Physical Chemistry , 1st ed.

The units L/mol cancel with the M unit in the denominator of the right side;

solving for t:

t65,100 s

which is over 18 hours.

b.For the initial amount [A] 0 0.00500 mol/L, if half of the initial amount

reacts, we will have [A]t0.00250 mol/L. Using the same equation and rate

constant, we have



0.00250

1

mol/L



0.00500

1

mol/L

3.07 10 ^4 

M

1

s

t

400.

m

L

ol

200.

m

L

ol

3.07 10 ^4 

M

1

s

t

200.

m

L

ol

3.07 10 ^4 

M

1

s

t

The molarity units cancel, and we solve for the time:

t651,000 s

which is more than seven days.

c.Notice that the time—the half-life—is much higher for the lower initial

amount. This shows that the half-life for a second-order reaction is not a con-

stant of the reaction, but rather depends on the initial amount.

The previous example shows that the term “half-life” can be applied to any

order of reaction, not just first-order reactions. However, only for first-order

reactions is the half-life independent of the initial amount, and a characteris-

tic of the reaction. For any other order of reaction, a half-life canbe defined,

but will always include the initial amount in the expression. For example, for

second-order reactions, the half-life t1/2can be defined as

t1/2

k

1

[A] 0

 (20.22)

This equation shows (as did the previous example) that as the initial amount

[A] 0 gets larger, the amount of time it takes for half of the reactant to react gets

smaller. Such relationships are useful to synthetic and industrial chemists who

are performing chemical processes.

There are a few other simple rate laws, and the integrations of those rate

laws follow the same type of steps we used to find integrated rate laws for first-

and second-order reactions. Rather than repeat such derivations, they will be

left to the student. The discussion will be confined to some more interesting

attributes of other rate laws. For example, a reaction following zeroth-order ki-

netics has a rate law of

d[

d

A

t

]k[A]^0 k (20.23)

That is, the rate of disappearance of A is a constant, the zeroth-order rate con-

stant. These types of reactions are rare, but they do occur: for example, con-

version of ingested ethyl alcohol (CH 3 CH 2 OH) into acetaldehyde (CH 3 CHO)

in the body follows zeroth-order kinetics.

We can consider equation 20.23 in several equivalent ways. Since the rate of

disappearance of reactant A is a constant, a plot of [A]tversus time is a straight

690 CHAPTER 20 Kinetics