Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation $\left(k=A e^{-E_{2} / R T}\right)$ b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

At equilibrium, we can express the equilibrium constant (K) as the ratio of the forward and reverse reaction rate constants, k_f and k_r: \(K = \frac{k_f}{k_r}\). We apply the Arrhenius equation to find k_f and k_r as \(k_f = A_{f}e^{-E_{f}/RT}\) and \(k_r = A_{r}e^{-E_{r}/RT}\), respectively. Expressing K as a ratio of k_f and k_r, we obtain \(K = e^{\frac{E_{r}-E_{f}}{RT}}\). Using the relationship between ΔG° and K, we derive the equilibrium constant K as a function of the forward and reverse reaction rate constants, proving the statement. For part b, the statement, "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction," is false as a catalyst lowers the activation energy for both the forward and reverse reactions, thereby increasing the rate of both reactions without altering the overall equilibrium constant K.