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Chapt 12 summary part 2 Today I want to discuss catalytic kinetics: Two key ideas for today: • Measure rate as a turnover number • Langmuir Hinshelwood kinetics 1 Turnover numbers Recall our discussion from before. CH3COOH CH3OH HI H2O CH3COI C O I I O C O I C H3C I C O Rh CH3I [Rh(CO)2I2]- CH3 Rh C O I I CO Figure 12.1 A schematic of the catalytic cycle for Acetic acid production via the Monsanto process. Printing press analogy • Reactants bind to sites on the catalyst surface • Transformation occurs • Reactants desorb 2 B +H2 H H O O O O O O - H 2O + 1/2 O 2 O O O O O O O +1/2 O 2 - H 2O H H O O O O O O O + H2 A H H O H H O Figure 5.10 Catalytic cycles for the production of water a) via disproportion of OH groups, b) via the reaction OH(ad)+H)ad)→H2O. B A B A B A B B A B B B A Langmuir-Hinshelwood A B A B A B A A Rideal-Eley B A B A B A B A B A B A A A Precursor Figure 5.20 Schematic of a) Langmuir-Hinshelwood, b) Rideal-Eley, c) precursor mechanism for the reaction A+B⇒AB and AB⇒A+B. 3 TN = RA NS (12.119) Physically, turnover number is the rate that the catalyst prints product. Most people discuss rates in terms of turnover numbers 2 10 Turnover Number, sec -1 Key plot: Dehydrogenation Hydrogenation Silicon Deposition 0 10 GaAs Deposition -2 10 -4 10 -6 10 Olefin Isomerization Alkane Hydrogenolysis Cyclization 200 400 600 800 1000 1200 Reaction Temperature, K 1400 Figure 12.28 Turnover numbers for some typical processes 4 Next topic why catalytic kinetics weird? 10 13 450 K 450 K 2 Rate, Molecules/cm -sec Recall from earlier in the semester catalytic rates weird: 10 12 440 K 440 K 425 K 410 K 10 11 10 415 K 390 K -8 -7 10 CO pressure, torr 10 -6 10 -8 -7 10 O 2 pressure, torr 10 -6 Figure 2.15 The influence of the CO pressure on the rate of CO oxidation on Rh(111). Data of Schwartz, Schmidt, and Fisher. 5 Rate, Molecules/cm 2-sec PO =2.5E-8 torr PCO=2.E-7 torr 2 1E+13 C 1E+12 F E D B 1E+11 A 400 600 800 400 600 800 Temperature, K Temperature, K Figure 2.18 The rate of the reaction CO + 2 O2 ⇒ CO2 on Rh(111). Data of Schwartz, Schmidt and Fisher[1986]. A) PCO = 2.5×10-8 torr, PO = 2.5×10-8 torr, B) 2 PCO = 1×10 -7 -8 -7 torr, PO = 2.5×10 torr, C) PCO = 8×10 torr, PO = 2.5×10-8 torr, D) 2 -7 2 -7 -7 -8 PCO = 2×10 torr, PO2 = 4×10 torr, E) PCO = 2×10 torr, PO2 = 2.5×10 torr, F) PCO = 2.5×10-8 torr, PO = 2.5×10-8 torr, 2 Today: Why the weird behavior? 6 Physical interpretation: Figure 12.33 Langmuir's model for the adsorption of gas on a solid catalyst. The light gray area represents the surface of the catalyst. The black dots are the sites on the catalyst are sites that are available to adsorb gas. The white ovals are the adsorbed A molecules. The dark ovals are adsorbed B molecules. Physical interpretation for A + B ⇒ AB • Catalysts have finite number of sites: • Initially rates increase because surface concentration increases. • Eventually A takes up so many sites than no B can adsorb. • Further increases in A decrease rate. 7 Can work out kinetics using steady state approximation: A⇒C (12.120) Mechanism 1 S + A º Aad 2 3 AAd º Cad 4 5 Cad º C + S 6 (12.121) Also have a species B 7 S + B B ad 8 (12.122) 8 Next use the steady state approximation to derive an equation for the production rate of Cad (this must be equal to the production rate of C). rC = k 3[A ad ] − k 4 [C ad ] (12.123) SS on [Aad] and [Cad] 0 = rA ad = k 1PA [S] − k 2 [A ad ] − k 3[A ad ] + k 4 [C ad ] (12.124) 0 = rC ad = k 6 PC [S] − k 5[C ad ] − k 4 [C ad ] + k 3[A ad ] (12.125) 9 People usually ignore reactions 3 and 4 since their rates very low rates compared to the other reactions. Dropping the k3 and k4 terms in equations 12.124 and 12.125 and rearranging yields: k [A ad ] = 1 PA [S] k2 (12.126) k6 [C ad ] = PC [S] k5 (12.127) Similarly for B k8 = PB [S] B [ ad ] k7 (12.128) 10 Rearranging Equations (12.126), (12.127) and (12.128) yields: [A ad ] k1 = PA [S] k 2 (12.129) [Bad ] k 8 = PB [S] k 7 (12.130) [C ad ] = k 6 Pc [S] k 5 (12.131) 11 Equations (12.129) and (12.130) imply that there is an equilibrium in the reactions: 1 A + S º Aad 2 8 B + S ºBad 7 6 C + S ºC ad 5 (12.132) 12 If sites are conserved One needs an expression for [S] to complete the analysis. One can get an expression for [S] by assuming that on any catalyst, there are a finite number of sites to hold the reactants. Each site can be bare or it can be covered by A, B or C. If we define S0 as the total number of sites in the catalyst, one can show: S 0 = [S] + [A ad ] + [Bad ] + [ C ad ] (12.133) Substituting Equations (12.126), (12.127) and (12.128) into Equation (12.133) and then solving for [S] yields: [S] = S0 k6 k8 k1 PC PB + 1+ PA + k5 k7 k2 (12.134) 13 Substituting Equation (12.134) into Equations (12.126) and (12.127) yields: [A ad ] = k 1+ 1 k2 k1 PA S 0 k2 k k PA + 8 PB + 6 PC k7 k5 (12.135) [C ad ] = k 1+ 1 k2 k6 PC S 0 k5 k PA + 8 PB + k7 k6 PC k5 (12.136) According to the analysis in Section 4.3, the equilibrium constant for the adsorption of A is given by: k1 KA = k2 (12.137) 14 Similarly, the equilibrium constants for the adsorption of B, and C are given by: k8 KB = k7 (12.138) k KC = 6 k5 (12.139) Substituting Equations (12.137), (12.138), and (12.139) into Equations (12.135) and (12.136) yields: [A ad ] = K A PA S 0 1 + K A PA + K B PB + K C PC (12.140) [C ad ] = K C PCS 0 1 + K A PA + K B PB + K C PC (12.141) 15 Substituting Equations (12.140) and (12.141) into Equation (12.123) yields: k 3 K A PA S 0 − k 4 K C PC S 0 r= 1 + K A PA + K B PB + K C PC (12.142) In the catalysis literature, Equation (12.142) is called the LangmuirHinshelwood expression for the rate of the reaction A⇒C, since we assumed that the reaction obeyed a Langmuir-Hinshelwood mechanism and the equation was first derived for surface reaction by Langmuir[1914]. The same equation is called the Michaelis-Menten equation in the enzyme literature; since it was also derived by Michaelis and Menten[1914]. 16 Rate, Molecules/cm 2 /sec Qualitative behavior for A + B ⇒ products: 2.5E+14 2.0E+14 1.5E+14 1.0E+14 5.0E+13 0.0E+0 0 10 20 30 40 50 PA 10 13 450 K 450 K 2 Rate, Molecules/cm -sec Figure 12.32 A plot of the rate calculated from equation (12.161) with K B PB = 1 0. 10 12 440 K 440 K 425 K 410 K 10 11 10 415 K 390 K -8 -7 10 CO pressure, torr 10 -6 10 -8 -7 10 O 2 pressure, torr 10 -6 Figure 2.15 The influence of the CO pressure on the rate of CO oxidation on Rh(111). Data of Schwartz, Schmidt, and Fisher. 17 Qualitative behavior for A⇒C Rate, Moles/cm 2 /sec 2.0E-8 PB =0 1.5E-8 1.0E-8 PB =25 5.0E-9 0.0E+0 0 10 20 30 40 50 Rate, Molecules/cm 2-sec PA 1E+20 1670 K 1E+19 1270 K 1070 K 1E+18 870 K 770 K 1E+17 1E+16 0.01 0.1 1 10 100 Ammonia pressure, torr Figure 2.16 The rate of the reaction NH 3 ⇒ 1 / 2N 2 + 3 / 2H 2 over a platinum wire catalyst. Data of Loffler and Schmidt[1976]. 18 Langmuir-Hinshelwood-Hougan-Watson rate laws Trick to simplify the rate equation. Hougan and watson's idea: • Identify rate determining step (RDS) • Assume all steps before RDS in equilibrium with recatants • All steps after RDS in equilibrium with products • Plug into site balance 19 Example: The reaction A + B ⇒ C obeys: S + A ⇒ A ad (1) S + B ⇒ Bad (2) A ad + Bad ⇒ C + 2S (3) (12.157) Derive an equation for the rate of formation of C as a function of the partial pressures of A and B. Assume that reaction (3) is rate determining. 20 My solution rC = k3[Aad][Bad] Assume reaction 1 in equilibrium [A ad ] = K1 S PA Similarly on reaction 2 [Bad ] = K2 S PB Combining 1,2 and 3 rC = K1K 2 k 3 PA PBS2 (1) (2) (3) (4) Need S to complete solution: get it from a site balance. (5) So= S + [Aad] + [Bad] Combining (2), (3) and (5) So= S + SK1PA + K2PB Solving (6) for S So S= 1 + K1PA + K 2 PB 21 (6) (7) Combining equations (4) and (7) K1K 2 k 3 PA PB (So ) rC = (1 + K1PA + K 2 PB )2 2 22 (8) Rate, Molecules/cm 2 /sec Next what does this look like? 2.5E+14 2.0E+14 1.5E+14 1.0E+14 5.0E+13 0.0E+0 0 10 20 30 40 50 PA 10 13 450 K 450 K 2 Rate, Molecules/cm -sec Figure 12.32 A plot of the rate calculated from equation (12.161) with K B PB = 1 0. 