On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)`
`H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ`
`H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ`
The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`

Options

A. `- 228.88 kJ`
B. `+ 228.88 kJ`
C. `- 22.88 kJ`
D. `- 343.52 kJ`

Answer

A

Explanation

According to thermodynamics, we have
`Delta_(r) H^(@) = sum a_(i) Delta_(f) H^(@)` )product) `- sum b_(i) Delta_(g) H^(@)` (reactants
Applying this relation ship of first equation, we have
`Delta_(r) H^(@) = [Delta_(f) H^(@) (H^(+), aq.) + Delta_(f) H^(@) (OH^(-), aq.)] - Delta_(f) H^(@) (H_(2) O, 1)`
`57.32 = [0 + Delta_(f) H^(@) (OH^(-) , aq.)]- (286.20)`
`:. Delta_(f) H^(@) (OH^(-), aq.) = 57.32 - 286.20 = - 228.88 kJ`