# Why does a catalyst increase the forward and reverse reactions equally

With reference to the picture below, how is it that when there is a lower activation energy (due to the alternate reaction pathway provided by the catalyst) that the catalyst will increase the forward and reverse reactions exactly the same amount.

It only makes intuitive sense that when the $$Ea$$ is lowered, the forward reaction increases more than the reverse reaction because the decrease was an absolute amount (that is, the same decrease for both forward and reverse) and thus proportionately, the forward $$Ea$$ has been reduced far more than the reverse $$Ea$$. So why should the reverse reaction instantly be raised such that it is still equal to the forward reaction (if from equilibrium)?

• Because the amount of particles in state "reactant" or "product" only depends on the energy difference, period. In QM, you have the Boltzmann distribution for that, in chemistry it's the same thing, just a bit more complicated because you usually have more than two species involved.
– Karl
Oct 27 '18 at 7:01
• Perhaps what you overlooked: There is a third thing in your system, the thermal energy reservoir $kT$. The relative energy difference from it to the product and educt state changes matters. The transition state only make a quick loan from the reservoir.
– Karl
Oct 27 '18 at 7:08
• Oct 27 '18 at 10:15
• Because of microscopic reversibility principle. Oct 27 '18 at 17:47
• chemistry.stackexchange.com/questions/58846/… Oct 27 '18 at 17:54

If the energy in your figure from reactants to transition state is $$E_a$$ and the difference in products to reactants $$G$$ and we lower the energy $$E_a$$ by $$E_0$$ then using Arrhenius type rate expressions the equilibrium constant is $$K_e=k_{forward}/k_{back}$$
$$\displaystyle K_e=\frac{\exp(-E_a/RT)}{\exp(-(E_a+G)/RT)}$$, and for the catalyst $$\displaystyle K_e^C= \frac{\exp(-(E_a-E_0)/RT)}{\exp(-(E_a-E_0+G)/RT)}$$ which is the same as $$K_e$$if you expand the terms in the exponentials.