Final answer:
Reducing the volume of a reaction mixture at equilibrium results in increased pressure and causes the equilibrium to shift towards the side with fewer moles of gas, following Le Chatelier's principle.
Step-by-step explanation:
When considering the effect of reducing the volume of a reaction mixture at equilibrium, it's important to look at Le Chatelier's principle and the stoichiometry of the reaction. If we have a reaction where decreasing the volume leads to an increase in pressure, and there are unequal numbers of moles of gases on each side of the equation, the equilibrium will shift to compensate. In this specific case, where the number of moles of products is different from the number of moles of reactants, reducing the volume would increase the pressure, causing the equilibrium to shift towards the side with fewer moles of gas to restore equilibrium.
For example, if a reaction has a greater number of moles on the reactant side compared to the product side, reducing the volume (and thus increasing pressure) would shift the equilibrium toward the product side with fewer moles of gas. This is to minimize the disturbance caused by the volume change, in line with Le Chatelier's principle. Conversely, if the product side had more moles of gas, the equilibrium would shift towards the reactants when the volume is reduced.