Question

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(B \longrightarrow A\) are both elementary processes with rate constants 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 for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})\) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Short answer

(a) The equilibrium constant (K) for the equilibrium A(g) ⇌ B(g) is 0.1. (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Step-by-step solution

Write down the given information

We are given the rate constants for the reactions: \(k_{1} = 2.5 \times 10^{-2} \mathrm{~min}^{-1}\) for the reaction A → B \(k_{2} = 2.5 \times 10^{-1} \mathrm{~min}^{-1}\) for the reaction B → A

Write down the relationship between rate constants and the equilibrium constant

For an elementary reaction, the equilibrium constant (K) can be determined from the rate constants of the forward and reverse reactions: \[K = \frac{k_{1}}{k_{2}}\]

Calculate the equilibrium constant K

Using the given rate constants, we can now determine the value of K: \[K = \frac{2.5 \times 10^{-2} \mathrm{~min}^{-1}}{2.5 \times 10^{-1} \mathrm{~min}^{-1}}\] \[K = \frac{1}{10} = 0.1\] The equilibrium constant (K) = 0.1.

Determine the partial pressures of A and B at equilibrium

At equilibrium, the ratio of partial pressures of the reactants and products is equal to the equilibrium constant K: \(\frac{P_{B}}{P_{A}} = K\) Given K < 1, this indicates that the reaction is more favorable in the reverse direction. Therefore: \(P_{B} < P_{A}\)

Summarize the findings

(a) The equilibrium constant (K) for the equilibrium A(g) ⇌ B(g) is 0.1. (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Key concepts

Gas-Phase Reactions

Gas-phase reactions occur when the reactants and products are all in the gaseous state. This type of reaction is often fast and can be influenced by various factors such as temperature and pressure due to the inherent high energy and movement of gas molecules. In the exercise example, we have a gas-phase reaction between A and B, denoted by \(A \longrightarrow B\) and \(B \longrightarrow A\). The beauty of gas-phase reactions is their simplicity, which makes them ideal for demonstrating fundamental chemical principles.

• They are often used in elementary reaction studies because of their straightforward behavior.
• Equilibrium in gas-phase reactions can be easily characterized using pressure measurements.

Understanding these reactions is crucial as they set the stage for more complex chemical studies, allowing us to predict how molecules will interact under various conditions.

Rate Constant

The rate constant, symbolized as \(k\), is a fundamental parameter in kinetics that gives us a measure of how quickly a reaction occurs. Its value indicates the speed of a reaction under specified conditions. In elementary reactions, each step has its own rate constant, which defines how fast that step proceeds.
In the exercise, two rate constants are considered:

• \(k_1 = 2.5 \times 10^{-2} \text{ min}^{-1}\) for the forward reaction \(A \rightarrow B\)
• \(k_2 = 2.5 \times 10^{-1} \text{ min}^{-1}\) for the reverse reaction \(B \rightarrow A\)

Having a higher rate constant generally means that the reaction goes faster. Here, the reverse reaction is faster since \(k_2 > k_1\). This difference in rate constants also helps determine the equilibrium state of the reactions.

Equilibrium Pressure

At equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration or, in gas-phase reactions, the partial pressures of the reactants and products.
The concept of equilibrium pressure is illustrated in the exercise by the relationship:\[\frac{P_B}{P_A} = K\]Where \(K\) is the equilibrium constant. This means that at equilibrium, the ratio of the partial pressure of product B to that of reactant A equals the equilibrium constant.
With \(K = 0.1\), a value less than 1 indicates that at equilibrium, the system favors the reactant A over the product B, thus \(P_A > P_B\). Understanding equilibrium pressure is critical for interpreting the chemical dynamics and predicting the composition of a gas system at equilibrium.

Elementary Reactions

Elementary reactions are single-step processes with a straightforward rate law that directly reflects the molecularity of the reaction. This means the rate is directly proportional to the concentration of the reactants involved.
In such reactions, the stoichiometric coefficients in the balanced equation provide us with the powers in the rate law. For the reactions presented:

• The reaction \(A \rightarrow B\) is elementary, and its rate is simply \(rate = k_1[A]\).
• The reverse reaction \(B \rightarrow A\) also follows an elementary rate law \(rate = k_2[B]\).

Elementary reactions are an excellent way of introducing kinetic principles since they skip intermediate steps and provide a cleaner insight into reaction dynamics.