We can combine the two constants on the right side of the equation into a sin-
gle variable, labeled k :
rate k [A]m
The constant k is called the pseudo rate constantfor the rate law. Under these
experimental conditions, it is much easier to determine the rate law and the
rate constant for this simpler rate expression.
Probably the most common usage of this tactic is where the order mequals
1, so that the rate of the reaction follows the approximate expression
rate k [A] (20.28)
Reactions that are intentionally set up this way are said to be following pseudo
first-order kinetics,and the constant k is the pseudo first-order rate constant.
Again, in reality a reaction might have some other, more specific rate law. Only
under special conditions [that is, with a high, almost unchanging concentration
of the other reactant(s)] will the reaction show pseudo first-order kinetics.
Note too that the pseudo first-order rate constant k has different units than
the true rate constant kof the same reaction. The units on k are always 1/s,
whereas the units on kdepend on the entire rate law. The specific numerical
value ofk depends on how high the concentration of the excess reactant is, as
the following example shows.Example 20.6
The following data are collected for a chemical reaction at constant temper-
ature, arbitrarily A B →products.[A] (M) [B] (M) Initial rate ( 10 ^7 M/s)
0.00636 0.00384 2.91
0.0108 0.00384 4.95
0.00636 0.00500 4.95a.Determine the rate law and the value of the rate law constant k.
b.Estimate the value ofk , the pseudo first-order rate law constant, if [B]
0.500 M and all other conditions are the same.Solution
a.By performing an analysis like the one shown in Example 20.2, you should
be able to see that the rate law is
rate k[A]^1 [B]^2
To get the order with respect to B, you have to use logarithms because the
concentrations and rates are not in exact (or even close!) whole-number ra-
tios. When the first and second trials are used, one gets^4
2.
.
9
9
5
1
1
1
0
0
7
7M
M
/
/
s
s(
(0
0
.
.
0
0
0
0
5
3
0
8
0
4
M
M
)
)
n
n1.701(1.302)n
In order to determine n, take the logarithm of both sides:
log (1.701) nlog (1.302)
0.2307n(0.1146)n
0
0
.
.
2
1
3
1
0
4
7
6
2.01 2
692 CHAPTER 20 Kinetics