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Statistical
Thermodynamics and
Chemical Kinetics
Lecture 9
May 12, 2003
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Chapter 9 Complex Reactions
9.1 Exact analytic solutions for complex reactions
9.1.1 Introduction
Most chemical processes are complex, i.e. they consist
of a number of coupled elementary reactions. These
complex reactions can be divided into several classes: (1)
opposing or reversible reactions, (2) consecutive
reactions, (3) parallel reactions, and (4) mixed reactions.
In this chapter, we examine methods for determining exact
analytic solutions for the time dependence of
concentrations of species involved in complex reactions.
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9.1.2 Reversible reactions
In reversible reactions or opposing reactions, the
products of the initial reaction can proceed to re-form the
original substances. A chemical example of this is the
cis-trans isomerization of 1,2-dichloroethylene:
Cl
(9-1)
Cl
Cl
Cl
As such, the simplest reversible reaction is of the form
kf
A2
(9-2)
A1
kr
and is first order in each direction. The differential
equation for this mechanism is
d [ A1 ]
(9-3)

 k f [ A1 ]  k r [ A2 ]
dt
d [ A2 ]

 k r [ A2 ]  k f [ A1 ]
dt
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(9-4)
If it is assumed that both A1 and A2 are present in the
system at time t=0, that is, [A1]=[A1]0 and [A2]=[A2]0,
then at any time afterwords the total amount of reactant
remaining and new product formed must equal the initial
amount of the reactants before reactions. Hence,
[A1]0 + [A2]0 = [A1] + [A2]
(9-5)
Solving for [A2], we obtain
[A2] = [A1]0 + [A2]0 - [A1]
(9-6)
and substituting this into equation (9-3) yields
d [ A1 ]
 k f [ A1 ]  k r ([ A1 ]0  [ A2 ]0  [ A1 ])
dt
d [ A1 ]

 k r ([ A1 ]0  [ A2 ]0 )  (k f  k r )[ A1 ]
dt
  k r ([ A1 ]0  [ A2 ]0 )

d [ A1 ]

 (k f  k r )
 [ A1 ]
State dt
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

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(9-7)
To find the solution of equation (9-7) we introduce a
variable m, defined as
 kr ([ A1 ]0  [ A2 ]0 )
m 
(k f  k r )
(9-8)
This allows us to rewrite equation (9-7) as
d [ A1 ]

 (k f  k r ) m  [ A1 ]
(9-9)
dt
which we may then integrate
d [ A1 ]

 m  [ A1 ]  (k f  kr ) dt (9-10)
The solution is
 k f [ A1 ]  k r [ A2 ] 
  (k  k )(t  t ) (9-11)
ln 
 k [A ]  k [A ] 
 f 1 0 r 2 0
f
t
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0
If only [A1] is present in the system initially, at t=0, then
the solution reduces to
 k f [ A1 ]  k r [ A2 ] 
  (k f  kt )t
ln 
(9-12)


k
[
A
]
f
1
0


which is just
[ A1 ]0
( k  k ) t
[ A1 ] 
[k r  k f e
]
(9-13)
( k f  kt )
Using the mass conservation constraint [A1]0=[A1] + [A2] ,
we can obtain the solution for [A2] :
f
[ A2 ] 
k f [ A1 ]0
(k f  kt )
[1  e
 ( k f  kt ) t
]
t
(9-14)
When equilibrium is reached, the individual reactions
must be balanced; in other words, the reaction AB must
occur just as frequently as the reverse reaction. The
forward and reverse reactions occur at the same rate.
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Consequently, for reaction (9-2) at equilibrium,
d [ A1 ]
d [ A2 ]

0
dt
dt
(9-15)
 k f [ A1 ]e  k r [ A2 ]e  0
and
(9-16)
and we have the following definition of the equilibrium
constant Keq expressed in terms of rate constants:
kf
[ A2 ]e

