Chapt 12 summary part 2
Today I want to discuss catalytic kinetics:
Two key ideas for today:
• Measure rate as a turnover number
• Langmuir Hinshelwood kinetics
1
Turnover numbers
Recall our discussion from before.
CH3COOH
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
CO
Figure 12.1 A schematic of the catalytic cycle for Acetic acid
production via the Monsanto process.
Printing press analogy
• Reactants bind to sites on the catalyst
surface
• Transformation occurs
• Reactants desorb
2
B
+H2
H H
O O O O O
O
- H 2O
+ 1/2 O 2
O O O O O
O O
+1/2 O 2
- H 2O
H H
O O O O O
O O
+ H2
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.
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
TN =
RA
NS
(12.119)
Physically, turnover number is the rate that
the catalyst prints product.
Most people discuss rates in terms of
turnover numbers
2
10
Turnover Number, sec
-1
Key plot:
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
Reaction Temperature, K
1400
Figure 12.28 Turnover
numbers for some typical
processes
4
Next topic why catalytic kinetics weird?
10
13
450 K
450 K
2
Rate, Molecules/cm -sec
Recall from earlier in the semester catalytic
rates weird:
10
12
440 K
440 K
425 K
410 K
10
11
10
415 K
390 K
-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.
5
Rate, Molecules/cm 2-sec
PO =2.5E-8 torr
PCO=2.E-7 torr
2
1E+13
C
1E+12
F
E
D
B
1E+11
A
400
600
800
400
600
800
Temperature, K
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) PCO = 2.5×10-8 torr, PO = 2.5×10-8 torr, B)
2
PCO = 1×10
-7
-8
-7
torr, PO = 2.5×10 torr, C) PCO = 8×10 torr, PO = 2.5×10-8 torr, D)
2
-7
2
-7
-7
-8
PCO = 2×10 torr, PO2 = 4×10 torr, E) PCO = 2×10 torr, PO2 = 2.5×10 torr, F) PCO =
2.5×10-8 torr, PO = 2.5×10-8 torr,
2
Today: Why the weird behavior?
6
Physical interpretation:
Figure 12.33 Langmuir's model for the
adsorption of gas on a solid catalyst. The light
gray area represents the surface of the catalyst.
The black dots are the sites on the catalyst are
sites that are available to adsorb gas. The white
ovals are the adsorbed A molecules. The dark
ovals are adsorbed B molecules.
Physical interpretation for A + B ⇒ AB
• Catalysts have finite number of sites:
• Initially rates increase because surface
concentration increases.
• Eventually A takes up so many sites than
no B can adsorb.
• Further increases in A decrease rate.
7
Can work out kinetics using steady state
approximation:
A⇒C
(12.120)
Mechanism
1
S + A º Aad
2
3
AAd º Cad
4
5
Cad º C + S
6
(12.121)
Also have a species B
7
S + B B ad
8
(12.122)
8
Next use the steady state approximation to
derive an equation for the production rate of
Cad (this must be equal to the production rate
of C).
rC = k 3[A ad ] − k 4 [C ad ]
(12.123)
SS on [Aad] and [Cad]
0 = rA ad = k 1PA [S] − k 2 [A ad ] − k 3[A ad ] + k 4 [C ad ]
(12.124)
0 = rC ad = k 6 PC [S] − k 5[C ad ] − k 4 [C ad ] + k 3[A ad ]
(12.125)
9
People usually ignore reactions 3 and 4
since their rates very low rates
compared to the other reactions.
