For the reaction A_2 + B_2 rightleftharpoons | vedclass

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(A) For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of th

Vedclass · Accepted Answer

The catalyst does not alter the Gibbs energy change of the reaction.

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(A) For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_f)$ and the backward reaction $(E_b)$: $\Delta H = E_f - E_b = 180 \ kJ \ mol^{-1} - 200 \ kJ \ mol^{-1} = -20 \ kJ \ mol^{-1}$. When a catalyst is added,it lowers the activation energy for both the forward and backward reactions by the same amount $(100 \ kJ \ mol^{-1})$: $E_{f(cat)} = 180 - 100 = 80 \ kJ \ mol^{-1}$ $E_{b(cat)} = 200 - 100 = 100 \ kJ \ mol^{-1}$ The new enthalpy change is $\Delta H_{cat} = 80 - 100 = -20 \ kJ \ mol^{-1}$. Since $\Delta H$ and $\Delta G$ are state functions,they depend only on the initial and final states of the reactants and products,not on the path taken. Therefore,a catalyst does not change the $\Delta H$ or $\Delta G$ of a reaction.

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