WAEC Chemistry 2019 Paper

Concept:This question is based on Le Chatelier's principle. A decrease in pressure favours the side with more moles of gas.Explanation:Consider the equilibrium: $$2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)$$ Count the total gaseous moles on each side. Reactant side: $2 + 1 = 3$ moles of gas. Product side: $2$ moles of gas. When the pressure is decreased, the system shifts to the side having a larger number of gas molecules. This is because a lower pressure favours the direction that increases volume. Therefore, the equilibrium shifts towards the reactants: $$2\mathrm{SO}_2 + \mathrm{O}_2 \leftarrow 2\mathrm{SO}_3$$ This means the equilibrium position shifts to the left.Answer:C. shift the equilibrium position to the left.