The standard enthalpy of neutralization of $K O H$ with $H C N$ and $H C l$ in dilute solution is -2480 $c a l m o l^{- 1}$ and -13.68 $k c a l m o l^{- 1}$ respectively. Find the enthalpy of dissociation of $H C N$ at the same temperature.

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Solution

When $K O H$ and $H C l$ react
$K O H + H C l ⟶ K C l + H_{2} O$
That is, $O H^{-} + H^{+} ⟶ H_{2} O Δ H_{n e u t r a l i s a t i o n} = - 1368 k c a l$
When $H C N$ reacts with $K O H$
$H C N ⟶ H^{+} + C N^{-} Δ H_{d i s s o c i a t i o n} O H^{-} + H^{+} ⟶ H_{2} O Δ H_{n e u t r a l i s a t i o n} = - 13.68 - --------------------------------------------------- - H C N + O H^{-} ⟶ C N^{-} + H_{2} O; Δ H$
$Δ H_{d i s s o c i a t i o n} = - 2480 + 1360 = 11, 200$ cal

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The standard enthalpy of neutralization of KOH with HCN and HCl in dilute solution is -2480 cal mol−1 and -13.68 kcal mol−1 respectively. Find the enthalpy of dissociation of HCN at the same temperature.

The standard enthalpy of neutralization of $K O H$ with $H C N$ and $H C l$ in dilute solution is -2480 $c a l m o l^{- 1}$ and -13.68 $k c a l m o l^{- 1}$ respectively. Find the enthalpy of dissociation of $H C N$ at the same temperature.