Thermodynamics 14

14.3.2 Illustrations of the Calculation of Equilibrium Compositions for Reacting Ideal Gas Mixtures It is often convenient to express Eq. 14.32 explicitly in terms of the number of moles that would be present at equilibrium. Each mole fraction appearing in the equation has the form yi 5 ni /n, where ni is the amount of component i in the equilibrium mixture and n is the total number of moles of mixture. Hence, Eq. 14.32 can be rewritten as K nnC C nnD D nnA A nnB B a p/pref n b nCnDnAnB (14.35) The value of n must include not only the reacting components A, B, C, and D but also all inert components present. Since inert component E has been assumed present, we would write n nA 1 nB 1 nC 1 nD 1 nE. Equation 14.35 provides a relationship among the temperature, pressure, and composition of an ideal gas mixture at equilibrium. Accordingly, if any two of temperature, pressure, and composition are known, the third can be found by solving this equation. suppose that the temperature T and pressure p are known and the object is the equilibrium composition. With temperature known, the value of K can be obtained from Table A-27. The n's of the reacting components A, B, C, and D can be expressed in terms of a single unknown variable through application of the conservation of mass principle to the various chemical species present. Then, since the pressure is known, Eq. 14.35 constitutes a single equation in a single unknown, which can be solved using an equation solver or iteratively with a hand calculator. b b b b b In Example 14.2, we apply Eq. 14.35 to study the effect of pressure on the equilibrium composition of a mixture of CO2, CO, and O2

14.3.3 Equilibrium Constant for Mixtures and Solutions The procedures that led to the equilibrium constant for reacting ideal gas mixtures can be followed for the general case of reacting mixtures by using the fugacity and activity concepts introduced in Sec. 11.9. In principle, equilibrium compositions of such mixtures can be determined with an approach paralleling the one for ideal gas mixtures. Equation 11.141 can be used to evaluate the chemical potentials appearing in the equation of reaction equilibrium (Eq. 14.26). The result is vA1gA RT ln aA2 vB1gB RT ln aB2 vC1g C RT ln aC2 vD1gD RT ln aD2 (14.36) where gi is the Gibbs function of pure component i at temperature T and the pressure pref 5 1 atm, and ai is the activity of that component. Collecting terms and employing Eq. 14.29a, Eq. 14.36 becomes ¢G RT ln a avC C avD D avA A avB B b (14.37) This equation can be expressed in the same form as Eq. 14.31 by defining the equilibrium constant as K a C vCavD D avA A avB B (14.38) TAKE NOTE... Study of Sec. 14.3.3 requires content from Sec. 11.9. 898 Chapter 14 Chemical and Phase Equilibrium Since Table A-27 and similar compilations are constructed simply by evaluating ¢G/RT for specified reactions at several temperatures, such tables can be employed to evaluate the more general equilibrium constant given by Eq. 14.38. However, before Eq. 14.38 can be used to determine the equilibrium composition for a known value of K, it is necessary to evaluate the activity of the various mixture components. Let us illustrate this for the case of mixtures that can be modeled as ideal solutions. IDEAL SOLUTIONS. For an ideal solution, the activity of component i is given by ai yi fi f i (11.142) where fi is the fugacity of pure i at the temperature T and pressure p of the mixture, and fi is the fugacity of pure i at temperature T and the pressure pref. Using this expression to evaluate aA, aB, aC, and aD, Eq. 14.38 becomes K 1yC fC /fC2vC 1yD fD /f D 2vD 1yA fA/f A 2vA1yB fB/f B2vB (14.39a) which can be expressed alternatively as K c 1fC /p2vC 1fD /p2vD 1fA/p2vA 1 fB/p2vB d c 1 f A/pref2vA1 fB/pref2vB 1 fC/pref2vC 1 f D /pref2vD d c yvC C yvD D yvA A yvB B a p prefb vCvD vAvB d (14.39b) The ratios of fugacity to pressure in this equation can be evaluated, in principle, from Eq. 11.124 or the generalized fugacity chart, Fig. A-6, developed from it. In the special case when each component behaves as an ideal gas at both T, p and T, pref, these ratios equal unity and Eq. 14.39b reduces to the underlined term, which is just Eq. 14.32.

Chemical Equilibrium In this part of the chapter, the equilibrium criterion dG]T, p 5 0 introduced in Sec. 14.1 is used to study the equilibrium of reacting mixtures. The objective is to establish the composition present at equilibrium for a specified temperature and pressure. An important parameter for determining the equilibrium composition is the equilibrium constant. The equilibrium constant is introduced and its use illustrated by several solved examples. The discussion is concerned only with equilibrium states of reacting systems, and no information can be deduced about the rates of reaction. Whether an equilibrium mixture would form quickly or slowly can be determined only by considering the chemical kinetics, a topic that is not treated in this text. 14.2 Equation of Reaction Equilibrium In Chap. 13 the conservation of mass and conservation of energy principles are applied to reacting systems by assuming that the reactions can occur as written. However, the extent to which a chemical reaction proceeds is limited by many factors. In general, the composition of the products actually formed from a given set of reactants, and the relative amounts of the products, can be determined only from experiment. Knowledge of the composition that would be present were a reaction to proceed to equilibrium is frequently useful, however. The equation of reaction equilibrium introduced in the present section provides the basis for determining the equilibrium composition of a reacting mixture. 14.2.1 Introductory Case Consider a closed system consisting initially of a gaseous mixture of hydrogen and oxygen. A number of reactions might take place, including 1H2 1 1 2O2 dS 1H2O (14.18) 1H2 dS 2H (14.19) 1O2 dS 2O (14.20) Let us consider for illustration purposes only the first of the reactions given above, in which hydrogen and oxygen combine to form water. At equilibrium, the system will consist in general of three components: H2, O2, and H2O, for not all of the hydrogen and oxygen initially present need be reacted. Changes in the amounts of these components during each differential step of the reaction leading to the formation of an equilibrium mixture are governed by Eq. 14.18. That is dnH2 dnH2O, dnO2 1 2dnH2O (14.21a) where dn denotes a differential change in the respective component. The minus signs signal that the amounts of hydrogen and oxygen present decrease when the reaction proceeds toward the right. Equations 14.21a can be expressed alternatively as dnH2 1 dnO2 1 2 dnH2O 1 (14.21b) which emphasizes that increases and decreases in the components are proportional to the stoichiometric coefficients of Eq. 14.18. Equilibrium is a condition of balance. Accordingly, as suggested by the direction of the arrows in Eq. 14.18, when the system is at equilibrium, the tendency of the hydrogen and oxygen to form water is just balanced by the tendency of water to dissociate into oxygen and hydrogen. The equilibrium criterion dG]T, p 5 0 can be used to determine the composition at an equilibrium state where the temperature is T and the pressure is p. This requires evaluation of the differential dG]T, p in terms of system properties.

