chemical-reaction-engineering-3ed-edition作者-octave

1、chemical-reaction-engineering-3ed-edition 作 者- octave-Levenspiel一课后习 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 REACTIONS1CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA5CHAPTER 4 INTRODUCT

2、ION TO REACTOR DESIGN18CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR21CHAPTER 6 DESIGN FOR SINGLE REACTIONS25CHAPTER 1。CHOOSING THE RIGHT KIND OF REACTOR32CHAPTER 11 BASICS OF NONJDEAL FLOW34CHAPTER 18 SOLID CATALYZED REACTIONS43Chapter 1 Overview of Chemical Reaction EngineeringLI Municipal waste wa

3、ter treatment plant Consider a municipal water treatment plant for a small community (Fig.PLl), Waste water, 32000 m3/day; flows through the treatment plant with a mean residence time of 8 hi; air is bubbled through the tanks, and microbes in the tank attack and break down the organic material(organ

4、ic waste) +O2CO2 + H2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg Oz/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Waste k WasteClean200 mgMeanZeroFigure PL1Solution:.-32000 X 1 ( n1' X dav) x (200

5、- 0). 1 吁 132g=1 3 day .L 1000mg mo/32000 x 1 dav!"他day 3 '3= S.15mol/(nr day) = 2Al x 10-4 wo/(nv - j)1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using tluiding bed combustors may be built some day (see Fig.PL2). These giants would be

6、fed 240 tons of coal/hr (90% C, 10%), 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 reacti

7、on within the11.1 beds, based on the oxygen used.Fluidized bedlOmxAmxlm240 torvhr, 90% C, 10% H2Fieure P1.2Solution:V =(20x4x1)x10 = 800/m3也=-240 x 103 x 0.5 x 0.9/hrA"' = 108 x IO' 极'=9000 mole /(bed - hr) kgcoalhrix800°°° = 1 1 .25kinolO21 (nr - hr)= 9000x 1 + J 200

8、。= 12000mol/(bed - hr) (It4T dO、_ 1.5x10" mol /(bed - hr)800=4. i7mol /(nr - s)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 elementaiy reaction or not, we

9、can't tell the index of the reaction.2.2 Given the reaction 2NO2 + 1/2 O2 = N2O5, what is the relation between the rates of formation and disappearance of the three reaction components?Solution: - rW2 = -4期=2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate e

10、xpression-fa = 2 C0.5 ACbWhat 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

11、 of disappearance of substrate is given by-”一What are the units of the two constants?Solution:mol _kAE. 版工一 6 + C"6 = CA = mol / Wmolnrmol I nt_ 1mol / m3 )(mol / m3) s2.5 For the complex reaction with stoichiometry A + 3B> 2R + S and with1second-orderrate expression-fa = kiAB are the reacti

12、on rates related as follows: fa= fb= fr? If the rates are not so related, then how are they related? Please account for the sings, + or Solution* 一 / = /r2.6 A certain reaction has a rate given by-fa = 0.005 C2 A, mol/cm3-ininIf the concentration is to be expressed in mol/liter and time in hours, wh

13、at would be the value and units of the rate constant?Solution:,、 mol /、 molE ”丁1 = F)xj- L hrcm -min-rA = L (r4) = 104 x6- r4 = 6x 104 x0.005C； = 300C；mol cm -min、mol 、mol(g)X = = (g)X Lcm.C=工7誓八=10七八mol cm.(_q) = 300C； =3OOx(1O-3C4)2 =3x10”：:.k =3x102.7 For a gas reaction at 400 K the rate is repor

14、ted as-= 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 is expressed as-fa =-=k C2 A, mol/m3-sV dtSolution: The unit of the rate constant is Iatm-hrV dtBecause it's a gas reaction occuring at the Hned

15、 terperatuse, so V=constant, andT=constant so the equation can be reduced toL生+华福P：考CM= (3.66RT)C； =ySo we can get that the value ofk = 3.66RT = 3.66 x 0.08205x 400 =120.12.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kj/mol.How much faster the decomposition at 650 than

16、at 500eC?Solution:r, 匕 _ E T 1 _300。/5厂守示(工一父=8.3140/(10%。/ . K)(173K1923K)=7.586生=1970.72.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot clays but were rather sluggish on cool clays. Checking his resul

