One mole of `H_(2)` and 2 moles of `I_(2) ` are taken initially in a two litre vessel. The number of moles of `H_(2)` at equilibrium is 0.2. Then the number of moles of `I_(2)` and `HI` at equilibrium is

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 chemical reaction and the changes in the number of moles of each substance involved.

### Step 1: Write the balanced chemical equation
The reaction between hydrogen (H₂) and iodine (I₂) can be represented as:
\[ H_2 + I_2 \rightleftharpoons 2 HI \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of \( H_2 = 1 \) mole
- Moles of \( I_2 = 2 \) moles
- Moles of \( HI = 0 \) moles

### Step 3: Define the change in moles
Let \( x \) be the amount of \( H_2 \) that reacts at equilibrium. Therefore, at equilibrium, the moles of each substance will be:
- Moles of \( H_2 = 1 - x \)
- Moles of \( I_2 = 2 - x \)
- Moles of \( HI = 2x \)

### Step 4: Use the information given about \( H_2 \)
We are told that at equilibrium, the number of moles of \( H_2 \) is 0.2 moles. Thus, we can set up the equation:
\[ 1 - x = 0.2 \]

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

### Step 6: Calculate the moles of \( I_2 \) at equilibrium
Using the value of \( x \) in the expression for \( I_2 \):
\[ \text{Moles of } I_2 = 2 - x = 2 - 0.8 = 1.2 \]

### Step 7: Calculate the moles of \( HI \) at equilibrium
Now, we can find the moles of \( HI \):
\[ \text{Moles of } HI = 2x = 2 \times 0.8 = 1.6 \]

### Conclusion
At equilibrium, the number of moles of \( I_2 \) is 1.2 moles, and the number of moles of \( HI \) is 1.6 moles.

### Final Answer
- Moles of \( I_2 \) at equilibrium: **1.2 moles**
- Moles of \( HI \) at equilibrium: **1.6 moles**

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