Equation 14.2 can be used to study equilibrium under various conditions. a process taking place in an insulated, constant-volume vessel, where dU 5 0 and dV 5 0, must be such that dS4U, V 0 (14.3) thermodynamic equilibrium Fixed U, V Smax S 14.1 Introducing Equilibrium Criteria 883 Equation 14.3 suggests that changes of state of a closed system at constant internal energy and volume can occur only in the direction of increasing entropy. The expression also implies that entropy approaches a maximum as a state of equilibrium is approached. This is a special case of the increase of entropy principle introduced in Sec. 6.8.1. b b b b b An important case for the study of chemical and phase equilibria is one in which temperature and pressure are fixed. For this, it is convenient to employ the Gibbs function in extensive form G H TS U pV TS Forming the differential dG dU p dV V dp T dS S dT or on rearrangement dG V dp S dT 1T dS dU p dV2 Except for the minus sign, the right side of this equation is the same as the expression appearing in Eq. 14.2. Accordingly, Eq. 14.2 can be written as dG V dp S dT 0 (14.4) where the inequality reverses direction because of the minus sign noted above. It can be concluded from Eq. 14.4 that any process taking place at a specified temperature and pressure (dT 5 0 and dp 5 0) must be such that dG4T, p 0 (14.5) This inequality indicates that the Gibbs function of a system at fixed T and p decreases during an irreversible process. Each step of such a process results in a decrease in the Gibbs function of the system and brings the system closer to equilibrium. The equilibrium state is the one having the minimum value of the Gibbs function. Therefore, when dG4T, p 0 (14.6) we have equilibrium. In subsequent discussions, we refer to Eq. 14.6 as the equilibrium criterion. Equation 14.6 provides a relationship among the properties of a system when it is at an equilibrium state. The manner in which the equilibrium state is reached is unimportant, however, for once an equilibrium state is obtained, a system exists at a particular T and p and no further spontaneous changes can take place. When applying Eq. 14.6, therefore, we may specify the temperature T and pressure p, but it is unnecessary to require additionally that the system actually achieves equilibrium at fixed T and fixed p. 14.1.1 Chemical Potential and Equilibrium In the present discussion, the Gibbs function is considered further as a prerequisite for application of the equilibrium criterion dG]T, p 5 0 introduced above. Let us begin by noting that any extensive property of a single-phase, single-component system is a function of two independent intensive properties and the size of the system. Selecting temperature and pressure as the independent properties and the number of moles n as the measure of size, the Gibbs function can be expressed in the form G 5 G(T, p, n). For a single-phase, multicomponent system, G may then be considered a function of temperature, pressure, and the number of moles of each component present, written G 5 G(T, p, n1, n2, . . . , nj)

For the present case, Eq. 14.10 giving the difference in the Gibbs function of the mixture between two states having the same temperature and pressure, but compositions that differ infinitesimally, takes the following form: dG4T, p mH2 dnH2 mO2 dnO2 mH2O dnH2O (14.22) The changes in the mole numbers are related by Eqs. 14.21. Hence, dG4T, p 11mH2 1 2mO2 1mH2O2 dnH2O At equilibrium, dG]T, p 5 0, so the term in parentheses must be zero. That is, 1mH2 1 2mO2 1mH2O 0 When expressed in a form that resembles Eq. 14.18, this becomes 1mH2 1 2mO2 1mH2O (14.23) Equation 14.23 is the equation of reaction equilibrium for the case under consideration. The chemical potentials are functions of temperature, pressure, and composition. Thus, the composition that would be present at equilibrium for a given temperature and pressure can be determined, in principle, by solving this equation. The solution procedure is described in Sec. 14.3. 14.2.2 General Case The foregoing development can be repeated for reactions involving any number of components. Consider a closed system containing five components, A, B, C, D, and E, at a given temperature and pressure, subject to a single chemical reaction of the form nAA 1 nBB dS nCC 1 nDD (14.24) where the n's are stoichiometric coefficients. Component E is assumed to be inert and thus does not appear in the reaction equation. As we will see, component E does influence the equilibrium composition even though it is not involved in the chemical reaction. The form of Eq. 14.24 suggests that at equilibrium the tendency of A and B to form C and D is just balanced by the tendency of C and D to form A and B. The stoichiometric coefficients nA, nB, nC, and nD do not correspond to the respective number of moles of the components present. The amounts of the components present are designated nA, nB, nC, nD, and nE. However, changes in the amounts of the components present do bear a definite relationship to the values of the stoichiometric coefficients. That is, dnA nA dnB nB dnC nC dnD nD (14.25a) where the minus signs indicate that A and B would be consumed when C and D are produced. Since E is inert, the amount of this component remains constant, so dnE 5 0. Introducing a proportionality factor d, Eqs. 14.25a take the form dnA nA dnB nB dnC nC dnD nD de from which the following expressions are obtained: dnA nA de, dnB nB de dnC nC de, dnD nD de (14.25b) extent of reaction The parameter is sometimes referred to as the extent of reaction.

