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:
\[ H_2 + I_2 \rightleftharpoons 2 HI \]

### Step 2: Set up the initial moles
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 number of moles of \( H_2 \) that dissociate at equilibrium. According to the stoichiometry of the reaction:
- \( H_2 \) will decrease by \( x \)
- \( I_2 \) will decrease by \( x \)
- \( HI \) will increase by \( 2x \)

At equilibrium, the moles will be:
- Moles of \( H_2 = 2 - x \)
- Moles of \( I_2 = 1 - x \)
- Moles of \( HI = 0 + 2x = 2x \)

### Step 4: Use the information given in the problem
We know that at equilibrium, the vessel contains 0.5 moles of \( HI \):
\[ 2x = 0.5 \]

### Step 5: Solve for \( x \)
From the equation \( 2x = 0.5 \):
\[ x = \frac{0.5}{2} = 0.25 \]

### Step 6: Calculate the degree of dissociation
The degree of dissociation \( \alpha \) is defined as the fraction of the initial amount that has dissociated. For \( H_2 \):
\[ \alpha = \frac{\text{moles dissociated}}{\text{initial moles}} = \frac{x}{2} \]
Substituting the value of \( x \):
\[ \alpha = \frac{0.25}{2} = 0.125 \]

### Step 7: Convert to percentage
To express the degree of dissociation as a percentage:
\[ \text{Percentage } \alpha = 0.125 \times 100 = 12.5\% \]

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

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