Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM

Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM | 793 I n Chapter 15 we analyzed combustion processes under the assumption that combustion is complete when there is sufficient time and oxygen. Often this is not the case, however. A chemical reaction may reach a state of equilib- rium before reaching completion even when there is sufficient time and oxygen. A system is said to be in equilibrium if no changes occur within the system when it is isolated from its surroundings. An isolated system is in mechanical equilibrium if no changes occur in pressure, in thermal equilibrium if no changes occur in temperature, in phase equilibrium if no transformations occur from one phase to another, and in chemical equilib- rium if no changes occur in the chemical composition of the system. The conditions of mechanical and thermal equilib- rium are straightforward, but the conditions of chemical and phase equilibrium can be rather involved. The equilibrium criterion for reacting systems is based on the second law of thermodynamics; more specifically, the increase of entropy principle. For adiabatic systems, chemical equilibrium is established when the entropy of the reacting system reaches a maximum. Most reacting systems encoun- tered in practice are not adiabatic, however. Therefore, we need to develop an equilibrium criterion applicable to any reacting system. In this chapter, we develop a general criterion for chemical equilibrium and apply it to reacting ideal-gas mixtures. We then extend the analysis to simultaneous reactions. Finally, we discuss phase equilibrium for nonreacting systems. Objectives The objectives of Chapter 16 are to: • Develop the equilibrium criterion for reacting systems based on the second law of thermodynamics. • Develop a general criterion for chemical equilibrium applicable to any reacting system based on minimizing the Gibbs function for the system. • Define and evaluate the chemical equilibrium constant. • Apply the general criterion for chemical equilibrium analysis to reacting ideal-gas mixtures. • Apply the general criterion for chemical equilibrium analysis to simultaneous reactions. • Relate the chemical equilibrium constant to the enthalpy of reaction. • Establish the phase equilibrium for nonreacting systems in terms of the specific Gibbs function of the phases of a pure substance. • Apply the Gibbs phase rule to determine the number of independent variables associated with a multicomponent, multiphase system. • Apply Henry’s law and Raoult’s law for gases dissolved in liquids. cen84959_ch16.qxd 5/11/05 10:08 AM Page 793 16–1 ■ CRITERION FOR CHEMICAL EQUILIBRIUM Consider a reaction chamber that contains a mixture of CO, O 2 , and CO 2 at a specified temperature and pressure. Let us try to predict what will happen in this chamber (Fig. 16–1). Probably the first thing that comes to mind is a chemical reaction between CO and O 2 to form more CO 2 : This reaction is certainly a possibility, but it is not the only possibility. It is also possible that some CO 2 in the combustion chamber dissociated into CO and O 2 . Yet a third possibility would be to have no reactions among the three components at all, that is, for the system to be in chemical equi- librium. It appears that although we know the temperature, pressure, and composition (thus the state) of the system, we are unable to predict whether the system is in chemical equilibrium. In this chapter we develop the neces- sary tools to correct this. Assume that the CO, O 2 , and CO 2 mixture mentioned above is in chemical equilibrium at the specified temperature and pressure. The chemical compo- sition of this mixture does not change unless the temperature or the pressure of the mixture is changed. That is, a reacting mixture, in general, has differ- ent equilibrium compositions at different pressures and temperatures. There- fore, when