A 1-litre container contains 2 moles of $P C l_{5}$ initially. If at equilibrium, $K_{c}$ is found to be 1, degree of dissociation of $P C l_{5}$ is

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-1

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\frac{1}{2}

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Solution

The correct option is D $\frac{1}{2}$
Initially:
$P C l_{5} 2 M ⇌ P C l_{3} 0 + C l_{2} 0$

At equilibrium:
$P C l_{5} 2 - x ⇌ P C l_{3} x + C l_{2} x$

A/q.

$K c = \frac{[P C l_{3}] [C l]}{[P C l_{5}]} = \frac{x^{2}}{2 - x} = 1$

$x^{2} + x - 2 = 0$

$x = \frac{- 1 \pm \sqrt{1 + 8}}{2} = \frac{3 - 2}{2} = 1$

$∴$ Degree of dissociation $= \frac{x}{C} = \frac{1}{2}$

Suggest Corrections

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