Problem 111 - General Chemistry: Principles and Modern Applications

Question

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

Step-by-Step Solution

Answer

The first reaction mechanism is consistent both with the stoichiometry of the overall reaction and with the rate law, but the second is only consistent with the stoichiometry of the overall reaction, not with the rate law.

Step 1: Checking the first mechanism

The first mechanism suggests that the reaction happens in two steps: 1. \(A + B \rightarrow I\) (slow) 2. \(I + B \rightarrow C + D\) (fast) The first step is the slow one, and thus will be the rate-determining step. So, the rate law is \(\text{rate} = k[I][B]\). Because \(I\) is an intermediate (formed in the first step and consumed in the second), it's more accurate to express the rate law in terms of reactants. According to the first step, the concentration of the intermediate is directly proportional to the concentration of \(A\) and \(B\). Thus, \([I] \propto [A][B]\). Replacing this in the rate law, we get \(\text{rate} = k'[A][B]\), which matches the given rate law. This confirms that this mechanism is consistent with both the stoichiometry of the overall reaction and with the rate law.

Step 2: Checking the second mechanism

The second mechanism has two steps: 1. \(2B \rightarrow B_2\) (fast) 2. \(A + B_2 \rightarrow C + D\) (slow) Even here, the slow step will determine the reaction rate. From the second step, the rate law will be \(\text{rate} = k'[A][B_2]\). However, the concentration of \(B_2\) is not directly proportional to \(A\) and \(B\). In fact, the equation suggests that a higher concentration of \(B\) forms more \(B_2\). Therefore, this rate law does not match the given rate law which requires the concentration of both \(A\) and \(B\). It's consistent with the overall stoichiometry of the reaction, but not with the rate law.

Key Concepts

Understanding the Reaction Rate Law

The reaction rate law is a mathematical expression that specifies how the rate of a chemical reaction depends on the concentration of its reactants. In the case of the reaction \( A + 2B \rightarrow C + D \) the rate law is given by the expression \(\text{rate} = k[A][B]\). This means that the reaction rate is directly proportional to the concentration of reactant A and the concentration of reactant B.

When attempting to understand the rate law of a given reaction, it's essential to identify the experimental conditions under which the reaction rate was determined. The coefficients in the rate equation reflect the influence of reactant concentrations on the reaction speed and not necessarily the stoichiometry of the overall reaction.

It's common for students to mistakenly assume the order of the reaction (the exponent applied to the reactant concentration in the rate law) equal to the stoichiometric coefficient of the reactant in the balanced equation. Hence, to improve students' understanding, emphasize that the reaction order is determined experimentally and that it may differ from the stoichiometric coefficients.

Role of the Rate-Determining Step

In a multi-step reaction mechanism, one step typically proceeds at a slower rate than the others and thus limits the overall reaction speed. This is known as the rate-determining step (RDS), and it's crucial for calculating the correct rate law. For the given reaction, the slow first step \( A + B \rightarrow I \) is the rate-determining step.

This concept can be challenging, so here's a helpful analogy: consider a factory production line where one station is much slower than all others. The overall production rate of the factory depends on the speed of the slowest station, not the average speed. Similarly, in chemical reactions, the rate-determining step sets the pace for the entire reaction.

To help students grasp this concept, focus on examples where they can practice identifying the rate-determining step from a mechanism and then derive the rate law accordingly. It is also useful to clarify that the rate-determining step might not involve all reactants but determines the expression for the reaction rate.

The Importance of Reaction Intermediates

Reaction intermediates are species that are formed and consumed during the course of a reaction mechanism. They are different from reactants and products because they do not appear in the overall balanced equation. As seen in the mechanism \( A + B \rightarrow I \) followed by \( I + B \rightarrow C + D \), the species \( I \) is an intermediate.

Understanding the role of intermediates is vital for students, as they are key to linking together the steps of a complex reaction mechanism. Intermediates can be tricky since they are not present at the start or end of the reaction but must be considered when determining the overall rate law.

One common student misconception is to include intermediates in the rate law as if they were reactants or products. It's essential to correct this misunderstanding and teach that the steady-state approximation or other methods are used to replace intermediates with reactants to match the experimentally determined rate law. Provide examples where students can practice writing rate laws with and without intermediates, emphasizing the need to substitute the intermediates with the initial reactants' concentrations wherever possible.