chemicalreactionengineering答案.doc

Corresponding Solutions for Chemical Reaction Engineering CHAPTER 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 1 Chapter 1 Overview of Chemical Reaction Engineering 1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for 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 + H2O microbes A typical entering feed has a 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 water 32,000 m3/day Waste water Treatment plant Clean water 32,000 m3/day 200 mg O2 needed/liter Mean residence time =8 hr t Zero O2 needed Figure P1.1 Solution: )/(107.2)/(75.18 3132/10)02()(3131204smoldaymol daymolgLgLdayVdtNrA 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%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. Find the rate of reaction within the beds, based on the oxygen used. 2 Solution: 3801)420(mV )/(99.501240 33 hrbedolchrkgckgcoalhrcoaltNc )/(25.18322 lOtVrccO )/(120490hrbedmoldt )/(17.48/5.1 32 2 smolltVrO 3 Chapter 2 Kinetics of Homogeneous Reactions 2.1 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. 2.2 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: 52224NONrr 2.3 A 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 = 2R + S Solution: 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/m3sA0617CE What are the units of the two constants? Solution: 603AAksmolr 3/molCsllsok1)/)(/( 333 2.5 For the complex reaction with stoichiometry A + 3B 2R + S and with second-order rate expression -rA = k1AB 4 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 , + or - . Solution: RBArr213 2.6 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)()(3 colrhLmolrAA 22443 305.10610in AA CclolhLr AAACcmolLC33 10)()(2422 )(03)(r 413k 2.7 For a gas reaction at 400 K the rate is reported as - = 3.66 p2 A, atm/hrdtpA (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 -rA = - = k C2 A , mol/m3sdtNVA1 Solution: (a) The unit of the rate constant is /1hratm (b) dtNVrAA1 Because its a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to 5 22)(6.3.)(1RTCPRTdtTdtPVRr AAAAA 2)6.3(AkC So we can get that the value of 1.204085.63. T 2.9 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)92317()10/(34.8)1(321212 KmolkJTREkLnr .9712r 2.11 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 find Running speed, m/hr 150 160 230 295 370 Temperature, 13 16 22 24 28 What activation energy represents this change in bustliness? Solution: RTETRTE ekatconsiconetraflticonetrafekr 00 )()(LA1 Suppose , TxnryA, so interceptREslopeLnk 6 )/(1hmrA150 160 230 295 370Ln -3.1780 - 3.1135 -2.7506 -2.5017 -2.2752CTo/ 13 16 22 24 28310 3.4947 3.4584 3.3881 3.3653 3.3206 -y = 5417.9x - 15.686R2 = 0.97 0 1 2 3 4 0.0033 0.00335 0.0034 0.00345 0.0035 1/T -Ln r -y = -5147.9 x + 15.686 Also , intercept = 15.686 ,KREslope9.7LnkmolJmolJ/80.42)/(31458.5 7 Chapter 3 Interpretation of Batch Reactor Data 3.1 If -rA = - (dCA/dt) =0.2 mol/litersec 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 cant answer this kind of question. 3.2 Liquid 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: Because the decomposition of A is a 1st-order reaction, so we can express the rate equation as: AkCr We know that for 1st-order reaction, ,tLnAo , 1ktCAo2ktnAo , o5.01 oC5.02 So equ(1)21)4()(1212 LnkLnknLktAAo equ(2)mi5)(11CntAo So mi512tt 3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd-order reaction, ,ktCA01 So we have two equations as follow: , equ(1)min5110 ktCAoAoA 8 , equ(2)212 3)(141ktCCAooAoA So , min1532t in01t 3.4 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?2 Solution: In a order reaction: ,15.0AAkCdtr After integration, we can get: ,5.01.2o So we have two equations as follow: , equ(1)min)10(.)4(5.05.05.0.15.0 ktCCAoAooAo , equ(2)min32. kt Combining these two equations, we can get: , but this means , which is 25.01ktAo 05.2AC impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted .1AX 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, monermonerkC And Loi)34(8.01in657.kmoermoner C)( 9 3.6 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 order reaction, , dissatisfied.43.151212 LnXnktA In 2nd order reaction, , satisfied.49/12.0.)