At temperature T, a 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 for `K_(p)` in terms of `alpha` and the total pressure `P_(T)` is

Options

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

Answer

To find the expression for \( K_p \) in terms of the degree of dissociation \( \alpha \) and the total pressure \( P_T \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The given reaction is:
\[ 2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g) \]

### Step 2: Define the initial conditions
Let the initial pressure of \( AB_2 \) be \( P_T \). Initially, there are no products, so:
- Pressure of \( AB_2 \) = \( P_T \)
- Pressure of \( AB \) = 0
- Pressure of \( B_2 \) = 0

### Step 3: Define the change in pressure due to dissociation
Let \( \alpha \) be the degree of dissociation of \( AB_2 \). Since \( 2 \) moles of \( AB_2 \) dissociate, the change in pressure can be expressed as:
- Pressure of \( AB_2 \) at equilibrium = \( P_T(1 - \alpha) \)
- Pressure of \( AB \) at equilibrium = \( 2P_T \alpha \) (since 2 moles of \( AB_2 \) produce 2 moles of \( AB \))
- Pressure of \( B_2 \) at equilibrium = \( P_T \alpha \) (since 2 moles of \( AB_2 \) produce 1 mole of \( B_2 \))

### Step 4: Write the expression for \( K_p \)
The expression for \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{AB})^2 \cdot P_{B_2}}{(P_{AB_2})^2} \]

### Step 5: Substitute the equilibrium pressures into the \( K_p \) expression
Substituting the equilibrium pressures we found:
\[ K_p = \frac{(2P_T \alpha)^2 \cdot (P_T \alpha)}{(P_T(1 - \alpha))^2} \]

### Step 6: Simplify the expression
Now we simplify the expression:
\[ K_p = \frac{(4P_T^2 \alpha^2) \cdot (P_T \alpha)}{P_T^2(1 - \alpha)^2} \]
\[ K_p = \frac{4P_T^3 \alpha^3}{P_T^2(1 - \alpha)^2} \]
\[ K_p = \frac{4P_T \alpha^3}{(1 - \alpha)^2} \]

### Step 7: Apply the approximation for small \( \alpha \)
Since \( \alpha \) is small compared to unity, we can approximate \( 1 - \alpha \approx 1 \):
\[ K_p \approx 4P_T \alpha^3 \]

### Final Result
Thus, the expression for \( K_p \) in terms of \( \alpha \) and \( P_T \) is:
\[ K_p = 4P_T \alpha^3 \]

---