Question

Suppose a reaction occurs with the following mechanism: (1) \(2 A \rightleftharpoons A_{2}\) \((\) fast \()\)(2) \(A_{2}+E \longrightarrow B+C\)(slow) in which the first step is a very rapid reversible reaction that can be considered to be essentially an equilibrium (forward and reverse reactions occurring at the same rate) and the second is a slow step. (a) Write the rate law for the forward reaction in step (1). (b) Write the rate law for the reverse reaction in step (1). (c) Write the rate law for the rate-determining step. (d) What is the chemical equation for the net reaction that occurs in this chemical change? Use the results of parts (a) and (b) to rewrite the rate law of the rate- determining step in terms of the concentrations of the reactants in the overall balanced chemical equation for the reaction.

Short answer

The rate laws are (a) Rate = k_forward [A]^2, (b) Rate = k_reverse [A2], (c) Rate = k_slow [A2][E], and the overall balanced chemical equation for the net reaction is 2A + E -> B + C. The rate law for the rate-determining step in terms of A and E is Rate = k_slow [(k_forward/k_reverse) [A]^2][E].

Step-by-step solution

Title - Write the Rate Law for the Forward Reaction in Step (1)

For the forward reaction of step (1), since it is fast and has reached equilibrium, the rate law is proportional to the concentration of reactant A raised to the power of its stoichiometric coefficient. The stoichiometric coefficient for A in the forward reaction is 2, so the rate law is: Rate = k_forward [A]^2, where k_forward is the rate constant for the forward reaction.

Title - Write the Rate Law for the Reverse Reaction in Step (1)

For the reverse reaction of step (1), the rate law is proportional to the concentration of the reactant A2 raised to the power of its stoichiometric coefficient, which is 1 in this case. Therefore, the rate law for the reverse reaction is: Rate = k_reverse [A2], where k_reverse is the rate constant for the reverse reaction.

Title - Write the Rate Law for the Rate-Determining Step

Step (2) is the slow, rate-determining step. The rate law for this step is determined by the concentration of reactants that are involved in this step. According to the stoichiometry, A2 and E react in a 1:1 ratio. Thus, the rate law for the rate-determining step is: Rate = k_slow [A2][E], where k_slow is the rate constant for the slow step.

Title - Write the Overall Chemical Equation

To obtain the net reaction, both steps must be added. Keep in mind that intermediates such as A2 which appear on both sides of the equation will cancel out. The net reaction of combining the two steps is obtained by cancelling out A2: 2A + E -> B + C. Thus, the equation for the net reaction is: 2A + E -> B + C.

Title - Rewrite the Rate Law of the Rate-Determining Step

To express the rate law of the rate-determining step in terms of the concentrations of the reactants (A and E), we have to eliminate the intermediate (A2). From step (1), we know that at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Thus, k_forward [A]^2 = k_reverse [A2]. We solve for [A2]: [A2] = (k_forward/k_reverse) [A]^2. Now we substitute [A2] in the rate law from step 3: Rate = k_slow [(k_forward/k_reverse) [A]^2][E].

Key concepts

Rate Law

Understanding the rate law is essential for analyzing how the concentration of reactants influences the speed of a chemical reaction. The rate law links the rate of a reaction to the concentration of its reactants raised to some power, which is generally determined empirically. For example, in a reaction mechanism with multiple steps, the rate law of the overall reaction can sometimes be deduced from the rate-determining (slowest) step.

In the given exercise, the rate-determining step is the second reaction, where the rate law is established as \( Rate = k_{slow} [A_2][E] \), with \( k_{slow} \) as the rate constant, and \( [A_2] \) and \( [E] \) as the concentrations of the reactants. This step is crucial as it effectively controls the overall rate at which the reaction proceeds, hence needing its own rate law. To further explicate, the concentration of the fast-reacting species \( A_2 \) in terms of the initial reactant \( A \) is derived from the equilibrium condition of the first step. The transformation leads to an expression where the rate law can now reflect the concentrations of the original reactants directly involved in the overarching reaction.

Reaction Mechanism

The reaction mechanism breaks down complex reactions into simpler steps, allowing us to analyze and understand the sequence of elementary reactions that occur.

In our exercise, we see a mechanism comprising two steps: a fast initial step reaching equilibrium quickly, and a subsequent slow step which is the rate-determining step. It’s important to highlight that not all intermediates, like \( A_2 \) in this case, are present in the final rate law for the overall reaction, as they are often transient species that do not appear in the net chemical equation. This simplification is vital for students to grasp, as it aids in focusing on the main reactants and products that should be considered when studying the progression of the reaction.

Reaction Rate

The reaction rate indicates how fast a reaction progresses. It is quantified by the change in concentration of reactants or products per unit time. Higher concentrations or more effective collisions usually mean faster reactions. However, the rate is not necessarily constant throughout the reaction; it may change as reactants are consumed and the concentration decreases.

In our exercise, the fast equilibrium in the first step suggests initial rapid consumption of reactant \( A \), while the slow rate of the second step governs the overall reaction pace. Understanding the distinction between instantaneous rates, which can vary at different points in time, and average rates, which are calculated over longer periods, can also enhance a student's comprehension of reaction rates in varying contexts.

Equilibrium Constant

The equilibrium constant (\( K_{eq} \)) quantifies the ratio of product concentrations to reactant concentrations at equilibrium, each raised to their respective stoichiometric coefficients. A large \( K_{eq} \) means the reaction favors products at equilibrium, while a small \( K_{eq} \) indicates that the reactants are favored.

In the initial fast step of our exercise, the equilibrium constant could be expressed as \( K_{eq} = [A_2]/[A]^2 \). However, to find the concentration of the intermediate \( A_2 \) via this constant, we would use the relationship \( k_{forward} / k_{reverse} \) as a stand-in for \( K_{eq} \) since \( k_{forward} [A]^2 = k_{reverse} [A_2] \) at equilibrium. This usage of the equilibrium concept helps to remove intermediates from the rate law of the rate-determining step, making it possible to link the kinetics of a reaction to its thermodynamic properties.