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Standard XII

Chemistry

Reaction Quotient

A sample of p...

Question

A sample of pure $P C l_{5}$ was introduced into an evacuated vessel at $473$ K. After equilibrium was attained, concentration of $P C l_{5}$ was found to be $0.5 \times 10^{- 1} m o l L^{- 1}$ .
$P c l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g), K_{C} = 8.3 \times 10^{- 3} m o l L^{- 1}$

0.08 M

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0.04 M

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0.2 M

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0.02 M

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Solution

The correct option is A $0.02 M$
Let $x M$ be the equilibrium concentration of $P C l_{3}$ .
The equilibrium reaction is shown below.
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
The equilibrium concentrations of $P C l_{5}, P C l_{3}$ and $C l_{2}$ are $0.5 \times 10^{- 2} M$ , $x$ and $x$ respectively.
The expression of the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{3}]}$ .
Substitute values in the above expression.
$8.3 \times 10^{- 3} = \frac{x^{2}}{0.5 \times 10^{- 1}}$
$x^{2} = 4.15 m \times 10^{- 4}$
$x ≃ 0.02 M$ .
Thus, option D is the correct answer.

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A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5×10−1mol L−1. Pcl5(g)⇌PCl3(g)+Cl2(g),KC=8.3×10−3 mol L−1

A sample of pure $P C l_{5}$ was introduced into an evacuated vessel at $473$ K. After equilibrium was attained, concentration of $P C l_{5}$ was found to be $0.5 \times 10^{- 1} m o l L^{- 1}$ .
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