Elements of Chemical Reaction Engineering 5th Edition

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

10-1 Solutions for Chapter 10 Catalysis and Catalytic Reactors P10-1 Individualized solution P10-2 (a) Example 10-1 (1) Pentane isomerization Pt nP iP!!" Maximum f = 5 molecules/site/sec Minimum f = 3e-3 molecukes/site/sec Maximum rate: 1 %100PPtPtr fD MW! "#=$ %& ' ( ) 41 15 0.5 1.28*10195 100Pmolrs

10-41 10-41 10-41 P10-18 (c) 10-42 10-43 P10-18 (d) 10-44 P10-18 (e) 10-45 10-45 10-45 P10-19(a) D da k dt

10-61 10-61 10-61 P10-25 10-62 10-62 10-62 P10-26 (a) 10-63 P10-26 (b) Using the same program we can see that the maximum conversion is 0.887

10-81 CDP10-I 10-82 CDP10-J CDP10-K CDP10-K CDP10-K 10-83 10-84 10-84 10-84 CDP10-L

10-93 CDP10-Q 10-94 10-94 10-94 CDP10-R 10-95 10-96 10-96 10-96

[3] Pa = y*Pao*(1-X)/(1+e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2) [8] rate = -ra [9] alpha = .03 [3] Pa = y*Pao*(1-X)/(1+e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2)

11-1 11-1 11-1 Solutions for Chapter 11 External Diffusion Effects on Heterogeneous Reactions P11-1 Individualized solution P11-2 (a) WB="2WAWA=cDAB1+yA( )dyAdz ="cDABdln 1+yA( )dz Integrating with yA=0 at z=" WA=cDAB" # z( )ln 1+yA( ) (1) at z = 0 yA=yA0 WA=cDAB"1+yA0( ) (2) WB="2WAWA=cDAB1+yA( )dyAdz

11-21 M=XAMA+XBMB CB=XBC VA=NA/CA VB=NB/CB=0 V*=NA+NBC=7.17 *10"3 V=nA+nB"=nA/" JA=CA(VA"V*)=NA"XANA=XBNA JB=CB(VB"V*)="XBNA="JA VA=NA/CA VB=NB/CB=0 V*=NA+NBC=7.17 *10"3 V=nA+nB"=nA/" JA=CA(VA"V*)=NA"XANA=XBNA JB=CB(VB"V*)="XBNA="JA VA=NA/CA VB=NB/CB=0 V*=NA+NB C=7.17 *10"3 V=nA+nB " =nA/ " JA=CA(VA"V*)=NA"XANA=XBNA JB=CB(VB"V*)="XBNA="JA (*105) (*103) z XB XA CB CA A A 0 0.329 0.671 1.3 2.65

12-1 Solutions for Chapter 12 Diffusion and Reaction in Porous Catalysts P12-1 Individualized solution P12-2 (a) (a) "=5 De~ DAB ~TPDex ~ De1 ~P1PT2T1" # $ % & ' 1.75 (lines and angles not to scale) De~ DAB ~TPDex ~ De1 ~P1PT2T1" # $ %

12-17 P12-10 (c) Lets try again with "0 = 10 "=1+10( )20.12#1[ ]=1#1020.99( ) "=1#99 =#99 not possible will be negative for any value of "0 greater than one. "=1+10( )20.12#1[ ]=1#1020.99( ) "=1#99 =#99 not possible will be negative for any value of "0 greater than one. "=1+10 ( ) 20.12#1 [

X=41%. X=41%. X =41%. 13-41 Maximum Mixedness ( ) ( )X F E C r d dX Ao A ! ! ! " += 1 Rate Law :2BAA CkCr =! ( )XCC AoA !=1 ( )XCC BoB !=1 ( )32 1XCkCr BoAoA != where k=175 dm6/mol2 min ( )32 1XkCCrBoAoA!= ( )zEdzdF != where z=14- Rate Law :2BAA CkCr =! (

