chemical reaction engineering 3ed edition作者 octave Levenspiel 課后習(xí)題答案

1、 Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING.1CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS.3CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA.7CHAPTER 4 INTRODUCTION TO REACTOR DESIGN.19CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR.22CHAPTE

2、R 6 DESIGN FOR SINGLE REACTIONS .26CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR .32CHAPTER 11 BASICS OF NON-IDEAL FLOW.34CHAPTER 18 SOLID CATALYZED REACTIONS.431Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant f

3、or a small community (Fig.P1.1). Waste water, 32000 m3/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) +O2 CO2 + H2Omicrobes A typical entering feed has a

4、BOD (biological oxygen demand) of 200 mg O2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks. Waste water32,000 m3/dayWaste waterTreatment plant Clean water 32,000 m3/day200 mg O2needed/literMean residencetime =8 hr tZero O2 neededF

5、igure P1.1Solution:)/(1017. 2)/(75.183132/100010001)0200()(313200031320001343333smmoldaymmoldaymolgmLmggLmgdaydaymdaydaymVdtdNrAA1.2 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). Thes

6、e giants would be fed 240 tons of coal/hr (90% C, 10%H2), 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. Fin

7、d the rate of reaction within the beds, based on the oxygen used.2Solution:380010) 1420(mV)/(9000101089 . 05 . 01024033hrbedmolchrkgckgcoalkgchrcoaltNc)/(25.111900080011322hrmkmolOtNVrrccO)/(12000412000190002hrbedmoldtdO)/(17. 4800)/(105 . 113422smmolhrbedmoldtdOVrO3Chapter 2 Kinetics of Homogeneous

8、 Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we dont know whether it is an elementary reaction or not, we cant tell the index of the reaction.2.2Given the reaction 2NO2 + 1/2 O2 = N2O5 , what is the relation between the rates of

9、 formation and disappearance of the three reaction components?Solution: 522224ONONOrrr2.3A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate expression-rA = 2 C0.5 ACB What is the rate expression for this reaction if the stoichiometric equation is written as A + 2B =

10、2R + SSolution: No change. The stoichiometric equation cant effect the rate equation, so it doesnt change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate is given by-rA = , mol/m3sA061760CEA What are the units of the two constants?Solution: 603AACEAksmmolr3

11、/6mmolCAsmmolmmolmmolsmmolk1)/)(/(/33332.5 For the complex reaction with stoichiometry A + 3B 2R + S and with second-order rate expression -rA = k1AB4 are the reaction rates related as follows: rA= rB= rR? If the rates are not so related, then how are they related? Please account for the sings , + o

12、r - .Solution: RBArrr21312.6 A certain reaction has a rate given by -rA = 0.005 C2 A , mol/cm3min If the concentration is to be expressed in mol/liter and time in hours, what would be the value and units of the rate constant?Solution:min)()(3cmmolrhrLmolrAA22443300005. 0106610)(minAAAAACCrrcmmolmolh

13、rLrAAAAACCcmmolmolLCcmmolCLmolC33310)()(24232103)10(300300)(AAAACCCr4103k2.7 For a gas reaction at 400 K the rate is reported as - = 3.66 p2 A, atm/hrdtdpA (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 is expressed as -

14、rA = - = k C2 A , mol/m3sdtdNVA1Solution:(a) The unit of the rate constant is /1 hratm(b)dtdNVrAA1Because its a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to 522)(66. 366. 3)(1RTCRTPRTdtdPRTdtdPVRTVrAAAAA22)66. 3(AAkCCRT So we can ge

15、t that the value of 1 .12040008205. 066. 366. 3RTk2.9 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(3211212KKKmolkJmolkJTTREkkLnrrLn7 .197012rr2.11 In the mid-nineteenth

16、 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 find Running speed, m/hr150160230295370Temperature, 1316222428 What activation energy represen

17、ts this change in bustliness?Solution:RTERTERTEekeaktconsionconcentratfletionconcentratfekr 00tan)()(RETLnkLnrA1Suppose , TxLnryA1,so intercept,REslopeLnk6)/(1hmrA150160230295370ALnr-3.1780-3.1135-2.7506-2.5017-2.2752CTo/13162224283101T3.49473.45843.38813.36533.3206-y = 5417.9x - 15.686R2 = 0.970123

