# 7-Heterogeneous Catalyst - Dicky Dermawan

```ITK-329 Kinetika & Katalisis
Chapter 7
HETEROGENEOUS
CATALYST & CATALYSIS
Dicky Dermawan
www.dickydermawan.net78.net
[email protected]
135
Steps in Heterogeneous Catalysis
1. External diffusion of reactant to catalyst
2. Internal diffusion of reactant in catalyst's
macro- and micro pores
3. Adsorption of reactant on catalyst surface
4. Surface Reaction
5. Desorption of products from catalyst surface
6. Internal diffusion of products through micro- and macro pores
7. External diffusion of product away from catalyst
136
Langmuir’s model for the adsoption of gas on a solid catalyst
permukaan active site dengan tekanan parsial zat itu pada temperatur
137
CO + S
CO +
CO + 2S
CO.S
C-O
CO +
CS + OS
C
O
138
H
H
H
C
C
A(g) + S
H
H
H
H
C
C
PA
Cv
ka
AS
k-a
CAS
H
Pt surface
 rs  k a  PA  C v  k a  CAS

CAS 
  rs  k a  PA  C v 

K
A 

At equilibrium : - rS  0, thus: PA  C v 
Site Balance:
CAS
KA
C AS  K A  PA  C v
Cv 
Ct  Cv CAS  Ct  Cv  KA  PA  Cv
ka
where K A 
 f(T )
k a
C AS 
Ct
1  K A  PA
K A  C t  PA
1  K A  PA
139
At low P:
C AS  K A  C t  PA
At high P:
C AS  C t
(1st
order)
(0th order)
CAS
K A  C t  PA
C AS 
1  K A  PA
PA
Estimation of the parameters:
C AS 
K A  C t  PA
1  K A  PA
1
1  K A  PA

C AS K A  C t  PA
PA 1 K A  PA

C AS
K A  Ct
PA
1
1

  PA
CAS K A  C t C t
PA
C AS
t an  
1
Ct
PA
140
CO + 2S
ka
k-a
CS + OS

C C 
 rA  k a  PCO  C v 2  CS OS 
KA 

At equilibrium:  rA  0  PCO  C v 2 
H H
H H
H H
CCS  COS
KA
CCS  COS , thus: PCO  C v
Site Balance:
H
2
H
Pt surface
COS2

KA
COS  K A1 / 2  PCO1 / 2  C v
Ct  Cv  CCS  COS  Cv  2  COS
Ct  Cv  2  K A1/ 2  PCO1/ 2  Cv
Cv 
Ct
1  2  K A1 / 2  PCO1 / 2
COS 
K A1 / 2  C t  PCO1 / 2

1  2  K A1 / 2  PCO1 / 2

141
Estimation of the parameters:
K A  C t  PCO1 / 2
COS 
1  2  K A  PCO1 / 2
1/ 2
1 1  2  K A  PCO

COS
K A  C t  PCO1 / 2
PCO1 / 2
COS

1  2  K A  PCO1 / 2
K A  Ct
PCO1/ 2
1
2

  PCO1/ 2
COS
K A  Ct Ct
PCO1 / 2
C AS
t an  
2
Ct
PCO1/2
142
144
Sample Problem S7-6
The following data were obtained at 70oC for the equilibrium
adsorption of n-hexane on silica gel particles.
Consider weather the adsorption is molecular or dissociative and
estimate the corresponding parameters.
Partial Pressure of
C6H14 in gas, atm
0.0020
0.0040
0.0080
0.0113
0.0156
0.0206
per gram gel
1.05E-04
1.60E-04
2.72E-04
3.46E-04
4.30E-04
4.73E-04
145
Sample Problem S7-7
The following data were obtained for the equilibrium adsorption
of n-hexane on the same silica gel particles at 110oC.
Consider weather the adsorption is molecular or dissociative and
estimate the corresponding parameters.
