Kinetics of Catalytic Reactions--Solutions Manual 0387259732, 9780387259734

Citation preview

Job #: 90652

Author name: vannice

Title of book: solutions manual for kinetics of catalytic reactions

ISBN number: 0387259732

Solutions Manual for

Kinetics of Catalytic Reactions

M. Albert Vannice

Solutions Manual for

Kinetics of Catalytic Reactions

~ Springer

M. Albert Vannice William H. Joyce Chaired Professor Department of Chemical Engineering The Pennsylvania State University University Park, PA 16802 [email protected]

ISBN-IO: 0-387-25973-2 ISBN-I3: 978-0387-25973-4

Printed on acid-free paper.

© 2005 Springer Science+Business Media, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, Inc., 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

Printed in the United States of America. 987 6 5 432 1 springeronline.com

(lBT)

Preface

This manual of solutions to the problems in "Kinetics of Catalytic Reactions" has been prepared to assist those who use this book in a teaching function. However, these solutions should also benefit those outside the classroom who want to apply the principles and concepts that are discussed in the book. By studying and observing the approaches used in solving these problems, it is very likely that similar applications can be envisioned in different kinetic problems that the investigator might face. Thus the availability of these solutions is a good learning tool for everyone. Additional details and insight about the solutions provided can be obtained by reading the cited references. I have tried to eliminate all errors, both conceptual and typographical, in these solutions; however, the probability is high that I have not succeeded completely. Should any errors of commission (or omission) be found, I would greatly appreciate being informed. I can be reached at this email address: [email protected], or mail can be sent to me at:

107 Fenske Laboratory, Department of Chemical Engineering, The

Pennsylvania State University, University Park, PA 16802.

Albert Vannice

v

Contents

Preface

v

Solutions to Problems Chapter 3 -

Catalyst Characterization

.

Chapter 4 -

Acquisition and Evaluation of Reaction Rate Data

Chapter 5 -

Adsorption and Desorption Processes

18

Chapter 6 -

Kinetic Data Analysis and Evaluation of Model Parameters for Uniform (Ideal) Surfaces

28

Chapter 7 -

Modeling Reactions on Uniform (Ideal) Surfaces

.41

Chapter 8 -

Modeling Reactions on Nonuniform (Nonideal) Surfaces

81

Chapter 9 -

Kinetics of Enzyme-Catalyzed Reactions

83

vii

7

Problem 3.1 Solution

o

0.1

0.2

0.4

0.3

p/po

N 2 BET Plot for BC-l From slope and intercept: TIm = 1120 umole N 2/g C == 1000

Am = [1120X10-6mole] g

fn C = qI - qL => qI RT

23 [6.023X10 mOleCUle] [16.2;\2 ] mole molecule

[IO-20m 2 ] = 110 m

= 6.9 (1.987 calJ (80 K)+ 1340 = 2440 cal mole-K

g mole

A 2

g

2

2

Problem 3.2(a) Solution

o

0.2

0.1

0.3

N2 Bet Plot for Si02 (Cab-O-Sil, Grade M5) From slope and intercept: n m = 1620 umole N 2/ g C=140 2 2 mOleCUle] [ 16.2 A ] [10-20 m2] = 158 m 2 umole molecule A g

Am = (1620 Jlmole N 2 ] (6.023XI0 g

17

.en C = ql - qL = 4.94 RT

qi = 4.94(1.987 cal/mole- K)(77 K)+ 1340 cal/mole = _2_10_0_ca_l mole

3

Problem 3.2(b) Solution

10

0.1

0.2

0.3

0.4

N2 BET Plot for Fresh, Reduced Fe203 From slope and intercept: n m = 10.00 umole N 2 /g C=47 6mole

Am =[10.00 xl0

2

N2

g

J(6.023X10 23 mOleCUle] [16.2A mole

] [10-20 m2] = 0.98 m 02 g molecule A

in C = ql - qL => ql = 3.85 RT ql = 3.85 (1.987 cal/mole- K)(SO K)+ 1340 cal/mole = 1950 cal/mole

2

4

Problem 3.2(c) Solution

0.2

~ '(I)

'0

§.

~

0.1

~

~I~

o

0.3

0.2

0.1

Ar Bet Plot for Used Bulk Iron From slope and intercept: n m = 1.60 umole Ar/g C =47 6

l

g

l

23

i

mOleCUle] [ 13.9 j·[10-20 m2] = 0.134 m mole molecule g

Am = (1.60xlO- mOleAr] ( 6.023xl0

fn C = ql - qL = 3.85 RT ql =3.85 (1.987 caIjmole.K)(77 K)+1550=2.14kcaIjmole

i

2

5 Problem 3.3 Solution

80

60 ...

40

20

o

100

200

300

400

P (Torr) Assume Had /Rh s = 1. The atomic weight of Rh is 103. The amount of hydrogen chemisorbed on the Rh is the intercept value of 50 umole H 2/g which is determined by the high-P region where the Rh surface is essentially saturated with H atoms.

(

50 JlmoleH2 g cat

J(2H J(lRh J ad

H2

s

1 Had

~----~-~~-~6

0.015 g ( g cat

RhJ (10 Jlmole RhJ 103 g Rh

= 0.69 = DM

6 Problem 3.4 Solution

.

.

80

t;-t

t

II

+

I

I

f

I;" i~:;~~

~.·.:-_:,~:~=-· -__:· -- ~:..~~~~ -~ :.: ~~:,~r~~: . ·'~_~ '·:-_~'_~I·.i _~_: :~::-._~ .~_'-'.:.:;~.;:.-,'~.,_-~:!-~;:-_:'~-+:~:__~:'. . .,~ : ~"r-t-:--+-;'-!-'-~I~I~~~~~~t±~::~~ +-:,-'"""~'+ ..... ~._.r-~ ~ _=-.-;---: .. _..',....

:::;.

:'.:_... ,:.-I-"-:.. ..... .. ..

.L....!L....::-t-r--:_.

+-: ..

I;

60 __ 1

I I

;

:

~ ~:~

;_~

; I

~;-:~-:-

ih'~

2-'--,.-.--:"';"'T-:-;

~ii:

I 1'1

I

:

:

I

•

I ,

I

:

r::-~

4°t==::t:::;::t:b::I±ttt±:@::;:qm:t:t:t;++:~::+tt:+++mm+++:;4±i::::t~4::::q

I I I

.

I

;

:: I

I I

-----

*-;..._-:.._~_.

'-.--' ': .

