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18.5: Multi-step Reaction

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

The student will:

• define multi-step reaction and reaction mechanism.
• identify the rate-determining step.
• draw a potential energy diagram for a multi-step reaction.

Vocabulary

• elementary step
• multi-step mechanism
• rate-determining step
• reaction mechanism

Introduction

In the last section, the uncatalyzed reaction between hydrogen and oxygen was compared to a catalyzed reaction between the same two reactants. Without a catalyst, all three particles needed to collide at the same time with sufficient energy to break all the bonds in all the molecules, giving the reaction a very high activation energy. The catalyzed reaction could occur because it allowed for a series of collisions where the bonds could be broken one at a time, thus requiring less activation energy.

Complicated reactions involving many bonds and many molecules do not occur in single collisions. These reactions instead occur in a series of collisions. Each collision in the series produces an intermediate product that undergoes further collisions until the final products are produced. This long series of collisions producing intermediate products is called a multi-step reaction.

The reaction below is an example of a multi-step reaction. Nitrogen dioxide and carbon monoxide react in a two-step process to form nitrogen monoxide and carbon dioxide. The two steps making up the reaction mechanism, which describes what occurs for each stage in the reaction, is shown below.

$\begin{array}{cl} \text{NO}_{2(g)} + \cancel{\text{NO}_{2(g)}} \rightarrow \cancel{\text{NO}_{3(g)}} + \text{NO}_{(g)} & (\text{step 1})\\\cancel{\text{NO}_{3(g)}} + \text{CO}_{(g)} \rightarrow \cancel{\text{NO}_{2(g)}} + \text{CO}_{2(g)} & (\text{step 2})\\\hline\text{NO}_{2(g)} + \text{CO}_{(g)} \rightarrow \text{NO}_{(g)} + \text{CO}_{2(g)} & (\text{overall reaction})\end{array}$

In this lesson, we will discuss multi-step reactions, as well as the individual reactions in the multi-step process.

Most Reactions Have Multi-Steps

In complicated reactions, the overall reaction will take place in a series of single steps, often called elementary steps. An elementary step is a single, simple step in a multi-step process. An elementary step almost always involves only two particles. The series of elementary steps outline the process of the reaction. Most reactions do not take place in one step but rather as a combination of two or more elementary steps. This series of steps is referred to as a multi-step reaction.

The rate of reaction is dependent on the reactants in the slowest step of the multi-step process. If we look at the reaction from the introduction again, it appears to be a two particle collision between $\mathrm{NO}_{2(g)}$ and $\mathrm{CO}_{(g)}$, yet it is known that the rate is only affected by the concentration of $\mathrm{NO}_{2(g)}$. This indicates that the reaction proceeds by way of a multi-step process.

$\mathrm{NO}_{2(g)} + \mathrm{CO}_{(g)} \rightarrow \mathrm{NO}_{(g)} + \mathrm{CO}_{2(g)}$

Example:

Which of the following reactions would most likely involve a multi-step reaction? Explain.

1. $2 \ \mathrm{I}^- + \mathrm{S}_2\mathrm{O}_8{^{2-}} \rightarrow \mathrm{I}_2 + 2 \ \mathrm{SO}_4{^{2-}}$
2. $\mathrm{Cl}_{2(g)} \rightarrow 2 \ \mathrm{Cl}^-_{(g)}$

Solution:

1. multi-step reaction; three reactant particles are present
2. elementary step; only one reactant particle is present

Each Step Has Its Own Activated Complex

When there is a single step reaction, we can draw potential energy diagrams like the ones we have seen earlier in this chapter. For a multi-step process where two or more elementary steps combine to form the net reaction, the potential energy diagram looks quite different. Look at the reaction below. This mechanism is involved in the depletion of the ozone layer.

$& \mathrm{NO}_{(g)} + \mathrm{O}_{3(g)} \rightarrow \mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)} && \text{(reaction 1)} && \text{(slow)}\\& \mathrm{NO}_{2(g)} + \mathrm{O}_{(g)} \rightarrow \mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} && \text{(reaction 2)} && \text{(fast)}$

The overall reaction is $\mathrm{O}_{3(g)} + \mathrm{O}_{(g)} \rightarrow 2 \ \mathrm{O}_{2(g)}$.

If we were to draw the potential energy diagram for this two-step process, it would look like the figure below. Notice that for each reaction in the multi-step process, there is an activation energy barrier. Therefore, $E_{a1}$ is the activation energy associated with reaction 1, and $E_{a2}$ is the activation energy associated with reaction 2. The slow step has an activation energy barrier that is higher than that of the faster reaction.

Each reaction also has its own activated complex. Remember that at the top of the activation energy barrier is the activated complex, the transition state between reactants and products that has the most potential energy. $AC_1$ is the complex created in the first reaction, while $AC_2$ is the activated complex created in the second reaction. Thus, for this two-step process, there are two activated complexes.

Example:

Draw the potential energy diagram for the following multi-step reaction $(\triangle H < 0)$. Properly label the diagram.

$& \mathrm{NO}_{2(g)} + \mathrm{F}_{2(g)} \rightarrow \mathrm{NO}_2\mathrm{F}_{(g)} + \mathrm{F}_{(g)} && \text{(reaction 1)} && \text{(slow)}\\& \mathrm{NO}_{2(g)} + \mathrm{F}_{(g)} \rightarrow \mathrm{NO}_2\mathrm{F}_{(g)} && \text{(reaction 2)} && \text{(fast)}$

Solution:

Rate of Reaction is Determined by Slowest Step

In a series of reactions that make up a multi-step reaction, each individual reaction step has its own reaction rate that is determined by the factors that have been discussed in this chapter. The overall reaction rate for the overall reaction (the sum of all the individual steps) can be determined from the rates of the individual steps. The relationship between the overall rate and the individual rates, however, is not what you might expect. The overall rate is not the sum or the average of the individual rates. In fact, the overall rate for the reaction is exactly the same as the rate of the slowest step. Let's look at an example from life to see how this occurs.

