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 \(I_{2}\) in a 2.30-L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

Short answer

\([\mathrm{I_2}] = 0.01936 \text{ M}; [\mathrm{I}] = 8.68 \times 10^{-4} \text{ M}\)

Step-by-step solution

Identify Initial Moles and Concentrations

We start with 0.0456 moles of \( \mathrm{I}_2(g)\) in a 2.30 L flask. Calculate the initial concentration of \( \mathrm{I}_2(g)\) using the formula: \[ \text{Concentration of } \mathrm{I}_2 = \frac{\text{Moles of } \mathrm{I}_2}{\text{Volume of flask}} = \frac{0.0456}{2.30} = 0.0198 \text{ M} \] The initial concentrations of \( \mathrm{I}(g)\) are zero because we are only starting with \( \mathrm{I}_2(g)\).

Set Up the Equilibrium Expression

For the equilibrium reaction \( \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{I}(g) \), the expression for the equilibrium constant \( K_c \) is:\[ K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]Where \([\mathrm{I}^2]\) is the concentration of iodine atoms and \([\mathrm{I}_2]\) is the concentration of iodine molecules at equilibrium.

Establish the Change and Equilibrium Concentrations

Let's denote \( x \) as the change in concentration for \( \mathrm{I}_2 \) that dissociates.- \( [\mathrm{I}_2] \) changes by \( -x \).- \( [\mathrm{I}] \) changes by \( +2x \).At equilibrium:- \( [\mathrm{I}_2] = 0.0198 - x \)- \( [\mathrm{I}] = 2x \)

Solve for x Using the Equilibrium Constant

Substitute the expressions for equilibrium concentrations into the \( K_c \) expression:\[ K_c = \frac{(2x)^2}{0.0198 - x} = 3.80 \times 10^{-5} \]This simplifies to:\[ \frac{4x^2}{0.0198 - x} = 3.80 \times 10^{-5} \]Solving this quadratic equation will give us the value of \( x \).

Approximating and Solving the Quadratic Equation

Assume \( x \) is very small compared to 0.0198, so \( 0.0198 - x \approx 0.0198 \). Hence:\[ \frac{4x^2}{0.0198} \approx 3.80 \times 10^{-5} \]\[ 4x^2 = (3.80 \times 10^{-5}) \times 0.0198 \] \[ 4x^2 = 7.524 \times 10^{-7} \]\[ x^2 = 1.881 \times 10^{-7} \]\[ x = \sqrt{1.881 \times 10^{-7}} \approx 4.34 \times 10^{-4} \]

Calculate Equilibrium Concentrations

Now use \( x \) to find equilibrium concentrations:- \( [\mathrm{I}_2] = 0.0198 - x \approx 0.0198 - 4.34 \times 10^{-4} = 0.01936 \text{ M} \)- \( [\mathrm{I}] = 2x = 2 \times 4.34 \times 10^{-4} = 8.68 \times 10^{-4} \text{ M} \)

Key concepts

Dissociation Reaction

In chemistry, a dissociation reaction involves breaking down a compound into two or more simpler substances. For molecular iodine (\(\mathrm{I}_2(g)\)), the reaction is: \[ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) \]Here, the molecule splits into two iodine atoms. This process is reversible and reaches a state of balance called equilibrium.Understanding dissociation is essential because it helps predict how molecules behave under different conditions. In this case, we start with only \(\mathrm{I}_2\). As the reaction proceeds, some \(\mathrm{I}_2\) becomes \(\mathrm{I}\) atoms.

Equilibrium Constant

The equilibrium constant \(K_c\) is a crucial factor in understanding chemical equilibrium. It measures the ratio of product concentration to reactant concentration at equilibrium.For the reaction \(\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{I}(g)\), the expression for the equilibrium constant \(K_c\) is:\[ K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]Here, \([\mathrm{I}]\) is the concentration of iodine atoms, and \([\mathrm{I}_2]\) is the concentration of iodine molecules.Knowing \(K_c\) allows us to predict the direction and extent of the reaction. At \(1000 \mathrm{~K}, K_c = 3.80 \times 10^{-5}\), indicating very low dissociation, meaning most \(\mathrm{I}_2\) remains as molecules.

Concentration Calculation

Calculating concentration involves converting moles and volume into molarity (M). Start with 0.0456 moles of \(\mathrm{I}_2\) in a 2.30 L flask to find its initial concentration:\[ \text{Concentration of } \mathrm{I}_2 = \frac{0.0456}{2.30} = 0.0198 \text{ M} \]At equilibrium, the concentration of \(\mathrm{I}\) atoms is determined by the change in concentration from dissociation.Assume the change is \(x\):

• \([\mathrm{I}_2] = 0.0198 - x\)
• \([\mathrm{I}] = 2x\)

Solving for \(x\) with the quadratic equation derived from \(K_c\) gives us the equilibrium concentrations of both \(\mathrm{I}_2\) and \(\mathrm{I}\).

Iodine Gas

Iodine gas \((\mathrm{I}_2)\) is a diatomic molecule, meaning it consists of two iodine atoms bonded together. It appears as a violet gas and is an important element in various chemical reactions.Understanding its behavior in gas form is essential for reactions such as the one we're examining. In our dissociation reaction, molecular iodine transitions between its gaseous form and individual iodine atoms. Monitoring this behavior helps us explore how molecules interact and change under specific conditions, such as temperature.