`X hArr nY, X` decomposes to give Y ( in one lietre vessel ) if degree of dissociation is `alpha` then `K_( C )` `:`

Options

A. `( (n alpha))/( ( 1+alpha))`
B. `( (n alpha))/( ( 1- alpha)) `
C. `(n alpha)/((1- alpha)) `
D. `((n alpha)^(n))/((1- alpha))`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction \( X \rightleftharpoons nY \) given the degree of dissociation \( \alpha \).

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction is given as:
\[ X \rightleftharpoons nY \]

2. **Define Initial Concentrations**:
Let's assume we start with 1 mole of \( X \) in a 1-liter vessel. Therefore, the initial concentration of \( X \) is:
\[ [X]_0 = 1 \, \text{mol/L} \]
The initial concentration of \( Y \) is:
\[ [Y]_0 = 0 \, \text{mol/L} \]

3. **Degree of Dissociation**:
The degree of dissociation \( \alpha \) represents the fraction of \( X \) that has dissociated. Thus, at equilibrium:
- The amount of \( X \) that dissociates is \( \alpha \) moles.
- The amount of \( Y \) produced is \( n\alpha \) moles.

4. **Calculate Equilibrium Concentrations**:
At equilibrium, the concentrations will be:
\[ [X] = 1 - \alpha \, \text{mol/L} \]
\[ [Y] = n\alpha \, \text{mol/L} \]

5. **Write the Expression for \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[Y]^n}{[X]} \]

6. **Substitute the Equilibrium Concentrations into the \( K_c \) Expression**:
Substituting the equilibrium concentrations into the expression for \( K_c \):
\[ K_c = \frac{(n\alpha)^n}{(1 - \alpha)} \]

7. **Final Expression for \( K_c \)**:
Thus, the final expression for the equilibrium constant \( K_c \) is:
\[ K_c = \frac{(n\alpha)^n}{(1 - \alpha)} \]

### Summary:
The value of \( K_c \) for the reaction \( X \rightleftharpoons nY \) when the degree of dissociation is \( \alpha \) is:
\[ K_c = \frac{(n\alpha)^n}{(1 - \alpha)} \]