17、ts with Oregon ants, I findRunning 15 16 23 29 37 speed, 00050 m/hrTemperatu 13 16 22 24 28 re, What activation energy represents this change in bustliness? Solution:E£r = koe R1 f (concentration)let f (concentration) = constant - -akoe KT = k e.1 E：.Lnr. = LnkT RSuppose y = LnrA = y ,Eso slope

18、 =, Intercept = Lnk小（”小）150 160 230 295 3705 rA-3.1 -3.1 -2.7 -2.5 -2.2780135 506 017 752T/T1316 22 2428947584 881 653 206o 1«0.00330.003350.00340.003450.00351/T-y = -5147.9 x + 15.686E.Also slope = - = -5147.9/f, intercept = Lnk = 15.686 ,E = 一 5147.9K x 8.3145J /（"?，/ K） = 42.80k J / mHC

19、hapter 3 Interpretation of Batch Reactor Data3.1 If -fa = - (clCA/dt) =0.2 mol/llter-sec when Ca= 1 mol/liter, what is the rate of reaction when Ca= 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 se

20、domposes 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: Because the decomposition of A is a ls,-order reaction, so we can express the rate equation as:CWe know that for ls,-order reaction, Ln= k

21、t,cc5L = k八,Ln3 = kf, C 川C.n-CA = 0.5g。， CA2 = 0 25equ(l)equ(2)一 5 幺)=-(Lji4 - bi2) = -L/i2Cu kk= Zv?(-) = Lii2 = 5 min k Cu kSo t1=t. = 5 nin33 Repeat the previous problem for second-order kinetics.Solution: We know that for 2nd-order reaction9 一=kt. g g。So we have two equations as follow:-=-=一=ktx

22、 = k5 min , equ(l)c川 c.lo g。 Ci。= - - - = 3() = 3kti = kt., equ(2)az心 Clo g, So t2 = 30 = 15 min , r2 -/)= 10 min3.4 A 10-minute experimental run shows that 75% of liquid reactant Is converted to product by a ； -order rate. What would be the fraction converted in a half-hour run? Solution:Ina - orde

23、r reaction: -r4 = - = kC5 924 dt 八ktAfter integration9 we can get: = C2So we have two equations as follow:"一；=4(*：；)=。"党=M = k(l。min), equ(l)琛_4=%(30min),equ(2)Combining these two equations, we can get: 1 .5C：； = kt?, but this means C% 。, which is impossible, so we can conclude that less t

24、han half hours, all the reactant is consumed up. So the fraction converted X A = 1.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 th

25、e monomer?Solution: The rate of reactant is independent of the initial concentration of monomers9 so we know the order of reaction is first-ordei;= kCmonotner monomercAnd Ln =(34 niin) k0.8Q% =0.00657 min 7=(0.00657 min t )Cmolumer3.6 After 8 minutes in a batch reactor, reactant (Cao = 1 mol/llter)

26、is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction.Solution:-Ln kLn5= 1.43, dissatisfied.In 1st order reaction, 2=k IIn 2nd order reaction,=当210According to the information, the reaction is a 2nd-or(ler reaction.3.7 nake-Eyes Magoo is a man of habi

27、t. For instance, his Friday evenings are all alikeinto the joint with his week's salaiy of $ 180, steady gambling at “2up” 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

28、his losses are also predictableat 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:nAo = 180，八=13, t = 2h,nA =135 , V = 3hy - rAaknALn(-)So we obtainbi- = kt, 

29、9;=“Att3.9 The first-order reversible liquid reactionA R , Cao = 0.5 mol/liter Cro=Otakes 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 sys

30、tem,Is* order reversible reaction, according to page56 eq. 53b, we obtaini= 1% =四=Ln&J。 J。匕一(自+七/八+k2 k(ki+k?)XA1t = 480sec = 8min, XA = 0.33, so we obtain eq(l)1女480 sec = 8 minLn! eq(l )kx +k2 自一(自+七)。33K( = = , M = 0, so we obtain eq(2)七以 i X. 以，2K 8 = J = 3 = 2,ki IF i 二3k = 2k2 eq(2)Combini

