Catalysts and Catalysis in Chemical Reaction Engineering
Explore the fascinating field of Chemical Reaction Engineering (CRE) that delves into the rates, mechanisms, and design of reactors for chemical reactions. Learn about catalysts, catalysis, active sites, and more through visual aids and insightful explanations.
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Presentation Transcript
Lecture 17 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.
Web Lecture 17 Class Lecture 22 Thursday 4/4/2013 Introduction to Catalysts and Catalysis Interstage cooling Noble Prize 2007 Catalytic steps 2 2
Catalysts and Catalysis A Catalyst is a substance that affects the rate of chemical reaction but emerges from the process unchanged. Catalysis is the occurrence, study, and use of catalysts and catalytic processes. Approximately 1/3 of the GNP of materials produced in the U.S. involves a catalytic process. 3
Catalysts and Catalysis Catalysts affect both selectivity and yield Different reaction paths 4 4
Catalysts and Catalysis Different shapes and sizes of catalyst. 5 5
Catalysts and Catalysis Catalytic packed-bed reactor, schematic. 6 6
Active Sites Reactions are not catalyzed over the entire surface but only at certain active sites or centers that result from unsaturated atoms in the surface. An active site is a point on the surface that can form strong chemical bonds with an adsorbed atom or molecule. 8 8
The Adsorption Step Vacant and occupied sites For the system shown, the total concentration of sites is Ct = Cv + CA.S + CB.S 10 10
The Adsorption Step + A S = A S = - v k / AD r k P C C k P C C K - A A A A S A A V A S A k = 1 - / [atm ] K k A A A = / = equilibriu @ m : 0 AD r C k P C AS A A V = 0 AD r k C k P C A AS A A V = + = + = + 1 ( ) C C C C K P C C K P t V A S V A A V V A A ?? ??= 1 + ???? 11 11
Langmuir Adsorption Isotherm ?? ??= 1 + ???? ?? ?= ?????? ???? 1+?????? ?? ?= ?? ? ?? ???? 1 + ???? = 12 12
Langmuir Adsorption Isotherm C K + P C = A S A K A P A S Increasing T 1 C C t A A T Slope=kA Langmuir Adsorption Isotherm P A 13
Desorption from the Surface for the Reaction A B+C C S C+S rDC= kDCC S PCC rDC= rADC (10-20) KDC 1 KDC= KC (10-21) rDC= kDCC S KCPCC
Steps in a Single-Site Catalytic Reactor C + A S A S = = Adsorption A S r r k P C A Ad Ad A v k A C = = A S B S Surface Reaction B S r r k C A S S A S k C + B S B S = = Desorption r r k C k P C A D D B S B B B Which step is the Rate Limiting Step (RLS)? 21 21
The Rate Rate Limiting Step: Which step has the largest resistance? Electrical analog to heterogeneous reactions 22 22
Collecting and Analyzing Data Collecting information for catalytic reactor design 23 23
Catalytic Reformers Normal Pentane Octane Number = 62 Iso-Pentane Octane Number = 95 25 25
Catalytic Reformers 0.75 wt% Pt n-pentane i-pentane Al2O3 +H2 Al2O3 -H2 n-pentane n-pentene i-pentene i-pentane Pt Pt Al2O3 n-pentene i-pentene N I 26 26
Catalytic Reformers Isomerization of n-pentene (N) to i-pentene (I) over alumina Al2O3 N I 1. Select a mechanism (Mechanism Single Site) + S S N N I S S S N I I S Adsorption on Surface: + Surface Reaction: Desorption: Treat each reaction step as an elementary reaction when writing rate laws. 27
Catalytic Reformers 2. Assume a rate-limiting step. Choose the surface reaction first, since more than 75% of all heterogenous reactions that are not diffusion- limited are surface-reaction-limited. The rate law for the surface reaction step is: S S N + + I S S C = = = ' I S r r r k C N I S S N S K S 28
Catalytic Reformers 3. Find the expression for the concentrations of the adsorbed species CN.S and CI.S. Use the other steps that are not limiting to solve for CN.S and CI.S. For this reaction: + N S N S r = S AD C P K C 0 : From N N N k A + I S I S r P C D = = 0 : I C K P C From I S I I k K D D 29
Catalytic Reformers 4. Write a Site Balance. = + + C C C C t N S I S 5. Derive the rate law. Combine steps 2, 3 and 4 to arrive at the rate law : k ( ) k C 1 K P P K = = s t + N N I P r r ( ) N S + K P K P N N I I ( K ) k + P P K = = N I P P r r ( ) N S + 1 P K N N I I 30
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas 1994 0.41 3.4 0.4 2004 0.125 3.4 0.4 2008 0.10 3.4 0.14 HC CO NO CO+ NO CO2+1 2N2
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas C + = = NO S NO S NO S ANO r k P C C K P C NO NO V NO S NO NO V K NO C = = CO S CO S CO S ACO r k P C C K P C CO CO V CO S CO CO V K CO + + + = CO S NO S CO N S S r k C C 2 S S CO S NO S ( ) g + + = = 2 N 2 V N S N S N 2 S r k C K P C C C K P 2 D D S N N N S V N N 2 2 2 32 32
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas rS= kSCNO SCCO S 2 rS= kSKNOKCOPNOPCOCV CT= CV+CNO S+CCO S+ CN S = CV+CVKNOPNO+ CVKCOPCO+ CV KN2PN2 33 33
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas C = t C V + + + 1 K P K P K P NO NO CO CO N N 2 2 k K 2 t k K C P P NO = = S NO + CO NO + CO K r r ( 1 ) S 2 + K P K P P NO NO CO CO N N 2 2 kP P NO = NO CO r ( 1 ) 2 + + + K P K P K P NO NO CO CO N N 2 2 34 34
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas kPNOPCO r NO= 1+KNOPNO+KCOPCO+ KN2PN2 ( ) 2 Neglect KN2PN2 kPNOPCO r NO= )2 1+KNOPNO+KCOPCO (
Catalytic Conversion of Exhaust Gas Catalytic Conversion of Exhaust Gas kPNOPCO r NO= )2 1+KNOPNO+KCOPCO ( Find optimum partial pressure of CO d rNO ( dPCO ) = 0 PCO=1+KNOPNO KCO
End of Web Lecture 17 37 37
