N₂ + 3H₂ → 2NH₃ ;ΔH = -46K.Cals. From the above reaction, heat of formation of ammonia is

Enthalpy of formation is the enthalpy change observed when one mole of a substance is formed from its constituent elements;

In the given reaction two moles of Ammonia is formed;

Thus ${∆}_{\mathrm{f}}^0 \mathrm{of} {\mathrm{NH}}_3=-\frac{46}{2}\mathrm{kCals}/\mathrm{mole}=-23 \mathrm{kCal}/\mathrm{mole};$