Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Marcus theory wikipedia , lookup
Solar air conditioning wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Heat transfer wikipedia , lookup
Electrolysis of water wikipedia , lookup
Thermomechanical analysis wikipedia , lookup
Spin crossover wikipedia , lookup
Glass transition wikipedia , lookup
Internal energy wikipedia , lookup
Transition state theory wikipedia , lookup
4 Thermodynamics: the Second Law Entropy 91 4.1 The direction of spontaneous change 91 4.2 Entropy and the Second Law 92 4.3 The entropy change accompanying expansion 94 4.4 The entropy change accompanying heating 95 4.5 The entropy change accompanying a phase transition 97 4.6 Entropy changes in the surroundings 99 4.7 The molecular interpretation of entropy 100 4.8 Absolute entropies and the Third Law of thermodynamics 101 4.9 The molecular interpretation of Third-Law entropies 103 4.10 The standard reaction entropy 105 4.11 The spontaneity of chemical reactions 105 The Gibbs energy 106 4.12 Focussing on the system 106 4.13 Properties of the Gibbs energy 107 109 CHECKLIST OF KEY CONCEPTS ROAD MAP OF KEY EQUATIONS QUESTIONS AND EXERCISES 9780199608119_C04.indd 90 109 110 Some things happen; some things don’t. A gas expands to fill the vessel it occupies; a gas that already fills a vessel does not suddenly contract into a smaller volume. A hot object cools to the temperature of its surroundings; a cool object does not suddenly become hotter than its surroundings. Hydrogen and oxygen combine explosively (once their ability to do so has been liberated by a spark) and form water; water left standing in oceans and lakes does not gradually decompose into hydrogen and oxygen. These everyday observations suggest that changes can be divided into two classes. A spontaneous change is a change that has a tendency to occur without work having to be done to bring it about. A spontaneous change has a natural tendency to occur. A non-spontaneous change is a change that can be brought about only by doing work. A non-spontaneous change has no natural tendency to occur. Non-spontaneous changes can be made to occur by doing work: gas can be compressed into a smaller volume by pushing in a piston, the temperature of a cool object can be raised by forcing an electric current through a heater attached to it, and water can be decomposed by the passage of an electric current. However, in each case we need to act in some way on the system to bring about the non-spontaneous change. There must be some feature of the world that accounts for the distinction between the two types of change. Throughout the chapter we shall use the terms ‘spontaneous’ and ‘non-spontaneous’ in their thermodynamic sense. That is, we use them to signify that a change does or does not have a natural tendency to occur. In thermodynamics the term spontaneous has nothing to do with speed. Some spontaneous changes are very fast, such as the precipitation reaction that occurs when solutions of sodium chloride and silver nitrate are mixed. However, some spontaneous changes are so slow that there may be no observable change even after millions of years. For example, although the 9/20/12 11:25 AM ENTROPY decomposition of benzene into carbon and hydrogen is spontaneous, it does not occur at a measurable rate under normal conditions, and benzene is a common laboratory commodity with a shelf life of (in principle) millions of years. Thermodynamics deals with the tendency to change; it is silent on the rate at which that tendency is realized. 91 Non-spontaneous Spontaneous Entropy A few moments’ thought is all that is needed to identify the reason why some changes are spontaneous and others are not. That reason is not the tendency of the system to move towards lower energy. This point is easily established by identifying an example of a spontaneous change in which there is no change in energy. The isothermal expansion of a perfect gas into a vacuum is spontaneous, but the total energy of the gas does not change because the molecules continue to travel at the same average speed and so keep their same total kinetic energy. Even in a process in which the energy of a system does decrease (as in the spontaneous cooling of a block of hot metal), the First Law requires the total energy of the system and the surroundings to be constant. Therefore, in this case the energy of another part of the world must increase if the energy decreases in the part that interests us. For instance, a hot block of metal in contact with a cool block cools and loses energy; however, the second block becomes warmer, and increases in energy. It is equally valid to say that the second block moves spontaneously to higher energy as it is to say that the first block has a tendency to go to lower energy! Fig. 4.1 One fundamental type of spontaneous process is the dispersal of matter. This tendency accounts for the spontaneous tendency of a gas to spread into and fill the container it occupies. It is extremely unlikely that all the particles will collect into one small region of the container. (In practice, the number of particles is of the order of 1023.) energy rather than that of matter. In a block of hot metal, the atoms are oscillating vigorously and the hotter the block the more vigorous their motion. The cooler surroundings also consist of oscillating atoms, but their motion is less vigorous. The vigorously oscillating atoms of the hot block jostle their neighbours in the surroundings, and the energy of the atoms in the block is handed on to the atoms in the surroundings (Fig. 4.2). The process continues until the vigour with which the atoms in the system are oscillating has fallen to that of the surroundings. The opposite flow of energy is very unlikely. It is highly improbable that there will be a net flow of energy into the system as a result of jostling from less vigorously oscillating molecules in the surroundings. In this case, the natural direction of change corresponds to the dispersal of energy. 4.1 The direction of spontaneous change We shall now show that the apparent driving force of spontaneous change is the tendency of energy to disperse and matter to become disordered. For example, the molecules of a gas may all be in one region of a container initially, but their ceaseless random motion ensures that they spread rapidly throughout the entire volume of the container (Fig. 4.1). Because their motion is so disorderly, there is a negligibly small probability that all the molecules will find their way back simultaneously into the region of the container they occupied initially. In this instance, the natural direction of change corresponds to the disorderly dispersal of matter. A similar explanation accounts for spontaneous cooling, but now we need to consider the dispersal of 9780199608119_C04.indd 91 Non-spontaneous Spontaneous Fig. 4.2 Another fundamental type of spontaneous process is the dispersal of energy (represented by the small arrows). In these diagrams, the yellow spheres represent the system and the purple spheres represent the surroundings. The double headed arrows represent the thermal motion of the atoms. 9/20/12 11:25 AM 92 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Hot source Heat Work Engine take entropy to be a synonym for the extent of disorderly dispersal, but shortly we shall see that it can be defined precisely and quantitatively, measured, and then applied to chemical reactions. At this point, all we need to know is that when matter and energy become dispersed, the entropy increases. That being so, we can express the basic principle underlying change as the Second Law of thermodynamics: The entropy of an isolated system tends to increase. Fig. 4.3 The Second Law denies the possibility of the process illustrated here, in which heat is changed completely into work, there being no other change. The process is not in conflict with the First Law, because the energy is conserved. The tendency of energy to disperse also explains the fact that, despite numerous attempts, it has proved impossible to construct an engine like that shown in Fig. 4.3, in which heat, perhaps from the combustion of a fuel, is drawn from a hot reservoir and completely converted into work, such as the work of moving an automobile. All actual heat engines have both a hot region, the ‘source’, and a cold region, the ‘sink’, and it has been found that some energy must be discarded into the cold sink as heat and not used to do work. In molecular terms, only some of the energy stored in the atoms and molecules of the hot source can be used to do work and transferred to the surroundings in an orderly way. For the engine to do work, some energy must be transferred to the cold sink as heat, to stimulate disorderly motion of its atoms and molecules. In summary, we have identified a single basic type of spontaneous physical process: Matter and energy tend to disperse. By ‘disperse’ we mean spread in a disorderly way through space or, if matter is confined to a particular region, for its structure to become disorganized, as when a solid melts. We must now see how this natural process results in some chemical reactions being spontaneous and others not. It may seem very puzzling that such a simple principle can account for the formation of such ordered systems as proteins and biological cells. Nevertheless, in due course we shall see that organized structures can emerge as energy and matter disperse. We shall see, in fact, that collapse into disorder accounts for change in all its forms. 