The standard enthalpy of formation of  ${\mathrm{NH}}_3\text{ is }-46.0\mathrm{kJ}{\mathrm{mol}}^{-1}$If the enthalpy of formation of H₂, from its atoms is $-436\mathrm{kJ}{\mathrm{mol}}^{-1}\text{ and that of }{\mathrm{N}}_2\text{ is }-712\mathrm{kJ}{\mathrm{mol}}^{-1}$the average bond enthalpy of N — H bond in NH₃, is

${\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3\mathrm{\Delta H}=2\times -46.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
Let x be the bond enthalpy of N — H bond then 
[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive.] $\mathrm{\Delta H}=\mathrm{\Sigma }\text{ Bond energies of reactants }-\mathrm{\Sigma }\text{ Bond energies }$of products 
$$\begin{gathered} 2\times -46=712+3\times \left(436\right)-6\mathrm{x} \\ -92=2020-6\mathrm{x} \\ 6\mathrm{x}=2020+92 \\ 6\mathrm{x}=2112 \\ \mathrm{x}=+352\mathrm{kJ}/\mathrm{mol} \end{gathered}$$