Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`

Options

A. `-57.3 kJ mol^(-1)`
B. `-114.6 kJ mol^(-1)`
C. `+57.3 kJ mol^(-1)`
D. `+114.6 kJ mol^(-1)`

Answer

C

Explanation

`NaOH+HCltoNaCl+H_(2)O`
or `H^(+)+OH^(-)toH_(2)O` , `DeltaH=-57.3 kJ`
` :. H_(2)O to H^(+)+OH^(-)` , `DeltaH=+57.3 kJ`
This is according to Laplace law.