solution manual essentials of chemical reaction engineering

1-2

Solutions for Chapter 1 Mole Balances

P1-1. This problem helps the student understand the course goals and objectives. Part (d) gives

hints on how to solve problems when they get stuck.

P1-2. 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. The results of this problem will appear in later chapters. Straight forward application of chapter

1 principles.

P1-10. Straight forward modification of the mole balance. Assigned for those who emphasize

bioreaction problems.

P1-11. 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 P1-14.

P1-12. 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.

P1-13. Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with

P1-6.

P1-14. I always assign this problem so that the students will learn how to use POLYMATH/MATLAB

before needing it for chemical reaction engineering problems.

P1-15 and P1-16. Help develop critical thinking and analysis.

CDP1-A Similar to problems 3, 4, and 10.

1-3

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 1-13 SF 15

P1-7 I SF 15

P1-8 S SF 15

P1-9 S SF 15

P1-10 O FSF 15

P1-11 I SF 1

P1-12 O FSF 30

P1-13 O SF 60

P1-14 AA SF 60

P1-15 O -- 30

P1-16 O FSF 15

CDP1-A AA FSF 30

Assigned

= Always assigned, AA = Always assign one from the group of alternates,

O = Often, I = Infrequently, S = Seldom, G = Graduate level

1-4

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

1-5

P1-1 Individualized solution.

P1-2

(b) The negative rate of formation of a species indicates that its concentration is decreasing as the

reaction proceeds ie. the species is being consumed in the course of the reaction.

A positive number indicates production of the particular compound.

(c) The general equation for a CSTR is:

A

AA

r

FFV 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 = 10dm

3 min-1

Substituting in the above equation we get:

3

A0 0 0 0 0

1

0

(1 0.1) (10 / min)(0.9)

0.1 (0.23min )(0.1)

A A

A A

C v C v C v dmV

kC kC

V = 391.304 m3

(d) k = 0.23 min-1

From mole balance: dNA

dtrA V

Rate law: rA k CA

rA kNA

V

Combine:

dNA

dtk NA

1-6

at 0t , NAO = 100 mol and t t , NA = (0.01)NAO

t = A

A

N

N

k

0ln1

min100ln23.0

1

t = 20 min




P1-6 Individualized solution

P1-7 (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.

1-7

P1- 7 (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.

P1-7(c)

The assumptions made in deriving the design equation of PFR are:

- Steady state.

- No radial variation in properties of the system.

P1-7 (d)

The assumptions made in deriving the design equation of PBR are:

- Steady state.

- No radial variation in properties of the system.

P1-7 (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 1-8

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.

1-8

'

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:

dVrFFdt

dNjjj

j

0

No accumulation and no spatial variation implies

dVrFF jjj00

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 FW

r

We can also just apply the general mole balance as

'

0( ) ( )j

j j j

dNF 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 FW

r

P1-9

Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 9), so

there is no cell growth and the nutrients are used in making product.

Lets do part c first.

1-9

[Flowrate In (moles/time)] penicillin + [generation rate (moles/time)]penicillin [ Flowrate Out(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 FV

r

Or,

,p out

p

FV

r

Similarly, for Corn Steep Liquor with FC = 0

C

C

C

CC

r

F

r

FFV 00

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

destroyed.

P1-10

Given

21010*2 ftA RTSTP 69.491 ftH 2000

31310*4 ftV T = 534.7 R PO = 1atm

Rlbmol

ftatmR

3

7302.0 yA = 0.02

3

1010*04.2ft

lbmolCS

1-10

C = 4*105 cars

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

hr

ftvA

3

3000 per car at STP

P1-10 (a)

Total number of lb moles gas in the system:

0PV

NRT

N = 13 3

3

1 (4 10 )

.0.73 534.69

.

atm ft

atm ftR

lbmol R

= 1.025 x 1011 lb mol

P1-10 (b)

Molar flowrate of CO into L.A. Basin by cars.

STP

AATAA

TR

PCvyFyF 0

carsft

lbmol

carhr

ftFT 400000

359

130003

3

(See appendix B)

FA = 6.685 x 104 lb mol/hr

P1-10 (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

P1-10 (d)

Molar flowrate of CO into basin from Sant Ana wind.

FS v0 CS

= 1.673 x 1013 ft3/hr 102.04 10 lbmol/ft3

= 3.412 x 103lbmol/hr

P1-10 (e)

1-11

Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO = COdN

dt

dt

dCVCvFF cocooSA (V=constant, VCN coco )

P1-10 (f)

t = 0 , coOco CC

0

co

coO

Ct

co

A S o coC

dCdt V

F F v C

cooSA

coOoSA

o CvFF

CvFF

v

Vt ln

P1-10 (g)

Time for concentration to reach 8 ppm.

8

0 32.04 10CO

lbmolC

ft, 8

3

2.0410

4CO

lbmolC

ft

From (f),

0

34 3 13 8

3 3

3 313 4 3 13 8

3

.ln

.

6.7 10 3.4 10 1.673 10 2.04 104

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 CVt

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

P1-10 (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

dCVCvF

tba cocoos

6sin

Now solving this equation using POLYMATH we get plot between Cco vs t

1-12

See Polymath program P1-10-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

1-13

(2) tf = 48 hrs sF = 0 dt

dCVCv

tba cocoo

6sin

Now solving this equation using POLYMATH we get plot between Cco vs t

See Polymath program P1-10-h-2.pol.

POLYMATH Results



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)


[1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V


[1] v0 = 1.67*10^12

[2] a = 35000

[3] b = 30000

[4] V = 4*10^13

1-14

(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.

P1-11 (a)

rA = k with k = 0.05 mol/h dm3

CSTR: The general equation is

A

AA

r

FFV 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

Substituting the values in the above equation we get,

05.0

10)5.0(01.010)5.0(00A0

k

vCvCV A

1-15

V = 99 dm3

FR: The general equation is

krdV

dFA

A , Now FA = CAv0 and FA0 = CA0v0 => kdV

vdCA 0

Integrating the above equation we get

VC

CA dVdC

k

v A

A 0

0

0

=> )( 00

AA CCk

vV

Hence V = 99 dm3

Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of

concentration.

