Calculate the free energy change in dissolving one mole of sodium chloride at `25^(@)C` . Lattic energy `= + 777.8 kJ mol^(-1)`. Hydration energy of `NaCl = - 774.1 kJ mol^(-1)` and `Delta S ` at `25^(@)C = 0.0 43kJ K^(-1) mol^(-1)`

Explanation

To calculate the free energy change (ΔG) for dissolving one mole of sodium chloride (NaCl) at 25°C, we will follow these steps:

### Step 1: Identify Given Data
- Lattice energy (ΔH_lattice) = +777.8 kJ/mol
- Hydration energy (ΔH_hydration) = -774.1 kJ/mol
- Temperature (T) = 25°C = 298 K
- ΔS (entropy change) = 0.043 kJ/K·mol

### Step 2: Calculate Enthalpy of Solution (ΔH_solution)
The enthalpy of solution can be calculated using the formula:
\[ \Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}} \]
Substituting the values:
\[ \Delta H_{\text{solution}} = 777.8 \, \text{kJ/mol} + (-774.1 \, \text{kJ/mol}) \]
\[ \Delta H_{\text{solution}} = 777.8 - 774.1 = 3.7 \, \text{kJ/mol} \]

### Step 3: Calculate Free Energy Change (ΔG)
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 3.7 \, \text{kJ/mol} - (298 \, \text{K} \times 0.043 \, \text{kJ/K·mol}) \]
Calculating \(T \Delta S\):
\[ T \Delta S = 298 \times 0.043 = 12.794 \, \text{kJ/mol} \]
Now substituting back into the ΔG equation:
\[ \Delta G = 3.7 \, \text{kJ/mol} - 12.794 \, \text{kJ/mol} \]
\[ \Delta G = 3.7 - 12.794 = -9.094 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta G \approx -9.094 \, \text{kJ/mol} \]

---