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?

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Chemistry

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9 July, 04:49

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?

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Chaz Greene · Answer 1

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.

Kinley Edwards · Answer 2

Activation energy (Ea) is the minimum energy required by the reactants so that they can overcome the energy barrier to form corresponding products.

The role of a catalyst is to lower this energy barrier and speed up the rate of the reaction. For a reaction at equilibrium, a catalyst tends to speed up both the forward and reverse reaction in the same ratio.

In this case, the Ea for the forward reaction decreases from 28.0 kcal/mol to 23.0 kcal/mol. The change in Ea for the forward reaction is: ΔEa = 5 kcal/mol. In a similar manner the Ea for the reverse reaction will also be lowered by 5 kcal/mol.