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 23 – 3/31/2011
 Interstage Cooling
 Noble Prize 2007
 Catalytic Steps
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 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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Different reaction Paths
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Different shapes and sizes of catalyst.
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Catalytic packed-bed reactor, schematic.
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 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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Vacant and occupied sites
For the system shown, the total concentration of sites is
Ct = Cv + CA.S + CB.S
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A  S  A S
rAD  k A PACv - k -A C AS  k A PACV  C AS / K A 
K A  k A / k A
[atm -1 ]
@ equilibriu m : rAD  0
rAD / k A  0
C AS  k A PACV
C AS  k A PACV
Ct  CV  C AS  CV  K A PACV  CV (1  K A PA )
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Ct
CV 
1  K A PACV
Ct
CV 
1  K A PACV
C A S  K A PACV
C A S
K A PA
Ct

1  K A PA
C A S
K A PA

1  K A PA
Ct
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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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A 
 B  C



C S 
CS


PC C  
rDC  k D CCS 

K DC 


(10-20)
rDC  rADC
K DC
1

KC
rDC  k D CCS  KC PC C  
(10-21)
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  k D CBS  k B PBCB 
Which step is the Rate Limiting Step (RLS)?
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Electrical analog to heterogeneous reactions
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Collecting information for catalytic reactor design
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 Catalytic Reforming
 Normal Pentane Octane Number = 62
 Iso-Pentane Octane Number = 95
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n-pentane
n-pentane
-H2
Pt
n-pentene
n-pentene
N
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0.75 wt% Pt
Al2O3
Al2O3
i-pentane
i-pentene
Al2O3
+H2
i-pentene
I
Pt
i-pentane
Isomerization of n-pentene (N) to i-pentene (I) over alumina
N
1. Select
I
a mechanism (Mechanism Single Site)
Adsorption on Surface:
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Al2O3
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.
2. Assume a rate-limiting step. Choose the surface
reaction first, since more than 75% of all heterogenous reactions
that are not diffusion-limited are surface-reaction-limited. The
rate law for the surface reaction step is
N S S  I S S

C IS 
'


 rN  rI  rS  k S  C NS 
KS 

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Find the expression for the concentrations of the
adsorbed species CN.S and CI.S. Use the other steps that are not
3.
limiting to solve for CN.S and CI.S. For this reaction,
N S  N S
rAD
From
 0:
kA
C NS  PN K N C
I S  I  S
rD
From
 0:
kD
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C IS
PI C 

 K I PI C 
KD
4. Write a Site Balance.
Ct  C  C NS  CIS
5. Derive the rate law. Combine steps 2, 3 and 4 to arrive at
the rate law:
k




k s C t 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 
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  CO2  N 2
2

NO  S
CO • S

NO • 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
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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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
 
rNO
Neglect



kPNOPCO
1 K NOPNO  KCOPCO 
K N 2 PN 2
K N 2 PN 2
 
rNO
kPNOPCO
2
1 K NOPNO  KCOPCO 

2
 
rNO
kPNOPCO
1 K NOPNO  KCO PCO 
Find optimum partial pressure of CO
drNO 
0
dPCO
PCO

2
1 K NOPNO

K CO
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