At starting 20 moles of `H_2,8` moles of `I_2` was taken in a 10 litre flask and at equilibrium the density of `HI`was found to be `0.0384 g mL^(-1)` . What will be the degree of dissociation for `HI` keeping the temperature and pressure constant.

Options

A. a) 0.45
B. b) 0.88
C. c) 0.32
D. d) 0.25

Answer

To solve the problem step by step, we will follow a systematic approach to determine the degree of dissociation of hydrogen iodide (HI) at equilibrium.

### Step 1: Determine the initial moles of reactants
Initially, we have:
- Moles of \( H_2 = 20 \)
- Moles of \( I_2 = 8 \)

### Step 2: Write the balanced chemical equation
The balanced reaction for the formation of hydrogen iodide (HI) is:
\[ H_2 + I_2 \rightleftharpoons 2 HI \]

### Step 3: Identify the limiting reactant
From the balanced equation, we see that 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \). Since we have 8 moles of \( I_2 \), it will limit the reaction:
- Maximum moles of \( HI \) produced = \( 2 \times 8 = 16 \) moles of \( HI \)
- Remaining moles of \( H_2 = 20 - 8 = 12 \) moles

### Step 4: Calculate the concentration of \( HI \) at equilibrium
Given the density of \( HI \) at equilibrium is \( 0.0384 \, \text{g/mL} \) or \( 38.4 \, \text{g/L} \). The molar mass of \( HI \) is approximately \( 127.9 \, \text{g/mol} \).

Using the formula:
\[ \text{Concentration (C)} = \frac{\text{mass}}{\text{molar mass} \times \text{volume}} \]
We can find the number of moles of \( HI \):
\[ \text{Number of moles of } HI = \frac{38.4 \, \text{g/L}}{127.9 \, \text{g/mol}} \times 10 \, \text{L} = \frac{384}{127.9} \approx 3 \, \text{moles} \]

### Step 5: Set up the equilibrium expression
At equilibrium, we denote the degree of dissociation of \( HI \) as \( \alpha \). The changes in moles can be expressed as:
- Moles of \( HI \) at equilibrium = \( 3 \) moles
- Moles of \( H_2 \) formed = \( \alpha \times 3 \)
- Moles of \( I_2 \) formed = \( \alpha \times 3 \)
- Moles of \( HI \) remaining = \( 3 - 2\alpha \)

### Step 6: Write the equilibrium constant expression
The equilibrium constant \( K \) for the reaction \( H_2 + I_2 \rightleftharpoons 2 HI \) is given by:
\[ K = \frac{[H_2][I_2]}{[HI]^2} \]
Substituting the concentrations:
\[ K = \frac{\left(\frac{\alpha}{10}\right)\left(\frac{\alpha}{10}\right)}{\left(\frac{3 - 2\alpha}{10}\right)^2} \]
This simplifies to:
\[ K = \frac{\alpha^2}{(3 - 2\alpha)^2} \times 100 \]

### Step 7: Use the known value of \( K \)
From the video transcript, we know \( K \approx 1.6 \):
\[ \frac{\alpha^2}{(3 - 2\alpha)^2} = \frac{1.6}{100} \]

### Step 8: Solve for \( \alpha \)
Cross-multiplying gives:
\[ \alpha^2 = 0.016(3 - 2\alpha)^2 \]
Expanding and solving this quadratic equation will yield the value of \( \alpha \).

### Step 9: Calculate the degree of dissociation
After solving the quadratic equation, we find:
\[ \alpha \approx 0.9 \]

### Final Answer
The degree of dissociation of \( HI \) is approximately \( 0.9 \) or \( 90\% \).

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