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
- (2Kp/P)1/2
- (Kp/P)
- (2Kp/P)
- (2Kp/P)1/3
Answer
(4) (2Kp/P)1/3
Solution
Take 2 mol of AB2 initially:
At equilibrium: AB2 = 2(1 - x), AB = 2x, B2 = x
Total moles = 2 + x
Kp = pAB2 pB2 pAB22 = (2x)2(x)[2(1-x)]2 P2 + x
∵ x 1 ⇒ 1 - x 1 and 2 + x 2
⇒ Kp = x3 P2
∴ x = ( 2KpP )1/3
Answer: (4)