Brian Vicente / Max Nori / H. Scott Fogler

-.. II!I!!III ..

Soludons Manual for Elements of Chemical Reaction Engineering Fourth Edition ~

Prentice Hallitaternational Series in the Phys!ptl and Chemical E.ngineering Sciences

A Complimentary Copy from

PRENTI CE FT Prentice Hall HALL

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~CD-ROM ~INClUDED

Solutions Manual for

Elements of Chemical Reaction Engineering Fourth Edition

Brian Vicente Max Nori H. Scott Fogler

~ PRENTICE

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ISBN 0-13-186383-5 Text printed in the United States at OPM in Laflin, Pennsylvania. First printing, November 2005 ______________ ... ________ . ________ :

**** CONFIDENTIAL ****

UNIVERSITY OF MICHIGAN INTERACTIVE COMPUTER MODULES FOR CHEMICAL ENGINEERING

CHEMICAL REACTION ENGINEERING MODULES

H. Scott Fogler, Project Director

M .. Nihat GUIllen, Project Manager (2002-2004) Susan Montgomery, Project Manager (1991-1993)

Department of Chemical Engineering University of Michigan

Ann Arbor, MI48109-2136

2005 Regents of the University of Michigan

- All Rights Reserved -

INTERPRETATION OF PERFORMANCE NUMBERS

Students should record their Performance Number for each program, along with the name of the program, and tum it in to the instructor. The Performance Number for each program is decoded as described in the following pages.

The official site for the distribution of the modules is http://www.engin.umich.edu/ -cre/icm

Please report problems to icm.support@umich.edu.

PerfOimance NumbeI InteIpIetation: eRE modules iii

**** CONFIDENTIAL ****

ICMs with Windows interface

Module Format

KINETIC CHALLENGE I CzBzzAzz

KINETIC CHALLENGE II CzBzzAzz

MURDER MYSTERY zzAzz

TIC TAC TOE zDzCzBzA

Interpretation

Score = 1.5 * AB.C z = random numbers Note: 75% constitutes mastery.

Score = 2.0 * ARC z = random numbers Note: 75% constitutes mastery.

A even: Killer and victim correctly identified

A odd: Killer and victim not identified

z = random numbers

Example

Perf. No. = 15.641~92 Score = 1.5 *( 62.7) = 94 %

Perf.. No. = Q3176.167 Score = 2.0*(47.0) = 94 %

Perf. No. = 50132 Score: No credit

Note: An even number for the middle digit constitutes mastery.

Score = 4.0 * AB.C z = random numbers

Configurations

Perf. No. = 718Q3281 Score = 4*(15 .. 0) = 60 configuration 7 completed

Note: Student receives 20 points for every square answered correctly. A score of 60 is needed for mastery of this module.

GREAT RACE zzzCzABz

PeIformance NumbeI Interpletation: eRE modules

Score = 6.0 * AB .. C z = random numbers

Perf No. = 777J.8Q18 Score = 6*(07.3) = 44

Note: A score of 40 is needed for mastery of this module.

iv

**** CONFIDENTIAL ****

ECOLOGY

STAGING

AzBCzaaD z = random numbers a = random characters

A gives info on rl\2 value of the student's linearized plot A=Y if rl\2 >= 0.9 A=A if 0.9 > rl\2 >= 0.8 A=X if 0..8> rl\2 >= 0.7 A=F if 0..7 > rl\2 A=Q if Wetland Analysis/Simulator portion has not been completed

B gives info on alpha B=l to 4 => student's alpha < (simulator's alpha 0.5) B=5 to 9 => student's alpha> (simulator's alpha 0.5) B=X if Wetland Analysis/Simulator portion has not been completed

C indicates number of data points deactivated during analysis C=number of deactivated data points if at least 1 point has been deactivated C=a randomly generated letter from A to Y if 0 points deactivated C=Z if Wetland Analysis/Simulator portion has not been completed

D gives info on solution method used by student D=l if polynomial regression was used D=2 if differential formulas were used D=3 if graphical differentiation was used D=4 to 9 if Wetland Analysis/Simulator portion has not been completed

zCBzAFzED z = random numbers

Final conversion = 2*ARC Final flow rate = 2*DE.F

Perf No .. = A7213DF2 1) A => 0.9 > rl\2 >= 0 .. 8 2) 2 => student's alpha < (simulator's

alpha 0 .. 5) 3) 1 => one data point was

deactivated 4) 2 => differential formulas were

used

Perf. No.. = 2125.:!8~913

conversion = 2*42 . .1 = 84.2 flow rate = 2*31.2 = 62.4

Please make a passlfiil criterion based on these values ..

Performance Number Interpretation: eRE modules v

**** CONFIDENTIAL ****

ICMs with Dos interface

Module Format

HETCAT

HEATFXl

HEATFX2

Interpretation

zzABzCD A=2,3,5,7: interaction done B=2,3,5,7: intra done C=2,3,5,7: review done D denotes how much they

did in the interaction:

D

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.,

PI-I. This problem helps the student understand the course goals and objectives.

PI-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.

PI-3. Helps the student understand critical thinking and creative thinking, which are two major goals of the course.

PI-4. Requires the student to at least look at the wide and wonderful resources available on the CD-ROM and the Web.

PI-S. The ICMs have been found to be a great motivation for this material.

PI-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 PI-IS.

PI-7. Straight forward modification of Example 1-1.

PI-S. Helps the student review and member assumption for each design equation.

PI-9 and PI-IO. The results of these problems will appear in later chapters. Straight forward application of chapter 1 principles.

PI-II. Straight forward modification of the mole balance. Assigned for those who emphasize bioreaction problems.,

PI-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.

16

PI-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 Montgomery's equipment module on the CD-ROM. See PI-17.

PI-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 (0 are printed out from the web. Part (g) is usually omitted.

PI-IS. Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with PI-6.

PI-16. Open-ended problem.

PI-17. I always assign this problem so that the students will learn how to use POLYMATHIMaLab before needing it for chemical reaction engineering problems.

PI-IS. Parts (a) and (b) are open-ended problem.

PI-19 and PI-20. Help develop critical thinking and analysis.

CDPI-A Similar to problems 3, 4, 11, and 12.

CDPI-B Points out difference in rate per unit liquid volume and rate per reactor volume.

Assigned PI--l AA

PI-2 I PI-3 0

PI-4 0 PI-5 AA PI-6 AA PI-7 I

PI-8 S PI-9 S Pl-IO S Pl-11 0

Summary

Alternates

1-15

1-7

Difficulty SF SF SF SF SF SF SF SF SF SF

FSF

Time (min) 60 30 30 30 30 15 15 15 15 15 15

PI-12 I - Read Only SF 30 PI-13 I SF 1 PI-14 0 FSF 30 PI-IS 0 SF 60 PI-16 S SF 15

PI-17 AA SF 60 PI-18 S SF 30 PI-19 0 30 PI-20 0 FSF 15 CDPI-A AA FSF 30 CDPI-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

Time

In problems that have a dot in conjunction with AA means that one of the problem, either the problem with a dot or anyone of the alternates are always assigned.

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 == CDPI-A.

Summary Table Ch-l ---------------,

Review of Definitions and Assumptions 1,5,6,7,8,9

Introduction to the CD-ROM 1,2,3,4

1-8

Make a calculation

Open-ended

PI-I Individualized solution. PI-2 Individualized solution. PI-3 Individualized solution. PI-4 Individualized solution.

6

8,16

PI-5 Solution is in the decoding algorithm given with the modules ..

PI-6 The general equation for a CSTR is:

V = F AO --FA -rA

Here rA is the rate of a first order reaction given by: fA=- kCA

Given: CA = O.lCAO , k = 0.23 min-I, Vo = 10dm3 min-I, FA = 5.0 moVhr And we know that FA = CAVo andFAo = CAOVo => CAO = F Aol Vo = 0.5 moVdm3 Substituting in the above equation we get:

v = CAOVO -CAVO = (0.5mol/dm 3 )(lOdm3 lmin)-0.1(0.5mq!ldm3 )(lOdm3 I min) kCA (0.23 min--I)(O. 1(0.5mol I dm 3

V =391.3 dm3

PI-7

t = INA -~- dNA -kNA

NAO k = 0.23 min-l

From mole balance: dNA -- = IA'V

dt

1-9

Rate law:

Combine:

at T:::: 0, NAo = 100 mol and T:::: T, NA = (O . .01)NAo

1 (N AO J -+ t= "kIn N A

1 :::: -In(lOO) min t = 20 min

0.23

Pl-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.

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

c) The assumptions made in deriving the design equation ofPFR are: Steady state .. No radial variation in properties of the system.

d) The assumptions made in deriving the design equation of PBR are: Steady state .. No radial variation in properties of the system.

e) For a reaction, A-7 B -rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=J moles/ (dm3 s)

1-10

P 1-9

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.

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. rA 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.

above:

Applying general mole balance we get: dN. --.L=Fo -F. + fr.dV dt J ] J

No accumulation and no spatial variation implies

o = Fo - F. + fr .dV J ] J Also Ij = Pb rj' and W = V Pb where Pb is the bulk density of the bed.

=> 0 = (FjO - F) + f r;CPbdV ) Hence the above equation becomes

Fo-F W=_l __ ,_l --rj

We can also just apply the general mole balance as dN j f' - = (FjO - Fj ) + r/dW) dt

Assuming no accumulation and no spatial variation in rate, we get the same form as

FO-F W = ._1_._, _1 -rj

1-11

PI-IO

Fj

F ,/ i' JO

Mole balance on species j is: v dN.

Po -P + fr.dV = __ 1 I I I dt

o

Let M j = molecular wt. of species j Then FjoM j = W jO = mass flow rate of j into reactor:

N jM j = m j = mass cif species j in the reactor Multiplying the mole balance on species j by M I

v dN FjOM j - FjM I +M I frjdV = M j . __ 1

o dt Now M I is constant:

F M. -PM. + vfM rdV = d0! j N j ) = d(m1l 10 I I I I I dt dt

o

v

w jO -- Wj + fM jrjdV = o

PI-II 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

Let's do pari c first

1-12

[In flowrate (moles/time)] penicillin + [generation rate (molesltime)]penicillin - [Out flowrate (molesltime)] penicillin = [rate of accumulation (moles/time)]penicillin

dNp Fp,in + Gp - Fp,out = dt Fp,in = 0 """ ''''', ...

v

Gp = fr~.dV Therefore,

v dNp f rp .dV - Fp,out = dt

,., .,(because no penicillin inflow)

Assuming steady state for the rate of production of penicillin in the cells stationary state,

dNp --=0

dt

And no variations

Or,

Similarly, fOI Corn Steep Liquor with Fe = 0

Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.

Pl-12 (a) Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below:

.--' Rank 2002 Rank 1995 Chemical 1 I H2SO4 2 2 N2 3 4 C2H4 4 3 O2 5 9 C3H6 6 - H2

1-13

7 6 NH3 8 10 Ch 9 - P20S 10

-C2H2Clz

-+ Cherrucals like H2 , P20 S , C2H2Cl2 has come III top 10 cherrucals and C3H6 has jumped to rank 5 now then rank 9 in 1995 .

