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. `-343.52 kJ`
C. `-22.88 kJ`
D. `-228.88 kJ`

Answer

D

Explanation

`H_(2)O_((l))rarr H^(+)(aq)+OH^(-)(aq) , Delta H=57.32`
`Delta H=[Delta H_(f)(H^(+))+Delta H_(f)(OH^(-)]-Delta H_(f)(H_(2)O)`
From eq. (ii) `Delta H_(f)(H_(2)O)=-286.20 kJ`
Given `Delta H_(f)(H^(+)) = 0`
`therefore 57.32 =0+Delta H_(f)(OH^(-))-(-286.20)`
`Delta H_(f)=-228.88 kJ`