Chapter 3: The Second Law: The Entropy of the Universe Increases PDF

Chapter 3 The Second Law: The Entropy of the Universe Increases Concepts Water does not flow spontaneously up a waterfall to enlarge a pool at the top, and a candle never assembles a column of wax out of thin air by burning in reverse. Expressing this intuition of the direction of natural processes...

Chapter 3 The Second Law: The Entropy of the Universe Increases Concepts Water does not flow spontaneously up a waterfall to enlarge a pool at the top, and a candle never assembles a column of wax out of thin air by burning in reverse. Expressing this intuition of the direction of natural processes in objective and precise language is the domain of the Second Law of Thermodynamics. The Second Law plays a crucial role in the discovery of a new variable of state, the entropy S, which is recognized as a measure of the disorder of the system, although the terms disorder and order do not capture the true meaning of entropy that was discovered by Ludwig Boltzmann (see sidebar). In an isolated system (which exchanges neither energy nor matter with the surroundings), every change of the system increases its entropy. If the system is not isolated from the surroundings, then the entropy of the system can decrease provided there is a larger increase in entropy of the surroundings. Thus, an elegant restatement of the Second Law emerges: the entropy of the universe—the system plus the surroundings—always increases. An additional variable of state, the Gibbs free energy G, can be defined as the enthalpy H minus the product TS. The Gibbs energy always decreases in a process that occurs spontaneously at constant temperature and pressure, important conditions for biochemical processes. By calculating a free-energy change for a reaction, we can tell if the reaction proceeds spontaneously at constant temperature and pressure. S = k lnW Inscribed on Ludwig Boltzmann’s tombstone in Vienna is the equation S = k ln W, meaning entropy is proportional to the logarithm of the number of different ways W a system can be rearranged without perceptibly changing it. Entropy had already been defined as a thermodynamic variable of state, but Boltzmann gave it a statistical meaning. Applications The First Law of Thermodynamics does not tell us whether a particular reaction will spontaneously occur; it just says that whichever way a reaction goes, the energy must be conserved. The Second Law of Thermodynamics specifies a criterion that predicts the direction of spontaneous change for a system. Living organisms maintain homeostasis—the stable and coordinated operation of all organismal processes such as respiration, etc.—by creating conditions that control the directions of useful chemical reactions. Consider that a voltage is established across a cell membrane to drive the synthesis of ATP. The dephosphorylation of this ATP may be coupled to a reaction that requires a certain minimum concentration of ATP to occur spontaneously in the cell. A knowledge of thermodynamics allows us to learn how the conditions can be changed to make impossible reactions possible. A critical limitation is that thermodynamics tells us a process is spontaneous, but not how fast it will actually occur, or how the rate will depend on the reaction conditions. For example, diamond at atmospheric pressure will convert spontaneously to graphite, but this 555 56 Chapter 3 | The Second Law: The Entropy of the Universe Increases (a) p1 , V1 I p2 , V2 Thot q1, w1 Thot IV q4, w4 q2, w2 II p4 , V4 q3, w3 p3 , V3 Tcold III Tcold (b) Thot 2.0 p1, V1 I p (bar) p2, V 2 IV Tcold Toward the Second Law: The Carnot Cycle In the early 1800s, steam engines were common but poorly understood. In 1824, Sadi Carnot published an analysis of an idealized heat engine that underwent a particular cycle of four steps to return to its original state. Unlike the operation of a real engine, such as a steam or automobile engine, all steps in this idealized engine are reversible. We will apply the Carnot cycle to an engine composed of an ideal gas in a piston cylinder (see figures 3.1 and 3.2). In step I, a hot ideal gas at Thot expands isothermally and reversibly, with work w1 and heat q1. The gas then expands adiabatically and reversibly in step II (q2 = 0). Recognize that in step II the engine does work (w2 is negative) without heat input, and so its energy drops and the temperature of the gas is reduced to Tcold. In step III, the gas is compressed isothermally and reversibly (w3, q3 ). In the last step, the gas is compressed adiabatically and reversibly (w4, q4 = 0) to return to its original condition. For one cycle, the total work is w = (w1 + w2 + w3 + w4 ), and the total heat absorbed is q = (q1 + q2 + q3 + q4 ). Since the initial and final states are the same for the cyclic process, we know that U = q + w = 0 and so q = -w. The First Law can be employed to analyze the heat and work for each step of the Carnot cycle. For the first step (an isothermal reversible expansion), V2 V2 . w 1 = - 3 p dV = -nRT hot ln V1 V1 II An ideal gas at constant temperature must have constant U (chapter 2). So the First Law for step I is q1 + w1 = 0, giving p4, V 4 1.0 process is so kinetically limited that it cannot be observed on human time scales. The rates of reactions are discussed in the chapters on kinetics. The application of thermodynamics is universal and is not limited to chemical reactions. For example, the Second Law is often framed as a constraint on the design of engines: no engine can be constructed, no matter how ingenious its mechanical design, which can convert heat into work without also discharging some heat to the surroundings. III q1 = -w 1 = nRT hot ln p3, V3 12 16 V (L) 20 FIGURE 3.1 (a) The Carnot cycle of an idealized heat engine. Each step is carried out reversibly. Steps I and III are carried out isothermally (step I at Thot and step III at Tcold) and steps II and IV adiabatically (q2 = q4 = 0). (b) A pV diagram for a Carnot cycle, illustrating the four reversible steps taken in moving between the states in (a). V2 . V1 The second step is an adiabatic reversible expansion, so q2 = 0 and the First Law gives w 2 + 0 = U = CV (T cold - T hot ), using the fact that the energy of an ideal gas depends only on temperature (Eq. 2.25). We will take a moment to find some additional helpful relationships for reversible adiabatic processes of ideal gases such as steps II and IV. Since dq = 0 and the differential of work is -pdV , we have dU = -pdV . Given that dU = CV dT is universally true, then C V dT = -pdV = - nRT dV . V Dividing both sides by T and integrating, we have for step II, Tcold CV 3 Thot CV ln V 3 dT dV = -nR 3 , T V2 V T cold V3 V2 = -nR ln = nR ln . T hot V2 V3 (3.1a) Toward the Second Law: The Carnot Cycle | 57 p2, V2, Thot p3, V3, Tcold p3, V3, Tcold p2, V2, Thot p4, V4, Tcold p1 , V1, Thot p1, V1, Thot Thot Tcold Thot Tcold I. Isothermal reversible expansion q1 is positive w1 is negative p4, V4, Tcold II. Adiabatic reversible expansion q2 = 0 w2 is negative III. Isothermal reversible compression q3 is negative w3 is positive FIGURE 3.2 A sketch of the four steps of the Carnot-cycle heat engine. The engine is in thermal contact with a hot heat reservoir at Thot (step I), a cold heat reservoir at Tcold (step III), or is thermally isolated (steps II and IV). Steps I and II are expansions in which