Chemical Reaction Engineering-chapter 1-01
chemicalreactionengineering3ededition作者octavelevenspiel课后习题答案
Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING .......................................... 错误!未定义书签。
CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS ........................................................ 错误!未定义书签。
CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA ..................................................... 错误!未定义书签。
CHAPTER 4 INTRODUCTION TO REACTOR DESIGN ............................................................... 错误!未定义书签。
CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR........................................................... 错误!未定义书签。
CHAPTER 6 DESIGN FOR SINGLE REACTIONS ....................................................................... 错误!未定义书签。
CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR ....................................................... 错误!未定义书签。
Chemical Reaction Engineering
Chemical Reaction Engineering Chemical reaction engineering is a crucial field in chemical engineering that deals with the design and optimization of chemical processes and reactors. It involves the study of the kinetics of chemical reactions, the mechanisms of chemical reactions, and the design of reactors to produce the desired products. This field is essential in various industries, including petrochemicals, pharmaceuticals, food and beverages, and environmental engineering. One of thekey challenges in chemical reaction engineering is the optimization of reaction conditions to maximize the production of desired products while minimizing the formation of undesired by-products. This requires a deep understanding of the kinetics of the chemical reactions involved, as well as the thermodynamics and transport phenomena that influence the reaction rates and product distributions. Engineers must also consider the economic and environmental implications of the reaction conditions, as well as the safety and operability of the reactor systems. From a theoretical perspective, chemical reaction engineering involves the application of principles from chemistry, physics, and mathematics to model and simulate chemical reactions and reactor systems. This requires a strong foundation in chemical kinetics, thermodynamics, and transport phenomena, as well as proficiency in numerical methods and computational tools for solving complex reaction engineering problems. Theoretical studies in chemical reactionengineering also contribute to the development of new reaction mechanisms and kinetic models, as well as the design of novel reactor systems for emerging chemical processes. On the other hand, from a practical perspective, chemical reaction engineering involves the design, operation, and optimization ofindustrial-scale reactors for the production of chemicals, fuels, and materials. This requires hands-on experience with various types of reactors, such as batch reactors, continuous stirred-tank reactors, packed bed reactors, and fluidized bed reactors. Engineers must also consider the scale-up of laboratory-scale reactions to industrial-scale production, as well as the integration of reaction systemswith downstream separation and purification processes. In addition, chemical reaction engineering plays a critical role in the development of sustainable and environmentally friendly processes. Engineers are constantly seeking ways toimprove the efficiency and selectivity of chemical reactions, as well as to minimize the use of energy and resources, and the generation of waste and emissions. This involves the use of advanced catalysts and reaction conditions, as well as the implementation of process intensification and integration strategies to improve the overall sustainability of chemical processes. Furthermore, chemical reaction engineering is also closely linked to the field of process safety, as the design and operation of chemical reactors must consider the potential hazards and risks associated with the handling of reactive chemicals and the release of toxic or flammable substances. Engineers must implement appropriate safety measures and risk assessment techniques to ensure the safe and reliable operation of reactor systems, as well as to prevent and mitigate the consequences of chemical accidents and incidents. In conclusion, chemical reaction engineering is a multidisciplinary field that encompasses theoretical and practical aspects of chemical kinetics, reactor design, and process optimization. It plays a crucial role in the development of new chemical processes and the improvement of existing ones, with a focus on efficiency, selectivity, sustainability, and safety. As the demand for new and improved chemical products continues to grow, the role of chemical reaction engineering in driving innovation and progress in the chemical industry will only become more significant.。
第一章 绪论
• <四>、按传热条件分类，分为 • 1． 等温反应器，整个反应器维持恒温， 这对传热要求很高。 • 2． 绝热反应器，反应器与外界没有热量 交换，全部反应热效应使物料升温或降 温。 • 3． 非等温、非绝热反应器，与外界有热 量交换，但不等温。
10
重 油 的 催 化 裂 化 流 化 床 反 应 器
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• 2．流体流动模型 活塞流模型
（1）理想流动模型
全混流模型
（2）非理想流动模型
多级串联模型 轴向扩散模型 离析流模型等
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五.反应器设计的基本任务（或内容）
• 1．选择合适的反应器型式
• 2．确定操作方式和最佳操作条件
• 3．针对所选的反应器型式和确定的操作 条件，计算完成规定的生产任务所需的 反应器有效体积。
化学反应工程（CRE） Chemical Reaction Engineering
使用教材：陈甘棠主编《化学反应工程》第三版 重点参考教材：郭锴等编《化学反应工程》第二版 主讲教师：化学化工学院 黄孟光
1
第一章 绪论
• 化学反应工程(CRE)学是一门研究涉及化 学反应的工程问题的学科,是化学工程的 一个重要分支，是高校化工类学生必修 的核心课程。 • 绪论阐述以下几个问题：
产物的分离与精制过程
3
原料的预处理 产物的分离精制
一般属物理过程，不发生化学反应
常用的方法是：蒸发、精馏、吸收、萃取、过 滤 、沉降、结晶、吸附等。属《化工原理》和 《分离工程》课程范畴。
所以CRE的研究对象是→工业规模的反应过程 （注：化学反应工程进行的场所是在工业反应器 中，因此工业反应器自然便成为CRE研究的核 心）
一：CRE的研究对象 四：工业反应器的研究方法 二：CRE的研究内容 五：反应器分析与设计的模型 三：工业反应器的分类 六：反应器设计的基本任务
Chemical-Reaction-Engineering (1)
