Two moles of HI when heated at `444^@C` until equilibrium is reached were found to be 22% dissociated. Calculate equilibrium constant for the reaction.

Explanation

To calculate the equilibrium constant for the dissociation of hydrogen iodide (HI), we can follow these steps:

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

### Step 2: Determine the initial moles of HI
We start with 2 moles of HI.

### Step 3: Calculate the amount dissociated
Since it is given that 22% of HI is dissociated, we can calculate the moles of HI that dissociate:
\[ \text{Moles of HI dissociated} = 22\% \text{ of } 2 \text{ moles} = 0.22 \times 2 = 0.44 \text{ moles} \]

### Step 4: Calculate the moles at equilibrium
At equilibrium, the moles of HI remaining can be calculated as:
\[ \text{Moles of HI at equilibrium} = 2 - 0.44 = 1.56 \text{ moles} \]

The stoichiometry of the reaction tells us that for every 2 moles of HI that dissociate, 1 mole of H2 and 1 mole of I2 are produced. Therefore, the moles of H2 and I2 produced at equilibrium will be:
\[ \text{Moles of H}_2 = \text{Moles of I}_2 = \frac{0.44}{2} = 0.22 \text{ moles} \]

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

### Step 6: Substitute the equilibrium concentrations into the Kc expression
To find the concentrations, we need to consider the total volume of the system. Assuming the reaction occurs in a 1 L container (for simplicity), the concentrations will be equal to the number of moles:
- \([\text{HI}] = \frac{1.56 \text{ moles}}{1 \text{ L}} = 1.56 \text{ M}\)
- \([\text{H}_2] = \frac{0.22 \text{ moles}}{1 \text{ L}} = 0.22 \text{ M}\)
- \([\text{I}_2] = \frac{0.22 \text{ moles}}{1 \text{ L}} = 0.22 \text{ M}\)

Now, substituting these values into the Kc expression:
\[ K_c = \frac{(0.22)(0.22)}{(1.56)^2} \]

### Step 7: Calculate Kc
Calculating the values:
\[ K_c = \frac{0.0484}{2.4336} \approx 0.0199 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at \( 444^\circ C \) is approximately \( 0.0199 \).

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