Pl-12 (b) Ranking of top 10 chemical companies in sales in yeru 2003 and 2002:

2003 2002 Company Chemical Sales

I----($ million 2003)

Dow Chemical 32632 1 1 2 2 DuPont 30249 3 3 ExxonMobil 20190 4 4 General Electric 8371

-

5 8 Chevron Phillips 7018 ---+-. 5 Huntsman Corp. 6990 --6 PPG Industries 6606

8 9 10

7 Equistar Chemicals 6545 11 Air Products 6029

--

9 Eastman Chemicals 5800

SOllIce: Chemical and Engineering News may 17,2004

-+ We have Chevron Phillips whichjumped to 5 rank in 2003 from 8th rank in 2002 and Air Products coming to 9th rank in 2003 flom 11 th in 2001. -+Chemical sales of each company has increased compared to year 2002 from 9%(Eastman Chemical) t028.2%(Chevron Phillips) but Huntsman Corp .. has a decrease by 2 .. 9%.

Pl-12 (C) Sulfuric acid is prime importance in manufacturing. It is used in some phase of the manufacture of neruly all industrial products It is used in production of every other strong acid. Because of its lruge number of uses, it's the most produced chemicaL Sulfuric acid uses are:

-+ It is consumed in production of fertilizers such as ammonium sulphate (NH4hS04 and superphosphate (Ca(H2P04)2) , which is formed when rock phosphate is treated with sulfUric acid .. -+Used as dehydrating agent. -+Used in manufacturing of explosives, dyestuffs, other acids, pruchment 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.

Pl-12 (d) Annual Production rate of ethylene for year 2002 is 5.21x 1010 lb/yeru Annual Production rate of benzene for yeru 2002 is 1.58 x 1010 lb/year

1-14

Annual Production rate of ethylene oxide for year 2002 is 7.6 x109 lb/year

Pl-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.

1-15

Pl-13 Type Characteristics Phases Usage Advantage Disadvantage

Batc All the reactants 1. Liquid 1, Small scale L High 1. High h fed into the phase pdn. Conversion per Operating

reactor. During 2" Gas 2. Used for lab unit volume. cost reaction nothing phase experimentation. 2. Flexibility of 2. Variable is added or 3" Liquid 3. using for product removed .. Easy Solid Pharmaceuticals multiple quality. heating or 4. Felmentation reactions" cooling. 3. Easy to clean

CST Continuous flow L Liquid L Used when 1 .. Continuous 1. Lowest R of reactants and phase agitation required. Operation" conversion

products .. 2. Gas- 2. Series 2.Good per unit Uniform liquid Configuration Temperature volume. composition 3. Solid -, possible for Control 2. By passing tluoughout liquid different 3 .. Two phase possible with

configuration reactions poor streams possible. agitation.

4.Good Control 3 High power 5. Simplicity of Input reqd. construction. 6. Low operating cost 7. Easy to clean

PFR One long reactor 1. 1. Large Scale 1. High 1. Undesired or number of Primarily pdn .. conversion per thelmal CSTR's in gas Phase 2 .. Fast reactions unit volume gradient series .. 3. Homogenous 2. Easy to 2. Poor No radial reactions maintain (No temperature variations .. 4, Heterogeneous moving parts) control Conc. changes reactions 3 .. low 3" Shutdown along the length, 5" Continuous operating cost and cleaning

pdn, 4. continuous expensive" operation

"" -

PBR Tubular reactor L Gas 1. Used plimarily 1. High L Undesired that is packed Phase in the conversion per thelmal with solid (Solid heterogeneous gas unit mass of gradient catalyst Catalyst) phase reaction catalyst 2" Poor particles" 2.Gas -- with solid catalyst 2" low temperature

solid e .. g Fischer operating cost control reactions. tropsch synthesis. 3., Continuous 3.

operation Channeling 4" Cleaning .~~pensiv:. __

_ .

116

Pl-14 Given A = 2 * 1010 ft 2

R = 0.7302 atm ft3 lbmolR

C = 4*105 cars

TSTP = 491.69R

T = 534 .. 7R

H = 2000ft

Po= latm

c S = 2.04 * 10-10 lbm301 ft

FS = CO in Santa Ana wind FA = CO emission from autos

ft3 V A = 3000-- per car at STP

hr

Pl-14 (a) Total number of Ib moles gas in the system:

POV N :=--

R-T

latmx(4xlO13 ft3) N = ( 3 J = L025 X 1011 lb mol

0.73 atmft_ x534.69R lbmol.R

Pl-14 (b) Molar flowrate of CO into L.A. Basin by cars ..

F - F - VA CPo A -YA T -YA Rr STP

F = ~~00ft3 X llbmol x400000 cars T hr car 359 ft3

FA = 6 .. 685 X 104 lb mollhr

Pl-14 (C) Wind speed through corridor is v = l.5mph W = 20 miles The volumetric flowrate in the conidor is Vo = v.W.H = (l5x5280)(20x5280)(2000) fe/ill

= L673 x lOl3 fe/ill

(See appendix B)

1-17

Pl-14 (d) Molar flowrate of CO into basin from Sant Ana wind FS:= vOCS

= 1.673 x 1013 ft3fhr x2.04xlO-1O Ibmolfft3 Fs = 3412 x 103lbmolflu

Pl-14 (e) dN

Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO = - co dt

Pl-14 (0 t = 0 , Ceo = Ccoo

Pl-14 (g) Time for concentration to reach 8 ppm,

= V dCco dt

C = 2.04 X 10-8 ~bmol C, = 3..:94 X 10-8 ~~mol coo ft3 ' co 4 ft3

From (f), _ V I (FA +Fs--vo.CcooJ t--- n

VO FA + Fs --VoCca

4ft3 = --In

1.673xlO13 ft' hr

t = 6.92 hr

(V =constant, Nco = C co V )

Pl-14 (h) (1) to 0 tl = 72lus

Ceo = 2.00E-1O IbmoUft3 a = 3.50E+041bmoUln v 0 L67E+ 12 fe flu b = 3,OOE+04 IbmoUln

1-18

Fs = 341.231bmol/hr

a + b Sin( JZ" ~ ) + Fs v = 4.0E+13 ft3

= vdCco

dt Now solving this equation using POLYMATH we get plot between Ceo vs

See Polymath program Pl-14h-l.pol.

POLYMATH Results Calculated values of the DEQ variahles Var'iable initial value minimal value maximal value t 0 0 C 2.0E-10 2 .. 0E-10 vO 1.67E+12 1 .. 67E+12 a 3.5E+04 3.5E+04 b 3 .. 0E+04 3 .. 0E+04 F 341..23 341 .. 23 V 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-vO*C)/V

Explicit equations as entered by the user [1] vO=1 .. 67*10A12 [2] a = 35000 [3] b = 30000 [4] F = 341 .. 23 [5] V = 4*10A13

3.0e-8,---------------

24e8

80

72 2.134E-08 1 .. 67E+12 3.5E+04 3 .. 0E+04 341..23 4.0E+13

(2) tf= 48lus F,=O a + bSin( JZ"~) Now solving this equation using POLYMATH we get plot between Ceo vs t

119

final value 72 1.877E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+13

v dCco dt

See Polymath program P 1-14-h--2.pol.

POL YMA TH Results Calculated values of the DEQ variables Variable initial value minimal value t 0 0 C 2 .. 0E-I0 2 .. 0E-I0 vO 1 .. 67E+12 1 .. 67E+12 a 3 .. 5E+04 3 .. 5E+04 b 3 .. 0E+04 3 .. 0E+04 V 4.0E+13 4 .. 0E+13

ODE Report (RKF45) Differential equations as entered by the user t 1] d(C)/d(t) = (a+b*sin(3 .. 14 *t/6)-vO*C)/V

Explicit equations as entered by the user [1] vO = 1.67*10A12 [2] a = 35000 [3] b = 30000 [4] V = 4*10A13

2 Oe-8 r-------- .----------,

50

(3)

maximal value final value 48 48 1 .. 904E-08 1 .. 693E-08 1 .. 67E+12 1.67E+12 3 .. 5E+04 3 .. 5E+04 3 .. 0E+04 3.0E+04 4 .. 0E+13 4.0E+13

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 Vo -+ As the value of Vo is increased the graph changes to a "shifted sin-curve" And as Vo is decreased graph changes to a smooth increasing curve ..

------------------ -----------

PI-IS (a) - rA = k with k = OD5 mol/h dm3

CSTR: The general equation is

1-20

v = FAO -FA -rA

Here CA = D"DICAo , Vo = 10 dm3/min, FA = 5 .. 0. mollhr Also we know that FA = CAVO and FAo = CAOVo, CAO = FAoI Vo = 0..5 mol/dm

3

Substituting the values in the above equation we get, V = CAOVO -CAvo = (0.5)10-0.01(0.5)10

k 0.05 -7 V=99dm3

PFR: The general equation is dFA dCAvO --=rA =k,NowFA=CAvoandFAo=CAovo=> =-k dV dV

Integrating the above equation we get V CA V V -~ JdCA = JdV => V =--.!l(CAG ,,- CA) k CAO 0. k

Hence V = 99 dm3 Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration.

PI-IS (b) - rA = kCA with k = 0.,,0.0.0.1 s"!

CSTR: We have already derived that

V = CAOVO -CAvo _ voC Ao (1-0.01) -rA kCA

k = D .. DDDls"! = 0. .. 0.0.0.1 x 360.0. hr"!= 0..36 hI'''!

-7 V = (10"dm3 / hr )(0.5mal / dny3 )(O.~?) => V = 2750 dm3 (0.36hr -1 )(0.01 * 0.5mal / dm 3 )

PFR: From above we already know that for a PFR

dCAvO - - kC -rA - A

dV Integrating

C dC V _VG f _A =- JdV k CAO C A 0.

Vo In CAG =V k CA

Again k = D.DDDls"! = 0..0.0.0.1 x 360.0. hI o != 0..36 ru-!

1-21

Substituing the values in above equation we get V = 127.9 dm3

PI-IS (c) - fA = kCA2 with k = 3 dm3/moLhr

CSTR:

-rA Substituting all the values we get

PFR:

V = (lOdm3 / hr)(O.5mol / dm3 )(O.99) (3dm3 / hr)(O.Ol * O.5mol / dm3 )2

dCAvO _ -kC 2 - rA - A dV

Integrating

=> V = 66000 dm3

V CA dC V vII --~ f _A =- fdV =>~-(----)=V k CAD C/o k C A C AO

=> V = 10dm3

I hr ( 1 _~_) = 660 dm3 3dm 3 lmol.hr O.OlCAO CAO

PI-IS (d)

tAO dN t- . - WA -- r V A

Constant Volume V=Vo

t = fAO dCA 1A - r

A

Zero order:

1 [ ] .999CAo t = - CAO -O.OOlCAO = ---= 9.99h k 0.05 FiIst Older:

1-22

t =~ln(CADJ =_I_ln(_I_) = 6908s k C A 0.001 .001

Second order:

1[1 1] 1[ 1 1] t = k C A - CAD ="3 0.0005 - 0.5 = 666 h

PI-16 Individualized Solution

PI17 (a) Initial number of rabbits, x(O) = 500 Initial number of foxes, y(O) = 200 Number of days = 500 dx dt = k1x-k2xy .................................... (1)

dy 'dt = k3 XY _. k4y .............. .

