Question

The lattice energy of solid $NaCl$ is $180kcal$ per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of $1kcal$ per mol. If the solvation energies of ${Na}^{+}$ and ${Cl}^{-}$ ions are in the ratio $6:5$, what is the enthalpy of hydration of sodium ion?

• A. $-85.6kcal/mol$
• B. $-97.5kcal/mol$
• C. $82.6kcal/mol$
• D. $+100kcal/mol$

Answer 1

Answer: (B)

$-97.5kcal/mol$

Solution:- (B) $-97.5kcal/mol$

Given that:-

${\Delta H}_{dissolution}=1kcal{mol}^{-1}$

${\Delta H}_{lattice}=180kcal{mol}^{-1}$

Let ${\Delta H}_{{Na}^{+}}$ and ${\Delta H}_{{Cl}^{-}}$ be the solvation energy of sodium and chloride ion respectively.

${\Delta H}_{hydration}=?={\Delta H}_{{Na}^{+}}+{\Delta H}_{{Cl}^{-}}$

Given that:-

$\frac{{\Delta H}_{{Na}^{+}}}{{\Delta H}_{{Cl}^{-}}}=\frac{6}{5}$

$\Rightarrow {\Delta H}_{{Cl}^{-}}=\frac{5}{6}{\Delta H}_{{Na}^{+}}$

$\therefore {\Delta H}_{hydration}={\Delta H}_{{Na}^{+}}+\frac{5}{6}{\Delta H}_{{Na}^{+}}$

$\Rightarrow {\Delta H}_{hydration}=\frac{11}{6}{\Delta H}_{{Na}^{+}}$

As we know that

${\Delta H}_{hydration}={\Delta H}_{lattice}-{\Delta H}_{dissolution}$

$\frac{11}{6}{\Delta H}_{{Na}^{+}}=180-1$

$\Rightarrow {\Delta H}_{{Na}^{+}}=\frac{6}{11}\times 179=97.63KJ{mol}^{-1}\approx 97.5Kcal{mol}^{-1}$

The sign will be $-ve$ as this energy is released during the process.

Hence the required answer is $-97.5Kcal{mol}^{-1}$.