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What is meant by the term average bond enthalpy ? Why is there difference in bond enthalpy of O - H bond in ethanol (C_(2) H_(5) OH) and water ? |
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Answer» Solution :The similar bonds in a molecule do not possess the same bond ENTHALPIES. e.g., in `H_(2)` O (H-O-H) molecule after the breaking of first O-H bond , the secon O-H bond undergoes some change because of changed CHEMICAL environment . Therefore, in polyatomic molecules the TERM mean or average bond enthalpy is used It is obtained by dividing total bond dissociation enthalpy by the number of bonds BROKEN. e.g. `H_(2)O_((g)) rarr H(g) + OH_((g)) ` `Delta_(a) H_(1)^(0) = 502 " kJ mol"^(-1) OH_((g)) rarr H + O_((g))`, `Delta_(a) H_(2)^(0) = 427 KJ mol^(-1)` Average 0-H bond enthalpy =`(502 + 427)/(2)` = 464.5 kJ `mol^(-1)` The bond enthalpies of O - H bond in `C_(2)H_(5)` OH and `H_(2)O` are different because of the different chemical (electronic) environment around OXYGEN atom. H - C - `underset(H)underset(|)overset(H)overset(|)(C) - underset(H)underset(|) overset(H)overset(|)(O) - H , H - O - H ` ` ""(C_(2) H_(5) OH) "" (H_(2) O)` |
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