Heat of neutralisation between `HCI` and `NaOH` is `13.7kcal` and between `HCN` and `NaOH` is `3`kcal at `45^(@)C`. Calculate the heat of ionisation of `HCN`

Explanation

`{:(HCI+,NaOHrarr,NaCI+,H_(2)O),(SA,SB,,):}`
`H^(o+) +overset(Theta)OH rarr H_(2)O, DeltaH_(1) =- 13.7 Kcal`
`HCN(W_(a)) +overset(Theta)OH(S_(B))rarrH_(2)O+CN^(Theta) , DeltaH_(2) = - 3kcal`
`HCN rarr H^(o+) +CN^(Theta)`,
`:. DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 3 -(-13.7) = 0.7 kcal`