Question

The ionisation 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 $N{a}^{+}$ and $C{l}^{-}$ ions are in the ratio $6:5,$ what is the enthalpy of hydration of sodium ion ?

• A. $-85.6kcal/mol$
• B. $-97.6kcal/mol$
• C. $82.6kcal/mol$
• D. $+100kcal/mol$

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

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

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

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

$=\frac{6}{11}x=\frac{6}{11}(-179)=-97.63kcalmo{l}^{-1}$