12.6 Reaction Mechanism

What do we mean by mechanism?

The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles (“collison”) 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.

For an example of a mechanism, consider the decomposition of nitrogen dioxide into nitric oxide and oxygen. The net balanced equation is

2 NO₂(g) → 2 NO(g) + O₂(g)

The mechanism of this reaction is believed to involve the following two elementary steps:

1)   2 NO₂ → NO₃ + NO

2)   NO₃ → NO + O₂

Note that the intermediate species NO₃ has only a transient existence and does not appear in the net equation.

A useful reaction mechanism consists of a series of elementary steps (defined below) that can be written as chemical equations, and whose sum gives the net balanced reaction equation; must agree with the experimental rate law; can rarely if ever be proved absolutely.

It is important to understand that the mechanism of a given net reaction may be different under different conditions. For example, the dissociation of hydrogen bromide 2 HBr(g) → H₂(g) + Br₂(g) proceeds by different mechanisms (and follows different rate laws) when carried out in the dark (thermal decomposition) and in the light (photochemical decomposition).Similarly, the presence of a catalyst can enable an alternative mechanism that greatly speeds up the rate of a reaction.

Elementary steps

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

Table 12.6

Elementary reactions differ from ordinary net chemical reactions in two important ways:

The rate law of an elementary reaction can be written by inspection. For example, a bimolecular process always follows the second-order rate law
k[A][B].

Elementary steps often involve unstable or reactive species that do not appear in the net reaction equation.

Some net reactions do proceed in a single elementary step, at least under certain conditions. But without careful experimentation, one can never be sure.

The gas-phase formation of HI from its elements was long thought to be a simple bimolecular combination of H₂ and I₂, but it was later found that under certain conditions, it follows a more complicated rate law.

Multi-step (consecutive) reactions

Mechanisms in which one elementary step is followed by another are very common.

step 1: A + B → Q

step 2: B + Q → C

net reaction: A + 2B → C

(As must always be the case, the net reaction is just the sum of its elementary steps.) In this example, the species Q is an intermediate, usually an unstable or highly reactive species.

If both steps proceed at similar rates, rate law experiments on the net reaction would not reveal that two separate steps are involved here. The rate law for the reaction would be

rate = k[A][B]²

(Bear in mind that intermediates such as Q cannot appear in the rate law of a net reaction.)

When the rates are quite different, things can get interesting and lead to quite varied kinetics as well as some simplifying approximations.

Some important simplifying approximations

When the rate constants of a series of consecutive reactions are quite different, a number of relationships can come into play that greatly simplify our understanding of the observed reaction kinetics.

The rate-determining step

The rate-determining step is also known as the rate-limiting step.

We can generally expect that one of the elementary reactions in a sequence of consecutive steps will have a rate constant that is smaller than the others. The effect is to slow the rates of all the reactions — very much in the way that a line of cars creeps slowly up a hill behind a slow truck.

The three-step reaction depicted here involves two intermediate species I₁ and I₂, and three activated complexes numbered X_1-3. Although the step I ₂ → products  has the smallest individual activation energy E_a3, the energy of X₃ with respect to the reactants determines the activation energy of the overall reaction, denoted by the leftmost vertical arrow . Thus the rate-determining step is  I₂ → products.

 Fig. 12.45

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Chemists often refer to elementary reactions whose forward rate constants have large magnitudes as “fast”, and those with forward small rate constants as “slow”. Always bear in mind, however, that as long as the steps proceed in single file (no short-cuts!), all of them will proceed at the same rate. So even the “fastest” members of a consecutive series of reactions will proceed as slowly as the “slowest” one.

What Limits the Rate?

• The overall reaction cannot occur faster than the slowest reaction in the mechanism.
• We call that the rate-determining step.

Fig. 12.46

Source: www.commons.wikimedia.org/

What is Required of a Plausible Mechanism?

• The rate law must be able to be devised from the rate-determining step.
• The stoichiometry must be obtained when all steps are added up.
• Each step must balance, like any equation.
• All intermediates are made and used up.
• Any catalyst is used and regenerated.

