The standard Gibbs free energy change `(DeltaG^(@))` at `25^(@)C` for the dissociation of `N_(2)O_(4)(g)` to `NO_(2)(g)` is (given, equilibrium constant = 0.15, R=8.314 `JK^(-) mol^(-1)`)

Options

A. 1.1 KJ
B. 4.7 KJ
C. 8.1 KJ
D. 38.2 KJ

Explanation

To find the standard Gibbs free energy change `(ΔG°)` for the dissociation of `N₂O₄(g)` to `NO₂(g)`, we can use the relationship between Gibbs free energy change and the equilibrium constant:

\[ ΔG° = -RT \ln K \]

where:
- \( R \) is the universal gas constant (8.314 J/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step-by-Step Solution:

1. **Convert the temperature to Kelvin**:
\[ T = 25°C + 273.15 = 298.15 K \]

2. **Substitute the values into the formula**:
Given that \( K = 0.15 \), we can substitute \( R \), \( T \), and \( K \) into the Gibbs free energy equation:
\[ ΔG° = - (8.314 \, \text{J/(K·mol)}) \times (298.15 \, \text{K}) \times \ln(0.15) \]

3. **Calculate \( \ln(0.15) \)**:
\[ \ln(0.15) \approx -1.897 \]

4. **Plug in the value of \( \ln(0.15) \)**:
\[ ΔG° = - (8.314) \times (298.15) \times (-1.897) \]

5. **Calculate the product**:
\[ ΔG° = 8.314 \times 298.15 \times 1.897 \approx 4.7 \, \text{kJ/mol} \]

6. **Final Result**:
\[ ΔG° \approx 4.7 \, \text{kJ/mol} \]

### Conclusion:
The standard Gibbs free energy change for the dissociation of `N₂O₄(g)` to `NO₂(g)` at `25°C` is approximately **4.7 kJ/mol**.