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.

  • $\begingroup$ I don't see a question here. Maybe provide a specific example? $\endgroup$ – Zhe Aug 10 '17 at 14:20
  • 2
    $\begingroup$ 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. $\endgroup$ – porphyrin Aug 10 '17 at 15:10

Energy diagram

As you can see from energy diagram of the reaction that 'Heat of Reaction' is

ΔH = Efwd- Erev.

So your expression is correct.

| improve this answer | |
  • $\begingroup$ No your equation is not correct; $\Delta H + E^{fwd} =E^{rev}$: you should not assume that $\Delta H $ is negative. $\endgroup$ – porphyrin Aug 11 '17 at 6:46
  • $\begingroup$ It is ...in that case Efwd > Erev and ΔH > 0 . $\endgroup$ – aks0854 Aug 12 '17 at 11:43
  • $\begingroup$ still no; in your diagram $\Delta H $ has the same sign as the $E_a$. $\endgroup$ – porphyrin Aug 12 '17 at 20:38
  • $\begingroup$ oh.. I just took the diagram from net..overall ΔH < 0 as Erev> Efwd. But expression will always hold as Efwd, Erev will always > 0 . their magnitude decides whether ΔH is positive or not. $\endgroup$ – aks0854 Aug 13 '17 at 21:45

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