elements of chemical reaction engineering 4th ed solution manual

1. Solutions Manual for Elements of Chemical Reaction Engineering Fourth Edition Brian Vicente Max Nori H. Scott Fogler

2. 1-1 Solutions for Chapter 1 Mole Balances Synopsis General: The goal of these problems are to reinforce the definitions and provide an understanding of the mole balances of the different types of reactors. It lays the foundation for step 1 of the algorithm in Chapter 4. P1-1. This problem helps the student understand the course goals and objectives. P1-2. Part (d) gives hints on how to solve problems when they get stuck. Encourages students to get in the habit of writing down what they learned from each chapter. It also gives tips on problem solving. P1-3. Helps the student understand critical thinking and creative thinking, which are two major goals of the course. P1-4. Requires the student to at least look at the wide and wonderful resources available on the CD-ROM and the Web. P1-5. The ICMs have been found to be a great motivation for this material. P1-6. Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the student an idea of things to come in terms of sizing reactors in chapter 4. An alternative to P1-15. P1-7. Straight forward modification of Example 1-1. P1-8. Helps the student review and member assumption for each design equation. P1-9 and P1-10. The results of these problems will appear in later chapters. Straight forward application of chapter 1 principles. P1-11. Straight forward modification of the mole balance. Assigned for those who emphasize bioreaction problems. P1-12. Can be assigned to just be read and not necessarily to be worked. It will give students a flavor of the top selling chemicals and top chemical companies. P1-13. Will be useful when the table is completed and the students can refer back to it in later chapters. Answers to this problem can be found on Professor Susan Montgomerys equipment module on the CD-ROM. See P1-17.

3. 1-2 P1-14. Many students like this straight forward problem because they see how CRE principles can be applied to an everyday example. It is often assigned as an in- class problem where parts (a) through (f) are printed out from the web. Part (g) is usually omitted. P1-15. Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with P1-6. P1-16. Open-ended problem. P1-17. I always assign this problem so that the students will learn how to use POLYMATH/MaLab before needing it for chemical reaction engineering problems. P1-18. Parts (a) and (b) are open-ended problem. P1-19 and P1-20. Help develop critical thinking and analysis. CDP1-A Similar to problems 3, 4, 11, and 12. CDP1-B Points out difference in rate per unit liquid volume and rate per reactor volume. Summary Assigned Alternates Difficulty Time (min) P1-1 AA SF 60 P1-2 I SF 30 P1-3 O SF 30 P1-4 O SF 30 P1-5 AA SF 30 P1-6 AA 1-15 SF 15 P1-7 I SF 15 P1-8 S SF 15 P1-9 S SF 15 P1-10 S SF 15 P1-11 O FSF 15 P1-12 I Read Only SF 30 P1-13 I SF 1 P1-14 O FSF 30 P1-15 O SF 60 P1-16 S SF 15 P1-17 AA SF 60 P1-18 S SF 30 P1-19 O -- 30 P1-20 O FSF 15

4. 1-3 CDP1-A AA FSF 30 CDP1-B I FSF 30 Assigned = Always assigned, AA = Always assign one from the group of alternates, O = Often, I = Infrequently, S = Seldom, G = Graduate level Alternates In problems that have a dot in conjunction with AA means that one of the problem, either the problem with a dot or any one of the alternates are always assigned. Time Approximate time in minutes it would take a B/B+ student to solve the problem. Difficulty SF = Straight forward reinforcement of principles (plug and chug) FSF = Fairly straight forward (requires some manipulation of equations or an intermediate calculation). IC = Intermediate calculation required M = More difficult OE = Some parts open-ended. ____________ * Note the letter problems are found on the CD-ROM. For example A CDP1-A. Summary Table Ch-1 Review of Definitions and Assumptions 1,5,6,7,8,9 Introduction to the CD-ROM 1,2,3,4 Make a calculation 6 Open-ended 8,16 P1-1 Individualized solution. P1-2 Individualized solution. P1-3 Individualized solution. P1-4 Individualized solution. P1-5 Solution is in the decoding algorithm given with the modules.

5. 1-4 P1-6 The general equation for a CSTR is: A AA r FF V ! ! = 0 Here rA is the rate of a first order reaction given by: rA = - kCA Given : CA = 0.1CA0 , k = 0.23 min-1 , v0 = 10dm3 min-1 , FA = 5.0 mol/hr And we know that FA = CAv0 and FA0 = CA0v0 => CA0 = FA0/ v0 = 0.5 mol/dm3 Substituting in the above equation we get: ))/5.0(1.0)(min23.0( min)/10)(/5.0(1.0min)/10)(/5.0( 31 3333 00A0 dmmol dmdmmoldmdmmol kC vCvC V A A ! ! = ! = V = 391.3 dm3 P1-7 1 0.23mink ! = From mole balance: dNA dt rA V! Rate law: rA! k CA" rA! k NA V " Combine: dNA dt k! NA" 0! = 0 t t1 ! " # d 1 k $ % & ' ( ) * NA0 NA NA 1 NA ! " " # d+ at 0! = , NAO = 100 mol and ! != , NA = (0.01)NAO t = !! " # $$ % & A A N N k 0 ln 1 ( ) min100ln 23.0 1 = t = 20 min

6. 1-5 P1-8 (a) The assumptions made in deriving the design equation of a batch reactor are: - Closed system: no streams carrying mass enter or leave the system. - Well mixed, no spatial variation in system properties - Constant Volume or constant pressure. P1-8 (b) The assumptions made in deriving the design equation of CSTR, are: - Steady state. - No spatial variation in concentration, temperature, or reaction rate throughout the vessel. P1-8 (c) The assumptions made in deriving the design equation of PFR are: - Steady state. - No radial variation in properties of the system. P1-8 (d) The assumptions made in deriving the design equation of PBR are: - Steady state. - No radial variation in properties of the system. P1-8 (e) For a reaction, A B -rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=] moles/ (dm3 .s). -rA is the rate of disappearance of species A per unit mass (or area) of catalyst [=] moles/ (time. mass of catalyst). rA is the rate of formation (generation) of species A per unit mass (or area) of catalyst [=] moles/ (time. mass catalyst). -rA is an intensive property, that is, it is a function of concentration, temperature, pressure, and the type of catalyst (if any), and is defined at any point (location) within the system. It is independent of amount. On the other hand, an extensive property is obtained by summing up the properties of individual subsystems within the total system; in this sense, -rA is independent of the extent of the system. P 1-9 Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per second. It is an Intensive property and the concentration, temperature and hence the rate varies with spatial coordinates. ' Ar on the other hand is defined as g mol of A reacted per gm. of the catalyst per second. Here mass of catalyst is the basis as this is what is important in catalyst reactions and not the reactor volume. Applying general mole balance we get: !+"= dVrFF dt dN jjj j 0 No accumulation and no spatial variation implies !+"= dVrFF jjj00

7. 1-6 Also rj = b rj` and W = Vb where b is the bulk density of the bed. => ' 00 ( ) ( )j j j bF F r dV!= " + # Hence the above equation becomes 0 ' j j j F F W r ! = ! We can also just apply the general mole balance as ' 0( ) ( ) j j j j dN F F r dW dt = ! + " Assuming no accumulation and no spatial variation in rate, we get the same form as above: 0 ' j j j F F W r ! = ! P1-10 Mole balance on species j is: ! =+" V j jjj dt dN dVrFF 0 0 Let jM = molecular wt. of species j Then 00 jjj wMF = = mass flow rate of j into reactor: jjj mMN = = mass of species j in the reactor Multiplying the mole balance on species j by jM ! =+" V j jjjjjjj dt dN MdVrMMFMF 0 0 Now jM is constant: ( ) ( ) dt md dt NMd dVrMMFMF j V jj jjjjjj ==+! "0 0

8. 1-7 dt dm dVrMww j V jjjj =+! "0 0 P1-11 Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 7), so there is no cell growth and the nutrients are used in making product. Lets do part c first. [In flowrate (moles/time)] penicillin + [generation rate (moles/time)]penicillin [Out flowrate (moles/time)] penicillin = [rate of accumulation (moles/time)]penicillin Fp,in + Gp Fp,out = dNp dt Fp,in = 0 .(because no penicillin inflow) Gp = . V pr dV! Therefore, . V pr dV! - Fp,out = dNp dt Assuming steady state for the rate of production of penicillin in the cells stationary state, dNp dt = 0 And no variations , ,p in p out p F F V r ! = ! Or, ,p out p F V r = Similarly, for Corn Steep Liquor with FC = 0 C C C CC r F r FF V ! = ! ! = 00 Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.

9. 1-8 P1-12 (a) Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below: Rank 2002 Rank 1995 Chemical 1 1 H2SO4 2 2 N2 3 4 C2H4 4 3 O2 5 9 C3H6 6 - H2 7 6 NH3 8 10 Cl2 9 - P2O5 10 - C2H2Cl2 Chemicals like H2 , P2O5 , C2H2Cl2 has come in top 10 chemicals and C3H6 has jumped to rank 5 now then rank 9 in 1995 . P1-12 (b) Ranking of top 10 chemical companies in sales in year 2003 and 2002: s o u r c e : C h e m i c al and Engineering News may 17,2004 We have Chevron Phillips which jumped to 5 rank in 2003 from 8th rank in 2002 and Air Products coming to 9th rank in 2003 from 11th in 2003. Chemical sales of each company has increased compared to year 2002 from 9%(Eastman Chemical) to28.2%(Chevron Phillips) but Huntsman Corp. has a decrease by 2.9%. P1-12 (c) Sulfuric acid is prime importance in manufacturing. It is used in some phase of the manufacture of nearly all industrial products .It is used in production of every other strong acid. Because of its large number of uses, its the most produced chemical. Sulfuric acid uses are: It is consumed in production of fertilizers such as ammonium sulphate (NH4)2SO4 and superphosphate (Ca(H2PO4)2) , which is formed when rock phosphate is treated with sulfuric acid. Used as dehydrating agent. Used in manufacturing of explosives, dyestuffs, other acids, parchment paper, glue , purification of petroleum and picking of metals. Used to remove oxides from iron and steel before galvanizing or electroplating. Used in non-ferrous metallurgy, in production of rayon and film. as laboratory reagent and etchant and in storage batteries. It is also general purpose food additive. 2003 2002 Company Chemical Sales ($ million 2003) 1 1 Dow Chemical 32632 2 2 DuPont 30249 3 3 ExxonMobil 20190 4 4 General Electric 8371 5 8 Chevron Phillips 7018 6 5 Huntsman Corp. 6990 7 6 PPG Industries 6606 8 7 Equistar Chemicals 6545 9 11 Air Products 6029 10 9 Eastman Chemicals 5800

10. 1-9 P1-12 (d) Annual Production rate of ethylene for year 2002 is 5.21x 1010 lb/year Annual Production rate of benzene for year 2002 is 1.58 x 1010 lb/year Annual Production rate of ethylene oxide for year 2002 is 7.6 x109 lb/year P1-12 (e) Because the basic raw material coal and petroleum for organic chemicals is very limited and their production is not increasing as production of raw material for inorganic chemicals. P1-13 Type Characteristics Phases Usage Advantage Disadvantage Batc h All the reactants fed into the reactor. During reaction nothing is added or removed. Easy heating or cooling. 1. Liquid phase 2. Gas phase 3. Liquid Solid 1. Small scale pdn. 2. Used for lab experimentation. 3. Pharmaceuticals 4. Fermentation 1. High Conversion per unit volume. 2. Flexibility of using for multiple reactions. 3. Easy to clean 1. High Operating cost. 2. Variable product quality. CST R Continuous flow of reactants and products. Uniform composition throughout. 1. Liquid phase 2. Gas liquid 3. Solid - liquid 1. Used when agitation required. 2. Series Configuration possible for different configuration streams 1. Continuous Operation. 2.Good Temperature Control 3. Two phase reactions possible. 4.Good Control 5. Simplicity of construction. 6. Low operating cost 7. Easy to clean 1. Lowest conversion per unit volume. 2. By passing possible with poor agitation. 3 High power Input reqd. PFR One long reactor or number of CSTRs in series. No radial variations. Conc. changes along the length. 1. Primarily gas Phase 1. Large Scale pdn. 2. Fast reactions 3. Homogenous reactions 4. Heterogeneous reactions 5. Continuous pdn. 1. High conversion per unit volume 2. Easy to maintain (No moving parts) 3. low operating cost 4. continuous operation 1. Undesired thermal gradient. 2. Poor temperature control 3. Shutdown and cleaning expensive.

