I Ii Mi IiIIiIS I III.,! I 'in. III. Is I mpty IV III IV Filled Ó8 / /// Figure 12.23 A rough diagram of the key MOs during reaction (12.101). During reaction (12.101) the system starts out with the orbitals in the left of Figure 12.23. There are four MOs labeled I, II, III, and IV. In the diagram, we have arbitrarily assigned the lobe on the R group to have a positive sign, and then considered all possible signs on the p hybrids on the carbon. In MO I, both the p's are positive; in MO II, the left p is negative, while the right p is positive. In the MO 111 the right p is negative and the left p is postitive, while in MO IV, both p's are negative. In MOs I and II, the 71 group has a bonding interaction with the carbon, while in III and IV, there is a sign change in moving from R to C and so the interaction is antibonding. As a result, orbitals I and II are bonding orbitals while orbitals III and IV are antibonding orbitals. In contrast, at the end of the reaction, the R group migrates from the right to the left of the molecule. In this case, orbitals I and III are bonding while orbitals 11 and IV arc antibonding. Now consider moving the R group. Notice that the R group needs to move across the molecule for reaction to occur. However, in order to move the positive orbital on the R group in orbital II, it will need to displace the negative nonbonding orbital on the carbon. In Chapter 10 we found that such orbital displacements have large barriers. The net effect is that 1,2 displacements have large barriers with neutral radicals. Notice that the repulsion occurs only because the electrons in the R group are pushing up against the nonbonding orbital in the molecule. If one modifies the charges on the molecule, one can remove the electrons from the nonbonding orbital. If one puts a +1 charge on the molecule, there will be no repulsions. Well, again on superacid catalysts the charge is nearly +1. That promotes easy isomer-ization. The net result is that the superacid catalyst is able to promote reaction (12.101). These results show that catalysts can modify the intrinsic barriers to reaction. These modifications allow very selective reactions to occur. These results show that catalysts can be designed to modify the changes on the reactants in a way that facilitates reaction. 12.12 CATALYSTS CAN BE DESIGNED TO BLOCK SIDE REACTIONS Another thing that catalysts can do is to block side reactions. The idea is simple. You design the catalyst so that it is shaped in such a way that the reactants can get together fiATAI V*T> n AN m i-r .H.i h ľ mhu", i. I II III Al / l't ......„„m,,.,,....................................•"".......'•;...... ^^rrrv^sÄR.,*................... ■titnylcne: CH3 nH2C=(IK'lll [-C-C-]n (12.103) Polypropylene undergoes free-radical orcalionic polymerization as discussed 111 ( lupiči "> 11 ih. free radical reaction occurs, the methyl group can go on (lie top 01 the bol tom 1 I hi molecule. If the methyl groups arc distributed randomly, the polymei hai pool .....linn, .il properties. As a result, the polymer cannot be used in many applii atioitl In ......1.1 il one can control the positions of the methyl groups, one can produce polymi IK llh much better mechanical properties. Such polymers are very valuable I heir are iwo key forms of oriented polypropylene: isotactic polypropylene, where ill ..1 1I1. nieiliyl groups are on one side HpH HCH HCH HCH h£h h£h iml syndiotactic polypropylene, where the methyl groups alternate from side......le u H H H HCHHCH HCHHCH V x x x /cXr XX X: H H HCH HfjH \^ v -cx-cx^ 1(1-1) 1 .insider making isotatic polyethylene. Figure 12.24 shows a diagram of a step during ih. production of isotactic polypropylene where a single propylene unit is added to i flowing polymer chain. Notice that one can add the propylene, with a methyl group 1.1. mg in the correct direction or in the wrong direction. If one would add (he propylene .miIi the methyl group facing the wrong way, polymerization would still occur, bin one would not end up with isotactic polypropylene. HqH h£h h£h H^H f£h h£h H»H V?" Cc _,XrX^ \ h£h h£h h£h r H^H X Methyl on wrong side h"h h"h h''h Figure 12.24 A rough diagram of one step during the production of isotatic polypropylene. Blocking _ group \ Ay/y Binding site Pocket \ Figure 12.25 A diagram of propylene polymerization in a Ziegler-Natta catalyst. When one runs these reactions industrially, one makes these polymers with what is called a Ziegler-Natta catalyst. The Ziegler-Natta catalyst is designed to produce only isotactic propylene. Figure 12.25 shows a diagram of the active site in a Ziegler-Natla catalyst. The titanium atom actually catalyzes the reaction. However, the titanium atom is surrounded by ligands that enclose the site. There are two pockets in the ligands. During the reaction, propylene squeezes into the pockets and reacts. The ligands prevent the propylene from twisting. That prevents the side reaction. The net result is that all of the methyl groups line up on the same side. This example shows that there can be an advantage to designing a catalyst with blocking groups to prevent side reactions. 12.13 CATALYSTS CAN BE DESIGNED TO DONATE AND ACCEPT ELECTRONS Another mode of catalytic action is for the catalyst to donate or accept electrons. Recall that in Section 2.3.2 we found that certain metal oxides are effective oxidation catalysts because they can store and give up lattice oxygen. Well, these catalysts work by a redox process where the metal is oxidized and reduced. For example, consider CO oxidation over a copper substrate. There are two stable copper oxides: Cu1+ oxide (i.e., Cu20) and Cu2+ oxide (i.e., CuO). During the reaction, the Cu1+ oxide is first oxidized to a Cu2+ oxide: Cu20 + \02 2CuO then the Cu2+ oxide reacts with CO to yield a Cul+ oxide: 2CuO + CO Cu20 + C02 (12.106) (12.107) Notice that in step (12.106) the copper donates an electron to the oxygen, while in step (12.107), the oxygen donates electrons back to the copper. The donation and release of electrons makes copper a good catalyst for CO oxidation. (Yrtin (IAN P i n mii in«ii nNi i w.i.i r i 1111 .mi inn ||U.„ „,e sover.il .x.in.....>» wIm-ii- .'le.l.nn Iriiiwlci process is „,..„• dual I in ninplo, Ihe electron liiinnli"i u-iit ll»" » It" I V < i , I ' IDS) i, .low because the +2 charge on (he iron repels the +3 charge on the vanadium I in11ii-1. the solvation shells on the iron and vanadium get in the way of each other. repulsion is less with C'u'1 because Cu 1 has only a +1 charge. Furthei coppei hai i much smaller solvation shell than either iron or nickel. As a result, coppei ions CM l,i< Dilate reaction (12.108). I he main mechanism of the reaction is electron exchange. First Cu ' reacts with V'' • Lulling an electron: Cul++V4+->Cu2+ + V,+ (12.109) I hen the Cu1+ is regenerated by reaction with Fe2+: Fe2++Cu2+-► Fe,+ +Cu1 + < i ' I 10) i loin-e that the copper is donating electrons in reaction (12.109) and accepting the electron I'.uk again in reaction (12.110). This is another example where the catalyst works by donating and accepting charge. Another application of this technique is in fuel cells. Fuel cells are commonly used In generate electricity. Figure 12.26 shows a diagram of a simple hydrogen fuel cell i ienerally, hydrogen reacts on a platinum catalyst on the anode of the fuel cell to produce pinions and electrons: H2=>2H++2e (12.111) where e~ is an electron. The protons are then transmitted through a polymer membrane In the fuel cell, where they react on the cathode: 02 + 4H+ + 4e~ 2H,0 ll.'.ll.'i Fuel cells can produce enough power to run an automobile. Notice that the catalyst on the mode is accepting electrons, while the catalyst on the cathode is giving up electrons. I nil is a further example showing that catalysts can work by accepting and donating electrons H, 2H+ + 2e~ Load 4H,0 Figure 12.26 A diagram of a polymer fuel cell. MAV, IIIAN'lllll.....1 1 .......,„.....,ANI WIMNAi AIAIV'U l!t I'MI »I N1 /:n> \:> II CAIAI YSIS CAN AC I A!. AN I I I l( .11 N I Ml ANS I Oil I NI.MGY TRANSFER There is one other mode of catalytic action that people do not talk alioul much in thl literature. Catalysts can be designed to act as an efficient means lor energy transfer. Recall] from Chapter 9 that the rate of most unimolecular reactions and bimolecular recombination reactions is limited by the rate at which energy enters or leaves the molecule. For example, the reaction Na + Cl-► NaCl (12.113)1 has a zero rate in the absence of a collision partner even though the reaction is 98 kcal/mol exothermic. If you add a collision partner, the reaction goes very quickly. In a sense, the collision partner is acting like a catalyst to carry heat away from the reactants. It happens that solid catalysts are often much better collision partners than arc gaseous species. For example, when Polanyi (1932) first examined reaction (12.113), he found that most of the reaction occurred on the walls of his vessel. The walls were acting as an efficient means of heat transfer, and thereby allowed reaction to occur. All solid catalysts or solvents have similar effects. The catalysts provide very efficient means for heat transfer. That allows reaction to occur. In practice, most unimolecular and bimolecular reactants occur much more rapidly in the presence of a liquid or a solid. Therefore, the reaction is being catalyzed even though the catalyst never binds strongly to the reactants. 12.15 MASS TRANSFER EFFECTS ARE MORE IMPORTANT WHEN A CATALYST IS PRESENT Finally, it is important to point out that mass transfer effects become more important when catalysts are present. Recall that, according to collision theory, the rate of reaction is determined by the rate that molecules collide, multiplied by a factor that accounts for the fact that not all molecules that collide react. In a solution, the collision rate is determined by mass transfer effects, while the fraction that reacts is determined by what happens after the reactants collide. A catalyst can modify the fraction of the molecules that react once the molecules collide. However, the catalyst cannot make the reactants collide more quickly. As a result, as you speed up a reaction with a catalyst, the rate of reaction will eventually be limited by the rate of collisions. The rate of collisions is determined by the rate at which the reactants diffuse together. The diffusion rate is, in turn, determined by the mass transfer rate. The net effect is that mass transfer became more important as you improve catalysts, so mass transfer controls the rate of reaction on the very best catalysts. When students first hear that mass transfer is controlling the rate, they think that the catalyst is not doing its job. However, catalysts are supposed to speed up rates of reactions, and if one speeds up a reaction enough, one always finds that diffusion starts to play a role. The best catalysts speed up a reaction so much that the reaction rate is instantaneous once the reactants collide. The result is that the mass transfer effects control the rate of reaction. People usually quantify the role of mass transfer on the basis of two parameters: the Thiele parameter, p; and what I call a mass transfer factor, n. The Thiele parameter is defined by reaction rate <»p = 7Tf-—r (12-114) diffusion rate I hi m.r.'i liunslei Im-toi in tliliiuil l>v in tun I reaction rule _ in in n. =3- (i... 119) reaclion mir il mass lianslei were instantaneous lid I......e people call the mass trunslei lacior an effectiveness factor We derive ........... loi i|>|. and i|c in the supplemental material. The result for a (irst-ordei rcu< lion mi I in)' m a spherical catalyst pellet is I r 1 q>|, p = yp il.1 1101 (12.1H) ii. 11 \, is ihe radius of the pellet, k| is the rate constant for the reaclion, and 11 1 thl ...... uiiy of the reactant in the pellet. I Igure 12.27 shows a plot of the mass transfer factor versus the Thiele paramelei foi Miii.inn into a solid catalyst pellet. Basically, the mass transfer factor is unity when iln I hide parameter is much less than one, and then tails off linearly when (he Thiele 1.......eter is greater than one. One can show that when 4>p > 10, the term in brackets in 1.....on 112.116) is between 0.9 and 1, under these circumstances: 0.95 ne « - (12 118) P In keep Figure 12.27 in perspective, there is some confusion created by calling h, ,m . ITcctiveness factor. The mass transfer factor does not measure the effectiveness 0! .1 Ullllyst. Good catalysts can have low mass transfer factors, while had catalysis can have hirh mass transfer factors. Generally, if a catalyst is not speeding up the reaction very .....rh, the Thiele parameter will be small, which means that, according to Figure 12.27, the mass transfer factor will be close to unity. Thus, a bad catalyst can have a large mass ii.msler factor. In contrast, if the catalyst speeds up a reaction by a considerable amount ,, will be large. Figure 12.27 shows that the mass transfer factor is reduced undei in Ii ■ Ircumstances. 10 CD 1 "eg 5 0 0 0.1 a <0 0.01 0.001 0.01 Figure 12.27 A plot of the mass l>ellet. 100 Thiele parameter, P parameter for diffusion in a porous catalyst transfer factor versus the Thiele .11II11111 I ..Ml Mi illl ii II'c il 11 AM I Wl II N A I AI Al Vil I*. I'lII M NI Ml All ol litis is a balance because you can also get a lar^e ,, il you have designed I he catalyst pellel wrong so thai I), is too small However, thai is rarely an issue with modern support materiuls. Usually, a small mass transfer factor is associated with a verj activi catalyst. Consequently, the mass transfer factor does not measure the effectiveness "I I catalyst even though the mass transfer factor is often called an effectiveness factor in the literature. I believe it is important to think of r)e as a mass transfer factor, not an effectiveness factor. Generally, when ne is small, you are near the mass transfer limit, so there is little-that you can do to improve the catalyst except speedup the mass transfer rate. On the other hand, when ne is large, the reaction rate is slow compared to the diffusion rale so that you can do to speed up the reaction. When 1 teach this material, I call n.,. a mass transfer factor. However, ne is still called an "effectiveness factor" in the literature. 