Problem 14 - Chemistry The Central Science In Si Units

Question

Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 K is \(K_{c}=5.4 \times 10^{-3}\). (a) Which species predominates at equilibrium, \(\mathrm{Br}_{2}\) or Br, assuming that the concentration of \(\mathrm{Br}_{2}\) is larger than \(5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?(\mathbf{b})\) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger numeric value of the rate constant, the forward or the reverse reaction?

Step-by-Step Solution

Answer

(a) Br₂ predominates. (b) The reverse reaction has a larger rate constant.

Step 1: Analyze the Equilibrium Expression

The equilibrium constant expression for the dissociation of \(\text{Br}_2\) into Br is given by \(K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]}\). This implies that at equilibrium, the ratio of the concentration of products to reactants should be equal to \(5.4 \times 10^{-3}\).

Step 2: Interpret Concentration Value

Given that \([\text{Br}_2] > 5.4 \times 10^{-3} \text{ mol/L}\), if [Br] is much smaller compared to [\text{Br}_2], then the ratio \(\frac{[\text{Br}]^2}{[\text{Br}_2]}\) will be less than \(5.4 \times 10^{-3}\). This means that \([\text{Br}_2]\) is larger at equilibrium than \([\text{Br}]\).

Step 3: Conclude Predominant Species at Equilibrium

Since the concentration of \(\text{Br}_2\) remains larger relative to \(5.4 \times 10^{-3}\) under the given conditions, the species \(\text{Br}_2\) predominates at equilibrium.

Step 4: Determine Rate Constants

The equilibrium constant \(K_c = \frac{k_f}{k_r}\), where \(k_f\) and \(k_r\) are the rate constants for the forward and reverse reactions, respectively. Since \(K_c < 1\), this implies \(k_r > k_f\), meaning the reverse reaction has a larger rate constant than the forward reaction.

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant ( \(K_{c}\) ) plays a crucial role. It defines the ratio between the concentrations of products and reactants at equilibrium. For the dissociation of molecular bromine, represented by \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) , the \(K_{c}\) is \(5.4 \times 10^{-3}\). This constant tells us how far the reaction proceeds before reaching equilibrium.
If \(K_{c}\) is less than one, like in our example, it indicates that the reactants are more prevalent at equilibrium. Here, a smaller equilibrium constant suggests that the dissociation does not significantly proceed in favor of forming \(\mathrm{Br}\) atoms. Instead, the equilibrium tends toward having more molecular bromine \(\mathrm{Br}_{2}\). This insight helps chemists understand reaction preferences right from the equilibrium constant value.

Reaction Rates

Understanding reaction rates in a chemical equation is pivotal. They give us insight into how quickly a reactant turns into a product and vice-versa. In our case, the equilibrium constant \(K_{c}\) is derived from the formula: \(K_{c} = \frac{k_f}{k_r}\), where \(k_f\) is the forward reaction rate constant and \(k_r\) is the reverse reaction rate constant.
Given \(K_{c} = 5.4 \times 10^{-3}\) , we know \(k_r > k_f\), suggesting that the reverse reaction, converting \(\mathrm{Br}\) back to \(\mathrm{Br}_{2}\), occurs more rapidly than the forward reaction. This insight into reaction rates not only helps predict the predominant species but also aids in designing conditions to control which reaction prevails by influencing factors like temperature or pressure.

Dissociation Reaction

A dissociation reaction describes the breakdown of a compound into two or more components. For molecular bromine \(\mathrm{Br}_{2}(g)\), this reaction is expressed as \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\).

• "Dissociation" indicates that \(\mathrm{Br}_{2}\) splits into two bromine atoms.
• Understanding dissociation is vital in predicting the behavior of reactions and the species present at equilibrium.

In our example, the \(K_{c}\) value indicates a reaction equilibrium that favors the non-dissociated form, \(\mathrm{Br}_{2}\). Dissociation reactions are all about balance, recognizing which conditions favor the breakdown and which conditions encourage recombination. This balance aids in identifying stable configurations in practical applications ranging from industrial processes to atmospheric chemistry studies.

Molecular Bromine

Molecular bromine ( \(\mathrm{Br}_{2}(g)\) ) is a diatomic molecule commonly encountered in chemistry, especially when studying halogens. It's essential to grasp this molecule's role in reactions, like the dissociation discussed here: \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\).
This is an example of a bimolecular process. Such reactions highlight how \(\mathrm{Br}_{2}\) can dissociate into two \(\mathrm{Br}\) atoms under specific conditions, primarily influenced by temperature.
Understanding molecular bromine's equilibrium behavior helps chemists anticipate how such molecules interact with their environment and affect various chemical processes. Its predominance in a reaction system signifies a certain reaction direction, which is vital for designing controlled chemical reactions in both laboratory and industrial settings.