11. 1-10 PBR Tubular reactor that is packed with solid catalyst particles. 1. Gas Phase (Solid Catalyst) 2.Gas solid reactions. 1. Used primarily in the heterogeneous gas phase reaction with solid catalyst e.g Fischer tropsch synthesis. 1. High conversion per unit mass of catalyst 2. low operating cost 3. Continuous operation 1. Undesired thermal gradient. 2. Poor temperature control 3. Channeling 4. Cleaning expensive. P1-14 Given 210 10*2 ftA = RTSTP 69.491= ftH 2000= 313 10*4 ftV = T = 534.7 R PO = 1atm Rlbmol ftatm R 3 7302.0= yA = 0.02 3 10 10*04.2 ft lbmol CS ! = C = 4*105 cars FS = CO in Santa Ana wind FA = CO emission from autos hr ft vA 3 3000= per car at STP P1-14 (a) Total number of lb moles gas in the system: N = 13 3 3 1 (4 10 ) . 0.73 534.69 . atm ft atm ft R lbmol R ! ! " # !$ % & ' = 1.025 x 1011 lb mol P1-14 (b) Molar flowrate of CO into L.A. Basin by cars. STP A ATAA TR PCv yFyF 0 !== cars ft lbmol carhr ft FT 400000 359 13000 3 3 !!= (See appendix B) FA = 6.685 x 104 lb mol/hr N P0 V R T :=

12. 1-11 P1-14 (c) Wind speed through corridor is v = 15mph W = 20 miles The volumetric flowrate in the corridor is vO = v.W.H = (15x5280)(20x5280)(2000) ft3 /hr = 1.673 x 1013 ft3 /hr P1-14 (d) Molar flowrate of CO into basin from Sant Ana wind. FS v0 CS!:= = 1.673 x 1013 ft3 /hr 10 2.04 10! " " lbmol/ft3 FS = 3.412 x 103 lbmol/hr P1-14 (e) Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO = COdN dt dt dC VCvFF co cooSA =!+ (V=constant, VCN coco = ) P1-14 (f) t = 0 , coOco CC = 0 co coO Ct co A S o coC dC dt V F F v C = + !" " !! " # $$ % & '+ '+ = cooSA coOoSA o CvFF CvFF v V t ln P1-14 (g) Time for concentration to reach 8 ppm. 8 0 3 2.04 10CO lbmol C ft ! = " , 8 3 2.04 10 4 CO lbmol C ft ! = " From (f), 0 3 4 3 13 8 3 3 3 3 13 4 3 13 8 3 . ln . 6.7 10 3.4 10 1.673 10 2.04 10 4 ln 1.673 10 6.7 10 3.4 10 1.673 10 0.51 10 A S O CO o A S O CO F F v CV t v F F v C lbmol lbmol ft lbmol ft hr hr hr ft ft lbmol lbmol ft lbmol hr hr hr hr ft ! ! " #+ ! = $ % + !& ' " # ( + ( ! ( ( ($ % $ %= $ % ( ( + ( ! ( ( ($ % & ' t = 6.92 hr

13. 1-12 P1-14 (h) (1) to = 0 tf = 72 hrs coC = 2.00E-10 lbmol/ft3 a = 3.50E+04 lbmol/hr ov = 1.67E+12 ft3 /hr b = 3.00E+04 lbmol/hr Fs = 341.23 lbmol/hr V = 4.0E+13 ft3 dt dC VCvF t ba co coos =!+" # $ % & ' + 6 sin ( Now solving this equation using POLYMATH we get plot between Cco vs t See Polymath program P1-14-h-1.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 72 72 C 2.0E-10 2.0E-10 2.134E-08 1.877E-08 v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12 a 3.5E+04 3.5E+04 3.5E+04 3.5E+04 b 3.0E+04 3.0E+04 3.0E+04 3.0E+04 F 341.23 341.23 341.23 341.23 V 4.0E+13 4.0E+13 4.0E+13 4.0E+13 ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)+F-v0*C)/V Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] F = 341.23 [5] V = 4*10^13 (2) tf = 48 hrs sF = 0 dt dC VCv t ba co coo =!" # $ % & ' + 6 sin ( Now solving this equation using POLYMATH we get plot between Cco vs t

14. 1-13 See Polymath program P1-14-h-2.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 48 48 C 2.0E-10 2.0E-10 1.904E-08 1.693E-08 v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12 a 3.5E+04 3.5E+04 3.5E+04 3.5E+04 b 3.0E+04 3.0E+04 3.0E+04 3.0E+04 V 4.0E+13 4.0E+13 4.0E+13 4.0E+13 ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] V = 4*10^13 (3) Changing a Increasing a reduces the amplitude of ripples in graph. It reduces the effect of the sine function by adding to the baseline. Changing b The amplitude of ripples is directly proportional to b. As b decreases amplitude decreases and graph becomes smooth. Changing v0 As the value of v0 is increased the graph changes to a shifted sin-curve. And as v0 is decreased graph changes to a smooth increasing curve. P1-15 (a) rA = k with k = 0.05 mol/h dm3 CSTR: The general equation is A AA r FF V ! ! = 0 Here CA = 0.01CA0 , v0 = 10 dm3 /min, FA = 5.0 mol/hr Also we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3

15. 1-14 Substituting the values in the above equation we get, 05.0 10)5.0(01.010)5.0(00A0 ! = ! = k vCvC V A V = 99 dm3 PFR: The general equation is kr dV dF A A == , Now FA = CAv0 and FA0 = CA0v0 => k dV vdCA !=0 Integrating the above equation we get !=! VC C A dVdC k v A A 0 0 0 => )( 0 0 AA CC k v V != Hence V = 99 dm3 Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration. P1-15 (b) - rA = kCA with k = 0.0001 s-1 CSTR: We have already derived that A A r vCvC V ! ! = 00A0 A A kC Cv )01.01(00 ! = k = 0.0001s-1 = 0.0001 x 3600 hr-1 = 0.36 hr-1 )/5.0*01.0)(36.0( )99.0)(/5.0)(/10( 31 33 dmmolhr dmmolhrdm V ! = => V = 2750 dm3 PFR: From above we already know that for a PFR AA A kCr dV vdC ==0 Integrating !"=! VC C A A dV C dC k v A A 0 0 0 V C C k v A A =00 ln Again k = 0.0001s-1 = 0.0001 x 3600 hr-1 = 0.36 hr-1 Substituing the values in above equation we get V = 127.9 dm3

16. 1-15 P1-15 (c) - rA = kCA 2 with k = 3 dm3 /mol.hr CSTR: ! V = CA0v0 " CAv0 "rA ! = v0CA0(1" 0.01) kCA 2 Substituting all the values we get ! V = (10dm3 /hr)(0.5mol/dm3 )(0.99) (3dm3 /hr)(0.01*0.5mol/dm3 )2 => V = 66000 dm3 PFR: ! dCAv0 dV = rA = kCA 2 Integrating ! v0 k dCA CA 2 CA 0 CA " = # dV 0 V " => ! v0 k ( 1 CA " 1 CA0 ) = V => ! V = 10dm3 /hr 3dm3 /mol.hr ( 1 0.01CA0 " 1 CA0 ) = 660 dm3 P1-15 (d) CA = .001CA0 ! t = dN "rAVNA NA 0 # Constant Volume V=V0 ! t = dCA "rA CA CA 0 # Zero order: ! t = 1 k CA0 " 0.001CA0[ ] = .999CAo 0.05 = 9.99h First order: ! t = 1 k ln CA0 CA " # $ % & ' = 1 0.001 ln 1 .001 " # $ % & ' = 6908s Second order: ! t = 1 k 1 CA " 1 CA0 # $ % & ' ( = 1 3 1 0.0005 " 1 0.5 # $% & '( = 666h

17. 1-16 P1-16 Individualized Solution P1-17 (a) Initial number of rabbits, x(0) = 500 Initial number of foxes, y(0) = 200 Number of days = 500 1 2 dx k x k xy dt = ! .(1) 3 4 dy k xy k y dt = ! ..(2) Given, 1 1 2 3 1 4 0.02 0.00004/( ) 0.0004/( ) 0.04 k day k day foxes k day rabbits k day ! ! = = " = " = See Polymath program P1-17-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 500 500 x 500 2.9626929 519.40024 4.2199691 y 200 1.1285722 4099.517 117.62928 k1 0.02 0.02 0.02 0.02 k2 4.0E-05 4.0E-05 4.0E-05 4.0E-05 k3 4.0E-04 4.0E-04 4.0E-04 4.0E-04 k4 0.04 0.04 0.04 0.04 ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(t) = (k1*x)-(k2*x*y) [2] d(y)/d(t) = (k3*x*y)-(k4*y) Explicit equations as entered by the user [1] k1 = 0.02 [2] k2 = 0.00004 [3] k3 = 0.0004 [4] k4 = 0.04

18. 1-17 When, tfinal = 800 and 3 0.00004/( )k day rabbits= ! Plotting rabbits Vs. foxes P1-17 (b) POLYMATH Results See Polymath program P1-17-b.pol.

19. 1-18 POLYMATH Results NLES Solution Variable Value f(x) Ini Guess x 2.3850387 2.53E-11 2 y 3.7970279 1.72E-12 2 NLES Report (safenewt) Nonlinear equations [1] f(x) = x^3*y-4*y^2+3*x-1 = 0 [2] f(y) = 6*y^2-9*x*y-5 = 0 P1-18 (a) No preheating of the benzene feed will diminish the rate of reaction and thus lesser conversions will be achieved. P1-18 (b) An interpolation can be done on the logarithmic scale to find the desired values from the given data. Now we can interpolate to the get the cost at 6000 gallons and 15000 gallons Cost of 6000 gal reactor = 1.905 x 105 $ Cost of 15000 gal reactor = 5.623 x 105 $ P1-18 (c) We are given CA is 0.1% of initial concentration CA = 0.001CA0 Also from Example 1.3, )ln( 00 A A C C k v V = Substituting v0 = 10 dm3 /min and k = 0.23 min-1 we get 3 300dmV = which is three times the volume of reactor used in Example 1-3. P1-18 (d) Safety of Plant. P1-19 Enrico Fermi Problem no definite solution P1-20 Enrico Fermi Problem no definite solution P1-21 Individualized solution. Cost vs Volume of reactor (log - log plot) 0 1 2 3 4 5 6 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 log volume (gallons)

20. 1-19 CDP1-A (a) How many moles of A are in the reactor initially? What is the initial concentration of A? If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite simple. We insert our variables into the ideal gas equation: 3 3 (20 )(200 ) 101.33 97.5 1. 8.3145 (500) PV atm dm kPa n moles RT atmkPa dm molK ! " = = =# $ ! " % & # $ % & Knowing the mole fraction of A (yAo) is 75%, we multiply the total number of moles (NTo) by the yA: 0.75 97.5 73.1AomolesA N= = ! = The initial concentration of A (CAo) is just the moles of A divided by the volume: 3 3 73.1 0.37 / 200 Ao Ao Nmoles moles C moles dm volume V dm = = = = CDP1-A (b) Time (t) for a 1st order reaction to consume 99% of A. A A dC r dt = Our first order rate law is: A Ar kC! = mole balance: A A dC kC dt = ! => 0 A Ao Ct A AC dC k dt C ! =" " ln A Ao C kt C ! " # = $ % & ' , knowing CA=0.01 CAo and our rate constant (k=0.1 min-1 ), we can solve for the time of the reaction: 1 1 4.61 ln(0.01) 46.1min 0.1min t k ! = ! = = CDP1-A (c) Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C. rate law: 2 A Ar kC! = mole balance: 2A A dC kC dt ! = " => 2 0 A Ao Ct A AC dC k dt C ! =" " => 1 1 A Ao kt C C ! = ! + We can solve for the time in terms of our rate constant (k = 0.7) and our initial concentration (CAo): 5 1 Ao Ao kt C C ! = ! +

21. 1-20 ( )( )3 3 4 4 15.4min. 0.7 / min 0.37 /Ao t kC dm mol mol dm = = = To determine the pressure of the reactor following this reaction, we will again use the ideal gas law. First, we determine the number of moles in the reactor: 0.25 0.2T I A B C To Ao B CN N N N N N N N N= + + + = + + + 0.8B C AoN N N= = (0.25) (0.2 0.8 0.8) 0.25(97.6) (1.8)(73.2) 156.1T To AoN N N moles= + + + = + = 3 3 (156.1 ) 0.082 (500 ) 32 200 T dm atm mole K molKN RT P atm V dm ! " # $ % &= = = CDP1-B Given: Liquid phase reaction in a foam reactor, A !B Consider a differential element, V! of the reactor: By material balance ( ) (1 )A A A AF F F r e V! + " = ! ! " Where, (1 )e V! " = fraction of reactor element which is liquid. or: (1 )A AF r e V! = ! ! " (1 )A A dF r e dV = ! Must relate ( )Ar! to AF , where, AF is the total (gas +liquid) molar flow rate of A. Ar! =rate of reaction (g mol A per cubic cm. of liquid per sec.); e = volume fraction of gas; AF = molar flow rate of A (g mol/sec.); V = volume of reactor

22. 2-1 Solutions for Chapter 2 - Conversion and Reactor Sizing Synopsis General: The overall goal of these problems is to help the student realize that if they have rA=f(X) they can design or size a large number of reaction systems. It sets the stage for the algorithm developed in Chapter 4. P2-1. This problem will keep students thinking about writing down what they learned every chapter. P2-2. This forces the students to determine their learning style so they can better use the resources in the text and on the CDROM and the web. P2-3. ICMs have been found to motivate the students learning. P2-4. Introduces one of the new concepts of the 4th edition whereby the students play with the example problems before going on to other solutions. P2-5. This is a reasonably challenging problem that reinforces Levenspiels plots. P2-6. Novel application of Levenspiel plots from an article by Professor Alice Gast at Massachusetts Institute of Technology in CEE. P2-7. Straight forward problem alternative to problems 8, 9, and 12. P2-8. To be used in those courses emphasizing bio reaction engineering. P2-9. The answer gives ridiculously large reactor volume. The point is to encourage the student to question their numerical answers. P2-10. Helps the students get a feel of real reactor sizes. P2-11. Great motivating problem. Students remember this problem long after the course is over. P2-12. Alternative problem to P2-7 and P2-9. CDP2-A Similar to 2-9 CDP2-B Good problem to get groups started working together (e.g. cooperative learning.

