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Thermodynamic Aspects of Heterogeneous Catalysis
Research
Raimund Horn
Fritz Haber Institute of the Max Planck Society Berlin
Emmy Noether Research Group „High Temperature Catalysis“
10/26/2012
Modern Methods in Heterogeneous Catalysis Research WS 2012/2013
1
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Introduction
2
One way of dealing
with thermodynamics!
or
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Content
3
1. Why does Thermodynamics matters in Catalysis?
2. Thermodynamic Quantities, Concepts and Tools
3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
4. Chemical Equilibrium: Conversion and Selectivity Calculations for
Multiple Reactions
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1. Why does Thermodynamics matters in Catalysis?
4
The best-known „catalytic reactor“:
One reaction catalyzed bythe catalytic converter:
In the gas phase
MCOMOCOOCOOCO
2
22
CO
2OCO
CO + ½ O2 CO2 rHo (298K) = -283 kJmol-1
OCO C
OO
OOC OOC
2CO 2CO
On Pt
Pt
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1. Why does Thermodynamics matter in Catalysis?
5
TemperatureEnergetics of CO oxidation
RTEr exp
E
E
In the gas phase
On Pt
Taken from: G. Ertl, Catalysis: Science and Technology Vol. 4 1983 p. 245ff, all enthalpies in kJmol-1
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1. Why does Thermodynamics matter in Catalysis?
6
Kinetics
TEquilibrium Transport
TCTkr AA
RTGK R
0
ln TCTDJ
AAB
A
0
0
00 T
Tpp
FFvv
T
T AKAKAads
ads
1
0
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1. Why does Thermodynamics matters in Catalysis?
7
Thermodynamics matters in Catalysis because
Thermodynamics determines the maximum extentof any given reaction
The reaction rate decreases withapproach to thermodynamic equilibrium
Heat release or uptake by chemical reactions leads tocomplex non-isothermal reactor behavior
Adsorption/Desorption thermodynamics determines theconcentration of adsorbed species and in turn the rate of surface reactions (e.g. Langmuir Hinshelwood rate expressions)
The kinetics of elementary steps are related to theirthermodynamics
macroscopic
microscopic
C
BAfA K
CCkrBA
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1. Why does Thermodynamics matter in Catalysis?
8Taken from: M. Salciccioli et al. Chem. Eng. Sci. 66 (2011) 4319
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9
Equilibrium Conversion and SelectivityCalculations for Multiple Reactions
(Examples)
Quantities, Concepts and Tools
First Law of TD, U, H, CV, Cp
Entropy S and Second Law of TD
Clausius Inequality and Gibbs Free Energy G
Chemical Potential µ
Chemical Equilibrium and Law of Mass Action
Chemical Conversion for a Single Reaction (Example)
Consequences for Reactor Design (Examples)
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2. Thermodynamic Quantities, Concepts and Tools?
10
2.1) Systems in Thermodynamics
System Surroundings Exchange of Matter
Exchange of Energy
e.g. a heatedflow reactor
e.g. an autoclave
e.g. a thermos bottleor the universe
open system closed system isolated system
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11
2.2) Processes
Irreversible Processes
2. Thermodynamic Quantities, Concepts and Tools?
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2.2) Processes
Reversible Processes
2. Thermodynamic Quantities, Concepts and Tools?
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13
Thermodynamic Quantity Symbol Intensive or Extensive? SI Unitmass m extensive kg
molar mass M intensive kgmol-1
temperature T intensive Kpressure P, p intensive Pafugacity f intensive Padensity intensive kgm-3
volume V extensive m3
molar volume Vm, v,… intensive m3mol-1
heat Q extensive Jwork W extensive J
inner energy U extensive Jenthalpy H extensive J
free energy, Helmholtz free energy F, A extensive Jfree enthalpy, Gibbs free energy G extensive J
