Do catalysts increase the rate of exothermic reaction more in a reversible reaction? 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?
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, $E_{a}-&gt;E_{a}+\Delta E$, 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.