A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2H`, the K is found to be 64. The amount of undreacted `I_(2)` at equilibrium is

Options

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

Explanation

To solve the problem step by step, we will follow the reaction and apply the principles of chemical equilibrium.

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

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

The initial concentrations are:
\[ [H_2] = \frac{0.3 \text{ moles}}{10 \text{ L}} = 0.03 \text{ M} \]
\[ [I_2] = \frac{0.3 \text{ moles}}{10 \text{ L}} = 0.03 \text{ M} \]
\[ [HI] = 0 \text{ M} \]

### Step 3: Define the change in concentration at equilibrium
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that reacts at equilibrium. Therefore, at equilibrium:
- \( [H_2] = 0.03 - x \)
- \( [I_2] = 0.03 - x \)
- \( [HI] = 2x \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(0.03 - x)(0.03 - x)} \]
Given \( K_c = 64 \), we can set up the equation:
\[ 64 = \frac{4x^2}{(0.03 - x)^2} \]

### Step 5: Rearranging the equation
Multiply both sides by \( (0.03 - x)^2 \):
\[ 64(0.03 - x)^2 = 4x^2 \]
Expanding the left side:
\[ 64(0.0009 - 0.06x + x^2) = 4x^2 \]
\[ 0.0576 - 3.84x + 64x^2 = 4x^2 \]
Rearranging gives:
\[ 60x^2 - 3.84x + 0.0576 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 60 \), \( b = -3.84 \), \( c = 0.0576 \)

Calculating the discriminant:
\[ b^2 - 4ac = (-3.84)^2 - 4 \times 60 \times 0.0576 \]
\[ = 14.7456 - 13.824 = 0.9216 \]

Now substituting into the quadratic formula:
\[ x = \frac{3.84 \pm \sqrt{0.9216}}{120} \]
Calculating \( \sqrt{0.9216} \approx 0.96 \):
\[ x = \frac{3.84 \pm 0.96}{120} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{4.8}{120} = 0.04 \)
2. \( x = \frac{2.88}{120} = 0.024 \)

### Step 7: Determine the amount of undreacted \( I_2 \)
Using \( x = 0.024 \) (the smaller value since it must be less than 0.03):
\[ [I_2]_{eq} = 0.03 - x = 0.03 - 0.024 = 0.006 \text{ moles} \]

### Final Answer
The amount of undreacted \( I_2 \) at equilibrium is:
\[ \text{Undreacted } I_2 = 0.006 \text{ moles} \]

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