A standard enthalpy of formation, written $\Delta H_f^\circ$, is the enthalpy change for forming exactly one mole of a substance from its constituent elements in their reference forms under specified standard-state conditions.

For example,

$$\mathrm{C(graphite)+O_2(g)\rightarrow CO_2(g)}$$

forms one mole of carbon dioxide from elemental carbon in its reference form and elemental oxygen, so its enthalpy change is $\Delta H_f^\circ(\mathrm{CO_2})$.

By convention, an element in its reference form has

$$\boxed{\Delta H_f^\circ=0}.$$

Examples include $\mathrm{O_2(g)}$ for oxygen and graphite for carbon under ordinary standard-state conditions. The zero is a reference convention, not a statement that the element contains no energy.

Because formation reactions provide a common thermochemical reference, Hess's law gives the standard enthalpy change of any reaction as

$$\boxed{\Delta H_{\mathrm{rxn}}^\circ =\sum_{\mathrm{products}}\nu_i\Delta H_{f,i}^\circ -\sum_{\mathrm{reactants}}\nu_i\Delta H_{f,i}^\circ},$$

where $\nu_i$ is the stoichiometric coefficient of species $i$ in the balanced reaction.

Example

For methane combustion,

$$\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)},$$

use approximate values

$$\Delta H_f^\circ(\mathrm{CH_4})=-74.8,\mathrm{kJ/mol},$$

$$\Delta H_f^\circ(\mathrm{CO_2})=-393.5,\mathrm{kJ/mol},$$

$$\Delta H_f^\circ(\mathrm{H_2O(l)})=-285.8,\mathrm{kJ/mol},$$

and $\Delta H_f^\circ(\mathrm{O_2})=0$.

Then

$$\begin{aligned} \Delta H_{\mathrm{rxn}}^\circ &=[-393.5+2(-285.8)]-[-74.8+2(0)]\ &\approx-890.3,\mathrm{kJ}. \end{aligned}$$

The negative sign indicates an exothermic reaction under the stated standard conditions.

Physical state must be matched correctly when using tables: $\Delta H_f^\circ$ for $\mathrm{H_2O(l)}$ differs from that for $\mathrm{H_2O(g)}$. Standard formation enthalpies therefore turn Hess's law into a reusable tabulated method rather than requiring a custom thermochemical cycle for every reaction.