Physical Chemistry , 1st ed.



d[

d

A

t

]

k 1 [A]tk 2 [A]t(k 1 k 2 ) [A]t (20.37)

Because equation 20.37 has only one variable (the concentration of A), it can

be integrated just like any other first-order rate law. In doing so (and in mak-

ing the same assumptions that the initial time is set to zero so that the variable

tstands for elapsedtime), we get

[A]t[A] 0 e(k^1 +k^2 )t (20.38)

This relates the concentration of A over time, and should be no great surprise.

The initial rates of appearances of the two products, B and C, are



d[

d

B

t

]

k 1 [A]t



d[

d

C

t

]

k 2 [A]t

Substituting for [A]tfrom equation 20.38, we get



d[

d

B

t

]

k 1 [A] 0 e(k^1 k^2 )t (20.39)



d[

d

C

t

]

k 2 [A] 0 e(k^1 k^2 )t (20.40)

These expressions can be integrated in order to determine the concentrations

of B and C over time. If it is assumed that the initial amounts of B and C are

zero at some initial time (that is, [B]t[C]t0 at t0), these two equa-

tions can be integrated to yield

[B]t

k

k

1

1





[A

k

]

2

^0 (1 e(k^1 k^2 )t) (20.41)

[C]t

k

k

2

1





[A

k

]

2

^0 (1 e(k^1 k^2 )t) (20.42)

Both of these concentrations depend on negative exponentials, but in these

cases the negative exponential is subtracted from 1. Therefore, as time increases

and the negative exponential gets smaller and smaller, the difference gets larger

and larger, and [B] and [C] increase as the elapsed time increases. Figure 20.10

shows the behavior of [A]t,[B]t, and [C]tfor a given set of rate constants.

20.5 Parallel and Consecutive Reactions 697

1.0

0

0

Time (s)

[A]

[C]

[B]

140

Fraction present

10080604020 120

0.8

0.6

0.4

0.2

Figure 20.10 Plots of [A]t, [B]t, and [C]t
versus time for two parallel reactions in which
k 1 0.005 s^1 and k 2 0.015 s^1. The initial
concentrations of B and C are dictated by the rel-
ative magnitudes of the two rate constants. For a
plot over a long period of time in which equilib-
ria are established, see Figure 20.11.