`K_p` for the reaction: `CO_2(g) + H_2 (g) hArr CO(g) + H_2O(g)` is found to be 16 at a given temperature. Originally equal number of moles of `H_2` and `CO_2` were placed in the flask. At equilibrium, the pressure of `H_2` is 1.20 atm. What is the partial pressure of CO and `H_2O` ?

Options

A. 1.20 atm each
B. 2.40 atm each
C. 4.80 atm each
D. 9.60 atm each

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2O(g) \]

### Step 2: Define the initial conditions
Let the initial number of moles of \( \text{CO}_2 \) and \( \text{H}_2 \) be \( n \). Since they are equal, we can say:
- Initial moles of \( \text{CO}_2 = n \)
- Initial moles of \( \text{H}_2 = n \)

### Step 3: Define the changes at equilibrium
Let \( x \) be the number of moles of \( \text{CO}_2 \) and \( \text{H}_2 \) that react at equilibrium. Therefore, at equilibrium, we have:
- Moles of \( \text{CO}_2 = n - x \)
- Moles of \( \text{H}_2 = n - x \)
- Moles of \( \text{CO} = x \)
- Moles of \( \text{H}_2O = x \)

### Step 4: Relate the pressures to the number of moles
Using the ideal gas law, we know that the pressure is directly proportional to the number of moles when volume and temperature are constant. Therefore, we can express the partial pressures as:
- \( P_{\text{CO}_2} = \frac{(n - x)RT}{V} \)
- \( P_{\text{H}_2} = \frac{(n - x)RT}{V} \)
- \( P_{\text{CO}} = \frac{xRT}{V} \)
- \( P_{\text{H}_2O} = \frac{xRT}{V} \)

Since \( R \) and \( T \) are constants and the volume \( V \) is fixed, we can simplify the pressures to:
- \( P_{\text{CO}_2} = P_{\text{H}_2} = P_0 - x \)
- \( P_{\text{CO}} = P_{\text{H}_2O} = x \)

### Step 5: Use the given pressure of \( H_2 \)
We know that at equilibrium, the pressure of \( H_2 \) is given as \( 1.20 \, \text{atm} \):
\[ P_{\text{H}_2} = 1.20 \, \text{atm} \]
Thus, we have:
\[ P_{\text{CO}_2} = 1.20 \, \text{atm} \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{\text{CO}} \cdot P_{\text{H}_2O}}{P_{\text{CO}_2} \cdot P_{\text{H}_2}} \]

Substituting the pressures we have:
\[ K_p = \frac{x \cdot x}{(1.20)(1.20)} = \frac{x^2}{1.44} \]

### Step 7: Set up the equation using the value of \( K_p \)
Given that \( K_p = 16 \):
\[ 16 = \frac{x^2}{1.44} \]

### Step 8: Solve for \( x \)
Multiplying both sides by \( 1.44 \):
\[ x^2 = 16 \times 1.44 \]
\[ x^2 = 23.04 \]
Taking the square root:
\[ x = \sqrt{23.04} \]
\[ x \approx 4.8 \, \text{atm} \]

### Step 9: Find the partial pressures of \( CO \) and \( H_2O \)
Since \( P_{\text{CO}} = x \) and \( P_{\text{H}_2O} = x \):
- \( P_{\text{CO}} \approx 4.8 \, \text{atm} \)
- \( P_{\text{H}_2O} \approx 4.8 \, \text{atm} \)

### Final Answer
The partial pressures at equilibrium are:
- \( P_{\text{CO}} = 4.8 \, \text{atm} \)
- \( P_{\text{H}_2O} = 4.8 \, \text{atm} \)