23. 2-2 CDP2-C Similar to problems 2-8, 2-9, 2-12. CDP2-D Similar to problems 2-8, 2-9, 2-12. Summary Assigned Alternates Difficulty Time (min) P2-1 O 15 P2-2 A 30 P2-3 A 30 P2-4 O 75 P2-5 O M 75 P2-6 S M 60 P2-7 AA 8,9,12 FSF 45 P2-8 S FSF 45 P2-9 AA 7,9,12 SF 45 P2-10 S SF 15 P2-11 AA SF 1 P2-12 AA 7,8,9 SF 60 CDP2-A O 9,B,C,D FSF 5 CDP2-B O 9,B,C,D FSF 30 CDP2-C O 9,B,C,D FSF 30 CDP2-D O 9,B,C,D FSF 45 Assigned = Always assigned, AA = Always assign one from the group of alternates, O = Often, I = Infrequently, S = Seldom, G = Graduate level Alternates In problems that have a dot in conjunction with AA means that one of the problems, either the problem with a dot or any one of the alternates are always assigned. Time Approximate time in minutes it would take a B/B+ student to solve the problem. Difficulty SF = Straight forward reinforcement of principles (plug and chug) FSF = Fairly straight forward (requires some manipulation of equations or an intermediate calculation). IC = Intermediate calculation required M = More difficult OE = Some parts open-ended. ____________ * Note the letter problems are found on the CD-ROM. For example A CDP1-A.

24. 2-3 Summary Table Ch-2 Straight forward 1,2,3,4,10 Fairly straight forward 7,9,12 More difficult 5,6,8 Open-ended 6 Comprehensive 4,5,6,7,8,9,12 Critical thinking P2-9 P2-1 Individualized solution. P2-2 Individualized solution. P2-3 Solution is in the decoding algorithm given with the modules. P2-4 (a) Example 2-1 through 2-3 If flow rate FAO is cut in half. v1 = v/2 , F1= FAO/2 and CAO will remain same. Therefore, volume of CSTR in example 2-3, 2.34.6 2 1 2 1 01 1 == ! = ! = A A A r XF r XF V If the flow rate is doubled, F2 = 2FAO and CAO will remain same, Volume of CSTR in example 2-3, V2 = F2X/-rA = 12.8 m3 P2-4 (b) Example 2-5 Now, FAO = 0.4/2 = 0.2 mol/s,

25. 2-4 New Table: Divide each term A A r F ! 0 in Table 2-3 by 2. X 0 0.1 0.2 0.4 0.6 0.7 0.8 [FAO/-rA](m3 ) 0.445 0.545 0.665 1.025 1.77 2.53 4 Reactor 1 Reactor 2 V1 = 0.82m3 V2 = 3.2 m3 V = (FAO/-rA)X ( )182.0 1 0 X r F XA A !! " # $$ % & ' = ( )22.3 2 0 X r F XA A !! " # $$ % & ' = By trial and error we get: X1 = 0.546 and X2 = 0.8 Overall conversion X0verall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673 P2-4 (c) Example 2-6 Now, FAO = 0.4/2 = 0.2 mol/s, 0201 202101 0 AA AA FF XFXF X + + = New Table: Divide each term A A r F ! 0 in Table 2-3 by 2. X 0 0.1 0.2 0.4 0.6 0.7 0.8 [FAO/-rA](m3 ) 0.445 0.545 0.665 1.025 1.77 2.53 4 V1 = 0.551m3 V2 = 1.614 m3 1 0 X AO A dX V F r = !" Plot FA0/-rA versus conversion. Estimate outlet conversions by computing the integral of the plotted function.

26. 2-5 X1 = 0.603 for V1 = 0.551m3 X2 = 0.89 for V2 = 1.614m3 Overall conversion X0 = (1/2)X1 + (1/2)X2 = (0.603+0.89)/2 = 0.746 P2-4 (d) Example 2-7 (1) For PFR, 0.2 2 0 3 0.222 AO A F V dX r m ! " = # $ %& ' = ( For first CSTR, X2 = 0.6, 3 1.32AOF m rA = ! , V1 = 32 1( ) .528AOF X X m rA ! = ! For second CSTR, X3 = 0.65, 3 2.0AOF m rA = ! , V3 = ( )3 2 3 0.1AOF X X m rA ! = ! (2) First CSTR remains unchanged For PFR: 0.5 0.2 AO A F V dX r ! " = # $ %& ' ( , Using the Levenspiel Plot VPFR = 0.22

27. 2-6 For CSTR, VCSTR2 = ( )3 2 3 0.3AOF X X m rA ! = ! (3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR. Conversion Original Reactor Volumes Worst Arrangement X1 = 0.20 V1 = 0.188 (CSTR) V1 = 0.23 (PFR) X2 = 0.60 V2 = 0.38 (PFR) V2 = 0.53 (CSTR) X3 = 0.65 V3 = 0.10 (CSTR) V3 = 0.10 (CSTR) For PFR, X1 = 0.2 1 1 0 X AO A F V dX r ! " = # $ %& ' ( Using trapezoidal rule, XO = 0.1, X1 = 0.1 ( ) ( ) ( ) [ ] 1 1 1 3 3 0.2 1.28 0.98 2 0.23 O O X X V f X f X rA m m ! = +" #$ %! = + = For CSTR, For X2 = 0.6, 3 1.32AOF m rA = ! , V2 = ( )2 1 AO A F X X r ! ! = 1.32(0.6 0.2) = 0.53 m3 For 2nd CSTR, For X3 = 0.65, 3 2AOF m rA = ! , V3 = 0.1 m3 P2-4 (e) Example 2-8 hrs5=! v0 = 1dm3 /min = 60dm3 /hr CA = 2.5 mol/dm3 X = 0.8 For CSTR, 0v V =! V = 300dm3

28. 2-7 (1) hrdmmol hrdmmol XC r A A 3 30 /4.0 / 5 8.05.2 = ! ==" # (2) V = 300dm3 (3)CA = CA0(1-X) = 0.5 mol/dm3 P2-5 X 0 0.1 0.2 0.4 0.6 0.7 0.8 FAO/-rA (m3 ) 0.89 1.08 1.33 2.05 3.54 5.06 8.0 V = 1.0 m3 P2-5 (a) Two CSTRs in series For first CSTR, V = (FAo/-rAX1) X1 => X1 = 0.44 For second CSTR, V = (FAo/-rAX2) (X2 X1) => X2 = 0.67 P2-5 (b) Two PFRs in series 1 2 10 X X Ao Ao A AX F F V dX dX r r ! " ! " = +# $ # $ % %& ' & ' ( ( By extrapolating and solving, we get X1 = 0.50 X2 = 0.74 P2-5 (c) Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 V = (0.5FAO/-rAX1) X1 Solving we get, Xout = 0.60 P2-5 (d) Two PFRs in parallel with the feed equally divided between the two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 By extrapolating and solving as part (b), we get Xout = 0.74

29. 2-8 P2-5 (e) A CSTR and a PFR are in parallel with flow equally divided Since the flow is divided equally between the two reactors, the overall conversion is the average of the CSTR conversion (part C) and the PFR conversion (part D) Xo = (0.60 + 0.74) / 2 = 0.67 P2-5 (f) A PFR followed by a CSTR, XPFR = 0.50 (using part(b)) V = (FAo/-rA-XCSTR) (XCSTR XPFR) Solving we get, XCSTR = 0.70 P2-5 (g) A CSTR followed by a PFR, XCSTR = 0.44 (using part(a)) ! " = PFR CSTR X X A AO dX r F V By extrapolating and solving, we get XPFR = 0.72 P2-5 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs, For PFR, XPFR = 0.50 (using part(b)) CSTR1: V = (FAo/-rA-XCSTR) (XCSTR XPFR) = 0.5 m3 XCSTR = 0.63 CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 XCSTR1) = 0.5 m3 XCSTR2 = 0.72 P2-6 (a) Individualized Solution P2-6 (b) 1) In order to find the age of the baby hippo, we need to know the volume of the stomach. The metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of what the mother eats then Fao (baby) = Fao (mother). The Levenspiel Plot is drawn for the baby hippo below.

30. 2-9 31.36 *0.34 0.23 2 baby A FaoX V m r = = = ! Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the babys stomach is half of an adult, then the baby hippo is half the age of a full grown hippo. 4.5 2.25 2 years Age years= = 2) If Vmax and mao are both one half of the mothers then 0 2 2 1 2 motherA Ao AM AM m m r r ! " # $ % &= ' ' and since max 2 A AM M A v C r K C ! = + then max 2 2 1 12 2baby mother A AM AM M A v C r r K C ! = = ! +

31. 2-10 2 2 2 1 2 1 2 Ao Ao Ao AM AM motherbaby AM mother mm m r rr ! " ! " # $ ! " = =# $ # $ # $# $% %& '# $& ' % & ' 2 Ao AM m r! will be identical for both the baby and mother. Assuming that like the stomach the intestine volume is proportional to age then the volume of the intestine would be 0.75 m3 and the final conversion would be 0.40 P2-6 (c) Vstomach = 0.2 m3 From the web module we see that if a polynomial is fit to the autocatalytic reaction we get: 0 1 A AM m r! = 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499 And since Vstomach = 0 1 A AM m r! X, solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3 Xstomach = .067. For the intestine: The Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178 Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.

32. 2-11 P2-6 (d) PFR CSTR PFR: Outlet conversion of PFR = 0.111 CSTR: We must solve V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499) X=0.42 Since the hippo gets a conversion over 30% it will survive. P2-7 Exothermic reaction: A B + C X r(mol/dm3 .min) 1/- r(dm3 .min/mol) 0 1 1 0.20 1.67 0.6 0.40 5 0.2 0.45 5 0.2 0.50 5 0.2 0.60 5 0.2 0.80 1.25 0.8 0.90 0.91 1.1 P2-7 (a) To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as given below. CSTR:

33. 2-12 Mole balance: ( )( ) ( )min./5 4.0min/300F V 3 A0 dmmol mol r X A CSTR = ! = => =>VCSTR = 24 dm3 PFR: Mole balance: ! " = X A APFR r dX FV 0 0 = 300(area under the curve) VPFR = 72 dm3 P2-7 (b) For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant over this conversion range. ! ! " = " = " = 6. 4. 6. 4. 0 6. 4. 00 X r F dX r F dX r F V A A A A A A PFR P2-7 (c) VCSTR = 105 dm3 Mole balance: A CSTR r X ! = A0F V moldm mol dm r X A min/35.0 min/300 105 3 3 == ! Use trial and error to find maximum conversion. At X = 0.70, 1/-rA = 0.5, and X/-rA = 0.35 dm3.min/mol Maximum conversion = 0.70

34. 2-13 P2-7 (d) From part (a) we know that X1 = 0.40. Use trial and error to find X2. Mole balance: ( ) 2 120 XA A r XXF V ! ! = Rearranging, we get 008.0 40.0 0 2 2 == ! ! AXA F V r X At X2 = 0.64, 008.0 40.0 2 2 = ! ! XAr X Conversion = 0.64 P2-7 (e) From part (a), we know that X1 = 0.40. Use trial and error to find X2. Mole balance: !! " = " == 22 40.040.0 0 30072 X A X A APFR r dX r dX FV At X2 = 0.908, V = 300 x (area under the curve) => V = 300(0.24) = 72dm3 Conversion = 0.908. P2-7 (f) See Polymath program P2-7-f.pol.

35. 2-14 P2-8 (a) S S r XF V ! = 0 FS0 = 1000 g/hr At a conversion of 40% g hrdm rS 3 15.0 1 = ! Therefore ( ) 3 60)40.0)(1000(15.0 dmV == P2-8 (b) At a conversion of 80%, g hrdm rS 3 8.0 1 = ! FS0 = 1000 g/hr Therefore ( ) 3 640)80.0)(1000(8.0 dmV == P2-8 (c) ! " = X S SPFR r dX FV 0 0 From the plot of 1/-rS Calculate the area under the curve such that the area is equal to V/FS0 = 80 / 1000 = 0.08 X = 12% For the 80 dm3 CSTR, S S r XF dmV ! == 03 80 X/-rs = 0.08. From guess and check we get X = 55%

36. 2-15 P2-8 (d) To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1/-rs. Next is a PFR with the necessary volume to achieve the 80% conversion following the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion For two CSTRs in series, the optimum arrangement would still include a CSTR with the volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1/-rs,first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR. P2-8 (e) SM CS s CK CkC r + =! and [ ] 001.01.0 0 +!= SSC CCC [ ]( ) SM SSS s CK CCkC r + +! =! 001.01.0 0 [ ]( )001.01.0 1 0 +! + =! SSS SM s CCkC CK r Let us first consider when CS is small. CS0 is a constant and if we group together the constants and simplify then CskCk CK r S SM s 2 2 1 1 + + =! since CS < KM CskCk K r S M s 2 2 1 1 + =! which is consistent with the shape of the graph when X is large (if CS is small X is large and as CS grows X decreases). Now consider when CS is large (X is small) As CS gets larger CC approaches 0: [ ] 001.01.0 0 +!= SSC CCC and 0SS CC ! If SM CS s CK CkC r + =! then CS SM s CkC CK r + =! 1 As CS grows larger, CS >> KM

37. 2-16 And CCS S s kCCkC C r 11 ==! And since CC is becoming very small and approaching 0 at X = 0, 1/-rs should be increasing with CS (or decreasing X). This is what is observed at small values of X. At intermediate levels of CS and X, these driving forces are competing and why the curve of 1/-rS has a minimum. P9-2 Irreversible gas phase reaction 2A + B 2C See Polymath program P2-9.pol. P2-9 (a) PFR volume necessary to achieve 50% conversion Mole Balance ! " = 2 1 )( 0 X X A A r dX FV Volume = Geometric area under the curve of (FA0/-rA) vs X) ( )5.01000005.0400000 2 1 !+" # $ % & ' !!=V V = 150000 m3 One of the points of this problem is for the students to recognize 150,000 m 3 is ridiculously reactor volume. P2-9 (b) CSTR Volume to achieve 50% conversion Mole Balance )( 0 A A r XF V ! = 1000005.0 !=V V = 50000m3 One of the points of this problem is for the students to recognize 150,000 m 3 is ridiculously reactor volume.

