Thermodynamics

800 | ThermodynamicsT, K100020003000400050006000H 2 → 2HP = 1 atmK P5.17 10 –182.65 10 –60.0252.54541.47267.7% mol H0.000.1614.6376.8097.7099.63FIGURE 16–9The larger the K P , the more completethe reaction.(a)(b)InitialcompositionEquilibriumcomposition at3000 K, 1 atm0.921 mol H 21 mol H 20.158 mol HK P = 0.02510.380 mol H 21 mol H 2 1.240 mol H1 mol N 21 mol N 2K P = 0.0251FIGURE 16–10The presence of inert gases does notaffect the equilibrium constant, but itdoes affect the equilibriumcomposition.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 sameK P value:H 2 1 2 O 2 ∆ H 2 Oat 1 atmH 2 1 2 O 2 ∆ H 2 Oat 5 atmH 2 1 2 O 2 3N 2 ∆ H 2 O 3N 2 at 3 atmH 2 2O 2 5N 2 ∆ H 2 O 1.5O 2 5N 2 at 2 atm2. 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, andthus the terms in the numerator move to the denominator and vice versa.Consequently, the equilibrium constant of the reverse reaction becomes1/K P . For example, from Table A–28,K P 0.1147 10 11 forH 2 1 2 O 2 ∆ H 2 Oat 1000 KK P 8.718 10 11 forH 2 O ∆ H 2 1 2 O 2 at 1000 K3. The larger the K P , the more complete the reaction. This is also apparentfrom Fig. 16–9 and Eq. 16–13. If the equilibrium composition consistslargely 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 andP B ), which results in a large value of K P . In the limiting case of a completereaction (no leftover reactants in the equilibrium mixture), K P approachesinfinity. Conversely, very small values of K P indicate that a reaction doesnot proceed to any appreciable degree. Thus reactions with very small K Pvalues at a specified temperature can be neglected.A reaction with K P 1000 (or ln K P 7) is usually assumed to proceedto completion, and a reaction with K P 0.001 (or ln K P 7) is assumednot to occur at all. For example, ln K P 6.8 for the reaction N 2 ∆ 2Nat 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 itdoes not affect the equilibrium constant K P ). This can be seen fromEq. 16–15, which involves the term P ∆n , where ∆n n P n R (the differencebetween the number of moles of products and the number of molesof reactants in the stoichiometric reaction). At a specified temperature, theK P value of the reaction, and thus the right-hand side of Eq. 16–15, remainsconstant. Therefore, the mole numbers of the reactants and the productsmust change to counteract any changes in the pressure term. The directionof the change depends on the sign of ∆n. An increase in pressure at a specifiedtemperature increases the number of moles of the reactants anddecreases the number of moles of products if ∆n is positive, have the oppositeeffect if ∆n is negative, and have no effect if ∆n is zero.5. The presence of inert gases affects the equilibrium composition (althoughit 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 numberof moles of the ideal-gas mixture at equilibrium, including inert gases. Thesign of ∆n determines how the presence of inert gases influences the equilibriumcomposition (Fig. 16–10). An increase in the number of moles ofinert gases at a specified temperature and pressure decreases the number of