CHBE 452 Lecture 31
Mass Transfer & Kinetics In Catalysis
1
Key Ideas For Today
Generic mechanics of catalytic reactions
 Measure rate as a turnover number
 Rate equations complex
 Langmuir Hinshelwood kinetics
2
Generic Mechanisms Of Catalytic
Reactions
B A
B
A
B
A
B
B
A
B
B
B
A
Langmuir-Hinshelwood
A
B
A
B
A
B
A
A
Rideal-Eley
B
A
B
A
B
A
B
A
B
A
B
A
A
A
Precursor
Figure 5.20 Schematic of a) Langmuir-Hinshelwood, b) Rideal-Eley, c) precursor mechanism for the reaction
A+BAB and ABA+B.
3
Turnover Number: Number Of Times Goes
Around The Catalytic Cycle Per Second
CH3COOH
Printing press analogy
CH3OH
HI
H2O
CH3COI
C
O
I
I
O
C
O
I
C
H3C
I
C
O
Rh
CH3I
[Rh(CO)2I2]-
CH3
Rh
C
O
I
I
•Reactants bind to sites on
the catalyst surface
•Transformation occurs
•Reactants desorb
CO
Figure 12.1 A schematic of the catalytic
cycle for Acetic acid production via the
Monsanto process.
4
Typical Catalytic Cycle
+ 1/2 O 2
O O O O O
O O
+1/2 O 2
H H
O O O O O
- H 2O
O O
B
H H
O O O O O
O
+ H2
+H2
- H 2O
A
H
H O
H
H O
Figure 5.10 Catalytic cycles for the production of water a) via disproportion of OH groups, b) via the
reaction OH(ad)+H)ad)H2O
5
Definition Of Turnover Number
RA
TN 
NS
(12.119)
Physically, turnover number is the rate that
the catalyst prints product per unit sec.
6
2
10
Turnover Number, sec
-1
Typical Turnover Numbers
Dehydrogenation
Hydrogenation
Silicon
Deposition
0
10
GaAs
Deposition
-2
10
-4
10
-6
10
Olefin
Isomerization
Alkane
Hydrogenolysis
Cyclization
200
400
600
800
1000
1200
1400
Reaction Temperature, K
7
10
13
450 K
450 K
2
Rate, Molecules/cm -sec
Next Topic Why Catalytic Kinetics
Different Than Gas Phase Kinetics
10
12
440 K
440 K
425 K
410 K
415 K
390 K
10
11
10
-8
-7
10
CO pressure, torr
10
-6
10
-8
-7
10
O 2 pressure, torr
10
-6
Figure 2.15 The influence of the CO pressure on the rate of CO oxidation on Rh(111). Data of
Schwartz, Schmidt, and Fisher.
8
Temperature Nonlinear
2
Rate, Molecules/cm -sec
PO2=2.5E-8 torr
PCO=2.E-7 torr
1E+13
C
1E+12
F
E
D
B
1E+11
A
400
600
Temperature, K
800
400
600
800
Temperature, K
Figure 2.18 The rate of the reaction CO + ½ O2  CO2 on Rh(111). Data of Schwartz, Schmidt and
Fisher[1986]. A) = 2.510-8 torr, = 2.510-8 torr, B) = 110-7 torr, = 2.510-8 torr, C) = 810-7 torr, =
2.510-8 torr, D) = 210-7 torr, = 410-7 torr, E) = 210-7 torr, = 2.510-8 torr, F) = 2.510-8 torr, =
2.510-8 torr,
9
Derivation Of Rate Law
For AC
Also have a species B
Mechanism
1
S +A
2
Aad
3
AAd  Cad
4
7
S+BBad
8
(12.122)
5
Cad C + S
6
(12.121)
10
Derivation:
Next uses the steady state approximation to
derive an equation for the production rate
of Cad (this must be equal to the
production rate of C).
11
Derivation Continued
rC  k 3[A ad ]  k 4 [C ad ]
(12.123)
SS on [Aad] and [Cad]
0  rA
 k 1PA [S]  k 2 [A ad ]  k 3 [A ad ]  k 4 [C ad ]
ad
(12.124)
0  rC ad  k 6 PC [S]  k 5 [C ad ]  k 4 [C ad ]  k 3 [A ad ]
(12.125)
People usually ignore reactions 3 and 4 since their rates very low
rates compared to the other reactions.
12
Dropping The k3 And k4 Terms In Equations
12.124 And 12.125 And Rearranging Yields:
 k1 
[A ad ]    PA [S]
 k2 
Similarly for B
(12.126)
k8
B

