I'm just learning chemistry for school and I've learnt that adding a catalyst reduces the activation energy for the forward and backward reaction by equal amounts, therefore the rate of forward and backward reaction increase by the same amounts so equilibrium stays the same but is reached more quickly.

I'm having some trouble understanding this. I know that the activation energy of the exothermic reaction is always smaller than that of the endothermic reaction. So thinking about the Maxwell Boltzmann distribution, a change in $E_a$ near to $0$ will have a greater effect on the rate of exothermic reaction than the endothermic reaction. Therefore, a catalyst should cause equilibrium to shift to the exothermic side?


1 Answer 1


Look at the form of the Arrhenius rate equation (this is fine since they mostly all have this form):

$$k = Ae^{-\frac{E_{a}}{RT}}$$

If you make the substitution, $E_{a}->E_{a}+\Delta E$, then you have:

$$k = Ae^{-\frac{E_{a} + \Delta E}{RT}} = Ae^{-\frac{E_{a}}{RT}}e^{-\frac{\Delta E}{RT}}$$

This means that you're just scaling the rate constant $k$ by another constant. This scaling applies to both the forward and reverse rate constants. If the concentrations were such that they were in equilibrium before, they're still in equilibrium now.

You'd have to prove this in general for complex reactions with complicated rate laws, but for simple reactions, this should suffice to help you see that the equilibrium is unaffected.

  • $\begingroup$ So the reaction rates don't increase by the same amounts, the increase by the same proportion? $\endgroup$
    – user43472
    Apr 4, 2017 at 20:14
  • $\begingroup$ Because rate dependence on energy is exponential, it is by proportion. In the end, the laws of chemistry work pretty well, so you can try to rationalize backwards from rate equivalence at equilibrium. $\endgroup$
    – Zhe
    Apr 4, 2017 at 20:16
  • $\begingroup$ Oh, I see what i'm missing. There have to be 2 distributions, one for the reactants and one for the products. This will allow for both areas to become the same $\endgroup$
    – user43472
    Apr 4, 2017 at 21:17
  • $\begingroup$ wait but then it should favour the side of the reaction with the greater concentration of reactants? $\endgroup$
    – user43472
    Apr 4, 2017 at 21:49
  • $\begingroup$ Again, chemistry works, so no. $\endgroup$
    – Zhe
    Apr 5, 2017 at 1:54

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