The degree of dissociation of $P C l_{5}$ at a certain temperature and $1 a t m$ pressure is $0.20$ . Calculate the value of equilibrium constant $K_{p}$ .

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Solution

Let the degree of dissociation be $α$ .
$P$ be the total pressure at equilibrium.
$P C l_{5} ⇌ P C l_{3} + C l_{2}$
Initial moles 1 0 0
At equi. $1 - α$ $α$ $α$
Total number of moles at equilibrium = $1 - α + α + α$ = $1 + α$
Now, we know $K_{p}$ = $\frac{P_{P C l_{3}} \times P_{C l_{2}}}{P C l_{5}}$
$P_{P C l_{5}}$ = $\frac{1 - α}{1 + α}$ P
$P_{P C l_{3}}$ = $\frac{α}{1 + α}$ P
$P_{C l_{2}}$ = $\frac{α}{1 + α}$ P
So, $K_{p}$ = $\frac{\frac{α}{1 + α} P \times \frac{α}{1 + α} P}{\frac{1 - α}{1 + α} P}$
$K_{p}$ = $\frac{α^{2}}{1 - α^{2}} P$
Given, P = 1 atm, $α$ = 0.2
$K_{p} = 0.042 a t m$

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The degree of dissociation of $P C l_{5}$ at a certain temperature and $1 a t m$ pressure is $0.20$ . Calculate the value of equilibrium constant $K_{p}$ .