For the system under present consideration, Eq. 14.10 takes the form dG4T, p mA dnA mB dnB mC dnC mD dnD mE dnE Introducing Eqs. 14.25b and noting that dnE 5 0, this becomes dG4T, p 1nAmA nBmB nCmC nDmD2 de At equilibrium, dG]T, p 5 0, so the term in parentheses must be zero. That is, nAmA nBmB nCmC nDmD 0 or when written in a form resembling Eq. 14.24 nAmA nBmB nCmC nDmD (14.26) For the present case, Eq. 14.26 is the equation of reaction equilibrium. In principle, the composition that would be present at equilibrium for a given temperature and pressure can be determined by solving this equation. The solution procedure is simplified through the equilibrium constant concept introduced in the next section. equation of reaction equilibrium 14.3 Calculating Equilibrium Compositions The objective of the present section is to show how the equilibrium composition of a system at a specified temperature and pressure can be determined by solving the equation of reaction equilibrium. An important part is played in this by the equilibrium constant. 14.3.1 Equilibrium Constant for Ideal Gas Mixtures The first step in solving the equation of reaction equilibrium, Eq. 14.26, for the equilibrium composition is to introduce expressions for the chemical potentials in terms of temperature, pressure, and composition. For an ideal gas mixture, Eq. 14.17 can be used for this purpose. When this expression is introduced for each of the components A, B, C, and D, Eq. 14.26 becomes nA ag A RT ln yAp pref b nB ag B RT ln yB p pref b nC ag C RT ln yC p pref b nD ag D RT ln yDp pref b (14.27) where gi is the Gibbs function of component i evaluated at temperature T and the pressure pref 5 1 atm. Equation 14.27 is the basic working relation for chemical equilibrium in a mixture of ideal gases. However, subsequent calculations are facilitated if it is written in an alternative form, as follows. Collect like terms and rearrange Eq. 14.27 as 1nCg C nDg D nAg A nBg B2 RT anC ln yC p pref nD ln yD p pref nA ln yA p pref nB ln yB p pref b (14.28) The term on the left side of Eq. 14.28 can be expressed concisely as DG8. That is, ¢G nC g C nD g D nAg A nBg B (14.29a) which is the change in the Gibbs function for the reaction given by Eq. 14.24 if each reactant and product were separate at temperature T and a pressure of 1 atm.

If each mole number is multiplied by , the size of the system changes by the same factor and so does the value of every extensive property. Thus, for the Gibbs function we may write aG1T, p, n1, n2, . . . , nj2 G1T, p, an1, an2, . . . , anj2 Differentiating with respect to while holding temperature, pressure, and the mole numbers fixed and using the chain rule on the right side gives G 0G 01an12 n1 0G 01an22 n2 . . . 0G 01anj2 nj This equation holds for all values of . In particular, it holds for 5 1. Setting 5 1, the following expression results: G a j i1 ni a 0G 0ni b T, p, nl (14.7) where the subscript nl denotes that all n's except ni are held fixed during differentiation. The partial derivatives appearing in Eq. 14.7 have such importance for our study of chemical and phase equilibrium that they are given a special name and symbol. The chemical potential of component i, symbolized by i, is defined as mi a 0G 0ni b T, p, nl (14.8) The chemical potential is an intensive property. With Eq. 14.8, Eq. 14.7 becomes G a j i1 nimi (14.9) The equilibrium criterion given by Eq. 14.6 can be written in terms of chemical potentials, providing an expression of fundamental importance for subsequent discussions of equilibrium. Forming the differential of G1T, p, n1, . . . , nj2 while holding temperature and pressure fixed results in dG4T, p a j i1 a 0G 0ni b T, p, nl dni The partial derivatives are recognized from Eq. 14.8 as the chemical potentials, so dG4T, p a j i1 mi dni (14.10) With Eq. 14.10, the equilibrium criterion dG]T, p 5 0 can be placed in the form a j i1 mi dni 0 (14.11) Like Eq. 14.6, from which it is obtained, this equation provides a relationship among properties of a system when the system is at an equilibrium state where the temperature is T and the pressure is p. Like Eq. 14.6, this equation applies to a particular state, and the manner in which that state is attained is not important. 14.1.2 Evaluating Chemical Potentials Means for evaluating the chemical potentials for two cases of interest are introduced in this section: a single phase of a pure substance and an ideal gas mixture.

