Consider the reaction:
P(s) + 3/2 Cl2(g)Doublearrow. GIFPCl3(g)

Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and bbelow:

a.) P(s) + 5/2 Cl2(g) Doublearrow. GIF PCl5(g) Ka
b.) PCl3(g) + Cl2(g) Doublearrow. GIF PCl5(g) Kb

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

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