# le Chatelier's Principle

I am really confused about how reducing partial pressure of a gas by a certain percentage would impact the behavior of an equilibrium system.

e.g $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ is at equilibrium. How would this system behave if the partial pressure of $\ce{Cl2}$ is reduced by 75%?

I understand the partial pressure is related to moles but having a hard time applying it here. My guess is that the reactant would be consumed to produce more of the $\ce{Cl2}$ since its pressure and thus moles are decreasing. However, my instructor said the system would react by consuming $\ce{POCl}$, the reason for which was not explained too well.

If I could get an explanation, it would be much appreciated.

• The partial pressure can be converted to moles and then you could use Kc. Or you could simply use Kp. However, I would think that you are correct: given no other changes, if products decrease, the system would act to create more products, until equilibrium – Andy Apr 14 '15 at 22:19
• There are several good questions that already address this. A change in partial pressure of a component means a change in concentration. This perturbs the system away from equilibrium which then adjusts to reach a new equilibrium. I agree with @Andy on the result: some POCl3 will dissociate to produce more products. chemistry.stackexchange.com/questions/4130, chemistry.stackexchange.com/questions/4262, chemistry.stackexchange.com/questions/9118 – Byron Wall Apr 16 '15 at 19:24
• I think what your instructor meant was that the reaction would involve the consumption of $\ce{POCl3}$ and that in turn leads to the formation of $\ce{POCl}$ & $\ce{Cl2}$. Thus resetting equilibrium. – Asker123 Jun 4 '15 at 14:15

Hence when the pressure of $\ce{Cl2}$ is reduced by 75% in the $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ system, the system naturally acts to counteract the change: by increasing the pressure of $\ce{Cl2}$.
Using the ideal gas law, $pv = nRT$ so pressure is directly proportionate to the number of moles of $\ce{Cl2}$ present. The position of equilibrium therefore shifts to the right, favouring the forward reaction that produces more $\ce{Cl2}$ (though not as much as to increase partial pressure of $\ce{Cl2}$ to its former value). This would mean it also consumes more $\ce{POCl3}$.
As for more $\ce{POCl}$ being consumed, if all other factors remain constant I'm not sure why this would happen.