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According to law of mass action for $C a C O_{3_{(s)}} \Leftrightarrow C a O_{(s)} + C O_{2_{(g)}}$ ( r_f= rate of forward and r_b= rate of backward reactions)

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$r_{f} / r_{b} = [C O_{2}]$

$r_{f} = k_{f} {[C a O]}^{2}$

$r_{b} = k_{b} {[C a C O_{3}]}^{2}$

$r_{f} = k_{b} [C O_{2}]$

answer is A.

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Detailed Solution

$r_{f} = k_{f} r_{b} = k_{b} [C O_{2}]$

At equilibrium r_f = r_b

$r_{f} = k_{b} [C O_{2}]$

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