For the reaction
`AB_(2)(g)hArrAB(g)+B(g)`
If `prop` is negligiable w.r.t `1` then degree of dissociaation `(prop)` of `AB_(2)` is proportional to :

Options

A. `1/P`
B. `1/V`
C. `1/sqrtP`
D. `sqrtV`

Answer

To solve the problem regarding the degree of dissociation (α) of the compound \( AB_2 \) in the reaction:

\[ AB_2 (g) \rightleftharpoons AB (g) + B (g) \]

we will follow these steps:

### Step 1: Write the expression for the degree of dissociation
Let the initial concentration of \( AB_2 \) be \( C \). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:

- Concentration of \( AB_2 \) at equilibrium: \( C - C\alpha = C(1 - \alpha) \)
- Concentration of \( AB \) at equilibrium: \( C\alpha \)
- Concentration of \( B \) at equilibrium: \( C\alpha \)

### Step 2: Write the equilibrium constant expression
The equilibrium constant \( K_c \) for the reaction can be expressed as:

\[ K_c = \frac{[AB][B]}{[AB_2]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 3: Assume \( \alpha \) is negligible with respect to 1
Since \( \alpha \) is negligible compared to 1, we can approximate \( 1 - \alpha \) as 1. Thus, the expression simplifies to:

\[ K_c \approx C\alpha^2 \]

### Step 4: Rearranging the equation
From the above equation, we can express \( \alpha^2 \):

\[ \alpha^2 = \frac{K_c}{C} \]

Taking the square root gives:

\[ \alpha = \sqrt{\frac{K_c}{C}} \]

### Step 5: Relate \( \alpha \) to volume and pressure
Since concentration \( C \) can be expressed as:

\[ C = \frac{n}{V} \quad \text{(where \( n \) is the number of moles and \( V \) is the volume)} \]

Substituting this into the equation for \( \alpha \):

\[ \alpha = \sqrt{\frac{K_c \cdot V}{n}} \]

At constant temperature, \( K_c \) is constant, and if the number of moles \( n \) is also constant, we find that:

\[ \alpha \propto \sqrt{V} \]

### Step 6: Relate \( \alpha \) to pressure
Since pressure \( P \) is inversely proportional to volume at constant temperature (from the ideal gas law), we have:

\[ P \propto \frac{1}{V} \]

Thus, we can relate \( \alpha \) to pressure:

\[ \alpha \propto \frac{1}{\sqrt{P}} \]

### Conclusion
Therefore, the degree of dissociation \( \alpha \) of \( AB_2 \) is proportional to both \( \sqrt{V} \) and \( \frac{1}{\sqrt{P}} \).