38. 2-17 P2-9 (c) Volume of second CSTR added in series to achieve 80% conversion )( )( 120 2 A A r XXF V ! ! = )5.08.0(5000002 !"=V V2 = 150000m3 Ridiculous volume P2-9 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion )3.0100000()3.0400000 2 1 ( !+!!=PFRV VPFR = 90000m3 Ridiculous volume P2-9 (e) For CSTR, V = 60000 m3 (CSTR) Ridiculous volume Mole Balance XX r XF V A A )500000800000(60000 )( 0 +!= ! = X = 0.463 For PFR, V = 60000 m3 (PFR) Mole balance ! " = X A A r dX FV 0 0 )( dXX X )100000800000(60000 0 ! +"= X = 0.134

39. 2-18 P2-9 (f) Real rates would not give that shape. The reactor volumes are absurdly large. P2-10 Problem 2-10 involves estimating the volume of three reactors from a picture. The door on the side of the building was used as a reference. It was assumed to be 8 ft high. The following estimates were made: CSTR h = 56ft d = 9 ft V = r2 h = (4.5 ft)2 (56 ft) = 3562 ft3 = 100,865 L PFR Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(11) = 3036 ft D = 1 ft V = r2 h = (0.5 ft)2 (3036 ft) = 2384 ft3 = 67,507 L Answers will vary slightly for each individual. P2-11 No solution necessary. P2-12 (a) The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in series and containing equal amounts of catalyst can be calculated from the figure below.

40. 2-19 The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR. This figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR. See Polymath program P2-12.pol. P2-12 (b) Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining the area of the shaded region in the figure below. The area of the rectangle is approximately 23.2 kg of catalyst. P2-12 (c) The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of the shaded rectangle shown in the figure below. The area of the rectangle is approximately 7.6 kg of catalyst.

41. 2-20 P2-12 (d) The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the shaded region in the figure below. The necessary catalyst weight is approximately 22 kg. P2-12 (e) The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating the area of the shaded region in the graph below. The necessary catalyst weight is approximately 13 kg.

42. 2-21 P2-12 (f) P2-12 (g) For different (-rA) vs. (X) curves, reactors should be arranged so that the smallest amount of catalyst is needed to give the maximum conversion. One useful heuristic is that for curves with a negative slope, it is generally better to use a CSTR. Similarly, when a curve has a positive slope, it is generally better to use a PBR. CDP2-A (a) Over what range of conversions are the plug-flow reactor and CSTR volumes identical? We first plot the inverse of the reaction rate versus conversion. Mole balance equations for a CSTR and a PFR: CSTR: A A r XF V ! = 0 PFR: ! " = X Ar dX V 0 Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical.

43. 2-22 i.e. CSTR A A A A A PFR V r XF dX r F r dX V = ! =" ! =" ! = 0 5.0 0 0 5.0 0 CDP2-A (b) What conversion will be achieved in a CSTR that has a volume of 90 L? For now, we will assume that conversion (X) will be less that 0.5. CSTR mole balance: A A A A r XCv r XF V ! = ! = 000 13 3 8 3 3 3 00 103 . 1032005 09.0 ! "= """ = ! = mol sm m mol s m m r Cv V X A A CDP2-A (c) This problem will be divided into two parts, as seen below: The PFR volume required in reaching X=0.5 (reaction rate is independent of conversion). 311000 1 105.1 m r XCv r XF V A A A A != " = " = The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).

44. 2-23 Finally, we add V2 to V1 and get: Vtot = V1 + V2 = 2.3 x1011 m3 CDP2-A (d) What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to raise the conversion to 90 % We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our CSTR mole balance equation: 311000 104.1 m r XCv r XF V A A A A CSTR != " # = " # = CDP2-A (e) If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40% conversion? Since there is no flow into or out of the system, mole balance can be written as: Mole Balance: dt dN Vr A A = Stoichiometry: )1(0 XNN AA !=

45. 2-24 Combine: dt dX NVr AA 0= From the stoichiometry of the reaction we know that V = Vo(1+eX) and e is 1. We insert this into our mole balance equation and solve for time (t): dt dX X N V r A A =+! )1( 0 0 ! +" =! X A A t Xr dX Cdt 0 0 0 )1( After integration, we have: )1ln( 1 0 XC r t A A + ! = Inserting the values for our variables: t = 2.02 x 1010 s That is 640 years. CDP2-A (f) Plot the rate of reaction and conversion as a function of PFR volume. The following graph plots the reaction rate (-rA) versus the PFR volume: Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.

46. 2-25 The volume required for 99% conversion exceeds 4*1011 m3 . CDP2-A (g) Critique the answers to this problem. The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactor of geological proportions (a CSTR or PFR approximately the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time. CDP2-B Individualized solution CDP2-C (a) For an intermediate conversion of 0.3, Figure below shows that a PFR yields the smallest volume, since for the PFR we use the area under the curve. A minimum volume is also achieved by following the PFR with a CSTR. In this case the area considered would be the rectangle bounded by X =0.3 and X = 0.7 with a height equal to the CA0/-rA value at X = 0.7, which is less than the area under the curve.

47. 2-26 CDP2-C (b) CDP2-C (c) CDP2-C (d) For the PFR,

48. 2-27

49. 2-28 CDP2-C (e) CDP2-D

50. 2-29 CDP2-D (a) CDP2-D (b) CDP2-D (c) CDP2-D (d)

51. 2-30 CDP2-D (e) CDP2-D (f) CDP2-D (g) CDP2-D (h)

52. 2-31 CDP2-E CDP2-F (a) Find the conversion for the CSTR and PFR connected in series. X -rA 1/(-rA) 0 0.2 5 0.1 0.0167 59.9 0.4 0.00488 204.9 0.7 0.00286 349.65 0.9 0.00204 490.19

53. 2-32 CDP2-F (b)

54. 2-33 CDP2-F (c) CDP2-F (d)

55. 2-34 CDP2-F (e) CDP2-F (f) CDP2-F (g) Individualized solution

56. 3-1 Solutions for Chapter 3 Rate Law and Stoichiometry P3-1 Individualized solution. P3-2 (a) Example 3-1 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 310 315 320 325 330 335 T (K) k(1/s) For E = 60kJ/mol For E1 = 240kJ/mol ! k =1.32 "1016 exp #60000J RT $ % & ' ( ) ! k1=1.32 "1016 exp #240000J RT $ % & ' ( ) E = 240 kj/mol 0 5E-23 1E-22 1.5E-22 2E-22 2.5E-22 3E-22 3.5E-22 0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325 1/T (1/K) k(1/s) E = 60 kj/mol 0 1000000 2000000 3000000 4000000 5000000 6000000 0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325 1/T (1/K) k(1/s) P3-2 (b) Example 3-2 Yes, water is already considered inert. P3-2 (c) Example 3-3 The solution to the example at a conversion of 20% would remain unchanged. For 90 % conversions of the caustic soda, the final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of caustic soda is possible.

57. 3-2 P3-2 (d) Example 3-4 ! A + b a B " c a C + d a D So, the minimum value of B = b/a = ! 1/3 1 = 0.33 P3-2 (e) Example 3-5 For the concentration of N2 to be constant, the volume of reactor must be constant. V = VO. Plot: 2 1 0.5(1 0.14 ) (1 )(0.54 0.5 )A X r X X ! = ! ! ! 1/(-ra) vs X 0 20 40 60 80 100 120 140 160 180 0 0.2 0.4 0.6 0.8 1 1.2 X 1/(-ra) The rate of reaction decreases drastically with increase in conversion at higher conversions. P3-2 (f) Example 3-6 For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor. Therefore the reverse reaction decreases. CT0 = constant and inerts are varied. ! N2O4 " 2NO2 A 2B Equilibrium rate constant is given by: ! KC = CB,e 2 CA,e Stoichiometry: ! " = yA0# = yA0(2 $1) = yA0 Constant volume Batch: ! CA = NA0(1" X) V0 = CA0(1" X) and ! CB = 2NA0X V0 = 2CA0X Plug flow reactor: ! CA = FA0(1" X) v0(1+ #X) = CA0(1" X) (1+ #X) and ! CB = 2FA0X v0(1+ "X) = 2CA0X (1+ "X)

58. 3-3 ! CAO = yAOPO RTO = yAO 0.07176( )mol/dm3 Combining: For constant volume batch: ! KC = CB,e 2 CA,e = 4CAo 2 X2 CAO (1" X) ! " Xe = KC (1# Xe) 4CA0 For flow reactor: )1)(1( 4 22 , 2 , XXC XC C C K AO Ao eA eB C !+" == 04 )1)(1( A eeC e C XXK X !+" =# See Polymath program P3-2-f.pol. POLYMATH Results NLES Report (safenewt) Nonlinear equations [1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao))^0.5 = 0 [2] f(Xef) = Xef - (kc*(1-Xef)*(1+eps*Xef)/(4*Cao))^0.5 = 0 Explicit equations [1] yao = 1 [2] kc = 0.1 [3] Cao = 0.07174*yao [4] eps = yao 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 y inerts EquilibriumConversion Batch Flow

59. 3-4 Yinert Yao Xeb Xef 0 1 0.44 0.508 0.1 0.9 0.458 0.5217 0.2 0.8 0.4777 0.537 0.3 0.7 0.5 0.5547 0.4 0.6 0.525 0.576 0.5 0.5 0.556 0.601 0.6 0.4 0.5944 0.633 0.7 0.3 0.6435 0.6743 0.8 0.2 0.71 0.732 0.9 0.1 0.8112 0.8212 0.95 0.05 0.887 0.89 0.956 0.044 0.893 0.896 P3-2 (g) No solution will be given P3-2 (h) A + 2 1 B 2 1 C Rate law: ! "rA = kACA 2 CB and ! kA = 25 1 s dm3 mol " # $ % & ' 2 ! rA "1 = rB "1/2 = rC 1/2 ! " 25CA 2 CB = kBCA 2 CB 1/2 = kCCA 2 CB #1/2 ! kC = kB =12.5 1 s dm3 mol " # $ % & ' 2 P3-2 (i) ! A + 3B " 2C Rate law: ! "rA = kACACB at low temperatures. At equilibrium, ! KC = CC,e CA,e 1/2 CB,e 3/2 ! ! CA 1/2 CB 3/2 " CC KC = 0 At equilibrium, -rA = 0 , so we can suggest that ! "rA = kA CA 1/2 CB 1/2 " CC KC # $ % & ' ( But at t = 0, CC = 0 So the rate law is not valid at t = 0. Next guess: ! KC 2 = CC,e 2 CA,eCB,e 3 , or ! CACB " CC 2 KC 2 CB 2 = 0 ! ! "rA = kA CACB " CC 2 KC 2 CB 2 # $ %% & ' (( which satisfies both the initial conditions and equilibrium rate law.

60. 3-5 Hence ! "rA = kA CACB " CC 2 KC 2 CB 2 # $ %% & ' (( is the required rate law. P3-3 Solution is in the decoding algorithm available separately from the author. P3-4 (a) Note: This problem can have many solutions as data fitting can be done in many ways. Using Arrhenius Equation For Fire flies: T(in K) 1/T Flashes/ min ln(flashe s/min) 294 0.003401 9 2.197 298 0.003356 12.16 2.498 303 0.003300 16.2 2.785 Plotting ln(flashes/min) vs 1/T, we get a straight line. See Polymath program P3-4-fireflies.pol. For Crickets: T(in K) 1/T x10 3 chrips/ min ln(chirps/ min) 287.2 3.482 80 4.382 293.3 3.409 126 4.836 300 3.333 200 5.298 Plotting ln(chirps/min) Vs 1/T, we get a straight line. Both, Fireflies and Crickets data follow the Arrhenius Model. ln y = A + B/T , and have the same activation energy. See Polymath program P3-4-crickets.pol. P3-4 (b) For Honeybee: T(in K) 1/T x10 3 V(cm/s) ln(V) 298 3.356 0.7 -0.357 303 3.300 1.8 0.588 308 3.247 3 1.098 Plotting ln(V) Vs 1/T, almost straight line. ln(V) = 44.6 1.33E4/T At T = 40o C(313K) V = 6.4cm/s At T = -5o C(268K) V = 0.005cm/s(But bee would not be alive at this temperature) See Polymath program P3-4-bees.pol.

