What is meant by the term average bond enthalpy? Why is there a difference in the bond enthalpy of the $O-H$ bond in ethanol $(C_{2}H_{5}OH)$ and water $(H_{2}O)$?

(N/A) The similar bonds in a molecule do not possess the same bond enthalpies. For example,in an $H_{2}O$ $(H-O-H)$ molecule,after the breaking of the first $O-H$ bond,the second $O-H$ bond undergoes some change because of the changed chemical environment.
Therefore,in polyatomic molecules,the term mean or average bond enthalpy is used. It is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken.
For example:
$H_{2}O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$; $\Delta_{a} H_{1}^{0} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)}$; $\Delta_{a} H_{2}^{0} = 427 \ kJ \ mol^{-1}$
Average $O-H$ bond enthalpy $= \frac{502 + 427}{2} = 464.5 \ kJ \ mol^{-1}$
The bond enthalpies of the $O-H$ bond in $C_{2}H_{5}OH$ and $H_{2}O$ are different because of the different chemical (electronic) environment around the oxygen atom.