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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 (1) to H ^(+) (aq) + O H ^(-) (aq), DeltaH = 57. 3 2 K J`
`H_(2) (g) + 1/2 O_(2) (g) to H_(2) O (I), DeltaH = -2 8 6 .2 0 K J` The value of enthalpy of formation of ` O H ^(-)` ion at `25^@C` is

A

`-228.88 KJ`

B

`+228.88 KJ`

C

`-343.52 KJ`

D

`-22.88 KJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`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`.
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