1. Surface Reactivity What determines the surface reactivity?
What is physisorption?
What is chemisorption?
What are the trends in Reactivity?
What is the underlying
fundament?

2. Physisorption By Taylor expansion Attractive part Repulsive part
(0,0,d)
(x,y,z) d + -
= (x,y,z),
By Taylor expansion
Attractive part
Repulsive part
Result:

3. Physisorption
Often is the less accurate Lennar-Jones potential used with n=12
Molecule
Enthalpy kJmol-1
CH4
-21 CO
-25
CO2 N2 O2
Remember:
Physisorption is only
polarizaion, there is no
exchange of electrons
Where do we utilize physisorption???

4. The Chemical Bond The simplest chemical bonding: Hydrogen molecule H2+
Saa= Sbb= 1 since and were already normalized.
Sab=S the overlap intergral

5. Simplified Approach Consider a simple two level system:
Consider the limit where S is small (small overlap)

6. Homonuclear system < 0 and S > 0 Strength of bond
b is proportional to the overlap

7. Homonuclear system < 0 and S > 0 Energy in the chemical bond
If E < 0 will the molecule be stable and the work required for dissociation will be Ediss= - E.
b is proportional to the overlap

8. The Chemical Bond So why is Cl2 much more reactive than O2
Ebond = eV 1s He
``He2´´
Ebond > 0 eV N N2
Ebond = eV 5s 1p 6s 2p O O2
Ebond = eV 2s
So why is Cl2 much
more reactive than O2
and why does Ne2 not
exist?

9. The heteronuclear system
s* E-
s E+
1sa E1s
1sb E1s Sb
b2/d

10. Molecules in general strong weak no bonding atomic molecular atomic
orbital orbital orbital
anti bonding
Big overlap
Small overlap
bonding

11. Electronic structure of the solid..
4 p
4 s
3 d
Energy
Atom
Metal
Density of states (DOS)

12. Trends in the periodic system
To the left the outer atomic orbital tends to be extended and will be
more localized as we go to the right where they become filled.
Thus the tendency for forming bond will decrease when going to the
left. From metallic to covalent to none.
When going down the periodic system the outer orbital increases and
becomes more extended and delocalized i.e. forms better bonding.
Eventually even Xe can form covalent bonding and the highest
melting point of metals have to be found in the third transition series
where there is an additional bonding effect.
gas
Filling =localized=less strong bonding
Increasing size= delocalization= increasing bonding
metals

13. What is the work functionEF
EVac
Sp-band
Core level 1
Core level 2
Metal
Adsorbed atom
Free atom
Work function
Vacuum level
Fermi level ea
Ionization
energy

14. Jellium model of a solid surface
metal vacuum
+ -
electrons
ionic cores
distance
density + -
dipole
Notice exponential decay
can be probed by STM

15. The solid surface The energy distribution of the electrons
will follow a Fermi-Dirac Distribution since
the electon gas is strongly degenerate, I.e.
many electons on few states:
where is the chemical potential of the electrons which at T= 0 K is
Notice how this expression becomes the usual Boltzman distribution for
large temperatues where the electron gas is getting diluted on many states

16. Free electron metals Typically for metals sp-bands like
Ca, Mg, Al, Cu, K, Cs; ect.
Evaporation of electrons
Increase T or lower F
When does metals melt?

17. The solid surface
k will be continous and they will occupy a sphere in k-space with radius kF and volume .

18. The Tight Binding Model
LCAO
Where F are the individual atomic
wave functions a e0
Core levels
Construct Block waves

19. The Tight Binding Model
Anti-bonding states
Bonding states

20. Metals vs. Insulators E Why is a metal a conductor?EF
Evac
Valence
band
Conduction
Free electron metal
Transition metal
Insulator
sp-band
d-band E
Why is a metal a conductor?
Why are metal reactive?
Why may insulatords break
down at elevated temperatures?
Band gap

21. Simple model of transition metalseF E
DOS
sp-band W
W/2
d-band
There will however always be a repulsive
Term proportional to the overlap:

22. Simple model of transition metals
There is an attractive term from the sp electrons roughly 5eV.
Why do we get a maximum?
Why does the interaction increase
with increasing d number?
Which metals have the highest
melting points?