Partial Pressure of
C6H14 in gas, atm
0.0005
0.0010
0.0020
0.0050
0.0100
0.0200
per gram gel
2.60E-05
4.50E-05
7.80E-05
1.70E-04
2.70E-04
4.00E-04
146
Sample Problem S7-10
147
Sample Problem S7-11
148
Sample Problem
The data below were recorded at constant temperature for the
adsorption of nitrogen on silica gel at -196oC. Consider weather
the adsorption is molecular or dissociative and estimate the
corresponding parameters.
Pressure, kPa
0.8
3.3
18.7
30.7
38.0
42.7
57.3
of sample (mL at STP)
6.1
12.7
17.0
19.7
21.5
23.0
27.7
149
Involving More Than One Species
Derive mathematical expression when 2 diatomic gases (say A2 & B2)
simultaneously adsorbed in the same active site at the surface of a
catalyst if:
k

a. Both A2 & B2 molecularly adsorbed, i.e.
A2  S
A2 S
a



k
a
k
B2  S
b


B2  S
kb
b. Both A2 & B2 dissociatively adsorbed, i.e.
k
A2  2 S
a



k
2 AS
a
k
B2  2 S
b


2 BS
kb
k
A2  S
a



k
A2 S
a
k
B2  2 S
b

150
2 BS

kb
4th Step: Surface Reaction
Langmuir – Hinshelwood
mechanism for
hydrogenation of ethylene
to ethane
151
Surface Reaction Mechanism
KS 
ks
k s
1. Langmuir - Hinshelwood: Single Site
AS
BS
A
B

C 
rS  k s   C AS  BS 
KS 

2. Langmuir - Hinshelwood: Dual Site
AS+S
AS + BS
S + BS
CS + DS
A
B
D
A B

C C 
rS  k s   C AS  C v  BS v 
KS 

D C

C C 
rS  k s   C AS  C BS  CS DS 
KS


C
3. Eley Rideal
AS + B
CS+ D
A
C
B and D do not adsorbed on catalyst’s surface active site

CCS  PD 

rS  k s   C AS  PB  152
KS 

5th Step: Desorption
CS
kd
k-d
C+S

P C 
rD  k DC   CCS  C v 
K DC 

K DC 
1
K AC
rD  k DC  CCS  KAC  PC  Cv 
153
Combining Ideas: Synthesizing Rate Law
Here, we assume that all diffusion processes (1,2,6,7)
are fast, so the overall processes will depend solely
on adsorption, surface reaction, and desorption.
As a base case, consider the decomposition of
cumene to form benzene and propylene:
catalyst
C6 H 5CH (CH 3 ) 2  
 C6 H 6  C3H 6
Mechanism:
Langmuir-Hinshelwood single site
mechanism
154
Combining Ideas: Synthesizing Rate Law
catalyst
C  
 B  P
KP 
PB  PP
PC
Mechanism:
C+S
kA
k-A
CS

C 
rA  k A   PC  C v  CS 
KC 

* Surface reaction
CS
ks
k-s
BS + P

C BS  PP 


rS  k s   CCS 
KS 

* Desorption
BS
kD
k-D
B+S
rB  k D  C BS  K B  C v  PB

KB 
1
KD
155
What if……..
C+S
kA
k-A
catalyst
C 
 B  P

CCS 


 rC  k A   PC  C v 
KC 

CS
Surface reaction will be in equilibrium:

C P 
rS  k s   CCS  BS P   0
KS 

CCS 
C BS  PP
KS
Desorption will be in equilibrium:
rB  k D  C BS  K B  C v  PB   0
C BS  K B  PB  C v
SITE BALANCE:
Ct  Cv  CBS  CCS
(K B  PB  Cv )  PP
Ct  Cv  K B  PB  Cv 
KS
Cv 
Ct
K
1  K B  PB  B  PB  PP
KS
156
catalyst
C 
 B  P
What if……..
