1 mol of hydrogen and 2 mol of iodine are taken initially in a 2L vessel. The number of moles of hydrogen at equilibrium is `0.2`. Then the number of moles of iodine and hydrogen iodide at equilibrium are

Options

A. `1.2, 1.6`
B. `1.8, 1.0`
C. `0.4, 2.4`
D. `0.8, 2.0`

Explanation

To solve the problem step by step, we will analyze the reaction and the initial conditions given.

### Step 1: Write the balanced chemical equation
The reaction between hydrogen (H₂) and iodine (I₂) to form hydrogen iodide (HI) can be represented as:
\[ \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \]

### Step 2: Identify initial moles
Initially, we have:
- Moles of H₂ = 1 mol
- Moles of I₂ = 2 mol

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the number of moles of H₂ that react at equilibrium. Therefore, the changes in moles can be represented as follows:
- Moles of H₂ at equilibrium = \( 1 - x \)
- Moles of I₂ at equilibrium = \( 2 - x \)
- Moles of HI produced = \( 2x \) (since 1 mole of H₂ reacts with 1 mole of I₂ to produce 2 moles of HI)

### Step 4: Use the information given in the problem
We are told that at equilibrium, the number of moles of hydrogen (H₂) is 0.2 mol. Therefore, we can set up the equation:
\[ 1 - x = 0.2 \]

### Step 5: Solve for \( x \)
From the equation:
\[ 1 - x = 0.2 \]
\[ x = 1 - 0.2 = 0.8 \]

### Step 6: Calculate moles of iodine (I₂) at equilibrium
Now that we have \( x \), we can find the moles of iodine at equilibrium:
\[ \text{Moles of I}_2 = 2 - x = 2 - 0.8 = 1.2 \, \text{mol} \]

### Step 7: Calculate moles of hydrogen iodide (HI) at equilibrium
Next, we can calculate the moles of hydrogen iodide produced at equilibrium:
\[ \text{Moles of HI} = 2x = 2 \times 0.8 = 1.6 \, \text{mol} \]

### Final Results
- Moles of I₂ at equilibrium = 1.2 mol
- Moles of HI at equilibrium = 1.6 mol

### Summary
At equilibrium, we have:
- Moles of H₂ = 0.2 mol
- Moles of I₂ = 1.2 mol
- Moles of HI = 1.6 mol