Given, kl = 0.02day--1 k2 = 0.00004/(dayx foxes) k3 = 0.0004/(dayx rabbits) k4 = 0.04day-1

. ............... (2)

See Polymath program P 1 .. 17 -a.pol.

POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value t 0 0 x SOO 2.9626929 Y 200 1.128S722 kl 0 .. 02 0 .. 02 k2 4 .. 0E-OS 4.0E-OS k3 4.0E-04 4.0E-04 k4 0 .. 04 0 .. 04

ODE Report (RKF45)

Differential equations as entered by the user [ 1 J d(x)/d(t) = (k1 *x)-(k2*x*y) [ 2 J d(y)/d(t) = (k3*x*y)-(k4 *y)

SOO S19.40024 4099.S17 0 .. 02 4.0E--OS 4.0E-04 0 .. 04

1-23

final value SOO 4.2199691 117.62928 0.02 4.0E-OS 4.0E-04 0.04

Explicit equations as entered by the user [1] k1 = 0 .. 02 [2] k2 = 0 .. 00004 [.3] k3 = 0.0004 [4] k4=0 .. 04

5000.------------------

4000

3000

2000

o 400 200 t 300 o 500

When, tfinal= 800 and k j = O.00004/(dayxrabbits)

160 320 t 480 640 800

Plotting rabbits Vs .. foxes

1-24

160lli-----------------

o 378 720 1061 rabddPl 1744 2086

PII7 (b) POL YMATII Results

See Polymath program Pl-17-b.pol.

POL YMA TH Results NLES Solution

Var~able Value f(x) X------- 2.3850387 --'253E-iI" y 3.7970279 1.72E-12

NLES Report (safenewt) Nonlinear equations ( 1] f(x) = xA3*y-4*yA2+3*x-1 = 0 [ 2 1 f(y) = 6*yJ\2-9*x*y-5 = 0

PI-IS (a)

Ini Guess 2 2

No preheatmg of the benzene feed wtli dlmmlsh the rate of reactIon and thus lesser converSIOns WIll be achIeved ..

PI-IS (b) An lDterpolation can be done on the logarithnnc scale to find the desued values from the gIven data. Now we can lDterpolate to the get the cost at 6000 gallons and 15000 gallons Cost of 6000 gal reactor = 1 905 x lOS $ Cost of 15000 gal reactor = 5 623 X 105 $

PI-IS (c) We are given CA IS 0 .. 1 % of minal concentration

~ CA = OOOlCAO

1-25

Coat va Volume of reactor (log - log plot)

.. --~-

. --

1 -

. , "

"--t-----~--

1011 volume (1I8110na) ..

Also from Example 1.3,

Substituting vo= 10 dm3/min and k = 0.23 min-! we get

v = 300dm 3

which is three times the volume of reactor used in Example 1-3.

PI-18 (d) Safety of Plant

PI-19 Enrico Fermi Problem - no definite solution PI-20 Enrico Fermi Problem -- no definite solution PI-21 Individualized solution ..

CDPI-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:

n = !.V = (20~tm)(200dm3) (lOL3~~a) = 97.5moles RT (8.3145 kPa.dm3 )(500) latm

molK Knowing the mole fraction of A (y Ao) is 75%, we multiply the total number of moles (N1o) by the Y A: molesA = N Ao = 0.75x97.5 = 73.1 The initial concentration of A (CAo) is just the moles of A divided by the volume:

C Ao = moles = N Ao = 73 .1mol~~ = 0.37 moles / dm3 volume V 200dm

CDPI-A(b) Time (t) for a 1 st order reaction to consume 99% of A.

dC r =_-A

A dt

OUf first order rate law is: -r =kC A A

1-26

mole balance: dC A =--kC dt A

-k tfdt = CfA dC A => C

o CAO A

-kt = In( CA ), knowing CA=O.OI CAo and our rate constant (k=O.1 min-I), we can solve CAO

for the time of the reaction: t = -. .!.In(O.OI) = 4.61 1 = 46. 1 min k O.lmin-

CDPI-A (c) Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.

rate law:

dC mole balance: :. _2. = --kC~

dt 1 1

=>-kt=--+-CA CAo

We can solve for the time in terms of our rate constant (k = 0.7) and our initial concentration (CAo): -kt = _~ +_1_

CAo CAO

t=_4_= 4 ----=15.4min.Todetermine kCAo (0.7dm3 / mol min)( 0.37moll dm3 )

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:

NB =Nc =0.8NAo NT = (0.25)Nyo + (0.2+0.8+0.8)NAo = 0.25(97.6) + (1.8)(73.2) = 156.1moles

(156. Imole) (0.082 dm3

atm] (500K) N RT molK p = _1_ = _ . ___ ---''---__ = 32atm

V 200dm3

CDPI-B Given: Liquid phase reaction in a foam reactor, A ~B Consider a differential element, ~ V of the reactor: By material balance

FA -(FA +~FA)=-rA(l-e)~V

1-27

Where, (1- e)~ V = fraction of reactor element which is liquid. or: -FA =-rA(1-e)~V

dF _A = r (I-e) dV A

Must relate (-rA) to FA ' where, FA is the total (gas +liquid) molar flow rate of A. -rA =rate of reaction (g mol A per cubic cm. of liquid per sec .. ); e = volume fraction of gas; FA = molar flow rate of A (g mol/sec .. ); V = volume of reactor

1-28

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 -r A=j{X) they can "design" or size a large number of reaction systems. It sets the stage for the algorithm developed in Chapter 4"

P2-l. 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-S. 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-S. 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-l0. Helps the students get a feel of real reactor sizes.

P2-ll. Great motivating problem. Students remember this problem long after the course is over.

P2-l2. Alternative problem to P2-7 and P2-9.

CDP2-A Similar to 2-9

2-1

CDP2-B Good problem to get groups started working together (e.g. cooperative learning.

CDP2-C Similar to problems 2-8, 2-9, 2-12.

CDP2-D Similar to problems 2-8,2-9,2-12.

Assigned P2-1 0

P2-2 A P2-3 A

P2-4 0 P2-.5 0 P2-6 S

P2-7 AA P2-8 S P2-9 AA P2-10 S

P2-11 AA P2-12 AA CDP2-A 0 CDP2-B 0 CDP2-C 0 CDP2-D 0

Summary

Alternates

8,9,12

7,9,12

7,8,9 9,B,C,D 9,B,C,D 9,B,C,D 9,B,C,D

Difficulty

M M

FSF FSF SF SF SF SF

FSF FSF FSF FSF

Time (min) 15 30 30 7.5 75 60 4.5 45 45 15 1

60 5 30 30 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 anyone of the alternates are always assigned.

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)

2-2

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 == CDPI-A.

Summary Table Ch-2

Straight forward

Fairly straight forward

More difficult

Open-ended

Comprehensive

Critical thinking

P2-l Individualized solution .. P2-2 Individualized solution.

-

1,2,3,4,10

7,9,12

5,6,8

6

4,5,6,7,8,9,12

P2-9

P2-3 Solution is in the decoding algorithm given with the modules ..

P2-4 (a) Example 2-1 through 2-3 If flow rate FAD is cut in half. VI = v/2, FI= FAd2 and CAD will remain same ..

Therefore, volume of CSTR in example 2-3, F;.X 1 FAOX 1 VI =----=---=-6.4=3.2 - rA 2 -r A 2

If the flow rate is doubled, F2 = 2F AD and CAD will remain same,

Volume ofCSTR in example 2-3, V2 = F2X1-rA = 12 .. 8 m3

P2-4 (b) Example 2-5

2-3

--

Faol2 Levenspiel Plot

45-.-------------------------------------

J--O+X overall

4 ~;==;==;==;===~ 35~~~

'" ';- 25 o '" 2 LL

15

Faol2 05

o+-~~r=~~~~~T=~-=~--~ o 02 04 06 08

Now, F AO = 0.,4/2 = 0...2 molls, Conversion

F New Table: Divide each term ~ in Table 2-3 by 2.

-- 'A cx:= ---.--/ 0.-10.-1 -.---, 0..2--'--, 0-.4 -.--, 0.-.6 ~-rAJ(m3) 0..445 TI545 0..665 1.0.25 1.77

Reactor 1 VI = 0.. 82m3 V = (F Aol-IA)X

0.82 = (_~~o J (Xl) A XI

By trial and error we get:

Reactor 2 V2 = 1.2 m3

XI = 0..546 and X2 = 0. .. 8 Overall conversion XOver.1I = (l/2)XI + (l/2)X2 = (0..546+0.._8)/2 = 0..673

P2-4 (C) Example 2-6 Now, FAO = 0..4/2 = 0..2 molls,

Fao/2

Faol2

X1

.... --+X overall

2-4

_ ,/ 0.---,,-.7 _Jj)4J.~J ~.53 k.-J

F New Table: Divide each term ~ in Table 2-3 by 2 ..

VI = 0.551m3

x dX V; = FAO I--

o -rA

-rA 0.1 0.2

0.665 0.4 0.6 1.025 1.77

Plot FAof-rA versus conversion. Estimate outlet conversions by computing the integral of the plotted function.

Levenspiel Plot

45 4

3.5 3

ca ... 25 "0 2 ca LL.

1.5 1

0.5 0

0 0 .. 2 0.4 06 0 .. 8

Conversion

[ XI = 0603 f~r VI = 0.551m3 "---X;-=-O .. 89 for Vz = L614m3 Overall conversion Xo= (l/2)XI + (1I2)Xz = (0 .. 603+0.89)/2 = 0.746

- -.~ ._----

P2-4 (d) Example 2-7 (1) ForPFR,

= 0.222m3

For first CSTR,

~ c5 '" IL

25

2

15

1

05

0

Levenspiel Plot

0 0.1 02 03 04

Convelsion

2-5

_J

05 06 07

FAO _ 3 X2 = 0..6, _. -1.32m , -rA

For second CSTR,

FAO 3 X3 = 0.65, -- = 2.0m , -rA

(2) First CSTR remains unchanged ForPFR:

05(F } V= f. _AO X' 02 rA

Using the Levenspiel Plot

VPPR = 0..22

ForCSTR,

(3) The worst arrangement is to put the PFR

F (X -X ) V - AO 3 2 = 01 3 3- . m

. -rA

Levenspiel Plot

25

2+-----------------~---

* 15+----------------------"0 .f 1 i=====~=iO;;;;;:~------~

o 5 +-------~----

o +---,---'-'-'---o 01 02 0.3 04 0.5

Conversion

first, followed by the larger CSTR and finally the smaller CSTR.