12.15.1 Shape-Selective Catalysis The other thing to recognize is that mass transfer limitations are not necessarily bad. In industrial practice, mass transfer limitations are used as design variables to improve the efficiency of catalysts. One of the problems in designing a catalyst is that most catalytic materials speed up a wide variety of reactions. The rate of the desired reaction is enhanced. However, often the rates of some undesirable reactions are also enhanced. Fortunately, there is a design strategy to get around this difficulty: . Use a solid catalyst. • Design the solid in such a way that only the desired product can get out of the catalyst. • Add enough catalytic components so that even if a side product is formed, the side product will be converted into the desired product. An example of this strategy comes in the production of para-xylene. Xylene can be made via the alkylation of toluene over an acid catalyst CH3C6H5 + CH^OH CH,Cf,H4CH3 + HzO (12.119) The alkylation reaction produces meta- and ortho-\y\ene in addition to para-xylene. People run the reaction in a catalyst that has acid sites in an interconnecting pore structure like that in Figure 12.28. The o-, m-, and p-xylene are formed at the acid sites, and then need to diffuse out of the pores before the xylene can leave the reactor. Note that if you align m- and o-xylene standing with the methyl facing up, the m-and o-xylene are wider than the para-xylene. As a result, if one designs a structure like that in Figure 12.28 with pores that just fit para-xylene, o- and m-xylene will not be able to diffuse down the pores, so the o- and m-xylene get trapped in the catalyst. One then adds an occasional isomerization site to the catalyst. The isomerization site converts the o- and m-xylene to p-xylene. The net result is that the catalyst produces mainly p-xylene, even though the alkylation reaction, reaction (12.119), produces similar amounts of £>-, m-, and p-xylene. People call this type of catalyst a shape-selective catalyst, because you are able to select molecules according to their shape and how easily they diffuse. Shape-selective catalysis is very important, because one can design very selective catalysts. Still, it is not I mure 12.28 An interconnecting pore structure that is selective for the formation of paraxylwm i i i ,y to make a catalyst with just the right size pores, so this is an area where there ll more art than science. fable 12.12 lists several different materials that are used as shape-selective catalysts. (|i ... ially, shape-selective catalysts are made out of materials called zeolites. Zeolitei ire silica aluminas, with very uniform pores. Figure 12.5 shows a single unit cell in I i olite. Generally, zeolites consist of small cages stacked on top of each other. There in small interconnecting pores between the cages. Table 12.12 shows the pore si/es m . une zeolites, while Table 12.13 shows the dimensions of some typical molecules. Notice thai you can choose the pore sizes so that only certain molecules can squeeze through the pores. Thus zeolite can be used to selectively produce a designed product, while nol producing a larger side product. /oolites can be used with a wide variety of catalyst materials, and can be used to promote selectivity in a wide number of reactions. Therefore, they are very useful for i alalyst design. Table 12.12 Pore sizes in some zeolites Size of Diffusion Size of Cavity, Zeolite Channel, A A Chabazite 3.6 x 3.7 5 Zeolite A 4.1 x 4.1 6.5 Erondite 3.6 x 5.2 11.6 Ferrierite 4.3 x 5.5 6.5 ZSM-5 5.5 x 5.6 10.5 Offretite 6.4 x 6.4 6.5 Mordenite 6.7 x 7.0 10.5 Faugasite 7.4 x 7.4 11.9 VFI 13 x 13 IIM I Ml M II II I II IN 11 I I ft I AI i ■ II'. KINI IK ' il i AIAI Ylli: III AC III IN'. 74:i Tulilo 12.13 Minimum diamolois ol 1,....... 111,,I,-, uli-. Minimum Molecule Diameter, A Linear alkane 4 Isoalkane 5.5 Benzene 5.1 Paraxylene 5.1 Orthoxylene 5.7 2-Methyl alkenes 5.1 Naphthalene 7.3 12.16 SUMMARY OF POTENTIAL CATALYST FUNCTIONS In summary, then, in the last few sections, we showed that catalysts can change rates in eight key ways: • Catalysts can he designed to help initiate reactions. • Catalysts can be designed to stabilize the intermediates of a reaction. • Catalysts can be designed to hold the reactants in close proximity. . Catalysts can be designed to hold the reactants in the right configuration to react. • Catalysts can be designed to block side reactions. • Catalysts can be designed to sequentially stretch bonds and otherwise make bonds easier to break. . Catalysts can be designed to donate and accept electrons. . Catalysts can be designed to act as an efficient means for energy transfer. It is also important to realize that . One needs a catalytic cycle to get reactions to happen. . Mass transfer limitations are more important when a catalyst is present. One of the questions I get when I teach this material is "Here are all of these effects, how do I know which effect will be the most important in a given catalyst system?" Unfortunately, at present we rarely know the answer to that question. As noted in Section 12.1, one would like the catalyst to do all of these things. For example, platinum is a wonderful catalyst for olefin hydrogenation. The platinum speeds up the initiation step during the hydrogenation process. The platinum stabilizes the intermediates of the reaction. The platinum holds the reactants in close proximity. The platinum also lowers the intrinsic barrier to hydrogenation. Platinum does everything so it is a great catalyst. It is not clear how to know that a priori. Generally, the most important effect of a catalyst is to stabilize intermediates and initiate reactions. However, all of the other effects also occur to some extent with any catalyst. Consequently, catalytic reactions are less well understood than, for example, gas-phase reactions. 11/ KINI IICS Ol (AIAI Y III III ACIIONS 1 we will be changing lupii •, in discuss ilie kinetics ol catalytic reactions. So far 111 llns ihiipli'i, we learned lhal most ciiliilyln tend.....s arc Ihe same. The readmits bind In linii ilvsi Then, the system noes through a catalytic cycle where intermediales arc formed nul iioatioyed Finally, Ihe product! leave the catalyst and the catalyst is regenerated Id ih, ncxl two sections, we will derive equations for the rales of catalytic reaction! Wi discussed many of the details in Section 5.10, but it is useful to repeal some of Ihe iiniliii)'s here: . First, all reactions go by a catalytic cycle. For example. Figure 12.2l) shows two dilTcrenl catalytic cycles for the production of water. The reactants adsorb, then rcuct. . rhere are three kinds of mechanisms as shown in Figure 12.30: a Langmuil Hinshelwood mechanism where all of the reactants are adsorbed; a Ridcal I In mechanism, where a gas-phase species collides with an adsorbed species and rem 1 and a precursor mechanism, where there is one strongly bound rcaclanl bound rcactanl and one reactant in a weakly bound precursor state. 1'hcre are several examples of these mechanisms in Section 5.10. Ihe readei ihould 1, \ icw Section 5.10 before proceeding. Next, we note that it is important to think about the rate of a catalytic reaction in lemis ill 1 lurnover number, TN. The turnover number is also called the turnover frequenc} (a) (b) / H H 00000 00000 www vwww -HoO H H 00 00 00000 I luure 12.29 Catalytic cycles for the production of water via (a) disproportion of OH groups, and (b) Urn reaction OH,ad) + H,ad> -» H20. B A 77 ~%777?l~V77777 w YZZZZ2 I A Y////1 B-A V////I (a) Langmuir-Hinshelwood A "A - V777žl~ V///// B-A A (b) Rideal-Eley \ B-A (c) Precursor "A Figure 12.30 Schematic of (a) Langmuir-Hinshelwood, (b) Rideal-Eley, (c) precursor mechanism lot UN inaction A + B => AB and AB => A + B. MNI III 'Kil CAIAI Ylli: Ml AC.IIIINII /r, In Chapter 1 we dcliucd 11 it- lumovci miinhi'i us ,., Ra 1 v = - Ns I 1 I -in where RA is the rate of reaction per unit area and Ns is the number of melal atoms pci unit area. Physically, the turnover number is the rate that the catalytic cycle occurs on each metal atom, measured in number of catalytic cycles per second. Figure 12.31 shows some typical turnover numbers for catalytic reactions. The fastest reactions occur at rates of 100/second, while the slowest occur at I0_4/second. There are I few examples of enzyme-catalyzed reactions that occur at 106/second. In my experience, in industrial reactions, turnover numbers are most commonly between 0.1 and 10/sccond. Faster reactions are usually mass-transfer-limited. Slower reactions are usually indicative of a catalyst needing improvement. 12.17.1 Langmuir Rate Laws As mentioned at the beginning of this section, next we want to discuss the kinetics of catalytic reactions. We briefly reviewed the kinetics of catalytic reactions in Section 2.7. Recall that we found that catalytic reactions show complex kinetics. Rates do not vary linearly with concentration. Arrhenius plots are curved. For example, Figures 12.32 and 12.33 show data for the reaction CO+ 5O2 => CO2. Notice that as we increase the CO partial pressure, the rate goes up, reaches a maximum, and then declines. In the remainder of this section we will derive an equation for the rate of a catalytic reaction to try to understand why the rates show such complex behavior. The derivation will be based on Langmuir's model of adsorption. Langmuir proposed that gases bind to fixed sites on a catalyst's surface. During reaction, reactants bind to the surface sites as illustrated in Figure 12.34. Some sites will be empty, and some sites will be covered by reactants A or B. Langmuir proposed that adsorption will occur only when a gas-phase A molecule reacts with an empty site. Consequently, the rate of adsorption will be proportional to S, the number of empty sites on the catalyst surface. 10° 10' i—i—i—i—i—i—i—i—i—i—i—i—r Dehydrogenation Hydrogenation - Olefin somerization Alkane hydrogenolysis -1 | cyclization J_I_I_I_I_I_I 200 400 600 800 1000 1200 1400 Reaction temperature, K Figure 12.31 Turnover numbers for some typical processes. I0 II) 108 io-7 to-" 10 CO pressure, torr the rate of CO oxidation on Rh(111). [Data of Schwartz 02 pressure, torr , ,„,„,, 12.32 The influence of the CO pressure on ,.l nl (l'IH(i).l 1E+13 E 0 1 1E+12 u QJ O E B 1E+11 co CC I........... -......i ' i ' I _. „ --------------................ \ c ■ S. B :1í _J§Í A i . i ■ i ■ - i . i , i ._i—i—i—i—: ■ . I I l 400 600 800 400 600 800 Temperature, K Temperature, K ......... „. ,md _ o y m-' torr Pn =4 x 10'7 torr; ePco = 2x10 torr, rc, ľ,: " ,rr2ľi 10° e torn Z =25 x 10- torr. [Data of Schwartz et al. (1986)., x 10 " tort, Po, - 2.5 x x io " torn I Innre „ mi face of the catalyst. The black dots are I he white ovals are the adsorbed A molecules. The dark ovals are adsorbed B molecules. ( diiMiiii the rati lion A-> C (12.121] and assume lhal ihc reaction occurs via ihc following mechanism: S +A AAd =F (12.122) ^ C + S In reaction (12.121), A is a gas-phase molecule, S is an empty site, Aad is an adsorbed A molecule, Cad is an adsorbed C molecule, and C is a gas-phase C molecule. We will assume that the reaction is being run in the presence of a species B, which adsorbs on the catalyst: S + B±=U:Bad (12.123) but does not participate in the reaction. During reaction (12.122), A reacts with a bare site on the catalyst, S, to yield an adsorbed complex. Then the adsorbed complex rearranges to form a new species, C, and then the C desorbs, regenerating a bare site. For purposes of derivation, it is useful to define So = the total concentration of sites available to adsorb gas (i.e., the total number of black dots in Figure 12.33), N/cm2 [S] = the concentration of empty sites, N/cm2 (i.e., the number of black dots that do not contain A or B) [Aad] = the concentration of adsorbed A molecules, N/cm2 [Bad] = the concentration of adsorbed B molecules, N/cm2 [Cad] = the concentration of adsorbed C molecules, N/cm2 PA = the partial pressure of A PB = the partial pressure of B Pc = the partial pressure of C Following the analysis in Section 4.3, rB, the net rate of formation of B, can be measured anywhere along the reaction path. If we consider reactions 3 and 4, we find that rc is given by: rc = k3[Aad]-k4lCadl (12.124) where ki and k4 are the rate constants for reactions 3 and 4, and [Aad] and [Cad] are the concentrations of adsorbed A and C. [Aad] and [Cad] have units of molecules/cm2 (or mol/cm2) for a solid catalyst, molecules/cluster for a metal cluster catalyst, or molecules/enzyme for an enzyme catalyst. (12.125) 11 ' i '''i KINI III ■.....A I Al......I ACIIONfl 747 • hu i .in i akuliili- |A,„i| mul |< ,„i| limit Ihr šinuly sliile approximation (I i,Vil k,l\|S| k.|A,„,| k,|A,„,| I K ,|t „., I 0=r(„, =k„l\|S| k,|C,„,l k,|('lu,| I k,|A,„,| I» li ' \,i and r(■ are the rales of formation of adsorbed A and B, I'a and P< lit iln i.....il pressures of A and C over the reactive surface, and |S| is the conceniiaiion nl . in|iis surface sites. In ihc literature, il is common to assume thai reactions 3 and t arc rale detenu.....ig i inI. i Mich circumstances, k, will be much smaller than k: and k,,, while k, will be .....I h iii.illii lhán ks and k|. Therefore the last two terms in equations (12.125) and (I.' I 16) Will he negligible. Under such circumstances 'k, |Aadl [Cad] = k«, PaISI PcIS] Similarly for B ,, .