6.6: Reaction Mechanisms

 Learning Objectives

• Identify the molecularity of elementary reactions (unimolecular, bimolecular, or termolecular).
• Write balanced chemical equations for overall reactions given a reaction mechanism.
• Write rate laws for elementary reactions and derive rate laws for overall reactions, considering slow steps and equilibrium approximations.
• Identify reaction intermediates within multi-step mechanisms.
• Assess the validity of a proposed mechanism by comparing its derived rate law to experimental data.
• Interpret and draw potential energy diagrams to illustrate reaction mechanisms, including transition states and rate-determining steps.

Elementary Reactions

A balanced chemical equation tells us which substances react, which substances are produced, and how the amounts of reactants and products are related. However, it does not show how the reaction actually proceeds at the molecular level. Most chemical reactions occur through a sequence of simpler reactions. A reaction mechanism is a step-by-step series of simpler reactions by which an overall reaction takes place.

For example, ozone in the stratosphere protects Earth’s surface from harmful ultraviolet radiation. Ultraviolet photons cause ozone molecules to decompose into oxygen molecules. The overall reaction equation is:

\[2\;\text{O}_3(g) \xrightarrow{h\nu} 3\;\text{O}_2(g) \nonumber \]

Although this equation indicates which substances are consumed and produced, it does not mean that two O₃ molecules collide and react directly. Instead, the reaction proceeds in two consecutive steps:

\[\begin{array}{rcl} &\text{step 1}:& \;\;\;\;\;\;\;\;\;\text{O}_3(g) &\xrightarrow{h\nu}& \text{O}_2(g) + \text{O}(g)\\[0.5em]&\text{step 2}:& \;\;\;\;\;\;\;\;\;\text{O}(g) + \text{O}_3(g) &\longrightarrow& 2\;\text{O}_2(g)\\[0.5em] &\text{overall}:& \;\;\;\;\;\;\;\;\;2\;\text{O}_3(g) &\longrightarrow& 3\;\text{O}_2(g)\end{array} \nonumber \]

Step 1 occurs when an O₃ molecule absorbs a UV photon, breaking a bond and producing an O₂ molecule and a free O atom:

Step 2 involves the newly formed O atom reacting with another O₃ molecule to produce two O₂ molecules:

 

 

 

Adding these steps together yields the overall reaction:

O₃(g) + O(g) + O₃(g) ⟶ O₂(g) + O(g) + 2O₂(g)

which simplifies to the overall reaction equation: 2 O₃(g) ⟶ 3 O₂(g).

A species such as O that appears as a product in one step but is consumed in a subsequent step is called a reaction intermediate. Intermediates are typically less stable than either the reactants or the products, but unlike transition states, they have fully formed bonds and can sometimes be isolated.

Each step in a mechanism is called an elementary reaction. An elementary reaction explicitly shows which particles collide, break apart, or rearrange to form the products and/or intermediates. For instance, step 2 above requires a collision between a single O atom and a single O₃ molecule, which then forms two O₂ molecules.

By contrast, the overall equation 2 O₃(g) ⟶ 3 O₂(g) does not imply a direct collision of two O₃ molecules. Instead, the overall equation is simply the net result of multiple elementary steps.

Because each elementary reaction has its own transition state (an energy maximum), a multi-step mechanism will involve a series of transition states. Figure \(\PageIndex{1}\) illustrates the reaction energy diagram for the ozone decomposition, showing two distinct peaks, each corresponding to a transition state.

Figure \(\PageIndex{1}\): Reaction energy diagram for the overall 2 O₃(g) → 3 O₂(g) reaction. This reaction proceeds via a mechanism consisting of two elementary reaction steps.

Even though one O₃ molecule does not react until step 2, it is included among the reactants in the energy diagram from the start as it is required in the second step. In a reaction energy diagram, all atoms and molecules that participate in the reaction are accounted for from the beginning to the end of the process.

In previous chapters we have emphasized that reaction stoichiometry cannot be used to predict the reaction orders of a rate law. This is true for overall reactions, but because elementary reactions describe an actual molecular event we can use their stoicheiometry to determine the reaction orders of the accompanying rate law. We will discuss this in more detail throughout this chapter.

Unimolecular Elementary Reactions

The molecularity of an elementary reaction is the number of reactant particles (atoms, molecules, or ions) involved in that step.  A unimolecular elementary reaction involves the rearrangement or decomposition of a single reactant molecule, for example:

A → products

At a given concentration of A ([A]) only a fraction of A molecules (those that can surpass the activation energy) will react at a given temperature. Doubling [A] doubles the number of A molecules capable of reacting, so the rate of the reaction doubles. Thus, for a unimolecular elementary reaction:

rate = k[A]

This means such a reaction is first-order overall. Because an elementary reaction directly represents the actual molecular event, we can determine its reaction order simply by looking at its stoichiometry.

