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Derivation of Kp and Kc

Prove that th...

Question

Prove that the pressure needed to obtain $50 %$ dissociation of $P C l_{5}$ at $250^{o} C$ is $3$ times of $K_{p}$

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Solution

For the reaction $P C l_{5} ⇌ P C l_{3} + C l_{2}$
$0.5$ moles $0.5$ $0.5$
Total= $15$ moles
$∴ P_{P C l_{5}} = P_{P C l_{3}} = P_{C l_{2}} = \frac{0.5}{1.5} P$
where $P$ is the required total pressure
Hence $K_{P} = \frac{P_{P C l_{3}} P_{C l_{2}}}{P_{P C l_{5}}} = \frac{(0.5 / 1.5)^{2} P^{2}}{(\frac{0.5}{1.5}) P} = \frac{1}{3} P \Rightarrow P = 3 K_{P}$
Pressure is $3$ times of $K_{P}$ .

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