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:

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

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- We start with 1 mole of HI.
- Initially, the moles of H2 and I2 are both 0.
- Therefore, we can summarize the initial moles as:
- HI: 1 mole
- H2: 0 moles
- I2: 0 moles

2. **Define Degree of Dissociation (α):**
- Let α (alpha) be the degree of dissociation of HI. This means that a fraction α of HI will dissociate into H2 and I2.

3. **Calculate Change in Moles:**
- From the balanced equation, we see that 2 moles of HI produce 1 mole of H2 and 1 mole of I2.
- If α moles of HI dissociate, then:
- Moles of HI that dissociate = 2α (since 2 moles of HI are needed for the reaction).
- Moles of H2 produced = α.
- Moles of I2 produced = α.

4. **Determine Remaining Moles of HI:**
- The remaining moles of HI after dissociation will be:
\[ \text{Remaining HI} = 1 - 2\alpha \]

5. **Calculate Total Moles at Equilibrium:**
- The total number of moles at equilibrium will be the sum of the remaining moles of HI and the moles of H2 and I2 produced:
\[ \text{Total moles} = (\text{Remaining HI}) + (\text{Moles of H2}) + (\text{Moles of I2}) \]
- Substituting the values we have:
\[ \text{Total moles} = (1 - 2\alpha) + \alpha + \alpha \]
- Simplifying this expression gives:
\[ \text{Total moles} = 1 - 2\alpha + 2\alpha = 1 \]

6. **Final Result:**
- Therefore, the total number of moles for the reaction starting with 1 mole of HI is:
\[ \text{Total moles} = 1 \]

### Conclusion:
The correct option is that the total number of moles for the reaction starting with 1 mole of HI is **1**.

---