Chemical Reaction Engineering Chemical reaction engineering is a fascinating field that lies at the intersection of chemistry, engineering, and mathematics. It involves studying the rates and mechanisms of chemical reactions and designing optimal processes for producing desired products on an industrial scale. From the production of pharmaceuticals and petrochemicals to the development of sustainable energy sources, chemical reaction engineering plays a crucial role in advancing technology and improving our quality of life. One of the key aspects of chemical reaction engineering is understanding the kinetics of reactions, which refers to the rates at which reactants are consumed and products are formed. This knowledge is essential for designing reactors that can operate efficiently and produce high yields of desired products. By studying the mechanisms of reactions and thefactors that influence their rates, chemical engineers can optimize reaction conditions such as temperature, pressure, and concentrations to achieve the desired outcomes. In addition to kinetics, another important consideration in chemical reaction engineering is thermodynamics. Thermodynamic principles govern the feasibility and direction of chemical reactions, influencing factors such as equilibrium constant, enthalpy, and entropy. By incorporating thermodynamic considerations into reactor design, engineers can ensure that reactions proceed in the desired direction and maximize the production of desired products. Furthermore, environmental sustainability is a growing concern in chemical reaction engineering. As the global population increases and natural resources become increasingly scarce, there is a pressing need to develop processes that are more energy-efficient, less wasteful, and produce fewer harmful byproducts. Chemical engineers are tasked with designing processes that minimize environmental impact while still meeting the demands of society for essential products. Moreover, the field of chemical reaction engineering is constantly evolving, driven by advances in technology, materials science, and computational modeling. Researchers are continuously exploring new catalysts, reactor designs, and process intensification techniques to improve the efficiency and selectivity of chemical reactions. By leveraging the power of computational tools such as computational fluid dynamics and quantum chemistry simulations, engineers can gain deeperinsights into reaction mechanisms and design more innovative processes. In conclusion, chemical reaction engineering is a dynamic and interdisciplinary field that plays a crucial role in shaping the future of technology and industry. By integrating principles from chemistry, engineering, and mathematics, chemical engineers can design processes that are efficient, sustainable, and environmentally friendly. As the demands of society continue to evolve, chemical engineers will play a key role in developing innovative solutions to address global challenges and improve the quality of life for future generations.。
chemical-reaction-engineering(标准答案)
Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (19)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (22)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (26)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (32)CHAPTER 11 BASICS OF NON-IDEAL FLOW (34)CHAPTER 18 SOLID CATALYZED REACTIONS (43)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for asmall community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material(organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, whilethe effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1 Solution: )/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day mol g m L mg g L mg day day m day day m VdtdN r AA ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MWelectrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within the beds, based on the oxygen used. Waste water32,000 m 3/day Clean water 32,000 m 3/day200 mg O 2needed/liter Zero O 2 neededSolution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hr kgckgcoal kgc hr coalt N c⋅-=⨯-=⨯⨯⨯-=∆∆)/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯=)/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction? Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the rates offormation and disappearance of the three reaction components?Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C0.5 AC BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate isgiven by-r A = A06]][[1760C E A + , mol/m 3·s What are the units of the two constants?Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and with second -order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related,then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C2 A , mol/cm 3·minIf the concentration is to be expressed in mol/liter and time in hours, what would bethe value and units of the rate constant? Solution:min )()(3'⋅⨯-=⋅⨯-cm mol r hr L mol r A A 22443'300005.0106610)(minA A A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ A A A A A C C cm mol mol L C cm mol C L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(A A A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as-dtdp A = 3.66 p2 A, atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = -dt dN V A 1 = k C2 A , mol/m 3·sSolution:(a) The unit of the rate constant is ]/1[hr atm ⋅(b) dt dN V r A A 1-=- Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=- 22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃? Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln 7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150 160 230 295 370 Temperature, ℃ 13 16 22 24 28What activation energy represents this change in bustliness? Solution:RT ERT ERT Ee k e ak t cons ion concentratf let ion concentrat f e k r ---=⋅⋅=⋅='00tan )()(RE T Lnk Lnr A 1'-=∴ Suppose T x Lnr y A 1,==, so ,R E slope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230295 370A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752C T o / 13 16 22 24 283101-⨯T 3.4947 3.45843.3881 3.36533.3206-y = -5147.9 x + 15.686 Also K R E slope 9.5147-=-=, intercept 'Lnk == 15.686 ,mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter?Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion? Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C Ln AAo =, 11kt C C Ln A Ao =, 22kt C C Ln A Ao = Ao A C C 5.01=, Ao A C C 25.02= So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics.Solution: We know that for 2nd -order reaction,kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C Ao Ao Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run? Solution: In a -21order reaction: 5.0A A A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao Ao A Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , which isimpossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =- And k C C Ln oo min)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction.Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Ln k X Ln k t t A A , dissatisfied. In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao AoAo Ao Ao Ao A Ao A C C C C C C C C k C C k t t , satisfied. According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise?Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n Ln A Ao=, ''')()(tn n Ln t n n Ln A Ao A Ao = 3135213180'Ao n Ln Ln =, 28'=A n 3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction.Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hr t ,A C ,mmol/LA Ao AAo C C C C Ln-AAo C C t-1 0.84 1.0897 6.252 0.68 1.2052 6.253 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 100.0065.146910.0604C A , millimol/liter 0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025 t,hr123456789101111 0.0025 6.0065 11.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graph9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:A −−→−enzymeR, -r A = min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10 mol/liter)into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.010)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A t C C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9℃: H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20t, min C 2H 5SO 4H, mol/liter t, min C 2H 5SO 4H, mol/liter 0 0 180 4.11 41 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t ,min L mol C R /,L mol C A /,A XA Ae A Ae Ae X X X X X Ln---)12( )1(AeA X XLn -0 0 5.5 0 0 0 41 1.18 4.91 0.1073 0.2163 -0.2275 48 1.38 4.81 0.1254 0.2587 -0.2717 55 1.63 4.685 0.1482 0.3145 -0.3299 75 2.24 4.38 0.2036 0.4668 -0.4881 96 2.75 4.125 0.25 0.6165 -0.6427 127 3.31 3.845 0.3009 0.8140 -0.8456 146 3.76 3.62 0.3418 1.0089 -1.0449 162 3.81 3.595 0.3464 1.0332 -1.0697 180 4.11 3.445 0.3736 1.1937 -1.2331 194 4.31 3.345 0.3918 1.3177 -1.3591 212 4.45 3.275 0.4045 1.4150 -1.4578 267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:0067.0)11(21=-=Ao AeC X k Slope , min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line:0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter 1 2 4 6 7 9 12 -r A , mol/liter·hr0.060.10.251.02.01.00.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃508427393356283k,cm 3/mol·s0.1059 0.00310 0.000588 80.9×10-6 0.942×10-6 Find the complete rate equation for this reaction. Use units of joules, moles, cm 3, and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.5670 4 8 12 16 20 02468101214Ca-1/RaE = 60851 J/mol k0 = 105556 cm3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e-60851/R T·C A2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400 K,π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcornto be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volume goesfrom 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. I f a space velocityof 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min. Whatspace-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be certain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double the bolume.For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) is to beconverted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a given enzymeconcentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R. Findthe size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flow reactor ata flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exit stream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC2 A1 ,-r A2=kC2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC CC -So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle of thetank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offered aboutthe first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①02112121021υV kC C C A A A =-=3/k=222221A A A kC C C - …… ② Combining equation ① and ② we obtain:C A21= 0.25C A0 ;C A22=0.25C A21=0161A C So it will help, and the expected activity of the exit stream is 1/16 of the feed.6.3 An aqueous reactant stream (4 mol A/liter) passes through a mixed flow reactorfollowed by a plug flow reactor. Find the concentration at the exit of the plug flow reactor if in the mixed flow reactor C A = 1 mol/liter. The reaction is second-order with respect to A, and the volume of the plug flow unit is three times that of the mixed flow unit.Solution: Constant volume system and 2nd-order reaction:υτmm V ==110A A A r C C --=2110A A A kC C C ->k 14-=3/k …… ① 03υυτmpp V V ===9/k= -⎰-fA A C C AA r dC 1= -⎰-Af C A AdC k C 12=)11(1-Af C k …… ②Combining equation. ① and ② we obtain:C Af = 0.1 mol/liter6.4 Reactant A (A → R,C A0=26 mol/m 3) passes in steady flow through four equal-sizemixed flow reactors in series (τtotal=2 min). When steady state is achieved theconcentration of A is found to be 11, 5, 2, 1 mol/m 3 in the four units. For this reaction,what must be τplug so as to reduce C A from C A0 = 26 to C Af = 1 mol/m3?Solution:4321m m m m m τττττ=====110A A A r C C --=221A A A r C C --=332A A A r C C --=443A A A r C C --C A0=26mol/liter, C A1=11 mol/liter, C A2=5 mol/liter, C A3= 2mol/liter, C A4=1mol/liter So we abtain: 15/(-r A1) = 6/(-r A2) = 3/(-r A3) = 1/(-r A4) We postalate the reaction rate is 1 unit when C A4=1 mol/liter So we obtain。
化学反应工程第一章
二 任务(Problems to be solved,
contents to be studied.) a. 改进和强化现有的反应; (Improvement) b. 开发新的反应技术和设备； c. 反应器的放大； (Design)
(Enlargement)
d. 选择最佳操作条件,实现最佳控制； (Optimum) e. 发展和完善反应工程理论。 (Development) 反应工程学： 研究工业规模下进行化学反应的规 律或工程反应器原理的一门学科。
时间分布 R T D
非均相+工业放大因素,使问题更复杂
三个分布（C，T，t）主要与反应器结构有关， 还与流动条件有关。
工业反应的影响因素：
T，C，t，类型（相态），传质，传热， 停留时间分布，反应器结构，操作条件等。
§ 1-3 研究方法——数学模型法 （Model Method）
1. 因次分析与相似论（只适用物理过程 eg:传质与传热）, 2. 经验放大 (速度较慢，周期较长) 实验室 —— 模拟实验 —— 小生产 —— 大生产 —— 工业化 3.实验技术 测试技术与计算机的发展给数学模型法提
非 均 相
Heterogeneou s
2.基元反应与非基元反应 基元反应（elementary reactions）： 反应速率式与化学式对应的反应
A B
dCA rA kC A dt
对于基元反应：反应级数=分子数 （整数）
K
非基元反应（non-elementary reactions）： 反应速率式与化学式并无 相对应关系的反应
化学反应工程学
Chemical Reaction Engineering 化学工艺教研室 主讲教师：王承学
第二章 反应动力学基础.
