Lecture 26 March 8, 2010 Heterogeneous Catalysis

Transcription

1 Lecture 26 March 8, 2010 Heterogeneous Catalysis Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Wei-Guang Liu Ted Yu Ch120a-Goddard-L21 copyright 2010 William A. Goddard III, all rights reserved 1

2 Remaining Course schedule Wednesday, Mar. 3 no class (wag at NIST/NAS review Maryland) Friday Mar. 5, 2pm, L24 (make up for March 3) TDAY Friday Mar 5 3pm L25 (catching up to March 5) Monday, Mar. 8 2pm L26, caught up, normal Wednesday, Mar. 10 no class (wag at conference in Chicago) Thursday Mar. 11, 2pm, L27 Last lecture (make up for Mar. 10) Final available Thursday March 11 Final due back Friday March 19 Ch120a-Goddard-L21 copyright 2010 William A. Goddard III, all rights reserved 2

3 Last time Ch120a-Goddard-L21 copyright 2010 William A. Goddard III, all rights reserved 3

4 Homogeneous CH 4 functionalization: how to best choose new metals ur QM mechanistic studies for a variety of complexes from AuIII to ReI show the continuum of charge transfer to methane Charge transfer Electron-rich methyl groups Electron-poor methyl groups HX + 4

5 CH activation and functionalization by nucleophilic d 6 metals M-CH 3 polarization based on C1s chemical shift The carbon 1s orbital energy is an excellent measure of the electron density on the methyl carbon. This illustrates the extremes of the polarity scale, which require very different functionalization mechanisms. 5

6 ngoing Work in Homogeneous CH 4 Functionalization Insertion We modeled bipyridine complexes of Ru II, s II and Re I to determine the dependence of ground states (protonation), H 3 C-H activation barriers (substitution and insertion) and functionalization barriers on metal and π-donating ligand substituents. Substitution Going forward, we are considering the kinetics of these steps using d 5 and d 4 metals and new coordinated bases (i.e. NH 2 ). 6

7 ngoing Work in Homogeneous CH 4 Functionalization metal substituent substitution insertion Ru II X=H n.a. n.a. =NH n.a. n.a. s II X=H =NH Re I X=H =NH (bpy) 2 Ru(H) 2 complexes do not participate in insertion mechanisms (i.e., the products are not a minimum on the potential energy surface), only in the substitution path. (bpy) 2 s(h) 2 complexes allow both pathways (each are identifiable saddle points). However, the insertion pathway is preferred. Electron-donating substituents labilize hydroxide, creating vacancies. Insertion barriers decrease with the electron-donating ability of the substituent. The catalyst s susceptibility to oxidation also increases with the C-H activation rate. After the resting state switches to Ru(H)(H 2 ), the substituents weakly effect substitution barriers. Insertion barriers can be tuned over an extreme range by varying the ligand and metal. Substitution barriers cannot be similarly tuned. 7

8 CH 4 functionalization with homogeneous catalysts Reductive functionalization mechanisms (red. elim., S N 2) well known for late metals (M-CH 3 δ+ ). With Periana we have sought complimentary mechanisms appropriate for electron rich metals: Baeyer-Villiger Nucleophilic attack Barrier Electrophilic attack HX + Periodic table Reductive elimination Transalkylation Going forward: Determine what combinations of Group 9 and 10 metals, ligands and nucleophiles will allow S N 2 functionalization with thermally accessible barriers. 8

9 Going forward in homogeneous CH 4 functionalization We explore functionalization mechanisms in which the oxidant is a higher oxidation state of the hypothetical CH activation catalyst: s VI + s IV L = (acac) 2 s VI + s III L = terpyridine 9

10 Catalytic cycle: Au in H 2 S 4 /H 2 Se 4 AuI to III Product. Act. CH 4 Act. CH 4 Cycle: oxidation CH activation SN 2 attack I AuI to III Problem: Inhibited by water Accessibility of both Au I and Au III oxidation states prevents deactivation due to oxidization of catalyst 1. CH activation by electrophilic substitution. 2. Functionalization by nucleophilic attack by HS 4-. Jones, Periana, Goddard, et al., Angew. Chem. Int Ed. 2004, 43, C, 27 bar CH 10 4, TF 10-3 s -1

