11.2 Forward Reactions with One Reactant 493
b.Find the half-life of the reaction of the previous example if the initial concentration is equal
to 0.1000 mol L−^1.If there is considerable experimental error it might be difficult to tell a first-order
reaction from a second-order reaction by inspection of a graph of the concentration ver-
sus time if the graph extends over only one half-life. Figure 11.2 shows two hypothetical
reactions with the same initial concentration: Substance A undergoes a first-order reac-
tion and substance B undergoes a second-order reaction with the same half-life. The
concentrations in the two cases differ only slightly for times up tot 1 / 2. For longer times
the difference is greater.[A] (first order)[B]
(second order)c(0)c(0)
c(0)0
t1/2 2 t1/2 3 t1/2Concentration2
4tFigure 11.2 Comparison of the
Concentrations of the Reactants
of a First-Order Reaction and a
Second-Order Reaction.Exercise 11.7
Find expressions for the time required for the concentration of the reactant in each of the reactions
of Figure 11.2 to drop to one-sixteenth of its original value, assuming that the reverse reaction
is negligible. Express these times in terms oft 1 / 2.nth-Order Reactions
The rate law for annth order reaction with a single reactant and negligible reverse
reaction isr−d[A]
dtkf[A]n (11.2-14)Consider the case that the ordernis not necessarily an integer but is not equal to unity
or zero. The variables can be separated by division by[A]nand multiplication bydt,
giving−
d[A]
[A]nkfdt (11.2-15)Assuming constant temperature, we perform a definite integration fromt0tott′:
∫[A]t′[A] 0d[A]
[A]n−kf∫t′0dt (11.2-16)The result is:1
n− 1[
1
[A]nt−^1−
1
[A]n 0 −^1]
kft (nth order, no reverse reaction) (11.2-17)where we omit the prime symbol ont. The half-life is found by substituting[A]t 1 / 2
[A] 0 /2 into Eq. (11.2-17). The result ist 1 / 2 2 n−^1 − 1
(n−1)kf[A]n 0 −^1(nth order, no reverse reaction) (11.2-18)These formulas are not valid ifn1orn0.