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`Delta_(f)H(H_(2)O) =- 68 kcal mol^(-1)` and `DeltaH` of neutralisation is `-13.7 kcal mol^(-1)`, then the heat of formation of `overset(Theta)OH` isA. `-68 kcal mol^(-1)`B. `-54.3 kcal mol^(-1)`C. `54.3 kcal mol^(-1)`D. `-71.7 kcal mol^(-1)` |
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Answer» `H^(o+) +overset(Theta)OH hArr H_(2)O` `DeltaH =- 13.7 kcal` `DeltaH^(Theta) = Delta_(f)H^(Theta) (H_(2)O) -[Delta_(f)H^(Theta)(H^(o+))+Delta_(f)H^(Theta) (overset(Theta)OH))]` `=- 68 - (0-13.7) =- 54.3 kcal mol^(-1)` `[Delta_(f)H^(Theta) (H^(o+)) =0 (convention)]` |
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