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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Kinetics > Rate Laws > Reaction Mechanisms > Reaction Mechanisms

Reaction Mechanisms

The mechanism of a chemical reaction is the sequence of events that take place as reactant molecules are converted into products.

Introduction

The study of kinetics includes very complex and sophisticated reactions that cannot be analyzed without a proposed mechanism, a series of steps that a reaction takes before reaching the final products.

Reaction mechanisms are step-by-step descriptions of what occurs on a molecular level in chemical reactions. Each step of the reaction mechanism is known as an elementary process, a term used to describe a moment in the reaction when one or more molecules changes geometry or is perturbed by the addition or omission of another interacting molecule. Collectively, an overall reaction and a reaction mechanism consist of multiple elementary processes. These elementary steps are the basic building blocks of a complex reaction, and cannot be broken down any further.

A reaction mechanism is only a guess at how a reaction proceeds. Therefore, even if a mechanism agrees with the experimental results of a reaction, it cannot be proven to be correct.

Example 1

Consider the chlorination reaction of methane, CH4:

\[CH_4 (g) + 2Cl_2(g) \longrightarrow CH_3Cl (g) + HCl (g) + Cl^- (g) \tag{overall reaction}\]

This reaction is proposed to occur via two successive elementary steps. Each step has its own characteristic reactants, product, rate law.

\[CH_4 (g) + Cl_2 (g) \longrightarrow CH_3 (g) + HCl (g) \tag{step 1 (slow)}\]

 with an elementary rate law of,

\[k_1 = [CH_4][Cl_2]\]

\[CH_3 (g) + Cl_2 (g) \longrightarrow CH_3 Cl (g) + Cl^- (g) \tag{step 2 (fast)}\]

with an elementary rate law of,

\[k_2 = [CH_4][Cl_2]\]

The steps combine to generate the final reaction equation,

\[CH_4 (g) + 2Cl_2(g) \longrightarrow CH_3Cl (g) + HCl (g) + Cl^- (g) \tag{overall reaction}\]

Example 2a

Consider the following reaction:

\[CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}\]

This reaction equation suggests that the carbon monoxide directly reacts with nitrogen dioxide to form the products. However, its reaction mechanism shows this is not the case. Consider the two elementary processes below:

\[2NO_2(g) \rightarrow NO_3(g) + NO(g) \tag{1}\]

\[CO(g) + NO_3(g) \rightarrow CO_2(g) + NO_2(g) \tag{2}\]

\[CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}\]

This reaction mechanism indicates that the compound NO3 performs a necessary role in allowing the reaction to take place, a concept not possible to confirm by looking at the overall equation.

Example 2b

Take the reaction \(2NO (g) + O_2(g) \rightarrow 2NO_2(g)\). A look at the overall reaction might suggest that two NO molecules must collide with an oxygen molecule to generate the product. However, termolecular processes are extremely rare; a further analysis of the reaction mechanism is required. The two elementary reactions below show:

\[2NO(g) \rightarrow N_2O_2(g) \tag{1}\]

\[ 2N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g) \tag{2} \]

\[2NO(g) + O_2(g) \rightarrow 2NO_2(g) \tag{overall reaction}\]

The reaction mechanism shows that, in fact, three molecules combining simultaneously to form a product does not occur in this case. Instead, the overall reaction is a compilation of two individual elementary processes that take place.

Description of a Reaction Mechanism

Because a reaction mechanism is used to describe what occurs at each step of a reaction, it also describes the transition state, or the state in which the maximum of potential energy is reached. A mechanism must show the order in which the bonds form or break and the rate of each elementary step. Also accounted for are the reaction intermediates, stable molecules that do not appear in the experimentally determined rate law because they are formed in one step and consumed in a subsequent step. Because a reaction cannot proceed faster than the rate of slowest elementary step, the slowest step in a mechanism establishes the rate of the overall reaction. This elementary step is known as the rate-determining step.

 A mechanism must satisfy the following two requirements:

  1. The elementary steps must add up to give the overall balanced equation for the reaction.
  2. The rate law for the rate-determining step must agree with the experimentally determined rate law.

Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collision") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an intermediate.

