A 10 litre box contains `O_3 and O_2` at equilibrium at 2000 k. `K_p=4.17xx10^(14) "for" 2O_3 hArr 3O_2` Assume that `P_(O_2) gt gt P_(O_3)` and if total pressure is 7.33 atm. Then partial pressure of `O_3` will be

Options

A. `9.71 xx 10^(-5)` atm
B. `9.71 xx 10^(-7)` atm
C. `9.71 xx 10^(-6)` atm
D. `9.71 xx 10^(-2)` atm

Answer

To find the partial pressure of \( O_3 \) in the given equilibrium reaction, we can follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium reaction is given as:
\[ 2O_3 \rightleftharpoons 3O_2 \]
The equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]

### Step 2: Substitute the known values
We know that:
- \( K_p = 4.17 \times 10^{14} \)
- Total pressure \( P_{total} = 7.33 \, \text{atm} \)
- Assume \( P_{O_2} \gg P_{O_3} \)

Since \( P_{O_2} \) is much greater than \( P_{O_3} \), we can approximate:
\[ P_{O_2} \approx P_{total} = 7.33 \, \text{atm} \]

### Step 3: Set up the equation
Using the equilibrium expression:
\[ 4.17 \times 10^{14} = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]
Substituting \( P_{O_2} \):
\[ 4.17 \times 10^{14} = \frac{(7.33)^3}{(P_{O_3})^2} \]

### Step 4: Calculate \( P_{O_3}^2 \)
Rearranging the equation gives:
\[ (P_{O_3})^2 = \frac{(7.33)^3}{4.17 \times 10^{14}} \]

Calculating \( (7.33)^3 \):
\[ (7.33)^3 = 393.66 \]

Now substituting this value:
\[ (P_{O_3})^2 = \frac{393.66}{4.17 \times 10^{14}} \]

Calculating the right side:
\[ (P_{O_3})^2 = 9.44 \times 10^{-13} \]

### Step 5: Find \( P_{O_3} \)
Taking the square root gives:
\[ P_{O_3} = \sqrt{9.44 \times 10^{-13}} \]
\[ P_{O_3} \approx 9.71 \times 10^{-7} \, \text{atm} \]

### Conclusion
The partial pressure of \( O_3 \) is approximately:
\[ P_{O_3} \approx 9.71 \times 10^{-7} \, \text{atm} \]

---