In this part of the chapter the equilibrium condition dG4T, p 0 introduced in Sec. 14.1 is used to study the equilibrium of multicomponent, multiphase, nonreacting systems. The discussion begins with the elementary case of equilibrium between two phases of a pure substance and then turns to the general case of several components present in several phases. For the first of these reactions, the form taken by the equilibrium constant when p 5 1 atm is K1 a3 1 2 11 a b241/2 11 a2 c 1 14 a2/2 d 11/21 a 1 a a 1 a b 4 a b 1/2 Similarly, the equilibrium constant for the second of the reactions is K2 b 3 1 2 11 a b241/2 3 1 2 11 b241/2 c 1 14 a2/2 d 11/21/2 2b 311 a b211 b241/2 At 3000 K, Table A-27 provides log10 K1 0.485 and log10 K2 0.913, giving K1 0.3273 and K2 0.1222. Accordingly, the two equations that must be solved simultaneously for the two unknowns a and b are 0.3273 a 1 a a 1 a b 4 a b 1/2 , 0.1222 2b 311 a b211 b241/2 The solution is a 5 0.3745, b 5 0.0675, as can be verified. The composition of the equilibrium mixture, in kmol per kmol of CO2 present initially, is then 0.3745CO, 0.0675NO, 0.6255CO2, 0.6535O2, 0.4663N2. ❶ If high enough temperatures are attained, nitrogen can combine with oxygen to form components such as nitric oxide. Even trace amounts of oxides of nitrogen in products of combustion can be a source of air pollution. ✓Skills Developed Ability to... ❑ apply Eqs. 14.54 and 14.55 to determine equilibrium composition given temperature and pressure for two simultaneous equilibrium reactions. ❑ retrieve and use data from Table A-27. Determine the mole fractions of the components of the equilibrium mixture. Ans. yCO 5 0.171, yNO 5 0.031, yCO2 5 0.286, yO2 5 0.299, yN2 5 0.213. Quick Quiz 14.5 Equilibrium between Two Phases of a Pure Substance Consider the case of a system consisting of two phases of a pure substance at equilibrium. Since the system is at equilibrium, each phase is at the same temperature and pressure. The Gibbs function for the system is G n¿g¿1T, p2 n-g-1T, p2 (14.57) where the primes 9 and 0 denote phases 1 and 2, respectively. Forming the differential of G at fixed T and p dG4T, p g¿ dn¿ g- dn- (14.58) Since the total amount of the pure substance remains constant, an increase in the amount present in one of the phases must be compensated by an equivalent decrease 14.5 Equilibrium between Two Phases of a Pure Substance 909 in the amount present in the other phase. Thus, we have dn0 5 2dn9, and Eq. 14.58 becomes dG4T, p 1g¿ g-2 dn¿ At equilibrium, dG4T, p 0, so g¿ g- (14.59) At equilibrium, the molar Gibbs functions of the phases are equal. CLAPEYRON EQUATION. Equation 14.59 can be used to derive the Clapeyron equation, obtained by other means in Sec. 11.4. For two phases at equilibrium, variations in pressure are uniquely related to variations in temperature: p psat1T2; thus, differentiation of Eq. 14.59 with respect to temperature gives 0g¿ 0T b p 0g¿ 0p b T dpsat dT 0g- 0T b p 0g- 0p b T dpsat dT With Eqs. 11.30 and 11.31, this becomes s¿ y¿ dpsat dT s- y- dpsat dT Or on rearrangement dpsat dT s- s¿ y- y¿ This can be expressed alternatively by noting that, with g h T s, Eq. 14.59 becomes h¿ T s¿ h- T s- or s- s¿ h- h¿ T (14.60) Combining results, the Clapeyron equation is obtained dpsat dT 1 T a h- h¿ y- y¿ b (14.61) An application of the Clapeyron equation is provided in Example 11.4. A special form of Eq. 14.61 for a system at equilibrium consisting of a liquid or solid phase and a vapor phase can be obtained simply. If the specific volume of the liquid or solid, y¿, is negligible compared with the specific volume of the vapor, y-, and the vapor can be treated as an ideal gas, y - R T/psat, Eq. 14.61 becomes dpsat dT h- h¿ RT 2 /psat or d ln psat dT h- h¿ RT 2 (14.62) which is the Clausius-Clapeyron equation. The similarity in form of Eq. 14.62 and the van't Hoff equation, Eq. 14.43b, may be noted. The van't Hoff equation for chemical equilibrium is the counterpart of the Clausius-Clapeyron equation for phase equilibrium.

In this part of the chapter, fundamental concepts are developed that are useful in the study of chemical and phase equilibrium. Among these are equilibrium criteria and the chemical potential concept. 14.1 Introducing Equilibrium Criteria A system is said to be in thermodynamic equilibrium if, when it is isolated from its surroundings, there would be no macroscopically observable changes. An important requirement for equilibrium is that the temperature be uniform throughout the system or each part of the system in thermal contact. If this condition were not met, spontaneous heat transfer from one location to another could occur when the system was isolated. There must also be no unbalanced forces between parts of the system. These conditions ensure that the system is in thermal and mechanical equilibrium, but there is still the possibility that complete equilibrium does not exist. A process might occur involving a chemical reaction, a transfer of mass between phases, or both. The objective of this section is to introduce criteria that can be applied to decide whether a system in a particular state is in equilibrium. These criteria are developed using the conservation of energy principle and the second law of thermodynamics as discussed next. Consider the case of a simple compressible system of fixed mass for which temperature and pressure are uniform with position throughout. In the absence of overall system motion and ignoring the influence of gravity, the energy balance in differential form (Eq. 2.36) is dU dQ dW If volume change is the only work mode and pressure is uniform with position throughout the system, W p dV. Introducing this in the energy balance and solving for Q gives dQ dU p dV Since temperature is uniform with position throughout the system, the entropy balance in differential form (Eq. 6.25) is dS dQ T ds Eliminating Q between the last two equations T dS dU p dV T ds (14.1) Entropy is produced in all actual processes and conserved only in the absence of irreversibilities. Hence, Eq. 14.1 provides a constraint on the direction of processes. The only processes allowed are those for which $ 0. Thus, T dS dU p dV 0 (14.2) Equation 14.2 can be used to study equilibrium under various conditions.