"._~

r---··--- I

J~_?!IfE~~j ~ o

100

300

P (Torr)

Assume no 02 adsorption on Si02 and the Oad/Ags ratio is 1 Adsorption on Ag = 50 umole 02/g cat as determined by the intercept determined in the high- P region of saturation. Dispersion = AgslAgtotal = OadlAg t

= (SO flIDole02/g)(2 Oad/02)(1 Ags/Oad) = 0.44

D M

6

0.0243 g AgJ (mole Ag J (10 JlmOle] ( g cat 108 gAg mole

7 Problem 4.1 Solution a)

For isothermal, 1st-order reaction: (Eq. 4.54)

ll Da o = mjkgaC o

(Eq. 4.50)

= ~ = 2nr + 2nrL = 18.75 cm " 2 2

a

V

(For a sphere, a

= A = ~ =18.75 cm") V

2r

=~ =

Co 80 2

for cylinder

Lnr

RT

= 9.58xl 0-7 mole 802

0.06 atm

J

(82 .06 atm· cm 3 (763.2 g mole-K

K)

cm '

See Table 1 for li"Dao and li" values. b)

The concentration drop can be determined from Eq. 4.52, i.e., so Co -C s

c.tc,

= l-li"Da o

= mjkga = dC

The ~C values are given in Table 1. The concentration gradients through the film thickness, 0, will be dC/O . From Eq. 4.44, kg =

D/o

so 0 =

Djk g •

From Eq. 4.75, the bulk diffusivity is: Db = VA 3 phase is (See Illustration 4.6): A_

1

_

The mean-free path in the gas

RT

- .fina 2 (N/V) - .fina 2 p for an ideal gas where 3

(

82.06 atm cm g mole- K

(J

= is the molecular cross-section area, so

J(763.2 K)(

1 mole 6.02x 10 23 molecule

J

..:.----------:..------.;~--------::...-

8

.fin[(4XI0- cml ](1 atm] molecule

5

= 1.5xl0- em

8

'1=

(8~~T

1/2

(Eq. 4.76)

J

v =[8 (1.38xIO-16 erg/K) (763.2 K)]1/2 = 5.0x10 4 em/s 1t

(64 amu] (1.66XIO-

24

g/amu)

Db = 1/3 ~.5xIO-5 em)(5.0xI0 4 em/s) = 0.25 em 2/s S 0, at hi19h SV :

~

u =

yk

D

2/s

g

= (

0.25 em )

=

4

0.020 em

\4.5xI0 cm/h (1 h/3600 s) 2

and at low SV: D=

e)

Is

0.25 em = 0.032 em ~.8xI04 em/h)(1 h/3600 s)

at high SV:

"'"C/o = 2.6xIO-

at low SV:

"'"C/o = 2.9xIO-

7

3

mole/em = 1.3x 10- 5 mole/ern" 0.020 em 7

3

mole/em 0.032 em

=9.0xlO-6 mole/ern"

Nonisothermal situation, assume C, == Co

and ~ = e-E/RT(I/t-l)

ko

(Eq. 4.55)

(Eq. 4.56)

~H~ = - 23.6 kcal/rnole S02

(CRC Handbook)

9 7

AI-' -

-

Then !1T = T, - To = 13 .11Da oTo , and the d)

3

3

- 23600 cal] (9.58X10- mole S02 J (9X10 cm KJ ( mole Sfr , cm' cal = 0.27 763.2 K

13 .11Da o

values are listed in Table 1.

The non-isothermal effectiveness factor is

'il' = ~s = : s = e-EjRTo(I/t-l) = (1- T(Da o ) e -EjRTo (- j3. T(Da o)(1 + j3. TjDaJ o

(Eq. 4.57)

0

where E/RT = - 30000 caI/mole = -19 78 . o (1.987caI/mole.K)(763.2K) The

'if'

values are also given in Table 1.

Table 1 in (mol/h-em")

0.215 0.204 0.194 0.153

kg

11Da o

(em/h)

4.5xl04 3.8xl04 3.4xl04 2.8xl04

0.27 0.30 0.32 0.30

11

~c

(isothermal)

(mole/ern")

13 .11Da o

0.73 0.70 0.68 0.70

2.6xl0· 7 2.9xl0·7 3.0xl0·7 2.9xl0·7

0.072 0.080 0.085 0.080

~T

(K)

55 61 65 61

11' (nonisothermal)

3.8 4.3 4.7 4.3

10 Problem 4.2 Solution Using the Thiele modulus:

Rp

= rp = 2 Vg = 2 ( Sg

3

0.42 em / g (420 m 2/ g) (100 em/m)2

J= 2.0x10- em 7

(Eq.4.76)

so

_¢- = 4.8 => ¢ = 4.8 ,and'll = 2- (1- _1_) = 0.495 == 0.5 tanh ¢

4.8

4.8

11

Problem 4.3 Solution

Use Weisz-Prater criterion:

Assume

c, = Co =~ =~ = V

RT

9t R 2 N w-r = - - p - ~ 0.3 C s Deff 2/3 (2 atm)

(82 .06 atm- em mole·K

maximum particle diameter = 1.3xl0-2 em

3

J( 673.2) K

= 2.4xlO-5

mo1

3e

em

12 Problem 4.4 Solution

Weisz-Prater criterion:

illR 2 N w-p = - - p -

c»:

ill = (0.0956 g mol S02/h· g cat) (1.6 Rp = 1.6 mm

=

< 0.3 => no significant pore diffusion effect

gl cm ' ) (h/3600 s)= 4.25x10-

5

0.16 em

Assume C, = Co = 9.58xI0-7 mole/em'

-r-..lo...--.------=----.---:.----~-----'--___._=

68 »

6

Mass transport limitations due to pore diffusion are definitely present.

mole/s. cm'

13 Problem 4.5 Solution

J(0.60 g eat J

in = (1.99 Jlmole Bz s g eat

C

s

= 1.2 Jlmole Bz

em 3

s em 3

= Co = P/RT = (SO Torr BZ)(l atm/760 Torr) = 1.9xlO-6 mo~e 82.06 atm· em [

3

J(413 K)

em

g moleK

D eff == D Kn beeause A »d p (averagepore diameter)

(Eq.4.76)

(Eq.4.78)

(a) With Pd in mesopores = 25 nm:

Worst case:

6

Best ease:

R -10 P -

4

3

R = SOO fl; N w-p = (1.2xlO- mole Bz/s em }(SOOxlOp (t.9XIO- 6 Bz/cm' )(2.8x10-2 cm' 6

.N

u;

w-p

3

4

eml = 0.OS6

Is)

eml -2 3 10-5

- P.2xlO- mole Bz s· em )(10xlO- (1.9xlO-6 mole Bz/em 3 ) (2.8xlO-2 crn'

Is) - . x

(b) With Pd in mieropores = 0.9 nm:

Worst ease:

Best ease:

R p

= SOO u ; N

w

_

p

= (1.2xlO-6 mole Bz/s. em 3 }(SOOxlO-4

eml = 1.6

(1.9xlO-6 mole Bz/cm' ) (1. Ox10- 3 cm'

Is)

14

Problem 4.6 Solution

C02

(2)

J

{200 Torr )( 1 atm =~= = 760 Torr = 4.4xl 0-6 mole CO 2 V RT ( 82.06 atm· em 3 (723 K) em 3 gmoleK

..!.-

( 1) C

v=

1/2

(8 k mT J B

1t

=

J

[

16

(.

