Question

The ionisation energy of solid $NaCl$ is $180 \,kcal$ per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of $1 \,kcal$ 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. $-85.6\, kcal / mol$
• B. $-97.6\, kcal / mol$
• C. $82.6\, kcal / mol$
• D. $+100\, kcal / mol$

Answer 1

Answer: (B)

$\Delta H_{\text {solution }}=\Delta H_{\text {ionization }}+\Delta H_{\text {hydration }}$

$\therefore 1=180+\Delta H_{h} \Rightarrow \Delta H_{h}=-179$ kcal mol $^{-1}$

Let $x$ be total heat of hydration $\left(\Delta H_{h}\right)$ but total

$\Delta H_{h}=\Delta H_{h\left( Na ^{+}\right)}+\Delta H_{h\left( Cl ^{-}\right)}$

$=\frac{6}{11} x+\frac{5}{11} x=x$

Thus, $\cdot \Delta H_{h\left( Na ^{+}\right)}$

$=\frac{6}{11} x=\frac{6}{11}(-179)=-97.63\, kcal\, mol ^{-1}$