What is the equilibrium constant of the reaction: 2HI(g) ⇌ H₂(g) + I₂(g), at 444°C? — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

The degree of dissociation α = 22.22% = 2/9. For the reaction 2HI ⇌ H₂ + I₂, initial moles of HI = 1. At equilibrium, moles of HI = 1 - α, H₂ = α/2, I₂ = α/2. K_c = [H₂][I₂] / [HI]² = (α/2 × α/2) / (1 - α)² = α² / 4(1 - α)². Substituting α = 2/9: K_c = (2/9)² / [4 × (7/9)²] = (4/81) / (4 × 49/81) = 4 / (4 × 49) = 1/49 ≈ 0.0204. Therefore, correct answer is B.