4.2 Entropy and the Second Law The measure of disorderly dispersal used in thermodynamics is called the entropy, S. Initially, we can 9780199608119_C04.indd 92 The Second Law The ‘isolated system’ may consist of a system in which we have a special interest (a beaker containing reagents) and that system’s surroundings: the two components jointly form a little ‘universe’ in the thermodynamic sense. To make progress and turn the Second Law into a quantitatively useful statement, we need to define entropy precisely. We shall use the following definition of a change in entropy for a system maintained at constant temperature: ∆S = qrev T Definition Entropy change (4.1) That is, the change in entropy of a substance is equal to the energy transferred as heat to it reversibly divided by the temperature at which the transfer takes place. We need to understand three points about the definition in eqn 4.1: the significance of the term ‘reversible’, why heat (not work) appears in the numerator, and why temperature appears in the denominator. • Why reversible? We met the concept of reversibility in Section 2.4, where we saw that it refers to the ability of an infinitesimal change in a variable to change the direction of a process. Mechanical reversibility refers to the equality of pressure acting on either side of a movable wall. Thermal reversibility, the type involved in eqn 4.1, refers to the equality of temperature on either side of a thermally conducting wall. Reversible transfer of heat is smooth, careful, restrained transfer between two bodies at the same temperature. By making the transfer reversible we ensure that there are no hot spots generated in the object that later disperse spontaneously and hence add to the entropy. • Why heat and not work in the numerator? Now consider why heat and not work appears in eqn 4.1. Recall from Section 2.3 that to transfer energy as heat we make use of the disorderly motion of molecules whereas to transfer energy as work we 9/20/12 11:25 AM ENTROPY make use of orderly motion. It should be plausible that the change in entropy—the change in the degree of disorder—is proportional to the energy transfer that takes place by making use of disorderly motion rather than orderly motion. • Why temperature in the denominator? The presence of the temperature in the denominator in eqn 4.1 takes into account the disorder that is already present. If a given quantity of energy is transferred as heat to a hot object (one in which the atoms have a lot of disorderly thermal motion), then the additional disorder generated is less significant than if the same quantity of energy is transferred as heat to a cold object in which the atoms have less thermal motion. The difference is like sneezing in a busy street (an environment analogous to a high temperature) and sneezing in a quiet library (an environment analogous to a low temperature). Brief illustration 4.1 Entropy changes The transfer of 100 kJ of energy as heat to a large mass of water at 0 °C (273 K) results in a change in entropy of DS = qrev 100 × 103 J = = + 366 J K −1 273 K T We use a large mass of water to ensure that the temperature of the sample does not change as heat is transferred. The same transfer at 100 °C (373 K) results in DS = 100 × 10 J = + 268 J K −1 373 K 3 93 undergoes a change of state is independent of how the change of state is brought about. Impact on technology 4.1 Heat engines, refrigerators, and heat pumps One practical application of entropy is to the discussion of the efficiencies of heat engines, refrigerators, and heat pumps. As remarked in the text, to achieve spontaneity—an engine is less than useless if it has to be driven—some energy must be discarded as heat into the cold sink. It is quite easy to calculate the minimum energy that must be discarded in this way by thinking about the flow of energy and the changes in entropy of the hot source and cold sink. To simplify the discussion, we shall express it in terms of the magnitudes of the heat and work transactions, which we write as |q | and |w |, respectively (so, if q = −100 J, |q| = 100 J). Maximum work— and therefore maximum efficiency—is achieved if all energy transactions take place reversibly, so we assume that to be the case in the following. Suppose that the hot source is at a temperature Thot. Then when energy |q | is released from it reversibly as heat, its entropy changes by −|q |/Thot. Suppose that we allow an energy |q′| to flow reversibly as heat into the cold sink at a temperature Tcold. Then the entropy of that sink changes by +|q′|/Tcold (Fig. 4.4). The total change in entropy is therefore Reduction of Increase of entropy of the entropy of the hot source cold sink DS total = |q | − Thot |q ′| + Tcold The engine will not operate spontaneously if this change in entropy is negative, and just becomes spontaneous as DStotal becomes positive. This change of sign occurs when DStotal = 0, which is achieved when The increase in entropy is greater at the lower temperature. A note on good practice The units of entropy are joules per kelvin (J K−1). Entropy is an extensive property. When we deal with molar entropy, an intensive property, the units will be joules per kelvin per mole (J K−1 mol−1). The entropy is a state function, a property with a value that depends only on the present state of the system.1 The entropy is a measure of the current state of disorder of the system, and how that disorder was achieved is not relevant to its current value. A sample of liquid water of mass 100 g at 60 °C and 98 kPa has exactly the same degree of molecular disorder—the same entropy—regardless of what has happened to it in the past. The implication of entropy being a state function is that a change in its value when a system 1 For a proof of this statement, see our Physical Chemistry (2010). 9780199608119_C04.indd 93 Thot Entropy q w = q – q’ q’ Tcold Entropy Fig. 4.4 The flow of energy in a heat engine. For the process to be spontaneous, the decrease in entropy of the hot source must be offset by the increase in entropy of the cold sink. However, because the latter is at a lower temperature, not all the energy removed from the hot source need be deposited in it, leaving the difference available as work. 9/20/12 11:25 AM 94 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW |q ′ | = Tcold × |q | Thot If we have to discard an energy |q′| into the cold sink, the maximum energy that can be extracted as work is |q | − |q′|. It follows that the efficiency, h (eta), of the engine, the ratio of the work produced to the heat absorbed by the engine, is h= work produced |q | − |q ′ | T |q ′ | = 1− = 1− cold = heat absorbed |q | Thot |q | This remarkable result tells us that the efficiency of a perfect heat engine (one working reversibly and without mechanical defects such as friction) depends only on the temperatures of the hot source and cold sink. It shows that maximum efficiency (closest to h = 1) is achieved by using a sink that is as cold as possible and a source that is as hot as possible. Brief illustration 4.2 The maximum efficiency The maximum efficiency of an electrical power station using steam at 200 °C (473 K) and discharging at 20 °C (293 K) is Tcold 293 293 K = 0.381 h = 1− = 1− 473 473 K Thot or 38.1 per cent. A refrigerator can be analysed similarly (Fig. 4.5). The entropy change when an energy |q| is withdrawn reversibly as heat from the cold interior at a temperature Tcold is −|q|/Tcold. The entropy change when an energy |q′| is deposited reversibly as heat in the outside world at a temperature Thot is +|q′|/Thot. The total change in entropy would be negative if |q′| = |q|, Thot Entropy w q Entropy Fig. 4.5 The flow of energy as heat from a cold source to a hot sink becomes feasible if work is provided to add to the energy stream. Then the increase in entropy of the hot sink can be made to cancel the entropy decrease of the cold source. 9780199608119_C04.indd 94 4.3 The entropy change accompanying expansion We can often rely on intuition to judge whether the entropy increases or decreases when a substance undergoes a physical change. For instance, the entropy of a sample of gas increases as it expands because the molecules get to move in a greater volume and so have a greater degree of disorderly dispersal. However, the advantage of eqn 4.1 is that it lets us express the increase quantitatively and make numerical calculations. For instance, as shown in the following Derivation, we can use the definition to calculate the change in entropy when a perfect gas expands isothermally from a volume Vi to a volume Vf , and obtain ∆S = nR ln Vf Vi Perfect gas Change in entropy on isothermal expansion (4.2) We have already stressed the importance of reading equations for their physical content. In this case: • If Vf > Vi, as in an expansion, then Vf /Vi > 1 and the logarithm is positive. Consequently, eqn 4.2 predicts a positive value for ∆S, corresponding to an increase in entropy, just as we anticipated (Fig. 4.6). q+w Tcold and the refrigerator would not work. However, if we increase the flow of energy into the warm exterior by doing work on the refrigerator, then the entropy change of the warm exterior can be increased to the point at which it overcomes the decrease in entropy of the cold interior, and the refrigerator operates. The calculation of the maximum efficiency of this process is left as an exercise (see Project 4.34). A heat pump is simply a refrigerator, but in which we are more interested in the supply of heat to the exterior than the cooling achieved in the interior. You are invited to show (see Project 4.34) that the efficiency of a perfect heat pump, as measured by the heat produced divided by the work done, also depends on the ratio of the two temperatures. • The change in entropy is independent of the temperature at which the isothermal expansion occurs. More work is done if the temperature is high (because the external pressure must be matched to a higher value of the pressure of the gas), so more energy must be supplied as heat to maintain the temperature. The temperature in the denominator of eqn 4.1 is higher, but the ‘sneeze’ (in terms of the analogy introduced earlier) is greater too, and the two effects cancel. 9/20/12 11:25 AM ENTROPY Change in entropy, ΔS/nR 5 4 3 2 1 0 0 20 40 60 80 100 Expansion, Vf /Vi Fig. 4.6 The entropy of a perfect gas increases logarithmically (as ln V ) as the volume is increased. The variation of the entropy of a perfect gas with volume We need to know qrev, the energy transferred as heat in the course of a reversible change at the temperature T. From eqn 2.6 we know that the energy transferred as heat to a perfect gas when it undergoes reversible, isothermal expansion from a volume Vi to a volume Vf at a temperature T is Vf Vi It follows that cancel T q nR T ln(Vf / Vi ) V DS = rev = = nR ln f T T Vi which is eqn 4.2. Brief illustration 4.3 The entropy change accompanying expansion When the volume occupied by 1.00 mol of any perfect gas molecules is doubled at any constant temperature, Vf /Vi = 2 and DS = (1.00 mol) × (8.3145 J K−1 mol−1) × ln 2 = +5.76 J K−1 Self-test 4.1 What is the change in entropy when the pressure of a perfect gas is changed isothermally from pi to pf? Answer: DS = nR ln(p i /pf) 9780199608119_C04.indd 95 Here is a subtle but important point. The definition in eqn 4.1 makes use of a reversible transfer of heat, and that is what we used in the derivation of eqn 4.2. However, entropy is a state function, so its value is independent of the path between the initial and final states. This independence of path means that although we have used a reversible path to calculate ∆S, the same value applies to an irreversible change (for instance, free expansion) between the same two states. We cannot use an irreversible path to calculate ∆S, but the value calculated for a reversible path applies however the path is traversed in practice between the specified initial and final states. You may have noticed that in Brief illustration 4.3 we did not specify how the expansion took place other than that it is isothermal. 