Elementary Processes

Elementary processes are usually either unimolecular or bimolecular. A unimolecular elementary process describes the dissociation of a single molecule. A bimolecular elementary process occurs when two molecules collide. A third process, called termolecular, is rare because it involves three molecules colliding at the same time. In the bimolecular and termolecular processes, the molecules may be different or the same.

Although a rate law for an overall reaction can only be experimentally determined, the rate law for each elementary step can be deduced from the chemical equation through inspection. A unimolecular elementary step has a first order rate law, whereas a bimolecular elementary step has a second order rate law. The table below summarizes the types of elementary steps and the rate laws that they follow. A, B, and C here represent the reactants or reaction intermediates.

 Elementary Steps and Rate Laws

Molecularity Elementary Step Rate Law for Elementary step
Unimolecular \[A \longrightarrow products\] \[\text{rate}= k[A]\]
Bimolecular \[A + B \longrightarrow products\] \[\text{rate}= k[A][B]\]
\[A + A \longrightarrow products\] \[\text{rate}= k[A]2\]
Termolecular \[A + A + B \longrightarrow products\] \[\text{rate}= k[A]2[B]\]
\[A + A + A \longrightarrow products\] \[\text{rate} = k[A]3\]
\[A + B + C \longrightarrow products\] \[\text{rate}= k[A][B][C]\]

Elementary processes are also reversible. Some may reach conditions of equilibrium, in which both the forward and reverse reaction rates are equal.

Bimolecular Elementary Processes

Example 2: Mechanism with a slow step followed by a fast step

Consider the reaction between the gases iodine monochloride and hydrogen produce two gases, iodine and hydrogen chloride as products.

\[H_2(g)+ 2 ICl (g) \longrightarrow I_2 (g)+ 2 HCl (g)\]

Through experiment, the rate law is found to be,

rate of reaction = k [H2] [ICl]

Using a proposed reaction mechanism:

(1) slow reaction:       H2+ ICl\(\longrightarrow\) HI + HCl             Rate(1):  k1 [H2] [ICl]

(2) fast reaction:         HI + ICl\(\longrightarrow\) I2 + HCl             Rate(2):k2  [HI] [ICl]

Overall reaction:         H2 + 2 ICl\(\longrightarrow\) I2+ 2 HCl

The mechanism chosen for the reaction above proposes that Step (1) is the rate-determining step, because it occurs more slowly. According to the proposed mechanism, the rate law agrees that the mechanism works for this reaction. The rate of Step (1) =  K1 [H2] [ICl] is in agreement with the experimentally determined rate law, so the mechanism matches the stoichiometry of the overall reaction. However, this does not mean that the mechanism is correct. It only shows that the chosen mechanism works perfectly for this reaction. This is a reasonable proposal.

When there is enough energy for the slow reaction to occur, the reaction moves to a faster step, Step (2). Step (2) occurs rapidly, and the mechanism suggests that HI is consumed in the second step as quickly as it is formed. HI is, therefore, the reaction intermediate.

The diagram below shows the reaction progress of the two-step mechanism proposed. Notice that the reaction intermediate and the transition states are short-lived, but the intermediate can be isolated due to its fully formed bonds.

Diagram of a Two-Step Reaction

final graph.jpg

 

Example 3: Mechanism with a fast step followed by slow step

Below is the equation of nitric oxide reacting with oxygen to produce nitrogen dioxide:

\[2 NO (g) + O_2 \longrightarrow 2 NO_2 (g)\]

which has an experimentally determined third order rate law:

\[\text{rate of reaction} = k [NO_2]^2[O_2]\]

Because a termolecular mechanism (a three-body reaction occurring in a single step) is extremely rare, a mechanism with two bimolecular elementary steps can be suggested:

\[ 2 NO (g) \rightleftharpoons N_2O_2 (g) \tag{Step 1 fast}\]

\[N_2O_2 (g)   +   O_2 (g)  \longrightarrow 2 NO_2 (g) \tag{Step 2 slow}\]

part 1 .jpg

part 2.jpg

The proposed mechanism has the same stoichiometry as the overall reaction. Step 1 is a reversible process in which equilibrium is reached quickly. Step 2 is therefore the rate-determining step. The relative rate of the Step 2 is \(k_3[N_2O_2][O_2]\), which does not seem to match with the experimental rate of \(k [NO_2]^2[O_2]\). However, N2O2 is actually a reaction intermediate, because it appears in both elementary steps but not in the equation of the overall reaction. Because an intermediate cannot appear in the rate law for the overall reaction, a value equivalent to N2O2 is required. Because Step 1 is a fast, reversible reaction, it can be assumed that:

\[k_1 [NO]^2 = k_2 [N_2O_2]\]

