# Why does reducing pressure cause the equilibrium to shift towards the side with less moles?

Take the example:

$$\ce{N2(g) + 3H2(g) <-> 2NH3(g)}$$

I understand that if I increase the pressure of the system, it'll shift towards the $\ce{NH3}$ side. This is because of the reaction rate increasing (increased collisions) more for the forward reaction than the reverse reaction. What I don't understand is why a decrease in pressure of the system would shift the equilibrium to $\ce{N2(g) + 3H2(g)}$. There is overall less collisions, but there's still more moles of reactants to collide and less molecules for $\ce{NH3}$ to collide and decompose.

• When you decrease the pressure, the reaction will try to keep the pressure up i.e try to increase the pressure. This is only possible if the reverse reaction takes place which is increasing the number of moles so that more collisions can occur thereby increasing pressure. – dr.drizzy Feb 26 '18 at 12:55