13-1 Solutions for Chapter 13 Distributions of Residence Times for Chemical Reactors P13-1 No solution will be given. P13-2 (a) The area of a triangle (h=0.044, b=5) can approximate the area of the tail :0.11 P13-2 (b) ~0.11 ~0.11 ~0.11 13-2 13-2

13-21 ( ) ( ) ( ) ( ) ( ) ( ) ! ! ! ! ! ! ! ! ! ! d F eE e C C d F E k d F E k Ao A"# $ % & ' ( ) * +* $ % & ' ( ) * +* $ $ $ % & ' ' ' ( ) * " *= " 1 1 1 ( )( ) ( )( )( )( )!!!!!!!!!!!dFeeEeCCdFEkdFEkAoA"#$%&'()***$%&'()*+*$$$%&'''()*"*="111 by definition ( ) ( ) ( )!!!dFdE = gives: ( )( ) ( ) ( )( )[ ] ( )( )!!!!!!Feee FLndFdFdFE"==#=#"""$%&'()""1111 ( )( ) ( )( )( )( )!!!!!!!!!!!dFeeEeCCdFEkdFEkAoA"#$%&'()***$%&'()*+*$$$%&'''()*"*="111 by definition

14-1 Solutions for Chapter 14 Models for Non-ideal Reactors P14-1 Individualized solution P14-2 (a) Approximated formula for Segregation Model (1st reaction) Errortkkte kt ++!=!2122 . The error is then o((kt)3). Approximating the !33ktError != Errortkkte kt ++!=!2122 . The error is then o((kt)3). Approximating the !33ktError

14-41 P14-17 The presence of a minimum (run2) imply the presence of a radial temperature profile that effects the reaction rate and determines the deviation from the concentration profile given by Eq. (14.51). The enthalpy balance allows the identification of the dimensionless thermal parameters: Defining 00

14-54 I. Variation of Da number Conversion Damkhler number/ Da Closed-vessel Open-vessel Parameter 0.69 0.432 0.378 8*U0 1.38 0.713 0.641 4*U0 2.76 0.959 0.911 2*U0 5.52 1 1 U0 11.04 / / U0/2 22.08 / / U0/4 44.16 / / U0/8 2.76 0.959 0.911 2*U0 5.52 1 1 U0 11.04

14-21 P14-7 Tubular Reactor 1st order, irreversible, pulse tracer test 2 ! = 65 2 s and m t For a 1st order reaction, PFR: X =1- e-k = 0.98 We need and k. There being no data for diffusivity (Schmidt number) Da and hence Per cannot be obtained using tubular flow correlations. Calculate V= 3

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 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. The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in

3-21 ( )( ) 30.305 1 0.39 0.19AmolCdm=!= ( )( ) 30.305 0.39 0.36CmolC= = ( )( ) 30.305 1 0.39 0.19AmolCdm=!= ( )( ) 30.305 0.39 0.36CmolC= = ( )( ) 3 0.305 1 0.39 0.19 A mol C dm =!= ( )( ) 3 0.305 0.39 0.36 C mol C = = dm P3-16 (d) Gas phase reaction in

3-1 3-1 3-1 Solutions for Chapter 3 Rate Law and Stoichiometry P3-1 Individualized solution. P3-2 (a) Example 3-1 00.0010.0020.0030.0040.005310 315 320 325 330 335T (K)k (1/s) 00.0010.0020.0030.0040.005310 315 320 325 330 335T (K)k (1/s) 0 0.001 0.002 0.003 0.004 0.005 310 315 320 325 330 335 T (K) k (1/s) 0.006 0.007 0.008 For E =

y 1 0.2366432 1 0.2366432 Dp 0.0075 0.0075 0.0075 0.0075 Q 0.5625 0.5625 0.5625 0.5625 Fao 5 5 5 5 alpha 0.00944 0.00944 0.00944 0.00944 Cao 0.207 0.207 0.207 0.207 kprime 2.9385672 2.9385672 2.9385672 2.9385672 ra -0.1259147 -0.1259147 -0.0012992 -0.0012992 y 1 0.2366432 1 0.2366432 Dp 0.0075 0.0075