)

8 1.38xl0- erg/K 723 K 24 gjamu 1t (44 amu] 1.66xlO-

]1/2

= S.9xl0 4 cmls

(Eq.4.76)

a)

b)

c)

4 7 2 2/s d) D Kn = 1/3(S.9xl0 cm/s){20nm)(lcm/l0 nm)=3.9xl0- cm 6 3 - (237xl0- mole/s· g) (1 g/cm ){0.01 cm)2 _ 0 14 (P ibl ) N w-r - ( OSSI e concern \4.4xl0- 6 mole/em." \3.9xl0- 2 cm 2/s

)f

) -.

15 Problem 4.7 Solution

n-1

R p k 112C2 s 'f/= Dl/2 A.

Thiele modulus

eff

= k' P H 2 = kC

Rate

C

=

H

2

E- = ~ = V

RT

(

H

= (1.99 !!mole BZJ(0.60 2

S.

(710/760 atm) = 2 76xl0-5 mole 8206em 3.atm ] ( ) em 3 • . 413K g mole-K

6

3

k = Rate = 1.19xl0- mole/s. em = 0.0432 C H2 2.76xlo- 5 mOle/em 3

(

A=

g/cm ' )= 1.19xlO-6 mole Bz s -cm '

g

S-1

82.06 em3 . atm](413K)( 1 mole ] gmole- K 6.02xl0 23 molecules

D eff ~ D Kn

r

fi 1t (2Axl 0-8 em (I atm total p)

= 1/3 vd p

where

dp

= pore

diameter

=2.2 x 10-5 em

16 Case 1: Smallest particle, all Pd in mesopores: D Kn r/J

= 1/3 ~.1 x 105 cm/s) (25xl 0 -9 m)(102 crn/ rn) = 0.175 em 2 / s

= (IOxlO-

6m}(102

cm/m~~.0432/sya =5.0xlO-4

(0.175

cm/s]

Case 2: Largest particle, all Pd in mesopores:

r/J =

6m)(102

(500xlO-

cm/ml~0.0432/sYI2 = 0.025

(0.175 cm/s]

Case 3: Largest particle, all Pd in micropores:

No concern - in all cases 11 is essentially unity.

17

Problem 4.8 Solution

Use Weisz-Prater criterion.

N w -p

mR 2

=- - p - «

0.6 (or 0.3)

c, D eff

(Eq. 4.78)

Assume 1 atm: C = C = ~ = ~ = (9 Torr)(l atm/760 Torr) = 2.93xlO- 7 mole CH 2 0

J

s

V

N

_ w p

RT

=JAXIO-

7

( 82.06 cm 3 ·atm (493 K) gmol-K 3

3

mole/em .sl[7A5xlO- emt = 2.25xlO-3

[2.93xl0- 7 mole/cm 3 ][1.18xl0- 2 cm 2/s]

Yes, it's free of diffusion effects.

cm '

«0.3

18

Problem 5.1 Solution

l*J= L -lo - *J-ls - *J= L - 3ls - *j K = (2 [S-*D2 [S-*] = PS02 [*]3

4 [S_*]3 PS02 (L - 3 [S- *]

K (L-3 [S-*]y = 4 [S_*]3/PS02 => K

y

1 3(L-3 /

1I3 LK 1I3pl/3 r S02 -3 r~S_*]K p1l3 S02 = 314 V~ ~S-*] 1I3 LK 1I3p1l3 r _ *] S02 = rl3 K1I3pS02 + 314]· V~ ~S [S_ *]-

-

LK1/3p1/3 _ L (K/4)1/3 PSIO/32 S02 Vir +3 K1/3p~~2 - 1+3 (K/4Yl3p~~2

[s-*D= Vir ~-*]/P~~2

19

Problem 5.2 Solution

. Ie-site . adsorptIon, . (Eq. 5.27) ,whereas to test To test sing p Iot -P = -I -P +n Kn m nm pIn 1 pU2 dual (double)-site adsorption, plot - = - + - (Eq. 5.28). U2 n K nm nm

The single-site

equation gives a reasonable linear fit, whereas the dual-site equation is curved with negative slopes (See Figures 1 and 2, respectively). From the slopes and intercepts of the plots in Figure 1:

T(K)

K (atm- I)

n m (umole g-I)

343 363 383 403

181 87 59 28

1380 1180 772 719

or

V m{cm~TP g-l )a 30.9 26.6 17.1 16.1

Use Eq. 5.38 and plot in K vs. 1fT (Figure 3) Slope =4.13x10 3 =-~H~d / R

Intercept =-6.85=~S~d / R

0 ~~ ~ , so ~Had =-8.2--or -34-mole mole

0 cal J , so ~S d =-14---or-59--a mole.K moleK

(a)

V

rn

= nRT/P = (n g mole) (82.06em

3

·atrn/g mole· K)(273.2 K) 1 atm

= n (2.242x104 em3 )

20

Benzene Adsorption On Silica Gel Single Site Langmuir Isotherm 1.Oe-4~----------------------

"0

E

=

8.0e-5

.......

C)

«

6.0e-5

....Eas

"as

4.0e-5

Z .......

a.

2.0e-5

O.Oe+O

-+------.-----__-----,r----~--___1 0.01

P (atm)

Figure 1

0.02

21

Benzene Adsortion On Silica Gel Double Site Langmuir Isotherm ..-.

2.0e-3

"0

E

=

x

""'-

C) -Ie

b.

70 (C)

0

90 (C)

[J

110 (C)

x

130 (C)

It)

0 < E ..,

x 1.0e-3 -

ca

[J

"'C

co

x

[J

)(

Z

[J It)

0


=> A

E

=

73.9

kcal mole

309~ mole

29 Problem 6.2 Solution

= k , = (k BT/ h) f;~~t

pre-exponential factor

because a N atom has no rotational or vibrational

f tr

= (k T/h) (aT)3/2 (bT) => k a Tl/2

modes. Therefore, k o

QCOOH2

c)

B

(cTy

0

has 3 x 5 = 15 modes, with 15 - 7 = 8 degrees of vibrational freedom; therefore

Q H2

~

Q Cl 2

= ft~

fr~t f vib; assume non-linear activated

vibrational freedom = 3N - 7 = 5, and

Q+

= QH2Cl2

=

f~

complex,

so degrees of

f:ot f;ib' Therefore,

30

Problem 6.3 Solution

1.

r= dl:rtkl[Br-][H2]+k2 [H.][BrJ-k_1[H.][HBr]

(1)

Use SSAonH· radical: d~Lo=kl [Br.][H 2]-k2[H.][Br2]-k_1[H.][HBr]

(2)

Use SSA

on Br· radical:

d[Br.] . way IS. : r =r - =0,= but an easier i t dt

[Br.] =(k i /kt )1/2 [Br2)1/2 From (1):

[H.]=

k1[Br.][H 2]

, then

k2[Br2]+k_I [HBr]

r=k l(k l/k t Y/2 [Br2 ]t/2 [H2]+(k2[Br2]-k_I[HBr])(kl (kjktr [Br2 J/2 [H 2 ] ] k2[Br2]+ k_I [HBr] =(k 2 [Br2]+ k_ I [HBr]) (k l (ki /kt )112[Br2 J/2 [H 2])+ (k 2 [Br2 ] - k_ l [HBrD(kI (ki /k )112 [Br2 J/2 [H 2 ] ) k2 [Br2 ]+k_l [HBr] t

or

31 Problem 6.4

at constant volume

32

Problem 6.5 Solution

a)

QN> Qo

. /N=O so N end down, t.e., Fe , n

1 ) QN 140 For (11 III : QoN = ~/ = ~ = 84 \2 - 1/n) \5/3)

=

3.

