Standard enthalpy and standard entropy change for the oxidation of `NH_(3)` at `298K` are `-382.64KJ mol^(-1)` and `145.6Jmol^(-1)` respectively. Standard free energy change for the same reaction at `298K` is

Options

A. `-221.1 KJ "mol"^(-1)`
B. `-339.3 KJ "mol"^(-1)`
C. `-439.3 KJ "mol"^(-1)`
D. `-523.2 KJ "mol"^(-1)`

Explanation

To find the standard free energy change (ΔG°) for the oxidation of NH₃ at 298 K, 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 (ΔH°) = -382.64 kJ/mol
- Standard entropy change (ΔS°) = 145.6 J/mol·K
- Temperature (T) = 298 K

2. **Convert ΔS° from J/mol·K to kJ/mol·K:**
- Since we need to keep the units consistent (kJ), we convert ΔS°:
\[ \Delta S° = 145.6 \, \text{J/mol·K} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 0.1456 \, \text{kJ/mol·K} \]

3. **Substitute the values into the Gibbs free energy equation:**
\[ \Delta G° = -382.64 \, \text{kJ/mol} - (298 \, \text{K} \times 0.1456 \, \text{kJ/mol·K}) \]

4. **Calculate the term involving temperature and entropy change:**
\[ 298 \, \text{K} \times 0.1456 \, \text{kJ/mol·K} = 43.3768 \, \text{kJ/mol} \]

5. **Now substitute this value back into the equation:**
\[ \Delta G° = -382.64 \, \text{kJ/mol} - 43.3768 \, \text{kJ/mol} \]

6. **Perform the final calculation:**
\[ \Delta G° = -382.64 \, \text{kJ/mol} - 43.3768 \, \text{kJ/mol} = -426.0168 \, \text{kJ/mol} \]

7. **Round the answer to three significant figures:**
\[ \Delta G° \approx -426.02 \, \text{kJ/mol} \]

### Final Answer:
The standard free energy change (ΔG°) for the oxidation of NH₃ at 298 K is approximately **-426.02 kJ/mol**.