-r) -( '[J_O~23J)-IIT53 ~

06 07

Conversion Original Reactor Volumes Xl = 0..20 VI = 0.188 (CSTR) X2 = 0..60 V2 = 0.,38 (PFR)

LX:.::.::..3 _=-=0..:.:.6:.::.5 ________ V.3 = 0.10 (CSTR)

Worst Ana"itgement_,, ___ ~ VI = 0..23 (PFR) V2 = 0.53 (CSTR)

_____ y3 = 0.10 (CSTR)

ForPFR, Xl = 0..2

V. = f( ~~J1X Using trapezoidal rule, Xo = 0,1, Xl = 0.1

2-6

~ = (X~:D)[f(XD)+f(Xl)J = 0.2 [1.28+0.98]m3

2 =0.23m3 ForCSTR,

FAD 3 For X2 = 0 .. 6, --= 1.32m , -rA

FAD () 3 V2 = -- X2 - Xl = 1.32(0.6- 0.2) = 0.53 ill -rA

For 2nd CSTR, FAD 3 For X3 = 0.65, --= 2m , -rA

P2-4 (e) Example 2-8 T = 5hrs Vo = 1dm3/min = 60dm3/hr CA = 2.5 mol/dm3X = 0.8

V For CSTR, T = .-

Vo V= 300dm3

(1) --r A= CAOX = 2.5xO.8 moll dm 3hr

T 5 = OAmol / dm 3 hr

(2) V = 300dm3 (3)CA = CAO(1-X)

= 0.5 mol/dm3

P2-5 -

0 0.1

m3 ) 0.8 1.0 9 8

0..2

13 3

-

OA 0.6 0..7 O. 8

2 .. 0 3.5 5.0 8.. 5 4 6 0

._.

Levenspiel plot

.............. - .... _ .. _ ... _._ ...... _ .. - .... _"

P2-5 (a) Two CSTRs in series For first CSTR, V = (F Ad-rAXl) XI =>Xl =OA4

For second CSTR, V = (FAd-rAX2) (X2 - XI) => X2 = 0.67

~5~-----------------------------~L----.~ g ~4~--------------------

01 02 03 04 05 06 07 08 09 x

P2-5 (b) Two PFRs in series

v ~ 1( ~~:}x +)( ~~:}X By extrapolating and solving, we get Xl = 0..50. X2 = 0.74

P2-5 (C)

4

3 +------------

F"o -r A 2

(m')

----f--------~

Two CSTRs in parallel with the feed, F AO, divided equally between two reactors, FANEW/-IAXI = o.5FAd-IAXI

1+-----------~~----------------~ .... .-..

----_ .. --_ .. _ ....

V = (o..5F Ad-rAXI) Xl Solving we get, X out = 0.,,60.

o +------,-----.------.------.----~ o 0,2 0,4 0.6

P2-5 (d) Two PFRs in parallel with the feed equally divided between the two reactors, FANEW/-rAXI = o.5FAOI-rAXI By extrapolating and solving as part (b), we get Xout = 0..74

P2-5 (e) A CSTR and a PFR are in parallel with flow equally divided

X 0,,8

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) 12 = 0.,67

P2-5 (0 A PFR followed by a CSTR, X PFR = 0...50. (using palt(b V = (F Ad-rA-XCSTR) (XcSTR - XPFR) Solving we get, X CSTR = 0.,70.

P2-5 (g) A CSTR followed by a PFR,

X CS1R = 0.,44 (using part(a

2-8

1

X pFR F V = f -.MLdX

-rA X CSTR By extrapolating and solving, we get

P2-5 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs, ForPFR, XPFR = 0..50. (using part(b

XPFR = 0. .. 72

CSTR1: V = (F Ad-rA-XCSTR) (XCSTR - XPFR) = 0..5 m3 XCSTR = 0..63 CSTR2: V = (F Ad-rA-XCSTR2) (XCSTR2 - XCSTRI) = 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) = Y2 Fao (mother). The Levenspiel Plot is drawn for the baby hippo below ..

Autocatalytic Reaction

5 ........................................ __ ... _ ...... - ................................. .,

4~------------------~--4

35~r----------------~----4

i 3 -l-\---------------+------1-'-------:-:--::-~ e

-.l. 2 5 t-~~--------------j~----I o ~ 2t\-~~---------~~-.I_~r~--~-~

1.5 t=~==~S;;;;;;;;o.."","'--~_r-----;

05

02 04

Conversion 06 08

2-9

Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the baby's stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.

4 . .5 years 2 25 Age = = . . years 2

2) IfVmax and mao are both one halfofthe mother's then

and since

-r AM2bllby

then

1 -vrnaxCA 1

..=2:....-.. __ = - - r KM + C

A . 2 AM2mothef

1 '-m

Catalytic Reaction

2

o JS= 0.4 Conversion

2 Ao -----

1 - 2 rAM2

( mAo J

-rAM2 baby (

mAo J = - rAM ~ mother

mother

m ~ will be identical for both the baby and mother -rAM2

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 040

P2-6 (C) VSlomach = 0.2 m3

From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:

m -- AO = 127X4 _ 172.J6X3 + 100.l8X2 - 28354X + 4.499 -rAMI

210

r.nAO And smce Vstomach = ---X, -rAMI

solve V= 127X5 - 172.36X4 + 1OG.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 ..

Autocatalytic Reaction

5

45 +--------

4+\----

35~--------------+_

i 3- n:t

----+---l !!: 25 fil

-f-..'~-----------j'-------i

E 2 1.5

1-----

05 I t-------.----... ---o -----,-----r-------.,..------/

o X~0067 02

P2-6 (d) PFR~ CSTR PFR:

04

Conversion

Outlet conversion of PFR = 0 .. 111

2

CSTR:

Catalytic Readion

o X= 0.111 Conversion

06 08

2

5

45

4

3.5 ,...

:E ca .. 2.5 -.!. 0 ca 2 E

1.5

05

0 0

2-11

Catalyti c Reaction

o X=0178 Con v ersi on

Autocatalytic Reaction

0.2 0.4

Conversion 0.6 08

We must solve

v = 0..46 = (X-QJ 1l)(127X4 - 172.36X3 + 100. J 8X2 - 28,354X + 4.499)

X=O,42

Since the hippo gets a conversion over 30.% it will survive"

P2-7 Exothermic reaction: A ~ B + C

r(mol/dm3 ,min 1/-r(dm3.min/mol) X )

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 hom the chart above. Second, use mole balance as given below,

CSTR:

Mole balance: V CSTR = FAo ! = (300mollmin)(0.4) =>

- 'A (Smol / dm 3 .min)

=> V CSTR = 24 dm3

PFR:

x dX VpFR = FAO f--

-r Mole balance: 0 A

P2-7 (b)

= 3QQ(area under the curve) VPFR = 72 dm3

2-12

0.4 0.6 X

For a feed stream that enters the reaction with a previous conversion of DAD 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.

P2-' (C) V CSTR = 105 dm3

FAOX Mole balance: V CSlR =--

-rA

0040

x ._. fA

105dm3 ---. = O.35dm3 mini mol

300mollmin

Use trial and error to find maximum conversion..

At X= 0..70., 1/-fA = 0..5, and X/-rA = 0..35 dm3 .. minlmol

Maximum conversion = 0..70.

P2-' (d) ..

From part (a) we know that Xl = DAn. ".

Use trial and error to find X2 .

2-13

CSTR 0.60

1 PFR II Xl

1

CSTR

--

.....

X2 -

Reananging, we get

X 2 - 0040 = ~ = 0.008 - rAlx2 FAD

X 2 -OAO At X2 = 0.64, I = 0.008

-rA X 2

Conversion = 0.64

P2-7 (e)

From part (a), we know that Xl = OAO. Use trial and error to find X2

X 2 dX X2 dX Mole balance: VPFR = 72 = FAO f --. = 300 f--

040 - r A 040 - rA At X2 = 0 .. 908, V = 300 x (area under the curve)

-lIrA

=> V = 300(024) = 72dm3 Conversion = 0..908 ..

P2-7 () See Polymath program P27fpoL

2-14

0.65

052

0.39

0.26

0.13

0.000 20

P2-8 (a) FsoX V=----rs

Fso = 1000 glhr

Q

40 V 60 80 100

1 dm 3hr At a conversion of 40% -- = 0.15---

-rs g Therefore V = (0.15)(1000)(0.40) = 60dm 3

P2-8 (b) 1 dm 3hr

At a conversion of 80%, -- = 0.8-----_. rs g

Fso = 1000 glhr

Therefore V = (0.8)(1000)(0.80) = 640dm 3

P2-8 (C) x dX

VPFR = Fso f--o -rs

6.0

5 .. 0

to GJ 3.0

2.0

1.0

0 .. 0 0 17 33 50V 67 83 100

From the plot of 1/-rs Calculate the area under the curve such that the area is equal to V IPso = 80/ 1000 = 0..08

X= 12%

F X For the 80 dm3 CSTR, V = 80 dm 3 =~

-fs XI-rs = 0.08.. From guess and check we get X = 55%

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 1I-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 CSTR's in series, the optimum arrangement would still include a CSTR with the volume to achieve a conversion of about 45%, or the conversion that cOllesponds to the minimum value of 1I-rs, first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR

-r l

_ kCs (O.l[Cso - Cs ]+ 0.001) KM + Cs

Let us first consider when Cs is small

Cso is a constant and if we group together the constants and simplify

since Cs < KM

1 KM - which is consistent with the shape of the graph when X is large (if Cs is small X is

r, klC~ + k2 Cs large and as Cs grows X decreases),

Now consider when Cs is large (X is small)

As Cs gets larger Cc appwaches 0:

kCsCc 1 KM + Cs If - r = then - - = ---''-'---l KM + Cs r, kCsCc

As Cs grows large!, Cs KM

2-16

And since Cc is becoming very small and approaching 0 at X = 0, 1/-1s 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.

P2-9 Irreversible gas phase reaction

2A+B ~2C See Polymath program P2"-9.poJ.

P2-9 (a) PFR volume necessary to achieve 50% conversion Mole Balance

X2 dX V=FAO f---

Xl (-rA ) Volume = Geometric area under the curve of (F AoI-rA) vsX)

V = GX400000XO.5) + (100000 X 0.5) V = 150000 m3

P2-9 (b) CSTR Volume to achieve 50% conversion

Mole Balance

V=!AO~_ (-rA )

V = 0.5 X 100000 V = 50000m3

P2-9 (c) Volume of second CSTR added in series to achieve 80% conversion

V __ l!.AO(X2- Xl). 2- (-rA ) V2 = 500000 X (0.8"- 0.5) V2 = 150000m3

2-17

x

x

x

-I- -+"-o.s 0

P2-9 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion

1 VPFR = (-x 400000 x 0.3) + (100000 X 0.3) 2 VPFR = 90000m3

P2-9 (e) ForCSTR, VI = 60000 m3 (CSTR) Mole Balance

V= FAOX (-rA )

x == 0.515

ForPFR, V2 = 60000 m3 Mole balance

X2 dX V=FAO f---

Xl (-rA ) Eqn .. Of 2nd line (sloping upwards)

y -100000 = 1.3 x106 (x - 0.5) X2

60000 = f (l.3x10 6 (x -- 0.51) + 100000)dX Xl

=> X2 = 0 . .746

P2-9 (0 Real rates would not give that shape. The reactor volumes are absurdly large ..

P2-10

x

x

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:

h = 56ft d = 9 ft

V = nr 2h = 11:(45 fti(56 ft) = 3562 ft' = 100,865 L

218

Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(II) = 3036 ft D=lft V = m 2h = n(O.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.

CSTR

PBR

Conversion~ X

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 ..

2-19

60

2 ,4 6 Conversion, X

8 10

The area of the rectangle is approximately 232 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.

