The standard enthalpy of formation of ${\mathrm{NH}}_3$ is $-46.0\mathrm{kJ}{\mathrm{mol}}^{-1}$. If the enthalpy of formation of ${\mathrm{H}}_2$ from its atom is $-436\mathrm{kJ}{\mathrm{mol}}^{-1}$ and that of ${\mathrm{N}}_2$ is $-712\mathrm{kJ}{{\mathrm{mol}}^{-}}^1$ , the average bond enthalpy of  $\mathrm{N}-\mathrm{H}$ bond in ${\mathrm{NH}}_3$ is 

Given that

(i)$\frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)⟶{\mathrm{NH}}_3\left(\mathrm{g}\right); \mathrm{\Delta }{H}^{\circ }=-46.0\mathrm{kJ}{\mathrm{mol}}^{-1}$

(ii) $2\mathrm{H}\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\left(\mathrm{g}\right); \mathrm{\Delta }{H}^{\circ }=-436\mathrm{kJ}{\mathrm{mol}}^{-1}$

(iii) $2\mathrm{N}\left(\mathrm{g}\right)⟶{\mathrm{N}}_2\left(\mathrm{g}\right) \mathrm{\Delta }{H}^{\circ }=-712\mathrm{kJ}{\mathrm{mol}}^{-1}$

We require to calculate

(iv) ${\mathrm{NH}}_3\left(\mathrm{g}\right)⟶\mathrm{N}\left(\mathrm{g}\right)+3\mathrm{H}\left(\mathrm{g}\right) \mathrm{\Delta }{H}_1^{\circ }=?$

This is obtained by the manipulation Eq. (i) $-\frac{3}{2}$ Eq. (ii) $-\frac{1}{2}$ (iii) 

Hence, $\mathrm{\Delta }{H}_1^{\circ }=\left(+46.0+\frac{3}{2}\times 436+\frac{1}{2}\times 712\right)\mathrm{kJ}{\mathrm{mol}}^{-1}=1056\mathrm{kJ}{\mathrm{mol}}^{-1}$

Since three  $\mathrm{N}-\mathrm{H}$ bonds are broken in Eq.(iv), we get ${\epsilon }_{\mathrm{N}-\mathrm{H}}=\frac{1}{3}\left(1056\mathrm{kJ}{\mathrm{mol}}^{-1}\right)=352\mathrm{kJ}{\mathrm{mol}}^{-1}$.