Book Cover Chemistry 8th Edition / Chang
Student Study Guide

Chapter 13: Chemical Kinetics


Index | 13.1 | 13.2 | 13.3 | 13.4 | 13.5 | 13.6 |

RATE LAWS (13.2)

STUDY OBJECTIVES
  1. Describe all the terms in "the rate law," and define the order of reaction.
  2. Apply the isolation method to derive the rate law for a reaction.

Effect of Concentration. The rate of a reaction is proportional to the reactant concentrations. For the reaction

NO + O2 NO2

the rate is proportional to the concentrations of NO and O2.

The rate law (or rate equation) for the reaction is:

rate = k[NO]x[O2]y

The proportionality constant k is called the rate constant. The value of k depends on the reaction and the temperature. x and y are exponents which are often, but not always 1 or 2.

The exponents x and y determine how strongly the concentration affects the rate. The exponent x is called the order with respect to NO, and y is the order with respect to O2. The sum x + y is the overall order. The values of x and y must be determined from experiment, and cannot be derived by any other means. We will discuss how to determine the order of reaction in the next section. For now we will just use the results. For the NO reaction with O2 experiments show that x = 2 and y = 1. Therefore, the rate law for this reaction is:

rate = k[NO]2[O2]

This reaction is second order in nitric oxide, and first order in oxygen. It is third order overall.

The fact that the reaction is first order in O2 means that the rate is directly proportional to the O2 concentration. If [O2] doubles or triples, the rate will double or triple also. We can show this mathematically. Consider two experiments. In expt 1 the concentration of O2 is c. In expt 2 the concentration of O2 is doubled from c to 2c. If the concentration of NO is the same in both experiments, it will have no effect on the rate. Use of the rate law allows us to write the ratio of the two rates:

As discussed, we see that doubling the concentration of a reactant that is first order will cause the rate to double.

If the concentration of O2 is held constant in two experiments and the concentration of NO doubles (from c to 2c), the rate law predicts that the rate will quadruple.

The fact that the reaction is second order in NO means that the rate is proportional to the square of the concentration of NO. Doubling or tripling of [NO] causes the rate to increase four- or nine-fold, respectively.

In general, if the concentration of one reactant is doubled while the other reactant concentration is unchanged, and the rate is:

  1. unchanged, the order of the reaction is zero order with respect to the changing reactant.
  2. doubled, the order of the reaction is first order with respect to the changing reactant.
  3. quadrupled, the order of the reaction is second order with respect to the changing reactant.

EXAMPLE Concentration Effect on the Rate

The reaction A + 2B products was found to have the rate law: rate = k[A][B]3. By what factor will the rate of reaction increase if the concentration of B is increased from c to 3c, while the concentration of A is held constant?

The Isolation Method. One procedure used to determine the rate law for a reaction involves the isolation method. In this method the concentration of all but one reactant is fixed, and the rate of reaction is measured as a function of the concentration of the one reactant whose concentration is varied. Any variation in the rate is due to the variation of this reactant's concentration. In practice the experimenter observes the dependence of the initial rate on the concentration of the reactant.

To determine the order with respect to A in the following chemical reaction

2A + B C

the initial rate would be measured in several experiments in which the concentration of A is varied and the concentration of B is held constant. To determine the order with respect to B, the concentration of A must be held constant and the concentration of B is varied in several experiments.

EXAMPLE Finding the Rate Law

The following rate data were collected for the reaction:

2NO + 2H2 N2 + 2H2O
Experiment [NO]0
(M) 		[H2]0
(M) 		[N2]/t
(M/h)
1 0.60 0.15 0.076
2 0.60 0.30 0.15
3 0.60 0.60 0.30
4 1.20 0.60 1.21

Remember to put subscripts in brackets and superscripts in braces -- x2 = x[2]; x3 = x{3}

a. Determine the rate law.
rate =

b. Calculate the rate constant.
k = /M2 h

•Comment

You should get the same value of k from all four experiments. Note that the units of k are those of a third-order rate constant. Taking a closer look at proving that x = 2. Write the ratio of rate laws for experiments 4 and 3.

4.0 = 2.0x

Therefore x = 2

OBJECTIVE CHECK

Complete the following questions to check your understanding of the material. Select the check button to see if you answered correctly.

  1. The rate law for the reaction 2A + B C was found to be rate = k[A][B]2. If the concentration of B is tripled and the concentration of A is unchanged, by how many times will the reaction rate increase?
  2. Use the following data to determine (a) the rate law and (b) the rate constant for the reaction

    3. 2A + B C

    Experiment [A]0 [B]0 Rate (M/s)
    1 0.25 0.10 0.012
    2 0.25 0.20 0.048
    3 0.50 0.10 0.024

    rate law:
    rate constant:


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