MCQ
The lattice energy of $NaCl$ is $-780\, kJ\, mol^{-1}$. The enthalpies of hydration of $Na_{\left( g \right)}^ + $ and $Cl_{\left( g \right)}^ - $ ions are $-406\, kJ\, mol^{-1}$ and $-364\, kJ\, mol^{-1}$. The enthalpy of solution of $NaCl_{(s)}$ is.....$kJ\, mol\,^{-1}$
  • A
    $738$
  • $10$
  • C
    $-10$
  • D
    $-822$

Answer

Correct option: B.
$10$
b
$\mathrm{NaCl}(\mathrm{s}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) ; \Delta \mathrm{H}_{1} =780 \,\mathrm{KJ} $

$ \mathrm{Na}^{+}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{Na}^{+}(\mathrm{aq})  \Delta \mathrm{H}_{2}=-406 \,\mathrm{KJ} $

$ \mathrm{Cl}^{-}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})  \Delta \mathrm{H}_{3}=-364 \,\mathrm{KJ} $

$ \mathrm{NaCl}(\mathrm{s})+\mathrm{aq} \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $

$ \Delta \mathrm{H}=\Delta \mathrm{H}_{1}+\Delta \mathrm{H}_{2}+\Delta \mathrm{H}_{3}$

$= 780-406-364 $

$=10 \,\mathrm{KJ\,mol}^{-1} $

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