• # question_answer 95)   What is meant by average bond enthalpy? Why is there difference in bond enthalpy of O - H bond in ethanol and water?

If a molecule contains more than one bond of the same type, then their bond enthalpies are not the same because of the presence of different neighbouring atoms. For example, four C-H bonds in $C{{H}_{4}}$ molecule have different bond enthalpies. In such cases, bond enthalpy is the average of the bond enthalpies of all these bonds. $\text{Average bond enthalpy =}\frac{\begin{array}{*{35}{l}} \text{Total enthalpies of all the} \\ \text{same bonds in the molecule} \\ \end{array}}{\text{Number of similar bonds}}$ Average bond energy of C-H bond in $C{{H}_{4}}$ is equal to one fourth of the energy of dissociation of methane into C+4H, i.e., $\frac{1663}{4}=416kJmo{{l}^{-1}}.$ Water has 2 O-H bonds. The average bond energy of O-H bond   is   $({{H}_{2}}O\to 2H+O,\Delta H=926kJ\,mo{{l}^{-1}})$ $\frac{926}{2}=463kJmo{{l}^{-1}}$ However, in enthanol, there is only one O-H bond and it is linked to ethyl group. Thus, in this molecule environment is different. The bond enthalpy is, therefore, different in both the compounds for the same O-H bond.