# Heat of reaction for a reversible reaction

If we have a reversible reaction

$$\ce{A <=> B}$$

and we have the activation energies of the forward and reverse reaction can we use

$$E_a^\mathrm{rev} = E_a^\mathrm{fwd} - \Delta H$$

to find the heat of reaction? $E_a^\mathrm{rev}$ and $E_a^\mathrm{fwd}$ are the activation energies of the reverse and forward reactions, respectively; $\Delta H$ is the heat of reaction.

I have tried using this but am unsure as to the layout of the equation.

• I don't see a question here. Maybe provide a specific example? – Zhe Aug 10 '17 at 14:20
• Suppose that the forward reaction is exothermic so that the product is $\Delta H$ below the reactant then $E_a^{fwd} +\Delta H= E_a^{rev}$. Draw a sketch. – porphyrin Aug 10 '17 at 15:10

• No your equation is not correct; $\Delta H + E^{fwd} =E^{rev}$: you should not assume that $\Delta H$ is negative. – porphyrin Aug 11 '17 at 6:46
• still no; in your diagram $\Delta H$ has the same sign as the $E_a$. – porphyrin Aug 12 '17 at 20:38