Question

Heat of neutralisation of $HCl$ by $NaOH$ is 13.7 kcal per equivalent and by $N{H}_{4}OH$ is 12.27 kcal. The heat of dissociation of $N{H}_{4}OH$ is:

• A. -25.97 kcal
• B. 25.97 kcal
• C. -1.43 kcal
• D. 1.43 kcal

Answer 1

The correct option is D 1.43 kcal

Heat of dissociation of $N{H}_{4}OH+$ heat of neutralisation by $N{H}_{4}OH=$ heat of neutralisation by $NaOH$
$\therefore \Delta H_{diss}+\left(12.27\right)=13.7$
$\therefore \Delta H_{diss}=13.7-12.27$
$\therefore \Delta H_{diss}=1.43$ kcal

Option D is correct.