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Law of Mass Action

At equilibriu...

Question

At equilibrium of the reaction
$P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ the concentrations of $P C l_{5} (g)$ and $P C l_{3} (g)$ are 0.2 and 0.1 moles/lit.respectively $K_{c}$ is 0.05. The equilibrium concentration of $C l_{2} (m o l e s . l i t^{- 1})$ is:

A

0.5

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B

0.1

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C

1.5

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D

0.75

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Solution

The correct option is A 0.1
The equilibrium reaction is $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ .
The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$ .
Substitute values in the above expression.
$0.05 = \frac{0.1 \times [C l_{2}]}{0.2}$
Thus $[C l_{2}] = 0.1$ .

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Similar questions

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

attains equilibrium. If the equilibrium concentration of

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P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

C O C l_{2} (g) ⇌ C O (g) + C l_{2} (g)

are simultaneously in equilibrium at constant volume. A few moles of

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Q. At a certain temperature, the reaction

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

has an equilibrium constant

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