For the reaction, `AB(g)hArr A(g)+B(g), AB" is "33%` dissociated at a total pressure of 'p' Therefore, 'p' is related to `K_(p)` by one of the following options

Options

A. `p=K_(p)`
B. `p=4K_(p)`
C. `p=3K_(p)`
D. `p=8K_(p)`

Answer

To solve the problem, we need to analyze the dissociation of the compound AB into its products A and B and relate the total pressure \( p \) to the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Initial Setup**:
- Consider 1 mole of \( AB \) at the start (time \( t = 0 \)).
- At \( t = 0 \):
- Moles of \( AB = 1 \)
- Moles of \( A = 0 \)
- Moles of \( B = 0 \)

2. **Dissociation Information**:
- Given that \( AB \) is 33% dissociated, we can express this as:
\[ \text{Dissociated moles of } AB = \frac{1}{3} \text{ moles} \]
- Remaining moles of \( AB \):
\[ \text{Remaining } AB = 1 - \frac{1}{3} = \frac{2}{3} \text{ moles} \]
- Moles of \( A \) and \( B \) formed:
\[ \text{Moles of } A = \frac{1}{3}, \quad \text{Moles of } B = \frac{1}{3} \]

3. **Total Moles Calculation**:
- Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } AB + \text{Moles of } A + \text{Moles of } B = \frac{2}{3} + \frac{1}{3} + \frac{1}{3} = \frac{4}{3} \]

4. **Mole Fraction Calculation**:
- Mole fraction of \( AB \):
\[ \text{Mole fraction of } AB = \frac{\text{Moles of } AB}{\text{Total moles}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2} \]
- Mole fraction of \( A \) and \( B \):
\[ \text{Mole fraction of } A = \text{Mole fraction of } B = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} \]

5. **Partial Pressure Calculation**:
- Total pressure \( p \) is given.
- Partial pressure of \( AB \):
\[ P_{AB} = p \times \text{Mole fraction of } AB = p \times \frac{1}{2} = \frac{p}{2} \]
- Partial pressure of \( A \) and \( B \):
\[ P_A = P_B = p \times \text{Mole fraction of } A = p \times \frac{1}{4} = \frac{p}{4} \]

6. **Equilibrium Constant \( K_p \) Calculation**:
- The expression for \( K_p \) is:
\[ K_p = \frac{P_A \times P_B}{P_{AB}} = \frac{\left(\frac{p}{4}\right) \times \left(\frac{p}{4}\right)}{\frac{p}{2}} \]
- Simplifying this:
\[ K_p = \frac{\frac{p^2}{16}}{\frac{p}{2}} = \frac{p^2}{16} \times \frac{2}{p} = \frac{p}{8} \]

7. **Relating \( p \) to \( K_p \)**:
- From the equation \( K_p = \frac{p}{8} \), we can express \( p \) in terms of \( K_p \):
\[ p = 8 K_p \]

### Conclusion:
The relationship between the total pressure \( p \) and the equilibrium constant \( K_p \) is:
\[ p = 8 K_p \]