61. 3-6 P3-4 (c) For ants: T(in K) 1/T x10 3 V(cm/s) ln(V) 283 3.53 0.5 -0.69 293 3.41 2 0.69 303 3.30 3.4 1.22 311 3.21 6.5 1.87 Plotting ln(V) Vs 1/T, almost straight line. See Polymath program P3-4-ants.pol. So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same. Insect Activation Energy Cricket 52150 Firefly 54800 Ant 95570 Honeybee 141800 P3-4 (d) There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would be helpful is the maximum and the minimum temperature that these insects can endure before death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless. P3-5 There are two competing effects that bring about the maximum in the corrosion rate: Temperature and HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the column and consequently the rate of corrosion should increase. However, the HCN concentrations (and the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion rate somewhere around the middle of the column. P3-6 Antidote did not dissolve from glass at low temperatures. P3-7 (a) If a reaction rate doubles for an increase in 10C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1. Then with k = Ae-E/RT in general, 1/ 1 E RT k Ae! = and 2/ 2 E RT k Ae! = , or

62. 3-7 2 1 1 1 2 1 E R T Tk e k ! " # #$ % & ' = or ( ) 2 2 1 1 1 2 1 22 1 ln ln 1 1 k k k kE T TR TTT T ! " ! " # $ # $ % & % &= ' = ' '! " '# $ % & Therefore: ( )( ) ( ) ( ) ( )( ) 2 1 1 1 11 2 1 ln 10 ln 2 10 10 k T T T Tk E R R T T ! " +# $ +% &= = ' ( )1 1 10 10 ln 2 E T T R + = which can be approximated by 0.5 10 ln 2 E T R = P3-7 (b) Equation 3-18 is E RT k Ae ! = From the data, at T1 = 0C, 1/ 1 E RT k Ae! = , and at T2 = 100C, 2/ 2 E RT k Ae! = Dividing gives 2 1 1 1 2 1 E R T Tk e k ! " # #$ % & ' = , or ( ) 2 1 1 2 2 1 2 1 2 1 ln ln 1 1 k R k RTT k E T T k T T ! " # $ ! "% &= ' = ' # $ '! " % &'# $ % & [ ][ ]1.99 273 373 .050 ln 7960 100 .001 cal K K mol K cal E K mol ! " # $ % &' (= =) * + , ( ) 1 3 1 1 1 7960 10 min exp 2100min 1.99 273 E RT cal molA k e cal K mol K ! ! ! " # $ % $ %= = = $ %& ' $ %( ) * +, - P3-7 (c) Individualized solution

63. 3-8 P3-8 When the components inside air bag are ignited, following reactions take place, 2NaN3 2Na + 3N2 ..(1) 10Na + 2KNO3 K2O + 5Na2O + N2 (2) K2O + Na2O + SiO2 alkaline silicate glass ...(3) ( ) ( ) ( ) ( )43215 rxnrxnrxnrxn =++! NaN3 + 0.2KNO3 + 0.1SiO2 0.4Na2O + 1.6N2 + complex/10(4) Stoichiometric table: Species Symbol Initial Change Final NaN3 A NA -NAX NA(1-X) KNO3 B NA B! -0.2X NA NA( B! - 0.2X) SiO2 C NA C! -0.1XNA NA( C! - 0.1X) Na2O D 0 0.4XNA 0.4XNA N2 E 0 1.6XNA 1.6XNA Given weight of NaN3 = 150g Mwt of NaN3 = 65 Therefore, no. of moles of NaN3 = 2.3 1 moles of NaN3 requires 0.2 mole of KNO3 => Moles of B, KNO3 = 0.2(2.3) = 0.46 moles Mwt of KNO3 = 101.1 Therefore, grams of KNO3 required = 0.46 x 101.1 = 46.5 g 1 moles of NaN3 requires 0.1 mole of SiO2. Moles of C, SiO2 = 0.1(2.3) = 0.23 moles Mwt of SiO2 = 60.08 Therefore, grams of SiO2 required = 0.23 x 60.08 = 13.8 g Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards: Store cars in cool, dry, ventilated areas. Avoid Physical damage of the bag in car. It is stable under ordinary conditions of storage. Decomposes explosively upon heating (over 221F or 105 C), shock, concussion, or friction. Conditions to avoid: heat, flames, ignition sources and incompatibles. P3-9 (a) From the web module we know that (1 ) dX k x dt = ! and that k is a function of temperature, but not a linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and therefore halve the cooking time. P3-9 (b) When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only be 100C. When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than double that of boiling water. P3-9 (c) No solution will be given

64. 3-9 P3-10 (a) 1) C2H6 C2H4 + H2 Rate law: -rA = 62HCkC 2) C2H4 + 1/2O2 C2H4O Rate law: -rA = 2/1 0242 CkC HC 3) (CH3)3COOC(CH3)3 C2H6 + 2CH3COCH3 A B + 2C Rate law: -rA = k[CA CBCC 2 /KC] 4) n-C4H10 I- C4H10 Rate law: -rA = k[ 104HnCC 104HiCC /Kc] 5) CH3COOC2H5 + C4H9OH CH3COOC4H9 + C2H5OH A + B C + D Rate law: -rA = k[CACB CCCD/KC] P3-10 (b) 2A + B C (1) -rA = kCACB 2 (2) -rA = kCB (3) -rA = k (4) -rA = kCACB -1 P3-10 (c) (1) C2H6 C2H4 + H2 Rate law: -rA = 62HCkC (2) H2 + Br2 2HBr Rate law: -rHBr = 2 22 2 2/1 1 Br HBr BrH C C k CCk + (3) H2 + I2 2HI Rate law: 222 IHH CkCr =! P3-11 (a) Liquid phase reaction, O CH2--OH CH2 - CH2 + H2O CH2--OH A + B C CAO = 1 lbmol/ft3 CBO = 3.47 lbmol/ft3 Stoichiometric Table: Species Symbol Initial Change Remaining Ethylene oxide A CAO=1 lbmol/ft3 - CAOX CA= CAO(1-X) = (1-X) lbmol/ft3 Water B CBO= 3.47 lbmol/ft3 , B! =3.47 -CAOX CB = CAO( B! -X) =(3.47-X) lbmol/ft3 Glycol C 0 CAOX CC = CAOX = X lbmol/ft3 Rate law: -rA = kCACB

65. 3-10 Therefore, -rA = k 2 AOC (1-X)( B! -X) = k(1-X)(3.47-X) At 300K E = 12500 cal/mol, X = 0.9, k = 0.1dm3 /mol.s =1.6018 slbmolft ./3 CSTR! = AO A C X r! = ( )( ) ( )( ) ( )( ) sec186.2 9.047.39.0116018.1 9.01 2 = !! At 350K, k2 = k exp((E/R)(1/T-1/T2))= 1.6018exp((12500/1.987)(1/300-1/350)) = 32.034 dm3 /mol.s Therefore, CSTR! = AO A C X r! = ( )( ) ( )( ) ( )( ) sec109.0 9.047.39.011034.32 9.01 2 = !! P3-11 (b) Isothermal, isobaric gas-phase pyrolysis, C2H6 C2H4 + H2 A B + C Stoichiometric table: Species symbol Entering Change Leaving C2H6 A FAO -FAOX FA=FAO(1-X) C2H4 B 0 +FAOX FB=FAOX H2 C 0 +FAOX FC=FAOX FTO=FAO FT=FAO(1+X) ! = yao ! = 1(1+1-1) = 1 v = vo(1+! X) => v = vo(1+X) CAO = yAO CTO = yAO P RT = ( )( ) ( ) 3 1 6 0.082 1100 . atm m atm K K kmol ! " # $ % & = 0.067 kmol/m3 = 0.067 mol/dm3 CA = ( ) ( ) 1(1 ) (1 ) 1 AOA AO O XF XF C v v X X !! = = + + mol/dm3 CB = ( ) ( ) (1 ) 1 AOB AO O F XF X C v v X X = = + + mol/dm3 CC = ( ) ( ) (1 ) 1 C AO AO O F F X X C v v X X = = + + mol/dm3 Rate law: -rA = kCA = kCAO ( ) ( ) 1 1 X X ! + =0.067 k ( ) ( ) 1 1 X X ! + If the reaction is carried out in a constant volume batch reactor, =>( ! = 0) CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm3 CC = CAO X mol/dm3 P3-11 (c)

66. 3-11 Isothermal, isobaric, catalytic gas phase oxidation, C2H4 + 1 2 O2 C2H4O A + 1 2 B C Stoichiometric table: Species Symbol Entering Change Leaving C2H4 A FAO -FAOX FA=FAO(1-X) O2 B FBO - B! FAOX FB=FAO( B! -X) C2H4O C 0 +FAOX FC=FAOX B! = 1 12 2 AO BO AO AO FF F F = = 2 3 AO AO AO TO AO BO F F y F F F = = = + 2 1 1 1 0.33 3 2 AOy! " # $ = = % % =& ' ( ) ( ) ( ) 33 62 0.092 3 . 0.082 533 . AO AO TO AO atmP mol C y C y RT dmatm dm K mol K = = = = ! " # $ % & ( ) ( ) ( ) ( ) ( ) ( ) 1 1 0.092 1 1 1 0.33 1 0.33 AO AOA A O F X C X XF C v v X X X! " " " = = = = + " " ( ) ( ) ( ) 0.046 12 1 1 0.33 AO B B B O X F XF C v v X X ! " # $ %& ' %( )= = = + % ( ) ( ) ( ) 0.092 1 1 0.33 C AO C O XF F X C v v X X! = = = + " If the reaction follow elementary rate law Rate law: 0.5 A A Br kC C! = ! ( ) ( ) ( ) ( ) 0.5 0.092 1 0.046 1 1 0.33 1 0.33 A X X r k X X ! "! "# #$ $$ $ # = % &% & # #$ $$ $' (' ( P3-11 (d) Isothermal, isobaric, catalytic gas phase reaction in a PBR C6H6 + 2H2 C6H10 A + 2B C Stoichiometric table: Species Symbol Entering Change Leaving Benzene A FAO -FAOX FA=FAO(1-X) H2 B FBO=2FAO -2FAOX FB=FAO( B! -2X) C6H10 C 0 FAOX FC=FAOX

67. 3-12 ( ) 2 2 1 3 1 2 1 2 1 3 3 BO AO B AO AO AO AO AO TO AO BO AO F F F F F F y F F F y ! " # = = = = = = + = = $ $ = $ =AOC ( ) 3 3 1 6 1 0.055 / 3 3. 0.082 443.2 . TO AO P atm CAO C y mol dm RT atm dm K mol K ! " ! " = = = =# $ # $ ! "% & % & # $ % & ( ) ( ) ( ) ( )1 1 0.055 1 2 21 1 1 3 3 AO AOA A O F X C X XF C v v X X X ! " " " = = = = + # $ # $ " "% & % & ' ( ' ( ( ) ( ) ( )2 0.11 1 21 1 3 AO BB B O F X XF C v v X X ! " # # = = = + $ % #& ' ( ) ( ) 0.055 2 21 1 1 3 3 C AO AO C O F F X C X X C v v X X X ! = = = = + " # " # $ $% & % & ' ( ' ( If the reaction follow elementary rate law. Rate law: ( ) 2 3 3 ' 1 ' 0.0007 2 1 3 A A B A r kC C X r k X ! = ! ! = " # !$ % & ' For a fluidized CSTR: 0 ' A A F X W r = ! ( ) 0 3 3 1 0.0007 2 1 3 AF X W X k X = ! " # !$ % & ' k = 53 3 min mol kgcat atm at 300K @ T = 170oC 1 1 1 1 80000 1 1 exp 53exp 8.314 300 443 E k k R T T ! "! " ! "! " = # = #$ %$ % $ %$ %$ % & '& '& '& ' = 1663000 3 min mol kgcat atm

68. 3-13 FA0 = CA0* v0 v0 = 5 dm3 /min ( ) 0 0 3 3 1 0.0007 2 1 3 AC v X W X k X = ! " # !$ % & ' at X = 0.8 3 104.2 ! "=W kg of catalyst @ T = 270oC 3 1 1 atmminkgcat mol 90787719 543 1 300 1 314.8 80000 exp53 11 exp !! ="" # $ %% & ' " # $ % & ' (=" " # $ % % & ' "" # $ %% & ' (= TTR E kk FA0 = CA0* v0 v0 = 5 dm3 /min ( ) 0 0 3 3 1 0.0007 2 1 3 AC v X W X k X = ! " # !$ % & ' at X = 0.8 5 104.4 ! "=W kg of catalyst P3-12 C2H4 + 1 2 O2 C2H4O A + 1 2 B C Stoichiometric table for the given problem will be as follows Assuming gas phase Species Symbol Entering Change Leaving C2H4 A FA0 - FA0X FA0(1-X) O2 B FB0 = BFA0 -1/2 FA0X FA0(B X/2) N2 I FI = IFA0 ------- FA0I C2H4O C 0 FA0X FA0X ! "B = 1 2 FA0 FA0 = 1 2 ! "I = FIO FAO ,FIO = 0.79 0.21 FBO ! " #I = #B 0.79 0.21 =1.88

69. 3-14 ! yA0 = FA0 FT 0 = 0.30, ! " = yA0# = $0.15 ! CA0 = yA0P RT = 0.041 mol dm3 ! CA = FA0 v = CA0 (1" X) (1+ #X) = 0.041(1" X) 1" 0.15X ! CB = FB v = CA0 ( 1 2 " 1 2 X) 1" 0.15X = 0.020(1" X) 1" 0.15X ! CC = FC v = CA0X 1" 0.15X = 0.041X 1" 0.15X P3-13 (a) Let A = ONCB C = Nibroanaline B = NH3 D = Ammonium Chloride A + 2B C + D - A A Br kC C !!" = P3-13 (b) Species Entering Change Leaving A FA0 - FA0X FA0(1-X) B FB0 = BFA0 =6.6/1.8 FA0 -2 FA0X FB= FA0(B 2X) C 0 FA0X FC= FA0X D 0 FA0X FD=FA0X P3-13 (c) For batch system, CA=NA/V -rA = kNANB/V2 P3-13 (d) - A A Br kC C= ( ) ( ) ( )0 0 0 0 0 0 1 1 , 1AA A A A A A A A NN N F F F X C X C C X V V V v v = = = ! = ! = = = ! ( ) ( ) ( )0 0 0 0 0 0 2 2 , 2AB B B B B A B B A B NN N F F X C X C C X V V V v ! ! != = = " = " = = " ( )( )2 0 1 2A A Br kC X X!" = " " 0 0 6.6 3.67 1.8 B B A C C ! = = =