60

10

~ _____ ~ __ -LI ____ ~ __ ___ 2 4 6 8 10

Conversion, X

The area of the rectangle is approximately 7,6 kg of catalyst

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,

2-20

6 Conversion, X

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.

60

1;; f 40 If

1.0

0 .. 8

0.6

04

0.2

0.0 0 .. 0 6.-1

~ ~

12..8 \V 19 .. 2 25 .. 6 32.0

P2-12 (g) For different (-rA) vs .. (X) curves, reactOIs 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.

10" 8 r A

/

/' / /

/

o , .. -......... -.. -.---.-,.-..... -... --.-------,--.... --.--.-----,-.... -.-... ----.. -..... .,.----.---,-.---.. o 0.2 OA ,)_6 0 .. 8 1.0

Conversion X

Mole balance equations for a CSTR and a PFR:

CSTR: V = FAOX -- rA

x dX PFR: V= S-

0-- rA

Until the conversion eX) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical.

2-22

05 dX F 05 F X . v f AO fdX - AO - V 1.e. PFR = --=-- - '- CSTR

o - rA - rA 0 - rA

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:

V = !AO X = voC AO X - rA - rA

V 0.09m3 X =---=. =3xlO-13 V C 3 l 3 o AO 5 m X 200 mo X 3 X 108 !'1 .~ - rA S m

3 mol CDP2-A (c) This problem will be divided into two parts, as seen below:

l()~ 8

-'A

(m3 '\ , -----1 ! mot 1 ,- )

0.2 {I.4 0,6 1.0 Conven,ion X

The PFR volume required in reaching X=O.5 (reaction rate is independent of conversion).

The PFR volume required to go from X=O..5 to X=o.,7 (reaction rate depends on convelsion).,

223

= 8*10lO m 3

Finally, we add V2 to VI and get:

Vtot = VI + V2 = 2.3 xlO11 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 %

10' 1/ r /\

l' 1113 ~,~ ruol f 2

I)

C;onversion X

We notice that the new inverse of the reaction rate (I/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:

CDP2-A (e) If the reaction is canied 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:

V _dNA Mole Balance: rA - --dt

Stoichiometry: N A = N AD (1 -- X)

2-24

dX Combine: rA V = N AO --dt

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):

Vo dX -rA -(1+X)=--N AO dt

tSdt - C XS __ d_X __ o - AO 0 - rA (1 + X)

After integration, we have:

1 t =--CAoln(l + X)

- rA

Inserting the values for our variables: t = 2.02 X 1010 s That is 640 year'S.

CDP2A (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: Ii' ~.. ~"~"".'" "''''''''''' " '" ""''''''''''''''' '"'''''' ' il 4 .. 0*10- Reaction Rate (--r A) ver'sus Reactor' Volume (V) 'I

II-'A 3O'10_9"--"--------~ 11[~o:s)

2 .. 0*10-9_

------

.'-..._ .. -...... _,"'_.","", .. 1 .. 0*10-9

o ----t---___ ----,>----_, --_, o 1 .. 0*10" 2 .. 0*10" 3 .. 0*10" 4 .. 0*10" 5 .. 0*1011

V (m3 )

Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.

225

Conversion (X) versus Reactor Volume (V) 1 .. 0

0 ... 8

x 0 ... 6

0 ... 4

0 .. 2

0--1

v .... so t/min o

v ... Vo I. 'Where

,0 .3 I ,. J_

o

*"

to t11(1 &::C& of tlu rcctan.;lc up to tho specified c;ouvcr:Jion.

eM I '" (O.45-0)

-rA (0.45-0)(37) - 16.55 ~in

:X .. O .1

So V ~ v I. (50 I/min}{16.65 min.) u 832.5 1 o

For thePFR,

fo.4S I .. o

4(48) ... 2(43) + 31)

..9.....Sli (10 ... 4(1.5 ... 3S ... 43 ... 48) ... 2(20 + 43 ... 50 + 43) ... ]1) ... :1.5. n .. 3

So V p v I - (50 I/min} {15.12 miu) - 186 1 o

There is also ~ solution at au X>O.1

Tr:r I. .. O.S

For the PFR

+ 4( 15) .... 33)

.!hl. UO ... 4' 20 +- SO +- 3Z .... 1.3) .... ~H43 .... 43 .... 1.1} ... 33) ... Z3.9 "" 3 \

For tl:a CSU

I (O.19)(30) - 13.7 min So V ... 'V t ... (SO 1/;:dn)(23.7 min) ... 11S! 1 Cdm~)

c

228

For t:.h.a PFR.

I - 23.9 - ~ (O.S-Q.79)(30+33) - 23.9 - 0.315 - 23.58 mi: ...

So V .. v I '" (50 I/mill)(23.SS mal -1179 1 o

o.~ diffc:rcu:u:o

CDP2-C (e)

x 0.0 0.1

C A0/-:r A {J:).inJ 10 20

-;(minJ 0.0 l.O

3CI

1$

20

o 0.1 02

v 7'00 1. 50 If::r.i.a. "" 14 m.in.

"= .-,.

0,,2

43

S .. t)

0.4 X

0.3

SO

15.0

2-29

0.4 0.5 0 .. 6

43 32 11

17.2- 16.0 10,.2

0.7

0.1 0.8

15 33

10 .. 5 2,6.4

CDP2-D Data taken at WI.;; kPa (10 atrn) and 227"C (500.2 K)

y/\O =0333 ~ y,,(}p (0.333XlO) 00811"-)'d-' c ;;;;; . __ ... -;;:;; ---------,- ==. J gInO I m Ao RT (0,082X500.2) ~

100000 ~"'--"""----'-.. -... ~,-- .. -------.~-~-. -l

60000

40000

zoooC)

!

I ... ------1

I I

----------------~i t l

() -~---.- .... -.-...... - ... --... ---.~ .. -----.. -.-.-.... -... -.---.-... --.-. o 0.2

x 0.4 0,6

CDP2-D (a) 30% conversion in PFR:

, 0.3 cLX -C pfR ;:;: C AO 5--....... :;::;: 4,664 .. 84 s

0'-' fA V - - (4 ~64 84 f 1 min }2 '/ .) 5 - - 3 =? - V o " - \' 0 . s, 6.:-.-- m mm = L ).) m Os ,

CDP2-D (b) 30 to 50% conversion in CSTR:

c (X,-X) 1:CS'IR ;;;:; .... _Au, .. : ............... !...: :;::;: 12,169.2 s =?

.... IAl "1 (12 169 ') {' 1 min ]2 3/..) 0 - 64 3 YCSIR :::::: ,._S -....... In IIlln =4 3. f m 60s

CDP2-D (C)

2-30

Total Volume:

VTouu == 155.5+405.6 == 561.1 mJ

CDP2-D (d) 60% conversion in PFR:

1:PFR ::-:. C At> J."~X.:;;;; 20.281.9 s o ""fA

80% conversion in PFR:

fA is not known for X>O.60 - can not do.

CDP2-D (e) 50 % in CSTR:

X 't' :::; C Ao ~- :::; 30,422.9 s

- fA.

V:::; v 0"' ;;;; (30,422.9 s { I ~~:~ Y. 2 m> Imin)-;;;:: 1014.1 m 3 \ DOS J

CDP2-D (1) 50 to 60% conversion in CSTR:

C.\o(X) -Xl) . t -;;;:: --.:...--. ....:::."" ..... - :::::: 8112.8

- fAl

V = v.' = (8112.8S{To':-}2 m'/min)= 270.4 m' CDP2-D (g)

2-31

Rate of Reaction vs. VolUl:oe IOE,05 ,"-""'.'--"--- ., '--1 7 5E 06 I ... --- "--, -,-- I

? 5 OE 061 .. _, ---, -- I 2 SE 06 -, - - ~

O.OEtOO L__,_._"~ ____ ,_~_,,,_J o 200 400 600 800

CDP2-D (h) Critique

Volnme(m3)

Answers are Valid:

1. Constant Temperature and Pressure No heat effects No pressure drop

Conversion vs. Volurne 06 ~--~-

0.4

0.2

0, o 200 400 600 800

Volume (m3)

2. Single interpolation to X .. :;;; 0.15, 030,0,45, and 0.50 allowable 3. tiuge volume (the size of the LA Basin)! Raise T? Raise P?

CDP2-E

For the CSTR : F X VI =: , .. ",::'l, .. .L ::;;;; F ...

o ( Area)

.,[' ....

] Alea == VI == 1200 drri

From the graph we can see thaI Xl :: 0.,60 For the PFR:

F"o(Xl - XI) ( d ) V7. :::: . --,"._, == F Ao Area un .er curve ''''''rA

J Area under curve == V, == 600 dm

From the graph We can see that Xz ::;;;; 0. .. 80

A ---- poiu:::t

4000

Fio 3000 orA 2000

1000

2-32

CDP2-F (a) Find the conversion for the CSTR and PFR connected in series

X ,rA lI(-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

400 L CSTR and 100 L PFR Feed is 41 % A, 41 % B, and 18% L

P ;:: 10 attn T :::;:; 2270 C = 500 K P 10 atm C = .. - ;;;:;.--.. ''., ... ----.-...... ,-----... --",.,.".,--- = 0_244 mollL.

' To RT (0.082 I.. . atmfmol " K)( 500 K) CA. ;;:: OAlC,o;:: 0.41(0.244 maUL)::::: 0.1 mol/I.. FAo :;;; t)OC AO ::::: 1 Us(O.l moUL)::::: 0.1 molls::: 6 mol/min

There are tv.'o possible arrangements of the system: 1. CSTR followed by the PFR 2. PFR followed by the CSTR Case 1: CSTR -.,. PFR

CSTR: VI = F .... )Area) V 400 Area =: """,l". = ' .... , , :;;:;: 66.67

FA., 6 From the graph ~ XI = 0.36

PFR : V 1 :;;:;: FA., (Area under curve) V, 100 Area under curve:;: ,-- ;;: -""".,.,:;:; 16.667 F: ... ., 6

From the graph .. X~ = 0.445

....... '.' .. ~' ...... '.~-'- .. ~-..-.,...-"."..,.,..-~--........ - .. -~ .. ~-

Case 1: CSTR > PFR 500r-----~-------.,,~_, 400 ~ 300

;;::; 200 100

o 0,0 0.2 0.4 0.6 Q,.8 1.0

X

Case 2: PFR --+ CSTR

PFR: Area under curve:: 1667 From the graph - Xl ::;; 0.259

CSTR. : /\rea::: 66.67 From the grapb . X 2 ::;; 0.515

CDP2-F (b) Two 400 L CSTR's in series.

CSTRl: V;;:;: FAQ(A.rea) mea;;:;; 66.67 From the graph - XI ::: 0.36

CSTR2 : Area = 66.67 From the graph . X z = 0.595

(b) Two CSTR's in Series 500~--------------r~ 400 ~ 300 .

::: 200 100

o 0,0 0.2 0.4 0.6 0.8 LO

X

CDP2-F (C) Two 400 L CSTR's in parallel.