developing a general criterion for chemical equilibrium, we consider a reacting system at a fixed temperature and pressure. Taking the positive direction of heat transfer to be to the system, the increase of entropy principle for a reacting or nonreacting system was expressed in Chapter 7 as (16–1) A system and its surroundings form an adiabatic system, and for such systems Eq. 16–1 reduces to dS sys Ն 0. That is, a chemical reaction in an adiabatic chamber proceeds in the direction of increasing entropy. When the entropy reaches a maximum, the reaction stops (Fig. 16–2). Therefore, entropy is a very useful property in the analysis of reacting adiabatic systems. When a reacting system involves heat transfer, the increase of entropy principle relation (Eq. 16–1) becomes impractical to use, however, since it requires a knowledge of heat transfer between the system and its surround- ings. A more practical approach would be to develop a relation for the equilibrium criterion in terms of the properties of the reacting system only. Such a relation is developed below. Consider a reacting (or nonreacting) simple compressible system of fixed mass with only quasi-equilibrium work modes at a specified temperature T and pressure P (Fig. 16–3). Combining the first- and the second-law relations for this system gives (16–2) dQ Ϫ P dV ϭ dU dS Ն dQ T ¶ dU ϩ P dV Ϫ T ds Յ 0 dS sys Ն dQ T CO ϩ 1 2 O 2 ¬ S ¬ CO 2 794 | Thermodynamics CO 2 CO O 2 O 2 O 2 CO 2 CO 2 CO CO FIGURE 16–1 A reaction chamber that contains a mixture of CO 2 , CO, and O 2 at a specified temperature and pressure. 100% products Violation of second law S Equilibrium composition 100% reactants dS = 0 dS > 0 dS < 0 FIGURE 16–2 Equilibrium criteria for a chemical reaction that takes place adiabatically. W b REACTION CHAMBER δ δ Control mass T, P Q FIGURE 16–3 A control mass undergoing a chemical reaction at a specified temperature and pressure. cen84959_ch16.qxd 5/11/05 10:08 AM Page 794 The differential of the Gibbs function (G ϭ H Ϫ TS) at constant tempera- ture and pressure is (16–3) From Eqs. 16–2 and 16–3, we have (dG) T,P Յ 0. Therefore, a chemical reac- tion at a specified temperature and pressure proceeds in the direction of a decreasing Gibbs function. The reaction stops and chemical equilibrium is established when the Gibbs function attains a minimum value (Fig. 16–4). Therefore, the criterion for chemical equilibrium can be expressed as (16–4) A chemical reaction at a specified temperature and pressure cannot proceed in the direction of the increasing Gibbs function since this will be a viola- tion of the second law of thermodynamics. Notice that if the temperature or the pressure is changed, the reacting system will assume a different equilib- rium state, which is the state of the minimum Gibbs function at the new temperature or pressure. To obtain a relation for chemical equilibrium in terms of the properties of the individual components, consider a mixture of four chemical components A, B, C, and D that exist in equilibrium at a specified temperature and pres- sure. Let the number of moles of the respective components be N A , N B , N C , and N D . Now consider a reaction that occurs to an infinitesimal extent during which differential amounts of A and B (reactants) are converted to C and D (products) while the temperature and the pressure remain constant (Fig. 16–5): The equilibrium criterion (Eq. 16–4) requires that the change in the Gibbs function of the mixture during this process be equal to zero. That is, (16–5) or (16–6) where the g – ’s are the molar Gibbs functions (also called the chemical poten- tials) at the specified temperature and pressure and the dN’s are the differen- tial changes in the number of moles of the components. To find a relation between the dN’s, we write the corresponding stoichio- metric (theoretical) reaction (16–7) where the n’s are the stoichiometric coefficients, which