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Presentation on theme: "CHEMISTRY 3030: CATALYSIS COURSE, 2011"— Presentation transcript:


Given the following reaction at equilibrium: A2(g) +2B2(g)  2AB2(g) Q: What will happen to the equilibrium position in such a reaction, if a small quantity of a catalyst were to be added? A:

3 INTRODUCTION HENCE: A catalyst is any substance (____________) which, when present in a reaction mixture is directly involved with the reaction sequence (mechanism), and that increases the reaction rate (_________) without altering the position of the thermodynamic equilibrium, but is itself not consumed or altered. WHAT THEN IS CATALYSIS? A:

4 = uncatalysed reaction
INTRODUCTION HOW DOES A CATALYST DO THIS? Reactants = uncatalysed reaction Energy Reaction co-ordinate Products Activation Energy Enthalpy change

ACTIVITY (A) This a measure of the _________ at which the catalyst is able to transform reactants into products. This speed is related to the rate constant ‘k’ i.e. Rate = –k [reactants]n The activity (A) of a catalyst is measured by the SI unit: katal (abbreviated to kat). If the activity of a catalyst is 1 kat : then it ‘enables’ the reaction rate to be___________

6 CONT. Often it is necessary to disperse (scatter) the catalyst (or ________) on a solid material which has a high surface area. This material is called a _________. Examples of supports are: Al2O3(s), TiO2(s), CaCO3(s), carbon nanotubes (CNTs), etc. In these cases, the _________ at which the reactions of supported catalysts are measured, have units of molecules converted/ surface area of exposed active phase (in cm2). If however the number of exposed catalyst sites have been determined experimentally (for later) then the rate has units of molecules converted/ exposed catalyst sites ____________________.

7 CONT. SELECTIVITY (S) Multiple products are often formed in a reaction when a catalyst is added. The catalyst thus has an activity for each reaction that leads to a different product. The catalyst selectivity is then just a ratio of the activity of one product over another (more about this later). The larger the ratio the higher the catalyst selectivity for that product. Alternatively selectivity can be viewed as the ability of a catalyst to _____________ the rate of _____________of the thermodynamically feasible reactions more than the others.

8 1) CO(g) + H2(g) → CH3OH(g) = 1.0 x 10-1 Kat
CONT. Consider the following reactions and then place them in decreasing order of the catalyst selectivity: 1) CO(g) + H2(g) → CH3OH(g) = 1.0 x 10-1 Kat 2) CO(g) + 2H2(g) → ½C2H5OH(g) = 2.5 x 10-2 Kat 3) CO(g) + 3H2(g) → CH4(g) + H2O(g) = 9.8 x 10-2 Kat ORDER: _____________ Cu/Zn/Al2O3 Cu/Zn/Al2O3 Cu/Zn/Al2O3

9 CoO nanoparticle Carbon nanotube DEACTIVATION
CONT. DEACTIVATION For a variety of reasons catalysts can lose their activity and hence their selectivity ___________. Some reasons: 1) SINTERING (_______________) Supported metal catalyst particles are oftentimes more stable when they are ____________or spread out on the support surface. CoO nanoparticle Carbon nanotube

10 CONT. ___________ occurs at high temperatures when the supported metal catalyst particles spontaneously migrate on the surface. They combine/coalesce with one another to form bigger particles. Hence the metal catalyst ________________. A.Binder et al. J. Phys. Chem. C 2010, 114, 7816–7821 (Pd on SiO2 or TiO2)

11 CONT. 2) POISONING Some elements/ions (e.g. Cl, S, C, etc.), when they build up in concentration, can block active sites on the surface of a catalyst and hence reduce the activity and selectivity of the catalyst. For example: ___________, in leaded petrol, can deactivate catalytic converters in cars.

12 Q: Which species is the catalyst and which the intermediate?
TYPES OF CATALYSIS Several types of catalysis: Homogeneous Catalysis When the reactants and the catalyst are in the ___________: e.g. O3(g) + A·(g) → O2(g) + AO(g) ….(1) AO(g) + O3(g) → A·(g) + O2(g) ….(2) Q: Which species is the catalyst and which the intermediate?

13 TYPES OF CATALYSIS Heterogeneous Catalysis
When the reactants and the catalyst are in _________________: For example: The photoreduction of carbon dioxide on titania Pt/TiO2(s) CO2(g) CO(g) + O2(g)

14 Examples of heterogeneous catalysis:
CO + ½ H2  CO2 (Co, Co-Ni) - Fuel cells. Selec Ox. of CO in H2 CH2CH2 + H2  CH3CH3 (Ni/Pd) Olefin Hydrogenation – Fuel industry. c.a. 119 million tonnes of C2H4 in 2010!) CHCH + 2H2  CH3CH3 (Pt/Pd) Removal of C2H2 from olefins by hydrogenation) CO + H2  -CH2n+2- (Fe, Co) Fischer-Tropsch. Fuels, waxes, etc. N2 + 3H2  2NH3 (Fe, Ru) Haber Process. Fertilizers, explosives.. C6H6 + 3H2  C6H12 (Cu -1925!, Ru,Ni) 90% of cyclohexane used for Nylon 6 and 66.