11 Predicted Pourbaix Diagram for Trans-(bpy) 2 Ru(H) 2 Black experimental data from Meyer, Red is from QM calculation (no fitting) using M06 functional, no explicit solvent Maximum errors: 200 mev, 2pH units Experiment: Dobson and Meyer, Inorg. Chem. Vol. 27, No.19,

12 Evaluating multi-oxidation state cycles for nucleophilic metals s V s III s IV 1 Volt 0.5 V s II (trpy)s(h n ) 3 Pipes and Meyer, Inorg. Chem. 1986, 25, Meyer, et al. Inorg. Chem. 1984, 23, (trpy)(bpy)s(h n ) xidation states VI II are present within ~0.5 V window. Aqua ligands stabilize many oxidation states. dd-electron oxidations are common. Ligands,anions influence the redox properties over a very wide range. 12

13 Industrially Important Catalytic Processes Many important organic chemicals are produced by catalytic processes 85% of industrial organic chemicals are produced from petroleum and natural gas 21% are produced by heterogeneous catalysis: allylic oxidation (acrolein, H 2 C=CHCH) and ammoxidation (acrylonitirle, H 2 C=CHCN), epoxidation, aromatic oxidation 13

14 Reaction Mechanism Study for Propane xidative Dehydrogenation on the Cubic V 4 10 cluster Manufacturing processes for lefins Title 1. Thermal Dehydrogenation of Alkanes C 3 H 8 C 3 H 6 + H 2 E= 38.0 kcal/mol Disadvantage: high temperature (>900 K); Low selectivity!! 2. xidative Dehydrogenation (DH) of Alkanes Catalyst C 3 H 8 + 1/2 2 C 3 H 6 + H 2 E= kcal/mol Advantage: very thermodynamically favorable Difficulty: the catalyst is still not understood 14

15 Unsupported Vanadium oxide (bulk V 2 5 ) Selectivity: % Supported Vanadium xide (V x /Al 2 3, V x /Zr 2 ) Selectivity: % Advantages: (a) higher mechanical strength (b) better thermal stability (c) larger surface area (d) highly dispersed V 4 unit 15

16 How model V= chemistry for V V? Want V= with 3 V- single bonds Also want each singly bonded to be bonded to two V Do not want edges with different chemistry Solution: periodicially infinite, but this is computationally complex How do finite system? V 4 10 cluster model 16

17 uster Model V 2 5 (001) surface V 4 10 cluster model R V-(1) =1.58 Å R V-(2) =1.78 Å R V-(3) =1.88 Å R V-(3 ) =2.02 Å (D-A bond) R V-(1) =1.57 Å R V-(2) =1.79 Å 17

18 Propene, H 2 out + H 2 CH H CH H 3 V V V V Start here V V V V propane in H V V V V Exper RDS 27.0 kcal/mol. (KIE) Theory gets 29.6 kcal/mol H 3 C CH 3 H Complete Catalytic Cycle on isolated V 2 5 propane + 2 propene H V V V V H 3 C CH3 H J. Phys. Chem. C, 111: 5115 (2007) propene out H 3 C H V V V V propane in CH3 H Net: 2 propane+ 2 2 propene + 2 H V V V V H 2 H H V V V V H 2 out H V V V V 2 H H 2 product cannot desorb from V 3+ site 2 in promotes desorption of H 2 while reactivating catalyst 18

19 Compare CH bonds propane H 3 C H 105 kcal/mol H 2 MeC H 100 kcal/mol HMe 2 C H 97 kcal/mol Me 3 C H 93 kcal/mol Thus H3C CH2 CH Expect to extract the H from the central C 19

20 Which of V410? Probe by bond H to them DH = - xx kcal/mol DH = - xx kcal/mol Thus much better to attack V= bond 20

21 Get propene product plus either ne V III -H 2 Eact = 11.8 kcal/mol or two V IV -H Eact = 12.2 kcal/mol Ultimately must get H2 to desorb the product H2 But H2 bond to V III very strong (xx kcal/mol) 21

