Question

Calculate the free energy change when 1 mole of NaCl is dissolved in water at 25°C. Lattice energy of NaCl = 777.8 kJ $mol^{-1}$ and ΔS for dissolution = 0.043 kJ $mol^{-1}$ and hydration energy of NaCl = –774.1 kJ $mol^{-1}$.

Answer 1

Given, when 1mole of NaCl dissolved in water at 25℃;

Lattice energy of NaCl = 777.8 KJ/mol

Hydration energy of NaCl = -774.1 KJ/mol

We know that, ${ \Delta H }_{ Disolution }\quad =\quad Lattice\quad energy\quad +\quad Hydration\quad energy$

$=\quad 777.8\quad { KJ }/{ mol }\quad -\quad 774.1\quad { KJ }/{ mol }$

${ \Delta H }_{ Disolution }\quad =\quad 3.75\quad { KJ }/{ mol }$

We know that, $\Delta G\quad =\quad \Delta H\quad -\quad T\Delta S$

Where, $T = 298 K; \Delta S = 0.043; and\quad \Delta H = 3.755 KJ/mol$

$=\quad 3.7\quad -\quad 298\quad \times \quad 0.043$

$\Delta G\quad =\quad -9.114\quad { KJ }/{ mol }$