23. The Newns-Anderson Model
What happens when an atom with an energy level Fa, ea approaches a
surface with valence electron at Yk, ek ?
Fa, ea EF Yk
The affinity level
becomes filled
when crossing EF
Fa, ea
na(e)

24. The Newns-Anderson model
Now we should try to explain what happens when an atom approaches
a metal surface
When a simple atomic level approaches a surface with a constant sp-band
will the level broaden:
Why is it broadened?

25. The Newns-Anderson model
When a simple atomic level approaches a surface with a simple sp-band
will the level broaden and be lowered in bonding energy, I.e. be
chemisorbed

26. The Newns-Anderson model
If we again have a narrow atomic level approaching a broad sp-band on
which there is superimposed a narrow d-band will we again see a
broadening and a lowering, but when the interaction to the d-band is
turned on will we have a splitting into bonding and anti-bonding levels.
Non-bonding orbitals

27. Summary of The Newns-Anderson model
Simple sp-band results in broadening and lowering:
Evac EF
Metal
Adsorbate levels
Distance from surface

28. Summary of The Newns-Anderson model
Narrow d-band on top of sp-band results in splitting:
Evac EF
Metal
Adsorbate levels
Distance from surface

29. Summary of The Newns-Anderson model
For molecules this may mean that the anti bonding orbitals get occupied
resulting in bonding to the surface, but weakening of the internal
bonding the essence of catalysis
sp-band
Evac EF
Metal
Adsorbate levels
Distance from surface
d-band s s*

30. Free electron metals Cl orbital is occupied and it will be Cl-
Li 2s is not populated so it
will be positively charged Li+

31. Free electron metals + - E + - E

32. Trend in atomic chemisorption energies

33. Trend in atomic chemisorption energies
Attractive part
Repulsive part

34. Trend in atomic chemisorption energies
will be fixed for fixed adsorbate and fixed site
and will reflect a proportionality for different
adsorbates x
Vad2 or b2 decreases with filling because d orbital becomes more
localized, while increases with n since they become more extended

35. Trend in atomic chemisorption energies
Remember Ebond=Esp+DEd-hyp
Since Esp~-5 eV
we will start out with strong
bonding that decreases with
incresing f (filling of d-band)
f=1, n=5
Why is Gold so noble?

36. Trend in molecular chemisorption energies

37. Trend in molecular chemisorption energies
For the late transition metals
(from Fe,Ru,Os)
Attractive part
Repulsive part
Not as easily interpreted as the atomic trends, but CO adsorbs stronger
when going to the left, at least to the middle part of the TM

38. Trends in surface reactivity
For the late transition metals the 2p is more important than the 5s
The bonding increases when
going to the left since the
d-band moves upwards
faster than f decreases
If we look at a fixed f then we can evaluate the influence of strain
or compression of a metal

39. Trends in surface reactivityEF Ed p*

40. The overall Picture Physisorption Molecular Chemisorption
Atomic Chemisorption
Reaction
Desorption?

41. Density Functional Theory (DFT)z
Repeat super cell
in xyz
Use a plane basis
set for describing
eletrons
Solve the Schødinger
equation in an
approximate manner
Accurate within eV x

42. N2 approaching Ru(0001) Ex: DFT Calculation
It costs 9.8 eV to dissociate
N2 in the gas phase while on
Ru(0001) it only cost 1.4 eV
i.e. Catalyst

43. Trends in Reactivity
The hight in the transition state will basically go as the molecular
bonding energy. For the late transition metals:
Cu, Ag, and Au cannot dissociate CO
Ni can barely, but Pd and Pt cannot. In principle they should be able
to do this by the final state is bonded to weakly (I.e. C and O)
So this is a final state effect.

