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.

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

Molecules must collide before they can react

This fundamental rule must guide any analysis of an ordinary chemical reaction mechanism.

This explains why termolecular processes are so uncommon. The kinetic theory of gases tells us that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction.

Consider a simple bimolecular step

A + B → products

Clearly, if two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. We call such an encounter a collision.

The frequency of collisions between A and B in a gas will be proportional to the concentration of each; if we double [A], the frequency of A-B collisions will double, and doubling [B] will have the same effect. So if all collisions lead to products, than the rate of a bimolecular process will be first-order in A and B, or second-order overall:

rate = k[A][B]

but...

Not all collisions are equal

When two billiard balls collide, they simply bounce off of each other. This is also the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, collisions must be sufficiently energetic (kinetic energy) to bring about this bond disruption. More about this further on.

And there is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N₂O must hit the nitrogen end of NO; reversing the orientation of either molecule prevents the reaction.

Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. But of course, the more critical this orientational requirement is, the fewer collisions will be effective.

Anatomy of a collision

Energetic collisions between molecules cause interatomic bonds to stretch and bend farther, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds.

A particular collision will typically excite a number of bonds in this way. Within about 10^–13 second this excitation gets distributed among the other bonds in the molecule in rather complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction.

Consider, for example, the isomerization of cyclopropane to propene which takes place at fairly high temperatures in the gas phase.

We can imagine the collision-to-product sequence in the following [grossly oversimplified] way:

Note that

• To keep things simple, we do not show the hydrogen atoms here. This is reasonable because C–C bonds are weaker then C–H bonds and thus less likely to be affected.
• The collision at will usually be with another cyclopropane molecule, but because no part of the colliding molecule gets incorporated into the product, it can in principle be a noble gas or some other non-reacting species;
• Although the C–C bonds in cyclopropane are all identicial, the instantaneous localization of the collisional energy can distort the molecule in various ways (), leading to a configuration sufficiently unstable to initiate the rearrangement to the product.

Unimolecular processes also begin with a collision

The cyclopropane isomerization described above is typical of many decomposition reactions that are found to follow first-order kinetics, implying that the process is unimolecular. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are really rather complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as pseudounimolecular. The details are beyond the scope of this course, but a good introduction can be found on this U. Arizona page.

Activation energy

Higher temperatures, faster reactions

It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.

It is not hard to understand why this should be. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelyhood of bond cleavages and rearrangemens as described above.

Activation energy diagrams

Most reactions involving neutral molecules cannot take place at all until they have acquired the energy needed to stretch, bend, or otherwise distort one or more bonds. This critical energy is known as the activation energy of the reaction. Activation energy diagrams of the kind shown below plot the total energy input to a reaction system as it proceeds from reactants to products.

In examining such diagrams, take special note of the following:

 

Gallery of activation energy plots

Activation energy diagrams can describe both exothermic and endothermic reactions:

... and the activation energies of the forward reaction can be large, small, or zero (independently, of course, of the value of ΔH):

 

Processes with zero activation energy most commonly involve the combination of oppositely-charged ions or the pairing up of electrons in free radicals, as in the dimerization of nitric oxide (which is an odd-electron molecule).

 

 

 

In this plot for the dissociation of bromine, the E_a is just the enthalpy of atomization

Br₂(g) → 2 Br· (g)

and the reaction coordinate corresponds roughly to the stretching of the vibrationally-excited bond. The "activated complex", if it is considered to exist, is just the last, longest "stretch". The reverse reaction, being the recombination of two radicals, occurs immediately on contact.

 

Where does the activation energy come from?

 

In most cases, the activation energy is supplied by thermal energy, either through intermoleculr collisions or (in the case of thermal dissocation) by thermal excitation of a bond-stretching vibration to a sufficiently high quantum level.

As products are formed, the activation energy is returned in the form of vibrational energy which is quickly degraded to heat.

 

 

 

It's worth noting, however, that other sources of activation energy are sometimes applicable:

• Absorption of light by a molecule (photoexcitation) can be a very clean and efficient, but it doesn't always work. It's not enough that the wavelength of the light correspond to the activation energy; it must also fall within the absorption spectrum of the molecule, and (in a complex molecule) enough of it must end up in the right part of the molecule, such as in a particular bond.
• Electrochemical activation. Molecules capable of losing or gaining electrons at the surface of an electrode can undergo activation from an extra potential (known as the overvoltage) between the electrode and the solution. The electrode surface often plays an active role, so the process is also known as electrocatalysis.

