Question

An equilibrium mixture of the following reaction has \(\left[\mathrm{I}_{2}\right]=0.0205 \mathrm{M}\) at \(1200{ }^{\circ} \mathrm{C}\). What is the concentration of \(\mathrm{I} ?\) $$ \begin{gathered} \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) \\ K_{\mathrm{eq}}=1.1 \times 10^{-2} \text { at } 1200^{\circ} \mathrm{C} \end{gathered} $$

Short answer

The concentration of \(\mathrm{I}\) at equilibrium is 0.015 M.

Step-by-step solution

State the Equilibrium Expression

For the reaction \(\mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g)\), the equilibrium expression based on the law of mass action is given as \(K_{eq} = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\).

Insert Known Values

Insert the known value of \(K_{eq}\) and the equilibrium concentration of \(\mathrm{I}_2\) into the equilibrium expression: \(1.1 \times 10^{-2} = \frac{[\mathrm{I}]^2}{0.0205}\).

Solve for \(\mathrm{I}\) Concentration

To find \(\mathrm{I}\), rearrange the formula to solve for \(\mathrm{I}\): \[\mathrm{I}]^2 = (1.1 \times 10^{-2}) \times 0.0205\.\] Then take the square root of both sides to find \(\mathrm{I}\): \[\mathrm{I} = \sqrt{(1.1 \times 10^{-2}) \times 0.0205}\.\]

Calculate the Concentration of \(\mathrm{I}\)

Perform the calculation: \[\mathrm{I} = \sqrt{1.1 \times 10^{-2} \times 0.0205} = \sqrt{2.255 \times 10^{-4}} = 0.015 \mathrm{M}\.\] The concentration of \(\mathrm{I}\) at equilibrium is 0.015 M.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is crucial in comprehending how chemical reactions reach a state of balance. It is a numerical value that represents the ratio between the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation.

For the reaction \( \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g) \) at equilibrium, the equilibrium constant \( K_{eq} \) is given by \( K_{eq} = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \). In this example, the value of \( K_{eq} \) is \(1.1 \times 10^{-2}\), which was determined experimentally at \(1200^\circ \mathrm{C}\).

This value indicates how far the reaction will proceed before reaching equilibrium. A low \( K_{eq} \) value suggests that, at equilibrium, the reactants are favored, while a high \( K_{eq} \) value implies that the products are favored. In this case, the value of \( K_{eq} \) indicates that at high temperature, the dissociation of iodine (\( \mathrm{I}_2 \) to \( \mathrm{I} \) atoms) is limited, favoring the formation of the reactant \( \mathrm{I}_2 \) over the product \( \mathrm{I} \) in the equilibrium state.

Law of Mass Action

The law of mass action describes the relationship between the rate of a chemical reaction and the concentration of the reactants. It states that at a constant temperature, the rate of a reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their coefficients in the balanced chemical equation.

When applied to the equilibrium condition for the reaction \( \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g) \), the law of mass action helps us derive the equilibrium expression, as shown in the exercise. By knowing the equilibrium constant \( K_{eq} \) and the concentration of one of the reactants or products, we can use the law of mass action to calculate the unknown concentrations at equilibrium.

The law of mass action is vital for constructing the equilibrium expression, which then allows us to solve for unknown concentrations, as demonstrated in the steps provided.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a chemical reaction when the reaction has reached equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of all reactants and products remain constant over time.

In our exercise, we are given the equilibrium concentration of \( \mathrm{I}_2 \) and asked to calculate that of \( \mathrm{I} \) using the equilibrium constant. The concentration of \( \mathrm{I}_2 \) at equilibrium is crucial information as it contributes to the equilibrium expression which we use to solve for \( \mathrm{I} \).

Calculating the equilibrium concentrations is a common task when studying chemical equilibrium, as it helps predict the amounts of substances present when a reaction is under specific conditions, such as the high temperature of \(1200^\circ \mathrm{C}\) in this example. As we've determined in the solution, the equilibrium concentration of \( \mathrm{I} \) is 0.015 M, which provides insights into how much of \( \mathrm{I} \) is formed at this state.