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To calculate the equilibrium constant \( K_p \) for this reaction, we use the general formula for the equilibrium constant in terms of partial pressures:

\[
K_p = \frac{(P_{\text{products}})^{\text{coefficients}}}{(P_{\text{reactants}})^{\text{coefficients}}}
\]

For the reaction:

\[
4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \rightarrow 4 \text{NH}_3(g) + 5 \text{O}_2(g)
\]

The expression for \( K_p \) is:

\[
K_p = \frac{(P_{\text{NH}_3})^4 \times (P_{\text{O}_2})^5}{(P_{\text{NO}})^4 \times (P_{\text{H}_2\text{O}})^6}
\]

From the table in the image, the equilibrium partial pressures of the compounds are:

- \( P_{\text{NO}} = 11.2 \, \text{atm} \)
- \( P_{\text{H}_2\text{O}} = 48.4 \, \text{atm} \)
- \( P_{\text{NH}_3} = 3.01 \, \text{atm} \)
- \( P_{\text{O}_2} = 66.4 \, \text{atm} \)

Substitute these values into the \( K_p \) expression:

\[
K_p = \frac{(3.01)^4 \times (66.4)^5}{(11.2)^4 \times (48.4)^6}
\]

Now let's calculate this step by step.

1. \( (3.01)^4 = 82.0 \)
2. \( (66.4)^5 = 1.47 \times 10^9 \)
3. \( (11.2)^4 = 15,746 \)
4. \( (48.4)^6 = 2.12 \times 10^9 \)

Now, multiply the numerators and the denominators:

Numerator:
\[
82.0 \times 1.47 \times 10^9 = 1.21 \times 10^{11}
\]

Denominator:
\[
15,746 \times 2.12 \times 10^9 = 3.34 \times 10^{13}
\]

Now divide:

\[
K_p = \frac{1.21 \times 10^{11}}{3.34 \times 10^{13}} = 3.6 \times 10^{-3}
\]

Thus, the equilibrium constant \( K_p \) is approximately:

\[
K_p = 3.6 \times 10^{-3}
\]

Would you like more details or have any questions?

### Related Questions:
1. How is \( K_p \) related to \( K_c \) in terms of the ideal gas law?
2. Why are partial pressures used in equilibrium constant calculations for gases?
3. What conditions affect the value of \( K_p \) in a chemical reaction?
4. How does changing temperature affect \( K_p \) for an exothermic reaction like this one?
5. Can you explain why ammonia production is favored or disfavored at high pressure?

**Tip:** Always ensure the units of pressure are consistent when calculating \( K_p \). Here, they are in atmospheres, which is appropriate for gas-phase equilibria.

Calculate the value of the equilibrium constant K_p for the reaction: 4NO(g) + 6H2O(g) -> 4NH3(g) + 5O2(g), given the equilibrium pressures of NO (11.2 atm), H2O (48.4 atm), NH3 (3.01 atm), and O2 (66.4 atm).

Learn how to calculate the equilibrium constant K_p for the reaction between nitrogen monoxide (NO) and water vapor (H2O) to form ammonia (NH3) and oxygen (O2), using the given partial pressures. Includes step-by-step calculation with relevant formulas.

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