10 12 440 K 440 K 425 K 410 K 10 11 10 415 K 390 K -8 -7 10 CO pressure, torr 10 -6 10 -8 -7 10 O 2 pressure, torr 10 -6 Figure 2.15 The influence of the CO pressure on the rate of CO oxidation on Rh(111). Data of Schwartz, Schmidt, and Fisher. 23 Discussion problem: consider a different reaction A⇒C (12.120) 1 S + A º Aad 2 3 AAd º Cad 4 5 Cad º C + S 6 (12.121) Derivation the same: rC Ad = k 3[A ad ] − k 4 [C ad ] (12.123) Derive rate equation 24 r= k 3K A PA S0 − k 4 K C PCS0 1 + K A PA + K B PB + K C PC (12.142) What does this look like? Rate, Moles/cm 2 /sec 2.0E-8 PB =0 1.5E-8 1.0E-8 PB =25 5.0E-9 0.0E+0 0 10 20 30 40 50 PA Figure 12.29 A plot of the rate of the reaction A⇒C calculated from Equation (12.142) with k4=0, PB = 0, 1, 2, 5, 10 and 25., KA = KB =1. 25 Rate, Molecules/cm 2-sec 0 1E+20 1670 K 1E+19 1270 K 1070 K 1E+18 870 K 770 K 1E+17 1E+16 0.01 0.1 1 10 100 Ammonia pressure, torr Figure 2.16 The rate of the reaction NH 3 ⇒ 1 / 2N 2 + 3 / 2H 2 over a platinum wire catalyst. Data of Loffler and Schmidt[1976]. 26 Summary: • Catalytic reactions follow a catalytic cycle reactants + S ⇒adsorbed reactants Adsorbed reactants ⇒products + S • Different types of reactions Langmuir Hinshelwood Rideal-Eley • Calculate kinetics via Hougan and Watson; • Identify rate determining step (RDS) • Assume all steps before RDS in equilibrium with recatants • All steps after RDS in equilibrium with products • Plug into site balance • Predicts non-linear behavior also seen experimentally 27 Next topic: Mass transfer in catalysts Mass transfer is important to catalytic reactions: Catalyst increases Reaction rate by 1020-1040 Mass transfer rate does not change. If you have a really active catalyst, mass transfer will control the rate. 28 Mass transfer control is good: - design pore structure to pass desired products much faster than undesired products. Cavity C H H H H C H H Diffusion Channel Figure Error! No text of specified style in document..1 An interconnecting pore structure which is selective for the formation of paraxylene. 29 Equations for mass transfer limitations Define Thiele Parameter Reaction Rate Φp = Diffusion Rate Define mass transfer factor (literature calls this effectiveness factor). ηe = Actual Reaction Rate Reaction Rate If Mass Transfer Was Instantaneous Derive equation: 30 Consider a spherical catalyst particle Edge of Pellet Center of Pellet Concentration 1 Gas Phase Conc 0.8 Avg conc in pellet Gas Phase 0.6 0.4 0.2 0 Conc in pellet 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Distance from the center of the pellet Figure 12.29 The concentration of species in a spherical catalyst pellet. The diffusion equation is d 2 C A 2 dC A rA + + =0 dy y dy D e Solution: y Sinh (3Φ P y/yP ) C A =C0A P y Sinh (3Φ P ) with 31 yP k A ΦP = 3 De Calculating the reaction rate: Mass Transfer Factor 1 1 1 ηe = − Φ P tanh (3Φ P ) 3Φ P 10 1 0.1 0.01 0.001 0.01 0.1 1 10 100 Thele Parameter, ΦP A plot of the mass transfer factor versus the Thiele parameter for diffusion in a porous catalyst pellet. 32 The mass transfer factor does not measure the effectiveness of a catalyst. Good catalysts can have low mass transfer factors while bad catalysts can have high mass transfer factors. Generally if a catalyst is not speeding up the reaction very much the Thiele parameter will be small, which means that according to the Figure the mass transfer factor will be close to unity. Thus, a bad catalyst can have a large mass transfer factor. In contrast, if the catalyst speeds up a reaction a lot, Φp will be large. The figure shows that the mass transfer factor is reduced under such circumstances. 33 Actual equations not useful - never know De, but ideas useful. 34