Following the reasoning in eqs. (2)–(5), we can write the equilibrium constant as \[ K_a = \dfrac{N_C}{N_AN_B}=\dfrac{q_C}{q_Aq_B}V \] This expression reflects that the equilibrium constant is related to the stoichiometrically scaled ratio of molecular partition functions per unit volume \(K_a = \prod_i(q_i/V)^{v_i}\). Then the standards binding free energy is determined by eq. (4). | 2,219 | 916 |
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