„anging equations (12.127)-( 12.129) yields [Bad]=^PBlSl [Aad! - Pa[S] " [Badl . PB[S] ' [Cadi PJS1 k2 ks k, k„ k, (12.127) (I2.I2K) (12.129) (12.130) (12.131) (12.1 i quations ( reactions: 12.130) and (12.131) imply that there is an equilibrium in the following A + S Aad 2 B + S ^ c + s^= Bad - Cad (12.133) [S] : neeas an ca^—------u - - number of sites to noic me by assuming that on any catalyst, there are a hn.te n ^ „ S0 = [S] + [Aad] + [Bad] + [Cad] (12.134) l'INI III '. I l| ( AIAI I IK III Al III HI'. M1> Substituting equations (12.127) (12,1 29) yields IM 1.....equation (12 134) ami ihm solving loi IS] S0 1 + ^Pa+S',, I S k2 k7 ks (12 1351 Substituting equation (12.135) into equations (12.127) and (12.128) yields [Aad] = - b1 K2 K7 k5 tCad] = PcSo (12.136) (12.137) K-A = — k2 According to the analysis in Section 4.3, the equilibrium constant for the adsorption of A is given bv (12.138) Similarly, the equilibrium constants for the adsorption of B, and C are given by (12.139) (12.140) KB = Kc = k7 k(, k5 Substituting equations (12.138)-(12.140) into equations (12.136) and (12.137) yields [A*,] = [Cad] = KAPAS0 1 + KAPA + KBPB + KCPC _ KcPcSp 1 + KAPA + KBPB + KCP^ Substituting equations (12.141) and (12.142) into equation (12.124) yields k3KAPAS0 - lqKcPcSo 1 + KAPA + KBPB + KCPC (12.141) (12.142) (12.143) exprLlnSe \ eqUat,0n (12143) iS Ca,led the Langmuir-Hinshelwood uZT Hin h °h rCtl0n A C' SinCC WC aSSUmed that the ™" obeyed a Ksssrs sm and the equation was first derived for su*ce —n by Langmuir (1915). The same equation is called the Michaelis-Menten equation in the enzyme hterature since it was also derived by Michaelis and Menten (1913) I,' I/',' I .1III|IIIIIII I liir.li. Iwnml IIoimi.iii W. ll •.(.II I t.ll< ■ I . I w. N, it, I wanl to talk aboul u trie) In ihi homework set, wc ask the readci i" derive i 11 I .111]'1111111 Hlnshclwood Michaells Menten rate equations One could follow the id i iv .iiHHi iii Section 12.17 l lowevei. Hougan and Watson (1943) found a trick that makes i hi in do the derivation, In ibis section, we will discuss the trick. I ris go back to the example in Section 12.17. Notice that ki is much smallei than •lull- k.| is much smaller Mian k-, in equations (12.125) and (12.126). Consequently, iii, i.i.i two terms in equations (12.125) and (12.126) are negligible. Consider equation I'm II I he last two terms in equation (12.125) arc negligible, Ihen k,Pa [S] = k2 [Aad] ( I ' I I 11 ■in r that the left side of equation (12.144) is the rate of reaction I, while the right nil- ill equation (12.144) is the rate of reaction 2. Therefore, the implication ol equation i I ' 144) is that when steps 3 and 4 are rate-determining, the rate of reaction I will almost • qua I the rate of reaction 2. A similar analysis shows that when reactions 3 ami I art i lie determining, the rate of reaction 5 must almost equal the rate of reaction 6. Note, however, that this is only an approximation. The rate of reaction 5 cannol be ■ uictly equal to the rate of reaction 6, because then there would be no nel production "I products. Similarly, if the rate of reaction 1 were exactly equal to the rate ol n in iion 2, there would be no net consumption of reactants. However, the implication nl the preceeding derivation is that one can calculate an accurate value of the surface i oncentration of each of the species attached to the catalyst by assuming that the rate "I reaction 1 approximately equals the rate of reaction 2 and the rate of reaction 5 approximately equals the rate of reaction 6, even though in reality, the rales are mil exactly equal. One can extend these ideas to other situations. Consider, for the moment, how the itrguments would change if reactions 5 and 6 were rate-determining in reaction (12.122). ll 5 and 6 were rate-determining, then a steady-state approximation on Cad would give 0 = rBad = k3 [Aad] - k4 [Cad] - k5 [Cad] As before, the last term in equation (12.145) is negligible: k3 [Aad] = k4 [Cad] (12.145) (12.146) I he left side of equation (12.146) is the rate of reaction 3, while the right side is the rate ol reaction 4. Therefore, the implication of equation (12.146) is that one can calculate ilie concentrations of adsorbed A and C by assuming that the rate of reaction 3 is equal in the rate of reaction 4. Now consider reactions 1 and 2. The steady-state approximation lor Aad implies 0 = rAad = k,PA [S] - k2 [Aad] + k4 [Cad] - k3 [Aad] (12.147) Note, however, that according to equation (12.146), the last two terms in equation (12.147) are approximately equal when reactions 5 and 6 are rate-determining. Therefore, when reactions 5 and 6 are rate-determining, one obtains k,PA [S] S k2 [Aad] (12.148) r ww 111 I MWI M M , I It IIM II M ,A I Al V'UM MNI IM MM I AIAIY1.....A.....N'l 781 Hie iiii side ol equation 112 148) li the rate <>i re* tion I and iiu righl Hide In the late ol reaction .'. Therefore, the implication <>i equation 11.' 148) is dial when reaction! j and 6 arc rate determining, the rate of reaction I is approximately equal io the rate a reaction 2. We note again that the rates are not exactly equal. However, the ke> resufi is that one can calculate accurate surface concentrations by assuming dial the rates an almost equal even though the rates are not exactly equal. One can generalize these results to say that when there is a simple rate-determinini step in a reaction, one can calculate an accurate rate equation by assuming that . All the steps before the rate-determining step are in equilibrium with the reactants, Consequently, the concentration of all of the species before the rate-determining step can be calculated via an equilibrium expression with the reactants. . All the steps after the rate-determining step are in equilibrium with the products. Consequently, the concentration of all of the species after the rate-determining step can be calculated via an equilibrium expression with the products. . Sites are conserved, so one can calculate the concentration of bare sites via a site balance. For example, if we have the reaction A + B C following the mechanisms A + S i=? Aad B2 + 2S ±=? 2Bad Aad + Bad -> C then the rate of production of C is given by rc = ki51[Aad][Bad] (12.149) (12.150) (12.151) (12.152) If step (12.151) is rate-determining, then it is okay to assume that steps (12.149) and (12.150) are in equilibrium: lAadl Pa[S] [Badj2 PB[S]2 KA (12.153) (12.154) In equations (12.152), (12.153), and (12.154), [S] is concentration of empty sites, PA and PB are the partial pressures of A and B2, [AadJ and [Bad] are the concentrations of adsorbed A and B, KA and KB are the equilibrium constants for the adsorption of A and B, and k150 is the rate constant for reaction (12.151). Some of the terms in equation (12.154) are squared because, according to (12.150), when B2 adsorbs, it produces two adsorbed B atoms. Solving equations (12.153) and (12.154) for [AadJ and [Bad] and substituting that result into equation (12.152) yields Ni*xl. we need mi expiration loi |S| Wi y i il lioin n site bnliincc let's dcllnc S,, us it......,ii number of silos available in ndmirh )■■>•■ tic.. Muck dots in lij-mc I.' it) Noiicc .....tin black dots can be empt) ot Ihej can hi covered bj A oi h Therefore S„ |S| I |Aad| I |H„,| (12 136) nil. hinting |A.„|| and |Had| from equations (12.153) and (12.154) into (12.156) yields S„ = |S| + KAPA|S| + (Kl!)"S(Pi>)"S|S| (I.MV) ...... equation (12.157) for |S| yields [SI + KAPA + (Kb)05(Pb)° (12 158) lib Hinting (12.158) into (12.155) yields the Langmuir-Hinshclwood-lloiigan Watson intc equation: k|5oKA(KB)05(S0)2PA(PB)0-5 (1 + KAPA + (KB)°-5(PB)0-5)2 (12.159) rc = kmKA(KBrPA(PBf[S] (12.155) Wi work out more examples in solved Examples 12.B and 12.C. The reader should inline those examples before proceeding. 12.17.3 Qualitative Features i we want to discuss the predictions of the model. Figure 12.35 shows a plol "I Mi. rate of the A => C reaction as a function of the reactant pressure calculated from . quation (12.143) with k* = 0. Notice that the rate of reaction goes up and then levell nil furthermore, as the partial pressure of B rises, the rate goes down even though we in . onsidering the reaction A => C, so B is not participating in the reaction. Noiicc that the trends in Figure 12.35 are quite different from the trends seen with gas phase reactions. In the gas phase, the rate would increase with increasing A concentration .mil never level off. A species such as B, which is not participating in the reaction, would Imve no effect on the rate. Il is useful to compare the results in Figure 12.35 to the experimental results in I'igure 2.14. Notice that the experimental results follow the trends expected from equation (12.143). In fact, the lines in Figure 2.14 were calculated from equation (12.143). i herefore, it seems that the experiments do follow the trends in equation (12.143). Clearly, iIk- kinetics of catalytic reactions look quite different from the kinetics of gas-phase reactions even though the mechanisms of the two reactions look quite similar. The reason why the rate shows this weird behavior is that the concentration of A on the catalyst does not have a simple relationship to the partial pressure of A. Figure 12.36 is a plot of [Aad], the concentration of A on the catalyst, calculated from the equation 112.141). Notice that the concentration of A on the catalyst increases as the partial pressure nl A increases, and then levels off. As we increase the partial pressure of B, the surface concentration of A on the catalyst decreases, even though there is no direct reaction between A and B. I' II II III .'M l 7MIM I in . Ill «V .III Mil O E GC ;'ni m «■ 1.5E-8 o 1 .OE-8 5.0E-9 O.OE+0 50 10 20 30 40 Pa 1 i92U5e To and 25^t Kb ""^ A * ° Ca'CU'ated 6qUati°n With * = °' P' 1.2E+15 1.0E + 15 B=0, * 8.0E + 14H o 0) 1 6.0E+14 2 4.0E + 14 CD > O ° 2.0E+14 0.0E+0 k=VkZ iap^=SfHt^STfrom the Lan9muir adsorption isotherm'equa,ion (12-141»'with Has and A and R T ^ * ^ °f sites °" the «° hold gas and A and B are compet.ng for those sites. At low pressures, there are plenty of bare sites so A sunply adsorbs. However, at higher A pressures, the surface wUl start o mi up with gas. In that case, when one increases the partial pressure of A in th gas phase one does not put that much more A onto the surface. S P ' i iinii ii useful io (nii'.iiiii j.....iii. i 1111 it 111 My, "\. defined by I A.,, | A s ( I ' Kill) n In ie HA is the fraction ol the sites on the catalyst that are covered with A. Substituting • ii mi ii hi (12.141) into equation (12,160) shows KaI'a .....lar derivation for B shows 1 + KAPA + KbP,, KbPb 1 + KAPA + KBPB (12 161) (12.162) liquation (12.161) is called the Langmuir adsorption isotherm. It was lirsi derived b) i imgmuir (1912). figure 12.37 is a plot of 0A as a function of PA for various values of P|(. Noiu e ili.il in 0A approaches 1.0, a change in the partial pressure of A does not change the sin l.u e • urn miration of A significantly, because one cannot squeeze more A molecules onto Ihc surface. Further, when PB increases at constant PA, the surface concentration ol A decreases! Molecule B takes up some of the surface sites and thereby partially blocks the idsorption of A. Consequently, as the partial pressure of B rises, the surface concentiaiion nl A decreases. Figure 12.37 A plot of the Langmuir adsorption isotherm. Nest il is iim'IiiI lo go li.uk and consider I ho implication ol these results lor the reaction: A^=>C (I2.l(>3) According to equation (12.124), when K4 = 0 k3 [Aad] (12.164) Therefore, the rate is proportional to the concentration of A on the catalyst. Figure 12.36 shows a plot of the concentration of A on the surface. Notice that at high A pressures, [Aad] is almost independent of PA. Consequently, according to (12.163), increases in PA do not significantly enhance the rate of reaction (12.163). Similarly, the presence of B reduces [Aad]. Consequently, the rate of reaction (12.163) goes down. The result is that the presence of B decreases the rate of the reaction even though B does not directly participate in the reaction. Notice, however, that according to equation (12.162), the fractional coverage of B goes to zero when PA is large. Therefore, B will not affect the rate at high enough A partial pressures. Experimentally, one usually observes such trends, which is why equation (12.143) has prove to be so useful. 12.17.4 Alternative Isotherms There is one other detail. When you fit real adsorption data, you sometimes have to use an equation for 0 other than the Langmuir adsorption isotherm. In the literature, there are a number of adsorption isotherms, where an adsorption isotherm is defined as an equation or plot of [Aad] versus PA. Masel (1996) lists several isotherms. The most important are the Toth equation KAPATo,h 1 + KAPX CTolh and the multisite model: XjKAPA 1 + KAPA (12.165) (12.166) In equations (12.165) and (12.166) i is a sum over different sites in a sample and C,0,h, X„ K'a are parameters that are adjusted to fit data as described in Chapter 4 of Masel (1996). The Toth equation arises because of attractive or repulsive interactions between molecules. The multisite model works because there can be different sites (i.e., places where reactants can adsorb) on a catalyst with different activities. For condensable substances (e.g., H2O), people sometimes use what is called the Brunauer-Emmett-Teller (BET) equation: [Aad] = ( pT 1-S Pi l+(cB n — (12.167) where [Aad] equals the concentration of A on the catalyst, S0 is the concentration of sites on the catalyst, PA is the partial pressure of A, and PA is the pressure you would have to ,.. m to condense toward A equation 11.' lo/l wises because, Willi .1 1.....lensuhle gits. 11 ■ in (jet what is called 11 imillllnvi 1 when- .nlililioii.il laycis ol g.is condense on lop ..I tin lusl layer of gas. I Extension to More Complex Reactions •tu Im. we have discussed only the reaction A C, but it is helpful to consider a slightly ....... 