Consider the following equilibrium in a closed container:
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
At a fixed temperature, the volume of the reaction container is halved. For this change which of the following statements holds true regarding the equilibrium constant `(K_(p))` and the degree of dissociation `(alpha)`?

Options

A. Neither `K_(p)` nor alpha changes
B. Both `K_(p)` and alpha change
C. `K_(p)` does not change but alpha changes
D. `K_(p)` changes, but alpha does not change

Answer

To solve the problem regarding the equilibrium of the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) when the volume of the container is halved, we will analyze the effects on the equilibrium constant \( K_p \) and the degree of dissociation \( \alpha \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The equilibrium reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
This reaction involves the conversion of 1 mole of \( N_2O_4 \) into 2 moles of \( NO_2 \).

2. **Effect of Volume Change on Pressure**:
When the volume of the container is halved, the pressure inside the container increases. According to the ideal gas law, \( PV = nRT \), if the volume \( V \) decreases while the temperature \( T \) is constant, the pressure \( P \) must increase.

3. **Applying Le Chatelier's Principle**:
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (like pressure), the equilibrium will shift in a direction that counteracts that change. In this case, since the pressure increases, the equilibrium will shift towards the side with fewer moles of gas.

- The left side (reactants) has 1 mole of \( N_2O_4 \).
- The right side (products) has 2 moles of \( NO_2 \).

Therefore, the equilibrium will shift to the left, towards \( N_2O_4 \), which has fewer moles of gas.

4. **Effect on Degree of Dissociation \( \alpha \)**:
The degree of dissociation \( \alpha \) refers to the fraction of the reactant that has dissociated into products. Since the equilibrium shifts to the left (favoring the formation of \( N_2O_4 \)), the amount of \( NO_2 \) will decrease, and hence the degree of dissociation \( \alpha \) will decrease.

5. **Effect on Equilibrium Constant \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
The value of \( K_p \) is only dependent on temperature. Since the temperature is fixed in this scenario, \( K_p \) will not change regardless of the changes in volume or pressure.

### Conclusion:
- The equilibrium constant \( K_p \) remains unchanged.
- The degree of dissociation \( \alpha \) decreases.

### Final Answer:
- \( K_p \) does not change, but \( \alpha \) decreases.