Therefore,

\[[N_2O_2] = \dfrac{k_1}{k_2} [NO]^2\]

Through substitution, k3[N2O2][O2] = k3(k1/k2) [NO]2[O2]. The rate constant is \(k = k_3 frac{k_1}{k_2}\), so the rate of reaction is \(k [NO]^2[O_2]\), matching the experimental rate law. This proposed mechanism fulfills both requirements for a possible mechanism.

Another Example of Reaction Mechanism

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The problem explained here gives a plausible reaction mechanism to approach the overall reaction.

The Steady-State Approximation

In some reaction mechanisms, more than one elementary step may control the rate of the reaction. In such mechanisms, no assumptions about the rates of the steps within the mechanism can be made. For example, take the reaction of nitric oxide with oxygen:

\[NO+ NO \longrightarrow N_2O_2 \tag{step 1}\]

\[N_2O_2 +O_2 \longrightarrow 2NO_2  \tag{step 2}\]

\[2NO + O_2 \longrightarrow 2NO_2 \tag{overall}\]

To find the overall reaction rate without knowledge of the rate-determining step, it must be assumed that the reaction rate of each elementary step is equal. To do this, select one of the elementary steps and write the reaction rate:

\[ \text{Reaction Rate} = k_2[N_2O_2][O_2]\]

However, the intermediate N2O2 still remains. To eliminate the reaction intermediate, we use the other elementary step, assuming that the [N2O2] reaches a steady-state condition. Using this assumption, [N2O2] can be written in terms of NO:

\[\text{rate of formation of } N_2O_2 = \text{Rate of disappearance of } N_2O_2\]

The rate of disappearance of N2O2 is separated into two parts, giving the equation:

\[ \text{rate of disappearance of } N_2O_2 = k_{-1}[N_2O_2] +  k_2[N_2O_2][O_2]\]

Now because of steady-state assumption, the rate of disappearance of N2Ois the following:

\[N_2O_2 = k_1[NO]^2\]

giving the equation:

\[k_1[NO]^2 = k_{-1}[N_2O_2] +  k_2[N_2O_2][O_2]\]

Rearranging this equation gives:

\[[N_2O_2] = \dfrac{k_1[NO]^2}{k_{-1}+k_2[O]}\]

Substituting this value of [N2O2] into our initial rate of reaction of the elementary step will provide us with the overall reaction of the mechanism:

\[\text{rate} = \dfrac{k_1k_2[O_2][NO]^2}{k_{-1}+k_2[O_2]}\]

Most overall reactions do not occur in a single step and involve mechanisms with successive elementary steps. Reaction mechanisms, however, can never be proven to be correct, but can be disproved (by showing that they do not fit the experimental data).

Problems

  1. When we find a reaction mechanism, according to the rate law, that perfectly works to solve a complex reaction, why can we not say that the reaction mechanism proves that the rate law is right?
     
  2. Which elementary process is bimolecular?
        (a) A\(\longrightarrow\) products
        (b) A + B\(\longrightarrow\) products
        (c) A + A + B\(\longrightarrow\) products
        (d) Both (b) and (c)
     
  3. Propose a mechanism for the reaction 2 NO(g) + O2 (g) \(\longrightarrow\) 2 NO2 (g)  and show that the chosen mechanism is consistent with the rate law: k[NO]2[O2]. 
     
  4. For the reaction, 2NO2(g) + F2(g)  \(\longrightarrow\)  2NO2F with a rate law = k[NO2][F2]:
        (a) write the steps for a bimolecular elementary process.
        (b) identify which elementary process is the rate determinant and which one is the fast reaction.
     
  5.  The decomposition of hydrogen peroxide is catalyzed by iodide ion:

      2 H202 (aq)\( \xrightarrow[ ]{I^-}\) 2H2O (l) + O2 (g)             Rate = k [H2O2] [I]

      Find a possible mechanism and identify the intermediate.
     