70. 3-15 0 3 1.8A kmol C m = ( ) ( )( ) 2 1.8 1 3.67 2Ar k X X! = ! ! P3-13 (e) 1) At X = 0 and T = 188C = 461 K min 0202.067.38.1 min 0017.0 3 2 3 3 2 00 m kmol m kmol kmol m kCr BAA =! " # $ % & ='=( 0 3 kmol 0.0202 m min Ar! = 2) At X = 0 and T = 25C = 298K ! ! " # $ $ % & !! " # $$ % & '= TTR E kk O O 11 exp min. 3 1003.2 298 1 461 1 . 987.1 11273 exp min. 0017.0 6 3 kmol m Kmol cal mol cal kmol m k ! "= # # # # $ % & & & & ' ( # $ % & ' ( != -rAO = kCAOCBO = 2.41 X 10-5 kmol/m3 min 3) 0 0 1 1 exp E k k R T T ! "# $ = %& '( ) & '* +, - 3 11273 m 1 1 0.0017 exp kmolmin 461 5611.987 cal molk cal K K mol K ! " # $% & = # ' $( ) # $* + # $, - 3 m 0.0152 kmolmin k = 0 0 0A A Br kC C! = 3 3 3 m 0.0152 1.8 6.6 kmolmin A kmol kmol r m m ! "! " # = $ %$ % & '& ' 3 kmol 0.1806 m min Ar! = P3-13 (f) rA = kCAO 2 (1-X)(B-2X)

71. 3-16 At X = 0.90 and T = 188C = 461K 1) at T = 188 C = 461 K ( ) ( )( ) min 00103.0 9.0267.39.018.1 min. 0017.0 3 2 3 3 m kmol m kmol kmol m rA = !!" # $ % & ' "" # $ %% & ' =! 2) At X = 0.90 and T = 25C = 298K ( ) ( )( ) min 1023.1 9.0267.39.018.1 min. 1003.2 3 6 2 3 3 6 m kmol m kmol kmol m rA ! ! "= !!# $ % & ' ( ## $ % && ' ( "=! 3) At X = 0.90 and T = 288C = 561K ( ) ( )( ) min 0092.0 9.0267.39.018.1 min. 0152.0 3 2 3 3 m kmol m kmol kmol m rA = !!" # $ % & ' "" # $ %% & ' =! P3-13 (g) FAO = 2 mol/min 1) For CSTR at 25o C -rA min 1023.1 3 6 m kmol! "= ( ) 3 3 3 9.0 60.162 min 1023.1 1.0min/2 1 m m mol mol r XF V XA AO = ! ! = " " = " =M 2)At 288o C, -rA min 0092.0 3 m kmol = ( ) 3 3 9.0 739.21 min 0092.0 1.0min/2 1 m m mol mol r XF V XA AO = ! = " " = =M P3-14 C6H12O6 + aO2 + bNH3 c(C4.4H7.3N0.86O1.2) + dH2O + eCO2 To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients a, b, c, d, and e. This can be done with mass balances on each element involved in the reaction. Once all the coefficients are found, you can then calculate the yield coefficients by simply assuming the reaction

72. 3-17 proceeds to completion and calculating the ending mass of the cells. P3-14 (a) Apply mass balance For C 6 = 4.4c + e For O 6 + 2a = 1.2c + d + 2e For N b = 0.86c For H 12 + 3b = 7.3c + 2d Also for C, 6(2/3) = 4.4c which gives c = 0.909 Next we solve for e using the other carbon balance 6 = 4.4 (0.909) + e e = 2 We can solve for b using the nitrogen balance b = 0.86c = 0.86* (0.909) b = 0.78 Next we use the hydrogen balance to solve for d 12 + 3b = 7.3c + 2d 12 + 3(0.78) = 7.3(0.909) + 2d d = 3.85 Finally we solve for a using the oxygen balance 6 + 2a = 1.2c + d + 2e 6 + 2a = 1.2(0.909) + 3.85 + 2(2) a = 1.47 P3-14 (b) Assume 1 mole of glucose (180 g) reacts: Yc/s= mass of cells / mass of glucose = mass of cells / 180 g mass of cells = c*(molecular weight) = 0.909 mol* (91.34g/mol) mass of cells = 83.12 g Yc/s = 83.12 g / 180 g Yc/s = 0.46 Yc/o2 = mass of cells / mass of O2 If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83.12 g of cells are produced. mass of O2 = 1.47 mol * (32 g/mol) mass of O2 = 47.04 g Yc/o2 = 83.12 g /47.04 g Yc/o2 =1.77 P3-15 (a) Isothermal gas phase reaction.

73. 3-18 2 2 3 1 3 2 2 N H NH+ ! Making H2 as the basis of calculation: 2 2 3 1 2 3 3 H N NH+ ! 1 2 3 3 A B C+ ! Stoichiometric table: Species Symbol Initial change Leaving H2 A FAO -FAOX FA=FAO(1-X) N2 B FBO= B! FAO -FAOX/3 FB=FAO( B! -X/3) NH3 C 0 +2FAOX/3 FC=(2/3)FAOX P3-15 (b) 2 1 2 1 3 3 3 2 1 0.5 3 3 AOy ! " ! # $ = % % = %& ' ( ) # $ = = * % = %& ' ( ) ( ) ( ) 3 16.4 0.5 . 0.082 500 . AO atm C atm dm K mol K = ! " # $ % & = 0.2 mol/dm3 ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 3 1 0.2 1 0.1 / 1 1 3 1 0.22 2 0.1 / 3 1 3 1 3 AO H A AO NH C C X X C C mol dm XX C X X C C mol dm XX ! ! " " = = = = + # $ "% & ' ( " = = ) = ) = + # $ "% & ' ( P3-15 (c) kN2 = 40 dm3 /mol.s (1) For Flow system: ( ) ( ) 2 2 2 2 31 2 2 31 2 2 2 1 13 40 1 1 3 3 N N N H AO r k C C X X C X X ! " ! "# = $ % $ % ! " ! "& ' #( )* + * +#, -* + * += & ' & '* + * +# #( ) ( )* + * +, - , -$ % $ %

74. 3-19 ! "rN2 =1.6 1" X( ) 1" X 3 # $ % & ' ( ) * + + + + , - . . . . 3 2 (2) For batch system, constant volume. ( ) ( ) [ ] 2 2 2 2 2 31 2 2 1 2 32 2 1 2 3 2 40 1 1 3 1.6 1 1 3 N N N H AO N r k C C X C X X r X ! " ! "# = $ % $ % ! "& ' = # #! "( )* + $ % , -$ % ! " # = # #* +$ % P3-16 (a) Liquid phase reaction assume constant volume Rate Law (reversible reaction): C A A B C C r k C C K ! " # = #$ % & ' Stoichiometry: ( )0 1A AC C X= ! , ( )0 1B AC C X= ! , 0C AC C X= To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe. A B C CC C K C= ( ) 22 0 01A e C A eC X K C X! = 2 0 1 2 1 0e e A C X X C K ! " # + + =$ % & ' 0.80eX = To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric relations. ( ) ( )0 3 3 1 2 1 0.80 0.4A A mol mol C C X dm dm = ! = ! = ( ) ( )0 3 3 1 2 1 0.80 0.4B A mol mol C C X dm dm = ! = ! = 0 3 3 2 *0.80 1.6A A mol mol C C X dm dm = = = P3-16 (b) Stoichiometry:

75. 3-20 ( )( )0 1 3 1 2Ay! "= = # = and 0C! = ( ) ( ) ( ) ( ) 0 0 0 1 1 1 1 2 AA A A N X XN C C V V X X! " " = = = + + ( ) ( ) 0 0 0 3 3 1 1 2 C A C A N N X X C C V V X X! = = = + + Combine and solve for Xe. ( ) ( ) ( ) 3 0 0 1 3 1 2 1 2 e e C A A e e X X K C C X X ! "# = $ % + +$ %& ' ( )( ) 2 2 3 01 1 2 27C e e A eK X X C X! + = 2 3027 4 3 1 0A e e C C X X K ! " # + + + =$ % & ' 0.58eX = Equilibrium concentrations: ( ) 0 0 33 0 10 0.305 400 0.082 A P atm mol C RT dmdm atm K mol K = = = ! " # $ % & ( ) ( )( ) 3 1 0.58 0.305 0.059 1 2 0.58 A mol C dm ! = = + ( )( ) ( )( ) 3 3 0.58 0.305 0.246 1 2 0.58 C mol C dm = = + P3-16 (c) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction. Stoichiometry: ( ) ( )0 0 0 1 1AA A A N XN C C X V V ! = = = ! 0 0 0 3 3C A C A N N X C C X V V = = = Combine and solve for Xe ( ) ( ) 3 0 01 3C A e A eK C X C X! = 0.39eX = Equilibrium concentrations

76. 3-21 ( )( ) 3 0.305 1 0.39 0.19A mol C dm = ! = ( )( ) 3 0.305 0.39 0.36C mol C dm = = P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction): 3 C A A C C r k C K ! " # = #$ % & ' Stoichiometry: ( )( )0 1 3 1 2Ay! "= = # = and 0C! = ( ) ( ) ( ) ( ) 0 0 0 1 1 1 1 2 AA A A N X XN C C V V X X! " " = = = + + ( ) ( ) 0 0 0 3 3 1 1 2 C A C A N N X X C C V V X X! = = = + + Combine and solve for Xe: ( ) 3 0 0 1 3 1 2 1 2 C A e A e e e K C X C X X X ! " # = $ % + +& ' 0.58eX = Equilibrium concentrations: ( ) ( ) 3 0.305 1 0.58 0.059 1 2 0.58 A mol C dm ! = = + ( )( ) ( ) 3 3 0.305 0.58 0.246 1 2 0.58 C mol C dm = = + P3-17 Given: Gas phase reaction A + B 8C in a batch reactor fitted with a piston such that V = 0.1P0 ( ) 2 3 2 1.0 sec ft k lbmol = 2 A A Br kC C! = NA0 = NB0 at t = 0 V0 = 0.15 ft3 T = 140C = 600R = Constant

77. 3-22 P3-17 (a) 0 0 0 0 0.5A A A B N y N N = = + 8 1 1 6! = " " = 0 3Ay! "= = Now ( )0 0 0 1 V P V X T P T != + " # $ % & ' and 0 1 T T = , 0 010P V= , and 10P V= Therefore ( ) 2 010 1 10 V V X V != + or ( )2 2 0 1V V X!= + [ ]0 1A AN N X= ! [ ]0B A BN N X!= " 0 0 1B B A N N ! = = [ ] ( ) 332 02 33 3 2 0 1 1 AA B A A B kN XkN N r kC C V V X! " " = = = + 0 0 0 0 A A y P N V RT ! " = # $ % & Therefore [ ] ( ) 3 0 0 3 2 1 1 A A Xy P r k RT X! "# $ " = % & ' ( + [ ] ( ) 3 9 3 3 2 1 5.03*10 sec 1 3 A X lbmol r ft X ! ! ! = + P3-17 (b) ( )2 2 0 1V V X!= + ( )2 2 0.2 0.15 1 X!= + 0.259X = 10 3 8.63*10 sec A lbmol r ft ! ! = P3-18 No solution will be given. P3-19 No solution will be given.

78. 3-23 P3-20 No solution will be given. CDP3-A CDP3-B Polanyi equation: E = C (-HR) We have to calculate E for the reaction CH3 + RBr CH3Br + R Given: HR = - 6 kcal/mol From the given data table, we get 6.8 = C (17.5) and 6.0 = C (20) => C = 12.4 KJ/mol and = 0.32 Using these values, and HR = - 6 kcal/mol, we get E = 10.48 KJ/mol CDP3-C (a) A B Rate law at low temperature: ! "rA = kCA The rate law t higher temperature must: 1) Satisfy thermodynamics relationships at equilibrium, and 2) Reduce to irreversible rate law when the concentration of one or more of the reaction products is zero. Also, We know, ! KC = CBe CAe Rearranging, we get ! CAe " CBe KC = 0

79. 3-24 So, lets assume rate law as ! "rA = kA CA " CB KC # $ % & ' ( Also when CB = 0, it satisfies the given rate law. Hence the proposed rate law is correct. CDP3-C (b) A + 2B 2D Rate law at low temperature: ! "rA = kCA 1/2 CB Here, ! KC = CDe 2 CAeCBe 2 ! CAeCBe 2 " CDe 2 KC = 0 ! "rA = kA CAeCBe 2 " CDe 2 KC # $ %% & ' (( But it does not satisfy the irreversible rate law at low temperatures. Hence it is not correct So, taking square root of KC ! KC = CDe CAe 1/2 CBe , ! CAe 1/2 CBe " CDe KC = 0 ! "rA = kA CAe 1/2 CBe " CDe KC # $ %% & ' (( Which satisfies the irreversible rate law. Hence it is the required rate law. CDP3-C (c) A + B C + D Irreversible rate law: ! "rA '= kPAPB 1+ KAPA + KBPB We know, ! KP = PC PD PBPA or ! PBPA " PC PD KP = 0 Hence assume rate law as: ! "rA = k PAPB " PC PD KP # $ % & ' ( 1+ KAPA + KBPB + KC PC + KDPD Which satisfies both the above mentioned conditions. CDP3-D

80. 3-25

81. 3-26

82. 3-27 CDP3-E

83. 3-28

84. 3-29 CDP3-F

85. 3-30 CDP3-G

86. 3-31 CDP3-H

87. 3-32

88. 3-33 CDP3-I

89. 3-34

90. 3-35

91. 3-36 CDP3-J Species Symbol Entering Change Leaving SiH4 A FA0 3FA0 FA0(1X) NH3 B FB0 = BFA0 4FA0X FA0(4X/3) Si3N4 C 0 FA0X FA0X/3 H2 D 0 12FA0X 4FA0X