31、ng eq(l) and eq(2), we obtaink, = 0.02888 nin t = 4.8 x IO_4 sec-1 一k、= 2k、= 0.05776 min " = 9.63 x 10-4 sec-1dC、So the rate equation is -rA = -= k1CA -k2(CAo -CA) dt= 4.8x10 sec-1 CA - 9.63 xlO-4 sec-1 (C10 一 CA)3.10 Aqueous A reacts to form R (A>R) and in the first niinute in a batch react

32、or its concentration drops from Cao = 2.03 mol/liter to Cxr = 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 obtain1 1 . t .=k. - 1 min, 1.97 2.0

33、31so k =0.015mol - minso the rate equation is - rA = (0.015 min 一 )C； 3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrosexitcraye* productsStarting with a sucrose concentration Cao = 1.0 inilllmol/llter and an enzyme concentrationCeo= 0.01 m

34、illimol/Iitei; the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):0.0.0.0.0.0.0.0.0.0.0.0010002/litert,hr 123456789 10 11Determine whether these data can be reasonably fitted by a knietic equation of theMichaelis-Menten type, ork

35、 c c.= 2- where Cm = Michaelis constantCi + cMIf the fit is reasonable, evaluate the constants ks and Cm. Solve by the integral method.Solution: Solve the question by the integral method:"Ct _ k4cA出 Ga + Gw 1 + k5ccA ,mmol/L0.841.08976.250.681.20526.25345678910110.530.380.270.160.090.040.0180.0

36、060.00251.35081.56061.79362.18162.64613.35304.09105.14696.00656.38306.45166.84937.14287.69238.33339.165010.060411.0276tCASuppose y=, x=-, thus we obtain such straight line graphCAo - CA C4, 一 CA01234567LiUCao/Ca)/( Cao-Ca)CkSlope = = = 0.9879 , intercept=上=5.0497*4后 3 c1 _ 0.9879一 % - 5.Q497=0.1956

37、(mmol / L),k _kCv _°1956Cl t 0.9879 xO.Ol= 19.8O/7r-1123.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:A Siwr, 第4= 2QQgC£o2 + Ca liter minIf we introduce enzyme (Ceo = 0.001 mol/liter) and reactant (Cao = 10 mol/llter) into a batch rector and let the reacti

38、on 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:1_ dt _2 + Ca _1015rA dCA 200 x 0.001 CA CA -Rearranging and integrating, we obtain:rqlO3 L JJ 0.025=1 OLn 蓝 + -八)=109.7

39、9 min3.20 M.Hellin and J.C. Jungers, Bulk soc chim. France, 386(1957), present the data inTable P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9 *C:H2SO4 + (CzHs)2sO4 - 2c2H5so4HInitial concentrations of H2SO4 and (C2HS)2s04 are each 5.5 mol/liter. Find a rate e

40、quation for this reaction.Table P3.20C2H5C2H5SO4HSO4Ht,t,min ' min ' mol/limol/literter1804.1143411.181944.31481.382124.45551.632674.86752.24962.753185.153685.321273.313795.351463.764105.421623.81 oo (5.80)Solution: It's a constanbvoluine system, so we can use Xa solving the problem: i)

41、We postulate it is a 2nd order reversible reaction system A + B 0 2RThe rate equation is: 一 rA =dt= kCACHC*=CB°=55mol/L, CA=CAo(-XA),= Ci，g=2C"X八When t =oo, CRc = 2CAoXAe = 5.8mol/L5.8So XLy7n 0.5273, 2x5.5CAe = CHe = CAo(- = 5.5x(1 -0.5273) = 2.6molILAfter integrating, we obtain5 X”2X _1)

42、X,= 2K故_心 eq(l)The calculating result is presented in following Table.：r in r i / r yX%(2乂. 一 1/八XAHl! CrjuoI/L C 2moi ILXA 5A Ae A AA Aen005.5411.184.910 0.10 73 0.12 54 0.14 82 0.20 36 0.25 0.30 09 0.34 18 0.34 64 0.37 36 0.39 18 0.40 45481.384.81551.634.685752.244.38962.754.12512 3.313.8457143.76

