Question

The dissociation of molecular iodine into iodine atoms is represented as: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a 2.30-L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

Short answer

[I2] = 0.0194 M, [I] = 8.66 × 10^-4 M.

Step-by-step solution

Initial Concentrations

First, calculate the initial concentration of \( \text{I}_2 \) using the formula \( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \). The volume is 2.30 L and moles of \( \text{I}_2 \) is 0.0456. Thus, the initial concentration \( [\text{I}_2]_0 = \frac{0.0456}{2.30} = 0.0198 \text{ M} \).

Set up ICE Table

Set up an ICE (Initial, Change, Equilibrium) table based on the dissociation reaction \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \). Initially, \([\text{I}]_0 = 0\, \text{M}\). At equilibrium, the change in \([\text{I}_2]\) is \(-x\), and \([\text{I}]\) will be \(+2x\) due to the stoichiometry of the reaction. Thus, \([\text{I}_2]\) at equilibrium is \(0.0198-x\) and \([\text{I}]\) at equilibrium is \(2x\).

Apply Equilibrium Constant Expression

The equilibrium constant expression for this reaction is \( K_c = \frac{[\text{I}]^2}{[\text{I}_2]} \). Substitute the equilibrium concentrations: \( K_c = \frac{(2x)^2}{0.0198-x} \).

Solve for 'x'

Given \( K_c = 3.80 \times 10^{-5} \), solve the equation \( \frac{4x^2}{0.0198-x} = 3.80 \times 10^{-5} \). After simplifying, we find the quadratic equation: \( 4x^2 + (3.80 \times 10^{-5})x - (7.52 \times 10^{-7}) = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \), yielding \( x \approx 4.33 \times 10^{-4} \).

Calculate Equilibrium Concentrations

Using the value of \( x = 4.33 \times 10^{-4} \), calculate the equilibrium concentrations of \([\text{I}_2]\) and \([\text{I}]\). Thus, \([\text{I}_2] = 0.0198 - 4.33 \times 10^{-4} \approx 0.0194 \text{ M} \) and \([\text{I}] = 2(4.33 \times 10^{-4}) \approx 8.66 \times 10^{-4} \text{ M} \).

Key concepts

Dissociation Reaction

A dissociation reaction occurs when a compound breaks down into its smaller components. In this exercise, we are focusing on the dissociation of molecular iodine (\( \mathrm{I}_2(g) \)) into iodine atoms (\( 2\mathrm{I}(g) \)).
In a dissociation reaction, the original molecule breaks up to form two or more ions or atoms, rearranging chemical bonds in the process.

• The reaction can reach a point where the rate of dissociation equals the rate of recombination, leading to a state of balance known as equilibrium.
• During this state, the concentrations of reactants and products remain constant over time, even though the reaction continues at the molecular level.

This understanding helps us analyze and predict the behavior of various chemical systems, such as the dissociation of gases in this problem.

Equilibrium Constant

The equilibrium constant (\( K_c \)) is a value that helps us understand the ratio of the concentrations of products to reactants at equilibrium for a given reaction.
For the reaction \( \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) \), the equilibrium constant expression is formulated as:\[ K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]

• This expression is derived from the balanced chemical equation, allowing us to plug in the concentrations of each species at equilibrium.
• The magnitude of \( K_c \) gives us insight into the position of equilibrium. A small \( K_c \) value, such as \( 3.80 \times 10^{-5} \), indicates that at equilibrium, the reaction mixture contains mostly the reactants, meaning that the \( \mathrm{I}_2 \) molecules are predominantly present, and only a small fraction has dissociated into \( \mathrm{I} \) atoms.

This concept is fundamental for determining the equilibrium concentrations of chemical species in any reversible reaction.

ICE Table

To systematically track the changes in concentrations of reactants and products during a chemical reaction, chemists use an ICE Table, which stands for Initial, Change, and Equilibrium.
For our dissociation reaction, we set up the ICE Table as follows:

• **Initial**: Calculate the initial concentration of \( \mathrm{I}_2 \) using the formula \( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \). In this case, \( [\text{I}_2]_0 = 0.0198 \text{ M} \), and \( [\text{I}]_0 = 0 \text{ M} \).
• **Change**: Define the change in concentration using variables. For \( \mathrm{I}_2 \), it's \(-x\), and for \( \mathrm{I} \), it's \(+2x\).
• **Equilibrium**: Express the equilibrium concentrations for each species. \( [\text{I}_2] = 0.0198 - x \) and \( [\text{I}] = 2x \)

The ICE Table serves as a visual aid to lay out what you know and what you need to solve for, simplifying the path to finding equilibrium concentrations.

Quadratic Formula

The quadratic formula is often employed in chemistry to solve for unknowns when the relationship between variables is quadratic in nature, particularly in equilibrium calculations.
In this scenario, we use it to solve for \( x \), the change in concentration during the dissociation of iodine.

• We arrive at a quadratic equation: \[ 4x^2 + (3.80 \times 10^{-5})x - (7.52 \times 10^{-7}) = 0 \]
• Solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Applying this formula helps find \( x \approx 4.33 \times 10^{-4} \) M, the concentration change necessary to determine final equilibrium values.

Mastering this tool allows students to tackle a range of equilibrium-related problems more effectively, as it provides a systematic method for finding solutions to complex equations.