Question

The Equilibrium Constant and Reaction Quotient \(K=5.6 \times 10^{-12}\) at \(500 \mathrm{K}\) for the dissociation of iodine molecules to iodine atoms. $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) $$ A mixture has \(\left[\mathrm{I}_{2}\right]=0.020 \mathrm{mol} / \mathrm{L}\) and \([\mathrm{I}]=2.0 \times 10^{-8}\) mol/L. Is the reaction at equilibrium (at \(500 \mathrm{K}\) )? If not, which way must the reaction proceed to reach equilibrium?

Short answer

Not at equilibrium; reaction must proceed to the right.

Step-by-step solution

Write the Expression for the Reaction Quotient

The reaction quotient, \( Q \), is calculated similarly to the equilibrium constant, but using the current concentrations (not equilibrium concentrations). For the reaction \(\text{I}_2(g) \rightleftharpoons 2 \text{I}(g)\), the expression for \( Q \) is:\[Q = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\]

Substitute Current Concentrations into the Expression

Substitute the given concentrations into the expression for \( Q \): \[Q = \frac{(2.0 \times 10^{-8})^2}{0.020}\]Calculate this value to determine \( Q \).

Calculate the Reaction Quotient

Calculate \( Q \):\[Q = \frac{4.0 \times 10^{-16}}{0.020} = 2.0 \times 10^{-14}\]

Compare \( Q \) and \( K \)

Compare the calculated \( Q \) with the given equilibrium constant \( K \) (\( K = 5.6 \times 10^{-12} \)).- If \( Q < K \), the reaction will proceed to the right to reach equilibrium.- If \( Q > K \), the reaction will proceed to the left to reach equilibrium.- If \( Q = K \), the reaction is at equilibrium.

Determine the Direction of Reaction

Since \( Q = 2.0 \times 10^{-14} \) and \( K = 5.6 \times 10^{-12} \), we see that \( Q < K \).This means that the reaction will proceed to the right (forward direction) to reach equilibrium.

Key concepts

Reaction Quotient

The reaction quotient, also known as \( Q \), helps us determine the current state of a chemical reaction. It uses the same formula as the equilibrium constant \( K \), but with the concentrations of the reactants and products at any given time, not just at equilibrium. For the dissociation reaction of iodine molecules

• \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \)

The formula for \( Q \) is:\[ Q = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]By substituting the concentrations of iodine atoms and molecules given in the problem, we can calculate \( Q \) to understand whether the reaction needs to proceed in the forward or reverse direction to achieve balance. Understanding this concept is crucial for predicting the behavior of chemical reactions under different conditions.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant, though not necessarily equal. For any given temperature, the reaction has an equilibrium constant \( K \). When we compare \( Q \) to \( K \), we can determine:

• If \( Q < K \), the reaction favors the forward direction, producing more products.
• If \( Q > K \), the reaction favors the reverse direction, forming more reactants.
• If \( Q = K \), the reaction is already at equilibrium and no shift is necessary.

In the iodine dissociation example, calculating and comparing \( Q \) and \( K \) allows us to decide the necessary direction of the reaction to reach equilibrium.

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into smaller molecules or atoms. In the exercise provided, iodine molecules (\( \text{I}_2 \)) dissociate into iodine atoms (\( \text{I} \)). This type of reaction is fundamental in understanding chemical equilibrium.

• At higher temperatures, such as 500 K in this scenario, molecules gain enough energy to break bonds and dissociate.
• The equilibrium constant \( K \) indicates the extent of this dissociation at a given temperature.
• In the case of \( \text{I}_2 \), a small \( K \) value implies that at equilibrium, there are fewer atoms of iodine and more molecules.

These reactions are powerful examples of how substances reach balance in a closed system, and analyzing their equilibrium helps us predict how systems behave under various conditions.