A unimolecular reaction can be part of a more complex mechanism (as with the ozone decomposition, where, \(\ce{O3 ⟶ O2 + O}\) is a unimolecular step), or it may represent the sole step in the reaction mechanism. In other words, an elementary reaction can also be an overall reaction.

Bimolecular Elementary Reactions

A bimolecular elementary reaction involves the collision of two reactant particles. The general forms are:

1. Collision between two different reactants:

A + B → products

rate = k[A][B]

This elementary step is first order in A and first order in B, making the reaction second-order overall.

2. Collision between two identical reactant molecules:

A + A → products or 2 A → products

rate = k[A]²

This elementary step is second order in A and second-order overall.

Figure \(\PageIndex{2}\) illustrates a reaction of the type A + B → products.  It demonstrates that the collision frequency (and thus the reaction rate) doubles as the concentration of either reactant doubles.  If both reactant concentrations double, the total number of collisions increases by a factor of four, consistent with a second-order rate law.

Figure \(\PageIndex{2}\) The effect of concentration on the collision frequency for a bimolecular reaction. (a) A single black molecule moving among 50 white molecules collides with 5 white molecules in 1 s. (Each white molecule that has been struck by a moving black molecule is shown in color.) (b) Doubling the number of white molecules doubles the collision frequency to 10 collisions/s.  c) Having two black molecules moving among 50 white ones also yields 10 collisions/s. The collision rate is proportional to both the concentration of white molecules and the concentration of black molecules.

These observations confirm that the collision rate—and thus the reaction rate—is proportional to the concentration of both reactants.

A real-world example of a bimolecular elementary reaction is Step 2 in the ozone decomposition mechanism discussed earlier:
O(g) + O₃(g) → 2O₂(g).

Termolecular Elementary Reactions

A termolecular elementary reaction involves three reactant particles colliding simultaneously.  Because the probability of three particles colliding at once is quite low, termolecular elementary reactions are rare (less than 0.1% of the probability of two particles colliding). More often, what appears to be a three-reactant reaction is actually a series of unimolecular and/or bimolecular steps. Nevertheless, a few established termolecular reactions exist. For instance, the oxidation of nitric oxide by oxygen appears to involve a termolecular step:

\[\ce{2NO + O2 ⟶ 2NO2}\\
\ce{rate}=k[\ce{NO}]^2[\ce{O2}] \nonumber \]

 

Relating Reaction Mechanisms to Rate Laws

A reaction mechanism consists of a series of elementary reactions and a valid mechanism for a multi-step reaction must meet the following criteria:

• It is composed of plausible elementary reactions (usually unimolecular or bimolecular).
• The sum of the elementary steps matches the overall balanced reaction.
• The mechanism agrees with the experimentally determined rate law.

For elementary reactions, rate laws can be written directly from the stoichiometric coefficients. However, the rate law of an overall multi-step reaction cannot be derived just by looking at the overall balanced equation. Instead, the overall rate law must be determined by combining the rate laws of the individual elementary steps in a way that reflects the reaction mechanism.

To evaluate a mechanism's validity, the theoretical rate law (derived from the mechanism) can be compared to the experimentally determined rate law. If the two agree, the proposed mechanism is consistent with experimental data and is a plausible mechanism to describe how the reaction proceeds. Additional experimental evidence, such as the detection of a proposed intermediate, can further support a proposed mechanism.

When proposing a reaction mechanism it is also necessary to define the relative energetics of each step. These energetics do not need to be quantitative with proposed activation energies for each step, they simply need to identify which step in the mechanism is the slowest step, known as the rate-determining step. The rate determining step determines the overall reaction rate. Even if all other steps are faster, the reaction cannot proceed faster than its slowest step. This is analogous to a production line, where the speed of the entire process is limited by the slowest machine. In this way, the rate law of the overall reaction corresponds to the rate law of the rate-determining step.

The next two sections will take you through the derivation of rate laws from reaction mechanisms. First we will discuss how to derive a rate law from a mechanism where the initial step is rate determining, then we will discuss how to derive a rate law from a mechanism where the initial step is not rate determining.

