Reaction Mechanisms

Synopsis

Remember the learning goal for this chapter? Now we are ready to determine the details of the trip from Atlanta to Chicago. The overall chemical reaction just gives the starting and ending points. Now we can attempt to determine the intermediate steps. These steps are called the mechanism of the reaction and they consist of elementary reactions, meaning reactions that themselves do not have a mechanism. For an elementary reaction, we can talk about the molecularity of the reaction, meaning the number of reactants involved in the elementary reaction. The term molecularity does not have any meaning for a reaction that is not an elementary reaction. An elementary reaction can be unimolecular, bimolecular, or termolecular. However termolecular reactions are very rare for we would have to have three reactants coming together at the same point in space, at the correct orientation, and with the correct energy for the reaction to occur and such a collision would be very rare.

The mechanism for an overall reaction is written as a series of elementary reactions. You will not be expected to develop a mechanism yourself, but you will need to be familiar with how to use a given mechanism. The rate law for an elementary reaction can be written using the stoichiometric coefficients. Remember that we said earlier that the stoichiometric coefficients were not related to the orders, but for elementary reactions they are. So for the elementary reaction aA + bB --> pP we can immediately write the rate law r = k[A]a[B]b. The molecularity of an elementary reaction is the same as the order. So the order for this elementary reaction would be n = a + b.

Now how do we know that our mechanism is the correct sequence of steps through which the reactants actually go to arrive at the product? Well, you can't prove a mechanism is the correct mechanism. You can prove that a mechanism is incorrect. We can be reasonably certain that the mechanism is correct by experimental observations, and often that level of certainty is enough to be of significant benefit in understanding the reaction. The first item that must match with experiment is the rate law. The rate law that we derive from the mechanism must be the same as the experimental rate law. The matching of the rate laws is a necessary, but not a sufficient condition for the reaction to be correct.

So let's see how this works out with an example. We will use the same example as in the text:

2NO2(g) + F2(g) --> 2FNO2(g), with the experimental rate law: r = k[NO2][F2]

The proposed mechanism for the reaction is

NO2(g) + F2(g) --> FNO2(g) + F(g), with the rate constant k1

NO2(g) + F(g) --> FNO2(g), with the rate constant k2

The first step has the rate law r = k1[NO2][F2] for the rate of formation of FNO2.

The second step has the rate law r = k2[NO2][F] for the rate of formation of FNO2.

You see that IF step one is very slow so that it is the rate-determining step, then the rate law matches that of experiment. Note that the product of step one is F, which is not the stable form of fluorine. Such unstable species are called reaction intermediates (they do not appear in the overall reaction), and they will rapidly try to form a stable species such as indicated by step two. So we expect step two to be very fast and step one to be the slow, rate-determining step. We would search for experimental evidence of the existence of the F atoms to partially verify the mechanism.

Now look at a little more complicated example:

2 NO(g) + O2(g) --> 2 NO2(g) with the experimental rate law r = k[NO]2[O2]

The fact that the orders are equal to the stoichiometric coefficients means that we must examine the possibility of a one-step termolecular reaction, but experiment rules it out. The proposed mechanism is

NO(g) + O2(g) --> OONO(g), with the rate constant k1

OONO(g) --> NO(g) + O2(g), with the rate constant k-1

NO(g) + OONO(g) --> 2 NO2(g), with the rate constant k3

Note that steps one and two constitute an equilibrium reaction, which we write as two distinct steps because the mechanism must be a series of elementary reactions. We can write the rate of formation of product (NO2) for the 3rd step as

r = k3[NO][OONO] 8

but now we see a problem. We cannot leave a rate law in this form because it contains a reaction intermediate. But remember that step 1 and step 2 constitute an equilibrium reaction, which means that the rates are equal:

k1[NO][O2] = k-1[OONO] ..9

As we will see in Chapter 16, k1/k-1 = K, the equilibrium constant.

So eq 9 becomes K[NO][O2] = [OONO]. Now if we substitute this equation for [OONO] into eq 8, we obtain r = k3K[NO]2[O2] = k'[NO]2[O2]

And this is the same form as the experimental rate law. We emphasize that having the theoretical and experimental rate laws equate does not mean that we have proven that the mechanism is the mechanism. However, we have proven that it is a possible mechanism.

Review Questions

  1. Work Exercise 15.13 on page 649..
  2. Work problems 82, 66, and 60 at the end of the chapter.
  3. Go to this site for more information about reaction mechanisms


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Web Author: Dr. Leon L. Combs
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