1
1.4 The first-order reactions (T = constant and V = constant)
Kinetic experiments usually yield concentrations of reacting
species as a function of time. The rate law that governs a
reaction is essentially a differential equation that gives dci/dt, the
rate of change in the concentrations of the reacting species. To
obtain the concentration vs. time dependencies predicted by the
rate law, one must of course be able to integrate rate laws.
In the following discussion, it is assumed that:
1) The reaction is carried out at constant temperature i.e.
T = constant and k = constant
2) The volume of the reaction mixture V is constant
3) The reaction is "irreversible" (no significant amount of
reverse reaction occurs). This will be true only if the equilibrium
constant Kc of the reaction is very large.
For example the decay of a radioactive isotope follows the
simple first-order kinetics. The rate law for the first-order
reaction, A→ products, is given by Eq. (1.24)
r = kcA
(1.24)
For the first-order reaction the degree of the advancement of
reaction is defined according to Eq. (1.6):
cA = cA0 - ξ
(1.25)
In Eq. (1.25) we have:
cA = concentration of A at the moment t
cA0 = the initial concentration of A at t = 0
ξ = ξ#/V = the degree of the advancement of reaction per
unit volume
2
By substituting cA from Eq. (1.25) into Eq. (1.24) and
representing the rate law according to Eqs. (1.9) and (1.10) we
obtain the differential equation
dξ
= k(c A0 − ξ)
dt
(1.26)
for the degree of the advancement of reaction, ξ. In fact we
should replace ξ in Eq.(1.26) by ξ#. However, as ξ# is an
extensive property it is more convenient to use ξ (which is an
intensive property) in discussing kinetics of chemical reactions
in general. If the volume V is constant we can with no
difficulties replace ξ# with ξ. To solve this differential equation
we rearrange it to separate ξ and t on the opposite sides of the
equation:
dξ
= kdt
(cAO − ξ)
(1.27)
dx 1
∫ X = a lnX (X = ax + b), where a ≡ -1, b ≡
cA0 and x ≡ ξ and integrating this equation, we get
Using the identity
- ln( c A0 − ξ) = kt + I
(1.28)
Setting for the initial conditions cA = cA0 at t = 0 and ξ = 0 we
get for the value of the integration constant I = - lncA0 and by
substituting this result into Eq. (1.28) we get for the solution of
Eq. (1.27):
⎡c − ξ ⎤
ln ⎢ A0
⎥ = − kt
c
⎣ A0 ⎦
(1.29)
The degree of the advancement of the first-order reaction ξ can
now be solved and we get:
3
ξ = c A0 [1 − exp(-kt )]
(1.30)
From Eq. (1.26) we now see that the solution of Eq. (1.25)
becomes
cA = cA0 exp(–kt)
(1.31)
⎡c ⎤
ln ⎢ A ⎥ = − kt
⎣ c A0 ⎦
(1.32)
or
It is evident from Eq. (1.32) that in order to determine the rate
constant k for a first-order reaction at any time (t ≠ 0) it is only
necessary to determine the ratio of the concentrations cA at two
different moments of time. Also any physical quantities
proportional to the respective concentrations may be substituted
for the concentrations in these equations since the
proportionality constants cancel.
The half-life time, t1/2 of a reaction is the time required for half
of the amount of reactant to disappear. For the first-order
reaction, A → B, Eq. (1.32) leads to:
⎡ c
⎤
ln ⎢ A0 ⎥ = k t 1/ 2
⎣ 0,5 c A0 ⎦
and
k t1/2 = 0,693
(1.33)
For any first-order reaction the half-life time is evidently
independent of the initial concentration. The relaxation time, τ
for the first-order reaction is equal to the reciprocal of the firstorder rate constant τ = 1/k. Thus Eq. (1.32) may as well be
written as
c A = c A0 exp( − t / τ)
(1.34a)
4
For the time dependency of the product concentration cB we
then obtain:
cB = cB0 + cA0[1 - exp(-kt)]
(1.34b)
The properties of first-order reactions can be interpreted by
graphical methods depicted in Fig. 1.1.
c
ln( c c )
0
c0
lnc=lnc 0- kt
c
2
lnc 0
slope = -k
t
half life-time:
- does not depend
t 1 = ln2
k
2
on initial concentration
t
Figure 1.1 Concentration vs. time curve (kinetic curve) for an
irreversible first-order reaction and its linear anamorphos.
The decay of a radioactive isotope follows the first-order
kinetics. The number of nuclei that decay, –dN (the number of
radioactive nuclei N is decreasing with time) during a small
period of time dt it is proportional to the number of the nuclei N
of the radioactive isotope currently present in the system and
also to the length of the small time interval:
dN
= − λN
dt
(1.35)
In Eq. (1.35) the proportionality constant λ is called the decay
constant. According to the first-order reaction rate theory Eq.
