Steam at pressure of 2 atm is passed through a furnace at 2000 K wherein the reaction `H_(2)O(g) hArr H_(2)(g)+(1)/(2)O_(2)(g), K_(p)=6.4xx10^(-5)` occurs. The percentage of oxygen in the exit steam would be

Options

A. 0.32
B. 0.08
C. 0.04
D. 0.16

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the reaction and the equilibrium constant expression
The reaction is:
\[ \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \]

The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{H}_2} \cdot P_{\text{O}_2}^{1/2}}{P_{\text{H}_2\text{O}}} \]

### Step 2: Define initial conditions
At the start (t = 0), we have:
- \( P_{\text{H}_2\text{O}} = 2 \, \text{atm} \)
- \( P_{\text{H}_2} = 0 \)
- \( P_{\text{O}_2} = 0 \)

### Step 3: Define changes at equilibrium
Let \( X \) be the degree of dissociation of water vapor. At equilibrium:
- \( P_{\text{H}_2\text{O}} = 2 - X \)
- \( P_{\text{H}_2} = X \)
- \( P_{\text{O}_2} = \frac{X}{2} \)

### Step 4: Write the total pressure at equilibrium
The total pressure at equilibrium is:
\[ P_{\text{total}} = P_{\text{H}_2\text{O}} + P_{\text{H}_2} + P_{\text{O}_2} = (2 - X) + X + \frac{X}{2} = 2 + \frac{X}{2} \]

### Step 5: Substitute into the equilibrium expression
Substituting the pressures into the \( K_p \) expression:
\[ 6.4 \times 10^{-5} = \frac{X \cdot \left(\frac{X}{2}\right)^{1/2}}{2 - X} \]

### Step 6: Simplify the equation
Since \( K_p \) is very small, we can assume that \( X \) is very small compared to 2, so \( 2 - X \approx 2 \):
\[ 6.4 \times 10^{-5} = \frac{X \cdot \left(\frac{X}{2}\right)^{1/2}}{2} \]
\[ 6.4 \times 10^{-5} = \frac{X \cdot \frac{\sqrt{X}}{\sqrt{2}}}{2} \]
\[ 6.4 \times 10^{-5} = \frac{X^{3/2}}{2\sqrt{2}} \]

### Step 7: Solve for \( X \)
Rearranging gives:
\[ X^{3/2} = 6.4 \times 10^{-5} \cdot 2\sqrt{2} \]
Calculating \( 2\sqrt{2} \approx 2.828 \):
\[ X^{3/2} = 6.4 \times 10^{-5} \cdot 2.828 \approx 1.81 \times 10^{-4} \]
Taking the cube root:
\[ X \approx (1.81 \times 10^{-4})^{2/3} \approx 0.0032 \]

### Step 8: Calculate the partial pressure of \( O_2 \)
The partial pressure of \( O_2 \) is:
\[ P_{\text{O}_2} = \frac{X}{2} = \frac{0.0032}{2} = 0.0016 \, \text{atm} \]

### Step 9: Calculate the total pressure at equilibrium
The total pressure at equilibrium is:
\[ P_{\text{total}} = 2 + \frac{X}{2} \approx 2 \, \text{atm} \]

### Step 10: Calculate the percentage of \( O_2 \) in the exit steam
The percentage of \( O_2 \) is given by:
\[ \text{Percentage of } O_2 = \left( \frac{P_{\text{O}_2}}{P_{\text{total}}} \right) \times 100 = \left( \frac{0.0016}{2} \right) \times 100 \approx 0.08\% \]

### Final Answer
The percentage of oxygen in the exit steam is approximately **0.08%**.

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