Question

Consider the following reaction at \(1600^{\circ} \mathrm{C}\) $$\mathrm{Br}_{2}(g) \rightleftarrows 2 \mathrm{Br}(g)$$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 -L flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Short answer

The equilibrium constant \( K_c \) is 0.000624.

Step-by-step solution

Determine initial concentrations

Calculate the initial concentration of \( \text{Br}_2 \) in the flask. Use the formula \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \). The initial moles of \( \text{Br}_2 \) is 1.05 and the volume is 0.980 L.\[ \text{Initial concentration of } \text{Br}_2 = \frac{1.05}{0.980} = 1.071 \text{ M} \]

Find moles of Br2 dissociated

Calculate the amount of \( \text{Br}_2 \) that dissociates: \( 1.20\% \) of the initial moles dissociates. \[ \text{Dissociated moles of } \text{Br}_2 = \frac{1.20}{100} \times 1.05 = 0.0126 \text{ moles} \]

Determine equilibrium concentrations

Calculate the equilibrium concentration of \( \text{Br}_2 \) and \( \text{Br} \).\[ \text{Equilibrium concentration of } \text{Br}_2 = 1.071 - \frac{0.0126}{0.980} = 1.058 \text{ M} \]\[ \text{Equilibrium concentration of } \text{Br} = \frac{2 \times 0.0126}{0.980} = 0.0257 \text{ M} \]

Write the expression for Kc

The equilibrium constant expression for the reaction \( \text{Br}_2(g) \rightleftharpoons 2\text{Br}(g) \) is:\[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \]

Calculate the equilibrium constant Kc

Substitute the equilibrium concentrations into the expression for \( K_c \).\[ K_c = \frac{(0.0257)^2}{1.058} = \frac{0.00066049}{1.058} = 0.000624 \]

Key concepts

Dissociation Reaction

At the heart of many chemical processes lies the concept of a dissociation reaction. This occurs when a compound breaks down into two or more different substances. For gas reactions like \( \text{Br}_2(g) \rightleftharpoons 2 \text{Br}(g) \), dissociation is a common occurrence.

In this exercise, we see that \( \text{Br}_2 \) - which exists as a diatomic molecule - dissociates into individual bromine atoms, \( \text{Br} \). This breakdown is represented by the reaction arrow pointing in both directions, indicating that the process is reversible.

Dissociation is often a partial process and in our example, only 1.20% of the initial \( \text{Br}_2 \) actually dissociates. Recognizing this incomplete dissociation is crucial for understanding how equilibrium concepts apply.

Initial Concentration

The initial concentration is a critical starting point when analyzing chemical reactions. It refers to the concentration of reactants right before any reaction occurs.

For our problem, calculate this using:

• the number of moles of a substance
• the volume of the container holding the substance

Initially, 1.05 moles of \( \text{Br}_2 \) are placed in a 0.980 L flask. By applying the formula \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \), we found the initial concentration of \( \text{Br}_2 \) to be 1.071 M.

This step is foundational, as it provides the baseline from which changes can be measured during the reaction.

Equilibrium Concentrations

Once a reaction begins, the concentrations of the substances involved will shift until they reach a state of equilibrium, where no further net change occurs. At this point, we refer to these as equilibrium concentrations.

Our task involves determining these values for both \( \text{Br}_2 \) and \( \text{Br} \). After partial dissociation, the remaining concentration of \( \text{Br}_2 \) and the newly formed \( \text{Br} \) can be calculated.

• Original concentration of \( \text{Br}_2 \) minus the dissociated part gives the equilibrium concentration of \( \text{Br}_2 \).
• The dissociated part doubles (since each \( \text{Br}_2 \) yields two \( \text{Br} \) atoms) yields the equilibrium concentration of \( \text{Br} \).

Here, this results in 1.058 M for \( \text{Br}_2 \) and 0.0257 M for \( \text{Br} \).

Understanding these concentrations is key to assessing the status of the reaction at equilibrium.

Kc Expression

The equilibrium constant, \( K_c \), is a representation of the relative concentrations of products and reactants at equilibrium. It can be tailored for different reactions.

For our reversible dissociation of \( \text{Br}_2 \) into \( \text{Br} \), the expression for \( K_c \) is derived as:

• \( K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \)

This equation captures the essence of the system at equilibrium.

By substituting the equilibrium concentrations (0.0257 M for \( \text{Br} \) and 1.058 M for \( \text{Br}_2 \)), we calculate \( K_c = 0.000624 \).

Such expressions are fundamental, as they provide insights into how far a reaction proceeds towards the products under specific conditions.