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If dissociation for reaction, ${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ is 20% at 1 atm pressure. Calculate $K_{C}$ .

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0.07

0.05

0.04

0.06

answer is B.

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Detailed Solution

$K_{C} = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]} = \frac{[\frac{20}{100}] \times [\frac{20}{100}]}{[\frac{80}{100}]}$

$= \frac{0.2 \times 0.2}{0.8} = \frac{0.04}{0.8} = 0.05$

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