4.4 The entropy change accompanying heating Derivation 4.1 qrev = nRT ln 95 We should expect the entropy of a sample to increase as the temperature is raised from Ti to Tf , because the thermal disorder of the system is greater at the higher temperature, when the molecules move more vigorously. To calculate the change in entropy, we go back to the definition in eqn 4.1 and, as shown in the following Derivation, find that, provided the heat capacity is constant over the range of temperatures of interest, ∆S = C ln Tf Ti Constant heat capacity Change in entropy on heating (4.3) where C is the heat capacity of the system; if the pressure is held constant during the heating, we use the constant-pressure heat capacity, Cp, and if the volume is held constant, we use the constant-volume heat capacity, CV. Once more, we interpret the equation: • When Tf > Ti, Tf /Ti > 1, which implies that the logarithm is positive, that ∆S > 0, and therefore that the entropy increases as the temperature is raised (Fig. 4.7). • The higher the heat capacity of the substance, the greater the change in entropy for a given rise in temperature. A high heat capacity implies that a lot of heat is required to produce a given change in temperature, so the ‘sneeze’ must be more powerful than for when the heat capacity is low, and the entropy increase is correspondingly high. 9/20/12 11:25 AM 96 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Change in entropy, ∆S/nCp 5 DS = Tf Ti C dT =C T Tf Ti T dT = C ln f Ti T 4 We have used the same standard integral as in Derivation 2.2, and evaluated the limits similarly. 3 2 Brief illustration 4.4 The entropy change upon heating 1 0 0 20 40 60 80 100 Temperature ratio, Tf /Ti Fig. 4.7 The entropy of a sample with a heat capacity that is independent of temperature, such as a monatomic perfect gas, increases logarithmically (as ln T ) as the temperature is increased. The increase is proportional to the heat capacity of the sample. When calculating entropy changes at constant volume, the heat capacity to be used in eqn 4.3 is CV,m. For example, the change in molar entropy when hydrogen gas is heated from 20 °C to 30 °C at constant volume is (assuming that CV,m = 22.44 J K−1 mol−1 is constant over this temperature range): 30 0°C Tf 303 K −1 −1 DSm = CV ,m ln = (22.44 J K mol ) × ln 293 K Ti 20 °C = +0.75 J K −1 −1 mol Derivation 4.2 The variation of entropy with temperature Equation 4.1 refers to the transfer of heat to a system at a temperature T. In general, the temperature changes as we heat a system, so we cannot use eqn 4.1 directly. Suppose, however, that we transfer only an infinitesimal energy, dq, to the system, then there is only an infinitesimal change in temperature and we introduce negligible error if we keep the temperature in the denominator of eqn 4.1 equal to T during that transfer. As a result, the entropy increases by an infinitesimal amount dS given by dS = C dT T Experimental basis of determining an entropy change (4.5) Tf Ti C dT T (4.4) For many substances and for small temperature ranges we may take C to be constant. This assumption is strictly true for a monatomic perfect gas. Then C may be taken outside the integral, and the latter evaluated as follows: 9780199608119_C04.indd 96 Derivation 4.3 The entropy change when the heat capacity varies with temperature In Derivation 4.2 we found, before making the assumption that the heat capacity is constant, that DS = The total change in entropy, DS, when the temperature changes from Ti to Tf is the sum (integral; see The chemist’s toolkit 2.1) of all such infinitesimal terms with T in general different for each of the infinitesimal steps: DS = ∆S = area under the graph of C/T plotted against T, between Ti and Tf dqrev T To calculate dq, we recall from Section 2.5 that the heat capacity C = q/DT, is where DT is macroscopic change in temperature. For the infinitesimal change dT brought about by an infinitesimal transfer of heat we write C = dq /dT. This relation also applies when the transfer of energy is carried out reversibly. It follows that dqrev = CdT and therefore that dS = When we cannot assume that the heat capacity is constant over the temperature range of interest, which is the case for all solids at low temperatures, we have to allow for the variation of C with temperature. As we show in the following very brief Derivation, the result is Tf C dT T T i This, which is eqn 4.4, is our starting point. All we need recognize is the standard result from calculus, illustrated in Derivation 2.2 and The chemist’s toolkit 2.1, that the integral of a function between two limits is the area under the graph of the function between the two limits. In this case, the function is C/T, the heat capacity at each temperature divided by that temperature. The process of calculating an entropy change is illustrated in Fig. 4.8: 9/20/12 11:25 AM (a) Ti (a) 97 Cp /T Heat capacity, Cp ENTROPY Temperature, T Tf (b) Temperature, T Fig. 4.8 The experimental determination of the change in entropy of a sample that has a heat capacity that varies with temperature involves: (a) measuring the heat capacity over the range of temperatures of interest, then plotting Cp /T against T and (b) determining the area under the curve (the tinted area shown here). The heat capacity of all solids decreases toward zero as the temperature is reduced. • First we measure and tabulate the heat capacity C throughout the range of temperatures of interest (Fig. 4.8a). • Then we divide each one of the values of C by the corresponding temperature, to get C/T at each temperature, and plot these C/T against T (Fig. 4.8b). • Lastly, we evaluate the area under the graph between the temperatures Ti and Tf (Fig. 4.8b). The only way to proceed reliably is to fit the data to a polynomial in T and then to use a computer to evaluate the integral. 4.5 The entropy change accompanying a phase transition We can suspect that the entropy of a substance increases when it melts and when it boils because its molecules become more disordered as it changes from solid to liquid and from liquid to vapour. The transfer of energy as heat occurs reversibly when a solid is at its melting temperature. If the temperature of the surroundings is infinitesimally lower than that of the system, then energy flows out of the system as heat and the substance freezes. If the temperature is infinitesimally higher, then energy flows into the system as heat and the substance melts. Moreover, because the transition occurs at constant pressure, we can identify the heat transferred per mole of substance with the enthalpy of fusion (melting). Therefore, the entropy of fusion, ∆fusS, the change 9780199608119_C04.indd 97 (b) Fig. 4.9 When a solid, depicted by the orderly array of spheres (a), melts, the molecules form a liquid, the disorderly array of spheres (b). As a result, the entropy of the sample increases. of entropy per mole of substance, at the melting temperature, Tf (with f now denoting fusion), is ∆ fus S = ∆ fus H(Tf ) Tf At the melting temperature Entropy of fusion (4.6) Notice how we must use the enthalpy of fusion at the melting temperature. To get the standard entropy of fusion, ∆fusSa, at the melting temperature we use the melting temperature at 1 bar and the corresponding standard enthalpy of fusion at that temperature. All enthalpies of fusion are positive (melting is endothermic: it requires heat), so all entropies of fusion are positive too: disorder increases on melting. The entropy of water, for example, increases when it melts because the orderly structure of ice collapses as the liquid forms (Fig. 4.9). Brief illustration 4.5 The entropy of fusion From eqn 4.6 and the information in Table 3.1, the entropy of fusion of ice at 0 °C is DfusH (Tf ) of water 103 J 6.01 kJ mol−1 −2 DfusS = = + 2.20 × 10 kJ K −1 mol−1 273.15 K Tf of water = +22.0 J K−1 mol−1 The entropy of other types of transition may be discussed similarly. Thus, the entropy of vaporization, ∆ vapS, at the boiling temperature, Tb, of a liquid is related to its enthalpy of vaporization at that temperature by ∆ vap S = ∆ vap H(Tb ) Tb At the boiling temperature Entropy of vaporization (4.7) To use this formula, we use the enthalpy of vaporization at the boiling temperature. For the standard value, ∆vapSa, we use data corresponding to 1 bar. 9/20/12 11:25 AM 98 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Because vaporization is endothermic for all substances, all entropies of vaporization are positive. The increase in entropy accompanying vaporization is in line with what we should expect when a compact liquid turns into a gas. Brief illustration 4.6 The entropy of vaporization From eqn 4.7 and the information in Table 3.1, the entropy of vaporization of water at 100 °C is DvapH (Tb ) of water 40.7 kJ mol−1 D vapS = = + 1.09 × 10−1 kJ K −1 mol−1 373.2 K Tb of water almost the same for all liquids at their boiling temperatures. The exceptions to Trouton’s rule include liquids in which the interactions between molecules result in the liquid being less disordered than a random jumble of molecules. For example, the high value for water implies that the H2O molecules are linked together in some kind of ordered structure by hydrogen bonding, with the result that the entropy change is greater when this relatively ordered liquid forms a disordered gas. The high value for mercury has a similar explanation but stems from the presence of metallic bonding in the liquid, which organizes the atoms into more definite patterns than would be the case if such bonding were absent. = +109 J K−1 mol−1 Brief illustration 4.7 Trouton’s rule Entropies of vaporization shed light on an empirical relation known as Trouton’s rule: The quantity ∆vapHa(Tb)/Tb is approximately the same (and equal to about 85 J K−1 mol−1) for all liquids except when hydrogen bonding or some other kind of specific molecular interaction is present. Trouton’s rule The data in Table 4.1 provide support for this rule. We know that the quantity ∆vapHa(Tb)/Tb, however, is the entropy of vaporization of the liquid at its boiling point, so Trouton’s rule is explained if all liquids have approximately the same entropy of vaporization at their boiling points. This near equality is to be expected because when a liquid vaporizes, the compact condensed phase changes into a widely dispersed gas that occupies approximately the same volume whatever its identity. To a good approximation, therefore, we expect the increase in disorder, and therefore the entropy of vaporization, to be Table 4.1 D vapS/(J K−1 mol−1) 9780199608119_C04.indd 98 DvapH - ≈ (332.4 K) × (85 J K−1 mol−1) = 28 kJ mol−1 The experimental value is 29 kJ mol−1. Self-test 4.2 Estimate the enthalpy of vaporization of ethane from its boiling point, which is −88.6 °C. Answer: 16 kJ mol−1 To calculate the entropy of phase transition at a temperature other than the transition temperature, we have to do additional calculations, as shown in the following Example. Example 4.1 Entropies of vaporization at 1 atm and the normal boiling point Ammonia, NH3 Benzene, C6H6 Bromine, Br2 Carbon tetrachloride, CCl4 Cyclohexane, C6H12 Hydrogen sulfide, H2S Mercury, Hg(l) Water, H2O We can estimate the enthalpy of vaporization of liquid bromine from its boiling temperature, 59.2 °C. No hydrogen bonding or other kind of special interaction is present, so we use the rule after converting the boiling point to 332.4 K: 97.4 87.2 88.6 85.9 85.1 87.9 94.2 109.1 Calculating the entropy of vaporization Calculate the entropy of vaporization of water at 25 °C from thermodynamic data and its enthalpy of vaporization at its normal boiling point. Strategy The most convenient way to proceed is to perform three calculations. First, calculate the entropy change for heating liquid water from 25 °C to 100 °C (using eqn 4.3 with data for the liquid from Table 2.1). Then use eqn 4.7 and data from Table 3.1 to calculate the entropy of transition at 100 °C. Next, calculate the change in entropy for cooling the vapour from 100 °C to 25 °C (using eqn 4.3 again, but now with data for the vapour from Table 2.1). Finally, add the three contributions together. The steps may be hypothetical. 