If the equilibrium constant of the reaction
`2HI(g)hArrH_(2)(g)+I_(2)(g)` is `0.25`, find the equilibrium constant of the reaction.
`(1)/(2)H_(2)+(1)/(2)I_(2)hArrHI(g)`

Explanation

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

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

given that the equilibrium constant for the reaction:

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

is \( K_c = 0.25 \).

### Step-by-Step Solution:

1. **Write the given equilibrium reaction and its constant:**
The first reaction is:
\[ 2 HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
The equilibrium constant for this reaction is given as:
\[ K_{c1} = 0.25 \]

2. **Reverse the reaction:**
When we reverse the reaction, we get:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]
The equilibrium constant for the reversed reaction is the reciprocal of the original:
\[ K_{c1}' = \frac{1}{K_{c1}} = \frac{1}{0.25} = 4 \]

3. **Multiply the reversed reaction by 1/2:**
Now, we need to find the equilibrium constant for the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g) \]
To obtain this reaction from the reversed reaction, we multiply the entire equation by \( \frac{1}{2} \):
\[ \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g) \]

4. **Relate the equilibrium constant:**
When we multiply a reaction by a factor, the equilibrium constant is raised to the power of that factor. Since we multiplied the reaction by \( \frac{1}{2} \), we have:
\[ K_{c2} = (K_{c1}')^{\frac{1}{2}} = (4)^{\frac{1}{2}} = 2 \]

5. **Final answer:**
Therefore, the equilibrium constant for the reaction:
\[ \frac{1}{2} H_2 + \frac{1}{2} I_2 \rightleftharpoons HI(g) \]
is:
\[ K_{c2} = 2 \]