The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxy`1` ion is

Options

A. `+228.52 kJ mol^(-1)`
B. `-114.26 kJ mol^(-1)`
C. `-228.52 kJ mol^(-1)`
D. `+114.2 kJ mol^(-1)`

Explanation

The process of neutralisation is
`H^(o+)(aq) + overset(Theta)OH (aq) rarr H_(2)O(l),DeltaH^(Theta) =- 57.32 kJ mol^(-1)`
`DeltaH_(reaction)^(Theta)=sum` Heat of formation of products `-sum`Heat of fromation of reactants
`= Delta_(f)H_(H_(2)O(l))^(Theta)-(Delta_(f)H_(H^(o+)(aq))^(Theta) +Delta_(f)H_(OH_((aq))^(Theta))^(Theta))`
`-57.32 =- 285.84 +57.32 =- 228.52 kJ`
`x =- 285.84 + 57.32 =- 228.52 kJ`
Thus, the entalpy of formation of hydroxy`1` ion is `-228.52 kJ`.