K B  PB  C v   PP


KS
 rC  k A   PC  C v 

KC


 rC '  k A 

CCS 


 rC '  k A   PC  C v 
K AC 

CCS 






Ct
K
1  K B  PB  B PB  PP
KS
Cv 


KB
 PC  
 PP  PB 
KS  K C


C BS  PP
KS
C BS  K B  PB  C v
Ct
K
1  K B  PB  B  PB  PP
KS
; at equilibrium:
PC,eq

KB
KS  K C
Consider the overall reaction as
equilibrium process, i.e. C  B + P
From thermodynamics:
PP,eq  PB,eq
G  R  T  ln(K P )
KB
1

KS  K C K P
Finally, the rate
equation becomes:
KP 
PB,eq  PP,eq
PC,eq
Thus:

P P 
k1   PC  P B 
KP 

 rC ' 
K
1  B  PB  PP  K B  PB
KS
k1  k A  Ct
157
What if……..
catalyst
C 
 B  P
Surface Reaction-controlling
CS
ks
k-s
BS + P

CCS 
  0
rA  k A   PC  C v 
KC 


C P 
 rC  rS  k s   CCS  BS P 
KS 

CCS  K C  PC  C v
Desorption will be in equilibrium:
rD  k D  C BS  K B  C v  PB   0
CBS  K B  PB  Cv
SITE BALANCE:
C t  C v  CBS  CCS
C t  C v  K B  PB  C v  K C  PC  C v
Ct
Cv 
1  K B  PB  K C 158
 PC
catalyst
What if……..
Surface Rxn-controlling (cont’)
Ct
1  K B  PB  K C  PC
ks  KC  Ct
 rC 
1  K B  PB  K C  PC
CBS  K B  PB  Cv
Cv 


K
 K C  PC  B  PB  PP 
KS


Ct
1  K B  PB  K C  PC
KB
1

KS  K C K P


KB
 PC 
 PB  PP 
KS  K C


Finally, the rate equation becomes:
 rC 

C P 
 rC  rS  k s   CCS  BS P 
KS 

CCS  K C  PC  Cv


K B  PB  C v   PP 

 rC  k s   K C  PC  C v 
KS


 rC  k s 
C 
 B  P

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC
k 2  k s  K C  Ct
159
What if……..
Desorption-controlling
Derive the rate expression, i.e. –rC’ for this case
if desorption of benzene is slow compared to all the
other steps
catalyst
C  
 B  P
C+S
Surface reaction CS
Desorption
BS
kA
k-A
ks
k-s
kD
k-D
CS
cepat
BS + P
cepat
B+S
lambat
160
What if……..
Desorption-controlling
BS
kD
k-D
B+S
catalyst
C 
 B  P
 rC '  rD  k D  C BS  K B  C v  PB


CCS 

rA  k A   PC  C v 
KC 

CCS  K C  PC  C v
Surface reaction will be in equilibrium:

C BS  PP 
  0
rS  k s   CCS 
KS 

C BS 
K S  CCS K S  K C  PC  C v 

PP
PP
SITE BALANCE:
C t  C v  CBS  CCS
C t  C v  KS  K C 
PC
 C v  K C  PC  C v
PP
Cv 
Ct
161
P
1  K S  K C  C  K C  PC
P
catalyst
C 
 B  P
What if……..
