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)`.

Enthalpy of neutralisation of acetic acid by `NaOH` is `-50.6 kJ mol^(

1.	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)`.
Answer» 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)`

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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)`.

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