1 (implicated case: A + B-> C (12 161) ......isc was lirst considered by Langmuir [19I9|. Langmuir assumed thai the rem...... followed a l.angmuir-Hinshelwood mechanism: S + A = S + B Aad + Bad Aad (D ■ Bad (2) . C + 2S (3) I I ' !<•'" 1.....-—- Harris äsä" rc = k3[Aad][Bad] d.' I /m r = kABOA0B w here HA and 0B are the fractional coverage of A and B, that is 9a_ So ' |Bad] (I.' I /1 1 (12.172) in,I kt(So)2 = kAB. Langmuir then assumed that the adsorption of A and B would follow .1 Langmuir adsorption isotherm from equations (12.161) and (12.162) to obtain kABKAKBPAPB , ,, rc = -r (12.173) (1+KAPA + KBPB)2 liquation (12.173) is the key result in this section. One should memorize it before proceeding. Figure 12.38 shows a plot of the rc/kAB as a function of PA calculated from equation (12.173) with KA = KB = 1. Notice that the rate goes up with increasing PA, reaches ,1 maximum, and then declines. The results are in qualitative agreement with the data 111 2 51 i M O.OE+0 O 10 20 30 40 50 Figure 12.38 A plot of the rate calculated from equation (12.173) with KBPB = 10. CD a CO ČĎ > O O Í Figure 12.39 The changes in 0A and 0B as a function of PA with KBPB = 10. Figure 2.17. In fact, the fit is quantitative. The lines in Figure 2.17 are actual fits to the data via equation (5.127). Just to keep this plot in perspective, note that for a gas-phase reaction, an increase in the partial pressure of the reactants always produces an increase in the reaction rate However, the data in Figure 2.17 and the plot in Figure 12.38 show that with a catalytic reaction, an increase in the partial pressure of the reactants may decrease, not increase the rate. 1IAI V III HI Al III IN' l'li\ mi .illy, whnl c Ili|'|m limp ľ, llhil in 111111111 j* in equation I I.' I • ■ i ■ when l\ llli M i.is. more A will nilsoih \ . .i n .nil. nA will incieasc llowevei, aicoidnifi In ■ I < ■ 1111 • 11 ( I.' Mi.'), whľii ľA im if, isis. il„ dri leases because llieie ale lewei vacaill siles In In ild It (see Figure I.'. Il>), Now i nnsulei (lie product ol CA and (*„ m equal ion ( I .M /1 I id ih.ii since oA goes up and "n gocN down, the product of the iwo can incream 01 i ic Willi increasing pressure Ol A. II the fractional decrease in It,, is smallei than the i. u llonal increase in 0A, the nel rale will increase. However, if the fractional dc..... m IIh exceeds the fractional increase in 0A, (he net rate will decrease Consequently, ID .....ease in the partial pressure of A will not necessarily increase die rale ol die icai i..... L. m een A and B, and it may decrease it. I.' 17.6 Further Extensions ' ine run extend these ideas to a wide variety of reactions. For example, considei the ( am where reaction (12.168) is run in the presence of a species D2 that dissocialively adsorb* hu die catalyst and blocks sites but does not otherwise participate in the reaction 1 >n< i mi wtile equilibrium constants for A and B as before: [A. Pa[S] [Bad] PB[S] (12.174) (12.175) Now let's assume that D2 dissociatively adsorbs to form two D atoms, that is, thai during die adsorption process, D follows D2 + 2S i= 2D„ (12.176) AI equilibrium: [Da, Pd2(S0)' (12.177) where [Dad] is the concentration of adsorbed D atoms, Pd2 is the partial pressure of I) , S(1 is the number of sites in the catalyst, and KD is the equilibrium constant for the adsorption of D. The site balance becomes So = [S] + [Aad] + [Bad] + [D,d] (12.178) Substituting equations (12.174), (12.175), and (12.177) into equation (12.178) and rearranging yields [S1= l+KAPA+KBSpB + (KDPD2)>/2 (l2-17,,) Substituting equation (12.179) into Equations (12.174) and (12.175), and then substiluiiii)' Ihe result into equation (12.170), yields Rc kAB K A Kb Pa Pb [I + KaPa + KbPb + ÍKdPdj)1/2)' (12.180) I mi IN I III II II II I II IN II I I A I Al , I . II ll'I'l I Ml N I ni MftllMlrtl row Equation ii.mkoi ii the rate equation foi reaction (I2.I6H) In the pretence ol.......aq species I), tli.il dissociates. Masel (1996, Chapter 7) provides many oilier examples of these methods. In thl problem set, we also ask the readers to do some similar examples, the derivations an-simple but amazingly powerful. One can lit kinetic data lor most catalytic reaciions, mosl enzyme reactions, and many reactions in electronic materials production using equation (12.180). 12.18 SUMMARY In summary, then, in this chapter, we described many different types of catalysts: Homogeneous catalysts: . Acids or bases . Metal salts • Enzymes • Radical initiators • Solvents Heterogeneous catalysts: • Supported metals . Metal oxides, carbides, and sulfides . Solid acids and bases . Immobilized enzymes and other polymer-bound species . Photocatalysts We found that all catalysts work in basically the same way: • Catalysts can be used to help initiate reactions. • Catalysts can be used to stabilize intermediates. . Catalysts can be used to hold the reactants in close proximity. i Catalysts can be used to hold the reactants in the right configuration to react. . Catalysts can be used to block side reactions. • Catalysts can be used to modify the forces between the reactants, which changes the intrinsic barriers to reactions. • Catalysts can be used to act as efficient means for energy transfer. It is also important to realize that . One needs a catalytic cycle to get a reaction to happen. • Mass transfer limitations are more important when a catalyst is present. . Most catalytic reactions also follow Langmuir-Hinshelwood rate laws. In the ncxl lew iliiiptn. «i will In ' B occurring in a spherical pellet of radius yp: i ^tanhiStfrp) p, w nli <|>p given by equation (12.117), or equivalently (for first order) yp(-r°A) (I2.IKI) (12 IK.') here yP is the diameter of the pellet (-r^) is the reaction rate (in the absence ol mass transfer), Dc is the effective diffusing into the pellet, and is the concentration ol the i. 1.1.mi outside the pellet. Equation (12.181) predicts that the actual rate of reaction is less than the rate in the absence of mass transfer resistances. The rate is reduced because the reactants are used up .is ihey diffuse into the pellet, as shown in Figure 12.40. The average concentration ni reactants in the pellet is less than the concentration in the gas phase. As a result, the rule is less than the rate one would have if the concentrations were equal to the gas phase . niiccntration. Center of pellet c B occurring in the spherical pellet shown in Figure 12.40. In this section, we derive an equation for the effectiveness factor assuming that all of the mass transfer resistance is associated with diffusion of reactants into the pellet. We will also assume that the diffusion process follows Fick's law with an effective diffusiity, I), First let us derive an expression of the concentration of A diffusing into the pellet. A mass balance on A shows 4;ry2De dy y+Ay 4jry2Dc dCs "~dy~ = 47ry2Ay(-rA) (12.184) where y is the distance from the center of the pellet, CA is the concentration of A at a distance y, Ay is the thickness of the slice, De is the effective diffusivity, and rA is the rate of the A => B reaction. The first term in equation (12.184) is the amount that diffuses into the slice. The second term in equation (12.184) is the amount that diffuses out. The last term is the amount that is used up. The negative sign in the first term expression arises because A is diffusing in the negative direction. Taking the limit as Ay goes to zero yields d2CA | 2dCA | rA dy2 y dy Dc (12.185) The boundary conditions on equation (12.185) are dCA —= 0 at y = 0 dy CA=C(> at y = yp (12.186) (12.187) Note that equation (12.186) arises because the concentration profile must be symmetric around y = 0, or else the second term in equation (12.185) will grow to infinity. In order to solve equation (12.187), it is useful to define a new variable, A, by yPcA (12.188) where CA is the concentration of A in the fluid and yp is the radius of the pellet. One can show CAyp d2,:A d2CA 2 dCA ._ajp—a_ a+__a. (12.189) y dy2 dy2 ' y dy Equation (12.185) becomes dy (12.190) I'm simplicity, we will immune rA - kA('A * I. i|t,.....g equations (12 ihm and (12 191) Into equation (12 190) yields d2 A kA dy2 I), . * a = 0 11 ' Di I il ' I').'I I In lolution of equation (12.192) is (12 193) I ,,inciting equation (12.193) to standard form yields yP sinh (3p) w here p is called the Thiele modulus defined by p (12 194) (12 l'>< I Next, we want to compute how much product is consumed by the pellet. There ait I wo ways to do that. One could integrate the rate over the volume of the pellet. However. , ( in also calculate how much A is diffusing into the pellet and note that at steady state il,, amount of A consumed by the pellet equals the amount of A that diffuses into the pellet. The algebra is simpler if we choose the latter derivation. From the definition ol ill,- diffusivity, we have dCA (A consumed) = -47rypDc-^- (12 I"m Next, let's define rA,r, the average rate of reaction per unit volume of pellet. From a mass balance 47T 3 (A consumed) = — yPrA (12.197) i ombining equations (12.196) and (12.197) yields 3De /dCA A " yp V dy , y=yp Substituting equation (12.194) into equation (12.198) yields 1111 _ r a — 3DeC'A yp 3p) yP sinh(3p) V(yp)2/ sinh(3P). (12.198) (12.199) II II VI Ii I XAMI'I 1 !U I Factoring oul l^./y,, fro.n.he hlg bnuke........,• Id, ol „|..uil......2.199) yield. r„,r= M\C<< ,/3,,\ / | , yp \ yP / V tanlK^p VyP/VkA Substituting equation (12.202) into equation (12.201) yields C = -[^)c°*P l I tanh (3P \tanh3p 3*F 11 • »00! (12.201) (12.202) (12.203) (12.204) (12.205) (12.206) Equation (12.206) was cited in Section 12 15 fiJltT ThT* °f WC thC reader 10 d° a Simiiar deriv^«n for diffusion i Hat plate of thickness 2 L. The result is into a Tie = tanh p (12.207) where De (12.208) Similarly, one can integrate equation (12.185) for other rate forms (see also Figure 12.41). rA=kCA (12.209) I ii Ii ii im, g 0.4 I 0.2 0.0 \ V Z hid o 1 Ii '1 cfir! ,i mi er Sec >nd c Hin 2 4 6 8 Thiele parameter in I If UN 12.41 A plot of the effectiveness factor for reaction in a flat plate where tin -». it. - lull, >w. in i...... 07) and (12.208). rA = k2CA d+k2CA)n ( I ' 'III! lii .ill cases equation (12.206) is an excellent approximation whenever the factor is gn uli i iii in about 2. i .•:'() SOLVED EXAMPLES I' Miinple 12.A Fitting Data to Langmuir-Hinshelwood and Michaelis-Mcnlm Kiilf Forms Steger and Masel (1998) examined the etching of copper wilh hexaflu iinipentanedione (CF3COCH2COCF3). The main reaction is IKF3COCH2COCF3) + 2Cu + 02 ==> 2Cu(CF3COCHCOCF3)2 + 2H20 (12.A.1) fable 12.A.1 shows some data for the rate of copper etching via the reaction. Fit the data in the the Langmuir-Hinshelwood expression: Re = l+K2Po-25 I I ' A 'I where Re is the etch rate, Po2 is the partial pressure of oxygen, and k] and K2 are constant! Table 12.A.1 The rate of copper etching measured by Steger and Masel Re Po2, Re Po2- Re Po2, lim/minute ton- |jm/minute torr (im/minute lorr 0.3 0.1 0.8 1.5 1.2 7.5 0.5 0.3 0.9 2.0 1.3 10.0 0.6 0.5 1.0 3.0 1.4 15.0 0.8 1.0 1.1 5.0 1.5 20.0 mil VI HI KAMI'I I II Solution (iencially I H'< i| ill Uy In solve piiililcius nl this l\pc l>\ li .ui.li n null)' equation ( 12.A. 2) hi .1 linear form anil then usuir a Icasl squaics approach Id hl the dala I'licic ,uc two key transformations: Ihe Linewcaver Hurke transformation and Ihe liadie llofslec transformation as discussed in example (A. liolh iransformalions weie developed hy biochemists to lit data to lit data to (he Michaelis-Menten equation and then were adopted by catalytic chemists. The Lineweaver-Burke transformation rearranges equation (12.A.2) by taking one over both sides: 1 K2 Therefore a plot of 1/Rc versus P0"5 should be linear. + lc7 (I2.A.3) The Eadie-Hofstee transformation can be derived by multiplying equation (12.A.2) 0.5-i D 1 D V - D °2 by d +P0f): Rc + ReK^o^ p0.5 02 (12.A.4) Dividing equation (12.A.4) by P1^ yields R^ pO.5 ťo2 RĽK2 (12.A.5) Therefore the plot of Re/Pof versus Rc should be linear. One can also rearrange the Lineweaver-Burke transformation as ~ k, + k- ^ 0.3 R. Therefore the plot of Pgf/Re versus Pu^ should be linear. If one had perfect data measured to several significant figures, one would get the same result, independent of whether one plotted the data as shown in equation (12.A.3), (12.A.5), or (12.A.6). However, in actual practice the three methods give slightly different results. Let's illustrate that point using the data in Table 12.A.1. Figure 12.A. I shows a plot of the data using three methods. Notice that all the plots look linear, which suggests that all three methods work. Lineweaver-Burke Eadie-Hofstee Equation(12.A.6) I In plots me mil i uinplrirls ic\rulnip ilmupli .n one needs numbers In i onipiiiv Ihe I haw- analyzed the dala Willi Hit help nl I In spreadsheet m fable I2.A.2 . In Ihe I......Khcel, column A is Ihe piessiue, column II is Ihe rale, column (' is one over Ihe ...... ion! ol Ihe pressure, column ll is one over Ihe rale, column li is minus Ihe rale, ..li.....i I is ihe rale over Ihe squaic loot of the pressure, column (i is I he square tool ol • It. pressure, and column II is ihe square rool ol Ihe pressure divided by Ihe rale I able I 2 A3 shows the numerical values in ihe spreadsheet. In the l ineweaver-Burke method, one tits the data to equation (I2.A.3). Coin..... • i.....I.I be linear with column D. In fact, il is, and the regression output (cells Did and 111 /1 show 0 879 \ (umpanson of equations (12.A.3) and (12.A.7) yields — = 0.879 k. k, = —— = 1137 1 0.879 ■......larly, comparison of equations (12.A.3) and (12.A.7) shows ^ = 0.477 k, K2 = 0.477 x k, = 0.477 x 1.137 = 0.544 ( I.' \ K I (I2.A.9) (I2.A.I0) (12.A.II) ......e Eadie-Hofstee method, one fits the ~*^<^™T " "" be linear as column F. The regression output 14 (cells F16 and rl snow. ^ = 1.166-0.577 x Re («2A12» Pol Therefore k, = 1.166 K2 = 0.577 (12.A.13) (12.A.14) Similarly one can fit the data to equation (12.A.6). Column G should be linear with