Modern Methods in Heterogeneous Catalysis Research WS 2012/2013

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

1

Introduction

One way of dealing with thermodynamics!

or

2

Content

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

3

1. Why does Thermodynamics matters in Catalysis?

The best-known „catalytic reactor“:

One reaction catalyzed by the catalytic converter: CO + ½ O2  CO2 rHo (298K) = -283 kJmol-1

 CO2  O

CO  O  M  CO2  M 

CO2

On Pt CO  O 2 OC O OC O

In the gas phase

CO  O2

CO2

CO O  C O

Pt

O O C 

4

1. Why does Thermodynamics matter in Catalysis?

Energetics of CO oxidation Temperature

E

 E   r  exp   RT 

In the gas phase

E

On Pt Taken from: G. Ertl, Catalysis: Science and Technology Vol. 4 1983 p. 245ff, all enthalpies in kJmol-1

5

1. Why does Thermodynamics matter in Catalysis?

Kinetics  rA  k T C A T 

Equilibrium  RG 0 ln K   RT K ads A0 A  1  K ads  A

  

T

 Transport JA   D AB T  C A T  v  v0

FT p 0 T FT 0 p T 0 6

1. Why does Thermodynamics matters in Catalysis? Thermodynamics matters in Catalysis because  Thermodynamics determines the maximum extent of any given reaction  C  The reaction rate decreases with A  B  rA  k f  C A  B KC  approach to thermodynamic equilibrium

  

macroscopic

 Heat release or uptake by chemical reactions leads to complex non-isothermal reactor behavior  Adsorption/Desorption thermodynamics determines the concentration of adsorbed species and in turn the rate of surface reactions (e.g. Langmuir Hinshelwood rate expressions)

microscopic

 The kinetics of elementary steps are related to their thermodynamics 7

1. Why does Thermodynamics matter in Catalysis?

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

8

Equilibrium Conversion and Selectivity Calculations for Multiple Reactions (Examples)

Consequences for Reactor Design (Examples) Chemical Conversion for a Single Reaction (Example) Chemical Equilibrium and Law of Mass Action Chemical Potential µ Clausius Inequality and Gibbs Free Energy G Entropy S and Second Law of TD First Law of TD, U, H, CV, Cp Quantities, Concepts and Tools

9

2. Thermodynamic Quantities, Concepts and Tools? 2.1) Systems in Thermodynamics

open system

System

closed system

Surroundings

e.g. a heated flow reactor

isolated system

Exchange of Matter

e.g. an autoclave

Exchange of Energy

e.g. a thermos bottle or the universe 10

2. Thermodynamic Quantities, Concepts and Tools? 2.2) Processes Irreversible Processes

11

2. Thermodynamic Quantities, Concepts and Tools? 2.2) Processes Reversible Processes

12

2. Thermodynamic Quantities, Concepts and Tools? 2.3) Important Quantities Thermodynamic Quantity

Symbol Intensive or Extensive?

SI Unit

mass

m

extensive

kg

molar mass

M

intensive

kgmol-1

temperature

T

intensive

K

pressure

P, p

intensive

Pa

fugacity

f

intensive

Pa

density



intensive

kgm-3

volume

V

extensive

m3

molar volume

Vm, v,…

intensive

m3mol-1

heat

Q

extensive

J

work

W

extensive

J

inner energy

U

extensive

J

enthalpy

H

extensive

J

free energy, Helmholtz free energy

F, A

extensive

J

free enthalpy, Gibbs free energy

G

extensive

J 13

2. Thermodynamic Quantities, Concepts and Tools? 2.4) Important Quantities Thermodynamic Quantity

Symbol

Intensive or Extensive

SI Unit

entropy

S

extensive

JK-1

molare inner energy

Um,u

intensive

Jmol-1

molar enthalpy

Hm,h

intensive

Jmol-1

molar free energy

Am, a, Fm, f

intensive

Jmol-1

molar free enthalpy

Gm, g

intensive

Jmol-1

chemical potential

µ

intensive

Jmol-1

molar entropy

Sm,s

intensive

JK-1mol-1

heat capacity at constant volume

CV

extensive

JK-1

heat capacity at constant pressure

CP

extensive

JK-1

specific heat capacity

c(V,p)

intensive

Jkg-1K-1

molar heat capacity

Cm(V,p), cm(V,p)

intensive

Jmol-1K-1

14

2. Thermodynamic Quantities, Concepts and Tools? 2.4.1) Internal Energy U in J

U  U trans  U vib  U rot  U chem  U nuc  U el  U magn  .....

