The standard enthalpy or the decomposition of `N_2 O_5 `to `NO_2` is 58.04 kJ and standard entropy of this reaction is 176.7 J/K. The standard free energy change for this reaction at `25^@`C is_____________.

Options

A. 539 kJ
B. `-539 kJ`
C. `-5.39 kJ`
D. 5.39 kJ

Answer

To find the standard free energy change (ΔG°) for the decomposition of \( N_2O_5 \) to \( NO_2 \), we can use the Gibbs free energy equation:

\[ \Delta G° = \Delta H° - T \Delta S° \]

### Step-by-step solution:

1. **Identify the given values**:
- Standard enthalpy change (\( \Delta H° \)) = 58.04 kJ
- Standard entropy change (\( \Delta S° \)) = 176.7 J/K
- Temperature (T) = 25°C

2. **Convert temperature to Kelvin**:
\[ T = 25°C + 273 = 298 \text{ K} \]

3. **Convert the entropy change from J/K to kJ/K**:
Since \( \Delta S° \) is given in J/K, we need to convert it to kJ/K:
\[ \Delta S° = 176.7 \text{ J/K} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 0.1767 \text{ kJ/K} \]

4. **Substitute the values into the Gibbs free energy equation**:
\[ \Delta G° = \Delta H° - T \Delta S° \]
\[ \Delta G° = 58.04 \text{ kJ} - (298 \text{ K} \times 0.1767 \text{ kJ/K}) \]

5. **Calculate \( T \Delta S° \)**:
\[ T \Delta S° = 298 \times 0.1767 = 52.6886 \text{ kJ} \]

6. **Calculate \( \Delta G° \)**:
\[ \Delta G° = 58.04 \text{ kJ} - 52.6886 \text{ kJ} = 5.3514 \text{ kJ} \]

7. **Round the answer**:
\[ \Delta G° \approx 5.35 \text{ kJ} \]

### Final Answer:
The standard free energy change for the reaction at \( 25°C \) is approximately **5.35 kJ**.