cnlumn H. The regression out put (cells H.6 and H17) shows. R 12 = 0.869 + 0.490 x Pl0'' Therefore 0 0.5 1 1.5 2 2.5 3 3.5 = 0.869 (12.A.15) (12.A.I6) Figure 12.A.1 A plot of the data in Table 12.A.1 according to equations (12.A.3), (12.A.5), and (12.A.6). Table 12.A.2 The formulas used to fit the rate data by the various methods A B C D E F G I H 01 Lineweaver-Burke Eadie-Hoffstee Equation (12.A.6) 02 pressure rate 1/p 0.5 1/rate -rate "rate/p"0.5 p".5 p"0.5/rate 03 0.1 0.3 = 1/A3"0.5 =1/B3 =-B3 =+B3/(A3"0.5) =+A3"0.5 =+G3/B3 04 0.3 0.5 =1/A4"0.5 =1/B4 =-B4 =+B4/(A4"0.5) =+A4"0.5 =+G4/B4 05 0.5 0.6 = 1 /A5"0.5 =1/B5s =-B5 =+B5/(A5"0.5) =+A5"0.5 =+G5/B5 06 1 0.8 = 1/A6"0.5 = 1/B6 = -B6 =+B6/(A6"0.5) =+A6"0.5 =+G6/B6 07 1 .5 0.8 = 1/A7"0.5 = 1 /B7 =-B7 =+B7/(A7"0.5) =+A7"0.5 =+G7 /B7 08 2 0.9 = 1/A8"0.5 = 1/B8 =-B8 =+B8/(A8"0.5) =+A8"0.5 =+G8/B8 09 3 1 = 1/A9"0.5 = 1 /B9 = -B9 =+B9/(A9"0.5) =+A9"0.5 =+G9/B9 10 5 1 .1 = 1/A10"0.5 = 1/B10 = -B10 =+B10/(A10"0.5) =+A10"0.5 =+G10/B10 11 7.5 1 .2 = 1/A11"0.5 = 1 /B11 = -B11 =+B11/(A11"0.5) =+A1 ro.5 =+G11 /B11 12 10 1 .3 = 1/A12"0.5 =1/B12 = -B12 =+B12/(A12"0.5) =+A12"0.5 =+G12/B12 13 15 1 .4 = 1/A13"0.5 = 1 /B13 = -B13 =+B13/(A13"0.5) =+A13"0.5 =+G13/B13 14 20 1 .5 = 1/A14"0.5 = 1/B14 = -B14 =+B14/(A14'0.5) =+A14"0.5 =+G14/B14 15 16 slope =SL0PE(D3:D14, C3:C14) slope =SL0PE(F3:F14, E3:E14) slope =SL0PE(H3:H"i. G3:G14) To" intercept =INTERCEPT(D3: D14,C3:C14) intercept =INTERCEPT(F3: F14,E3:E14) intercept =INTERCEPT(H3: H14,G3:G14) 18 r2 =RSQ(D3:D14, C3:C14) r2 =RSQ(F3:F14, E3:E14) r2 =RSQ(H3:H14; 33:G14) Table 12.A.3 The numerical values in the spreadsheet used to fit the rate data by the various methods 01 02 03 04 05 06 07 0.1 0.3 0.5 1 .5 7.5 10 15 20 rate 0.3 0.5 0.6 0.8 0.8 0.9 1 .1 1 .2 Lineweaver-Burke 1/p"0.5 3.162278 1.825742 1.414214 1 0.816497 0.707107 0.57735 0.447214 1 .3 1 .4 1 .5 0.365148 0.316228 0.258199 0.223607 slope intercept r2 1 / rate 3.333333 1.666667 1 .25 1 .25 1.111111 1 Eadie-Hoffstee -rate -0.3 -0.5 -0.6 -0.8 -0.8 0.909091 0.833333 0.769231 0.714286 0.666667 0.879058 0.477853 0.994711 -0.9 -1 -1 .1 -1 .2 -1 .3 -1 .4 "rate/p"0.5 G Equation (12.A.6) 0.948683 0.912871 0.848528 0.8 0.653197 0.636396 0.57735 0.491935 0.438178 0.411096 -1 .5 slope intercept r2 0.361478 0.33541 0.576817 p\5 0.316228 0.547723 0.707107 1.224745 1.414214 1.732051 2.236068 2.738613 3.162278 3.872983 4.472136 slope 1.165903 0.961547 intercept r2 2.432Í 2.981424 0.489952 0.86925- 0.992587 111 >i vl l> l "Wi l n f Similarly I 150 = 0.490 ||, ii- is ilio spivmlshcrl I '.rl H|' '•" 'I" ' 'il''"""'- or K2 = 0.490 x k, = 0.490 x 1.150 = 0.564 (I2.A.18) (I2.A.I9) Table I2.A.4 summarizes these results. Notice that the three methods give different value of the coefficients although the differences are not large. Figure 12.A.2 shows a replot of the data on a semilog scale. In this particular case there is almost no difference between the methods. In the homework set we will ask the reader to show that the Lineweaver-Burke plot is sensitive to errors. Physically, when one uses the Lineweaver-Burke method, one is fitting to 1/Rc rather than Re. 1/Re is very sensitive to noise at low concentrations. As a result, Lineweaver-Burke plots sometimes fail. If you try to derive your own method, be careful to compare errors on Re not 1/Rc, or else you will do a poor job of fitting real data. In my experience, it is better to avoid the linear methods and use nonlinear least-squares method instead. The nonlinear least-squares method can be done using the solver function in Microsoft Excel or Lotus 1-2-3. Table 12.A.4 Fits of the data to equations (12.A.3), (12.A.5), and (12.A.6) Lineweaver-Burke Eadie-Hofstee Equation (12.A.6) Slope Intercept Regression coefficient ki K2 0.879 0.477 0.9994711 1.137 0.544 0.577 1.166 0.961547 1.166 0.577 0.490 0.869 0.992587 1.150 0.564 1 10 Pressure, torr Figure 12.A.2 A plot of the data in Table 12.A.1 Eadie-Hofstee method. 100 fit by using the Lineweaver-Burke method and the I I I ]|.....iStll'l! .11 r rate calculated from < equation(12.A.2) srror"2 lbs (in i in ) 0 i 0.3 =C$19'($A6'05)/(1+C$20 "($A6"0.5)) =(C6-$B6)"2 =ABS(C6 $B6) 1 0.3 0.5 =C$19($A7"0.5)/(1+C$20 •($A7"0.6)) = (C7-$B7)"2 =ABS(C7 ii.i. 0.6 =C$19'($A8"0.5) / (1+C$20 *($A8"0.5)) = (C8-$B8)"2 =ABS(C8 $BB) 1 0.8 =C$19'($A9"0.5)/(1+C$20 "($A9"0.5)) = (C9-$B9)"2 =ABS(C9-$B9) 1 1.5 0.8 =C$19'($A10"0.5)/(1+C$20 *($A10"0.5)) = (C10-$B10)"2 =ABS(C10-$B10) 1 2 0.9 =C$ 19* ($A11" 0.5) / (1 +C$20 *($A11 "0.5)) = (C11 -$B11 )"2 =ABS(C11 $H1I) 1 =C$19"($A12"0.5)/(1+C$20 •($A12"0.5)) = (C12-$B12)"2 =ABS(C12-$B12) 3 5 1 .1 =C$19"($A13"0.5)/(1+C$20 "($A13"0.5)) = (C13-$B13)"2 =ABS(C13-$B13) 4 7.5 1 .2 =C$19*($A14"0.5)/(1+C$20 •($A14"0.5)) = (C14-$B14)"2 =ABS(C14 $B14) 15 10 1 .3 =C$19'($A15"0.5)/(1+C$20 ■($A15"0.5)) = (C15-$B15)"2 =ABS(C15-$B1b) 1 15 1 .4 =C$19'($A16"0.5/)/(1+C$20 "($A16"0.5)) = (C16-$B16)"2 =ABS(C16-$B16) 1 20 1 .5 =C$19"($A17"0.5)/(1+C$20 "($A17"0.5)) = (C17-$B17)"2 =ABS(C17-$B17) 1 a k_1 1 .15667844795306 (Calculated via solver) =SUM(D6:D17) =S0M(E6:E17) i K_2 0.570642862949246 (Calculated via solver) In the spreadsheet, cl9 = k|, c20 = K2, column C, contains the rates were calculated from equation (12.A.2). Column D contains the individual errors squared. Cell dl9 is the sum of the squares of the errors. When I set up the spreadsheet, I guessed k| = I ami K: = 0.5 and used the solver function in Excel to minimize the dl9 by varying cl9 and c20. The result is that Excel calculated that the best ki is 1.157 and the best K2 is 0.571 There is one subtlety in the calculation. Notice that I calculated the error squared in column D of the spreadsheet. If you suspect that you have bad points in the calculation, il is better to instead minimize the absolute value of the error. Column E in the spreads heel shows the absolute value of the error. Again, one can use the solver function in Ext el to minimize the d!9 i>y varying cl9 and c20 The resuii is ihm lixi el i al, ulaied that the best k| is 1.131 and the best k, is 0.554, Additional details are m Problem * A Example 12.1i Simple Eangimiir Derivation The reaction A I Ii >('obeys NH + H2 + X-► NH3 + X Derive an equation for the rate of formation of C as a function of the partial pressures oj A and B. Assume that reaction 3 is rate-determining. Solution Ml M VI IM X AMI'I I 'I iii rc = k3[AadJ[Bad] Assume that reaction 1 is in equilibrium: [Aad] SPA Similarly for reaction 2 = K. [Bad] SPR K2 Combining (12.B. 1)-(12.B.3) yields rc = K,K2k3PAPBS2 I need S to complete solution; I get it from a site balance: S„ = S + [Aad] + [Bad] Combining (12.B.2), (12.B.3), and (I2.B.5) yields S0 = S + SK1PA + K2PB Solving (12.B.6) for S, we obtain S=_S2_ 1 + K,PA + K2PB Combining equations (12.B.4) and (12.B.7) yields rc = K|K2k3PAPB(S0)2 (1 +K,PA + K2PB)2 (12.B.I) (12.B.2) (12.B.3) (12.B.4) (12.B.5) (12.B.6) (12.B.7) (12.B.8) Example 12.C Derivation of a Langmuir-Hinshelwood rate equation Tanaka (1960) proposed that the decomposition of silane, SiH4, on a silicon wafer followed the following mechanism: SiH4 + S SiH4ad (12.C.1) SiH4ad + S-> SiH3ad + Had (12.C.2) SiH3(ad) Si + S+ §H2 (12.C.3) 2H(ad) ±=7 H2 + 2S (12.C.4) lletlve a l.migmuu llinshiluiiiiil ||oil|imi Wnlsoii uile i'i|iiiilloll loi llie iniclion, I i.......| iliai reaction (I2.C2) I* rad d< i...........y. mul irieversible. Solution II reaclion ( It' .') is inie determining, then the rale ol silicon dcposilion, i.„. In luven by is, k. |Sill......„I |S| (12.C.5) I ni lowing llougan and Watson (1943), we will assume that reactions (12.('. I), l I.' < I), mil 11 ' < 'I) are in equilibrium. Sill., [SlPSiH4 [SiH3(ad,] = K Sill, Ksii [S]P^[Si] ill (ad) I -VP) — ISJPh7, i I ' ( ■ in (I2.C.7) (12.C.8) where KSjh4, Ks,h3 and Kh are the equilibrium constants for reactions (12.C. I),(I.'< t) iiiid (I2.C.4), respectively. i ombining equations (12.C.5) and (12.C.6) yields rsi = kzKsiHaPs;^ [S]2 SVc gel I SI from a site balance: S,, = [S] + [SiH4(ad)] + [SiH3(ad)] + [H(ad)] (12.C.9) (12.(10) where S0 is the total number of surface sites. Note that if you have a growing water, linn, lifter you deposit a silicon atom, the silicon atom is a site for further adsorption. Adsorbed iluon does not use up sites. Therefore, there is no [Siad] term in equation (12.C.10) Instead (Si] = S0. Substituting equations (12.C.6)-(12.C.8) into equation (12.C.10) yields So = [S] + KSiH4 [S] PsiH4 + KsiH, [S] P^2 [Si] + [Kh] [S] P£ Noting that [Sj] = S0 and solving for [S] yields So ,1/2 [S] 1 + KSiH4PsiH4 + KSjh,Ph2 So + KHP£ Substituting equation (12.C.12) into equation (12.C.9) yields k2KsiH4PsiH4So + KSiH4PsiH4 + KsíHjPh^So + KhPh22^ (12.C.I2) (12.C.13) liquation (12.C.13) is the Langmuir-Hinshelwood-Hougan-Watson rate equation for silicon deposition. ii* in mi iniii in in M 11 AIAI Y\r. Villi ""Mil II I I >1 I \ani|)lo 12.1) Disiingiiisliing itciwtTii Mechanlimi Assume thai iIn- dccompoi....... of silane instead obeys Nubstltutlng 11.' i > 3i iniii 11 11 > /1 mill rearranging ihowi SiH4 + 4S-► Si + 4H(ull) 2H(ad) -► H2 + 2S i I ' I > I i (I'l) I Reaction (12.D.2) is rate-determining. How would you tell the difference between dir. mechanism and the mechanism in Example 12.C: (a) spectroscopically or (b) on the basil of kinetic analysis. Solutions There are two different solutions: a spectroscopic solution and a kim-iu solution. (a) The spectroscopic solution is to do spectroscopy of the surface, and see whal species are present. For example, if reaction (12.C.2) is rate-determining, then there should be SiH4 on the surface; if reaction (12.D.2) is rate-determining, adsorbed hydrogen atoms should be seen. There are some uncertainties because the concentrations could be low. However, one can often find conditions where the reactants in the rate-determining step are on the surface of the water. One can then use IR (infrared) to see which species are present. If silane is present, reaction (12.C.2) is rate-determining. If hydrogen atoms are present, then (12.D.2) is rate-determining. If both are present, no reaction is rate-determining. (b) You can try to determine which mechanism is better by seeing whether equation (12.C.13) or the equivalent expression, reactions (12.D.1) and (12.D.2), work better. First let us derive an equation for the rate, assuming that reaction (12.D.2) is rate-determining. According to the analysis in Chapter 4, r12c 2, the rate of reaction (12.D.2), is given by F12.C.2 = k2C[H(ad)] If reaction 12.D.1 is in equilibrium, then [H(ad)]4 [S]4Ps,H4 K, where K12C.1 is the equilibrium constant for reaction (12.D.1). Rearranging equation (12.D.4) shows [H(ad)] = [S](Kl2.c.l)°-25(PsiH4)0'25 Substituting (12.D.5) into (12.D.3) yields rl2.c.2 = k2c[S]2(Kl2.cl)0'5P^H4 Again, we calculate [S] from the balance: S0 = [SI + [H(ad)] where So is the total number of sites on the surface. (I2.D.3) (12.D.4) (12.D.5) (12.D.6) (12.D.7) |S| I I (Km , >" ^(I's.ii., )" 1 I ' I I K I ......1.......equations (I2.l).(>) and 11.' D.8) yields k, Sg(K12,c,1)1/2(Ps1H4)0-5 CI2.C.2 = (I + (K12.C.I )"'25(PsiH4)°'25)2 (12.1).1)) low the question is how to do experiments to distinguish between equations (12.C.13) mill (I 2.D.9). The easier way to find out is to put both rale equations in a spreadsheet and i| for differences. I in the purposes here, it is useful to consider a case where PH, is negligible In thi ill nominator of equation (12.C.13). In that case (I2.C.13) becomes rSi = k2KSiH4PsiH4So (1 +KsiH4 +PSiH4)2 (I2.D.I0) fable 12.D.1 shows a spreadsheet that I used to illustrate the difference between liquations (I2.C.14) and (12.D.9). In the spreadsheet I set column A to the pressure, I olumn B to the rate calculated from equation (I2.C.14), and column C to be the rate I111111 equation (12.D.9). I picked S0 = 1, k2 = 1 KsiH4 = 1. I then calculated an em..... ■ iilniiiii D, and used the solver function of the spreadsheet to find values of k2<- and K, ■, , in minimize cell D$4, the total difference between the values of the rate was calculated from equations (12.D.9) and (12.C.14). Figure 12.D.1 shows a plot of the data. Notice that the two curves look similar ai low pressure, but there are significant differences at high pressure. Equation (12.D.9) risei over the range of pressure shown. One can show that the rate eventually saturates at high pressures. In contrast, equation (12.C.14) reaches a maximum and then declines. Figure 12.D.1 tells us how to run experiments to distinguish between the mechanisms ( me runs the reaction over a wide range of pressure, and looks to see how the rate varies II the rate reaches a maximum and declines, then equation (12.C.14) will be better. II the 0.01 0.1 1 Pressure Figure 12.D.1 A plot of the rate calculated via equations (12.C.14) and (12.D.9). I.llll.' Ii*I) 1 The formulas In the soreadahMt-----■ _______. qinwi onod to crente Flguro 12.D.1 SO KSiH4*$A9 /(1+KSiH4'$A9r2 = k2s0" KSiH4*$A10 /(1+KSiH4'$A10r2 k2sOC*(ABS(K12.c.1)*$A8r0 5 /(l + tABSfK^.c.lj-SASro^S)^ k2s0C (ABS(K12.c.1)'$A9)"0 5 /(1 + (ABS(K12.c.1)-$A9)'0,25r2 =k2sOC* (ABS (K12.c. 1) *$A10)"0 = /(1 + (ABS(K12.c.1)-$A10)-0.25r2 =ABS(B9-C9)"2 =ABS(B10-C10)"2 11 10"A6 12 =k2sO*KSiH4"$A11 /(1+KSiH4*$A11 )"2 = k2s0 KSiH4*$A12 /(1+KSiH4'$A12r2 =k2s0 KSiH4*$A13 / (1+KSiH4'$A13r2 k2sOC'(ABS(K12.c.1)-$A11)-0 5 / (1 + (ABS (K12.C.1 ) -$A11 ro.25 V? =k2sOC (ABS(K12.c.1)*$A12)"0 5 /J1 + (ABS(K12.c.1 ) *$A12 )"0.25 =ABS(B11-C11)"2 =ABS(B12-C12)"2 14 15 16 17 18 :k2sO"KSiH4*$A14 /(1+KSiH4*$A14)_2 k2 = 10 A11 ^10*A12 k2sOC'(ABS(K12.c,1)-$A13)-0 5 / (1 + (ABS(K12.C. 1 ) -$A13ro.PS VP =ABS(B13-C13)"2 *2s0(T (ABS(K12.c.1 )*$A14)"0 5 / (1 + (ABS (K12.c. 1 ) -$A14 T0.25 )"2 :ABS(B14-C14)"2 SO KSiH4* $A15 /(1+KSiH4'$A15)"2 K2S0 KS1H4'$A16| =k2s0C' (ABS(K12.c. 1) 'SAlffot /(1+KSiH4-$A16)"2 / (1 + (ABS (K12.c. 1) '$A16)"0.25 )"2 k2sOC (ABS(K12.c.1)*$A15r0 5 /(1 + (ABS(K12.c.ir$A15 fO.25 )~2 ABS(B15-C15)"2 19 = 10"A13 :10*A14 = k2s0"KSiH4*$A17 /(1+KSiH4'$A17)'2 = k2s0 KSiH4"$A18 /(1+KS1H4'$A18)"2 =k2s0 KSiH4"$A19 =ABS(B16-C16)"2 =k2s0C* (ABS(K12.c.1)$A17yY75 / (1 + (ABS (K12.C. 1) *$A17)"0 25 )"2 =k2sOC (ABS(K12.C.1) '$A18)"0 5 /(H-(ABS(K12.c.1)-$A18)-Q 25V2 =ABS(B17-C17)"2 ABS(B18-C18)"2 20 21 10*A15 = 10*A16 /(1+KSiH4'$A19)'2 =k2sOKSiH4*$A20 /(1+KSiH4,$A20)'2 =K2S0 KSiH4"$A21 k2s0C' (ABS(K12 / (1 + (ABS(K12.c.1 =k2sOC (<\BS(K12. / (1 + (ABS(K12.c.1 ) c.1 )'$A19)"0.5 *$A19)-0.25)"2 ABS(B19-C19)"2 c.1 )*$A20)"0.5 '$A20)"Q.25r2 ABS(B20-C20)"2 /(1+KSiH4*$A21)"2 k2s0C (ABS(K12 /(1 + (ABS(K12.c.1) c.1 ) $A21 )"0.5 "$A21 )"0.25)"2 ABS(B21 -C21 )"2 rate goes up and saturates, then equation (I2.D.9) will be better. If the rate merely rises then one cannot stmply distinguish between the rate equations Y ' On occasion one can still use the F test in Fyamr.1^ t r ,n a-—s. However, wtthout a ^^^^^Zi STl^E I't f°nStrUCti"8 Sachtler-Fahrenfort and Tanaka-Tamaru P,otS c^tlJoTTt^ r3te °f ethyle"e W™^™ on a number metals at 0 C. Next, let s construct a Sachtler-Fahrenfort plot of the data. i it 'itn nr. i i I I t I II It I II It III ■■! Ml I' | 1,1,1.' 