P1-11 (b)

- rA = kCA with k = 0.0001 s-1

CSTR:

We have already derived that

A

A

r

vCvCV 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

dmmolhrdmV => V = 2750 dm3

PFR:

From above we already know that for a PFR

AAA kCr

dV

vdC 0

Integrating

VC

C A

A dVC

dC

k

v A

A 0

0

0

VC

C

k

v

A

A00 ln

Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1

1-16

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

P1-11 (c)

- rA = kCA2 with k = 3 dm3/mol.hr

CSTR:

VCA0v0 CAv0

rA

v0CA0(1 0.01)

kCA2

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

dVrA kCA

2

Integrating

v0

k

dCA

CA2

CA0

CA

dV

0

V

=>v0

k(1

CA

1

CA0) V

=> V10dm3 /hr

3dm3 /mol.hr(

1

0.01CA0

1

CA0) = 660 dm3

P1-11 (d)

CA = .001CA0

tdN

rAVNA

NA 0

Constant Volume V=V0

tdCA

rACA

CA 0

Zero order:

t1

kCA0 0.001CA0

.999CAo

0.059.99h

First order:

1-17

t1

klnCA0

CA

1

0.001ln

1

.0016908 s

Second order:

t1

k

1

CA

1

CA0

1

3

1

0.0005

1

0.5666h

P1-12 (a)

Initial number of rabbits, x(0) = 500

Initial number of foxes, y(0) = 200

Number of days = 500

1 2

dxk x k xy

dt .(1)

3 4

dyk 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 P1-12-a.pol.

POLYMATH Results



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

1-18

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)


[1] d(x)/d(t) = (k1*x)-(k2*x*y)

[2] d(y)/d(t) = (k3*x*y)-(k4*y)


[1] k1 = 0.02

[2] k2 = 0.00004

[3] k3 = 0.0004

[4] k4 = 0.04

When, tfinal = 800 and 3 0.00004 /( )k day rabbits

1-19

Plotting rabbits Vs. foxes

P1-12 (b)

POLYMATH Results

See Polymath program P1-12-b.pol.

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

P1-13 Enrico Fermi Problem no definite solution

P1-14

Mole Balance:

A0 A

A

F FV =

r

1-20

Rate Law :

2

A Ar kC

Combine:

A0 A

2

F FV =

AkC

3

0 0 3

2 63 .A Adm molA molA

F v Cs dm s

3

0 3

0.1 0.33 .A Adm molA molA

F v Cs dm s

3

32

3

(6 0.3)

V = 19,000

(0.03 )(0.1 ).

mol

s dmdm mol

mol s dm

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Solutions for Chapter 2 - Conversion and Reactor Sizing

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. Straight forward problem alternative to problems 7, 8, and 11. P2-7. To be used in those courses emphasizing bio reaction engineering. P2-8. The answer gives ridiculously large reactor volume. The point is to encourage the student to

question their numerical answers. P2-9. Helps the students get a feel of real reactor sizes. P2-10. Great motivating problem. Students remember this problem long after the course is over. P2-11. Alternative problem to P2-6 and P2-8. P2-12. Novel application of Levenspiel plots from an article by Professor Alice Gast at Massachusetts

Institute of Technology in CEE. CDP2-A Similar to 2-8 CDP2-B Good problem to get groups started working together (e.g. cooperative learning). CDP2-C Similar to problems 2-7, 2-8, 2-11. CDP2-D Similar to problems 2-7, 2-8, 2-11.

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 AA 7,8,11 FSF 45 P2-7 S FSF 45 P2-8 AA 6,8,11 SF 45 P2-9 S SF 15 P2-10 AA SF 1 P2-11 AA 6,7,8 SF 60 P2-12 S M 60 CDP2-A O 8,B,C,D FSF 5 CDP2-B O 8,B,C,D FSF 30 CDP2-C O 8,B,C,D FSF 30 CDP2-D O 8,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.

Summary Table Ch-2

Straight forward 1,2,3,4,9

Fairly straight forward 6,8,11

More difficult 5,7, 12

Open-ended 12

Comprehensive 4,5,6,7,8,11,12 Critical thinking P2-8


P2-2 (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,

2.34.62

1

2

1 011

A

A

A r

XF

r

XFV

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

P2-2 (b) Example 2-4

Now, FAO = 0.4/2 = 0.2 mol/s,

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 m

3

V = (FAO/-rA)X

182.0

1

0 Xr

F

XA

A 22.3

2

0 Xr

F

XA

A

By trial and error we get:

X1 = 0.546 and X2 = 0.8

Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673

P2-2 (c) Example 2-5

(1) For first CSTR,

at X=0 ;

Levenspiel Plot

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.2 0.4 0.6 0.8 1

Conversion

Fa

o/-

ra

0A

A

F

r1.28m3

at X=0.2 ; 0A

A

F

r.94 m3

From previous example; V1 ( volume of first CSTR) = .188 m3

Also the next reactor is PFR, Its volume is calculated as follows

0.5

2

0.2

30.247

AO

A

FV dX

r

m

For next CSTR,

X3 = 0.65, 32AO

Fm

rA, V3 =

33 2( ) .3AOF X X

mrA

(2)

Now the sequence of the reactors remain

unchanged.

But all reactors have same volume.

First CSTR remains unchanged

Vcstr = .1 = (FA0/-rA )*X1

=> X1 = .088

Now

For PFR:

2

0.088

X

AO

A

FV dX

r

,

By estimation using the levenspiel plot

X2 = .183

For CSTR,

VCSTR2 =3 2 30.1

AOF X Xm

rA

=> X3 = .316

(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

FV dX

r

Using trapezoidal rule,

XO = 0.1, X1 = 0.1

1

1 1

3

3

0.21.28 0.98

2

0.23

O

O

X XV f X f X

rA

m

m

For CSTR,

For X2 = 0.6, 31.32AO

Fm

rA, V2 = 2 1

AO

A

FX X

r= 1.32(0.6 0.2) = 0.53 m3

For 2nd CSTR,

For X3 = 0.65, 32AO

Fm

rA, V3 = 0.1 m

3


P2-4 Solution is in the decoding algorithm given with the modules.

