Consider the exothermic reaction

A(g) + B(g) ⇌ C(g)          ΔH < 0

The following table illustrates the changes expected.

Change	Rate of Reaction	Position of Equilibrium	Change to K
Increasing P by decreasing V	Overall rate increases	Shifts to the right	None
Reducing P by increasing V	Overall rate decreases	Shifts to the left	None

Again, the overall rate increases if we decrease V. This is because there is a higher probability of favourable interactions between molecules, increasing both the forward and backward rate of reaction. As with the case for temperature, the rate of forward reaction increases more than the increase in rate of backward reaction, as the forward reaction (to form lesser moles of gas) is more favourable.

Since temperature is not a factor here, it does not matter if the reaction is exothermic or endothermic.

What happens, however, if we keep one quantity constant, and vary the other? The answer is that there will be no change to the equilibrium position. Why?

The answer has to do with addition of inert substances, or removal of reactants. It is physically impossible to have one of pressure and volume constant while varying the other, without either addition or removal. Similar to when we discussed Hess’s Law, substances are either added in equal ratio to both sides, or removed from both sides, of the equation. This ensures that the position of equilibrium does not shift.