Two moles of HI were heated in a sealed tube at `440^@ C` till the equilibrium was reached. HI was found to be 22% decomposed.The equilibrium constant for disssociation is :

Options

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

Answer

To find 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 as:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Step 2: Determine the initial moles and the change at equilibrium
Initially, we have 2 moles of HI. Given that HI is 22% decomposed, we can calculate the amount of HI that has decomposed:
- Decomposed moles of HI = 22% of 2 moles = \( 0.22 \times 2 = 0.44 \) moles

### Step 3: Calculate the moles at equilibrium
At equilibrium:
- Moles of HI remaining = Initial moles - Decomposed moles = \( 2 - 0.44 = 1.56 \) moles
- Since 2 moles of HI produce 1 mole of H2 and 1 mole of I2, the moles of H2 and I2 produced will be:
- Moles of H2 = Moles of I2 = \( \frac{0.44}{2} = 0.22 \) moles

### Step 4: Set up the equilibrium concentrations
Assuming the volume of the container is 1 liter (for simplicity):
- Concentration of HI = \( \frac{1.56 \text{ moles}}{1 \text{ L}} = 1.56 \text{ M} \)
- Concentration of H2 = \( \frac{0.22 \text{ moles}}{1 \text{ L}} = 0.22 \text{ M} \)
- Concentration of I2 = \( \frac{0.22 \text{ moles}}{1 \text{ L}} = 0.22 \text{ M} \)

### 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 expression
Substituting the values we calculated:
\[ K_c = \frac{(0.22)(0.22)}{(1.56)^2} \]

### Step 7: Calculate the value of Kc
Calculating the numerator:
\[ 0.22 \times 0.22 = 0.0484 \]

Calculating the denominator:
\[ (1.56)^2 = 2.4336 \]

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

### Conclusion
The equilibrium constant \( K_c \) for the dissociation of HI at 440°C is approximately \( 0.0199 \).

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