Ch 203 Problem 14.64

14.64 What is the molecularity of each of the following elementary processes? Write the rate law for each.

An elementary step (or process) in a reaction mechanism is a step which takes place as a result of a single collision. To this point in our studies, we have been concerned with substances which are present at the start of a reaction (the reactants) and at the end of a reaction (the products); we have learned that various substances react in characteristic ways (e.g., metal oxide + water -----> metal hydroxide), but we haven't yet examined how chemical reactions occur. This topic is examined through the study of reaction mechanisms: we take a complicated reaction and try to break it down into a series of elementary steps which add up to give the reaction we're interested in. If the elementary steps in the proposed mechanism add up to give the correct reaction, and if the mechanism yields a rate law which agrees with experiment, then we accept the mechanism.

The rate law for any elementary step is based directly on the molecularity of the step. The molecularity of an elementary step is defined by the number of reactant molecules which participate in the step. If there is one reactant molecule, the step is unimolecular; if there are two reactant molecules, the step is bimolecular, and if there are three reactant molecules, the step is termolecular. Termolecular elementary steps are rare (why?)

Elementary steps have another important property: the rate law of any elementary step is based directly on its molecularity. Thus, unimolecular elementary steps yield first-order rate laws, and bimolecular elementary steps yield second-order rate laws.

Based on the above discussion, we can determine the molecularity and write the rate laws for the following elementary steps:

(a) 2NO(g) ----> N2O2(g)

This elementary step is bimolecular (two reactant molecules: NO and NO); its rate law is therefore second order: rate = k[NO]2

(b)

This step is unimolecular (one reactant molecule); its rate law is first order:

rate = k[C3H6]

(c) SO3(g) ----> SO2(g) + O2(g)

Here, we have a unimolecular step (one reactant molecule); it has a first order rate law:

rate = k[SO3].

Note that you cannot predict a rate law from looking at a balanced chemical equation, but you can predict the rate law for an elementary step from the molecularity of the step. A balanced chemical equation is the sum of a number of elementary steps, and the rate law for the slowest elementary step will determine the rate law for the overall chemical reaction. You will always be told when a reaction is an elementary step.

DAC 8/5/08