Temperature and kinetic energy

A review of the principles of gas molecular velocities and the Boltzmann distribution can be found on the "KMT-classic" page.

In the vast majority of cases, we depend on thermal actvation, so the major factor we need to consider is what fraction of the molecules possess enough kinetic energy to react at a given temperature.

According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.

It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.

  

 

Take a moment to focus on the meaning of this equation, neglecting the A factor for the time being.

First, note that this is another form of the exponential decay law we discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –E_a /RT. And what is the significance of this quantity? If you recall that RT is the average kinetic energy, it will be apparent that the exponent is just the ratio of the activation energy E^_a to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign.) This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. And because these terms occur in an exponent, their effects on the rate are quite substantial.

The two plots below show the effects of the activation energy (denoted here by E^‡) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10⁸.

 

Looking at the role of temperature, we see a similar effect.

(If the x-axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.)

 

 

 

Determining the activation energy

The Arrhenius equation

can be written in a non-exponential form which is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields

which is the equation of a straight line whose slope is –E_a /R. This affords a simple way of determining the activation energy from values of k observed at different temperatures; we just plot ln k as a function of 1/T.

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Thus for the isomerization of cyclopropane to propene

the following data were obtained (calculated values shaded in pink):

From the calculated slope, we have

– (E_a/R) = –3.27 × 10⁴ K

E_a=– (8.314 J mol^–1 K^–1) (–3.27 × 10⁴ K) = 273 kJ mol^–1

Comment: This activation energy is rather high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature.

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You don't always need a plot

(... if you are willing to live a bit dangerously!) Since the ln k-vs.-1/T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that

(... in which we have made the ln-A term disappear by subtracting the expressions for the two ln-k terms.) Solving the expression on the right for the activation energy yields

 

 

 

 

The pre-exponential factor

It is now time to focus in on the pre-exponential term A in the Arrhenius equation. We have been neglecting it because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. But since A multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate.

Recall that the exponential part of the Arrhenius equation expresses the fraction of reactant molecules that possess enough kinetic energy to react, as governed by the Maxwell-Boltzmann law. This fraction can run from zero to nearly unity, depending on the magnitudes of E_a and of the temperature.

If this fraction were unity, the Arrhenius law would reduce to

k = A

In other words, A is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded E_a — admittedly, an uncommon scenario.

It's all about collisions

So what would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor Z.

In some reactions, the relative orientation of the molecules at the point of collision is important, so we can also define a geometrical or steric factor (commonly denoted by ρ (Greek lower case rho).

In general, we can express A as the product of these two factors:

A = Zρ

Values of ρ are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which A is assumed to be the same as Z.

Direction makes a difference

The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other.

We showed one example of this near the top of the page, but for another, consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane.

Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below.

The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive.

The double bond of ethene consists of two clouds of negative charge corresponding to the σ (sigma) and π (pi) molecular orbitals. The latter, which extends above and below the plane of the C₂H₄ molecule, interacts with and attracts the HCl molecule.

If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to bounce away from each other before any reaction can take place. The same thing happens in ; the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom.

Now that you know what it takes to get a reaction started, you are ready for the next lesson that describes their actual mechanisms.

 

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.

• Explain the meaning of a reaction mechanism and define elementary step and intermediate.
• Describe the role of collisions in reaction mechanisms, and explain why not all collisions lead to the formation of products.
• Sketch out activation energy diagrams for simple reactions that are endothermic or exothermic,
• Explain how an activated complex differs from an intermediate.
• Define catalyst, and sketch out an activation energy diagram that illustrates how catalysts work.
• Explain the significance of the various terms that appear in the Arrhenius Law.
• Sketch out a typical Arrhenius Law plot for a hypothetical reaction at higher and lower temperatures.
• Explain how the activation energy of a reaction can be determined experimentally.
• Explain the significance of the various terms that appear in the pre-exponential factor of the Arrhenius equation.