# Concentration and Rate

Synopsis

Usually there is proportionality between the rate of a reaction and the concentration of the reactants. We can convert a proportionality to an equality by the use of a proportionality constant so for the reaction

aA + bB --> pP

we can write that the rate of the reaction, r, is given by the following rate equation, or rate law:

r = k[A]a[B]b .........(2)

where "a" and "b" are the order of the reaction with respect to A and B, respectively. The sum of "a" and "b" is n, the overall order of the reaction. These orders have nothing to do with the stoichiometric coefficients except sometimes fortuitously. These rate constants have to be determined experimentally and can be zero, positive or negative, and integers or fractions. The proportionality constant, k, is called the rate constant for the reaction and it is a function of temperature.

Let's consider a simple example, the hypothetical reaction

2A --> pP

The rate law would then be r = k[A]a. The order has to be determined experimentally, and one method used to obtain the order is by experimentally determining the initial rate of the reaction. The initial rate is the rate of the reaction immediately after the reaction begins. So now we experimentally measure the initial rate as a function of the initial concentration of A, and let's suppose that we obtain the following data (no units):

 [A] Initial Rate [A]o 2 2[A]o 4

You see that by doubling the initial concentration of A, the rate doubled. For the doubling of the rate to match the doubling of the initial concentration of A, " must be one so that the rate law is r = k[A]. How do we get this result? Let the initial rate be ro = k[[A]o], and then double the initial concentration of A to have r = k[2[A]o] = 2k[[A]o] and you see that r = 2ro, which is the desired result.

But suppose that instead of the above data, we experimentally obtain the following data:

 [A] Initial Rate [A]o 2 2[A]o 8

Now the experimental data tells us that if the initial concentration of A doubles, the rate quadruples. So we apply the same procedure as for the first set of data, and we see that " must be equal to two, so that the rate law would be r = k[A]2.

If there were two reactants, we would need similar data involving both reactants. Consider the hypothetical reaction

aA + bB --> pP

Now you would need to plan your experiment carefully so that you would be able to use the experimental data to determine the order with respect to A and B. You would need to hold the initial concentration of A constant for two runs of the experiment with the concentration of B doubled in the second run. You would then need to hold the initial concentration of B constant for two runs with the concentration of A doubled in the second run. The minimal set of data below would accomplish your goal:

 [A] [B] Initial Rate [A]o [B]o 1 [A]o 2[B]o 2 2[A]o [B]o 4

We see that from this hypothetical set of data, the rate law for this reaction would be r = k[A]2[B]. To be sure of our conclusions we would probably want to use more variation in initial concentrations.

Now how do we obtain the rate constant? After you have used the experimental data to obtain the rate law, just put any set of the data into the rate law and solve for k. If there were no experimental error, then each of the data sets will give the same value for k. Because we do expect experimental error, it is best to use each set of data to calculate k and then to average the results. Of course the rate constants will have units, and the units will vary depending upon the overall order of the reaction.

Review Questions

1. Work though all of the examples to make sure that you understand the procedures of this section.
2. Work Exercise 15.3 on page 613.
3. Suppose that you have three different reactants involved in a chemical reaction. Set up a table similar to the above for the minimum number of experiments that you would have to perform to determine the order with respect to each reactant.

Web Author: Dr. Leon L. Combs