One of the reaction that takes plece in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and `CO_(2)`.
`FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g), K_(p)=0.265` atm at `1050 K`
What are the equilibrium partial pressure of `CO` and `CO_(2)` at `1050 K` if the partical pressure are: `p_(CO)=1.4 atm` and `p_(CO_(2))=0.80 atm`?

Explanation

To solve the problem, we need to find the equilibrium partial pressures of CO and CO₂ at 1050 K given the initial partial pressures and the equilibrium constant \( K_p \).

### Step 1: Write the balanced chemical equation and the expression for \( K_p \)

The reaction is:
\[ \text{FeO(s)} + \text{CO(g)} \rightleftharpoons \text{Fe(s)} + \text{CO}_2(g) \]

The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{CO}_2}}{P_{\text{CO}}} \]

### Step 2: Calculate the reaction quotient \( Q_p \)

Given:
- \( P_{\text{CO}} = 1.4 \, \text{atm} \)
- \( P_{\text{CO}_2} = 0.80 \, \text{atm} \)

Calculate \( Q_p \):
\[ Q_p = \frac{P_{\text{CO}_2}}{P_{\text{CO}}} = \frac{0.80}{1.4} \approx 0.571 \]

### Step 3: Compare \( Q_p \) with \( K_p \)

Given:
- \( K_p = 0.265 \, \text{atm} \)

Since \( Q_p (0.571) > K_p (0.265) \), the reaction will shift to the left (backward direction) to reach equilibrium.

### Step 4: Set up the equilibrium expression

Let \( x \) be the change in pressure at equilibrium. The equilibrium pressures will be:
- For CO: \( P_{\text{CO}} = 1.4 + x \)
- For CO₂: \( P_{\text{CO}_2} = 0.80 - x \)

At equilibrium, we can write:
\[ K_p = \frac{P_{\text{CO}_2}}{P_{\text{CO}}} = \frac{0.80 - x}{1.4 + x} \]

### Step 5: Substitute \( K_p \) and solve for \( x \)

Substituting the value of \( K_p \):
\[ 0.265 = \frac{0.80 - x}{1.4 + x} \]

Cross-multiplying gives:
\[ 0.265(1.4 + x) = 0.80 - x \]

Expanding and rearranging:
\[ 0.371 + 0.265x = 0.80 - x \]
\[ 0.265x + x = 0.80 - 0.371 \]
\[ 1.265x = 0.429 \]
\[ x \approx 0.339 \, \text{atm} \]

### Step 6: Calculate the equilibrium partial pressures

Now, substitute \( x \) back into the expressions for \( P_{\text{CO}} \) and \( P_{\text{CO}_2} \):
- \( P_{\text{CO}} = 1.4 + 0.339 = 1.739 \, \text{atm} \)
- \( P_{\text{CO}_2} = 0.80 - 0.339 = 0.461 \, \text{atm} \)

### Final Answer

The equilibrium partial pressures at 1050 K are:
- \( P_{\text{CO}} \approx 1.739 \, \text{atm} \)
- \( P_{\text{CO}_2} \approx 0.461 \, \text{atm} \)

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