Let us return to the discussion of Sec. 14.2 and consider the possibility of more than one reaction among the substances present within a system. For the present application, the closed system is assumed to contain a mixture of eight components A, B, C, D, E, L, M, and N, subject to two independent reactions: (1) nAA 1 nBB dS nCC 1 nDD (14.24) (2) nA9A 1 nLL dS nMM 1 nNN (14.46) As in Sec. 14.2, component E is inert. Also, note that component A has been taken as common to both reactions but with a possibly different stoichiometric coefficient (A9 is not necessarily equal to A). Analysis: The ionization of cesium to form a mixture of Cs, Cs, and e is described by Cs S 11 z2Cs z Cs ze where z denotes the extent of ionization, ranging from 0 to 1. The total number of moles of mixture n is n 11 z2 z z 1 z At equilibrium, we have Cs dS Cs1 1 e2, so Eq. 14.35 takes the form K 1z21z2 11 z2 c p/pref 11 z2 d 111 a z2 1 z2 b a p prefb (a) Solving for the ratio p/pref and introducing the known value of K p pref 115.632 a 1 z2 z2 b For pref 1 atm and z 0.95 (95%), p 1.69 atm. Using an equation solver and plotting package, the following plot can be constructed: z (%) 100 95 90 85 80 75 70 0 2 4 6 8 10 p (atm) Fig. E14.8 Figure E14.8 shows that ionization tends to occur to a lesser extent as pressure is raised. Ionization also tends to occur to a greater extent as temperature is raised at fixed pressure. ✓Skills Developed Ability to... ❑ apply Eq. 14.35 to determine the extent of ionization of cesium given temperature and pressure. Solving Eq. (a) for z, determine the percent of ionization of Cs at T 5 28808R (K 5 0.78) and p 5 1 atm. Ans. 66.2%. Quick Quiz 906 Chapter 14 Chemical and Phase Equilibrium The stoichiometric coefficients of the above equations do not correspond to the numbers of moles of the respective components present within the system, but changes in the amounts of the components are related to the stoichiometric coefficients by dnA nA dnB nB dnC nC dnD nD (14.25a) following from Eq. 14.24, and dnA nA¿ dnL nL dnM nM dnN nN (14.47a) following from Eq. 14.46. Introducing a proportionality factor de1, Eqs. 14.25a may be represented by dnA nA de1, dnB nB de1 dnC nC de1, dnD nD de1 (14.25b) Similarly, with the proportionality factor de2, Eqs. 14.47a may be represented by dnA nA¿ de2, dnL nL de2 dnM nM de2, dnN nN de2 (14.47b) Component A is involved in both reactions, so the total change in A is given by dnA nA de1 nA¿ de2 (14.48) Also, we have dnE 5 0 because component E is inert. For the system under present consideration, Eq. 14.10 is dG4T, p mA dnA mB dnB mC dnC mD dnD mE dnE mL dnL mM dnM mN dnN (14.49) Introducing the above expressions giving the changes in the n's, this becomes dG4T, p 1nA mA nBmB nCmC nDmD2 de1 1nA¿mA nLmL nMmM nNmN2 de2 (14.50) Since the two reactions are independent, d1 and d2 can be independently varied. Accordingly, when dG4T, p 0, the terms in parentheses must be zero and two equations of reaction equilibrium result, one corresponding to each of the foregoing reactions: nAmA nBmB nCmC nDmD (14.26b) nA¿mA nLmL nMmM nNmN (14.51) The first of these equations is exactly the same as that obtained in Sec. 14.2. For the case of reacting ideal gas mixtures, this equation can be expressed as a ¢G RT b 1 ln c ynC C ynD D ynA A ynB B a p prefb nCnD nAnB d (14.52) Similarly, Eq. 14.51 can be expressed as a ¢G RT b 2 ln c ynM M ynN N ynA¿ A ynL L a p prefb nM nN nA¿nL d (14.53) In each of these equations, the DG8 term is evaluated as the change in Gibbs function for the respective reaction, regarding each reactant and product as separate at temperature T and a pressure of 1 atm. From Eq. 14.52 follows the equilibrium constant K1 ynC C ynD D ynA A yvB B a p prefb nCnD nAvB (14.54) 14.4 Further Examples of the Use of the Equilibrium Constant 907 and from Eq. 14.53 follows K2 yvM M yvN N yvA¿ A yvL L a p prefb vMvNvA¿vL (14.55) The equilibrium constants K1 and K2 can be determined from Table A-27 or a similar compilation. The mole fractions appearing in these expressions must be evaluated by considering all the substances present within the system, including the inert substance E. Each mole fraction has the form yi 5 ni/n, where ni is the amount of component i in the equilibrium mixture and n nA nB nC nD nE nL nM nN (14.56) The n's appearing in Eq. 14.56 can be expressed in terms of two unknown variables through application of the conservation of mass principle to the various chemical species present. Accordingly, for a specified temperature and pressure, Eqs. 14.54 and 14.55 give two equations in two unknowns. The composition of the system at equilibrium can be determined by solving these equations simultaneously. This procedure is illustrated by Example 14.9. The procedure discussed in this section can be extended to systems involving several simultaneous independent reactions. The number of simultaneous equilibrium constant expressions that results equals the number of independent reactions. As these equations are nonlinear and require simultaneous solution, the use of a computer is usually required.

SINGLE PHASE OF A PURE SUBSTANCE. An elementary case considered later in this chapter is that of equilibrium between two phases involving a pure substance. For a single phase of a pure substance, Eq. 14.9 becomes simply G nm or m G n g (14.12) That is, the chemical potential is just the Gibbs function per mole. IDEAL GAS MIXTURE. An important case for the study of chemical equilibrium is that of an ideal gas mixture. The enthalpy and entropy of an ideal gas mixture are given by H a j i1 ni hi 1T2 and S a j i1 nisi 1T, pi2 where pi 5 yip is the partial pressure of component i. Accordingly, the Gibbs function takes the form G H TS a j i1 ni hi 1T 2 T a j i1 ni si 1T, pi2 a j i1 ni 3hi 1T 2 T si 1T, pi24 1ideal gas2 (14.13) Introducing the molar Gibbs function of component i gi1T, pi2 hi1T2 T si1T, pi2 1ideal gas2 (14.14) Equation 14.13 can be expressed as G a j i1 nigi 1T, pi2 1ideal gas2 (14.15) Comparing Eq. 14.15 to Eq. 14.9 suggests that mi gi 1T, pi2 1ideal gas2 (14.16) That is, the chemical potential of component i in an ideal gas mixture is equal to its Gibbs function per mole of i, evaluated at the mixture temperature and the partial pressure of i in the mixture. Equation 14.16 can be obtained formally by taking the partial derivative of Eq. 14.15 with respect to ni, holding temperature, pressure, and the remaining n's constant, and then applying the definition of chemical potential, Eq. 14.8. The chemical potential of component i in an ideal gas mixture can be expressed in an alternative form that is somewhat more convenient for subsequent applications. Using Eq. 13.23, Eq. 14.14 becomes

