Chemistry Reference
In-Depth Information
0
50
100
150
2
0
0
1.0
1.0
n
A
0.8
0.8
n
C
k
A
= 10 s
-1
0.6
0.6
0.4
0.4
0.2
0.2
k
B
= 100 s
-1
k
B
= 20 s
-1
n
B
0.0
0.0
0
50
100
150
200
Reaction time in ms
FIGURE 3.25
Example of consecutive chemical reactions of first order. Shown is the con-
sumption of the particles A with
k
A
10 s
−
1
, and the formation the final product C via the
intermediate product B for two selected reaction coefficients,
k
B
=
=
100 s
−
1
and
k
B
=
20 s
−
1
.
The rate equations are solved elementary with initial conditions
n
A
(
0
)
and
n
B
(
0
)
=
n
C
(
0
)
=
0.
exp
−
t
.
k
(
1
)
A
n
A
(
t
)
=
n
A
(
0
)
·
·
(3.187)
k
(
1
)
A
k
(
1
)
A
·
exp
−
t
−
exp
−
t
.
k
(
1
)
A
k
(
1
)
B
n
B
(
t
)
=
n
A
(
0
)
·
·
·
(3.188)
k
(
1
)
B
−
1
t
.
k
(
1
)
B
·
k
(
1
)
exp
−
t
−
k
(
1
A
exp
−
1
k
(
1
)
A
k
(
1
)
B
n
C
(
t
)
=
n
A
(
0
)
·
+
·
·
·
B
k
(
1
)
A
−
(3.189)
3.
Second-order chemical reaction:
The product formation depends on two
reactant concentrations.
dn
A
dt
=
dn
B
dt
=−
k
(
2
)
AB
−→
k
(
2
)
AB
A
+
B
AB
·
n
A
·
n
B
.
(3.190)
exp
[
n
A
(
t
k
(
2
)
AB
(
n
A
(
0
)
−
n
B
(
0
))
·
0
)
−
n
B
(
0
)
]
·
·
n
A
(
t
)
=
n
A
(
0
)
·
n
A
(
exp
(
t
−
)
. (3.191)
k
(
2
)
AB
0
)
·
n
A
(
0
)
−
n
B
(
0
))
·
·
n
B
(
0
In the case of the reaction between identical particles it follows:
dn
A
dt
=−
k
(
2
)
AA
−→
k
(
2
)
AA
A
+
A
A
2
2
·
·
n
A
.
(3.192)