2.3) Important Quantities
2. Thermodynamic Quantities, Concepts and Tools?
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14
Thermodynamic Quantity Symbol Intensive orExtensive SI Unit
entropy S extensive JK-1
molare inner energy Um,u intensive Jmol-1
molar enthalpy Hm,h intensive Jmol-1
molar free energy Am, a, Fm, f intensive Jmol-1
molar free enthalpy Gm, g intensive Jmol-1
chemical potential µ intensive Jmol-1
molar entropy Sm,s intensive JK-1mol-1
heat capacity at constant volume CV extensive JK-1
heat capacity at constant pressure CP extensive JK-1
specific heat capacity c(V,p) intensive Jkg-1K-1
molar heat capacity Cm(V,p), cm(V,p) intensive Jmol-1K-1
2.4) Important Quantities
2. Thermodynamic Quantities, Concepts and Tools?
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2.4.1) Internal Energy U in J
..... magnelnucchemrotvibtrans UUUUUUUU
VC (internal pressure )0 for ideal gases
WQU First Law of Thermodynamics:
2. Thermodynamic Quantities, Concepts and Tools?
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16
2.4.2) Internal Energy U in J
WQU
The First Law of Thermodynamics is important for describing non-isothermalreactors (energy balance) but says nothing about the direction of a physical orchemical process!
2. Thermodynamic Quantities, Concepts and Tools?
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17
2.4.3) Enthalpy H in J
weight
area
....
),....,,,,(
1,,1,,
21
1
dnnHdp
pHdT
THdH
nnnpTHH
jjj nTpnTnp
k
pC (isothermer drosseleffekt in m3) 0 for ideal gases
2. Thermodynamic Quantities, Concepts and Tools?
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18
2.4.3) Enthalpy H in J
2. Thermodynamic Quantities, Concepts and Tools?
iiiii
DCBA nndndDCBA 0, 1 extent of reaction
H
Tp,
H
enthalpy of reaction i
ii HH
322 23
21 NHHN example:
3221
23
21
NHHN HHHH
The standard enthalpy of formation Hf° or standard heat of formation of acompound is the change of enthalpy that accompanies the formation of 1mole of the compound from its elements, with all substances in their standardstates. The standard state of a gas is the (hypothetical) ideal gas at 1bar and298.15K. For liquids and solids it is the pure substance at 1bar and 298.15K.
32231
23
21)( 3 NHHNfNH HHHNHHH
Hess Law
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19
2.4.4) Heat capacities CV, Cp in J/K RCC mVmp ,,for ideal gases
2. Thermodynamic Quantities, Concepts and Tools?
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2.4.4) Heat capacities CV, Cp in J/K
2. Thermodynamic Quantities, Concepts and Tools?
KmolJCC
molkJNHHN
ipiipr
186.22 9.45H 23
21
r322
22 23
21 HN
barpKT
115.298
31NHbarp
KT1
15.298
molkJ9.45
22 23
21 HN
barpKT
1400
31NHbarp
KT1400
molkJ2.48
2
1
r12 )(H)(HT
Tprr dTCTT
Kirchhoff‘s law
pCT
Hr
p
r
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21
2.4.5) EntropyExample: 3 Molecules (A,B,C) distributed among 4 energy states(0=0, 1= 1, 2=21, 3=31) with a total energy of total=31 (isolated system)
