The dissociation equilibrium of a gas `AB_2` can be represented as, `2AB_2(g) hArr 2AB (g) +B_2(g)`. The degree of disssociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant `k_p` and total pressure P is

Options

A. `((K_p)/P)`
B. `((2k_p)/P)`
C. `((2K_p)/P)^(1//3)`
D. `((2K_p)/P)^(1//2)`

Answer

To derive the expression relating the degree of dissociation (x) with the equilibrium constant \( K_p \) and total pressure \( P \) for the reaction:

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

we will follow these steps:

### Step 1: Define the Initial Conditions
Let the initial pressure of \( AB_2 \) be \( P \). The initial pressures of \( AB \) and \( B_2 \) are both zero.

### Step 2: Define the Change in Pressure
Let the degree of dissociation of \( AB_2 \) be \( x \). At equilibrium:
- The change in pressure for \( AB_2 \) will be \( -2xP \) (since 2 moles of \( AB_2 \) dissociate).
- The pressure of \( AB \) will increase by \( 2xP \) (2 moles of \( AB \) are formed).
- The pressure of \( B_2 \) will increase by \( xP \) (1 mole of \( B_2 \) is formed).

### Step 3: Write the Equilibrium Pressures
At equilibrium, the pressures will be:
- Pressure of \( AB_2 \): \( P_{AB_2} = P - 2xP = P(1 - x) \)
- Pressure of \( AB \): \( P_{AB} = 2xP \)
- Pressure of \( B_2 \): \( P_{B_2} = xP \)

### Step 4: Write the Expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:

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

Substituting the equilibrium pressures:

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

### Step 5: Simplify the Expression
Now, simplify the expression:

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

This simplifies to:

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

### Step 6: Cancel Out \( P^2 \)
Cancelling \( P^2 \) from the numerator and the denominator gives:

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

### Step 7: Approximate \( 1 - x \)
Since \( x \) is small compared to 1, we can approximate \( 1 - x \approx 1 \):

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

### Step 8: Solve for \( x \)
Rearranging the equation to solve for \( x \):

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

Taking the cube root gives:

\[ x \approx \left(\frac{K_p}{4P}\right)^{1/3} \]

### Final Expression
Thus, the expression relating the degree of dissociation \( x \) with the equilibrium constant \( K_p \) and total pressure \( P \) is:

\[ x \approx \left(\frac{K_p}{4P}\right)^{1/3} \]