Phosphorus pentachloride dissociates as follows in a closed reaction vessel.
${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of ${PCl}_{5}$ is x. the partial pressure of ${PCl}_{3}$ will be :

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$(\frac{x}{1 - x}) P$

$(\frac{2 x}{1 - x}) P$

$(\frac{x}{x + 1}) P$

$(\frac{x}{x - 1}) P$

answer is A.

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Detailed Solution

For the standard reaction,

$\begin{matrix} a A + b B ⇌ c C + d D \\ K_{p} = \frac{{(P_{C})}^{c} {(P_{D})}^{d}}{{(P_{A})}^{a} {(P_{B})}^{b}} \end{matrix}$

Let the degree of dissociation for ${PCl}_{5}$ is x.

$\underset{1 - x}{{PCl}_{5}} \underset{}{\underset{x}{⇌ {PCl}_{3}} (g)} + \underset{x}{{Cl}_{2} (g)}$

Therefore,

${PCl}_{3} = (\frac{x}{1 + x}) \times P$

Hence, option A is correct.

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