chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案.doc

Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING1CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS3CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA7CHAPTER 4 INTRODUCTION TO REACTOR DESIGN19CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR22CHAPTER 6 DESIGN FOR SINGLE REACTIONS26CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR32CHAPTER 11 BASICS OF NON-IDEAL FLOW34CHAPTER 18 SOLID CATALYZED REACTIONS43Chapter 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 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 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 water32,000 m3/dayWaste waterTreatment plant Clean water 32,000 m3/day200 mg O2needed/literMean residencetime =8 hr Zero O2 neededFigure P1.1Solution: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.Solution: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 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: 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 + 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/m3s What are the units of the two constants?Solution: 2.5 For the complex reaction with stoichiometry A + 3B 2R + S and with second-order rate expression -rA = k1AB 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: 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:2.7 For a gas reaction at 400 K the rate is reported 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 -rA = - = k C2 A , mol/m3sSolution:(a) The unit of the rate constant is (b)Because its a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to So we can get that the value of 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: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/hr150160230295370Temperature, 1316222428 What activation energy represents this change in bustliness?Solution:Suppose , so intercept150160230295370-3.1780-3.1135-2.7506-2.5017-2.275213162224283.49473.45843.38813.36533.3206-y = -5147.9 x + 15.686Also , intercept = 15.686 ,Chapter 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 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:We know that for 1st-order reaction, , , So equ(1) equ(2)So 3.3 Repeat the previous problem for second-order kinetics.Solution: We know that for 2nd-order reaction, ,So we have two equations as follow:, equ(1), equ(2)So , 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?Solution:In a order reaction: ,After integration, we can get: ,So we have two equations as follow:, equ(1), equ(2)Combining these two equations, we can get:, but this means , which is impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted .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,And 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.In 2nd order reaction, , 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 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 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: , , , , So we obtain , , 3.9 The first-order reversible 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 reaction, according to page56 eq. 53b, we obtain , , so we obtain eq(1) eq(1), , so we obtain eq(2), eq(2)Combining eq(1) and eq(2), we obtain So the rate equation is 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 ,so so the rate equation is 3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as follows: Aucrose products 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):CA, millimol/liter0.840.680.530.380.270.160.090.040.0180.0060.0025 t,hr1234567891011Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -rA = where CM = Michaelis constant If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method. Solution: Solve the question by the integral method:, ,mmol/L10.841.08976.2520.681.20526.2530.531.35086.383040.381.56066.451650.271.79366.849360.162.18167.142870.092.64617.692380.043.35308.333390.0184.09109.1650100.0065.146910.0604110.00256.006511.0276Suppose y=, x=, thus we obtain such straight line graph, intercept=So ,3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A R, -rA = 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: Rearranging and integrating, we obtain:3.20 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. 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 postulate it is a 2nd order reversible reaction system The rate equation is: , , , When , So , After integrating, we obtain eq (1)The calculating result is presented in following Table., min005.5000411.184.910.10730.2163-0.2275481.384.810.12540.2587-0.2717551.634.6850.14820.3145-0.3299752.244.380.20360.4668-0.4881962.754.1250.250.6165-0.64271273.313.8450.30090.8140-0.84561463.763.620.34181.0089-1.04491623.813.5950.34641.0332-1.06971804.113.4450.37361.1937-1.23311944.313.3450.39181.3177-1.35912124.453.2750.40451.4150-1.45782674.863.070.44181.7730-1.81973185.152.9250.46822.1390-2.18863685.322.840.48362.4405-2.49183795.352.8250.48642.5047-2.55644105.422.790.49272.6731-2.72545.82.60.5273Draw t plot, we obtain a straight line:,When approach to equilibrium, ,so So the rate equation isii) We postulate it is a 1st order reversible reaction system, so the rate equation is After rearranging and integrating, we obtain eq (2)Draw t plot, we obtain another straight line:,So So the rate equation isWe 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/literhr0.060.10.251.02.01.00.5We 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 CA0 = 10 mol/liter to CAf = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002468101214Ca-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,508427393356283k,cm3/mols0.10590.003100.00058880.910-60.94210-6Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds.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:From the figure, we get slope = E/R = 7319.1 intercept = - In(k0) = -11.567E = 60851 J/mol 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 TCA2Chapter 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 gaseous feed, , , , find , , , 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, , , , find , , Aqueous reaction system, so When , When , So , , which is impossible.So , 4.3 Given a gaseous feed, CA0 =200, CB0 =100, A +B R, CA = 50. Find XA, XB, CB.Solution: Given a gaseous feed, , , .find , , , which is impossible.So 4.4 Given a gaseous feed, CA0 = CB0 =100, A +2B R, CB = 20. Find XA, XB, CA.Solution: Given a gaseous feed, ，, ,Find , , , , , 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, , , , , , , , find , , ., , According to eq page 87, 4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn to