The per cent dissociation of H₂S(g) if 0.1 mole of H₂S is kept in 0.4 L vessel at 1000 K for the rea — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Initial concentration of H₂S = 0.1 / 0.4 = 0.25 M. Let the dissociated concentration of H₂S be 2x. At equilibrium: [H₂S] = 0.25 - 2x, [H₂] = 2x, [S₂] = x. K_c = ([H₂]²[S₂]) / [H₂S]² = (2x)²(x) / (0.25 - 2x)². Since K_c is very small, 0.25 - 2x ≈ 0.25. 4x³ / (0.25)² = 1.0 × 10⁻⁶ => 4x³ / 0.0625 = 10⁻⁶ => 4x³ = 6.25 × 10⁻⁸ => x³ = 15.625 × 10⁻⁹ => x = 2.5 × 10⁻³ M. Dissociated concentration = 2x = 5.0 × 10⁻³ M. Per cent dissociation = (5.0 × 10⁻³ / 0.25) × 100 = 2%. Therefore, correct answer is 2.