chemicalreactionengineering答案.doc

Corresponding Solutions for Chemical Reaction Engineering CHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING1 CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS.3 CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA7 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 REACTIONS43 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 + H2Omicrobes 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: )/(1017 . 2 )/(75.18 3 1 32 / 1000 1000 1 )0200()( 3 1 32000 3 1 32000 1 343 3 3 3 smmoldaymmol day mol g m L mg g L mg day day m day day m Vdt dN 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%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: 3 80010) 1420(mV )/(9000101089 . 05 . 010240 33 hrbedmolc hr kgc kgcoal kgc hr coal t Nc )/(25.11 1 9000 800 11 3 2 2 hrmkmolO t N V rr c cO )/(12000 4 12000 19000 2 hrbedmol dt dO )/(17 . 4 800 )/(105 . 11 3 4 2 2 smmol hrbedmol dt dO V rO 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. 2.2Given the reaction 2NO2 + 1/2 O2 = N2O5 , what is the relation between the rates of formation and disappearance of the three reaction components? Solution: 5222 24 ONONO rrr 2.3A 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 = 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/m3s A 0 6 1760 C EA What are the units of the two constants? Solution: 6 0 3 A A C EAk sm mol r 3 /6mmolCA smmolmmol mmol sm mol k 1 )/)(/( / 33 3 3 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: RBA rrr 2 1 3 1 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 cm mol r hrL mol r AA 2244 3 300005 . 0 106610)( min AAAAA CCrr cm mol mol hrL r AAA AA CC cm mol mol L C cm mol C L mol C 3 3 3 10 )()( 2 4232 103)10(300300)( AAAA CCCr 4 103 k 2.7 For a gas reaction at 400 K the rate is reported as - = 3.66 p2 A, atm/hr dt dpA (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/m3s dt dN V A 1 Solution: (a) The unit of the rate constant is /1 hratm (b) dt dN V r A A 1 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 )( 66 . 3 66 . 3 )( 1 RTC RT P RTdt dP RTdt dP VRT V r AA AA A 22 )66 . 3 ( AA kCCRT So we can get that the value of 1 . 12040008205 . 0 66 . 3 66 . 3 RTk 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 ) 923 1 173 1 ( )10/(314 . 8 /300 ) 11 ( 3 211 2 1 2 KKKmolkJ molkJ TTR E k k Ln r r Ln 7 . 1970 1 2 r r 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/hr150160230295370 Temperature, 1316222428 What activation energy represents this change in bustliness? Solution: RT E RT E RT E ekeaktconsionconcentratfletionconcentratfekr 00 tan)()( R E T LnkLnrA 1 Suppose , T xLnry A 1 , so intercept, R E slope Lnk 6 )/( 1 hmrA 150160230295370 A Lnr -3.1780- 3.1135 -2.7506-2.5017-2.2752 CT o / 1316222428 3 10 1 T 3.49473.45843.38813.36533.3206 -y = 5417.9x - 15.686 R2 = 0.97 0 1 2 3 4 0.00330.003350.00340.003450.0035 1/T -Ln r -y = -5147.9 x + 15.686 Also , intercept = 15.686 ,K R E slope 9 . 5147 Lnk molkJKmolJKE/80.42)/(3145 . 8 9 . 5147 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. 3.2Liquid 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: Because the decomposition of A is a 1st-order reaction, so we can express the rate equation as: AA kCr We know that for 1st-order reaction, ,kt C C Ln A Ao , 1 1 kt C C Ln A Ao 2 2 kt C C Ln A Ao , AoA CC5 . 