If for `2A_(2)B(g)hArr2A_(2)(g)+B_(2)(g),K_(P)="TOTAL PRESSURE" ("at equilibrium")` and starting the dissociation from" 4 "mol of" A_(2)B` then:

Options

A. degree of dissociation of `A_(2)B "will be" (2//3)`
B. total no. of moles at equilibrium will be `(14//3)`.
C. at equilibrium the no. of moles of `A_(2)B` are no equal to the no. of moles of `B_(2)`
D. at equilibrium the no. of moles of `A_(2)B` are equal to the no. of moles of `A_(2)`

Answer

To solve the problem step by step, we will analyze the reaction and calculate the degree of dissociation, total moles at equilibrium, and compare the moles of different species at equilibrium.

### Step 1: Write the Reaction
The given reaction is:
\[ 2A_2B(g) \rightleftharpoons 2A_2(g) + B_2(g) \]

### Step 2: Initial Moles
We start with 4 moles of \( A_2B \):
- Initial moles of \( A_2B = 4 \)
- Initial moles of \( A_2 = 0 \)
- Initial moles of \( B_2 = 0 \)

### Step 3: Define Degree of Dissociation
Let \( \alpha \) be the degree of dissociation of \( A_2B \). Since 2 moles of \( A_2B \) dissociate to form 2 moles of \( A_2 \) and 1 mole of \( B_2 \), we can express the changes in moles as follows:
- Change in moles of \( A_2B = -2\alpha \cdot 4 = -8\alpha \)
- Change in moles of \( A_2 = +2\alpha \cdot 4 = +8\alpha \)
- Change in moles of \( B_2 = +\alpha \cdot 4 = +4\alpha \)

### Step 4: Moles at Equilibrium
At equilibrium, the moles will be:
- Moles of \( A_2B = 4 - 2\alpha \)
- Moles of \( A_2 = 8\alpha \)
- Moles of \( B_2 = 4\alpha \)

### Step 5: Total Moles at Equilibrium
The total moles at equilibrium can be calculated as:
\[ \text{Total moles} = (4 - 2\alpha) + (8\alpha) + (4\alpha) = 4 + 10\alpha \]

### Step 6: Partial Pressures
The partial pressures can be expressed in terms of total pressure \( P \):
- Partial pressure of \( A_2B = \frac{(4 - 2\alpha)}{(4 + 10\alpha)}P \)
- Partial pressure of \( A_2 = \frac{8\alpha}{(4 + 10\alpha)}P \)
- Partial pressure of \( B_2 = \frac{4\alpha}{(4 + 10\alpha)}P \)

### Step 7: Equilibrium Constant Expression
The equilibrium constant \( K_p \) can be expressed as:
\[ K_p = \frac{(P_{A_2})^2 \cdot (P_{B_2})}{(P_{A_2B})^2} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{8\alpha}{(4 + 10\alpha)}P\right)^2 \cdot \left(\frac{4\alpha}{(4 + 10\alpha)}P\right)}{\left(\frac{(4 - 2\alpha)}{(4 + 10\alpha)}P\right)^2} \]

### Step 8: Simplifying the Expression
After simplification, we find:
\[ K_p = \frac{4\alpha^3 P^2}{(4 + 10\alpha)^2} \]
Given that \( K_p = P \), we set:
\[ \frac{4\alpha^3 P^2}{(4 + 10\alpha)^2} = P \]

### Step 9: Solving for \( \alpha \)
Rearranging gives:
\[ 4\alpha^3 = (4 + 10\alpha)^2 \]
Expanding and simplifying leads to:
\[ 4\alpha^3 - 100\alpha^2 - 80\alpha - 16 = 0 \]
Solving for \( \alpha \) yields \( \alpha = \frac{1}{3} \).

### Step 10: Degree of Dissociation
The degree of dissociation of \( A_2B \) is:
\[ \text{Degree of dissociation} = 2\alpha = 2 \times \frac{1}{3} = \frac{2}{3} \]

### Step 11: Total Moles at Equilibrium
Substituting \( \alpha \) back into the total moles:
\[ \text{Total moles} = 4 + 10 \times \frac{1}{3} = 4 + \frac{10}{3} = \frac{22}{3} \]

### Final Results
- Degree of dissociation of \( A_2B \) = \( \frac{2}{3} \)
- Total moles at equilibrium = \( \frac{22}{3} \)