# If reaction A-> B is exothermic, how does the activation energy for the forward reaction compare with the activation energy for the reverse reaction, B -> A?

May 17, 2015

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

$\text{A" -> "B" + "heat}$

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 cse for $\text{A}$, to a lower energy level, as is the case for $\text{B}$.

The difference in energy between $\text{A}$ and $\text{B}$ will the enthalpy change of the reaction, $\Delta H$. In the case of the reverse reaction, you're going to have

$\text{heat" + "B" -> "A}$

You're going to have to supply energy to the more stable $\text{B}$ in order to get it to reform the *less stable", i.e. higher in energy, $\text{A}$.

The difference between the activation energy of the forward reaction and the activation energy of the reverse reaction will be $\Delta H$.

You need to supply to $\text{B}$ the exact amount of energy needed to g et it to go back to $\text{A}$. 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. ${E}_{\text{a forward}}$.

This means that you have

${E}_{\text{a reverse" = DeltaH + E_"a forward}}$

For exothermic reactions, you'll always have a bigger ${E}_{a}$ for the reverse reaction than you had for the forward reaction.