Chapter 18, Problem 66AP

For reactions carried out under standard-state conditions, Equation 18.10 takes the form $\Delta Gº=\Delta Hº-T\Delta Sº$.(a) Assuming $\Delta Hº\text{ and }\Delta S\text{º}$ are independent of temperature, derive the equation:

$\text{ln}\frac{K_2}{K_1}=\frac{\Delta Hº}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Where $K_1\text{ and }{K}_{\text{2}}$ are the equilibrium constants at $T_1\text{ and }{T}_{\text{2}}\text{,}$ respectively. (b) Given that at $\text{25ºC }{K}_{\text{c}}\text{ is 4}\text{.63 }\times {\text{ 10}}^{\text{-3}}$ for the reaction:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)\rightleftarrows 2{\text{NO}}_{\text{2}}\left(g\right)\Delta H\text{º=58}\text{.0kJ/mol}$

Calculate the equilibrium constant at $\text{65ºC}$.