Suppose you and two of your friends organize a car wash. You set up an assembly line operation where the cars at station 1 are wetted with a hose, cars at station 2 are washed with soapy water and rinsed, and cars at station 3 are towel dried.

Suppose your job is to wet the cars, which takes $3 \ \mathrm{minutes}$ to accomplish. The job of washing at station 2 takes $18 \ \mathrm{minutes}$, and the drying job requires $14 \ \mathrm{minutes}$. When the car is done at a station, the car gets in line for the next station. In order to evaluate the efficiency of your assembly line, you count the minutes between the finished cars coming off the end of the line. The time lapse between completed cars is your reaction rate.

Regardless of what appears in the picture above, you should realize that there will be no cars in line for station 3. Each car requires $18 \ \mathrm{minutes}$ at station 2, and when the cars move to station 3, another car immediately goes to station 2. The drying of the first car and the washing of the second car begin at exactly the same time. Since it takes $18 \ \mathrm{minutes}$ to wash and $14 \ \mathrm{minutes}$ to dry, the car at station 3 is always finished and gone $4 \ \mathrm{minutes}$ before the car at station 2. Therefore, the worker at station 3 is always standing around and waiting for $4 \ \mathrm{minutes}$ before the next car is ready to be dried. The time between cars coming off the line will be the $4\;\mathrm{minutes}$ the station 3 worker waits plus the $14\;\mathrm{minutes}$ required to dry, so the time between cars will be $18\;\mathrm{minutes.}$ The reaction rate for this arrangement will be $18\;\mathrm{minutes.}$ You should note that this overall reaction rate is exactly the same as the slowest step in the process, namely the wash step at station 2.

Suppose you bring in another person to work on your car wash and you assign that person to station 2 so that you have two people washing cars. The time to wash a car now becomes only $9 \ \mathrm{minutes}$ since you have doubled the work force. The wash process now will finish before the drying process, so cars will back up in line for station 3. The drying stage is now the slowest step. Therefore, when the worker at station 3 finishes drying a car, another car immediately enters station 3 for the worker to begin drying. With this new organization, the time lapse between cars coming out will now be $14 \ \mathrm{minutes}$. The overall reaction rate is faster. The important point is that the slowest step takes $14 \ \mathrm{minutes}$, and the overall rate is $14 \ \mathrm{minutes}$.

Let's look at another important point about this concept. We will go back to the original set up with a $3-\mathrm{minute}$ wetting (station 1), an $18-\mathrm{minute}$ washing (station 2), and a $14-\mathrm{minute}$ drying (station 3). Suppose when the first extra worker was brought in, you assigned that person to help with the drying so that the drying time is now $7 \ \mathrm{minutes}$. What would the overall rate be? The overall rate is exactly the same as the slowest step, and the slowest step is still the $18-\mathrm{minute}$ washing time. Therefore, you have increased the rate of the drying step, but you have not affected the overall reaction rate. The only way you can alter the overall reaction rate is to increase the rate of the slowest step. Increasing the rate of steps other than the slowest step does nothing to the overall rate.

In chemical reactions, the speed of the other steps is so much faster than the slowest step that the slowest step is referred to as the rate-determining step. It is the speed of this slowest step that determines the rate of the overall reaction, so changing the concentrations of the reactants in this step will change the rate.

Example:

In the multi-step reaction below, identify the rate-determining step and write the overall reaction.

$& \mathrm{H}_2\mathrm{O}_{2(aq)} + \mathrm{I}^-{_{(aq)}} \rightarrow \mathrm{IO}^-{_{(aq)}} + \mathrm{H}_2\mathrm{O}_{(l)} && \text{(fast)} \\& \mathrm{H}_2\mathrm{O}_{2(aq)} + \mathrm{IO}^-{_{(aq)}} \rightarrow \mathrm{I}^-{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)} && \text{(slow)}$

Overall reaction: $2 \ \mathrm{H}_2\mathrm{O}_{2(aq)} \rightarrow 2 \ \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)}$

The overall reaction is found by adding the two elementary steps together and canceling identical species that appear on both sides of the chemical equation, which in this case are $\mathrm{I}^-{_{(aq)}}$ and $\mathrm{IO}^-{_{(aq)}}$.

Solution:

The slow step (reaction 2) is the rate-determining step. Whatever the reaction rate is for reaction 2, the overall rate will be exactly the same.

Lesson Summary

• A multi-step reaction is a combination of two or more elementary steps.
• An elementary step is a single, simple step involving one or two particles.
• The rate-determining step is the slowest step in a multi-step reaction, and the overall reaction rate will be exactly the same as the rate of the slowest step.

The following is a video lecture on reaction mechanism.

Review Questions

1. Why do most reactions take place in more than one step?
2. The overall rate of a reaction depends on
1. the temperature
2. the surface area
3. the pressure
4. the slowest step
3. Suppose a reaction takes place according to the following reaction mechanism. Determine which step in the mechanism is the rate-determining step.

$& X + Y \rightarrow Z && \text{(fast)} \\& X + Z \rightarrow F && \text{(slow)}$

1. If you wanted to increase the overall rate of the reaction in Question #3, would you increase the concentration of $X$ or $Y$? Explain
2. Consider the following equation for the formation of ammonia. Explain why this equation is not likely to represent the reaction mechanism.

$\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightarrow 2 \ \mathrm{NH}_{3(g)}$

Date Created:

Feb 23, 2012

Jan 07, 2015
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