The pressure necessary to obtain $50 %$ dissociation of $P C l_{5}$ at $400 K$ is numerically equal to how many times the value of the equilibrium constant $K_{p}$

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Solution

$P C l_{5} will dissociate as,$
$P C l_{5} \to P C l_{3} + C l_{2}$
$t = 0100$
$t = t_{e q} 0.5 0.5 0.5$
So, total moles at equilibrium = $0.5 + 0.5 + 0.5 = 1.5$
$Let the total pressure be P$
Again,
$K_{p} = \frac{P_{P C l_{3}} \times P_{C l_{2}}}{P_{P C l_{5}}}$
$\Rightarrow K_{p} = \frac{\frac{0.5}{1.5} P \times \frac{0.5}{1.5} P}{\frac{0.5}{1.5} P}$
$\Rightarrow K_{p} = \frac{P}{3}$
$\Rightarrow 3 K_{p} = P$

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