The equilibrium constant for the reaction $C a C O_{3} (s) ⇋ C a O (s) + C O_{2} (g)$ is:

K_{c} = \frac{[C a O] [C O_{2}]}{[C a C O_{3}]}

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K_{c} = \frac{[C a C O_{3}]}{[C a O] [C O_{2}]}

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K_{c} = [C O_{2}]

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K_{c} = \frac{1}{[C O_{2}]}

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Solution

The correct option is C $K_{c} = [C O_{2}]$
The equilibrium reaction is as follows:
$C a C O_{3} (s) \Leftrightarrow C a O (s) + C O_{2} (g)$ .

The expression for the equilibrium constant is $K_{c} = \frac{[C a O] [C O_{2}]}{[C a C O_{3}]}$

The concentration of solids is considered as unity.

Substituting $[C a C O_{3}] = 1$ and $[C a O] = 1$ in the expression for the equilibrium constant, we get-

$K_{c} = [C O_{2}]$

Hence, option $C$ is the correct answer.

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