Transcript che 551 lectures - Classnotes For Professor Masel's Classes

CHBE 553 Lecture 26
How Do Surfaces Change
Activation Barriers
1
Objective
• Describe how surfaces changes rates of
reaction
– What are key mechanisms
– Start to give examples
2
Background: Surfaces Can Change Rates By A
Large Amount
Reaction
Catalyst
Rate
Temperature
Enhancement
Ortho H2  Para H2
Pt (solid)
1040
300K
2NH3N2 + 3H2
Mo (solid)
1020
600K
C2 H4 + H2  C2 H6
Pt (solid)
1042
300K
H2 +Br2  2HBr
Pt (solid)
1  108
300K
2NO + 2H2 N2 + 2H2 O Ru (solid)
3  1016
500K
CH3COH  CH4 + CO
I2 (gas)
4  106
500K
CH3CH3  C2H4 +H2
NO2 (gas)
1  109
750K
(CH3)3 COH 
(CH3)2CH2CH2+H2O
HBr (gas)
3  108
750K
3
Today: Mechanisms Of Catalyst Action
• Catalysts can be designed to help
initiate reactions.
• Catalysts can be designed to stabilize
the intermediates of a reaction.
• Catalysts can be designed to hold the
reactants in close proximity.
• Catalysts can be designed to hold the
reactants in the right configuration to
react.
4
Mechanism Of Catalysts,
Continued
• Catalysts can be designed to block side
reactions.
• Catalysts can be designed to sequentially
stretch bonds and otherwise make bonds
easier to break.
• Catalysts can be designed to donate and
accept electrons.
• Catalysts can be designed to act as
efficient means for energy transfer.
5
Mechanism Of Catalysis
Continued
Is is also important to realize that:
• One needs a catalytic cycle to get
reactions to happen.
• Mass transfer limitations are more
important when a catalyst is present.
6
Today: The Role Of Catalysts In Initiating
Reactions, Stabilizing Intermediates
• Catalysts initiate reactions by help creating
active centers (i.e. a radical or ion).
– Active center could be catalyst itself
– Could be a radical R-O.
• Catalysts stabilize intermediates.
– Catalyst binds to intermediate, lowering the free
energy of the reactive intermediates.
• Raises intermediate concentration.
• Intermediates less reactive.
7
How Could Catalysts Change H2+Br22HBr
50
+H2
A
H+HBr
Br
B
H+HBr
+Br2
+Br2
1/2Br2
1/2Br2
Br+2HBr
0
0
Br+2HBr
Br
1/2 Br2+2HBr
Free energy, kcal/mole of bromine atoms
Mechanism
Br2→2Br
Br+H2→HBr+H
H+Br2→HBr+Br
2Br→Br2
50
+H2
1/2 Br2+2HBr
Initiate Reaction
Gas Phase
-50
-50
Reaction Progress
Reaction Progress
50
50
C
Pt
D pt
Sulfided
+H2
1/2Br2
+H2
1/2Br2
0
0
+Br2
Br
+Br2
Br
H+HBr
H+HBr
Br+2HBr
Br+2HBr
1/2 Br2+2HBr
Modify Intrinsic
Barriers
Stabilize Intermediates
-50
1/2 Br2+2HBr
-50
Reaction Progress
Reaction Progress
8
Key Principles Of Catalytic
Mechanisms
• Catalysts bind intermediates at distinct
sites.
• Mechanism often the same in gas
phase & on catalyst.
• Initiation much faster – in effect do not
need initiation reacion.
9
Example Of Catalysts
