Standard enthalpy and standard entropy changes for the oxidation of ammonia at `298 K` are `-382.64 kJ mol^(-1)` and `-145.6 jK^(-1)mol^(-1)` respectively. Standard Gibbs energy change for the same reaction at `298 K` is

Options

A. `-523.2 kJ mol^(-1)`
B. `-22.1 kJ mol^(-1)`
C. `-339.3 kJ mol^(-1)`
D. `-439.3 kJ mol^(-1)`

Explanation

To calculate the standard Gibbs energy change (ΔG°) for the oxidation of ammonia at 298 K, we can use the Gibbs free energy equation:

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

Where:
- ΔG° = standard Gibbs energy change
- ΔH° = standard enthalpy change
- T = temperature in Kelvin
- ΔS° = standard entropy change

### Step 1: Identify the values
From the question, we have:
- ΔH° = -382.64 kJ/mol (we need to convert this to J/mol for consistency with ΔS°)
- ΔS° = -145.6 J/K·mol
- T = 298 K

### Step 2: Convert ΔH° to J/mol
Since ΔH° is given in kJ/mol, we convert it to J/mol:
\[ \Delta H° = -382.64 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -382640 \text{ J/mol} \]

### Step 3: Substitute the values into the Gibbs free energy equation
Now, we can substitute the values into the Gibbs free energy equation:
\[ \Delta G° = -382640 \text{ J/mol} - (298 \text{ K} \times -145.6 \text{ J/K·mol}) \]

### Step 4: Calculate TΔS°
Calculate \( T \Delta S° \):
\[ T \Delta S° = 298 \text{ K} \times -145.6 \text{ J/K·mol} = -43488.8 \text{ J/mol} \]

### Step 5: Substitute TΔS° back into the equation
Now substitute \( T \Delta S° \) back into the Gibbs free energy equation:
\[ \Delta G° = -382640 \text{ J/mol} - (-43488.8 \text{ J/mol}) \]
\[ \Delta G° = -382640 \text{ J/mol} + 43488.8 \text{ J/mol} \]
\[ \Delta G° = -339151.2 \text{ J/mol} \]

### Step 6: Convert ΔG° back to kJ/mol (if needed)
To convert back to kJ/mol:
\[ \Delta G° = -339151.2 \text{ J/mol} \div 1000 = -339.15 \text{ kJ/mol} \]

### Final Answer
The standard Gibbs energy change for the oxidation of ammonia at 298 K is:
\[ \Delta G° = -339.15 \text{ kJ/mol} \]