(1212 AoAoAo CCkt 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 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 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: , , ,180Aon3Aht2 , , 5 t; Aknr So we obtain , ktnLAo)()(tLtAoAo , 3152 80Ao28n 3.9 The first-order reversible liquid reaction A R , CA0 = 0.5 mol/liter, CR0=0 takes 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 obtain 10 12121021 )()( AXAt XkLnkkddA , , so we obtain eq(1) mi8sec40t3.0A eq(1).)(in2121kLk , , so we obtain eq(2)AeAecXMCkK1R21 0AoRC ,23121AecXkK eq(2)21 Combining eq(1) and eq(2), we obtain 1412 sec08.min08. k1 63957 So the rate equation is )(21AoAA CkdtCr )(sec1063.9sec08.40144 AC 3.10 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 from 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 ,min103.297.1k so min015.olLk so the rate equation is 21)in5.0(AACr 3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as follows: Aucrose products sucrae 11 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 measurements): Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -rA = where CM = Michaelis constantMECk03 If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method. Solution: Solve the question by the integral method: ,AAEoAkdtr5431 , MEoCk345AoAoLnkt451hrt, ,mmol/LACAoCAot 1 0.84 1.0897 6.25 2 0.68 1.2052 6.25 3 0.53 1.3508 6.3830 4 0.38 1.5606 6.4516 5 0.27 1.7936 6.8493 6 0.16 2.1816 7.1428 7 0.09 2.6461 7.6923 8 0.04 3.3530 8.3333 9 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 11 0.0025 6.0065 11.0276 CA, 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,hr 1 2 3 4 5 6 7 8 9 10 11 12 Suppose y= , x= AoC Ln , thus we obtain such straight line graphAoCt y = 0.9879x + 5.0497 R2 = 0.998 0 2 4 6 8 10 12 0 1 2 3 4 5 6 7Ln(Cao/Ca)/(Cao-Ca) t/(C ao-C a) , intercept=987.034EoMkSlope 049.5k So ,)/(16.4.51Lmol13 80.9.987.0hrCkEoM 3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A R, -rA = enzym min200lteroCE If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 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: 510.201AAACdCtr Rearranging and integrating, we obtain: 1025.025.10 )(51)( AoAoAt Lnd 13 min79.10)(502.1AoCLn 3.20 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: H2SO4 + (C2H5)2SO4 2C 2H5SO4H Initial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter. Find a rate equation for this reaction. Table P3.20 t, min C2H5SO4H, mol/liter t, min C2H5SO4H, 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 5.42 162 3.81 (5.80) Solution: Its a constant-volume system, so we can use XA solving the problem: i) We postulate it is a 2nd order reversible reaction system RB2 The rate equation is: 21RBAACkdtCr , , LmolCBoA/5.)(oX , AAXAR2 When , t Lleo/8.5Re So ,73.0528.Ae LmolXCAeoB /6.2)73.01(.)1( After integrating, we obtain eq (1)tCXkLnAoeAeA )1()2(1 14 The calculating result is presented in following Table. , t min LmolCR/,olA/,AXAeAXLn)12( )(AeXLn 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 410 5.42 2.79 0.4927 2.6731 -2.7254 5.8 2.6 0.5273 Draw t plot, we obtain a straight line:AeAXLn)12( y = 0.0067x - 0.0276 R2 = 0.9988 0 0.5 1 1.5 2 2.5 3 0 100 200 300 400 500t Ln 15 ,067.)1(21AoeCXkSlopmin)/(94.5)273.0(6.1 lLk When approach to equilibrium, ,BeAcCkK2R21 so min)/(0364.8.5.0794.622Re12 olLCkBA So the rate equation is in)/()10364.10794.6( 2LolCCr RBAA ii) We postulate it is a 1st order reversible reaction system, so the rate equation is RAAkdtr21 After rearranging and integrating, we obtain eq (2)tkXLnAee1)1( Draw t plot, we obtain another straight line:)1(AeXLn -y = 0.0068x - 0.0156 R2 = 0.9986 0 0.5 1 1.5 2 2.5 3 0 100 200 300 400 500 x -Ln ,68.1AeXkSlop 16 So 131 min0586.273.068. k 13Re2 i7.15CA So the rate equation is in)/()1067.10586.3( 33 LmolCr RAA We find that this reaction corresponds to both a 1st and 2nd 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 =0.5273 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/liter 1 2 4 6 7 9 12 -rA, mol/literhr 0.06 0.1 0.25 1.0 2.0 1.0 0.5 We 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 CA0 = 10 mol/liter to CAf = 2 mol/liter. Solution: By using graphical integration method, we obtain that the shaped area is 50 hr. 0 4 8 12 16 20 0 2 4 6 8 10 12 14Ca -1/Ra 3.31 The thermal decomposition of hydrogen iodide 2HI H 2 + I2 is reported by M.Bodenstein Z.phys.chem.,29,295(1899) as follows: T, 508 427 393 356 283 17 k,cm3/mol s 0.1059 0.00310 0.000588 80.910-6 0.94210-6 Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds. According to Arrhenius Law, k = k0e-E/R T transform it, - In(k) = E/R(1/T) In(k0) Drawing the figure of the relationship between k and T as follows: y = 7319.1x - 11.567 R2 = 0.9879 0 4 8 12 16 0.001 0.002 0.003 0.0041/T -Ln (k) From the figure, we get slope = E/R = 7319.1 intercept = - In(k0) = -11.567 E = 60851 J/mol k0 = 105556 cm3/mols From the unit k we obtain the thermal decomposition is second-order reaction, so the rate expression is - rA = 105556e-60851/R TCA2 18 Chapter 4 Introduction to Reactor Design 4.1 Given a gaseous feed, CA0 = 100, CB0 = 200, A +B R + S, XA = 0.8. Find XB,CA,CB. Solution: Given a gaseous feed, , , 10Ao20BoS , find , , XABC , BA 20.1)(AoXC4.208Bob16.)1( 4.2 Given a dilute aqueous feed, CA0 = CB0 =100, A +2B R + S, CA = 20. Find XA, XB, CB. Solution: Given a dilute aqueous feed, , 10BoA , , find , , SRBA22AXB Aqueous reaction system, so 0 When , 0AXV When , 1 So , 2A41AoBbC ,8.012AoCX , which is impossible.6.0812BoABabX So , 1B0oC 4.3 Given a gaseous feed, CA0 =200, CB0 =100, A +B R, CA = 50. Find XA, XB, CB. Solution: Given a gaseous feed, , ,20Ao1Bo 19 , .find , , RBA50ACAXBC ,75.21AoX , which is impossible.5.BoCb So 10BoC 4.4 Given a gaseous feed, CA0 = CB0 =100, A +2B R, CB = 20. Find XA, XB, CA. Solution: Given a gaseous feed, ， , ,10BoAR220o Find , , AXBA , 0B201V 155R , .20B 5.012.A , 84.05.1BX 421.08.AX3.721.51AAoC 4.6 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, , , , Ko1atm5010Ao20Bo , , , , find , , .RBA5KT40atm2AXB , , 136AAoBBbC5.0410T According to eq page 87, 20 81.0512.100TCXAoAA 49.8.BoaXb1308.012)2(1)(0 AAoBXTC 4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn to be operated 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. Solution: , , 30A.1uaCAo .281uaCAo%5.46301AoAX 21 Chapter 5 Ideal Reactor for a single Reactor 5.1 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 and mean residence time or holding time of fluid in the plug flow reactor. Solution: ,min1s Varying volume system, so cant be found.t 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 0A According to eq.4 on page 92, min130XAordCt Eq.13, , cant be certain.AoAoRFM.RFM. Eq.17, , so XAoRFPrdC0. in13.RBFPt 5.4 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 by A R, -rA = kC1.5 A Solution: Liquid reaction system, so 0 , AoArXCFV 5.1)()(AoAoXCkr Now we know: , , , 27.0 So we obtain 22 5.15.15.15.1 )(2)(2AoAoAo XkCXkCFV .8)7.0()(515.1A94. 5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) is to be converted to product in a plug flow reactor. The kinetics of the reaction is represented by A +B R, -rA = 200CACB minltero Find the volume of reactor needed for 99.9% conversion of A to product. Solution: Aqueous reaction system, so 0A According to page 102 eq.19, AfAf XXoAo rdrCtFV001 , ,fXAod0 min/4lteoLrdXrVAXAoAf 3.12401.9.0 5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a given enzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (CA0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given by A R , -rA = enzym litermoCAn5.01 Solution: P.F