15 TYPES OF CATALYSIS Enzyme Catalysis Enzymes are polymeric molecules which regulate the majority of __________ reactions that take place in living organisms. In the main they are proteins which are made up of amino acid building blocks or ____________. Enzymes are ___________and have extremely ________________(typically between 10 to 103 molecules converted/enzyme/s).

16 Q: How does an enzyme catalyse a reaction?
CONT. Q: How does an enzyme catalyse a reaction? A: The reactant molecules that interact with an enzyme are called ___________. Each enzyme has a specific site (_________) where only certain shaped substrates can fit into or bind (______________). When the substrate ___________binds to the active site the enzyme changes shape (___________) to form the ___________. The reaction then takes place, product/s formed and released, enzyme returns to its original shape.

17 CONT.

18 Polymer Supported Catalysis
TYPES OF CATALYSIS Polymer Supported Catalysis “These are catalyst systems comprising a polymer support (often based on _________________in the form of ___________to pack in a reactor) in which catalytically active species are ___________through chemical bonds or weaker interactions such as hydrogen bonds or donor–acceptor interactions and can be used repeatedly.” PAC, 2004, 76, 889 (Definitions of terms relating to reactions of polymers and to functional polymeric materials (IUPAC Recommendations 2003)) on page 896

19 CONT. Example: Pd nanoparticles (PdNPs) immobilised on microporous Poly(amidoamine) (PAMAM) dendrimers. Shin Ogasawara and Shinji Kato J. AM. CHEM. SOC. 2010, 132, 4608–4613

20 Suzuki-Miyaura reaction in water.
CONT. Suzuki-Miyaura reaction in water. Advantages: Water as a solvent for organic rxn! GREEN rxn! In bio-active compounds synthesized the catalyst is easily retrievable : No ___________ of the product! Reduced costs (Pd=$$$); catalyst ___________ Shin Ogasawara and Shinji Kato * J. AM. CHEM. SOC. 2010, 132, 4608–4613

Catalyst Type: Examples: Reactions: Metals Ni, Pd, Fe, Pt, Ag Hydrogenation Dehydrogenation Hydrogenolysis Oxidation Semiconducting oxides /sulphides NiO, ZnO, MnO2, Cr2O3, Bi2O3-MoO3, WS2 Desulphurisation

Catalyst Type: Examples: Reactions: Insulating oxides Al2O3, SiO2, MgO Dehydration Acids SiO2-Al2O3, Zeolites Polymerisation, Isomerisation, cracking, alkylation

23 Catalysts and Surfaces
A reactant must react with the surface atoms of a catalyst. Hence the more atoms on the surface, the more reactants can be transformed into products. Thus the expectation that high surface areas lead to ___________. Surface Areas fall into three categories: 10 m2 g-1  Small (e.g. _______________) 200 m2 g-1  Normal (e.g. _______________) 1200 m2 g-1  Large (e.g. _______________)

24 Catalysts and Surfaces
The rate might be expected to be __________ to the number of surface atoms BUT not all surface atoms are the _______. This gives rise to the concept of an _____________(Taylor 1925). Q: How then can the surface area be maximized? A: _______________ For spherical particles on a hemispherical support, the total surface area is give by SA (in m2)= ____________________ Where M = total mass of catalyst,  = density and r = average particle radius, Vpart= volume of particle.

25 Catalysts and Surfaces
Q: A batch of hemispherical catalysts (support and active nanoparticles) weighs 1.23 g and has a density of 3.14 g/ml. What is the total surface area of the catalysts if they are loaded with spherical nanoparticles with diameters of 50 nm? A: SA = _______________

26 Inorganic support Organic support Good _______ OK
Q: How then can the surface area be maximized? A2: __________ Factor: Inorganic support Organic support Effect of Temperature Good _______ Heat transfer Chemical reactivity OK

27 Q: How then can the surface area be maximized?
Promoters are substances that increase the _______ _______, even though they are not catalysts by themselves. In addition they “allow the active phase to function at its _______________” (Bond pp 76) e.g. Co or Ni in WS2 catalyst for desulphurisation† † C. Roukoss et al. / C. R. Chimie 12 (2009) There are two types of promoters _________

28 Types of promoters: Structural
A structural promoter ___________by separating the surface ___________. For e.g. The active phase for the NH3 synthesis catalyst is Fe, but its promoters may include: _________, ________, _____ and ______*. These inhibit Fe crystallites from coalescing. *I. Siminiceanu, I. Lazau, Z. Ecsedi, L. Lupa*, C. Burciag.Chem. Bull. "POLITEHNICA" Univ. (Timisoara) Volume 53(67), 1-2, 2008.