are evaluated easily once the reaction is specified. The stoichiometric reaction plays an impor- tant role in the determination of the equilibrium composition of the reacting n A A ϩ n B B ∆ n C C ϩ n D D g Ϫ C¬ dN C ϩ g Ϫ D¬ dN D ϩ g Ϫ A¬ dN A ϩ g Ϫ B¬ dN B ϭ 0 1dG 2 T,P ϭ a 1dG i 2 T,P ϭ a 1g Ϫ i¬ dN i 2 T,P ϭ 0 dN A A ϩ dN B B ¬ ¡ ¬ dN C C ϩ dN D D 1dG 2 T,P ϭ 0 ϭ dU ϩ P dV Ϫ T dS ϭ 1dU ϩ P dV ϩ V dP2Ϫ T dS Ϫ S dT 1dG 2 T,P ϭ dH Ϫ T dS Ϫ S dT Chapter 16 | 795 100% products Violation of second law G Equilibrium composition 100% reactants dG = 0 dG < 0 dG > 0 FIGURE 16–4 Criteria for chemical equilibrium for a fixed mass at a specified temperature and pressure. REACTION CHAMBER T, P N A moles of A N B moles of B N C moles of C N D moles of D dN A A + dN B B → dN C C + dN D D FIGURE 16–5 An infinitesimal reaction in a chamber at constant temperature and pressure. → 0 → 0 cen84959_ch16.qxd 5/11/05 10:08 AM Page 795 mixtures because the changes in the number of moles of the components are proportional to the stoichiometric coefficients (Fig. 16–6). That is, (16–8) where e is the proportionality constant and represents the extent of a reac- tion. A minus sign is added to the first two terms because the number of moles of the reactants A and B decreases as the reaction progresses. For example, if the reactants are C 2 H 6 and O 2 and the products are CO 2 and H 2 O, the reaction of 1 mmol (10 Ϫ6 mol) of C 2 H 6 results in a 2-mmol increase in CO 2 ,a 3-mmol increase in H 2 O, and a 3.5-mmol decrease in O 2 in accordance with the stoichiometric equation That is, the change in the number of moles of a component is one-millionth (e ϭ 10 Ϫ6 ) of the stoichiometric coefficient of that component in this case. Substituting the relations in Eq. 16–8 into Eq. 16–6 and canceling e,we obtain (16–9) This equation involves the stoichiometric coefficients and the molar Gibbs functions of the reactants and the products, and it is known as the criterion for chemical equilibrium. It is valid for any chemical reaction regardless of the phases involved. Equation 16–9 is developed for a chemical reaction that involves two reactants and two products for simplicity, but it can easily be modified to handle chemical reactions with any number of reactants and products. Next we analyze the equilibrium criterion for ideal-gas mixtures. 16–2 ■ THE EQUILIBRIUM CONSTANT FOR IDEAL-GAS MIXTURES Consider a mixture of ideal gases that exists in equilibrium at a specified temperature and pressure. Like entropy, the Gibbs function of an ideal gas depends on both the temperature and the pressure. The Gibbs function val- ues are usually listed versus temperature at a fixed reference pressure P 0 , which is taken to be 1 atm. The variation of the Gibbs function of an ideal gas with pressure at a fixed temperature is determined by using the defini- tion of the Gibbs function and the entropy-change relation for isothermal processes . It yields Thus the Gibbs function of component i of an ideal-gas mixture at its partial pressure P i and mixture temperature T can be expressed as (16–10) g Ϫ i 1T, P i 2ϭ g Ϫ * i 1T 2ϩ R u T ln P i 1¢g Ϫ 2 T ϭ ¢h Ϫ Ϫ T 1¢ s Ϫ 2 T ϭϪT 1¢ s Ϫ 2 T ϭ R u T ln P 2 P 1 3¢ s Ϫ ϭϪR u ln 1P 2 >P 1 24 1g Ϫ ϭ h Ϫ Ϫ Ts Ϫ 2 n C g Ϫ C ϩ n D g Ϫ D Ϫ n A g Ϫ A Ϫ n B g Ϫ B ϭ 0 C 2 H 6 ϩ 3.5O 2 ¬ S ¬ 2CO 2 ϩ 3H 2 O dN A ϭϪen A dN C ϭ en C dN B ϭϪen B dN D ϭ en D 796 | Thermodynamics 0.1H 2 → 0.2H H 2 → 2H H 2 = 1 H = 2 0.01H 2 → 0.02H 0.001H 2 → 0.002H n n FIGURE 16–6 The changes in the number of moles of the components during a chemical reaction are proportional to the stoichiometric coefficients regardless of the extent of the reaction. → 0 cen84959_ch16.qxd 5/11/05 10:08 AM Page 796 where g – i * (T) represents the Gibbs function of component i at 1 atm pres- sure and temperature T, and P i represents the partial pressure of component i in atmospheres. Substituting the Gibbs function expression