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 neutrlisation of a strong acid by a strong base is represented by:
`H^(o+) (aq) +overset(Theta)OH(aq) 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`
Operate: `:. (ii) -(i)`
`:. DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 50.6 -(-55.9) = 5.3 kJ mol^(-1)`