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If heat of neutralisation is -13.7kcal a...

If heat of neutralisation is `-13.7kcal` at `25^(@)C` and `H_(f(H_(2)O))^(@)=-68kcal`, then standard enthalpy of `OH^(-)` would be `:`

A

`54.3kcal`

B

`-54.3kcal`

C

`71.3kcal`

D

`-71.3kcal`

Text Solution

Verified by Experts

The correct Answer is:
B
`H^(+)+OH^(-)rarrH_(2)O,DeltaH=-13.7kcal`
Also`DeltaH=H_(f H_(2)O)^(@)-[H_(H^(+))^(@)+H_(OH^(-))^(@)]`
`:. -13.7=-68.0-[0+H_(HO^(-))^(@)][:'H_(H^(+))^(@)=0]`
`:. H_(OH^(-))^(@)=-68.9+13.7=-54.3kcal`
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