Initiating Reactions
C2H6C2H4+H2
(12.41)
Gas phase mechanism
C2H6 2CH3
(12.42)
CH3+C2H6C2H5+CH4
(12.43)
C2H5C2H4+H
H+C2H6C2H5+H2
(12.45)
2CH3C2H6
(12.46)
2c2H5C4H10
(12.47)
(12.44)
CH3C2H5 C3H8
(12.48)
10
Catalysis By NO2
Consider adding NO2
NO2  C2 H 6  C2 H5  HNO2
(12.50)
Catalysts can initiate reactions. The mechanisms are
similar to the mechanisms without a catalyst, but the
initiation process is much faster with the catalyst.
Effect 109 (small for catalysis)
11
Some Examples Of Reactions Initiated By
Catalysts
Reaction
Catalyst
Mechanism of Initiation
CH3 CH3  C2H4 +H2
NO2
NO2 + CH3CH3  HNO2 + CH3CH2
CH3COH  CH4 + CO
I2
X+I2  2I+X
I + CH3COH  HI + CH3CO
Ethylene  polyethylene
R OOR
R OOR  2R O
R O+CH2=CH2R OCH2CH2
H2 + Br2  2HBr
metalic
platinum
Br2 + 2S  2Brad
Propylene 
Polypropylene
Ti+
Tl+ + propylene  CH3CHTiCH2+
C2H5OH  C2H4 + H2O
H+
C2H5OH + H+  [C2H5OH2]+
[C2H5OH3]+  [C2H5]+ + H2O
2O3  O2
Cl
[C2H5]+  C2H4 + H+
O3 + Cl  O2 + ClO
12
Gas Phase
50
+H2
H+HBr
Br
+Br2
0
1/2Br2
Br+2HBr
1/2 Br2+2HBr
-50
Reaction Progress
Enthalpy, kcal/mole of bromine atoms
Enthalpy, kcal/mole of bromine atoms
Initiation Often Not Enough. Also Need To
Stabilize Intermediates
Surface
50
0
1/2Br2
+H2
Br(ad)
H(ad)+HBr
+Br2
1/2 Br2+2HBr
Termination
-50
Br(ad)+2HBr
Reaction Progress
Figure 12.7 The enthalpy changes during the
Figure 12.8 The enthalpy changes during the
gas phase reaction H2 + Br2  2 HBr assuming Rideal-Eley surface reaction H2 + Br2  2 HBr
that the reaction terminates after one cycle
on Pt(111) assuming that the reaction
terminates after one cycle
13
Stabilization Of Ionic
Intermediates
RHC  CRH  RRC  CHH
(12.63)
Possible gas phase mechanism
X + RHC  CRH  RHC  CH  R   X
(12.64)
R  RHC  CRH  X  R2 HCCRH   X
(12.65)
R2 HCCH   X  R2 C  CH  H   X
(12.66)
Note big barrier to first step
14
Acid Catalyzed Reaction
RHC  CRH  H   RHC  CRHH 
(12.67)
RHC  CRHH   RRHC  CHH 
(12.68)
RRHC  CHH   RRC  CHH  H 
(12.69)
15
Stabilization Of Intermediates. Can We Have Too
Much of a Good Thing?
When we stabilize intermediates we
increase the intermediate
concentration. We also decrease the
reactivity of the intermediates.
Which wins? (recall lecture 21)
16
Experimental Evidence
HCOOHH2+CO2
 H
HCOOH  HCOO(ad)
ad
  CO  H
H (ad)  HCOOad
2
2
(12.75)
350
Ir
Temperature For
50% Conversion
Pt
400
Ru
Pd
Rh
450
Cu
Co
W
Au
550
Fe
Ni
Ag
500
600
50
60
70
80
90
100
110
120
Heat Of Formation Of Formate
17
Sabatier’s Principle
The best catalysts are substances
which bind the reactants strongly, but
not too strongly.
18
Consider H2+Br2HBr
Rideal Eley mechanism
Br2  2S  2Brad
Brad  H 2  HBr
H ad  Br2  HBr  Br
 H ad
Complicated derivation (see
text)
1
rHBr  2 k Br exp (1   p,2 ) H ad /  B T [H 2 ][S][Br2 ]2