CV

(internal pressure ) 0 for ideal gases

First Law of Thermodynamics:

U  Q  W 15

2. Thermodynamic Quantities, Concepts and Tools? 2.4.2) Internal Energy U in J The First Law of Thermodynamics is important for describing non-isothermal reactors (energy balance) but says nothing about the direction of a physical or chemical process!

U  Q  W

16

2. Thermodynamic Quantities, Concepts and Tools? 2.4.3) Enthalpy H in J weight area

H  H (T , p, n1 , n2 ,...., nk )  H   H   H    dp   dH   dn1  ....  dT    T  p ,n j  p T ,n j  n1  p ,T ,n j1 Cp

(isothermer drosseleffekt  in m3) 0 for ideal gases

17

2. Thermodynamic Quantities, Concepts and Tools? 2.4.3) Enthalpy H in J

1  A A   B B   C C   D D d  dni ni  ni ,0  i i

 H     H    p,T example:

enthalpy of reaction

H   i H i

extent of reaction Hess Law

i

1 3 N 2  H 2  NH 3 2 2

1  3 H    H N 2      H H 2  1 H NH 3 2  2

The standard enthalpy of formation Hf° or standard heat of formation of a compound is the change of enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states. The standard state of a gas is the (hypothetical) ideal gas at 1bar and 298.15K. For liquids and solids it is the pure substance at 1bar and 298.15K.

H NH 3

1  3  H ( NH 3 )    H N 2      H H 2  1 H NH 3 2  2  f

18

2. Thermodynamic Quantities, Concepts and Tools? 2.4.4) Heat capacities CV, Cp in J/K

for ideal gases

C p ,m  CV ,m  R

19

2. Thermodynamic Quantities, Concepts and Tools? 2.4.4) Heat capacities CV, Cp in J/K

1 3 kJ J  N 2  H 2  NH 3  r H  45.9  r C p   i C pi   22.186 2 2 mol mol  K i Kirchhoff‘s law T  298.15 K p  1bar

1 3 N2  H2 2 2

T  298.15 K p  1bar 1NH 3

kJ 45.9 mol

   r H       r Cp  T  p

T2

T  400 K p  1bar 1 3 N2  H2 2 2

 r H (T2 )   r H (T1 )    r C p dT T1

T  400 K p  1bar 1NH 3

 48.2

kJ mol

20

2. Thermodynamic Quantities, Concepts and Tools? 2.4.5) Entropy Example: 3 Molecules (A,B,C) distributed among 4 energy states (0=0, 1= 1, 2=21, 3=31) with a total energy of total=31 (isolated system) Energy State

Energy

Macrostate

S  k  ln W

I 31

2

21

C

B

C

A

B

A

1

11

B

C

A

C

A

B

0

01

A

A

B

B

C

C

BC

B

III

3

# of Microstates (statistical weight W)

A

II

AC 3

C

AB

6

ABC

1

21

2. Thermodynamic Quantities, Concepts and Tools? 2.4.5) Entropy

Second Law of Thermodynamics 22

2. Thermodynamic Quantities, Concepts and Tools? 2.4.5) Entropy irreversible process

  1

Tlow Thigh

Sadi Carnot

reversible process

23

2. Thermodynamic Quantities, Concepts and Tools? 2.4.5) Entropy

Clausius Inequality

Rudolph Clausius

dQ dS  T 24

2. Thermodynamic Quantities, Concepts and Tools? 2.4.6) Helmholtz Free Energy (Free Energy) and Gibbs Free Energy (Free Enthalpy)