Desorption-controlling (cont’)
 rC '  k D  C BS  K B  C v  PB
 rC '  k D  CBS  K B  Cv  PB 
C BS  K S  K C 

P
 rC '  k D   K S  K C  C  C v  K B  C v  PB
PP

 rC '  k D 
Ct
P
1  K S  K C  C  K C  PC
PP
KS  K C  Ct
 rC '  k D 
P
1  K S  K C  C  K C  PC
PP
Finally, the rate
equation becomes:
 rC ' 



Cv 
 PC
KB
 

 PB
P
K

K
S
C
 P




PB PP
KP
Ct
P
1  K S  K C  C  K C  PC
PP
KB
1

KS  K C K P




KB
  PC 
 PB  PP
K

K
S
C

k 3  PC 
PC
 Cv
PP
CCS  K C  PC  Cv

P
  K S  K C  C  K B  PB
PP

KS  K C  Ct
 rC '  k D 
PP  K S  K C  PC  K C  PC  PP





PP  K S  K C  PC  K C  PC  PP
k3  k D  KS  KC  Ct
162
catalyst
Summary for C 
 B  P
(kinetic term)  (driving force term)
rate 
C+S
Surface reaction CS
Desorption
BS
 rC ' 

k1  PC 
1
KB
KS
PP PB
KP

 PB  PP  K B  PB
If surface reaction is rate-limiting:
 rC ' 

k 2  PC 
PB PP
KP

k 2  k s  K C  Ct
k3  k D  KS  KC  Ct
CS
k-A
ks
k-s
kD
k-D
BS + P
B+S
PB,eq  PP,eq
PC,eq
KS  K C
KP 
KB
G  R  T  ln( K P )
1  K B  PB  K C  PC
If desorption is rate-limiting:
 rC ' 
k1  k A  Ct
KP 
kA

k 3  PC 
PB PP
KP

PP  K S  K C  PC  K C  PC  PP
Actually, which one is
the rate-limiting step?
163
The Use of Initial-Rate Method
catalyst
C 
 B  P
Carried out experiments using pure C at various PC,0
Thus…….Initially PB = PP = 0
Read the rate of reaction at t = 0, i.e. –rC,0’
-rC’
 rC ' 

k1  PC 
PP PB
KP

K
1  KB  PB  PP  K B  PB
 rC,0 '  k1   PC,0 
PC,0
 rC '
S
If surface reaction is rate-limiting:
 rC ' 

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC
 rC,0 ' 
k 2  PC,0

1  K C  PC,0
PC,0
-rC,0’
If desorption is rate-limiting:
 rC ' 

k 3  PC 
PB PP
KP

PP  K S  K C  PC  K C  PC  PP
 rC ' 
k 3  PC 
 k3 '
K S  K C  PC
164
PC,0
The Use of Initial-Rate Method
catalyst
C 
 B  P
Experimental data:
 rC,0 ' 
Thus…..
k 2  PC,0
1  K C  PC,0
Experiment shows surface reaction is rate-limiting:
 rC ' 

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC

165
Dual Site Mechanism
This is a molecular adsorption,Langmuir-Hinshelwood dual site mechanism
Mechanism:
kA
C+S
k-A
Surface reaction CS + S
Desorption
BS
PS
ks
k-s
kD1
k-D1
kD2
k-D2
CS
BS + PS
If Surface Reaction is controlling:
B+S
P+S
 rC ' 

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC  K P  PP  2
166
What if (a) adsorption (b) desorption is the rate-limiting step?
Dual Site Mechanism
Mechanism:
catalyst
C 
 B  P
C+2S
kA
Surface reaction C’S +C”S
If Surface Reaction is controlling:
 rC ' 

k 2 ' PC 
1 K B  PB 
PB PP
KP

K C  PC  K P  PP
Desorption
BS
PS

C’S + C”S
k-A
kD1
k-D1
kD2
k-D2
ks
k-s
BS + PS
B+S
P+S
2
167
What if (a) adsorption (b) desorption is the rate-limiting step?
P10-3A
168
Adsorbed inert will use some of the vacant site which should be used to
adsorbed the reactants, thus reducing the available site for the reaction.
This will retard the rate of overall reaction.
