In the dissociation of $P C l_{5}$ as
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
if the degree of dissociation is a at equilibrium pressure P, then the equilibrium constant for the reaction is :

K_{p} = \frac{α^{2}}{1 + α^{2} P}

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K_{p} = \frac{α^{2} P^{2}}{1 - α^{2}}

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K_{p} = \frac{P^{2}}{1 - α^{2}}

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K_{p} = \frac{α^{2} P}{1 - α^{2}}

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Solution

The correct option is B $K_{p} = \frac{α^{2} P}{1 - α^{2}}$
$P C l_{5} I n i t i a l 1 F i n a l 1 - α P a r t i a l p r e s s u r e = \frac{1 - α}{1 + α} \cdot P ⇌ P C l_{3} 0 α \frac{α}{1 + α} \cdot P + C l_{2} 0 α \frac{α}{1 + α} \cdot P$
Total mole $= 1 = α + α + α = 1 + α$
$K_{p} = \frac{\frac{α}{1 + α} P \cdot \frac{α}{1 + α} \cdot P}{\frac{1 - α}{1 + α} P} = \frac{α^{2} P}{1 - α^{2}}$

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