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 = 1.6 m3

P2-5 (a) Two CSTRs in series

For first CSTR,

V = (FAo/-rAX1) X => X1 = 0.53

For second CSTR,

V = (FAo/-rAX2) (X2 X1)

=> X2 = 0.76

P2-5 (b)

Two PFRs in series

1 2

10

X X

Ao Ao

A AX

F FV dX dX

r r

By extrapolating and solving, we get

X1 = 0.62 X2 = 0.84

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 = 0.68

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 = 0.88

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 dXr

FV

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

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

P2-6 (a)

To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as given

below.

CSTR:

Mole balance: min./5

4.0min/300F V

3

A0

dmmol

mol

r

X

A

CSTR =>

=>VCSTR = 24 dm3

PFR:

Mole balance:

X

A

APFRr

dXFV

0

0

= 300(area under the curve)

VPFR = 72 dm3

P2-6 (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 Xr

FdX

r

FdX

r

FV

A

A

A

A

A

A

PFR

P2-6 (c)

VCSTR = 105 dm3

Mole balance: A

CSTRr

XA0F V

moldmmol

dm

r

X

A

min/35.0min/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

P2-6 (d)

From part (a) we know that X1 = 0.40.

Use trial and error to find X2.

Mole balance:

2

120

XA

A

r

XXFV

Rearranging, we get

008.040.0

0

2

2AXA

F

V

r

X

At X2 = 0.64, 008.040.0

2

2

XAr

X

Conversion = 0.64

P2-6 (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

APFRr

dX

r

dXFV

At X2 = 0.908, V = 300 x (area under the curve)

=> V = 300(0.24) = 72dm3

Conversion = 0.908.

P2-6 (f)

See Polymath program P2-6-f.pol.

P2-7 (a)

S

S

r

XFV 0

FS0 = 1000 g/hr

At a conversion of 40% g

hrdm

rS

3

15.01

Therefore 360)40.0)(1000(15.0 dmV

P2-7 (b)

At a conversion of 80%, g

hrdm

rS

3

8.01

FS0 = 1000 g/hr

Therefore 3640)80.0)(1000(8.0 dmV

P2-7 (c) X

S

SPFRr

dXFV

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

XFdmV 0380

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

P2-7 (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.

P2-7 (e)

SM

CS

sCK

CkCr and 001.01.0 0 SSC CCC

SM

SSSs

CK

CCkCr

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

rS

SM

s 2

2

1

1

since CS < KM

CskCk

K

rS

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

sCK

CkCr then

CS

SM

s CkC

CK

r

1

As CS grows larger, CS >> KM

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.

P2-8

Irreversible gas phase reaction

2A + B 2C

See Polymath program P2-8.pol.

P2-8 (a) PFR volume necessary to achieve 50% conversion Mole Balance

2

1)(

0

X

X A

Ar

dXFV

Volume = Geometric area under the curve of (FA0/-rA) vs X)

5.01000005.04000002

1V

V = 150000 m3

.

P2-8 (b) CSTR Volume to achieve 50% conversion Mole Balance

)(

0

A

A

r

XFV

1000005.0V V = 50000m3

P2-8 (c) Volume of second CSTR added in series to achieve 80% conversion

)(

)( 1202

A

A

r

XXFV

)5.08.0(5000002V

V2 = 150000m3

P2-8 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion

)3.0100000()3.04000002

1(PFRV

VPFR = 90000m3

P2-8 (e) For CSTR, V = 60000 m3 (CSTR) Mole Balance

XX

r

XFV

A

A

)500000800000(60000

)(

0

X = 0.463 For PFR, V = 60000 m3 (PFR) Mole balance

X

A

Ar

dXFV

0

0)(

dXXX

)100000800000(600000

X = 0.134

P2-8(f)

Real rates would not give that shape. The reactor volumes are absurdly large.

P2-9

Problem 2-9 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 = r2h = (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 = r2h = (0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L

Answers will vary slightly for each individual.

P2-10 No solution necessary.

P2-11 (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.

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 P2-11.pol.

P2-11 (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.

P2-11 (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.

P2-11 (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.

P2-11 (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.

P2-11 (f)

P2-11 (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.

P2-12 (a) Individualized Solution

P2-12 (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 shown:

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8

ma

o/-

ra

M1

Conversion

Autocatalytic Reaction

Mother

Baby

31.36 *0.34 0.232

baby

A

FaoXV 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.

4.52.25

2

yearsAge years

2) If Vmax and mao are both one half of the mothers then

0

2 2

1

2 motherA

Ao

AM AM

mm

r r

and since

max2

AAM

M A

v Cr

K C then

max

2 2

112

2baby mother

A

AM AM

M A

v C

r rK C

2 22

1

21

2

AoAo Ao

AM AM motherbaby AMmother

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

P2-12 (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

rX,

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.

P2-12 (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.

P2-13

For a CSTR we have :

V = X0

|f

A

A X X

F

r

So the area under the 0A

A

F

r versus X curve for a CSTR is a rectangle but the height of rectangle

corresponds to the value of 0A

A

F

r at X= Xf

But in this case the value of 0A

A

F

r is taken at X= Xi and the area is calculated.

Hence the proposed solution is wrong.

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

XFV 0 PFR:

X

Ar

dXV

0

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

i.e. CSTRA

A

A

A

A

PFR Vr

XFdX

r

F

r

dXV 0

5.0

0

05.0

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

XFV 000

13

38

3

3

3

00

103.

1032005

09.0

mol

sm

m

mol

s

m

m

r

Cv

VX

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 mr

XCv

r

XFV

A

A

A

A

The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).

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 mr

XCv

r

XFV

A

A

A

ACSTR

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

dNVr AA

Stoichiometry: )1(0 XNN AA

Combine: dt

dXNVr 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

dXX

N

Vr

A

A )1(0

0

X

A

A

t

Xr

dXCdt

00

0 )1(

After integration, we have:

)1ln(1

0 XCr

t AA

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%.