rA kcA c B
A B
2.2 反应速率方程
强调：对非基元反应，须根据反应机理推导动力学方程
举例1： A → P + D 的反应机理
A A* P
A* → D
c A* c p
cA
K1
c A* K1c A
cP
rA rA* k 2 c A*
rA k 2 K1c A / cP kcA / cP
r f (c, T )
r k f ( X A ) k g( X A )
温度
浓度
化学计量数？速率控制步骤出现的次 数。
例2-2 等温条件下进行醋酸 (A)和丁醇(B) 的酯化反应： CH3COOH+C4H9OH=CH3COOC4H9+H2 O 醋酸和丁醇的初始浓度分别为 0.2332 和 1.16kmolm-3 。测得不同时间下醋酸转化 量如表所示。
2.2 反应速率方程
目前，绝大多数化学反应的机理还不清楚，因 此主要是根据实验结果来确定速率方程。方程的 一种形式是幂函数型。
i A B r kc c ... k c i A B
i 1 N
对于可逆反应
r k ci k ci
i
i 1 i 1
N
N
i
Rate Equation
可见，非基元反应的速率方程不能根据质量作用定律写出。
2.2 反应速率方程
举例2：一氧化氮氧化反应 2NO+O2→ 2NO2
反应机理（1）： 反应机理（2）：
NO+NO
（NO）2
NO + O2
NO3
（NO）2+O2→ 2NO2
化学反应工程
2 3 10
化学反应工程(Chemical
Reaction Engineering)
西南科技大学
§2-3 复合反应
二、平行反应
1)速率方程积分式 为简化讨论，假定主、副反应均为一级不可逆反应，
其微分速率方程分别为：
化学反应工程(Chemical
Reaction Engineering)
§2-3 复合反应
二、平行反应
根据选择性的定义可得：
aP aA,2 k1 SP aS aA,1 k2
2 3 26
将式(2-3-17)分别代入式(2-3-20)和(2-3-23)中可得：
CP CPO
a p k1 1 exp k1 k2 t C AO a k k A,1 1 2
2 3 1
西南科技大学
Reaction Engineering)
§2-3 复合反应
一、基本概念及术语
２）如果几个反应是依次发生的，这样的复合反应称为
串联反应，如下式所示： A B P R S
2 3 2
３）此外，还有由平行和串联反应组合在一起的复合反 应，如下两式所示：
摩尔数与反应掉的着眼反应组份Ａ的摩尔数之比值。
P
4. 得率
np npo nAO nA
2 3 5
得率，以符号Xp记之，它表示生成的目标产物P的摩尔 数与着眼反应物姓的起始摩尔数之比，即：
X P n p n po / nAO
化学反应工程(Chemical
2 3 6
rA.1 k1CA 主
化学反应工程(ChemicalReactionEngineering)
§1-1化学反应工程
第一节 化学反应工程 一、化学反应工程的研究对象
化学反应工程是化学工程学科的一个重要分支，主要包括 两个方面的内容，即反应动力学和反应器设计分析。
反应动力学--研究化学反应进行的机理和速率，以获得工 业反应器设计与操作所需的动力学知识和信息，如反应模式、 速率方程及反应活化能等。其中速率方程可表示为：
化学反应工程(Chemical Reaction Engineering) 西南科技大学
§1-1化学反应工程
例如： so2 o2 为钒 一 气固催s化o反3 应
化学反应工程(Chemical Reaction Engineering) 西南科技大学
§1-1化学反应工程
三、反应过程的举例
化学反应工程(Chemical Reaction Engineering) 西南科技大学
§1-1化学反应工程
无论对于放热过程，还是吸热过程，催化剂与 反应物气体存在温差。 就整个反应器而言，如反应器内的浓度和温度 随位置变化，需将化学反应与传递现象综合起 来考虑。
四、化学反应工程作用
对于化学产品和加工过程的开发、反应器 的设计放大起着重要的作用。运用化学反应工程 知识可以： 提高反应器的放大倍数，减少试验和开发周期。
Chapter Ⅰ绪 论 Chapter Ⅰ绪 论 Chapter 1 Introduction
§1-1 化学反应工程 §1-2 转化率、收率和选择性 §1-3 化学反应器的类型 §1-4 反应器的操作方式 §1-5 反应器的设计与基本过程 §1-6 工业反应器的放大
化学反应工程(Chemical Reaction Engineering) 西南科技大学
chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案
chemical-reaction-engineeri ng-3ed-edition作者-octave-Levenspiel-课后习题答案Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (20)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (23)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (27)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (34)CHAPTER 11 BASICS OF NON-IDEAL FLOW (36)CHAPTER 18 SOLID CATALYZED REACTIONS (45)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1Solution:)/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day molgm L mg g L mg day day m dayday m VdtdN r A A ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within theWaste Waste Clean200 mgMean residenZerobeds, based on the oxygen used.Solution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hrkgckgcoal kgc hr coal t N c ⋅-=⨯-=⨯⨯⨯-=∆∆ )/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯= )/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the ratesof formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C 0.5 A C BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance ofsubstrate is given by-r A =A06]][[1760C E A + , mol/m 3·sWhat are the units of the two constants? Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and withsecond-order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C 2 A , mol/cm 3·min If the concentration is to be expressed in mol/liter and time in hours, what wouldbe the value and units of the rate constant?Solution:min)()(3'⋅⨯-=⋅⨯-cm molr hr L mol r A A 22443'300005.0106610)(minAA A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ AA A A A C C cmmol mol L C cmmolC L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(AA A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as -dtdp A= 3.66 p 2 A , atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = - dtdN V A1 = k C2 A , mol/m 3·s Solution:(a) The unit of the rate constant is ]/1[hr atm ⋅ (b) dtdN V r AA 1-=-Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=-22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃?Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150160230295370Temperatu re, ℃13 16 22 24 28 What activation energy represents this change in bustliness? Solution:RTE RTE RTE ek eak t cons ion concentrat f let ion concentrat f ek r ---=⋅⋅=⋅='00tan )()(RET Lnk Lnr A 1'-=∴ Suppose Tx Lnr y A 1,==, so ,REslope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230 295 370 A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752CT o / 13 16 22 24 28 3101-⨯T3.4947 3.4584 3.3881 3.3653 3.3206-y = 5417.9x - 15.686R2 = 0.9712340.00330.003350.00340.003450.00351/T-L n r-y = -5147.9 x + 15.686Also K REslope 9.5147-=-=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter? Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C LnAAo=, 11kt C C LnA Ao =, 22kt C CLn A Ao = Ao A C C 5.01=, Ao A C C 25.02=So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd -order reaction, kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C AoAo Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run?Solution: In a-21order reaction: 5.0AA A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao AoA Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , whichis impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =-And k C C Lnoomin)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Lnk X Ln k t t A A , dissatisfied.In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao Ao Ao Ao Ao Ao A Ao A C C C C C C C C k C C k t t, satisfied.According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n LnAAo=, ''')()(tn n Ln t n n Ln AAo A Ao= 3135213180'Ao n Ln Ln =, 28'=An3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction. Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or-r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hrt ,AC ,mmol /L A Ao AAo C C C C Ln-AAo C C t -1 0.84 1.0897 6.25 20.681.20526.25C A , millimol /liter0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 1 2 3 4 5 6 7 8 9 10 113 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 110.00256.006511.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graphy = 0.9879x + 5.0497R 2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t /(C a o -C a )9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A −−→−enzyme R, -r A =min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.0100)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A tC C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9℃:H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20 t, minC 2H 5SO 4H , mol/li ter t, minC 2H 5SO 4H , mol/li ter1804.1141 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t,mi nLmol C R /,Lmol C A /,AXAAe AAe Ae X X X X X Ln---)12()1(AeAX X Ln -0 0 5.5 0 0 041 1.18 4.91 0.10730.2163 -0.227548 1.38 4.81 0.12540.2587 -0.271755 1.63 4.685 0.14820.3145 -0.329975 2.24 4.38 0.20360.4668 -0.488196 2.75 4.125 0.25 0.6165 -0.642712 7 3.31 3.8450.30090.8140 -0.845614 6 3.76 3.620.34181.0089 -1.044916 2 3.81 3.5950.34641.0332 -1.069718 0 4.11 3.4450.37361.1937 -1.233119 4 4.31 3.3450.39181.3177 -1.359121 2 4.45 3.2750.40451.4150 -1.4578267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:y = 0.0067x - 0.0276R 2= 0.998800.511.522.530100200300400500tL n0067.0)11(21=-=Ao AeC X k Slope ,min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line: -y = 0.0068x - 0.0156R 2 = 0.998600.511.522.530100200300400500x-L n0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter1 2 4 6 7 9 12-r A , mol/liter·hr0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002 4 68 10 12 14Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃ 508427 393 356 283k,cm 3/mol·s0.10590.003100.00058880.9×10-60.942×10-6Find the complete rate equation for this reaction. Use units of joules, moles, cm 3,and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R 2= 0.987904812160.0010.0020.0030.0041/T-L n (k )From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.567E = 60851 J/mol k 0 = 105556 cm 3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e -60851/R T ·C A 2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400K, π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-literpopcorn to be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volumegoes from 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a spacevelocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min.What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be cert ain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double thebolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC 1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) isto be converted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a givenenzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R.Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flowreactor at a flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exitstream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC 2 A1 ,-r A2=kC 2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC C C - So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle ofthe tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offeredabout the first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①。
化学反应工程Chemical+Reaction+Engineering总复习
rA m ax C A C M
平推流最优
CA
1 rA
xA
C E 0 m ax
2、竞争性和非竞争性抑制
k3CE 0CA CA CM (1 NCB0 ) 竞争性 k3 ( )CE 0C A 1 LC B 0 rA 非竞争性 1 NC B 0 CM ( ) CA 1 LC B 0 rA
比较a1与a2、b1与b2大小可以确定A或B的加料 方式及所采用的反应器型式。
Chapter 8 Potpourri of Multiple Reactions
1、串联反应 A→R → S 目的产物 选择性提高→平推流最佳(n>0)
Chapter 8 Potpourri of Multiple Reactions
u
r 2 ur u0 1 R
2 、E(t)
0 E (t ) t 2t 3
2
t t 2 t t 2
Chapter 15 The Convection Model for Laminar Flow
3、反应转化率计算
xA 0 xA (t ) E(t )dt
r
T↑，r↓
可逆放热反应：T↑
T
(2)平衡转化率： 可逆吸热反应：T↑ ，xe↑ 可逆放热反应：T↑ ，xe ↓ (3)选择性： T↑对E大的反应有利
Chapter 9 Temperature and Pressure Effects
(4)最佳温度曲线 吸热和不可逆放热反应， Topt Tmax 可逆放热反应：
Chapter 6 Design for Single Reactions 5、组合反应器
1 rA
《反应过程》李绍芬版 第三章釜式反应器
3.2.2 最优反应时间