PB S
 ad 
k7
 k6 
[C ad ]    PC [S]
 k5 
(12.128)
(12.127)
13
Rearranging Equations (12.126),
(12.127) And (12.128) Yields:
[A ad ]  k1 
 
PA [S]  k 2 
(12.129)
[Bad ]  k 8 
 
PB [S]  k 7 
(12.130)
C ad    k 6 
 
Pc S  k 5 
(12.131)
14
Derivation Continued
Equations (12.129) and (12.130) imply that
there is an equilibrium in the reactions:
[A ad ]  k1 
 
PA [S]  k 2 
Aad
A + S
1
2
(12.129)
B + S
Bad
B
7
8
[Bad ]  k 8 
 
PB [S]  k 7 
(12.130)
C ad    k 6 
 
Pc S  k 5 
(12.131)
6
Cad
C + S
5
(12.132)
15
Site Balance To Complete The
Analysis
If we define S0 as the total number as sites n the
catalyst, one can show:
S0  [S]  [A ad ]  [Bad ]  Cad 
(12.133)
Pages of Algebra
K A PA S 0
[A ad ] 
1  K A PA  K B PB  K C PC
(12.140)
K C PC S 0
[C ad ] 
1  K A PA  K B PB  K C PC
(12.141)
16
Substituting Equations (12.140) And
(12.141) Into Equation (12.123) Yields:
k 3 K A PA S 0  k 4 K C PCS 0
r
1  K A PA  K B PB  K C PC
(12.142)
In the catalysis literature, Equation
(12.142) is called the LangmuirHinshelwood expression for the rate of the
reaction AC, also called Michaele’s
Menton Equation.
17
2.5E+14
10
13
450 K
2
2.0E+14
Rate, Molecules/cm -sec
Rate, Molecules/cm 2 /sec
Qualitative Behavior For Bimolecular
Reactions (A+Bproducts)
1.5E+14
1.0E+14
5.0E+13
0.0E+0
0
10
20
30
40
PA
50
10
12
440 K
410 K
390 K
10
11
10
Figure 12.32 A plot of the rate calculated
from equation (12.161) with KBPB=10.
-8
-7
10
CO pressure, torr
10
-6
18
Physical Interpretation Of
Maximum Rate For A+BAB




Catalysts have finite
number of sites.
Initially rates increase
because surface
concentration increases.
Eventually A takes up so
many sites that no B can
adsorb.
Further increases in A
decrease rate.
19
Rate, Moles/cm 2/sec
2.0E-8
PB =0
1.5E-8
1.0E-8
PB =25
5.0E-9
0.0E+0
0
10
20
PA
30
40
50
Rate, Molecules/cm 2-sec
Qualitative Behavior For Unimolecular
Reactions (AC)
1E+20
1670 K
1E+19
1270 K
1070 K
1E+18
870 K
770 K
1E+17
1E+16
0.01
0.1
1
10
100
Ammonia pressure, torr
20
Langmuir-Hinshelwood-Hougan-Watson
Rate Laws: Trick To Simplify The Algebra
Hougan and Watson’s Method:
 Identify rate determining step (RDS).
 Assume all steps before RDS in equilibrium
with reactants.
 All steps after RDS in equilibrium with
products.
 Plug into site balance to calculate rate
equation.
21
Example:
The reaction A + B  C obeys:
S
S
A ad
 A

