Correct Answer – A
According to thermodynamics, we have
`Delta_(r) H^(@) = sum a_(i) Delta_(f) H^(@)` )product) `- sum b_(i) Delta_(g) H^(@)` (reactants
Applying this relation ship of first equation, we have
`Delta_(r) H^(@) = [Delta_(f) H^(@) (H^(+), aq.) + Delta_(f) H^(@) (OH^(-), aq.)] – Delta_(f) H^(@) (H_(2) O, 1)`
`57.32 = [0 + Delta_(f) H^(@) (OH^(-) , aq.)]- (286.20)`
`:. Delta_(f) H^(@) (OH^(-), aq.) = 57.32 – 286.20 = – 228.88 kJ`

Correct Answer – D
`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`

Correct Answer – 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`.

Correct Answer – B
`H_(2)+1/2 O_(2) rarr H^(+)+OH^(-) Delta H=-228.88 kJ`

Correct Answer – A
`H_(2)O(l) rarr H^(+) (aq) +OH^(-)(aq), DeltaH=57.32 kJ” “`….(i)
`H_(2)(g) +1/2 O_(2)(g) rarr H_(2)O(l), DeltaH=-286.20 KJ” “`…..(ii)
For the reaction
`H_(2)(g)+1/2 O_(2) (g) rarr H^(+)(aq) +OH^(-)(aq)” “DeltaH=?`
By adding equation (i) and (ii)
`H_(2)(g)+1/2 O_(2)(g) rarr H^(+) (aq) +OH^(-) (aq)” “DeltaH=57.32 kJ -286.20 kJ`
`DeltaH=-280.88″ kJ/mol”`

Correct Answer – A