22 Instead, react 2 with V III site 2 has S=1 and V III has S=1, thus combining them leads to S=0, 1, 2 Form ground state cyclic peroxide singlet Goes through V-- biradical with unpaired spin on V and on outer This is initially triplet but crosses with singlet and then forms cyclic peroxide Now H 2 desorbs easily (xx kcal/mol) Cyclic V can also react with propane to form new V-H plus i-pr 22

23 V V H Cyclic V react with propane to form new V- H plus i-pr The V--HiPr then goes to H V---iPr Which rearranges to position 2 nd H above H Then transer this H to form H2 and V= plus propene 23

24 Potential Energy Surface Methylene C-H bond of propane is activated in the rate-determining step with the barrier of 27.0 kcal/mol. (kinetic isotopic effects) Theory says 29.6 kcal/mol 24

25 Now want to study Reactive Dynamics CH 3 H + 2 H 2 C= + H 2 over V 2 5 Problem: QM methods not practical to determine mechanism and describe catalysis for mixture of reactants at experimental T and pressure 25

26 Problem QM methods not practical to determine mechanism and describe catalysis for mixture of reactant at experimental T and pressure Solution: ReaxFF reactive force field Describes reaction mechanisms (transition states and barriers) at nearly the accuracy of QM at computation costs nearly as low as ordinary force fields Val Coul E = E + E + E VdW Valence energy Electrostatic energy short distance Pauli Repulsion + long range dispersion (pairwise Morse function) Allow bonds to break and form smoothly, describe barriers for reactions. Allows charges to change continuously as reactions proceed (QEq method) All parameters from quantum mechanics (no empirical data) ReaxFF describes reactive processes (from oxidation to combustion to catalysis to shock induced chemistry) for 1000s to millions atoms Applied to catalysis for realistic temperatures and pressures 26

27 Critical element ReaxFF: charges flow as geometry changes Self-consistent Charge Equilibration (QEq) Describe charges as distributed (Gaussian) Thus charges on adjacent atoms shielded (interactions constant as R 0) and include interactions over ALL atoms, even if bonded (no exclusions) Allow charge transfer (QEq method) atomic interactions J ij E { χ iq ηij η kl i ( q ) = + i} J ij ( qi, q j, rij ) i< j E int Erf r η k l ij + η ij kl k l ( r, Q, Q ) = Q Q ij i j r ij i k 1/r ij i 1 + J 2 i q Charge Equilibration for MD Simulations; Rappé, Goddard J. Phys. Chem. 95, 3358 (1991) 2 i Keeping: q i = Q Hardness (IP-EA) Electronegativity (IP+EA)/2 I i r i 0 + r j 0 Three universal parameters for each element: r ij 1991: use experimental IP, EA, R i ; ReaxFF get from fitting QM χ o i, J, o i R c i 27

28 Bond order nd Essential element of ReaxFF Bond distance bond order forces General analytic form. Get parameters from fitting QM bond breaking for systems with single, double, and triple bonded. All parameters from σ σ σ π π ππ QM Bond order Bond order (uncorrected) Sigma bond Pi bond Double pi bond E bond = D B f ( B ) Distance (A) Interatomic distance (Å) Interatomic distance (Å) Valence Terms (E Val ) based on Bond rder: dissociates smoothly Forces depend only on geometry (no assigned bond types) Allows angle, torsion, and inversion terms (where needed) Describes resonance (benzene, allyl) Describes forbidden (2 s + 2 s ) and allowed (Diels-Alder) reactions Atomic Valence Term (sum of Bond rders gives valency) e ij Bond energy (kcal/mol) Bond energy ij D e B ij D Sigma energy Pi energy e Double pi energy Total bond energy B ππ ij Distance (A) Adri van Duin 28

29 ReaxFF Development: Propane propene on V 4 10 ReaxFF QM V propane Cheng, Chenoweth, xgaard, van Duin, Goddard JPC-C 2007, 111, ReaxFF reproduces the QM energies for the entire reaction pathway Kimberly Chenoweth MSC, Caltech Binding of 2 displaces propene product V H propene 29

