The energies of activation for the forward and reverse directions of the reversible reaction are and — Chemical Kinetics Chemistry Question
Question
The energies of activation for the forward and reverse directions of the reversible reaction $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \text{ kJ mol}^{-1}$ and $200 \text{ kJ mol}^{-1}$ respectively. The introduction of a positive catalyst lowers the activation energy of both the forward and reverse reactions by exactly $100 \text{ kJ mol}^{-1}$. What will be the precise overall enthalpy change ($\Delta H$) of the reaction in the presence of this catalyst?
💡 Solution & Explanation
Enthalpy change ($\Delta H$) is a thermodynamic state function independent of the path taken, defined kinetically as $\Delta H = E_{a(forward)} - E_{a(backward)}$. Initially, $\Delta H = 180 - 200 = -20 \text{ kJ mol}^{-1}$. Since a catalyst lowers both activation barriers by the exact same magnitude ($100 \text{ kJ mol}^{-1}$), the difference between them remains absolutely constant. Thus, the enthalpy change remains $-20 \text{ kJ mol}^{-1}$.