Chemical Kinetics-Reaction Mechanisms and Complex Reactions(Notes)
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Transcript of Chemical Kinetics-Reaction Mechanisms and Complex Reactions(Notes)
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Reaction Mechanisms and Complex
Reactions
Reversible first-order reaction
Ak1
k1B
Reversible has a different meaning in kinetics than in
thermodynamics!
All chemical reactions are reversible to some extent,
but ifK>
100, this fact can usually be ignored for pur-
poses of chemical kinetics.
d[A]
d t= fAvf+bAvb
f
A=1, vf= k1[A]bA=+1, vb= k1[B]d[A]
d t=k1[A]+k1[B]
PChem BA 8.1
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d[B]
d t= fBvf+bBvb
d[B]
d t= k1[A]k1[B]
Add the two rate equations:
d[A]
d t +d[B]
d t =k1[A]+k1[B]+k1[A]k1[B]d
d t
[A]+ [B]
= 0
[A](t)+ [B](t)= const= [A](0)+ [B](0)=C[B](t)=C [A](t)d[A]
d t=k1[A]+k1{C [A](t)}
d[A]
d t= k1C (k1+k1)[A]
equilibrium:
d[A]eq
d t= 0
PChem BA 8.2
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0=k1[A]eq+k1[B]eq
[B]eq[A]eq
= k1k1
=Kc
C [A]eq[A]eq
= k1k1
=Kc
[A]eq
=
k1
k1+k1C
d[A]
d t= k1C (k1+k1)[A]
=(k1+k1)
[A] k1k1+k1
C
d[A]
d t =(k1+k1)[A] [A]eq[A](t)[A]0
d[A]
[A] [A]eq=(k1+k1)t
ln
[A](t) [A]eq[A]0
[A]eq
=(k1+k1)t
[A](t) [A]eq[A]0 [A]eq
= exp(k1+k1)t
PChem BA 8.3
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[A](t)= [A]eq+
[A]0 [A]eq
exp
(k1+k1)t
relaxation methods: relaxation = return of a system
to equilibrium
Relaxation methods are used to measure rate con-
stants. Apply a sudden perturbation to the equilib-
rium, e.g., a temperature jump or a pressure jump,
and monitor the relaxation.
Let x be the deviation from equilibrium. Then for a re-
versible first-order reaction A B, our results showthat
x(t)= [A](t) [A]eq= x(0)exp(t/)with
= 1k1+k1
i.e., after a temperature jump at t= 0, the system re-laxes exponentially with a rate 1/ = k1+ k1. Com-bine the measurement of the equilibrium constant
PChem BA 8.4
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Kc = k1/k1 with a measurement of the relaxationtime to obtain the forward and backward rate con-
stants k1 and k1.
Parallel first-order reactions
Assume the reactions are irreversible for the moment.
Ak1
BA
k2Cd[A]
d t=k1[A]k2[A]
d[B]
d t = k1[A]d[C]
d t= k2[A]
add the three equations:
dd t
[A]+ [B]+ [C]= 0
[A](t)+ [B](t)+ [C](t)= const
PChem BA 8.5
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= [A](0)+ [B](0)+ [C](0)= [A]0
if initially only the reactant A is present
conversion of reactant into products:
[A](t)
=[A]0 exp{
(k1+
k2)t}
d[B]
d t= k1[A]= k1[A]0 exp{(k1+k2)t}
[B](t)= k1[A]0k1+k2
1exp[(k1+k2)t]
[C](t)
=k2[A]0
k1+k2 1exp[(k1+k2)t]yield (t)
YB
irr
=[B]()
[A]0 =k1
k1+k2Y
Cirr =
[C]()[A]0
= k2k1+k2
PChem BA 8.6
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selectivity
Sirr= [B](t)[C](t)
= k1k2
for all times
What happens if the reactions are reversible?
