Answer in General Chemistry for Akash #156268

For the reaction, A → B, is an exothermic reaction. The value of activation energy is 50 kJ/mol. Draw an energy vs. reaction path diagram for the above reaction and calculate the activation energy for the reverse reaction B to A when the reaction enthalpy corresponds to 12 kJ.

In the case of an exothermic reaction, the activation energy of the forward reaction will always be smaller than the activation energy of the reverse reaction.

In your case, the forward reaction will look like this A →B + heat

Since heat is released during the reaction, the product will have a lower energy than the reactant. You go from a higher energy level, as was the case for A, to a lower energy level, as is the case for B.

The difference in energy between A and B will the enthalpy change of the reaction, ΔH

In the case of the reverse reaction, you’re going to have heat + B → A

You’re going to have to supply energy to the more stable B in order to get it to reform the “less stable”, i.e. higher in energy, A.

The difference between the activation energy of the forward reaction and the activation energy of the reverse reaction will be ΔH.

You need to supply to B the exact amount of energy needed to get it to go back to A. More specifically, you need to provide it with the energy released during the forward reaction plus the energy needed to get the forward reaction going, i.e. E_a forward.

This means that you have

E_areverse = ΔH + E_aforward = 50 kJ/mol + 12 kJ = 62 kJ/mol