Answer:
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.
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