Equilibrium question

Question

Te disassociation equilibrium of a gas AB2 can be represented as: 2AB2(g)---------2AB(g) + B2(g) . The degree of disassociation is 'x' and is small compared to 1. The expression relating the degree of disassociation (x) with equilibrium constant Kp and total pressure is?

Answer 1

2AB₂(g) < ==== > 2AB(g) + B₂(g)

Initial: AB₂ = 1; AB = 0; B₂ = 0

Equilibrium: AB₂ = 2(1-x); AB = 2x; B₂ = x

Total moles = 2-2x + 2x + x = (2 + x)

Since the K_p is related to the partial pressures of the gasses at equilibrium, and partial pressure can be found from mole fraction * P_total (P), let us find partial pressures:

P_AB = 2xP/(2+x)

P_B2 = xP/(2+x)

P_AB2 = 2(1-x)P/(2+x)

K_p expression can be written as Kp = (P_AB)²(P_B2) / (P_AB)²

Substitute the above values for partial pressures into K_p expression and solve for x. Hopefully, you will find x = (2K_p/P)^1/3 or cube root of (2K_p/P)

You could also express the answer as K_p = Px³/2, I would think.