QNO = Q

and

Q~N oN

7+

D

(Table 6.5) NO

QNO = (0.6)2 (140)2/[(0.6)(140)+ 151]= 30 kcal mole."

SO

o II

(l]IIlZ): QNO = (O.6Y (140)zj [(O.6)~140) + 151] = 37

For

N

kcal mole"

=>

F:-ie (preferred)

b)

Orientation is 11 1111 with 0 end down, n = 4. 115 -1 Qoo = (7/4) = 65.7 kcal mole

QH 0 2

c)

=

Q~o

=

Q oO + D H20

(

Y

(65.7 65.7 + 220

Table 6.5

=15

This is a symmetric molecule with A

D H- C = = 230+ 81 = 311

= Dc=c

)

kcal mole

(Table 6.7a)

= CH, i.e.,

+ D C- H

(9/2)(90)2 QHC=CH = ( ) ( ) = 13 kcal mole 3 90 + 8 311

(Eq. 6.46)

33

d)

Q2 QCH2 == Q + ~ C

for "strong" chemisorption (adsorbs in hollow site) CH2

use Qc = 160 kcal mole-Ion Pd(111) , so Q

160)2

CH2

( = 160 = + 183

75 kcal mole "

34

Problem 6.6 Solution

a)

N ad > Oad regarding bond strength, so N end down. This is "weak" chemisorption, n

= 4.

o II For TJ\l2 =>

£N::N"i

so n' = 2,

Q oN

(Table 6.6), so

QNO

QoN + 151 2

b)

(4/7)2 (135)2 -1 = ( )() = 31 kcal mole 4/7 13 5 + 151

2

V\!

This is "weak" chemisorption, n = 3,

Ju..&.L&.LI\ Pd--Pd

(Table 6.7b)

H

c)

This is "weak" chemisorption, n

= 3,

N end down, i.e.,

I

H-lf-H Fe

QNH3

so

=

Q~A

Q OA +D AB

QNH3

,

from ref. 25, D NH3

= 279 kcal mole -1 ,

(0.6)2 (140)2 -1 )() =19 kcal mole 0.60 140 + 279

= (

35 d)

This is "strong" chemisorption

QNH = 2

Q N

Q~

+D NH2

so Q

=

NH2

(140)2

. = 63 kcal mole." 140+ 169

Q~

=

Q NH

Q N

, from ref.25, D NH 2 = 169kcal mole-I,

+D NH

, from ref. 25, D NH =81kcalmole

-1

,SOQNH

(140)2 1 =---=89 kcal mole140+81

If "intermediate" chemisorption is chosen for NH2, then

Q

=

NH 2

~[ 2

r

(0.6)2(140

(0.6)(140) +169 2

+

(140)2 ] =48 kcal mole-1 if n'=2

140+169

=46 kcal mole-1 if n'=1

36 Problem 6.7 Solution

For acetone, (CH3)2 CO [or

kcal/mole [25]; therefore,

31"

DTotal

6l

+

~:=

0

l

the enthalpy offormation is -217.3 kJ/mole or -51.9

= D(CH3) 2 CO = L~H~i (atoms)-~H~

+

so, based on:

l

3 C(s) + 3 H2(g) + ~ 02(g) and from reference [25]: 3 (C

~

C·)

3 (H2 ~ 2 H') ~ (02 ~ 2 0') ~H~ DTotal

3 (171.3 kcal/mole) (3) (2) (51.2 kcal/rnole) (0.5) (119.2 kcaljmole) -(-51.9 kcal/rnole) 938.0 kcal/rnole

From Table 6.5, on Pt: Qc = 150 kcal/rnole and Qo = 85 kcal/rnole. (a)

For on-top adsorption

(111~1)' use equation 6.35 with n =

3 for Pt(III). Molecule will

bond O-end down; so

Qoo = (3/5) (85) and from Table 6.7b, the bond energy associated with each C-CH3 group is 376 kcaljmole, so

DAB = Dc=o

= 938.0 - 2 (376) = 186 kcal/rnole.

Then, for "weak" chemisorption, use equation 6.41 :

37

(b)

C~;=H3

For di-e-adsorption, i.e.,

;C-~

Pt---Pt equation 6.42 mustbe used, keepingin mindthatatomsA andB can have othergroups attached to alterthe A-B bondenergy,thus frompart (a) DAB = Dc=o = 186 kcal/rnole, From Eq. 6.43, a

= Q~(' (Qoc' + 2 roO) where

Qoc' represents the heat of

Qoc' +Q oo

chemisorption of the quasi-atomic (CH3)2 C· fragment. For this relatively large species, a reasonable way to calculate its heat of chemisorption is to againuse eq. 6.41 where DAB now represents the totalbond energyof all the bonds formed by the secondaryC atom, thus fromTable 6.7b:

so Q ,= «0.6)2 ((150))2 ) = 31.6 kcal/rnole then oC

(0.6) (150) + 166

' 5

a= (31.6)2[31.6+2 (51.0)L l.33xl0 =19.6 kcal/rnole.and

(31.6+51.0)2 from Eq, 6.44,

b = (51.0

6.82x10 3

Y[51.0+ 2

5

(31.6)L 2.97x10 = 43.5 kcal/mole

(51.0 + 31.6)2

6.82x10 3

-

(19.6) (43.5) (19.6 + 43.5) + (186) (19.6 - 43.5)2 = 1.60xl0 = 12.7 ~ 13 kcal/rnole 5

(19.6) (43.5) + (186) (19.6 + 43.5)

1.26x104

38 Problem 6.8 Solution

Rate constant for bimolecular surface rxn or desorption = A b < 10- 4 cm 2/s a.

Collision theory, r=P

ZAA

2-dimensional gas

[Aad]2 = Ab [Aadf , Probability P= 1 , # collisions =

ZAA = cry = o 8:~T

(

ZAA

1/2

J

where o = molecular diameter (molecule sweeps out an area)

(given in ref. 11, p. 68) b.

Absolute rate theory, immobile adsorption:

2 A -S

~

'--y---J

S- A2 - S ~ A2+2 S

site pair

where

L, = density of site pairs

c.

Absolute rate theory,

= 2/2 L => (Z == 4) ,

2-dimensional gas:

2 Aad

then

~

A2 ad

~

A2,g

39 Problem 6.9 Solution

Figure a. Curves for ActivationEnergy,Enthalpy, Entropy Determination on activated carbon

--k -

...

u

R= 0.89098

=535.11* e(-175.1sm

••• - K ...