18、40.00330.003350.00340.003450.00351/T-Ln r-y = -5147.9 x + 15.686Also , intercept = 15.686 ,KREslope9 .5147LnkmolkJKmolJKE/80.42)/(3145. 89 .51477Chapter 3 Interpretation of Batch Reactor Data3.1 If -rA = - (dCA/dt) =0.2 mol/litersec when CA = 1 mol/liter, what is the rate of reaction when CA = 10 mo

19、l/liter?Note: the order of reaction is not known.Solution: Information is not enough, so we cant answer this kind of question.3.2Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A is converted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution

20、: Because the decomposition of A is a 1st-order reaction, so we can express the rate equation as:AAkCr We know that for 1st-order reaction, ,ktCCLnAAo, 11ktCCLnAAo22ktCCLnAAo, AoACC5 . 01AoACC25. 02So equ(1)21)24(1)(11212LnkLnLnkCCLnCCLnkttAAoAAo equ(2)min521)(111LnkCCLnktAAoSo min5112ttt3.3Repeat t

21、he previous problem for second-order kinetics.Solution: We know that for 2nd-order reaction, ,ktCCAA011So we have two equations as follow:, equ(1)min511211101kktCCCCCAoAoAoAA8, equ(2)2123)1(31411ktktCCCCCAoAoAoAoASo , min15312 ttmin1012tt3.4 A 10-minute experimental run shows that 75% of liquid reac

22、tant is converted to product by a -order rate. What would be the fraction converted in a half-hour run?21Solution:In a order reaction: ,215 . 0AAAkCdtdCrAfter integration, we can get: ,5 . 015 . 02AAoCCktSo we have two equations as follow:, equ(1)min)10(5 . 0)41(15 . 05 . 05 . 05 . 015 . 0kktCCCCCAo

23、AoAoAAo, equ(2)min)30(25 . 025 . 0kktCCAAoCombining these two equations, we can get:, but this means , which is 25 . 05 . 1ktCAo05 . 02ACimpossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted .1AX3.5 In a hmogeneous isothermal liquid poly

24、merization, 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

25、 reaction is first-order,monomermonomerkCrAnd kCCLnoomin)34(8 . 01min00657. 0kmonomermonomerCr)min00657. 0(193.6 After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction.Solution:In 1st ord

26、er reaction, , dissatisfied.43. 1511111111212LnLnXLnkXLnkttAAIn 2nd order reaction, , satisfied.49/4/912 . 0111 . 01)11(1)11(11212AoAoAoAoAoAoAoAAoACCCCCCCCkCCkttAccording to the information, the reaction is a 2nd-order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evening

27、s are all alikeinto the joint with his weeks salary of 180, steady gambling at “2-up” for two hours, then home to his family leaving 45 behind. Snake Eyess betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictableat a rate proport

28、ional 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: , , ,180Aon13Anht2, , 135Anht3;AAknrSo we obtain , ktnnLnAAo)()(tnnLntnnLnAAoAAo, 3135213180AonLnLn28An3.9 The first-order rever

29、sible liquid reaction A R , CA0 = 0.5 mol/liter, CR0=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 react

30、ion, according to page56 eq. 53b, we obtain 10121112102110)(1)(AXAAtXkkkkLnkkXkkkdXdttA, , so we obtain eq(1) min8sec480t33. 0AX eq(1)33. 0)(1min8sec480211121kkkkLnkk, , so we obtain eq(2)AeAeAecXXMCCkkK1Re210AoRoCCM,232132121AeAecXXkkK eq(2)212kk Combining eq(1) and eq(2), we obtain 1412sec108 . 4m

31、in02888. 0k14121sec1063. 9min05776. 02 kkSo the rate equation is )(21AAoAAACCkCkdtdCr )(sec1063. 9sec108 . 401414AAACCC3.10 Aqueous A reacts to form R (AR) and in the first minute in a batch reactor its concentration drops from CA0 = 2.03 mol/liter to CAf = 1.97 mol/liter. Find the rate equation fro

32、m the reaction if the kinetics are second-order with respect to A.Solution: Its a irreversible second-order reaction system, according to page44 eq 12, we obtain ,min103. 2197. 111kso min015. 01molLkso the rate equation is 21)min015. 0(AACr3.15 At room temperature sucrose is hydrolyzed by the cataly

33、tic action of the enzyme sucrase as follows: Aucrose products sucrase11 Starting with a sucrose concentration CA0 = 1.0 millimol/liter and an enzyme concentration CE0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation mea

