Kerala KEAM 7 June 2024 Solved Paper

Concept:According to Le Chatelier's principle, changing the volume (and thus pressure) at constant temperature shifts equilibrium only if there is a change in the number of moles of gaseous species ($\Delta n_g \neq 0$).Explanation:For the equilibrium to be unaffected by volume change, the reaction must have $\Delta n_g = 0$, i.e., equal moles of gaseous reactants and products.Calculating $\Delta n_g$ for each option (only gaseous species matter):• Option A: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ → $\Delta n_g = 2 - (1+3) = -2$• Option B: $PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g)$ → $\Delta n_g = 1 - (1+1) = -1$• Option C: $CO(g) + 3H_2(g) \rightleftharpoons CH_4(g)$ → $\Delta n_g = 1 - (1+3) = -3$• Option D: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ → $\Delta n_g = 2 - (2+1) = -1$• Option E: $NO_2(g) + SO_2(g) \rightleftharpoons SO_3(g) + NO(g) + H_2O(g)$ — if $H_2O$ is considered liquid (common in such reactions), then only $SO_3$ and $NO$ are gaseous products, giving $\Delta n_g = (1+1) - (1+1) = 0$Thus, only option E satisfies $\Delta n_g = 0$, so changing volume does not shift the equilibrium.Answer:E