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 degree of dissociation `(alpha)`?

Options

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

Answer

To solve the problem, we need to analyze the equilibrium reaction and the effects of halving the volume of the container on the equilibrium constant \( K_p \) and the degree of dissociation \( \alpha \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The equilibrium reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

2. **Understand the Effect of Volume Change**:
When the volume of the reaction container is halved, the pressure of the gases in the container increases (since pressure is inversely proportional to volume, according to Boyle's Law).

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

4. **Count the Moles**:
- On the left side (reactants), there is 1 mole of \( N_2O_4 \).
- On the right side (products), there are 2 moles of \( NO_2 \).
Since there are more moles of gas on the product side, the equilibrium will shift to the left (towards \( N_2O_4 \)) when the volume is halved.

5. **Effect on Degree of Dissociation (\( \alpha \))**:
The degree of dissociation \( \alpha \) refers to the fraction of the original \( N_2O_4 \) that dissociates into \( NO_2 \). As the equilibrium shifts to the left, less \( N_2O_4 \) will dissociate, resulting in a decrease in \( \alpha \).

6. **Effect on Equilibrium Constant (\( K_p \))**:
The equilibrium constant \( K_p \) is only affected by temperature. Since the temperature is constant in this scenario, \( K_p \) remains unchanged despite the change in volume.

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

### Final Answer:
- \( K_p \) is constant, and \( \alpha \) decreases.