chapter 10 lecture 1
TRANSCRIPT
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Chemical Reaction Engineering
Week 10 (S1-2012)Fogler Chapter 10
Catalysis and Catalytic Reactors
Lecture 1
Gia Hung Pham
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Objectives
- Catalysts and catalytic reactor design
- Steps in a heterogeneous reaction
- Reaction mechanism
- Rate law
- Catalyst decay
- Reactor types
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Catalysts
Definition
- Fogler: A substance that affects the rate of a reaction but
emerges from the process unchanged.
- Ostwald (1895): A catalyst accelerates a chemicalreaction without affecting the position of the equilibrium.
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Phenomenon
Reaction A P
Catalyst changes reaction pathway.
A
Cat Cat-A
Chemical bonding
Cat-PP
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- Positive catalyst increases reaction rate.
- Negative catalyst decreases reaction rate.
Importance of catalysis
- About 75% of all chemicals are produced by catalytic process.
Catalyst can be:- Gas (NOx)
- Liquid (H2SO4)
- Solid (Ni)
Note: In this chapter only solid catalysts will be considered.
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Reaction energetic
Energy
A
P
Ea (without cat)
H
Ea (with cat)Cat-A
Cat-P
adsorption
reaction
desorption
Reaction progress
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Heterogeneous catalyst
PBR
Reactants (gas or liquid)
Solid catalyst
Products
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Heterogeneous catalyst
PBR
Catalyst Bed
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Heterogeneous catalyst
PBR
Catalyst Bed Catalyst pellet
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Heterogeneous catalyst
PBR
Catalyst Bed Catalyst pelletCatalyst pore
Active site
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Catalyst surface area
- Specific surface area of silica-alumina cracking catalyst is
300 m2/g.
- External surface area of 1mm alumina sphere is
0.0015m2
/g.- A catalyst that has large area resulting from pores
(internal surface area) is called a porous catalyst.
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Film theory
Fogler. Page 773
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Individual steps in heterogeneous catalysis ( A P )
CAb
CAs
Cat-A
Cat-P
A
A
AA
PP
P
4
Boundary layer
P
Catalyst pellet
Catalyst pore
1
2 3
56
7
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Individual steps in heterogeneous catalysis (Mass transfer and
chemical reaction)
Step 1. Diffusion of the reactant A from the fluid to the catalystsurface (External mass transfer).
Step 2. Diffusion of the reactant A from the pore mouth into thepore (Internal mass transfer or pore diffusion).
Step 3. Adsorption of the reactant onto the surface of the pore
(Reactant adsorption).Step 4. Chemical reaction on the catalyst surface to form product
P (Surface reaction).
Step 5. Desorption of the product P from the catalyst surface(Product desorption).
Step 6. Diffusion of the product P out of the pore (Internal masstransfer or pore diffusion).
Step 7. Diffusion of the product from the pore mouth through theboundary layer into the bulk fluid (External mass transfer).
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Catalytic reaction Kinetics
Mass transfer
and Effective reaction raterA,eff
Catalytic reaction
Reaction kinetics without mass transfer effects.
We choose the conditions, under which the mass transfers do not affect the effective
reaction rate.
For example: High feed flow rate and /or small catalyst particle size
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No mass transfer limiting CA,b = CA,s = CA
Conc.
CAs
Catalyst pore
bulk
CAb CA
Concentration of A in catalyst pore does NOT change
with pore length L.
L
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Assumptions: - Uniform surface
- Active site has the same attraction
Conc.
CAs
Catalyst pore
s s s s s
bulk
Active site
Catalyst surface
CAb CA
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Steps in a heterogeneous catalytic without mass transfer effects
- Step 3 : Adsorption of reactants- Step 4 : Surface reaction catalytic chemical reaction
- Step 5 : Desorption of products
Reaction A P
s s s s s
A P
Catalyst surface
3
45
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Adsorption Isotherms
Concentration of active site
Active site balance: Ct = Cv + CAs + CPs
s s s s s s
PA
Occupied sitesVacant sites
Total active si te per gram catalyst (mol/gcat)
Vacant site per gram catalyst (mol/gcat)
Occupied sites by species A and B (mol/gcat)
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Two adsorption models
- Model 1: Molecular or nondissociated adsorption.
Example: Adsorption of CO on Ni-catalyst
CO
- Model 2: Dissociative adsorption
Example: Adsorption of CO on Fe-catalyst
ss s s s
CO
Molecule COS = Ni
CO + S COS
CO
s s s s
C O
Atom C and O
S = Fe
CO + 2S C-S +OS
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Model 1
Rate law for the reaction
Rate of attachment = kA PCO Cv
Rate of detachment = k-A CCOS
Net rate of adsorption rAD = kAPCOCv k-A CCOS
or
with (adsorption equilibrium constant)
CO + S COS
kA
k-A
A
A
A
k
kK
A
SCO
vCOAAD
K
CCPkr
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Site balance : Ct = CV + CCOSAt equilibrium rAD = 0 we have CCOS = KACVPCO
This adsorption isotherm equation is called Langmuir Isotherm.
