Question

Lattice enthalpy and enthalpy of solution of NaCl are $788\mathrm{kJ}{{\mathrm{mol}}^{–1}}^{}$, and $4\mathrm{kJ}{{\mathrm{mol}}^{–1}}^{}$, respectively. The hydration enthalpy of NaCl is

• A. –780 kJ mol^–1
• B. 784 kJ mol^–1
• C. –784 kJ mol^–1
• D. 780 kJ mol^–1

Answer 1

The correct option is C

–784 kJ mol^–1

The explanation for the correct answer:

(C)–784 kJ mol^–1

Given, that Lattice enthalpy NaCl= $788\mathrm{kJ}{\mathrm{mol}}^{-1}$and enthalpy of solution of NaCl= $4\mathrm{kJ}{{\mathrm{mol}}^{–1}}^{}$

then hydration enthalpy of NaCl is,

$$\begin{gathered} \mathrm{\Delta Hsol}=\mathrm{L}.\mathrm{E}+{\mathrm{\Delta H}}_{\mathrm{hyd}} \\ 4=788+{\mathrm{\Delta H}}_{\mathrm{hyd}} \\ {\mathrm{\Delta H}}_{\mathrm{hyd}}=–784\mathrm{KJ}/\mathrm{mol} \end{gathered}$$

Hence, option (c) is correct.