Dropping the k3 and k4 terms in
equations 12.124 and 12.125 and
rearranging yields:
k 
[A ad ] =  1  PA [S]
 k2 
(12.126)
 k6 
[C ad ] =   PC [S]
 k5 
(12.127)
Similarly for B
k8
=
PB [S]
B
[ ad ]
k7
(12.128)
10
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)
11
Equations (12.129) and (12.130) imply that
there is an equilibrium in the reactions:
1
A + S º Aad
2
8
B
+ S ºBad
7
6
C + S ºC
ad
5
(12.132)
12
If sites are conserved
One needs an expression for [S] to complete
the analysis. One can get an expression for
[S] by assuming that on any catalyst, there
are a finite number of sites to hold the
reactants. Each site can be bare or it can be
covered by A, B or C. If we define S0 as the
total number of sites in the catalyst, one can
show:
S 0 = [S] + [A ad ] + [Bad ] + [ C ad ]
(12.133)
Substituting Equations (12.126), (12.127)
and (12.128) into Equation (12.133) and
then solving for [S] yields:
[S] =
S0
k6
k8
k1
PC
PB +
1+
PA +
k5
k7
k2
(12.134)
13
Substituting
Equation
(12.134)
into
Equations (12.126) and (12.127) yields:
[A ad ] =
k
1+ 1
k2
 k1 
  PA S 0
 k2 
k
k
PA + 8 PB + 6 PC
k7
k5
(12.135)
[C ad ] =
k
1+ 1
k2
 k6 
  PC S 0
 k5 
k
PA + 8 PB +
k7
k6
PC
k5
(12.136)
According to the analysis in Section 4.3, the
equilibrium constant for the adsorption of A
is given by:
k1
KA =
k2
(12.137)
14
Similarly, the equilibrium constants for the
adsorption of B, and C are given by:
k8
KB =
k7
(12.138)
k
KC = 6
k5
(12.139)
Substituting Equations (12.137), (12.138),
and (12.139) into Equations (12.135) and
(12.136) yields:
[A ad ] =
K A PA S 0
1 + K A PA + K B PB + K C PC
(12.140)
[C ad ] =
K C PCS 0
1 + K A PA + K B PB + K C PC
(12.141)
15
Substituting Equations (12.140) and
(12.141) into Equation (12.123) yields:
k 3 K A PA S 0 − k 4 K C PC S 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, since we assumed that the
reaction obeyed a Langmuir-Hinshelwood
mechanism and the equation was first
derived
for
surface
reaction
by
Langmuir[1914]. The same equation is
called the Michaelis-Menten equation in the
enzyme literature; since it was also derived
by Michaelis and Menten[1914].
16
Rate, Molecules/cm 2 /sec
Qualitative behavior for A + B ⇒ products:
2.5E+14
2.0E+14
1.5E+14
1.0E+14
5.0E+13
0.0E+0
0
10
20
30
40
50
PA
10
13
450 K
450 K
2
Rate, Molecules/cm -sec
Figure 12.32
A plot of the rate
calculated from equation (12.161) with
K B PB = 1 0.
10
12
440 K
440 K
425 K
410 K
10
11
10
415 K
390 K
-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.
17
Qualitative behavior for A⇒C
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
PA
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
Figure 2.16 The rate of the reaction NH 3 ⇒ 1 / 2N 2 + 3 / 2H 2 over a platinum
wire catalyst. Data of Loffler and Schmidt[1976].
18
Langmuir-Hinshelwood-Hougan-Watson
rate laws
Trick to simplify the rate equation.
Hougan and watson's idea:
• Identify rate determining step (RDS)
• Assume all steps before RDS in
equilibrium with recatants
• All steps after RDS in equilibrium
with products
• Plug into site balance
19
Example:
The reaction A + B ⇒ C obeys:
S
+
A
⇒
A ad
(1)
S
+
B
⇒
Bad
(2)
A ad
+ Bad
⇒ 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.
20
My 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)
Need S to complete solution: get it from a
site balance.
(5)
So= S + [Aad] + [Bad]
Combining (2), (3) and (5)
So= S + SK1PA + K2PB
Solving (6) for S
So
S=
1 + K1PA + K 2 PB
21
(6)
(7)
Combining equations (4) and (7)
K1K 2 k 3 PA PB (So )
rC =
(1 + K1PA + K 2 PB )2
2
22
(8)
Rate, Molecules/cm 2 /sec
Next what does this look like?
