Lecture 24
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.
Today’s lecture
 Normal Pentene  Iso-Pentane
 Langmuir Adsorption Isotherm
 Catalytic Converters for Automobiles
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Last Lecture
 Catalytic Reforming Reaction (Gasoline)
 Normal Pentane Octane Number=62
 Iso-Pentane Octane Number=90
 Normal Pentane  Iso-Pentane
3
4
Steps in a Catalytic Reaction
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Steps in a Catalytic Reaction
AB
Adsorption on Surface
Surface Reaction
Single Site
A  S  B S
Dual Site
A  S  S  B S  S
Desorption from Surface
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A  S  A S
B S  B  S
n-pentane
n-pentane
-H2
Pt
n-pentene
n-pentene
N
7
0.75 wt% Pt
Al2O3
Al2O3
i-pentane
i-pentene
Al2O3
+H2
i-pentene
I
Pt
i-pentane
k




2
k SC1 K N PN  P1 / K P 
 rN  rS 
2
1  K N PN  K1P1 
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Isomerization of n-pentene (N) to i-pentene (I) over alumina
N
Al2O3
I
1. Select a mechanism (Mechanism Dual Site)
Adsorption on Surface:
Surface Reaction:
Desorption:
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N  S  N S
N S  S  I S S
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


C
C
1

S


rN  rS  k S  C C NS 
KS 

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3. Find the expression for concentration of the adsorbed species
CI.S. Use the other steps that are not limiting to solve for CI.S
(e.g., CN.S and CI.S). For this reaction,
rAD
From
 0:
kA
rD
From
 0:
kD
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CNS  PN K NC
C IS
PI C 

 K I PI C 
KD
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




2
k s C I K N PN  PI K P 

 rN  rS 
2
1  K N PN  K I PI 
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13
14
15
Automotive Catalytic Converters
Placed inline with exhaust system (~$500)
NOx plays big role in smog formation
CO 2
CO
NO
'
 rNO
CO  NO  inert products(CO2  1 / 2N2 )
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Automotive Catalytic Converters
1) Propose a mechanism
NO  S  NO  S
CO  S  CO  S
CO  S  NO  S  CO 2  N  S  S
rADNO  k NO PNO C V  C NO S / k NO 
rADCO  k CO PCO C V  C COS / k NO 
rS  k S C COS N NO S 
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Automotive Catalytic Converters
(from graph data, we see CO2 is not on the surface and that the
surface reaction is irreversible)
N  S  N  S  N2  2S

rD  k C
2
NS
 PN2 CVk N2

rADNO / k NO  0  C NO S  k NO PNO CV
rADCO / k CO  0  CCOS  k CO PCO CV
rD / k N 2  0  C NS  CV k N PN 2
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Automotive Catalytic Converters
'
 rNO
 rS  kSk NO k CO PCO PNO CV
CT  CV  CNOS  CCOS  CNS
CT
CV 
1  k NO PNO  PCO k CO  k N2 PN2
r
'
NO
19

1  k
k Sk NO k CO PCO PNO C
2
T
P  PCO k CO  k N 2 PN 2
NO NO

2
Automotive Catalytic Converters
Is this consistent with the given data?
-No CO2 in rate law
-rate law decreases as PN2 increases
The contribution of PN2 is very small and can be neglected
r
'
NO
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kPCO PNO

2
1  k NO PNO  k CO PCO 
Automotive Catalytic Converters
rNO
optimum
@ optimum :
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drNO
0
dPCO
PCO
opt
1 k NO PNO

kCO
N  S  N S
N S  I S
I S  I  S
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End of Lecture 24
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