work is done by the engine. Steps III and IV are compressions in which work is done on the engine. Thus, we have a relation between the temperature and volume changes for an adiabatic reversible expansion (or compression) of an ideal gas that applies for steps II and IV. The third and fourth steps are similar to the first and second. For the isothermal step III, V4 . q3 = -w 3 = nRT cold ln V3 For the adiabatic step IV, we have q4 = 0 and w 4 = C V (Thot - Tcold ) and also CV ln T hot V4 = nR ln . T cold V1 (3.1b) The total heat absorbed is the sum over all steps qcycle = q1 + q2 + q3 + q4 = nRT hot ln V2 V4 + 0 + nRT cold ln + 0, V1 V3 and the total work done by the engine is also summed to give -wcycle = - (w1 + w2 + w3 + w4 ) = nRThot ln V2 V4 + nRTcold ln , V1 V3 since w2 and w4 cancel. It is a good check to see that the cycle conforms to the First Law (e.g., Ucycle = qcycle + w cycle = 0). Notice that unlike U, neither q nor w is zero for this cyclic path, a reminder that heat and work are not variables of state (chapter 2). Carnot made an important finding that although the sum of qi for all the steps of this reversible cycle is not zero, the sum of the quantities qrev >T over the path is zero. The subscript “rev” is used to emphasize that the heat exchange occurs reversibly. In general, qrev >T can be evaluated by integrating dqrev >T , with dqrev representing an infinitesimal heat change along a reversible path and T being the temperature at which dqrev is exchanged.* For the isothermal steps I and III, *Although we use the symbol dq to express a differential change in heat, we should keep in mind that there is an important difference between dq and the differential of a state function, such as dU. For any cyclic path from state 1 back to state 1, the integral dU is always zero. But dq is dependent on the particular path, and its integral is generally not zero for a cyclic path. Mathematically, we say that dU is an exact differential, but dq (and dw as well) is not. IV. Adiabatic reversible compression q4 = 0 w4 is positive 58 Chapter 3 | The Second Law: The Entropy of the Universe Increases dqrev q1 1 3 T = T hot 3 dqrev = T hot , (Step I) q3 dqrev 1 3 T = T cold 3 dqrev = T cold . (Step III) For steps II and IV, the temperature is not constant, but dqrev is zero (adiabatic), giving dqrev 3 T = 0. (Steps II and IV) Then 3 dqrev >T for the cyclic path reduces to the contributions from steps I and III, q3 q1 V2 V4 V 2V 4 + = nR ln + nR ln = nR ln . T hot T cold V1 V3 V 1V 3 Now add Eqs. 3.1a and 3.1b, to find a useful identity for the Carnot cycle, nR ln V 2V 4 = 0, V 3V 1 and therefore, q1 q3 + = 0. T hot T cold (3.2) Equation 3.2 is an important finding. Our system (an ideal gas in a cylinder) has gone through a cycle and returned to its original state. Carnot recognized that the sum of the quantities 3 dqrev >T for this cyclic path is zero and could represent a state function, just as we have learned that state functions such as U and H exhibit no change for a cyclic path. But before we get too excited about this realization, we must ask two questions. First, is Eq. 3.2 just a special result of choosing an ideal gas or can it apply to all matter? Second, is Eq. 3.2 only an outcome of the Carnot cycle, or is it true for any reversible cycle? The First Law of Thermodynamics cannot answer these questions. In deducing that Eq. 3.2 is general for all reversible cycles, Carnot was the first to clearly specify the Second Law of Thermodynamics. A New State Function, Entropy Carnot had a superb intuition for his model steam engine, but there is some mystery about how much he knew as he withheld equations. In modern terms, we can tackle the preceding questions about Eq. 3.2 by analyzing the efficiency of the Carnot cycle as the total work done by the engine in one complete cycle -w divided by the heat absorbed at the hot heat reservoir qhot. For the Carnot engine, -w -w efficiency = = . (3.3) qhot q1 The rest of the energy is discharged as heat qcold at the cooler heat reservoir (qcold = q3 for the Carnot engine) and is wasted, as far as the efficiency for converting heat into work is concerned. With a little algebra, we can solve for efficiency in terms of the temperatures of the heat reservoirs. From Eq. 3.2, -q3 q1 = . (3.4) T hot T cold From the First Law, the total work done by the engine in one complete cycle is -w = q1 + q3 . A New State Function, Entropy | 59 Thus, q1 + q3 -w -w = = , qhot q1 q1 q3 = 1 + . q1 efficiency = (3.3a) It follows from Eq. (3.4) that q3 T cold = , q1 T hot giving the final result in terms of temperature, efficiency = 1- T cold . T hot (3.5) For any heat engine that operates reversibly at all steps, the engine can also be operated in the reverse direction as a refrigerator or heat pump. In the forward direction as a heat engine, there is a net transfer of heat from the hot reservoir to the cold one, and the system does work on the surroundings. In the reverse direction, as a heat pump, the surroundings perform work on the system, and there is a net flow of heat from the cold reservoir to the hot one. Carnot recognized the impossibility of a certain kind of perpetual motion, and he concluded that all heat engines operating with reversible cycles between Thot and Tcold must have the same efficiency. We will illustrate the argument in modern terms with a specific example (figure 3.3). Suppose Thot and Tcold are 1200 and 300 K, respectively. The efficiency of a Carnot engine is therefore 1 - 300>1200, or 0.75 (75%). If the engine takes up 100 kJ of heat in step I (qhot = 100 kJ), 25 kJ would be discharged in step III (qcold = -25 kJ), and the engine does 75 kJ of work (w = -75 kJ) in each cycle. Suppose another hypothetical engine also operates on a cycle of reversible steps but has a lower efficiency of 50%. To get 75 kJ of work out of it, 150 kJ would have to be absorbed at Thot and 75 kJ would be discharged at Tcold. Although this second engine is an inferior heat engine compared with the Carnot engine, it becomes a superior heat pump when operated in reverse: an input of 75 kJ of work (w = 75 kJ) would lead to the absorption of 75 kJ of heat at the cooler heat reservoir (qcold = 75 kJ) and the delivery of a total of 150 kJ of heat to the hotter heat reservoir (qhot = -150 kJ). If the work output of the Carnot engine is used to drive the hypothetical heat pump, we can see that the total qhot for the combined engine is 100 kJ - 150 kJ = -50 kJ, and the total qcold for the combined engine is -25 kJ + 75 kJ = 50 kJ. The combined effect of the two reversible engines is a net flow of 50 kJ of heat per cycle from the cooler reservoir to the hotter reservoir, with no change in the surroundings. If any two heat engines operating with reversible cycles between Thot and Tcold differ in their efficiencies, we can always operate the less efficient heat engine in reverse as a heat pump and use the more efficient Carnot Hypothetical 1200 K 1200 K q=+100 kJ 1200 K q=-150 kJ w=-75 kJ η=75% + w=+75 kJ η=50% q=-25 kJ q=-50 kJ = Impossible Engine q=+75 kJ 300 K 300 K q=+50 kJ 300 K “the maximum of [work] resulting from the employment of steam is also the maximum of [work] realizable by any means whatever” —S. Carnot, 1824* Carnot published his extraordinary findings in 1824 in a popular science format, with few equations. His work was largely unnoticed in its time, but has since been recognized as the first realization and application of the Second Law. *Note that [work] is substituted for Thurston’s translation of motive power (Carnot S. 1890. Reflections on the Motive Power of Heat, transl. R. H. Thurston. Macmillan and Co.) FIGURE 3.3 If two reversible heat engines (grey circles) each operate between the same heat baths, but with different efficiencies, then they form a new engine whose only result is the transfer of heat from a cold bath to a hot bath with no work on or by the surroundings, a violation of the Second Law of Thermodynamics. Thus all reversible heat engines must have the same efficiency. 