(12.85)
(12.76)
   E 0   H  TS  /  T
k Br  2k 0
exp
p,2 O
ad 
B 
2
  a,2
(12.86)
19
For Unlimited Sites
1E+26
2
Rate, Molecules/Cm Sec
1E+31
1E+21
1E+16
1E+11
1E+6
-40
-20
0
20
40
60
80
Heat Of Formation Of Intermediate, Kcal/mole
Figure 12.10 The rate of HBr formation as calculated from Equation (12.85), with [S] =
1e14/cm2 and p= 0.5, T = 500K, PH 2  PBr2 1atm.
20
For Finite Number Of Sites
So
[S] 
1  K Br PBr  K H PH
2
2
2
2
(12.87)
1E+14
1E+12
2
Rate, Molecules/Cm Sec
1E+16
1E+10
1E+8
1E+6
1E+4
1E+2
1E+0
- 40
- 20
0
20
40
60
80
Heat Of Formation Of Intermediate, Kcal/mole
Figure 12.11 The rate of HBr formation calculated from Equation (12.85), with [S] from
Equation (12.87) and p= 0.5, T = 500K, PH PBr 1atm.
2
2
21
Common Plots In Literature
Sachtler-Frahenfort
Tanaka-Tamaru plots:
plots:
Use heat of oxidation
Use heat of oxidation
per mole of metal as
per mole of oxygen
surrogate for heat of
as surrogate for
formation of product.
heat of formation of
product.
22
Comparison Of Sachtler-Frahrenfort And
Tanaka-Tamaru
Sachtler-Frahrenfort
Ru
1E+18
Ni
1E+14
W
2
Rh
Pd Co
Ir
Fe
1E+16 Pt
Ta
1E+12
Cu
1E+10
-10 -30 -50 -70 -90 -110
0 -20 -40 -60 -80 -100
Heat of formation of oxide
per mole of oxygen, Kcal/mol
Tanaka-Tamaru
1E+20
Rate, Molecules/cm /sec
2
Rate, Molecules/cm /sec
1E+20
Ru
1E+18 Rh
Pd Ir Co
1E+16
Fe
Pt
Ni
1E+14
W
Ta
1E+12
Cu
1E+10
0
-100
-50
-200
-150
-300
-250
Heat of formation of oxide
per mole of metal, Kcal/mol
Figure 12.13 A Sachtler-Frahrenfort and Tanaka-Tamaru plot for the
hydrogenation of ethylene.
23
Summary So Far
• Catalysts work by initiating reactions,
stabilizing intermediates.
• Leads to 1020 increase in rates – need
other effects to get to 1040.
• Can stabilize too much.
24
Example: Constructing Sachtler-Frauhenfort
And Tanaka-Tamaru Plots
Table 12.E.1 some
data for the rate of
ethylene
hydrogenation on a
number of metals 0
C. Next lets
construct a
Sachtler-Fahrenfort
plot of the data.
Table 12.E.1 The spreadsheet for
example 12.E
A
4 Metal
5
6
7
8
9
10
11
12
13
14
15
16
17
Pt
Pd
Ir
Rh
Ru
Cu
Cu
Co
Ni
Fe
Fe
W
Ta
B
rate
1.0E+16
3.0E+16
3.0E+16
3.0E+17
9.0E+18
8.3E+10
8.3E+10
3.0E+16
1.0E+14
1.0E+16
1.0E+16
3.0E+13
3.0E+13
C
D
E
Hf Of Oxygens Metal
Oxide in oxide atoms
in oxide
-9.7
1
1
-20.4
1
1
-40.1
2
1
-21.7
1
1
-52.5
2
1
-39.8
1
2
-37.1
1
1
-57.2
1
1
-58.4
1
1
-197.5
3
2
-63.7
1
1
-136
2
1
-499.9
5
2
F
Hf per mole
of oxide
G
Hf per mole of
metal
=$C5/D5
=$C6/D6
=$C7/D7
=$C8/D8
=$C9/D9
=$C10/D10
=$C5/E5
=$C6/E6
=$C7/E7
=$C8/E8
=$C9/E9
=$C12/D12
=$C13/D13
=$C14/D14
=$C11/E11
=$C12/E12
=$C13/E13
=$C14/E14
=$C16/D16
=$C17/D17
=$C16/E16
=$C17/E17
25
Solution
A
4 Metal
5
6
7
8
9
10
11
12
13
14
15
16
17
Pt
Pd
Ir
Rh
Ru
Cu
Cu
Co
Ni
Fe
Fe
W
Ta
B
rate
1.0E+16
3.0E+16
3.0E+16
3.0E+17
9.0E+18
8.3E+10
8.3E+10
3.0E+16
1.0E+14
1.0E+16
1.0E+16
3.0E+13
3.0E+13
C
D
E
Hf Of Oxygens Metal
Oxide in oxide atoms
in oxide
-9.7
1
1
-20.4
1
1
-40.1
2
1
-21.7
1
1
-52.5
2
1
-39.8
1
2
-37.1
1
1
-57.2
1
1
-58.4
1
1
-197.5
3
2
-63.7
1
1
-136
2
1
-499.9
5
2
F
Hf per mole
of oxide
G
Hf per mole of
metal
=$C5/D5
=$C6/D6
=$C7/D7
=$C8/D8
=$C9/D9
=$C10/D10
=$C5/E5
=$C6/E6
=$C7/E7
=$C8/E8
=$C9/E9
=$C12/D12
=$C13/D13
=$C14/D14
=$C11/E11
=$C12/E12
=$C13/E13
=$C14/E14
=$C16/D16
=$C17/D17
=$C16/E16
=$C17/E17
26
Frahrenfort And TanakaTamaru
Sachtler-Frahrenfort
Ru
1E+18
Ni
1E+14
W
2
Rh
Pd Co
Ir
Fe
1E+16 Pt
Ta
1E+12
Cu
1E+10
-10 -30 -50 -70 -90 -110
0 -20 -40 -60 -80 -100
Heat of formation of oxide
per mole of oxygen, Kcal/mol
Tanaka-Tamaru
1E+20
Rate, Molecules/cm /sec
2
Rate, Molecules/cm /sec
1E+20
Ru
1E+18 Rh
Pd Ir Co
1E+16
Fe
Pt
Ni
1E+14
W
Ta
1E+12
Cu
1E+10
0
-100
-50
-200
-150
-300
-250
Heat of formation of oxide
per mole of metal, Kcal/mol
Figure 12.13 A Sachtler-Frahrenfort and Tanaka-Tamaru plot for the
hydrogenation of ethylene.
27
Stabilizing Intermediates Not
Entire Effect
• Leads to 1020 increases in rates –
-need other effects to get to 1040
• Does not lead to selectivity
28
Summary
Catalysts work by initiating reactions,
stabilizing intermediates.

Leads to 1020 increase in rates – need
other effects to get to 1040.


Can stabilize too much.

Other effects change selectivity.
29