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The problem asks you to calculate the equilibrium constant \( K_p \) for the given reaction at equilibrium:

\[
4 \, \text{HCl}(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{Cl}_2(g) + 2 \, \text{H}_2\text{O}(g)
\]

The pressures of the gases at equilibrium are provided:

- \( \text{HCl} \): 72.5 atm
- \( \text{O}_2 \): 57.3 atm
- \( \text{Cl}_2 \): 34.1 atm
- \( \text{H}_2\text{O} \): 61.6 atm

The equilibrium constant \( K_p \) is calculated using the expression:

\[
K_p = \frac{(P_{\text{Cl}_2})^2 (P_{\text{H}_2\text{O}})^2}{(P_{\text{HCl}})^4 (P_{\text{O}_2})}
\]

Where:
- \( P_{\text{Cl}_2} \) is the equilibrium pressure of chlorine
- \( P_{\text{H}_2\text{O}} \) is the equilibrium pressure of water vapor
- \( P_{\text{HCl}} \) is the equilibrium pressure of hydrogen chloride
- \( P_{\text{O}_2} \) is the equilibrium pressure of oxygen

Substitute the given pressures:

\[
K_p = \frac{(34.1)^2 \times (61.6)^2}{(72.5)^4 \times (57.3)}
\]

Let's calculate:

1. \( (34.1)^2 = 1162.81 \)
2. \( (61.6)^2 = 3794.56 \)
3. \( (72.5)^4 = 27631406.25 \)
4. Multiply the numerator: \( 1162.81 \times 3794.56 = 4412280.61 \)
5. Multiply the denominator: \( 27631406.25 \times 57.3 = 1583533554.625 \)
6. Now divide: 

\[
K_p = \frac{4412280.61}{1583533554.625} \approx 0.002785
\]

Rounded to two significant digits:

\[
K_p \approx 0.0028
\]

Let me know if you'd like any clarifications or further details!

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Here are 5 related questions:
1. How does the equilibrium constant \( K_p \) change with temperature?
2. What is the difference between \( K_c \) and \( K_p \)?
3. How would adding more \( \text{O}_2 \) affect the equilibrium of this reaction?
4. Why are pressure values used instead of concentrations in this \( K_p \) calculation?
5. How does Le Chatelier's principle apply to this reaction?

**Tip**: Always double-check units when calculating equilibrium constants, as incorrect units can lead to wrong results.

Hydrogen chloride and oxygen react to form chlorine and water: 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g). A chemist finds that at a certain temperature the equilibrium mixture has the following pressures: HCl = 72.5 atm, O2 = 57.3 atm, Cl2 = 34.1 atm, H2O = 61.6 atm. Calculate the equilibrium constant Kp.

This example shows how to calculate the equilibrium constant Kp for the reaction 4HCl(g) + O2(g) ⇌ 2Cl2(g) + 2H2O(g), given equilibrium partial pressures of HCl (72.5 atm), O2 (57.3 atm), Cl2 (34.1 atm), and H2O (61.6 atm). Suitable for undergraduate chemistry students.

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