The reaction; O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products. ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g) -107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0). ∴2BE of O3(g) = 594.9kJ/mol Average bond enthalpy = 594.9kJ/mol/2 =297.45kJ/mol ∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.