If the equilibrium constant of the reaction `2HIhArr H_(2)+I_(2)` is 0.25, then the equilibrium constant for the reaction, `H_(2)(g)+I_(2)(g)hArr 2HI(g)` would be

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step-by-Step Solution:

1. **Identify the given reaction and its equilibrium constant**:
The given reaction is:
\[ 2HI \rightleftharpoons H_2 + I_2 \]
The equilibrium constant \( K_1 \) for this reaction is given as:
\[ K_1 = 0.25 \]

2. **Write the equilibrium expression for the given reaction**:
The equilibrium constant expression for the reaction \( 2HI \rightleftharpoons H_2 + I_2 \) is:
\[ K_1 = \frac{[H_2][I_2]}{[HI]^2} \]

3. **Reverse the reaction**:
To find the equilibrium constant for the reverse reaction \( H_2 + I_2 \rightleftharpoons 2HI \), we need to reverse the original reaction. The equilibrium constant for a reversed reaction \( K_2 \) is the reciprocal of \( K_1 \):
\[ K_2 = \frac{1}{K_1} \]

4. **Calculate the new equilibrium constant**:
Substituting the value of \( K_1 \):
\[ K_2 = \frac{1}{0.25} = 4 \]

5. **Conclusion**:
Therefore, the equilibrium constant for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is:
\[ K_2 = 4 \]

### Final Answer:
The equilibrium constant for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is **4**.

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