44. Sabatier´s Principle
Perfect
Too Reactive
Too Noble

45. Separating of electronic and geometric effects
Geometrical

46. Active Sites on Ruthenium
Surface
Particle

47. Size Distribution Before and After Annealing
B5 sites, (110) geometry
B5 sites, (311) geometry
600 particles analyzed
TEM
PDSP01 Ru/MgAl O 2 4
P: 50 bar ; Temp.: 475 °C
gas: H :N
= 3:1 2 2
170 hours at 500 °C.
6,0
2½ hours at
C. H. Jacobsen, S. Dahl,
P. L. Hansen, E. Toernqvist,
H. Topsøe, D. V. Prip,
P. B Møenshaug, and
I. Chorkendorff, J. Mol.
Catal. A: Chemical
163 (2000) 19.
5,0
500 °C
Activity [mmol/(g*s]
4,0
3,0 50
100
150
200
250
300
Time [h]

48. Designing catalysis by alloying?
Ni, Cu / Fe
Cu / Co
Cu, Pd, Ag / Ru
Cu, Pd, Ag / Rh
Ag / Pd
Cu, Pd, Ag, Pt,
Au / Ir
Ag, Au / Pt
Ru, Ir / Co
Fe, Ru, Rh, Ir, Pt / Ni
Ni, Pt / Cu
Fe, Co / Rh
Fe, Co, Ni, Cu / Pd
Co, Ni, Cu, Rh, Pd, Pt / Ag
Fe, Co, Ru / Ir
Fe, Co, Ni, Cu, Ru, Rh / Pt
Fe, Co, Ni, Cu, Rh, Pd, Pt / Au
Fe, Co / Cu
Ru, Rh, Ir / Pd
Fe, Ru, Ir / Ag
Ru, Ir / Au
Co, Rh, Pd/Fe
Ag, Ir, Pt, Au / Fe
Rh, Pd, Ag, Pt, Au / Co
Cu, Pd, Ag, Au / Ni
Ag, Au / Cu
Fe, Co, Ni, Pt, Au / Ru
Ni, Pt / Rh
Au / Pd
Curve shape +
alloy -
phase-separation
substrate atoms segregate to the surface
adsorbate atoms segregate to the surface
Designing reactivity

49. Thermal Experiment: CH4 Dissociation at 530 K
2 min annealing to 1100 K
3e-7
Initial sticking probability
700 K flash
2e-7
1e-7 1 2 3 4
Ni Coverage [ML]
R. C. Egeberg and I. Chorkendorff, Catal.Lett. In press 2001.

50. Universality in Heterogeneous Catalysis
Brøndsted-Evans-Polany (BEP)
relationship: There is a linear
relation between bonding energy
and activation energy barrier.
Why is it structure dependent?
For a given substrate and site the TS
all look alike
Why is it adsorbate independent?
In TS the molecules lose indentity
Why is the relationship between the
activation energy and the adsorption
energy linear?
When the molecule have lost identity in
TS the energy is determined by the products

51. The Industrial Ammonia synthesis
The Real Ammonia reactor
The schematic reactor
inlet
outlet A B D C
The equilibrium curve
The optimum operating line

52. The Industrial Ammonia synthesis
The equilibrium curve
The optimum operating line
C. H. Jacobsen J. Catal. 205 (2002) 382

53. The Concept of Optimal Catalyst Curve
2:1 80bar 420oC
3:1 200bar 450oC 5% 5%
Defines optimal Catalyst
90%
90%
High ammonia conc. 90%
requires low bonding energy
2:1 80bar 420oC
3:1 200bar 450oC
Low ammonia conc. 5%
requires higher bonding energy
Claus Jacobsen et al. J. Catal. 205 (2002) 382