34、surements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -rA = where CM = Michaelis constantMAEACCCCk03 If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method. Solution: Solve the question by the integra

35、l method:,AAMAAEoAACkCkCCCCkdtdCr5431, MEoCCkk34MCk15AAoAAoAAoCCCCLnkkkCCt4451hrt,mmol/LACAAoAAoCCCCLnAAoCCt10.841.08976.2520.681.20526.2530.531.35086.383040.381.56066.451650.271.79366.849360.162.18167.142870.092.64617.692380.043.35308.333390.0184.09109.1650100.0065.146910.0604110.00256.006511.0276C

36、A, millimol/liter0.840.680.530.380.270.160.090.040.0180.0060.0025 t,hr123456789101112Suppose y=, x=AAoAAoCCCCLn, thus we obtain such straight line graphAAoCCty = 0.9879x + 5.0497R2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t/(Cao-Ca), intercept=9879. 0134EoMCkCkSlope0497. 545kkSo , )/(1956. 00497.

37、59879. 015LmmolkCM14380.1901. 09879. 01956. 0hrCCkkEoM3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A R, -rA = enzymemin22000litermolCCCAEA If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 10 mol/liter) into a batch rector and let the reaction pr

38、oceed, 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. 020021AAAAACCCdCdtrRearranging and integrating, we obtain:10025. 0025. 0100)(510)510(AAoAAoAAtCCCCLndCCdtt13min79.10

39、9)(5025. 01010AAoCCLn3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data inTable P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9: H2SO4 + (C2H5)2SO4 2C2H5SO4HInitial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter

40、. Find a rate equation for this reaction.Table P3.20t, minC2H5SO4H, mol/litert, minC2H5SO4H, mol/liter001804.11411.181944.31481.382124.45551.632674.86752.243185.15962.753685.321273.313795.351463.764105.421623.81(5.80)Solution: Its a constant-volume system, so we can use XA solving the problem:i) We

41、postulate it is a 2nd order reversible reaction system RBA2The rate equation is: 221RBAAACkCCkdtdCr, , LmolCCBoAo/5 . 5)1 (AAoAXCC, AAAoBoBCXCCCAAoRXCC2When , tLmolXCCAeAo/8 . 52ReSo ,5273. 05 . 528 . 5AeX LmolXCCCAeAoBeAe/6 . 2)5273. 01 (5 . 5)1 (After integrating, we obtain eq (1)tCXkXXXXXLnAoAeAA

42、eAAeAe) 11(2) 12(114The calculating result is presented in following Table., tminLmolCR/,LmolCA/,AXAAeAAeAeXXXXXLn) 12()1 (AeAXXLn005.5000411.184.910.10730.2163-0.2275481.384.810.12540.2587-0.2717551.634.6850.14820.3145-0.3299752.244.380.20360.4668-0.4881962.754.1250.250.6165-0.64271273.313.8450.300

43、90.8140-0.84561463.763.620.34181.0089-1.04491623.813.5950.34641.0332-1.06971804.113.4450.37361.1937-1.23311944.313.3450.39181.3177-1.35912124.453.2750.40451.4150-1.45782674.863.070.44181.7730-1.81973185.152.9250.46822.1390-2.18863685.322.840.48362.4405-2.49183795.352.8250.48642.5047-2.55644105.422.7

44、90.49272.6731-2.72545.82.60.5273Draw t plot, we obtain a straight line:AAeAAeAeXXXXXLn) 12(y = 0.0067x - 0.0276R2 = 0.998800.511.522.530100200300400500tLn15,0067. 0) 11(21AoAeCXkSlopemin)/(10794. 65 . 5) 15273. 01(20067. 041molLkWhen approach to equilibrium, ,BeAecCCCkkK2Re21so min)/(10364. 18 . 56

45、. 210794. 642242Re12molLCCCkkBeAeSo the rate equation ismin)/()10364. 110794. 6(244LmolCCCrRBAAii) We postulate it is a 1st order reversible reaction system, so the rate equation is RAAACkCkdtdCr21After rearranging and integrating, we obtain eq (2)tkXXXLnAeAeA11)1 (Draw t plot, we obtain another str