Vr cR FR t t0 dcR cR dt t t0 AT at a0t 0 a f Vr cR
dA T 0 dt
3.3 等温 BR 的计算
反应时间的计算
复合反应
3.3.1 平行反应 A P rP k1c A 对各组分作物料衡算(恒容条件)： A Q rQ k 2c A 对A：(k1 k2 )c A 对P：
等温CSTR 的计算
复合反应－平行反应 对关键组分A有： Vr
Q0c A0 X Af (k1 k2 )c A
Q0c p k1c A ， 设c p 0 0
对目的产物P有： Vr
对副产物Q有：
Vr
Q0cQ k2c A
， 设cQ0 0
三式中有两式独立，可解Vr、XA、YP三者关系
全混流反应器图
等温CSTR 的计算
对稳态操作，有：
dni dt
M
0
i 1,2, K
单一反应
则物料衡算通式变为：
Q0 ci 0 Qci Vr
j 1
ij r j
连续釜式反应器物料衡算式
Vr
Q0 c A0 X Af rA ( X Af )
Q0 (c A0 c A ) rA
3.1 釜式反应器设计方程
物料衡算：
3.2 等温 BR 的t计算
间歇反应器Batch Reactor 间歇操作的充分搅拌槽式反
应器。
用于液相反应。 在反应过程中没有进出料。 反应器内物料充分混合，器
内各点温度浓度相同。
间歇操作，需要辅助生产时
间。
3.2.1 等温 BR 的Vr计算
化学反应工程-第一章_绪论
62
研究方法
• 化学反应工程中，有化
学过程，又有物理过程。 实验的纳果，只能还原应 用于该实验条件本身，既 没有普遍性，也不能转用 于实验条件之外的其它情 况。 •小实验的结果，不能应 用于大型装置。
兰州石油化工
63
化学反应工程的基本研究方法
研究方法：理想化和模型化。 理想化：对于实际反应器，通过理想化处理， 抽象出某几种具有代表性的典型反应器，当 然它们是理想反应器。理想的反应器主要有 三种： （1）理想间隙搅拌釜式反应器 （2）理想连续搅拌釜式反应器 （3）理想管式或塔式反应器
58
工业反应器的参数
（2）浓度：在气液反应中，浓度指液相
A
中B的浓度和溶解于液相中的A的浓度。在 同一时刻在反应器的不同部位，A的浓度 不同。 （3）温度：在反应器中心处与釜壁处温 度存在差异。
（4）时间： 在连续操作的反应器，平均 的停留时间是否等于反应时间呢? B
C
59
知识准备
处理反应工程的问题需要具备三 个方面的知识：
28
其他型式
釜式反应器 又称反应釜，搅拌反应器 特征: 反应器高度与直径相当或稍高。 釜内设有搅拌装置和挡板。 常带夹套或釜内放置蛇管，传 热以维持釜内所需温度。 适用于液相均相反应、气液反 应、液液反应、液固反应、气 液固三相反应。
29 29
30 30
各种搅拌桨的形式
31 31
搅拌釜的各种换热形式
化学反应工程 ——化学工程学科的一个 分支，以工业反应过程为主要研究对象， 以反应技术的开发、反应过程的优化和 反应器设计为主要目的的一门新兴工程 学科。 研究对象——工业反应过程，研究过程 速率及其变化规律、传递规律及其对化 学反应的影响。
chemical_reaction_engineering
c h e m i c a l_r e a c t i o n_e n g i n e er i n g(答案)(总44页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING ..................... 错误!未定义书签。
CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS ............................. 错误!未定义书签。
CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA .......................... 错误!未定义书签。
CHAPTER 4 INTRODUCTION TO REACTOR DESIGN ................................ 错误!未定义书签。
CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR ............................ 错误!未定义书签。
CHAPTER 6 DESIGN FOR SINGLE REACTIONS ................................... 错误!未定义书签。
CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR ........................... 错误!未定义书签。
CHAPTER 11 BASICS OF NON-IDEAL FLOW ..................................... 错误!未定义书签。
Chemical Reaction Engineering
Chemical Reaction EngineeringChemical Reaction Engineering is a branch of engineering that dealswith the design and optimization of chemical reactors for industrial processes. It is a field that combines the principles of chemistry, physics, and engineering to develop efficient and cost-effective processes for the production of chemicals, fuels, and other products. Chemical reactions areat the heart of many industrial processes, and the ability to control and optimize these reactions is crucial for the success of many industries.One of the main challenges in chemical reaction engineering isdesigning a reactor that can produce the desired product with high yieldand purity. This requires a deep understanding of the kinetics of the reaction, as well as the thermodynamics and transport phenomena that govern the behavior of the reactants and products. The choice of reactor type, size, and operating conditions can have a significant impact on the performance of the process, and careful consideration must be given tothese factors during the design phase.Another important aspect of chemical reaction engineering is safety. Many chemical reactions are exothermic, meaning they release heat, and if not properly controlled, they can lead to runaway reactions that can result in explosions or fires. Designing a reactor that can safely handle the reaction conditions is therefore critical, and safety must be given top priority during the design and operation of chemical processes.In addition to safety, environmental considerations are also importantin chemical reaction engineering. Many industrial processes generate waste products or emit pollutants, and minimizing the environmental impact of these processes is crucial for sustainability. Engineers must consider the entire life cycle of a process, from raw material extraction to product disposal, and strive to minimize the environmental impact at every stage.One of the key tools used in chemical reaction engineering is mathematical modeling. By developing mathematical models that describe the behavior of the reactants and products in a reactor, engineers can simulate different operating conditions and optimize the process for maximum efficiency and yield. These models can also be used to predict the behavior of the reactor under different scenarios, such as changes in feed composition or temperature, and can help engineers design a reactor that can handle a wide range of conditions.Finally, chemical reaction engineering is a field that is constantly evolving. New technologies and materials are being developed all the time, and engineers must stay up-to-date with the latest advances in their field. This requires a commitment to lifelong learning and a willingness to embrace new ideas and approaches. By staying on the cutting edge of technology, chemical reaction engineers can continue to develop innovative solutions to the world's most pressing challenges.。
