If the equilibrium constant of the reaction `2HI hArrH_(2) + I_(2)` is `0.25` , then the equilibrium constant of the reaction `H_(2) + I_(2)hArr2HI` would be

Options

A. `1.0`
B. `2.0`
C. `3.0`
D. `4.0`

Explanation

To find the equilibrium constant for the reaction \( H_2 + I_2 \rightleftharpoons 2HI \) given that the equilibrium constant for the reaction \( 2HI \rightleftharpoons H_2 + I_2 \) is \( K = 0.25 \), we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:
\[ 2HI \rightleftharpoons H_2 + I_2 \]
The equilibrium constant for this reaction is:
\[ K_1 = 0.25 \]

### Step 2: Write the reaction for which we need to find the equilibrium constant
We need to find the equilibrium constant for the reverse reaction:
\[ H_2 + I_2 \rightleftharpoons 2HI \]
Let's denote the equilibrium constant for this reaction as \( K_2 \).

### Step 3: Relate the equilibrium constants of the forward and reverse reactions
The relationship between the equilibrium constants of a reaction and its reverse is given by:
\[ K_2 = \frac{1}{K_1} \]
This means that if you reverse a reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

### Step 4: Substitute the value of \( K_1 \)
Now we can substitute the value of \( K_1 \) into the equation:
\[ K_2 = \frac{1}{0.25} \]

### Step 5: Calculate \( K_2 \)
Calculating \( K_2 \):
\[ K_2 = \frac{1}{0.25} = 4 \]

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

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