(1.35) can as well be rewritten by Eq. (1.36).
5
dc N
= −k NcN
dt
(1.36)
where kN = λ. After integration Eq. (1.35) gives:
N = N0 exp(- λ t)
(1.37)
where N0 is the number of radioactive nuclei present at t = 0.
1.5 The second-order reactions (T = constant and V =
constant)
There exists two forms of second-order reaction rate laws given
below:
For the reaction, 2A → products, where νA = -2, we have the
rate law
r = k cA2
(1.38)
For the reaction, A + B → products, we have the rate law
r = k cAcB
(1.39)
1.5.1 The second order reaction with the rate law r = kcA2
For the second-order reaction described in Eq. (1.38) the degree
of advancement of the reaction is defined according to Eq. (1.6)
cA = cA0 - 2ξ
(1.40)
6
By substituting cA from Eq. (1.40) into Eq. (1.38) and
representing the rate law of the reaction according to Eqs. (1.9)
and (1.10) we obtain the differential equation:
dξ
= k(c A0 − 2ξ) 2
dt
(1.41)
To solve this equation we rearrange it and separate ξ and t on the
opposite sides of the equation. Then we have
dξ
(c A0 − 2 ξ )
2
= kdt
(1.42)
1
X n+1 where X = b + ax,
a(n + 1)
a = -2, b = cA0, x = ξ and n = – 2 to integrate this equation, we
get:
Using the identity
n
∫ X dx =
1
(cA0 − 2ξ) −1 = k t + I
2
(1.43)
Setting for the initial conditions cA = cA0 at t = 0 and ξ = 0 we
1
get for the value of the integration constant I =
. By
2c A0
substituting this result into Eq. (1.43) we get for the solution of
differential equation, Eq. (1.42):
1⎛
1
1 ⎞
−
⎜
⎟ = kt
2 ⎝ c A0 − 2ξ c A0 ⎠
(1.44)
As cA0 -2ξ = cA Eq. (1.44) becomes
1
1
−
= 2kt
c A c A0
(1.45)
7
The half-life time of this reaction is given by Eq. (1.46) and it is
not independent of the initial concentration of A, cA0
t1 =
2
1
2kc A0
(1.46)
If νA = -1, which means that the reaction is written as
A → products then Eq. (1.45) becomes
1
1
−
= kt
c A c A0
(1.47)
The general form for the degree of the advancement of the
second order reaction is obtained by representing ξ according to
Eq. (1.44)
k t c 2A0
ξ=−
1 + 2 k t c A0
(1.48)
Substituting this result into Eq. (1.6) we get for time dependency
of the concentration of A
2 k t c 2A0
c A = c A0 −
1 + 2 k t c A0
(1.49)
The properties of second-order reaction obeying the rate law
given by Eq. (1.38) can be interpreted by the graphical methods
depicted in Fig. 1.2
8
c
1
c0
c
1
1
=
+ kt
c
c0
c
2
slope = k
t
t 1 = 1 - is proportional to (1 c0 )
2
kc0
1
c0
t
Figure 1.2 Concentration vs. time curve (kinetic curve) for the
irreversible second-order reaction 2A → products and its linear
anamorphos.
1.5.2 The second-order reaction with the rate law r = kcAcB
A completely different integrated rate law for a second-order
reaction is obtained if the stoichiometry for the chemical
reaction is A + B → products, and the rate law obeys Eq. (1.39).
The differential equation for this kind of second-order reaction is
according to Eq. (1.39) given by
dξ
= k(c A0 − ξ)(c B0 − ξ)
dt
(1.50)
By rearranging and separating the variables ξ and t we get
dξ
= k dt
(cA0 − ξ)(c B0 − ξ)
Using the identity
(1.51)
9
dx
1 fx + g
=
∫ (b + ax)(g + fx) Δ ln ax + b
where ξ = x, a = -1, f = -1, cA0 = b, cB0 = g and Δ = cb0 - cA0, to
integrate Eq. (1.51) we get:
⎡c − ξ ⎤
1
=kt+ I
ln ⎢ B0
c B0 − c A0 ⎣ c A0 − ξ ⎥⎦
(1.52)
Setting for the initial conditions cA = cA0 and cB = cB0 at t = 0 and
ξ = 0 we get for the value of the integration constant I:
I=
1
c
ln B0
c B0 − c A0 c A0
(1.53)
By substituting I back to Eq. (1.52) we get for the solution of
Eq. (1.51)
c B0
⎡ (c − ξ )c A0 ⎤
1
ln ⎢ B0
=kt
− c A0 ⎣ (c A0 − ξ )c B0 ⎥⎦
(1.54)
As cB0 - ξ = cB and cA0 - ξ = cA Eq. (1.54) becomes
1
[c c ]
ln B B0 = kt
c B0 − c A0 [c A c A0 ]
(1.55)
Thus for the second-order reaction, A + B → products, in which
the initial concentrations of reactants A and B are nonstoichiometric the plot, ln (cA/cB) vs. t is linear as indicated in
Fig. 1.3.
