The value of equilibrium constant of the reaction. `HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g)` is `8.0` The equilibrium constant of the reaction. `H_2(g)+I_2(g)hArr2HI(g)` will be

Options

A. (a) `(1)/(16)`
B. (b) `(1)/(64)`
C. (c) `16`
D. (d) `(1)/(8)`

Explanation

To solve the problem, we need to find the equilibrium constant \( K_2 \) for the reaction:

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

given that the equilibrium constant \( K_1 \) for the reaction:

\[ HI(g) \rightleftharpoons \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \]

is \( 8.0 \).

### Step-by-Step Solution:

1. **Identify the Given Equilibrium Constant**:
- The equilibrium constant \( K_1 \) for the reaction \( HI(g) \rightleftharpoons \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \) is given as \( K_1 = 8.0 \).

2. **Write the Expression for \( K_1 \)**:
- The expression for the equilibrium constant \( K_1 \) can be written as:
\[ K_1 = \frac{[H_2]^{1/2} [I_2]^{1/2}}{[HI]} \]

3. **Determine the Relationship Between \( K_1 \) and \( K_2 \)**:
- The second reaction is the reverse of the first reaction multiplied by 2:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
- The equilibrium constant for this reaction, \( K_2 \), can be related to \( K_1 \) using the following relationships:
- When a reaction is reversed, the equilibrium constant is 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:
\[ K_2 = (K')^2 \]

4. **Calculate \( K_2 \)**:
- First, find \( K' \):
\[ K' = \frac{1}{K_1} = \frac{1}{8.0} = 0.125 \]
- Now, square \( K' \) to find \( K_2 \):
\[ K_2 = (K')^2 = (0.125)^2 = \frac{1}{64} \]

5. **Final Answer**:
- Therefore, the equilibrium constant \( K_2 \) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is:
\[ K_2 = \frac{1}{64} \]

### Conclusion:
The value of the equilibrium constant \( K_2 \) is \( \frac{1}{64} \).