30 Derive one FF for V to describe all coordinations in the metal and oxide and all oxidation states Metal FCC, BCC,HCP, A15, SC, Diamond Metal oxides Heat of formation (kcal/unit) 4, 6, 8 xygen coordination V 2 5 V 2 V 2 3 QM V V(bcc) QM ReaxFF QM: SeqQuest (SNL Gaussianbased periodic DFT) Energy difference for oxidation changes is in good agreement with QM data Indicates that ReaxFF is able to capture energetics of redox reactions V 2 V 2 5 V ReaxFF V(bcc) V

31 ReaxFF Reactive Dynamics Simulation CH 3 H + ½ 2 H 2 C= + H 2 over V layer V 2 5 (001) periodic slab 30 methanol molecules Total number of atoms = 684 Slab Temp = 650K CH 3 H Temp = 2000K Time step = 0.25 fs Temperature damping = 100 fs Total simulation time = 250ps 31

32 ReaxFF RD Simulation Methanol xidative DeHydrogenation (DH)/V H 3 CH Methanol Formaldehyde H2CH Radical Water thers Surface species 5 0 H 2 C= H Time (ps) methanol converts to formaldehyde with production of water thers include the number of hydrocarbons bound to the surface Expt.: Major product is formaldehyde (also H 2, C x ) 32

33 ReaxFF Mechanistic Details Formation of H 2 C-H radical 3.45ps 8.80ps Formation of formaldehyde H-abstraction by surface vanadyl groups 33

34 Desorption of Water from catalyst Snapshots from simulation showing atoms within 5.5Å of V bound to H Water bound to V III, bond very strong, will not desorb 2 nd layer has V V = pointing at V III of top layer 2 nd layer bonds to top V get V IV --V IV H 2 bonds weakly to V IV now desorbs 34

35 Bi III 1/2 2 or [] ~ ~ Bi ~ ~ G 673 ~ 37.0 ~ ~ allyl, H propene propene H allyl ~ IV Bi ~ ~ 2 allyl V Summary: xidation Propene + 2 Acrolein + H 2 G 673 = kcal/mol CH 2 CHCH 2 CHCHCH 2 H ~ Mo VI ~ ~ Mo VI ~ ~ Mo VI ~ ~ Mo V ~ ~ Mo V ~ ~ Mo IV ~ G 673 = 11.4 CH 2 CHCH 2 G 673 = 9.2 H CHCHCH 2 ~ Mo VI ~ ~ Mo VI ~ ~ Mo VI ~ ~ Mo V ~ ~ Mo V ~ ~ Mo IV ~ H 2 G 673 ~ G 673 = acrolein 1/2 2 or [] H 2 or 2[] H ~ Mo V ~ ~ Mo VI ~ ~ Mo V ~ ~ Mo IV ~ All in agreement with experiment, except for the role of Bi V H Mo V ~ ~ Mo VI ~ ~ G 673 ~ H Mo V ~ ~ ~ Mo IV ~ 35

36 Calculation: Allyl adsorption Spectator Mo= Mo VI CH 2 =CHCH 2 Mo V G 673 = 5.7 NH Mo VI G 673 = -4.0 CH 2 CH=CH 2 NH Mo V Spectator Mo=NH NH CH 2 =CHCH 2 NH HN NH HN CH 2 CH=CH 2 NH Mo VI G 673 =12.8 Mo V Mo VI G 673 = 2.9 Mo V insertion N insertion Spectator effect: Mo= > Mo=NH by 7 kcal/mol Consistent with k I >>k I Chemisorption on Mo=NH is easier than on Mo= by 10 kcal/mol Consistent with the assumption k NI >> k I 36

