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 of determining the percentage of oxygen in the exit steam when steam at a pressure of 2 atm is passed through a furnace at 2000 K, we will follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The reaction is given as:
\[ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g) \]

Initially, we have:
- Pressure of \( H_2O(g) = 2 \, \text{atm} \)
- Pressure of \( H_2(g) = 0 \, \text{atm} \)
- Pressure of \( O_2(g) = 0 \, \text{atm} \)

### Step 2: Define Changes at Equilibrium
Let \( x \) be the change in pressure due to the dissociation of \( H_2O \) at equilibrium. Therefore, at equilibrium:
- Pressure of \( H_2O(g) = 2 - x \)
- Pressure of \( H_2(g) = x \)
- Pressure of \( O_2(g) = \frac{x}{2} \)

### Step 3: Write the Expression for Total Pressure
The total pressure at equilibrium can be expressed as:
\[ P_{\text{total}} = (2 - x) + x + \frac{x}{2} = 2 + \frac{x}{2} \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given as:
\[ K_p = \frac{(P_{H_2}) (P_{O_2})^{1/2}}{P_{H_2O}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{x \left(\frac{x}{2}\right)^{1/2}}{(2 - x)} \]

### Step 5: Substitute \( K_p \) Value
Given \( K_p = 6.4 \times 10^{-5} \):
\[ 6.4 \times 10^{-5} = \frac{x \left(\frac{x}{2}\right)^{1/2}}{(2 - x)} \]

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

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

### Step 8: Calculate the Pressure of Oxygen
The pressure of oxygen at equilibrium is:
\[ P_{O_2} = \frac{x}{2} = \frac{0.0032}{2} = 0.0016 \, \text{atm} \]

### Step 9: Calculate Total Pressure
The total pressure at equilibrium is:
\[ P_{\text{total}} = 2 + \frac{x}{2} \approx 2 + 0.0016 \approx 2 \, \text{atm} \]

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

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