2.5E+14
2.0E+14
1.5E+14
1.0E+14
5.0E+13
0.0E+0
0
10
20
30
40
50
PA
10
13
450 K
450 K
2
Rate, Molecules/cm -sec
Figure 12.32
A plot of the rate
calculated from equation (12.161) with
K B PB = 1 0.
10
12
440 K
440 K
425 K
410 K
10
11
10
415 K
390 K
-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.
23
Discussion problem:
consider a different reaction
A⇒C
(12.120)
1
S + A º Aad
2
3
AAd º Cad
4
5
Cad º C + S
6
(12.121)
Derivation the same:
rC Ad = k 3[A ad ] − k 4 [C ad ]
(12.123)
Derive rate equation
24
r=
k 3K A PA S0 − k 4 K C PCS0
1 + K A PA + K B PB + K C PC
(12.142)
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.
25
Rate, Molecules/cm 2-sec
0
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
Figure 2.16 The rate of the reaction NH 3 ⇒ 1 / 2N 2 + 3 / 2H 2 over a platinum
wire catalyst. Data of Loffler and Schmidt[1976].
26
Summary:
• Catalytic reactions follow a catalytic
cycle
reactants + S ⇒adsorbed reactants
Adsorbed reactants ⇒products + S
• Different types of reactions
Langmuir Hinshelwood
Rideal-Eley
• Calculate kinetics via Hougan and
Watson;
• Identify rate determining step
(RDS)
• Assume all steps before RDS in
equilibrium with recatants
• All steps after RDS in equilibrium
with products
• Plug into site balance
• Predicts non-linear behavior also seen
experimentally
27
Next topic: Mass transfer in catalysts
Mass transfer is important to catalytic
reactions:
Catalyst increases Reaction rate by 1020-1040
Mass transfer rate does not change.
If you have a really active catalyst, mass
transfer will control the rate.
28
Mass transfer control is good:
- design pore structure to pass desired
products much faster than undesired
products.
Cavity
C
H
H
H
H
C
H
H
Diffusion
Channel
Figure Error! No text of specified style in document..1 An interconnecting pore structure
which is selective for the formation of paraxylene.
29
Equations for mass transfer limitations
Define Thiele Parameter
Reaction Rate
Φp =
Diffusion Rate
Define mass transfer factor (literature calls
this effectiveness factor).
ηe =
Actual Reaction Rate
Reaction Rate If Mass Transfer Was Instantaneous
Derive equation:
30
Consider a spherical catalyst particle
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
0
Conc in pellet
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Distance from the center of the pellet
Figure 12.29 The concentration of species
in a spherical catalyst pellet.
The diffusion equation is
d 2 C A 2 dC A rA
+
+ =0
dy y dy D e
Solution:
y Sinh (3Φ P y/yP )
C A =C0A P
y Sinh (3Φ P )
with
31
yP k A
ΦP =
3 De
Calculating the reaction rate:
Mass Transfer
Factor
1 
1
1 
ηe = 
−
Φ P  tanh (3Φ P ) 3Φ P 
10
1
0.1
0.01
0.001
0.01
0.1
1
10
100
Thele Parameter, ΦP
A plot of the mass transfer factor versus the Thiele parameter for diffusion in a porous catalyst
pellet.
32
The mass transfer factor does not
measure the effectiveness of a catalyst.
Good catalysts can have low mass transfer
factors while bad catalysts can have high
mass transfer factors. Generally if a catalyst
is not speeding up the reaction very much
the Thiele parameter will be small, which
means that according to the Figure the mass
transfer factor will be close to unity. Thus, a
bad catalyst can have a large mass transfer
factor.
In contrast, if the catalyst speeds up a
reaction a lot, Φp will be large. The figure
shows that the mass transfer factor is
reduced under such circumstances.
33
Actual equations not useful - never know
De, but ideas useful.
34