Some additional aspects of the use of the equilibrium constant are introduced in this section: the equilibrium flame temperature, the van't Hoff equation, and chemical equilibrium for ionization reactions and simultaneous reactions. To keep the presentation at an introductory level, only the case of ideal gas mixtures is considered. 14.4.1 Determining Equilibrium Flame Temperature In this section, the effect of incomplete combustion on the adiabatic flame temperature, introduced in Sec. 13.3, is considered using concepts developed in the present chapter. We begin with a review of some ideas related to the adiabatic flame temperature by considering a reactor operating at steady state for which no significant heat transfer with the surroundings takes place. Let carbon monoxide gas entering at one location react completely with the theoretical amount of air entering at another location as follows: CO 1 2O2 1.88N2 S CO2 1.88N2 As discussed in Sec. 13.3, the products would exit the reactor at a temperature we have designated the maximum adiabatic flame temperature. This temperature can be determined by solving a single equation, the energy equation. At such an elevated temperature, however, there would be a tendency for CO2 to dissociate CO2 S CO 1 2O2 Since dissociation requires energy (an endothermic reaction), the temperature of the products would be less than the maximum adiabatic temperature found under the assumption of complete combustion. When dissociation takes place, the gaseous products exiting the reactor would not be CO2 and N2 but a mixture of CO2, CO, O2, and N2. The balanced chemical reaction equation would read CO 1 2O2 1.88N2 S z CO 11 z2CO2 z 2 O2 1.88N2 (14.40) where z is the amount of CO, in kmol, present in the exiting mixture for each kmol of CO entering the reactor. Accordingly, there are two unknowns: z and the temperature of the exiting stream. To solve a problem with two unknowns requires two equations. One is provided by an energy equation. If the exiting gas mixture is in equilibrium, the other equation is provided by the equilibrium constant, Eq. 14.35. The temperature of the products may then be called the equilibrium flame temperature. The equilibrium constant used to evaluate the equilibrium flame temperature would be determined with respect to CO2 Sd CO 1 2O2 . Although only the dissociation of CO2 has been discussed, other products of combustion may dissociate, for example, H2O Sd H2 1 2O2 H2O Sd OH 1 2H2 O2 Sd 2O H2 Sd 2H N2 Sd 2N When there are many dissociation reactions, the study of chemical equilibrium is facilitated by the use of computers to solve the simultaneous equations that result. equilibrium flame temperature 900 Chapter 14 Chemical and Phase Equilibrium Simultaneous reactions are considered in Sec. 14.4.4. The following example illustrates how the equilibrium flame temperature is determined when one dissociation reaction occurs.

The dependence of the equilibrium constant on temperature exhibited by the values of Table A-27 follows from Eq. 14.31. An alternative way to express this dependence is given by the van't Hoff equation, Eq. 14.43b. The development of this equation begins by introducing Eq. 14.29b into Eq. 14.31 to obtain on rearrangement RT ln K 31nChC nD hD nAhA nB hB2 T1nC s C nD s D nAs A nB sB 24 (14.41) Each of the specific enthalpies and entropies in this equation depends on temperature alone. Differentiating with respect to temperature RT d ln K dT R ln K cnC a dhC dT T d s C dT b nD a dhD dT T d s D dT b nA a d hA dT T d s A dT b nBa d hB dT T d s B dT b d 1nC sC nD sD nAsA nBsB2 From the definition of s1T2 (Eq. 6.19), we have d s/dT cp /T. Moreover, dh/dT cp. Accordingly, each of the underlined terms in the above equation vanishes identically, leaving RT d ln K dT R ln K 1nCs C nD s D nAsA nB sB 2 (14.42) Using Eq. 14.41 to evaluate the second term on the left and simplifying the resulting expression, Eq. 14.42 becomes d ln K dT 1nC hC nDhD nAhA nBhB2 RT 2 (14.43a) or, expressed more concisely, d ln K dT ¢H RT 2 (14.43b) which is the van't Hoff equation. In Eq. 14.43b, ¢H is the enthalpy of reaction at temperature T. The van't Hoff equation shows that when ¢H is negative (exothermic reaction), K decreases with temperature, whereas for ¢H positive (endothermic reaction), K increases with temperature. van't Hoff equation 904 Chapter 14 Chemical and Phase Equilibrium The enthalpy of reaction ¢H is often very nearly constant over a rather wide interval of temperature. In such cases, Eq. 14.43b can be integrated to yield ln K2 K1 ¢H R a 1 T2 1 T1 b (14.44) where K1 and K2 denote the equilibrium constants at temperatures T1 and T2, respectively. This equation shows that ln K is linear in 1/T. Accordingly, plots of ln K versus 1/T can be used to determine ¢H from experimental equilibrium composition data. Alternatively, the equilibrium constant can be determined using enthalpy data. 14.4.3 Ionization The methods developed for determining the equilibrium composition of a reactive ideal gas mixture can be applied to systems involving ionized gases, also known as plasmas. In previous sections we considered the chemical equilibrium of systems where dissociation is a factor. For example, the dissociation reaction of diatomic nitrogen N2 dS 2N can occur at elevated temperatures. At still higher temperatures, ionization may take place according to N dS N1 1 e2 (14.45) That is, a nitrogen atom loses an electron, yielding a singly ionized nitrogen atom N1 and a free electron e2. Further heating can result in the loss of additional electrons until all electrons have been removed from the atom. For some cases of practical interest, it is reasonable to think of the neutral atoms, positive ions, and electrons as forming an ideal gas mixture. With this idealization, ionization equilibrium can be treated in the same manner as the chemical equilibrium of reacting ideal gas mixtures. The change in the Gibbs function for the equilibrium ionization reaction required to evaluate the ionization-equilibrium constant can be calculated as a function of temperature by using the procedures of statistical thermodynamics. In general, the extent of ionization increases as the temperature is raised and the pressure is lowered. Example 14.8 illustrates the analysis of ionization equilibrium.

