Derive the expressions $K_{c}$ and $K_{p}$ for decomposition of $P C l_{5}$ .

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Solution

The dissociation equilibrium of $P C l_{5}$ in gaseous state is written as $P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}$
Let 'a' moles of $P C l_{5}$ vapour be present in V, liters initially. If $x$ moles of $P C l_{5}$ dissociate to $P C l_{3}$ and $C l_{2}$ gases at equilibrium then, the molar concentrations of $P C l_{5}$ , $P C l_{3}$ and $l_{2}$ gases at equilibrium will be $\frac{a - x}{V}, \frac{x}{V}$ and $\frac{x}{V}$ respectively.
$∴ K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$
$= \frac{\frac{x}{V} \times \frac{x}{V}}{\frac{a - x}{V}}$
$= \frac{x^{2}}{V^{2}} \times \frac{V}{a - x}$
$K_{c} = \frac{x^{2}}{(a - x) V}$
$x =$ degree of dissociation
$x = \frac{N u m b e r o f m o l e s d i s s o c i a t e d}{T o t a l n u m b e r o f m o l e s p r e s e n t i n i t i a l l y}$
$K_{c} = \frac{x^{2}}{(1 - x) V}$
$[∵$ initially the no. of $P C l_{5}$ present $= 1$ mole $]$
$= \frac{\cdot x^{2} P}{(1 - x) R T}$ $[∵ P V = R T]$
In terms of partial pressure of $P C l_{5}, P C l_{3}$ and $C l_{2}$ , then
$K_{p} = \frac{p P C l_{3} \cdot p C l_{2}}{p P C l_{5}}$ atm
$K_{p} = \frac{x^{2} P}{1 - x^{2}}$
atm.

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