Question

The free energy change when 1 mole of $NaCl$ is dissolved in water at 298 K is $-xkJ$. Find the value of 'x'.
Given:-
$\text{Lattice energy of}NaCl=778kJmo{l}^{-1}$
$\text{Hydration energy of}NaCl=-775kJmo{l}^{-1}$
$\text{Entropy change (at 298 K) =}40Jmo{l}^{-1}$

• A. x = 9
• B. x = -9
• C. x = 9000
• D. x = -9000

Answer 1

The correct option is A x = 9

For the dissolution of a compound, we have the following relation
$\Delta H_{\text{dissolution}}=\Delta H_{\left(\text{lattice energy}\right)}+\Delta H_{\left(\text{hydration}\right)}$

$\Delta H_{\left(\text{lattice energy}\right)}=778kJ\Delta H_{\left(\text{hydration}\right)}=-775kJ$

Putting the above values, we get;
$\Delta H_{\text{dissolution}}=778-775=3kJ{\text{mol}}^{-1}=3000J{\text{mol}}^{-1}$

Given,
$\Delta S_{\text{dissolution}}=40J{\text{mol}}^{-1}$

Also we know,
$△G=\Delta H-T\Delta S=3000-300\times 40=-9000J$
$\Delta G=-9kJ=-xkJ\to x=9$