0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 ($\ce{4HCl + O2}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.