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. −228.88 kJ
C. +228.88 kJ
D. −343.52kJ

Answer

B

Explanation

`H_2(g) + 1/2O_2(g) to H_2O (l) , DeltaH = -286.20 kJ`
`DeltaH_r = DeltaH_f (H_2O,l) - DeltaH_f (H_2,g) - 1/2 DeltaH_f (O_2,g)`
` - 286.20 = DeltaH_f (H_2O(l))`
So, `DeltaH_f (H_2O,l) = -286.20`
`H_2O(l) to H^(+) (aq) + OH^(-) (aq), DeltaH = 57.32 kJ`
`57.32 = 0 + DeltaH_f^@ (OH^(-),aq) - (-286.20)`
`DetlaH_f^@ (OH^(-), aq) = 57.32 - 286.20 = -288.88kJ`