Transcript che452lect31

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
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Definition Of Turnover Number
RA
TN 
NS
(12.119)
Physically, turnover number is the rate that
the catalyst prints product per unit sec.
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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,
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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).
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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)
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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)
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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.
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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.
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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.
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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.
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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)
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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)
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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.
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Introduction

In a supported catalyst




reactants first diffuse into the catalyst,
then they react
The products diffuse out
Gives opportunity for catalyst design
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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
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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
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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
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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 
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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