for each com- ponent into Eq. 16–9, we obtain For convenience, we define the standard-state Gibbs function change as (16–11) Substituting, we get (16–12) Now we define the equilibrium constant K P for the chemical equilibrium of ideal-gas mixtures as (16–13) Substituting into Eq. 16–12 and rearranging, we obtain (16–14) Therefore, the equilibrium constant K P of an ideal-gas mixture at a specified temperature can be determined from a knowledge of the standard-state Gibbs function change at the same temperature. The K P values for several reactions are given in Table A–28. Once the equilibrium constant is available, it can be used to determine the equilibrium composition of reacting ideal-gas mixtures. This is accom- plished by expressing the partial pressures of the components in terms of their mole fractions: where P is the total pressure and N total is the total number of moles present in the reaction chamber, including any inert gases. Replacing the partial pressures in Eq. 16–13 by the above relation and rearranging, we obtain (Fig. 16–7) (16–15) where Equation 16–15 is written for a reaction involving two reactants and two products, but it can be extended to reactions involving any number of reac- tants and products. ¢n ϭ n C ϩ n D Ϫ n A Ϫ n B K P ϭ N C n C N D n D N A n A N B n B a P N total b ¢n P i ϭ y i P ϭ N i N total P K P ϭ e Ϫ¢G*1T2>R u T K P ϭ P C n C P D n D P A n A P B n B ¢G* 1T 2ϭϪR u T 1n C ln P C ϩ n D ln P D Ϫ n A ln P A Ϫ n B ln P B 2ϭϪR u T ln P C n C P D n D P A n A P B n B ¢G* 1T2ϭ n C g Ϫ * C 1T 2ϩ n D g Ϫ * D 1T 2Ϫ n A g Ϫ * A 1T 2Ϫ n B g Ϫ * B 1T 2 Ϫn A 3g Ϫ * A 1T 2ϩ R u T ln P A 4Ϫ n B 3g Ϫ * B 1T 2ϩ R u T ln P B 4ϭ 0 n C 3g Ϫ * C 1T 2ϩ R u T ln P C 4ϩ n D 3g Ϫ D * 1T 2ϩ R u T ln P D 4 Chapter 16 | 797 (1) In terms of partial pressures K P = P C P D CD P A P B AB (3) In terms of the equilibrium composition K P = N C N D ( N A N B A B ( N total P ∆ (2) In terms of ∆G*(T) K P = e –∆ G*(T)/R T C D u nn nn nn nn n FIGURE 16–7 Three equivalent K P relations for reacting ideal-gas mixtures. cen84959_ch16.qxd 5/11/05 10:08 AM Page 797 798 | Thermodynamics EXAMPLE 16–1 Equilibrium Constant of a Dissociation Process Using Eq. 16–14 and the Gibbs function data, determine the equilibrium constant K P for the dissociation process N 2 → 2N at 25°C. Compare your result to the K P value listed in Table A–28. Solution The equilibrium constant of the reaction N 2 → 2N is listed in Table A–28 at different temperatures. It is to be verified using Gibbs func- tion data. Assumptions 1 The constituents of the mixture are ideal gases. 2 The equi- librium mixture consists of N 2 and N only. Properties The equilibrium constant of this reaction at 298 K is ln K P ϭ Ϫ367.5 (Table A–28). The Gibbs function of formation at 25°C and 1 atm is O for N 2 and 455,510 kJ/kmol for N (Table A–26). Analysis In the absence of K P tables, K P can be determined from the Gibbs function data and Eq. 16–14, where, from Eq. 16–11, Substituting, we find or The calculated K P value is in agreement with the value listed in Table A–28. The K P value for this reaction is practically zero, indicating that this reaction will not occur at this temperature. Discussion Note that this reaction involves one product (N) and one reactant (N 2 ), and the stoichiometric coefficients for this reaction are n N ϭ 2 and n N 2 ϭ 1. Also note that the Gibbs function of all stable elements (such as N 2 ) is assigned a value of zero at the standard reference state of 25°C and 1 atm. The Gibbs function values at other temperatures can be calculated from the enthalpy and absolute entropy data by using the definition of the Gibbs func- tion, , where . h Ϫ 1T 2ϭ h Ϫ * f ϩ h Ϫ T Ϫ h Ϫ 298 K g Ϫ * 1T 2ϭ h Ϫ 1T 2Ϫ Ts Ϫ * 1T 2 K P Х 2 ؋ 10 ؊160 ϭϪ367.5 ln K P ϭϪ 911,020 kJ>kmol 18.314 kJ>kmol # K21298.15 K2 ϭ 911,020 kJ>kmol ϭ 12 21455,510 kJ>kmol2Ϫ 0 ¢G* 1T2ϭ n N g Ϫ * N 1T 2Ϫ n N 2 g Ϫ * N 2 1T 2 K P ϭ e Ϫ¢G*1T2>R u T EXAMPLE 16–2 Dissociation Temperature of Hydrogen Determine the temperature at which 10 percent of diatomic hydrogen (H 2 ) dissociates into monatomic hydrogen (H) at a pressure of 10 atm. Solution The temperature at which 10 percent of H 2 dissociates into 2H is to be determined. Assumptions 1 The constituents of the mixture are ideal gases. 