Chapter 2 Introduction to Kinetic Processes in Materials 41 Chapter 3 Second-Order and Multistep Reactions 61 Chapter 4 Temperature Dependence of the Reaction Rate Constant 95
3.5 PARALLEL AND SERIES REACTIONS .1 i ntROductiOn
for both the exact and a spreadsheet numerical solutions to Equation 3.32 with Δt = 0.5. The two solu- tions are pretty close and get closer at longer times because, in this case, the derivative is getting smaller. Clearly, the numerical solution is the easier and faster way to solve this relatively simple problem and numerical solutions become mandatory as the kinetics become more complex, as indeed they do, even for simple reactions.
3.5 PARALLEL AND SERIES REACTIONS
and therefore
A A e k k t
[ ]
=[ ]
0 −(1+ 2). (3.35)Substitution of Equation 3.35 in the rate expression for B gives d B
dt k A k A e k k t
[ ]
= 1[ ]
= 1[ ]
0 −(1+ 2)and integrating
∫
0[ ]Bd B[ ]
=k A1[ ]
0∫
0te−(k1+k t2)dt to giveB k
k k A e k k t
[ ]
= 1+1 2[ ]
0{
1− −(1+ 2)}
. (3.36)Similarly, d C
dt
[ ]
=k A2[ ]
=k A e2[ ]
0 −(k1+k t2) and is integrated to giveC k
k k A e k k t
[ ]
= 1+2 2[ ]
0{
1− −( 1+ 2)}
(3.37)which clearly shows that the faster reaction—the larger k—controls the rate of reaction and determines which product is preferentially created, as shown in Figure 3.8. This is a very general result: for paral- lel reactions the faster (fastest if more than two products) controls the overall rate of the reaction. For the k-values used to plot Figure 3.8, [C] is always about 20% of [B].
0 20 40 60 80 100
0.0 0.2 0.4 0.6 0.8 1.0 [A]0
k1= 0.02 k2 = 0.005
Time (s) [A]
[B]
[C]
Concentration
FIGURE 3.8 Concentrations as a function of time for parallel reactions that go to completion: A → B and A → C. For parallel reactions, the faster (fastest) of the rates, k1, determines the rate of the overall reaction and the relative amount of the major product, B.
3.5.3 s
eRiesR
eactiOns3.5.3.1 Exact Solution
Consider a series reaction A →k1 B →k2 C in which the slower of the two reactions—the smaller of k1 and k2—determines the overall rate, or [C](t). The rates of reactions are given by
d A
dt k A d B
dt k A k B d C
dt k B
[ ]
= −[ ] [ ]
=[ ]
−[ ] [ ]
=[ ]
1
1 2
2 .
(3.38)
The solutions to these three simultaneous ordinary differential equations are given in many physical chemistry texts without the mathematical details that require some techniques for solving differen- tial equations (but given in Appendix A.3):
A A e k t
[ ]
=[ ]
0 −1 (3.39)B A k
k k e k t e k t
[ ]
=[ ]
−0 1
(
− − −)
2 1
1 2 (3.40)
C A
k k k e k t k e k t
[ ]
=[ ]
+ −(
−)
− −
0 2 1
1 2
1 1 2 1 . (3.41)
Figure 3.9 shows a case where k1k2 so A quickly forms B, which reacts more slowly to form C.
Figure 3.10 shows the opposite case where k2 ≫ k1 so that hardly any B is present at any time because it quickly reacts to form C. In Appendix A.3.5, it is shown that it is the smaller of the two “ks”—the slower reaction—controls the overall reaction rate. Note that [C](t) does not depend on which of the two rates is the faster, which is obvious from Equation 3.41 because the reversal of k1 and k2 gives the same rate equation for [C]. For example, with k1 = 0.05 and k2 = 0.3 as in Figure 3.10, Equation 3.41 becomes
0 20 40 60 80 100
0.0 0.2 0.4 0.6 0.8 1.0 [A]0
Time (s) [A]
[C]
[B]
Concentration
k2= 0.05 s−1 k1= 0.3 s−1
FIGURE 3.9 Series reaction in which the reaction of A to B, k1, is considerably faster than the reaction of B to C, k2. Because the rate of reaction of A to B is fast, A quickly reacts to form B and it is the slow reaction of B going to C that controls the reaction. As a result, the slower controls the rate of the overall reaction to C.