60 Chapter 3 | The Second Law: The Entropy of the Universe Increases FIGURE 3.4 Two views are depicted of the same Carnot cycle. Since U = 0 for the cycle, the First Law requires wcycle = - qcycle. Then the shaded areas bounded by the reversible pathways represented in (a) pV or (b) TS coordinates must be equal. The integral symbol with the circular arrow represents summing the integral over the whole cycle. (a) (b) Thot I p I Thot IV T Tcold IV II Tcold III II III V S = p dV T dS one as a heat engine to achieve the result of figure 3.3. The combined actions of these two engines would contradict vast experience that heat flows spontaneously strictly only from a hotter reservoir to a cooler reservoir without changing the surroundings. Therefore, the only way to avoid the impossible engine in figure 3.3 is to conclude that all reversible engines have the same efficiency. If all reversible engines have the same efficiency, then Eq. 3.4 and therefore Eq. 3.2 cannot be a consequence of using an ideal gas in our engine or of the particular paths chosen. In other words, we have defined a new state function S, entropy: dqrev S = 3 , T which for isothermal processes is then S = qrev . T (3.6) Entropy is an extensive variable of state; S depends on the initial and final states and the amount of the system. From Eq. 3.6, one sees that the dimensions of entropy are energy/ temperature; we will use units of J K - 1. We have “proven” that entropy is a state function because we accepted as true that heat flows spontaneously from a hot body to a cooler one. We can now see an alternative depiction of the four reversible steps of the Carnot cycle by representing the pathway through T and S coordinates (figure 3.4). It is especially clear in this depiction that steps II and IV are isentropic. The Second Law of Thermodynamics When we calculate the entropy changes of both the system and the surroundings for simple processes that can occur spontaneously—such as the flow of heat from a hot body to a cooler one, the expansion of an ideal gas into a vacuum, or the flow of water down a hill—we find that the sum of the entropy changes is not zero. Unlike energy, entropy is not conserved. The generalization of a great deal of experience, which is the Second Law of Thermodynamics, is that the sum of the entropy changes of the system and the surroundings is always positive. Even zero values can be approached only as a limit, and negative values are never found. If there is a decrease in entropy in a system, there must be an equal or larger increase in entropy in the surroundings: S (system) + S (surroundings) Ú 0 . (3.7) For an isolated system, since there is no energy or material exchange between such a system and the surroundings, there is no change in the surroundings. Therefore, S (isolated system) Ú 0 . (3.8) The Second Law of Thermodynamics | 61 The Ú sign means greater than or equal to; the former applies for irreversible processes and the latter for reversible ones. There are many different ways of stating the Second Law. That heat spontaneously flows from a hot body to a cold body, for example, can be taken as a statement of this law. We express the Second Law by Eqs. 3.7 and 3.8 because these choices are more convenient for problems in which chemists and bioscientists are interested. E X A M P L E 3 .1 One mole of an ideal gas initially at p1 = 2 bar, T, and V1 expands to p2 = 1 bar, T, and 2V1. Consider two paths: (a) irreversible expansion into a vacuum, as shown next, and (b) the expansion is reversible (not depicted). Calculate qirrev, S (system) and S (surroundings) for (a) and qrev, S (system), and S (surroundings) for (b). 2 bar V1, T Vacuum Open 1 bar V1, T 1 bar V1, T valve V2 = 2V1 SOLUTION Recall that S (system) is independent of path and must be the same for paths (a) and (b) because the initial and final states of the system are the same. However, q depends on the path, and the change in the state of the surroundings is different for (a) and (b). Thus, the values of S (surroundings) will be different. Path a: Begin with the First Law analysis for this path, w = 0 (no work done against surroundings) U = 0 (U for an ideal gas is independent of volume) qirrev = U - w = 0 Since no heat or work exchange between the system and surroundings occurs, the system is isolated and so the surroundings undergoes no change at all. Then, S (surroundings) = 0 . Since S (system) must be computed over a reversible path, we can choose the path of (b), in which the gas expands isothermally and reversibly, V2 V2 w rev = - 3 pdV = -RT 3 V1 V1 V2 dV = -RT ln V V1 = -RT ln 2 . Since U = 0 for both paths, S (system) = qrev U - w rev 0 - ( -R ln 2) = = = R ln 2 . T T T So, for the spontaneous, irreversible process in part (a), S (system) + S (surroundings) = R ln 2 + 0 7 0 , consistent with Eqs. 3.7 and 3.8. 62 Chapter 3 | The Second Law: The Entropy of the Universe Increases Path b: We have already calculated for this path that qrev = RT ln 2 , S (system) = R ln 2 . For path b, an amount qrev of heat is transferred reversibly into the system from the surroundings. This could occur if the surrounding temperature is higher than T by an infinitesimal amount, so that the surrounding temperature is still T. For the surroundings, the heat input is -qrev and -qrev S (surroundings) = = -R ln 2 . T Note that for this reversible process, S (system) + S (surroundings) = 0 , consistent with Eq. 3.7. Molecular Interpretation of Entropy The Second Law tells us that the entropy of the universe is increasing. This does not mean that a small part of the universe (the system) cannot have its entropy decrease. When we freeze liquid water to solid ice in an ice tray in the freezer, we are decreasing the entropy inside the ice tray. However, the large amount of heat dissipated outside the freezer compartment (remember that this is a heat pump) leads to a greater increase of entropy in the surroundings. The total entropy change of the ice tray and the surroundings is therefore positive. Many biological processes involve decreases of entropy for the organism that are always coupled to other processes that increase the entropy by a greater amount elsewhere. In photosynthesis in green plants, the large decrease of entropy for converting simple gases like CO2 and H2O into a very complex organism is offset by greater increases in entropy occurring elsewhere, such as in the Sun, which produces the light that drives photosynthesis. It is a pessimistic, but accurate, view that anything we do will always increase the entropy of the universe. Feynman (1963) explained entropy as the number of ways the insides can be arranged so that from the outside the system looks the same. For example, 1 mol of ice has lower entropy (informally: is more ordered) than 1 mol of water at the same temperature, because water molecules in the liquid may have many different arrangements but still have the properties of liquid water. Water molecules in solid ice can have only the periodic arrangement of the crystal structure of ice to have the properties of ice. The molar entropies of water at 1 bar are the following: Entropy mol-1 H2O (s) (273.15 K) 41.3 J K - 1 H2O (l) (298.15 K) 69.9 J K - 1 H2O (g) (298.15 K) 188.9 J K - 1 These values are consistent with the fact that the water molecules in ice are confined to positions in the crystalline lattice, whereas in water vapor the molecules are free to move about in a relatively large space. The liquid state is intermediate, but it is more like the solid than the gas in terms of entropy; consider that liquids and solids have similar densities so that the molecules of a liquid are still confined to about the same volume as in the corresponding solid. For monatomic elements, the entropy can be qualitatively related to Molecular Interpretation of Entropy | 63 the hardness of an element; in hard solids the molecules are more rigidly confined to lattice positions and have lower entropies than in soft solids. Contrast graphitic carbon, which is arranged in two-dimensional planes that can easily slide between each other, with the more rigid three-dimensionally network-bonded structure of diamond (see sidebar, 298 K, 1 bar). All entropies increase as the temperature is raised since increasing molecular motion increases the possible positions and configurations of the molecules. Interpreting entropy as a measure of disorder helps us to understand and predict entropy changes in reactions. In the gas phase, if the number of product molecules is less than the number of reactant molecules, entropy decreases. If the number of gas molecules increases, entropy increases. Two examples at 25°C and 1 bar are shown as follows. The superscript in  r S specifies that the values are for reactions in which all reactants and products are in their standard states, which we will define later. Phase Entropy Graphite 5.74 J K - 1mol - 1 Soft graphite  r Sm(J K - 1 mol - 1) Reaction 2H2 (g) + O2 (g) S 2H2 O(g) -88.8 PCl5 (g) S PCl3 (g) + Cl2 (g) 170.3 c c Reaction  r Sm (J K - 1 mol - 1 ) CO32 - (aq) + H + (aq) S HCO3- (aq) +148.1 NH 4+ (aq) + CO32 - (aq) S NH3 (aq) + HCO3- (aq) +146.0 NH 4+ (aq) + HCO3- (aq) S CO2 (aq) + H2O(l) + NH3 (aq) +94.2 + NH3 (aq) + H (aq) S HC2O4- (aq) - +2.1 NH4+ (aq) + OH (aq) S C2O42 - (aq) + H2O(l) CH4 (aq) S CH4 (CCl4 ) -23.1 +75 For the first four reactions, neutralization of charges occurs. Because a charged species tends to orient the water molecules around it, charge neutralization results in disorientation of some of the solvent molecules and an increase in entropy. The next two reactions involve transfer of charge but no neutralization. The entropy changes are small as a consequence. The last reaction involves the transfer of a nonpolar methane (CH4) molecule from a polar solvent (H2O) to a nonpolar solvent (CCl4). The large positive S is primarily the result of the ordering of water molecules around a nonpolar molecule. Removal of the nonpolar molecule results in the randomization of the water molecules and hence a positive S. The ordering of water molecules around nonpolar moieties is important for many reactions in aqueous media, such as the binding of a nonpolar substrate to the nonpolar region of an enzyme or the folding of a protein. Boltzmann derived the result that the entropy is proportional to the logarithm of the number of microscopic states of a system: S = kB lnW , (3.9) c c c +80.8 c c c If the number of molecules is unchanged by the reaction, a small change in entropy—either positive or negative—is expected. For reactions in solution, it is more difficult to predict the entropy change because of the potentially large effects of the solvent. Two solute molecules may react to form one product molecule, but unless we know what is happening to the solvent—whether it is being ordered or disordered—we cannot predict the change in entropy. However, we can rationalize and interpret the measured results. A few examples of reactions carried out at 25°C, 1 bar are as follows: OH - (aq) + H + (aq) S H2O(l) Property Diamond 2.38 J K - 1mol - 1 Hard c c diamond 64 Chapter 3 | The Second Law: The Entropy of the Universe Increases where kB is Boltzmann’s constant (1.381 * 10 - 23 J K - 1 ) and W is the number of microscopic states of the system. Note that the gas constant R (8.314 J K - 1 mol - 1) is proportional to kB by Avogadro’s number R = k B N A . The number of microscopic states is the number of different ways the inside of the system can be rearranged without changing its properties. For a simple molecular system, we can calculate its number of microstates and hence its entropy; for a complex system, it becomes harder to calculate the entropy from the structures of the molecules and their interactions. We can also use the measured entropy to calculate W, the number of microscopic states for any system. Solving for W in Eq. 3.9, we obtain W = eS>k B . (3.10) It is possible, in principle, to calculate the entropy and other thermodynamic properties of complex systems from molecular structures and interactions; the methods used are described in books on statistical mechanics. Fluctuations As we saw in example 3.1, if we assume that a system will spontaneously reach uniform pressure, we arrive at the conclusion that the sum of entropy changes of the system and surroundings is positive (the Second Law). The quantitative statement of the Second Law in Eqs. 3.7 and 3.8 was based on a wealth of observations and can now be used to predict the direction of a spontaneous process. So the Second Law predicts that the system shown in figure 3.5(a) will spontaneously tend to reach uniform pressure. Similarly, starting with the Second Law, we predict that the system shown in figure 3.5(b) will tend to reach uniform composition spontaneously. That is, at constant pressure, gases originally separated in two halves of a system will tend to mix. Both of these processes can take place without changing the surroundings. The reverse of these reactions—unmixing gases, or going from a uniform pressure to unequal pressures—can only be done if the surroundings are also changed and thus will not occur spontaneously. The Second Law was based on observations of relatively large systems and measurements averaged over time. Consider instead a system of 10 N2 molecules contained within a rigid box, whose volume is not very important for this thought experiment. It is evident that there will not be a uniform number of collisions of the gas particles with any wall so that pressure will be a fluctuating and poorly defined property (as will density). Similarly, if we watched a system containing 10 N2 molecules and 10 O2 molecules, we may occasionally notice that molecules segregate spontaneously so that all 10 molecules of N2 are on one side of the system and all 10 molecules of O2 are on the other side. We might wonder if this spontaneous decrease of entropy with no effects in the surroundings disproves the Second Law. FIGURE 3.5 Two flasks at the same temperature are connected through a valve that can be turned to the “Open” position (right-side drawings) to permit the passage of the molecules from