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

Answer

To solve the problem, we need to relate the total pressure \( p \) to the equilibrium constant \( K_p \) for the reaction:

\[ AB(g) \rightleftharpoons A(g) + B(g) \]

Given that \( AB \) is 33% dissociated at a total pressure of \( p \), we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with \( C \) moles of \( AB \). Initially, the moles of \( A \) and \( B \) are both 0.

### Step 2: Calculate Moles at Equilibrium
Since \( AB \) is 33% dissociated, we can calculate the number of moles dissociated:
- Degree of dissociation \( \alpha = 0.33 \)
- Moles of \( AB \) dissociated = \( 0.33C \)

At equilibrium:
- Moles of \( AB \) remaining = \( C - 0.33C = 0.67C \)
- Moles of \( A \) formed = \( 0.33C \)
- Moles of \( B \) formed = \( 0.33C \)

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = \text{Moles of } AB + \text{Moles of } A + \text{Moles of } B = 0.67C + 0.33C + 0.33C = 1.33C \]

### Step 4: Calculate Partial Pressures
To find the partial pressures, we need to calculate the mole fractions:
- Mole fraction of \( AB \):
\[ \text{Mole fraction of } AB = \frac{0.67C}{1.33C} = \frac{0.67}{1.33} \]

- Mole fraction of \( A \):
\[ \text{Mole fraction of } A = \frac{0.33C}{1.33C} = \frac{0.33}{1.33} \]

- Mole fraction of \( B \):
\[ \text{Mole fraction of } B = \frac{0.33C}{1.33C} = \frac{0.33}{1.33} \]

Now, we can express the partial pressures in terms of the total pressure \( p \):
- Partial pressure of \( AB \):
\[ P_{AB} = \left(\frac{0.67}{1.33}\right) p \]

- Partial pressure of \( A \):
\[ P_A = \left(\frac{0.33}{1.33}\right) p \]

- Partial pressure of \( B \):
\[ P_B = \left(\frac{0.33}{1.33}\right) p \]

### Step 5: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} \]

Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left(\frac{0.33}{1.33} p\right) \cdot \left(\frac{0.33}{1.33} p\right)}{\left(\frac{0.67}{1.33} p\right)} \]

### Step 6: Simplify the Expression
\[ K_p = \frac{\left(\frac{0.33^2}{1.33^2} p^2\right)}{\left(\frac{0.67}{1.33} p\right)} = \frac{0.33^2 p}{0.67 \cdot 1.33} \]

### Step 7: Relate \( p \) to \( K_p \)
Rearranging gives:
\[ p = \frac{K_p \cdot 0.67 \cdot 1.33}{0.33^2} \]

### Step 8: Calculate the Numerical Value
Calculating the numerical value:
\[ p = \frac{0.67 \cdot 1.33}{0.33^2} K_p \]
Calculating \( \frac{0.67 \cdot 1.33}{0.33^2} \) gives approximately \( 8.1 \).

Thus, we can conclude:
\[ p \approx 8 K_p \]

### Final Answer
The relation between total pressure \( p \) and \( K_p \) is:
\[ p = 8 K_p \]