If `alpha` is degree of dissocliation, then the total number of moles for the reaction starting with 1 mole of HI `2HI= H_2 +I_2` will be

Options

A. 1
B. `1-alpha`
C. 2
D. `2-alpha`

Answer

To solve the problem, we need to determine the total number of moles at equilibrium for the reaction:

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

starting with 1 mole of HI and given that the degree of dissociation is represented by \( \alpha \).

### Step-by-Step Solution:

1. **Initial Moles**:
- Initially, we have 1 mole of HI, and no moles of H2 or I2.
- Therefore, the initial amounts can be represented as:
- \[\text{HI} = 1 \text{ mole}\]
- \[\text{H}_2 = 0 \text{ moles}\]
- \[\text{I}_2 = 0 \text{ moles}\]

2. **Degree of Dissociation**:
- The degree of dissociation \( \alpha \) indicates the fraction of the initial moles of HI that dissociate.
- For the reaction \( 2 \text{HI} \rightarrow \text{H}_2 + \text{I}_2 \), if \( \alpha \) is the degree of dissociation, then:
- Moles of HI dissociated = \( 2\alpha \)
- Moles of H2 formed = \( \alpha \)
- Moles of I2 formed = \( \alpha \)

3. **Moles at Equilibrium**:
- After dissociation, the moles of each species will be:
- Moles of HI remaining = \( 1 - 2\alpha \)
- Moles of H2 = \( \alpha \)
- Moles of I2 = \( \alpha \)

4. **Total Moles at Equilibrium**:
- To find the total number of moles at equilibrium, we sum the moles of all species:
\[ \text{Total moles} = (\text{Moles of HI remaining}) + (\text{Moles of H2}) + (\text{Moles of I2}) \]
\[ \text{Total moles} = (1 - 2\alpha) + \alpha + \alpha \]
\[ \text{Total moles} = 1 - 2\alpha + 2\alpha \]
\[ \text{Total moles} = 1 \]

### Conclusion:
The total number of moles at equilibrium is **1 mole**.