Question

I understand that when you increase pressure for a reaction at equilibrium it shifts to the side that makes less moles of product and when you decrease pressure, it shifts to the side that makes more moles. Why does the reaction shift in these directions?

Answer 1

Try thinking of it in terms of the equilibrium constant definition. For the general reaction:

`nu_A A(g) + nu_B B(g) rightleftharpoons nu_C C(g) + nu_D D(g)`

`K_P = Pi_i P_i^(nu_i)`

`= (P_C^(nu_C)P_D^(nu_D))/(P_A^(nu_A)P_B^(nu_B))`

where `P_i` is the partial pressure of gas `i`, and `nu_i` is the stoichiometric coefficient of gas `i`.

Now, let's say it was a specific reaction, like this simple one:

`2NO_2(g) rightleftharpoons N_2O_4(g)`

For this:

`K_P = P_(N_2O_4)/(P_(NO_2)^2)`

If you increase the total pressure `P`, note that the partial pressure is defined as `chi_iP = P_i`, where `chi_i = (n_i)/(n_1 + n_2 + . . . + n_N)` is the `"mol"` fraction of gas `i`.

Thus, if you increase the total pressure, you increase the partial pressure of everything in the equilibrium mixture. However, the influence on the equilibrium constant is weighted by the exponent on the partial pressure.

When `P` increases, `P_(NO_2)` and `P_(N_2O_4)` increase. However, since `nu_(NO_2) > nu_(N_2O_4)` (the exponent is larger), the result is that `K_P` decreases, because the denominator increases by more than the numerator.

To undo the shift in equilibrium, Le Chatelier's principle states that we undo what just happened.

We just increased the pressure, so the easiest way to decrease the pressure (i.e. undo the increase) is to react reactants and shift towards the products.

Thus, the reaction shifts right when the mols of gas are larger on the reactants side and the pressure is increased.

In general, it means the reaction shifts towards the side with fewer mols of gas when the total pressure is increased.

You can follow the same logic for a decrease in pressure. If it helps, plug in actual numbers.