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On the basis of the following thermochem...

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 `:`

A

`-22.88kJ`

B

`-228.88kJ`

C

`+228.88kJ`

D

`-343.52kJ`

Text Solution

Verified by Experts

The correct Answer is:
B

`H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-), DeltaH=57.32kJ …..(1)`
`H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((l)), DeltaH=-286.20kJ ...(ii)`
By adding `eqs. (i)` and `(ii)`,
`H_(2(g))+_(1)/(2)O_(2(g))rarr H_((aq.))^(+)+OH_((aq.))^(-), DeltaH=-228.88kJ`
`:' Delta_(f)H^(@) of H_((aq.))^(+)=0`
`:. Delta_(f)H^(@) of OH^(-)=-228.88kJ`
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Calculate enthalpy of ionisation of OH^(-) ion. Given: H_(2)O_((l)) rarr H_((aq))^(+)+OH_((aq))^(-), ? H^(0) =57.32 kJ H_(2_((g)))+1/2 O_(2_((g))) rarr H_(2)O_((l)), ? H^(0) =-285.83 kJ

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