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 hydroxyl 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)`

Answer

C

Explanation

The process of neutralization is
`H^(+)(aq.)+OH^(-)(aq.)toH_(2)O(l),DeltaH^(@)=-57.32kJ" "mol^(-1)`
`DeltaH_("reaction")=sum` heat of foramtion of product
`-sum` heat of formation of reactants
`=DeltaH_(f" "H_(2)O(l))^(@)-[DeltaH_(f" "H^(+)(aq.))^(@)+DeltaH_(f" "OH^(-)(aq.))^(@)`
`=57.32=-285.84-(0+x)`
`x=-285.84+57.32`
`=-228.52kJ`