For the equilibrium `AB(g) hArr A(g)+B(g)`. `K_(p)` is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium reaction is given as:
\[ AB(g) \rightleftharpoons A(g) + B(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of \( AB = 1 \)
- Moles of \( A = 0 \)
- Moles of \( B = 0 \)

### Step 3: Define the change in moles at equilibrium
Let \( \alpha \) be the amount of \( AB \) that dissociates at equilibrium. Therefore, at equilibrium:
- Moles of \( AB = 1 - \alpha \)
- Moles of \( A = \alpha \)
- Moles of \( B = \alpha \)

### Step 4: Calculate the total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 5: Write the expression for \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}} \]
Where \( P_A \), \( P_B \), and \( P_{AB} \) are the partial pressures of \( A \), \( B \), and \( AB \) respectively.

### Step 6: Express partial pressures in terms of total pressure
The partial pressures can be expressed as:
- \( P_A = \frac{\alpha}{1 + \alpha} P \)
- \( P_B = \frac{\alpha}{1 + \alpha} P \)
- \( P_{AB} = \frac{1 - \alpha}{1 + \alpha} P \)

### Step 7: Substitute into the \( K_p \) expression
Substituting the expressions for partial pressures into the \( K_p \) equation gives:
\[ K_p = \frac{\left(\frac{\alpha}{1 + \alpha} P\right) \cdot \left(\frac{\alpha}{1 + \alpha} P\right)}{\frac{1 - \alpha}{1 + \alpha} P} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 P^2}{(1 - \alpha)(1 + \alpha) P} \]
\[ K_p = \frac{\alpha^2 P}{1 - \alpha} \]

### Step 8: Use the given condition for \( K_p \)
According to the problem, \( K_p = 4P \). Therefore, we can set up the equation:
\[ \frac{\alpha^2 P}{1 - \alpha} = 4P \]

### Step 9: Simplify the equation
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ \frac{\alpha^2}{1 - \alpha} = 4 \]

### Step 10: Rearranging the equation
Rearranging gives:
\[ \alpha^2 = 4(1 - \alpha) \]
\[ \alpha^2 + 4\alpha - 4 = 0 \]

### Step 11: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 4, c = -4 \):
\[ \alpha = \frac{-4 \pm \sqrt{16 + 16}}{2} \]
\[ \alpha = \frac{-4 \pm \sqrt{32}}{2} \]
\[ \alpha = \frac{-4 \pm 4\sqrt{2}}{2} \]
\[ \alpha = -2 \pm 2\sqrt{2} \]

### Step 12: Choose the positive root
Since \( \alpha \) must be positive, we take:
\[ \alpha = -2 + 2\sqrt{2} \]

### Step 13: Calculate the number of moles of \( A \)
The number of moles of \( A \) formed at equilibrium is:
\[ \text{Moles of } A = \alpha = -2 + 2\sqrt{2} \]

### Final Answer
Calculating \( -2 + 2\sqrt{2} \):
Approximately, \( \sqrt{2} \approx 1.414 \):
\[ \alpha \approx -2 + 2 \times 1.414 \approx -2 + 2.828 \approx 0.828 \]

Thus, the number of moles of \( A \) formed is approximately \( 0.828 \).