 K eq
k r [ A1 ]e
(9-17)
The same argument can be extended to a reversible
reaction that occurs in multiple stages.
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9.1.2.1 First-order reversible reactions
involving two steps
Reversible reactions may be distinguished by number
of stages and the number of initial reactants involved in
the reaction. Here, we consider the complete derivation
for first order reversible reactions involving only two
stages, that is,
k1
k2
A2
A3
A1
k-1
k-2
(9-18)
The kinetic equations for the system are
d [ A1 ]
(9-19)
 k 1[ A2 ]  k1[ A1 ]
dt
d [ A2 ]
 k1[ A1 ]  k 1[ A2 ]  k 2 [ A2 ]  k  2 [ A3 ]
dt
d [ A3 ]
 k 2 [ A2 ]  k 2 [ A3 ]
State
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(9-20)
(9-21)
We assume that at t=0, [A1]=[A1]0, and [A2]0=[A3]0 =0
and that the amounts of A1, A2 and A3 which have
reacted at a later time satisfy the equation:
[A1]0=[A1] + [A2] + [A3]
(9-22)
By the principle of detailed balance, we have
k1 [ A2 ]e
k 2 [ A3 ]e
and
(9-23,24)


k 1
[ A1 ]e
k2
[ A2 ]e
Using equations (9-22)-(9-24) gives
So
 k1 k2

[ A1 ]0  

 1[ A2 ]e
 k1 k2 
k1k  2
[ A2 ]e 
[ A1 ]0
k 1k  2  k1k 2  k1k  2
(9-25)
(9-26)
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Substituting equation (9-26) into equation (9-19) we
obtain the new first-order differential equaion:
(9-27)
d [ A1 ]
k1k 1k 2
dt

k 1k  2  k1k 2  k1k  2
[ A1 ]0  k1[ A1 ]
Solving this euation using standard methods, we have
(9-28)
[ A1 ]0
k t
[ A1 ] 
k 1k  2  k1k 2  k1k  2
[k 1k  2  (k1k 2  k1k  2 )e
1
]
Substituting equation (9-22),(9-26) and (9-28) into this
expression, we have, upon simplification,
[ A1 ]0
[ A3 ] 
[k1k 2  (k1k 2  k1k  2 )e  k1t ]
k 1k  2  k1k 2  k1k  2
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(9-29)
As t,
k 1k  2 [ A1 ]0
k 1k 2  k1k 2  k1k  2
k1k  2 [ A1 ]0
[ A2 ] 
k 1k  2  k1k 2  k1k  2
k1k 2 [ A1 ]0
[ A3 ] 
k 1k  2  k1k 2  k1k  2
[ A1 ] 
(9-30)
(9-31)
(9-32)
and the system is then in a state of equilibrium.
The misuse of the principle arises when the intermediate A2 is difficult to detect, so that an experimentalist
might think that A1A3 is an elementary reaction. Thus
the equilibrium constant for [A3]e/[A1]e might be taken to
be k1/k-2. However, since A1A3 is not an elementary
reaction, this conclusion is incorrect.
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To determine K = [A3]e/[A1]e correctly, the equilibrium of
each elementary reaction must be considered, that is,
equations (9-23) and (9-24) must be used. The correct
expression of K can be found:
[A ]
kk
(9-33)
K 3 e  1 2
[ A1 ]e
k 1k  2
Another example of a first-order reversible reaction
involving only two stages is the cyclic reaction
A1
A2
A3
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9.1.2.2 First and second order
reversible reactions.
A reversible reaction may be of mixed order, such as,
k1
A1
k-1
A2
+
A3
An example is N2O4  2NO2. The rate expression for
this type of reaction is
d [ A1 ]
(9-34)
 k 1[ A2 ][ A3 ]  k1[ A1 ]
dt
To find the solution, we introduce a progress variable,
x = [A1]0 - [A1]
(9-35)
Equation (9-34) thus becomes
dx
 k1 ([ A1 ]0  x)  k 1 ([ A2 ]0  x)([ A3 ]0  x)
dt
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dx
 k1[ A1 ]0  k 1[ A2 ]0 [ A3 ]0  (k1  k 1[ A3 ]0  k 1[ A2 ]0 ) x  k 1 x 2
dt
Or
We let
  k1[ A1 ]0  k1[ A2 ]0 [ A3 ]0
  (k1  k1[ A3 ]0  k1[ A2 ]0 )
  k1
and
dx
So that
   x  x 2
dt
Then
dx
   x  x
2
  dt
(9-36)
(9-37)
(9-38)
(9-39)
The solution of equation(9-39) is
 t0  (   q1/ 2 ) / 2 
 t  (   q1/ 2 ) / 2 
1/ 2
ln 

ln

q
(t  t0 )  