A
k1
k1 B
Ak2
k2C
d[A]
d t=k1[A]+k1[B]k2[A]+k2[C]
d[B]d t
= k1[A]k1[B]d[C]
d t= k2[A]k2[C]
[A](t)+ [B](t)+ [C](t)= const= [A]0
can eliminate on variable, say [C]:
[C](t)= [A]0 [A](t) [B](t)
PChem BA 8.7
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Consider the case t. The reaction reaches equi-librium, i.e., the concentrations no longer change with
time:
d[A]
d t=k1[A]+k1[B]k2[A]+k2[C]= 0
d[B]
d t =k1[A]
k
1[B]
=0
d[C]
d t= k2[A]k2[C]= 0
k1[A]eq= k1[B]eqk2[A]eq= k2[C]eq[B]
eq
[A]eq= k1
k1=Kc,1
[C]eq
[A]eq= k2
k2=Kc,2
[A]eq+ [B]eq+ [C]eq= [A]0
[A]eq+Kc,1[A]eq+Kc,2[A]eq= [A]0[A]eq=
[A]0
1+Kc,1+Kc,2
PChem BA 8.8
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[B]eq=Kc,1[A]0
1+
Kc,1+
Kc,2
[C]eq=Kc,2[A]0
1+Kc,1+Kc,2yield (t)
Y
Brev=
[B]eq
[A]0 =Kc,1
1+Kc,1+Kc,2
YC
rev=[C]eq
[A]0= Kc,2
1+Kc,1+Kc,2selectivity
Srev=[B]eq
[C]eq= Kc,1
Kc,2= k1
k1 k2
k2= k1
k2 k2
k1
Srev=Sirr k2k1
The selectivity
S = [B][C]
= k1k2=Sirr
PChem BA 8.9
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will be observed for short times, as along as the back-
ward reactions are negligible: The reaction is un-
der kinetic control.
For long times, the selectivity approaches the equilib-
rium value
S=
[B]
[C] tKc,1
Kc,2 =S
rev
The reaction is under thermodynamic control.
Example: k1 = 1 s1, k1 = 0.01s1, k2 = 0.1s1, k2 =0.0005s1
Initially B is formed ten times faster than C, however
Kc,1=1 s1
0.01s1= 100
Kc,2= 0.1s10.0005s1
= 200
and C is the more stable product.
PChem BA 8.10
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1000 2000 3000 4000 5000 6000 7000
t
0.2
0.4
0.6
0.8
1.
conc
Figure 1: Parallel first-order reversible reactions. [A]:
solid line, [B]: dashed line, [C]: dot-dash line.
PChem BA 8.11
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1 2 3 4 5 6 7 8 9 10
t
0.2
0.4
0.6
0.8
1.
conc
Figure 2: Parallel first-order reversible reactions: Ini-
tial stage. [A]: solid line, [B]: dashed line, [C]: dot
dash line.
PChem BA 8.12
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0.01 0.1 1 10 100 1000 10000
t
0.2
0.4
0.6
0.8
1.
conc
Figure 3: Parallel first-order reversible reactions: con-
centrations versus logt. [A]: solid line, [B]: dashed
line, [C]: dot-dash line.
In the above parallel-reaction mechanism C can be
converted into B only via A. What happens if there isthe additional reaction
Ck3
k3B
(An example are the isomerization reactions be-
tween ortho-xylene, meta-xylene, and para-xylene:
o-xylene m-xylene, m-xylene p-xylene, p-xyleneo-xylene.) Let also assume that the three steps are
PChem BA 8.13
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the mechanism of the reaction, i.e., all reactions are
elementary.
Is it then possible to establish equilibrium by the
cyclic reaction mechanism below?
A
B
C!
