For the reaction,`SO_(2)(g)+1//2O_(2)(g)hArrSO_(3)(g)`
`DeltaH_(298)^(@)=-98.32KJ//mol e,DeltaS_(298)^(@)=-95.0J//mol e-K`. Find the `K_(p)` for this reaction at `298K. `

Answer

To find the equilibrium constant \( K_p \) for the reaction

\[ SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g) \]

at 298 K, we will use the Gibbs free energy change (\( \Delta G^\circ \)) and its relation to \( K_p \).

### Step-by-step Solution:

1. **Identify the Given Values**:
- \( \Delta H^\circ_{298} = -98.32 \, \text{kJ/mol} \)
- \( \Delta S^\circ_{298} = -95.0 \, \text{J/mol K} \)

2. **Convert \( \Delta H^\circ \) to Joules**:
Since \( \Delta H^\circ \) is given in kJ, we convert it to J:
\[ \Delta H^\circ = -98.32 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -98320 \, \text{J/mol} \]

3. **Use the Gibbs Free Energy Equation**:
The Gibbs free energy change is given by:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Plugging in the values:
\[ \Delta G^\circ = -98320 \, \text{J/mol} - 298 \, \text{K} \times (-95.0 \, \text{J/mol K}) \]

4. **Calculate \( T \Delta S^\circ \)**:
\[ T \Delta S^\circ = 298 \times (-95.0) = -28310 \, \text{J/mol} \]

5. **Calculate \( \Delta G^\circ \)**:
\[ \Delta G^\circ = -98320 + 28310 = -70010 \, \text{J/mol} \]

6. **Relate \( \Delta G^\circ \) to \( K_p \)**:
The relationship between \( \Delta G^\circ \) and \( K_p \) is given by:
\[ \Delta G^\circ = -RT \ln K_p \]
Rearranging gives:
\[ \ln K_p = -\frac{\Delta G^\circ}{RT} \]

7. **Substituting Values**:
Using \( R = 8.314 \, \text{J/mol K} \) and \( T = 298 \, \text{K} \):
\[ \ln K_p = -\frac{-70010}{8.314 \times 298} \]

8. **Calculate \( \ln K_p \)**:
\[ \ln K_p = \frac{70010}{2477.572} \approx 28.24 \]

9. **Calculate \( K_p \)**:
To find \( K_p \), we take the exponential:
\[ K_p = e^{28.24} \approx 1.8 \times 10^{12} \]

### Final Answer:
\[ K_p \approx 1.8 \times 10^{12} \]