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# 18.3: Reaction Mechanisms

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Describe how the elementary steps of a chemical reaction combine to produce the overall reaction mechanism.
• Be able to identify intermediates and catalysts when they appear in a reaction mechanism.
• Write a rate law for an elementary step, and identify the molecularity of the reaction.
• Identify the rate-determining step of a reaction.
• Analyze the potential energy diagram for a multi-step reaction.

## Lesson Vocabulary

• elementary step
• intermediate
• molecularity
• rate-determining step
• reaction mechanism

### Recalling Prior Knowledge

• What is a rate law?
• What is the effect of using a catalyst on the rate of a reaction?

Chemical reactions rarely occur in one simple step. More commonly, many small steps combine in a specific sequence, resulting in the overall reaction. In this lesson, you will be introduced to reaction mechanisms and how they are related to rate laws.

## Elementary Steps

The overall balanced equation for a chemical reaction does not always tell us how a reaction actually proceeds. In many cases, the overall reaction takes place in a series of small steps. An elementary step (or elementary reaction) is one step in a series of simple reactions that show the progress of a reaction at the molecular level. A reaction mechanism is the sequence of elementary steps that together comprise an entire chemical reaction. As an analogy, consider the route that you might take while driving to the grocery store. That route may consist of several turns, similar to the elementary steps. The overall reaction specifies only the beginning point (your house) and the final destination (the store), with no information about the specifics in between.

The reaction mechanism concept can be illustrated by the reaction between nitrogen monoxide and oxygen to form nitrogen dioxide.

$\mathrm{2NO}{(g)} \mathrm{+O_{2}}{(g)} \mathrm{\rightarrow 2NO_{2}}{(g)}$

It may seem as though this reaction would occur as the result of a collision between two NO molecules and one O2 molecule. However, a careful analysis of the reaction has detected the presence of N2O2 during the reaction. A proposed mechanism for this reaction consists of two elementary steps:

Step 1: $\mathrm{2NO}{(g)} \mathrm{\rightarrow N_2O_2}{(g)}$
Step 2: $\mathrm{N_2O_2}{(g)} \mathrm{+O_2}{(g)} \mathrm{\rightarrow 2NO_2}{(g)}$

In the first step, two molecules of NO collide to form a molecule of N2O2. In the second step, that molecule of N2O2 collides with a molecule of O2 to produce two molecules of NO2. The overall chemical reaction is the sum of the two elementary steps:

$\mathrm{2NO}{(g)} \rightarrow \cancel{\mathrm{N_2O_2}{(g)}} & \\\cancel{\mathrm{N_2O_2}{(g)}}+\mathrm{O_2}{(g)} \rightarrow \mathrm{2NO_2}{(g)} & \\\hline\mathrm{2NO}{(g)}+\mathrm{O_2}{(g)} \rightarrow \mathrm{2NO_2}{(g)} &$

The N2O2 molecule is not part of the overall reaction. It was produced in the first elementary step, then reacts in the second elementary step. An intermediate is a species which appears in the mechanism of a reaction, but not in the overall balanced equation. An intermediate is always formed in an earlier step of the mechanism and then consumed in a later step.

### Molecularity of a Reaction

The molecularity of an elementary step is the total number of reactant molecules in that step. In both steps of the reaction mechanism shown above, two reactant molecules collide with one another. These are both bimolecular reactions. Notice that the colliding molecules may be the same (as in step 1 above) or different (as in step 2 above). A unimolecular step is one in which only one molecule is present as a reactant. A termolecular reaction involves three reacting molecules in one elementary step. Termolecular steps are relatively rare because they require the simultaneous collision of three molecules with sufficient energy in the correct orientation, which is a rare event. When termolecular reactions do occur, they tend to be very slow.

## Rate Laws and Mechanisms

The rate law for a reaction can be determined from knowledge of the reaction mechanism. Consider the following unimolecular elementary reaction:

A → products

Because it occurs in a single elementary step, the rate of product formation will increase linearly with the concentration of A, making the rate of this reaction first-order with respect to A.

rate=k[A]

A bimolecular elementary reaction could be one of two types. Either a molecule of A could react with a molecule of B, or two molecules of A could react with each other. In either case, the rate of reaction depends on how frequently the collisions between reactant molecules occur.

$\mathrm{A+B} & \rightarrow \mathrm{products} & & \mathrm{rate=k[A][B]} \\\mathrm{2A} & \rightarrow \mathrm{products} & & \mathrm{rate=k[A]^2}$

The reaction order for each reactant in an elementary step is equal to its stoichiometric coefficient in the equation for that step. In the first equation above, each coefficient is a 1, and so the reaction is first-order with respect to A and first-order with respect to B. In the second equation, the coefficient of 2 means the reaction is second-order with respect to A.

The determination of a reaction mechanism can only be made in the laboratory. When a reaction occurs in a sequence of elementary steps, the overall reaction rate is governed by whichever one of those steps is the slowest. The rate-determining step is the slowest step in the reaction mechanism. To get an idea of how one step is rate determining, imagine driving on a one-lane road where it is not possible to pass another vehicle. The rate of flow of traffic on such a road would be dictated by whichever car is traveling at the lowest speed. The decomposition of hydrogen peroxide is discussed below and illustrates how reaction mechanisms can be determined through experimental studies.

