(b) 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=Σ Bond energies of reactants −Σ Bond energies of products 2×−46=712+3×(436)−6x−92=2020−6x 6x=2020+92 6x=2112 x=+352kJ∕mol