Physical Chemistry , 1st ed.

Since 0.25 is 4 ^1 and 0.50 is ^12 , it is seen by inspection that the exponent nequals

2. (You should verify this.) Using the first and third sets of experimental data,
   we can construct a similar comparison:

^5

2

.

.

3

6

7

9

1

1

0

0





3

3

M

M

/

/

s

s



Units and k’s cancel to give

2 (0.50)m

Since 0.50 is equal to ^12 , we can see that m1. (You should verify this also.)

Finally, to determine the value of the rate constant k, we can substitute any

set of data into the rate law. It should not matter which data set is used for

this. Using the first set:

5.37 10 ^3 M

s

k(1.44 M)^1 (0.70 M)^2

Solving:

5.37 10 ^3 

M

s

k0.340 M

k1.59 10 ^2 s^1

The units on kare appropriate to give a unit of M/s on the rate. Verify that

the second and third sets of experimental data give a similar value for k.

Students should be aware that in real cases the numerical treatment does
not work out as perfectly as it did in the above example. Some level of math-
ematical sophistication, and the application of logarithms, may be necessary to
determine the orders in the rate law. Example 20.6 shows one such case where
use of logarithms is necessary to determine the order of the reaction.

20.3 Characteristics of Specific Initial Rate Laws

There are several simple forms of rate laws, so it is common to discuss them
and their particular characteristics. Although many rate laws depend on the
amounts of more than one component, we will focus on rate laws of the type

rate k[A]n (20.8)

in this section. What is the behavior of the rate when the order with respect to
A has various common whole number values?
To apply calculus to equation 20.8, we need to rewrite the quantity “rate” in
terms of a change in amount over some period of time. We do this by using the
forms in equation 20.4. If we are focusing on a reactant species A, the rate is

rate 

d[

d

A

t

]

 (20.9)

where the [A] term can be an amount in moles or in concentration units. We
will presume a concentration unit—molarity—at this point. The general equa-
tion whose properties we hope to understand is



d[

d

A

t

]

k[A]n (20.10)

k(1.44 M)m(0.35 M)n

k(2.89 M)m(0.35 M)n

20.3 Characteristics of Specific Initial Rate Laws 685