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Introduction to Reversible and Irreversible Reactions

PCl5 dissocia...

Question

$P C l_{5}$ dissociation a closed container as:
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
If the total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $P C l_{5}$ is $α$ . Then partial pressure of $P C l_{5}$ will be:

P . [\frac{1 - α}{α + 1}]

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P \cdot [\frac{2 α}{1 - α}]

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P \cdot [\frac{α}{1 - α}]

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P \cdot [\frac{α}{1 + α}]

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Solution

The correct option is A $P . [\frac{1 - α}{α + 1}]$
The equilibrium reaction is $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ .
The value of the equilibrium constant is $K_{p} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$ .
At the start of the reaction, the number of moles are $n_{P C l_{5}} = 1 M, n_{P C l_{3}} = 0 M$ and $n_{C l_{2}} = 0 M$
At the equilibrium, the number of moles are
$n_{P C l_{5}} = 1 - α, n_{P C l_{3}} = α$ and $n_{C l_{2}} = α$ .
At the equilibrium, the partial pressures are
$P_{P C l_{5}} = \frac{1 - α}{1 + α} \times P,$
$P_{P C l_{3}} = \frac{α}{1 + α} \times P$ and
$P_{C l_{2}} = \frac{α}{1 + α} \times P$

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