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. `22.88 KJ`
B. `228.88KJ`
C. `+228.88kJ`
D. `-343. 52KJ`

Answer

B

Explanation

Consider the heat of formation of `H_2O`
`H_2 (g) + 1/2 O_(2)(g) rarr H_(2O(l)`,
`Delta H = -286.20 kJ`
`Delta H_(r ) = Delta H_(f) (O_(2),g) = -286.20 = Delta H_(f) (H_(2)O,l) - 0 - 0`
`Delta H_(f) (H_2O,l) = -286.20`
Now, consider the ionization of `H_2O`
`H_2O (l) to H^(+) (aq) + OH^(-) (aq)`
`Delta H = 57.32 kJ`
`Delta H_(r) = Delta H_(f) (H^(+), aq) +Delta H_(f) (OH^(-),aq) - Delta H_(f) (H_2O, l)`
`57.32 = 0 + Delta H_(f) (OH^(-) , aq)- (286.20)`
Thus, `Delta H_(f) (OH^(-) , aq) = 57.32 - 286.20 = -228.80 kJ`.