Le Chatelier's Principle - Pressure Effect

Question

For the reaction CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g), what will happen to the equilibrium position when the total pressure is increased at constant temperature by reducing the volume?

RankGuru · Accepted Answer

Le Chatelier's principle for pressure changes states that when pressure is increased by decreasing volume, the equilibrium shifts toward the side with fewer moles of gas. For CO(g) + 3H_2(g) ightleftharpoons CH_4(g) + H_2O(g): moles of gaseous reactants = 1 + 3 = 4; moles of gaseous products = 1 + 1 = 2. Since the product side has fewer moles (2 vs 4), increasing pressure shifts equilibrium to the right, favouring CH_4 and H_2O formation. Option B is wrong because the reverse direction would have 4 moles of gas, which would make the pressure problem worse, not better; Le Chatelier's principle opposes the disturbance, not amplifies it. Option C is partially true — K_c is indeed independent of pressure — but the position of equilibrium (not K_c) does shift with pressure when \Delta n_g 
eq 0. Option D is incorrect; temperature and pressure effects are independent, and pressure alone can shift equilibrium position when \Delta n_g 
eq 0. This concept is fundamental in NCERT and tested in JEE Main for industrial processes like methanation. Plausibility check: \Delta n_g = 2 - 4 = -2 < 0, confirming that increased pressure favours the forward (lower mole) direction.

Le Chatelier's Principle - Pressure Effect

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