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. 0.87
B. 0.13
C. 0.435
D. 0.16

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and apply the concepts of equilibrium constants and percentage dissociation.

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

Let:
- Initial moles of \( AB \) = 1 mole
- Moles of \( A \) and \( B \) formed at equilibrium = \( \alpha \) (the degree of dissociation)

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

Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 3: Use the total pressure to relate to moles
Given that the total pressure at equilibrium is 1 atm, we can express the partial pressures in terms of \( \alpha \):
- Total pressure \( P = 1 \) atm

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} \]

Where:
- \( P_A = \frac{\alpha}{1 + \alpha} \cdot P \)
- \( P_B = \frac{\alpha}{1 + \alpha} \cdot P \)
- \( P_{AB} = \frac{1 - \alpha}{1 + \alpha} \cdot P \)

Substituting these into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{\alpha}{1 + \alpha} \cdot P\right) \cdot \left(\frac{\alpha}{1 + \alpha} \cdot P\right)}{\frac{1 - \alpha}{1 + \alpha} \cdot P} \]

### Step 5: Simplify the expression
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Substitute known values
We know:
- \( K_p = 2.56 \times 10^{-2} \) atm
- \( P = 1 \) atm

Substituting these values:
\[ 2.56 \times 10^{-2} = \frac{\alpha^2 \cdot 1}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Approximate \( 1 - \alpha \)
Assuming \( \alpha \) is small, \( 1 - \alpha \approx 1 \). Thus:
\[ 2.56 \times 10^{-2} = \frac{\alpha^2}{1 + \alpha} \]

### Step 8: Further approximation
Since \( \alpha \) is small, we can also approximate \( 1 + \alpha \approx 1 \):
\[ 2.56 \times 10^{-2} \approx \alpha^2 \]

### Step 9: Solve for \( \alpha \)
Taking the square root:
\[ \alpha = \sqrt{2.56 \times 10^{-2}} \]
\[ \alpha \approx 0.16 \]

### Step 10: Calculate percentage dissociation
Percentage dissociation is given by:
\[ \text{Percentage dissociation} = \left(\frac{\alpha}{1} \times 100\right) = 0.16 \times 100 = 16\% \]

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

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