Given

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 $\mathrm{NH}_{3(g)}$

$=1 / 2 \Delta_{r} H^{\theta}$

$=1 / 2\left(-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$=-46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$