化学反应工程 第一章 绪论
2001年
6、Octave Levenspiel Chemical Reaction Engineering 2002 化学工业出版社（第一版）
第一章
绪
论
Chapter 1 Preface and Overview of Chemical Reaction Engineering
典型的化工过程（ Typical Chemical Process ）
系统工程：反应过程的动态特性与反应系统的控制及
相关的最优化问题；
化学反应工程与相关学科的联系：
化学热力学：确定物系的各种物性常数，反应的可行性及可能达到
的程度。
反应动力学：反应器的选型、设计及操作方式选择都依赖于对反应
动力学特性(characteristics of reaction kinetics)的认识。 例如： ★裂解制乙烯； ★乙烯氧化制环氧乙烷； ★合成氨催化 剂的开发等。
§1.2 化学反应工程学的基本方法
化学反应工程研究方法：模型方法（Modelling Method)。
模型方法：是用数学模型来分析和研究化学反应工程问题。
数学模型（Mathematical Model)：用数学语言来表达过程中各种变量
之间的关系。数学模型的建立应从应用的角度着眼，使之适用和能用，
而不是力图将过程中的一切关系不分主次地罗列出来。 数学模型的分类： ◆机理模型(Experience Model)----从过程机理出发推导得到的； ◆经验模型(Mechanism Model)----从实验数据归纳得到的。
而且产生质的变化，这样一定模型过程，需要做不同规模的反应器试验，反复 将模型进行检验，不断修正。可以用下图表示：
Lab. Experiment
Chemical Reaction Engineering
Chemical Reaction Engineering Chemical reaction engineering is a crucial field within chemical engineering that focuses on the design and optimization of chemical reactions to produce desired products efficiently and safely. It involves understanding the kineticsand thermodynamics of reactions, as well as the design of reactors and processesto achieve specific goals. As a chemical reaction engineer, one must consider various factors such as reaction rates, selectivity, heat transfer, mass transfer, and reactor design to develop effective and sustainable processes. One of the key challenges in chemical reaction engineering is balancing the competing factors of reaction rate and selectivity. Faster reaction rates often lead to higher productivity, but they can also result in unwanted byproducts or decreased selectivity for the desired product. On the other hand, high selectivity may require slower reaction rates, which can reduce overall productivity. Finding the optimal balance between these factors is essential for designing efficient andcost-effective processes. Another important aspect of chemical reaction engineering is reactor design. Different types of reactors, such as batch reactors, continuous stirred-tank reactors, and packed bed reactors, have unique advantages and limitations. The choice of reactor type depends on various factors, including the reaction kinetics, heat and mass transfer requirements, and desired product specifications. By selecting the appropriate reactor design, chemical engineerscan maximize the efficiency and performance of the reaction process. In addition to reaction kinetics and reactor design, chemical reaction engineers must also consider the environmental and safety implications of their processes. Sustainable development and green chemistry principles are becoming increasingly important in the field, as society seeks to reduce its environmental impact and minimize waste generation. Engineers must strive to design processes that are not only efficient and economical but also environmentally friendly and safe for workers and the surrounding community. Furthermore, advancements in technology and computational tools have revolutionized the field of chemical reaction engineering. Computer simulations and modeling techniques allow engineers to predict reaction behavior, optimize process conditions, and troubleshoot potential issues before they occurin the lab or plant. These tools enable faster and more cost-effective processdevelopment, leading to improved product quality and reduced time to market. Overall, chemical reaction engineering is a dynamic and multidisciplinary field that plays a crucial role in the development of new materials, chemicals, and energy sources. By understanding the fundamental principles of reaction kinetics, thermodynamics, and transport phenomena, engineers can design innovative processes that meet the demands of a rapidly changing world. Through collaboration with scientists, researchers, and industry partners, chemical reaction engineers continue to push the boundaries of what is possible, driving progress and innovation in the field of chemical engineering.。
Chemical Reaction Engineering
Chemical Reaction Engineering Chemical Reaction Engineering is a field of study that deals with the design and optimization of chemical processes. It is a crucial aspect of the chemical industry, which involves the production of various chemicals, fuels, and materials. The goal of chemical reaction engineering is to develop efficient and sustainable processes that can produce high-quality products while minimizing the environmental impact. In this essay, we will discuss the various aspects of chemical reaction engineering, including its importance, applications, challenges, and future prospects. Chemical reaction engineering plays a vital role in the chemical industry as it provides the necessary tools and techniques to design and optimize chemical processes. It involves the study of