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${PCl}_{5}$ decomposes as ${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$ . If at equilibrium, total pressure is P and density of gaseous mixture is d at temperature T then degree of dissociation $(α)$ is :
Molecular wt. of ${PCl}_{5}$ $= M$

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$α = \frac{d R T}{P M} - 1$

$α = 1 - \frac{d R T}{P M}$

$α = \frac{P M}{d R T} - 1$

$α = 1 - \frac{P M}{d R T}$

answer is A.

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Detailed Solution

For the given reaction,

${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$

The standard formula for the degree of dissociation is,

$α = \frac{D - d}{d} α = \frac{M_{t} - M_{o}}{M_{o}}$

$P (\frac{M}{1 + α}) = d \cdot R \cdot T \Rightarrow α = \frac{P \cdot M}{d \cdot R \cdot T} - 1$

Hence, the correct option is a.

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