Cooking food (effect of temperature), removing of stains with bleach (effect of bleach conc.), dissolution of sugar in coffee or tea. Cooking food (effect of temperature), removing of stains with bleach (effect of bleach conc.), dissolution of sugar in coffee or tea. Cooking food (effect of temperature), removing of stains

4-74 4-75 CDP4-I (a) CDP4-I (b) 4-76 CDP4-I (c)

[2] d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2^2 [3] d(Ca3)/d(t) = (Ca2 - Ca3)/tau -k*Ca3^2 Explicit equations as entered by the user [1] k = 0.025 [2] Cao = 2 [3] tau = 10 [4] X = 1 - 2*Ca3/Cao From Polymath, the steady state conversion of A is approximately

Now using: ( )( )( )2 22 201 1C CZ X Z XV f X Vz zX XX XK K! " ! "=#= = $% & % &' ( ' () * ) *$ $ $ $+ , + ,- . - . where 1 1expO Ok Ezk R T

P5-1 (d) Individualized solution P5-1 (e) Individualized solution P5-1 (f) Individualized solution P5-1 (g) Individualized solution P5-1 (h) Example 5-1 P5-1 (d) Individualized solution P5-1 (e) Individualized solution P5-1 (f) Individualized solution P5-1 (g) Individualized solution P5-1 (h) Example 5-1 P5-1 (d) Individualized solution P5-1 (e) Individualized solution P5-1 (f)

5-18 60 S g MW gmol = 6 3 2.32 2.32 *10 S g g ml m ! = = (Handbook of Chemistry and Physics, 57th ed., p.B-155) 631 10g gml m!"= 20FgMW gmol= 3 30.5 0.0005V dm m= = 91.2986 *10!"= ( ) ( ) ( )1.775639 31.7752 6360101.2986 *10 0.000520 0.2 2.3*10 100ggkgmolmmggmgmol m!" #" # $ %$

constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is slow and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the concentration profiles for k1 = 0.10 h-1 and k2 = 0.8 h-1. Note that

The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3 The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3 The concentration of B is highest at t = 100 sec, where its value is 0.1363

[2] E1 = 20000 [3] R = 1.987 [4] k2 = 0.71*exp(E2/R*(1/946-1/T)) [5] tau = 10 [6] cao = 0.05*0.75 [7] k1 = 0.22*exp(E1/R*(1/946-1/T)) [8] ca = cao/(1+tau*(k1+k2)) [9] cb = tau*k1*ca [10] cm = tau*k1*ca [11] cp = tau*k2*ca [2] E1 = 20000 [3] R = 1.987

2321 AAACkCkkddC !!!=" 1kddC X=! ABCkddC2=! 23AYCkddC =! In PFR with = V/vo = 783 sec we get X = 0.958 and SB/XY = 0.624 also at = 560.24 sec SB/XY is at its maximum value of 0.644 2321 AAACkCkkddC !!!=" 1kddC X=! ABCkddC2=! 23AYCkddC =! In PFR

6-61 P6-21 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst. A B + C 1C& 'A D '2 2D D Ar k C= 2C + D 2E ' 23 3E E C Dr k C C= ( )3324.6 0.6 /0.082 / . (500 )AOP atmC

7-21 P7-10 (a) E+Sk1 " # " ES ESk2 " # " E+S ESk3 " # " P+E P+Ek4 " # " ES rP=k3ES[ ]"P[ ][ ]KC$ % ' ( , KC=k3/k4 rES=k1E[ ]S[ ]"k2ES[ ]"k3ES[ ]+k4P[ ]E[ ] ES[ ]=k1E[ ]S[ ]+k4P[ ]E[ ]k2+k3=E[ ]k1S[ ]+k4P[ ]( )k2+k3 Et=E+ES=E1+k1S[ ]+k4P[ ]" %

7-41 C g dC r dt = Substrate Balance: 0 0 0SS S S gCdC v C v C Y rdt =! ! max S CgM SC CrK C=+ This would result in the Cell concentration growing exponentially. This is not realistic as at some point there will be too many cells to fit

P7-1 (b) Example 7-2 For t = 0 to t = 0.35 sec, PSSH is not valid as steady state not reached. And at low temperature PSSH results show greatest disparity. See Polymath program P7-1-b.pol. P7-1 (b) Example 7-2 For t = 0 to t = 0.35 sec, PSSH is

7-59 CDP7-C (e) From part (c), we have: 7-60 7-60 7-60 7-61 CDP7-D No solution will be given.

