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Characteristics of Equilibrium Constant

For the follo...

Question

For the following reaction,
$P C l_{5} (g) ⇌ P C l_{2} (g) + C l_{2} (g)$
0.4 Mole of $P C l_{5}, 0.2$ mole of $P C l_{3}$ and 0.6 mole of $C l_{2}$ are taken in 1 litre flask. if $K_{c}$ is 0.2 then, predict the direction in which reaction proceeds.

A

forward

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B

backward

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C

Reaction will remain at equilibrium

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D

None of these

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Solution

The correct option is B backward
the given reaction is
$P C l_{5} (g) ⇌ P C l_{2} (g) + C l_{2} (g) K_{c} = 0.2$
volume = 1 litre, $n_{P C l_{5}} = 0.4, n_{P C l_{3}} = 0.2, n_{C l_{2}} = 0.6$
the concentrations of different species becomes
$[P C l_{5}] = \frac{0.4}{1} = 0.4 M, [P C l_{3}] = \frac{0.2}{1} = 0.2 M$
$[C l_{2}] = \frac{0.6}{1} = 0.6 M s i n c e c o n c e n t r a t i o n = \frac{n u m b e r o f m o l e}{v o l u m e o f s o l u t i o n i n l i t r e}$
Hence, $Q = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.2 \times 0.6}{0.4} = 0.3 M > K_{c} (= 0.2)$
Since, $Q > K_{c}$ , the reaction proceeds in the backward direction, i.e., more $P C l_{5}$

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