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Let's break down the problem step by step to determine the equilibrium constants \( K_c \) and \( K_p \) for the given reaction.

### Reaction:
\[
4 \text{Fe}(s) + 3 \text{O}_2(g) \rightleftharpoons 2 \text{Fe}_2\text{O}_3(s)
\]

### Given Data:
- The reaction occurs in a 2.0 L vessel.
- At equilibrium, the mixture contains:
  - \( 1.2 \, \text{mol Fe} \)
  - \( 1.0 \times 10^{-3} \, \text{mol O}_2 \)
  - \( 2.0 \, \text{mol Fe}_2\text{O}_3 \)
- The temperature is \( 2000.0^\circ C \).

### Step 1: Write the Expression for \( K_c \)

The equilibrium constant \( K_c \) for the reaction only includes gaseous species (since solids like Fe and Fe₂O₃ are not included in the expression). Thus, \( K_c \) is written in terms of the concentration of the gaseous species \( \text{O}_2 \) (oxygen).

\[
K_c = \frac{[\text{Fe}_2\text{O}_3]^2}{[\text{O}_2]^3}
\]

However, both Fe and Fe₂O₃ are solids, so they do not appear in the expression for \( K_c \). The final expression for \( K_c \) is:

\[
K_c = \frac{1}{[\text{O}_2]^3}
\]

### Step 2: Calculate Concentration of \( \text{O}_2 \)

The concentration of \( \text{O}_2 \) is found by dividing the moles of \( \text{O}_2 \) by the volume of the reaction vessel:

\[
[\text{O}_2] = \frac{\text{moles of } \text{O}_2}{\text{volume of vessel}} = \frac{1.0 \times 10^{-3} \, \text{mol}}{2.0 \, \text{L}} = 5.0 \times 10^{-4} \, \text{mol/L}
\]

### Step 3: Solve for \( K_c \)

Now, substitute the concentration of \( \text{O}_2 \) into the \( K_c \) expression:

\[
K_c = \frac{1}{(5.0 \times 10^{-4})^3}
\]

\[
K_c = \frac{1}{1.25 \times 10^{-10}} = 8.0 \times 10^9
\]

### Step 4: Use \( K_c \) to Find \( K_p \)

The relationship between \( K_c \) and \( K_p \) is given by the equation:

\[
K_p = K_c \cdot (RT)^{\Delta n}
\]

Where:
- \( R \) is the gas constant (0.0821 L·atm/mol·K),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in moles of gas between products and reactants.

For this reaction:
\[
\Delta n = (0 \text{ moles of gas on the product side}) - (3 \text{ moles of gas on the reactant side}) = -3
\]

First, convert the temperature to Kelvin:

\[
T = 2000.0^\circ C + 273.15 = 2273.15 \, \text{K}
\]

Now, calculate \( K_p \):

\[
K_p = K_c \cdot (RT)^{\Delta n}
\]

Substitute the values:

\[
K_p = 8.0 \times 10^9 \cdot (0.0821 \times 2273.15)^{-3}
\]

\[
K_p = 8.0 \times 10^9 \cdot (186.59415)^{-3}
\]

\[
K_p = 8.0 \times 10^9 \cdot 1.537 \times 10^{-6}
\]

\[
K_p \approx 1.23 \times 10^4
\]

### Final Answers:
- \( K_c = 8.0 \times 10^9 \)
- \( K_p \approx 1.23 \times 10^4 \)

Would you like further details or have any questions about this process? Here are 5 related questions to expand your understanding:

1. How does the presence of solids affect the equilibrium constant expression?
2. Why is the \( \Delta n \) important in converting between \( K_c \) and \( K_p \)?
3. How does temperature affect the equilibrium constant \( K_p \)?
4. What assumptions do we make about ideal gases when using the ideal gas law in equilibrium calculations?
5. Could you solve for \( K_p \) if \( \Delta n \) were zero? Why?

**Tip:** Always check the units and ensure they cancel out correctly in equilibrium constant calculations!

Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 2.0 L reaction vessel contained a mixture of 1.2 mol Fe, 1.0 x 10^-3 mol O2, and 2.0 mol of Fe2O3 at 2000.0 °C. What are the values of Kc and Kp for this reaction? 4 Fe(s) + 3 O2(g) ⇄ 2 Fe2O3(s)

Learn how to calculate the equilibrium constants Kc and Kp for a reaction involving Fe, O2, and Fe2O3. This example includes key formulas for Kc and Kp, stoichiometric considerations, and an analysis of gas behavior at high temperatures. Suitable for high school chemistry students.

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