If `alpha` is the fraction of HI dissociated at equilibrium in the reaction, `2HI(g)hArrH_2(g)+I_2(g)` starting with the 2 moles of HI. Then the total number of moles of reactants and products at equilibrium are

Options

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

Explanation

To solve the problem, we need to analyze the dissociation of hydrogen iodide (HI) in the given reaction:

\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

We start with 2 moles of HI and let \( \alpha \) be the fraction of HI that dissociates at equilibrium.

### Step-by-Step Solution:

1. **Initial Moles**:
- Initially, we have 2 moles of HI and 0 moles of H2 and I2.
- So, at the start (before any reaction occurs):
- Moles of HI = 2
- Moles of H2 = 0
- Moles of I2 = 0

2. **Change in Moles**:
- At equilibrium, \( \alpha \) fraction of HI dissociates. Since we started with 2 moles of HI, the amount that dissociates is:
\[ \text{Moles of HI dissociated} = 2\alpha \]
- Therefore, the moles of HI remaining at equilibrium will be:
\[ \text{Moles of HI at equilibrium} = 2 - 2\alpha \]

3. **Formation of Products**:
- For every 2 moles of HI that dissociate, 1 mole of H2 and 1 mole of I2 are formed. Thus, the moles of H2 and I2 formed at equilibrium will be:
\[ \text{Moles of H2 at equilibrium} = \alpha \]
\[ \text{Moles of I2 at equilibrium} = \alpha \]

4. **Total Moles at Equilibrium**:
- Now, we can calculate the total number of moles at equilibrium by summing the moles of all species:
\[ \text{Total moles} = \text{Moles of HI} + \text{Moles of H2} + \text{Moles of I2} \]
\[ \text{Total moles} = (2 - 2\alpha) + \alpha + \alpha \]
\[ \text{Total moles} = 2 - 2\alpha + 2\alpha = 2 \]

Thus, the total number of moles of reactants and products at equilibrium is **2**.

### Final Answer:
The total number of moles of reactants and products at equilibrium is **2**.