Therefore, the overwhelming majority of atoms A is in the unbound state; it is clear
that if the pressure is greatly increased, the picture may become the opposite.
One can estimate the time after which d[A]/dt = 0 and d[(A…B)
# ]/dt = 0. It has
been already shown that in steady-state conditions r A = k 4.10 [(A…B)
# ] ss [M]. This
allows to calculate [(A…B)
# ] ss = r A /k 4.10 [M] = 10
6 cm
−3 and [A] ss = 10
3
Á[(A…
B)
# ] ss = 10
9 cm
−3 . This number of atoms can be accumulated within *10
–5 s.
Although this value is not equal to the time of establishment of the steady-state
concentrations, it may indicate the order of time for the establishment of it.
The rate at which steady-state conditions are reached can be judged by the
magnitudes of the coefficients in the rfs of (2.4.11, 2.4.12) for [A] and [(A…B)
# ]. In
the case under discussion, this is k 4.9 [B] = 10
7 s
−1 and k −4.9 = 10
10 s
−1 . If they are
so large, then it can be assumed that at times not exceeding the time of accumulation of the stationary concentration of active intermediate products (in our case A
and (A…B)
# ) the steady-state conditions have have occurred. Of course, in
unobvious cases, it is not bad to solve a system of differential equations, and
calculate the time to establish stationary concentrations. Only then the steady-state
method can be used.
And this method is very, very useful. Indeed, one replaces the system of differential equations, often non-linear, using a system of algebraic equations. Using
this method, one can often easily get the dependences of the rate of a change of the
final product concentrations on the experimental conditions and, from the dependencies observed in the experiment, obtain the values of the rate constants or any of
their combinations (ratios, etc.). One should only remember the expression of one
of the authors of the book [2], Professor V.N. Kondratyev: “Kinetic proof of the
mechanism is the weakest proof!”.
But where the steady-state method is merely irreplaceable, is in understanding
the mechanism of the process. The reader behind these algebraic exercises can
practically understand, in the first approximation, the mechanism of atom and
molecule recombination. The reader will see below how to get the recombination
rate constant using the rate constants of elementary processes and what essential
conclusions can be drawn from this.
Using (2.4.13, 2.4.18), one gets:
r 4:8 ¼
k 4:9 k 4:10 A
½ Á B
½ Á M
½
k À4:9 þ k 4:10 M
½
ð2:4:20Þ
A very simple and elementary equation. Analyze it using the above values of
constants and concentrations. The most simple are the limits of small and large
pressures, which can be formulated as:
k À4:9 ) k 4:10 ½M
ð 2:4:21Þ
and
2.4 Complex Reactions. Consecutive Reactions. Steady-State Method
33
that if the pressure is greatly increased, the picture may become the opposite.
One can estimate the time after which d[A]/dt = 0 and d[(A…B)
# ]/dt = 0. It has
been already shown that in steady-state conditions r A = k 4.10 [(A…B)
# ] ss [M]. This
allows to calculate [(A…B)
# ] ss = r A /k 4.10 [M] = 10
6 cm
−3 and [A] ss = 10
3
Á[(A…
B)
# ] ss = 10
9 cm
−3 . This number of atoms can be accumulated within *10
–5 s.
Although this value is not equal to the time of establishment of the steady-state
concentrations, it may indicate the order of time for the establishment of it.
The rate at which steady-state conditions are reached can be judged by the
magnitudes of the coefficients in the rfs of (2.4.11, 2.4.12) for [A] and [(A…B)
# ]. In
the case under discussion, this is k 4.9 [B] = 10
7 s
−1 and k −4.9 = 10
10 s
−1 . If they are
so large, then it can be assumed that at times not exceeding the time of accumulation of the stationary concentration of active intermediate products (in our case A
and (A…B)
# ) the steady-state conditions have have occurred. Of course, in
unobvious cases, it is not bad to solve a system of differential equations, and
calculate the time to establish stationary concentrations. Only then the steady-state
method can be used.
And this method is very, very useful. Indeed, one replaces the system of differential equations, often non-linear, using a system of algebraic equations. Using
this method, one can often easily get the dependences of the rate of a change of the
final product concentrations on the experimental conditions and, from the dependencies observed in the experiment, obtain the values of the rate constants or any of
their combinations (ratios, etc.). One should only remember the expression of one
of the authors of the book [2], Professor V.N. Kondratyev: “Kinetic proof of the
mechanism is the weakest proof!”.
But where the steady-state method is merely irreplaceable, is in understanding
the mechanism of the process. The reader behind these algebraic exercises can
practically understand, in the first approximation, the mechanism of atom and
molecule recombination. The reader will see below how to get the recombination
rate constant using the rate constants of elementary processes and what essential
conclusions can be drawn from this.
Using (2.4.13, 2.4.18), one gets:
r 4:8 ¼
k 4:9 k 4:10 A
½ Á B
½ Á M
½
k À4:9 þ k 4:10 M
½
ð2:4:20Þ
A very simple and elementary equation. Analyze it using the above values of
constants and concentrations. The most simple are the limits of small and large
pressures, which can be formulated as:
k À4:9 ) k 4:10 ½M
ð 2:4:21Þ
and
2.4 Complex Reactions. Consecutive Reactions. Steady-State Method
33