Chapter 18: Kinetics

18.6 Reaction Mechanisms

Learning Outcomes

  • Distinguish net reactions from elementary reactions (steps)
  • Identify the molecularity of elementary reactions
  • Write a balanced chemical equation for a process given its reaction mechanism
  • Derive the rate law consistent with a given reaction mechanism
    • Use rate laws to assess the validity of a reaction mechanism

Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.

The decomposition of ozone, for example, appears to follow a mechanism with two steps:

[latex]\begin{array}{l}{\ce{O}}_{3}\left(g\right)\rightarrow{\ce{O}}_{2}\left(g\right)+\ce{O}\\ \ce{O}+{\ce{O}}_{3}\left(g\right)\rightarrow 2{\text{O}}_{2}\left(g\right)\end{array}[/latex]

We call each step in a reaction mechanism an elementary reaction. Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Elementary reactions add up to the overall reaction, which, for the decomposition, is:

[latex]\ce{2O3}(g)\rightarrow \ce{3O2}(g)[/latex]

Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediate.

While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one [latex]\ce{O3}[/latex] decomposes to yield [latex]\ce{O2}[/latex] and an oxygen atom, and a second [latex]\ce{O3}[/latex] molecule subsequently reacts with the oxygen atom to yield two additional [latex]\ce{O2}[/latex] molecules.

Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction’s equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.

Unimolecular Elementary Reactions

The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the rearrangement of a single reactant species to produce one or more molecules of product:


The rate equation for a unimolecular reaction is:


A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:


illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism. However, some unimolecular reactions may have only a single reaction in the reaction mechanism. (In other words, an elementary reaction can also be an overall reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, [latex]\ce{C4H8}[/latex], to ethylene, [latex]\ce{C2H4}[/latex], is represented by the following chemical equation:

In this figure, structural formulas are used to illustrate a chemical reaction. On the left, a structural formula for cyclobutane is shown. This structure is composed of 4 C atoms connected with single bonds in a square shape. Each C atom is bonded to two other C atoms in the structure, leaving two bonds for H atoms pointing outward above, below, left, and right. An arrow points right to two identical ethane molecules with a plus symbol between them. Each of these molecules contains two C atoms connected with a double bond oriented vertically between them. The C atom at the top of these molecules has H atoms bonded above to the right and left. Similarly, the lower C atom has two H atoms bonded below to the right and left.

This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first-order behavior:

[latex]\text{rate}=-\dfrac{\Delta\left[{\ce{C}}_{4}{\ce{H}}_{8}\right]}{\Delta t}=k\left[{\ce{C}}_{4}{\ce{H}}_{8}\right][/latex]

This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.

Bimolecular Elementary Reactions

A bimolecular reaction involves two reactant species, for example:

[latex]\begin{array}{c}A+B\rightarrow\text{products}\\ \text{and}\\ 2A\rightarrow\text{products}\end{array}[/latex]

For the first type, in which the two reactant molecules are different, the rate law is first-order in A and first order in B (second-order overall):

[latex]\text{rate} = k[A][B][/latex]

For the second type, in which two identical molecules collide and react, the rate law is second order in A:

[latex]\text{rate} = k[A][B] = k[A]^{2}[/latex]

Some chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:


This figure provides an illustration of a reaction between two H I molecules using space filling models. H atoms are shown as white spheres, and I atoms are shown as purple spheres. On the left, two H I molecules are shownwith a small white sphere bonded to a much larger purple sphere. The label, “Two H I molecules,” appears below. An arrow points right to a similar structure in which the two molecules appear pushed together, so that the purple spheres of the two molecules are touching. Below appears the label, “Transition state.” Following another arrow, two white spheres are shown vertically oriented and bonded together with the label, “H subscript 2” above. The H subscript 2 molecule is followed by a plus sign and two purple spheres bonded together with the label, “I subscript 2” above. Below these structures is the label, “Hydrogen iodide molecules decompose to produce hydrogen H subscript 2 and iodine I subscript 2.”
Figure 18.6.1. The probable mechanism for the reaction between [latex]\ce{NO2}[/latex] and [latex]\ce{CO}[/latex] to yield [latex]\ce{NO}[/latex] and [latex]\ce{CO2}[/latex].

Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:


Termolecular Elementary Reactions

An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:

[latex]\begin{array}{l}\ce{2NO}+{\ce{O}}_{2}\longrightarrow 2{\ce{NO}}_{2}\\ \text{rate}=k{\left[\ce{NO}\right]}^{2}\left[{\ce{O}}_{2}\right]\end{array}[/latex]

Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:

[latex]\begin{array}{l}\text{2NO}+{\text{Cl}}_{2}\longrightarrow 2\text{NOCl}\\ \text{rate}=k{\left[\text{NO}\right]}^{2}\left[{\text{Cl}}_{2}\right]\end{array}[/latex]

Relating Reaction Mechanisms to Rate Laws

A photo is shown of cattle passing through a narrow chute into a holding pen. A person directs them through the gate with a long white and red pole.
Figure 18.6.2. A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)

It’s often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction.

As described above, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example:


For temperatures above 225 °C, the rate law has been found to be:


The reaction is first order with respect to [latex]\ce{NO2}[/latex] and first-order with respect to [latex]\ce{CO}[/latex]. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.

At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to [latex]\ce{NO2}[/latex]:


This is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step:

[latex]\begin{array}{l}{\ce{NO}}_{2}\left(g\right)+{\ce{NO}}_{2}\left(g\right)\longrightarrow{\ce{NO}}_{3}\left(g\right)+\ce{NO}\left(g\right)\qquad\left(\text{slow}\right)\\ {\ce{NO}}_{3}\left(g\right)+\ce{CO}\left(g\right)\longrightarrow{\ce{NO}}_{2}\left(g\right)+{\ce{CO}}_{2}\left(g\right)\qquad\left(\text{fast}\right)\end{array}[/latex]

The rate-determining step gives a rate law showing second-order dependence on the [latex]\ce{NO2}[/latex] concentration, and the sum of the two equations gives the net overall reaction.

In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an equilibrium reaction, the rate law for the overall reaction may be more difficult to derive.

a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species [latex]\ce{N2O2}[/latex]. When this reaction is at equilibrium:

[latex]\begin{array}{l}\ce{NO}+\ce{NO}\rightleftharpoons{\ce{N}}_{2}{\ce{O}}_{2}\\ {\text{rate}}_{\text{forward}}={\text{rate}}_{\text{reverse}}\\ {k}_{1}{\left[\ce{NO}\right]}^{2}={k}_{-1}\left[{\ce{N}}_{2}{\ce{O}}_{2}\right]\end{array}[/latex]

This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:


Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.

Example 18.6.1: Deriving A Rate Law From a Reaction Mechanism

The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:

Step 1: [latex]\ce{NO}(g)+\ce{Cl2}(g)\rightleftharpoons\ce{NOCl2}(g)\qquad\text{fast}[/latex]

Step 2: [latex]\ce{NOCl2}(g)+\ce{NO}(g)\longrightarrow\ce{2NOCl}(g)\qquad\text{slow}[/latex]

Use this mechanism to derive the equation and predicted rate law for the overall reaction.

Show Solution

The equation for the overall reaction is obtained by adding the two elementary reactions:


To derive a rate law from this mechanism, first write rates laws for each of the two steps.

[latex]\begin{array}{ccc}\hfill {\text{rate}_{1}}& =& {k}_{1}\left[{\ce{NO}}\right]\left[\ce{Cl}_{2}\right]\qquad\text{for the forward reaction of step 1}\hfill \\ \hfill{\text{rate}_{-1}}& =& {k}_{-1}\left[{\ce{NOCl}}_{2}\right]\qquad\text{for the reverse reaction of step 1}\hfill \\ \hfill{\text{rate}_{2}}& =& {k}_{2}\left[\ce{NOCl}_{2}\right]\left[\ce{NO}\right]\qquad\text{for step 2}\hfill \end{array}[/latex]

Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [latex]\ce{[NOCl2]}[/latex]. To remedy this, use the first step’s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.

Assuming Step 1 is at equilibrium:





Substituting this expression into the rate law for Step 2 yields:



Check Your Learning

Key Concepts and Summary

The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.