A Mechanism With a Slow Initial Step

• Overall equation: NO₂ + CO → NO + CO₂
• Rate law: Rate = k [NO₂]²
• If the first step is the rate-determining step, the coefficients on the reactants side are the same as the order in the rate law!
• So, the first step of the mechanism begins:

                        NO₂ + NO₂ → ?

A Mechanism With a Slow Initial Step (continued)

• The easiest way to complete the first step is to make a product:

            NO₂ + NO₂ → NO + NO₃

• We do not see NO₃ in the stoichiometry, so it is an intermediate, which needs to be used in a faster next step.

            NO₃ + CO → NO₂ + CO₂

A Mechanism With a Slow Initial Step (completed)

• Since the first step is the slowest step, it gives the rate law.
• If you add up all of the individual steps (2 of them), you get the stoichiometry.
• Each step balances.
• This is a plausible mechanism.

• Equation for the reaction:

            2 NO + Br₂  ⇌  2 NOBr

• The rate law for this reaction is found to be

            Rate  =  k [NO]² [Br₂]

• Because termolecular processes are rare, this rate law suggests a multistep mechanism.

• The rate law indicates that a quickly established equilibrium is followed by a slow step.
• Step 1:             NO  +  Br₂  ⇌  NOBr₂
• Step 2:             NOBr₂  +  NO  →  2 NOBr

What is the Rate Law?

• The rate of the overall reaction depends upon the rate of the slow step.
• The rate law for that step would be

Rate = k₂[NOBr₂] [NO]

• But how can we find [NOBr₂]?

[NOBr₂] (An Intermediate)?

• NOBr₂ can react two ways:
  • With NO to form NOBr.
  • By decomposition to reform NO and Br₂.
• The reactants and products of the first step are in equilibrium with each other.
• For an equilibrium (as we will see in the next chapter):

Rate_f = Rate_r

The Rate Law (Finally!)

• Substituting for the forward and reverse rates:

            k₁ [NO] [Br₂]  =  k₋₁ [NOBr₂]

• Solve for [NOBr₂], then substitute into the rate law:

            Rate = k₂ (k₁/k₋₁) [NO] [Br₂] [NO]

• This gives the observed rate law!

            Rate = k [NO]² [Br₂]

SUPPLEMENTAL

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Rapid equilibria

In many multi-step processes, the forward and reverse rate constants for the formation of an intermediate Q are of similar magnitudes and sufficiently large to make the reaction in each direction quite rapid. Decomposition of the intermediate to product is a slower process.

This is often described as a rapid equilibrium in which the concentration of Q can be related to the equilibrium constant K = k₁/k_–1 (This is just the Law of Mass Action.). It should be understood, however, that true equilibrium is never achieved because Q is continually being consumed; that is, the rate of formation of Q always exceeds its rate of decomposition. For this reason, the steady-state approximation described below is generally preferred to treat processes of this kind.

The steady-state approximation

Consider a mechanism consisting of two sequential reactions

in which Q is an intermediate. The time-vs-concentration profiles of these three substances will depend on the relative magnitudes of k₁ and k₂, as shown in the following diagrams.

Construction of these diagrams requires the solution of sets of simultaneous differential equations, which is [fortunately!] beyond the scope of this course.

Fig. 12.47

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The steady-state approximation is usually not covered in introductory courses, although it is not particularly complicated mathematically. For a nice introduction, see this U. Waterloo tutorial.

In the left-hand diagram, the rate-determining step is clearly the conversion of the rapidly-formed intermediate into the product, so there is no need to formulate a rate law that involves Q. But on the right side, the formation of Q is rate-determining, but its conversion into B is so rapid that [Q] never builds up to a substantial value. (Notice how the plots for [A] and [B] are almost mutually inverse.) The effect is to maintain the concentration of Q at an approximately constant value. This steady-state approximation can greatly simplify the analysis of many reaction mechanisms, especially those that are mediated by enzymes in organisms.