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. `-343.52 KJ`
D. `-22.88 KJ`

Answer

A

Explanation

`H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l)" " =-286.20 KJ`
`DeltaH_(r)=DeltaH_(f)(H_(2)O,l) = DeltaH_(f)(H_(2),g) -(1)/(2)DeltaH_(r) (O_(2),g)`
`-286.20 = DeltaH_(f)(H_(2)O(l))`
So `DeltaH-=_(f) (H_(2)O,l) =-286.20 KJ//"mole"`
`H_(2)O(l) rarr H^(+) (aq) + OH^(-)(aq)" "DeltaH = 57.32 KJ`
`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)`
`DeltaH_(f)^(@)(OH^(-), aq) = 57.32 - 286.20 =-228.88 KJ`.