Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$

(N/A) No,the $\Delta _{r}H^{o}$ for the given reaction is not the same as $\Delta _{f}H^{o}$.
The standard molar enthalpy of formation,$\Delta _{f}H^{o}$,is defined as the standard enthalpy change for the formation of $1 \ mol$ of a compound from its constituent elements in their most stable reference states.
The reaction for the formation of $CaCO_{3(s)}$ is:
$Ca_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} \to CaCO_{3(s)}$
In the given reaction,$CaO_{(s)} + CO_{2(g)} \to CaCO_{3(s)}$,the product is formed from compounds ($CaO$ and $CO_{2}$) rather than from elements in their standard states.
Therefore,$\Delta _{r}H^{o} \neq \Delta _{f}H^{o}$.