Increasing mC increases the slope of R(T) and will increase the exit concentration of cells. Increasing Ta will lower the exit concentration of cells. Increasing mC increases the slope of R(T) and will increase the exit concentration of cells. Increasing Ta will lower the exit concentration of cells. Increasing mC

8-96 T 550 225.22683 550 225.22683 Cao 0.064 0.064 0.064 0.064 To 550 550 550 550 k 1.949E+05 5.047E-05 1.949E+05 5.047E-05 Kc 1.01E+06 1841.4832 1.01E+06 1841.4832 f 1 1 2.4419826 2.4419826

8-41 9 P0 1.013E+06 1.013E+06 1.013E+06 1.013E+06 10 CA0 270.8283 270.8283 270.8283 270.8283 11 CA 270.8283 258.1071 270.8283 258.1071 12 kr 0.2 0.076962 0.2 0.076962 13 CC 0 0 15.77362 15.77362 14 CB 0 0 15.77362 15.77362 15 rA -36.02017 -36.02017 -0.0062421 -0.0062421 11 CA 270.8283 258.1071 270.8283 258.1071 12

8-21 8-22 P8-5 2A B C+! A B C Fio lb mole hr ! " # $ % & ' 10 10 0.0 Tio(F) 80 80 - ~PioBtuClb mole F! "# $% & 51 44 47.5 ,lbMWlb mol! "# $% & 128 94 222

8-81 8-81 8-81 P8-25 Mole balance: C A B A B C dF dF dF r r r dW dW dW = = = Rate Laws: 1 231 123 3B A BC AA AB B BA Cr r rr rr k Cr k Cr k C=!+=!!=!=!= 1 231 123 3B A BC AA AB B BA Cr

8-1 8-1 8-1 Solutions for Chapter 8 Steady-State Nonisothermal Reactor Design P8-1 Individualized solution P8-2 (a) Example 8-1 For CSTR X=$k1+$k=$Ae"E RT1+Ae"E RTOne equation, two unknowns Adiabatic energy balance T=T0"#HRx XCPA In two equations and two unknowns In Polymath form the solution f X( )=X"#Ae"E RT1+Ae"E RTf T(

9-21 P9-10 (c) The unsteady energy balance for a batch reactor is showed in equation 9-11. ( ) ( ) ! "#"$#++ = ii ARxs CpN VrHWQ dt dT & & Adiabatic operation ( Q & =0), neglecting s W & ( ) ( )( ) ( ) ( )( )[ ]( )( )( ) ( ) ( )( )[ ]!"!"!"!"!"!"000100000001TTTTTTTTCkCpHTTTTTTTTCkCCpHVrCpNHdtdTBfBffAiiRxfBfBfAAiiRxAiiARx#$%+%#$#%%#&&'())*+$,%=%#$%+%#$#%####&&'())*+$,%=#%$,%=+%+++--- (

The new T0 of 20 F (497 R) gives a new HRn and T. With T=497+89.8X the polymath program of example 9-1 gives t= 8920 s for 90 % conversion. The new T0 of 20 F (497 R) gives a new HRn and T. With T=497+89.8X the polymath program of

tau 0.12 0.12 0.12 0.12 Ca 0.479966 0.0241085 0.479966 0.0241085 Cp 0.75 0.75 0.75 0.75 Gt 2.404E+05 2.404E+05 5.711E+06 5.711E+06 Rt -5.25E+05 -5.25E+05 6.825E+06 6.825E+06 tau 0.12 0.12 0.12 0.12 Ca 0.479966 0.0241085 0.479966 0.0241085 Cp 0.75 0.75 0.75 0.75 Gt 2.404E+05 2.404E+05 5.711E+06 5.711E+06 Rt -5.25E+05 -5.25E+05