Derive an expression for the degree of dissociation, α, in terms of Kp

For the homogeneous gaseous reaction
2 AB2 (g) ⇌ 2 AB (g) + B2 (g)
Derive an expression for the degree of dissociation, α, in terms of Kp and total pressure P assuming that α << 1. Given reaction is: 2 AB2 (g) ⇌ 2 AB (g) + B2 (g)
Number of moles at Equilibrium:
AB2 = 2(1 – α)
AB = 2 α
B2 = α
Total number of moles at equilibrium = 2(1 – α) + 2α + α = 2 + α
Kp = p2 (AB) X p (B2) / p2 (AB2)
P (AB2) = 2(1 – α) p / 2 + α
P (AB) = 2α p / 2 + α
P (B2) = αp / 2 + α
Kp reaction

= α3 p / (2 + α) (1 – α) 2
As α << 1, it can be neglected in the denominator and we will get: Kp = α3 p / 2
Hence, α = (2 Kp / p) 1/3

Category: Free Energy and Chemical Equilibria

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