Problem 11 - Chemistry: The Central Science

Question

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate con- stants of \(3.8 \times 10^{-2} \mathrm{~s}^{-1}\) and \(3.1 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Explain.

Step-by-Step Solution

Answer

The equilibrium constant for the given gas-phase reaction is \(K = 0.123\). At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Step 1: Define the rate laws for the forward and reverse reactions

For an elementary reaction, the rate law can be directly obtained from the balanced chemical equation. We have the following reactions: Forward reaction (A → B): \(k_1 = 3.8 \times 10^{-2} s^{-1}\) Reverse reaction (B → A): \(k_2 = 3.1 \times 10^{-1} s^{-1}\)

Step 2: Determine the equilibrium expression for the overall reaction

The overall chemical equation for this reaction is: \[A(g) \rightleftharpoons B(g)\] At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This means that the ratio of the rate constants is equal to the equilibrium constant (K): \[K = \frac{k_1}{k_2}\]

Step 3: Calculate the equilibrium constant, K

Substitute the rate constants provided in the exercise to find K: \[K = \frac{3.8 \times 10^{-2}s^{-1}}{3.1 \times 10^{-1}s^{-1}} = 0.123\]

Step 4: Determine whether the partial pressure of A or B is greater at equilibrium

At equilibrium, the partial pressures of A and B can be represented as follows: \[[A]_{eq} = P_A \\ [B]_{eq} = P_B\] We can use the equilibrium expression to find the relationship between the partial pressures: \[K = \frac{P_B}{P_A}\] Substituting the value of K, we get: \[0.123 = \frac{P_B}{P_A}\] Since the value of K is less than 1, this indicates that the partial pressure of A is greater than the partial pressure of B at equilibrium: \[P_A > P_B\] Therefore, the partial pressure of A is greater at equilibrium. In conclusion, the equilibrium constant for the given gas-phase reaction is 0.123, and at equilibrium, the partial pressure of A is greater than the partial pressure of B.

Key Concepts

Equilibrium Constant

In the realm of chemical equilibrium, the equilibrium constant, often symbolized as \(K\), plays a pivotal role. It provides insight into the ratio of the concentrations of products to reactants at equilibrium. For the elementary reaction \(A(g) \rightleftharpoons B(g)\), we can express \(K\) as \[K = \frac{k_1}{k_2}\]\
where \(k_1\) and \(k_2\) represent the forward and reverse reaction rate constants, respectively. In our example, \(k_1 = 3.8 \times 10^{-2} \text{s}^{-1}\) and \(k_2 = 3.1 \times 10^{-1} \text{s}^{-1}\). By substituting these values, we find \(K = 0.123\).
This value of \(K\) tells us that, at equilibrium, the ratio of products to reactants favors the reactants, since \(K < 1\). This means that the concentration of A (reactants) is greater than B (products) under equilibrium, providing a quantitative measure of the position of equilibrium.

Rate Laws

Rate laws are mathematical expressions that describe the rate of a chemical reaction. They are derived based on the concentration of reactants and their respective rate constants. For the reactions \(A \to B\) and \(B \to A\), the rate laws can be written succinctly:

• Forward reaction: \(\text{Rate} = k_1[A]\)
• Reverse reaction: \(\text{Rate} = k_2[B]\)

A fundamental aspect of chemical reactions is that, at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This creates a balance where the concentrations of A and B remain constant over time, even though both reactions are still occurring. It is this balance that allows us to define the equilibrium constant \(K\) in terms of rate constants \(k_1\) and \(k_2\).
Rate laws thereby serve as the cornerstone in understanding how chemical reactions progress and how equilibrium is established in reversible reactions.

Partial Pressure

The concept of partial pressure is crucial when dealing with gases in a chemical equilibrium. It refers to the pressure that a gas in a mixture would exert if it occupied the whole volume independently. For a gas-phase reaction like \(A(g) \rightleftharpoons B(g)\), the equilibrium aspect focuses on the partial pressures of these components.
Using the equilibrium expression \(K = \frac{P_B}{P_A}\), we gain insight into the relative quantities of gases present at equilibrium.
In our scenario, with \(K = 0.123\), we have the equation \(0.123 = \frac{P_B}{P_A}\). This implies that \(P_A > P_B\) as \(K < 1\), meaning that A has a higher partial pressure compared to B at equilibrium. This relationship is fundamental in predicting the direction of reaction shifts when external conditions change, according to Le Chatelier's principle.Understanding partial pressures helps to predict how gases will distribute in a reaction at equilibrium and is an essential tool in gas dynamics within a chemical system.