Calculate `DeltaG^(@)` for the reaction
`NO(g) iff 1/2 N_(2)(g) + 1/2 O_(2)(g)` if `K = 1.55 xx 10^(15)` at 298 K.

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the reaction

\[ \text{NO(g)} \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \]

given that the equilibrium constant (K) is \( 1.55 \times 10^{15} \) at 298 K, we can use the following equation that relates ΔG° to K:

\[ \Delta G° = -RT \ln K \]

Where:
- R is the universal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (298 K)
- K is the equilibrium constant

### Step-by-Step Solution:

1. **Identify the values**:
- R = 8.314 J/(mol·K)
- T = 298 K
- K = \( 1.55 \times 10^{15} \)

2. **Convert K to natural logarithm**:
We need to calculate \( \ln K \):
\[ \ln(1.55 \times 10^{15}) = \ln(1.55) + \ln(10^{15}) = \ln(1.55) + 15 \ln(10) \]
Using \( \ln(10) \approx 2.303 \):
\[ \ln(1.55) \approx 0.439 \]
Therefore,
\[ \ln(1.55 \times 10^{15}) \approx 0.439 + 15 \times 2.303 \approx 0.439 + 34.545 = 34.984 \]

3. **Substitute values into the ΔG° equation**:
\[ \Delta G° = -RT \ln K = - (8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times (34.984) \]

4. **Calculate ΔG°**:
\[ \Delta G° = - (8.314 \times 298 \times 34.984) \approx - 86,700 \, \text{J/mol} \approx - 86.7 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G° \approx -86.7 \, \text{kJ/mol} \]