If a catalyst changes the activation energy of a forward reaction from 28.0 kcal/mol to 23.0 kcal/mol, what effect does it have on the reverse reaction?

Explanation:

For an equilibrium reaction, the rate of forward reaction is equal to the rate of backward reaction.

As in the given reaction, change in activation energy is as follows.

                 (28.0 - 23.0) kcal/mol

                 = 5 kcal/mol

Therefore, when activation energy is increasing 5 kcal/mol for the given reaction in the forward direction then for the backward reaction there will be be a decrease of 5 kcal/mol.

Thus, we can conclude that if a catalyst changes the activation energy of a forward reaction from 28.0 kcal/mol to 23.0 kcal/mol, then it reduces the energy of activation of the reverse reaction by 5.0 kcal/mol.

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