### Decomposition of Hydrogen Peroxide

Recall that a catalyst is a substance that increases the rate of a chemical reaction without being consumed. Catalysts lower the overall activation energy for a reaction by providing an alternative mechanism for the reaction to follow. One such catalyst for the decomposition of hydrogen peroxide is the iodide ion (I).

$\mathrm{2H_2O_2}{(aq)} \mathrm{\overset{I^-}{\rightarrow}} \mathrm{2H_2O}{(l)}+ \mathrm{O_2}{(g)}$

By experiment, the rate of this reaction is found to be first-order with respect to both H2O2 and I, which makes it second-order overall.

rate = k[H2O2][I-]

Based on this experimental rate law, we know that the reaction cannot occur in a single elementary step. If it did, the reaction would be second-order with respect to H2O2, since the coefficient of H2O2 in the balanced equation is 2. A reaction mechanism can be constructed which accounts for the rate law and for the detection of the IO ion as an intermediate. It consists of two bimolecular elementary steps.

$& \text{Step 1:} & \mathrm{H_2O_2}{(aq)} \mathrm{+I^-}{(aq)} & \rightarrow \mathrm{H_2O}{(l)} \mathrm{+IO^-}{(aq)} \\& \text{Step 2:} & \mathrm{H_2O_2}{(aq)} \mathrm{+IO^-}{(aq)} & \rightarrow \mathrm{H_2O}{(l)} \mathrm{+O_2}{(g)} \mathrm{+I^-}{(aq)}$

If step 1 is the rate-determining step, then the rate law for that step will be the rate law for the overall reaction.

rate = k[H2O2][I-]

The rate law for the slow step of the proposed mechanism agrees with the overall experimentally determined rate law. IO is present as an intermediate in the reaction. The iodide ion catalyst also appears in the mechanism. It is a reactant in the first elementary step and a product in the second step. By definition, catalysts are not used up in a reaction, so if they are consumed by one step in a mechanism, they must be regenerated by a subsequent step.

Hydrogen peroxide is decomposing all the time, but the addition of solid potassium iodide catalyzes the decomposition of hydrogen peroxide into water and oxygen gas. The effect of adding potassium iodide can be seen in the following video: http://www.youtube.com/watch?v=mtjbg6ZF5K4 (2:06).

### Potential Energy Diagrams

A potential energy diagram can illustrate the mechanism for a reaction by showing each elementary step of the reaction with its own distinct activation energy (Figure below).

The potential energy diagram shows an activation energy peak for each of the elementary steps of the reaction. The valley between these peaks represents the intermediate for the reaction.

The reaction, whose potential energy diagram is shown in the figure above, is a two-step reaction. The activation energy for each step is labeled (Ea1 and Ea2). Additionally, each elementary step has its own activated complex (AC1 and AC2). In between the two activated complexes is a potential energy well that corresponds to the intermediates in this reaction. Unlike activated complexes, intermediates can exist for long enough to be detected and characterized. Note that the overall enthalpy change of the reaction (ΔH) is unaffected by the individual steps, since it depends only on the initial and final states.

## Lesson Summary

• Chemical reactions often consist of a sequence of simpler reactions, called elementary steps. All of the elementary steps of a reaction combine to produce the reaction mechanism.
• Reaction mechanisms must be experimentally determined, and the rate law for the rate-determining step of the mechanism must agree with the rate law for the overall reaction.
• Mechanisms show the presence of intermediates and catalysts, neither of which are present in the balanced equation for the overall reaction.
• A potential energy diagram can show the mechanism of a reaction by a series of peaks and valleys corresponding to each elementary step.

## Lesson Review Questions

### Reviewing Concepts

1. How is an elementary step different from an overall chemical reaction?
2. Why can a rate law be written based on the equation for an elementary step?
3. The overall rate for a reaction depends most on which step of the mechanism? Explain.
4. Classify the following elementary reactions as unimolecular, bimolecular, or termolecular.
1. $\mathrm{SO+O_2 \rightarrow SO_2+O}$
2. $\mathrm{CH_3NC \rightarrow CH_3CN}$
3. $\mathrm{H_2+2I \rightarrow 2HI}$
5. The equation for the production of ammonia from nitrogen and hydrogen gases is: Why is it unlikely that this reaction proceeds in a single elementary step?

### Problems

1. Given the reaction and experimentally determined rate law below:
1. Why must the mechanism of the reaction consist of more than one step?
2. The mechanism below has been proposed for the reaction.

If this mechanism is correct, which step is the rate-determining step? Explain.

2. Consider the following two-step reaction mechanism. Write the overall reaction. Which substance is an intermediate in the reaction? Which substance is a catalyst?
3. Given the overall reaction below: By experiment it is found that doubling the concentration of A2 doubles the rate. Doubling the concentration of B also doubles the rate. Doubling the concentration of C has no effect on the rate.
1. Write the rate law for the reaction.
2. How can changing the concentration of C have no effect on the rate?
3. Propose a reaction mechanism that is consistent with the rate law. Indicate the relative speed (fast or slow) of each step.

## Points to Consider

Many chemical reactions do not proceed entirely to completion. In other words, after some of the products are formed in the reaction, they can recombine to form the reactants again. Such reactions are said to be reversible.

• How can a reversible reaction be shown in a chemical equation?
• What is the relationship between the rate of the forward reaction and the rate of the reverse reaction?

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