Clausius Inequality

dQ dS   0  dQ  TdS T

Q V  dU Q  p  dH

spontaneous process at constant volume

dU  TdS  0

spontaneous process at constant pressure

dH  TdS  0

Helmholtz Free Energy (Free Energy):

Gibbs Free Energy (Free Enthalpy):

25

2. Thermodynamic Quantities, Concepts and Tools? 2.4.7) Dependencies of G by Combining the First and Second Law of Thermodynamics

dU  Q  W  Qrev  Wrev  TdS  pdV H  U  pV  dH  dU  pdV  Vdp dH  TdS  pdV  pdV  Vdp  TdS  Vdp dG  dH  TdS  SdT  TdS  Vdp  TdS  SdT  VdP  SdT

 G   G     V    S  T  p  p T 26

2. Thermodynamic Quantities, Concepts and Tools? 2.4.8) Gibbs Free Energy in Mixtures and the Chemical Potential

 G   G   G     dG  VdP  SdT   dn1   dn2  ....   dni  n1  p ,T ,n j1  n2  p ,T ,n j2  ni  p ,T ,n ji  G    i    ni  p,T ,n ji

T,p=const.

chemical potential of i

dG   i , Adni , A   i , B dni , B   i , B   i , A dni , B  0

27

2. Thermodynamic Quantities, Concepts and Tools? 2.4.8) Gibbs Free Energy in Mixtures and the Chemical Potential

  S     i     S m ,i      T  p  ni  p ,T ,n ji   i   p

 V     T  ni

   Vm ,i  p ,T ,n ji

Example: pure ideal gas

pV  nRT  Vm 

V RT  n p

28

2. Thermodynamic Quantities, Concepts and Tools? 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 and liquid mixtures

for real systems

in any case 29

2. Thermodynamic Quantities, Concepts and Tools? 2.5)

Chemical Equilibrium

 A A   B B   C C   D D ni  ni ,0  i

extent of reaction

in equilibrium

30

2. Thermodynamic Quantities, Concepts and Tools? 2.5)

Chemical Equilibrium

G G

TT=const. = k o n s t. pp=const. = k o n s t.

dG=0 G = 0



R e a k tio n s k o o r d in a te

31

2. Thermodynamic Quantities, Concepts and Tools? 2.5)

Chemical Equilibrium

law of mass action

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

Kp

32

2. Thermodynamic Quantities, Concepts and Tools? 2.5)

Chemical Equilibrium

K  f ( p)

Kp i i   K    x i p  p   x i    i i  i



K p f ( p) i i    p p  K x    p   p 

K x f ( p)

piV  ni RT  pi  ci RT K   c i RT 

i

i

i  i  RT    p   ci     i  p  

 i

 p K c     p 

K c  f ( p)

 i

33

2. Thermodynamic Quantities, Concepts and Tools? 2.5)

Chemical Equilibrium

Temperature Dependance of K: Gibbs Helmholtz Equation

van‘t Hoff Equation

34

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction Example: Calculate the maximum NH3 yield of the Haber-Bosch-Process for 25°C    600 °C and 1bar  p  500 bar! Assume a stoichiometric feed and ideal gases for simplicity!

1 3 N 2  g   H 2 ( g )  NH 3 ( g ) Equation: 2 2 Thermodynamic Data (NIST Chemistry Webbook, CRC Handbook, PC books...): Species