Mechanism where the Initial Step is Rate Determining

Let’s revisit the decomposition of ozone:

\[\begin{array}{rcl} &\text{step 1}:& \;\;\;\;\;\;\;\;\;\text{O}_3(g) &\xrightarrow{h\nu}& \text{O}_2(g) + \text{O}(g)\;\text{  slow}\\[0.5em]&\text{step 2}:& \;\;\;\;\;\;\;\;\;\text{O}(g) + \text{O}_3(g) &\longrightarrow& 2\;\text{O}_2(g)\\[0.5em] &\text{overall}:& \;\;\;\;\;\;\;\;\;2\;\text{O}_3(g) &\longrightarrow& 3\;\text{O}_2(g)\end{array} \nonumber \]

Since Step 1 is the slow step, it determines the overall reaction rate. The slow step produces O atoms, which are rapidly consumed in the faster second step. Because the reaction cannot proceed faster than its slowest step, the rate law for the overall reaction is the same as the rate law for Step 1:

\(\text{rate}=k[O_3]\)

The reaction energy diagram in Figure \(\PageIndex{1}\) illustrates the relative activation energies of the two steps, showing that Step 1, the rate-determining step, has a larger activation energy than Step 2, \(E_{a1} > E_{a2}\).

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 the first step.

Example \(\PageIndex{1}\)

Consider the following mechanism:

\[\ce{NO2}(g)+\ce{NO2}(g)⟶\ce{NO3}(g)+\ce{NO}(g)\:\ce{(slow)}\\
\ce{NO3}(g)+\ce{CO}(g)⟶\ce{NO2}(g)+\ce{CO2}(g)\:\ce{(fast)} \nonumber \]

a) What is the overall reaction?

b) List any reaction intermediates.

c) At temperatures below 225 °C, measurements show that the reaction is second order with respect to NO₂.  Is the mechanism consistent with this experimental observation?

Solution

a) \[\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber \]

b) NO₃ is an intermediate.  It is made in Step 1 and consumed in Step 2.

c) Step 1 is the rate-determining step, so the rate law for that step is the rate law for the overall reaction:

rate = k[NO₂]².  This is consistent with the rate law determined experimentally.

 

Mechanism where the Initial Step is not Rate Determining

The rate law for the overall reaction is always governed by the rate determining step, or slow step, of the mechanism. When it's not the first step, the rate law for the slow step will include at least one reaction intermediate. Reaction intermediates cannot be included in a rate law because they are transient species whose concentration depends on the concentration of the initial reactants and the rate of subsequent steps. For this reason, rate laws are always expressed in terms of the concentration of the reactants and products in the overall reaction. 

To describe the rate law of the rate determining step in terms of only reactant and product concentrations, we often use the equilibrium assumption which assumes that the preceding fast steps of the mechanism are reversible and quickly reach equilibrium. At equilibrium, the rate of product formation equals the rate of its conversion back to reactants, allowing us to express the concentration of any intermediates in terms of the concentration of the reactant and products.

Consider the following multi-step reaction as an example:

2 NO(g) + Cl₂(g) ⟶ 2 NOCl(g)

The currently accepted mechanism for this reaction is:

\[\begin{array}{rcl} &\text{Step 1}:& \text{NO}(g) + \text{Cl}_2(g) &\xrightleftharpoons[k_{-1}]{k_1}& \text{NOCl}_2(g)\;\;\;\;\;\text{fast}\\[0.5em]&\text{Step 2}:& \text{NOCl}_2(g) + \text{NO}(g) &\xrightarrow{k_2}& \text{2NOCl}(g)\;\;\;\;\;\text{slow}\\[0.5em]&\text{Overall}:& \text{2NO}(g) + \text{Cl}_2(g) &\longrightarrow& \text{2NOCl}(g)\end{array} \nonumber \]

As seen in Figure \(\PageIndex{3}\), the activation energy for the second step is greater than that of the first step, making it the rate-determining step.

Figure \(\PageIndex{3}\): Reaction energy diagram for the 2 NO(g) + Cl₂(g) → 2 NOCl(g) reaction

Since Step 2 is slowest step, its rate law determines the overall reaction rate:

rate₂ = k₂ [NOCl₂] [NO],

However, this expression contains [NOCl₂], the concentration of an intermediate, which is not possible to properly define, and so [NOCl₂] needs to be expressed in terms of concentrations of reactants and/or products.

To do this we need to turn to Step 1, which we assume is a fast and reversible reaction, meaning NO and Cl₂ quickly form NOCl₂, and NOCl₂ just as quickly breaks down back into NO and Cl₂. Because Step 2 is slower than Step 1, most NOCl₂ molecules decompose back into reactants rather than proceeding to Step 2. This allows us to assume that Step 1 is at equilibrium.