0 1 AoA CC25 . 0 2 So equ(1)2 1 )24( 1 )( 1 12 12 Ln k LnLn kC C Ln C C Ln k tt A Ao A Ao equ(2)min52 1 )( 1 1 1 Ln kC C Ln k t A Ao So min5 112 ttt 3.3Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd-order reaction, ,kt CC AA 0 11 So we have two equations as follow: , equ(1)min5 11211 1 01 kkt CCCCC AoAoAoAA 8 , equ(2) 21 2 3) 1 (3 1411 ktkt CCCCC AoAoAoAoA So , min153 12 ttmin10 12 tt 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 1 Solution: In a order reaction: , 2 1 5 . 0 A A A kC dt dC r After integration, we can get: , 5 . 0 1 5 . 0 2 AAo CC kt So we have two equations as follow: , equ(1)min)10(5 . 0) 4 1 ( 1 5 . 05 . 05 . 05 . 0 1 5 . 0 kktCCCCC AoAoAoAAo , equ(2)min)30( 2 5 . 0 2 5 . 0 kktCC AAo Combining these two equations, we can get:, but this means , which is 2 5 . 0 5 . 1ktCAo0 5 . 0 2 A C 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, monomermonomer kCr And k C C Ln o o min)34( 8 . 0 1 min00657 . 0 k monomermonomer Cr)min00657 . 0 ( 1 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 . 1 5 1 1 11 1 11 1 2 1 2 Ln Ln X Ln k X Ln k t t A A In 2nd order reaction, , satisfied. 4 9 /4 /9 1 2 . 0 1 1 1 . 0 1 ) 11 ( 1 ) 11 ( 1 1 2 1 2 Ao Ao AoAo AoAo AoA AoA C C CC CC CCk CCk t t 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: , , ,180 Ao n13 A nht2 , , 135 A nht3 ; AA knr So we obtain , kt n n Ln A Ao )()( t n n Ln t n n Ln A Ao A Ao , 3 135 2 13 180 Ao n LnLn 28 A n 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 1 211 1 21 0 211 0 )( 1 )( A X A A t Xkkk k Ln kkXkkk dX dtt A , , so we obtain eq(1) min8sec480t33 . 0 A X eq(1) 33 . 0 )( 1 min8sec480 211 1 21 kkk k Ln kk , , so we obtain eq(2) Ae Ae Ae c X XM C C k k K 1 Re 2 1 0 Ao Ro C C M ,2 3 2 1 3 2 1 2 1 Ae Ae c X X k k K eq(2) 21 2kk Combining eq(1) and eq(2), we obtain 141 2 sec108 . 4min02888 . 0 k 141 21 sec1063 . 9 min05776 . 0 2 kk So the rate equation is )( 21AAoA A A CCkCk dt dC r )(sec1063 . 9 sec108 . 4 0 1414 AAA CCC 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 ,min1 03 . 2 1 97 . 1 1 1 k so min 015 . 0 1 mol L k so the rate equation is 21) min015 . 0 ( AA Cr 3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as follows: Aucrose products sucrase 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 constant MA EA CC CCk 03 If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method. Solution: Solve the question by the integral method: , A A MA AEoA A Ck Ck CC CCk dt dC r 5 43 1 , M Eo C Ck k 3 4 M C k 1 5 AAo A Ao AAo CC C C Ln kk k CC t 44 5 1 hrt,mmol/L A C AAo A Ao CC C C Ln AAo CC t 10.841.08976.25 20.681.20526.25 30.531.35086.3830 40.381.56066.4516 50.271.79366.8493 60.162.18167.1428 70.092.64617.6923 80.043.35308.3333 90.0184.09109.1650 100.0065.146910.0604 110.00256.006511.0276 CA, millimol/liter0.840.680.530.380.270.160.090.040.0180.0060.0025 t,hr1234567891011 12 Suppose y=, x= AAo A Ao CC C C Ln , thus we obtain such straight line graph AAo CC t y = 0.9879x + 5.0497 R2 = 0.998 0 2 4 6 8 10 12 01234567 Ln(Cao/Ca)/(Cao-Ca) t/(Cao-Ca) , intercept=9879 . 0 1 34 Eo M Ck C k Slope0497 . 5 4 5 k k So , )/(1956 . 0 0497 . 5 9879 . 0 1 5 Lmmol k CM 14 3 80.19 01 . 