Changing the volume of the container with a reaction mixture is essentially only possible for a reaction involving gases.
First, let's look at Le Chatelier's principle. If we mechanically decrease the volume of a container of gases the pressure inside the container will increase. Le Chatelier's principle tells us the reaction will re-achieve equilibrium by shifting to counteract this change. Since the change we made was to increase the pressures the reaction will shift in such a way to decrease the pressures. This can be achieved by moving towards the side of the reaction with fewer gas molecules.
Two key ideas to note. First, this is only an effect for reactions that involve gases. Second, the volume only affects reactions that have different numbers of moles of gas in the products and reactants. Why is this? It is again Q vs K. It is easier to see if we look at an example reaction. Let's take
\[{\rm 2NO_2(g) \rightleftharpoons N_2O_4(g)}\]
For this reaction
\[Q = {P_{\rm N_2O_4} \over P_{\rm NO_2}^2}\]
So, if you increase the pressure by decreasing the volume, the partial pressures will increase. Since the reactants have two moles of gas, the pressures of the reactants are squared. This means that the effect will be larger for the reactants. Dividing by a bigger number will make Q smaller and you'll find that after increasing the pressures Q < K. Then equilibrium is towards the products side. This is the side with fewer molecules.
Similarly if we mechanically increase the volume of a container, the pressure of the gas inside the container will decrease. Now the system will re-establish equilibrium by shifting to side of the reaction with the greatest number of moles of gas.
If the reaction has the same number of moles of gas in the reactants and products (\(K_{\rm p} = K_{\rm c}\)) then volume (pressure) will have no effect.
Finally, sometimes the volume is increased by adding an inert gas to the mixture at constant pressure. In this case the total pressure will stay constant, but the partial pressures of the reactants and products will decrease. This is because the total pressure now includes the pressure resulting from the inert gas. This has the effect of diluting the mixture and is identical to simply mechanically increasing the volume.
Increasing the pressure by adding an inert gas at constant volume has no effect. This is because what matters for the equilibrium are the partial pressures of the reactants and products. Adding an inert component to a system at constant volume will change the total pressure but not the partial pressures of the compounds of interest.
For reactions in the solution, you can have identical effects by changing the total volume of the solution. Adding water will effectively dilute all the concentrations. The reaction will then shift to counter this by moving toward the side of the reaction with more aqueous species in solution. Conversely concentration the solution lowers the volume and raises the concentrations. This will cause the reaction to shift to the side with fewer aqueous species. It follows the identical reasoning to the Q vs K examples given above except that the mass action quotient has concentrations instead of partial pressures.
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