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 above 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.

CDP2-C (b)

CDP2-C (c)

CDP2-C (d)

For the PFR,

CDP2-C (e)

CDP2-D

CDP2-D (a)

CDP2-D (b)

CDP2-D (c)

CDP2-D (d)

CDP2-D (e)

CDP2-D (f)

CDP2-D (g)

CDP2-D (h)

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

CDP2-F (b)

CDP2-F (c)

CDP2-F (d)

CDP2-F (e)

CDP2-F (f)

CDP2-F (g) Individualized solution

3-1







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these materials .

3-2

Solutions for Chapter 3 Rate Laws


P3-2 (a) Example 3-1

For, E = 60kJ/mol For, E = 240kJ/mol

16 600001.32 10 expJ

kRT

16 2400001.32 10 expJ

kRT

T (K) k (1/sec) 1/T ln(k)

310 1023100 0.003226 13.83918

315 1480488 0.003175 14.2087

320 2117757 0.003125 14.56667

325 2996152 0.003077 14.91363

330 4194548 0.00303 15.25008

335 5813595 0.002985 15.57648

T (K) k (1/sec) 1/T ln(k)

310 4.78E-25 0.003226 -56.0003

315 2.1E-24 0.003175 -54.5222

320 8.77E-24 0.003125 -53.0903

325 3.51E-23 0.003077 -51.7025

330 1.35E-22 0.00303 -50.3567

335 4.98E-22 0.002985 -49.0511

3-3

P3-2 (b) No solution will be given

P3-2 (c)

A + 2

1B

2

1C

Rate law: rA kACA2CB and kA 25

1

s

dm3

mol

2

rA

1

rB

1/2

rC

1/2 25CA

2CB

kBCA2CB

1/2

kCCA2CB

1/2

kC kB 12.51

s

dm3

mol

2

P3-3(a) Refer to Fig 3-3 The fraction of molecular collisions having energies less than or equal to 50 Kcal is given by the area

under the curve, f(E,T)dE from EA = 0 to 50 Kcal.

P3-3(b) The fraction of molecular collisions having energies between 10 and 20 Kcal is given by the area under

the curve f(E,T) from EA = 10 to 20 Kcal.

3-4

P3-3(c) The fraction of molecular collisions having energies greater than the activation energy EA= 30 Kcal is

given by the area under the curve f(E,T) from EA =30 to 50 Kcal.

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(flashes/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

chirps/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 similar 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.

3-5

ln (V) = 44.6 1.33E4/T

At T = 40oC (313K) V = 6.4cm/s

At T = -5oC (268K) V = 0.005cm/s(But bee would not be alive at this temperature)

See Polymath program P3-4-bees.pol.

P3-4 (c) For Ants:

T(in K) 1/T x103 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,

We get almost a 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.

3-6

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, 1/1E RTk Ae and 2/2

E RTk Ae , or

2 1

1 1

2

1

E

R T Tke

k or

2 2

1 1

1 2

1 22 1

ln ln

1 1

k k

k kE

T TR

TTT T

Therefore:

21 1

1 11

2 1

ln 10ln 2 10

10

kT T

T TkE R R

T T

1 1

1010

ln 2

ET T

R

which can be approximated by 0.510

ln 2

ET

R

P3-7 (b)

Equation 3-18 is E

RTk Ae

From the data, at T1 = 0C, 1/

1

E RTk Ae , and at T2 = 100C, 2/

2

E RTk Ae

Dividing gives 2 11 1

2

1

E

R T Tke

k, or

2

1 1 2 2

1 2 1

2 1

ln

ln1 1

kR

k RTT kE

T T k

T T

1.99 273 373.050

ln 7960100 .001

calK K

mol K calE

K mol

3-7

1 3 1 1

1

7960

10 min exp 2100min

1.99 273

E

RT

cal

molA k ecal

Kmol K

P3-7 (c) Individualized solution

P3-8

From the given data -rA(dm

3/mol.s) T(K)

k= -rA/(4*1.5)

1/T ln (k)

0.002 300 0.00033333

0.003333 -8.00637

0.046 320 0.00766667

0.003125 -4.87087

0.72 340 0.12

0.002941 -2.12026

8.33 360 1.38833333

0.002778 0.328104

Plotting ln(k) vs (1/T), we have a straight line:

Since, ln k = ln A - ( , therefore ln A = 41.99 and E/R = 14999

3-8

(a) Activation energy (E), E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol

(b) Frequency Factor (A), ln A = 41.99 A = 1.72X1018

(c) k = 1.72X1018 exp(- (1)

Given T0 = 300K

Therefore, putting T = 300K in (1), we get k(T0) = 3.33X10-4

Hence,

k(T) = k(T0)exp[

P3-9 (a)

From the web module we know that (1 )dX

k xdt

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.

P3-10

1) C2H6 C2H4 + H2 Rate law: -rA = 62HC

kC

2) C2H4 + 1/2O2 C2H4O Rate law: -rA = 2/1

0242CkC HC

3) (CH3)3COOC(CH3)3 C2H6 + 2CH3COCH3

A B + 2C

Rate law: -rA = k[CA CBCC2/KC]

4) n-C4H10 I- C4H10 Rate law: -rA = k[104HnC

C 104HiC

C /Kc]

5) CH3COOC2H5 + C4H9OH CH3COOC4H9 + C2H5OH

A + B C + D

Rate law: -rA = k[CACB CCCD/KC]

3-9

P3-11 (a) 2A + B C

(1) -rA = kCACB2

(2) -rA = kCB

(3) -rA = k

(4) -rA = kCACB-1

P3-11 (b)

(1) H2 + Br2 2HBr Rate law: -rHBr =

2

22

2

2/1

1

Br

HBr

BrH

C

Ck

CCk

(2) H2 + I2 2HI Rate law: -rA = 2 21 H I

k C C

P3-12 a)

we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium

condition. If we assume the rate law for the reverse reaction (B->A) is =

then:

