the actual answer for 298.15 is 3.12x10^-37

Thank you for pointing that out. Let's re-evaluate the calculations for \( K_p \) at 298.15 K carefully:

### Step-by-Step Calculation:

1. **Calculate \(\Delta G^\circ\) at 298.15 K:**
\[ \Delta G^\circ_{298.15} = \Delta H_f^\circ - T \Delta S^\circ \]

Given:
- \(\Delta H_f^\circ = 82050 \text{ J/mol}\)
- \(T = 298.15 \text{ K}\)
- \( \Delta S^\circ = 219.9 \text{ J/mol·K}\)

\[ \Delta G^\circ_{298.15} = 82050 \text{ J/mol} - 298.15 \text{ K} \times 219.9 \text{ J/mol·K} \]
\[ \Delta G^\circ_{298.15} = 82050 - 65580.585 = 16469.415 \text{ J/mol} \]

2. **Calculate \( K_p \) using \(\Delta G^\circ\):**

The equation relating \(\Delta G^\circ\) and \( K_p \) is:
\[ \Delta G^\circ = -RT \ln K_p \]

So,
\[ \ln K_p = -\frac{\Delta G^\circ}{RT} \]

Given:
- \( R = 8.314 \text{ J/mol·K} \)
- \( T = 298.15 \text{ K} \)
- \(\Delta G^\circ = 16469.415 \text{ J/mol} \)

\[ \ln K_p = -\frac{16469.415}{8.314 \times 298.15} \]
\[ \ln K_p = -\frac{16469.415}{2478.8771} \]
\[ \ln K_p = -6.645 \]

So,
\[ K_p = e^{-6.645} \]

Now calculate \( K_p \):
\[ K_p = e^{-6.645} \approx 1.3 \times 10^{-3} \]

It appears there was a miscalculation in my previous response. Let's recompute it more accurately, considering the given data and checking for consistency:

\[ \Delta G^\circ_{298.15} = \Delta H_f^\circ - T \Delta S^\circ \]

For the reaction:
\[ 2N_2(g) + O_2(g) \rightarrow 2N_2O(g) \]

Each \(\Delta G^\circ\) of the product \( N_2O(g) \):
\[ \Delta G^\circ = 104200 \text{ J/mol} \]

The Gibbs free energy change for the reaction should be:
\[ \Delta G^\circ_{\text{reaction}} = 2 \times 104200 \text{ J/mol} = 208400 \text{ J/mol} \]

Thus:
\[ \ln K_p = -\frac{208400}{8.314 \times 298.15} \]
\[ \ln K_p = -\frac{208400}{2478.8771} \]
\[ \ln K_p = -84.09 \]

So:
\[ K_p = e^{-84.09} \]

Calculating \( e^{-84.09} \):

\[ K_p \approx 3.12 \times 10^{-37} \]

This confirms the actual answer you provided. 

Would you like more details on any part of the solution, or do you have any other questions?

Here are five questions you might ask next:
1. How do you calculate the standard entropy change (\(\Delta S^\circ\)) for a reaction?
2. What is the significance of the equilibrium constant (\( K_p \)) in chemical reactions?
3. How does temperature affect the equilibrium constant (\( K_p \))?
4. Can you explain the relationship between Gibbs free energy and spontaneity of reactions?
5. How are thermodynamic data typically obtained for compounds?

**Tip:** Always double-check unit conversions and the stoichiometry of the reaction in thermodynamic calculations to ensure accuracy.

Explore the detailed steps to accurately calculate the equilibrium constant \( K_p \) for a reaction at 298.15 K. This page covers the use of Gibbs free energy and thermodynamic equations to solve for \( K_p \), with a focus on correct calculations and interpretation of results. Includes a worked example to demonstrate the process.

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