Physical Chemistry , 1st ed.

equation 20.7 is written,mis called the order with respect to A,and nis the or-

der with respect to B,and so on, if other terms exist. Orders are usually small

positive whole numbers, but they may be negative whole numbers, zero, or

even fractions. The sum of all orders of the algebraic expression is the overall

orderof the reaction.

The complete expression in equation 20.7 is called a rate law.Rate laws must

be determined experimentally.Sometimes they are simple; sometimes they can

be very complicated. But in all cases, they must be determined by setting up a

chemical reaction under different conditions of initial concentrations, mea-

suring the initial rates by some experimental method, and algebraically de-

ducing the individual orders and the numerical value of the rate constant k.

The rate constant kshould also have units that will give a proper unit for the

overall rate (which is usually mol/s or M/s). Although problems involving rate

laws may already be familiar to you, the following example illustrates the math-

ematical tactic for determining a simple rate law from experimental data.

Example 20.2

For a general reaction “aA bB →products,” the following initial rates are

determined experimentally when reactions are set up with the initial amounts

indicated in units of molarity, M.

[A] (M) [B] (M) Initial rate (M/s)

1.44 0.35 5.37 10 ^3

1.44 0.70 2.15 10 ^2

2.89 0.35 2.69 10 ^3

Assuming that the rate law can be written as

rate k[A]m[B]n

determine the values ofm,n, and k.

Solution

The tactic in determining orders is to set up two rate law expressions using

two different sets of numbers, and then divide one expression by the other.

Select the sets so that the concentration of one of the species cancels (as will

k). Then, the order with respect to the other species can be determined by in-

spection or by taking the logarithm of the resulting equation.

Using the first and second sets of data, we can substitute into the general

form of the rate law to get two equations:

5.37 10 ^3 M

s

k(1.44 M)m(0.35 M)n

and

2.15 10 ^2 M

s

k(1.44 M)m(0.70 M)n

Dividing the first equation by the second, we have

^5

2

.

.

3

1

7

5

1

1

0

0





3

2

M

M

/

/

s

s



The units on the left cancel, as do the k’s on the right. Notice that the (1.44 M)m

also cancels on the right side no matter what value m is .We get

0.25 (0.50)n

k(1.44 M)m(0.35 M)n



k(1.44 M)m(0.70 M)n

684 CHAPTER 20 Kinetics