`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` is

Options

A. `-68 kcal mol^(-1)`
B. `-54.3 kcal mol^(-1)`
C. `54.3 kcal mol^(-1)`
D. `-71.7 kcal mol^(-1)`

Explanation

`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)]`