At equilibrium, the forward and reverse reaction rates of Step 1 are equal:

rate₁ = rate₋₁

Since Step 1 has a forward rate constant k₁​ and a reverse rate constant k₋₁​, we can express these rates as:

k₁[NO][Cl₂] = k_-1[NOCl₂]

Rearranging for [NOCl₂]:

\[[\text{NOCl}_2] = \left(\dfrac{k_1}{k_{-1}}\right)[\text{NO}][\text{Cl}_2] \nonumber \]

Substituting this into the rate law for Step 2:

\[\text{rate}_2 = k_2[\text{NOCl}_2][\text{NO}] = k_2\left(\dfrac{k_1}{k_{-1}}\right)[\text{NO}][\text{Cl}_2][\text{NO}] = \left(\dfrac{k_1k_2}{k_{-1}}\right)[\text{NO}]^2[\text{Cl}_2] \nonumber \]

Defining a new overall rate constant:

\[k = \dfrac{k_1k_2}{k_{-1}} \nonumber \]

we obtain the final rate law:

rate = k [NO]²[Cl₂]

This rate law can now be directly compared to experimental data to determine whether the proposed mechanism is consistent with observed reaction kinetics.

Example \(\PageIndex{2}\): Deriving the Overall Rate Law Expression for a Multistep Reaction

Consider the following mechanism:

\[\begin{array}{rcl} &\text{Step 1}:& \text{NO}(g) + \text{O}_2(g) &\xrightleftharpoons[k_{-1}]{k_1}& \text{NO}_3(g)\;\;\;\;\;\text{fast}\\[0.5em]&\text{Step 2}:& \text{NO}_3(g) + \text{NO}(g) &\xrightarrow{k_2}& \text{2NO}_2(g)\;\;\;\;\;\text{slow}\end{array} \nonumber \]

a) What is the overall reaction?

b) List any intermediates.

c) Determine the overall rate law.

Solution

a) For the overall reaction, sum the two steps, cancel intermediates, and combine like formulas:

\[\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\nonumber \]

b) NO₃ is the intermediate.

c) Write the rate law for the slow step:

rate = k₂[NO₃][NO]

Since NO₃ is an intermediate, we need to replace it in the rate law.

Using the fast equilibrium from step 1:

rate₁ = rate_-1

k₁[NO][O₂]=k_-1[NO₃]

\( \ce{[NO3]}=\dfrac{k_1}{k_{−1}}\ce{[NO][O2]} \)

Now substitute this expression for [NO3] into the overall rate law expression:

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

Example \(\PageIndex{3}\) : Nitrogen Oxide Reacting with Molecular Hydrogen

Assume the reaction between \(\ce{NO}\) and \(\ce{H_2}\) occurs via a three-step process:

\(\textrm{step 1}\)	\(\mathrm{NO}+\mathrm{NO}\xrightleftharpoons[k_{-1}]{k_1}\mathrm{N_2O_2}\)	\(\textrm{(fast)}\)
\(\textrm{step 2}\)	\(\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}\)	\(\textrm{(slow)}\)
\(\textrm{step 3}\)	\(\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}\)	\(\textrm{(fast)}\)

Write the balanced chemical equation for the overall reaction, identify any intermediates, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction, \(\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \)

Answer

The overall reaction is \(\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber \)

The intermediates are N₂O₂ and N₂O.

Rate Determining Step : #2

From Step #2, \(rate = k_2[\ce{N_2O_2}][\ce{H_2}]\).  Since N₂O₂ is an intermediate, we need to replace it in the rate law.

\(k_1[\ce{NO}]^2=k{-1}[\ce{N_2O_2}]\).

\([\ce{N_2O_2}] = =\dfrac{k_1}{k_{−1}}\ce{[NO]}^2 \)

Substituting into the rate law for step 2 gives:

\(rate = \dfrac{k_1k_2}{k_{-1}}[\ce{NO}]^2[\ce{H_2}]\) which matches the experimentally derived rate law.

Summary

A reaction mechanism is the sequence of elementary steps by which reactants are converted to products.  Each elementary step has its own transition state, shown as a maximum on a reaction energy diagram.

Elementary reactions depict actual molecular events and the stoichiometry of an elementary reaction can be used to define reaction orders in a theoetical rate law.

Reaction intermediates are species formed in one step and consumed in a later step. They have fully formed bonds but are typically unstable.

A valid mechanism must (1) sum to the overall balanced equation, (2) consist of plausible (generally unimolecular or bimolecular) steps, and (3) have a rate law consistent with experimental measurements.

The rate-determining step is the slowest step in the mechanism and controls the overall reaction rate.   When a fast step precedes the rate-determining step, we often use the equilibrium approximation to eliminate intermediates from the overall rate law.

Understanding the step-by-step mechanism and identifying the rate-determining step allows chemists to explain why a reaction follows a particular rate law and predict how changes in conditions will affect the reaction rate.