9/20/12 11:25 AM ENTROPY Answer From eqn 4.3 with data for the liquid from Table 2.1: T DS1 = C p,m(H2O,l) ln f Ti A typical resting person heats the surroundings at a rate of about 100 W. Estimate the entropy you generate in the surroundings in the course of a day at 20 °C. 373 K 298 K = +16.9 J K−1 mol−1 Strategy We can estimate the approximate change in entropy from eqn 4.8 once we have calculated the energy transferred as heat. To find this quantity, we use 1 W = 1 J s−1 and the fact that there are 86 400 s in a day. Convert the temperature to kelvins. From eqn 4.7 and data from Table 3.1: D vap H(Tb) Tb = 4.07 × 104 J K −1 mol−1 373 K = +109 J K−1 mol−1 Solution The heat transferred to the surroundings in the course of a day is From eqn 4.3 with data for the vapour from Table 2.1: DS1 = C p,m(H2O,g) ln Tf Ti qsur = (86 400 s) × (100 J s−1) = 86 400 × 100 J = (33.58 J K −1 mol−1) × ln = −7.54 J K −1 Example 4.2 Estimating the entropy change of the surroundings = (75.29 J K −1 mol−1) × ln DS 2 = 99 The increase in entropy of the surroundings is therefore 298 K 373 K DSsur = −1 mol The sum of the three entropy changes is the entropy of transition at 25 °C: DvapS (298 K) = DS1 + DS2 + DS3 = +118 J K−1 mol−1 q sur 86 400 × 100 J = = + 2.95 × 104 J K −1 T 293 K That is, the entropy production is about 30 kJ K−1. Just to stay alive, each person on the planet contributes about 30 kJ K−1 each day to the entropy of their surroundings. The use of transport, machinery, and communications generates far more in addition. Self-test 4.3 Calculate the entropy of vaporization of benzene at 25 °C from the following data: Tb = 353.2 K, DvapH -(Tb) = 30.8 kJ mol−1, Cp,m(l) = 136.1 J K−1 mol−1, Cp,m(g) = 81.6 J K−1 mol−1. −1 Answer: 96.4 J K mol −1 Self-test 4.4 Suppose a small reptile operates at 0.50 W. What entropy does it generate in the course of a day in the water in the lake that it inhabits, where the temperature is 15 °C? Answer: +150 J K−1 4.6 Entropy changes in the surroundings We can use the definition of entropy in eqn 4.1 to calculate the entropy change of the surroundings in contact with the system at the temperature T: ∆Ssur = qsur T The surroundings are so extensive that they remain at constant pressure regardless of any events taking place in the system, so qsur,rev = ∆Hsur. The enthalpy is a state function, so a change in its value is independent of the path and we get the same value of ∆Hsur regardless of how the heat is transferred. Therefore, we can drop the label ‘rev’ from q and write ∆Ssur = qsur T Entropy change of the surroundings in terms of heating of the surroundings (4.8) This formula can be used to calculate the entropy change of the surroundings regardless of whether the change in the system is reversible or not. 9780199608119_C04.indd 99 Equation 4.8 is expressed in terms of the energy supplied to the surroundings as heat, qsur. Normally, we have information about the energy supplied to or escaping from the system as heat, q. The two quantities are related by qsur = −q. For instance, if q = +100 J, an influx of 100 J, then qsur = −100 J, indicating that the surroundings have lost that 100 J. Therefore, at this stage we can replace qsur in eqn 4.8 by −q and write ∆Ssur = − q T Entropy change of the surroundings in terms of heating of the system (4.9) This expression is in terms of the properties of the system. Moreover, it applies whether or not the process taking place in the system is reversible. To gain insight into eqn 4.9, let’s consider two scenarios: • Suppose a perfect gas expands isothermally and reversibly from Vi to Vf. The entropy change of 9/20/12 11:25 AM 100 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW the gas itself (the system) is given by eqn 4.2. To calculate the entropy change in the surroundings, we note that q, the heat required to keep the temperature constant, is given in Derivation 4.1. Therefore, ∆Ssur = − q nRT ln(Vf /Vi ) =− T T cancel T = − nR ln Vf Vi The change of entropy in the surroundings is therefore the negative of the change in entropy of the system, and the total entropy change for the reversible process is zero. • Now suppose that the gas expands isothermally but freely (pex = 0) between the same two volumes. The change in entropy of the system is the same, because entropy is a state function. However, because ∆U = 0 for the isothermal expansion of a perfect gas and no work is done, no heat is taken in from the surroundings. Because q = 0, it follows from eqn 4.9 (which, remember, can be used for either reversible or irreversible heat transfers), that ∆Ssur = 0. The total change in entropy is therefore equal to the change in entropy of the system, which is positive. We see that for this irreversible process, the entropy of the universe has increased, in accord with the Second Law. they draw on information that we have not yet encountered. However, it is possible to understand the basis of the approach that we use there, and see how it illuminates what we have achieved so far. The fundamental equation that we need is the Boltzmann formula, which was originally proposed by Ludwig Boltzmann towards the end of the nineteenth century (and is carved as his epitaph on his tombstone): Brief illustration 4.8 The weight of a configuration Suppose we had a tiny system of four molecules A, B, C, and D that could occupy three equally spaced levels of energies 0, e, and 2e, and we know that the total energy is 4e. The 19 arrangements shown in Fig. 4.10 are possible, so W = 19 and the system has 19 microstates. Self-test 4.5 Constant pressure Entropy change of the surroundings (4.10) This enormously important expression will lie at the heart of our discussion of chemical equilibrium. We see that it is consistent with common sense: if the process is exothermic, ∆H is negative and therefore ∆Ssur is positive. The entropy of the surroundings increases if heat is released into them. If the process is endothermic (∆H > 0), then the entropy of the surroundings decreases. 4.7 The molecular interpretation of entropy We have referred frequently to ‘molecular disorder’ and have interpreted the thermodynamic quantity of entropy in terms of this so far ill-defined concept. The concept of disorder, however, can be expressed precisely and used to calculate entropies. The procedures required will be described in Chapter 22, for 9780199608119_C04.indd 100 (4.11) where k = 1.381 × 10−23 J K−1 is Boltzmann’s constant (Foundations 0.11). The quantity W is the number of ways that the molecules of the sample can be arranged yet correspond to the same total energy and formally is called the ‘weight’ of a ‘configuration’ of the sample. Each way to arrange the molecules in a system (under the constraint of keeping the total energy constant) is also called a ‘microstate’ of the system. If a chemical reaction or a phase transition takes place at constant pressure, we can identify q in eqn 4.9 with the change in enthalpy of the system, and obtain ∆Ssur = − ∆H T Boltzmann formula for the entropy S = k ln W How many arrangements are possible for three molecules with a total energy 3e that could occupy equally spaced levels of energies 0, e, and 2e? Answer: 7 2ε ε 0 2ε ε 0 2ε ε 0 C D B D B C A D A C A B A B A C A D B C B D C D D B C A C B D A B C D A A C D B D A C B C A D B D A B C B A D C A B D C C A B D B A C D A B C D 2ε ε A B C D 0 Fig. 4.10 The 19 arrangements of four molecules (represented by the blocks) in a system with three energy levels and a total energy of 4e. 9/20/12 11:25 AM ENTROPY (b) Energy (a) Fig. 4.11 As a box expands, the energy levels of the particles inside it come closer together. At a given temperature, the number of arrangements corresponding to the same total energy is greater when the energy levels are closely spaced than when they are far apart. The entropy calculated from the Boltzmann formula is sometimes called the statistical entropy. We see that, if W = 1, which corresponds to one microstate (only one way of achieving a given energy; all molecules in exactly the same state), then S = 0 because ln 1 = 0. However, if the system can exist in more than one microstate, then W > 1 and S > 0. These are important results, to which we shall return in Section 4.9. For now, we note that if the molecules in the system have access to a greater number of energy levels, then there may be more ways of achieving a given total energy. That is, there are more microstates for a given total energy, W is greater, and the entropy is greater than when fewer levels accessible. Therefore, the statistical view of entropy summarized by the Boltzmann formula is consistent with our previous statement that the entropy is related to the dispersal of energy. For instance, when a perfect gas expands, the available translational energy levels get closer together (Fig. 4.11; this is a conclusion from quantum theory that we verify in Chapter 12), so it is possible to distribute the molecules over them in more ways than when the volume of the container is small and the energy levels are further apart. Therefore, as the container expands, W and therefore S increase. It is no coincidence that the thermodynamic expression for ∆S (eqn 4.2) is proportional to a logarithm: the logarithm in Boltzmann’s formula turns out to lead to the same logarithmic expression (see Chapter 22). Brief illustration 4.9 The Boltzmann formula Suppose that a large, flexible polymer can adopt 1.0 × 1031 different conformations of the same energy. It follows from eqn 4.11 that the entropy of the polymer is 9780199608119_C04.indd 101 101 k W S = (1.381× 10−23 J K −1) × ln(1.0 × 1031) = 9.9 × 10−22 J K −1 Multiplication by Avogadro’s constant gives the molar entropy: Sm = 6.0 × 102 J K−1 mol−1. The Boltzmann formula also illuminates the thermodynamic definition of entropy (eqn 4.1) and in particular the role of the temperature. Molecules in a system at high temperature can occupy a large number of the available energy levels, so a small additional transfer of energy as heat will lead to a relatively small change in the number of accessible energy levels. Consequently, the number of microstates does not increase appreciably and nor does the entropy of the system. In contrast, the molecules in a system at low temperature have access to far fewer energy levels (at T = 0, only the lowest level is accessible), and the transfer of the same quantity of energy by heating will increase the number of accessible energy levels and the number of microstates significantly. Hence, the change in entropy upon heating will be greater when the energy is transferred to a cold body than when it is transferred to a hot body. This argument suggests that the change in entropy should be inversely proportional to the temperature at which the transfer takes place, as in eqn 4.1. 