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. `(1)/(16)`
B. `(1)/(64)`
C. 16
D. `(1)/(8)`

Explanation

To find the equilibrium constant for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), we can use the relationship between the equilibrium constants of related reactions.

### Step-by-Step Solution:

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

2. **Write the Second Reaction**:
The second reaction we need to analyze is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

3. **Relate the Two Reactions**:
Notice that the second reaction is the reverse of the first reaction, but it also involves doubling the coefficients of the first reaction.

4. **Adjust the First Reaction**:
If we multiply the first reaction by 2, we get:
\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
The equilibrium constant for this reaction, \( K_2 \), is related to \( K_1 \) by:
\[ K_2 = K_1^2 \]
Thus, substituting the value of \( K_1 \):
\[ K_2 = (8.0)^2 = 64.0 \]

5. **Take the Inverse**:
Since the reaction \( 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \) is the reverse of the reaction we need, the equilibrium constant for the desired reaction will be the inverse of \( K_2 \):
\[ K = \frac{1}{K_2} = \frac{1}{64.0} \]

6. **Final Result**:
Therefore, the equilibrium constant for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is:
\[ K = \frac{1}{64.0} \approx 0.015625 \]

### Summary:
The equilibrium constant for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is approximately \( 0.015625 \).