1/ 2
1/ 2
 t  (   q ) / 2 
 t0  (   q ) / 2 
(9-40)
2
1/ 2
1/ 2
q



4

;


ln((


q
)
/(


q
))
Where
t  0;
 t  (   q1/ 2 ) / 2 
ln 
  q t 
1/ 2
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 t  (   q ) / 2 
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(9 - 41)
9.1.3 Consecutive Reactions
• Irreversible reactions can be defined as those
which start with an initial reactant and produce
products or intermediates generally in only one
direction.
• Consecutive reactions are sequential
irreversible reactions.
• There are two classes of consecutive reactions:
Those which are first order and those which are
mixed first order and second order.
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9.1.3.1 First-order consecutive reactions.
9.1.3.1.1 First-order with two steps.
Consider the consecutive first-order reaction
involving two stages: A1 k1 A2 k2 A3
(9-42)
This mechanism can be described by the following set of
rate expressions:
(9-43)
d [ A1 ]
 k1[ A1 ]
dt
d [ A2 ]
 k1[ A1 ]  k 2 [ A2 ]
dt
d [ A3 ]
 k 2 [ A2 ]
dt
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(9-44)
(9-45)
The concentration of A1 is obtained after integration as
[ A1 ]  [ A1 ]0 e  k t
(9-46)
The concentration of A2 is computed from equation (9-44)
d [ A2 ]
 k 2 [ A2 ]  k1[ A1 ]0 e  k t
(9-47)
dt
Solving this equation, we obtain the time dependence of
[A2]:
k1[ A1 ]0  k t
k t
[ A2 ]  [ A2 ]0 e 
(e  e  k t )
k 2  k1
(9-48)
k1[ A1 ]0  k1t
If [A2]0=0 at t=0, then
[ A2 ] 
(e  e  k 2 t )
k 2  k1
(9-49)
1
1
2
1
2
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The concentration of A3 can be determined from
conservation of mass: [A1]0 = [A1] + [A2] + [A3] (9-50)
Hence we have
k1
k t
[ A3 ]  [ A1 ]0 [1  e 
(e  k t  e  k t )]
k 2  k1
(9-51)
Simplifying this expression gives
k2
k1
k t
(9-52)
[ A3 ]  [ A1 ]0 [1 
e 
e k t ]
k 2  k1
k 2  k1
1
1
1
2
2
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9.1.3.1.2 First-order with three steps.
The system of differential equations for a first-order
consecutive reaction involving three steps, viz.:
k1
k2
k3
(9-53)
A1
A2
A3
A4
can be integrated in a way similar to the two-step case.
The differential equations for A1 and A2 and their
solutions do not differ from those obtained in the two-step
case, while the differential equation for A3 is
d [ A3 ]
(9-54)
 k 2 [ A2 ]  k3 [ A3 ]
dt
Substituting equation (9-49) for [A2] into equation (9-54)
gives
 k1k2 [ A1 ]0  k t k t
d[ A3 ]
 k3[ A3 ]  
(9-55)
(e  e )
dt
 k2  k1 
1
2
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Integrating this linear equation subject to the initial
condition that [A3] =0 at t=0, we obtain

  k1t 
  k 2t
k1k 2 [ A1 ]0
k1k 2 [ A1 ]0
[ A3 ]  
e  
e
 (k 2  k1 )( k3  k1 ) 
 (k1  k 2 )( k3  k 2 ) 

  k 3t
k1k 2 [ A1 ]0

e
 (k1  k3 )( k 2  k3 ) 
(9-56)
Then


k2 k3e  k1t
k1k2e  k2t
k1k2e  k3t
[ A4 ]  [ A1 ]0 



(
k

k
)(
k

k
)
(
k

k
)(
k

k
)
(
k

k
)(
k

k
)
1
2
3
2
1
3
2
3 
 2 1 3 1
(9-57)
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9.1.3. 2 Higher order consecutive reactions.
Some higher order consecutive reactions may be first
order in one step and second order the in second, such as
k1
k2
(9-58)
A1
A2
A1 + A2
A3
Or both steps may be second order consecutive reactions,
as in
k1
k2
A1 + A2
A3
A1 + A3
A4
(9-59)
The differential equations for most of these systems of
reactions are nonlinear and generally they have no exact
solutions. Nevertheless, analytic solutions for such
systems can be obtained if time is eliminated as a variable.
To illustrate this method, consider the reaction sequence
described by the system of equations (2-58):
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d [ A1 ]
 k1[ A1 ]  k 2 [ A1 ][ A2 ]
dt
d [ A2 ]
 k1[ A1 ]  k 2 [ A1 ][ A2 ]
dt
d [ A3 ]
 k 2 [ A1 ][ A2 ]
dt
(9-60)
(9-61)
(9-62)
To solve this system of equations we divide equation (961) by (9-60)
(9-63)
 k1  k2 [ A2 ] 
d [ A2 ] k1[ A1 ]  k2 [ A1 ][ A2 ]
d [ A1 ]


