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 \( K_p \) for this reaction is given by:
\[ K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]

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

### Step 3: Assume \( P_{O_3} \) is negligible
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 4: Substitute \( P_{O_2} \) into the equilibrium expression
Using the equilibrium expression:
\[ K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} \]
Substituting \( P_{O_2} \):
\[ 4.17 \times 10^{14} = \frac{(7.33)^3}{(P_{O_3})^2} \]

### Step 5: Calculate \( (7.33)^3 \)
Calculating \( (7.33)^3 \):
\[ (7.33)^3 = 7.33 \times 7.33 \times 7.33 = 389.017 \]

### Step 6: Rearranging the equation to find \( P_{O_3} \)
Now we rearrange the equation to solve for \( P_{O_3}^2 \):
\[ P_{O_3}^2 = \frac{(7.33)^3}{K_p} = \frac{389.017}{4.17 \times 10^{14}} \]

### Step 7: Calculate \( P_{O_3}^2 \)
Calculating \( P_{O_3}^2 \):
\[ P_{O_3}^2 = \frac{389.017}{4.17 \times 10^{14}} = 9.34 \times 10^{-13} \]

### Step 8: Take the square root to find \( P_{O_3} \)
Taking the square root to find \( P_{O_3} \):
\[ P_{O_3} = \sqrt{9.34 \times 10^{-13}} \approx 9.67 \times 10^{-7} \, \text{atm} \]

### Final Answer
The partial pressure of \( O_3 \) is approximately:
\[ P_{O_3} \approx 9.67 \times 10^{-7} \, \text{atm} \]

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