From Appendix C we know that for a reaction at equilibrium: KC

At equilibrium, rnet 0, so:

Solving for KC gives:

3-10

b)

If we assume the rate law for the reverse reaction is

then:




c)

If we assume the rate law for the reverse reaction is

then:

3-11




P3-13 The rate at which the beetle can push a ball of dung is directly proportional to its rate constant, therefore

-rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant

k = A exp(

From the data given

-rA T(K)

1/T ln k

6.5 300

0.003333 1.871802

13 310

0.003226 2.564949

18 313

0.003195 2.890372

3-12

Refer to P3-8 (similar procedure)

Therefore, A = 1.299X1011

E = 59195.68 J/mol

k = 1.299X1011 exp(-7120/T)

Now at T = 41.5 C = 314.5 K

k = 19.12 cm/s

Therefore, beetle can push dung at 19.12 cm/s at 41.5 C

P3-14 Since the reaction is homogeneous, which means it involves only one phase.

Therefore,

So, option (4) is correct.

P3-15 Assuming the reactions to be elementary:

2 Anthracene -> Dimer

2( )DimerA AnthraceneC

Cr k C

K where, Kc = k+/k-

Similarly for the second reaction:

3-13

Norbornadiene Quadricyclane

( )Quadricyclane

Norbornadiene Norbornadiene

C

Cr k C

K where, Kc = k+/k-

P3-16 The mistakes are as follows:

1. For any reaction, the rate law cannot be written on the basis of the stoichiometric equation. It

can only be found out using experimental data.

2. In the evaluation of the specific reaction rate constant at 100o C the gas constant that should

have been used was 8.314 J K-1 mol -1.

3. In the same equation, the temperatures used should have been in K rather than oC.

4. The units for calculated k(at 100oC) are incorrect.

5. The dimension of the reaction rate obtained is incorrect. This is due to the fact that the rate law

that has been taken is wrong.

CDP3-A 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

4-1














these materials .

4-2

Solutions for Chapter 4 Stoichiometry

P4-1 Individualized solution

P4-2 (a) Example 4-1 Yes, water is already considered inert.

P4-2 (b) Example 4-2 For 20% Conversion

CD = same as example & CB = CAo (B XA/3) = 10 ( - ) = 2.33 mol/dm3

For 90 % Conversion

CD = same as example & CB = CAo (B XA/3) = 10 ( ) = 0 mol/dm3

The final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of

caustic soda is possible.

P4-2 (c) Example 4-3 For the concentration of N2 to be constant, the volume of reactor must be constant. V = VO.

Plot: 21 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.

4-3

P4-2 (d) Example 4-4

POLYMATH Report No Title Ordinary Differential Equations 05-May-2009

Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 epsilon -0.14 -0.14 -0.14 -0.14

2 FA0 1000. 1000. 1000. 1000.

3 k 9.7 9.7 9.7 9.7

4 K 930. 930. 930. 930.

5 KO2 38.5 38.5 38.5 38.5

6 KSO2 42.5 42.5 42.5 42.5

7 PO2 2.214 0.19639 2.214 0.1963901

8 PSO2 4.1 0.0115348 4.1 0.011535

9 PSO20 4.1 4.1 4.1 4.1

10 PSO3 0 0 4.754029 4.754029

11 rA -0.0017387 -0.0063092 5.915E-08 -1.659E-09

12 w 0 0 5.0E+05 5.0E+05

13 x 0 0 0.9975796 0.9975795

Differential equations

1 d(x)/d(w) = -rA/FA0

Explicit equations

1 FA0 = 1000

mol.hr-1

2 PSO20 = 4.1

3 K = 930

atm-1/2

4 KSO2 = 42.5

5 KO2 = 38.5

atm-1

6 k = 9.7

7 epsilon = -0.14

8 PSO3 = PSO20*x/(1+epsilon*x)

9 PO2 = PSO20*(1.08-x)/2/(1+epsilon*x)

10 PSO2 = PSO20*(1-x)/(1+epsilon*x)

11 rA = -k*(PSO2*PO2^0.5 - PSO3/K)/(1+(PO2*KO2)^0.5+PSO2*KSO2)^2

mol/hr.g.cat

General

4-4

Total number of equations 12

Number of differential equations 1

Number of explicit equations 11

Elapsed time 0.000 sec

Solution method RKF_45

Step size guess. h 0.000001

Truncation error tolerance. eps 0.000001

4-5

Therefore, catalyst weight required for 30% conversion is 1.611x10

5 g = 161.1 kg

4-6

Therefore, catalyst weight required for 60% conversion is 2.945x105 g = 294.5 kg

Therefore, catalyst weight required for 99% conversion is 4.041x10

5 g = 404.1 kg

P4-2 (e) Example 4-5 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: KCCB,e

2

CA,e

Stoichiometry: yA0 yA0(2 1) yA0

Constant volume Batch:

CANA0(1 X)

V0CA0(1 X) and CB

2NA0X

V02CA0X

Plug flow reactor:

CAFA0(1 X)

v0(1 X)

CA0(1 X)

(1 X) and CB

2FA0X

v0(1 X)

2CA0X

(1 X)

4-7

CAOyAOPO

RTOyAO 0.07176 mol /dm

3

Combining: For constant volume batch:

KCCB,e

2

CA,e

4CAo2 X 2

CAO (1 X) Xe

KC (1 Xe)

4CA0

For flow reactor:

)1)(1(

4 22

,

2

,

XXC

XC

C

CK

AO

Ao

eA

eB

C 04

)1)(1(

A

eeC

eC

XXKX

See Polymath program P4-2-e.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

4-8

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

P4-3 Solution is in the decoding algorithm available separately from the author.