'
eee
d[A]
d t=k1[A]+k2[C]
d[B]
d t= k1[A]k3[B]
d[C]
d t =k3
[B]
k2
[C]
[A](t)+ [B](t)+ [C](t)= const= [A]0stationary state:
k1
[A]=
k2
[C]=
[C]
[A] =k1
k2
k1[A]= k3[B] =[B]
[A]= k1
k3
PChem BA 8.14
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k3[B]= k2[C]
which follows from the previous two equations
[B]= k1k3
[A]
[C]= k1k2
[A]
[A]+ k1k3
[A]+ k1k2
[A]= [A]0
[A]= [A]01+ k1
k3+ k1
k2
We have determined the stationary concentrations
[A], [B], and [C] in terms of [A]0 and the three rateconstants.
This stationary state is not an equilibrium state!! It
violates the principle of detailed balance.
The principle of microscopic reversibility:
In a system at equilibrium, any molecular pro-
cess and the reverse of that process occur, on
the average, at the same rate.
PChem BA 8.15
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implies the principle of detailed balance:
In a system at equilibrium, each collision has itsexact counterpart in the reverse direction, so
that the rate of every chemical process is ex-
actly balanced by that of the reverse process.
=In a system at equilibrium, every elementary forward
reaction is balanced by its own backward reaction.
In other words, in a system at equilibrium each ele-
mentary reaction must be individually at equilibrium.
Principle of detailed balance implies that all elemen-
tary reactions must be reversible.
Application to the scheme
Ak1
k1B
Ak2
k2C
Ck3
k3B
PChem BA 8.16
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equilibrium:
k1[A]eq= k1[B]eq = Kc,1= [B]eq[A]eq
= k1k1
k2[A]eq= k2[C]eq = Kc,2 =[C]eq
[A]eq= k2
k2
k3
[C]eq=
k3
[B]eq=
Kc,3=
[B]eq
[C]eq =k3
k3
The product of the second and third equilibrium con-
stants is
Kc,2Kc,3=[C]eq
[A]eq
[B]eq
[C]eq =[B]eq
[A]eq =Kc,1
=
k1
k1= k2
k2
k3
k3
or
k1k2k3= k1k2k3PChem BA 8.17
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This condition on the six rate constants has been de-
rived by considering chemical equilibrium, but rate
constants are concentration independent. Therefore,
this relation is general for this mechanism; it also
holds for all nonequilibrium conditions. We see that
the rate constants can never be chosen independent-
ly, and the cyclic mechanism is impossible, since it
requires k1=
k2
=k3
=0, which is forbidden by the
relation.
Consecutive first-order reactions (series reactions)
Ak1B k2C
Assume the reactions are first order and irreversible.
d[A]
d t=k1[A]
d[B]
d t = k1[A]k2[B]d[C]
d t= k2[B]
PChem BA 8.18
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Assume again [A](0) = [A]0, [B](0) = 0, [C](0) = 0, andadd the three equations:
[A](t)+ [B](t)+ [C](t)= const= [A]0[A](t)= [A]0 exp(k1t)d[B]
d t= k1[A]0 exp(k1t)k2[B]
[B](t)= k1[A]0k2k1
{exp(k1t)exp(k2t)}
[C]= [A]0 [A](t) [B](t)
= [A]0
1+ 1k2k1
[k1 exp(k2t)k2 exp(k1t)]
Consider the case: k2 k1. Then
exp(k2t) exp(k1t)
and
k2k1 k2
PChem BA 8.19
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[C] [A]01+1
k2 (k2)exp(k1t)
= [A]0{1exp(k1t)}
The formation ofC depends onlyon the rate constant
of the first step A
B.
AB is the rate-determining step.
Motivation for the
Steady-State Approximation: After an initial induction
period, the concentration and the rate of change of
all reaction intermediates are negligibly small.