92. 3-37 ! "B = FB0 FA0 =1 yA0 = 0.5 # = yA0$ = 0.5 12 % 3% 4( ) = 2.5 CA0 = yA0CT 0 = 0.5 P0 RT0 & ' ( ) * + = 0.5 1Pa 8314 Pa, dm3 mol, k ,973K & ' ( ( ( ( ) * + + + + = 6.18,10%8 CA = CA0 1% X( ) 1+ #X = 6.18,10%8 1% X( ) 1+ 2.5X CB = 1.68,10%8 1% 4X 3 & ' ( ) * + 1+ 2.5X CC = 2.06,10%8 X 1+ 2.5X CD = 2.47,10%7 X 1+ 2.5X CDP3-K

93. 3-38

94. 3-39

95. 4-1 Solutions for Chapter 4 - Isothermal Reactor Design P4-1 Individualized solution. P4-2 (a) Cooking food (effect of temperature), removing of stains with bleach (effect of bleach conc.), dissolution of sugar in coffee or tea. P4-2 (b) Example 4-1 There would be no error! The initial liquid phase concentration remains the same. P4-2 (c) Example 4-2 For 50% conversion, X = 0.5 and k = 0.311min-1 FAO = 6.137 12.27 / min 0.5 CF lbmol X = = Also, 3 12.27 min AO AO AO F ft v C = = vo = vAo + vBo = 2vAo and also, vAo = vBo vo= 24.52 ft3 /min Using Mole Balance, 24.52 0.5 78.93 (1 ) 0.311(0.5) 592 ov X V k X gal ! = = = " = This is less volume than Example 4-2 because the rate is higher. P4-2 (d) Example 4-3 For P = 60atm, CAO = 0.0415 lbmol/ft3 ( 198073.0 60 ! == O OAO AO RT Py C ) Using equation E-4-3.6, for X = 0.8 We see that the only thing that changes is CA0 and it increases by a factor of 10, therby decreasing the volume by a factor of 10. ! V " 1 P P4-2 (e) Example 4-4 New Dp = 3D0/4 Because the flow is turbulent ! "o # 1 Dp "1 = "2 Dp2 Dp1 = 0.0775 1 0.75 = 0.1033

96. 4-2 ( ) 1 2 1 12 2 2 0.103 60 2 1 1 0.24 10o atm ft L ft y P atm ! " # $ $% &" # % &= ' = ' = '% & % &( ) % & ( ) Now 0 2 1 !" o o P L# , so too much pressure drop P = 0 and the flow stops. P4-2 (f) Example 4-5 For without pressure drop, conversion will remain same as example X = 0.82. With Pressure drop, 1 0037.0 10 ! == kgo new " " For decrease in diameter, o = 20.01244[266.92+12920.8] = 334.75lbf/ft3 = 334.75(lbf/ft3 )1/144(ft2 /in2 )1/14.7(atm/( lbf/in2 )) = 0.15814atm/ft = 52.71 kPa/m For turbulent flow: ! " # 1 Dp and $ # 1 P0 % $2 = $1 P01Dp1 P02Dp2 1 2 4.6(0.949) 1 4.6(0.949) 1 1 2 AO o AO o kC W W v X kC W W v ! ! " # $% & ' ( = = +" # + $% & ' ( X = 0.8136 virtually the same (2) Optimum diameter would be larger 152.71 0.037 0.0756 25.8 kg! " = # = if 2 1 1 2 P pD D= 1 2 1 1 2 2 p p D D ! ! != = 2 11 1 1 (0.037)(27.5) 0.0175 2 W W ! !" = " = " = " Now 1-W < 0, too much pressure drop due to higher superficial velocity.

97. 4-3 P4-2 (g) Example 4-6 For turbulent flow ! " # 1 Dp and " # 1 P0 011 2 1 1 1 2 02 1 1 1 5 5 P P PD D P ! ! ! ! " # $ %" #" # " # = = =$ %$ %$ % $ % & '& '& ' $ % & ' Therefore there is no change. P4-2 (h) Exmple 4-7 For pressure doubled and temperature decrease CTO = 2*Po/RT and T = 688K See Polymath program P4-2-h.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 1.0E-04 1.0E-04 Fa 2.26E-04 1.363E-05 2.26E-04 1.363E-05 Fb 0 0 2.124E-04 2.124E-04 Fc 0 0 1.062E-04 1.062E-04 E 2.4E+04 2.4E+04 2.4E+04 2.4E+04 T 688 688 688 688 Cto 0.573773 0.573773 0.573773 0.573773 Ft 2.26E-04 2.26E-04 3.322E-04 3.322E-04 Ca 0.573773 0.0236075 0.573773 0.0236075 k 213.40078 213.40078 213.40078 213.40078 ra -70.254837 -70.254837 -0.1189309 -0.1189309 Fao 2.26E-04 2.26E-04 2.26E-04 2.26E-04 rb 70.254837 0.1189309 70.254837 0.1189309 vo 3.939E-04 3.939E-04 3.939E-04 3.939E-04 rc 35.127419 0.0594654 35.127419 0.0594654 X 0 0 0.9395277 0.9395277 Tau 0 0 0.253044 0.253044 rateA 70.254837 0.1189309 70.254837 0.1189309 ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc

98. 4-4 Explicit equations as entered by the user [1] E = 24000 [2] T = 688 [3] Cto = 2*1641/8.314/T [4] Ft = Fa+Fb+Fc [5] Ca = Cto*Fa/Ft [6] k = 0.29*exp(E/1.987*(1/500-1/T)) [7] ra = -k*Ca^2 [8] Fao = 0.000226 [9] rb = -ra [10] vo = Fao/Cto [11] rc = -ra/2 [12] X = 1-Fa/Fao [13] Tau = V/vo [14] rateA = -ra P4-2 (i) Example 4-8 Individualized solution. P4-2 (j) Example 4-9 Using trial and error, we get maximum feed rate of B = 0.0251dm3 /s to keep concentration of B 0.01mol/dm3 . See Polymath program P4-2-j.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 500 500 ca 0.05 0.0063485 0.05 0.0063485 cb 0 0 0.009981 0.009981 cc 0 0 0.0078965 0.0078965 cd 0 0 0.0078965 0.0078965 k 0.22 0.22 0.22 0.22 v00 0.0251 0.0251 0.0251 0.0251 cb0 0.025 0.025 0.025 0.025 v0 5 5 5 5 ca0 0.05 0.05 0.05 0.05 rate 0 0 3.91E-05 1.394E-05 v 5 5 17.55 17.55 x 0 0 0.5543321 0.5543321

99. 4-5 Differential equations as entered by the user [1] d(ca)/d(t) = -k*ca*cb-v00*ca/v [2] d(cb)/d(t) = -k*ca*cb+v00*(cb0-cb)/v [3] d(cc)/d(t) = k*ca*cb-v00*cc/v [4] d(cd)/d(t) = k*ca*cb-v00*cd/v Explicit equations as entered by the user [1] k = .22 [2] v00 = 0.0251 [3] cb0 = 0.025 [4] v0 = 5 [5] ca0 = 0.05 [6] rate = k*ca*cb [7] v = v0+v00*t [8] x = (ca0*v0-ca*v)/(ca0*v0) If the concentration of A is tripled the maximum feed rate becomes 0.064 dm3 /s P4-2 (k through r) Individualized solution. P4-3 Solution is in the decoding algorithm given with the modules. P4-4 We have to find the time required to cook spaghetti in Cuzco, Peru. Location Elevation (km) Pressure (mm Hg) Boiling Point (C) Time (min) Ann Arbor 0.21 739 99.2 15 Boulder 1.63 625 94.6 17 Cuzco 3.416 493 88.3 ? Assume reaction is zero order with respect to spaghetti conversion: E ART A dC r k Ae dt ! ! = = = ! so that 0 A A E C RT C C Ae t ! " # ! = $ % & ' For complete conversion (i.e.: well cooked) CA = 0 at time t. Therefore 0 0 E RT A E A RT C tAe C te A ! ! = = 0 1 ln ln lnA b C E k t A R T ! " = = #$ % & '

100. 4-6 Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to Tb = 88.3C = 361.45 K yields t = 21 minutes. P4-5 (a) ( )( )22 0 1 10 AV kC X X ! =

101. 4-7 2 2 0 0 (1 )AOA A A kC XrdX dV F F !! = = 2 0 01 A A kC VX X F = ! X=0.85 So, considering the above results, we will choose a CSTR. P4-5 (b) P4-5 (c) P4-5 (d) 1) CSTR and PFR are connected in series: 3 3 3 2 2 (200 )(0.07 / .min)(1 / ) (1 ) 10 / min CSTR dm dm mol mol dm X X mol ! = Solving the quadratic equation, XCSTR = 0.44 For PFR, 0A A dX F r dV = !

102. 4-8 3 2 (0.07 / / min) (1 )(1 ) 10 / min AOdm mol C X X dX dV mole ! ! = 3 3 2 3 3 0.44 (0.07 / / min)(1 / ) (800 ) (1 ) 10 / min X dX dm mol mol dm dm X mole = !" 0.736X = 2) when CSTR and PFR are connected in parallel, 3 3 3 2 2 (200 )(0.07 / .min)(1 / ) (1 ) 5 / min. CSTR dm dm mol mol dm X X mol ! = XCSTR = 0.56 For PFR, 3 3 2 3 2 0 (0.07 / .min)(1 / ) (800 ) (1 ) 5 / min X dX dm mol mol dm dm X mol = !" XPFR = 0.92 Hence, final conversion X = 0.56 0.92 2 + =0.74 P4-5 (e) To process the same amount of species A, the batch reactor must handle 3 5 60min 24 2 14400 min dm h mol M hr day day ! " ! "! " =# $ # $# $ % &% &% & If the reactants are in the same concentrations as in the flow reactors, then 3 3 1 14400 14400 mol dm dm V day mol day ! "! " = =# $# $ % &% & So the batch reactor must be able to process 14400 dm3 every 24 hours. Now we find the time required to reach 90% conversion. Assume the reaction temperature is 300K. ( )2 0 0 0 1AA A A kC X Vr VdX dt N N !! = = 0 2 0 1 A R A N X t VkC X = ! , and since 0 0 A A N C V = 3 0 3 1 1 0.9 2.14 1 0.1 4.2 *1 R A X t hr dm molkC X mol hr dm = = = ! " # $ % & ' (

103. 4-9 Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tf = 3 hours ttotal = tR + tf ttotal = 2.14hours + 3 hours = 5.14 hours. Therefore, we can run 4 batches in a day and the necessary reactor volume is 3 314400 3600 4 dm dm= Referring to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be approximately $85,000 for the reactor. P4-5 (f) The points of the problem are: 1) To note the significant differences in processing times at different temperatures (i.e. compare part (b) and (c)). 2) That the reaction is so fast at 77C that a batch reactor is not appropriate. One minute to react and 180 to fill and empty. 3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black. P4-6 (a)

104. 4-10 P4-6 (b) To increase conversion, use PFR, higher temperature, or use better catalyst. P4-6 (c) P4-6 (d)

105. 4-11 P4-6 (e) P4-6 (f) Individualized solution P4-6 (g) Individualized solution P4-6 (h) Individualized solution

106. 4-12 P4-7 (a) Elementary gas phase reaction. P4-7 (b) P4-7 (c) For = 0.001dm-3 See Polymath program P4-7-c.pol.

107. 4-13 POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value v 0 0 500 500 x 0 0 0.656431 0.656431 y 1 0.1721111 1 0.1721111 Co 0.3 0.3 0.3 0.3 esp 2 2 2 2 alfa 0.001 0.001 0.001 0.001 C 0.3 0.0077768 0.3 0.0077768 k 0.044 0.044 0.044 0.044 r -0.0132 -0.0132 -3.422E-04 -3.422E-04 fo 2.5 2.5 2.5 2.5 Differential equations as entered by the user [1] d(x)/d(v) = -r/fo [2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y) Explicit equations as entered by the user [1] Co = 0.3 [2] esp = 2 [3] alfa = 0.001 [4] C = Co*(1-x)*y/(1+esp*x) [5] k = 0.044 [6] r = -k*C [7] fo = 2.5 At V = 500, x = 0.66, y = 0.17 P4-7 (d) Individualized solution P4-7 (e) A B + 2C

108. 4-14 Using these equations in Polymath we get the volume to be 290 dm3 . P4-7 (f) Using Polymath to solve the differential equation gives a volume of 290 dm3 See Polymath program P4-7-f.pol.