102

=5.95*105 * e(-1539ff)

R= 0.27915

_ ... _H20... _ ...•-- ... _ ... _ ... _ ...

•• • •••

K

N0 2

-,.,.

= 1.01*107 * e(-39S4.4m R= 0.98668

........

······ ..·······a····· .... _-

. ••••••••••• •• 51

3.05

3.1

3.15

3.2

3.25

3.3

3.35

Iff x 1000(K-l) From slope & intercept: 1)

For N0 2

~H~d = + 7.9 kcal mole" ~S~d

= + 32.1 e.u.

~H~d

= + 0.4 kcal mole" = + 12.5 e.u.

~S~d

(1 e.u. =1cal/mole- K)

Not consistent

~

both

~H

&

~S

values are positive

40

Figureb. Curvesfor Activation Energy, Enthalpy and Entropy Determination on Si0 2 . - - - k = 5.21*10-5 * e(4656.1/11 R= 0.99901 ...•....... K ~ 2.9139*10.7 :II e(5784.3{f> R= 0.99483 N0 2

-

... - K

H20

= 6.6838*10·i 2 III e(9889.8tr) R= 0.99991 ".

....

.... ~

."..,....

."."

3

3.05

3.1

3.15

3.2

3.25

3.3

3.35

Iff x 1000 (K-l)

1)

E,

2)

For N02

=

-9.3 kcal mole"

=>

negative and inconsistent

=>

inconsistent

~H~d = -11.5 kcal mole." ~S~d

= - 29.9 e.u.

S~ =

~H~d ~S~d

57 e.u.

= -19.7 kcaI mole" =- 51.1 e.u.

S~ = 45 e.u. < 1- 51.1 e.u.1

41 Problem 7.1 Solution

(a)

SSA statesthat d [0 ]jdt

(b)

Cannot assume Quasi-Equilibrated Adsorption (No Langmuir isotherm) r = d [C0 2] ; @ steady state, so r\ = r2 , Site Balance: L = [S- 0]+ [S] dt

r=k 2 [CO][S-0]=Lk 1k 2 [CO] [N 20]j(k l [N20]+k 2 [CO])

= d [cn.] lCH 4 jdt

r][ ]

(c)

r

1.

Literature solutionfor reaction as written

= k 2 lCH 2ad H 2

&

[C2H2ad][H2F K=---

[C 2H 6][S]

[C2H 2ad] = K [C2H 6][S]I [H2]2 = K[C 2H 6 ] (L - [C2H 2ad])

[H 2 F

[C2H2ad](1+K[C 2H6]1[H2]2)=LK [C2H 6 ] => [C H d]=

[HJ2

@ SteadyState

2 2 a

LK [C2H 6 ]

r)

[H2]2 (I+K[C 2H 6 ] / [H 2

d lCH 2ad J de 1 d lCH 2ad J r ][ ] =0 : - == k, lC2H2ad H 2 dt dt dt 2

42

II.

Alternate solution for reaction written with well defined sites, S, i.e.,

K

C2 H6(g)+ S ~ C2H2-S + 2 H 2(g)

(1) (2)

k

1 C 2H2-S + H2(g) + S -=---t2 CH2-S

k2 2 [CH 2-S + H 2 (g) ----=r CH4 (g)]

(3)

C2H6 + H2 ====} 2 CH4 Note that a CH 2 group almost certainly requires a site for itself as shown in step 2.

If [C 2 H 2 - S] is the MARl, then L

[C2H2 -

s]~ + K

[C2H6 ] /

[C H -S]2 2 @S tea d y state..

-

=

[C2H 2 - S] + [S] for the site balance, and

[H 2]2) = LK[C2H6 V[H2]2

&

LK [C 2H6 ] [H 2]2{I + K [C2H6 ]/ [H2J2 )

d[CH 2 - S] -_ O & dt

2-S]_k -dc_l - - d[CH - 1 2 dt dt

lc H -S][H 2][S]& ~ 2 2

43

2k l [C 2H 2 -S][H 2][S]=k 2 [CH 2 -S][H 2 ]=> [CH 2 -S]=2(k l/k 2)[C 2H 2 -S][S] [S]= L[H 2 ]2 I([H 2 F+ K[C 2H 6 ] )

so

d[CH 4 ] =kz[CH z-S] [H z]= (.!:-J 2Lzk[K [C ZH 6][H zP1([Hzf + K[C ZH 6 ]Y dt

2L

= k' [C 2H 6 ] [H 2 ]3 =

k' [C 2H 6 ] =r [Hz]Z + K[c zH6 1f [Hzl(l + K [C 2H6]]2 [H 2]2

44

Problem 7.2 Solution

r=-

dlAt dlBt k [B-S] dt

Site balance: (a)

dt

3

L = [S]+ [A - S]+ [B- S] If [B- S] is the MARl, then L = [S]+ [B-S] and

The mathematical forms are identical.

45 Problem 7.3 Solution

Write balanced sequence:

R NO 4 [NO + * .~ NO *]

(1)

K1

NO*+ * ~ N*+O* K2 CH4+O* ~CH2*+H20

(2) (3) (4)

CH2* + NO*

~

CHNO* + H*

K3

(5)

H* + CHNO* + 2 NO*~ C02 + N2+ H20 + N* + 3*

(6)

2N* ~ N2+2*

1/KN2

Step 4 is RDS

where K

NO = K IK 2K I f2

K N2

46

and

r=

47 Problem 7.4 Solution

r= -d[H 20]=d[H 2]=d[C0 2 ] dt dt dt

(Site balance) Steady-state approximations:

48

Problem 7.5 Solution

(a)

A+S

k :

reversible RDS

A·S -1

K2 B+S~ A·S+B·S

K3 ~

B·S

C·S +D·S

Kj D·S ~ D+S A+B

~

C+D

,

or

[DoS]= PD[S] K s

'

For an ideal gas

49

substitution gives

Lk1(PA -PcPn/K PB) r = ~(l-+-K-A-Pc-P-n-/K:.......:..P....:...:B=--+-K....;:;..B~P-:-B+-K~cP-c-+-K-n-P~n)

(b)

K1

A+ST=@==A'S

K2 B+S T=@== B·S

K3 A·S+B·S T=@== C'S +D·S

Reversible RDS

50

r

=

LksKIK2K3K4PAPB/PC - Lk_SPO

---~~~--:........;~;;..;..........~--.-;~--

(1 + KIPA +K 2PB +PC/K 4 + KIK2K3K4PA PB/PC)

substitution gives r

=

(c)

Lk_sKPAPB/PC - Lk_SPO

-----'-----.;;.,;;'---=-'-...::...--...::...-=---

(1 + KAPA + KBPB + KcP c + KOKPAPB/PC) k1 k

A2 + 2 S

2 A· S

reversible RDS

-1

K2 2[B+S~ B'S]