29 Types of promoters: Electronic
Electronic promoters are effective due to their ______________. The most widely used electronic promoters belong to _______, _______ and the _______. For Groups 1A and 2A, their ability to promote is inversely proportional to their electronegativity. Examples: Cs > K > Na (Group 1A)+ Ba > Ca > Mg (Group 2A)++ + S. Murata, K. Aika, T. Onishi, Chem. Lett., 1990, p ++ S. R.Tennison, in: J.R. Jennings (Ed.), Catalytic Ammonia Synthesis, Fundamentals and Practice, Plenum Press, New York, 1991, p. 303.

30 Electronic and structural promoter?
When Cs is added: No Caesium ruthenates form. The electronic properties change. RuO2 clusters reduced to metallic Ru. Increased Activ. Y.V. Larichev. Effect of Cs+ Promoter in Ru/MgO Catalysts. J. Phys. Chem. C 2011, 115, 631–635

31 Pauling’s Electronegativity value
Problem Given the following data for the lanthanides, place them in order of increasing ability to act as electronic promoters: ______________ Element Pauling’s Electronegativity value Samarium 1.198 Lutetium 1.201 Lanthanum 1.101

32 Sol-gel process Q: How then can the surface area be maximized?
A3: ______/ _______ maximization Sol-gel process TEOS = Tetraethyl orthsilicate Polymers include: polyvinyl alcohol and polyethylene glycol with different molec.masses S SATO, T MURAKATA, T SUZUKI, T OHGAWARA. JOURNAL OF MATERIALS SCIENCE 25 (1990)

33 Q: How then can the surface area be maximized?
A3: _______/ _______maximization No polymer Polymer of low molec. mass Polymer of higher molec. mass Mean pore sizes increase from 3 nm to 7 nm S SATO, T MURAKATA, T SUZUKI, T OHGAWARA. JOURNAL OF MATERIALS SCIENCE 25 (1990)

The atom surface concentration can be determined by the _____________ Assume that the bulk density is 1 g/cm3 then the molecular density will be 5 x 1022 molecules per cm3. The surface concentration (molecules per cm2) is proportional to _______if one assumes cube like packing. This gives a value of _______molecules per cm2.

35 For very small particles D = 1
DISPERSION The fraction of the atoms on the surface is referred to as _______. Mathematically, dispersion (D) is the ratio of the number of surface atoms (NS) to the TOTAL no. of atoms (NT): i.e. _______ For very small particles D = 1 However, as the particle grows the number of surface atoms will _______. For a cube of 100 Å, D = 10-3 !!!!

36 DISPERSION 1.0 Total number of atoms Dispersion (D)

37 Surface Atoms Consider a cube of metal (or metal oxide). The surface atoms rest on the bulk atoms and so must reflect this situation. Previously you learned about how atoms can pack and the way in which structures were built up. Example: _______ metallic_structures.htm

38 Surface Atoms Consider a cube of atoms with the _____structure. We can take slices through this structure and this will yield different faces with Miller Indices: (100), (111), (100), etc. M Bowker, The Basis and Applications of Heterogeneous Catalysis, Oxford University Press, 1998., pp 12.

39 Surface Atoms It is quite difficult to get surfaces with only one type of face. Most surfaces have many faces and contain _______, _______, _______ and _______. Thus surfaces are generally not _______. This has implications for the reactant molecules (see example with ammonia synthesis that follows). Different arrangements of surface atoms have different _______ _______. Generally, surfaces with _______ coordination number have the _______ surface free energy (are the most reactive).

40 Adatom Step Kink edge Terrace
A. N. Chaika, S. I. Bozhko, A. M. Ionov, A. N. Myagkov, and N. V. Abrosimov. Semiconductors, 2007, Vol. 41, No. 4, pp. 431–435

41 Spectroscopy in Catalysis: An Introduction, Third Edition, J. W
Spectroscopy in Catalysis: An Introduction, Third Edition, J. W. Niemantsverdriet Copyright WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim, ISBN:

42 Top plane co-ordination Adatom Kink edge 3 Step 4 Terrace 6
Atom type Top plane co-ordination Adatom Kink edge 3 Step 4 Terrace 6 Thus reactivity order is: _______ > _______ > _______ > _______ M Bowker, The Basis and Applications of Heterogeneous Catalysis, Oxford University Press, 1998., pp 13.