one flask to another. The Second Law states that (a) pressures tend to become uniform and (b) composition tends to become uniform. (a) pA > p2 pB < p2 N2 O2 Open valve p2 p2 (b) Open valve N2 + O2 N2 + O2 Chemical Reactions | 65 Fluctuations will be important only for very small systems and very short times. For large numbers of molecules and for small numbers of molecules observed for a long time, the Second Law always applies. This discussion on fluctuations is a reminder that thermodynamics is a macroscopic theory; it applies to matter in bulk and does not acknowledge molecules in any way, although we can and should connect thermodynamic insights to molecular structures and interactions. We can estimate the probability that a process will occur that is not consistent with the Second Law. From Eq. 3.9, we can show that the probability of observing a violation of the Second Law depends exponentially on the number of molecules N in the system. The probability that a change contrary to the Second Law occurs in a system containing N molecules is e -N . We thus find that for 5 molecules, the probability of contradicting the Second Law is about 1/100; for 10 molecules, 1/104; for 20 molecules, 1/108; and for 100 molecules, 1/1043. Furthermore, observing five molecules for a long time is equivalent to observing many molecules for a short time. Scold = Measurement of Entropy qin Tcold > qin = S hot Thot The experimental measurement of entropy changes is based on the definition of entropy, Eq. 3.6. The entropy change when a system changes from state 1 to state 2 is state 2 state 2 S = S 2 - S 1 = 3 dS = 3 state 1 state 1 dqrev . T (3.11) We must be sure to understand entropy and its relation to the reversible heat given by Eq. 3.11. Remember that entropy depends only on the initial state (1) and the final state (2). It does not matter how we get from state 1 to state 2. However, the entropy difference can be measured as in example 3.1 by finding a reversible path between states 1 and 2, measuring the heat change, and using Eq. 3.11. For an irreversible path, the entropy change is not equal to the heat absorbed divided by T; furthermore, the entropy change is always greater than the irreversible heat divided by the temperature: dqirrev S = S 2 - S 1 7 3 . T S hot Thot qin (3.12) EX E R C I S E 3 .1 Show that example 3.1(a) is consistent with Eq. 3.12. Before we explore the calculation of entropy, notice that the inverse dependence of the entropy change on T in Eq. 3.11 conveys a remarkable insight: a fixed qrev changes the entropy of colder bodies more than of hotter bodies (figure 3.6). In other words, this property illustrates how the entropy changes of the two nonadiabatic steps of the Carnot cycle can have the same magnitude even when they involve different amounts of heat transfer. Chemical Reactions For a general reaction n AA + n BB h n CC + n DD at a chosen T and p, the entropy change is the difference in the entropies of the reactants and products:  rS = nCS m,C + nDS m,D - nAS m,A - nBS m,B . Scold Tcold qin FIGURE 3.6 Consider performing reversible, isothermal expansions of equal molar amounts of an ideal gas at temperatures Tcold and Thot such that each gas receives the same qin. (pV is a unique constant for each process.) Then the entropy of the cold gas increases more than that of the hot gas. 66 Chapter 3 | The Second Law: The Entropy of the Universe Increases where S m, A is the entropy per mol of compound A and S m, B is the entropy per mol of compound B, and so on, at T and p. (A similar expression was given in chapter 2 for  r H of chemical reactions; see Eq. 2.67.) The entropy of each reactant or product depends on T and p; thus,  rS of a given reaction will depend on T and p. A reference state with a temperature of 25°C (298 K) and a pressure of 1 bar is often picked, and the entropy per mol of a substance at this reference state is termed the standard molar entropy, S°m. We use the symbol  r Sm for the molar entropy change of a reaction at 1 bar. Standard molar entropies of compounds can be obtained from tables (see appendices). Unlike the convention for the standard enthalpies of formation, the standard molar entropies of the elements are not equal to zero at 25°C and 1 bar. Rather, the Third Law of Thermodynamics will be seen to define an absolute scale for entropy. EXAMPLE 3.2 Calculate the entropy change at 25°C and 1 bar for the decomposition of 1 mol of liquid water to H2 and O2 gas. SOLUTION The reaction is H2O(l) S H2 (g) + 21O2 (g). The entropy change at 25°C and 1 bar is 1  r S m (25C) = Sm (O2 (g) ) + Sm (H2 (g) ) - Sm (H2O(l) ) 2 1 = (205.25) + 130.79 - 69.91 2 = 163.51 J mol - 1K - 1 The entropy change is positive, consistent with our expectations for a reaction that involves the formation of two gases from one liquid. There is an increase in disorder. Third Law of Thermodynamics The Third Law of Thermodynamics states that the entropy of any pure, perfect crystal is zero at 0 K (absolute zero), S A (0 K) K 0 , (3.13) where A is any pure, perfect crystal. The Third Law is an experimental one, as the first two laws are, but we can understand it in terms of the relation of entropy and disorder. Near 0 K, the disorder of a substance can approach zero; the number of microstates approaches 1. For this to happen, the substance must be pure; in a mixture, the entropy could be reduced by separating the components. So, for a perfect crystal of any pure compound, we know the entropy at absolute zero. We can obtain the entropy at any other temperature if we know how entropy changes with temperature. Temperature Dependence of Entropy The entropy change for heating or cooling a system is easy to calculate. The heating or cooling can be done essentially reversibly so that the entropy change is T2 S2 - S1 = 3 T 1 T2 dqrev CdT = 3 . T T T1 Third Law of Thermodynamics | 67 At constant p, T2 S = 3 Cp dT (3.14) T T1 = Cp ln T2 , T1 (3.15) if Cp, the heat capacity at constant p, is independent of temperature and thus can be placed outside of the integral (figure 3.7). At constant V, T2 CV dT T T1 T2 = CV ln , T1 S = 3 (3.16) (3.17) where C V, the heat capacity at constant V, is assumed independent of temperature. Because Cp and CV are always positive, Eqs. 3.15 and 3.17 show that raising the temperature will always increase the entropy; this is expected from the increase in disorder. Ideal Gas (p, 298 K) Ideal Gas (p, 310 K) 0.04 C p,m T 0.03 0.02 0.01 298 310 Temperature / K Area = Sm= 310 C p,mdT 298 T = C p,m ln T2 T1 FIGURE 3.7 The area under the curve described by Cp,m/ T is the change in entropy for expanding an ideal gas at constant pressure (from 298 K to 310 K in this example). For an ideal gas, Cp,m is independent of temperature, an assumption which can also be applied when the temperature range is very small. For finding S over large temperature ranges, Cp,m/ T must be measured by calorimetric methods and the integral evaluated by curve fitting. 