Energy State Energy Macrostate
I II III
3 31 A B C
2 21 C B C A B A
1 11 B C A C A B ABC
0 01 BC AC AB A A B B C C
# of Microstates(statistical weight W) 3 6 1
WkS ln
2. Thermodynamic Quantities, Concepts and Tools?
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22
2.4.5) Entropy
Second Law of Thermodynamics
2. Thermodynamic Quantities, Concepts and Tools?
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23
2.4.5) Entropy
irreversibleprocess
high
low
TT
1
reversibleprocess
2. Thermodynamic Quantities, Concepts and Tools?
Sadi Carnot
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24
2.4.5) Entropy Rudolph Clausius
Clausius Inequality
2. Thermodynamic Quantities, Concepts and Tools?
TdQdS
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25
2.4.6) Helmholtz Free Energy (Free Energy) andGibbs Free Energy (Free Enthalpy)
Clausius Inequality dHQ
dUQ
p
V
spontaneous process at constant volume
spontaneous process at constant pressure 00
TdSdHTdSdU
Helmholtz Free Energy (Free Energy):
Gibbs Free Energy (Free Enthalpy):
2. Thermodynamic Quantities, Concepts and Tools?
TdSdQTdQdS 0
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2.4.7) Dependencies of G by Combining the First and Second Law of Thermodynamics
pdVTdSWQWQdU revrev
VdppdVdUdHpVUH
VdpTdSVdppdVpdVTdSdH
SdTVdPSdTTdSVdpTdSSdTTdSdHdG
STGV
pG
pT
2. Thermodynamic Quantities, Concepts and Tools?
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27
2.4.8) Gibbs Free Energy in Mixtures and the Chemical Potential
inTpinTpnTp
dnnGdn
nGdn
nGSdTVdPdG
ijjj
,,
2,,2
1,,1
....21
ijnTpii n
G
,,
chemical potential of i
0 ,,,,,,,
BiAiBi
BiBiAiAi
dndndndG
T,p=const.
2. Thermodynamic Quantities, Concepts and Tools?
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28
2.4.8) Gibbs Free Energy in Mixtures and the Chemical Potential
im
nTpip
i Sn
ST
ij
,,,
imnTpiT
i VnV
pij
,,,
Example: pure ideal gas
2. Thermodynamic Quantities, Concepts and Tools?
pRT
nVVnRTpV m
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2.4.8) Gibbs Free Energy in Mixtures and the Chemical Potential
Membrane permeable for i (e.g. Pd and i=H2)
for ideal gaseous andliquid mixtures
for real systems
in any case
2. Thermodynamic Quantities, Concepts and Tools?
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2.5) Chemical Equilibrium
iiiDCBA nnDCBA 0, extent of reaction
2. Thermodynamic Quantities, Concepts and Tools?
in equilibrium
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31
G
R e a k tio n s k o o rd in a te
T = k o n s t.p = ko n s t.
d G = 0
G T=const.p=const.
G=0
2.5) Chemical Equilibrium
2. Thermodynamic Quantities, Concepts and Tools?
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32
standard state for all gases is the pure ideal gas at 1bar
activity of an ideal gas
pK
2.5) Chemical Equilibrium
2. Thermodynamic Quantities, Concepts and Tools?
law of mass action
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33
pK
2.5) Chemical Equilibrium
)( pfK p
)( pfK
ii
iii
ppK
ppxppxK x
ii
ii
RTcpRTnVp iiii
ii
iii
ppK
pRTcpRTcK c
ii
ii
)( pfK x
)( pfK c
2. Thermodynamic Quantities, Concepts and Tools?
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34
2.5) Chemical Equilibrium
2. Thermodynamic Quantities, Concepts and Tools?
Temperature Dependance of K:
Gibbs Helmholtz Equation
van‘t Hoff Equation
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
35
Example: Calculate the maximum NH3 yield of the Haber-Bosch-Process for25°C 600 °C and 1bar p 500 bar! Assume a stoichiometric feed andideal gases for simplicity!
