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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.

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  • $\begingroup$ 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. $\endgroup$ – dr.drizzy Feb 26 '18 at 12:55
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Actually, the shift of reaction towards left on decreasing pressure (and towards right on increasing pressure) is due to Le Chatelier's Principle, which states that if a change is brought in the equilibrium conditions of a reaction, the reaction will proceed in such a manner that it counteracts the change.

In case of increasing pressure the reaction shifts to right due to lesser number of moles on right. And according to gas equation, lesser moles means lesser pressure. The opposite happens when the pressure is decreased.

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    $\begingroup$ So, it is essentially less collisions = less pressure and vice versa. Initially when the pressure is decreased, does this mean that the reverse reaction occurs faster than the forward reaction until equilibrium is reestablished? $\endgroup$ – CipherBot Feb 26 '18 at 13:19
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    $\begingroup$ @CipherBot Yes, definitely. It no longer remains in equilibrium, until a new one is reestablished. Also in new equilibrium, the concentration of the reactant would be greater than that in old one. $\endgroup$ – Kirti Agrawal Feb 26 '18 at 13:25

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