4.8 Absolute entropies and the Third Law of thermodynamics The graphical procedure summarized by Fig. 4.8 for the determination of the difference in entropy of a substance at two temperatures has a very important application. If Ti = 0, then the area under the graph between T = 0 and some temperature T gives us the value of ∆S = S(T) − S(0). We are supposing that there are no phase transitions below the temperature T. If there are any phase transitions (for example, melting) in the temperature range of interest, then the entropy of each transition at the transition temperature is calculated using an equation like eqn 4.6. In any case, at T = 0, all the motion of the atoms has been eliminated, and there is no thermal disorder. Moreover, if the substance is perfectly crystalline, with every atom in a well-defined location, then there is no spatial disorder either. We can therefore suspect that at T = 0, the entropy is zero. When we have done some quantum mechanics, we shall see that molecules cannot lose all their vibrational energy, so they retain some motion even at T = 0. However, they are then all in the same state (their lowest energy state), and so in this sense lack any thermal disorder. 9/20/12 11:25 AM ? 0 (a) Rhombic 369 Temperature, T/K Area Monoclinic ? 0 (b) Entropy, S ? +1.09 J K–1 mol–1 Cp / T Monoclinic Molar entropy, Sm/(J K–1 mol–1) Molar entropy, Sm/(J K–1 mol–1) 102 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Rhombic 369 Temperature, T/K Temperature, T The entropies of all perfectly crystalline substances are the same at T = 0. The Third Law For convenience (and in accord with our understanding of entropy as a measure of disorder), we take this common value to be zero. Then, with this convention, according to the Third Law, S(0) = 0 for all perfectly ordered crystalline materials. The Third-Law entropy at any temperature, S(T), is equal to the area under the graph of C/T between T = 0 and the temperature T (Fig. 4.13). If there are any phase transitions (for example, melting) in the temperature range of interest, then the entropy of each transition at the transition temperature is 9780199608119_C04.indd 102 Melt Boil Cp /T One example of the thermodynamic evidence for the conclusion that S(0) = 0 is as follows. Sulfur undergoes a phase transition from rhombic to monoclinic at 96 °C (369 K) and the enthalpy of transition is +402 J mol−1. The entropy of transition is therefore ∆S = (+402 J K−1 mol−1)/(369 K) = +1.09 J K−1 mol−1 at this temperature. We can also measure the molar entropy of each phase relative to its value at T = 0 by determining the heat capacity from T = 0 up to the transition temperature (Fig. 4.12). At this stage, we do not know the values of the entropies at T = 0. However, as we see from the illustration, to match the observed entropy of transition at 369 K, the molar entropies of the two crystalline forms must be the same at T = 0. We cannot say that the entropies are zero at T = 0, but from the experimental data we do know that they are the same. This observation is generalized into the Third Law of thermodynamics: Temperature, T Fig. 4.13 The absolute entropy (or Third-Law entropy) of a substance is calculated by extending the measurement of heat capacities down to T = 0 (or as close to that value as possible), and then determining the area of the graph of Cp /T against T up to the temperature of interest. The area is equal to the absolute entropy at the temperature T. Solid (a) Liquid Gas Tb Tf ∆vapS Entropy, S Fig. 4.12 (a) The molar entropies of monoclinic and rhombic sulfur vary with temperature as shown here. At this stage we do not know their values at T = 0. (b) When we slide the two curves together by matching their separation to the measured entropy of transition at the transition temperature, we find that the entropies of the two forms are the same at T = 0. ∆fusS (b) 0 Tf Tb Temperature, T Fig. 4.14 The determination of entropy from heat capacity data. (a) Variation of Cp /T with the temperature of the sample. (b) The entropy, which is equal to the area beneath the upper curve up to the temperature of interest plus the entropy of each phase transition between T = 0 and the temperature of interest. calculated like that in eqn 4.6 and its contribution added to the contributions from each of the phases, as shown in Fig. 4.14. The Third-Law entropy, which is commonly called simply ‘the entropy’, of a substance depends on the pressure; we therefore select a standard pressure (1 bar) and report the standard molar entropy, S a m, the molar entropy of a substance in its standard state at the temperature of interest. Some values at 298.15 K (the conventional temperature for reporting data) are given in Table 4.2. 9/20/12 11:25 AM ENTROPY Table 4.2 Standard molar entropies of some substances at 298.15 K* Substance = Sm /(J K−1 mol−1) 103 That is, the molar entropy at the low temperature T is equal to one-third of the constant-pressure heat capacity at that temperature. The Debye T 3 law strictly applies to Cp, but CV and Cp converge as T → 0, so we can use it for estimating CV too without significant error at low temperatures. Gases Ammonia, NH3 Carbon dioxide, CO2 Helium, He Hydrogen, H2 Neon, Ne Nitrogen, N2 Oxygen, O2 Water vapour, H2O 192.5 213.7 126.2 130.7 146.3 191.6 205.1 188.8 Derivation 4.4 Entropies close to T = 0 Once again, we use the general expression, eqn 4.4, for the entropy change accompanying a change of temperature deduced in Derivation 4.2, with DS interpreted as S(Tf ) − S(Ti ), taking molar values, and supposing that the heating takes place at constant pressure: Liquids Benzene, C6H6 Ethanol, CH3CH2OH Water, H2O Sm(Tf ) − Sm(Tf ) = 173.3 160.7 69.9 C p,m T Ti dT If we set Ti = 0 and Tf some general temperature T, we transform this expression into 0 Sm(T ) − Sm(0) = Solids Calcium oxide, CaO Calcium carbonate, CaCO3 Copper, Cu Diamond, C Graphite, C Lead, Pb Magnesium carbonate, MgCO3 Magnesium oxide, MgO Sodium chloride, NaCl Sucrose, C12H22O11 Tin, Sn (white) Sn (grey) Tf 39.8 92.9 33.2 2.4 5.7 64.8 65.7 26.9 72.1 360.2 51.6 44.1 TC 0 p ,m T dT According to the Third Law, S(0) = 0, and according to the Debye T 3 law, Cp,m = aT 3, so Sm(T ) = T T T aT 3 dT = a 0 T 2 At this point we can use the standard integral x dx = 2 1 3 x 3 + constant to write T *See the Data section for more values. T T 2dT = ( 13 T 3 + constant) 0 Heat capacities can be measured only with great difficulty at very low temperatures, particularly close to T = 0. However, as remarked in Section 2.10, it has been found that many non-metallic substances have a heat capacity that obeys the Debye T 3 law: At temperatures close to T = 0, Cp,m = aT 3 Debye T 3 law 9780199608119_C04.indd 103 Entropy at low temperatures 0 = ( 13 T 3 + constant) − constant = 13 T 3 We can conclude that Sm(T ) = 13 aT 3 = 13 Cp,m(T ) as in eqn 4.12b. (4.12a) where a is an empirical constant that depends on the substance and is found by fitting eqn 4.12a to a series of measurements of the heat capacity close to T = 0. With a determined, it is easy to deduce, as we show in the Derivation below, the molar entropy at low temperatures: At temperatures close to T = 0, Sm(T) = 13 Cp,m(T) dT 0 (4.12b) 4.9 The molecular interpretation of Third-Law entropies We can readily verify that the Boltzmann formula (eqn 4.11) agrees with the Third-Law value S(0) = 0. When T = 0, all the molecules must be in the lowest possible energy level. Because there is just a single arrangement of the molecules, W = 1, and as ln 1 = 0, eqn 4.11 gives S = 0 too. We can also see that the Boltzmann formula is consistent with the entropy of 9/20/12 11:25 AM Energy Energy 104 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW (a) (b) Fig. 4.15 The arrangements of molecules over the available energy levels determines the value of the statistical entropy. (a) At T = 0, there is only one arrangement possible: all the molecules must be in the lowest energy state. (b) When T > 0 several arrangements may correspond to the same total energy. In this simple case, W = 3. a substance always being positive (because W ≥ 1, the natural logarithm term in eqn 4.11 is never negative) and increasing with temperature. When T > 0, the molecules of a sample can occupy energy levels above the lowest one, and now many different arrangements of the molecules will correspond to the same total energy (Fig. 4.15). That is, when T > 0, W > 1 and according to eqn 4.11 the entropy rises above zero (because ln W > 0 when W > 1). It is worth spending a moment to look at the values in Table 4.2 to see that they are consistent with our molecular interpretation of entropy. For example, the standard molar entropy of diamond (2.4 J K−1 mol−1) is lower than that of graphite (5.7 J K−1 mol−1). This difference is consistent with the atoms being linked less rigidly in graphite than in diamond and their thermal disorder being correspondingly greater. The standard molar entropies of ice, water, and water vapour at 25 °C are, respectively, 45, 70, and 189 J K−1 mol−1, and the increase in values corresponds to the increasing disorder on going from a solid to a liquid and then to a gas. The Boltzmann formula also provides an explanation of a rather startling observation: the entropy of some substances is greater than zero at T = 0, apparently contrary to the Third Law. When the entropy of carbon monoxide gas is measured thermodynamically (from heat capacity and boiling point data, and assuming from the Third Law that the entropy is zero a at T = 0), it is found that Sm (298 K) = 192 J K−1 −1 mol . However, when Boltzmann’s formula is used and the relevant molecular data included, the standard molar entropy is calculated as 198 J K−1 mol−1. One explanation might be that the thermodynamic calculation failed to take into account a phase transition in solid carbon monoxide, which could have 9780199608119_C04.indd 104 Fig. 4.16 The positional disorder of a substance that accounts for the residual entropy of molecules that can adopt either of two orientations at T = 0 (in this case, CO). If there are N molecules in