Following equilibria are established on mixing two gases A₂ and C: 3A₂(g) ⇌ A₆(g); K_p = 1.6 atm⁻² and A₂(g) + C(g) ⇌ A₂C(g); K_p = 'x' atm⁻¹. When A₂(g) and C(g) are mixed in 2:1 molar ratio, the tot

Question

Following equilibria are established on mixing two gases A₂ and C: 3A₂(g) ⇌ A₆(g); K_p = 1.6 atm⁻² and A₂(g) + C(g) ⇌ A₂C(g); K_p = 'x' atm⁻¹. When A₂(g) and C(g) are mixed in 2:1 molar ratio, the total pressure of gases at equilibrium is found to be 1.4 atm and partial pressure of A₆(g), 0.2 atm. The value of '4x' is

Prashant Kotian · Accepted Answer

Correct answer: 6. From the first equilibrium: K_p = P_A₆ / (P_A₂)³ = 1.6. With P_A₆ = 0.2 atm, 0.2 / (P_A₂)³ = 1.6 => (P_A₂)³ = 1/8 => P_A₂ = 0.5 atm. Total pressure = P_A₆ + P_A₂ + P_C + P_A₂C = 1.4 atm. Thus, 0.2 + 0.5 + P_C + P_A₂C = 1.4 => P_C + P_A₂C = 0.7 atm. The initial molar ratio of A₂ to C is 2:1. Using mass balance via partial pressures (since V, T are constant): Initial A₂ ∝ P_A₂ + 3P_A₆ + P_A₂C = 0.5 + 0.6 + P_A₂C = 1.1 + P_A₂C. Initial C ∝ P_C + P_A₂C. Ratio = (1.1 + P_A₂C) / (P_C + P_A₂C) = 2. Since P_C + P_A₂C = 0.7, 1.1 + P_A₂C = 2(0.7) = 1.4 => P_A₂C = 0.3 atm. Then P_C = 0.7 - 0.3 = 0.4 atm. For the second equilibrium: x = P_A₂C / (P_A₂ × P_C) = 0.3 / (0.5 × 0.4) = 0.3 / 0.2 = 1.5. Thus, 4x = 4(1.5) = 6. Therefore, correct answer is 6.

💡 Solution & Explanation

Practice 22,000+ questions like this