 
 k1[ A1 ]  k2 [ A1 ][ A2 ]
 k1  k2 [ A2 ] 
If we let K=k1/k2, then
d [ A2 ]
K  [ A2 ]

d [ A1 ]
K  [ A2 ]
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(9-64)
Equation (9-64) can be rearranged to
( K  [ A2 ]) d [ A2 ] 2 Kd[ A2 ]


 d [ A1 ]
K  [ A2 ]
K  [ A2 ]
2 Kd [ A2 ]
 d [ A2 ] 
 d [ A1 ]
K  [ A2 ]
(9-65)
(9-66)
Integrating equation(9-66) over the respective limits, i.e.,
[A ]
[ A ] d ( K  [ A ])
[A ]
(9-67)
2

d[ A ]  2K

d[ A ]

2
[ A2 ] 0
2

2
[ A2 ] 0
K  [ A2 ]

1
[ A1 ]0
gives A

 [ A2 ]   [ A1 ]0  [ A1 ]
2
 2 ln 1 
 1


K
K   K  [ A1 ]0 

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1
(9-68)
9.1.4 Parallel reactions
• Parallel reactions are defined as two or more
processes in which the same species participate in
each reaction step.
• The most common cases of parallel reactions are:
(1) those in which the initial reactant decomposes
into several different products; (2) those in which
the initial reactants are different, but yield the
same products; and (3) those in which a substance
reacts with two or more initial reactants.
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9.1.4.1 First order decay to different products
Consider the mechanism
k2
k3
k4
(9-69)
At t=0 the initial concentrations of the four components
are [A1]= [A1]0 ; [A2]= [A3] = [A4] = 0
(9-70,71)
The differential equation for the system can be written, as
exemplified for [A1],
d [ A1 ]
 k 2 [ A1 ]  k3 [ A1 ]  k 4 [ A1 ]  (k 2  k3  k 4 )[ A1 ]  kT [ A1 ] (9-72)
A1
A2 ; A1
A3 ; A1
A4
dt
Integrating equation (9-72) yields
[ A1 ]  [ A1 ]0 e
 kT t
 [ A1 ]0 e
 ( k 2  k3  k 4 ) t
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(9-73)
To solve for [A2] we substitute equation (9-73) into
equation (9-74),
d [ A2 ]
 k 2 [ A1 ]
(9-74)
dt
And upon integrating we obtain
k 2 [ A1 ]0
(9-75)
 kT t
[ A2 ] 
[1  e ]
kT
Similarly we obtain
k3 [ A1 ]0
 kT t
[ A3 ] 
[1  e ]
(9-76)
kT
k 4 [ A1 ]0
[ A4 ] 
[1  e  kT t ]
kT
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(9-77)
It follows that the relative rate constants can be
determined by measuring the relative product yields:
(9-78)
[ A3 ] k3
[ A4 ]

k4
This ratio defines the branching ratio for the reaction;
note that this branching ration is independent of time.
The above example can be easily generalized to
reactions of a single reactant into n different products. We
n
thus have

t
k
 kT t

i 2 i
[ A1 ]  [ A1 ]0 e  [ A1 ]0 e
(9-79)
k n [ A1 ]0
 kT t
(9-80)
[ An ] 
[1  e ]
kT
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
9.1.4.2 First order decay to the same products.
Consider the reaction sequence
k1
k3
(9-81)
The rate expressions for the disappearance of A1 and A3
and the appearance of A2 are
d [ A1 ]
(9-82)
A1

A2 ; A3
A2
 k1[ A1 ]
dt
d [ A2 ]
 k1[ A1 ]  k3[ A3 ]
dt
d [ A3 ]