109. 4-15 POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value X 0 0 0.47 0.47 V 0 0 290.23883 290.23883 Kc 0.025 0.025 0.025 0.025 Fao 2.5 2.5 2.5 2.5 Cao 0.3 0.3 0.3 0.3 k 0.044 0.044 0.044 0.044 e 2 2 2 2 ra -0.0132 -0.0132 -9.391E-04 -9.391E-04 ODE Report (RKF45) Differential equations as entered by the user [1] d(V)/d(X) = Fao/(-ra) Explicit equations as entered by the user [1] Kc = .025 [2] Fao = 2.5 [3] Cao = .3 [4] k = .044 [5] e = 2 [6] ra = -(k*Cao/(1+e*X))*((1-X)-(4*Cao^2*X^3)/((1+e*X)^2*Kc)) PFR with pressure drop: Alter the Polymath equations from part (c). See Polymath program P4-7-f-pressure.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value v 0 0 500 500 x 0 0 0.5077714 0.5077714 y 1 0.3181585 1 0.3181585 Kc 0.025 0.025 0.025 0.025 alfa 0.001 0.001 0.001 0.001 Cao 0.3 0.3 0.3 0.3 k 0.044 0.044 0.044 0.044 esp 2 2 2 2 fo 2.5 2.5 2.5 2.5 r -0.0132 -0.0132 -1.846E-04 -1.846E-04

110. 4-16 ODE Report (STIFF) Differential equations as entered by the user [1] d(x)/d(v) = -r/fo [2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y) Explicit equations as entered by the user [1] Kc = .025 [2] alfa = 0.001 [3] Cao = 0.3 [4] k = 0.044 [5] esp = 2 [6] fo = 2.5 [7] r = -(k*Cao/(1+esp*x))*(4*Cao^2*x^3/((1+esp*x)^2*Kc)) At V = 500 dm3 X = 0.507 and y = 0.381 P4-7 (g) Membrane reactor: A B + 2C CA = COFA/FT CB = COFB/FT CC = COFC/FT FT = FA + FB + FC and -rA = rB = rC/2 Using polymath, For PFR, See Polymath program P4-7-g.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value v 0 0 1040 1040 Fa 2.5 1.3231889 2.5 1.3231889 Fb 0 0 0.3635477 0.0684325 Fc 0 0 2.3536223 2.3536223 Kc 0.025 0.025 0.025 0.025 Ft 2.5 2.5 3.7452437 3.7452437 Co 0.3 0.3 0.3 0.3 K 0.044 0.044 0.044 0.044 kc 0.08 0.08 0.08 0.08 ra -0.0132 -0.0132 -3.827E-04 -3.827E-04 X 0 0 0.4707245 0.4707245 Differential equations as entered by the user [1] d(Fa)/d(v) = ra [2] d(Fb)/d(v) = -ra - kc*Co*Fb/Ft [3] d(Fc)/d(v) = -2*ra Explicit equations as entered by the user [1] Kc = 0.025 [2] Ft = Fa+ Fb+ Fc [3] Co = 0.3 [4] K = 0.044 [5] kc = .08 [6] ra = - (K*Co)*(Fa/Ft- Co^2*Fb*Fc^2/(Kc*Ft^2)) [7] X = 1 - Fa/2.5 Solving for when X = 0.47, we get V = 1040 dm3

111. 4-17 P4-8 (a) The blades makes two equal volumes zones of 500gal each rather than one big mixing zone of 1000gal.

112. 4-18 P4-8 (b) A CSTR is been created at the bend due to backmixing, so the effective arrangement is a PFR is in series with a CSTR. A B k = 5 min-1 vo = 5 dm3 /min. Xexpected = 0.632 Xactual = 0.586 0 1 ln 1 X AO o A Expected F dX v V r k X ! " = = # $ % %# $& ' ( = 5 1 ln 1.0 5 1 .632 ! " =# $ %& ' Now, For PFR, 1 1 ln 1 PV X ! " = # $ %& ' .1 For CSTR, ( ) ( ) 11( ) 1 actualAO actual C A actual X XF X X V r X !! = = ! ! 2 CSTR zone created due to backmixing

113. 4-19 Also, 1CP VV V V + = ..3 Solving 1, 2 and 3 by using polymath, See Polymath program P4-8-b.pol. POLYMATH Results NLES Solution Variable Value f(x) Ini Guess X1 0.350949 3.148E-10 0 Vc 0.567756 -3.297E-14 1 V 1 X2 0.586 Vp 0.432244 NLES Report (safenewt) Nonlinear equations [1] f(X1) = ln(1/(1-X1)) -Vp = 0 [2] f(Vc) = (X2-X1)/(1-X2) - Vc = 0 Explicit equations [1] V = 1 [2] X2 = .586 [3] Vp = V - Vc VC = 0.57 dm3 ; VP = 0.43 dm3 ; X1 = 0.35 P4-8 (c) CAO = 2 mol/dm3 A !B Assuming 1st order reaction, For CSTR, AO A C X r ! = " -rA = kCA = kCAO(1-X) => 0.4 0.67 1 0.6 X k X ! = = = " For PFR, ( )0 00 1 X A A dX V F kC X = !" , 0 1 X dX k X ! = "# => 1 exp( )PFRX k!= " " =1-exp(-0.67) = 0.486 Now assuming 2nd order reaction, For CSTR, Now, assuming 2nd order reaction, For CSTR, AO A C X r ! = " 2 A Ar kC! = = ( ) 22 1AOkC X! => ( ) 2 2 0.4 1.111 0.61 AO X kC X ! = = = " For PFR, ( ) 2 0 1 1 11 X AO AO dX X kC kC XX ! = = "" #

114. 4-20 => 1 1 1 1 .526 1 2.111AO X kC! = " = " = + So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second order reaction. P4-8 (d) A graph between conversion and particle size is as follows: Originally we are at point A in graph, when particle size is decreased by 15%, we move to point B, which have same conversion as particle size at A. But when we decrease the particle size by 20%, we reach at point C, so a decrease in conversion is noticed. Also when we increase the particle size from position A, we reach at point D, again there is a decrease in the conversion. P4-9 A ! B To = 300K KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3 Mole balance: A AO r XF V ! = Rate law: ! " # $ % & '=' C B AOA K C Ckr 2 2 Stoichiometry: ( )XCC AOA != 1 and XCC BOB = ! ( ) ! " # $ % & '' = C AOO AO K X X X Ck F V 2 2 2 1 ( ) 2 2 2 2 2 1 3785.4 0.4 0.6 0.4 3 CAO O AO X V X KF Z k C X ! " # #$ % ! "& '= = = #$ % & ' 3 2902.2Z dm= , ! Z = k ko = exp E R 1 T0 " 1 T # $ % & ' ( ) * + , - .

115. 4-21 Now using: ( ) ( ) ( ) 2 2 2 2 0 1 1 C C Z X Z X V f X V z zX X X X K K ! " ! " = # = = $% & % & ' ( ' () * ) * $ $ $ $+ , + , - . - . where 1 1 exp O O k E z k R T T ! "# $ = = %& '( )& ' * +, - and ! ! " # $ $ % & ' ( ) * + , - . = TTR H KK O RX COC 11 exp Solving using polymath to get a table of values of X Vs T. See Polymath program P4-9.pol. POLYMATH Results NLE Solution Variable Value f(x) Ini Guess X 0.4229453 3.638E-12 0.5 To 300 T 305.5 z 2902.2 V 3785.4 E 1.5E+04 R 2 y 1.5684405 Kco 3 Hrx -2.5E+04 Kc 1.4169064 NLE Report (safenewt) Nonlinear equations [1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0 Explicit equations [1] To = 300 [2] T = 305.5 [3] z = 2902.2 [4] V = 3785.4 [5] E = 15000 [6] R = 2 [7] y = exp(E/R*(1/To-1/T)) [8] Kco = 3 [9] Hrx = -25000 [10] Kc = Kco*exp(Hrx/R*(1/To-1/T)) 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 295 300 305 310 315 320 Temperature X

116. 4-22 T(in K) X 300 0.40 301 0.4075 303 0.4182 304 0.4213 305 0.4228 305.5 0.4229 305.9 0.4227 307 0.421 310 0.4072 315 0.3635 We get maximum X = 0.4229 at T = 305.5 K. P4-10 (a) For substrate: Cell: voCC = rgV FSO FS + rsV = 0 (CSO CS)vO = rg VYS/C = VYS/C max S C M S C C K C ! " # $ +% & P4-10 (b) [ ]SSOSCC CCYC != / ( ) / 0MAX S SO S O S C C M S C C C v VY C K C ! " # # =$ % +& ' ! ( ) ( )/ 0MAX S SO S O C S SO S M S C C C v VY C C K C ! " # # # =$ % +& ' ( ) ( ) 0.5 30 5 25 0.8 30 0 5 S S S S C C C C ! "# $ $ # # $ =% & +' ( Solving we get CS = 5.0 g/dm3 or 30 g/dm3 . if CS = CS0 no reaction has occurred so the only valid answer is CS = 0.5 g/dm3 . P4-10 (c) CC = YC/S(CSO - CS) = 0.8(30 5.0)g/dm3 = 20 g/dm3 . P4-10 (d) vnew = vo/2 = 2.5 dm3 /h Using equation from above, we get CS = 1.67 g/dm3 and CC = 22.67 g/dm3

117. 4-23 P4-10 (e) Vnew = Vo/3 = 25/3 dm3 Using equation from above, we get CS = 3.0 g/dm3 and CC = 21.6 g/dm3 P4-10 (f) For batch reactor: CSO = 30 g/dm3 CCO = 0.1 g/dm3 CC = CC0 + YC/S(CSO - CS) V = 10 dm3 See Polymath program P4-10-f.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 15 15 Cs 30 0.0382152 30 0.0382152 Cso 30 30 30 30 Ycs 0.8 0.8 0.8 0.8 Km 5 5 5 5 Umax 0.5 0.5 0.5 0.5 Cco 0.1 0.1 0.1 0.1 Cc 0.1 0.1 24.069428 24.069428 rg 0.0428571 0.0428571 5.4444349 0.0912841 rs -0.0535714 -6.8055436 -0.0535714 -0.1141052 negative_ 0.0535714 0.0535714 6.8055436 0.1141052 ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = rs Explicit equations as entered by the user [1] Cso = 30 [2] Ycs = 0.8 [3] Km = 5 [4] Umax = 0.5 [5] Cco = 0.1 [6] Cc = Cco+Ycs*(Cso-Cs) [7] rg = (Umax*Cs/(Km+Cs))*Cc [8] rs = -(1/Ycs)*rg [9] negative_rs = -rs

118. 4-24 P4-10 (g) Graphs should look the same as part (f) since reactor volume is not in the design equations for a constant volume batch reactor. P4-11 Gaseous reactant in a tubular reactor: A B 1 0.0015min at 80 F A Ar kC k ! ! = = o 25,000 cal E g mol = 0.90X = 1000B lb M hr = 58A B lb MW MW lbmol = = 1 inch (I.D.)tD = 10L ft= 132 146.7P psig psia= = 260 720T F R= =o o number of tubestn = 1000 17.21 58 B lb lbmolhrF lb hr lbmol = = 0 17.21 19.1 0.9 B A lbmol F lbmolhrF X hr = = = For a plug flow reactor: 0.92 0 0 4 t t A A n D L dX V F r ! = = "# 1 1 0! = " = 1.0Ay = 0Ay! "= = ( )0 1A Ar kC X! = ! 0 A A P P C RT RT = = ( ) 0.9 0.9 0 0 0 0 0 00 0 1 ln ln10 1 1 0.9 A A A A A A A F F RTdX dX V F F r kC X kC kP ! " = = = =# $ % % %& ' ( ( At T2 = 260F = 720R, with k1 = 0.0015 min-1 at T1 = 80F = 540R, 1 2 1 1 2 1 1 25000 1 1 exp 0.0015exp 53.6min 1.104 540 720 E k k R T T ! " #" # " #" # = ! = ! =$ %$ % $ %$ %$ % & '& '& '& ' 1 1 2 53.6min 3219k hr! ! = = ( ) ( )( ) 3 0 1 19.1 10.73 720 ln10 ln10 3219 146.7 A lbmol ft psia R hr lbmol RF RT V kP hr psia! " #" # $ %$ % & '& '= = o o 3 0.72V ft= 2 4 t tn D L V ! =

119. 4-25 ( ) ( ) 3 22 4 0.724 13.2 1 10 12 t t ftV n D L ft ft ! ! = = " # $ % & ' Therefore 14 pipes are necessary. P4-12 A B/2 AO AO TOC y C= 2 2 2 2 (1 1/ 4) 0.8 2( 1/ 4)(1 1/ 4)ln(1 0.8) ( 1/ 4) 0.8 2.9 1 0.8 PFR AO TO AO V ky C F ! = ! ! ! + ! + = ! .(6)

120. 4-26 2 2 2 ' ' ' PFR AO TO AO V ky C F = ( ) 2 2 8 8 2.9 2.58 9 9 PFR AO TO AO V ky C F = = = P4-13 Given: The metal catalyzed isomerization BA ! , liquid phase reaction 1 B A A eq C r k C K ! " # = #$ %$ % & ' with Keq = 5.8 For a plug flow reactor with yA = 1.0, X1 = 0.55 Case 1: an identical plug flow reactor connected in series with the original reactor. Since yA = 1.0, B = 0. For a liquid phase reaction ( )A A0C C 1 X= ! and B A0C C X= ( )0 1A A eq X r kC X K ! " # = # #$ %$ % & ' For the first reactor, ( ) 1 1 1 0 0 0 0 0 1 X X A A A A eq dX dX V F F r X kC X K = = ! " # ! !$ %$ % & ' ( ( or 1 1 0 1 0 0 0 1 1 ln 1 1 11 11 1 X X A A eq eq eq kC V dX X F K X KK ! "# $ = = % % +& '( )( )# $ & '* ++ , -% +( )( ) * + .

121. 4-27 ( )0 1 1 0 1 1 ln 1 1 0.853ln .355 0.883 1 1 A A eq eq kC V X F K K ! "# $ = % % + = % =& '( )( )& '* ++ , - Take advantage of the fact that two PFRs in series is the same as one PFR with the volume of the two combined. VF = V1 + V2 = 2V1 and at VF X = X2 0 0 0 0 1 1 ln 1 1 11 11 1 F F X X A F A eq eq eq kC V dX X F K X KK ! "# $ = = % % +& '( )( )# $ & '* ++ , -% +( )( ) * + . 0 0 1 2 0 0 1 1 2 ln 1 1 1 1 A F A A A eq eq kC V kC V X F F K K ! "# $ = = % % +& '( )( )& '* ++ , - ( )0 1 0 2 2 0.883 1.766A A kC V F = = 2 1 1 1.766 ln 1 1 1 5.81 5.8 X ! "# $ = % % +& '( ) * +, -+ X2 = 0.74 Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor, The analysis for the first reactor is the same as for case 1. 0 1 1 0 1 1 ln 1 1 1 1 A A eq eq kC V X F K K ! "# $ = % % +& '( )( )& '* ++ , - By performing a material balance on the separator, FA0,2 = FA0(1-X1)

122. 4-28 Sinc