An amount of 0.2 mole of each A₂(g) and B₂(g) is introduced in a sealed flask and heated to 2000 K w — Chemical Equilibrium Chemistry Question
Question
An amount of 0.2 mole of each A₂(g) and B₂(g) is introduced in a sealed flask and heated to 2000 K where following equilibrium is established: A₂(g) + B₂(g) ⇌ 2 AB(g). At equilibrium, moles of AB is 0.3. At this stage, 0.1 mole of C₂(g) is added and a new equilibrium is also established as: A₂(g) + C₂(g) ⇌ 2 AC(g). At the new equilibrium, the moles of AB becomes 0.24. What is the equilibrium constant for the second reaction?
💡 Solution & Explanation
Eq 1: A₂ + B₂ ⇌ 2AB. Let 2x = 0.3 => x = 0.15. Eq moles: A₂ = 0.05, B₂ = 0.05, AB = 0.3. K_c1 = (0.3)² / (0.05 × 0.05) = 36. C₂ is added, rxn 1 shifts back. New AB = 0.24, meaning total A₂ and B₂ converted to AB is 0.12 each. Moles of B₂ = 0.2 - 0.12 = 0.08. Since K_c1 is constant: 36 = (0.24)² / ([A₂] × 0.08) = 0.0576 / (0.08 × [A₂]) = 0.72 / [A₂]. So [A₂] = 0.02. Total A₂ consumed = 0.2 (initial) - 0.02 (left) = 0.18. A₂ in AB = 0.12, so A₂ in AC = 0.18 - 0.12 = 0.06. Moles of AC = 2 × 0.06 = 0.12. C₂ consumed = 0.06. C₂ left = 0.1 - 0.06 = 0.04. K_c2 = [AC]² / ([A₂][C₂]) = (0.12)² / (0.02 × 0.04) = 0.0144 / 0.0008 = 18. Therefore, correct answer is 0018.