 k3[ A3 ]
dt
At t=0, [A1]=[A1]0, [A3]=[A3]0, and [A2]=0.
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
(9-83)
(9-84)
The equations describing the time dependence for each
component are [ A1 ]  [ A1 ]0 e  k1t
(9-85)
(9-86)
[ A3 ]  [ A3 ]0 e  k3t
Since d [ A2 ]
 k1[ A1 ]0 e  k1t  k3 [ A3 ]0 e  k3t
(9-87)
dt
Upon integration we obtain
(9-88)
[ A2 ]  [ A1 ]0  [ A1 ]0 e  k1t  [ A3 ]0  [ A3 ]0 e  k3t
or [ A2 ]  ([ A1 ]0  [ A3 ]0 )  ([ A1 ]0 e  k1t  [ A3 ]0 e  k3t ) (9-89)
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
9.1.4.3 Parallel second-order reactions
In the case of parallel second-order reactions,
k1
k2
(9-90)
A1 + A2
A5
A4 ; A1 + A3
Two differential equations can be written,
d [ A2 ]
(9-91)
 k [ A ][ A ]
1
1
2
dt
d [ A3 ]
 k 2 [ A1 ][ A3 ]
dt
(9-92)
Conservation of mass demands
[A4] = [A2]0 - [A2]
(9-93)
[A5] = [A3]0 - [A3]
(9-94)
[A1]0 - [A1] = [A4] +[A5]=[A2]0-[A2] + [A3]0-[A3] (9-95)
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
Eliminating time as a variable, yields
(9-96)
d [ A3 ] k 2 [ A3 ]

d [ A2 ] k1[ A2 ]
which can be integrated under the initial condition that
[A3]=[A3]0 and [A2] = [A2]0 at time t=0. This gives
[ A3 ]  [ A2 ] 

 
[ A3 ]0  [ A2 ]0 
k 2 / k1
(9-97)
From the mass conservation relation we can determine
[A5] by substituting equation (9-94) into equation (9-97).
Rearranging then gives
k /k


  [ A2 ] 


[ A5 ]  [ A3 ]0 1  
 (9-98)
[A ]
2



2 0
1

State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室


The expression for [A1] is obtained from equation (9-95)
[A1] = [A1]0 - [A2]0 - [A3]0 +[A2] + [A3]
(9-99)
Substituting equation (9-96) into equation (9-99) gives
 [ A2 ] 

[ A1 ]  [ A1 ]0  [ A2 ]0  [ A3 ]0  [ A2 ]  [ A3 ]0 
 [ A2 ]0 
k 2 / k1
(9-100)
Using this expression for [A1], we obtain a differential
equation for [A2]:
k 2 / k1




d [ A2 ]
[ A2 ]



 k1[ A2 ][ A1 ]0  [ A2 ]0  [ A3 ]0  [ A2 ]  [ A3 ]0 

dt


 [ A2 ]0 


d [ A2 ]

 k1[ A2 ]   [ A2 ]   [ A2 ]k2 / k1
dt
  [ A1 ]0  [ A2 ]0  [ A3 ]0 ;   [ A2 ]  [ A3 ]([ A2 ])  k2 / k1


State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
(9-101)
(9-102)
Integrating (9-102), we obtain
[ A2 ]
t
d [ A2 ]
[ A2 ]0 [ A2 ]  [ A2 ]   [ A2 ]k2 / k1   0 k1dt
(9-103)
There is no explicit solution to this integral, but we can
consider some limiting cases.
Case I. k1 >> k2
Under this condition the ration k2/k1 is approximately
zero, thus equation (9-103) reduces to the simple form
[ A2 ]
t
d [ A2 ]
(9-104)
[ A2 ]0 [ A2 ]  [ A2 ]     0 k1dt
[ A2 ]0   [ A2 ]0   
(9-105)
 ln
 (   )k1t
[ A2 ]t   [ A2 ]t   
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
Rearranging and solving for [A2]t yields
[ A2 ]t  [ A2 ]0

(9-106)
(    [ A2 ]0 )e (   ) k1t  [ A2 ]0
Case II. k1 = k2
Under this condition the ration k2/k1 =1, thus equation (9103) can be integrable, viz.
[ A2 ]t
t
d [ A2 ]
(9-107)
[ A2 ]0 [ A2 ]  [ A2 ](1   )  0 k1dt
 [ A2 ]t  [ A2 ]0

(  (1   )[ A2 ]0 )e k1t  (1   )[ A2 ]0 (9-108)
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室
Assignments:
1. Consider the reaction
A1
k1
A2
k2
A3
..... An-2
kn-2
kn-1
An-1
An
at t=0, [A1]=[A1]0, [A2]=[A3]=…= [An-1]=[An]=0. Derive an
expression that will describe the concentration of the
intermediates existing at any time t during the reaction. [E.
Abel, Z. Phys. Chem. A56, 558(1906).]
A
2.Consider the first-order cyclic reaction
B
C
Assume that initially at t=0, [A]=[A]0, [B]=[C]=0, and the
amounts of A,B and C always follow [A]0=[A] + [B] + [C]
Using the detailed balanced method, solve for [A],[B] and [C].
State Key Laboratory for Physical Chemistry of Solid Surfaces
厦门大学固体表面物理化学国家重点实验室