A 10 litre box contains `O_3` and `O_2` at equilibrium at 2000 K. `K_p=4xx10^(14)` atm for `2O_3(g)hArr 3O_2(g)`
Assume that `P_(O_2)gtgtP_(O_3)` and if total pressure is 8 atm, then partial pressure of `O_3` will be :

Options

A. `8xx10^(-5)` atm
B. `11.3xx10^(-7)` atm
C. `9.71xx10^(-6)` atm
D. `9.71xx10^(-2)` atm

Answer

To find the partial pressure of \( O_3 \) in a 10-litre box containing \( O_3 \) and \( O_2 \) at equilibrium, we can follow these steps:

### Step 1: Understand the Reaction and Given Information
The equilibrium reaction is:
\[ 2 O_3(g) \rightleftharpoons 3 O_2(g) \]

We are given:
- \( K_p = 4 \times 10^{14} \) atm
- Total pressure \( P_{total} = 8 \) atm
- \( P_{O_2} \gg P_{O_3} \)

### Step 2: Set Up the Equation for Total Pressure
Since the pressure of \( O_2 \) is much greater than that of \( O_3 \), we can approximate:
\[ P_{total} \approx P_{O_2} \]
Thus, we can assume:
\[ P_{O_2} \approx 8 \, \text{atm} \]
And consequently, the pressure of \( O_3 \) can be considered negligible.

### Step 3: Write the Expression for \( K_p \)
The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]

Substituting the known values:
\[ 4 \times 10^{14} = \frac{(8)^3}{(P_{O_3})^2} \]

### Step 4: Calculate \( P_{O_2}^3 \)
Calculate \( 8^3 \):
\[ 8^3 = 512 \]

### Step 5: Substitute and Rearrange the Equation
Now we can substitute this back into the equation:
\[ 4 \times 10^{14} = \frac{512}{(P_{O_3})^2} \]

Rearranging gives:
\[ (P_{O_3})^2 = \frac{512}{4 \times 10^{14}} \]

### Step 6: Calculate \( (P_{O_3})^2 \)
Calculating the right side:
\[ (P_{O_3})^2 = \frac{512}{4} \times 10^{-14} = 128 \times 10^{-14} = 1.28 \times 10^{-12} \]

### Step 7: Take the Square Root to Find \( P_{O_3} \)
Taking the square root gives:
\[ P_{O_3} = \sqrt{1.28 \times 10^{-12}} \]

Calculating this:
\[ P_{O_3} \approx 1.13 \times 10^{-6} \, \text{atm} \]

### Final Answer
Thus, the partial pressure of \( O_3 \) is approximately:
\[ P_{O_3} \approx 1.13 \times 10^{-6} \, \text{atm} \]

---