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Reaction Mechanisms

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

Introduction

As we study kinetics, we encounter very complex and sophisticated reactions that we are not able to proceed analyzing without proposing a mechanism. In order to analyze the kinetics of complex reactions, we may need to devise a series of steps that a reaction  occurs  through before reaching the final products. As a result, scientists have developed a proposal that can help us see this process, step by step, in a reaction. We call this the reaction mechanism.

A reaction mechanism is considered our best 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. A reaction mechanism is defined as a proposed set of elementary steps, which account for the overall features of the reaction. Each of the reactions that comprises the mechanism is called an elementary step. We believe it is elementary because it takes place in a single reactive encounter between the reactants involved. These elementary steps are the basic building blocks of a complex reaction and cannot be broken down any further. The elementary processes in a reaction mechanism describe the molecular reaction and the changes it undergoes during the reaction.

Example 1

Consider the chlorination reaction of methane (\(CH_4\))

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

This reaction is proposed to occur via two sucessive elementary steps. Each step has its own characteristic reactants and product and characteristic 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

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

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 when the maximum of potential energy is reached, in a reaction. A mechanism must show the order that the bonds form or break and the rate of each elementary step. Also accounted for in a mechanism are the reaction intermediates, which are 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. Since 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:

  • The elementary steps must add up to give the overall balanced equation for the reaction.
  • 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 — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation.

Step by step...

A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called elementary reactions) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as

unimolecular A → by far the most common
bimolecular A + B →  
termolecular A + B + C → very rare

Elementary Processes

Elementary processes are usually either unimolecular or bimolecular. A unimolecular elementary process is when a single molecule dissociates. A bimolecular elementary process is when two molecules collide with each other. 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 where both the foward and reverse reactions 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, since it occurs slower. According to our 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 in any manner that the mechanism is proven to be 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 react, the reaction moves to a faster step, Step (2). Step (2) occurs rapidly and our mechanism suggests that HI is consumed in the second step as fast as it is being formed. HI is, therefore, the reaction intermediate.

The diagram below shows the reaction progress of the two-step mechanism proposed. Notice how the reaction intermediate and the transition states do not last very long, 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

Here 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

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

Since 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:

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

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

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 then the rate determinant, the step with a slow rate. The relative rate of the rate determinant is \(k_3[N_2O_2][O_2]\), which does not seem to match with the experimental rate of k [NO2]2[O2]. However, \([N_2O_2]\) is actually a reaction intermediate, because it appears in both elementary steps but not in the equation of the overall reaction. Since an intermediate cannot be in the rate law for the overall reaction, we want to have a value that is equivalent to \([N_2O_2]\) to substitute into the rate equation. Since Step 1 is a fast, reversible reaction, we can make the assumption that:

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

so

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

Through substitution, k3[N2O2][O2] = k3(k1/k2) [NO]2[O2]. The rate constant \(k = k_3 frac{k_1}{k_2}\), so rate of reaction is \(k [NO]^2[O_2]\), thus matching the experimental rate law. This proposed mechanism fulfills both requirements needed to be considered 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, we make no assumptions about the rates of the steps within the mechanism. 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, we assume that the reaction rate of each elementary step is equal. To do this, we select one of the elementary steps to begin with and write the reaction rate:

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

However, the intermediate \(N_2O_2\) 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, we are able to write \([N_2O_2]\) 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 \([N_2O_2]\) is separated into two parts, giving an equation of:

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

Now because of steady-state assumption, the rate of disappearance of

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

giving the equation:

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

Rearranging this equation will give us

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

Plugging in this value of \([N_2O_2]\) into our initial rate of reaction of the elementary step will provide us with the overall reaction of the mechanism:

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

Once again, reaction mechanisms only give us a guess at what is happening within the overall reaction. 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 (i.e., does 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. Since 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

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08:57, 13 Apr 2014

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