Question

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

• A. $-8.5kcalmo{l}^{-1}$
• B. $-97.5kcalmo{l}^{-1}$
• C. $+82.6kcalmo{l}^{-1}$
• D. $+100kcalmo{l}^{-1}$

Answer 1

Answer: (B)

$-97.5kcalmo{l}^{-1}$

The enthalpy change for dissolution is the sum of the lattice enthalpy and the enthalpy change for hydration

$\Delta H_{sol}=\Delta H_{lattice}+\Delta H_{hyd}$

Substitute values in the above expression,

$1=180+\Delta H_h$

$\Delta H_h=-179kcal$

The enthalpy of hydration for sodium chloride is the sum of the enthalpies of hydration for sodium ions and chloride ions

$\Delta H_{N{a}^{+}\left(h\right)}+\Delta H_{C{l}^{-}\left(h\right)}=-179kcal$

The hydration energies of sodium ion and chloride ions are in the ratio 6:5.

$\therefore \Delta H_{N{a}^{+}\left(h\right)}=\frac{6\times (-179)}{11}=-97.64kcal$