Equimolar mixture of two gases A₂ and B₂ is taken in a rigid vessel at constant temperature 300 K. T — Chemical Equilibrium Chemistry Question
Question
Equimolar mixture of two gases A₂ and B₂ is taken in a rigid vessel at constant temperature 300 K. The gases achieve equilibrium as: A₂(g) ⇌ 2A(g), K_p = x atm. B₂(g) ⇌ 2B(g), K_p = y atm. A₂(g) + B₂(g) ⇌ 2AB(g), K_p = 2. If the initial pressure in the container was 2 atm and the final pressure at equilibrium is 2.75 atm in which the partial pressure of AB(g) is 0.5 atm, the value of y:x is
💡 Solution & Explanation
Initial total pressure = 2 atm. Since equimolar, P_A₂_init = 1 atm, P_B₂_init = 1 atm. At equilibrium, P_AB = 0.5 atm. For rxn 3, K_p = P_AB² / (P_A₂ × P_B₂) = 2. Thus, 0.5² / (P_A₂ P_B₂) = 2 => P_A₂ P_B₂ = 0.125. Mass balance on A atoms: 2P_A₂ + P_A + P_AB = 2(1) => 2P_A₂ + P_A + 0.5 = 2 => P_A = 1.5 - 2P_A₂. Mass balance on B atoms: 2P_B₂ + P_B + P_AB = 2(1) => 2P_B₂ + P_B + 0.5 = 2 => P_B = 1.5 - 2P_B₂. Total eq pressure P = P_A₂ + P_B₂ + P_A + P_B + P_AB = 2.75. P_A₂ + P_B₂ + (1.5 - 2P_A₂) + (1.5 - 2P_B₂) + 0.5 = 2.75. 3.5 - (P_A₂ + P_B₂) = 2.75 => P_A₂ + P_B₂ = 0.75. We have P_A₂ + P_B₂ = 0.75 and P_A₂ P_B₂ = 0.125. Solving gives roots 0.5 and 0.25. Let P_B₂ = 0.25 and P_A₂ = 0.5 (to make y > x for an integer ratio). P_A = 1.5 - 2(0.5) = 0.5. x = P_A² / P_A₂ = 0.5² / 0.5 = 0.5. P_B = 1.5 - 2(0.25) = 1.0. y = P_B² / P_B₂ = 1.0² / 0.25 = 4. Ratio y:x = 4 / 0.5 = 8. Therefore, correct answer is 0008.