Thermodynamic Aspects of Heterogeneous Catalysis · PDF fileThermodynamic Aspects of...

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

Introduction

2

One way of dealing

with thermodynamics!

or

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


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

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


6

Kinetics

TEquilibrium Transport

TCTkr AA

RTGK R

0

ln TCTD

J

AAB

A

0

0

00 T

Tpp

FFvv

T

T AK

AKAads

ads

10


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


8Taken from: M. Salciccioli et al. Chem. Eng. Sci. 66 (2011) 4319

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)

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

11

2.2) Processes

Irreversible Processes


12

2.2) Processes

Reversible Processes


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


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


15

2.4.1) Internal Energy U in J

..... magnelnucchemrotvibtrans UUUUUUUU

VC (internal pressure )

0 for ideal gases

WQU First Law of Thermodynamics:


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!


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


18

2.4.3) Enthalpy H in J


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

19

2.4.4) Heat capacities CV, Cp in J/K RCC mVmp ,,for ideal gases


20

2.4.4) Heat capacities CV, Cp in J/K


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

pC

TH

rp

r

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


22

2.4.5) Entropy

Second Law of Thermodynamics


23

2.4.5) Entropy

irreversibleprocess

high

low

TT

1

reversibleprocess


Sadi Carnot

24

2.4.5) Entropy Rudolph Clausius

Clausius Inequality


TdQdS

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):


TdSdQTdQdS 0

26

2.4.7) Dependencies of G by Combining the First and Second Law of Thermodynamics

pdVTdSWQWQdU revrev

VdppdVdUdHpVUH

VdpTdSVdppdVpdVTdSdH

SdTVdPSdTTdSVdpTdSSdTTdSdHdG

STGV

pG

pT


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.


28


im

nTpip

i Sn

ST

ij

,,,

imnTpiT

i VnV

pij

,,,

Example: pure ideal gas


pRT

nVVnRTpV m

29


Membrane permeable for i (e.g. Pd and i=H2)

for ideal gaseous andliquid mixtures

for real systems

in any case


30

2.5) Chemical Equilibrium

iiiDCBA nnDCBA 0, extent of reaction


in equilibrium

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



32

standard state for all gases is the pure ideal gas at 1bar

activity of an ideal gas

pK



law of mass action

33

pK


)( pfK p

)( pfK

ii

iii

ppK

ppxppxK x

ii

ii

RTcpRTnVp iiii

ii

iii

ppK

pRTcpRTcK c

ii

ii

)( pfK x

)( pfK c


34



Temperature Dependance of K:

Gibbs Helmholtz Equation

van‘t Hoff Equation


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


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


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


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


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


40

T dependance of K: T

Tp TdTCTHTH

1

1

323

12

111

31

321

211

21

3232

32

1

TCTBTATCTBTATH

TTCTTBTTATH

TdTCTBATHTHT

T

20 TCTBATC p

T

Tp

TdTR

THTKTKRT

HT

K

1

212 lnlnln


41

T dependance of K:

31

2111

22

3231

2111

32.

32.

3232

TCTBTATHconst

TRC

RB

RTA

RTconst

RTTH

TCTBTATCTBTATHTH

21

21

111

21

62ln11.lnln

32.lnln

1

TTRCTT

RB

TT

RA

TTRconstTKTK

TdTRC

RB

RTA

TRconstTKTK

T

T


42

T dependance of K:

2/32/1

1

22/32/3

22/12/1

2)(lnexp

pp

pp

pp

TK

21

21

111 62

ln11.lnln TTRCTT

RB

TT

RA

TTRconstTKTK

Solve

numerically or graphically with p and T as parameters gives = (p,T)!


43

Solution: Numerical Solution (e.g. with Matlab, Mathematica....)

KT /barp /

typical process conditions forindustrial ammonia synthesis


44

Thermodynamics says nothing about the rate at which a process proceeds!


45

molkJH

SOOSO

r

CSOKOV

/99

2/1 3600400,22 4252

Te

Ta


46

molkJKHOHCHHCO r / 8.90300 2 32

Linde Isothermal Reactor, e.g. for

methanol synthesis

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions?

47

4.1 Theory min,.....,, 21, NpT nnnGG

but 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


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


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 H2

Gf° (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:


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: 01000314.8

2ln 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.


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


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.

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


4.3 Another Example: Equilibrium Calculations with CHEMKIN


4.4 Take home message: Some months in the Lab can save you one day in the Library!!!

typical transient experiment


4.4 Take home message: Some months in the Lab can save you one day in the Library!!!

tr/ss = 98%

tr/ss = 99%

Thermodynamics matters in Catalysis!!!

56

Kinetics

TEquilibrium Transport

TCTkr AA

RTGK R

0

ln TCTD

J

AAB

A

0

0

00 T

Tpp

FFvv

T

T AK

AKAads

ads

10

Maybe the better way to deal with Thermodynamics?

57

Thank you very much for your attention!!!

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