The equilibrium constant of the reaction `H_(2(g)) +I_(2(g)) hArr 2HI_((g))` is 64. If the volume of the container is reduced to one fourth of its original volume, the value of theequilibrium constant will be

Options

A. 16
B. 32
C. 64
D. 128

Explanation

To solve the problem, we need to understand how the equilibrium constant (K) behaves when the volume of the container changes.

### Step-by-Step Solution:

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

2. **Understand the Effect of Volume Change on Concentration**:
When the volume of the container is reduced to one-fourth of its original volume, the concentration of each gas will increase. The concentration of a gas is inversely proportional to the volume of the container. Therefore, if the volume is reduced to \( \frac{1}{4} \), the concentration of each gas will increase by a factor of 4.

3. **Calculate the New Concentrations**:
Let the initial concentrations of \( H_2 \), \( I_2 \), and \( HI \) at equilibrium be \( [H_2] \), \( [I_2] \), and \( [HI] \), respectively. After reducing the volume, the new concentrations will be:
\[ [H_2]_{new} = 4[H_2] \]
\[ [I_2]_{new} = 4[I_2] \]
\[ [HI]_{new} = 4[HI] \]

4. **Substitute New Concentrations into the Equilibrium Expression**:
The equilibrium constant expression for the reaction is:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the new concentrations into the expression gives:
\[ K_{new} = \frac{(4[HI])^2}{(4[H_2])(4[I_2])} \]

5. **Simplify the Expression**:
Simplifying the right-hand side:
\[ K_{new} = \frac{16[HI]^2}{16[H_2][I_2]} = \frac{[HI]^2}{[H_2][I_2]} = K \]
Thus, \( K_{new} = K \).

6. **Conclusion**:
Since the equilibrium constant does not change with changes in concentration or volume, the value of the equilibrium constant remains 64 even after the volume is reduced.

### Final Answer:
The value of the equilibrium constant remains **64**.