43 Single crystal faces of Fe for NH3 synthesis
G.A. Somorjai, N. Materer / Surface structures in ammonia synthesis. Topics in Catalysis 1 (1994)

44 Molecular Heterogeneous Catalysis
Metal cluster chemistry can assist as an important model to illustrate the interconnectedness between _______ chemistry and _______ chemistry. Consider the _____________ reaction of formic acid catalyzed on different metal surfaces: H-COO-H (aq)  CO2(g) + H2(g) Metal Plotting the _______ of each metal surface versus the standard enthalpy of formation (ΔHof) of each metal formate, gives what is called a ‘______________’ :

45 Molecular Heterogeneous Catalysis
_____________ _____________ Speed of decomposition of surface intermediate is low _______ ______________Speed of formation of surface intermediate is low.

46 Molecular Heterogeneous Catalysis
The shape of the ‘_______’ plot is consistent with the notion that surface intermediates closely resemble bulk intermediates (formates in this case). This is based on the _______ _______. This implies that: Reaction rate = f(_______, _______). Thus the optimum catalytic performance does not relate to a specific _______ _______ but to a balance of interaction & desorption. These are the elementary steps in _______ _______.

47 Molecular Heterogeneous Catalysis
Wolfgang Sachtler showed that when reacting molecules _______ onto a surface they form _______ _______. Furthermore, these complexes result in the partial destruction of metal-metal bonds and lead to a ‘______________’ of the surface. This was later termed ‘_______ _______’. Somorjai & Muetterties showed that the _______ _______ in a catalytic reaction, as well as the surface complexes are similar to homogeneous _______ complexes and reactions.

48 Molecular Heterogeneous Catalysis
Comparison of some Somorjai &Muetterties surface complexes with known organic complexes:

49 ADSORPTION Molecular and/or atomic species have essentially two ways in which they can attach or adsorb onto a surface: _______ or _______. Physical Adsorption or ___________ Physisorption often occurs in any liquid/solid or gas/solid system where the molecular/ atomic species attach to the solid surface through _______ _______ _______ (van der Waals forces). The elementary step in physisorption from a gas phase does not involve an _______ _______.

50 Physisorption (cont.) The typical binding energy of these physisorbed species on a surface is between ___________. (No chemical specificity). The process is _______ _______ _______, with little energy. See example of helium gas on metal surfaces afterwards. The adsorption enthalpy can range between _______ and _______. For physisorption, under appropriate conditions, gas phase molecules can form _______ adsorption.

51 Example: The physisorption profiles of He on various metal surfaces.
E. Zaremba and W. Kohn (1977). "Theory of helium adsorption on simple and noble-metal surfaces". Phys. Rev. B 15 (4): doi: /PhysRevB   Retrieved from ""

52 Chemical Adsorption or _____________
Chemisorption adsorption occurs when molecular/atomic species chemically attach to a surface through _______ _______. New species are formed. _______ _______ _______ _______ The typical binding energy of chemisorbed species on a surface is between _______. Binding is usually chemically specific. The process is less easy to reverse and often takes a lot of energy to do so. Sometimes the process is _______ i.e. due to _______ of the species. Example: _______ _______ _______ _______ _______ _______ _______ _______ _______.

53 Chemisorption (cont.) The elementary step in a chemisorption process from the gas phase often involves an _______ _______. (Recall dissociation of oxygen on metal.) In chemisorption, because the molecular/atomic species are adsorbed on the surface by covalent bonds, they often only form a single or _______ adsorption. In chemisorption, the adsorption enthalpy can range between _______ and _______. Q: Why are the adsorption enthalpies of physisorption and chemisorption negative?

54 ADSORPTION ISOTHERMS Q: Why is it important to obtain a relationship between the quantity of a substance adsorbed on the surface of a solid and its gas phase pressure? The _______ of gas coverage on the surface can be _______ determined. The _______ of the adsorption of the molecules can be quantitatively determined (i.e. chemi or physisorption). The _______ _______ of the solid can be quantitatively determined. The _______ of the catalytic system can be modelled.

55 ADSORPTION ISOTHERMS Q: What then is an adsorption isotherm? A: “The relationship between the _______ of gas adsorbed on a surface and the _______ with which it is in equilibrium, at a _______ _______, is called an adsorption isotherm” (Bond pg 15). Suppose the maximum surface that could be covered was ____ and that it was covered by an amount __, then the ratio or fraction (___) that the surface is covered is: _______ _______, then an adsorption isotherm, based on various assumptions can be developed.