Consider: C catalyst

 B  P
C+S
Surface reaction CS
Desorption
BS
I+S
kA
k-A
ks
k-s
kD
k-D
kI
k-I
CS
BS + P
B+S
IS

C 
rA  k A   PC  C v  CS 
KC 


C P 
rS  k s   CCS  BS P 
KS 

rB  k D  C BS  K B  C v  PB

C 
rI  k I   PI  C v  IS 
KI 

Use Langmuir – Hinshelwood approach to verify the abovementioned
conclusion if the reaction is:
b. Desorption limiting
c. Surface reaction limiting
169

C+S
catalyst
C 
 B  P
kA
k-A

CCS 


 rC  k A   PC  C v 
KC 

CS
Surface reaction will be in equilibrium:

C P 
rS  k s   CCS  BS P   0
KS 

CCS 
Desorption will be in equilibrium:
rB  k D  C BS  K B  C v  PB   0
C BS  K B  PB  C v
C BS  PP
KS
Inert will be in equilibrium:

C 
rI  k I   PI  C v  IS   0
KI 

C IS  K I  PI  C v
SITE BALANCE:
Ct  Cv  CBS  CCS  CIS
Ct  Cv  K B  PB  Cv 
Cv 
Ct
1  K B  PB 
(K B  PB  Cv )  PP
 K I  PI  Cv
KS
KB
 PB  PP  K I  PI
KS
170
catalyst
C 
 B  P

CCS 


 rC '  k A   PC  C v 
K AC 

CCS 
C BS  PP
KS
C BS  K B  PB  C v
CIS  K I  PI  Cv
Cv 
K B  PB  C v   PP


KS
 rC '  k A   PC  C v 

KC










KB
 PC  
 rC '  k A 
 PP  PB 
K
KS  K C

1  K B  PB  B PB  PP  K I  PI 
KS
 rC ' 
1  K B  PB 
KB
 PB  PP  K I  PI
KS
KP 
Ct
Finally, the rate
equation becomes:
Ct

P P 
k1   PC  P B 
KP 

K
1  B  PB  PP  K B  PB  K I  PI
KS
PB,eq  PP,eq
PC,eq
G  R  T  ln(K P )
k1  k A  Ct
171
Summary
for the Effect of
Mechanism:
Desorption
C 
 B  P
 rC ' 
1
KB
KS
PP PB
KP

 PB  PP  K B  PB
 rC ' 
kD
I+S

k-D
kI
k-I
PP PB
KP
CS
BS + P
B+S
IS

K
1  KB  PB  PP  K B  PB  K I  PI
S

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC
 rC ' 
If desorption is rate-limiting:
 rC ' 
k-s
k1  PC 
If surface reaction is rate-limiting:
 rC ' 
ks
BS

k-A
Surface reaction CS
catalyst
k1  PC 
kA
C+S

k 3  PC 
PB PP
KP

PP  K S  K C  PC  K C  PC  PP
 rC ' 

k 2  PC 
PB PP
KP

1  K B  PB  K C  PC  K I  PI

k 3  PC 
PB PP
KP

172
PP  K S  K C  PC  K C  PC  PP  K I  PI  PP
Reversible vs Irreversible Surface
Reactions
Compare the rate equations for hypothetical Langmuir – Hinshelwood singlesite mechanism cases when surface reactions are rate-controlling.
Case A: Irreversible Reaction
catalyst
A  
 B
Mechanism:
kA
A+S
AS
k-A
Surface reaction AS
Desorption
BS
ks
kD
k-D
BS
B+S
Case B: Reversible Reaction
catalyst
A  
 B
Mechanism:
kA
A+S
AS
k-A
Surface reaction AS
Desorption
BS
ks
BS
k-s
kD
B+S
k-D
Using the usual Hougen - Watson technique, it can be found that:
k s  K A  C t   PA 
 rA '  rS 
1  K A  PA  K B  PB
 rA '  rS 

P
k s  K A  C t  PA  KB
P
1  K A  PA  K B  PB

173
Langmuir – Hinshelwood Single Site & Dual Site
Mechanism
Consider hypothetical case when irreversible surface reactions are ratecontrolling:
Langmuir - Hinshelwood: Dual Site
Langmuir - Hinshelwood: Single Site
catalyst
A 
 B
catalyst
A 
 B
Mechanism:
A+S
Surface reaction AS
Desorption
BS
kA
k-A
ks
kD
k-D
AS
BS
B+S
Mechanism:
A+S
Surface reaction AS +S
Desorption
BS
kA
k-A
ks
kD
k-D
AS
S + BS
B+S
Using the usual Hougen - Watson technique, it can be found that:
k s  K A  C t   PA 
 rA '  rS 
1  K A  PA  K B  PB
 rA '  rS 
k s  K A  C t   PA 
1  K A  PA  K B  PB 2
174
What if (a) adsorption (b) desorption is the rate-limiting step?
