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

Answer

To solve the problem, we need to determine how the equilibrium constant \( K_c \) changes when the volume of the container is reduced to half its original volume for the reaction:

\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

### 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 \( K_c \) for this reaction is given as 64.

2. **Calculate the Change in Moles (\( \Delta N_g \)):**
- To find \( \Delta N_g \), we need to count the moles of reactants and products.
- **Products:** 2 moles of \( HI \).
- **Reactants:** 1 mole of \( H_2 \) + 1 mole of \( I_2 \) = 2 moles total.
- Therefore, \( \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \).

3. **Understand the Effect of Volume Change on \( K_c \):**
- According to Le Chatelier's principle, if the volume of the container is reduced, the system will shift to counteract the change.
- However, since \( \Delta N_g = 0 \), there is no shift in the equilibrium position when the volume is halved.
- The equilibrium constant \( K_c \) is only affected by changes in temperature, not by changes in volume.

4. **Conclusion:**
- Since \( K_c \) does not change with a change in volume when \( \Delta N_g = 0 \), the value of \( K_c \) remains 64.
- Therefore, the final answer is \( K_c = 64 \).

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

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