In a 10 litre box 2.5 mole hydroiodic acid is taken. After equilibrium
`2HI

Options

A. 2.4
B. 0.15
C. 1.5
D. `7.5 xx 10^(-2)`

Answer

To solve the problem step by step, we will analyze the equilibrium reaction and the given information.

### Step 1: Write the balanced chemical equation
The balanced reaction is:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Step 2: Identify initial moles and concentrations
Initially, we have:
- Moles of HI = 2.5 moles
- Moles of H₂ = 0 moles
- Moles of I₂ = 0 moles

The volume of the box is 10 liters.

### Step 3: Calculate initial concentrations
The initial concentration of HI can be calculated as:
\[ \text{Initial concentration of HI} = \frac{\text{moles}}{\text{volume}} = \frac{2.5 \, \text{moles}}{10 \, \text{L}} = 0.25 \, \text{mol L}^{-1} \]
The initial concentrations of H₂ and I₂ are both 0 mol L⁻¹.

### Step 4: Set up the equilibrium expression
Let \( x \) be the amount of HI that dissociates at equilibrium. The changes in concentration can be expressed as:
- Concentration of HI at equilibrium = \( 0.25 - 2x \)
- Concentration of H₂ at equilibrium = \( x \)
- Concentration of I₂ at equilibrium = \( x \)

### Step 5: Use the given equilibrium concentration of HI
We are given that the concentration of HI at equilibrium is \( 0.1 \, \text{mol L}^{-1} \). Therefore, we can set up the equation:
\[ 0.25 - 2x = 0.1 \]

### Step 6: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 0.25 - 0.1 \]
\[ 2x = 0.15 \]
\[ x = \frac{0.15}{2} = 0.075 \]

### Step 7: Calculate the concentration of H₂ at equilibrium
The concentration of H₂ at equilibrium is given by:
\[ \text{Concentration of H}_2 = \frac{x}{10} = \frac{0.075}{10} = 0.0075 \, \text{mol L}^{-1} \]

### Final Answer
The concentration of H₂ at equilibrium is:
\[ \text{Concentration of H}_2 = 0.0075 \, \text{mol L}^{-1} \text{ or } 7.5 \times 10^{-3} \, \text{mol L}^{-1} \]