Chem1 General Chemistry Virtual Textbook → dynamics → mechanisms
How chemical reactions take place
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We are now ready to open up the "black box" that lies between the reactants and products of a net chemical reaction. What we find inside may not be very pretty, but it is always interesting because it provides us with a blow-by-blow description of how chemical reactions take place.
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
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.
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 reactions differ from ordinary net chemical reactions in two important ways:
Some net reactions do proceed in a single elementary step, at least under certain conditions. But without careful experimenation, 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 followes a more complicated rate law.
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.
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.
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 X1-3. Although the step I 2 → products has the smallest individual activation energy Ea3, 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.
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:
A Q B
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.
Consider a mechanism consisting of two sequential reactions
A Q B
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.
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.
One possible mechanism might involve two intermediates Q and R:
|1 A + B Q||(slow, rate-determining)|
|2 Q + A R||(fast)|
|3 R + B C||(fast)|
The rate law corresponding to this mechanism would be that of the rate-determining step: rate = k1[A][B].
An alternative mechanism in which the rate-determining step involves one of the intermediates would display third-order kinetics:
|1 A + B Q||(rapid equilibrium)|
|2 Q + A R||(slow, rate-determining)|
|3 R + B C||(fast)|
Since intermediates cannot appear in rate law expressions, we must express [Q] in the rate-determining step in terms of the other reactants. To do this, we make use of the fact that Step 1 involves an equilibrium constant K1:
Solving this for [Q], we obtain [Q] = K1[A][B].
We can now express the rate law for 2 as
rate = k2K1[A][B][A] = k[A]2[B]
in which the constants k2 and K1 have been combined into a single constant k.
Application of the "chemical intuition" mentioned in the above box would lead us to suspect that any process that involves breaking of the strong F–F bond would likely be slow enough to be rate limiting, and that the resulting atomic fluorines would be very fast to react with another odd-electron species:
|1 F2 + 2 NO2 NO2F + F + NO2||(slow, rate-determining)|
|2 F + NO2 NO2F||(very fast)|
If this mechanism is correct, then the rate law of the net reaction would be that of the rate-determining step:
rate = k1[F2][NO2]2
Ozone is an unstable alltrope of oxygen that decomposes back into ordinary dioxygen according to the net reaction
2 O3 → 3 O2
A possible mechanism would be the simple one-step bimolecular collision suggested by the reaction equation, but this would lead to a second-order rate law which is not observed. Instead, experiment reveals a more complicated rate law:
rate = [O3]2[O2]–1
What's this? It looks as if O2 actually inhibits the reaction in some way. The generally-accepted mechanism for this reaction is:
|1 O3 O2+ O||(rapid equilibrium)|
|2 O + O3 2 O2||(rate-determing)|
Does this seem reasonable? Note that
To translate this mechanism into a rate law, we first write the equilibrium constant for Step 1 and solve it for the concentration of the intermediate:
We substitute this value of [O] into the rate expression [O][O3] for Step 2, which yields the experimentally-obtained rate law
2 NO + O2 → 2 NO2
The gas-phase oxidation of nitric oxide, like most third-order reactions, is not termolecular but rather a combination of an equilibrium followed by a subsequent bimolecular step:
|1 NO + NO N2O2||(equilibrium, eq. const. K)|
|2 N2O2 + O2 2 NO2||(rate-limiting)|
Since the intermediate N2O2 may not appear in the rate equation, we need to express its concentration in terms of the reactant NO. As in the previous example, we do this through the equilibrium constant of Step 1:
K = [N2O2] / [NO]2 [N2O2] = K [NO]2
rate = k2 [N2O2 ][O2] = k2 K [NO]2 [O2]
Many important reaction mechanisms, particularly in the gas phase, involve intermediates having unpaired electrons, commonly known as free radicals. Free radicals are often fairly stable thermodynamically and may be quite long-lived by themselves, but they are highly reactive, and hence kinetically labile.
The "atomic" forms of many elements that normally form diatomic molecules are free radicals; H·, O·, and Br· are common examples. The simplest and most stable (ΔG = +87 kJ/mol) molecular free radical or "odd-electron molecule" is nitric oxide, NO·.
The most important chemical property of a free radical is its ability to pass the odd electron along to another species with which it reacts. This chain-propagation process creates a new radical which becomes capable of initiating another reaction. Radicals can, of course, also react with each other, destroying both ("chain termination") while creating a new covalent-bonded species.
The synthesis of hydrogen bromide from its elements illustrates the major features of a chain reaction. The figures in the right-hand column are the activation energies per mole.
|1 Br2 + M → 2 Br·||chain initiation (ΔH° = +188 kJ)|
|2 Br· + H2 → H· + HBr||chain propagaton (+75 kJ)|
|3 H· + Br2 → HBr + Br·||chain propagation (–176 kJ)|
|4 H· + HBr → H2 + Br·||chain inhibition (–75 kJ)|
|5 2 Br· + M → M* + Br2||chain termination (–188 kJ)|
Note the following points:
The rate laws for chain reactions tend to be very complex, and often have non-integral orders.
The gas-phase oxidation of hydrogen has been extensively studied over a wide range of temperatures and pressures.
H2(g) + ½ O2(g) → H2O(g) ΔH° = –242 kJ/mol
This reaction does not take place at all when the two gases are simply mixed at room temperature. At temperatures around 500-600°C it proceeds quite smoothly, but when heated above 700° or ignited with a spark, the mixture explodes.
As with all combustion reactions, the mechanism of this reaction is extremely complex (you don't want to see the rate law!) and varies somewhat with the conditions. Some of the major radical formation steps are
|1 H2 + O2 → HO2· + H·||chain initiation|
|2 H2 + HO2· → HO· + H2O||chain propagaton|
|3 H· + O2 → HO· + O·||chain propagaton + branching|
|4 O· + H2 → HO· + H·||chain propagation + branching||5 HO· + H2 → H2O + H·||chain propagation|
Reactions 3 and 4 give birth to more radicals than they consume, so when these are active, each one effectively initiates a new chain process causing the rate overall rate to increase exponentially, producing an explosion.
An explosive reaction is a highly exothermic process that once initiated, goes to completion very rapidly and cannot be stopped. The distructive force of an explosion arises from the rapid expansion of the gaseous products as they absorb the heat of the reaction.
There are two basic kinds of chemical explosions:
Thermal explosions occur when heat is released by a reaction more rapidly than it can escape from the reaction space. This causes the reaction to proceed more rapidly, releasing even more heat, resulting in an ever-accelerating runaway rate.
Chain-branching explosions occur when the number of chain carriers increases exponentially, effectively seeding new reaction centers in the mixture. The process then transforms into a thermal explosion.
Whether or not a reaction proceeds explosively depends on the balance between formation and destruction of the chain-carrying species. This balance depends on the temperature and pressure, as illustrated here for the hydrogen-oxygen reaction.
Upper and lower explosion limits for several common fuel gases are shown below. [source]
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.