Question

Enthalpy of neutralization of $C{H}_{3}COOH$ by $NaOH$ is $-50.6kJ/mol$ and the heat of neutralization of a strong acid with $NaOH$ is $-55.9kJ/mol$. The value of $\Delta H$ for the ionisation of $C{H}_{3}COOH$ is:

• A. $3.5kJ/mol$
• B. $4.6kJ/mol$
• C. $5.3kJ/mol$
• D. $6.4kJ/mol$

Answer 1

Answer: (C)

$5.3kJ/mol$

Enthalpy of neutralisation of $C{H}_{3}COOH$ by $NaOH$ is $-50.6kJ/$mole

For strong acid and base, this value is $55.9kJ/$mole. The heat evolved in the first case is less as some heat is used up in ionisation of $C{H}_{3}COOH$.

So, $\Delta H$ for ionisation of $C{H}_{3}COOH=55.9-50.6=5.3kJ$/mole.

Hence, the correct option is $\text{C}$