Hf° / kJmol-1

S° / Jmol-1K-1

Cp°/ Jmol-1K-1 A

B / 10-3 K-1

C / 10-6 K-2

N2

0

191.6

24.98

5.912

-0.3376

H2

0

130.7

29.07

-0.8368

2.012

NH3

-45.9

192.8

25.93

32.58

-3.046

C p / J  K 1  mol 1  A  B  T  C  T 2 35

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction

G   H   T  S 

Calculate K for 25°C: Species

i

Hf° / kJmol-1 S° / Jmol-1K-1

Cp°/ Jmol-1K-1 A

B / 10-3 K-1

C / 10-6 K-2

N2

-1/2

0

191.6

24.98

5.912

-0.3376

H2

-3/2

0

130.7

29.07

-0.8368

2.012

NH3

1

-45.9

192.8

25.93

32.58

-3.046

kJ kJ H   1 / 2  0  (3 / 2)  0  1  (45.9)  45.9 mol mol J J S    1 / 2 191.6  (3 / 2) 130.7  1192.8  99.05 mol  K mol  K kJ kJ kJ G   45.9  298.15 K   99.05 10 3  16.37 mol mol  K mol  G    16.37 103 J  mol  K    exp   738 K  exp   RT   mol  8.314J  298.15 K  





36

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction ni

Calculate X from K:

Specie s

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



1

K   ai i  i

  



1 d  dni   dni   i  d  ni  ni 0  i i ni 0 0

n N 2, 0  n N 2 X n N 2, 0  N 2  (1 / 2)    nN 2,0 1/ 2 1

 p NH 3    p         p   2    p   1/ 2 3/ 2 1/ 2 3/ 2  1 / 2  1 / 2  p   3 / 2  3 / 2  p  p N 2   pH 2                p   p  p   2    p   2    37

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction 1

  p       2   p   738  1/ 2 3/ 2  1 / 2  1 / 2  p   3 / 2  3 / 2  p          p   2    p   2   

T=298.15 K and p=1 bar

Solution: Graphical Solution (e.g. with Excel, Origin....)

38

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction T dependance of K:    ln K  H  H  T     ln K T   ln K T1    dT     2 2 RT   T  p RT T1 T

T

H  T   H  T1    C p T dT  T1

Species i

Hf° / kJmol-1

Cp°/ Jmol-1K-1

S° / Jmol-1K-1 A

B

C

N2

-1/2

0

191.6

24.98

5.912

-0.3376

H2

-3/2

0

130.7

29.07

-0.8368

2.012

NH3

1

-45.9

192.8

25.93

32.58

-3.046

C 0p T   A  B  T  C  T 2 A   i Ai .... i

C T   30.17  30.9 10  T  5.90 10  T 0 p

3

6

2

39

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction T

H  T   H  T1    C p T dT 

T dependance of K:

T1

C 0p T   A  B  T  C  T 2 T





H  T   H  T1    A  B  T   C  T 2 dT  T1

B 2 C 3 2 3  H T1   AT  T1   T  T1  T  T1 2 3 B 2 C 3 B 2 C 3   H T1   A  T1  T1  T1  A  T  T  T 2 3 2 3 









   ln K  H  H  T     ln K T   ln K T1    dT     2 2 RT   T  p RT T1 T

40

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction

T dependance of K:

B 2 C 3 B 2 C 3 H T   H T1   A  T1  T1  T1  A  T  T  T 2 3 2 3 H  T  const. A B C     T 2 2 RT RT RT 2 R 3 R B 2 C 3  const.  H T1   A  T1  T1  T1 2 3 



T

 const . A B C   ln K T   ln K T1        T  dT 2 RT  RT 2 R 3 R  T1  const .  1 1  A T B C 2 2   ln K T   ln K T1     ln  T  T  T  T 1 1   R  T1 T  R T1 2 R 6R



41



3. Chemical Equilibrium: Reactor Conversion for a Single Reaction T dependance of K:

const .  1 1  A T B C 2 2   ln K T   ln K T1     ln  T  T  T  T 1 1   R  T1 T  R T1 2 R 6R



Solve 1

  p     2    p   expln K (T )   1/ 2 3/ 2  1 / 2  1 / 2  p   3 / 2  3 / 2  p          p   2    p   2    numerically or graphically with p and T as parameters gives = (p,T)! 42



3. Chemical Equilibrium: Reactor Conversion for a Single Reaction Solution: Numerical Solution (e.g. with Matlab, Mathematica....)



