The Foundations of Chemistry

Figure 16-4 (a) Plot of concentration versus time for a first-order reaction. During the
first half-life, 1.73 seconds, the concentration of A falls from 1.00 Mto 0.50 M.An
additional 1.73 seconds is required for the concentration to fall by half again, from 0.50 M

to 0.25 M,and so on. For a first-order reaction, t1/2; t1/2does not depend

on the concentration at the beginning of that time period. (b) Plot of concentration versus
time for a second-order reaction. The same values are used for a,[A] 0 , and kas in part (a).
During the first half-life, 2.50 seconds, the concentration of A falls from 1.00 Mto 0.50 M.
The concentration falls by half again from 2.50 to 7.50 seconds, so the second half-life is
5.00 seconds. The half-life beginning at 0.25 Mis 10.00 seconds. For a second-order

reaction, t1/2 ; t1/2is inversely proportional to the concentration at the beginning

of that time period.

1



ak[A] 0

0.693



ak

ln 2



ak

For tt1/2, we have [A]^12 [A] 0 , so

akt1/2

Simplifying and solving for t1/2, we obtain the relationship between the rate constant and
t1/2.

t1/2


In this case t1/2depends on the initial concentration ofA. Figure 16-4 illustrates the different
behavior of half-life for first- and second-order reactions.

EXAMPLE 16-7 Half-Life: Second-Order Reaction

Compounds A and B react to form C and D in a reaction that was found to be second order
in A and second order overall. The rate constant at 30°C is 0.622 liter per mole per minute.
What is the half-life of A when 4.10 10 ^2 MA is mixed with excess B?

A
B88nC
D ratek[A]^2

second order in A,

second order overall

1



ak[A] 0

1



[A] 0

1



^12 [A] 0

[A] (

M

)

1.0

0.8

0.6

0.4

0.2

0510

0

t (seconds)

(a)

[A] (

M

)

1.0

0.8

0.6

0.4

0.2

0510

0

t (seconds)

(b)

15 20

First-order reaction

a  1; [A] 0  1.0 M

k  0.40 s^1

Second-order reaction

a  1; [A] 0  1.0 M

k  0.40 M^1 ^ s^1

You should carry out the algebraic

steps to solve for t1/2.