68 Chapter 3 | The Second Law: The Entropy of the Universe Increases EXAMPLE 3.3 Calculate the change in entropy at constant p when 1 mol of liquid water at 100°C is brought in contact with 1 mol of liquid water at 0°C. Assume that the heat capacity of liquid water is independent of temperature and is equal to 75 J mol - 1K - 1. No heat is lost to the surroundings. SOLUTION As we bring equal amounts of the water in contact, C p,m is constant and no heat is lost; the First Law tells us that the final temperature of the mixture will be the average of the two temperatures, 50°C. We use Eq. 3.15 to calculate the change in entropy of the hot water and of the cold water and add the results: Hot water: S 1 = S (50C) - S (100C) = (1 mol)Cp,m ln 323 373 = -10.79 J K - 1 . Cold water: S 2 = S (50C) - S (0C) = (1 mol)Cp,m ln 323 273 = 12.61 J K - 1 . (1 mol) H2O(100C) + (1 mol)H2O(0C) h (2 mol)H2O(50C): S 1 + S 2 = 1.82 J K-1. The entropy change is positive, as it must be for a spontaneous process in an isolated system. Temperature Dependence of the Entropy Change for a Chemical Reaction To calculate the entropy change for a chemical reaction at 1 bar and some temperature other than 25°C, we can use Eq. 3.15. We consider a cycle in which products and reactants are heated or cooled to the new temperature and then add the entropy changes for the cycle: A(T2) A(25°C) ∆rS°(T2) ∆rS°(25oC) 298  rS(T2) =  rS(25C) + 3 T2 B(T2) B(25°C) T 2 dT dT Cp(A) + 3 Cp(B) . T T 298 Third Law of Thermodynamics | 69 We can generalize the result into a more compact form by using the identity b a 3 = -3 , a b T2  r S(T2 ) =  r S(T1 ) + 3  r Cp T1 dT , T (3.18) where  r Cp = Cp (products) - Cp (reactants) . EXAMPLE 3.4 If a spark is applied to a mixture of H2 (g) and O2 (g), an explosion occurs and water is formed. The gaseous water is cooled to 100°C. Calculate the entropy change when 2 mol of gaseous H2O is formed at 100°C and 1 bar from H2 (g) and O2 (g) at the same temperature and each at a partial pressure of 1 bar. SOLUTION The reaction for 2 mol of H2O is 2H2 (g) + O2 (g) h 2H2O(g) . The entropy change at 25°C, 1 bar, is  r S(25C) = 2Sp,m,H2O(g) - Sp,m,O2(g) - 2Sp,m,H2(g) = 2(188.93) - 205.25 - 2(130.79) = -88.97 J K - 1 mol - 1 . To find  r S (100°C, 1 bar), we need to use Eq. 3.18; therefore, we need to know the heat capacities of H2 (g), O2 (g), and H2O(g). They can be taken as constants over the temperature range 25°C to 100°C (apply Eq. 2.48 to the table on Pg. 24, and use Table 2.2 also)  rCp = 2Cp,m,H2O(g) - Cp,m,O2(g) - 2Cp,m,H2(g) = 2(36.5) - 29.4 - 2(28.8) = -14.0 J K - 1 mol - 1 . From Eq. 3.18, 373  r S(100C) =  r S(25C) + 3  r Cp 298 =  r S(25C) +  r Cp ln dT T 373 298 = -88.97 - (14.0)(0.224) = -92.11 J K - 1 mol - 1 . The entropy of the system decreased; the Second Law thus requires that there be a greater increase in the entropy of the surroundings. The reaction is exothermic, and heat is lost to the surroundings at constant temperature. 70 Chapter 3 | The Second Law: The Entropy of the Universe Increases Entropy Change for a Phase Transition On heating many compounds from 0 K to room temperature, various phase transitions, such as melting (fusion) and boiling (vaporization), may occur. The reversible heat absorbed divided by the equilibrium temperature of the transition gives the entropy change for the phase transition. It is important to stress that entropy should be calculated at the equilibrium conditions of phase changes. Otherwise, the transition is not reversible, and the heat absorbed is not the reversible heat. For a phase transition, at constant p and T, qp = qrev =   H  S =  H T (2.43) (3.19) , where   S is the entropy of the phase transition and   H is the enthalpy of the phase transition at the equilibrium temperature T. Third-Law entropies or absolute entropies, such as those in tables A.5 -7 in the appendix, are obtained by adding together (1) the entropy of the substance at 0 K (which will be zero if it is a perfect crystal), (2) the entropy increase associated with increasing the temperature, and (3) the entropy changes associated with any phase changes. For example, to obtain the Third-Law entropy of a liquid compound at 25°C and 1 bar, we would use the equation 298  fus H dT dT + + 3 Cp,m(l) T Tfus T Tfus 0 where Tfus is the melting temperature at 1 bar and we have assumed that there are no solid–solid transitions. If there are, we would add (  H >T  ) for each transition and use an appropriate C p,m for each solid phase. Sm (25C, 1 bar) = Sm(0 K) + 3 Tfus Cp,m(s) Pressure Dependence of Entropy For solids and liquids, we generally ignore the direct effect of pressure on entropy: S = S (p2 ) - S (p1 ) ⬵ 0 . (3.20) For gases, we approximate the effect of pressure at constant temperature by that on an ideal gas. The change in entropy dS for a small change dp in pressure is dqrev dS = T dU - dw rev . = T For an ideal gas that changes its pressure at constant temperature, dU is zero because U is independent of p at constant temperature. Thus, pdV -dw rev = . T T Because pV = nRT is a constant at constant temperature, dS = d(pV ) = pdV + V dp = 0 , (constant temperature) and so pdV = -V dp for an ideal gas at constant temperature. Therefore, dS = -V dp -nRdp pdV = = , p T T Third Law of Thermodynamics | 71 and p2 S = -nR 3 p1 dp p2 = -nR ln . p p1 (3.21) We notice that in Eq. 3.21 if either p1 or p2 is zero, the entropy of n mol of gas becomes infinite. This should not surprise us; zero pressure means infinite volume, and infinite volume implies infinite disorder of the molecules in the volume. The point to remember is that as the pressure of a gas decreases at constant temperature, the entropy will increase. Equations 3.20 and 3.21 can be used to calculate the entropy change of a substance when the pressure is changed. A useful application of Eq. 3.21 is in the calculation of the entropy change of mixing two gases. Two gases, nA mol of A and nB mol of B, are initially in two separate flasks connected by a tube with a valve in it, similar to the arrangement illustrated in figure 3.5b. It is convenient to solve the special case with the two gases at the same pressure (pA = pB = p) and temperature T. If the volumes of the flasks are the same then nA = nB. After opening the valve, the two gases will eventually become completely mixed, as both our experience and the Second Law of Thermodynamics tell us. Because we start with two gases at the same pressure and we also assume that the two gases behave like ideal gases and do not react with each other, there will be no change in the total pressure p after mixing. The partial pressures of A and B after mixing, however, become pA = p[nA >(nA + nB )] = x A p, and pB = p[nB >(nA + nB )] = x B p , respectively. [x A = nA >(nA + nB ) and x B = nB >(nA + nB ) are the mole fractions of A and B, respectively.] From Eq. 3.21, the entropy change accompanying a change of the pressure of A from p to xAp is, using the ideal gas approximation, S A = -nA R ln a xA p b = -nA R ln x A . p S B = -nB R ln a xB p b = -nB R ln x B . p Similarly, The total entropy change for mixing the two ideal gases is therefore  mixS = SA + SB = -R[nA ln xA + nB ln xB] . (3.22) It is interesting that if the two gases are identical, there is no entropy change upon mixing the two sides. The entropy increases when the valve is opened only if the two gases are distinguishable. One gas can be an isotope of the other, but as long as we can tell them apart, we calculate the same  mixS . If the two are indistinguishable,  mixS is zero. This conclusion is entirely consistent with the molecular interpretation of entropy. For two indistinguishable gases, swapping molecules between the two flasks does not alter the total number of microscopic states. Spontaneous Chemical Reactions It is natural to ask if graphite could