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^(-6)`
B. `22.62xx10^(-7)`
C. `9.71xx10^(-6)`
D. `9.71xx10^(-2)`

Explanation

To solve the problem, we need to find the partial pressure of \( O_3 \) at equilibrium given the total pressure and the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Write the Reaction and Equilibrium Expression**:
The given reaction is:
\[ 2 O_3(g) \rightleftharpoons 3 O_2(g) \]
The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]

2. **Given Values**:
- \( K_p = 4 \times 10^{14} \, \text{atm} \)
- Total pressure \( P_{total} = P_{O_2} + P_{O_3} = 8 \, \text{atm} \)
- Assume \( P_{O_2} \gg P_{O_3} \)

3. **Neglect \( P_{O_3} \)**:
Since \( P_{O_2} \) is much greater than \( P_{O_3} \), we can approximate:
\[ P_{O_2} \approx P_{total} = 8 \, \text{atm} \]
Thus, we can write:
\[ P_{O_3} = P_{total} - P_{O_2} \approx 8 - P_{O_3} \]

4. **Substituting into the Equilibrium Expression**:
Substitute \( P_{O_2} \) into the \( K_p \) expression:
\[ K_p = \frac{(8)^3}{(P_{O_3})^2} \]
Therefore,
\[ 4 \times 10^{14} = \frac{512}{(P_{O_3})^2} \]

5. **Rearranging to Solve for \( P_{O_3} \)**:
Rearranging gives:
\[ (P_{O_3})^2 = \frac{512}{4 \times 10^{14}} \]
\[ (P_{O_3})^2 = \frac{512}{4} \times 10^{-14} = 128 \times 10^{-14} \]
\[ P_{O_3} = \sqrt{128 \times 10^{-14}} = \sqrt{128} \times 10^{-7} \]

6. **Calculating \( \sqrt{128} \)**:
\[ \sqrt{128} = \sqrt{64 \times 2} = 8 \sqrt{2} \approx 8 \times 1.414 \approx 11.31 \]
Therefore:
\[ P_{O_3} \approx 11.31 \times 10^{-7} \, \text{atm} \]

### Final Answer:
The partial pressure of \( O_3 \) is approximately:
\[ P_{O_3} \approx 11.3 \times 10^{-7} \, \text{atm} \]