2 The equi- librium mixture consists of H 2 and H only. cen84959_ch16.qxd 5/11/05 10:08 AM Page 798 A double arrow is used in equilibrium equations as an indication that a chemical reaction does not stop when chemical equilibrium is established; rather, it proceeds in both directions at the same rate. That is, at equilibrium, the reactants are depleted at exactly the same rate as they are replenished from the products by the reverse reaction. 16–3 ■ SOME REMARKS ABOUT THE K P OF IDEAL-GAS MIXTURES In the last section we developed three equivalent expressions for the equilib- rium constant K P of reacting ideal-gas mixtures: Eq. 16–13, which expresses K P in terms of partial pressures; Eq. 16–14, which expresses K P in terms of the standard-state Gibbs function change ∆G*(T); and Eq. 16–15, which expresses K P in terms of the number of moles of the components. All three relations are equivalent, but sometimes one is more convenient to use than the others. For example, Eq. 16–15 is best suited for determining the equi- librium composition of a reacting ideal-gas mixture at a specified tem- perature and pressure. On the basis of these relations, we may draw the following conclusions about the equilibrium constant K P of ideal-gas mixtures: 1. The K P of a reaction depends on temperature only. It is independent of the pressure of the equilibrium mixture and is not affected by the presence of inert gases. This is because K P depends on ∆G*(T), which depends on Chapter 16 | 799 Analysis This is a dissociation process that is significant at very high tem- peratures only. For simplicity we consider 1 kmol of H 2 , as shown in Fig. 16–8. The stoichiometric and actual reactions in this case are as follows: Stoichiometric: Actual: reactants products (leftover) A double-headed arrow is used for the stoichiometric reaction to differentiate it from the actual reaction. This reaction involves one reactant (H 2 ) and one product (H). The equilibrium composition consists of 0.9 kmol of H 2 (the leftover reactant) and 0.2 kmol of H (the newly formed product). Therefore, N H 2 ϭ 0.9 and N H ϭ 0.2 and the equilibrium constant K P is determined from Eq. 16–15 to be From Table A–28, the temperature corresponding to this K P value is Discussion We conclude that 10 percent of H 2 dissociates into H when the temperature is raised to 3535 K. If the temperature is increased further, the percentage of H 2 that dissociates into H will also increase. T ϭ 3535 K K P ϭ N H n H N H 2 n H 2 a P N total b n H Ϫn H 2 ϭ 10.2 2 2 0.9 a 10 0.9 ϩ 0.2 b 2Ϫ1 ϭ 0.404 H 2 ¡ 0.9H 2 ϩ 0.2H H 2 ¬ ∆ ¬ 2H ¬ 1thus n H 2 ϭ 1 and n H ϭ 22 1 kmol H 2 10 atm Initial composition 0.9H 2 0.2H Equilibrium composition FIGURE 16–8 Schematic for Example 16–2. 123 123 cen84959_ch16.qxd 5/11/05 10:08 AM Page 799 temperature only, and the ∆G*(T) of inert gases is zero (see Eq. 16–14). Thus, at a specified temperature the following four reactions have the same K P value: at 1 atm at 5 atm at 3 atm at 2 atm 2. The K P of the reverse reaction is 1/K P . This is easily seen from Eq. 16–13. For reverse reactions, the products and reactants switch places, and thus the terms in the numerator move to the denominator and vice versa. Consequently, the equilibrium constant of the reverse reaction becomes 1/K P . For example, from Table A–28, 3. The larger the K P , the more complete the reaction. This is also apparent from Fig. 16–9 and Eq. 16–13. If the equilibrium composition consists largely of product gases, the partial pressures of the products (P C and P D ) are considerably larger than the