${\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3;\Delta \mathrm{H}=-46\mathrm{ }\mathrm{K}$. Cals. Heat of formation of ammonia is

 Thermodynamic properties - Heat of formation Detailed solution:

- Reacting 1 mole nitrogen with 3 mole hydrogen produces 2 moles of ammonia with heat of formation $-46\mathrm{ }\mathrm{K}$.cal

- $\Delta H_f=-46\mathrm{ }\mathrm{K}$. cal for 2 moles ammonia

- Then for one mole ammonia the heat of formation is

- $\Delta H_f=-\frac{46}{2}=-23\mathrm{ }\mathrm{K}.\mathrm{cal}$

Hence the correct option is $\left(\mathrm{C}\right)$.