To each CSTR goes half of the feed

F Ao :::;:; 6/2 ;:::: 3 moUmin V::: FAo(Area)

V 400 Area ::: .. _,-- :: ------ = 13 3...3 FA" 3

From the graph: X = 0.52

2-34

(c) Two CSTR's in Parallel 500 1"'----------:Jr-. 400 ~ 30G .

::.; 200 100

o 0.0 0 .. 2 0.4 0.6 0 .. 8 1.0

X

CDP2-F (d) PFR : V ::: F Ao (Area under curve) From the graph we can find the area under the curve for a conversion of 0.60:

Area == (0 .. 60)(300) :.::: 90 2

V = (2 mol1min)(90) = 180 L ~--~.-.. -------................... -.-,-----.. ".,.--~ ............ ---....... -.-.. ~ .. , .. " .......... .

(d) Single .PFR 500 T-----------.---.---.--~._, 400

.::: 300 .

- 200 1010

o 0.0 0.2 OA 0.6 0.8 1..0

X

CDP2-F (e) Pressure reduced by a factor of 10.

A decrease in pressure would cause a decrease in the overall concentration which would in turn cause a decrease in C Ao and F.-\,o" By looking at the design equation:

v::: !.",,,X "'[A

it is apparent that to compensate for the decrease in F .Aa there would be and increase in X.

CDP2-F (1) Use the graph of 1/.,( ... VS. X to find values for all volumes. (Assume a flow rate of 1 mol/min.) Generate the following table and graphs:

X -r" V

...... "

0 0.2 0 --;:, ........ -

0.1 ""'0.'01'67 -;--3 .. 494 " ... ,,-

0.4 0.0048S-- -"4i984 i----

=+ ... 0.00286 0 .. 7 125.878 0 .. 9 225 .. 088 0.00204

2-35

1"""'"""" .. """'-'----,.,-'"'-'-'-"'- ""'---~-~~---'--" -------, t ! ! (f) Conversion "'S. ''flolume !

~ i - _",~ _____

0& 0.,6

0 .. 4

0,2. I o .,.,,_,_.,.,_ ... , __ ,-..-J

I 0 100 V 200 300 L-_, _____ , __ , ,_,~ __ "' ____ ,J CDP2-F (g) Individualized solution

2-36

(0-1."" 'IS. Volume DO! -,--.---'-------~

0.015

.; 0,01

0005

o , .. ___ ,_ .. ,._ .. ~ __ , ___ ..J

Q 100 V ZOO 300

Solutions for Chapter 3 - Rate Law and Stoichiometry

P3-1 Individualized solution.

P3-2 (a) Example 3-1

0 .. 008

0 .. 007 -t-------------

0 .. 006 -t-----------------------#----j

0 .. 005 ~ ~ 0 .. 004 +------------------F .iOi

0 .. 003

0 .. 002 +-----------~~-------

0 .. 001 -t---------= ....... ~---

o 310

. T ..... " ..... ,.-++ .... _ ............... ., 315 320

T(K)

For E = 60kJ/moi

k = 1.32 X 1016 expC-- 6~~OJ ) E = 240 kj/mol

35E-22

3E-22 t---+------------i 25E-22 t---~._---------___j

g 2E-22 t----\--------------I :;; 1 5E'22 t------\-----

1E-22 t-------'b-------5E-23 t--------"'c--~

325

o 00295 0003 0 00305 0 0031 000315 00032 000325 1fT (11K)

330 335

For E1 = 240kJ/moi

kl = 1.32 X 1016 exp( - 2~~OOJ) E=60 kj/mol

6000000 . . ............................... -_ .

5000000

4000000 t----~-=--

"' 2. 3000000 t--------"..:::-:--------j -'"

2000000 t------

1000000 t---.----------~-__l

o 000295 0003 0.00305 00031 000315 0.0032 000325

1fT (11K)

P3-2 (b) Example 3-2 Yes, water is alJeady considered inert.

3-1

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.

bed P3-2 (d) Example 3-4 A + - B ~ -c + - D a a a

113 So, the minimum value of @B = b/a = -- = 0.33

1 P3-2 (e) Example 3-5 For the concentration of N2 to be constant, the volume of reactor must be constant. V = Vo.

Plot:

180

160

140

120

f"100

::!:-... 80

60

40

20

1 0.5(1-0.14x)2 -rA (I-X)(0.54--0.5X)

1/(-ra) vs X

02 04 06

X 08 12

The rate of reaction decreases drastically with increase in conversion at higher conversions.

P3-2 (0 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 .. Cro = constant and inerts are varied.

N 20 4 *"72N02 A ~ 2B

C 2 Equilibrium rate constant is given by: K c =.~

CA,e

Stoichiometry: = Y A06 = Y AO (2 -1) = Y AO Constant volume Batch:

3-2

C - 2NAO X -2C X B - - AO Vo

Plug flow reactor:

C = _~AO(l- X) = C Ao(l- X) and CB

= 2FAO X = 2CAO X A vo(1 + eX) (l + EX) vo(l + eX) (1 + eX)

CAO = Y;o;0 = YAo(O.07176)mol! dm 3 o

Combining: For constant volume batch: C 2 4C2 X2

K -~- Ao . c- -CAe C AO (1 - X) 4CAO

For flow reactor: C 2 4C2 X2

K =_._~= Ao K C (1 - X e )(1 + eX e) C CAe C AO (1 - X)(1 + eX)

See Polymath program P32-LpoL

POLYMA TH Results NLES Report (safenewt)

Nonlinear equations [1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*CaoI\() .. 5 = 0 [2] f(Xef) = Xef - (kc*(1-Xef)*(1 +eps*Xef)/(4*CaoA0 .. 5 = 0

Explicit equations [l) yao = 1 [2] kc = 0.1 [3] Cao = 0 .. 07174*yao [4] eps = yao

1 ........................

4CAO

0 .. 9 +-------------------------:.7-~---i

Yinert Yao 0

0.1 0.2 0.3 0.4 0.5

-

0.6 0.7 0.8 0.9

0.95 0.956

P3-2 (g) No solution will be given

P3-2 (h) 1 1

A+-B-7-C 2 2

Xeb 1 0.44

0.9 0.458 0.8 0.4777 0.7 0.5 0.6 0.525 0.5 0.556 0.4 0.5944 0.3 0.6435 0.2 0.71 0.1 0.8112

0.05 0.887 0.044 0.893

Rate law: - rA = kACA 2CS 1 dm 3

and kA = 25- --s mol ( J

2

~=.~.= rc -1 -112 112

1 (dm3J2 kc =kB =12.5- -s mol

P3-2 (i) A+3B -7 2C

Rate law: - rA = k A C A C B at low temperatures. At equilibrium,

CCe K = .. ---- ' C C I12C 3/2

A,e B,e

Xef 0.508

0.5217 0.537

0.5547 0.576 0.601 0.633

0.6743 0.732

0.8212 0.89

0.896

At "Iuil;b,;um, -, A = 0 , so we can sugge" th,t - r A = k A ( C A 1/2 C B 112 - ~ ~ J But at t = 0, Cc = 0 So the rate law is not valid at t = 0 ..

3-4

Next guess:

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 11T --- r;:::-_. Flashes/ In(flash K) min es/min) 294 0 .. 00340 9

1 298 0.00335 12.16

6 303 0.00330 16.2

0 -

Plotting In(flashes/min) vs lIT, we get a straight line ..

2.197

2.498

2 .. 785

See Polymath program P3-4-firetlics.pol.

For Crickets:

T(in K) 11T chrips/ In(chirPsi x103 min min) ~2 3.482 80 4 .. 382

293 .. 3 3.409 126 4.836 300 3.333 200 5.298 Plotting In(chups/min) Vs lIT, we get a straight line .. -+ Both, Fireflies and Crickets data follow the Arrhenius Model.

--

In y = A + BIT , and have the same activation energy. See Polymath program P3-4-crickcts.pol.

3-5

2.66

2.52

2.38

2.24

210L---~----~----~--~ 3.30E-:3 .332E-3 3.34E1h 3.36E-3 3.38E-3 3.40E-3

5.3

5.1

49

~.5

4..3 '---~--~--~--3.HE-3 1.36E-3 3.39Eih 1.42E-3 3.45E-3 3.48E-.3

P3-4 (b) F H b or oney! ee: T(in K) 1rr

x103 298 3.356 303 3.,300

rsoa- 3,,247

V(cm/s) In(V)

0.7 -0.357 1.8 0 .. 588 3 1.098

Plotting In(V) Vs liT, almost straight lme. In(V) = 44.,6 - 1.33E4/T At T = 40C(313K) V = 6Acmls At T = -5C(268K) V = 0,005cmls(But bee would not be alive at this temperatme) See Polymath program P3-4-bces.pol.

P3-4 (c) For ants' T(in K) 1rr x1Q" V(cm/s) In(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 In(V) Vs liT, 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

20 ,----------------,

1.4

o .. s

0.2

-0.4

-1.0 L-__ ~ __ ~ __ ~ __ ~ __ __' 3.25E-3 ,'.2'7E-3 3.29E1lr HIE-, 3.3.3E-3 3.36E-3

20 r---------------,

1.4

o.S

0.2

-10L--~------3.22E-J .uSE-J J,J,tE1tr 3.41E-J 3.47E-33.53E-3

increase in temperatme, Activation energies fOl fireflies and crickets are almost the same,

---:-c- --Insect Activation Energy Cricket 52150 Firefly 54800

---

Ant 95570 Honeybee 141800

P3-4 (d) There is a limit to temperature for which data for anyone of he insect can be extrapolate, Data which would be helpful is the maximum and the minimum temperatme that these insects can endme before death., Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperatme it could be useless.,

P3-5 There are two competing effects that bring about the maximum in the corrosion rate: Temperatme and HCN-H2S04 concentration. The corrosion rate increases with increasing temperatme and increasing concentration of HCN-H2S04 complex, The temperatme increases as we go from top to bottom of the column and consequently the rate of conosion should increase" However, the HCN concentr'ations (and the

3-6

HCN-H2S04 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 cOIrosion 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 lOoC, at T = T 1 let k = kl and at T = T 2 = T 1+ 10, let k = k2 = 2kl Then with k = Ae-EIRT in general, kl = Ae-E1R1i and k2 = Ae-E'RT2 , or

k2 _ -~(*-*) E --e or kl R

Therefore:

In (k2 J (T. (T. + 10)) kIll (In 2)(7; (7; +10))

E = R-- = R---,- " (7;-7;) 10 T. (T. + 10) = lOE

1 1 RIn2 !OED 5

which can be approximated by T = ----RIn2

P3-7 (b) E

Equation 3-18 is k = Ae RT

E( 1 I) Dividing gives ~2, = e -Ii Tz""-"1;" , or

1

3-7

[1.99 cal ] [ 273 K][ 373 K] E = mol K In(050) = 7960 cal

lOOK .001 mol

E

A = k eRT, = 10-3 min -I exp . 1

( 1.99_Cal ](273K)

molK

P3-7 (c) Individualized solution

P3-8

= 2lO0min-1

When the components inside air bag are ignited, following reactions take place, 2NaN3 --7 2Na + 3N2 ... .. .,' , ............. ' .. ' ,,(1) lONa + 2KN03 --> K20 + 5Na20 + N2 , , ,(2) K20 + Na20 + Si02 --7 alkaline silicate glass" ",.,.,(3)

5x rxn(l) + rxn(2) + rxn(3) = rxn(4) NaN3 + 0.2KN03 + 01Si02 -7 04Na20 + 1.6N2 + complex/l0., ... , ,(4) Stoichiometric table:

r--'--- ,--_. -

Species Symbol Initial Change Final ---

NaN3 _ A NA -NAX NA(1-X) KN03 B NA~B -O.2XNA NA( BB - 02X)

--

Si02 C NABe OIXNA NA( Be - 01X) Na20 D 0 O.4XNA O.4XNA N2 E 0 1.6XNA 1.6XNA

Given weight of NaN3 = 150g Therefore, no., of moles of NaN3 = 2 .. 3

Mwt of NaN3 = 65

1 moles of NaN3 requires 02 mole of KN03 => Moles ofB, KN03 = 02(2..3) = 0.46 moles Mwt ofKN03 = lOLl Therefore, grams ofKN03 required = 046 x lOLl = 46..5 g

1 moles ofNaN3 requires 01 mole of Si02, Moles of C, Si02 = 0.1(2.3) = 023 moles Mwt of Si02 = 60,08 Therefore, grams of Si02 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 cal. It is stable under ordinary conditions of storage" Decomposes explosively upon heating (over

221 0 For 105 0 C), shock, concussion, or friction,

3-8

Conditions to avoid: heat, flames, ignition sources and incompatibles.

