Problem 13 - Chemistry: An Atoms-Focused Approach

Question

Suppose the forward rate constant of the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) is greater than the rate constant of the reverse reaction at a given temperature. Is the valuc of the cquilibrium constant less than, greater than, or cqual to \(1 ?\)

Step-by-Step Solution

Answer

Answer: The equilibrium constant is greater than 1.

Step 1: Understand the relationship between rate constants and the equilibrium constant

The reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) is reversible, meaning it proceeds in both the forward direction (\(\mathrm{A} \rightarrow \mathrm{B}\)) and the reverse direction (\(\mathrm{B} \rightarrow \mathrm{A}\)). The forward and reverse rate constants, \(k_f\) and \(k_r\), describe the rates at which the reaction proceeds in these two directions, respectively. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant \(K\) can be defined as the ratio of the forward rate constant and the reverse rate constant: \(K = \frac{k_f}{k_r}\).

Step 2: Determine the direction of the reaction with a larger rate constant

We are given that the forward rate constant of the reaction is greater than the rate constant of the reverse reaction at a given temperature: \(k_f > k_r\). This means that the reaction proceeds more readily in the forward direction at this temperature.

Step 3: Determine the relationship between the equilibrium constant and \(1\)

Since \(k_f > k_r\), then the ratio of the forward rate constant and the reverse rate constant is greater than \(1\): \(\frac{k_f}{k_r} > 1\). Since \(K = \frac{k_f}{k_r}\), the equilibrium constant \(K\) is greater than \(1\) at the given temperature. Therefore, the equilibrium constant is greater than \(1\).

Key Concepts

Rate Constant

In chemical reactions, the concept of a rate constant is a key factor in determining how fast a reaction occurs. The rate constant, often represented by the symbol \(k\), quantifies the speed at which reactants are converted into products in a chemical reaction under a constant temperature. This value is essential for calculating the rate of both forward and reverse reactions in a chemical system.

• The rate constant is specific to a particular reaction at a given temperature.
• It is influenced by factors such as the nature of the reactants, the presence of a catalyst, and the temperature.

The greater the rate constant, the faster the reaction proceeds. In the exercise above, the given information that the forward rate constant \(k_f\) is greater than the reverse rate constant \(k_r\) informs us that the reaction occurs more readily in the forward direction, leading to a greater production of products (B from A) at the specified temperature.

Forward Reaction

The forward reaction is the process where reactants are converted into products. In reversible reactions such as \( \mathrm{A} \rightleftharpoons \mathrm{B} \), the forward reaction is denoted as \(\mathrm{A} \rightarrow \mathrm{B}\). This direction represents the conversion from A to B.

• In chemical equilibrium, the rate of the forward reaction becomes equal to the rate of the reverse reaction.
• The rate constant for the forward reaction, \(k_f\), indicates how quickly this transformation occurs.

If \(k_f\) is greater than \(k_r\), it indicates that the forward reaction is favored, meaning more product (B) will be present at equilibrium than reactant (A). As seen in the solution, because \(k_f > k_r\), the equilibrium constant \(K\) results in being more than 1, suggesting a higher concentration of products than reactants.

Reverse Reaction

In the context of reversible reactions, the reverse reaction is when products revert back into reactants. For the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\), the reverse direction is \(\mathrm{B} \rightarrow \mathrm{A}\). This process is just as important as the forward reaction in systems at equilibrium, where the rates of the forward and reverse reactions are equal.

• The rate constant of the reverse reaction is \(k_r\), which determines how quickly products can revert to reactants.
• An important observation is that if \(k_r\) is less than \(k_f\), the reverse process is less favored.

In our given example, since \(k_r < k_f\), the reverse reaction is slower, meaning less of the product B is converted back to reactant A over time. Thus, this contributes to an equilibrium state where there is a predominantly higher concentration of B compared to A, reflected by the equilibrium constant being greater than 1.