Solution

Average Bond Enthalpy is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.

All the identical bonds in a molecule do not have the same bond enthalpies, e.g., in water ($$H_{2}O$$), there are two O−H bonds but breaking of first O−H bond, the second O−H bond undergoes some change because of charge. Therefore, in polyatomic molecules average bond enthalpy is used and calculated by dividing total bond dissociation enthalpy by the number of bonds broken.

$$ H_{2}O \rightarrow H + OH ; \Delta H_{1} =502\ kJ/mol$$

$$ OH \rightarrow H + O ; $$ $$\Delta H_{2} =472\ kJ/mol$$

Average bond enthalphy $$= \dfrac{502 + 472}{2} = 464.5\ kJ/mol$$

The bond enthalpies of $$O−H$$ in $$C_{2}H_{5}OH $$ and $$H_{2}O $$ are different because of the different electronic environments around the oxygen atom.

$$CH_{3}CH_{2}OH $$,$$H-O-H$$

In ethanol, −OH is attached to carbon and in water O−H is attached to the hydrogen atom.