For reactions carried out under standard-state conditions, Equation (17.10)
takes the form \(\Delta G^{\circ}=\Delta H^{\circ}\) \(-T \Delta S^{\circ}\). (a)
Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) 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 \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and
\(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C},
K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction
$$
\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad
\Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol}
$$
calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\).