Consider the following equilibrium `PCl_(5(g)) Leftrightarrow PCl_(3(g))+Cl_(2(g))` in a closed container. 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) and degree of dissociation (a ) ?

Options

A. Neighter `K_(p)` nor a changes
B. Both `K_(p)` and a change
C. `K_(p)` changes, but a does not change
D. `K_(p)` does not change, but `alpha` changes

Explanation

To solve the problem regarding the equilibrium of the reaction \( PCl_5(g) \leftrightarrow PCl_3(g) + Cl_2(g) \) when the volume of the reaction container is halved at a fixed temperature, we will analyze the effects on the equilibrium constant \( K_p \) and the degree of dissociation \( \alpha \).

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constant**:
The equilibrium constant \( K_p \) for the reaction at a given temperature is defined as:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
where \( P \) denotes the partial pressures of the gases involved in the reaction.

2. **Effect of Volume Change on Partial Pressures**:
When the volume of the container is halved, the partial pressures of the gases will change. According to the ideal gas law, if the volume is halved, the pressure (and hence the concentration) of the gases will double, assuming the temperature remains constant.

3. **Analyzing the Change in Concentrations**:
Let’s denote the initial concentration of \( PCl_5 \) as \( C_0 \). When the volume is halved, the new concentrations will be:
- \( [PCl_5] = 2C_0 \)
- \( [PCl_3] = 2x \) (where \( x \) is the degree of dissociation)
- \( [Cl_2] = 2x \)

4. **Revising the Equilibrium Expression**:
The new expression for \( K_p \) after halving the volume becomes:
\[ K_p = \frac{(2x)(2x)}{2C_0} = \frac{4x^2}{2C_0} = \frac{2x^2}{C_0} \]

5. **Understanding the Degree of Dissociation \( \alpha \)**:
The degree of dissociation \( \alpha \) is defined as the fraction of the original \( PCl_5 \) that has dissociated. If the volume is halved, the system will shift to the right to counteract the change (according to Le Chatelier's principle), which means that \( \alpha \) will increase.

6. **Conclusion**:
- The equilibrium constant \( K_p \) remains unchanged because it is dependent only on temperature, which is constant in this case.
- The degree of dissociation \( \alpha \) increases due to the halving of the volume, leading to a shift in equilibrium towards the products.

### Final Answer:
- The equilibrium constant \( K_p \) does not change.
- The degree of dissociation \( \alpha \) increases.