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!

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Answer 1 · 2018-01-05T05:06:37

$E_text(a) = "40 kJ·mol"^"-1"$

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

ProgReaction
(Adapted from BBC)

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

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

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

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