CHBE 452 Lecture 31 Mass Transfer & Kinetics In Catalysis 1 Key Ideas For Today Generic mechanics of catalytic reactions Measure rate as a turnover number Rate equations complex Langmuir Hinshelwood kinetics 2 Generic Mechanisms Of Catalytic Reactions 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+BAB and ABA+B. 3 Turnover Number: Number Of Times Goes Around The Catalytic Cycle Per Second CH3COOH Printing press analogy 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 •Reactants bind to sites on the catalyst surface •Transformation occurs •Reactants desorb CO Figure 12.1 A schematic of the catalytic cycle for Acetic acid production via the Monsanto process. 4 Typical Catalytic Cycle + 1/2 O 2 O O O O O O O +1/2 O 2 H H O O O O O - H 2O O O B H H O O O O O O + H2 +H2 - H 2O 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 5 Definition Of Turnover Number RA TN NS (12.119) Physically, turnover number is the rate that the catalyst prints product per unit sec. 6 2 10 Turnover Number, sec -1 Typical Turnover Numbers 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 1400 Reaction Temperature, K 7 10 13 450 K 450 K 2 Rate, Molecules/cm -sec Next Topic Why Catalytic Kinetics Different Than Gas Phase Kinetics 10 12 440 K 440 K 425 K 410 K 415 K 390 K 10 11 10 -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. 8 Temperature Nonlinear 2 Rate, Molecules/cm -sec PO2=2.5E-8 torr PCO=2.E-7 torr 1E+13 C 1E+12 F E D B 1E+11 A 400 600 Temperature, K 800 400 600 800 Temperature, K Figure 2.18 The rate of the reaction CO + ½ O2 CO2 on Rh(111). Data of Schwartz, Schmidt and Fisher[1986]. A) = 2.510-8 torr, = 2.510-8 torr, B) = 110-7 torr, = 2.510-8 torr, C) = 810-7 torr, = 2.510-8 torr, D) = 210-7 torr, = 410-7 torr, E) = 210-7 torr, = 2.510-8 torr, F) = 2.510-8 torr, = 2.510-8 torr, 9 Derivation Of Rate Law For AC Also have a species B Mechanism 1 S +A 2 Aad 3 AAd Cad 4 7 S+BBad 8 (12.122) 5 Cad C + S 6 (12.121) 10 Derivation: Next uses the steady state approximation to derive an equation for the production rate of Cad (this must be equal to the production rate of C). 11 Derivation Continued rC k 3[A ad ] k 4 [C ad ] (12.123) SS on [Aad] and [Cad] 0 rA k 1PA [S] k 2 [A ad ] k 3 [A ad ] k 4 [C ad ] ad (12.124) 0 rC ad k 6 PC [S] k 5 [C ad ] k 4 [C ad ] k 3 [A ad ] (12.125) People usually ignore reactions 3 and 4 since their rates very low rates compared to the other reactions. 12 Dropping The k3 And k4 Terms In Equations 12.124 And 12.125 And Rearranging Yields: k1 [A ad ] PA [S] k2 Similarly for B (12.126) k8 B PB S ad k7 k6 [C ad ] PC [S] k5 (12.128) (12.127) 13 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) 14 Derivation Continued Equations (12.129) and (12.130) imply that there is an equilibrium in the reactions: [A ad ] k1 PA [S] k 2 Aad A + S 1 2 (12.129) B + S Bad B 7 8 [Bad ] k 8 PB [S] k 7 (12.130) C ad k 6 Pc S k 5 (12.131) 6 Cad C + S 5 (12.132) 15 Site Balance To Complete The Analysis If we define S0 as the total number as sites n the catalyst, one can show: S0 [S] [A ad ] [Bad ] Cad (12.133) Pages of Algebra K A PA S 0 [A ad ] 1 K A PA K B PB K C PC (12.140) K C PC S 0 [C ad ] 1 K A PA K B PB K C PC (12.141) 16 Substituting Equations (12.140) And (12.141) Into Equation (12.123) Yields: k 3 K A PA S 0 k 4 K C PCS 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 AC, also called Michaele’s Menton Equation. 