For the equilibrium `AB(g) hArr A(g)+B(g)` at a given temperature, the pressure at which one-third of AB is dissociated is numerically equal to

Options

A. `8` times `K_(p)`
B. `16` times `K_(p)`
C. `4` times `K_(p)`
D. `9` times `K_(p)`

Explanation

To solve the problem regarding the equilibrium \( AB(g) \rightleftharpoons A(g) + B(g) \) and to find the pressure at which one-third of \( AB \) is dissociated, we can follow these steps:

### Step 1: Define the degree of dissociation
Let the degree of dissociation \( \alpha \) be equal to \( \frac{1}{3} \) since one-third of \( AB \) is dissociated.

### Step 2: Set up the initial and equilibrium conditions
Assume we start with 1 mole of \( AB \):
- Initial moles:
- \( AB = 1 \)
- \( A = 0 \)
- \( B = 0 \)

At equilibrium, if \( \alpha \) moles of \( AB \) dissociate, then:
- Moles of \( AB \) at equilibrium = \( 1 - \alpha \)
- Moles of \( A \) at equilibrium = \( \alpha \)
- Moles of \( B \) at equilibrium = \( \alpha \)

### Step 3: Substitute the value of \( \alpha \)
Since \( \alpha = \frac{1}{3} \):
- Moles of \( AB \) at equilibrium = \( 1 - \frac{1}{3} = \frac{2}{3} \)
- Moles of \( A \) at equilibrium = \( \frac{1}{3} \)
- Moles of \( B \) at equilibrium = \( \frac{1}{3} \)

### Step 4: Calculate the total moles at equilibrium
Total moles at equilibrium = Moles of \( AB \) + Moles of \( A \) + Moles of \( B \)
\[ \text{Total moles} = \frac{2}{3} + \frac{1}{3} + \frac{1}{3} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

### Step 5: Write the expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} \]
Where:
- \( P_A = \frac{\alpha}{\text{Total moles}} \cdot P \)
- \( P_B = \frac{\alpha}{\text{Total moles}} \cdot P \)
- \( P_{AB} = \frac{1 - \alpha}{\text{Total moles}} \cdot P \)

### Step 6: Substitute the values into the \( K_p \) expression
Substituting the values:
\[ P_A = \frac{\frac{1}{3}}{\frac{4}{3}} \cdot P = \frac{1}{4} P \]
\[ P_B = \frac{\frac{1}{3}}{\frac{4}{3}} \cdot P = \frac{1}{4} P \]
\[ P_{AB} = \frac{\frac{2}{3}}{\frac{4}{3}} \cdot P = \frac{1}{2} P \]

Now substituting into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{1}{4} P\right) \cdot \left(\frac{1}{4} P\right)}{\frac{1}{2} P} = \frac{\frac{1}{16} P^2}{\frac{1}{2} P} = \frac{1}{8} P \]

### Step 7: Solve for pressure \( P \)
From the equation \( K_p = \frac{1}{8} P \), we can rearrange to find \( P \):
\[ P = 8 K_p \]

### Final Answer
The pressure at which one-third of \( AB \) is dissociated is numerically equal to \( 8 K_p \).