1/1 I I III' ',|III>,I.I',I|IM,| till I .....I|l|ll|.' I I I A B C II E F 0 Mill .i 1 rule \n, in Oxide Oxytliins III n . 1 111 • Metal atoms III ox 1 tic AHf per tun I ii of oxide All, pill' llllllll ol ill.' 1 ,l 1 IM I'll 1.OE+16 -9.7 1 ~~i $Cb/l)b $05/1 !> 3.0E+16 20.4 1 1 =$C6/D6 $C(i/l (. $C7/I 7 I r 3.OE+16 -40.1 2 1 =$C7/D7 Hh 3.0E+17 -21 .7 1 1 =$C8/D8 $C8/I ll Hit 9.0E+18 -52.5 2 I =$C9/D9 $C9/I 'l Cll 8.3E+10 -39.8 1 2 $C10/I)I() $C 1 1 / I II $012/1 i;1 (.11 8.3E+10 -37.1 1 1 Co 3.OE+16 b7.2 1 1 =$C12/D12 N 1 1 .0E+14 -58.4 1 1 =$C13/D13 $ci:i/i ri Fe 1 .OE+16 -197.5 3 2 =$C14/D14 $014/1 M 1 Fe 1.OE+16 -63.7 1 1 w 3.0E+13 -136 2 1 =$C16/D16 $Ct(>/l II. T, 3.0E+13 -499.9 5 2 =$C17/D17 =$C17/E17 Solutions There are two types of plots: Sachtler-Fahrenfort plots and Tanaku Tumuru plots. Sachtler-Fahrenfort plots are graphs of the rate of reaction on a series nl mi Ml ■ ulalysts as a function of the heat of formation of the metal per mole of oxygen, whll* I unaka—Tamaru plots are graphs of the rate of reaction on a series of metal catalysis as a I unction of the heat of formation of the metal per mole of metal. If we wanted to construct a Sachtler-Fahrenfort plot of the data, we would calculate Ihe heat of formation of the oxide per mole of oxygen. A spreadsheet to do the calculations Is given in Table 12.E.1. Column A lists the metal; column B lists the rale. Column C in ihe heat of formation of the oxide per mole from Table 12.11, while column I) is ilic number of oxygens in the corresponding oxide. The heat of formation per mole of oxygen is given in column F. Similarly, column E gives the number of metal atoms in the oxide, while column G gives the heat of formation of the oxide per mole of metal. Ihe one subtlety in the analysis comes when there are two oxides of the same matei lal For example, copper is available as both Cu20 and CuO. The heat of formation ol CthO per mole of oxygen is greater that the heat of formation of CuO per mole ol oxygen. Consequently, one uses the heat of formation of Cu20 when constructing I Sachtler-Fahrenfort plot. In contrast, the heat of formation of CuO per mole of COppei is larger than the heat of formation of Cu20 per mole of copper. Therefore, one uses ihe heat of formation of CuO when constructing the Tamaru-Tanaka plot. The Sachtler-Fahrenfort plot is a graph of the rate data (i.e., column A) versus column T in the spreadsheet. The Tanaka-Tamaru plot is a graph of the rate data (i.e., column A I versus column G in the spreadsheet. The actual plot is given in Figure 12.14 in the chaptei 12.21 SUGGESTIONS FOR FURTHER READING Introductory books on catalysis include: M. Boudart, Kinetics of Heterogeneous Catalytic Reactions, Princeton University Press. Princeton NJ, 1984. (i. C. Bond, Heterogeneous Catalysis, Principles and Applications, Clarendon Press, Oxford, UK 1987. It (' (Jules, Calalvlu fhrmiMiv. Wiley. New Ynik, I')'».' More advanced treatments include: H. Pines, The Chemistry of Catalytic Hydrocarbon Conversions, Academic Press, San Diego, 1991. G. V. Smith and F. Notheisz, Heterogeneous Catalysis in Organic Chemistry, Academic Press, San Diego, CA, 1999. J. M. Thomas and W. J. Thomas, Principles and Practice of Heterogeneous Catalysis, Wiley, New York, 1996. J. R. Anderson and M. Boudart eds., Catalysis.Science and Technology, Elsevier, Amsterdam, 1984-1998. Mass transfer effects are well described in: R. 1. Wijngaarden and K. R. Westerterp, Industrial Catalysts, New York, VCH, 1998. Enzyme catalysis is covered in K. Faber, Biotransformations in Organic Chemistry, 4th ed, Springer, NY 2000. A. Fersht, Structure and Mechanism in Protein Science, 3rd ed, W. H. Freeman 1999. 12.22 PROBLEMS 12.1 Define the following terms: (a) Homogeneous catalyst (b) Heterogeneous catalyst (c) Volcano plot (d) Isomerization (e) Alkylation (f) Cracking (g) Aldol condensation (h) Ziegler-Natta catalyst (i) Wilkinson catalyst (j) Wacker process (k) Hydroformylation (1) Monsanto carbonylation process (m) Oxidoreductases (n) NADH peroxidase (o) Hydrolases (p) Lyases (q) Isomerase (r) Langmuir adsorption isotherm (y) Fischer-Tropsch synthesis (z) Hydrogenolysis (aa) Hydrodesulfurization (bb) Selective catalytic reduction (cc) Aminoxidation (dd) Water-gas shift (ee) Faugasite (ff) zeolite (gg) Superacid (hh) ZSM-5 (ii) Mordenite (jj) Sodalite (kk) Photocatalysts (11) Langmuir-Hinshelwood kinetics (mm) Sabatier's principle (nn) Adsorption (oo) Adsorbate is) 1 lie i iiwiiiIm im It) Michaelis Menleii Kineln •. tul Transferase* (v) l.igases (w) Menscluitkin reactions (x) Supported metals I'liiiiiiiM'i 777 (p|l) Mulct ill.II adsinplniii ( and precursor mechanisms. How would you distinguish between them experimentally 12.4 Describe in your own words why the kinetics of surface reactions are so different from the kinetics of gas-phase reactions. What makes the kinetics different? 12.5 Summarize the various types of catalysts described in the first half of this chaptei Provide an illustration of each type of catalyst and an explanation of how ihe catalyst works. 12.6 Visit the Websites of some of the large catalyst manufacturers: (a) Johnson Matthey (http://www.rnatthey.com/divisions/catalytic.hliii), Fngel hard (http://www.engelhard.com/), W. R. Grace (http:/Avww.gracedavison com/). United Catalysts Sud-chemie (http://www.unitedcatalysts.com/). Find a list of catalysts that they offer and describe what the catalysts do. Then go to the Shell and Exxon sites, search the word catalyst, and see what you find. (b) Visit the Websites of the major enzyme manufacturers: Novozymei (http://www.novozymes.dk/), Genencor (http://www.genencor.com/). Royal Gist Brochades (http://www.gist-brocades.com/). What do you find? 12.7 Describe in your own words when Sabatier's principle works. (a) When do you get a maximum as in Figure 12.12? (b) When does the rate continue to increase as in Figure 12.11? (c) How would changes in the intrinsic barriers to reaction affect a volcano plot? 12.8 Find five examples of catalytic reactions in your home or your car. Be sure to consider reactions in your oven, your washing machine, your stomach, your intestines, and so on. 12.9 Diagram the catalytic cycle for (a) Ethylene hydrogenation catalyzed by the Wilkinson catalyst. (b) Platinum-catalyzed ethylene hydrogenation (c) The copper-catalyzed reaction: Fe2+ + V4+ => Fe3+ + V3+ (d) The dehydration of ethanol (e) Hydrogen oxidation on a fuel cell (f) Free-radical polymerization of styrene (g) Photocatalysis of CO oxidation I' III 12.11 Voiii stoiiKn.il contains ,i solution ol hydroi hlorii tu nl ili.u is buttered to ,i pll ol l.o. On the basis ol the results iii this i haptei (a) What would you expect the acid to do to? (1) Proteins (2) Fats (3) Sugars and starches (4) Invading bacteria (b) Your saliva contains the enzyme lingual amylase. Look in the proton database or a biochemistry book and find out what each enzyme docs in the substances listed in items (l)-(4) (above). What is the main mode ol enzyme action? (c) Your intestines contain a-amylase, trypsin, chymotrypsin, and elastase. Look in the protein database and find out what each enzyme does to the substances in items (l)-(4). What is the main mode of enzyme action? Enzymes are widely used in detergents. The enzymes catalyze the hydrolysis of stains. (a) Look in the protein database and find enzymes that would be active for the hydrolysis of (1) Proteins (2) Fats (3) Sugars and starches (4) Cellulose (b) Detergents have a pH of about 8.5. Would any of the compounds you found be stable at a pH of 8.5? (c) Look at some of the Websites for the commercial enzyme manufacturers: Novozymes (http://www.novoz.ymes.dk/), Genencor (http://www.genencor. corn/). What do they tell you about their enzymes? (d) The following is a noninclusive list of some of the major commercial detergent enzymes: Alcalase, Esperase, Maxatase, Optimase, Durazym, Kazusase, Opticlean, Savinase, Celluzyme, Duramyl. Look up each of the enzymes and see what they do. 12.12 In Example 12.A we used a variety of methods to fit Steger and Masel's data to rate equations. (a) Set up the spreadsheets yourself and reproduce the findings (b) Assume that Steger and Masel misrecorded the data so the rate was 0.03 at a pressure of 0.1 torr. Reanalyze the data. How well does each of the methods do? (c) What do the results in this problem tell you about the sensitivity of the various methods to errors? 12.13 Ethylene hydrogenation was one of the first catalytic reactions to be studied in detail. Langmuir and Hinshelwood (1918) proposed that the reaction went via a Langmuir-Hinshclwood reaction: H2 + 2S 2Ha, (P12.13.1) .....IIS III < II \ 'II,,, ......Nl.ll d'l.' 13,2) (P12.13 ii Rideal and Eley proposed i Rldeal Eley mechanism: II, I 2S H2C=CH2 ■+ 11,,, CILClLia.i) + HiK| 21 l,„i CH3CH2(«i) C2H<, (P12 13.4) (P12.13.S) (PI 2.1 Id) Houriti and Polanyi proposed a third mechanism: 2H;, H2 + 2S H2C=CH2 + S H2C—CH2 4- Haj CHjCH^ajj + Had -► C2H6 (a) Derive a rate equation for each mechanism. =p H2C—CH2(a,D =? CHjCH^aj) (PL' W.I) (PI 2.1 t HI (P12 l 1.9) (P12.13 mi (b) What experiments would you do to distinguish between the mechanisms? For the purpose of this problem, assume that you are working in 1920 and do not know how to measure the surface concentration of species. (c) Repeat part (b) assuming that you can measure surface concentrations 12.14 The synthesis of ethylene (C2H4) is one of the largest chemical processes in Ununited States. (a) Find a reasonable mechanism for the conversion of ethane (CILCII,) into ethylene and hydrogen. (b) What other products would you expect? (c) Use the steady-state approximation to derive an expression for the rate "I the overall reaction. (d) Estimate the preexponentials and activation barriers for each of the reactions (e) How could you catalyze the reaction? Pick one homogeneous catalyst and one heterogeneous catalyst and explain how they would work. Do not pick the catalysts from part (0 or (g) (in this problem) or Problem 12.6. (f) In industrial practice, people use water and a zeolite (a solid catalyst) as catalysts. The water dissociates on the zeolite to yield OH and H. The water also reacts with carbon deposited on the pores of the zeolite. Explain how the water/zeolite mixture would catalyze the reaction. (g) In Section 12.1.1 we noted that I2 can catalyze the decomposition of ethane How would that work? (h) Find a reasonable mechanism for the decomposition of ethane in the present ( of iodine. (i) Estimate all of the preexponentials and rate constants for the reaction in part (h). IHK Hill M', 7H1 12.15 (J) Use 11n* siľiuly slali" approximation to dcrlvt un oi|inilit>iI loi the uile of rea< tion foi the mechanism in pari (h) (k) Show that the rate is highei than in the case in part (a). Repeat Problem 12.14. pails (g) (k) usiiiü acclaklehyile as a catalyst. Could ethanol also be used as a catalyst. (Hint: What products are formed when clhanol decomposes?) 12.16 What type of catalyst would you use for the following reaction: 12.17 (a) Hydrogenation of corn oil to form margarine (b) Oxidation of carbon monoxide in your home furnace (c) Hydrogenation of furfurylamine (d) Hydrolysis of protein stains on clothes (e) Production of mclhylamine from methanol and ammonia (f) Transformation of 2,4-pentanedione into 3-methyl-2-cyclopentanone (g) Partial oxidation of Cj-Qj olefins (h) Softening of a pair of jeans by hydrolyzing the celluose. Jennings (1991) summarized the available data for ammonia synthesis on a number of catalysts. Table P12.17 summarizes the data. (a) How well do these data follow a Sachtler-Fahrenfort plot? (b) On the basis of your findings, what other materials would be good catalysis for ammonia synthesis? (c) In reality, lead and other substances you would guess from Table 6.5 are not good catalysts for the reaction. What does that tell you? 12.18 Sinfelt and co-workers did a considerable amount of data on ethane hydrogenol-ysis. Table PI2.18 gives a selection of their results. (a) How well do these data follow a Sachtler-Fahrenfort plot and Tanaka-Tamaru plot? (b) Are there any points way off of the curve? What do you make of that? (c) On the basis of your plots and the data in Table 12.11, what other substances would you expect to be active for this reaction? Table P12.17 The relative rate of ammonia synthesis over a number of catalysts Metal Relative Rate Heat of Adsorption of Nitrogen, kcal/mol of nitrogen atoms Ru 1.0 -30 Fe 0.01 -42 Rh 0.005 -27 Re 1 x 10~5 -99 Pi 1 x 10~7 -71 Note: Heat of adsorption of N2 = 3* energy in Table 6.5. Source: Data from Jennings (1991). l.ililnlMSMII Die. intn of nlll...... liviliiiuminlVhl« nl f.