Principles of catalysis and use of nanoparticles

1. Selected Topics in Inorganic
Chemistry Part 1/3
(Course 242434)
Catalysis and Nanoparticles
Dr.Christoph UP 2018
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2. 2

3. What is a catalyst ?
A substance that increases the speed of a
reaction without being used up itself.
(https://www.youtube.com/watch?v=m_9bpZep1QM) 3

4. What is a catalyst ?
A catalyst opens an alternative pathway for a chemical
reaction – with lower activation energy
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5. Catalyst Parameters
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6. 6

7. Two kinds of catalysis
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8. What kind in these industrial applications ?
http://www.ncert.nic.in/NCERTS/l/lech105.pdf
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9. A catalyst does NOT change the free enthalpy G and therefore
NOT the equilibrium constant K !
Remember: G = - RT ln K
It only lowers the activation energy Ea and therefore the SPEED
G
Transition
state
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10. Activation Energy and Speed
The relation between activation energy and speed is given by:
http://www.chemguide.co.uk/physical/basicrates/arrhenius.html 10

11. Calculate the speed constant when we can lower the
activation energy to 25 kJ/mol
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12. 12

13. How does it work ?
When we want to increase the reaction speed, normally we heat
up the mixture.
In this way we create more molecule collisions.
BUT: in an exothermic reaction, increasing the heat reduces the
yield :
=> Higher speed BUT lower yield ! (Le Chatelier’s Principle)
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14. How a system a equilibrium reacts
to changes in temperature
(Le Chatelier’s Prinicple)
“A chemical system at equilibrium always responds to
changes in order to maintain that equilibrium”
Examle: an exothermic reaction with H = - 75 kJ/mol
https://www.youtube.com/watch?v=XhQ02egUs5Y
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15. An INCREASE in temperature would favor the endothermic
back-reaction:
So the equilibrium is shifted to the left side !
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16. Demonstration
https://www.youtube.com/watch?v=LgJGTH5Xp0o
https://www.youtube.com/watch?v=ZOYyCTvLa9E
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17. Experiment
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18. How about heat ?
We can include heat release in an exothermic reaction equation:
A + B <===> C + heat (H < 0)
When we add heat from outside, we shift the equilibrium
left or right ?
How about an endothermic reaction:
A + B + heat <===> C
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19. HETEROGENEOUS CATALYSIS
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20. Heterogeneous catalysis
Advantages:
• Easy separation between catalyst and product
• Catalyst can be re-used many times
• Higher selectivity towards a product
Disadvantages:
• Lower activity than homogeneous (surface !)
• Less specific for one reaction pathway
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21. Important industrial applications
http://www.ncert.nic.in/NCERTS/l/lech105.pdf
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22. 1. Porous catalytic particles
For example cracking of hydrocarbons
https://www.youtube.com/watch?v=jeUW6h2oVEY 22

23. Typical example of supported catalyst
Most often transition metals, finely distributed on a
support.
“Inert” porous materials which
carry the metal atoms or ions.
Most often used are
• Aluminumoxide
• Titaniumoxide
• Active Charcoal
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24. Supported catalysts
The catalyst is fixed on the big surface of a support material
(like alumina, charcoal) 24

25. 3. Monolithic catalysts
example: Haber Process
N2 + 3 H2 == 2 NH3 H = - 92 kJ/mol
The process needs high pressure
– explain from the entropy !
http://www.chemguide.co.uk/physical/equilibria/haber.html
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26. Demonstration
https://www.youtube.com/watch?v=NWhZ77Qm5y4
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27. Application of Le Chatelier’s Principle
https://www.youtube.com/watch?v=Tt6diRJli6Q
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28. Steps in catalytic reactions
https://www.youtube.com/watch?v=Gg5drDzrWwo
The steps 3 – 5
define the rate
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29. (1) Adsorption Isotherms
(1) Langmuir: molecules occupy step by step the
whole surface, but do not sit upon each other
Typical for gas
molecules on a
metal surface
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30. (2) Freundlich Isotherm:
more common in liquids
How could we design an experiment to find the Freundlich parameters of
the adsorption of a dye on charcoal ?
(max 80 mg dye could be a adsorbed by 1g of charcoal)
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31. Practical example
How could we design an experiment to find the Freundlich parameters of
the adsorption of a dye on charcoal ?
(max 80 mg dye could be a adsorbed by 1g of charcoal)
We can re-write the relation: to:
So we can measure the solution conc. C at different initial concentrations
after adsorption took place:
Example: we can put 10 mg dye in 1000 ml + 1 g Charcoal
=> 8 mg should be adsorbed x/m = 8 mg/g
=> we measure C = 2 ppm = 2 mg/L
Adsorbed
mg/g
Equilibrium
conc. mg/L
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32. Example data
Initial C eq C x/m
(1) 2 ppm 1.8 ppm 0.2 mg/g
(2) 6 ppm 5.7 ppm 0.3 mg/g
(3) 12 ppm 11.6 ppm 0.4 mg/g
(4) 20 ppm 19.5 ppm 0.5 mg/g
Calculate log x/m and log C and draw a graph.
What is k and n ?
from
preparation
measure
this
Calculate as
difference
amount
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33. Result
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34. (2) Chemical reaction step
Single site mechanism
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35. Dual site mechanism - Langmuir
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36. (3) Desorption
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37. Summary
The slowest step is the rate-determining step ! 37

38. HOMOGENEOUS CATALYSIS
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39. Catalytic Steps
http://web.iitd.ac.in/~sdeep/Elias_Inorg_lec_7.pdf
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40. 40

41. Reductive Elimination
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42. 42

43. 43

44. 44

45. Problem solving
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46. Wilkinson Catalyst
Hydrogenation of Alkenes
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47. 47

48. Increase catalyst efficiency
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49. Problem solving
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50. Practical examples
https://www.youtube.com/watch?v=q-dSFBduLnw 50

51. Bond formation by oxidative addition
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52. 52

53. Introduce F-ligands
for pharmaceuticals
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54. Pd catalysed way
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55. Fluorination by Cu catalyst
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56. Fluorination by Cu catalyst
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57. 57

58. Use Fluorine(18) for proton imaging
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59. Detection of cancer cell which consume most glucose
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