Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are `-582.64 kJ mol^(-1) and -150.6 J mol^(-1) k^-1` , respectively. Standard Gibb's energy change for the same reaction at 298 K is

Options

A. `-221.1 kJ mol^(-1)`
B. `-537.7 kJ mol^(-1)`
C. `-439.3 kJ mol^(-1)`
D. `-523.2 kJ mol^(-1)`

Answer

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

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

### Step-by-step Solution:

1. **Identify the given values**:
- Standard enthalpy change (ΔH) = -582.64 kJ/mol
- Standard entropy change (ΔS) = -150.6 J/(mol·K)
- Temperature (T) = 298 K

2. **Convert ΔS from J/(mol·K) to kJ/(mol·K)**:
Since ΔH is given in kJ, we need to convert ΔS to the same unit:
\[ \Delta S = -150.6 \, \text{J/(mol·K)} = \frac{-150.6}{1000} \, \text{kJ/(mol·K)} = -0.1506 \, \text{kJ/(mol·K)} \]

3. **Substitute the values into the Gibbs energy equation**:
Now we can substitute the values into the Gibbs energy equation:
\[ \Delta G = -582.64 \, \text{kJ/mol} - (298 \, \text{K} \times -0.1506 \, \text{kJ/(mol·K)}) \]

4. **Calculate the second term**:
Calculate \(T \Delta S\):
\[ T \Delta S = 298 \times -0.1506 = -44.8788 \, \text{kJ/mol} \]
Since we have a negative sign in front of ΔS, this will be positive when subtracted.

5. **Combine the terms**:
Now, we can calculate ΔG:
\[ \Delta G = -582.64 \, \text{kJ/mol} + 44.8788 \, \text{kJ/mol} \]
\[ \Delta G = -582.64 + 44.8788 = -537.7612 \, \text{kJ/mol} \]

6. **Round to appropriate significant figures**:
Rounding to two decimal places, we get:
\[ \Delta G \approx -537.76 \, \text{kJ/mol} \]

### Final Answer:
The standard Gibbs energy change for the oxidation of ammonia at 298 K is approximately \(-537.76 \, \text{kJ/mol}\).

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