10
[A]t
ln
[B]t
[A]t=0
[B]t=0
slope = -k([A]
t=0
-[B]
t=0
time
Figure 1.3 The linear anamorphos of the irreversible secondorder reaction of the type A + B → products.
The concept of reaction half-life time is inapplicable to Eq.
(1.55). For example when cA = 0.5 cA0, cB will not equal 0.5 cB0
unless the reactants are present initially at stoichiometric
concentrations and in this special case Eq. (1.55) is clearly not
applicable since the difference cB0 -cA0 becomes zero.
Kinetics of most of the elementary reactions taking place in
homogeneous liquid or gas phase are described by the secondorder rate law. Typical example is the formation of HI given in
Eq. (1.22).
1.6 The third-order reactions (T = constant and V =
constant)
One can imagine the following third-order reactions:
3A → products
2A + B → products
)
11
A + B + C → products
with the rate laws, r = kcA3, r = kcA2cB and r = kcAcBcC,
respectively.
For the simplest reaction, 3A → products , the rate law over the
extent of the reaction is described by the differential equation:
dξ
3
= k(c A0 − 3ξ)
dt
(1.56)
To solve this equation we rearrange it and separate ξ and t on
opposite sides of equation:
dξ
= k dt
(c A0 − 3ξ) 3
(1.57)
Using the same mathematical identity as for the second-order
rate law, Eq. (1.42), with n = –3 we get
1
6(c A0 − 3ξ)
2
= kt + I
(1.58)
Setting the initial conditions as before i.e. cA = cA0 at ξ = 0 and
1
t = 0 we get the value for the integration constant I = 2 . By
6c A0
substituting I back to Eq. (1.58) we get for the solution of Eq.
(1.57)
1
(cA0 − 3ξ)
2
−
1
= 6kt
c 2A0
(1.59)
If we write the reaction equation in the form A → products, we
have νA = -1 and cA = cA0 - ξ . Then Eq. (1.59) becomes:
12
1
1
−
= 2kt
c 2A c 2A0
(1.60)
For the half-life time t1/2 of the third-order reaction, 3A →
products, we get:
t 1/ 2 =
1
2 k c 2A0
(1.61)
For the third-order reaction, A → products, we obtain the plot, in
which 1 / c A2 0 versus t is a linear function as indicated in Fig. 1.4.
c
1 c2
c0
c
2
slope = 2k
t
t1 = 3 2
2 2kc 0
1 c2
0
t
Figure 1.4 Concentration vs. time curve (kinetic curve) for the
irreversible third order reaction 3A → products and its linear
anamorphos.
The other two third-order rate laws are considered next. For the
reaction, 2A + B → products, which obeys rate law r = k cA2 cB
we have νA = -2 and νB = -1. For the reaction A + B + C →
products which obeys the rate law r = k cA cB cD we have νA = -1,
νB = -1 and νC = -1. Both rate laws yield complicated differential
equations. Using a good handbook of integrals we can get the
13
solutions for these differential equations. For the third order rate
law r = kcA2cB, we get the solution
⎛ 1 1
νB
c /c ⎞
1
ln B B0 ⎟ = kt (1.62)
⎜ − +
νAcB0 − νBcA0 ⎝ cA0 cA νAcB0 − νBcA0 cA / cA0 ⎠
For the third order rate law r = k cA cB cD, we get the solution
c
c
c
νA
νB
νC
ln D = kt
ln B +
ln A +
Δab Δad cA0 ( −Δab )Δ bd cB0 Δac Δ bd cD0
(1.63)
where Δab ≡ νA cB0 – νB cA0 , Δad ≡ νA cD0 – νD cA0 and Δbd ≡
νB cD0 – νD cB0
Only very few elementary reactions are described by third order
kinetics. One of them is
2NO + O2 → 2NO2
(1.64)
Also the gas phase recombination reaction of hydrogen atoms
given by Eq. (1.65) is a third order reaction
H + H + M → H2 +M
(1.65)
Here M is a third body adsorbing the energy liberated in the
hydrogen bond formation between two hydrogen atoms.
1.7 General n-th order reactions
No elementary reactions where more than three molecules take
part in are known because of the extremely low probability of
simultaneous collision between more than three molecules.
