For the reaction o (g) + o2 (g) → o3 (g) δh o = - 107.2 kj/mol given that the bond enthalpy in o2 (g) is 498.7 kj/mol, calculate the average bond enthalpy in o3.

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.