Problem 42 - Chemistry The Molecular Science

Question

Suppose \(0.086 \mathrm{~mol} \mathrm{Br}_{2}\) is placed in a 1.26-L flask. The flask is sealed and heated to \(1756 \mathrm{~K}\), a temperature at which the \(\mathrm{Br}_{2}\) dissociates to atoms $$ \mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Br}(\mathrm{g}) $$ If \(\mathrm{Br}_{2}\) is \(3.7 \%\) dissociated at this temperature, calculate \(K_{c}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_c \) is approximately 0.00038.

Step 1: Calculate initial concentration of \( \mathrm{Br}_2 \)

The initial concentration \( [\mathrm{Br}_2]_0 \) can be calculated using the formula:\[[\mathrm{Br}_2]_0 = \frac{\text{moles of } \mathrm{Br}_2}{\text{volume of the flask}}\]Substitute the values:\[[\mathrm{Br}_2]_0 = \frac{0.086 \text{ mol}}{1.26 \text{ L}} \approx 0.0683 \text{ M}\]This is the initial concentration of \( \mathrm{Br}_2 \).

Step 2: Determine change in concentration due to dissociation

Since \( \mathrm{Br}_2 \) is 3.7% dissociated, the change in concentration \( x \) can be calculated as:\[x = 0.037 \times [\mathrm{Br}_2]_0 = 0.037 \times 0.0683 \approx 0.0025 \text{ M}\]This is the amount of \( \mathrm{Br}_2 \) that has dissociated into \( \mathrm{Br} \) atoms.

Step 3: Find equilibrium concentrations

At equilibrium, the concentration of \( \mathrm{Br}_2 \) is decreased by \( x \), and \( \mathrm{Br} \) is produced at twice the rate of \( x \) since each \( \mathrm{Br}_2 \) dissociates into 2 \( \mathrm{Br} \).\[[\mathrm{Br}_2]_{eq} = 0.0683 - x = 0.0683 - 0.0025 = 0.0658 \text{ M}\]\[[\mathrm{Br}]_{eq} = 2x = 2(0.0025) = 0.0050 \text{ M}\]These are the equilibrium concentrations.

Step 4: Calculate \( K_c \) using equilibrium concentrations

The equilibrium constant \( K_c \) for the reaction is given by:\[K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\]Substitute the equilibrium concentrations:\[K_c = \frac{(0.0050)^2}{0.0658} = \frac{0.000025}{0.0658} \approx 0.00038\]This is the equilibrium constant \( K_c \) at the given temperature.

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \( K_c \), is a critical value to understand when studying chemical reactions reaching equilibrium. It helps us know how far a reaction proceeds before reaching equilibrium. The constant is specific to a particular reaction at a given temperature and is calculated using the concentrations of the reactants and products once equilibrium is achieved. For the dissociation of \( \text{Br}_{2} \), the equilibrium expression is derived from the balanced equation and is given by:\[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_{2}]} \]This expression tells us the proportion of products to reactants at equilibrium. A small \( K_c \) value, like the 0.00038 from our exercise, indicates that at equilibrium, the reactants are favored, meaning most of the \( \text{Br}_{2} \) remains undissociated.

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into its constituent elements or simpler compounds. For example, in the given exercise, \( \text{Br}_{2} \) gas dissociates into two \( \text{Br} \) atoms. The reaction can be represented as:\[ \text{Br}_{2}(g) \rightleftharpoons 2 \text{Br}(g) \]In this reversible reaction, the double arrows indicate that the reaction can proceed in both forward (dissociation) and backward (association) directions. The balance between these two directions at a given temperature eventually establishes an equilibrium state. Understanding this dynamic process is crucial because it shows how reactions can adjust themselves to achieve a stable state where the rates of the forward and reverse reactions are equal.

Concentration Calculations

Calculating concentrations is key when analyzing chemical equilibria as it allows us to determine how much of each substance is present at different stages of the reaction. To calculate the initial concentration of \( \text{Br}_{2} \), we use the formula:\[[\text{Br}_{2}]_0 = \frac{\text{moles of } \text{Br}_{2}}{\text{volume of the flask}}\]Substituting the provided values gives us an initial concentration. When 3.7% of \( \text{Br}_{2} \) dissociates, we calculate the change in concentration, termed \( x \), using:\[ x = 0.037 \times [\text{Br}_{2}]_0 \]At equilibrium, we see a decrease of \( x \) in \( [\text{Br}_{2}] \), and an increase in \( [\text{Br}] \) by twice \( x \) because each \( \text{Br}_{2} \) produces two \( \text{Br} \) atoms.These concentration changes help us derive the equilibrium concentrations used for calculating \( K_c \), providing insight into the distribution of substances at equilibrium.