Illustrate for the consecutive scheme:
[B](t)
[A](t) =
k1[A]0
k2k1{exp(k1t)exp(k2t)}
[A]0 exp(k1t)= k1
k2k1{1exp[(k2k1)t]}
PChem BA 8.20
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For k2 k1:
[B](t)
[A](t) k1
k2{1exp[k2t]
0 for t1/k2
}= k1k2
[B](t)= k1k2
[A](t) [A](t)
d[B]
d t =k1
k2
d[A]
d t d[A]
d t
If we apply directly the steady-state approximation
(SSA), also known as the quasisteady-state approxi-
mation (QSSA), to the intermediate B, we obtain
0 d[B]d t
= k1[A]k2[B]
=
[B]=
k1
k2[A]
d[C]
d t= k2[B]= k2
k1
k2[A]= k1[A]
PChem BA 8.21
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= k1[A]0 exp(k1t)
[C](t)= [A]0{1exp(k1t)} (same as above)
Application: decomposition of ozone
2 O3 3 O2
empirical rate law (M is an inert gas):
v=12
d[O3]
d t= 1
3
d[O2]
d t= k[O3]
2[M]
k[O2][M]+ [O3]
proposed mechanism:stoichiometric number
O3+Mk1O2+O+M 2
O2+O+Mk1O3+M 1
O+
O3
k2
2 O
21
2 O3+2 M+O2+O+M+O+O3
PChem BA 8.22
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2 O2+2 O+2 M+O3+M+2 O2
2 O3 3 O2
intermediate O: steady-state approximation
d[O]
d t= 0
Intermediates do not appear in the rate law!!
d[O]
d t= k1[O3][M]k1[O2][O][M]k2[O][O3]= 0
[O]
=
k1[O3][M]
k1[O2][M]+k2[O3]d[O3]
d t=k1[O3][M]+k1[O2][O][M]k2[O][O3]
=k1[O3][M]+
k1[O2][M]k2[O3]
[O]
=k1[O3][M]++
k1[O2][M]k2[O3] k1[O3][M]
k1[O2][M]+k2[O3]
PChem BA 8.23
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= 1k
1[O2][M]
+k2[O3]
k1[O3][M]k1[O2][M]k1[O3][M]k2[O3]+
+k1[O2][M]k1[O3][M]k2[O3]k1[O3][M]
= 2k1k2[O3]2[M]
k1[O2][M]+k2[O3]=2 k1[O3]
2[M]k1k2
[O2][M]+ [O3]=2v with k= k1, k= k1/k2
empirical rate law
If initially we have pure ozone, then k1[O2][M] k2[O3] and
d[O3]
d t=2k1[O3][M]
= the first step is rate-determiningThe reaction is self-inhibiting in O2, v as [O2]
PChem BA 8.24
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Another approximation method
Pre-equilibrium or fast-equilibrium approximation
A+B k1k1
Xk2C
k1 k2, k1 k2
A, B, and Xare in equilibrium:
k1[A][B]= k1[X]
[X]= k1k
1
[A][B]=K[A][B]
(Note: K=Kc!!)d[C]
d t= k2[X]= k2K[A][B]= k[A][B]
k= k2K=k2k1
k1
PChem BA 8.25
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enzyme reactions
Michaelis-Menten mechanism
E+S k1k1
(ES)k2 P+E
d[P]
d t= v= k2[(ES)]
Apply the steady-state approximation:
d[(ES)]
d t= k1[E][S]k1[(ES)]k2[(ES)]= 0
[(ES)]
=
k1[E][S]
k1+k2 =
[E][S]
KM
KM=k1+k2
k1Michaelis constant
d[P]
d t= k2
KM[E][S]
conservation of total enzyme concentration:
[E]0= [E]+ [(ES)]
PChem BA 8.26
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= [E]+ [E][S]KM
= [E]
1+ [S]KM
[E]= [E]01+ [S]
KM
d[P]
d t =k2
KM [E][S]
= k2KM [E]0
1+ [S]KM
[S]
d[P]
d t= v= k2[E]0[S]
KM+
[S]
Michaelis-Menten rate law
[Figure: rate versus substrate concentration, Fig. 8.13]
At high substrate concentrations, [S]KM,
d[P]d t
k2[E]0[S][S]
= k2[E]0,
the reaction is zero order, and the rate reaches its
PChem BA 8.27
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maximum:
vmax= k2[E]0
The Michaelis-Menten rate law can be written in terms
of vmax:
d[P]
d t= v= vmax
1+KM/[S]
We rearrange this expression into a form that is
amenable to data analysis by linear regression:
1
v= 1
vmax+
KM
vmax
1
[S]
A Lineweaver-Burk plot is a plot of the reciprocal of
the initial velocity v0 versus the reciprocal of the ini-
tial substrate concentration [S]0 at fixed enzyme con-
centration. If the enzyme obeys Michaelis-Menten ki-
netics, then the plot should yield a straight line, with
PChem BA 8.28
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slope KM/vmax, that intercepts the ordinate axis at
1/vmax and the abscissa at
1/KM.