  6. NO2 (g) + CO (g)\(\longrightarrow\) NO (g) + CO2 (g)
    The experimental rate = k[NO2]2
    Explain why the following mechanism cannot be used. Propose a plausible mechanism and identify the rate-determining step.

 

single step.jpg

Solutions

  1. A reaction mechanism is only a rationalization of how a reaction progresses. It cannot be proven to be true. There may be more than one possible proposed mechanism, in which the slow elementary process is considered the rate-determining step.
     
  2. (b) \( A + B \longrightarrow \text{products}\) is a bimolecular process because two molecules are involved.
     
  3. Refer back to the example under "A Mechanism with a Fast Reversible First Step Followed by a Slow Step."
     
  4. (a)    Step (1):             NO2 + F2\(\longrightarrow\) NO2F + F          Rate (1): k1[NO2][F2]
            Step (2):             NO2+ F\(\longrightarrow\) NO2F                  Rate (2): k2[NO2][F]
           _________________________________________________________________________
            Overall Reaction:   2 NO+ F2\(\longrightarrow\) 2NO2F
       
    Step (1) and (2) add up to an overall reaction that matches the stoichiometry of the given equation. The rate of Step (2) is consistent with the experimental rate law. Therefore, the proposed mechanism works.

    (b)    The slowest elementary process gives the rate of the overall reaction. Because the rate of Step (2) matches the experimental rate law, Step (2) is the rate determinant. Step (1) has a faster reaction.
     
  5. Step (1):             H2O2 + I-\(\longrightarrow\) H2OH + IO-     (Intermediate)                 Rate (1) = k1 [H2O2][I-]
    Step (2):             H2O2 +IO-\(\longrightarrow\) H2O + O2 + I-        (Catalyst)                 Rate (2)= k2 [H2O2][IO-]
    _______________________________________________________
    Overall Reaction: 2 H202 (aq)\( \xrightarrow[ ]{I^-}\) 2H2O (l) + O2 (g)                                    Rate = k [H2O2] [I]

    Although I- appears in both elementary steps, it is a catalyst in the reaction, so it can occur in the rate law, unlike the intermediate, IO-.
     
  6. Although the one-step mechanism proposed in this problem satisfies the molecular equation, it's rate of reaction (k [NO2][CO]) does not agree with the experimentally determined rate law.

    Step (1):                NO2+ NO2\(\longrightarrow\) NO3 + NO                    Rate (1)= k1 [NO2]2     
    Step (2):                NO3 + CO\(\longrightarrow\) NO2 +CO2                        Rate (2) = k2 [NO3][CO]
    ___________________________________________________
    Overall Reaction:     2 NO2 + CO\(\longrightarrow\) NO + CO2

    NO2 and NO3 are intermediates. Step (1) has the same rate as the experimental rate law, so it is the rate-determining step.

References

  1. Fábián, István. Reaction Kinetics, Mechanisms and Catalysis. Springer, 2009.
  2. Kinetics and Mechanism Concept Test Questions: http://www.jce.divched.org/JCEDLib/QBank/collection/ConcepTests/kinetics.html
  3. Petrucci, Ralph. General Chemistry, Principles and Modern Applications. Ninth Edition. Chemical Kinetics, Reaction Mechanism, pg 596-602.
  4. Kemp, T. J. Progress in Reaction Kinetics & Mechanism. Science Reviews, 2000 Ltd. 
  5. Upadhyay, Santosh K. Chemical Kinetics and Reaction Dynamics. Springer, 2007.
  6. Zumdahl, Steven S. Chemistry. Fifth edition. Chemical Kinetics, Reaction Mechanism, pg 583-586.
  7. Petrucci, Ralph. General Chemistry, Principles and Modern Applications. Custom Edition. Chemical Kinetics, Reaction Mechanism, pg 1093-1097

Contributors

  • Abel Silva, Marisol Ahumada

Viewing 1 of 1 comments: view all
Hello! I wanted to note that there's an error within Example 3, the first step in the mechanism is labeled as the "fast" step, and the second is labeled as the "slow" step, however, in the following commentary contradicts these labels.
Posted 21:44, 26 Jul 2014
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