Transcript Chapter 27 Molecular Reaction Dynamics

Chapter 24. Molecular Reaction
Dynamics
Purpose: Calculation of rate constants for simple
elementary reactions.
For reactions to take place:
1. Reactant molecules must meet.
2. Must hold a minimum energy.
Gas phase reactions:
Collision theory.
Solution phase reactions: Diffusion controlled.
Activation controlled.
24.1 Collision theory
• Consider a bimolecular elementary reaction
A + B → P
v = k2[A][B]
The rate of v is proportional to the rate of collision, and
therefore to the mean speed of the molecules,
v   (T / M )1/ 2 N A N B   (T / M )1/ 2[ A][B]
• Because a collision will be successful only if the kinetic energy
exceeds a minimum value. It thus suggests that the rate constant
should also be proportional to a Boltzmann factor of the form, e  E / RT .
a
k2   (T / M )1 / 2 e  Ea / RT
• Consider the steric factor, P,
k2  P (T / M )1 / 2 e  Ea / RT
• Therefore, k2 is proportional to the product of steric requirement x
encounter rate x minimum energy requirement
Collision rate in gases
•
Collision density, ZAB, is the number of (A, B) collisions in a region of the
sample in an interval of time divided by the volume of the region and the
duration of the interval.
1/ 2
Z AB
 8kT 


 u 
N A2 [ A][B]
where σ =  d2
d = ½(dA + dB)
and u is the reduced mass
u
•
m AmB
m A  mB
when A and B are the same, one gets
1/ 2
Z AA
•
 4kT 

 (1 / 2) 
 m A 
N A2 [ A]2
The collision density for nitrogen at room temperature and pressure, with d
= 280 pm, Z = 5 x 1034 m-3s-1.
The energy requirement
•
For a collision with a specific relative speed of approach vrel
d [ A]
  ( )v rel N A[ A][ B ]
dt

d [ A]
    ( )v rel f ( )d  N A[ A][B]
dt
 0

•
reorganize the rate constant as


k2  N A    ( )v rel f ( )d 
 0

•
Assuming that the reactive collision cross-section is zero below εa
 a 
 ( )  1  
 

for    a

k2  N A c rel e  Ea / RT
The steric effect
• Steric factor, P,
• Reactive cross-section, σ*,
•
σ* = P σ
•
1/ 2
 8kT 
k 2  P 


u


N Ae  Ea / RT
• Harpoon mechanism: Electron transfer preceded
the atom extraction. It extends the cross-section for the
reactive encounter.
• K and Br2 reaction
Example 24.1 Estimate the steric factor for the reaction
H2 + C2H4 -> C2H6 at 628K
given that the pre-exponential factor is 1.24 x 106 L mol-1 s-1.
Solution:
Calculate the reduced mass of the colliding pair
u
m Am B
 3.12 1027 kg
m A  mB
1/ 2
 8kT 


 u 
 2.66 103 m s 1
From Table 24.1 σ(H2) = 0.27 nm2 and σ(C2H4) = 0.64 nm2, given a
mean collision cross-section of σ = 0.46 nm2.
1/ 2
 8kT 
A 

 u 
N A  7.37 1011 Lmol1 s 1
P = 1.24 x 106 L mol-1 s-1/7.37 x 1011 L mol-1s-1
= 1.7 x 10-6
• Example 24.2: Estimate the steric factor for the
reaction: K + Br2 → KBr + Br
Solution: The above reaction involves electron flip
K + Br2 → K+ + Br2Three types of energies are involved in the above process:
(1) Ionization energy of K, I
(2) Electron affinity of Br2, Eea
(3) Coulombic interaction energy:
 e2
4 0 R
Electron flip occurs when the sum of the above three energies changes sign
from positive to negative
24.2 Diffusion-controlled reactions
• Cage effect: The lingering of one molecule near another on account
of the hindering presence of solvent molecules.
• Classes of reaction
Suppose that the rate of formation of an encounter pair AB is
first-order in each of the reactants A and B:
A + B →AB
v = kd[A][B]
The encounter pair, AB, has the following two fates:
AB → A + B
v = kd’[AB]
AB → P
v = ka[AB]
• The net rate of change of [AB]:
d [ AB ]
dt
= kd[A][B] - kd’[AB] - ka[AB]
• Invoking steady-state approximation to [AB]
[ AB ] 
kd [ A][B]
k d '  ka
• The net rate of the production:
k k [ A][B]
d[ P ]
 ka [ AB ]  a d
 k2 [ A][B]
dt
k d '  ka
k2 
k a kd
k d '  ka
• When kd’<< ka
k2 = kd
• When kd’>> ka
reaction)
k2 
ka kd
kd'
(This is diffusion-controlled limit)
(This is activation-controlled
Reaction and Diffusion
kd  4R* DN A
•
where R* is the distance between the reactant
molecules and D is the sum of the diffusion coefficients of the two
reactant species.
DA 
kT
6RA
DB 
kT
6RB
where η is the viscosity of the medium. RA and RB are
the hydrodynamic radius of A and B.
•
If we assume RA = RB = 1/2R*
kd 
8RT
3
24.3 The material balance equation
(a) The formulation of the equation
[ J ]
 2 [J ]
[ J ]
D
v
2
t
x
x
the net rate of change due to chemical reactions
[ J ]
  k[ J ]
t
the over rate of change
[ J ]
 2 [J ]
[ J ]
D
v
 k[ J ]
2
t
x
x
the above equation is called the material balance equation.
(b) Solutions of the equation
[ J ]
 2 [J ]
D
 k[ J ]
2
t
x
t
[J ]  k  [J ]e kt dt  [J ]e kt
*
0
[J ] 
n0e
 x 2 / 4 Dt
A(Dt )1 / 2