Transcript che 377 lectures - Classnotes For Professor Masel's Classes

ChE 553 Lecture 14
Catalytic Kinetics
1
Objective
• Provide an overview of catalytic kinetics
– How do rates vary with concentration
– Simple modles: langmuir
2
Key Ideas For Today
Generic mechanisms of catalytic
reactions
• Measure rate as a turnover number
• Rate equations complex
• Langmuir Hinshelwood kinetics
3
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
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.
4
A
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)2I 2]-
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.
5
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
6
Definition Of Turnover
Number
RA
TN 
NS
(12.119)
Physically, turnover number is the rate that
the catalyst prints product per unit sec.
7
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
8
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.
9
More Typical Behavior
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 + 2 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,
10
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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Derivation Of Rate Law
For AC
Mechanism
1
S + A  Aad
2
3
AAd  Cad
4
 Rate Determining Step
5
Cad  C + S
6
(12.121)
12
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).
13
Derivation
1

S +A
2
Aad
3
AAd

4
Cad
5
Cad  C + S
6
(12.121)
rC  k 3[A ad ]  k 4 [C ad ]
SS approximation on Aad and Cad
0  rA
0  rC
ad
 k1PA [S]  k 2 [A ad ]  k 3[A ad ]  k 4 [Cad ]
ad
 k 6 PC [S]  k 5[Cad ]  k 4 [Cad ]  k 3[A ad ]
People usually ignore reactions 3 and 4 since their rates very low
rates compared to the other reactions.
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Dropping The k3 And k4 Terms In Equations 12.124
And 12.125 And Rearranging Yields:
 k1 
[A ad ]    PA [S]
 k2 
(12.126)
k 
[C ad ]   6  PC [S]
 k5 
(12.127)
15
Rearranging Equations (12.126),
(12.127) And (12.128) Yields:
[A ad ]  k1 
 
PA [S]  k 2 
C ad    k 6 
 
Pc S  k 5 
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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 
C ad    k 6 
 
Pc S  k 5 
A+S
C+S


Aad
Cad
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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 ]  C ad 
Pages of Algebra
K A PAS0
[A ad ] 
1  K A PA  K C PC
K C PCS0
[C ad ] 
1  K A PA  K C PC
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Substituting Equations (12.140) And (12.141)
Into Equation (12.123) Yields:
k 3K A PAS0  k 4 K C PCS0
r
1  K A PA  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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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
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
PA
r
k 3 K A PA S 0  k 4 K C PC S 0
1  K A PA  K B PB  K C PC
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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.
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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 + SK1PA + K2PB
Solving (6) for S
So
S
1  K1PA  K 2 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
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
25
-6
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.
26
Another Example:
Consider a different reaction
AC
(12.120)
1
 Aad
S +A
2
3
AAd

4
Cad
5
Cad

6
C+S
(12.121)
27
Derivation The Same
rC
 k 3[A ad ]  k 4 [C ad ]
Ad
(12.123)
Derive rate equation
k 3K A PA S0  k 4 K C PCS0
r
1  K A PA  K B PB  K C PC
(12.142)
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What Does This Look Like?
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
30
40
50
PA
Figure 12.29 A plot of the rate of the reaction AC calculated from
Equation (12.142) with k4=0, PB = 0, 1, 2, 5, 10 and 25., KA = KB =1.
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Summary
• Catalytic reactions follow a catalytic
cycle
reactants + S adsorbed reactants
Adsorbed reactants products + S
• Different types of reactions
Langmuir Hinshelwood
Rideal-Eley
30
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
31
Key Predictions
Unimolecular reactions
• Rate increases with pressure, levels
• Rate always increases with temperature
• Very sensitive to poisons
Bimolecular reactions
• Rate rises reaches a maximum at finite temp
and pressure, then drops
• Sensitive to poisons
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