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The heat released on neutralisation of a strong alkali say `NaOH` with strong acid is `13.7kcal//mol.` The heat released on neutralisation of `NaOH` with `HF` if `-16.4 kcal//mol.` Calculate `DeltaH^(@)` of ionisation of `HF` in water. |
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Answer» `OH_((aq.))^(-)+H_((aq.))^(+)rarr H_(2)O_((aq.)), ....(1)` `DeltaH^(@)=-13.7kcal` `OH_((aq.))^(-)+HF_((l)) rarr H_(2)O+F_((aq.))^(-), ....(2)` `DeltaH^(@)=-16.4kcal` By `eq. (2)-(1) HF rarr H_((aq.))^(+)+F_((aq.))^(-),` `DeltaH^(@)=-2.7kcal` The negative value is due to high heat of hydration of `F^(-)` ion on account of smaller size. |
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