P3-9 (a) dX

From the web module we know that -- = k (1 - .x) and that k is a function of temperature, but not a dt

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

P3-10 (a) 1) C2H6 ~ C2H4 + H2 Rate law: -fA = kCC2H6 2) C2H4 + 11202 -+ C2H40 Rate law: -fA = kCC2H4 C~:2 3) (CH3)3COOCCCH3)3 ~ C2H6 + 2CH3COCH3

A ~ B + 2C Rate law: -fA = k[CA - CBCc2IKcJ

4) n-C4HIO +-+ 1- C4HlO Rate law: orA = k[ C C H --C IKe] n 4 10 iC4H lO 5) CH3COOC2Hs + C4H90H +-+ CH3COOC4H9 + C2HsOH

A + B +-+ C + D

P3-10 (b) 2A + B-- C (1) (2) (3) (4)

P3-10 (C)

orA = kCACB2 orA = kCB orA = k orA = kCACB-1

(1) C2H6 ~ C2H4 + H2

Rate law: -fA = k[CACB - CcCnlKcJ

3-9

(2) H2 + Br2 ----> 2HBr

(3) H2 + 12 ----> 2HI

P3-11 (a) Liquid phase reaction,

o /"-

CH2--OH I

CH2 - CH2 + H20~ CHT-OH A +B--~ C

CAO = 1 Ibmol/ft3 CBO = 3.47 lbmol/fe

Stoichiometric Table Species

Ethylene oxide

Water

f--Glycol

'----

Rate law: Therefore, At 300K

Symbol Initial A CAo=1Ibmol/ft3

--

CBO= 347lbmol/ft1, ----

B

BB=347 C 0

--------_ .. _-----

-rA = kCACB -rA = k C~o (1-X)( BB -X) = k(l-X)(347-X) E = 12500 cal/mol, X = 0 . .9,

Change Remaining -CAOX CA= CAO(l-X)

= (lX) Ibmol/ft3 -CAOX CB = CAO( BB -X)

=(3.47-X) lbmol/fe CAOX Cc = CAOX

= X Ibmol/ft3

3 0.1 3 3 k = O . .ldm ImoLs =--x35.315ft Ilbmol.s = 0.0035ft Ilbmol.s

1000 CAOX (1)(0.9) -1

'rCSTR = --- = --2--------- = 1000 . .56 s -rA (0.0035)(1) (1- 0.9)(3.47 - 0.9)

At 350K, k2 = k expE/R)(lIT-lIT 2= O.0035exp125001l.987)(11T-IIT2= 0.071 dm3/moLs Therefore,

CAOX (1)(0.9) _-1 'rCSTR = -- = 2 - 49.3s

-rA (0.071)(1) (1- 0.9)(3.47 - 0.9)

P3-11 (b) Isothermal, isobaric gas-phase pyrolysis,

CZH6 C2H4 + H2 A ~ B + C

Stoichiometric table:

3-10

.... -

e =Yaob= 1(1+1-1)= 1 v = vo(1+ eX)

P

C2H4 B H2 C

=> v = vo(l+X)

CAO = YAO CTQ = YAO -RT

0 +FAOX FB=FAoX 0 +FAoX Fc=FAOX FTO=FAO F~FAO(1+X)

(I)(6atm) = = 0..067 kmol/m3 = 0..067 mol/dm3

( 0.082 m3atm J(1100K) K.kmol CA = FA = FAO(l- X) = C (1- X) mol/dm3

v v 0 (l + X) AO (1 + X) FB FAO(X) X 3 CB= -=- =CAO--- mol/dm v vo(l+X) (I+X)

Fe FAO(X) X 3 Cc = - = = CAO .--- mol/dm

v vo(l+X) (I+X)

Rate law:

(I-X) (I-X) =kCAO--- =o..o.67k---(I+X) (I+X)

-fA = kCA

If the reaction is carried out in a constant volume batch reactor, =>( e = 0.) CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm3 Cc = CAOX mol/dm3

P3-11 (C) Isothermal, isobaric, catalytic gas phase oxidation,

1 C2H4 + - O2 ~ C2~0 2

1 A + -B ~ C

2 Stoichiometric table:

Species C2~ O2 C2H4O

Symbol A B

C

---.-::::------,--"------Entering Change Leaving FAO -FAOX FA=FAO(l-X) FBo

-BBFAOX FB=FAO( BB -X) 0. +FAOX Fc=FAoX

3-11

_ _ P _ 2 ( 6atm ) _ mol CAO - YAOCro - YAO-- ( 3) -0.092--3

RT 3 0.082 atm.dm (533K) dm mol.K

FA FAo(1-X) CAO (1-X) 0.092(1-X) C - - - - - --~---.:.. A - V - Vo (1 + eX) - (1-0.33X) - (1-0.33X)

FB FAo(eB-~) 0.046(1-X) C - - - - --'-----'-

B- V - vo(1+eX) - (1-0.33X) Fe FAoX 0.092(X) C = -- = = ---'---"-

e V vo(1+eX) (1-0.33X) If the reaction follow elementary rate law

_ {0.092(1- X)}{0.046(1- X)}O.5 => -r -k .

A (1-0.33X) (1-0.33X) P3-11 (d) Isothermal, isobaric, catalytic gas phase reaction in a PBR

C6H6 + 2H2 ~ C6HlO A + 2B ~ C

Stoichiometric table: Species Symbol Entering Change Benzene A FAO -FAOX H2 B FBO=2FAO -2FAOX

----

C6HlO C 0 FAOX

1 2 e=y J=-(1-2-1)=-,-

, AO 3 3

--------

Leaving FA=FAO(1-X) FB=F AO( BB -2X) Fc=FAOX

-,-

P (1) 6atm ( 1 ) 3 CAO = CroYAO = RT - = ( 3) - = 0.055mol / dm 3 0.082 atm.dm (443.2K) 3

mol.K _ FA _ FAo(1-X) _ CAo (1-X) _ 0.055(1-X) CA --- - ------

V vo(1+eX) (1-~X) (1-~X)

3-12

C = FB = FAO (BB -2X) = O.l1(I-X) B v Vo (1+X) (1-~X)

C = Fe = FAoX __ = CAOX = 0.055X _

c V vo (1+X) (1-~X) (1-~X) If the reaction follow elementary rate law. Rate law: -rA '= kCAC;

(1 X)3 -rA '= 0.0007k-----3 (1- ~ X) For a fluidized CSTR:

W = FAoX. -r ' A

W = FAoX _ (1- X)3

0.0007k--"-3 (1-~X ) mol

k=53 - "-- at300K kgcat min atm3

k = kj exp(E(l.. __ !.J) = 53exp(80~9.Q_(_I_""_-I-_)J= 1663000 mol R ~ T 8.314 300 443 kgcatminatm3 FAO = CAO* Vo Vo = 5 dm3/min

atX= 0.8

W = 0.0024 kg of catalyst

3-13

P3-12 1

A + -B -7 C 2

Stoichiometric table for the given problem will be as follows

A h ssummg gas pl ase Species Symbol Entering Change C2H4 A FAO -FAOX O2 B FBo = eBFAO -112 FAOX

-

N2 I FI = elFAO -----_. C2H4O C 0 FAOX

1 2 FAO 1 8B =---=- e =_FIO F = 0.79 F I FAD' 10 0.21 BD FAO 2 F

Y AO = -.:iQ.. = 0.30, FTO

C = YAO P = 0.041!!101 AO RT dm 3

(1-X) F C =_AO =C A V 'AO (1 + EX)

0.041(1- X) 1-- 0.15X

(!--~-X) _ FB _ 2 2 0.020(1---- X) CB ---CAO ----------

V 1-----0.15X 1-0.15X C

c =J:c = CAO~ 0.041X

V 1- 0.15X 1- 0.15X

P3-13 (a) C = Nibroanaline

Leavin~ FAoCI-X) F AOceB - x/2) FAOel FAOX

=? e = e 0.79 = 1.88 I B 0.21

Let A=ONCB B=NH3 D = Ammonium Chloride

A+2B ~ C+D

3-14

P3-13 (b) Species

--

A B

c--:::-------C D ---

P3-13 (C) For batch system,

CA=NAN

P3-13 (d) -rA = kCACB

Entering Change Leaving FAO -FAOX FAO(1-X) FBO = 0 BFAO -2 FAOX FB=

=6.6/1.8 FAO FAOC0B -2X) 0 FAOX Fc=FAOX 0 FAOX Fo=FAOX

---

FA = _~A = _~A = ~:O (1- X) = C AO (1- X) , C A = !A = FA = C AO (1- X) Vo Vo V Vo

N N N F FB =-; =-v,B =- T:O(BB-2X)=CAO(BB-2X), CB =-1L=CAo (BB- 2X )

o Vo Vo -rA = kC~o (1- X)( BB -2X) BB = _~BO_ = 6.6 = 3.67

CAO 1.8

C = 1 8i'!!lol AO . 3 m

FrA =k(1.8r(1-X)(3.67--~~)] P3-13 (e) 1) At X = 0 and T = 188C = 461 K

2 m3

( kmOl)( kmOl) -rAO = kCAoBB = kACAOCBO = 0.0017 . 1.8--3- 6.6-3-kmolnun m m

E krnol -rAO = 0.0202 3 . In nun - -

2) At X = 0 and T = 25C = 298K

3-15

k ~ ko exp( !(:o + nJ [

112735!i 1 m 3 l 1 1 k =0.0017 exp mo -+-kmol.min 1.987 .~ (461 298)

mol.k

= 2.12xlO-6 m3 kmol.min

-lAO = kCAOCBO = 2.52 X W 5 kmoJ/m3min

3)

k ~koexp[ !( t-nJ m3 k =O.0017---exp

kmolmin 11273~ (1 1 J

1.987 cal 461K 561K molK

m 3 k=0.0152---kmolmin

-rAO = kCAOCBO

-rA =0.0152-~~-. (1.8 kmol)(6.6 kmol) kmolrmn m3 m3

I kmol

-rA = 0.1806-3-.-m rmn '---------

P3-13 (f) lA = kCAO\1-X)(eB-2X)

At X = 0.90 and T = 188C = 461K l)att=188C

--_. rA = (0.0017 --.!!!~-J(1.8 km~!_)2 (1- 0.9 )(3 .. 67 - 2(0.9)) kmol.rmn m 2)

= 0.00103 kmol m

3 min

At X = 0.90 and T = 25C = 298K

3-16

- rA = (2.12XlO-6 -m~J(1.8 km~1)2 (1-0.9)(3.67 - 2(0.9)) kmol.rmn m 3)

= 1.28 X 10-6 kmol m

3min

At X = 0..90 and T = 288C = 561K

- rA = (0.0152 m3. J(1.8 km~1)2 (1-- 0.9 )(3.67 - 2(0.9)) kmol.rmn m = 0.00333 __ kmol

m3

min P3-13 (g) Vo = 2m3/min = 0..002m3/min

kmol 1)Fof CSTR at 25C -fA = = 1.28 X 10-6 3 .

mmm

- rA:X =09

.9.0.02m3 /m~nX1.8km~!/m3 xO.l = 281.25m3 1.28xlO-6 _ kmol_

m3 min kmol

2)At 288C, -rA = 0.00333- 3 . m rmn

v = v oCAo (I- ~) - rA:X =O.9

V = 0.002!113 /m~?

