The equilbrium constant of the reaction `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 50. If the volume of the container is reduced to half of its original value, the value of equilibrium constant will be

Options

A. 25
B. 50
C. 75
D. 100

Answer

To solve the problem, we need to understand the concept of the equilibrium constant (K) and how it is affected by changes in conditions such as volume.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Information**:
The reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
The equilibrium constant \( K \) for this reaction is given as 50.

2. **Understand the Effect of Volume on Equilibrium Constant**:
The equilibrium constant \( K \) is defined by the expression:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
where \([HI]\), \([H_2]\), and \([I_2]\) are the molar concentrations of the respective gases at equilibrium.

3. **Consider the Change in Volume**:
The problem states that the volume of the container is reduced to half of its original value. While this change will affect the concentrations of the gases, it does not affect the equilibrium constant \( K \).

4. **Reasoning**:
The equilibrium constant is dependent only on temperature. Changes in volume or concentration will shift the position of equilibrium (according to Le Chatelier's principle) but will not change the value of \( K \).

5. **Conclusion**:
Since the equilibrium constant \( K \) is independent of volume changes, the value of the equilibrium constant remains the same:
\[ K = 50 \]

### Final Answer:
The value of the equilibrium constant will remain **50**.

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