K3

2[A·S+B·S

~

2[C·S

~

C'S + S]

K4

r = kIPA2 [S]2 - k.,

[ ] =K 2PBlS~ ] B·S

,

fA· S]2

C+S]

, L=

lc. S]= Pc [S]

II

K4

[A. S]+ [B. S]+ [C. S]+ [S] , K =~ [C. S][S] 3 lA,SJlB,SJ

(from site balance)

51

r=

(Z/2L)r}k l PA2 - (Z/2L)r}k_1 (pc /K 2K 3K 4PB Y

-~--""------------,---

with site-pair probability of

(1 + Pc /K 2K 3K 4PB + K 2PB + Pc /K 4 )2 where Z

= coordination number

or, because

KKK _Kl/2jKl/2 2 3 4 I

r

Lki(PA2 -p~jKP~) =....,.-------"'------.;.--.---

(d)

{1+K:

/2

1

PC/K !2 PB +KBPB +Kcpcf

K1

A2 + 2S T=@=:= 2A· S

K2 2 [B + S T=@=:=

B'S]

K3 2 [A' S + B· S T=@=:=

C'S + S]

k4 2 [C' S

A2+ 2 B

~4

-----

(reversible RDS)

C+ S]

2C

r = k 4[C. S]- k_ 4PC[S] , [A .Sp = KIP A2 [SP

,

[B .S]= K 2PB[S]

-----rsr-

K = [C.S][S] => rC. S] = K 3 [A . s I B . s ] =K I/2K K pl/ 2p rS] 3 lA.sIB.S] ~ I 2 3 A2 B~

so

(Z/2L)

,

52

Now with

r -

L(Zj2)k-4 (K l/2pl/2 p - P ) \ A2 B C

- (1 + Kl/A22pl/2 +K P +KK pl/2p ) A2 B B C As B

so

53 Problem 7.6 Solution

(a)

Use steady-state approximation for total surface carbon atoms to get another equation for BC2HS ' i.e., dec (total) = O. Then, because steps 2 and 4 are very rapid: dt

~

and

but e C2 H4 can be rewritten to give:

so that

which is notation consistent with reference 55.

54

k1 C2H6 + 2 S;;:=:-

(b)

~l

C2Hs . S + H . S

r=- d[C2H 6]=k dt

K2

~

C2H, . S + H . S

C2H4 • S + S + H2

lc 2 H 4 .s]rs]p ~ H2

3~

Steady-state approximation on all surface C atoms gives:

rC

H .S](l + K k /k ) = 4 2 3 -1

~ 2

k K P H (k K /k)P H 1 2 C2 6 => rC H . S] = 1 2 -1 C2 6 k P ~ 2 4 1 + K k /k -1 H2

2

3

IP

-1

H2

55

Now, site balance to get [8] is:

Agreement with (a) is achieved only if

[8] =L, i.e., the surface is essentially free of all adsorbed

species, i.e., 8 H ,8 C2Hs ,8 C2H4 ,8 CH3 «1 . This a questionable assumption.

56

Problem 7.7 Solution

Plotting in rate vs. in Pi using a power rate law gives the following reaction orders:

T(K)

Reaction Order

623 653 673

lliQ

lli

lli

0.08 0.24 0.31

-0.31 -0.12 -0.07

0 0 0

The simplest L-H model would be for unimolecular decomposition: KN20 2 [N20 + * ~ N20

(1) (2)

2 [N20 *

-l

k

- - + N2 + 0*]

(3)

r

=~ dN =~(dNdt m

N 2oJ=k[N

N2

mmdt

Site balance gives:

consequently,

L

0*] 2

= [*] + [N 20*] + [0*]

thus

57 Arrhenius plots of the fitting parameters listed in Table 2 provide the following values:

For K 02 : ~H~d For k: E RD S

=- 25 kcal mole"

= 57 kcal

and ~S~d

=- 35 e.u.

mole "

The enthalpy and entropy values for adsorption fulfill all the guidelines in Table 6.9, thus they are consistent. From either a linear extrapolation of the high- P portions of the two isotherms in Figure 1 or using the difference between the two at 100 Torr CO pressure, the irreversible uptake is 580 umole CO g~~t. The dispersion ofCu is: D eu = CUs/Cu tot ,and with COad/Cu s

D

=

cu

580 Jlmole CUs

(0.0456 g Cu

g~~t

g~~t)(mole Cu/63.55 g CU)(106 Jlmole/mole)

= 1,

= 0.81

Under differential reaction conditions, P02 ::::: 0 and can be ignored; therefore, an easy way is to choose a known differential rate and correct for temperature, for example: r k e-36200 caI/mole/(1.987 cal/mole-K)(823K) 823K - 823K -139 -r- - k - e-36200caI/mole/(1.987caI/mole.K)(673K) 673K 673K

TOF

thus

= r823K = (12.6 Jlmole/s· g)(13 atm- 1 )(0.0666 atm)(139) = 1.4 S-l

823K

CUs

~ + ~3 atm-l ) (0.0666 atm)] (580

umole CUs /g)

58

Problem 7.8 Solution

Step 2 defines the rate:

=~m dNdt =~(dNdt m

N 2oJ=k[N

N2

r m

2

0*]

Assuming all surface species are included, a site balance gives L

= [*]+ lN 2 0 *J+ lo *J

To remove the unknown

[0 *], the SSA must be used:

and

Rearranging and squaring each side gives:

and

59

The solution for this quadratic expression is:

Substituting back and using the positive root gives:

H= kKN20PN20/kl + 2L(1 + KN20PN20)+ ~2~_1 Po2/k1 ~(I+KN20PN2of]

Because the rate is:

rm

= k[N 20 *] = kKN20PN20 [*], the final rate expression is:

dP N20 {I + (a + C)PN20 + [aPN20 + (a2 + aC~N20 + bPo

J1/2}

=---'--------:-------:------------'--

r m

(1 + CPN20

f -bPo2

parameters can be combined in various ways to give the values in Table 1. An Arrhenius plot of the adsorption equilibrium constant, K N20, gives ~H~d =- 25.2 kcal mole- l and ~S~d =- 33 e.u., which satisfy the guidelines in Table 6.9. The rate constant Lk represents a unimolecular decomposition reaction on the surface, so guideline 2 in Table 6.10 can be applied:

60

or, with correction for activation energy: LAde-34400/1.987.843

= 2xl0 12 = 1.2x10-9

LAd

Both values are below 1028 molecule/scm 2 .

and

LAd

= 2xl0 21 molecule/s cm s

61

Problem 7.9 Solution

One has dissociative H2 adsorption, and to get a negative dependence on a reactant (acetone), the denominator must be squared, which indicates a bimolecular surface reaction as a RDS with competitive adsorption between acetone (Ace) molecules and H atoms for active sites, so:

KH2

(1)

H2+ 2S ~ 2H-S

(2)

Ace + S ~ Ace-S

(3)

Ace-S + H-S ~ AceH-S + S

KAce K3

AceH-S + H-S

(4)

k4

~

IPA-S + S

llK IPA IPA-S ~ IPA+S

(5)

Ace + H2 ~ IPA (Isopropyl alcohol) To get a 1st-order dependence on H 2, step 4 and not step 3 must be the RDS. The overall site balance appears to be simplified to: L = [S]+ [Ace] = [S]+ K Ace PAce [S]

(1)

because there is no dependence on IPA at these differential conversions and the dependence on PH2

rm and

must be as close to unity as possible, so [Ace- S] is the MARI. Therefore,

=~ dN IPA m

dt

=k 4 [AceH - S][H - S]

(2)

62

[H-S]=K~~PU~[S] fromstepl.