partial pressures of the reactants (P A and P B ), which results in a large value of K P . In the limiting case of a complete reaction (no leftover reactants in the equilibrium mixture), K P approaches infinity. Conversely, very small values of K P indicate that a reaction does not proceed to any appreciable degree. Thus reactions with very small K P values at a specified temperature can be neglected. A reaction with K P Ͼ 1000 (or ln K P Ͼ 7) is usually assumed to proceed to completion, and a reaction with K P Ͻ 0.001 (or ln K P Ͻ Ϫ7) is assumed not to occur at all. For example, ln K P ϭϪ6.8 for the reaction N 2 ∆ 2N at 5000 K. Therefore, the dissociation of N 2 into monatomic nitrogen (N) can be disregarded at temperatures below 5000 K. 4. The mixture pressure affects the equilibrium composition (although it does not affect the equilibrium constant K P ). This can be seen from Eq. 16–15, which involves the term P ∆n , where ∆n ϭ ͚ n P Ϫ ͚ n R (the dif- ference between the number of moles of products and the number of moles of reactants in the stoichiometric reaction). At a specified temperature, the K P value of the reaction, and thus the right-hand side of Eq. 16–15, remains constant. Therefore, the mole numbers of the reactants and the products must change to counteract any changes in the pressure term. The direction of the change depends on the sign of ∆n. An increase in pressure at a speci- fied temperature increases the number of moles of the reactants and decreases the number of moles of products if ∆n is positive, have the oppo- site effect if ∆n is negative, and have no effect if ∆n is zero. 5. The presence of inert gases affects the equilibrium composition (although it does not affect the equilibrium constant K P ). This can be seen from Eq. 16–15, which involves the term (1/N total ) ∆n , where N total is the total number of moles of the ideal-gas mixture at equilibrium, including inert gases. The sign of ∆n determines how the presence of inert gases influences the equi- librium composition (Fig. 16–10). An increase in the number of moles of inert gases at a specified temperature and pressure decreases the number of K P ϭ 8.718 ϫ 10 Ϫ11 ¬¬ for ¬¬¬ H 2 O ∆ H 2 ϩ 1 2 O 2 ¬ at 1000 K K P ϭ 0.1147 ϫ 10 11 ¬¬ for ¬ H 2 ϩ 1 2 O 2 ∆ H 2 O ¬¬¬ at 1000 K H 2 ϩ 2O 2 ϩ 5N 2 ∆ H 2 O ϩ 1.5O 2 ϩ 5N 2 H 2 ϩ 1 2 O 2 ϩ 3N 2 ∆ H 2 O ϩ 3N 2 H 2 ϩ 1 2 O 2 ∆ H 2 O H 2 ϩ 1 2 O 2 ∆ H 2 O 800 | Thermodynamics 4000 5000 1000 2000 3000 T, K P = 1 atm 6000 5.17 ϫ 10 –18 2.65 ϫ 10 –6 2.545 41.47 0.025 267.7 76.80 97.70 0.00 0.16 14.63 99.63 K P % mol H H 2 → 2H FIGURE 16–9 The larger the K P , the more complete the reaction. 1 mol H 2 Initial composition Equilibrium composition at 3000 K, 1 atm (a) (b) 1 mol N 2 1 mol H 2 K P = 0.0251 0.158 mol H 0.921 mol H 2 K P = 0.0251 1.240 mol H 0.380 mol H 2 1 mol N 2 FIGURE 16–10 The presence of inert gases does not affect the equilibrium constant, but it does affect the equilibrium composition. cen84959_ch16.qxd 5/11/05 10:08 AM Page 800 moles of the reactants and increases the number of moles of products if ∆n is positive, have the opposite effect if ∆n is negative, and have no effect if ∆n is zero. 6. When the stoichiometric coefficients are doubled, the value of K P is squared. Therefore, when one is using K P values from a table, the stoichio- metric coefficients (the n’s) used in a reaction must be exactly the same ones appearing in the table from which the K P values are selected. Multiply- ing all the coefficients of a stoichiometric equation does not affect the mass balance, but it does affect the equilibrium constant calculations since the stoichiometric coefficients appear as exponents of partial pressures in Eq. 16–13. For example, For But for 7. Free electrons in the equilibrium composition can be treated as an ideal gas. At high temperatures (usually above 2500 K), gas molecules start to dissociate into unattached atoms (such