chemical kinetics, thermodynamics, and transport phenomena to understand the behavior of chemical reactions and their interaction with the environment. The knowledge gained from chemical reaction engineering is used to develop new processes, improve existing ones, and troubleshoot problems that arise during operation. One of the most significant applications of chemical reaction engineering is in the production of fuels. The demand for energy is increasing rapidly, and the chemical industry is playing a crucial role in meeting this demand. Chemical reaction engineering is used to design and optimize processes for the production of various fuels,including gasoline, diesel, and biofuels. The efficiency and sustainability of these processes are critical, as they have a significant impact on the environment and the economy. Another important application of chemical reaction engineeringis in the production of chemicals and materials. The chemical industry produces a wide range of products, including polymers, plastics, fertilizers, and pharmaceuticals. Chemical reaction engineering is used to develop processes that can produce these products efficiently and sustainably. The design of these processes requires a deep understanding of the chemical reactions involved, the reaction kinetics, and the thermodynamics of the system. Despite its importance, chemical reaction engineering faces several challenges. One of the main challenges is the complexity of chemical reactions. Chemical reactions involve multiple steps, and the kinetics of each step can vary significantly. The design of efficient and sustainable processes requires a deep understanding of these reactions, which canbe challenging. Another challenge is the optimization of processes. Chemical processes involve multiple variables, and the optimization of these variables can be complex. The design of efficient and sustainable processes requires the optimization of these variables, which can be time-consuming and costly. The future of chemical reaction engineering is promising, as new technologies and techniques are being developed to overcome these challenges. One of the most promising technologies is the use of artificial intelligence and machine learning. These technologies can be used to analyze large amounts of data and develop models that can predict the behavior of chemical reactions. This can significantly reduce the time and cost required for process design and optimization. Another promising technology is the use of renewable energy sources. The chemical industry is one of the largest energy consumers, and the use of renewable energy sources can significantly reduce the environmental impact of chemical processes. In conclusion, chemical reaction engineering is a crucial aspect of the chemical industry, which involves the design and optimization of chemical processes. The knowledge gained from chemical reaction engineering is used to develop efficient and sustainable processes that can produce high-quality products while minimizing the environmental impact. Chemical reaction engineering has several applications, including the production of fuels, chemicals, and materials. Despite its importance, chemical reaction engineering faces several challenges, including the complexity of chemical reactions and the optimization of processes. However, the future of chemical reaction engineering is promising, as new technologies and techniques are being developed to overcome these challenges.。
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An experimental rate measurement on the decomposition of A is made with a particular catalyst (see perinent data listed below).
(a)Is it likely that film resistance to mass transfer influences the rate?
(b)Could this run have been made in the regime of strong proe diffusion?
(c)Would you expect to have temperature variations within the pellet or across the gas film?
Data
For the spherical particle:
d p=2.4 mm or L=R/3=0.4 mm=4×10-4 m cat
D e=5×10-5 m3/hr·m cat (effective mass conductivity)
k eff=1.6kJ/hr·m cat·K (effective thermal conductivity)
For the gas film surrounding the pellet (from correlations in the literature):
H=160kJ/hr·m2cat·K (heat transfer coefficient)
K g=360 m3/hr·m2cat (mass transfer coefficient)
For the reaction:
∆H r=-160kJ/mol A (exothermic)
C Ag=20 mol/ m3 (at 1 atm and 336°C)
=105 mol/hr·m3 cat
-r''',obs
A
Assume that the reaction is first order.
No.2
−4R
The catalytic reaction A−→
is run at 3.2 atm and 117o C in a plug flow reactor which contains 0.01kg of catalyst and uses a feed considering of the partially converted product of 20 liters/hr of pure unreacted A. The results
Find a rate equation to represent this reaction.
No.3 (To be continued)
Directly from this data, and without using a rate equation, find the sixe of packed bed needed to
ε=3) to 35% conversion, all at 3.2 treat 2000 mol/hr of pure A at 117o C (or C A0=0.1 mol/liter,
A。