The dissociation equilibrium of a gas AB 2 can be represented as, 2 AB 2 (g) ⇌ 2 AB(g) + B 2 (g) The degree of dissociation is x and is small as compared to 1 . The expression relating the degree of dissociation (x) with equilibrium constant K P and total pressure P is:

Step-by-step solution

2 AB ⁡ 2 g ⇌ 2AB ⁡ g + B ⁡ 2 g ⁡ At equilibrium 2(1 − x) 2x x Total moles = (2 + x) Partial pressures P AB 2 = 2(1 − x) 2 + x p, P AB = 2x 2 + x p, P B 2 = x 2+x p K p = p AB 2 pB 2 ( pAB 2 ) 2 = [ ( 2 x 2 + x ) p ] 2 [ x p 2 + x ] [ 2 ( 1 - x ) 2 + x p ] 2 K p = 4 x 3 P 4 2 + x ( 1 − x ) 2 2 + x → 2 ( x is small) 1 − x → 1 K P = 4 x 3 p 8 = x 3 p 2 x = 2 kp p 3