)()(23
21
322 gNHgHgN Equation:
Thermodynamic Data (NIST Chemistry Webbook, CRC Handbook, PC books...):
Species Hf° / kJmol-1 S° / Jmol-1K-1 Cp°/ Jmol-1K-1
A B / 10-3 K-1 C / 10-6 K-2
N2 0 191.6 24.98 5.912 -0.3376H2 0 130.7 29.07 -0.8368 2.012NH3 -45.9 192.8 25.93 32.58 -3.046
211/ TCTBAmolKJC p
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
36
Calculate K for 25°C: STHG
73815.298314.8
1037.16expexp
37.161005.9915.2989.45
05.998.19217.130)2/3(6.1912/1
9.45)9.45(10)2/3(02/1
3
3
KJmolKmolJ
RTGK
molkJ
KmolkJK
molkJG
KmolJ
KmolJS
molkJ
molkJH
Species i Hf° / kJmol-1 S° / Jmol-1K-1 Cp°/ Jmol-1K-1
A B / 10-3 K-1 C / 10-6 K-2
N2 -1/2 0 191.6 24.98 5.912 -0.3376H2 -3/2 0 130.7 29.07 -0.8368 2.012NH3 1 -45.9 192.8 25.93 32.58 -3.046
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
37
Calculate X from K:
Species i ni0/mol ni/mol xi
N2 -1/2 1/2 1/2-1/2 (1/2-1/2)/(2-)H2 -3/2 3/2 3/2-3/2 (3/2-3/2)/(2-)NH3 1 0 /(2-) 2 2- 1
iiii
n
nii
i
nnddndndi
i
000
1
2/1)2/1(
0,2
2
0,2
20,2
N
N
N
NN
n
nnn
X
2/32/1
1
2/3
2
2/1
2
13
22/32/3
22/12/1
2
pp
pp
pp
pp
pp
pp
aKHN
NH
ii
i
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
38
Solution: Graphical Solution (e.g. with Excel, Origin....)
T=298.15 K and p=1 bar
2/32/1
1
22/32/3
22/12/1
2738
pp
pp
pp
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
39
T dependance of K:
T
Tp
TdTR
THTKTKRT
HT
K
1
212 lnlnln
T
Tp TdTCTHTH
1
1
Speciesi
Hf° / kJmol-1
S° / Jmol-1K-1
Cp°/ Jmol-1K-1
A B CN2 -1/2 0 191.6 24.98 5.912 -0.3376H2 -3/2 0 130.7 29.07 -0.8368 2.012NH3 1 -45.9 192.8 25.93 32.58 -3.046
2630
20
1090.5109.3017.30
....
TTTC
AATCTBATC
p
iiip
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
40
T dependance of K: T
Tp TdTCTHTH
1
1
32312111
31
321
211
21
3232
32
1
TCTBTATCTBTATH
TTCTTBTTATH
TdTCTBATHTHT
T
20 TCTBATC p
T
Tp
TdTR
THTKTKRT
HT
K
1
212 lnlnln
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
41
T dependance of K:
312111
22
3231
2111
32.
32.
3232
TCTBTATHconst
TRC
RB
RTA
RTconst
RTTH
TCTBTATCTBTATHTH
212111
1
21
62ln11.lnln
32.lnln
1
TTRCTT
RB
TT
RA
TTRconstTKTK
TdTRC
RB
RTA
TRconstTKTK
T
T
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
42
T dependance of K:
2/32/1
1
22/32/3
22/12/1
2)(lnexp
pp
pp
pp
TK
212111
1 62ln11.lnln TT
RCTT
RB
TT
RA
TTRconstTKTK
Solve
numerically or graphically with p and T as parameters gives = (p,T)!
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
43
Solution: Numerical Solution (e.g. with Matlab, Mathematica....)
KT /barp /
typical process conditions forindustrial ammonia synthesis
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
44
Thermodynamics says nothing about the rate at which a process proceeds!
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
45
molkJH
SOOSO
r
CSOKOV
/99
2/1 3600400,22 4252
Te
Ta
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3. Chemical Equilibrium: Reactor Conversion for a Single Reaction
46
molkJKHOHCHHCO r / 8.90300 2 32 Linde Isothermal Reactor, e.g. for
methanol synthesis
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
47
4.1 Theory min,.....,, 21, NpT nnnGGbut the ni cannot vary independently because they have to fulfill thematerials balances
N
ikiki bna
1
number of atoms of element k in species i
moles of species i in the reaction mixture
total moles of atoms of elementk in the reaction mixtures
N
i
M
k
N
ikiki
N
ikikikiki bnabnabna
1 1 1k
1k 0 0 0
Lagrange Multipliers
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
48
M
k
N
ikikipT bnaGF
1 1k,
Lagrange Function
01
k1
k,,,,
M
kkii
M
kki
npTinpTi
aanG
nF
ijij
0ln1
0
M
kkikii aaRTG
01
k
N
ikiki bna
set of N equations(1 for each of the N species)
set of M equations(1 for each of the M elements)
N + M unknowns (n - mole numbers + m - lambdas)
4.1 Theory
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
49
4.2 Example: Calculate the equilibrium composition of a steam reformingmixture consisting of CH4, H2O, CO, CO2 and H2 at T=1000K and p=1bar. n0,CH4=2mol, n0,H2O=3mol. Ideal gases can be assumed.