be converted into diamonds. One way to answer this question is to learn if the entropy of the universe increases when the reaction occurs. We can easily learn from tables A.5 –7 in the appendix whether the entropy of the system increases, but it is not so easy to learn about the change in entropy of the surroundings. For processes that occur in an isolated system, however, the surroundings do not change, and 72 Chapter 3 | The Second Law: The Entropy of the Universe Increases we can limit our attention to the system: the sign of the entropy change tells whether the reaction can occur. Very few reactions occur at constant energy and volume and thus do not affect the surroundings. Conversion of an optically active molecule to a racemic mixture is an example. The two enantiomers have identical energies, volumes, and entropies, but the mixture has a larger entropy and nearly the same energy and volume. The conversion reaction is spontaneous. Other examples are the mixing of two ideal gases or the transfer of heat from a hot object to a cold one. Gibbs Free Energy For most chemical reactions, there are large changes in energy and enthalpy, and we are interested in whether a reaction will occur at constant T and p. These are the conditions under which most experiments are done in the laboratory: the system is free to exchange heat with the surroundings to remain at room temperature, and it can expand or contract in volume to remain at atmospheric pressure. Constant temperature and pressure are also common conditions for life processes. We need a criterion of spontaneity that applies to the system for these conditions. A new thermodynamic variable of state, the Gibbs free energy, is useful in this situation. The Gibbs free energy, G, is defined as a combination of enthalpy, temperature, and entropy: G K H - TS . (3.23) Because G is specified by the state variables H, T, and S and because both H and TS are quantities that increase in proportion to the amounts of materials, the Gibbs free energy is an extensive variable of state and has the same units as enthalpy or energy. We will see in the next two sections that if G for a process at constant T and p is negative, the process can occur spontaneously; if G at constant T and p is positive, the process will not occur spontaneously; and if G at constant T and p is zero, the system is at equilibrium. G and a System’s Capacity to Do Nonexpansion Work To see how the change in Gibbs free energy is related to the spontaneity of a process, we first examine the relation between G and a system’s capacity to do work. From the definition of G, for a closed system a small change dG in the Gibbs free energy is related to changes in the other thermodynamic parameters by dG = dH - TdS - SdT (Eq. 3.23) = dU + pdV + V dp - TdS - SdT (H = U + PV ) = dqrev + dw rev + pdV + V dp - TdS - SdT (First Law, reversible) = TdS + dw rev + pdV + V dp - TdS - SdT (definition of S) = dw rev + pdV + V dp - SdT. (3.24) We express dwrev as the sum of two types of work: the pressure–volume work, -pdV owing to a change dV in the volume of the system, and all other forms of work, dw other, dw rev = -pdV + dw other . (3.25) dG = V dp - SdT + dw other . (3.26) Equation 3.24 then becomes At constant temperature and pressure, dp = dT = 0 and dG = dw other (3.27) G = w other . (3.28) or, upon integration, Gibbs Free Energy | 73 Equation 3.28 states that G of a process at constant pressure and temperature is equal to w other or that - G is equal to -w other. In any process, such as in a chemical reaction, a system must do the necessary pV work associated with the change of the system from its initial state to its final state. Therefore, the non-pV work that can be obtained from a reversible process at constant temperature and pressure is -w other, also termed the nonexpansion work. If a reversible process does electrical work, then -w other is the electrical work that the process is capable of doing. Because the work done by a system is maximal when a process is carried out reversibly, -w other = - G is the maximum amount of nonexpansion work a system can do on the surroundings by a process at constant temperature and pressure. Spontaneous Processes at Constant T and p Equation 3.28 provides a criterion for processes that can spontaneously occur at constant pressure and temperature. Our experience tells us that if a system undergoes a spontaneous process, it is capable of performing useful work: water flowing down a fall can be utilized to generate electricity, and heat flowing from a hot to a cold reservoir can be utilized to power a heat engine. Conversely, for a nonspontaneous process, such as water flowing uphill or heat flowing from cold to hot, the surroundings must do work on the system. Thus, -w other, the maximum nonexpansion work a system can do on the surroundings, is positive for a spontaneous process. From Eq. 3.28, it follows that - G = -w other 7 0 for a spontaneous process at constant temperature and pressure, or G 6 0 (spontaneous process at constant T and p) . (3.29) If a process is not spontaneous, w other must be positive and G 7 0 (nonspontaneous process at constant T and p) . (3.30) A system at equilibrium cannot perform any work. Thus, w other = 0 and G = 0 (system at equilibrium at constant T and p) . (3.31) The preceding discussion can help us explore the meaning of the catchphrase “Conserve energy!”—a thermodynamically misleading statement since the First Law of Thermodynamics says that energy is always conserved. At constant temperature and pressure, Eq. 3.28 shows that the decrease in Gibbs free energy corresponds to the maximum non-pV work a system can do. This is the maximum energy that could be extracted, for example, from a fuel cell. And this is also the energy that is available to drive nonspontaneous reactions in a cell. Generally, life processes take place at constant T and p, while pV work is often neglected in cells so that G is the only free energy to be recovered from a given process.† Therefore, we are especially interested to learn the change in free energy at 1 bar and 310 K for biological reactions. A compromise catchphrase that has both social and biological appeal might be, “Conserve free energy!” Calculation of Gibbs Free Energy The Gibbs free-energy change for a reaction at constant temperature can be obtained from the enthalpy and entropy changes,  rG =  rH - T  rS . (3.32) Tables A.5–7 in the appendix give values of  fH  and S for various substances at 25°C and 1 bar. These values can then be used to calculate  rH  and  rS, and hence  rG, for chemical reactions at 25°C and 1 bar. †Cell volumes can certainly change, generally in response to environmental conditions, but cells do attempt to maintain constant volume. And pV work can be performed by organisms as the collective motions of major organs (e.g., lungs). 