According to law of mass action, for CaCO3⇌CaO + CO2 (Rf = Rate of forward and Rb = Rate of backward reactions)

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According to law of mass action, for CaCO₃ $⇌$ CaO + CO₂ (R_f = Rate of forward and R_b = Rate of backward reactions)

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\frac{R_{f}}{R_{b}} = [{CO}_{2}]

R_{b} = K_{b} [{CO}_{2}]

R_{b} = K_{b} {[{CaCO}_{3}]}^{2}

R_{f} = K_{f} [CaO]

answer is A.

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$\large CaC{O_3}(s)\, \rightleftharpoons \,CaO(s)\, + \,C{O_2}(g)$

At equilibrium applying law of mass action to backward reaction.

r_b = k_b [CO₂]

Note: CaCO, CaO are solids. Their concentration changes are negligible and law of mass aciton cannot be applied.

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