A vessel of 2.5 L was filled with 0.01 mole of Sb₂S₃ and 0.01 mole of H₂ to attain equilibrium at 44 — Chemical Equilibrium Chemistry Question
Question
A vessel of 2.5 L was filled with 0.01 mole of Sb₂S₃ and 0.01 mole of H₂ to attain equilibrium at 440°C as: Sb₂S₃(s) + 3H₂(g) ⇌ 2Sb(s) + 3H₂S(g). After equilibrium, the H₂S formed was analysed by dissolving it in water and treating with excess of Pb²⁺ to give 1.19 g of PbS as precipitate. The value of K_c of the reaction at 440°C is (Pb = 206)
💡 Solution & Explanation
Molar mass of PbS = 206 (Pb) + 32 (S) = 238 g/mol. Moles of PbS = 1.19 / 238 = 0.005 mol. This implies 0.005 mol of H₂S was present at equilibrium. From the reaction stoichiometry, change in H₂ is also 0.005 mol. Equilibrium moles of H₂ = 0.01 (initial) - 0.005 (reacted) = 0.005 mol. Since Δn_g = 3 - 3 = 0, volume cancels out. K_c = [H₂S]³ / [H₂]³ = (0.005)³ / (0.005)³ = 1. Therefore, correct answer is 1.