Question

Heat of neutralisation between $HCl$ and $NaOH$ is $13.7kcal$ and between $HCN$ and $NaOH$ is $3Kcal$ at ${45}^{\circ }C$.

Calculate the heat of ionisation of $HCN$ :

• A. $+10.7Kcal$.
• B. $+16.7Kcal$.
• C. $-10.7Kcal$.
• D. None of these

Answer 1

The correct option is A $+10.7Kcal$.

$\underset{SA}{HCl}+\underset{SB}{NaOH}⟶NaCl+H_{2}O$ or

$H^{\oplus }+\stackrel{\ominus }{O}H⟶H_{2}O,\Delta H_1=-13.7Kcal$.

$HCN\left(W_A\right)+\stackrel{\ominus }{O}H\left(S_B\right)⟶H_{2}O+C{N}^{\ominus };\Delta H_2=-3Kcal$

$HCN⟶H^{\oplus }+C{N}^{\ominus }$; $\therefore \Delta H=\Delta H_2-\Delta H_1$ $=-3-(-13.7)$ $=10.7Kcal$.

Hence, the heat of ionization of HCN is $10.7kcal$.