Explain the standard enthalpy of formation.

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called standard molar enthalpy of formation. Its symbol is $\Delta_{f} H^{\ominus}$.
Elements are considered in their most stable states of aggregation. For example,$H_{2}$ and $O_{2}$ are in their gaseous state at $298 \ K$ temperature and $1 \ bar$ pressure.
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} ; \Delta_{f} H^{\ominus} = -285.8 \ kJ \ mol^{-1}$
$C_{\text{(graphite,s)}} + 2H_{2(g)} \rightarrow CH_{4(g)} ; \Delta_{f} H^{\ominus} = -74.81 \ kJ \ mol^{-1}$
In these examples,one mole of a compound is formed from its constituent elements.
In contrast,consider the enthalpy change for the following exothermic reaction:
$CaO_{(s)} + CO_{2(g)} \rightarrow CaCO_{3(s)} ; \Delta_{r} H^{\ominus} = -178.3 \ kJ \ mol^{-1}$
This is not an enthalpy of formation of calcium carbonate,since calcium carbonate is formed from other compounds,not from its constituent elements.
Similarly,the following reaction is not the standard enthalpy of formation for $HBr_{(g)}$ because two moles are formed:
$H_{2(g)} + Br_{2(l)} \rightarrow 2HBr_{(g)} ; \Delta_{r} H^{\ominus} = -72.8 \ kJ \ mol^{-1}$
By convention,the standard enthalpy of any element in its most stable state is taken as zero.
Calculation of heat needed for the decomposition of $CaCO_{3}$:
$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} ; \Delta_{r} H^{\ominus} = ?$
$\Delta_{r} H^{\ominus} = \sum a_{i} \Delta_{f} H^{\ominus} \text{(products)} - \sum b_{i} \Delta_{f} H^{\ominus} \text{(reactants)}$
$\Delta_{r} H^{\ominus} = \Delta_{f} H^{\ominus} [CaO_{(s)}] + \Delta_{f} H^{\ominus} [CO_{2(g)}] - \Delta_{f} H^{\ominus} [CaCO_{3(s)}]$
$= [1(-635.1) + 1(-393.5)] - [-1206.9]$
$= 178.3 \ kJ \ mol^{-1}$
Thus,the decomposition of $CaCO_{3(s)}$ is an endothermic process.