Question

How much heat is evolved using neutrilization of $HCN$ by a strong base? Heat of ionization of $HCN$ $10.8$ Kcal.

Answer 1

Solution:-

As we know that heat of neutralization of a strong acid by a strong base is the heat of formation of water.

Therefore,

$H^{+}+{OH}^{-}⟶H_{2}O\Delta H=-13.6kcal.....\left(1\right)$

Given that heat of ionisation of $HCN$ is $10.8kcal$.

$HCN⟶H^{+}+{CN}^{-}\Delta H=10.8kcal.....\left(2\right)$

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

$H^{+}+HCN+{OH}^{-}⟶H^{+}+{CN}^{-}+H_{2}O\Delta H=-13.6+10.8$

$HCN+{OH}^{-}⟶{CN}^{-}+H_{2}O\Delta H=-2.8kcal.....\left(2\right)$

Hence the heat evolved using neutralisation of $HCN$ by a strong base is $-2.8kcal$.