Question

The heat of neutralization of $HCl$ by $NaOH$ is $-55.9$ kJ/mole. If the heat of neutralisation of $HCN$ by $NaOH$ is $-12.1$ kJ/mole, the energy of dissociation of $HCN$ is:

• A. $43.8$ kJ
• B. $-43.8$ kJ
• C. $68$ kJ
• D. $-68$ kJ

Answer 1

The correct option is A $43.8$ kJ

Solution:- (A) $43.8kJ$

$NaOH+HCl⟶NaCl+H_{2}O\Delta H=-55.9kJ$

As we know that heat of neutralisation of a strong acid by a strong base is heat of fomation of water.

Therefore,

$H^{+}+{OH}^{-}⟶H_{2}O\Delta H=-55.9kJ$

$H_{2}O⟶H^{+}+{OH}^{-}\Delta H=55.9kJ.....\left(1\right)$

Given that heat of neutralisation of $HCN$ is $-12.1kJ$

$HCN+{OH}^{-}⟶{CN}^{-}+H_{2}O\Delta H=-12.1kJ.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$H_{2}O+HCN+{OH}^{-}⟶H^{+}+{OH}^{-}+{CN}^{-}+H_{2}O\Delta H=55.9+(-12.1)$

$HCN⟶H^{+}+{CN}^{-}\Delta H=43.8kJ$

Hence the heat of dissociation of $HCN$ is $43.8kJ$.