$A$ balanced chemical equation together with the value of its $\Delta_{r} H$ is called a thermochemical equation.
It is necessary to remember the following conventions regarding thermochemical equations:
$(1)$ The coefficients in a balanced thermochemical equation refer to the number of moles of the substances involved in the reaction.
$(2)$ The numerical value of $\Delta_{r} H^{\ominus}$ refers to the number of moles of substances specified by the equation. Standard enthalpy change $\Delta_{r} H^{\ominus}$ has units of $kJ \ mol^{-1}$.
Example: $Fe_{2}O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$
Given standard enthalpies of formation:
$\Delta_{f} H^{\ominus}(H_{2}O) = -285.83 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\ominus}(Fe_{2}O_{3}) = -824.2 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\ominus}(Fe) = 0$ and $\Delta_{f} H^{\ominus}(H_{2}) = 0$
Then,$\Delta_{r} H^{\ominus} = [3(-285.83)] - [1(-824.2)] = -33.3 \ kJ \ mol^{-1}$.
If the equation is balanced differently,e.g.,$\frac{1}{2} Fe_{2}O_{3(s)} + \frac{3}{2} H_{2(g)} \rightarrow Fe_{(s)} + \frac{3}{2} H_{2}O_{(l)}$,then $\Delta_{r} H^{\ominus} = -16.6 \ kJ \ mol^{-1} = \frac{1}{2} \Delta_{r} H_{1}^{\ominus}$. This shows that enthalpy is an extensive property.
$(3)$ When a chemical equation is reversed,the sign of $\Delta_{r} H^{\ominus}$ is reversed.
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}, \Delta_{r} H^{\ominus} = -91.8 \ kJ \ mol^{-1}$
$2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}, \Delta_{r} H^{\ominus} = +91.8 \ kJ \ mol^{-1}$