46、aight line:)1 (AeAXXLn-y = 0.0068x - 0.0156R2 = 0.998600.511.522.530100200300400500 x-Ln,0068. 01AeXkSlope16So 131min10586. 35273. 00068. 0k133Re12min10607. 18 . 56 . 210586. 3CCkkAeSo the rate equation ismin)/()10607. 110586. 3(33LmolCCrRAAWe find that this reaction corresponds to both a 1st and 2n

47、d order reversible reaction system, by comparing eq.(1) and eq.(2), especially when XAe =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 XAe =0.5.(The data that we use just have XAe =0

48、.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 CA alone determines this rate:CA,mol/liter12467912-rA, mol/literhr0.060.10.251.02.01.00.5We plan to run this re

49、action 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 CA0 = 10 mol/liter to CAf = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002468101

50、214Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI H2 + I2 is reported by M.Bodenstein Z.phys.chem.,29,295(1899) as follows:T,50842739335628317k,cm3/mols0.10590.003100.00058880.910-60.94210-6Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds.

51、According to Arrhenius Law,k = k0e-E/R Ttransform it,- In(k) = E/R(1/T) In(k0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R2 = 0.987904812160.0010.0020.0030.0041/T-Ln(k)From the figure, we get slope = E/R = 7319.1 intercept = - In(k0) = -11.567E = 60851 J/mo

52、l k0 = 105556 cm3/molsFrom the unit k we obtain the thermal decomposition is second-order reaction, so the rate expression is- rA = 105556e-60851/R TCA218Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, CA0 = 100, CB0 = 200, A +B R + S, XA = 0.8. Find XB,CA,CB.Solution: Given a gase

53、ous feed, , , 100AoC200BoCSRBA, find , , 0AXBXACBC, 0BA202 . 0100)1 (AAoAXCC4 . 02008 . 01001BoAAoBCXbCX1206 . 0200)1 (BBoBXCC4.2 Given a dilute aqueous feed, CA0 = CB0 =100, A +2B R + S, CA = 20. Find XA, XB, CB.Solution: Given a dilute aqueous feed, , 100BoAoCC, , find , , SRBA 220ACAXBXBCAqueous

54、reaction system, so 0BAWhen , 0AX200VWhen , 1AX100VSo , 21A41AoBoABbCC,8 . 01002011AoAACCX, which is impossible.16 . 11008 . 010012BoAAoBCXCabXSo , 1BX100BoBCC4.3 Given a gaseous feed, CA0 =200, CB0 =100, A +B R, CA = 50. Find XA, XB, CB.Solution: Given a gaseous feed, , ,200AoC100BoC19, .find , , R

55、BA50ACAXBXBC,75. 02005011AoAACCX, which is impossible.15 . 1BoAAoBCXbCXSo 100BoBCC4.4 Given a gaseous feed, CA0 = CB0 =100, A +2B R, CB = 20. Find XA, XB, CA.Solution: Given a gaseous feed, ，, ,100BoAoCCRBA 220BoCFind , , AXBXAC, 0BX200100100BAV , 1BX15010050RAV, 25. 0200200150B5 . 01002110025. 0A,

56、842. 02025. 010020100BX421. 0100842. 010021AX34.73421. 05 . 01421. 0110011AAAAoAXXCC4.6 Given a gaseous feed, T0 =1000 K, 05atm, CA0=100, CB0=200, A +B5R,T =400 K, =4atm, CA =20. Find XA, XB, CB.Solution: Given a gaseous feed, , , , KTo1000atm50100AoC200BoC, , , , find , , .RBA5KT400atm420ACAXBXBC,

57、, 1300300600A2AoBoABbCCa5 . 041000540000TTAccording to eq page 87, 20818. 05 . 010020115 . 0100201110000TTCCTTCCXAoAAAoAA409. 0200818. 0100BoAAoBaCXbCX130818. 011200)818. 0100200(1)(00AAAoAAoBoBXCTTXabCCC4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn to be ope

58、rated in steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/min of mixed exit stream. Independent tests show that when raw corn pops its volume goes from 1 to 31. With this information determine what fraction of raw corn is popped in the unit.Soluti

59、on: , , 301131A.1 uaCAo.281281uaCCAoA%5 .462813012811AAAoAAoACCCCX21Chapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A R + 2S with unknown kinetics. If a space velocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time

60、and mean residence time or holding time of fluid in the plug flow reactor.Solution: ,min11sVarying volume system, so cant be found.t5.2 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 fl