For $\large PC{l_{5\left( g \right)\,}}\, \rightleftharpoons \,PC{l_{3\left( g \right)}}\, + \,C{l_{2\left( g \right)}}$ at equilibrium, $\large {K_p}\, = \,\frac{P}{3}$ , where P is equilibrum pressure. Then degree of dissociation of PCl₅ at that temperature is

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0.75

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answer is D.

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Detailed Solution

Let initial moles of PCl₅ = 1

Degree of decomposition of PCl₅ = α

Equilibrium pressure = P

$\large {K_P} = \frac{P}{3}$

	$\large PC{l_{5\left( g \right)\,}}$	$\large \rightleftharpoons$	$\large \,PC{l_{3\left( g \right)}}\, +$	$\large C{l_{2\left( g \right)}}$
Initial moles	1		_____	_____
	1 - α		α	α

Total moles at equilibrium = 1 - α + α + α = 1 + α

$\large {X_{PC{l_3}}} = {X_{C{l_2}}} = \frac{\alpha }{{1 + \alpha }}$

$\large {X_{PC{l_5}}} = \frac{\alpha }{{1 + \alpha }}$

$\large {K_P} = \frac{{{P_{PC{l_3}}}.{P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}$

$\large {K_P} = \frac{{\left( {\frac{\alpha }{{1 + \alpha }}P} \right)\left( {\frac{\alpha }{{1 + \alpha }}P} \right)}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}P} \right)}}$