Problem 71 - Chemistry The Central Science In Si Units

Question

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?

Step-by-Step Solution

Answer

The equilibrium constant \( K \) at \( 25^{\circ} \mathrm{C} \) is \( 1.505 \times 10^{-39} \). Reactants are more plentiful at equilibrium.

Step 1: Understand the Relationship Between Rate Constants and Equilibrium Constant

For a reversible reaction, the equilibrium constant \( K \) is the ratio of the rate constant of the forward reaction \( k_f \) to that of the reverse reaction \( k_r \). This means: \[ K = \frac{k_f}{k_r} \]

Step 2: Identify Given Rate Constants

From the given problem, the rate constant for the forward reaction \( k_f \) is \( 1.4 \times 10^{-28} \, \text{M}^{-1} \text{s}^{-1} \) and for the reverse reaction \( k_r \) is \( 9.3 \times 10^{10} \, \text{M}^{-1} \text{s}^{-1} \).

Step 3: Calculate the Equilibrium Constant

We use the equation for equilibrium constant:\[ K = \frac{k_f}{k_r} = \frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}} \]Calculate \( K \) by dividing these values, which leads to \( K = 1.505 \times 10^{-39} \).

Step 4: Analyze the Magnitude of Equilibrium Constant

If \( K \) is much less than 1 (as it is here, \( K = 1.505 \times 10^{-39} \)), it signifies that the concentration of reactants is far greater than that of the products at equilibrium.

Key Concepts

Equilibrium Constant

An equilibrium constant, denoted by \( K \), is a fundamental concept when dealing with chemical equilibriums. It provides insight into the composition of a reaction mixture at equilibrium. The value of \( K \) is derived from the ratio of the rate constants of the forward and reverse reactions. For the reaction given, \( K \) can be calculated using the formula:

• \( K = \frac{k_f}{k_r} \)

where \( k_f \) is the rate constant of the forward reaction, and \( k_r \) is that of the reverse reaction.
In this particular exercise, \( K \) has been calculated as \( 1.505 \times 10^{-39} \) at \( 25^{\circ} \text{C} \).
A \( K \) value much less than 1 suggests that at equilibrium, the reactants significantly outnumber the products. This indicates that the forward reaction is not favored under these conditions.

Rate Constant

The rate constant is a crucial factor in determining the speed of a reaction. It is represented by \( k \) and varies with temperature. In reversible reactions, there are usually two rate constants involved:

• \( k_f \): Rate constant for the forward reaction
• \( k_r \): Rate constant for the reverse reaction

For the provided reaction, the forward rate constant \( k_f \) is \( 1.4 \times 10^{-28} \text{ M}^{-1} \text{s}^{-1} \), while the reverse rate constant \( k_r \) is \( 9.3 \times 10^{10} \text{ M}^{-1} \text{s}^{-1} \).
Comparing these values shows a stark difference, indicating that the reverse reaction is much faster than the forward reaction. This discrepancy in rate constants gives us a deeper understanding of why the products are less abundant at equilibrium.

Forward and Reverse Reactions

In chemical reactions at equilibrium, both the forward and reverse reactions occur simultaneously. The forward reaction consumes reactants to form products, while the reverse reaction does the opposite.The coexistence of these reactions at equilibrium means that over time, their rates become equal, stabilizing the concentrations of reactants and products. For the reaction:

• Forward Reaction: \( \text{CO}(g) + \text{Cl}_2(g) \rightarrow \text{COCl}(g) + \text{Cl}(g) \)
• Reverse Reaction: \( \text{COCl}(g) + \text{Cl}(g) \rightarrow \text{CO}(g) + \text{Cl}_2(g) \)

The extremely low equilibrium constant seen in the original exercise reflects a preference for the reactants, chiefly because the reverse reaction occurs at a much quicker rate compared to the forward reaction. As a result, equilibrium is established with mostly reactants present.