How the concentration of the chlorine would be affected if the volume of the container is increased? $$\ce{4 HCl + O2 <=> 2 Cl2 + 2 H2O + $Q$}$$

My initial thought was that the total and partial pressures were decreased on increasing the volume of the container. But I did not account for the fact that the $\ce{HCl}$ did not play a role in this pressure.

So, according to my previous process, the reaction should proceed in the backward direction, concentration of $\ce{Cl2}$ should decrease. On the other hand, if we exclude the $\ce{HCl}$ factor, the reaction should proceed in forward direction increasing the concentration if $\ce{Cl2}$.

Please help me understand what will be the actual effect.


1 Answer 1


According Le Chatelier's Principle, when a change in concentration or pressure is applied to an equilibrium, the equilibrium changes to counteract the change.

In your equilibrium, the volume of the container is increased, causing a decrease in pressure. To counteract this, the side of the equilibrium with the highest number of moles is favoured. I.o.w. the equilibrium will move somewhat to the left ($\ce{4HCl + O2}$).


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