Example 20.10
What is the expected rate law of the equation below, assuming that it is an el-
ementary process? What is the molecularity with respect to each reactant?
H 2 O 2 →2OHSolution
According to the statement in the paragraph before the example, the rate
law for this elementary process is given directly by the stoichiometry of the
reaction. Therefore, we can simply look at the reaction (that is, we can tell “by
inspection”) and state that
rate k[H 2 ]^1 [O 2 ]^1
Since the exponent 1 is typically not written explicitly, the rate law is usually
written as
rate k[H 2 ][O 2 ]
The molecularity with respect to H 2 is 1, and the molecularity with respect
to O 2 is also 1.There is an additional consideration: is the rate law of this first elementary
process consistent with the experimentally determinedrate law of the overall re-
action? Unfortunately, we don’t know this yet, and the determination of this
will be left to the next section. But you should at least be getting an idea of the
factors involved in determining the mechanism of a reaction.
What are other possible elementary processes for this reaction? Below, we
list some possible elementary processes. Note that the lack of charges on the
species is intentional: many of the intermediates from the elementary processes
are radicals.
H 2 O 2 →2OH (from previously)
OHH 2 →H 2 O H
HO 2 →OHO
OHH 2 →H 2 O H
HO→OH
OHH 2 →H 2 O H
and so on
The combination of all of these processes, for all of the billions of billions of
molecules in any sample, is
2H 2 O 2 →2H 2 O
As it should be.
Another example of elementary processes comes from halogenation of
alkanes. For the gas-phase chlorination of methane, the major overall reaction
(ignoring unwanted products) is
CH 4 Cl 2 →CH 3 Cl HCl708 CHAPTER 20 Kinetics