# A reaction is endothermic with H=100 kJ/mol. If the activation enthalpy of the forward reaction is 140 kJ/mol, what is the activation enthalpy of the reverse reaction? I know the answer is 40 kJ/mol but I’m not quite sure how to get there, thanks!

Jan 5, 2018

${E}_{\textrm{a}} = \text{40 kJ·mol"^"-1}$

#### Explanation:

Here's a schematic progress of reaction diagram for an endothermic reaction.

We see that Δ_text(‡)H("fwd") (from reactants to transition state) is $\text{140 kJ·mol"^"-1}$.

We also see that Δ_text(rxn)H (from reactants to products is $\text{100 kJ·mol"^"-1}$.

Thus, Δ_text(‡)H("rev") (products to transition state) must be

Δ_text(‡)H("rev") = "140 kJ·mol"^"-1" - "100 kJ·mol"^"-1" = "40 kJ·mol"^"-1"