Given
$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \longrightarrow 2 \mathrm{NH}_{3(g)} ; \Delta_{r} H^{\theta}=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
What is the standard enthalpy of formation of NH3 gas?
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of NH3(g),
$\frac{1}{2} \mathrm{~N}_{2(g)}+\frac{3}{2} \mathrm{H}_{2(g)} \longrightarrow \mathrm{NH}_{3(g)}$
$\therefore$ Standard enthalpy of formation of $\mathrm{NH}_{3(g)}$
$=1 / 2 \Delta_{r} H^{\theta}$
$=1 / 2\left(-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
$=-46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$