In the reaction `AB(g) hArr A(g) + B(g)` at `30^(@)C, k_(p)` for the dissociation equilibrium is `2.56xx10^(-2) atm`. If the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is

Options

A. `87%`
B. `13%`
C. `43.5%`
D. `16%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced reaction and define the variables
The reaction is given as:
\[ AB(g) \rightleftharpoons A(g) + B(g) \]

Let:
- Initial moles of \( AB = 1 \)
- Let \( \alpha \) be the degree of dissociation of \( AB \).

### Step 2: Express the moles at equilibrium
At equilibrium:
- Moles of \( AB = 1 - \alpha \)
- Moles of \( A = \alpha \)
- Moles of \( B = \alpha \)

### Step 3: Write the expression for total moles at equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Use the total pressure to relate to moles
Given that the total pressure at equilibrium is \( 1 \, \text{atm} \):
\[ P_{\text{total}} = (1 + \alpha) \cdot P_{0} \]
Where \( P_{0} \) is the pressure contribution from each mole. Since the total pressure is given as \( 1 \, \text{atm} \), we can set:
\[ 1 = (1 + \alpha) \cdot P_{0} \]

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} \]
At equilibrium:
- \( P_A = \alpha \cdot P_{0} \)
- \( P_B = \alpha \cdot P_{0} \)
- \( P_{AB} = (1 - \alpha) \cdot P_{0} \)

Thus:
\[ K_p = \frac{(\alpha \cdot P_{0}) \cdot (\alpha \cdot P_{0})}{(1 - \alpha) \cdot P_{0}} \]

### Step 6: Simplify the expression for \( K_p \)
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P_{0}}{1 - \alpha} \]

### Step 7: Substitute \( K_p \) and solve for \( \alpha \)
Given \( K_p = 2.56 \times 10^{-2} \, \text{atm} \):
\[ 2.56 \times 10^{-2} = \frac{\alpha^2 \cdot P_{0}}{1 - \alpha} \]

Since \( P_{0} = 1 \) atm (from total pressure):
\[ 2.56 \times 10^{-2} = \frac{\alpha^2}{1 - \alpha} \]

Assuming \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):
\[ 2.56 \times 10^{-2} \approx \alpha^2 \]

### Step 8: Solve for \( \alpha \)
Taking the square root:
\[ \alpha \approx \sqrt{2.56 \times 10^{-2}} \]
\[ \alpha \approx 0.16 \]

### Step 9: Calculate the percentage dissociation
Percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.16 \times 100 = 16\% \]

### Final Answer
The percentage dissociation of \( AB \) is **16%**.

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