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

Answers (1)

(i) Average bond enthalpy is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.

(ii) 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 the charge.

Therefore, in polyatomic molecules average bond enthalpies is used and calculated by dividing total bond dissociation enthalpy by the number of bonds broken.

H_{2}O\; (g)\; \rightarrow H(g)+OH(g);\Delta _{a}H{_{1}}^{o}=502\; kJ\; mol^{-1}

OH\; (g)\; \rightarrow H(g)+O(g);\Delta _{b}H{_{2}}^{o}=427\; kJ\; mol^{-1}

Therefore, Average O-H bond enthalpy = 502+\frac{427}{2} = 464.5\; kJ\; mol-1

The bond enthalpies of O-H in C_{2}H{_{5}}^{2}OH.H_{2}O are different because of different electronic environment around 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 hydrogen atom.

 

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