Question #44069

The gas-phase reaction \ce{A(g) + 2B(g) <=> C(g)} has an equilibrium constant $K_c = 2500 \text{ M}^{-2}$ at $600 \text{ K}$. The activation energy for the forward uncatalyzed reaction is $80 \text{ kJ mol}^{-1}$.

To improve the rate of product formation, a solid catalyst is introduced, which alters the reaction pathway to the following mechanism: Step 1: \ce{A(g) + B(g) <=> I(g)} (Fast equilibrium) Step 2: \ce{I(g) + B(g) <=> C(g)} (Slow, rate-determining step)

The catalyst lowers the activation energy of the forward reaction by $11.5 \text{ kJ mol}^{-1}$. Assuming ideal gas behavior and that the concentration of the intermediate \ce{I(g)} is negligible at macroscopic equilibrium, what is the value of the equilibrium constant $K_p$ for the catalyzed reaction at $600 \text{ K}$?

(Given: $R \approx \frac{1}{12} \text{ L bar K}^{-1} \text{ mol}^{-1}$ for pressure calculations, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ for energy calculations, and $e^{2.3} \approx 10$)