37 ne-center or Multi-center? /NH Insertion G 673 = -4.8 CHCH=CH 2 HN NH 2 Mo di-imido di-oxo CH 2 =CHCH 2 Mo G 673 =18.6 G 673 =14.5 Mo G 673 = 4.6 CHCH=CH 2 Mo Mo H H CHCH=CH 2 Mo H CHCH=CH 2 Mo CH 2 =CHCH 2 Multi-center for di-oxo Multi-center for oxo-imido May be one or two center for di-imido All in agreement with experiment CH 2 =CHCH 2 NH Mo NH NH Mo G 673 = 4.3 HN Mo NH G 673 = -6.5 G 673 ~ 4.4 G 673 ~ 14.5 Mo NH G 673 = -9.2 HN HN Mo HN CHCH=CH 2 Mo Mo H HN NH 2 CHCH=CH 2 Mo H H CHCH=CH 2 NH 2 Mo HN NH HN CHCH=CH 2 Mo NH oxo-imido CHCH=CH 2 Mo 37

38 Breakthrough 1995 (Mitsubishi, BP-America). Mixed Metal xide Catalysts (MoVNbTaTex +xxx) for propane ammoxidation CH 3 -CH 2 -CH 3 + NH CH 2 =CH-CN + 4 H 2 Progress: now know that 2 phases (M1 and M2) are involved in MM Catalysis. Have powder xray structures.. But 13 years of R&D still not commercial. No real idea about how atomic level structure affects the chemistry model for critical sites in M1 and M2 phases M To make rapid progress NEED THERY T UNDERSTAND Mechanism M2 M4 M5 Te1 M4 Need to model critical sites in M1 and M2 phases. But must use huge unit cell (>3000 atoms) because of partial occupation M3 Te2 M4 M4 M3 Not practical for QM M5 Te1 38

39 Experimental ccupations in M1 Phase of MoVNbTe x Tot Mo : V 7.3: : :1.5 1 Mo : V 0.46: : :0.2 2 Mo : V 0.63: : :0.2 3 Mo : V 0.46: : :0.5 4 Mo : V 0.88: : :0.0 5 Mo : V 0.80: : :0.0 6 Mo Mo : V 0.62: : :0.5 8 Mo Nb Mo Mo Te Te H. Murayamai et al. Applied Catalysis A 318 (2007) 137. P. Desanto et al. Z. Kristallogr. A 219 (2004) 152. HP. DeSanto et al. Topics in Catalysis 23 (2003)

40 General problem with Crystal structures of Mixed Metal xide catalysts: partial occupancy M1 phase of the MoVTeNb x catalyst To get mechanism, need whole atoms at each site. Use ReaxFF to find stable whole atom structures for large supercell Trigonal Mo 3 V x catalyst 40

41 Compare Calculated x-ray powder diffraction intensities from final ReaxFF structure with experimental peaks Crystal-x-ray From Predicted Structure Annealed at 300K Conclusion: predicted structure with resolved partial occupations is consistent with experimental x-ray diffraction intensities Peak at 22 is indicative of the layered structure. 41

42 Final Configuration trigonal Mo 3 V x catalyst from ReaxFF-MC-RD C7 1 metals in highest oxidation state C7 2 and C7 3 containing some reduced metals C6 1 and C6 2 contain some reduced metal sites. C7 2 C6 2 C7 1 C7 3 C6 1 vanadyl oxygens are shown in blue. C6 2 C7 3 C7 1 C6 1 C7 2 42

43 Vanadium Coordination M3 site, V= vanadyl groups align along the c-axis, similar to bulk V 2 5 M1 2 site has V= Vanadyl groups pointing into the C7 2 channel 43

44 Donor-acceptor Network a) Network of donoracceptor interactions (Mo VI =---Mo VI ) in the Mo 3 V x catalyst where b) Mo VI (purple) facilitates the continuation of the network while other metal sites disrupt the network. c) Network is more complete in the c- direction compared to the a-b plane due the chains of Mo VI. 44

45 Final Configuration from ReaxFF- RD of Mo 3 Vx with Propane Initial Configuration final Configuration final Configuration Top view Mo = purple V = green = red propane molecules Yellow: in channel blue-gray: exterior 3 propane molecules all go into the C7 2 channel 45

46 Cross-section final configuration from the propane/mo3vx ReaxFF-RD 3 Propane moved into C7 2 Heptagonal Channel Average channel radius = 4.6 Å length ~ 18Å Channel C7 1 is smaller with average radius = 4.1Å and remains empty 46