0 9879 . 0 1956 . 0 hr C Ck k Eo M 3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A R, -rA = enzyme min2 200 0 liter mol C CC A EA 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: 5 10 001 . 0 200 21 AA A AA CC C dC dt r Rearranging and integrating, we obtain: 10 025 . 0 025 . 0 100 )(510)5 10 ( AAo A Ao A A t CC C C LndC C dtt 13 min79.109)(5 025 . 0 10 10 AAo CCLn 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 2C2H5SO4H 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 001804.11 411.181944.31 481.382124.45 551.632674.86 752.243185.15 962.753685.32 1273.313795.35 1463.764105.42 1623.81(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 RBA2 The rate equation is: 2 21RBA A A CkCCk dt dC r , , LmolCC BoAo /5 . 5)1 ( AAoA XCC , AAAoBoB CXCCC AAoR XCC2 When , tLmolXCC AeAo /8 . 52 Re So ,5273 . 0 5 . 52 8 . 5 Ae X LmolXCCC AeAoBeAe /6 . 2)5273 . 0 1 (5 . 5)1 ( After integrating, we obtain eq (1)tC X k XX XXX Ln Ao AeAAe AAeAe ) 1 1 (2 ) 12( 1 14 The calculating result is presented in following Table. , t min LmolCR/,LmolCA/, A X AAe AAeAe XX XXX Ln ) 12( )1 ( Ae A X X Ln 005.5000 411.184.910.10730.2163-0.2275 481.384.810.12540.2587-0.2717 551.634.6850.14820.3145-0.3299 752.244.380.20360.4668-0.4881 962.754.1250.250.6165-0.6427 1273.313.8450.30090.8140-0.8456 1463.763.620.34181.0089-1.0449 1623.813.5950.34641.0332-1.0697 1804.113.4450.37361.1937-1.2331 1944.313.3450.39181.3177-1.3591 2124.453.2750.40451.4150-1.4578 2674.863.070.44181.7730-1.8197 3185.152.9250.46822.1390-2.1886 3685.322.840.48362.4405-2.4918 3795.352.8250.48642.5047-2.5564 4105.422.790.49272.6731-2.7254 5.82.60.5273 Draw t plot, we obtain a straight line: AAe AAeAe XX XXX Ln ) 12( y = 0.0067x - 0.0276 R 2 = 0.9988 0 0.5 1 1.5 2 2.5 3 0100200300400500 t Ln 15 ,0067 . 0 ) 1 1 (2 1 Ao Ae C X kSlope min)/(10794 . 6 5 . 5) 1 5273 . 0 1 (2 0067 . 0 4 1 molLk When approach to equilibrium, , BeAe c CC C k k K 2 Re 2 1 so min)/(10364 . 1 8 . 5 6 . 210794 . 6 4 2 24 2 Re 1 2 molL C CCk k BeAe So the rate equation is min)/()10364 . 1 10794 . 6 ( 244 LmolCCCr RBAA ii) We postulate it is a 1st order reversible reaction system, so the rate equation is RA A A CkCk dt dC r 21 After rearranging and integrating, we obtain eq (2)tk XX X Ln AeAe A 1 1 )1 ( Draw t plot, we obtain another straight line:)1 ( Ae A X X Ln -y = 0.0068x - 0.0156 R2 = 0.9986 0 0.5 1 1.5 2 2.5 3 0100200300400500 x -Ln ,0068 . 0 1 Ae X k Slope 16 So 13 1 min10586 . 3 5273 . 0 0068 . 0 k 13 3 Re 1 2 min10607 . 1 8 . 5 6 . 210586 . 3 C Ck k Ae So the rate equation is min)/()10607 . 1 10586 . 3 ( 33 LmolCCr 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/liter12467912 -rA, mol/literhr0.060.10.251.02.01.00.5 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 02468101214 Ca -1/Ra 3.31 The thermal decomposition of hydrogen iodide 2HI H2 + I2 is reported by M.Bodenstein Z.phys.chem.,29,295(1899) as follows: T,508427393356283 17 k,cm3/mol s 0.10590.003100.00058880.910-60.94210-6 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 R 2 = 0.9879 0 4 8 12 16 0.0010.0020.0030.004 1/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, , , 100 Ao C200 Bo CSRBA , find , , 0 A X B X A C B C , 0 BA 202 . 0100)1 ( AAoA XCC