Suppose the value of the equilibrium constant for the given reaction `H_2(g) + I_(2)(g) hArr 2HI(g)` at 717 K is 64, then calculate the same (eq. const.) for the reaction `HI hArr 1/2 H_2 + 1/2 I_2`.

Options

A. 64
B. 8
C. `1/64`
D. `1/8`

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ \text{HI} \rightleftharpoons \frac{1}{2} \text{H}_2 + \frac{1}{2} \text{I}_2 \]

given that the equilibrium constant for the reaction:

\[ \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \]

is \( K_1 = 64 \) at 717 K.

### Step-by-Step Solution:

1. **Identify the Equilibrium Constants**:
- For the first reaction (let's call it Reaction 1):
\[ K_1 = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 64 \]
- For the second reaction (let's call it Reaction 2):
\[ K_2 = \frac{[\text{H}_2]^{1/2}[\text{I}_2]^{1/2}}{[\text{HI}]} \]

2. **Relate the Two Reactions**:
- Reaction 2 can be derived from Reaction 1 by reversing it and halving the coefficients:
\[ \text{HI} \rightleftharpoons \frac{1}{2} \text{H}_2 + \frac{1}{2} \text{I}_2 \]
- When a reaction is reversed, the equilibrium constant becomes the reciprocal:
\[ K' = \frac{1}{K_1} \]
- When the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. In this case, since we are halving the coefficients, we take the square root:
\[ K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} \]

3. **Calculate \( K_2 \)**:
- Substitute \( K_1 = 64 \):
\[ K_2 = \frac{1}{\sqrt{64}} = \frac{1}{8} = 0.125 \]

### Final Answer:
The equilibrium constant \( K_2 \) for the reaction \( \text{HI} \rightleftharpoons \frac{1}{2} \text{H}_2 + \frac{1}{2} \text{I}_2 \) is \( 0.125 \).

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