ľ.'l K iiiul 10 nim ovni mi miiillini »I ..ii|i|il x 10'' (» x 10" 2 x 1013 '.nil .i.ini e Rh IM Os Ii PI Miili-c iili-s/li 111 ....."Ii 2 x ur 2 x 10" 2 x It)1" 5 x 10" 2 x 10'' ....... Data "I I II. Sinfelt, Adv Catalysis, 23, 91 (1973a,b), J. Catalysis, 29, 308 (1973) (d) Experimentally, none of these other metals actually shows significant catalytic activity. What do you conclude from that? i ' r> The reaction 2CH3CH2OH =» CH3CH2CH=CH2 + 2H20 is being run in an U Id catalyst. (a) Find a feasible mechanism for the reaction. (b) What side products do you expect? (c) If you run the reaction in a zeolite, could you eliminate some of the lide products? (d) What zeolite would you choose? 12.20 Imbert, Gnep, and Guisset, (1997) J. Catalysis. 172, 307 examined the isomei ization of creosol, CH3(C6H4)OH. (a) Find a feasible mechanism for the reaction (b) What side products do you expect? (c) If you run the reaction in a zeolite, could you eliminate some of the lldt products? (d) What zeolite would you choose? (e) Imbert et al. actually ran the reaction in ZSM-5. They found thai initially they got mainly para-creosol. However, after they ran the reaction for a long time so that there was plenty of time for all species to diffuse in and out of the catalyst, meřa-creosol was the major product. How can you accounl tin these observations? 12.21 Ilao, Yamamoto, Segawa (1996). Journal of Catalysis. 161, 20 examined the reaction between methanol and ammonia to yield methylamine. (a) What general class of reaction would the addition of a methyl group to ammonia fall under? (b) What type of catalyst would you use? (c) What side products would you expect? (d) What would you do to avoid the formation of dimethyl and trimethyl amine'1 (e) Look at the lists of catalysts in this chapter. Which one is properly sized to avoid the formation of dimethyl and trimethyl amine? (f) How do your predictions in a), b), c) compare to Ilao, Yamamoto, Scgau a findings? mi mii i mm 12.22 lluupl'cur, Olson and Sthniidl (1994) Journal of thr Ein inn linniial Society 141. 1043 examined I lit- kinetics ol Si<>, deposition limn loll nolhy loithoiiliCkt*. (( Ml II ()),Si also called I I ON 1 he in......r.u.......s (CH,CH20)4Si SiQ2 + 4CH2CH2 + 2H20 (a) Derive a Langmuir-Hinshelwood rale equation for the reaction, assuming that the mechanism of the reaction is TEOS ^ TEOS(aj) —> products. (b) Haupfear, et al. found that the rate of decomposition is reduced when ethylene is fed into the reactor even though the reaction is irreversible. Show that your rate equation predicts that the ethylene will slow down the reaction, (c) Would the ethylene affect the activation barrier for the reaction. In other words, would a plot of log rate versus 1/T change? Try to be quantitative. (d) What effect would you expect acetylene to have on the reaction rate? How about ammonia? 12.23 In Table 2.9 we said that the hydrogenation of ethylene over a nickel catalyst obeys klS(|K2K,PH2Pc2H4 ,pl2 „ , (1 + K2PH2+K3Pc2h4)2 Tc,h- The objective of this problem is to work out the behavior of equation (P12.23.1). Assume k| = 10'Vsecond x exp(-l8 kcal/mol)/kBT, K2 = 10"/atm x exp(16 kcal/mol)kBT, K3 = ICTIH/atm x exp(14 kcal/mol)kBT), S„ = 1015 molecules/cm2 (a) Plot the rate as a function of the C2H4 pressure at 300 K. Assume an H2 pressure of I atm and consider C2H4 pressures from 0.5 to 10 atm. (b) Why does the rate decline at high C2H4 pressures? (c) Plot the rate as a function of temperature for temperatures between 200 and 500 K. Assume a C2H4 pressure of 1 atm and a H2 pressure of 1 atm. (d) Why does the rate decline at high pressures? 12.24 The decomposition of urea in the presence of an enzyme called urease follows the following rate law: kiK2[urea] Turca = - j - K2[urea] Table P12.24 shows rate data for the reaction. (a) Fit the data to the rate law using a Lineweaver-Burke plot, the Eadie-Hofstee plot, and a nonlinear fit to the data. (Hint: Look back to Examples 3.A and 12.A.) (b) Why do the different methods give you such different numbers? 12.25 Vannice and Poondi (1997). /. Catalysis. 169, 166 examined benzyl alcohol hydrogenation over on a number of supported platinum catalysts. They found that the main reaction was hydrogenation of the benzyl alcohol to toluene. Initio I'll' II.it' il.il-i In' I'Milllnni '-' I lieu ('< >ik i-nl i ill it l teiiclioii iissui.....r ih.u ir.nn.hi (I'l. .'<> I) is rale determining. (b) Derive equations tor the rale ol reaction assuming tti.it reaction ll'12.26.2)| is rate-determining. (c) Derive equations for the rate of reaction assuming thai reaction (PI2.26..1) is rate-determining. (d) Table P12.26 shows some rate data extrapolated from Saucr and Olis' results. Use an F test as described in Example 3.B to determine which rate equation fits best. 12.27 In Section 12.13 we presented the mechanism of hydrogen oxidation in a fuel cell. However, that mechanism works with only one kind of fuel cell: a fuel sell with a polymer electrolyte. The electrolyte is active at 60°C. H2 02 + 4H+ + 4e" > 2H+ + 2e~ => 2H20 However, under other conditions, alternative mechanisms occur: With an alkaline electrolyte at 120°C H2 + 20H 02 + 2H20 + 4e~ 2H20 + 2e" => 40H- With a molten carbonate at 600°C H2 + 2C032" 02 + 2C02 + 4e" With a molten salt at 900°C H2 + O2-02 + 4e~ H20 + C02 + 2e" h2 2CO,2" H20 + 2e~ 202- Table P12.26 Rate data for Problem 12.26 Acetone Concentration, mg/m3 Rate, mg/(cm3 ■minute) Acetone Concentration, mg/m3 Rate, mg/(cm3 ■minute) Acetone Concentration, mg/m1 Rate, mg/(cm3 minute) 0 0 25 0.08 150 0.25 3 0.01 30 0.10 170 0.26 4 0.02 4(1 0.12 200 0.27 5 0.02 50 0.14 250 0.29 7 0.03 70 0.17 300 0.3 III 0.04 100 0.21 350 0.032 15 0.06 130 0.23 400 0.32 (u) (iné.', .i I. . ■.. 11,1.....i Illinium loi cm Ii ol die relictions on n platinum i alnlvsl tllmi The ii-aitions above aie not elementary reactions, Find elementary reactions that on in on platinum llieic should be a dilletenl mechanism on the anode and cathode See I iguic YIO lot ideas.) (Ii) Diagram the catalytic cycle lot each reaction. He sure lo constdet the production ami destruction of the catalytic site, {iiuu in each caae then is one catalytic cycle on the anode and one on (he cathode I (c) Mow is electricity generated? (d) I .ook up Hie half cell potential loi each of these reactions. How mueli VOlttgC will you produce? 12.28 The production of indendiol is one of the key steps in the manufacture ol m AIDS drug called Trixovan (sec Figure PI2.28). The objective of tins projei i ll for you to propose ways to make indendiol. (a) One scheme to make the product is to oxidize indenc with hydrogen peroxide. Search the protein database to find an enzyme that would . atalyn the simultaneous addition of two OH groups. (b) An alternative scheme is to add iodine across the double bond .......lent to form a indene diiodate and then hydrolyze the product (replace lodlni with water). What kind of catalyst would you use to catalyze die add....... of iodine? (c) What type of catalyst would you use to catalyze the hydrolysis reaction? (Hint: What type of catalyst works for the reverse reaction, i.e., dehydration ') (d) Repeat (b) and (c) when the first step is to phosphorylating the indene 12.29 The reaction CH3CH2OH + NAD+ => CH3CHO + NADH + H 1 is critical to the destruction of alcohol in the human body. (a) Look up NAD and NADH in a biochemistry book. What are they? (b) Look in the protein database and find an enzyme that would cataly/e die reaction in your liver. (c) Assume that the reaction follows the mechanism: NAD+ + S NAD+d) (PI2.29.I) (P12.29.2) NAD+d) + CH3CH2OH [CH3CH2OH:NAD](<,d| [CH3CH2OH:NAD]+d)-► NADH + [CH3CH20]+d) (PI2.."M i [CH3CH20]+d) ±=? CH3CHO + H+ + S (PI2.29.4) Indene Indendiol Inden-di-iodate Figure P12.28 Structures of some of the intermediates used in the production of Trixovan. where 11 11 i I I 1 'I I N AI is a NAD. cllianol rumples mul S is a site mi the enzyme Derive an equation foi the rate ol the reaction assuming thai reaction (PI2.29.3) is rate-determining. (d) How do your kinetics differ from Langmuir lliushelwood kinetics? (e) Look up a biochemistry book, and find mil how much ol the NAD' is available in a typical human being? (f) Does the amount of NAD+ vary with the weight of the individual? (g) Use your findings to suggest a possible hangover cure. (h) Would a vitamin work? 12.30 The US Chemical Industry Statistical Handbook for 1997 gives a chart showing the feedstocks used to produce all of the major chemicals in the United Stales Figure PI2.30 shows part of the benzene chain, specifically, the chemicals produced starting with benzene. The chart includes the major organic chemicals used in the process, but not hydrogen, oxygen. CO, or water. (a) Look at each step (i.e., arrow) and write a balanced reaction for the step, adding hydrogen oxygen, water, or CO as needed. (b) Use your knowledge from Chapter 5 to suggest a possible mechanism for each reaction. (c) Suggest a possible catalyst or series of catalysts for each step. (d) Look in the chemical processes handbook. How do your guesses for catalysts compare to the ones actually used? 12.31 In the human body, cells use the reaction Glucose + ATP > glucose-6-phosphate + ADP to trap glucose in cells. The kinetics of the reaction varies between various tissues. Your heart and brain use an enzyme called hexokinase. Your liver uses a different enzyme called glucokinase. Ethylene Benzene Ethyl benzene Styrene Dodecylbenzene Detergents Maleic anhydride -Lj A Resins Cumene Phenol Acetone Cyclohexane Cyclohexanol Cyclohexanone Adiptic acid Caprolactam Figure P12.30 A part of the benzene chain for the production of industrial chemicals. [Adapted from The U.S. Chemical Industry Statistical Handbook for 1997 and Dr D. L Burdick and Dr W. L Leffler Petrochemical Chart.] Mihi« IM/ :il Ilm nun Uvily ol hnxoMiuinn mul t|lii< oklnimii lni |ihoii|ilioiyliillnii Olu....... II loud Kule Willi Katr Willi lil.....1 Kule Willi Rail * illi 1 IlUCOMi HexoUni ■• (ihn <>kimisi-. (IlUOOM, i lexoklneMi i Um ni in i i IIIMIill/llU'1 iiiiiini/iiiiiii iiiniiil/l min inniol/litci ........11.....' imiiiil/liiiiii nos 0.3 0.02 3.0 0.97 0.7 lil 0.5 0.03 in 1 o 0.9 o ! 0.75 0.10 5.0 1 li i o 0.5 O.S 0.15 6.0 1 o i i i o 0.9 0.3 7.0 1 i) 1.2 ' o 0.95 0.5 10.0 1.0 1 1 (a) Fit the data in fable PI2.3I to Michaelis Menten kinetics. (b) How do the results above help your brain and heart to survive famine? W hal happens as the blood glucose level drops? Where does the glucose go? To your heart or to your liver? (c) Now assume that there is an excess of glucose in your bloodstream, What happens? (Note: 5 mmol/liter is a normal glucose level in humans.) 12.32 In Section 12.5 we found that N02 can catalyze the decomposition ol ethani N02 catalysis occurs at low temperature, where there arc not many radical I in the system. However, at higher temperatures, N02 can instead inhibit the decomposition of ethane. The main mechanism is for the N02 to trap the radii all in the system (i.e., N02 reacts with the radicals to form stable species). The objective of this problem is to develop a criterion to see whether N02 will enhani e or inhibit ethane decomposition. (a) Write a complete mechanism for ethane decomposition including the l.ui that N02 can initiate the mechanism as discussed in Section 12.5 and thai N02 can trap the hydrogen atoms and ethyl radicals. (b) Use the steady-state approximation to develop a rate equation for the reaction. (c) Use the results in Chapters 5, 7, and 11 to guess a preexponcntial and .111 activation barrier for each step. (d) Use your results to calculate the rate of ethane decomposition at 500, 800, and 1200 K, assuming that you have a reactor filled with 1 atm of ethane and 0.01 atm of N02. (e) How do your results compare to the case in Figure 12.9? 12.33 Blaser, Jalett, Garland, Studer, Thies, and Wirthtijani (1998). J. Catalysis. 173, 282 examined the kinetics of the hydrogenation of ethyl pyruvate (CHjCOCOOCills) to ethyl lactate (CH3CHOHCOOC2H5) on a 5% Pt/Al203 catalyst in toluene In both presence and absence of a chiral modifier. (a) What is the stoichiometry of the reaction? Provide a balanced overall reaction. (b) Propose a feasible mechanism for the reaction. (c) Blaser et al. propose that in the absence of the chiral modifier, the rate determining step in the reaction is the addition of an adsorbed hydrogen to an adsorbed ethyl pyruvate. Derive a Langmuir-Hinshelwood rate equation for the reaction, assuming that this mechanism is correct. I in Mil i M'. 7IM) (d) lllnsci el ill also propose llml in the piescnce ol the »lni.il modifier, the addition ol the Aral hydrogen to the ethyl pyruvate rate Is rapid, whila the addition oi a second hydrogen to the adsorbed ethyl pyruvate is rate determining. Derive a Langmuir llinshclwood rale equation fol the reaction assuming that this mechanism is correct. (e) What types of experiments would you do to distinguish between yom i.inequations in c) and d)? (f) Which equation fits the data in Table P 12.33 better? 12.34 Schiirch, et al. (1998). J. Catalysis. 173, 187 examined the role of cerlam amines in modifying the stereospecificity of ethyl pyruvate (CH3COCOOC2H,) hydrogenation to ethyl lactate (CH3CHOHCOOC2H5) on a 5% Pt/Al203 catalyst, (a) If you were trying to modify the stereospecificity of a reaction, what would you do? (Hint: Draw an analogy to the production of isotactic polyethylene 1 (b) Amines bind moderately strongly to platinum. How could the amine work? (c) If you were designing the modifiers, what other species would you choose'.' Look specifically at thiols since they bind even more strongly to platinum. (d) Would you anticipate side reactions? (Hint: What other reactions would the platinum do? What other reactions would the alumina do?) 12.35 Keane, (1997). Journal of the Chemical Society-Faraday Transactions. 