Question

What is meant by the term average bond enthalpy? Why is there difference in bond enthalpy of $O-H$ bond in ethanol $\left(C_2{H}_{5}OH\right)$ and water ?

Answer 1

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.

$H_{2}O\to H+OH;\Delta H_1=502kJ/mol$

$OH\to H+O;$ $\Delta H_2=472kJ/mol$

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

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.

$C{H}_{3}C{H}_{2}OH$,$H-O-H$

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