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If a slow step precedes a fast step in a two-step mechanism, do the substances in the fast step appear in the overall rate law? Explain.

Short Answer

Expert verified
Substances in the fast step do not appear in the overall rate law because the slow step determines the reaction rate.

Step by step solution

01

Understand the Reaction Mechanism

In a two-step reaction mechanism, there are typically two elementary steps: one slow step and one fast step. The overall rate of the reaction depends largely on the slow step, which is known as the rate-determining step.
02

Identify the Rate-Determining Step

The slow step is the rate-determining step. This means that the rate of the overall reaction is governed by the rate at which this slow step occurs. The substances involved in this step are what primarily influence the overall rate law.
03

Analyze the Influence of the Fast Step

Since the fast step follows the slow step, the substances formed or consumed in the fast step do not directly influence the overall rate law because the slow step's rate has already determined the reaction rate. Essentially, the substances in the fast step reach equilibrium quickly and do not appear in the rate law.
04

Construct the Overall Rate Law

The overall rate law is constructed based on the rate-determining step. Only the reactants (and sometimes intermediates if they appear in the slow step) of this slow step will appear in the rate law. Substances purely involved in the fast step do not contribute to the rate law.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

reaction mechanism
In chemical reactions, a reaction mechanism is a detailed step-by-step description of how reactants transform into products. Most reactions occur through a series of intermediate steps, rather than a single event. These steps are called elementary steps. Understanding the mechanism helps chemists figure out which step is the slowest, and therefore, which one limits the overall rate of the reaction. This slowest step is called the rate-determining step. By knowing the detailed mechanism, we can better predict and control the reaction's behavior. In our example, we are dealing with a two-step mechanism where one step is slower (rate-determining) and one is faster.
elementary steps
Each elementary step in a reaction mechanism represents a single event at the molecular level. These steps can involve the breaking and making of chemical bonds. For a given mechanism, chemists can write out each elementary reaction, which together represent the full reaction. These steps are not usually observable directly but are inferred from experimental data and the reaction's overall behavior. Each elementary step has its own distinct rate law which only depends on the species involved in that step. In our example, the two elementary steps include a slow step and a fast step. The slow step controls the reaction rate, while the fast step quickly reaches equilibrium and does not directly affect the rate law.
overall rate law
The overall rate law of a reaction describes how the rate of the reaction depends on the concentration of reactants. It is derived from the rate-determining step of the reaction mechanism. Reactants and intermediates that are involved in this slow step appear in the rate law equation. For two-step mechanisms where the slow step precedes the fast step, only the concentrations of species in the slow step will appear in the overall rate law. The substances from the fast step do not directly influence the rate because the fast step quickly reaches equilibrium. Thus, for our example, the substances in the rate-determining (slow) step determine the overall rate law.
reaction kinetics
Reaction kinetics is the study of the rates at which chemical reactions occur. It involves understanding how different factors such as concentration, temperature, and the presence of catalysts affect the reaction rate. By examining the rate laws and mechanisms, chemists can predict the reaction rate under various conditions. Studying kinetics provides insights into the efficiency and speed of reactions, which is crucial for industries relying on chemical processes. In our two-step mechanism example, the reaction kinetics is primarily determined by the rate-determining step. Understanding this can help in optimizing the conditions to achieve the desired reaction rate efficiently.

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Most popular questions from this chapter

What is the central idea of collision theory? How does this model explain the effect of concentration on reaction rate?

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ \mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g}) $$ When [B] is decreasing at \(0.5 \mathrm{~mol} / \mathrm{L}\) -s, how fast is \([\mathrm{A}]\) decreasing?

Give the individual reaction orders for all substances and the overall reaction order from the following rate law: $$ \text { Rate }=k\left[\mathrm{BrO}_{3}^{-}\right]\left[\mathrm{Br}^{-}\right]\left[\mathrm{H}^{+}\right]^{2} $$

Like any catalyst, palladium, platinum, or nickel catalyzes both directions of a reaction: addition of hydrogen to (hydrogenation) and its elimination from (dehydrogenation) carbon double bonds. (a) Which variable determines whether an alkene will be hydrogenated or dehydrogenated? (b) Which reaction requires a higher temperature? (c) How can all-trans fats arise during hydrogenation of fats that contain some double bonds with a cis orientation?

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both \(\mathrm{NOBr}\) and \(\mathrm{Br}_{2}\) are reddish brown: $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ Use the data below to answer the following: (a) Determine the average rate over the entire experiment. (b) Determine the average rate between 2.00 and \(4.00 \mathrm{~s}\). (c) Use graphical methods to estimate the initial reaction rate. (d) Use graphical methods to estimate the rate at \(7.00 \mathrm{~s}\). (e) At what time does the instantaneous rate equal the average rate over the entire experiment? $$ \begin{array}{cc} \text { Time (s) } & \text { [NOBr] (mol/L) } \\ \hline 0.00 & 0.0100 \\ 2.00 & 0.0071 \\ 4.00 & 0.0055 \\ 6.00 & 0.0045 \\ 8.00 & 0.0038 \\ 10.00 & 0.0033 \end{array} $$

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