(3)

Now steps 2 and 3 can be added to get

(6) so

Consequently, from equation 1:

ls J= L/(l + K

Ace PAce)

and

(Qad

~

Add later)

(5)

63 Problem 7.10 Solution

(a)

The rate, defined by N2 formation, is r

= d[N 2] =.::..!. d[NO] = k 2[N *]2 dt

2

Fromthe 3 quasi-equilibrated steps:

Site balancegives:

L= [*]+ [NO *]+ [N -l-[0 -l so

for site pairs: ~ is the probability factor, so 2L

(b) (c)

(d) (e)

dt

64

(f)

(g)

[N *]nearsaturation (»

[*D~ [N *]= L ~ r = ~k2L2 = Lk~ 2L

Reject: (e) - Cannot give reaction order on PNobelow 2 (g) - Observed reaction order is not zero order

65

Problem 7.11 Solution

The rate can be defined as rm

CH4 + 2 02

~

dN N 2 =-1 ( - ___ dN c H 4 =-1 __ m

dt

m

K02 02 +2 * ~ 20*

(2)

CH4+ *

(3)

CH4* + 0*

--------::,.

(4)

CH3* + 0*

~

K CH

4

~ CH4*

kl

CH3* + OH*

K2

CH20* +H*

(5)

K3 CH20* + 0* ~ HCO* + OH*

(6)

HCO* +0* ~ CO* +OH*

(7)

H* +OH* ~ H20* + *

(8)

CO* + 0 * ~ C02* + *

K4

Kj

K6

(10)

lIK C0 2 C02* ~ C02+*

K7 20H* ~ H20* + 0*

If.KH

0 2

(11)

(12)

Jfor the reaction

C02 + 2 H20. With dissociative 02 adsorption:

(1)

(9)

dt

2 [H20* ~ H20 + *] llK co CO* ~ CO+*

66

Step 12 allows for CO desorption as a product. The rate based on this sequence is given by step 3 (the RDS):

Thus

K K 1I2P p1l2 [*]2 r m -k 1 CH4 02 CH4 02 -

The site balance with only adsorbed reactants and products is:

Steps 9, 11, and 12 give, respectfully:

and

From Arrhenius plots of K CH4 & K 0 2

'

For CH 4 := L\H~d = - 20 kcal mole-I, L\S~d = -13 e.u.; For 02 : L\H~d

=- 30 kcal mole-I, L\S~d =- 25 e.u.

(1 e.u. = 1 cal mole" K-1)

67 Problem 7.12 Solution

(a)

r fiNO

1 dN -1 dN NO = - -i - = - - m v.dt

2m

dt

respectively. SSA on HNO* gives:

Site balance on * sites: L.

= [*]+[NO*] and K

_ [NO*] PNol~

so

NO -

Site balance on S sites:

[S]=+--~

,

r., = [S]+[H - S] and K H2

[ S] eH =~= L s

(b)

K

II 2 p H2 H2

= [H

-~S]]:

so

PH2LS

II 2

1 KII2 pIl2 + H2 H2

=K II 2pIl2 H2 H2

if

eH «1

68

8N20 =

Lsko8N08H + k-3PNz08* (k2 + k 3)

LskoKNOK~~PNOPU~

k- 3PNZO

= (1

N K k P (L-[N*])2 -k p 3/2 l *J(L - lN *D 2k [N*]2 =0 1 2 NH3 -2 H2 KKK 3/2 6 3

4

5

2 Solution for quadratic equation: ax + bx + c = 0 is

- b ± ~b2 - 4ac 2a

so

70

Divide numerator and denominator by -2 kt; and use positive root:

Inside [ ---- ] 1/2 after factoring Lout:

so

71

~(- 2Klk2PNH3 4

k6

[N*]=~------'---=--------=~

and, after introducing probability factor Z/2L for neighbor sites:

where

72

Problem 7.14 Solution

k l K N20PN208 y k2KeoPeo8y

An Arrhenius plot of'Lk, gives EI

=

26 kcal mole" while similar plots for K N20 and

Keo give:

for CO:

~H~d

= -13

kcal mole" and ~S~d

=- 29

cal mole" K-

~H~d

= -10

kcal mole" and ~S~d

=- 23

e.u.

1

(e.u.)

73 All these thermodynamic values satisfy the criteria in Table 6.9. The TOF at 573K 2100flmoleN2g-1 S-I I (769 umole Cu g- ) (0.91) Aoe-26000/1.987 (573)

30 .

-I

s.

Th

=

h . If . . 30-1 us t e pre-exponentia tactor per site IS: • s =

and A o = 2.4 x 1010 S-I which is less than 1013 S-I and satisfies criterion 2

in Table 6.10 for a unimolecular reaction. Also, the heat of adsorption of 10 kcal mole" for CO on Cu is consistent with values of7-10 kcal mole" reported in the literature.

74 Problem 7.15 Solution

from step (6): [CH 20 *] = K 6 [CH 2 *][OH *]/ [H *]

from step (8):

[CO *] = KgPco [*]

SSA on surface CH 2

* (and CH 20 *) species:

therefore

75

from step (6) :

Site balance with [CH 20 *]as MARl: L

= [*]+ [CH 20 *] =

so

76

.l".

Th e TOF lor step

(5.35 umole 8-

7.

IS:

1

g-1 )

=.'1 1 s -1

~.4 umole H 2 g-1 )(2 Jlmole Ni s J umole H 2

Rate per site: TOF A = 1.1/3.25 6.10)

X

=

1.1 S-1

10-12 = 3.4

X

=

Ae-EIRT

1011

S-I,

= Ae-38000/1.987 (723) =

A (3.25 x 10-12) .

Therefore

which is less than 1013 (See Criterion 2 in Table

77 Problem 7.16 Solution

(a)

1 d~ 1 dN -1 dN NO 1 dN N2 dN 2 -1 dN NO = - - = - -i = - - =---=>=---

r m

_ dN NO dt

(b)

m dt

= 2k

m

vi dt

[NO *]2 }

2m

dt

m

*]

dt

[NO K NO =~ => PNol*J

dt

2

dt

[ ] =KNOPNO[*] NO*

In the site balance: k 3 [N20 *] = k, [NO *]2 and and

[N20 *] = (k}/k 3 )KNOP NO[*]2

P and

k 2[*(NO)2 *] = k, [NO *

so L = H+ 2klKkNOPNO H2 + klK~oPNO 2

[*(NO)2 *] = (k}/k 2 )KNOP NO[*]2

H +KNOPNOH+K~pb~H

3

Consequently, a quadratic expression for l*J is obtained and the solution for such an equation is complicated.