as H 2 ∆ 2H), and at even higher temperatures atoms start to lose electrons and ionize, for example, (16–16) The dissociation and ionization effects are more pronounced at low pres- sures. Ionization occurs to an appreciable extent only at very high tempera- tures, and the mixture of electrons, ions, and neutral atoms can be treated as an ideal gas. Therefore, the equilibrium composition of ionized gas mixtures can be determined from Eq. 16–15 (Fig. 16–11). This treatment may not be adequate in the presence of strong electric fields, however, since the elec- trons may be at a different temperature than the ions in this case. 8. Equilibrium calculations provide information on the equilibrium compo- sition of a reaction, not on the reaction rate. Sometimes it may even take years to achieve the indicated equilibrium composition. For example, the equilibrium constant of the reaction at 298 K is about 10 40 , which suggests that a stoichiometric mixture of H 2 and O 2 at room temperature should react to form H 2 O, and the reaction should go to com- pletion. However, the rate of this reaction is so slow that it practically does not occur. But when the right catalyst is used, the reaction goes to comple- tion rather quickly to the predicted value. H 2 ϩ 1 2 O 2 ∆ H 2 O H ∆ H ϩ ϩ e Ϫ 2H 2 ϩ O 2 ∆ 2H 2 O ¬¬ K P 2 ϭ P 2 H 2 O P 2 H 2 P O2 ϭ 1K P 1 2 2 H 2 ϩ 1 2 O 2 ∆ H 2 O K P 1 ϭ P H 2 O P H 2 P 1>2 O 2 Chapter 16 | 801 EXAMPLE 16–3 Equilibrium Composition at a Specified Temperature A mixture of 2 kmol of CO and 3 kmol of O 2 is heated to 2600 K at a pres- sure of 304 kPa. Determine the equilibrium composition, assuming the mix- ture consists of CO 2 , CO, and O 2 (Fig. 16–12). Solution A reactive gas mixture is heated to a high temperature. The equi- librium composition at that temperature is to be determined. 2 kmol CO Initial composition x CO 2 Equilibrium composition at 2600 K, 304 kPa y CO z O 2 3 kmol O 2 FIGURE 16–12 Schematic for Example 16–3. where N total = N H + N H + + N e – K P = H → H + + e – ∆ = H + + e – – H = 1 + 1 – 1 = 1 N H + N e – ∆ N total ( ( N H H + e – P H n nn n nn n n FIGURE 16–11 Equilibrium-constant relation for the ionization reaction of hydrogen. cen84959_ch16.qxd 5/11/05 10:08 AM Page 801 802 | Thermodynamics Assumptions 1 The equilibrium composition consists of CO 2 , CO, and O 2 . 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are as follows: Stoichiometric: Actual: products reactants (leftover) C balance: O balance: Total number of moles: Pressure: The closest reaction listed in Table A–28 is CO 2 ∆ CO ϩ 1 – 2 O 2 , for which ln K P ϭϪ2.801 at 2600 K. The reaction we have is the inverse of this, and thus ln K P ϭϩ2.801, or K P ϭ 16.461 in our case. Assuming ideal-gas behavior for all components, the equilibrium constant relation (Eq. 16–15) becomes Substituting, we get Solving for x yields Then Therefore, the equilibrium composition of the mixture at 2600 K and 304 kPa is Discussion In solving this problem, we disregarded the dissociation of O 2 into O according to the reaction O 2 → 2O, which is a real possibility at high temperatures. This is because ln K P ϭϪ7.521 at 2600 K for this reaction, which indicates that the amount of O 2 that dissociates into O is negligible. (Besides, we have not learned how to deal with simultaneous reactions yet. We will do so in the next section.) 1.906CO 2 ؉ 0.094CO ؉ 2.074O 2 z ϭ 3 Ϫ x 2 ϭ 2.047 y ϭ 2 Ϫ x ϭ 0.094 x ϭ 1.906 16.461 ϭ x 12 Ϫ x 213 Ϫ x>22 1>2 a 3 5 Ϫ x>2 b Ϫ1>2 K P ϭ N CO 2 n CO 2 N CO n CO N O 2 n O 2 a P N total b n CO 2 Ϫn CO Ϫn O 2 P ϭ 304 kPa ϭ 3.0 atm N total ϭ x ϩ y ϩ z ϭ 5 Ϫ x 2 8 ϭ 2x ϩ y ϩ 2z ¬ or z ϭ 3 Ϫ x 2 2 ϭ x ϩ y ¬ or y ϭ 2 Ϫ x 2CO ϩ 3O 2 ¡ xCO 2 ϩ yCO ϩ zO 2 CO ϩ 1 2 O 2 ∆ CO 2 1thus n CO 2 ϭ 1, n CO ϭ 1, and n O 2 ϭ 1 2 2 123 1552553 cen84959_ch16.qxd 5/11/05 10:08 AM Page 802 [...]