In case of H2O molecule, the enthalpy needed to break the two O - H bonds is not the same, i.e. H2O(g)⟶H(g)+OH(g);ΔaH1⊖=502 kJ mol-1OH(g.)⟶H(g)+O(g);ΔaH2⊖=427 kJ mol-1What should be the mean bond enthalpy of O-H bonds in case of H2O molecule?

In case of H2O molecule, the enthalpy needed to break the two O-H bonds is not the same,i.e.  H2O(g)⟶H(g)+OH(g);ΔaH1⊖=502 kJ mol-1OH(g)⟶H(g)+O(g);ΔaH2⊖=427 kJ mol-1The mean or average bond enthalpy is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule.Average bond enthalpy = 502+4272 =  464.5kJ mol-1