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Law of Mass Action

In a 500 ml...

Question

In a $500$ ml capacity vessel $C O$ and $C l_{2}$ are mixed to form $C O C l_{2}$ at equilibrium, it contains 0.2 moles of $C O C l_{2}$ and 0.1 mole each of $C O$ and $C l_{2}$ . The equilibrium constant $K_{c}$ for reaction
$C O + C l_{2} ⇋ C O C l_{2}$ , is :

A

20

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B

15

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C

10

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D

5

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Solution

The correct option is C 10
500 ml of solution contains 0.1 moles of CO, 0.1 moles of $C l_{2}$ and 0.2 moles of $C O C l_{2}$ .
Thus the equilibrium concentrations of CO, $C l_{2}$ and $C O C l_{2}$ are 0.2M, 0.2M and 0.4 M respectively.
The expression for the equilibrium constant becomes $K_{c} = \frac{[C O C l_{2}]}{[C O] [C l_{2}]} = \frac{0.4}{0.2 \times 0.2} = 10$ .

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