P4-4

A1

2B

1

2C

yA01

2

1

2

1

21

1

2

(a) CB0 CA0 0.1mol

dm3

(b) CA CA01 X

1 X0.1

1 0.25

11

2X

0.1 0.75

0.8750.086

mol

dm3

0 3

0.25

0.1 0.1252 20.1 0.0143

11 0.8751

2

C A

X

molC C

X dmX

(c) CB CA0

11

2X

1 X

CA0 11

2X

11

2X

CA0 0.1mol

dm3

(d) CB 0.1mol

dm 3

4-9

(e) rA

4

rC

2, rA 2rC 4 mol dm

3 min

P4-5 (a) Liquid phase reaction,

O CH2--OH

CH2 - CH2 + H2O CH2--OH

A + B C

CAO = 16.13mol/dm3 CBO = 55.5 mol/dm

3

Stoichiometric Table:

Species Symbol Initial Change Remaining

Ethylene

oxide

A CAO=16.13 mol/dm3 - CAOX CA= CAO(1-X)

= (1-X) lbmol/ft3

Water B CBO= 55.5 mol/dm3,

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

Therefore, -rA = k 2

AOC (1-X) ( B -X) = k (16.13)2(1-X) (3.47-X)

At 300K E = 12500 cal/mol, X = 0.9,

k = 0.1dm3/mol.s

CSTR = AO

A

C X

r =

2

16.13 0.92.186sec

0.1 16.13 1 0.9 3.47 0.9 and, V = X FA0 = 2.186 X 200 liters = 437.2 liters

At 350K,

k2 = k exp((E/R)(1/T-1/T2))= 0.1exp((12500/1.987)(1/300-1/350))

= 19.99 dm3/mol.s

Therefore,

CSTR = AO

A

C X

r =

2

16.13 0.90.109sec

19.99 16.13 1 0.9 3.47 0.9,

and, V = X FA0 = 0.109 X 200 liters = 21.8 liters

4-10

P4-5 (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 atmK

K kmol

= 0.067 kmol/m3 = 0.067 mol/dm3

CA = 1(1 )

(1 ) 1

AOAAO

O

XF XFC

v v X X mol/dm3

CB = ( )

(1 ) 1

AOBAO

O

F XF XC

v v X X mol/dm3

CC = ( )

(1 ) 1

C AOAO

O

F F X XC

v v X X mol/dm3

Rate law:

-rA = kCA= kCAO1

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/dm

3 CC = CAO X mol/dm3

P4-5 (c) Isothermal, isobaric, catalytic gas phase oxidation,

C2H4 + 1

2O2 C2H4O

A + 1

2B C

4-11


Species Symbol Entering Change Leaving

C2H4 A FAO -FAOX FA=FAO(1-X)

O2 B FBO -

1

2FAOX

FB=FAO( B -X/2)

C2H4O C 0 +FAOX FC=FAOX

B =

112

2

AOBO

AO AO

FF

F F

2

3

AO AOAO

TO AO BO

F Fy

F F F

2 11 1 0.33

3 2AOy

33

620.092

3 .0.082 533

.

AO AO TO AO

atmP molC y C y

RT dmatm dmK

mol K

1 1 0.092 1

1 1 0.33 1 0.33

AO AOAA

O

F X C X XFC

v v X X X

0.046 12

1 1 0.33

AO B

BB

O

XF

XFC

v v X X

0.092

1 1 0.33

C AOC

O

XF F XC

v v X X

If the reaction follow elementary rate law

Rate law: 0.5A A Br kC C

0.5

0.092 1 0.046 1

1 0.33 1 0.33A

X Xr k

X X

P4-5 (d) Isothermal, isobaric, catalytic gas-phase reaction in a PBR

C6H6 + 2H2 C6H10

A + 2B C

Given:

min/50 30 dm

Stoichiometric Table:

Species Symbol Entering Change Leaving

C6H6 A FA0 -FA0X FA=FA0(1-X)

H2 B FB0=2FA0 -2FA0X FB=FA0(B-2X)

C6H10 C 0 FA0X Fc=FA0X

4-12

22

0

0

0

0

A

A

A

B

BF

F

F

F

3

1

00

0

0

0

0

BA

A

T

A

AFF

F

F

Fy

3

2)121(

3

10Ay

3000055.0

3

1

)443(*0821.0

6

3

13

dm

mol

K

atm

RT

PyCC

KmolatmdmATA

)1(

)1(

)1(

)1(

320

0

0

X

XC

X

XFFC A

AAA

)1(

)1(*2

)1(

)22(

)1(

)2(

320

320

0

0

X

XC

X

XC

X

XFFC AA

BABB

)1()1(320

0

0

X

XC

X

XFFC A

AC

C

Rate Law:

NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of

concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or

concentration units are used. See below for an example. 2

' BAA PkPr

2

3**

minminatmatm

atmkgcat

mol

kgcat

mol

Notice that if you use concentrations in this rate law, the units will not work out.

RTCRTCyPyP AAAA *)( 00

3

3

32

33

0

322 )()1(

)1(4)( RT

X

XkCRTCkCPkPr ABABAA

Design Equation for a fluidized CSTR:

'

0

A

A

r

XFW

333

0

3

32

0

)()1(4

)1(

RTXkC

XXFW

A

A

332

0

3

32

0

)()1(4

)1(

RTXkC

XXW

A

Evaluating the constants:

Katatmkgcat

molk 300

min53

3

At 170C (443K),

4-13

atmkgcat

mol

KK

Kmol

Jmol

J

TTR

Ekk A

min1663000

443

1

300

1

314.8

80000

5311

443300

300443

Plugging in all the constants into the design equation:

X = 0.8

kgcatK

WKmolatmdm

dm

mol

atmkgat

mol

dm7

332

min

3

32

min 1025.5)4430821.0()8.01()055.0(16630004

)8.01(8.0503

33

3

At 270C (543K),

atmkgcat

mol

KK

Kmol

Jmol

J

TTR

Ekk A

min9079000

543

1

300

1

314.8

80000

5311

543300

300543 Plu

gging in all the constants into the design equation:

X = 0.8

kgcatK

WKmolatmdm

dm

mol

atmkgat

mol

dm8

332

min

3

32

min 1022.5)5430821.0()8.01()055.0(90790004

)8.01(8.0503

33

3

P4-6 (a) Let A = ONCB C = Nibroanaline

B = NH3 D = Ammonium Chloride

A + 2B C + D

- A A Br kC C

P4-6 (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

P4-6 (c) For batch system,

CA=NA/V -rA = kNANB/V2

P4-6 (d)

4-14

- A A Br kC C

00 0

0 0 0

1 1 , 1AA A A AA A A ANN N F F

F X C X C C XV V V v v

00 0

0 0 0

2 2 , 2AB B BB B A B B A BNN N F

F X C X C C XV V V v

2

0 1 2A A Br kC X X

0

0

6.63.67

1.8

BB

A

C

C

0 31.8A

kmolC

m

21.8 1 3.67 2Ar k X X

P4-6 (e) 1) At X = 0 and T = 188C = 461 K

min0202.067.38.1

min0017.0

3

2

3

32

00m

kmol

m

kmol

kmol

mkCr BAA

0 3

kmol0.0202

m minAr

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

TTR

Ekk

O

O

11exp

min.

31003.2

298

1

461

1

.987.1

11273

expmin.

0017.0

6

3

kmol

m

Kmol

calmol

cal

kmol

mk

-rAO = kCAOCBO = 2.41 X 10-5 kmol/m3min

3)

0

0

1 1exp

Ek k

R T T

4-15

3 11273m 1 10.0017 exp

kmol min 461 5611.987

cal

molkcal K K

mol K

3m0.0152

kmolmink

0 0 0A A Br kC C

3

3 3

m0.0152 1.8 6.6

kmolminA

kmol kmolr

m m

3

kmol0.1806

m minAr

P4-6 (f) rA = kCAO

2(1-X)(B-2X)

At X = 0.90 and T = 188C = 461K

1) at T = 188 C = 461 K

min00103.0

9.0267.39.018.1min.

0017.0

3

2

3

3

m

kmol

m

kmol

kmol

mrA

2) at X = 0.90 and T = 25C = 298K

min1023.1

9.0267.39.018.1min.

1003.2

3

6

2

3

36

m

kmol

m

kmol

kmol

mrA

3) at X = 0.90 and T = 288C = 561K

min0092.0

9.0267.39.018.1min.

0152.0

3

2

3

3

m

kmol

m

kmol

kmol

mrA

P4-6 (g) FAO = 2 mol/min

1) For CSTR at 25oC -rA min

1023.13

6

m

kmol

4-16

3

3

3

9.0

60.162

min1023.1

1.0min/2

1

m

m

mol

mol

r

XFV

XA

AO

2)At 288oC, -rA min

0092.03m

kmol

3

3

9.0

739.21

min0092.0

1.0min/2

1

m

m

mol

mol

r

XFV

XA

AO

P4-7 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 proceeds to completion and calculating the ending mass of the cells.

P4-7 (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

4-17

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

P4-7 (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

P4-8 (a) Isothermal gas phase reaction.

2 2 3

1 3

2 2N H NH

Making H2 as the basis of calculation:

2 2 3

1 2

3 3H N NH

1 2

3 3A B C


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

4-18

P4-8 (b)

2 1 21

3 3 3

2 10.5

3 3AOy

3

16.40.5

.0.082 500

.

AO

atmC

atm dmK

mol K

= 0.2 mol/dm3

2

3

3

3

1 0.2 10.1 /

11

3

0.22 20.1 /

3 1 31

3

AO

H A

AO

NH C

C X XC C mol dm

XX

C X XC C mol dm

XX

P4-8 (c)

kN2 = 40 dm3/mol.s

(1) For Flow system:

2 2 2 2

312 2

312 2

2

113

40

1 13 3

N N N H

AO

r k C C

X

XC

X X

rN2 1.61 X

1X

3

32

(2) For batch system, constant volume.

4-19

312 2

2 2 2 2

2

2

12

32

2

12

32

0

0 0

0

0

0

2

0

1

1

3

13

40 1 13

1.6 1 13

N N N H

AA AA

A A

A B

BB

H O

N A

r k C C

N XN NC

V V V

C C X

XN

NC

V V

XC

Xr C X

XX

P4-9 (a) Liquid phase reaction assume constant volume

Rate Law (reversible reaction):

CA A B

C

Cr 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

12 1 0e e

A C

X XC K

0.80eX

To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric

relations.

0 3 31 2 1 0.80 0.4A A

mol molC C X

dm dm

0 3 31 2 1 0.80 0.4B A

mol molC C X

dm dm

4-20

0 3 32 *0.80 1.6A Amol mol

C C Xdm dm

P4-9 (b) Stoichiometry:

0 1 3 1 2Ay and 0C

0

0

0

1 1

1 1 2

AAA A

N X XNC C

V V X X

00

0

3 3

1 1 2

C AC A

N N X XC C

V V X X

Combine and solve for Xe.

3

0 0

1 3

1 2 1 2

e eC A A

e e

X XK C C

X X

2 2 3

01 1 2 27C e e A eK X X C X

230274 3 1 0A e e

C

CX X

K

0.58eX

Equilibrium concentrations:

00 33

0

100.305

400 0.082

A

P atm molC

RT dmdm atmK

mol K

3

1 0.580.305 0.059

1 2 0.58A

molC

dm

3

3 0.58 0.3050.246

1 2 0.58C

molC

dm

P4-9 (c) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.