COA
tCOA
SCO PK
CPKC
1
CCOS
PCO
Linear
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Model 2
CO + 2S CS + OS
Net reaction rate
At equilibrium rAD
= 0
Site balance Ct = CV + COS + CCS
kA
K-A
SCSOAVCOAADCCkCPkr
2
A
SOSC
VCOAAD
K
CCCPkr
2
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2/1
2/1
21 COA
COAt
SO
PK
PKCC
COS
PCO
parabolic
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Inverse of both sides and multiply2/1
COP
t
CO
AtOS
CO
C
P
KCC
P2/1
2/1
2/121
2/1
COP
OS
CO
C
P2/1
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Adsorption isotherm equation of A in the presence of adsorbate B
s s s s s
A B
Ct = CV + CAS + CBS
BBAA
tAA
SA
PKPK
CPKC
1
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Surface reaction
* Model 1: Single site
A S BS
s s
A B
kS
k-S
S
SB
SASs
K
CCkr
ks
k-s
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* Model 2: Dual site
AS + S BS + S
AB
S
VSB
VSASs
K
CCCCkr
ks
k-s
ks
k-s
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* Model 3: Eley Rideal
AS + B(g) CS
B(g)
A C
S
SC
BSASs
K
CPCkr
s s
ks
k-s
ks
k-s
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Desorption
CS C + S
With KDC = 1/KC
s s
CC
DC
VC
SCDDC K
CPCkr
VCCSCDDC
CPKCkr
kD
k-DkD
k-D
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Synthesizing a rate law
Mechanism and rate limiting step
Foglers Example (page 671): Reaction
C6H5CH(CH3)2 C6H6 + C3H6Cumene Benzene PropyleneC B P
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Reaction mechanism
C + S CS (Adsorption)
CS BS + P(gas) (Surface reaction)
BS B + S (Desorption)
kA
k-A
kS
k-S
kD
k-D
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Reaction rate
Adsorption
rAD = kA Pc Cv k-A CCS
Surface reaction
rS = kS C CS k-S PP CBS
Desorption
rD = kDCBS k-D PB CV
S
SBPSCSS
K
CPCkr
VBBSBDD CPKCkr
C
SC
VCAAD
K
C
CPkr
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Case 1
The reaction rate of three discussed reactions are equal
DSADC rrrr '
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Case 2
Reaction rate limiting
One or two of three discussed reactions has/have very slow reaction rate.
We assume the adsorption of cumene is rate limiting (slowest step).
What is the rate law if it is true?
Adsorption rate
We need to express Cv and CCS in terms of PC, PB and PP
C
SC
VCAAD
K
CCPkr
Unknown
Measurable
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kA : small
kS : large or rS/kS 0
kD : large or rD/kD 0
Surface reaction
S
SBP
SCSS
KCPCkr
S
PSB
SC
K
PC
C
unknown
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Desorption VBBSBDD
CPKCkr VBBSB
CPKC
V
S
PB
BSCC
K
PPKC
CS
VPBB
VCAAD
KK
CPPKCPkr
Overall reaction C B + P
B
CS
PK
KKK
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Site balance Ct = CV + CCS + CBS
Solving for CV we have
VBBVPB
S
B
VtCPKCPP
K
KCC
BBBP
S
B
t
V
PKPP
K
K
CC
1
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Adsorption control rate law
At t = 0 min we have Pc = Pc,0
PP = 0 (no product)
PB = 0 (no product)
BPBP
S
B
P
PB
CAt
AD
PKPPK
K
K
PPPkC
r
1
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Adsorption control rate law at t = 0 min
Plot initial rate versus
.
0,0,
'
, CCAtADOC PkPkCrr '
,OCr
0,CP
'
,OCr
0,CP
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We assume the surface reaction is rate limiting (slowest step).
What is the rate law if it is true?
Surface reaction
* kS : small
* kA : large or rAD/kA 0
* kD : large or rD/kD 0
S
SBP
SCSS K
CPCkr
We get these concentrations from
adsorption and desorption reaction.
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Adsorption reaction
Desorption reaction
C
SC
VCAAD K
CCPkr
VCCSCCPKC
VBBSBDD
CPKCkr
VBBSBCPKC
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Surface reaction rate limiting
V
P
BP
CCSSC
K
PPPKkr
We get this one from site balance
CCBBt
V
PKPK
CC
1
CCBB
B
PB
CtCS
S
PKPK
K
PPPCKk
r
1
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At t = 0 min
At low partial pressure of C we have 1 >> KC PC,0
Initial rate increase linearly with the partial pressure of C
0,
0,,
0,
1 CC
C
SC
PK
Pkrr
tCSCKkkwith
0,
,
0, CSC Pkrr
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At high initial partial pressure of C we have KC PC,0 >>1
tconsK
krr
CSC tan
,
0,
'
,OCr
0,CP
k
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We assume the desorption of Benzene reaction is rate limiting (slowest step).
What is the rate law if it is true?
Desorption reaction
* kD : small
* kA : large or rAD/kA 0
* kS : large or rS/kS 0
VBBSBDD
CPKCkr
Not measureable but we can get this one from adsorption and surface reaction.
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Surface reaction
Adsorption reaction
Site balance
P
SC
SSBP
C
KC
,
VCCSCCPKC
CC
P
C
SC
t
V
PK
P
PKK
CC
1
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Desorption limiting
At t= 0 min
PCCCSCP
B
PB
C
tSCDDC
PPKPKKP
K
PP
P
CKKkrr,
0,
tconsCkrtDC
tan,
0,
'
,OCr
0,CP
Foglers algorithm for determining reaction mechanism and rate
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Fogler s algorithm for determining reaction mechanism and rate
limiting step.
1. Select a mechanism
2. Assume a rate-limiting step
3. Find the expression for concentration of the
adsorbed species Ci,S
4. Write a site balance
5. Derive the rate law
6. Compare with data
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In-class exercise
The following reaction is carried out over a solid catalyst, the reactionmechanism is believed to be
We assume step 3 is the slowest step, derive the rate law for thereaction.
SDSD
SCSC
SDSCSBSA
SBSB
SASA
)5
)4
)3
)2
)1