Langmuir - Hinshelwood and Eley-Rideal
Mechanism
Consider hypothetical case when irreversible surface reactions are ratecontrolling:
Eley – Rideal:
Langmuir - Hinshelwood:
catalyst
A  D  
 B
catalyst
A  D 
 B
Mechanism:
Mechanism:
A+S
D+S
kA1
k-A1
kA2
k-A2
ks
Surface reaction AS+DS
Desorption
BS
kD
k-D
AS
DS
A+S
kA
Surface reaction AS + D(g)
BS + S
Desorption
BS
B+S
AS
k-A
kD
k-D
ks
BS
B+S
Using the usual Hougen - Watson technique, it can be found that:
 rA '  rS 
k s  K A  C t   PA  Pd 
1  K A  PA  K B  PB  K D  PD 2
 rA '  rS 
k s  K A  C t   PA  PD 
1  K A  PA  K B  PB
175
What if (a) adsorption (b) desorption is the rate-limiting step?
Behavior of Single Site, Dual Site, and EleyRideal Mechanism (cont’)
k s  K A  C t   PA 
 rA '  rS 
1  K A  PA  K B  PB
Pt wire
NH 3   21 N 2  32 H 2
 rA '  rS 
k s  K A  C t   PA 
1  K A  PA  K B  PB 2
Rh(111)
CO  21 O 2   CO 2
176
Integration
0,75 wt Pt , Al O
n  pen tan e    23  i  pen tan e
catalyst
N  
 I
 rN ' 
2,14 10 4   PN  0,954  PI 
1  1,39  PN  1,08  PI 
2
Atm/detik
How long would it takes to achieve 30% conversion in a batch reaction
starting with pure N at PN0 = 10 atm?
What if we start with PN0 = 10 atm & PH0 = 1 atm.
H = n-hexane, considered as inert, KH= 1.4
177
Deduction of Reaction Mechanism
from Rate Equation
(kinetic term)  (driving force term)
rate 
catalyst
A  D  
 B  C
 rA ' 
1  0,49  P
2,087 10 3   PA  PD 
1/ 2
A
 2,76  PB  7,21  PD  5,36  PE
From driving force term:
- Irreversible reaction
- A, B, D, & E adsorbed in catalyst’ surface
-B, D, & E molecularly adsorbed
-Dual site mechanism
-Surface reaction control
178

2
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
Hidrodealkylation of toluene:
catalyst
C6 H 5CH 3  H 2  
 C6 H 6  CH 4
8,7 10 4   PT  PH  gmol toluene
 rT ' 
1  1,39  PB  1,038  PT kg cat  min
catalyst
T  H  
 B  M
From driving force term:
- Irreversible reaction
- B & T adsorbed in catalyst’ surface, both molecularly
- M & H do not OR weakly adsorbed
- Single site mechanism
- Surface reaction control
kA
T+S
k-A
Surface reaction TS + H
Desorption
BS
ks
kD
k-D
TS
BS + M
B+S
fast
slow
fast
179
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
Hidrodealkylation of toluene:
catalyst
C6H5CH3  H 2 
 C6H6  CH4
T  H 
 B  M What if:  rT ' 
catalyst
8,7 10 4   PT  PH 
1  1,39  PB  1,038  PT 2
gmol toluene
kg cat  min
From driving force term: - Irreversible reaction
- B & T adsorbed in catalyst’ surface, both molecularly
- M & H do not OR weakly adsorbed
- Dual site mechanism
- Surface reaction control
kA
T+S
k-A
Surface reaction TS + S
Desorption
BS
MS
ks
kD
k-D
kD
k-D
TS
BS + MS
B+S
M+S
fast
slow
fast
fast
180
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
Isomerization of pentane:
catalyst
N 
 I
0,75wt Pt, Al O
n  pen tan e 23  i  pen tan e
2,14 10 4   PN  0,954  PI 
 rN ' 
1  1,39  PN  1,08  PI 2
