Enthalpy of neutralisation of acetic acid by `NaOh` is `-50.6 kJ mol^(-1)`. Calculate `DeltaH` for ionisation of `CH_(3)COOH`. Given. The heat of neutralisation of a strong acid with a strong base is `-55.9 kJ mol^(-1)`.

Explanation

The neutralisation of a strong acid by a strong base is represented by:
`H^(o+)(aq) +overset(Theta)OH rarr H_(2)O(l) DeltaH =- 55.9 kJ…..(i)`
We have to calculate:
`CH_(3)COOH rarr CH_(3)COO^(Theta) +H^(o+) DeltaH = ?`
Given:
`CH_(3)COOH + overset(Theta)OH rarr CH_(3)COO^(Theta) +H_(2)O, DeltaH_(2) =- 50.6 ...(ii)`
Substract equation (i) from equation (ii), we get
`DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 50.6 -(-55.9) = 5.3 kJ mol^(-1)`