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Derivation of Kp and Kc

P Cl 5 ⇌ P Cl...

Question

$P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}$ in the reversible reaction the moles of $P {C l}_{5}, P {C l}_{3}, {C l}_{2}$ are $a, b$ and $c$ respectively and total pressure is $P$ then value of $K_{p}$ is:

\frac{b c}{a} R T

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\frac{b}{(a + b + c)} P

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\frac{b c P}{(a + b + c)}

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\frac{c}{(a + b + c)} P

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Solution

The correct option is A $\frac{b c}{a} R T$
$P C l_{5} ⇌ P C l_{3} + C l_{2}$
Total pressure is $P$ .
Moles of $P C l_{5}, P C l_{3}$ and $C l_{2}$ are $a, b$ and $c$ respectively.
$K_{p} = \frac{P_{P C l_{3}} \cdot P_{C l_{2}}}{P_{P C l_{5}}} = \frac{(\frac{P \cdot P}{1 + P}) \cdot (\frac{P \cdot P}{1 + P})}{\frac{(1 - P) \cdot P}{(1 + P)}}$
$= \frac{P^{2} P}{(1 - P)^{2}}$

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