Physical Chemistry , 1st ed.

Example 20.7

Kinetics of consecutive reactions are easily applicable to nuclear decay

processes, in which a parent isotope produces a radioactive daughter isotope

that also decays. (In fact, in the early twentieth century, such sequential

processes were a major complicating factor in trying to understand this new

phenomenon.) One such example is

210 t1/2,1 t1/2,2

83 Bi →^21084 Po →^20682 Pb

which are the last two steps in the radioactive decay series starting with^23892 U

and ending in the nonradioactive isotope of Pb. (It is sometimes called the

4n 2 seriesbecause all of the mass numbers of the isotopes involved can be

represented by that general equation.) The half-lives,t1/2,1and t1/2,2, are 5.01

days and 138.4 days, respectively. Comment on the relative amounts of^210 Bi,

(^210) Po, and (^206) Pb over time.
Solution
The presentation of this example is potentially misleading because the prob-
lem gives half-lives, not rate constants! Equation 20.17 should be consulted
for the relationship between t1/2and k:

t1/2

0.6

k

93



Using this, and keeping track of two different half-lives and their related rate

constants using subscripts, we get

t1/2,15.01 d 4.33 105 s

t1/2,2138.4 d 1.196 107 s

(where the half-lives have been converted into standard units). Therefore

k 1 

4.33

0.

69

1

3

05 s

1.60 10 ^6 s^1

k 2 

1.196

0.6

93

107 s

5.79 10 ^8 s^1

20.5 Parallel and Consecutive Reactions 701

1.0

0

0

Time (s)

[A]

[C]

[B]

2500

Fraction present

0.8

0.6

0.4

0.2

500 1000 1500 2000

Figure 20.13 For consecutive reactions in which k 1 is much smaller than k 2 , there is very
little initial buildup of the intermediate product, B. The final product, C, is formed almost im-
mediately. In this plot,k 1 0.0005 s ^1 and k 2 0.01 s^1. Compare this to Figure 20.12.