Question

What is the heat of solution of sodium chloride if hydration energy of $N{a}^{+}$ is $-389.4{\text{kJ mol}}^{-1}$, hydration energy of $C{l}^{-}$ is $-382.3{\text{kJ mol}}^{-1}$ and lattice energy of $NaCl=+776{\text{kJ mol}}^{-1}$?

• A. $+8.6{\text{kJ mol}}^{-1}$
• B. $+4.3{\text{kJ mol}}^{-1}$
• C. $-4.3{\text{kJ mol}}^{-1}$
• D. $-8.6{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $+4.3{\text{kJ mol}}^{-1}$

$\Delta {\text{H}}_{\text{sol}}=\text{Lattice energy + Hydration energy}$

Hydration energy of $NaCl=(-389.4-382.3){\text{kJ mol}}^{-1}$
$=-771.7{\text{kJ mol}}^{-1}$

$\Delta {\text{H}}_{\text{sol}}=776-771.7=4.3{\text{kJ mol}}^{-1}$