Calculate `DeltaG^@` for the reaction
`NO hArr 1/2N_2(g) +1/2O_2(g)`
If `K=1.55xx10^15` at 298 K

Explanation

To calculate \(\Delta G^\circ\) for the reaction

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

given that \(K = 1.55 \times 10^{15}\) at 298 K, we can use the following formula:

\[ \Delta G^\circ = -2.303 \cdot R \cdot T \cdot \log K \]

where:
- \(R\) is the universal gas constant, which is \(8.314 \, \text{J/(mol K)}\),
- \(T\) is the temperature in Kelvin, which is \(298 \, \text{K}\),
- \(K\) is the equilibrium constant.

### Step-by-Step Solution:

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

2. **Calculate \(\log K\)**:
\[ \log K = \log(1.55 \times 10^{15}) \approx \log(1.55) + \log(10^{15}) = 0.190 + 15 = 15.190 \]

3. **Substitute the values into the \(\Delta G^\circ\) equation**:
\[ \Delta G^\circ = -2.303 \cdot 8.314 \cdot 298 \cdot 15.190 \]

4. **Calculate the product**:
- First calculate \(2.303 \cdot 8.314 \cdot 298\):
\[ 2.303 \cdot 8.314 \cdot 298 \approx 5730.57 \]

5. **Now multiply by \(\log K\)**:
\[ \Delta G^\circ = -5730.57 \cdot 15.190 \approx -86973.72 \, \text{J/mol} \]

6. **Convert to kJ/mol**:
\[ \Delta G^\circ = -86973.72 \, \text{J/mol} \div 1000 = -86.97372 \, \text{kJ/mol} \approx -86.97 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G^\circ \approx -86.97 \, \text{kJ/mol} \]