## General Chemistry: Principles and Modern Applications (10th Edition)

(a) $\Delta n = 0$, so the volume does not affect the equilibrium constant. (b) $K_c = K_p = 0.659$
(a) There are 2 moles of gaseous reactants and 2 moles of gaseous products. $\Delta n = 0$, therefore, the change in volume will not affect the equilibrium. Mathematically: $$K_c = \frac{[CO][H_2O]}{[CO_2][H_2]}$$ Since molarity is number of moles divided by volume: $$K_c = \frac{\frac{n_{CO}}{V}{}\frac{n_{H_2O}}{V}}{\frac{n_{CO_2}}{V}\frac{n_{H_2}}{V}}$$ Simplifying: $$K_c = \frac{{n_{CO}}{}{}{n_{H_2O}}{}}{{n_{CO_2}}{}{n_{H_2}}{}}$$ The Volume do not appear on the equilibrium constant expression. (b) Substitute: $$K_c = \frac{(0.224)(0.224)}{(0.276)(0.276)} = 0.659$$ $$K_p = K_c(RT)^{\Delta n} = K_c(RT)^0 = K_c(1) = K_c = 0.659$$