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 of finding the pressure at which one-third of AB is dissociated in the equilibrium reaction \( AB(g) \rightleftharpoons A(g) + B(g) \), we will follow these steps:

### Step 1: Define the Initial Conditions
Let the initial amount of \( AB \) be \( C \) (in moles or concentration). At equilibrium, one-third of \( AB \) is dissociated, which means:

\[ \text{Dissociated amount} = \frac{1}{3}C \]

### Step 2: Calculate the Amounts at Equilibrium
At equilibrium, the amounts of the substances will be:

- Amount of \( AB \) remaining:
\[ C - \frac{1}{3}C = \frac{2}{3}C \]

- Amount of \( A \) formed:
\[ \frac{1}{3}C \]

- Amount of \( B \) formed:
\[ \frac{1}{3}C \]

### Step 3: Calculate the Total Moles at Equilibrium
The total number of moles at equilibrium is the sum of the moles of \( A \), \( B \), and remaining \( AB \):

\[ \text{Total moles} = \frac{2}{3}C + \frac{1}{3}C + \frac{1}{3}C = \frac{2}{3}C + \frac{2}{3}C = \frac{4}{3}C \]

### Step 4: Calculate the Partial Pressures
Using Dalton's Law of Partial Pressures, we can express the partial pressures in terms of total pressure \( P \):

- Partial pressure of \( A \) (\( P_A \)):
\[ P_A = \left(\frac{\frac{1}{3}C}{\frac{4}{3}C}\right)P = \frac{1}{4}P \]

- Partial pressure of \( B \) (\( P_B \)):
\[ P_B = \left(\frac{\frac{1}{3}C}{\frac{4}{3}C}\right)P = \frac{1}{4}P \]

- Partial pressure of \( AB \) (\( P_{AB} \)):
\[ P_{AB} = \left(\frac{\frac{2}{3}C}{\frac{4}{3}C}\right)P = \frac{1}{2}P \]

### Step 5: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} = \frac{\left(\frac{1}{4}P\right) \cdot \left(\frac{1}{4}P\right)}{\frac{1}{2}P} \]

### Step 6: Simplify the Expression
Substituting the values into the equation:

\[ K_p = \frac{\frac{1}{16}P^2}{\frac{1}{2}P} = \frac{1}{16}P^2 \cdot \frac{2}{1} = \frac{1}{8}P \]

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

\[ P = 8K_p \]

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