The balanced chemical equations used to describe a reaction are a description of the overall process, but usually doesn’t give an accurate picture of what actually happens. There may be several steps represented by the overall equation, which are important to understand the process. Kinetics can help determine what these smaller steps are.

** Elementary reactions:**

Take the example:

NO_{2(g)} + CO_{(g)} → NO_{(g)} + CO_{2(g)}

From looking at the overall equation, it would appear that two NO_{2} molecules collide with a CO molecule and BAM! you get a reaction. In reality, a successful tri-molecular collision is fairly unlikely and this reaction would sit far to the left. However, it is believed that this reaction actually occurs in two smaller steps, known as elementary reactions:

NO_{2} + NO_{2} → NO_{3} + NO

NO_{3} + CO → NO_{2} + CO_{2}

An elementary reaction is a single step, involving one molecular interaction that results in a reaction. The sum of all the elementary steps gives the overall equation. In this case, two NO_{2} molecules collide to give the intermediate NO_{3} and NO. The intermediate, NO_{3} is then immediately consumed in the second collision to give the product CO_{2} and NO_{2}.

We can use the elementary reactions to derive the overall rate law, or use the overall rate law to confirm a mechanism. For an elementary reaction, (but NOT for an overall reaction), we can assume that the rate is proportional to concentration of the reactants, raised to their stoichiometric values.

The following example is modified from Ebbing’s *General Chemistry* 5^{th} edition.

2NO_{2} + F_{2} → 2NO_{2}F

The above reaction is believed to occur via two elementary steps:

Adding the two equations together gives you the overall reaction, with F as an intermediate, so that satisfies that requirement. However, the mechanism must also be in agreement with the experimental rate law. In this case, the rate law was found to be:

Rate = k[NO_{2}][F_{2}]

To show that the mechanism is in agreement, we look at the **rate determining step**, which is the slowest step. In this case, the first step is shown to be slower, so that is the reaction that will determine the overall rate, (you can only move as fast as the slowest step – for a practical demonstration of this try to hurry down Bourke St during the Christmas shopping season). We assume that the overall rate is the same as that of the rate determining step and can write:

Rate = k_{1}[NO_{2}][F_{2}]

This is the same as the experimentally determined step, and so we can say that this mechanism is consistent.

Of course, it isn’t always that easy. Sometimes, the first step is the fast step and we need to invoke the **steady state approximation** to find the rate law.

For example, let’s look at the destruction of ozone:

2O_{3} → 3O_{2}

Rate = k[O_{3}]^{2}[O_{2}]^{-1}

The following mechanism is proposed:

We can see that the two add together to give the overall reaction, and that the rate determining step is the second step.

Rate = k_{2}[O_{3}][O]

The problem with this is that there is an intermediate in the rate law, and we can’t have that. So we apply the steady state approximation – assume that the concentration of the intermediate remains constant.

This means that we are creating it at the same rate as we are using it:

Rate production = rate consumption

Rate production – rate consumption = 0

k_{1}[O_{3}] – k_{1}^{’}[O_{2}][O] – k_{2}[O_{3}][O] = 0

rearrange for [O]:

We can then plug this into the rate equation to get a rate law without the intermediate:

Because the second step is much slower than the first, we can assume that k_{1}^{’ }>> k_{2}, so the denominator simplifies:

and we assume that

We get the experimental rate law!

Some links:

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadys2.html

http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum04/handout_Lecture_06.pdf