43、3.626163.813.5952184.113.4450194.313.3454214.453.2752000.2163-0.22750.2587-0.27170.3145-0.32990.4668-0.48810.6165-0.64270.8140-0.84561.0089-1.04491.0332-1.06971.1937-1.23311.3177-1.35911.4150-1.45782674.863183685.155.323795.35415.42 00.443.07180.462.925820.482.84360.482.825640.492.79270.522.6731.773

44、02.13902.44052.50472.6731-1.8197-2.1886-2.4918-2.5564-2.7254Drawt plot, we obtain a straight line:1Slope = 2k(-1)C =0.0067,0.0067=6.794 x 10-4L/(/7/t?/-min)2(0.5273-l)x5.5When approach to equilibrium, K(=-, k? C A/C Beso k、= k'C，= 6.794*0 二2.忆=j 364x 10-4L/(,/. min) 豆5"So the rate equation

45、is-rA = (6.794 x 10CACB -1.364 x 107C；)mol /(L - nin)ii) We postulate it is a 1st order reversible reaction system, so the rate equation isAfter rearranging and integrating, we obtain沙(1-意)= -%，eq DrawAe X Ae) t plot, we obtain another straight line:Slope = -0.0068,So k = -0.0068 x 0.5273 = -3.586 x

46、 10-3 nin皿=-2.6 = _1 607 xl0-3min-. Ge5.8So the rate equation is一 rA = (-3.586 x 10-3CA + 1.607 xlO-3CR )mol /(L - min)We find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(l) and eq(2), especially when X.e =0.5 , the two equations are identi

47、caL 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 Xac =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueou

48、s reactant A is converted to product at the following rates, and Ca alone determines this rate:CA,mol/liter 12467912-rA,0.060.10.251.02.01.00.5mol/litere hrWe 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 lo

49、wer the concentration of A from Cao = 10 mol/liter to CAf = 2 mol/liter.Solution: By using graphical integration method9 we obtain that the shaped area is 50 hr.3.31 The thermal decomposition of hydrogen iodide2Hl -> H2 + 12is reported by M.Bodenstein Z.phys.chem.,29,295( 1899) as follows:T/C 50

50、427 393 356 283k,cm3 0.1/mob 050.0 0.00 80.9 0.942031 058 xlO- xlO6Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds.According to Arrhenius9 Law,k = koe-E/RT transform it,-In(k) = E/R-(l/T) In(ko)Drawing the figure of the relationship between k and T as

51、follows:1612y = 7319.1x- 11.567R2 = 0.98798win4 /o0 ''10.0010.0020.0030.0041/TFrom the figure, we getslope = E/R = 7319.1 intercept = - In(ko) = -11.567E = 60851 J/mol ko = 105556 cm3/mobsFrom the unit k we obtain the thermal decomposition is second-order reaction, so the rate expression is-

52、fa= 105556e-608S,/RT-CA2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, Cao = 100, Cbo = 200, A +B> R + S, Xa = 08 Find Xb,Ca,Cb.Solution: Given a gaseous feed, CAo = 100 , C刖=200 , A + B - R + SX/|=0,fliid Xb , g, Cbj =% = 0, CA =C4o(l-X4) = 100x0.2 = 20GjxlOOxO.8Cb0 200Cw = CH

53、 (1 -) = 200 x 0.6 = 120nttO 'n z4.2 Given a dilute aqueous feed, Cao = Cbo =100, A +2B> R + S, Ca = 20. Find Xa, Xb, Cb.Solution: Given a dilute aqueous feed, CA(, = CHo = 100,4 + 28 -R + S, g =20,findCflAqueous reaction system, so sA =幺 =0When X、=0, V = 200When X八=1, V = 100So £ . = ,

54、£( = 0.8,2 bCAo 4Xa=1 Cb.2 100x0.8-x1100= 1.6 > which is impossible.So XH=, CH = C =1004.3 Given a gaseous feed, Cao =200, Cbo =100, A+Bt R, Ca = 50. Find Xa, Xb, Cb.Solution: Given a gaseous feed, CAo = 200 , CBo = 100,A + 8 -R, CA = 50.And X, CHx31 £l = T200=0.75 ,X“A"X& = 1.5 > 1, which is impossible.CboSo C B = C BO = 1004.4 Given a gaseous feed, Cao = Cbo =100, A +2B&