56 LANGMUIR ISOTHERM Suppose that a solid surface has a gas (of a certain pressure) _______ _______ on its surface and at dynamic equilibrium then: 1 2 3 1 2 3

57 If the ratio or fraction that the surface is
If the ratio or fraction that the surface is covered is θ, then Langmuir made the following assumptions: Adsorption isotherms do not exceed a _______ _______. Thus molecular/ atomic species can only maximally fill the surface, then no further adsorption occurs i.e. _______ _______ _______ All sites on the surface are _______ and _______. This implies that the surface is _______ _______ and that the _______ _______ would be equivalent throughout the surface.

58 When molecules adsorb onto a surface,
When molecules adsorb onto a surface, they _______ _______ or _______ one another or incoming molecules that will adsorb. i.e. molecules will adsorb onto a site ___________ from whether or not a site next to them is _______ or _______. In the light of what you have already learned, take some time to critically analyze Langmuir’s assumptions.

1) Mono/single site adsorption Consider a gas A that can reversibly _______ and ______ on an active site *, then: Where kA = ________________________ Where kD = ________________________ the number of molecules colliding with the surface in unit time is proportional to the ____________of gas A i.e. ____________

60 Suppose the surface has a total of _____
________and at some time the ________ ________ is , then: Fraction of unoccupied sites = ________ Number of unoccupied sites = ________ Rate of adsorption  PA But: Rate of adsorption  no of unoccupied sites  ________ So: Rate of adsorption  PA × N(1-) Rate of adsorption = _____________, (where kA = _____________________)

61 But, the rate of desorption  amount of adsorbed gas
 ________ Rate of desorption = ________ (where kD = _____________________) At equilibrium: Rate ___________ = Rate ___________ thus: kA× PA × N(1-) = kD × N Where___ and ___ are the equilibrium values of pressure and surface coverage.

62 Cancelling like terms and rearranging gives:
(kA× PA ) – ( × kA× PA ) = kD ×  So:  = (kA× PA ) (kD + kA× PA ) Then let _______ or the ______________ _________ then you obtain the following: L ___ This is the ______________for _____ site adsorption.

63 Langmuir adsorption isotherm
Thus: ________ Recall: b = kA/kD

64 If V = volume actually covered
And Vm = monolayer coverage then: θ = ________ then since ___ Then rearranging and substituting gives: This is in the form Y= mX + C , where Y = ____ m = ____ , X = __ C = ____.This is allows the monolayer coverage to be calculated

65 Since ________ and kA = AA×e–{EA/RT} with kD = AD×e–{ED/RT} ____ Where ________ Thus b is a function of ________ and ________ at temperature ________. When b is ________ then ________ is _______ bonded, conversely when b is ________ it is ________ bonded.

66 2) Dual site adsorption Consider a gas A that must strike the surface at a location where there are _____ adjacent active sites *, then: Up to fairly high fractional coverage () it can be assumed that the adsorption of _____ _________ will depend on the fraction of vacant sites or ____, and that the adsorption of the other fragment will also depend on this ____________.

67 2) Dual site adsorption (cont.)
Then: Rate of adsorption = _____________ Rate of desorption = ________ At equilibrium: Rate of ads= Rate of desorp, So: k’APA[N(1 - )]2 = k’D(N)2 If b = k’A/k’D, Then by subsitution and Rearrangement:  = (bPA)½ /(1 + (bPA)½ ) For the monolayer coverage (Vm), again, let  = V/Vm, then This is in the form Y= mX + C , where Y =______, m = ____ , X = ___ C = _______. Thus a plot of _______ against ___ will give a straight line.

68 Measured amounts adsorbed of the pure gases CH4 ( ),
CO2 ( ), and N2 ( ) on AC Norit R1 at T D 298 K. Simultaneous fit of all data with the generalized dual-site Langmuir isotherm (—) F. DREISBACH, R. STAUDT AND J.U. KELLER Adsorption 5, 215–227 (1999)

69 3) Non-competitive adsorption
It is possible that ___________(gas A, and gas B), may be in the same container and will adsorb on different sites. The adsorption is _________________. Then the isotherm for each gas is simply a Langmuir isotherm for ________ gas. i.e.  = bPA/(1 + bPA) for gas A, and  = bPB/(1 + bPB) for gas B

70 4) Adsorption of more than one species on the
same surface Consider the reaction of _______ gases on a surface (i.e. the adsorption is ____________). with rate constants ka(A) and kd(A) with rate constants ka(B) and kd(B) Then A = bAPA/(1 + bAPA + bBPB) And B = bBPB/(1 + bAPA + bBPB)

