Problem 13 - Chemistry The Central Science In Si Units

Question

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(B \longrightarrow A\) are both elementary processes with rate con- stants of \(2.5 \times 10^{-2} \mathrm{~min}^{-1}\) and \(2.5 \times 10^{-1} \mathrm{~min}^{-1}\), respectively. (a) What is the value of the equilibrium constant (b) Which is greater for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) ? at equilibrium, the partial pressure of A or the partial pressure of B?

Step-by-Step Solution

Answer

(a) The equilibrium constant \(K\) is 0.1. (b) The partial pressure of A is greater at equilibrium.

Step 1: Write the Reaction Rates

The reaction rates are given as \(A \rightarrow B\) with a rate constant \(k_1 = 2.5 \times 10^{-2} \; \text{min}^{-1}\) and \(B \rightarrow A\) with a rate constant \(k_2 = 2.5 \times 10^{-1} \; \text{min}^{-1}\).

Step 2: Determine Equilibrium Constant Expression

Since these are elementary processes, the equilibrium constant \(K\) for the reaction can be expressed as the ratio of the rate constants: \[ K = \frac{k_1}{k_2} \] Substitute the given values: \[ K = \frac{2.5 \times 10^{-2}}{2.5 \times 10^{-1}} \] Simplify to find \(K\).

Step 3: Calculate the Equilibrium Constant

Simplify the expression for \(K\): \[ K = \frac{2.5 \times 10^{-2}}{2.5 \times 10^{-1}} = \frac{1}{10} = 0.1 \]

Step 4: Analyze Relative Partial Pressures

The small value of \(K = 0.1\) indicates that the reaction favors the reactant side, \(A\) over \(B\) at equilibrium. Therefore, the partial pressure of \(A\) must be greater than the partial pressure of \(B\) at equilibrium.

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, denoted as \( K \), plays a crucial role in understanding the balance between reactants and products. It represents the ratio of the concentration of products to reactants when a reaction is at equilibrium. For elementary reactions, such as in our exercise, this ratio can be directly tied to the rate constants of the forward and reverse reactions.

• The equilibrium constant provides insight into which side of the reaction is favored—whether reactants or products are more dominant.
• A large value of \( K \) means more products at equilibrium, while a small \( K \) indicates more reactants.

In the given exercise, the equilibrium constant is calculated by dividing the rate constant of the forward reaction \( (A \rightarrow B) \) by that of the reverse reaction \( (B \rightarrow A) \). With rate constants of \( 2.5 \times 10^{-2} \; \text{min}^{-1} \) and \( 2.5 \times 10^{-1} \; \text{min}^{-1} \), we find \( K = 0.1 \), suggesting that either reactants or products have a relatively stronger presence at equilibrium.

Reaction Rate

The rate of a chemical reaction measures how quickly reactants are converted into products. This is an essential concept when understanding chemical equilibrium, as different reactions can occur at different speeds depending on various conditions.

• Reaction rates can be expressed in terms of rate constants \( k \), which are intrinsic to each reaction pathway.
• For a given reaction, knowing the rate constant can help predict how fast a reaction will reach equilibrium.

In our exercise, the rate constants for the reactions \( A \rightarrow B \) \((k_1 = 2.5 \times 10^{-2} \; \text{min}^{-1})\) and \( B \rightarrow A \) \((k_2 = 2.5 \times 10^{-1} \; \text{min}^{-1})\) are provided. These values indicate the speed at which each process takes place. The larger rate constant for the backward reaction \( B \rightarrow A \) shows that this process happens faster compared to its forward counterpart, influencing the equilibrium position to favor the reactants.

Elementary Reaction

An elementary reaction is a single-step process that describes the direct transformation from reactants to products. Understanding elementary reactions is crucial because they represent the simplest form of a chemical reaction without any intermediate steps.

• Elementary reactions have a straightforward stoichiometry, and their rate laws can be derived directly from their molecularity.
• Each step in a reaction mechanism can often be characterized as an elementary reaction, and the mechanism's overall rate is determined by the slowest step known as the rate-determining step.

For the exercise at hand, both \( A \rightarrow B \) and \( B \rightarrow A \) are considered as elementary reactions. This simplifies the calculation of the equilibrium constant since no intermediate steps or additional complexities need to be considered. With this direct approach, understanding how these reactions proceed becomes more manageable, allowing for straightforward predictions about system behavior at equilibrium.