93, 2001 examined the hydrogenation of methylacetoacetate (MAA) (CH3COCH2COO CH3) to methyl 3-hydroxybutyrate (CH3CHOHCH2COOCH3) over a Ni/Si02 catalyst. (a) What is the reaction? Provide a balanced equation. (b) Derive a Langmuir-Hinshelwood rate equation for the reaction assuming that the addition of the first hydrogen is rate-determining (c) Derive a Langmuir-Hinshelwood rate equation for the reaction assuming that the addition of the second hydrogen is rate-determining. (d) Which equation fits the data in Table PI 2.35 better? (e) Why is there a maximum rate at intermediate MAA concentrations? 12.36 Many different catalysts can decompose H202: catalyse from potatoes, Mn02, iron, and iron oxide. Put a 30% solution of H202 into a Petri dish, than add Table P12.33 Rate data for problem 12.33 - rates/gram of catalyst Ethyl Pyruvate Rate.mmol/ H2 Pressure, Concentration, Rate mmol/ H2 Pressure, (gm hour) bar mol/liter (gmhour) bar 0 2I 0 17 10 20 21 0.3 24 20 20.5 21 1 30 30 25 21 2.5 34 40 24 21 3 42 60 23 21 4.2 48 80 18 21 6.8 54 100 15 21 9.0 68 160 Ethyl Pyruvate Concentration, mol/liter Initio P 12.35 li.,1. .I..I.I loi I'liihlitin in MAA MAA Win,-, mmol/ < mi,, nil.nun. II; pressure, Rate, mmol/ ( Hill ľ II 11 111 IOI1. it, pri " 1 in, 1 minute) ......ol/lltei aim iiiii-i minute) llllllol/llH'l .ihn 2.1 1 1 0 2.0 1 0.5 / 'i 5 1 0 7.2 5 0.5 11.5 10 1.0 10.0 10 0.5 13.5 30 i 0 11.0 1(1 0.5 12.8 40 1.0 10.2 ID 0.3 20 50 1.0 y.4 50 0.5 10.4 /o 10 8.0 70 0.5 8.5 100 1.0 6.4 100 n The 0.5-atm dala were made up lor this problem. Keane actually measured al only a slice of potato, some Mn02, a clean nail, and a rusty nail. Which subsi,urn-decomposes the H202 the quickest? 12.37 Consider a first-order reaction occurring in a 5-mm-diametcr spherical caiah.i pellet. (a) Calculate the effectiveness factor for the reaction. Data: k, = 8/second, De = 0.8 cm2/second. (b) How would the effectiveness factor change if you changed to 2-min diametei spherical catalyst pellets? (c) How would the effectiveness factor change for 4-mm-diameler pellets with a ke of 1/min? 12.38 As noted in Problem 12.11, enzymes are commonly added to detergents to eliminate stains. The detergents work best in cold water. The object of this problem is to quantify the idea that the enzymes work best in cold water. (a) Consider removing a food stain from your shirt. Food stains contain proteins that take a week to set. Guess at a kinetic expression for the selling of the stain assuming the activation barrier to be 26 kcal/mol, that is, the value from Figure 2.8, assuming that the half-life is a week. (b) Now consider the reaction between the stain and the enzyme. Develop a kinetic expression for the reaction between the enzyme and the stain. Assume thai (1) The reaction between the enzyme is catalyzing the hydrolysis of the protein in the stain, (2) The reaction follows a Rideal-Eley mechanism; that is, the enzyme binds to the stain, then water is added without binding to the enzyme, (3) The binding energy (i.e., AGad) between the protein and the enzyme is 25 kcal/mol, (4) The reaction has a preexponential of 1013/second and an activation barrier of 18 kcal/mol. (5) S0 = 1 enzyme site/1000 linkages in the stain. (c) At this point, you have two parallel reactions. Integrate the rate equations to calculate the concentration of the stain, the portion of the stain that is set, and the portion of the stain that is hydrolyzed as a function of time. I'IK Hli I M'. 701 Ml) Plot yum icsulls as u Illinium ol nine al T 2/1. .">(), 1(H). Ill) K Wlml Fraction «>i die stain is gel in eai ii i Me ' 12.39 In the supplemental material, we derived an equation foi the effectiveness factoi lor diffusion into a Hat plate. The objective of this problem is to derive an expression for the diffusion into a Hal slab of thickness 2L. Assume that the top and bottom of the slab are surrounded by gas and that the slab is infinite in the x and y directions. (a) Use a shell balance to show that the slab obeys dz2 + rA = 0 (PI 2.39.1) where z is the distance from the center of the slab, Ca is the concentration of the reactant, and rA is the rate of formation of the reaclant A. (b) Substitute in a first-order rate law to derive a differential equation for the concentration in the slab. (c) What are the boundary conditions for your rate equation? (d) Show that the solution of the differential equation is CA = Cj cosh(P) / (P12.39.2) with (P12.39.3) (e) Now use your results to derive an equation for the effectiveness factor. (f) Make a plot of the effectiveness factor as a function of the Thiele modulus. How do your results compare to those for a spherical catalyst pellet? 12.40 Chen, Lu Pradier, Paul and Flodstrom (1997). J. Catalysis. 177, 3, examined the effect of S02 on the reduction of NO by butene over a platinum catalyst. They found that small additions of S02 enhanced the rate of reaction, while large amounts of S02 stopped the reaction. (a) Use a Langmuir-Hinshelwood rate expression to show how large amounts of S02 could stop the reaction. (b) Now consider small amounts of S02. Small amounts of S02 act as a poison to weaken the bonds between the reactants and the platinum. How could that increase the rate of reaction? (c) S02 also creates acid sites. How would that affect the reaction? More Advanced Problems 12.41 Gouverneur, Houk, Pascual-Teresa, Beno, Janda, and Lerner (1993). Science 262, 204 produced catalytic antibodies to control the percentage of cis and trans isomers in a series of Diels-Alder reactions. Their technique is to calculate the transition state geometry, find a "hapten" that has a geometry that closely mimics the transition sliilc, and then euate antibodies to lite hapten I he iinlllinily sliiinyly binds 1 tu- transition state and so ails like a cnlnlysl lot the leaelion (n) In the papei they use KIII/3 21(1 methods to calculate the transition stale geometry for the reaction. Recalculate the transition slale geometry al lite MP2/6 IIG" level. Iii Simplify things, replace the p.ul ol the (liene pail the nitrogen with a hydrogen. How much does the geometry change? Nee l.ongcltarich. Brown and Houk ( I'W)), ./. Oryanic (hem. 54. I 129 foi ....... details of the transition state calculation. (b) How would the changes in the geometry ol the transition slale alleel the conclusions in the paper? Is the real transition stale geometry elnsei 01 farther away from the hapten geometry? 12.42 Korrc, Klein, and Quann (1997). Industrial & Engineering Chemistry Resean h 36, 2041 say that they used "inhibition studies" to examine "the kinetics of naphthalene and phenanthrene hydrocracking over a presulfided NiW/USV catalyst in an 1-L batch autoclave at PH2 = 68.1 atm and T = 350 degrees ( (a) What is the overall reaction? Write a balanced equation for the real tanta md major products. (b) Describe the catalyst. What do you know about the structure of the catalyil from the properties given in the paper? (c) How was selective inhibition used to help determine kinetics? (d) Korre, et al. conclude that the reaction occurs on two sites. What is the evidence for this conclusion? Do you agree with the conclusion? 12.43 Prins, Jian, and Flechsenhar (1997a,b). Polyhedron. 16, 3235, (1997). Journal oj Catalysis. 168, 491 examined the kinetics and mechanism of aniline hydrodem trogenation. They find that the data are not easily fit by a Langmuir-Hinshelwood rate expression, but a multisite model works. (a) Use the multisite model to derive an expression for the hydrodenitrogenation of analine. (b) Read Prins' paper. What evidence does he provide for the multisite model' Is the evidence convincing? 12.44 HJ. Bart W. Kaltcnbrunner and H. Landschutzer, examined the kinetics ol the esterification of acetic acid with propyl alcohol catalyzed by the ion exchange resin Dowex Monosphere at 650°C. The mixture is very nonideal. (a) How do Bart et al. account for the nonideal mixing properties in their kineiie model? Do the changes make sense? In other words, does it make sense to modify the kinetic equations in this way? 12.45 Read the following papers and report on the findings; and then devise a possible homework problem based on the results. (a) M. C. Ilao, H. Yamamoto, and K. Segawa, Shape-selective methylamine synthesis over small-pore zeolite catalysts. J. Catal. 161(1); 20-30 (l')')(i> (b) H. U. Blaser, H. P. Jalett, Garland, M., Studer, M., Thies, H., and A. Wirthtijan, Kinetic studies of the enantioselective hydrogenation of ethj I pyruvate Catalyzed by a cinchona modified PT/AL20;, catalyst. SO J. Catal 173(2); 282-294 (1998). I'I H Hli I M' If) Megalofonos, S K , and ľiipttyimriiikiis, N (I Kinetics ol caliilylii reaction ol median and hydrogen sulphide ovei M( )S... 4/«/'/ (iilnl. A 165(1 .'); 249-258 (1997). (d) Mahajani, S. M.. and Slianna, M. M. Reaction of glyoxal willi aliphatic alcohols using cationic exchange resins as catalysts. Org. Process Res. Dev. 1(2); 97-105 (1997). (e) Olsbye, U., Wurzel, T., and Mleczko, L. Kinelics and reaction engineering studies of dry reforming of methane over a NI/LA/AL203 catalyst. Ind. ľni;. Chem. Res. 36(12); 5180-5188 (1997). (f) Mark, M. F., Mark, F. and Maier, W. F., Reaction kinelics of the C02 reforming of methane. Chem. Eng. Technol. 20(6); 361-370 (1997). (g) Lechert, H. The mechanism of faujasite growth studied by crystallization kinetics. Zeolites 17(5-6); 473-482 (1996). (h) Bond, G. C, Hooper, A. D., Slaa, J. C, and Taylor, A. O. Kinetics of metal-catalyzed reactions of alkanes and the compensation effect. J. Catal. 163(2); 319-327 (1996). (i) Yu, Y. S., Bailey, G. W., and Jin, X. C. Application of a lumped, nonlinear kinetics model to metal sorption on humic substances. J. Environ. Qual. 25(3); 552-561 (1996). (J) Loumagne, F., Langlais, F., and Naslain, R. Reactional mechanisms of the chemical vapour deposition of sic-based ceramics from CH3SICL3/H-2 gas precursor. /. Crys. Growth 155(3-4); 205-213 (1995). (k) Kim, D. S., and Lee, Y. H., Growth mechanism of room-temperature deposited A-SIC-H films by ion-assisted RF glow discharge. /. Electrochem. Soc. 142(10); 3493-3504 (1995). (1) Holm, D. R., Hill, C. G., and Conner, A. H., Kinetics of the liquid phase hydrogenation of furan amines. Ind. Eng. Chem. Res. 34(10); 3392-3398 (1995). (m) Kim, E. J., and Gill, W. N. Low pressure chemical vapor deposition of silicon dioxide films by thermal decomposition of tetra-alkoxysilanes. J. Electrochem. Soc. 142(2); 676-682 (1995). (n) Sauer, M. L., and Ollis, D. F. Acetone oxidation in a photocatalytic monolith reactor. J. Catal. 149(1); 81-91 (1994). (o) Creighton, J. R. The surface chemistry and kinetics of tungsten chemical vapor deposition and selectivity loss. Thin Solid Films 241(1-2); 310-317 (1994). (p) Edwards, M. E., Villa, C. M., Hill, C. G., Jr., and Chapman, T.W. Effectiveness factors for photocatalytic reactions occurring in planar membranes. Ind. Eng. Chem. Res. 35(3), 712-720 (1996). (q) Haupfear, E. A., and Schmidt, L. D. Kinetics of boron deposition from BBr3 plus H2. Chem. Eng. Sci. 49(15), 2467 (1994). (r) Kuhne, H., How hydrogen influences axial growth rate distribution during silicon deposition from silane. Semicond. Sci. Technol. 8(11) 2018-2022 (1993). (s) Garant, H., and Lynd, L. Applicablity of competitive and noncompetitive kinetics to the reductive dechlorination of chlorinated ethenes. Biotechnol. Bioeng. 57, 751 (1998). (I) Islukawii, I . Yoihliniiui, K , Sukni, K., Iiiinoi. M , Tiikcdu, I , and siugeoka. s. Moiii iii.ii characterization ami physiological roll "i a glyoxyiOttM bound MCOlbata peroxidase from spinach. I'luni (ill Physiol 39(1); 23 34 (I9'»K) in) Leaf, T, A.. Srienc, P Metabolli modeling of polyhydrozybutyrate biosyn thesis. Bioti-clmol. Bioeng. 57(5); 557-570(1998). (v) Xorko, M„ l'ooga, M, Saar. K.. Rczaei, K.. and l.angel, U. Differential regulation of gtphase activity by mastoparan and galparan. Arch, Biochem Biophys. 349(2); 321-328 (1998). (w) Hoh, C. Y., and Cordruwisch, R. Experimental evidence foi the need ol thermodynamic considerations in modeling of anaerobic environmental bioprocesses. Water Sci. Technol. 36(10); 109-115 (1997). (x) Fadiloglu, S., and Soylcmez, Z. Kinetics of lipase-catalyzed hydrolysl ol olive oil. Food Res. Int. 30(3-4); 171-175 (1997). (y) Jones. A. W.. Jonsson. K. A., and Kechagias. S. Effect "I 111j■ I• l.u lnrli protein, and high-carbohydrate meals on the pharmacokinetics >>i a in.ill dose of ethanol. Br. J. Clin. Pharmaco. 44(6); 521-526 (1997). (z) Iwuoha, E. I., Devillaverde, D. S., Garcia, N. P., Smyth. M. R., Pingarron, J. M., Reactivities of organic phase biosensors. 2. The amperometrli behaviour of horseradish peroxide immobilized on a platinum electrodi modified with an electrosynthetic polyaniline film. Biosensors Biolectron 12(8); 749-761 (1997). (aa) Egger, D., Wehlje, E., and Adlercreutz, P. Characterization and optimization of phospholipase A(2) catalyzed synthesis of phosphatidylcholine. Huh him Biophys. Acta 1343, 76-84 (1997). (bb) Aoyama, T., Yamamoto, K., Kotaki, H., Sawada, Y., and lgo, T. Pharmaco dynamic modeling for change of locomotor activity by methylphenidate In rats. Pharm. Res. 14(11); 1601-1606 (1997). (cc) Davies, R.R., and Distcfano, M.D., A semisynthetic mctallocn/.ymc based on a protein cavity that catalyzes the enantioselective hydrolysis of estei and aminde substrates. J. Am. Chem. Soc. 119(48); 11643-11652 (1997) (dd) Franssen, O., Vanooijen, R.D., Deboer, D., Maes, R.A.A., Herron, J.N . and Hennink, W.E. Enzymatic degradation of methacrylated dcxtrans. Macro molecules 30(24); 7408-7413 (1997). (ee) Turhan, M., and Mutly, M. Kinetics of kappa-casein/chymosin hydrolysis. Milchwissenschaft 52(10); 559-563 (1997). (ff) Deka, R.C., and Vetrivel, R., Adsorption sites and diffusion mechanism of alkylbenzenes in large pore zeolite catalysts as predicted by moleculai modeling techniques. J. Catal 174(1); 88-97 (1998). 12.46 Look at the following journals: Journal of Catalysis, Applied Catalysis A. Applied Catalysis B, Biotechnology & Bioengineering. Find an interesting article and report on its findings.