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

Answer

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

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

Given:
- The equilibrium constant \( K_c = 37.6 \times 10^{-6} \) at \( 1000 \, K \)
- Initial moles of \( I_2 = 1.0 \, \text{mole} \)
- Volume of the flask = \( 1.0 \, \text{litre} \)

### Step 1: Calculate the initial concentration of \( I_2 \)

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

\[ \text{Initial concentration of } I_2 = \frac{1.0 \, \text{mole}}{1.0 \, \text{litre}} = 1.0 \, \text{M} \]

**Hint:** Remember that concentration is calculated as moles divided by volume in litres.

### Step 2: Set up the equilibrium expression

The equilibrium constant expression for the reaction is given by:

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

Where:
- \([I]\) is the concentration of iodine (I) at equilibrium.
- \([I_2]\) is the concentration of diiodine (I2) at equilibrium.

### Step 3: Define the change in concentration at equilibrium

Let \( x \) be the change in concentration of \( I_2 \) that dissociates at equilibrium. Therefore, at equilibrium:

- The concentration of \( I_2 \) will be \( 1.0 - x \)
- The concentration of \( I \) will be \( 2x \) (since 2 moles of I are produced for every mole of \( I_2 \) that dissociates)

### Step 4: Substitute into the equilibrium expression

Substituting these values into the equilibrium expression gives:

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

Substituting the value of \( K_c \):

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

### Step 5: Solve for \( x \)

To solve for \( x \), we can rearrange the equation:

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

Expanding this gives:

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

Rearranging leads to:

\[ 4x^2 + 37.6 \times 10^{-6} x - 37.6 \times 10^{-6} = 0 \]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \) where:
- \( a = 4 \)
- \( b = 37.6 \times 10^{-6} \)
- \( c = -37.6 \times 10^{-6} \)

### Step 6: Use the quadratic formula

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting the values of \( a \), \( b \), and \( c \):

\[ x = \frac{-37.6 \times 10^{-6} \pm \sqrt{(37.6 \times 10^{-6})^2 - 4 \cdot 4 \cdot (-37.6 \times 10^{-6})}}{2 \cdot 4} \]

Calculating the discriminant and solving for \( x \) will yield the equilibrium concentrations.

### Step 7: Analyze the equilibrium concentrations

Once \( x \) is found, we can determine the equilibrium concentrations:

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

### Step 8: Compare concentrations

Given that \( K_c \) is very small (\( 37.6 \times 10^{-6} \)), it indicates that the equilibrium position favors the reactants. Therefore, we can conclude that:

\[ [I_2] > [I] \]

**Final Conclusion:** The concentration of \( I_2 \) is significantly greater than the concentration of \( I \) at equilibrium.

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