the sample, there are 2N possible arrangements with the same energy. contributed the missing 6 J K−1 mol−1. An alternative explanation is that the CO molecules are disordered in the solid, even at T = 0, and that there is a contribution to the entropy at T = 0 from positional disorder that is frozen in. This contribution is called the residual entropy of a solid. We can estimate the value of the residual entropy by using the Boltzmann formula and supposing that at T = 0 each CO molecule can lie in either of two orientations (Fig. 4.16). Then the total number of ways of arranging N molecules is (2 × 2 × 2. . . .)N times = 2N. Then S = k ln 2N = Nk ln 2 (We used ln xa = a ln x; see The chemist’s toolkit 2.2.) The molar residual entropy is obtained by replacing N by Avogadro’s constant: Sm = NAk ln 2 = R ln 2 This expression evaluates to 5.8 J K−1 mol−1, in good agreement with the value needed to bring the thermodynamic value into line with the statistical value, for instead of taking Sm(0) = 0 in the thermodynamic calculation, we should take Sm(0) = 5.8 J K−1 mol−1. Brief illustration 4.10 Residual entropy Ice has a residual entropy of 3.4 J K−1 mol−1. This value can be traced to the positional disorder of the location of the H atoms that lie between neighbouring molecules. Thus, although each H2O molecule has two short O—H covalent bonds and two long O. . . H—O bonds, there is a randomness in which bonds are long and which are short (Fig. 4.17). When the statistics of the disorder are analysed for a sample that contains N molecules, it turns out that W = (3/2)N. It follows that the residual entropy is expected to be S(0) = k ln (3/2)N = Nk ln 3/2, and therefore the molar residual entropy is Sm(0) = R ln 3/2, which evaluates to 3.4 J K−1 mol−1, in agreement with the experimental value. 9/20/12 11:25 AM ENTROPY 105 (H2O,l) − {2Sm (H2,g) + Sm (O2,g)} DrS- = 2Sm = 2(70 J K−1 mol−1) − {2(131 J K−1 mol−1) + (205 J K−1 mol−1)} = −327 J K−1 mol−1 A note on good practice Do not make the mistake of setting the standard molar entropies of elements equal to zero: they have nonzero values (provided T > 0), as we have already discussed. Fig. 4.17 The origin of the residual entropy of ice is the randomness in the location of the hydrogen atom in the O—H. . . O bonds between neighbouring molecules. Note that each molecule has two short O—H bonds and two long O. . . H hydrogen bonds. This schematic diagram shows one possible arrangement. Self-test 4.7 (a) Calculate the standard reaction entropy for N2(g) + 3 H2(g) → 2 NH3(g) at 25 °C. (b) What is the change in entropy when 2 mol H2 reacts? Answer: (a) (Using values from Table 4.2) −198.7 J K−1 mol−1; (b) −132.5 J K−1 Self-test 4.6 Calculate the residual molar entropy of a solid in which the molecules can adopt six orientations of equal energy at T = 0. Answer: Sm(0) = 14.9 J K−1 mol−1 4.10 The standard reaction entropy Now we move into the arena of chemistry, where reactants are transformed into products. When there is a net formation of a gas in a reaction, as in combustion, we can usually anticipate that the entropy increases. When there is a net consumption of gas, as in photosynthesis, it is usually safe to predict that the entropy decreases. However, for a quantitative value of the change in entropy, and to predict the sign of the change when no gases are involved, we need to make an explicit calculation. The difference in molar entropy between the products and the reactants in their standard states is called the standard reaction entropy, ∆rSa. It can be expressed in terms of the molar entropies of the substances in much the same way as we have already used for the standard reaction enthalpy: a ∆rSa = ∑vSm (products) a (reactants) − ∑vSm The standard reaction entropy (4.13) where the v are the stoichiometric coefficients in the chemical equation. Brief illustration 4.11 Standard reaction entropy For the reaction 2 H2(g) + O2(g) → 2 H2O(l) we expect a negative entropy of reaction as gases are consumed. To find the explicit value we use the values in the Data section to write 9780199608119_C04.indd 105 4.11 The spontaneity of chemical reactions The result of the calculation in Brief illustration 4.11 should be rather surprising at first sight. We know that the reaction between hydrogen and oxygen is spontaneous and, once initiated, that it proceeds with explosive violence. Nevertheless, the entropy change that accompanies it is negative: the reaction results in less disorder, yet it is spontaneous! The resolution of this apparent paradox underscores a feature of entropy that recurs throughout chemistry: it is essential to consider the entropy of both the system and its surroundings when deciding whether a process is spontaneous or not. The reduction in entropy by 327 J K−1 mol−1 in the reaction 2 H2(g) + O2(g) → 2 H2O(l) relates only to the system, the reaction mixture. To apply the Second Law correctly, we need to calculate the total entropy, the sum of the changes in the system and the surroundings that jointly compose the ‘isolated system’ referred to in the Second Law. It may well be the case that the entropy of the system decreases when a change takes place, but there may be a more than compensating increase in entropy of the surroundings so that overall the entropy change is positive. The opposite may also be true: a large decrease in entropy of the surroundings may occur when the entropy of the system increases. In that case we would be wrong to conclude from the increase of the system alone that the change is spontaneous. Whenever considering the implications of entropy, we must always consider the total change of the system and its surroundings. To calculate the entropy change in the surroundings when a reaction takes place at constant pressure, we use eqn 4.10, interpreting the ∆H in that expression as the standard reaction enthalpy, ∆rHa. 9/20/12 11:25 AM 106 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Brief illustration 4.12 Entropy change in the surroundings resulting from a chemical reaction Consider the water formation reaction, 2 H2(g) + O2(g) → 2 H2O(l), for which DrH - = −572 kJ mol−1. The change in entropy of the surroundings (which are maintained at 25 °C, the same temperature as the reaction mixture) is DrSsur DH(− 572 × 103 J mol−1) =− r =− T 298 K = +1.92 × 103 J K−1 mol−1 Now we can see that the total entropy change is positive: DrStotal = (−327 J K−1 mol−1) + (1.92 × 103 J K−1 mol−1) = +1.59 × 103 J K−1 mol−1 This calculation confirms that the reaction is spontaneous under standard conditions. In this case, the spontaneity is a result of the considerable disorder that the reaction generates in the surroundings: water is dragged into existence, even though H2O(l) has a lower entropy than the gaseous reactants, by the tendency of energy to disperse into the surroundings. Self-test 4.8 Calculate the entropy change in the surroundings for the reaction N2(g) + 3 H2(g) → 2 NH3(g) at 25 °C, given the standard enthalpy of reaction DrH - = −92.2 kJ mol−1 and standard entropy of reaction DrS - = −199 J K−1 mol−1 at this temperature. Answer: +111 J K−1 mol−1 The Gibbs energy One of the problems with entropy calculations is already apparent: we have to work out two entropy changes, the change in the system and the change in the surroundings, and then consider the sign of their sum. The great American theoretician J.W. Gibbs (1839–1903), who laid the foundations of chemical thermodynamics towards the end of the nineteenth century, discovered how to combine the two calculations into one. The combination of the two procedures in fact turns out to be of much greater relevance than just saving a little labour, and throughout this text we shall see consequences of the procedure he developed. 4.12 Focussing on the system The total entropy change that accompanies a process is ∆Stotal = ∆S + ∆Ssur, where ∆S is the entropy change 9780199608119_C04.indd 106 for the system; for a spontaneous change, ∆Stotal > 0. If the process occurs at constant pressure and temperature, we can use eqn 4.10 to express the change in entropy of the surroundings in terms of the enthalpy change of the system, ∆H. When the resulting expression is inserted into this one, we obtain ∆ Stotal = ∆ S − ∆H T Constant temperature and pressure Total entropy change (4.14) The great advantage of this formula is that it expresses the total entropy change of the system and its surroundings in terms of properties of the system alone. The only restriction is to changes at constant pressure and temperature. Now we take a very important step. First, we introduce the Gibbs energy, G, which is defined as G = H − TS Definition Gibbs energy (4.15) The Gibbs energy is commonly referred to as the ‘free energy’ and the ‘Gibbs free energy’. Because H, T, and S are state functions, G is a state function too. A change in Gibbs energy, ∆G, at constant temperature arises from changes in enthalpy and entropy, and is ∆G = ∆H − T∆S Constant temperature Gibbs energy change (4.16) By comparing eqns 4.14 and 4.16 we obtain ∆G = −T∆Stotal Constant temperature and pressure Gibbs energy change (4.17) We see that at constant temperature and pressure, the change in Gibbs energy of a system is proportional to the overall change in entropy of the system plus its surroundings. The difference in sign between ∆G and ∆Stotal implies that the condition for a process being spontaneous changes from ∆Stotal > 0 in terms of the total entropy (which is universally true) to ∆G < 0 in terms of the Gibbs energy (for processes occurring at constant temperature and pressure). That is, in a spontaneous change at constant temperature and pressure, the Gibbs energy decreases (Fig. 4.18). It may seem more natural to think of a system as falling to a lower value of some property. However, it must never be forgotten that to say that a system tends to fall to lower Gibbs energy is only a modified way of saying that a system and its surroundings jointly tend towards greater total entropy. The only criterion of spontaneous change is the total entropy of the system and its surroundings; the Gibbs energy merely contrives a way of expressing that total change in terms of the properties of the system alone, and is valid only for processes that occur at constant 9/20/12 11:25 AM THE GIBBS ENERGY chemical reaction to produce an electric current, like those used on the space shuttle—then up to 237 kJ of electrical energy can be generated for each mole of H2O produced. This energy is enough to keep a 60 W light bulb shining for about 1.1 h. If no attempt is made to extract any energy as work, then 286 kJ (in general, DH ) of energy will be produced as heat. If some of the energy released is used to do work, then up to 237 kJ (in general, DG) of non-expansion work can be obtained. Equilibrium Total entropy G or S 107 Gibbs energy Progress of change Fig. 4.18 (a) The criterion of spontaneous change is the increase in total entropy of the