Problem 111 - Chemistry The Central Science In Si Units

Question

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: A+B \longrightarrow C+X \\ \text { Step } 2: A+X \longrightarrow C+D \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Step-by-Step Solution

Answer

(a) Rate = k_1 [A][B]; (b) Rate = \frac{k_2 k_1}{k_{-1}} [A]^2 [B]; (c) Option (iv).

Step 1: Understanding Reaction Steps

The overall reaction is composed of two elementary steps. In these steps, the intermediate 'X' is formed in Step 1 and used up in Step 2.

Step 2: Predicted Rate Law for Step 1 as Rate Determining

If Step 1 is the rate-determining step, the rate of the reaction is governed by this step. The rate law can be directly written from the reaction stoichiometry of Step 1 as it is elementary. Thus, for Step 1: \[\text{Rate} = k_1 [A][B]\] where \(k_1\) is the rate constant for Step 1.

Step 3: Predicted Rate Law for Step 2 as Rate Determining

If Step 2 is the rate-determining step, we write the rate law based on the stoichiometry of Step 2. However, it involves the intermediate 'X', which has to be expressed in terms of reactants (since intermediates are not included in the overall rate law). From Step 1 which is fast, we use the steady state approximation or pre-equilibrium condition: \[\text{From Step 1, } [X] = \frac{k_1[A][B]}{k_{-1}}\]Substituting \([X]\) into the rate law for Step 2, \[\text{Rate} = k_2 [A][X] = k_2 \left(\frac{k_1[A][B]}{k_{-1}}\right)[A] = \frac{k_2 k_1}{k_{-1}} [A]^2 [B]\].

Step 4: Understanding Surprising Rate Law Features

For part (b), the rate law \( \text{Rate} = \frac{k_2 k_1}{k_{-1}} [A]^2 [B] \) includes only positive integer reaction orders for the reactants. There are no product concentrations or negative reaction orders in the rate law, making option (iv) correct: 'Neither reasons (i) nor (ii).'

Key Concepts

Reaction Mechanism

In chemistry, a reaction mechanism is a detailed step-by-step description of the pathway that a chemical reaction follows. Each step, known as an elementary step, represents a single molecular event. These combine to form the overall reaction.
For the hypothetical reaction given, the mechanism involves two steps. In the first step, reactants A and B interact to form an intermediate product X and a molecule of C:

• Step 1: A + B → C + X
• Step 2: A + X → C + D

The existence of intermediate X is crucial here. Reaction mechanisms help chemists understand how a reaction progresses and what species are temporarily formed along the way.

Rate-Determining Step

In a multi-step reaction mechanism, the rate-determining step is the slowest step. This step acts like a bottleneck, restricting the overall speed of the reaction. Since it slows everything down, it determines the overall reaction rate.
If Step 1 is the rate-determining step in our given reaction, the rate of the reaction depends on the concentrations of A and B. Thus, the rate law is \[ \text{Rate} = k_1 [A][B]\]. However, if Step 2 is the rate-determining step, the situation is more complex because it involves intermediate X. Nonetheless, it ultimately simplifies to \[ \text{Rate} = \frac{k_2 k_1}{k_{-1}} [A]^2 [B]\] , showing how much the rate depends on the reactants' concentrations.

Steady State Approximation

The steady-state approximation is a concept used to handle intermediate species in complex chemical reactions. It assumes that the concentration of an intermediate remains relatively constant during the reaction, despite being consumed and produced.
In our reaction example, when Step 2 is the rate-determining step, the concentration of intermediate X appears. To eliminate X from the rate law, we use the steady-state approximation.

• We assume that \([X]\) is stabilized by the fast step, Step 1.
• We express this as \( [X] = \frac{k_1[A][B]}{k_{-1}} \)

This allows us to link the intermediate back to concentrations of reactants, which are easier to measure and use in kinetic equations.

Intermediate Species

Intermediate species are molecules formed and consumed during a reaction mechanism. They don't appear in the overall balanced equation for the reaction but play a crucial role in the steps of the mechanism.
In the reaction provided, X is an intermediate species. It is formed in Step 1 and consumed in Step 2. Intermediates are typically unstable and fleeting, making them difficult to detect.
Understanding how intermediates work helps chemists to refine their understanding of a reaction's mechanism, as well as predict reaction kinetics and final products.