For the homogeneous gaseous reaction

2 AB_{2} (g) ⇌ 2 AB (g) + B_{2} (g)

Derive an expression for the degree of dissociation, α, in terms of K_{p} and total pressure P assuming that α << 1.
Given reaction is:
2 AB_{2} (g) ⇌ 2 AB (g) + B_{2} (g)

Number of moles at Equilibrium:

AB_{2} = 2(1 – α)

AB = 2 α

B_{2} = α

Total number of moles at equilibrium = 2(1 – α) + 2α + α = 2 + α

K_{p} = p^{2} (AB) X p (B_{2}) / p^{2} (AB_{2})

P (AB_{2}) = 2(1 – α) p / 2 + α

P (AB) = 2α p / 2 + α

P (B_{2}) = αp / 2 + α

= α^{3} p / (2 + α) (1 – α) ^{2}

As α << 1, it can be neglected in the denominator and we will get:
K_{p} = α^{3} p / 2

Hence, α = (2 K_{p} / p) ^{1/3}