system and its surroundings. (b) Provided we accept the limitation of working at constant pressure and temperature, we can focus entirely on properties of the system, and express the criterion as a tendency to move to lower Gibbs energy. temperature and pressure. Every chemical reaction that is spontaneous under conditions of constant temperature and pressure, including those that drive the processes of growth, learning, and reproduction, are reactions that change in the direction of lower Gibbs energy, or—another way of expressing the same thing—result in the overall entropy of the system and its surroundings becoming greater. As well as providing a criterion of spontaneous change, another feature of the Gibbs energy is that the value of ∆G for a process gives the maximum non-expansion work that can be extracted from the process at constant temperature and pressure. By non-expansion work, w′, we mean any work other than that arising from the expansion of the system. It may include electrical work, if the process takes place inside an electrochemical or biological cell, or other kinds of mechanical work, such as the winding of a spring or the contraction of a muscle (we saw examples in Chapter 2). To demonstrate this property, we need to combine the First and Second Laws, and as shown in the following Derivation we find ∆G = w′max Gibbs energy and nonexpansion work (4.18) Brief illustration 4.13 Non-expansion work Experiments show that for the formation of 1 mol H2O(l) at 25 °C and 1 bar, DH = −286 kJ and DG = −237 kJ. It follows that up to 237 kJ of non-expansion work can be extracted from the reaction between hydrogen and oxygen to produce 1 mol H2O(l) at 25 °C. If the reaction takes place in a fuel cell—a device for using a 9780199608119_C04.indd 107 Maximum non-expansion work We need to consider infinitesimal changes, because dealing with reversible processes is then much easier. Our aim is to derive the relation between the infinitesimal change in Gibbs energy, dG, accompanying a process and the maximum amount of non-expansion work that the process can do, dw ′. We start with the infinitesimal form of eqn 4.16: At constant temperature: dG = dH − TdS where, as usual, d denotes an infinitesimal difference. A good rule in the manipulation of thermodynamic expressions is to feed in definitions of the terms that appear. We do this twice. First, we use the expression for the change in enthalpy at constant pressure (eqn 2.13; dH = dU + pdV ), and obtain At constant temperature and pressure: dG = dH − TdS = dU + pdV − TdS 4.13 Properties of the Gibbs energy Constant temperature and pressure Derivation 4.5 Then we replace dU in terms of infinitesimal contributions from work and heat (dU = dw + dq): dG = dU + pdV − TdS = dw + dq + pdV − TdS A note on good practice Recall from Chapter 2 that, because heat and work are processes and the internal energy is a property, the symbol d carries a different meaning when it is attached to q, w, and U. Namely, dU denotes a very small change in the internal energy of a system whereas dq and dw denote very small transfers of energy as heat and work. The work done on the system consists of expansion work, −pexdV, and non-expansion work, dw ′. Therefore, dG = dw + dq + pdV − TdS = −pexdV + dw ′ + dq + pdV − TdS This derivation is valid for any process taking place at constant temperature and pressure. Now we specialize to a reversible change. For expansion work to be reversible, we need to match p and pex, in which case the first and fourth terms on the right cancel. Moreover, because the heat transfer is also reversible, we can replace dq by TdS, in which case the third and fifth terms also cancel: dG = −pdV + dw ′ + TdS + pdV − TdS We are left with At constant temperature and pressure, for a reversible process: dG = dw rev ′ 9/20/12 11:25 AM 108 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Maximum work is done during a reversible change (Section 2.4), so another way of writing this expression is At constant temperature and pressure: dG = dw max ′ Because this relation holds for each infinitesimal step between the specified initial and final states, it applies to the overall change too. Therefore, we obtain eqn 4.18. Example 4.3 Estimating a change in Gibbs energy Suppose a certain small bird has a mass of 30 g. What is the minimum mass of glucose that it must consume to fly to a branch 10 m above the ground? The change in Gibbs energy that accompanies the oxidation of 1.0 mol C6H12O6(s) to carbon dioxide and water vapour at 25 °C is −2828 kJ. Strategy First, we need to calculate the work needed to raise a mass m through a height h on the surface of the Earth. As we saw in eqn 0.11, this work is equal to mgh, where g is the acceleration of free fall. This work, which is non-expansion work, can be identified with DG. We need to determine the amount of substance that corresponds to the required change in Gibbs energy, and then convert that amount to a mass by using the molar mass of glucose. Solution The non-expansion work to be done is w ′ = mgh = (30 × 10−3 kg) × (9.81 m s−2) × (10 m) = 3.0 × 9.81 × 1.0 ×10−1 J (because 1 kg m2 s−2 = 1 J). The amount, n, of glucose molecules required for oxidation to give a change in Gibbs energy of this value given that 1 mol provides 2828 kJ is n= The great importance of the Gibbs energy in chemistry is becoming apparent. At this stage, we see that it is a measure of the non-expansion work resources of chemical reactions: if we know ∆G, then we know the maximum non-expansion work that we can obtain by harnessing the reaction in some way. In some cases, the non-expansion work is extracted as electrical energy. This is the case when the reaction takes place in an electrochemical cell, of which a fuel cell is a special case, as we see in Chapter 9. In other cases, the reaction may be used to build other molecules. This is the case in biological cells, where the Gibbs energy available from the hydrolysis of ATP (adenosine triphosphate) to ADP is used to build proteins from amino acids, to power muscular contraction, and to drive the neuronal circuits in our brains. Indeed, simple arguments involving entropy and the Gibbs energy lead easily to an explanation of why life is consistent with the Second Law (see Impact 4.2). This interpretation can be expressed in terms of ∆G = ∆H − T∆S. Thus, we saw in Section 2.9 that in the absence of non-expansion work ∆H can be identified with the energy transferred as heat (at constant pressure). But in a reversible process T∆S is also equal to the energy transferred as heat even if other processes, including non-expansion work, are taking place. Therefore, the difference between these two quantities must represent the energy transfer due to these other processes, and specifically to the nonexpansion work that the system does. 3.0 × 9.81× 1.0 × 10−1 J 2.828 × 106 J mol−1 Impact on biology 4.2 can ncel J 3.0 × 9.81× 1.0 × 10−7 = mol 2.828 Life and the Second Law Therefore, because the molar mass, M, of glucose is 180 g mol−1, the mass, m, of glucose that must be oxidized is ⎛ 3.0 × 9.81× 1.0 × 10−7 ⎞ mol⎟ × (180 g mol−1) m = nM = ⎜ ⎜ ⎟ 2 . 828 ⎝ ⎠ cancel mol = 1.9 × 10−4 g That is, the bird must consume at least 0.19 mg of glucose for the mechanical effort (and more if it thinks about it). Self-test 4.9 A hard-working human brain, perhaps one that is grappling with physical chemistry, operates at about 25 W (1 W = 1 J s−1). What mass of glucose must be consumed to sustain that power output for an hour? Answer: 5.7 g 9780199608119_C04.indd 108 Every chemical reaction that is spontaneous under conditions of constant temperature and pressure, including those driving the processes of growth, learning, and reproduction, is a reaction that proceeds in the direction of lower Gibbs energy, or—another way of expressing the same thing— results in the overall entropy of the system and its surroundings becoming greater. With these ideas in mind, it is possible to explain why life, which can be regarded as a collection of biological processes taking place in a highly organized body, proceeds in accord with the Second Law of thermodynamics. Conditions in a cell ensure that many of the reactions that lead to the breakdown of food in organisms are spontaneous. For instance, the dissociation of large molecules, such as sugars and lipids, into smaller molecules leads to the dispersal of matter in the cell. Energy is also dispersed, as it is released upon reorganization of bonds in foods when they are oxidized. More difficult to rationalize is life’s requirement of organization of a very large number of molecules into biological cells, which in turn assemble into organisms. To be 9/20/12 11:25 AM ROAD MAP OF KEY EQUATIONS sure, the entropy of the system—the organism—is very low because matter becomes less dispersed when molecules assemble to form cells, tissues, organs, and so on. However, the lowering of the system’s entropy comes at the expense of an increase in the entropy of the surroundings. To understand this point, we need to know that cells grow and act by converting energy from the Sun or oxidation of foods partially into work. The remaining energy is released as 109 heat into the surroundings, so qsur > 0 and DSsur > 0. As with any process, life is spontaneous and organisms thrive as long as the increase in the entropy of the organism’s environment compensates for decreases in the entropy arising from the assembly of the organism. Alternatively, we may say that DG < 0 for the overall sum of physical and chemical changes that we call life. Checklist of key concepts 1 A spontaneous change is a change that has a tendency to occur without work having to be done to bring it about. 2 Matter and energy tend to disperse. 3 The Second Law states that the entropy of an isolated system tends to increase. 4 In general, the entropy change accompanying the heating of a system is equal to the area under the graph of C/T against T between the two temperatures of interest. 5 The Third Law of thermodynamics states that the entropies of all perfectly crystalline substances are the same at T = 0 (and may be taken to be zero). 6 The residual entropy of a substance is its entropy at T = 0 due to any positional disorder that remains. 7 At constant temperature and pressure, a system tends to change in the direction of decreasing Gibbs energy. 8 At constant temperature and pressure, the change in Gibbs energy accompanying a process is equal to the maximum non-expansion work the process can do. Road map of key equations Isothermal expansion, perfect gas C constant Heating, ∆S = q rev /T Entropy change Low T, non-metallic solid ∆S = nR ln(Vf /Vi) ∆S = C ln(Tf /Ti) ∆G = w'max Sm(T ) = 1/ 3Cp,m(T ) Constant pressure ∆G = ∆H − T∆S Phase transition ∆ trsS a(T trs) = ∆trsH a(Ttrs)/Ttrs Chemical reaction ∆ rS a = ΣnS ma (products) − ΣnS ma(reactants) Constant temperature S = k ln W Standard reaction entropy Boltzmann formula ∆Ssurr = –q/T Constant pressure ∆Ssurr = −∆H/T Entropy change of surroundings G = H − TS Gibbs energy ∆Stotal = ∆S − ∆H/T Total entropy change A blue box indicates a relation valid only for perfect gases. 