B
 Bad
 A ad (1)
 Bad (2)
 C  2S (3)
(12.157)
Derive an equation for the rate of formation
of C as a function of the partial pressures of
A and B. Assume that reaction (3) is rate
determining.
22
Solution
rC = k3[Aad][Bad]
Assume reaction 1 in equilibrium
[A ad ]
 K1
S PA
Similarly on reaction 2
[Bad ]
 K2
S PB
Combining 1,2 and 3
rC  K1K 2 k 3 PA PBS2
(1)
(2)
(3)
(4)
23
Solution Continued
Need S to complete solution: get it from a
site balance.
So = S + [Aad ] + [Bad ]
(5)
Combining (2), (3) and (5)
So = S + SK1 PA + SK2 PB
Solving (6) for S
So

S
1  K1 PA  K2 PB
Combining equations (4) and (7)
K1K 2 k 3 PA PB So 
rC 
1  K1PA  K 2 PB 2
(6)
(7)
2
(8)
24
Background: Mass Transfer
Critical to Catalyst Design
Cavity
C
H
H
H
H
H
H
C
Diffusion
Channel
Figure 12.27 An interconnecting pore structure which is
selective for the formation of paraxylene.
25
Introduction

In a supported catalyst




reactants first diffuse into the catalyst,
then they react
The products diffuse out
Gives opportunity for catalyst design
26
Distance
Reactant Concentration Drops
Moving Into the Solid
Reactant Concentration
Edge of
Pellet
Center of
Pellet
Concentration
1
Gas Phase Conc
0.8
Avg conc in pellet
Gas
Phase
0.6
0.4
0.2
Conc in pellet
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Distance from the center of the pellet
27
Thiele Derivation for Diffusion
In Catalysis
Distance
Assume:
 Constant effective diffusivity
 First Order reaction per unit volume
 Irreversible reaction – Rate of diffusion of
products out of catalyst does not affect rate
Reactant Concentration
28
Define “Effectiveness Factor”
Actual rate for a catalyst pellet with mass transfer limitations
e 
Rate in the absence of mass transfer limitations
29
Derivation: Mass Balance On
Differential Slice
Y


 dC S  
 dC A  
2
2
  4 y D e  
 4 y D e  


 dY   y  y 
 dy   y

 4 y 2 y  rA 
Spherical pellet
Taking the limit as Δy → 0
2
d C A 2 dC A rA


0
dy y dy D e
30
Long Derivation
y P Sinh 3y/y P 
C A C
y Sinh 3P 
0
A
yP
P 
3
kA
De
1 
1
1 
e 



 P  tanh3 P 3 P 
31
Mass Transfer Factor
Thiele Plot
1.0
0.8
Zero Order
0.6
0.4
0.2
0.0
0
First Order
Second Order
2
4
6
8
10
Thele Parameter
32
Issues With “Effectiveness Factor”

Best catalysts have low “effectiveness
factors”



Effectiveness goes down as rate goes up
 High rate implies low selectivity
Often want mass transfer limitations for
selectivity
I prefer “mass transfer factor” not
“effectiveness factor”
33
Issues, Continued:
Effective Diffusivity, De, Unknown
Figure 14.3 A cross sectional diagram of a typical
catalyst support.
34
Knudsen Diffusion


Diffusion rate in gas phase controlled by
gas – gas collisions
Diffusion rate in small poles controlled by
gas – surface collisions
35
Knudsen Diffusion
From kinetic theory
1
Dgas  vλ
3
1
D k  V2a 
3
applies when
(λ  mean free path)
a  pore diameter 
2a  gas
36
Knudsen Diffusion
DK  1.68x10
1
2
 a  T  2  AMU
 
 

sec  1R  300K   M 
3 CM
1
2
100 Å pore DK  0.12cm / sec for H2
2
compared to 0.86 in the gas phase
37
Simple Models Never Work
Notice one molecule interferes with diffusion of second molecule
38
Summary


Catalytic reactions follow a catalytic cycle
reactants + S adsorbed reactants
Adsorbed reactants products + S
Different types of reactions
Langmuir Hinshelwood
Rideal-Eley
39
Summary

Calculate kinetics via Hougan and
Watson;





Identify rate determining step (RDS)
Assume all steps before RDS in equilibrium
with reactants
All steps after RDS in equilibrium with
products
Plug into site balance
Predicts non-linear behavior also seen
experimentally
40
Query

What did you learn new in this lecture?
41