10.11 Kinetic Theories for Chemical Reactions
509
For simplicity in this example, we will take the initial concentration of B to match
that of A. The first two equations for the change of A and B with time insures that
A remains equal to B. Similarly, the concentration of D will match that of C. We
also have dC/dt = −dA/dt, so that C = A o − A. In the following, we will take
κ > λ.
With these assumptions, the solution for A, B, C, and D can be found from the
solution of
dA
dt
= −κA
2
+ λ(A o − A)
2 .
(10.284)
This equation can be expressed in terms of a known (logarithmic) integral:
t =
A
A o
dx
−κx 2 + λ(A o − x) 2 .
(10.285)
Solving for A gives
A = A o
√
λκ
1 + e −2A o
√
λκt
κ +
√
λκ
−
κ −
√
λκ
e −2A o
√
λκt
.
(10.286)
A plot of the concentrations of A and C as a function of time is given in Fig. 10.6.
After a time much longer than 1/(A o
√
λκ), the concentration of A molecules
reaches
A(∞) = A o
1
1 +
√
κ/λ
,
(10.287)
as does the concentration of B molecules, while C and D follow C(∞) = A o −
A(∞)
C(∞) = A o
√
κ/λ
1 +
√
κ/λ
.
(10.288)
The equilibrium constant for the reaction is [C][D]/([A][B]) = κ/λ.
In biological reactions, evolution has optimized the production of useful products, so that many biochemical reactions can be taken as ‘one directional’, proceeding to completion of the products. In these cases, the kinematical differential
equations for the reactions become far easier to solve. For the reaction A + B
C + D treated above, the reverse reaction will be negligible as λ/κ → 0. In this
case,
A(t) = A o
1
1 + A o κt
→ 0