T/K

p / bar

typical process conditions for industrial ammonia synthesis 43

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction

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

44

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction

Te SO 2 1 / 2O2 V   SO3 2 O5 , K 2 SO4  400  600 C   r H  99kJ / mol

Ta

45

3. Chemical Equilibrium: Reactor Conversion for a Single Reaction

CO  2 H 2  CH 3OH  r H 300 K   90.8 kJ / mol Linde Isothermal Reactor, e.g. for methanol synthesis

46

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.1 Theory

G T , p  G n1 , n2 ,....., nN   min

but the ni cannot vary independently because they have to fulfill the materials balances moles of species i in the reaction mixture N

 aki ni  bk i 1

total moles of atoms of element k in the reaction mixtures

number of atoms of element k in species i N

 N  a ki ni  bk  0 k   a ki ni  bk   0  i 1  i 1 

M

 N  k   a ki ni  bk   0  k 1  i 1 

Lagrange Multipliers 47

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.1 Theory

M

 N  F  G T , p   k   a ki ni  bk  k 1  i 1 

Lagrange Function

 F   ni

M M   G        k a ki   i   k a ki  0 k 1 T , p ,n j i  ni T , p ,n j i k 1 M

Gi0  RT ln a i   k a ki  0 k 1

 N  k   a ki ni  bk   0  i 1 

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) 48

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.2 Example: Calculate the equilibrium composition of a steam reforming mixture 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) / Jmol-1

19475

-192603

-200281

-395865

0

From the species formulas we obtain:

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 49

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.2 Example: Formulating the equations

G 0f ,i RT

M

k aki  0 k 1 RT

 ln a i  

ai 

pi ni 1bar ni p  x   i p p   ni 1bar  ni

 n 19475  ln  CH 4 CH4: 8.314 1000   ni  i

  C 4 H   0  8 . 314  1000 8 . 314  1000  

  nH 2O  192603  ln H2O:  8.314 1000   ni  i   nCO  200281  ln CO:  8.314 1000   ni  i

  O 2 H  0  8 . 314  1000 8 . 314  1000     C O  0   8.314 1000 8.314 1000 

 n  395865  ln  CO 2 CO2: 8.314 1000   ni  i

  C 2O  0  8 . 314  1000 8 . 314  1000  

H2:

  n ln  H 2   ni  i

  2 H 0  8 . 314  1000  

i

i

starting values n0,CH4 = 2 mol n0,H2O = 3 mol. atom balance on C nCH 4  nCO  nCO 2  2  0 atom balance on O nH 2O  nCO  2nCO 2  3  0 atom balance on H 4nCH 4  2nH 2 O  2nH 2  14  0 50

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.2 Example: Solving the system of nonlinear equations e.g in Matlab (fsolve)

nCH 4  0.175

xCH 4  0.0202

nH 2O  0.856

x H 2O  0.0990

nCO  1.507

xCO  0.1742

nCO 2  0.319

xCO 2  0.0368

nH 2  5.795

x H 2  0.6698

n

x

i

i

 8.651

i

1

i

The most difficult thing is to find starting values that work. Make educated guesses based on physical or chemical knowledge (0  xi  1), e.g.

  nH 2 ln   ni  i

  2 H  8.314 1000  ln 0.5  0  H ,0   2881  2  8.314 1000  51

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.3 Another Example: Equilibrium Calculations with CHEMKIN

Methan Oxidation on Rh and Pt Coated Foam Catalysts (T  1000°C) CH4 + O2

 ~ 10-3 s p, H const. H2, CO, H2O, CO2 52

4. Chemical Equilibrium: Conversion and Selectivity Calculations for Multiple Reactions? 4.3 Another Example: Equilibrium Calculations with CHEMKIN 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? 4.4 Take home message: Some months in the Lab can save you one day in the Library!!! typical transient experiment

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%

Thermodynamics matters in Catalysis!!!

Kinetics  rA  k T C A T 

Equilibrium  RG 0 ln K   RT K ads A0 A  1  K ads  A

  

T

 Transport JA   D AB T  C A T 

v  v0

FT p 0 T FT 0 p T 0 56

Maybe the better way to deal with Thermodynamics?

Thank you very much for your attention!!! 57