Chapter 15, Problem 15.122QP

For reactions earned out under standard-state conditions, Equation 14.10 takes the form ΔG° = ΔH° − TΔS°. (a) Assuming ΔH° and ΔS° are independent of temperature, derive the equation

$\ln \frac{K_2}{k_1}=\frac{\Delta H^{\circ }}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

where K₁ and K₂ are the equilibrium constants at T₁ and T₂ respectively. (b) Given that at 25°C K_c is 4.63 x 10⁻³ for the reaction

$\begin{array}{cc}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)\rightleftarrows {\text{2NO}}_{\text{2}}\left(g\right)& \Delta H^{\circ }\end{array}=\text{58}\text{.0kJ/mol}$

calculate the equilibrium constant at 65°C.