Q. The dissociation equilibrium of a gas AB₂ can be represented as, 2AB₂(g) ⇌ 2AB(g) + B₂(g) The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
A) (Kp/P)
B) (2Kp/P)
C) (2Kp/P)^(1/3)
D) (2Kp/P)^(1/2)
Explanation
[Solution from source] Starting with 1 mol AB₂: AB₂ = 1 − x, AB = x, B₂ = x/2, total = 1 + x/2 ≈ 1. Kp = (xP)²(xP/2)/(P)² = x³P/2, so x = (2Kp/P)^(1/3). The square-root form would hold for a reaction where x enters only as x², e.g. AB₂ ⇌ AB + B. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.