Lecture 17
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Web Lecture 17
Class Lecture 22–Thursday 4/4/2013
Introduction to Catalysts and Catalysis
 Interstage cooling
 Noble Prize 2007
 Catalytic steps
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Catalysts and Catalysis
 A Catalyst is a substance that affects the rate of
chemical reaction but emerges from the process
unchanged.
 Catalysis is the occurrence, study, and use of
catalysts and catalytic processes.
Approximately 1/3 of the GNP of materials
produced in the U.S. involves a catalytic process.
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Catalysts and Catalysis
Catalysts affect both selectivity and yield
Different reaction paths
4
Catalysts and Catalysis
Different shapes and sizes of catalyst.
5
Catalysts and Catalysis
Catalytic packed-bed reactor, schematic.
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Steps in a Catalytic Reaction
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Active Sites
 Reactions are not catalyzed over the entire
surface but only at certain active sites or centers
that result from unsaturated atoms in the surface.
 An active site is a point on the surface that can
form strong chemical bonds with an adsorbed
atom or molecule.
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Active Sites – Ethylidyne on Platinum
The Adsorption Step
Vacant and occupied sites
For the system shown, the total concentration of sites is
Ct = Cv + CA.S + CB.S
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The Adsorption Step
A  S  A S
rAD  k A PACv - k -AC AS  k A PACV  C AS / K A 
K A  k A / k A
-1
[atm ]
@ equilibrium : rAD  0
rAD / k A  0
C AS  k A PACV
C AS  k A PACV
Ct  CV  CAS  CV  K A PACV  CV (1  K A PA )
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𝐶𝑡
𝐶𝑉 =
1 + 𝐾𝐴 𝑃𝐴
Langmuir Adsorption Isotherm
𝐶𝑡
𝐶𝑉 =
1 + 𝐾𝐴 𝑃𝐴
𝐶𝐴∙𝑆 = 𝐾𝐴 𝑃𝐴 𝐶𝑉
𝐶𝐴∙𝑆 =
𝐾𝐴 𝑃𝐴
𝐶𝑡
1+𝐾𝐴 𝑃𝐴
𝐶𝐴∙𝑆
𝐾𝐴 𝑃𝐴
=
𝐶𝑡
1 + 𝐾𝐴 𝑃𝐴
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Langmuir Adsorption Isotherm
C AS
CT
Increasing T
Slope=kA
C AS
K A PA

Ct
1  K A PA
Langmuir Adsorption
Isotherm
PA
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The Surface Reaction Step
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The Surface Reaction Step
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The Surface Reaction Step
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The Surface Reaction Step
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The Surface Reaction Step
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Steps in a Catalytic Reaction
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Desorption from the Surface for the Reaction
A 
 B  C


C S 
CS

rDC


PC C  
 k D CCS 

K DC 

(10-20)
rDC  rADC
K DC
1

KC
rDC  k D CCS  K C PC C  
(10-21)
Steps in a Single-Site Catalytic Reactor
Adsorption
A  S  A S
Surface Reaction A  S  B  S
Desorption
BS  B S

C AS 
 rA  rAd  k Ad  PACv 

k
A 


CBS 
 rA  rS  k S C AS 

k
C 

 rA  rD  kD CBS  kB PBCB 
Which step is the Rate Limiting Step (RLS)?
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The Rate Limiting Step:
Which step has the largest resistance?
Electrical analog to heterogeneous reactions
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Collecting and Analyzing Data
Collecting information for catalytic reactor design
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Collecting and Analyzing Data
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Catalytic Reformers
 Normal Pentane Octane Number = 62
 Iso-Pentane Octane Number = 95
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Catalytic Reformers
n-pentane
0.75 wt% Pt
i-pentane
Al2O3
n-pentane
-H2
n-pentene
Al2O3
+H2
i-pentene
Pt
Pt
n-pentene
N
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i-pentane
Al2O3
i-pentene
I
Catalytic Reformers
Isomerization of n-pentene (N) to i-pentene (I) over alumina
N
Al2O3
I
1. Select a mechanism (Mechanism Single Site)
Adsorption on Surface:
N S  N S
Surface Reaction:
N S  IS
Desorption:
IS  I S
Treat each reaction step as an elementary reaction when writing rate laws.
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Catalytic Reformers
2. Assume a rate-limiting step.
Choose the surface reaction first, since more than 75%
of all heterogenous reactions that are not diffusionlimited are surface-reaction-limited. The rate law for the
surface reaction step is:
N SS  ISS


C
I

S

 rN   r  rS  k S  C NS 
KS 

'
I
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Catalytic Reformers
3. Find the expression for the concentrations of
the adsorbed species
CN.S and CI.S. Use the other steps that are not limiting to
solve for CN.S and CI.S. For this reaction:
N S  N S
rAD
From
 0:
kA
CNS  PN K NC
I S  I  S
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rD
From
 0:
kD
C IS
PI C 

 K I PI C 
KD
Catalytic Reformers
4. Write a Site Balance.
Ct  C  CNS  CIS
5. Derive the rate law. Combine steps 2, 3 and 4 to
arrive at the rate law :
k




k s Ct K N PN  PI K P 

 rN  rS 
1  K N PN  K I PI 
 rN   rS 
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k PN  PI K P 
1  K N PN  K I PI 
Catalytic Conversion of Exhaust Gas
HC
CO
1994
0.41
3.4
2004
0.125
3.4
2008
0.10
3.4
NO
0.4
0.4
0.14
1
CO  NO  CO 2  N2
2

Catalytic Conversion of Exhaust Gas

NO  S NO • S

CO • S

CO • S



C
rANO  k NO PNO C V  NO •S  C NO •S  K NO PNO C V
K NO 


C 
rACO  k CO PCO C V  CO•S 
K CO 

C CO•S  K CO PCO C V
CO • S  NO • S  CO 2  N • S  S rS  k S C CO•S C NO •S 
N •S N •S
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
N 2 g   2S


rD  k D C 2N•S  K N 2 PN 2 C 2V

C N•S  C V K N PN 2
Catalytic Conversion of Exhaust Gas
rS  k S C NO•SCCO•S 
rS  k SK NOKCO PNOPCO C 2V
C T  C V  C NO•S  CCO•S  C N•S
 C V  C V K NOPNO  C V KCO PCO  C V K N 2 PN 2
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Catalytic Conversion of Exhaust Gas
CV 
Ct
1  K NO PNO  K CO PCO  K N 2 PN 2
  rS 
 rNO
 
 rNO
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1  K
1  K
k



k S K NO K CO C 2t PNO PCO
NO
PNO  K CO PCO  K N 2 PN 2
kPNO PCO
NO
PNO  K CO PCO  K N 2 PN 2

2

2
Catalytic Conversion of Exhaust Gas
 
rNO
Neglect
kPNOPCO
1 KNOPNO  KCO PCO 
K N 2 PN 2
 
rNO



K N 2 PN 2
kPNOPCO
2
1 K NOPNO  KCO PCO 

2
Catalytic Conversion of Exhaust Gas
 
rNO
kPNOPCO
2
1 K NOPNO  KCO PCO 
Find optimum partial pressure of CO
drNO 
0
dPCO
PCO 

1 K NOPNO
K CO
End of Web Lecture 17
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