17 2.5E+14 10 13 450 K 2 2.0E+14 Rate, Molecules/cm -sec Rate, Molecules/cm 2 /sec Qualitative Behavior For Bimolecular Reactions (A+Bproducts) 1.5E+14 1.0E+14 5.0E+13 0.0E+0 0 10 20 30 40 PA 50 10 12 440 K 410 K 390 K 10 11 10 Figure 12.32 A plot of the rate calculated from equation (12.161) with KBPB=10. -8 -7 10 CO pressure, torr 10 -6 18 Physical Interpretation Of Maximum Rate For A+BAB Catalysts have finite number of sites. Initially rates increase because surface concentration increases. Eventually A takes up so many sites that no B can adsorb. Further increases in A decrease rate. 19 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 PA 30 40 50 Rate, Molecules/cm 2-sec Qualitative Behavior For Unimolecular Reactions (AC) 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 20 Langmuir-Hinshelwood-Hougan-Watson Rate Laws: Trick To Simplify The Algebra Hougan and Watson’s Method: Identify rate determining step (RDS). Assume all steps before RDS in equilibrium with reactants. All steps after RDS in equilibrium with products. Plug into site balance to calculate rate equation. 21 Example: The reaction A + B C obeys: S S A ad A B Bad A ad (1) Bad (2) 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. 22 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) 23 Solution Continued Need S to complete solution: get it from a site balance. So = S + [Aad ] + [Bad ] (5) Combining (2), (3) and (5) So = S + SK1 PA + SK2 PB Solving (6) for S So S 1 K1 PA K2 PB Combining equations (4) and (7) K1K 2 k 3 PA PB So rC 1 K1PA K 2 PB 2 (6) (7) 2 (8) 24 Background: Mass Transfer Critical to Catalyst Design Cavity C H H H H H H C Diffusion Channel Figure 12.27 An interconnecting pore structure which is selective for the formation of paraxylene. 25 Introduction In a supported catalyst reactants first diffuse into the catalyst, then they react The products diffuse out Gives opportunity for catalyst design 26 Distance Reactant Concentration Drops Moving Into the Solid Reactant Concentration 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 Conc in pellet 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Distance from the center of the pellet 27 Thiele Derivation for Diffusion In Catalysis Distance Assume: Constant effective diffusivity First Order reaction per unit volume Irreversible reaction – Rate of diffusion of products out of catalyst does not affect rate Reactant Concentration 28 Define “Effectiveness Factor” Actual rate for a catalyst pellet with mass transfer limitations e Rate in the absence of mass transfer limitations 29 Derivation: Mass Balance On Differential Slice Y dC S dC A 2 2 4 y D e 4 y D e dY y y dy y 4 y 2 y rA Spherical pellet Taking the limit as Δy → 0 2 d C A 2 dC A rA 0 dy y dy D e 30 Long Derivation y P Sinh 3y/y P C A C y Sinh 3P 0 A yP P 3 kA De 1 1 1 e P tanh3 P 3 P 31 Mass Transfer Factor Thiele Plot 1.0 0.8 Zero Order 0.6 0.4 0.2 0.0 0 First Order Second Order 2 4 6 8 10 Thele Parameter 32 Issues With “Effectiveness Factor” Best catalysts have low “effectiveness factors” Effectiveness goes down as rate goes up High rate implies low selectivity Often want mass transfer limitations for selectivity I prefer “mass transfer factor” not “effectiveness factor” 33 Issues, Continued: Effective Diffusivity, De, Unknown Figure 14.3 A cross sectional diagram of a typical catalyst support. 34 Knudsen Diffusion Diffusion rate in gas phase controlled by gas – gas collisions Diffusion rate in small poles controlled by gas – surface collisions 35 Knudsen Diffusion From kinetic theory 1 Dgas vλ 3 1 D k V2a 3 applies when (λ mean free path) a pore diameter 2a gas 36 Knudsen Diffusion DK 1.68x10 1 2 a T 2 AMU sec 1R 300K M 3 CM 1 2 100 Å pore DK 0.12cm / sec for H2 2 compared to 0.86 in the gas phase 37 Simple Models Never Work Notice one molecule interferes with diffusion of second molecule 38 Summary Catalytic reactions follow a catalytic cycle reactants + S adsorbed reactants Adsorbed reactants products + S Different types of reactions Langmuir Hinshelwood Rideal-Eley 39 Summary Calculate kinetics via Hougan and Watson; Identify rate determining step (RDS) Assume all steps before RDS in equilibrium with reactants All steps after RDS in equilibrium with products Plug into site balance Predicts non-linear behavior also seen experimentally 40 Query What did you learn new in this lecture? 41