Problem 15 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the dissociation of iodine molecules to iodine atoms $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) $$ is \(3.76 \times 10^{-3}\) at \(1000 \mathrm{K}\). Suppose 0.105 mol of \(\mathrm{I}_{2}\) is placed in a \(12.3-\mathrm{L}\). flask at \(1000 \mathrm{K}\). What are the concentrations of I \(_{2}\) and I when the system comes to equilibrium?

Step-by-Step Solution

Answer

At equilibrium, \([\mathrm{I}_2] = 0.00799 \text{ M}\) and \([\mathrm{I}] = 0.001106 \text{ M}\).

Step 1: Write the equilibrium expression

For the reaction \( \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \), the equilibrium constant expression (\( K_c \)) is given by:\[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\]The value of \( K_c \) at \( 1000 \mathrm{K} \) is \( 3.76 \times 10^{-3} \).

Step 2: Calculate initial concentrations

The initial concentration of \( \mathrm{I}_2 \) is calculated by dividing the moles by the volume of the flask:\[[\mathrm{I}_2]_0 = \frac{0.105 \text{ mol}}{12.3 \text{ L}} = 0.00854 \text{ M}\]Initially, \( [\mathrm{I}] = 0 \text{ M} \) because no iodine atoms are present.

Step 3: Set up the expression for changes in concentration

Let the change in concentration of \( \mathrm{I}_2 \) be \( -x \). Thus, for \( \mathrm{I} \), the change will be \( +2x \) because two moles of \( \mathrm{I} \) are formed for each mole of \( \mathrm{I}_2 \) dissociated. At equilibrium, we have:- \( [\mathrm{I}_2] = 0.00854 - x \)- \( [\mathrm{I}] = 2x \)

Step 4: Substitute into equilibrium expression

Substitute the expressions for equilibrium concentrations into the equilibrium expression:\[K_c = \frac{(2x)^2}{0.00854 - x} = 3.76 \times 10^{-3}\]

Step 5: Solve the equation for x

Simplify and solve the equation \( \frac{4x^2}{0.00854 - x} = 3.76 \times 10^{-3} \) to find \( x \). Rearranging gives:\[4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x)\]This simplifies to a quadratic equation. Solving the quadratic equation via factoring or the quadratic formula gives:\[ x \approx 0.000553 \text{ M} \]

Step 6: Calculate equilibrium concentrations

Using the value of \( x \), calculate the equilibrium concentrations:- \( [\mathrm{I}_2] = 0.00854 - 0.000553 = 0.00799 \text{ M} \)- \( [\mathrm{I}] = 2 \times 0.000553 = 0.001106 \text{ M} \)

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant (\( K_c \)) plays a crucial role in understanding reactions at equilibrium. It provides a ratio of the concentrations of products to reactants for a reaction that has reached a stable state. Each equilibrium has its unique constant, indicating the extent of the reaction. For gas-phase reactions, like the dissociation of iodine molecules into atoms, the expression involves concentrations as \( M \) (molarity).

The formula is set up by placing the product concentrations in the numerator and the reactant concentrations in the denominator, each raised to the power of their stoichiometric coefficients from the balanced chemical equation. Essentially, \( K_c \) helps predict the direction and position of the equilibrium state:

• If \( K_c \) > 1: Products are favored.
• If \( K_c \) < 1: Reactants are favored.

Understanding \( K_c \) allows us to calculate unknown concentrations once the equilibrium state is established.

Dissociation

Dissociation describes the process by which a compound breaks down into simpler molecules or ions. For molecular iodine (\( \mathrm{I}_2 \)), dissociation involves splitting into individual iodine atoms—2 \( \mathrm{I} \).

When a substance dissociates, the relationship between its initial concentration and how much it changes is crucial. It's expressed in an ICE table (Initial, Change, Equilibrium) that tracks concentration shifts from the initial state to equilibrium.

• Initial: Start concentration before reaction has occurred.
• Change: Amount that dissociates, represented by variable \( x \).
• Equilibrium: Result after dissociation reaches equilibrium.

For iodine molecules, the reaction \( \mathrm{I}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \) signals that dissociating one mole of \( \mathrm{I}_2 \) yields two moles of atoms, making understanding and calculating precise concentrations imperative for equilibrium predictions.

Iodine Molecules

Iodine molecules are diatomic, meaning they contain two iodine atoms bonded together (\( \mathrm{I}_2 \)). They can exist as gases under high temperatures or specific conditions, such as in a closed flask.

The concentration of iodine molecules initially is calculated by dividing the number of moles by the volume, using the formula:

• \( [\mathrm{I}_2]_0 = \frac{\text{moles of } \mathrm{I}_2}{\text{volume of flask}} \)

This calculation sets the stage for understanding how much \( \mathrm{I}_2 \) will dissociate and how the reaction progresses within the closed system. It's the starting point from where changes due to dissociation are measured.

Concentration Calculations

Calculating concentrations involves a few simple mathematical operations. You start by determining initial concentrations and track changes that occur during the reaction to find equilibrium concentrations.

For reactions like \( \mathrm{I}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \), changes occur in accordance with the stoichiometry. Initially, only \( \mathrm{I}_2 \) might be present, hence:

• Initial Concentration: \( [\mathrm{I}_2] = 0.00854 \text{ M} \); \( [\mathrm{I}] = 0 \text{ M} \)
• Change: Represented with \( x \) for reactants and \( 2x \) for products due to stoichiometry.
• Equilibrium: \( [\mathrm{I}_2] = 0.00854 - x \), \( [\mathrm{I}] = 2x \)

After setting up the equilibrium expression, algebraic manipulation (in this case solving a quadratic equation) allows us to find \( x \), eventually calculating the equilibrium concentrations of iodine molecules and atoms. This systematic approach ensures accuracy in dynamic systems like chemical equilibria.