256 g of HI were heated in a sealed bulb at 444°C till the equilibrium was attained. The acid was found to be 22% dissociated at equilibrium. Calculate the equilibrium constant for the reaction `2HI(g) hArr H_2(g) +I_2 (g) `

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_c \) for the reaction:

\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

### Step 1: Calculate the number of moles of HI

First, we need to find the number of moles of HI using the formula:

\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} \]

The molecular weight of HI is calculated as follows:

\[ \text{Molecular weight of HI} = 1 \, (\text{H}) + 127 \, (\text{I}) = 128 \, \text{g/mol} \]

Now, substituting the values:

\[ \text{Number of moles of HI} = \frac{256 \, \text{g}}{128 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 2: Calculate the amount dissociated

We are given that the dissociation of HI is 22%. Therefore, the number of moles that dissociate is:

\[ \text{Moles dissociated} = 2 \, \text{moles} \times \frac{22}{100} = 0.44 \, \text{moles} \]

### Step 3: Calculate remaining moles of HI

The remaining moles of HI after dissociation can be calculated as:

\[ \text{Remaining moles of HI} = \text{Initial moles} - \text{Moles dissociated} = 2 - 0.44 = 1.56 \, \text{moles} \]

### Step 4: Calculate moles of products formed

From the stoichiometry of the reaction, for every 2 moles of HI that dissociate, 1 mole of H2 and 1 mole of I2 are formed. Therefore, the moles of H2 and I2 formed are:

\[ \text{Moles of H}_2 = \text{Moles of I}_2 = \frac{0.44}{2} = 0.22 \, \text{moles} \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

### Step 6: Calculate concentrations

Assuming the volume of the container is \( V \):

- Concentration of H2:
\[ [\text{H}_2] = \frac{0.22}{V} \]

- Concentration of I2:
\[ [\text{I}_2] = \frac{0.22}{V} \]

- Concentration of HI:
\[ [\text{HI}] = \frac{1.56}{V} \]

### Step 7: Substitute concentrations into the \( K_c \) expression

Substituting the concentrations into the equilibrium constant expression:

\[ K_c = \frac{\left(\frac{0.22}{V}\right) \left(\frac{0.22}{V}\right)}{\left(\frac{1.56}{V}\right)^2} \]

This simplifies to:

\[ K_c = \frac{0.22 \times 0.22}{1.56^2} \]

### Step 8: Calculate \( K_c \)

Calculating the values:

\[ K_c = \frac{0.0484}{2.4336} \approx 0.02 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) is:

\[ K_c \approx 0.02 \]

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