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

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

given that the equilibrium constant for the reaction:

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

is 0.25.

### Step-by-Step Solution:

1. **Write the Equilibrium Expression for the First Reaction:**
The equilibrium constant \( K \) for the reaction \( 2HI \rightleftharpoons H_2 + I_2 \) can be expressed as:
\[ K = \frac{[H_2][I_2]}{[HI]^2} \]
Given that \( K = 0.25 \), we have:
\[ 0.25 = \frac{[H_2][I_2]}{[HI]^2} \]

2. **Write the Equilibrium Expression for the Second Reaction:**
For the reverse reaction \( H_2 + I_2 \rightleftharpoons 2HI \), the equilibrium constant \( K_1 \) can be expressed as:
\[ K_1 = \frac{[HI]^2}{[H_2][I_2]} \]

3. **Relate the Two Equilibrium Constants:**
The relationship between the equilibrium constants of a reaction and its reverse is given by:
\[ K_1 = \frac{1}{K} \]
Therefore, substituting the value of \( K \):
\[ K_1 = \frac{1}{0.25} \]

4. **Calculate \( K_1 \):**
\[ K_1 = 4 \]

5. **Conclusion:**
The equilibrium constant for the reaction \( H_2 + I_2 \rightleftharpoons 2HI \) is \( 4 \).

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

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