71 Other Non-Langmuir Isotherms
5) The General expression We can work with 1,2,3,4, ….. up to i gases. Each gas can be expressed by a Langmuir Isotherm: A = bAPA/(1 +biPi) Other Non-Langmuir Isotherms i) ____________ Isotherm This assumes that a __________ decrease of the enthalpy of ____________ occurs with fractional coverage. ( = kP1/n where k and n are constants with n > 1)

72 L Zhang, S Hong, J He, F Gan, Y-S Ho
L Zhang, S Hong, J He, F Gan, Y-S Ho. Clean – Soil, Air, Water 2010, 38 (9), 831–836. Freundlich isotherms obtained using linear and nonlinear regression methods for the adsorption of phosphorus onto Al2O3 at temperature of 308 K.

73 ii) __________ Isotherm
Assumption here is that the ________ heat of adsorption falls off _________ with coverage: θ= k’ln(k”bPA) where b and PA have been defined previously.


75 POROUS MATERIALS Many materials are _______. Information about the pore ______, pore ________, pore ________ as well as the _________ can be obtained from two different types of adsorption experiments: 1) Multilayer gas adsorption Here we use an _________ _________ or _______e.g. N2, Ar, Kr to physisorb onto the material. 2) Mercury Porosimetry Liquid mercury is forced _______ into the pores of the material (e.g. _______ gives information on ______pore radius)

76 Uniform/cylindrical Blind pore Through pore Funnel shaped
PORE SHAPES/TYPES Uniform/cylindrical Blind pore Through pore Funnel shaped Porous network Ink bottle shaped Closed pore

77 Pores are classified according to size _________ < 2 nm
POROUS MATERIALS Pore Size Pores are classified according to size _________ < 2 nm _________ 2 nm < x < 20 nm _________ > 20 nm Yun Wu, Xianfeng Du, Honghua Ge and Zhen Lv Starch/ Stärke 2011, 00, 1–9 DOI /star Microprous starch

78 IUPAC Classification of porous solids
There are _________ of adsorption isotherms that have been observed. Each gives information about the types of pores contained in a solid as well as the capacity of the solid to adsorb a gas.

79 IUPAC Classification of porous solids

80 IUPAC Classification of porous solids (Cont.)
In the previous figures the _________ of the curve gives information about the solid – its _________. Let us examine these a bit closer: Microporous solids (see I) At ______pressure: adsorption in _________ first At _________ pressure: then coverage of _________ surface takes place. Mesoporous solids (see IV) At _____ pressure: ________coverage (plateaus) At _________ pressure: adsorption in _________. After the pores are filled adsorption occurs on the external surface.

81 Macroporous solids (see II) At _____ pressure: __________ coverage
At _______ pressure: _______ coverage until condensation occurs. There tends to be overlap between the 2 regions Uniform ultra-microporous solids (see VI) If all sites the same: _________ coverage If not: _________ isotherm for _________ Example: Zeolites Range of pore sizes Zeolite (small size range)

82 HYSTERESIS LOOPS Evaporation from a pore takes place at a _________ than condensation thus the path of _________ differs from _________. Four types of Hystereses have been identified and classified (IUPAC) P/P* Adsorbed volume The four _________ shapes of adsorption isotherms typically associated with N2 adsorption Type H1 Type H2 Type H3 Type H4

H1/H2 TYPE Particles with _________ or aggregates of _________ particles H1: uniform size/shape H2: non-uniform size/shape i.e. different size pore mouth and pore body e.g. ink bottle type pores H3/H4 TYPE Aggregates with _________ pores H4: uniform size/shape H3: non-uniform size/shape e.g. zeolites, carbons

84 KELVIN EQUATION Lord Kelvin noted that the evaporation of condensed gas molecules from a surface with very fine pores is more difficult than their condensation. This is because there is a greater probability, as compared to a _________ _________, that the molecules which evaporate from a ____________ meniscus will _________. Using the Kelvin equation, it is possible to measure a pore radius at a given P/P*: _________ _________, where V = molar volume of liquid,  = surface tension r = pore radius,  = contact angle (usually = zero) R=gas constant, T = temperature

85 PROBLEM P/P* Adsorbed volume ml/mol Consider a zeolite material onto which nitrogen was adsorbed and desorbed. If the P/P* was 0.25 at –183.15oC and the surface tension of nitrogen was 1.0 x 10-1 Nm-1 (and the contact angle was zero), then use the graph to calculate what the pore radius of the zeolite was.