47 Speculations about M1 selective oxidation from ReaxFF RD Simulations We believe that the migration of propane into the heptagonal channels found in the RD plays an important role in the selectivity. It has V= chain, just like V 2 5 and VP that can break CH bond (E act ~ 28 kcal/mol) After the activation, a 2 nd H can be transferred to any oxo group or ether group to form propene But this propene is in a protected site inside the channel where the V= chain has already been de-activated so that it can undergo selective activation of allylic CH bond followed by trapping on a M= bond to form M--CH2-CH-CH2 and then it continues the same as for propene selective oxidation in BiMox etc We think that the unselective oxidation to C 2 occurs at the surfaces and grain boundaries, where there may be multiple V= sites, leading to rapid oxidation Thus to obtain increased selectivity want to poison the surface V= sites but not the channel V= chains. This might be done with bulky groups 47

48 New material 48

49 Hemoglobin Blood has 5 billion erythrocytes/ml Each erythrocyte contains 280 million hemoglobin (Hb) molecules Each Hb has MW=64500 Dalton (diameter ~ 60A) Four subunits (α1, α2, β1, β2) each with one heme subunit Each subunit resembles myoglobin (Mb) which has one heme Hb Mb 49

50 The action is at the heme or Fe-Porphyrin molecule Essentially all action occurs at the heme, which is basically an Fe-Porphyrin molecule The rest of the Mb serves mainly to provide a hydrophobic envirornment at the Fe and to protect the heme 50

51 The heme group The net charge of the Fe-heme is zero. The VB structure shown is one of several, all of which lead to two neutral N and two negative N. Thus we consider that the Fe is Fe 2+ with a d 6 configuration Each N has a doubly occupied sp 2 σ orbital pointing at it. 51

52 Energies of the 5 Fe 2+ d orbitals x 2 -y 2 z 2 =2z 2 -x 2 -y 2 yz xz xy 52

53 Exchange stabilizations 53

54 Ferrous Fe II x 2 -y 2 destabilized by heme N lone pairs z 2 destabilized by 5 th ligand imidazole or 6 th ligand C y x 54

55 Four coordinate Fe-Heme High spin case, S=2 or q The 5 th axial ligand will destabilize q2 since dz2 is doubly occupied A pi acceptor would stabilize q1 wrt q2 Bonding 2 to 5 coordinate will stabilize q3 wrt q1 Future discuss only q1 and denote as q 55

56 Four coordinate Fe-Heme Intermediate spin, S=1 or t 56

57 Four coordinate Fe-Heme Low spin case, S=0 or s 57

58 ut of plane motion of Fe 4 coordinate 58

59 Add axial base N-N Nonbonded interactions push Fe out of plane is antibonding 59

60 Summary 4 coord and 5 coord states 60

61 Free atom to 4 coord to 5 coord Net effect due to five N ligands is to squish the q, t, and s states by a factor of 3 This makes all three available as possible ground states depending 61 on the 6 th ligand

62 Bonding of 2 with to form ozone 2 has available a pσ orbital for a σ bond to a pσ orbital of the atom And the 3 electron π system for a π bond to a pπ orbital of the atom 62

63 Bond 2 to Mb Simple VB structures get S=1 or triplet state In fact Mb 2 is singlet Why? 63

64 change in exchange terms when Bond 2 to Mb 2 pσ 2 pπ Assume perfect VB spin pairing Then get 4 cases Thus average K dd is ( )/4 = K dd 5*4/2 up spin down spin 7 K dd 4*3/2 + 2*1/2 7 K dd 4*3/2 + 2*1/2 6 K dd 3*2/2 + 3*2/2 64

65 Bonding 2 to Mb Exchange loss on bonding 2 65

66 Modified exchange energy for q state But expected t binding to be 2*22 = 44 kcal/mol stronger than q What happened? Binding to q would have H = = + 11 kcal/mol Instead the q state retains the high spin pairing so that there is no exchange loss, but now the coupling of Fe to 2 does not gain the full VB strength, leading to bond of only 8kcal/mol instead of 33 66

67 Bond C to Mb H 2 and N 2 do not bond strongly enough to promote the Fe to an excited state, thus get S=2 67

68 compare bonding of C and 2 to Mb 68