MCQ
The heat of neutralization of $HCl$ by $NaOH$ under certain condition is $-55.9\,\, kJ$ and that of $HCN$ by $NaOH$ is $-12.1\,\, kJ$. the heat of ionization of $HCN$ is .............. $\mathrm{kJ \,mol}^{-1}$
  • A
    $ - 68$
  • B
    $ - 43.8$
  • C
    $68$
  • $43.8$

Answer

Correct option: D.
$43.8$
d
$HCl + NaOH \longrightarrow NaCl + H _2 O \quad \Delta H =-55.9 KJ / mol$

$NaCl + H _2 O \longrightarrow NaOH + HCl \quad \triangle H =55.9 KJ / mol \quad-------(1)$

$HCN + NaOH \longrightarrow NaCH + H _2 O \quad \triangle H =-12.1 KJ / mol \quad-----(2)$

Adding $(1)$ and $(2) $

$HCN + NaCl \longrightarrow NaCN + HCl \longrightarrow \Delta H=55.9-12.1$

$=43.8 KJ / mol$

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