86 SURFACE AREA Porous Materials I) Internal surface area
If the pores are _______, _________ uniform cylinders then: r = radius of pore, S = internal surface area II) Total Surface Area This is given by: S = nmLm, where: nm = moles gas adsorbed in __________ m = area of ________ adsorbed molecule L (or N) = Avagadro constant S = 2Vp/r Vp =pore volume

87 SURFACE AREA X Slope Y The Brunauer, Emmett and Teller (BET) Method
(Used on type II adsorption curves for multiple layer physisorption - see IUPAC classification of porous solids) Here: Which can be rearranged to give the ____________: Vm = volume of gas, monolayer coverage, C = _________ _________ constant V = θ = CP Vm (P* –P) (C–1)P P* X P P P* = (C-1) P* Vm 1 – P C.Vm C. Vm P* Y Y – intercept. Thus calculate Vm Slope

88 _________ _________ _________
P/P* P/P* / 10-1 cm-3 Vm (1-P/P*)

Single point method This is a _________ method. It arises because the slope > intercept (which tends to zero when P/P* is in the 0.2 to 0.3 region see (type II)). Hence assume the BET plot passes through the origin. i.e. assume Y-intercept at the origin. Then slope = (C-1)/C.Vm and thus Vm can be calculated. An error bar of 5% is acceptable in these experiments.

90 V P/P* Single point method (cont.)
Alternatively, if a simple extrapolation is made from the _________ of a set of data, then the molar volume can be obtained. This method delivers a rough estimate of the molar volume of within 10% (Bowker pp 57). P/P* V

The constant C in the BET equation is related to the following equation: C = e[(Ha – H1)]/RT, where Ha = enthalpy of _________ of the _________ and H1 = enthalpy liberated from the second and subsequent layers (similar to the _________ of the gas). Thus C helps give an estimate of ΔHads and it influences the ______ of the adsorption isotherm: C=10 000 V 1 P/P* C=10 C=2 C=1

92 SURFACE AREA Gases used for analyses:
Gas Area/10-20m2 Saturation P (torr) N at 77K Ar at 77K O at 77 K There is a limit on using nitrogen: _______ pore corresponds to 5 molecule width. Hence we use Ar or Kr for low surface area measurements (_________ ). 

93 KINETICS Kinetic measurements and the interpretation of the kinetic data lies at the heart of catalysis. Kinetic analysis allows for: Reactor design Correlating and rationalising catalytic activity Mechanism determination Rates of reaction, Order, Effect of Temperature Consider the reaction __A + __B  __C Rate (r) given by (rate of form = rate of consump): r = – (1/__)dA/dt = – (1/__)dB/dt = +(1/__)dC/dt Units of r = _________

94 KINETICS In a heterogeneous reaction the rate will depend on the _____________area available to the reactants. This is expressed as the ____________ (TOF). TOF = no of molecules converted per unit of time _______________ . Gas Phase Reaction (Homogeneous reactions) We know that the reaction rate can be expressed in terms of _________ _________ . Reaction : aA + bB  cC Rate = k(PA)a(PB)b(PC)c ….This is referred to as a ________________

95 Given: Rate = k(PA)a(PB)b(PC)c ,
a,b,c = orders of the reactants (don’t have to = integers). k = rate constant Rate can be expressed per _________ : r = mrm, where m = mass of catalyst OR If the total surface area (S) of the catalyst has been determined, then rate can be expressed ______________ : r = Srs Recall the Arrhenius equation: k = Acatexp(–Ecat/RT) Where Ecat = Activation Energy And Acat = pre-exponential factor

We will now look at two different mechanisms proposed for adsorption: (A) Langmuir-Hinshelwood (B) Eley-Rideal

97 Two types to consider: LANGMUIR-HINSHELWOOD (L-H) MODEL
Three general assumptions to this model: Adsorption is _____ and _________from the gas phase The reaction of the adsorbed molecules is the rate determining step (RDS) i.e. the surface chemical reaction = RDS i.e. k very small The _________ of an _________ species is determined by the appropriate Langmuir Isotherm. Two types to consider:

98 (A) L-H model for unimolecular reactions
Example: E(g)  E (Ads) → C(g) θE fast Surface rxn =RDS Molecule E at PE Product Molecule C E C If bE or PE are _______, then: Rate= k bE PE …i.e. __________ If bE or PE are _______, then: Rate → k …..i.e. ____________

99 Things to note about L-H model for unimolecular reactions
This type of kinetics is not specific to catalysis Mainly applies in: 3 step reactions Pre-equilibrium systems e.g. Michaelis Menton equation for enzymatic catalysis in Biochemistry

100 (B) L-H model for bimolecular reactions
A(g)  A (Ads) B(g)  B (Ads) And Surface rxn =RDS Fast A(Ads) + B(Ads) → AB(Ads) → AB(g) Two extra assumptions to this model: Molecules A and B are adsorbed on _______ sites with _____________ Product molecule AB is very ___________and comes off the surface fast.

