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${PCl}_{5}$ dissociates as follows in a closed reaction vessel ${PCl}_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}$ . 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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$[\frac{x}{x-1}] p$

$[\frac{x}{1-x}] p$

$[\frac{x}{x+1}] p$

$[\frac{2x}{1-x}] p$

answer is C.

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Detailed Solution

$P C l_{5} ⇌ P C l_{3} + C l_{2}$
$t = 0 : 1 0 0$
$t = t_{\sec} : 1 - x x x$

Total number of moles $= 1 - x + x + x = 1 + x$
$\frac{x}{1 + x} \times P \Rightarrow$ partial pressure of $P C l_{3}$

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