C A e t e t A
[ ]
=[ ]
0 + −(
− − −)
=[ ]
+0 3 0 05
1 1 0
0 3 0 05 0 05 0 3 1 4 0
. . . . . .
( (
.005e−0 3. t−0 3. e−0 05. t) )
.Reversing k1 and k2 so that k1 = 0.3 and k2 = 0.05, the equation becomes
C A e t e t A
[ ]
=[ ]
0 + −(
− − −)
=[ ]
+0 05 0 3
1 1 0
0 05 0 3 0 3 0 05 1 4 0
. . . . . .
( (
.005e−0 3.t−0 3. e−0 05. t) )
which—not too surprisingly—is the same as the previous equation and [C](t) is the same in both Figures 3.9 and 3.10. However, [B](t) depends strongly on which is the faster reaction.
In the limit of k2 ≫ k1, Equations 3.31 and 3.32 become respectively:
B A k
k e k
k A
[ ]
≅[ ]
0 1 −k t=[ ]
2
1 2
1 (3.42)
C A e k t
[ ]
≅[ ]
0(
1− − 1)
. (3.43)These results clearly show that the rate of reaction that determines the concentrations of A, B, and C depend on the smaller reaction rate constant, k1: that is, for a series reaction, the slower (slowest) reaction controls the rate of the overall reaction.
3.5.3.2 Steady-State Assumption
An approach to get an approximate solution to these series reaction equations, Equation 3.38, is to make the steady-state assumption for the concentration of B. It is extremely important to note that steady-state and equilibrium have two very different meanings in reactions even though the terms are often used—incorrectly—interchangeably. The term equilibrium means that the reaction has stopped. In contrast, steady-state implies that concentrations are no longer changing with time but the reaction is still taking place and may be a long way from equilibrium.
The steady-state assumption is particularly useful for analyzing complex reactions having mul- tiple series and parallel steps (Laidler 1987). That is,
0 20 40 60 80 100
0.0 0.2 0.4 0.6 0.8 1.0 [A]0
k1= 0.05 s−1 k2= 0.3 s−1
Time (s) [A]
[C]
[B]
Concentration
FIGURE 3.10 Same series reaction as in Figure 3.9 only the reaction rates are reversed. Now B converts quickly to C and it is the slower rate of A reacting to B that controls the overall rate of reaction: the formation of C. Note also that the concentration of C, [C], as a function of time in both this figure and Figure 3.9 are the same because the slower reaction controls the overall rate, the concentration of C as a function of time.
d B
dt k A k B
[ ]
= 1[ ]
− 2[ ]
=0 (3.44)this leads immediately to Equations 3.42 and 3.43 and is, of course, tantamount to the assump- tion that k2 ≫ k1, for which, as Equation 3.42 shows, [B] does not vary rapidly with time because it depends on e−k t1 and k1 is small. Of course, Equations 3.42 and 3.44 are mathematically inconsistent, but the result is a reasonable approximation if k2 ≫ k1.
3.5.3.3 Example: Nuclear Decay
In a nuclear reactor, the roughly 23592U enriched to 4% or 5% of the uranium present undergoes fission generating energy and several neutrons. Some of these neutrons are captured by the 96% 23892U to give
23992U, which decays to 23993Np and then to 23994Pu by the following reactions:
23992 23 5 23993
23993 2 35 23994
1 28
U Np MeV
Np d P
. min
.
→ + + .
→
β−
uu MeV
Pu yr U He MeV
+ +
→ + +
β− 0 23 5 23
23994 24 000 23592
24
.
. .
,
(3.45)
If a reactor is operating at a constant power level for a long period of time, typically several years, then the amount of 23992U will stay at some small time-independent value as will the amount of 23993Np because of their relatively short half-lives. On the other hand, the amount of 23994Pu in the reactor core will continually increase. However, it too will reach a steady-state concentration because 23994Pu is also fissionable and can be used as fuel in nuclear power reactors, as it is in France. Nevertheless, it would be expected that the relative steady-state concentrations of 23992U and23993Np, which from Equations 3.44 and 3.45, would be 23993Np / 23992U =2 35 23 5. d/ . min=720.
Of course, the concentration in the atmosphere of radioactive 146C used for radiocarbon dating is a perfect example of a steady-state concentration between its formation by cosmic rays and its own nuclear decay.