In the following equilibrium reaction involving the dissociation of $C a C O_{3}$ , the equilibrium constant is given by:
$C a C O_{2} (s) ⇌ C a O (s) + C O_{2} (g)$

\frac{P_{C a O} \times P_{C O_{2}}}{P_{C a C P_{3}}}

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C_{C a O} \times \frac{P_{C O_{2}}}{C_{C a C O_{3}}}

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\frac{P_{C a O}}{P_{C a C O_{3}}}

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P_{C O_{2}}

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Solution

The correct option is D $P_{C O_{2}}$
The equilibrium reaction is:
$C a C O_{3} (s) \Leftrightarrow C a O_{s} + C O_{2} (g)$
The expression for the equilibrium constant is $K_{p} = \frac{[P_{C a O}] [P_{C O_{2}}]}{[P_{C a C O_{3}}]}$
The pressure of solids are considered as unity.
Substitute $[P_{C a C O_{3}}] = 1$ and $[P_{C a O}] = 1$ in the expression for the equilibrium constant.
$K_{p} = [P_{C O_{2}}]$

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