`HI` was heated in a sealed tube at `400^(@)C` till the equilibrium was reached. HI was found to be `22%` decomposed. The equilibrium constant for dissociation is

Options

A. `1.99`
B. `0.0199`
C. `0.0796`
D. `0.282`

Explanation

To find the equilibrium constant for the dissociation of hydrogen iodide (HI) at 400°C, we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of hydrogen iodide can be represented by the following balanced equation:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Step 2: Set up the initial concentrations
Let’s assume the initial concentration of HI is 2 M (molar). Initially, the concentrations of H₂ and I₂ are 0 M since they are products of the reaction.

- Initial concentrations:
- [HI] = 2 M
- [H₂] = 0 M
- [I₂] = 0 M

### Step 3: Determine the change in concentrations
According to the problem, HI is 22% decomposed at equilibrium. This means that 22% of the initial concentration of HI has reacted.

- Amount of HI decomposed = 22% of 2 M = \( \frac{22}{100} \times 2 = 0.44 \) M

Since the stoichiometry of the reaction shows that 2 moles of HI produce 1 mole of H₂ and 1 mole of I₂, we can express the changes in concentration as follows:

- Change in [HI] = -0.44 M (since it is decomposing)
- Change in [H₂] = +0.22 M (since 0.44 M of HI produces 0.22 M of H₂)
- Change in [I₂] = +0.22 M (same as H₂)

### Step 4: Write the equilibrium concentrations
Now we can calculate the equilibrium concentrations:
- [HI] at equilibrium = Initial [HI] - Change in [HI] = \( 2 - 0.44 = 1.56 \) M
- [H₂] at equilibrium = 0 + 0.22 = 0.22 M
- [I₂] at equilibrium = 0 + 0.22 = 0.22 M

### Step 5: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by the formula:
\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Now substitute the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{(0.22)(0.22)}{(1.56)^2} \]

### Step 7: Calculate \( K_c \)
Calculating the values:
- Numerator: \( 0.22 \times 0.22 = 0.0484 \)
- Denominator: \( (1.56)^2 = 2.4336 \)

Now, substituting these values into the equation:
\[ K_c = \frac{0.0484}{2.4336} \approx 0.0199 \]

### Conclusion
The equilibrium constant \( K_c \) for the dissociation of HI at 400°C is approximately \( 0.0199 \).

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