From driving force term:
- Reversible reaction with KP = 1/0.954
- N & I adsorbed in catalyst’ surface, both molecularly
- Dual site mechanism
- Surface reaction control
kA
N+S
k-A
Surface reaction NS + S
Desorption
IS
ks
kD
k-D
NS
S + IS
I+S
fast
slow
fast
181
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
Chemical Vapor Deposition:
GeCl 2(g )  H 2(g ) 
 Ge(s )  2 HCl(g)
rDep " 

k1  PGeCl2  PH 2



From driving force term:
1  k 2  PGeCl2  k 3  PH 2 3
- Irreversible
- Triple site mechanism
- Surface reaction control
GeCl2 + S
H2 + 2 S
kA
k-A
kA
k-A
Surface reaction GeCl2.S + 2 HS
Desorption
GeCl2.S
2 H.S
ks
fast
fast
Ge + 2 HCl + 3 S slow
182
none
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
183
Sample Problem for
Deduction of Reaction Mechanism from
Rate Equation
184
P10-7B
Sample Problem for
Deduction of Reaction Mechanism from Rate Equation
185
P10-11B
Deduction of Reaction Mechanism from Rate Equation
186
Interpretation of Kinetic Data
Mekanisme
Metode Least Square

PercoBaan
Data Kinetik
2
V
Bentuk Persamaan
Regresi – Estimasi Parameter
Penyederhanaan
187
P10-5B
Interpretation of Kinetic Data
188
P10-5B
Interpretation of Kinetic Data
189
P10-8B
Interpretation of Rate Data
190
P10-9C
Interpretation of Rate Data
191
P10-10B
Interpretation of Rate Data
192
INTERPRETASI DATA KINETIK
OH
+ H2O
A
Rx 105
B + C
PA
PB
PC
3.3
1
1
1
1.05
5
1
1
0.565
10
1
1
1.826
2
5
1
1.49
2
10
1
1.36
3
0
5
1.08
3
0
10
0.862
1
10
10
0
0
5
8
1.37
3
3
3
193
Catalyst Deactivation
at 
laju saat katalis berumur t
laju saat katalisnyamasih baru
Types of Decay
1.) Sintering
2.) Coking
3.) Poisoning
4.) Slow Decay
Temperature-Time Trajectories
5.) Moderate Decay
Moving Bed
6.) Rapid Decay
STTR
194
Catalyst Deactivation
Separable Kinetics:
Decay reaction order n (empirical): 
da
 kd  a n
dt
n = 0 (linear)
: Conversion of p-hidrogen on tungsten
poisoned with oxygen
n = 1 (exponential)
: & ethylene hydrogenation on Cu
poisoned with CO
& paraffin dehydrogenation on Cr.Al2O3
& cracking of gas oil
& vinyl chloride monomer formation
n = 2 (hyperbolic)
: % vinyl chloride monomer formation
% cyclohexane deydrogenation on Pt/Al2O3
% isobutylene hydrogenation on Ni
 Ed 
k

A

exp


Decay constant usually follow Arrhenius form: d
d
 RT 
195
Integration of Decay Reaction Order n
1st order decay:
a
t
da
 a  k d . dt
1
1
a
ln    k d  t
1
a  e  k D t
2nd order decay:

da
a
2
 k d   dt
1
. 1  kd  t
a
1
 1 kd  t
a
1
a
1 kd  t
196
Investigating Rate of Decay
t, hour
0
1
2
X,%
3
2.3
2.0
3/
2.3/
2.0/
a
1st order decay:
3
3
3
2nd order decay:
a  ek D t
a
-ln a
1
1 kd  t
1
a
Set Intersept = 0
t
1
Set Intersept = 1
t
197
Example P10-25C
T = 500 oC
T,hari
0
20
40
60
10
120
X,%
1
0.7
0.56
0.45
0.38
0.29
a
T = 550 oC
T,hari
0
5
10
15
20
30
40
X,%
2
1,2
0,89
0,69
0,57
0,42
0,33
a
198
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