Q.A catalyst lowers the activation energy of a reaction from $100\ \text{kJ mol}^{-1}$ to $80\ \text{kJ mol}^{-1}$ at $300\ \text{K}$, without changing the pre-exponential factor $A$. Calculate the ratio of the catalyzed to the uncatalyzed rate constant.

[!TLDR] The ratio $k_{cat}/k_{uncat} = e^{\Delta E_a/RT}$, where $\Delta E_a$ is the drop in activation energy. [!ANSWER] $k_{cat}/k_{uncat} \approx 3.0\t …

$\Delta E_a = 100 - 80 = 20\ \text{kJ mol}^{-1} = 20000\ \text{J mol}^{-1}$. Since $A$ is unchanged, $\dfrac{k_{cat}}{k_{uncat}} = \dfrac{Ae^{-E_{a,cat}/RT}}{Ae^{-E_{a,uncat}/RT}} = e^{(E_{a,uncat}-E_{a,cat})/RT} = e^{\Delta E_a/RT}$. $\dfrac{\Delta E_a}{RT} = \dfrac{20000}{8.314\times 300} = \dfrac{20000}{2494.2} = 8.019$. $e^{8.019} = 10^{8.019/2.303} = 10^{3.482} \approx 3.04\times 10^{3}$. [!ANSWER] The ratio of catalyzed to un …

Take the ratio of the two Arrhenius expressions (the unchanged $A$ cancels), leaving $e^{\Delta E_a/RT}$ with $\Delta E_a$ the D …

A sign error is common here -- since the catalyzed path has the LOWER $E_a$, the ratio $k_{cat}/k_{uncat}$ must come out greater t …