However, in practice it is possible to obtain a general rate
equation for the reaction (-νA)A → products, described by the
rate law
14
r = kcAn
(1.66)
The degree of the advancement of the n-th order reaction is
described by
dξ
(cA0 + νA ξ)
n
= k dt
(1.67)
Using the same identity as for solving Eq. (1.42) we get for the
solution of Eq. (1.67)
n
∫ X dx =
1
X n +1
a( n + 1)
(1.68)
Where X = cA0 + νAξ . It is evident that necessarily n ≠ 1. The
integration gives for the n-th order reaction
1
(cA0 + νA ξ) − n +1 = kt + I
νA ( − n + 1)
(1.69)
Setting the initial conditions in the usual way, we get the for the
− n +1
c A0
. By substituting
value of the integration constant I =
ν A ( − n + 1)
the integration constant I back into Eq. (1.69) and rearranging
we get for the solution of Eq. (1.67)
⎡ 1
1
1 ⎤
−
= kt
n −1 ⎥
νA ( − n + 1) ⎢⎣ c An−1 c A0
⎦
(1.70)
For the half-life time we get by setting cA = 0,5 cA0 at t = t1/2
t 1/ 2
2 n −1 − 1
=
n −1
νA (1 − n) k c A0
(1.71)
15
Sometimes Eq. (1.71) is used to determine the reaction order n
from the experimental half-life time results. Eqs. (1.70) and
(1.71 are universal and valid for reactions with any values of n
except for n = 1.
1.8 The 0-th order reactions
A reaction is of zeroth order if the rate is independent of the
concentration of the reactant. This can be the case if e.g. the rate
is limited by the concentration of a catalyst. In this case k is in
fact a pseudo reaction rate constant proportional to the
concentration of the catalysts.
Zeroth order reactions can also be detected in photochemical
reactions if the rate is determined by the light intensity. From
Eq. (1.70) it is possible to obtain solution for the zeroth order
reaction rate equation
c A − c A0
= kt
νA
(1.72)
If νA = –1 then Eq. (1.72) becomes
c A0 − c A = kt
(1.73)
The half-life time of the zeroth order reaction is respectively
t 1/ 2 =
c A0
2k
1.9 Reactions where T = constant but V ≠ constant
(1.74)
16
It is important from the viewpoint of engineering to notice that
many industrial reactions run under constant pressure rather
than under constant volume conditions. As the rate of reaction
was defined in Eq. (1.9) by
1 dξ #
r=
V dt
(1.75)
we must reconsider the case where the volume of the reaction
mixture is not constant. As the concentration of species A taking
part in the reaction is given by
n A n A 0 + νA ξ #
cA =
=
V
V
(1.76)
we get for the first order reaction, A → products, for which νA =
-1:
1 dξ #
n A0 − ξ #
=k
V dt
V
(1.77)
and it is seen that for example with the initial condition ξ# = 0 at
t = 0 we get the solution
ξ # = n A0 [1 − exp( − kt )]
(1.78)
regardless of the volume of the reaction mixture.
However, for reactions with orders other than unity we must
more carefully consider how the volume of the reaction mixture
depends on the advancement of the reaction. In most general
form the volume of reaction mixture is some function of the
amounts of the components in reaction mixture:
V = V(n1, n2, ...) = V(nj)
(1.79)
17
If the rate of the reaction can be expressed with Eq. (1.17)
r = k Π cβi i
(1.80)
i
we get for the rate expression the general form
⎡ ni ⎤
1 dξ #
= kΠ⎢
⎥
V( n j ) dt
i ⎣ V( n j ) ⎦
βi
(1.81)
In practice it is often more convenient to use in data analysis the
Levenspiel´s approximation given by Eq. (1.82)
V ≈ V0(1 + δAfA)
(1.82)
In this equation fA is the fraction conversion of the reactant A,
V0 is the volume of the reaction mixture at fA = 0 (at the
beginning of the reaction), and the Levenspiel's constant δA is
defined in Eq. (1.83)
δA =
V(at fA = 1) − V(at fA = 0)
V(at fA = 0)
(1.83)
Now the concentrations for any species i in the reaction mixture
are with good accuracy given by
νi
n A 0 fA
ni
νA
ci = =
V
V0 (1 + δ A fA )
ni0 −
and the rate of reaction can be expressed by
⎛ n A 0 dfA ⎞
1 dξ #
1
r=
=
⎜−
⎟
V dt
V0 (1 + δ A fA ) ⎝ νA dt ⎠
(1.84)
18
=−
cA 0
1 dfA
(1 + δ A fA ) νA dt
(1.85)
In most cases it is more convenient to use the approximate Eq.
(1.85) that to express the reaction rate with an exact equation.
Of course expressing the reaction rate r in Eq. (1.80) by Eq.
(1.85) and then the concentrations ci by Eq. (1.84) will lead to a
differential equation, which can in most cases be solved only by
means of numerical methods.