[Figure: Lineweaver-Burk plot, Fig. 8.14]
At low substrate concentrations, [S]KM,
d[P]
d t
k2[E]0[S]
KM = k[S],
the reaction is first order.
The turnover number or catalytic constant of an
enzyme, kcat, is the number of catalytic cycles
(turnovers) performed by the active site in a given
time interval divided by the duration of that inter-
val. It has units of a first-order rate constant, and for
the Michaelis-Menten mechanism is given by k2, the
rate constant for release of product from the enzyme-
substrate complex.
kcat= k2=vmax
[E]0
PChem BA 8.29
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The catalytic efficiency, , of an enzyme is given by
= kcatKM
The higher the value of , the more efficient is the
enzyme. For the Michaelis-Menten mechanism
= k2k1+k2
k1
= k1k2k1+k2
max= k1 if k2 k1
= rate of formation of the enzyme-substrate complex;
diffusion-limited: < 108 109 M1 s1 for enzyme-sizedmolecules at room temperature
catalase (decomposition of hydrogen peroxide) =4.0
108 M1 s1: catalytic perfection
Diffusion control
reactions in solutions
PChem BA 8.30
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reaction between two solute molecules A and B: the
two molecules approach by diffusion to form an en-
counter pair: {AB}
[Figure: processes in involved in solution phase reactions]
kinetic scheme:
A+Bkd
kd {AB}ka
products
kd: rate constant for diffusive approach; kd: rateconstant for separation of the encounter pair; ka: rate
constant for conversion of encounter pair into prod-
ucts
apply steady-state approximation to the encounter
pair
d[{AB}]
d t= kd[A][B] (kd+ka)[{AB}]= 0
[{AB}]= kd[A][B]kd+ka
PChem BA 8.31
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reaction velocity
v= ka[{AB}]= kakd[A][B]kd+ka
thus the rate constant of the bimolecular reaction
A
+B
k2
products
is given by
k2=v
[A][B]= kakd
kd+ka
two-limiting cases:
(i) diffusion-controlled limit: ka kd, separation ofA and B is relatively difficult, e.g., the solvent has a
large viscosity , or the reaction has a small activa-
tion energy = rate-determining step is the diffusiveapproach of the reactants; once they are in the sol-
vent reaction cage, reaction is assured
k2= kdPChem BA 8.32
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analysis of diffusion of molecules in liquids and use of
the Einstein relation and Stokes law:
k2= kd=8RT
3
for small molecules in aqueous solutions at room tem-
perature, kd
1091010 M1 s1
(ii) activation-controlled limit: ka kd, reactionswith large activation energies; Ea 20kJmol1 for re-actions in water,
k2= kakd
kd
the second factor is the equilibrium constant KAB for
the formation of the encounter pair; thus
k2= kaKAB
rate is determined by the equilibrium concentration
of encounter pairs and the rate of passage over the
activation energy barrier
PChem BA 8.33
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Gas phase reactions
kinetic theory of Gases
average speed
c=
8RT
M
1/2
R=NAk, k Boltzmann constant
R
M= NAk
NAm= k
m
c
= 8kT
m1/2
m mass of a particle
elastic collisions
determine how often a specific particle collides with
other particles in the gas
since this is a representative particle, we can assume
that it moves with the average speed c
PChem BA 8.34
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we can replace the moving collision partners by sta-
tionary particles, if we replace the average speed c
