At temperature ,T, a compound `AB_2(g)` dissociates according to the reaction `2AB_2 (g) ltimplies 2AB(g)+B_2(g)` with a degree of dissociation, x, which is small compared with unity.Deduce the expression for `K_P`, in terms of x and the total pressure , P.

Options

A. `(Px^3)/2`
B. `(Px^2)/3`
C. `(Px^3)/3`
D. `(Px^2)/2`

Answer

To derive the expression for \( K_P \) in terms of the degree of dissociation \( x \) and the total pressure \( P \) for the reaction:

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

we will follow these steps:

### Step 1: Define Initial Conditions
Assume the initial pressure of \( AB_2 \) is \( P_0 \) and the initial pressures of \( AB \) and \( B_2 \) are both 0.

- Initial pressures:
- \( P(AB_2) = P_0 \)
- \( P(AB) = 0 \)
- \( P(B_2) = 0 \)

### Step 2: Define Changes at Equilibrium
Let \( x \) be the degree of dissociation of \( AB_2 \). Since 2 moles of \( AB_2 \) dissociate to form 2 moles of \( AB \) and 1 mole of \( B_2 \), we can express the changes in pressure at equilibrium:

- Change in pressure for \( AB_2 \): \( -2P_0x \)
- Change in pressure for \( AB \): \( +2P_0x \)
- Change in pressure for \( B_2 \): \( +P_0x \)

### Step 3: Write Equilibrium Pressures
At equilibrium, the pressures will be:

- \( P(AB_2) = P_0 - 2P_0x = P_0(1 - 2x) \)
- \( P(AB) = 2P_0x \)
- \( P(B_2) = P_0x \)

### Step 4: Total Pressure at Equilibrium
The total pressure \( P \) at equilibrium can be expressed as:

\[ P = P(AB_2) + P(AB) + P(B_2) = P_0(1 - 2x) + 2P_0x + P_0x \]

Simplifying this gives:

\[ P = P_0(1 - 2x + 2x + x) = P_0(1 + x) \]

### Step 5: Solve for \( P_0 \)
From the equation \( P = P_0(1 + x) \), we can express \( P_0 \):

\[ P_0 = \frac{P}{1 + x} \]

### Step 6: Write the Expression for \( K_P \)
The expression for \( K_P \) is given by:

\[ K_P = \frac{(P(AB))^2 \cdot P(B_2)}{(P(AB_2))^2} \]

Substituting the equilibrium pressures:

\[ K_P = \frac{(2P_0x)^2 \cdot (P_0x)}{(P_0(1 - 2x))^2} \]

### Step 7: Substitute \( P_0 \)
Now substituting \( P_0 = \frac{P}{1 + x} \):

\[ K_P = \frac{(2 \cdot \frac{P}{1 + x} \cdot x)^2 \cdot (\frac{P}{1 + x} \cdot x)}{(\frac{P}{1 + x}(1 - 2x))^2} \]

### Step 8: Simplify the Expression
After substituting and simplifying, we get:

\[ K_P = \frac{4P^2x^2 \cdot \frac{P}{1 + x}x}{\frac{P^2(1 - 2x)^2}{(1 + x)^2}} \]

This simplifies to:

\[ K_P = \frac{4Px^3}{(1 - 2x)^2} \]

### Step 9: Final Expression
Since \( x \) is small compared to 1, we can approximate \( (1 - 2x) \approx 1 \):

\[ K_P \approx 4Px^3 \]

### Conclusion
Thus, the final expression for \( K_P \) in terms of \( x \) and \( P \) is:

\[ K_P = \frac{4Px^3}{1} \]