For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 1000K . If 1.0 mole of `I_(2)` is introduced into a 1.0 litre flask at 1000K, at equilibrium

Options

A. `[I_(2)]gt gt [I]`
B. `[I_(2)] = [I]`
C. `[I_(2)] lt [I]`
D. Unpredictable

Explanation

To solve the problem, we need to analyze the equilibrium of the reaction:

\[ I_2(g) \rightleftharpoons 2I(g) \]

Given:
- \( K_c = 37.6 \times 10^{-6} \) at \( 1000 \, K \)
- Initial amount of \( I_2 = 1.0 \, \text{mole} \) in a \( 1.0 \, \text{litre} \) flask

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant expression for the reaction is given by:

\[ K_c = \frac{[I]^2}{[I_2]} \]

### Step 2: Determine initial concentrations

Since we have 1 mole of \( I_2 \) in a 1.0 litre flask, the initial concentration of \( I_2 \) is:

\[ [I_2]_{initial} = \frac{1.0 \, \text{mole}}{1.0 \, \text{litre}} = 1.0 \, \text{M} \]

The initial concentration of \( I \) is \( 0 \, \text{M} \) because no iodine atoms are present initially.

### Step 3: Set up the change in concentrations at equilibrium

Let \( x \) be the amount of \( I_2 \) that dissociates at equilibrium. Therefore, at equilibrium, we have:

- Concentration of \( I_2 \): \( [I_2] = 1.0 - x \)
- Concentration of \( I \): \( [I] = 2x \)

### Step 4: Substitute into the equilibrium expression

Substituting these values into the \( K_c \) expression gives:

\[ K_c = \frac{(2x)^2}{(1.0 - x)} = \frac{4x^2}{1.0 - x} \]

### Step 5: Set up the equation with the given \( K_c \)

Now we can set this equal to the given \( K_c \):

\[ 37.6 \times 10^{-6} = \frac{4x^2}{1.0 - x} \]

### Step 6: Solve for \( x \)

Assuming \( x \) is small compared to 1.0 (which is reasonable given the small value of \( K_c \)), we can simplify the equation:

\[ 37.6 \times 10^{-6} \approx \frac{4x^2}{1.0} \]

This simplifies to:

\[ 4x^2 = 37.6 \times 10^{-6} \]

\[ x^2 = \frac{37.6 \times 10^{-6}}{4} \]

\[ x^2 = 9.4 \times 10^{-7} \]

\[ x = \sqrt{9.4 \times 10^{-7}} \approx 9.7 \times 10^{-4} \]

### Step 7: Calculate equilibrium concentrations

Now substituting \( x \) back to find the equilibrium concentrations:

- \( [I_2] = 1.0 - x \approx 1.0 - 9.7 \times 10^{-4} \approx 0.999 \, \text{M} \)
- \( [I] = 2x \approx 2 \times 9.7 \times 10^{-4} \approx 1.94 \times 10^{-3} \, \text{M} \)

### Conclusion

At equilibrium, the concentrations are approximately:

- \( [I_2] \approx 0.999 \, \text{M} \)
- \( [I] \approx 1.94 \times 10^{-3} \, \text{M} \)