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Derivation of Kp and Kc

PCl3 ⇌ PCl3 +...

Question

$P C l_{3} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction the moles of $P C l_{3}$ and $C l_{2}$ 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{a}{a + b + c}

. P

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

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

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Solution

The correct option is C $\frac{b c . P}{a (a + b + c)}$
Given
Moles of $P C l_{5}$ =a
Moles of $P C l_{3}$ =b
Moles of $C l_{2}$ =c
Total pressure =P
Solution
$P C l_{5} = P C l 3 + C l_{2}$
$K p = \frac{p (P C l_{3}) * p (C l_{2})}{p (P C l_{5})}$
Total no. of moles = a+ b+ c

$p (P C l_{5}) = \frac{a}{a + b + c} * P$

$p (P C l_{3}) = \frac{b}{a + b + c} * P$

$p (C l_{2}) = \frac{c}{a + b + c} * P$
Putting values
$K_{s p} = \frac{[\frac{b}{a + b + c}] * [\frac{c}{a + b + c}]}{\frac{a}{a + b + c}} * P$

$K_{s p} = \frac{b c}{a (a + b + c)} * P$

The correct option is C

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