by the relative average speed crel:
c=
8kT
m
1/2, crel=
8kT
1/21
1
mA+
1
mB
, = reduced mass
for identical particles
1
= 2
m= = m
2
crel=2c
[Figure: collision tube]
number of stationary particles inside the collision
tube: N0 crelt, N= N/V number density, 0 =d
2
= elastic collision cross-section, for collisions be-tween identical particles, d= 2rcollision frequency z = average number of collisions
PChem BA 8.35
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36/41
of a particle per unit time
z=N0 creltt
=N0crelz=
2N0c
z=
2N
V0c
z=2 PkT
0c; (PV=nRT= NNA
RT=N kT)
mean free path l = the average distance a molecule
travels between two successive collisions
l= c 1z= c
2N0c= 1
2(N/V)0= kT
2P0
total number of collisions per unit volume per unit
time for identical particles:
ZAA=1
2zNA=
0c2NA
2=0
4kT
mA
1/2NA
2
PChem BA 8.36
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for mixture of gases A and B, total number of colli-
sions per unit volume per unit time for dissimilar par-
ticles:
ZAB=0
8kT
1/2NANB
for dissimilar particles: 0
=d2 with d
=rA
+rB
if all collisions were reactive, then the kinetic equa-
tion for the number density of Afor the reaction
A+B P
would be
dNA
d t=ZAB
and for the molarity
d[A]d t
=ZAB/NA
where NA is Avogadros number
PChem BA 8.37
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so we obtain
v= kAB[A][B]=08kT
1/2NA[A][B]
kAB=0
8kT
1/2NA
most collisions will not be reactive, since collisionsmust occur with sufficient energy in the gas phase to
give rise to a reactive event
[Figure: collisions in the gas phase, Fig. 7.11]
before molecules can get close enough to react, they
must overcome an energy barrier a
only molecules with sufficient kinetic energy along
the line of centers AB to surmount this energy barrier
will react
[Figure: line-of-centers energy, Fig. 7.12]
fraction of reactive collisions
f= exp(Ea/RT), Ea=NAa
PChem BA 8.38
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v=08kT
1/2
NAexp(Ea/RT)[A][B]
kAB=0
8kT
1/2NAexp(Ea/RT)
the predicted rate constant follows the Arrhenius law
kAB= Aexp(Ea/RT)
the pre-exponential factor A is given by
Ath=08kT
1/2
NA
collision theory predicts that the pre-exponential fac-
tor is weakly temperature dependent
Ath
T
for many reactions, this temperature dependence is
swamped by the strong temperature dependence of
the exponential term
PChem BA 8.39
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the following Table (from M. J. Pilling and P. W. Seakins,
Reaction Kinetics, Oxford University Press, Oxford,
1995) compares the predicted and experimental val-
ues of A for some reactions, the quantity P is defined
as
P= AexpAth
Rxn T Ea 1011Aexp 10
11Ath P
1 600 0 10 2.1 4.8
2 300 0 0.24 1.1 0.22
3 470 102 0.094 0.59 0.16
4 800 180 1.24105 7.3 1.7106
T is in K, Ea in kJmol1
, A in Lmol1 s1
1: K+Br2KBr+Br2: CH3
+CH3
C2H6
3: 2NOCl 2 NO+Cl24: H2+C2H4C2H6PChem BA 8.40
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except for Rxn 1, the theoretical values are too large,
for Rxn 4 by more than a factor of 105
P< 1 indicates that the relative orientation of the mol-ecules is important in reactive collisions; P is known
as the steric factor and is generally several orders of
magnitude smaller than 1
kAB= P0 8kT
1/2 NAexp(Ea/RT)orientation does not explain values of P > 1, whichwould seem to imply that the molecules react faster
than they collide
P> 1: long range attractive interactions