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. `-22.88KJ`
B. `-22.88KJ`
C. `+228.88KJ`
D. `-343.52KJ`

Answer

B

Explanation

Consider the heat of formation of `H_(2)O`
`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),`
`DeltaH=-286.20KJ`
`DeltaH_(r)=DeltaH_(f)(H_(2)O,l)-DeltaH_(f)(H_(2),G)`
`-(1)/(2)DeltaH_(f)(O_(2),g)=-286.20=DeltaH_(f)(H_(2)O,l)-0-0`
`DeltaH_(f)(H_(2)O,l)=-286.20`
Now, consider the ionization of `H_(2)O`
`H_(2)O(l)rarrH^(+)(aq)+)H^(-)(aq)`
`DeltaH=57.32KJ`
`DeltaH_(r)=DeltaH_(f)(H^(+),aq)+DeltaH_(f)(OH^(-),aq)-DeltaH_(f)(H_(2)O,l)`
`57.32 =0_ DeltaH_(f)(OH^(-),aq)-(-286.20)`
Thus, `DeltaH_(f)(OH^(-),aq)=57.32-286.20=-228.80KJ`