Apply mass balance For C 6 = 44c + e For N b = o..86c

ForO ForH

Also for C, 6(2/3) = 4,4c which gives c = 0.909

Next we solve for e using the other carbon balance 6 = 44 (0.,,90.9) + e e=2

We can solve for b using the nitrogen balance b = o..86c = 0..86* (0.,90.9) 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.,,90.9) + 2d d = 3.85

Finally we solve for a using the oxygen balance 6 + 2a = 1.2c + d + 2e 6 + 2a = L2(o..9o.9) + 3.85 + 2(2) a = 1.47

P3-14 (b) Assume 1 mole of glucose (180. g) reacts:

6 + 2a = L2c + d + 2e 12 + 3b = 7.,3c + 2d

Y e/s= mass of cells / mass of glucose = mass of cells /180. g

mass of cells = c*(molecular weight) == 0.,90.9 mol* (9L34g1mol)

mass of cells = 83,.12 g

Y e1s = 81.12 g /180. g Yels = 0.46

Y e/02 = 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 = L47 mol * (32 glmol)

mass of O2 = 47,0.4 g

Y e/02 = 83.12 g /47,0.4 g Ye/o2 =1.77

3-18

P3-15 (a) Isothermal gas phase reaction. 1 3 --N +-H ---7NH 2 2 2 23 Making Hz as the basis of calculation:

1 2 H2 +-Nz -'?-NH3 3 3

1 2 A+-B-'?-C 3 3

Stoichiometric table' Species Symbol H2 A N2 B

'-----1----NH3 C

P3-15 (b) 6=(~ -i-1)=--~ e= YAo6 =0.5X( - ~)=-i

Initial change FAO -FAoX FBO=BBFAO -FAoXl3

0 +2FAoXl3

_ (16.4atm) 3 C AO - 0.5 ( J = 0..2 mol/dm

0.082 atm.dm~ (500K) mol.K

C = C = CAO (~_~~_Xl = 0.2(1- X) = O.lmol/ dm3 H2 A (l+eX) (1- })

2 CAo (l-X) 2 O.2(X) 3 C = C =--x =-x =O.lmol/ dm NH3 C 3 (l+eX) 3 (1-{)

P3-15 (C) kNZ = 40 dm3/mol.s

(1) For Flow system:

3-19

Leaving FA=FAO(1-X) FB=F AO( 0B -Xl3)

--

Fc=(2/3)F AOX

(1- X)

(I-X) (1--})

P3-16 (a) Liquid phase reaction -? assume constant volume Rate Law (reversible reaction):

-r =k[C C -~] A A B K _ c

Stoichiometry: CA =CAO (1-X), CB =CAO (1-X), Cc =CAOX To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe

CACBKc =Cc C~o (1- Xe r Kc = CAOXe X: -(2+ 1 JXe +1=0

CAoKC Xe =0.80

To find the equiliblium concentrations, substitute the equilibrium conversion into the stiochiometric relations.,

3-20

mol mol CA = CAO (I-X) = 2-3 (1-0.80) =0.4-3 dm dm mol ) mol CB =CAo (I-X)=2-3 (1-0.80 =0.4-3 dm dm

CA = CAOX = 2 mo! *0.80 = 1.6 mo!. dm dm

P3-16 (b) Stoichiometry: e= YAot5=(1)(3-1)=2 and Be =0

NAN AO (I- X) (1- X) C --- -C A - V - V

o(1+eX) - AO (1-t-2X)

C = Ne = __ .3NAO X = C 3X e V Vo (1 +eX) 'AO (1+2X)

Combine and solve for x.,.

K C .Q..-- Xe) =[c 3Xe ]3 C AO (1+ 2Xe) AO (1+ 2Xe)

Kc (1- Xe)(1 + 2Xe)2 = 27C~oX: (

27C~o J 3 - 4+--- X +3X +1=0 K e e

e

Xe = 0 . .58

Equilibrium concentrations:

Po lOatm mol C =--=--------------=0.305---AO RT (d 3 J d 3 o (400K) 0.082 m atm m

molK _ (1-0 . .58) _ mol

CA - 0.305 ( ( )) - 0.0.59--3 1+2 0.58 dm 3(0 . .58)(0.30.5) mol

C = =0246- e (1+2{0.58)) . dm3

P3-16 (C) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.,

3-21

Stoichiometry:

C =NA =NAO (1-X)=C (I-X) A V V AO

o

C =Nc =3NAO X =3C X c V V AO

o

Combine and solve for Xe

KCCAO (1- Xe) = (3CAo X e)3 Xe = 0.39

Equilibrium concentrations mol CA = (0.30.5)(1-0.39) = 0.19-3 dm

mol Ce = (0.30.5)(0.39) = 0.36--3 dm

P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction):

-r =k[C ._ C~] A A K

e

Stoichiometry:

= YAoO=(I)(3-1)=2 and Be =0 NAN AO (1- X) (1 - X ) C --- -C

A - V - Vo(1+X) - AO (1+2X)

C = Nc =_l!'AOX --=3C X c V Vo(1+X) AO (1+2X)

Combine and solve for Xe:

KCCAO (1- Xe) = [3CAO X e]3 1+ 2Xe 1+ 2Xe

Xe = 0.58

Equilibrium concentrations:

3-22

0.305(1-0.58) mol C = =0.059--A 1+2(0.58) dm3

3(0.305)(0.58) mol C = =0.246-c 1+2(0.58) dm3

P3-17 Given: Gas phase reaction A + B -7 8C in a batch reactor fitted with a piston such that V = O.IPo

(ft3f k=l.O---

lbmol 2 sec -rA =kC~CB NAO = NBo at t = 0 Vo = 0.15 ft3 T = 140C = 6000 R = Constant

P3-17 (a) Y = N AO =0.5 AO N AO +NBO 8=8-1-1=6 e = YAo8=3

Now T

and .-.- = 1, Po = lOVo' and P = lOV 1'0

Therefore

Therefore

v = lOVo2 (1 + eX) lOV

or

N = ( Y AOPo ) v: AO RT 0

3-23

[1 X]3 -r

A = 5.03*10-9 - 3 Ib

3mol

(1+3X)2 ft sec

P3-17 (b) V2 =Vo2(1+eX) 0.22 = 0.152 (1 + eX) X = 0.259

-rA = 8.63*10- 10 }b3mol

ft sec

P3-18 No solution will be given. P3-19 No solution will be given. P3-20 No solution will be given .. --_ ... _---------

CDP3-A

3-24

w InW r 6.5 13 18

1871802 2.564949 2.890372

300 310 313

1fT' 0.003333 0.003226 0 .. 003195

.... ...... -.--.. ---.--.--.-..i~w;;1rr .. ---- __ .. 1 3.5 ,..---.. - .. _ ... -._. ...... _ .... _- I

From the graph: E= 7120

3

2 . .5

~ 2 e - 1.S

0.5

y '" -7120.4x + 2:5.593 Fr.O.S8$3

0+-,----.-- '--.--.. "' .. __ . .,-.,-----1 o..(XX:I1S 0.0002 0..0032S

1trK ~." ....... ~,- .. " , .. "" ..... , "".,. ~ ...... -~~ ................... -,--"- " ......... ".".~ .... -"--"-,,~ .......... , .... _ ............... , .... 7"" .............. -. __ ........ .

In W = --7120.4 *(_._ .. -! .. ,,,.--) 1- 25.593 == 2.95 41.5+273

W(4L5C) = 1 9.2cm I s

I

---------------------------_.-_. __ ._--------

CDP3B Polanyi equation: E = C - a(-.1HR) We have to calculate E for the reaction

CH30 + RBI' -7 CH3Br + Ro Given: dHR = - 6 kcaUmol From the given data table, we get

6.8 = C - a(17.5) and 6,.0 = C -- a(20) => C = 12.4 KJ/mol and a = 0..32 Using these values, and .1HR = - 6 kcaUmol, we get E = 10.48 KJ/mol

CDP3-C (a) A-7B

Rate law at low temperature: - rA = kC A The rate law t higher temperature must: 1) Satis(y thermodynamics relationships at equilibrium, and 2) Reduce to irleversible rate law when the concentration of one or more of the reaction products is zero.

K __ CBe Also, We know, C -

Reauanging, we get

CAe

C - CBe =0 Ae K

c

3-25

So, lets assume rate law as -rA =kA( CA - ~:J

Also when CB = 0, it satisfies the given rate law, Hence the proposed rate law is conect.

CDP3-C (b) A+2B72D

Rate law at low temperature:

Here,

1/2 -rA =kCA CB

But it does not satisfy the irreversible late law at low temperatures" Hence it is not COlrect So, taking square root of Kc

fj{ CDe -V 1\' C = - 112'

CAe C Be

C 112 C _ C De = 0 Ae Be ~Kc

k (c 1I2C CDe J - rA = A Ae Be - ~ K C Which satisfies the irreversible rate law, Hence it is the required rate law.,

CDP3-C (C) A+B-7C+D

. ,kPAPB IrreversIble rate law: -, rA = . ..::::---1 + KAPA + KBPB PCPD PCPD We know, K p = or PBPA - = 0 PBPA Kp

Hence assume rate law as:

k PP -'--. A B K

p ( PCPDJ

- rA = . ."-----1 + KAPA + KBPB + KcPc + KDPD Which satisfies both the above mentioned conditions,

3-26

CDP3-D

a)

b)

c)

lJ

y - 0.15 NlIJ O

A$$umi~1 ideal gas law

C eo

- ~Q ... ~ .. - 8.2 Itra ' .. 0.2 ,mol/l Va H:ro 0.082 latln

1 OKCSOOK)

. ,:no

CNH .