Try It

  1. Nitrogen(II) oxide, [latex]\ce{NO}[/latex], reacts with hydrogen, [latex]\ce{H2}[/latex], according to the following equation:
    [latex]\ce{2NO}+\ce{2H2}\rightarrow\ce{N2}+\ce{2H2O}[/latex]What would the rate law be if the mechanism for this reaction were:
    [latex]\begin{array}{l}2\ce{NO}+{\ce{H}}_{2}\rightarrow{\ce{N}}_{2}+{\ce{H}}_{2}{\ce{O}}_{2}\left(\text{slow}\right)\\ {\ce{H}}_{2}{\ce{O}}_{2}+{\ce{H}}_{2}\rightarrow 2{\ce{H}}_{2}\ce{O}\left(\text{fast}\right)\end{array}[/latex]
  2. Experiments were conducted to study the rate of the reaction represented by this equation[1].
    Initial concentrations and rates of reaction are given here.

    Experiment Initial Concentration [latex]\ce{[NO]}[/latex] (mol/L) Initial Concentration, [latex]\ce{[H2]}[/latex] (mol/L) Initial Rate of Formation of [latex]\ce{N2}[/latex] (mol/L min)
    1 0.0060 0.0010 1.8 × 10−4
    2 0.0060 0.0020 3.6 × 10−4
    3 0.0010 0.0060 0.30 × 10−4
    4 0.0020 0.0060 1.2 × 10−4

    Consider the following questions:

    1. Determine the order for each of the reactants, [latex]\ce{NO}[/latex] and [latex]\ce{H2}[/latex], from the data given and show your reasoning.
    2. Write the overall rate law for the reaction.
    3. Calculate the value of the rate constant, k, for the reaction. Include units.
    4. For experiment 2, calculate the concentration of [latex]\ce{NO}[/latex] remaining when exactly one-half of the original amount of [latex]\ce{H2}[/latex] had been consumed.
    5. The following sequence of elementary steps is a proposed mechanism for the reaction.

    Step 1: [latex]\ce{NO}+\ce{NO}\rightleftharpoons\ce{N2O2}[/latex]
    Step 2: [latex]\ce{N2O2}+\ce{H2}\rightleftharpoons\ce{H2O}+\ce{N2O}[/latex]
    Step 3: [latex]\ce{N2O}+\ce{H2}\rightleftharpoons\ce{N2}+\ce{H2O}[/latex]

    Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

Show Selected Solutions
  1. The answers are as follows:
    1. Doubling [latex]\ce{[H2]}[/latex] doubles the rate. [latex]\ce{[H2]}[/latex] must enter the rate equation to the first power. Doubling [latex]\ce{[NO]}[/latex] increases the rate by a factor of 4. [latex]\ce{[NO]}[/latex] must enter the rate law to the second power.
    2. The rate law is Rate = k [latex]\ce{[NO]^2[H2]}[/latex].
    3. 1.8 × 10−4 mol/L/ min = k[0.0060 mol/L]2[0.0010 mol/L], k = 5.0 × 103 mol−2 L−2 min−1
    4. The reaction has consumed 0.0010 mol/L of [latex]\ce{H2}[/latex]. The amount of [latex]\ce{NO}[/latex] consumed is the same, 0.0010 mol/L of [latex]\ce{NO}[/latex]. Thus 0.0060 − 0.0010 = 0.0050 mol/L remains.
    5. Step II is the rate-determining step. If step I gives [latex]\ce{N2O2}[/latex] in adequate amount, steps 1 and 2 combine to give [latex]\ce{2NO}+\ce{H2}\rightarrow\ce{H2O}+\ce{N2O}.[/latex] This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.


bimolecular reaction: elementary reaction involving the collision and combination of two reactant species

elementary reaction: reaction that takes place precisely as depicted in its chemical equation

intermediate: molecule or ion produced in one step of a reaction mechanism and consumed in another

molecularity: number of reactant species (atoms, molecules or ions) involved in an elementary reaction

rate-determining step: (also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction

reaction mechanism: stepwise sequence of elementary reactions by which a chemical change takes place

termolecular reaction: elementary reaction involving the simultaneous collision and combination of three reactant species

unimolecular reaction: elementary reaction involving the rearrangement of a single reactant species to produce one or more molecules of product

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