Stoichiometry:

0

0

0

11

AAA A

N XNC C X

V V

4-21

00

0

33C AC A

N N XC 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

30.305 1 0.39 0.19A

molC

dm

30.305 0.39 0.36C

molC

dm

P4-9 (d) Gas phase reaction in a constant pressure, batch reactor

Rate law (reversible reaction):

3

CA A

C

Cr k C

K

Stoichiometry:

0 1 3 1 2Ay and 0C

0

0

0

1 1

1 1 2

AAA A

N X XNC C

V V X X

00

0

33

1 1 2

C AC A

N N X XC C

V V X X

Combine and solve for Xe:

3

0 01 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 0.580.059

1 2 0.58A

molC

dm

4-22

3

3 0.305 0.580.246

1 2 0.58C

molC

dm

P4-10 Given: Gas phase reaction A + B 8C in a batch reactor fitted with a piston such that

V = 0.1P0 2

3

21.0

sec

ftk

lbmol

2

A A Br kC C

NA0 = NB0 at t = 0

V0 = 0.15 ft3

T = 140C = 600R = Constant

P4-10 (a)

00

0 0

0.5AAA B

Ny

N N

8 1 1 6

0 3Ay

Now 0 0

0

1V P

V XT

PT

and 0

1T

T, 0 010P V , and 10P V

Therefore 2

010 110

VV X

V or

2 2

0 1V V X

0 1A AN N X 0B A BN N X 0

0

1BBA

N

N

332

02

333

20

1

1

AA BA A B

kN XkN Nr kC C

VV X

0 00 0

AA

y PN V

RT

Therefore

4-23

3

0 0

3

2

1

1

AA

Xy Pr k

RTX

3

9

3 3

2

15.03*10

sec1 3

A

X lbmolr

ftX

P4-10 (b)

2 2

0 1V V X

2 20.2 0.15 1 X

0.259X

10

38.63*10

secA

lbmolr

ft

P4-11

(1) For any reaction ,we cannot write the rate law on the basis of the stochiometric

equation. The rate law is to be obtained from the experimental data.

It has been mentioned as an elementary reaction in the problem statement but in the

proposed solution the rate law is based on the reaction equation that has been divided

by stoichiometric coefficient of A.

(2) The value of calculated is incorrect.

= yA0 = 0.6 (3+5-2-3) = 1.8 is the correct value.

(3) The expression for CA and CB will therefore be,

0 1

1

A

A

C XC

X

0

21

3

1

A

B

C X

CX

(4) According to the system of units being used in the calculations, R = 0.0821 atm. Liter/

mol & Temperature = 700 K should be used.

5-1














these materials .

5-2

Solutions for Chapter 5: Isothermal Reactor Design- Conversion


P5-2 (a) Example 5-1 There would be no error! The initial liquid phase concentration remains the same.

P5-2 (b) Example 5-2: For n equal sized CSTRs (volume Vi each) with equal feed FAi0 = FA0/n

CSTR mole balance on any reactor:

Since FAi0 = FA0/n

rA= kCA= kCAO(1-X)

(1)

For single CSTR with volume nVI and feed rate FA0

CSTR mole balance:

rA= kCA= kCAO(1-X)

(2)

From equation (1) and (2), Xi = X.

Therefore, we can generalize that the conversion achieved in any one of the parallel reactors with volume V i and feed

rate of FA0/n will be identical to the conversion achieved in one large CSTR of volume nVi and feed rate FA0.

P5-2 (c) Example 5-3 For 50% conversion, X = 0.5 and k = 3.07 sec-1 at 1100 K (from Example 5-3)

5-3

FB = 200X106 / (365 X 24 X 3600 X 28) lbmol/sec = 0.226 lbmol/sec

FAO = 0.026

0.452 / sec0.5

BF lbmolX

Also, 30.00415 /AOC lbmol ft

Now, we have from the example

0

0

11 ln

1

A

A

FV X

kC X

1 3

0.452 / sec 11 1 ln 1 0.5

3.07sec 0.00415 / 1 0.5

lbmolX

X lbmol ft

= 35.47 X 0.886 ft3

= 31.44 ft3

Now,

n = 31.44 ft3/0.0205 ft2 X40 ft = 38.33

So, we see that for lower conversion and required flow rate the volume of the reactor is reduced.

P5-2 (d) Example 5-4 New Dp = 3D0/4

Because the flow is turbulent

o

1

Dp

1 2

Dp2

Dp10.0775

1

0.750.1033

1

21

12

2

2 0.103 602

1 1 0.2410o

atmX X ft

L fty

P atm

Now, 02

1o

o

P

L, so too much pressure drop P = 0 and the flow stops.

P5-2 (e) Example 5-5 For P = 60atm,

CAO = 0.0415 lbmol/ft3 (

198073.0

60

O

OAO

AORT

PyC )

5-4

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.

V1

P

P5-2 (f) Example 5-6 For turbulent flow

1

Dp and

1

P0

0112 1 1 1

2 02

1 1

1 5

5

P

P

PD

D P

Therefore there is no change.

P5-2 (g) Example 5-3, Using ASPEN, we get (Refer to Aspen Program P5-2g from polymath CD)

(1) At 1000K, for the same PFR volume we get only 6.2% conversion. While at 1200K, we

get a conversion of nearly 100%. This is because the value of reaction constant k

varies rapidly with reaction temperature.

(2) Earlier for an activation energy of 82 kcal/mol we got approx. 81% conversion. For

activation energy of 74 kcal/mol keeping the PFR volume the same we get a

conversion of 71.1%. While for an activation energy of 90 kcal/mol we get a

conversion of 89.93%.

(3) On doubling both flow rate and pressure we find that the conversion remains the

same.

P5-2 (h) Individualized solution.

P5-2 (i) Individualized solution.

P5-3 Solution is in the decoding algorithm given with the modules.

5-5

P5-4 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

ARTA

dCr k Ae

dt

so that

0

A

A

EC

RT

CC Ae t

For complete conversion (i.e.: well cooked) CA = 0 at time t.

Therefore

0

0

E

RTA

E

A RT

C tAe

Cte

A

0 1ln ln lnA

b

C Ek t

A R T

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.

5-6

P5-5 (a) The blades makes two equal volumes zones of 500gal each rather than one big mixing zone of 1000gal.

So, we get 0.57 as conversion instead of 0.5.

5-7

P5-5 (b) A CSTR is been created at the bend due to back mixing, 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 = 0.618

0

1ln

1

X

AO o

A Expected

F dX vV

r k X=

5 1ln 1.0

5 1 .632

Now,

For PFR, 1

1ln

1PV

X.(1)

For CSTR, 11( )

1

actualAO actualC

A actual

X XF X

solution manual essentials of chemical reaction engineering

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