At certain temperature compound `AB_(2)(g)` dissociates according to the reaction
`2AB_(2)(g) hArr2AB (g)+B_(2)(g)`
With degree of dissociation `alpha` Which is small compared with unity, the expression of `K_(p)` in terms of `alpha` and initial pressure P is :

Options

A. (a) `(Palpha^(3))/(2)`
B. (b) `(Palpha^(2))/(3)`
C. (c) `(Palpha^(3))/(3)`
D. (d) `(Palpha^(2))/(2)`

Answer

To solve the problem, we need to derive the expression for the equilibrium constant \( K_p \) for the dissociation reaction of the compound \( AB_2(g) \):

\[ 2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g) \]

Given that the degree of dissociation is \( \alpha \) (which is small compared to unity) and the initial pressure is \( P \), we can follow these steps:

### Step 1: Define Initial Conditions
Initially, we have:
- The pressure of \( AB_2 \) is \( P \).
- The pressures of \( AB \) and \( B_2 \) are both 0.

### Step 2: Define Changes at Equilibrium
When \( AB_2 \) dissociates, the changes in pressure can be defined as follows:
- The pressure of \( AB_2 \) at equilibrium will be \( P(1 - \alpha) \).
- The pressure of \( AB \) at equilibrium will be \( 2\alpha P \) (since 2 moles of \( AB \) are produced for every 2 moles of \( AB_2 \) that dissociate).
- The pressure of \( B_2 \) at equilibrium will be \( \alpha P \) (since 1 mole of \( B_2 \) is produced for every 2 moles of \( AB_2 \) that dissociate).

### Step 3: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is expressed in terms of the partial pressures of the products and reactants:

\[ K_p = \frac{(P_{AB})^2 (P_{B_2})}{(P_{AB_2})^2} \]

Substituting the equilibrium pressures:
- \( P_{AB} = 2\alpha P \)
- \( P_{B_2} = \alpha P \)
- \( P_{AB_2} = P(1 - \alpha) \)

The expression becomes:

\[ K_p = \frac{(2\alpha P)^2 (\alpha P)}{(P(1 - \alpha))^2} \]

### Step 4: Simplify the Expression
Now, we simplify the expression:

\[ K_p = \frac{4\alpha^2 P^2 \cdot \alpha P}{P^2(1 - \alpha)^2} \]

This simplifies to:

\[ K_p = \frac{4\alpha^3 P^3}{(1 - \alpha)^2} \]

### Step 5: Approximate for Small \( \alpha \)
Since \( \alpha \) is small compared to unity, we can approximate \( (1 - \alpha) \) as approximately 1. Thus, we have:

\[ K_p \approx 4\alpha^3 P^3 \]

### Final Expression
Thus, the final expression for \( K_p \) in terms of \( \alpha \) and the initial pressure \( P \) is:

\[ K_p \approx \frac{4\alpha^3 P^3}{1} \]