Problem 14 - Chemistry: The Central Science in SI Units, Global Edition

Question

Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 \(\mathrm{K}\) 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 \(5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?\) (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) The \(\mathrm{Br}_2\) (molecular bromine) predominates at equilibrium. (b) The reverse reaction has a larger numeric value of the rate constant.

Step 1: Write the expression for the equilibrium constant

According to the given equation: \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) We can write the equilibrium constant \(K_c\) expression in terms of the concentrations of the species as: \[K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\] We are given \(K_c=5.4 \times 10^{-3}\), and the concentration of \(\mathrm{Br}_2\) is larger than \(5.4 \times 10^{-3}\ \mathrm{mol} / \mathrm{L}\).

Step 2: Compare the concentrations of Br₂ and Br at equilibrium

Since we know that the concentration of \(\mathrm{Br}_2\) is larger than \(5.4 \times 10^{-3}\ \mathrm{mol}\ /\ \mathrm{L}\), we have the following inequality: \[ [\mathrm{Br}_2] > 5.4 \times 10^{-3}\ \mathrm{mol}\ /\ \mathrm{L}\] Using the expression for the equilibrium constant, we can write: \[5.4 \times 10^{-3} = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\] We know that \([\mathrm{Br}_2]>5.4 \times 10^{-3}\), so the fraction \(\frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\) must be less than 1, which means that: \[ [\mathrm{Br}]^2 < [\mathrm{Br}_2]\] Taking the square root of both sides: \[ [\mathrm{Br}] < \sqrt{[\mathrm{Br}_2]}\] From the above inequality, we can conclude that the concentration of \(\mathrm{Br}_2\) predominates at equilibrium. Therefore, the answer to question (a) is Br₂.

Step 3: Determine which reaction has a larger numeric value of the rate constant

For an elementary reaction, the rate constant for the forward reaction (\(k_f\)) and the reverse reaction (\(k_r\)) are related to the equilibrium constant by: \[K_c = \frac{k_f}{k_r}\] We are given: \[K_c=5.4 \times 10^{-3}\] Since the value of the equilibrium constant \(K_c\) is less than 1, it means that \(k_f < k_r\). Therefore, the reverse reaction has a larger numeric value of the rate constant. Now we have the answers to both parts of the question: (a) The \(\mathrm{Br}_2\) (molecular bromine) predominates at equilibrium. (b) The reverse reaction has a larger numeric value of the rate constant.

Key Concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), plays a crucial role in chemical reactions. It's a value that tells us the ratio of the concentration of products to reactants at equilibrium. When a reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, meaning that the concentrations of the reactants and products remain constant over time.

For the dissociation of molecular bromine, \( \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) \), the equilibrium constant is given by the formula:

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

This expression shows how the concentration of the bromine atoms (Br) compared to molecular bromine (\( \mathrm{Br}_2 \)) dictates the direction in which the reaction favors: products or reactants.

If \( K_c < 1 \), the equilibrium is shifted towards the reactants because the denominator (reactants) is larger in value. In other words, more \( \mathrm{Br}_2 \) remains compared to \( \mathrm{Br} \). This is exactly what's happening in our exercise example where \( K_c = 5.4 \times 10^{-3} \).

Understanding \( K_c \) allows you to predict which species will predominate at equilibrium without having to measure all concentrations directly. It's an indispensable tool for chemists!

Reaction Rates

Reaction rates tell you how quickly a reaction proceeds. They're determined by the concentrations of reactants and a rate constant, often denoted as \( k \). For elementary reactions, the rate of the forward reaction is proportional to the concentration of the reactants.

In our bromine dissociation reaction,

• The forward reaction rate depends on the concentration of \( \mathrm{Br}_2 \).
• The reverse reaction rate relies on the concentration of \( \mathrm{Br} \).

When we talk about the rate constants in this context, \( k_f \) is the rate constant for the forward reaction, and \( k_r \) is for the reverse. The relationship between these constants and the equilibrium constant is given by:

• \( K_c = \frac{k_f}{k_r} \).

If the equilibrium constant is small, \( K_c < 1 \), it suggests that \( k_f < k_r \). This indicates the reverse reaction is faster than the forward reaction, meaning that products revert to reactants more readily than reactants form products.

In summary, reaction rates and their constants give us a dynamic picture of how equilibrium is achieved and maintained.

Elementary Reactions

Elementary reactions are the simplest forms of chemical reactions. They occur in a single step, with no intermediate steps, obviously influencing the rate law directly from the stoichiometry of the balanced equation.

In the given reaction, \( \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{Br}(g) \), both the forward and reverse reactions are described as elementary. This means that the reaction rates directly depend on the reactant concentrations as they appear in the chemical equation:

• For the forward reaction, rate \( = k_f \times [\mathrm{Br}_2] \)
• For the reverse reaction, rate \( = k_r \times [\mathrm{Br}]^2 \)

Understanding these concepts helps clarify why the equilibrium constant is related to the rate constants for elementary reactions—there are no intermediate steps, simplifying the relationship between \( K_c \), \( k_f \), and \( k_r \).

Thus, knowledge of whether a process is elementary provides chemists insight into reaction mechanisms and the simplicity or complexity of establishing equilibrium.