The equilibrium of systems that may involve several phases, each having a number of components present, is considered in this section. The principal result is the Gibbs phase rule, which summarizes important limitations on multicomponent, multiphase systems at equilibrium. 14.6.1 Chemical Potential and Phase Equilibrium Figure 14.1 shows a system consisting of two components A and B in two phases 1 and 2 that are at the same temperature and pressure. Applying Eq. 14.10 to each of the phases dG¿4T,p mA¿ dnA¿ mB¿ dn¿ B dG-4T,p m- A dn- A m- B dn- B (14.63) where as before the primes identify the two phases. When matter is transferred between the two phases in the absence of chemical reaction, the total amounts of A and B must remain constant. Thus, the increase in the amount present in one of the phases must be compensated by an equivalent decrease in the amount present in the other phase. That is, dnA- dnA¿ , dn- B dnB¿ (14.64) With Eqs. 14.63 and 14.64, the change in the Gibbs function for the system is dG4T,p dG¿4T,p dG-4T,p 1mA¿ m- A2 dnA¿ 1mB¿ m- B2 dnB¿ (14.65) Since nA9 and nB9 can be varied independently, it follows that when dG4T, p 0, the terms in parentheses are zero, resulting in mA¿ mA- and m¿ B mB- (14.66) At equilibrium, the chemical potential of each component is the same in each phase. The significance of the chemical potential for phase equilibrium can be brought out simply by reconsidering the system of Fig. 14.1 in the special case when the chemical potential of component B is the same in both phases: mB¿ mB-. With this constraint, Eq. 14.65 reduces to dG4T,p 1mA¿ mA- 2 dnA¿ Any spontaneous process of the system taking place at a fixed temperature and pressure must be such that the Gibbs function decreases: dG4T, p 0. Thus, with the above expression we have 1m¿ A m- A2dnA¿ 0 Accordingly, c when the chemical potential of A is greater in phase 1 than in phase 2 1m¿ A m- A2, it follows that dnA¿ 0. That is, substance A passes from phase 1 to phase 2. c when the chemical potential of A is greater in phase 2 than in phase 1 1mA- mA¿ 2, it follows that dnA¿ 0. That is, substance A passes from phase 2 to phase 1. At equilibrium, the chemical potentials are equal 1mA¿ mA- 2, and there is no net transfer of A between the phases. With this reasoning, we see that the chemical potential can be regarded as a measure of the escaping tendency of a component. If the chemical potential of a component is not the same in each phase, there will be a tendency for that component to pass from Fig. 14.1 System consisting of two components in two phases. Phase 2 Component A, n´´A, ´´A Component B, n´´B, ´´B µ µ Phase 1 Component A, n´ A, ´A Component B, n´ B, ´ µ B µ 14.6 Equilibrium of Multicomponent, Multiphase Systems 911 the phase having the higher chemical potential for that component to the phase having the lower chemical potential. When the chemical potential is the same in both phases, there is no tendency for a net transfer to occur from one phase to the other. In Example 14.10, we apply phase equilibrium principles to provide a rationale for the model introduced in Sec. 12.5.3 for moist air in contact with liquid water.

The requirement for equilibrium of a system consisting of two components and two phases, given by Eqs. 14.66, can be extended with similar reasoning to nonreacting multicomponent, multiphase systems. At equilibrium, the chemical potential of each Integrating from p1 to p2 at fixed temperature yf1p2 p12 RT lna pv,2 pv,1 b (b) Alternatively, pv,2 pv,1 expa yf1T21p2 p12 RT b (c) Eq. (c) gives the change in partial pressure pv accompanying a change in total pressure, as required in part (a). A special case faciliates part (b): Since Eq. (c) is obtained assuming the water vapor and liquid water phases are in equilibrium, it applies in particular when no dry air is present initially. For that case, p1 pv,1 psat(T). Eq. (c) then becomes pv psat expa yf1T21p psat1T22 RT b (d) where p denotes the total pressure and pv is the accompanying partial pressure of the water vapor. (b) At 708F, Table A-2 gives psat = 0.3632 lbf/in.2 and f = 0.01605 ft3 /lb. Then, when p = 1 atm, Eq. (d) gives pv psat exp ° 0.01605 ft3 /lb114.696 0.363221lbf/in.2 2144 in.2 /ft2 a 1545 ft lbf 18.02 lb Rb1530R2 ¢ 1.00073 ❹ Accordingly, for the specified conditions the departure of pv from psat owing to the presence of the dry air is negligible. ❶ The term moist air refers to a mixture of dry air and water vapor in which the dry air is treated as if it were a pure component (Sec. 12.5.1). ❷ For phase equilibrium there would be a small, but finite, concentration of air in the liquid water phase. However, this small amount of dissolved air is ignored in the present development. ❸ To increase the pressure, we think of adding dry air while keeping temperature constant. ❹ The departure of pv from psat is negligible at the specified conditions. This suggests that at normal temperatures and pressures the equilibrium between the liquid water phase and the water vapor is not significantly disturbed by the presence of the dry air. Accordingly, the partial pressure of the water vapor can be taken as equal to the saturation pressure of the water at the system temperature. This model, introduced in Sec. 12.5.3, is used extensively in Chap. 12. ✓Skills Developed Ability to... ❑ apply the concept of phase equilibrium to an air-water vapor mixture in equilibrium with liquid water. Using the methods of Sec. 12.5.2, determine the humidity ratio, v, of the air-water vapor mixture of part (b). Ans. 0.01577 lb(vapor)/ lb(dry air). Quick Quiz 14.6 Equilibrium of Multicomponent, Multiphase Systems 913 component must be the same in all phases. For the case of N components that are present in P phases we have, therefore, the following set of N(P 2 1) equations: P 1 N d m 1 1 m 2 1 m 3 1 . . . m P 1 m 1 2 m 2 2 m 3 2 . . . m P 2 . . . m 1 N m 2 N m 3 N . . . m P N (14.67) where mj i denotes the chemical potential of the ith component in the jth phase. This set of equations provides the basis for the Gibbs phase rule, which allows the determination of the number of independent intensive properties that may be arbitrarily specified in order to fix the intensive state of the system. The number of independent intensive properties is called the degrees of freedom (or the variance). Since the chemical potential is an intensive property, its value depends on the relative proportions of the components present and not on the amounts of the components. In other words, in a given phase involving N components at temperature T and pressure p, the chemical potential is determined by the mole fractions of the components present and not the respective n's. However, as the mole fractions add to unity, at most N 2 1 of the mole fractions can be independent. Thus, for a system involving N components, there are at most N 2 1 independently variable mole fractions for each phase. For P phases, therefore, there are at most P(N 2 1) independently variable mole fractions. In addition, the temperature and pressure, which are the same in each phase, are two further intensive properties, giving a maximum of P(N 2 1) 1 2 independently variable intensive properties for the system. But because of the N(P 2 1) equilibrium conditions represented by Eqs. 14.67 among these properties, the number of intensive properties that are freely variable, the degrees of freedom F, is F 3P1N 12 24 N1P 12 2 N P (14.68) which is the Gibbs phase rule. In Eq. 14.68, F is the number of intensive properties that may be arbitrarily specified and that must be specified to fix the intensive state of a nonreacting system at equilibrium. let us apply the Gibbs phase rule to a liquid solution consisting of water and ammonia such as considered in the discussion of absorption refrigeration (Sec. 10.5). This solution involves two components and a single phase: N 2 and P 1. Equation 14.68 then gives F 3, so the intensive state is fixed by giving the values of three intensive properties, such as temperature, pressure, and the ammonia (or water) mole fraction. b b b b b The phase rule summarizes important limitations on various types of systems. For example, for a system involving a single component such as water, N 1 and Eq. 14.68 becomes F 3 P (14.69) c The minimum number of phases is one, corresponding to P 1. For this case, Eq. 14.69 gives F 5 2. That is, two intensive properties must be specified to fix the intensive state of the system. This requirement is familiar from our use of the steam tables and similar property tables. To obtain properties of superheated vapor, say, from such tables requires that we give values for any two of the tabulated properties, for example, T and p. c When two phases are present in a system involving a single component, N 1 and P 2. Equation 14.69 then gives F 1. That is, the intensive state is determined degrees of freedom Gibbs phase rule μ 914 Chapter 14 Chemical and Phase Equilibrium by a single intensive property value. For example, the intensive states of the separate phases of an equilibrium mixture of liquid water and water vapor are completely determined by specifying the temperature. c The minimum allowable value for the degrees of freedom is zero: F 0. For a single-component system, Eq. 14.69 shows that this corresponds to P 3, a threephase system. Thus, three is the maximum number of different phases of a pure component that can coexist in equilibrium. Since there are no degrees of freedom, both temperature and pressure are fixed at equilibrium. For example, there is only a single temperature 0.018C (32.028F) and a single pressure 0.6113 kPa (0.006 atm) for which ice, liquid water, and water vapor are in equilibrium. The phase rule given here must be modified for application to systems in which chemical reactions occur. Furthermore, the system of equations, Eqs. 14.67, giving the requirements for phase equilibrium at a specified temperature and pressure can be expressed alternatively in terms of partial molal Gibbs functions, fugacities, and activities, all of which are introduced in Sec. 11.9. To use any such expression to determine the equilibrium composition of the different phases present within a system at equilibrium requires a model for each phase that allows the relevant quantities— the chemical potentials, fugacities, and so on—to be evaluated for the components present in terms of system properties that can be determined. For example, a gas phase might be modeled as an ideal gas mixture or, at higher pressures, as an ideal solution.