From thermodynamic tables (e.g. CRC Handbook) we extract:
Species CH4 H2O CO CO2 H2Gf° (1000K) / Jmol-1 19475 -192603 -200281 -395865 0
Species CH4 H2O CO CO2 H2# of C atoms aC,CH4=1 aC,H2O=0 aC,CO=1 aC,CO2=1 aC,H2=0# of O atoms aO,CH4=0 aO,H2O=1 aO,CO=1 aO,CO2=2 aO,H2=0# of H atoms aH,CH4=4 aH,H2O=2 aH,CO=0 aH,CO2=0 aH,H2=2
From the species formulas we obtain:
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
50
4.2 Example: Formulating the equations
0ln1
0,
M
kki
ki
if aRT
aRTG
ii
i
ii
ii
ii n
nbarbar
nn
ppx
ppa
11
01000314.8
41000314.8
ln1000314.8
19475 4
HC
ii
CH
nn
01000314.81000314.8
2ln1000314.8
192603 2
OH
ii
OH
nn
01000314.81000314.8
ln1000314.8
200281
OC
ii
CO
nn
01000314.8
21000314.8
ln1000314.8
395865 2
OC
ii
CO
nn
0224 COCOCH nnn
032 22 COCOOH nnn
014224 224 HOHCH nnn
CH4:
H2O:
CO:
CO2:
H2: 01000314.82ln 2
H
ii
H
nn
atom balance on C
atom balance on O
atom balance on H
starting valuesn0,CH4 = 2 moln0,H2O = 3 mol.
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
51
4.2 Example: Solving the system of nonlinear equations e.g in Matlab (fsolve)
ii
H
CO
CO
OH
CH
nnnnnn
651.8795.5319.0
507.1856.0175.0
2
2
2
4
ii
H
CO
CO
OH
CH
xxxxxx
16698.00368.0
1742.00990.00202.0
2
2
2
4
The most difficult thing is to find starting values that work. Make educatedguesses based on physical or chemical knowledge (0 xi 1), e.g.
28812
5.0ln1000314.801000314.8
2ln 0,2
HH
ii
H
nn
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
52
4.3 Another Example: Equilibrium Calculations with CHEMKIN
Methan Oxidation on Rh and Pt Coated Foam Catalysts (T 1000°C)
CH4 + O2
H2, CO, H2O, CO2
~ 10-3 s
p, H const.
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5 wt% Rh on 80ppi -Al2O3 foam, feed 5 ln/min, C/O = 1.0, Ar/O2 = 79/21, 1 bar
R. Horn*, K. A. Williams, N. J. Degenstein, A. Bitsch-Larsen, D. Dalle Nogare, S. A. Tupy, L. D. Schmidt J. Catal. 249 (2007) 380-393
4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
4.3 Another Example: Equilibrium Calculations with CHEMKIN
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
4.4 Take home message: Some months in the Lab can save you one day in the Library!!!
typical transient experiment
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4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?
4.4 Take home message: Some months in the Lab can save you one day in the Library!!!
tr/ss = 98%
tr/ss = 99%
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Thermodynamics matters in Catalysis!!!
56
Kinetics
TEquilibrium Transport
TCTkr AA
RTGK R
0
ln TCTDJ
AAB
A
0
0
00 T
Tpp
FFvv
T
T AKAKAads
ads
1
0
-
Maybe the better way to deal with Thermodynamics?
57
Thank you very much for your attention!!!