101 θB θA B A Product molecule BA Molecule B at PB Molecule A at PA fast
RDS Molecule A at PA Product molecule BA θB B A .. Langmuir-Hinshelwood Equation

102 Things to note about L-H model for bimolecular reactions
If molecules A and B are _______ adsorbed, bA and bB are _______ , then: Rate = k’ PAPB ….i.e. _______, (Where k’= kbA bB) If molecule ____is _______ adsorbed and molecule B is strongly adsorbed i.e. _____________then: Rate = k’’ PA/ PB , (Where k’’= kbA/bB). This is an indication that B poisons the surface. Similarly the rate is affected if the _______ _______of molecule A and B are varied:

103 θB>>θA θA >> θB B B
The effect of surface concentration on the rate in bimolecular reactions PA rate For constant PB Rate limited by surface Concentration of A θB>>θA A B Rate limited by surface concentration of B θA >> θB A B

104 II) Eley-Rideal (E-R) model for bimolecular reactions
Unlike the L-H model, in the E-R model it is proposed that an adsorbed molecule may react _______with an _______ gas molecule by a collisional mechanism. The surface reaction is still the RDS i.e. PA fast RDS PB A B θA

105 Eley-Rideal bimolecular surface reactions
Note: For constant PA, the rate is always first order wrt PB θA = 1 PA Rate For constant PB _________ bAPA >> 1 _________ Thus: Rate = k PB …zero order in A bAPA << 1 Thus: Rate = k bA PA PB …….. first order in A _____ _____

106 How can the two bimolecular reaction models be distinguished from one another?
Experimentally if the reaction rate is measured as a function of the surface coverage of A i.e. θA, then the rate will initially increase for both mechanisms at ___________________. There will be slight to no difference between the two proposed models observed. BUT: If the reaction proceeds by the model proposed by Eley-Rideal then at ___________________(i.e. θA  1) the _________________until the surface is covered by A. If the reaction proceeds by the model proposed by Langmuir-Hinshelwood then the ______________ ________________and then gradually decreases to zero as θA  1.

107 Q: Will the rate of E-L model increase ad infinitum?
The effect of surface coverage on rate for the Eley-Rideal and Langmuir-Hinshelwood models θA Rate 1 Eley-Rideal model Langmuir-Hinshelwood Q: Will the rate of E-L model increase ad infinitum?

THE EFFECT OF TEMPERATURE ON KINETICS: CURVATURE IN ARRHENIUS PLOTS In heterogeneous systems we will still assume for the reaction: A  C Rate = –dPA/dt = kA = kbAPA. BUT in this equation BOTH ___ and _____ are functions of temperature. For instance: If bA is ________ and _____ is ________, then k will be the only variable that is a function of T. i.e. k = A·exp(-Etrue/RT) If bA is small then  tends to _____, then from the van’t Hoff Isochore we have: (dlnbA)/dT = HθA/RT2 , then ∫(dlnbA) = ∫ (HθA/RT2) dT

109 Thus: lnbA = –HθA /RT + C OR bA = C·exp(–HθA/RT) (where C = integration constant & –HθA is the standard molar enthalpy of adsorption of reactant A) Now we have rate = kA and A = bAPA We can substitute into the equation for both k and bA i.e. Rate = –dPA/dt = kA = kbAPA = k[C·exp(–HθA/RT)](PA) = A·exp(–Etrue/RT)[C·exp(–HθA/RT)](PA) Re-writing this out: = PA·A·C·exp(–Etrue–HθA)/RT = PA·A·C·exp(–Eapp)/RT Where: (–Eapp) = (–Etrue–HθA), hence Etrue = Eapp–Hθa and since HθA is always negative, then Etrue > Eapp

110 Arrhenius plot for a catalysed reaction over a range of temperatures gives:
III II I Log 10 Rate 1/T I II III Surface coverage 1 1>>0 ~ 0 Order,n 1>n>0 Slope x 2.3R Etrue ---- Eapp

Diffusion or mass-transport limited reactions are those in which the transport of the ________ /or ____________ the catalyst influence the rate: Rate  [catalyst]n and n< 1 vs ______________ Rate  stirring rate vs ____________________ Eact about 10 – 15 kJ mol–1 vs ______________ Rate  (temperature)1/2 vs _________________ An Arrhenius plot showing the onset of diffusion limitation can be drawn:

112 R Q’ Q R’ Log 10 Rate 1/T A’A: Surface reaction is rate-limiting B’B: Reaction is diffusion limited A ____________ for diffusion limited reactions the RAQ’ curve (purple curve) is detected, because the slower of the two processes controls the rate.

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