The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be

Options

A. `+ 28`
B. 64
C. 32
D. 16

Explanation

To solve the problem, we need to understand how the equilibrium constant \( K_c \) behaves when the volume of the container is changed.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
The equilibrium constant \( K_c \) for this reaction is given as 64.

2. **Expression for \( K_c \)**:
The expression for the equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \([HI]\), \([H_2]\), and \([I_2]\) are the equilibrium concentrations of hydrogen iodide, hydrogen, and iodine, respectively.

3. **Effect of Volume Change**:
When the volume of the container is reduced to half, we need to analyze how this affects the concentrations:
- If the original volume is \( V \), then the new volume is \( \frac{V}{2} \).
- The concentrations will change as follows:
\[ [H_2] = \frac{n_{H_2}}{V/2} = \frac{2n_{H_2}}{V} \]
\[ [I_2] = \frac{n_{I_2}}{V/2} = \frac{2n_{I_2}}{V} \]
\[ [HI] = \frac{n_{HI}}{V/2} = \frac{2n_{HI}}{V} \]

4. **Calculating New \( K_c \)**:
Substituting these new concentrations into the \( K_c \) expression:
\[ K_c' = \frac{(2[HI])^2}{(2[H_2])(2[I_2])} \]
Simplifying this gives:
\[ K_c' = \frac{4[HI]^2}{4[H_2][I_2]} = \frac{[HI]^2}{[H_2][I_2]} = K_c \]

5. **Conclusion**:
The value of the equilibrium constant \( K_c \) remains unchanged at 64, even after the volume of the container is halved.

### Final Answer:
The value of the equilibrium constant \( K_c \) remains 64.