If 0.3 moles of hydrogen gas and 2.0 moles of sulphur solid are heated to 87°C in a 2.0 L vessel, what will be the partial pressure of H₂S gas at equilibrium? (Given: R = 0.08 l-atm/K-mol)
H₂(g) +

Question

If 0.3 moles of hydrogen gas and 2.0 moles of sulphur solid are heated to 87°C in a 2.0 L vessel, what will be the partial pressure of H₂S gas at equilibrium? (Given: R = 0.08 l-atm/K-mol)
H₂(g) + S(s) ⇌ H₂S(g); K_c = 0.08

Prashant Kotian · Accepted Answer

Correct answer: A. K_c = [H₂S] / [H₂] = 0.08. Let x moles of H₂S form. [H₂S] = x/2, [H₂] = (0.3 - x)/2. Thus x / (0.3 - x) = 0.08 => x = 0.024 - 0.08x => 1.08x = 0.024 => x = 0.0222 mol. Partial pressure of H₂S = (nRT)/V = (0.0222 × 0.08 × 360) / 2.0 = 0.32 atm. Therefore, correct answer is A.

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