This expression can be written alternatively in terms of specific enthalpies and entropies as ¢G nC1hC T sC 2 nD1hD T s D 2 nA1hA T s A 2 nB1hB T s B2 1nC hC nD hD nAhA nBhB2 T1nC sC nD s D nAs A nB s B 2 (14.29b) Since the enthalpy of an ideal gas depends on temperature only, the h's of Eq. 14.29b are evaluated at temperature T. As indicated by the superscript 8, each of the entropies is evaluated at temperature T and a pressure of 1 atm. Introducing Eq. 14.29a into Eq. 14.28 and combining the terms involving logarithms into a single expression gives ¢G RT ln c ynC C ynD D ynA A ynB B a p prefb nCnD nAnB d (14.30) Equation 14.30 is simply the form taken by the equation of reaction equilibrium, Eq. 14.26, for an ideal gas mixture subject to the reaction Eq. 14.24. As illustrated by subsequent examples, similar expressions can be written for other reactions. Equation 14.30 can be expressed concisely as ¢G RT ln K1T2 (14.31) where K is the equilibrium constant defined by K1T2 ynC C ynD D ynA A ynB B a p prefb nCnD nAnB (14.32) Given the values of the stoichiometric coefficients, nA, nB, nC, and nD and the temperature T, the left side of Eq. 14.31 can be evaluated using either of Eqs. 14.29 together with the appropriate property data. The equation can then be solved for the value of the equilibrium constant K. Accordingly, for selected reactions K can be evaluated and tabulated against temperature. It is common to tabulate log10K or ln K versus temperature, however. A tabulation of log10K values over a range of temperatures for several reactions is provided in Table A-27, which is extracted from a more extensive compilation. The terms in the numerator and denominator of Eq. 14.32 correspond, respectively, to the products and reactants of the reaction given by Eq. 14.24 as it proceeds from left to right as written. For the inverse reaction nCC nDD dS nAA nBB, the equilibrium constant takes the form K* ynA A ynB B ynC C ynD D a p prefb nAnBnCnD (14.33) Comparing Eqs. 14.32 and 14.33, it follows that the value of K* is just the reciprocal of K: K* 5 1/K. Accordingly, log10K* log10K (14.34) Hence, Table A-27 can be used both to evaluate K for the reactions listed proceeding in the direction left to right and to evaluate K* for the inverse reactions proceeding in the direction right to left. Example 14.1 illustrates how the log10K values of Table A-27 are determined. Subsequent examples show how the log10K values can be used to evaluate equilibrium compositions.

where pref is 1 atm and yi is the mole fraction of component i in a mixture at temperature T and pressure p. The last equation can be written compactly as mi gi RT ln yip pref 1ideal gas2 (14.17) where g i is the Gibbs function of component i evaluated at temperature T and a pressure of 1 atm. Additional details concerning the chemical potential concept are provided in Sec. 11.9. Equation 14.17 is the same as Eq. 11.144 developed there.

Thermodynamics 14

Kaugnay na mga set ng pag-aaral

Accountability - Fundamentals

HW #9 The Hip & Pelvic Joint

Exam 2

FIN 300 UKY Exam 1

Spanish Day Of The Dead Quiz

Assessment 2

Csci 161 test 1

Chapter 7 Survey Research

305 chap 1

Almost, Maine - Daniel (Man) and Hope

Communities Develop and Montana Becomes a Territory

A&P Ch. 12: Membrane Potential and the Action Potential

Physics momentum review

Pysch. Ch. 1-2

chapter 1: dynamic environment of hrm

Old Business Tests

Anatomy Terms Unit 1

Macro Final (CH 22 review)

Server OS Ch 3 Prototype

Penn Foster, Computer Applications