74 Chapter 3 | The Second Law: The Entropy of the Universe Increases Values of  f G , the standard free energy of formation, of various substances are also given in tables A.5–7 in the appendix. The molar standard free energy of formation is defined as the free energy of formation of 1 mol of any compound at 1 bar from its elements in their standard states at 1 bar. As we did with the enthalpy of formation, we arbitrarily assign the elements in their most stable state at 1 bar to have zero free energy. EXAMPLE 3.5 Calculate the Gibbs free-energy change for the following reaction at 25°C and 1 bar. Will the reaction occur spontaneously? H2O(l) h H2 (g) + 1 O (g) 2 2 SOLUTION  rG(25C) =  f Gm,H2(g) + 1  G -  f Gm,H2O(l) 2 f m,O2(g) = 0 + 0 - ( -237.1)kJ mol - 1 = 237.1 kJ mol - 1 . The Gibbs free-energy change is the negative of the Gibbs free energy of formation of liquid water. We can also calculate H and S from table A.5 (appendix): 1  rH = Hm,H2(g) + Hm,O2(g) - Hm,H2O(l) 2 = 0 + 0 - ( -285.83) = 285.83 kJ mol - 1  rS = Sm,H2(g) + 1 Sm,O2(g) - Sm,H2O(l) 2 1 (205.25) - 69.91 2 =163.51 J K - 1 mol - 1 . =130.79 + Now  rG can be computed:  rG(25C) = 285.83 kJ - (298.15)(163.51 * 10 - 3 ) kJ = 237.08 kJ mol - 1 . The reaction will not occur spontaneously because  rG  is positive. EXAMPLE 3.6 We wish to know whether proteins in aqueous solution are unstable with respect to their constituent amino acids. As an example, let’s calculate the standard free energy of hydrolysis for the dipeptide glycylglycine at 25°C and 1 bar in dilute aqueous solution. SOLUTION The reaction is + H3NCH2CONHCH2COO - (aq) + H2O(l) h 2 + H3NCH2COO - (aq) glycylglycine glycine Gibbs Free Energy | 75 The standard free-energy change when solid glycine dissolves has been measured and is small. We will assume that this is also true for solid glycylglycine. Therefore, the free-energy values from tables A.5–7 (appendix) for solid glycine and glycylglycine will be a good approximation to their aqueous values:  rG (25C) = 2 f G m(glycine, s) -  f G m(glycylglycine, s) -  f G m(H2O, l) = 2( -377.69) - ( -490.57) - ( -237.13) = -27.68 kJ mol-1 The reaction is spontaneous, but it normally occurs slowly. Catalysts such as proteolytic enzymes can cause the reaction to occur rapidly (e.g., to break down proteins in food). There are many such enzymes in living organisms whose myriad tasks include regulating cellular functions of various proteins and the programmed death of cells. A proposed method of storing solar energy is to use sunlight to overcome the large positive free energy to split water into hydrogen and oxygen gases that in turn can be used as fuel. Notably, such a feat has been achieved by inorganic catalysts by Prof. Daniel Nocera and coworkers, whose invention has been dubbed an “artificial leaf” (Science 334, 2011: 645–648). The light-harvesting reactions of green plant photosynthesis overcome a positive free-energy change, which is almost exactly the same as for the production of oxygen gas by the direct decomposition of water. However, green plants do not make hydrogen gas; instead, they reduce carbon dioxide to carbohydrates and other products. Temperature Dependence of Gibbs Free Energy For reactions carried out at 25°C and 1 bar, the Gibbs free-energy change  rG can often be calculated from tabulations of enthalpy of formation and entropy data. If  rG of a reaction is known at one temperature such as 25°C, how do we obtain its value at a different temperature T? It is clear from the definition of Gibbs free energy that G (and therefore  rG) depends explicitly on T. Here are three common approaches to estimate  rG at one temperature given thermodynamic information for the reaction at some other reference temperature such as 25°C. (i) Since the values of  rH and  rS for reactions do not vary significantly for temperatures close to 25°C, Eq. 3.32 is a simple way to calculate  rG at other temperatures:  rG(T) ⬵  rH(25C) - T rS(25C) . (3.33) For example, values of  r H and  r S at 25°C from tables A.5–7 in the appendix can be used to calculate the approximate free energy of the reaction at a “physiological temperature” of 37°C. Another useful equation that is easily derived from Eq. 3.32 is  rG(T) -  rG(25C) ⬵ - (T - 298) rS(25C) , (3.34) which shows that the sign of  rS indicates how  rG will change with temperature. If  rS is negative, then  rG increases with increasing temperature. (ii) A more general expression of the temperature dependence of  rG at constant pressure can be obtained by considering a reversible path involving only pV work, such that Eq. 3.26 becomes dG = V dp - SdT . (3.35) Although we derived Eq. 3.35 for a reversible path, notice that all variables in this equation are state functions that do not depend on a particular path. Therefore, this equation is 76 Chapter 3 | A(T2) The Second Law: The Entropy of the Universe Increases ΔG(T2) ΔG1 B(T2) generally applicable to all paths and not just to the particular path chosen for its derivation. At constant pressure, Eq. 3.35 becomes dG p = -SdT, ΔG3 A(T1) ΔG2=ΔG(T1) B(T1) (3.36) which can be integrated to give G at constant pressure: T2 Gp = G(T 2 ) - G(T 1 ) = - 3 SdT . (3.36a) T1 T1 G1 = - 3 S A dT T2 Eqs. 3.36 and 3.36a can also be written in terms of the Gibbs free energy Gm and entropy Sm per mole of a substance: G 2 = G(T1) dG p,m = -S mdT (3.37) T2 G3 = - 3 S B dT T2 Gm = Gm (T 2 ) - Gm (T 1 ) = - 3 S mdT T1 (3.37a) T1 G(T2) = G 1 + G 2 + G 3 To calculate the temperature dependence of  rG of a chemical reaction nAA + nBB h nCC + nDD T2 = G(T 1) - 3 (S B - S A ) dT T1 T2 we use the same approach for the temperature dependence of H (Eq. 2.73). The free-energy change of the reaction is = G(T 1) - 3 S dT  rG = nCGm,C + nDGm,D - nAGm,A - nBGm,B , T1 (3.38) where Gm,C, Gm,D are the molar free energies of compounds C, D, and so on. [Here we consider the reactants and products as pure substances that are present as separate phases (solid zinc pellets dropped into a dilute hydrochloric acid solution, steam passed over hot charcoal, etc.). A more general definition of the molar free energies will be presented in chapter 4.] Taking the derivative of Eq. 3.38 with respect to temperature gives nC dGm,C nDdGm,D nAdGm, A nBdGm,B d rG = + . dT dT dT dT dT At constant pressure, then, Eq. 3.37 can be substituted to give d rG = -nCSm,C - nDSm,D + nASm,A + nBSm, B = -  r S , dT (3.39) since  r S = nCSm,C + nDSm, D - nASm,A - nBSm,B is the entropy change of the reaction. Integrating Eq. 3.39 gives rG(T2) 3 rG(T1) T2 d rG = - 3  rSdT , T1 T2  rG(T 2 ) -  rG(T 1 ) = - 3  rSdT . (3.39a) T1 If  rS is assumed constant between T1 and T2, then Eq. 3.39a becomes  rG(T 2 ) -  rG(T 1 ) = -  rS (T 2 - T 1 ) , (3.39b) which is just a general form of Eq. 3.34. The preceding discussion is an important preview of molar free energies. An alternate route to derive Eq. 3.39b is to sum the free-energy changes over a convenient reversible path, as shown in the sidebar. Gibbs Free Energy | 77 (iii) Another useful equation for the temperature dependence of  rG at constant pressure when we have knowledge of the enthalpy change is T2  rG(T 2 )  rG(T 1 )  rH(T ) = -3 dT , T2 T1 T2 T1 (3.40) which is called the Gibbs–Helmholtz equation. If we assume that  r H(T) is independent of the temperature range, then  r H(T) can be selected at the midpoint of T1 and T2,  rG(T 2)  rG(T 1) 1 1 =  rH c d. T2 T1 T2 T1 (3.41) Depending on available data, one can also use  r H(T1 ) in Eq. 3.41 when the temperature range is very narrow. This equation is derived at the end of this chapter. EXAMPLE 3.7 What is the free energy of hydrolysis of glycylglycine at 37°C and 1 bar? SOLUTION We can use Eq. 3.39b to find  rG(37°C) if we assume that  rS is independent of temperature, or we can use Eq. 3.41 if we assume that  r H is independent of temperature. Both will be nearly identical, since  r H and  r S should be constant over the small temp