A mixture of 0.3 moles of `H_2 and 0.3` moles of `l_2` is allowed to react in a 10 litre evacuated flask at 500°C. The reaction is `H_2 + I_2 = 2HI, the K is found to be 64. The amount of unreacted 'l_2' equilibrium is

Options

A. 0.15 mole
B. 0.06 mole
C. 0.03 mole
D. 0.2 mole

Answer

To solve the problem step by step, we will analyze the reaction and apply the equilibrium constant expression.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ H_2 + I_2 \rightleftharpoons 2HI \]

### Step 2: Set up initial conditions
We start with:
- Initial moles of \( H_2 = 0.3 \) moles
- Initial moles of \( I_2 = 0.3 \) moles
- Volume of the flask = 10 liters

### Step 3: Define the change in moles
Let \( x \) be the number of moles of \( H_2 \) (and \( I_2 \)) that react. At equilibrium:
- Moles of \( H_2 \) remaining = \( 0.3 - x \)
- Moles of \( I_2 \) remaining = \( 0.3 - x \)
- Moles of \( HI \) formed = \( 2x \)

### Step 4: Write the equilibrium concentrations
The equilibrium concentrations can be calculated as:
- Concentration of \( H_2 = \frac{0.3 - x}{10} \)
- Concentration of \( I_2 = \frac{0.3 - x}{10} \)
- Concentration of \( HI = \frac{2x}{10} \)

### Step 5: Write the equilibrium constant expression
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2x}{10}\right)^2}{\left(\frac{0.3 - x}{10}\right)\left(\frac{0.3 - x}{10}\right)} \]

### Step 6: Simplify the expression
This simplifies to:
\[ K_c = \frac{4x^2}{(0.3 - x)^2} \]
Given \( K_c = 64 \), we can set up the equation:
\[ 64 = \frac{4x^2}{(0.3 - x)^2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 64(0.3 - x)^2 = 4x^2 \]
Expanding the left side:
\[ 64(0.09 - 0.6x + x^2) = 4x^2 \]
This leads to:
\[ 5.76 - 38.4x + 64x^2 = 4x^2 \]
Rearranging gives:
\[ 60x^2 - 38.4x + 5.76 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 60 \), \( b = -38.4 \), and \( c = 5.76 \):
1. Calculate the discriminant:
\[ b^2 - 4ac = (-38.4)^2 - 4 \times 60 \times 5.76 \]
\[ = 1474.56 - 1382.4 = 92.16 \]
2. Calculate \( x \):
\[ x = \frac{38.4 \pm \sqrt{92.16}}{120} \]
\[ = \frac{38.4 \pm 9.6}{120} \]
This gives two possible values for \( x \):
\[ x = \frac{48}{120} = 0.4 \text{ (not possible since it exceeds initial moles)} \]
\[ x = \frac{28.8}{120} = 0.24 \]

### Step 9: Calculate unreacted \( I_2 \)
The amount of unreacted \( I_2 \) at equilibrium is:
\[ \text{Unreacted } I_2 = 0.3 - x = 0.3 - 0.24 = 0.06 \text{ moles} \]

### Final Answer
The amount of unreacted \( I_2 \) at equilibrium is **0.06 moles**.

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