9780199608119_C04.indd 109 9/20/12 11:25 AM 110 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Questions and exercises Discussion questions 4.1 Explain the single criterion for a physical process to be classified as spontaneous. 4.2 Explain in molecular terms why the entropy of a gas increases (a) with volume, (b) with temperature. 4.3 Justify Trouton’s rule. What are the sources of discrepancies? 4.4 Justify the identification of the statistical entropy with the thermodynamic entropy. 4.5 Under what circumstances may the properties of the system alone be used to identify the direction of spontaneous change? 4.6 The evolution of life requires the organization of a very large number of molecules into biological cells. Does the formation of living organisms violate the Second Law of thermodynamics? State your conclusion clearly and present detailed arguments to support it. Exercises Assume all gases are perfect unless stated otherwise. All thermochemical data are for 298.15 K. 4.1 A goldfish swims in a bowl of water at 20 °C. Over a period of time, the fish transfers 120 J to the water as a result of its metabolism. What is the change in entropy of the water, assuming no losses to the environment? 4.2 Suppose you put a cube of ice of mass 100 g into a glass of water at just above 0 °C. When the ice melts, about 33 kJ of energy is absorbed from the surroundings as heat. What is the change in entropy of (a) the sample (the ice), (b) the surroundings (the glass of water)? 4.3 A sample of aluminium of mass 1.00 kg is cooled at constant pressure from 300 K to 250 K. Calculate the energy that must be removed as heat and the change in entropy of the sample. The molar heat capacity of aluminium is 24.35 J K−1 mol−1. 4.4 What is the change in entropy when (a) 100 g of ice is melted at 0 °C, (b) 100 g of water is heated from 0 °C to 100 °C, and (c) 100 g of water is vaporized at 100 °C? Use the values for each step to calculate the change in entropy when 100 g of ice at is transformed to water vapour at 100 °C? Suppose that the changes are brought about by a heater that supplies energy at a constant rate, and sketch a graph showing (a) the change in temperature of the system, (b) the enthalpy of the system, (c) the entropy of the system as a function of time. 4.5 Calculate the change in molar entropy when a sample of nitrogen gas expands isothermally from 1.0 dm3 to 5.5 dm3. 4.6 A sample of carbon dioxide that initially occupies 15.0 dm3 at 250 K and 1.00 atm is compressed isothermally. Into what volume must the gas be compressed to reduce its entropy by 10.0 J K−1? 4.7 Whenever a gas expands isothermally—when we exhale, when a flask is opened, and so on—the gas under- 9780199608119_C04.indd 110 goes an increase in entropy. A sample of methane gas of mass 15 g at 260 K and 105 kPa expands isothermally and (a) reversibly, (b) irreversibly until its pressure is 1.50 kPa. Calculate the change in entropy of the gas. 4.8 What is the change in entropy of 100 g of water when it is heated from room temperature (20 °C) to body temperature (37 °C)? Use Cp,m = 75.5 J K−1 mol−1. 4.9 Calculate the change in entropy of 1.0 kg of lead when it cools from 500 °C to 100 °C. Take Cp,m = 26.44 J K−1 mol−1. 4.10 Calculate the change in molar entropy when a sample of argon is compressed from 2.0 dm3 to 500 cm3 and simultaneously heated from 300 K to 400 K. Take CV,m = 32 R. 4.11 A monatomic perfect gas at a temperature Ti is expanded isothermally to twice its initial volume. To what temperature should it be cooled to restore its entropy to its initial value? Take CV,m = 32 R. 4.12 In a certain cyclic engine (technically, a Carnot cycle), a perfect gas expands isothermally and reversibly, then adiabatically (q = 0) and reversibly. In the adiabatic expansion step the temperature falls. At the end of the expansion stage, the sample is compressed reversibly first isothermally and then adiabatically in such a way as to end up at the starting volume and temperature. Draw a graph of entropy against temperature for the entire cycle. 4.13 Estimate the molar entropy of potassium chloride at 5.0 K given that its molar heat capacity at that temperature is 1.2 mJ K−1 mol−1. 4.14 Calculate the change in entropy when 100 g of water at 80 °C is poured into 100 g of water at 10 °C in an insulated vessel given that Cp,m = 75.5 J K−1 mol−1. 4.15 The enthalpy of the graphite → diamond phase transition, which under 100 kbar occurs at 2000 K, is +1.9 kJ mol−1. Calculate the entropy of transition at this temperature. 9/20/12 11:25 AM QUESTIONS AND EXERCISES 4.16 The enthalpy of vaporization of chloroform (trichloromethane), CHCl3, is 29.4 kJ mol−1 at its normal boiling point of 334.88 K. (a) Calculate the entropy of vaporization of chloroform at this temperature. (b) What is the entropy change in the surroundings? 4.17 Calculate the entropy of fusion of a compound at 25 °C given that its enthalpy of fusion is 36 kJ mol−1 at its melting point of 151 °C and the molar heat capacities (at constant pressure) of the liquid and solid forms are 33 J K−1 mol−1 and 17 J K−1 mol−1, respectively. 4.18 Octane is typical of the components of gasoline. Estimate (a) the entropy of vaporization, (b) the enthalpy of vaporization of octane, which boils at 126 °C at 1 atm. 4.19 Suppose that the weight of a configuration of N molecules in a gas of volume V is proportional to V N. Use Boltzmann’s formula to deduce the change in entropy when the gas expands isothermally from Vi to Vf. 4.20 An FClO3 molecule can adopt four orientations in the solid with negligible difference in energy. What is its residual molar entropy? 4.21 Without performing a calculation, estimate whether the standard entropies of the following reactions are positive or negative: trypsin → (a) Ala–Ser–Thr–Lys–Gly–Arg–Ser ⎯⎯⎯⎯ Ala–Ser–Thr–Lys + Gly–Arg (b) N2(g) + 3 H2(g) → 2 NH3(g) (c) ATP4−(aq) + 2 H2O(l) → ADP3−(aq) + HPO42−(aq) + H3O+(aq) 4.22 Use the information on standard molar entropies in the Data section to calculate the standard reaction entropy at 298 K of (a) 2 CH3CHO(g) + O2(g) → 2 CH3COOH(l) (b) 2 AgCl(s) + Br2(l) → 2 AgBr(s) + Cl2(g) (c) Hg(l) + Cl2(g) → HgCl2(s) (d) Zn(s) + Cu (aq) → Zn (aq) + Cu(s) 2+ 2+ (e) C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l) 4.23 Suppose that when you exercise, you consume 100 g of glucose and that all the energy released as heat remains in your body at 37 °C. What is the change in entropy of your body? 4.24 Calculate the standard reaction entropy and the change in entropy of the surroundings (at 298 K) of the reaction N2(g) + 3 H2(g) → 2 NH3(g). 111 nonlinear gaseous molecules are approximately 4R. Estimate the change in standard reaction entropy of the following two reactions when the temperature is increased by 10 K at constant pressure: (a) 2 H2(g) + O2(g) → 2 H2O(g) (b) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 4.26 Use the information you deduced in Exercise 4.24 to calculate the standard Gibbs energy of reaction of N2(g) + 3 H2(g) → 2 NH3(g). 4.27 In a particular biological reaction taking place in the body at 37 °C, the change in enthalpy was −135 kJ mol−1 and the change in entropy was −136 J K−1 mol−1. (a) Calculate the change in Gibbs energy. (b) Is the reaction spontaneous? (c) Calculate the total change in entropy of the system and the surroundings. 4.28 The change in Gibbs energy that accompanies the oxidation of C6H12O6(s) to carbon dioxide and water vapour at 25 °C is −2828 kJ mol−1. How much glucose does a person of mass 65 kg need to consume to climb through 10 m? 4.29 Fuel cells are being developed that make use of organic fuels; in due course they might be used to power tiny intravenous machines for carrying out repairs on diseased tissue. What is the maximum non-expansion work that can be obtained from the metabolism of 1.0 mg of sucrose to carbon dioxide and water? 4.30 The formation of glutamine from glutamate and ammonium ions requires 14.2 kJ mol−1 of energy input. It is driven by the hydrolysis of ATP to ADP mediated by the enzyme glutamine synthetase. (a) Given that the change in Gibbs energy for the hydrolysis of ATP corresponds to DG = −31 kJ mol−1 under the conditions prevailing in a typical cell, can the hydrolysis drive the formation of glutamine? (b) How many moles of ATP must be hydrolysed to form 1 mol glutamine? 4.31 The hydrolysis of acetyl phosphate has DG = −42 kJ mol−1 under typical biological conditions. If acetyl phosphate were to be synthesized by coupling to the hydrolysis of ATP, what is the minimum number of ATP molecules that would need to be involved? 4.32 Suppose that the radius of a typical cell is 10 mm and that inside it 106 ATP molecules are hydrolysed each second. What is the power density of the cell in watts per cubic metre (1 W = 1 J s−1)? A computer battery delivers about 15 W and has a volume of 100 cm3. Which has the greater power density, the cell or the battery? (For data, see Exercise 4.31.) 4.25 The constant-pressure molar heat capacities of linear gaseous molecules are approximately 72 R and those of 9780199608119_C04.indd 111 9/20/12 11:25 AM 112 CHAPTER 4 THERMODYNAMICS: THE SECOND LAW Projects The symbol ‡ indicates that calculus is required. 4.33‡ Equation 4.3 is based on the assumption that the heat capacity is independent of temperature. Suppose, instead, that the heat capacity depends on temperature as C = a + bT + c/T 2. Find an expression for the change of entropy accompanying heating from Ti to Tf. Hint: See Derivation 4.2. 4.34 Here we explore the thermodynamics of refrigerators and heat pumps. (a) Follow the arguments used in the definition of the efficiency of a heat engine in Impact on technology 4.1, to show that the best coefficient of cooling performance, ccool, the ratio of the energy extracted as heat at Tcold to the energy supplied as work in a perfect refrigerator, is ccool = Tcold /(Thot − Tcold). What is the maximum rate of extraction of 9780199608119_C04.indd 112 energy as heat in a domestic refrigerator rated at 200 W operating at 5.0 °C in a room at 22 °C? (b) Show that the best coefficient of heating performance, cwarm, the ratio of the energy produced as heat at Thot to the energy supplied as work in a perfect heat pump, is cwarm = Thot /(Thot − Tcold). What is the maximum power rating of a heat pump that consumes power at 2.5 kW operating at 18.0 °C and warming a room at 22 °C? 4.35‡ The temperature dependence of the heat capacity of non-metallic solids is found to follow the Debye T 3 law at very low temperatures, with Cp,m = aT 3. (a) Derive an expression for the change in molar entropy on heating for such a solid. (b) For solid nitrogen a = 6.15 × 10−3 J K−4 mol−1. What is the molar entropy of solid nitrogen at 5 K? 9/20/12 11:25 AM