2 mole of `H_(2) and "1 mole of "I_(2)` are heated in a closed 1 litre vessel. At equilibrium, the vessel contains 0.5 mole HI. The degree of dissociation of `H_(2)` is

Options

A. `12.5%`
B. `60%`
C. `79.1%`
D. `71.9%`

Answer

To find the degree of dissociation of \( H_2 \) in the given reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and iodine can be represented as:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of \( H_2 = 2 \)
- Moles of \( I_2 = 1 \)
- Moles of \( HI = 0 \)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of \( H_2 \) that dissociates at equilibrium. Therefore, the changes in moles can be expressed as:
- Moles of \( H_2 \) at equilibrium = \( 2 - x \)
- Moles of \( I_2 \) at equilibrium = \( 1 - x \)
- Moles of \( HI \) formed = \( 2x \)

### Step 4: Use the information given about equilibrium
At equilibrium, it is given that the vessel contains 0.5 moles of \( HI \):
\[ 2x = 0.5 \]
From this, we can solve for \( x \):
\[ x = \frac{0.5}{2} = 0.25 \]

### Step 5: Calculate the degree of dissociation of \( H_2 \)
The degree of dissociation \( \alpha \) is defined as the number of moles that dissociate divided by the initial number of moles:
\[ \alpha = \frac{x}{\text{initial moles of } H_2} \]
Substituting the values:
\[ \alpha = \frac{0.25}{2} = \frac{1}{8} \]

### Step 6: Convert to percentage
To express the degree of dissociation as a percentage:
\[ \text{Percentage } \alpha = \alpha \times 100 = \frac{1}{8} \times 100 = 12.5\% \]

### Final Answer
The degree of dissociation of \( H_2 \) is \( 12.5\% \).

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