N2+3H2→2NH3 ∆H=2×−46.0kJmol−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. ] ∆H=∑ Bond energies of products −∑ Bond energies of reactants 2×−46=712+3×(436)−6x −92=2020−6x 6x=2020+92 ⇒6x=2112 ⇒x=+352kJ∕mol