... liquid phase to the vapor phase, and the two phases are in phase equilibrium The conditions of phase equilibrium change, however, if the temperature or the pressure is changed Therefore, we examine phase equilibrium at a specified temperature and pressure Phase Equilibrium for a Single-Component System T, P VAPOR mg LIQUID mf FIGURE 16 18 A liquid–vapor mixture in equilibrium at a constant temperature and. .. data and (b) KP data Phase Equilibrium 16 56C Consider a tank that contains a saturated liquid– vapor mixture of water in equilibrium Some vapor is now allowed to escape the tank at constant temperature and pressure Will this disturb the phase equilibrium and cause some of the liquid to evaporate? 16 57C Consider a two -phase mixture of ammonia and water in equilibrium Can this mixture exist in two phases... Sons, 1986 6 K Wark and D E Richards Thermodynamics 6th ed New York: McGraw-Hill, 1999 cen84959_ch16.qxd 5/11/05 10:08 AM Page 817 Chapter 16 | 817 PROBLEMS* KP and the Equilibrium Composition of Ideal Gases 16 1C Why is the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy? 16 2C Is a wooden table in chemical equilibrium with the air? 16 3C Write three different... and 70 percent oxygen in the liquid phase and 66 percent nitrogen and 34 percent oxygen in the vapor phase Notice that y f,N2 ϩ y f,O2 ϭ 0.30 ϩ 0.70 ϭ 1 (16 21a) y g,N2 ϩ y g,O2 ϭ 0.66 ϩ 0.34 ϭ 1 (16 21b) Therefore, once the temperature and pressure (two independent variables) of a two-component, two -phase mixture are specified, the equilibrium composition of each phase can be determined from the phase. .. criterion for phase equilibrium 16 63 Show that a saturated liquid–vapor mixture of refrigerant-134a at Ϫ10°C satisfies the criterion for phase equilibrium 16 64 Consider a mixture of oxygen and nitrogen in the gas phase How many independent properties are needed to fix the state of the system? Answer: 3 16 65 In absorption refrigeration systems, a two -phase equilibrium mixture of liquid ammonia (NH3) and water... interface 16 73 A wall made of natural rubber separates O2 and N2 gases at 25°C and 500 kPa Determine the molar concentrations of O2 and N2 in the wall 16 74 Consider a glass of water in a room at 27°C and 97 kPa If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium, determine (a) the mole fraction of the water vapor in the air and (b) the... Will the equilibrium constant KP change? (b) Will the number of moles of CO2, CO, and O2 change? How? 16 7C Consider a mixture of NO, O2, and N2 in equilibrium at a specified temperature and pressure Now the pressure is tripled (a) Will the equilibrium constant KP change? (b) Will the number of moles of NO, O2, and N2 change? How? 16 8C A reaction chamber contains a mixture of CO2, CO, and O2 in equilibrium. .. and N2 that exit at 1200 K and 2 atm, determine CO Combustion 1200 K CO2 H2O chamber 2 atm O2 N2 FIGURE P16–32 16 33 Reconsider Prob 16 32 Using EES (or other) software, investigate if it is realistic to disregard the presence of NO in the product gases? 16 34E A steady-flow combustion chamber is supplied with CO gas at 560 R and 16 psia at a rate of 12.5 ft3/min and with oxygen (O2) at 537 R and 16. .. contents are ignited Determine the final temperature and pressure in the tank when the combustion gases are H2O, H2, and O2 cen84959_ch16.qxd 5/11/05 10:08 AM Page 819 Chapter 16 Simultaneous Reactions | 819 16 47 16 38C What is the equilibrium criterion for systems that involve two or more simultaneous chemical reactions? 16 39C When determining the equilibrium composition of a mixture involving simultaneous... temperatures 807 cen84959_ch16.qxd 5/11/05 10:08 AM Page 808 808 | Thermodynamics 16 6 ■ PHASE EQUILIBRIUM We showed at the beginning of this chapter that the equilibrium state of a system at a specified temperature and pressure is the state of the minimum Gibbs function, and the equilibrium criterion for a reacting or nonreacting system was expressed as (Eq 16 4) 1dG2 T,P ϭ 0 FIGURE 16 17 Wet clothes hung