78 Problem 7.17 Solution

The series of elementary steps is:

K1

(1)

MCH+* ~

MCH*

K2 MCH* ~

(2)

MCX* +H2

K3 MCX* ~

(3)

MCD* +H2

K4 (4)

MCD* ~

(5)

TOL*

~

MCH

===}

k,5

TOL* + H2 TOL+*

(RDS)

TOL

or

K.. .

MCH+*

~

TOL*

~

k,5

TOL*+3H2 TOL+*

Step 5 is RDS, so r = k s [TOL *J

K' _ [TOL *]P~2

-~

L = [*]+ [MCH

and

-l- [MCX -l- [MCD *]+ [TOL -l

This is not consistent with the behavior because it contains a P~2 term.

79 Problem 7.18 Solution

A. B.

C.

[HCO

2

2 PH2CO[0 - sF -s] -- K 1/2K p1/2 lo _S]1/2 f

H20 H20 ~

~S]I/2

--

K2PH2CO [0 - S]3/2 [S]-112

K~~oP~~O (Balance on O-S species)

D.

At steady-state:

E.

K P k 3/2 p3/2 fS]S/2 Substitute D in C: [HCO _ S] = 2 H2 CO 1 02 ~ 2 k3/2 [HCO _ S]3/2 K 1/2 p1/2 3 2 H20 H20

F.

G.

f - S]-

rO

-

k p2/sfS171/S pl/S 1 02 ~ Jl'-H20 H2 0 k 2/5K2/5k 3/S p2/5 3 2 1 H2CO

H.

1.

Assume PH20 is ~ constant, plus power of 0.2 is low

OA pO.6 K,"rnOAp-OA \2 Then, r = Lk 1 P02 /(1 + K ",nrH2CO 02 + r0 2 H2CO J

80 If 8 Hc 0 2« 1 and 8 0 «1, then r = kP0 2 If 8 0 is MASI, i.e., 8 0 » 8 Hc 0 2

then

(Note typographical error in power dependence on H 20 in ref. 62)

81

Problem 8.1 Solution

a= 2/3

kO a(t-to) k l= Ie

= kOe2/3(t-to) k 1 '-1

= k" e- 1I3 (t- to ) k = kOe-1I3 (t- to ) k = k? e 2/3(t-t o) -1 '2 2 '-2-2

Substitute into r:

r=

kfk~[AI] [A2]-k~Ik~2[Bl] B 2] E e 1l3(t- to ) + E2e-2/3(t-to)

where E I = kf [AI]+ k~2 [B2] and

I

E 2 =k~I[BI]+k~[A2]

Let D represent the denominator, then

[SI] __ 1/2 ---... lS2j

~

-,r

2~S] ~S] ~I =~2

and

e_ [S2] _ [S2] - ~ -~-~lSIj+ lS2j

Ij2 lS2 j+ lS2j

3

82 Problem 8.2 Solution

thus

83

Problem 9.1 Solution

A+E

~

A·E

~

P+E

is

A·E

~2

A =====}

P

SSA on A· E: kl[A][E]+ k_ 2 [P][E]-k_l [A. E]-k 2 [A .E]= 0

Active site balance: L, = [E]+ [A .E]

[A. E]= k] [A][E]+ k_2 [p ][E] k_ l +k 2

L = [E]+ k j [A][E]+ k_2 [p][E e

t

k_ l +k 2

(k-l + k 2 )Le r = (k_l + k 2 )+ k, [A k_ 2 [P]'

l-

(k_1 +k 2 1E]+k j [A][E]+ k_2 [p][E] k_ l +k 2

(k 1k 2 [A]- k_ 1k_2 [p D_ L e(k 1k 2 [A]- k_ 1k_2 [p D (k_l + k 2 ) - (k_l + k 2 )+ k, [AJ+ k_ 2 [P]

84

also of Michaelis-Menten form.

85 Problem 9.2 Solution

Useeq.9.9,whichis:

vr=(KmJ(_[l]J+_l- andplot! vs._[l] rmax A rmax r A

25 Y = 0.0808 X + 7625

M

(:)

~

20

X

...-.-. (1)

0 E ........ ..J

15 10

e

E

"'-'"

... ........ ~

5 0 0

50

100

150 3

1/A (L/mole) x10-

From the slope and the intercept: Intercept = _1_ = 7620 so rmax = 1.31xl0-4 mol~ rmax Lv min Slope

= 0.0808 min = K m r max

=>

K m = (0.0808 min) (1.31XIO- 4 mole/L· min)= 1.06xlO-5 mole L

200

86

Problem 9.3 Solution

r

= d[P] = k[ES] dt

Steps 1 and 2 are quasi-equilibrated, so

K = [ES] s

1EJlS]

Balance on enzyme:

and

K' = [ESS] s

1ESJlS]

r., = lEJ+ lESJ+ lESS J

Note: by convention, steps 1 and 2 are represented by a dissociation equilibrium constant, i.e.,

K1

ES

~ E+S

written in terms of dissociation constants. The mathematical form is identical to the above rate equation.

87 Problem 9.4 Solution

(a)

Steps 1 and 2 arein quasi-equilibrium:

(1)

k1 S+E~

(ES)1

~l

k2 (2)

(3)

(ES)1 ~ (ES)2 ~2 (ES)2 ~ P+E S ---=r P

r

= d[P] =k[(ES)2 ]=kK 2[(ES1]=kKlK 2[S][E] dt

Active site balance: L,

= E + (ES)1 + (ES)2

K = 1

[(ES)lt~ [S][E]

k.,

2 [(ES)2t~

K =

[(ES)1]

k_2

88

(b)

Steps 1 and 2 are reversible, so use SSA:

From (2) :

[(ES)2 ](k_2+ k) = k2[(ES)I] => [(ES)2] = k2[(ES )111(k-2 + k)

From (1):

kl[S][E]=.k2~-2[~E:)1].-(k_l +k2)[(ESh]=0 -2

[E] =

L e [k_1(k 2+k)+ k2k]

k_1 (k_2+k)+k2k + k,(k 2+k_2+kls]

=:>

89

r

=k[(ES)2]= kk2[(ES)1 t (k_2 + k)

kk\k2[S][E]

=

lk_ l (k_2 + k)+ k 2kJ

Lekk 2[S] jk

k., (k_2 + k)+ k 2k + k, (k 2 + k_ 2 + k

To get Michaelis-Menten form, divide numerator and denominator by

so rmax ( apparent) =

Lkk e

2

k 2 +k_ 2 -i-k

Is]

90 Problem 9.5 Solution

Y= rEt r., y

-2

1

(l+KJ[W ]+K\K

2/[H+

n

and

rE-21 1 - ~ - ----------.........---

- r., - (1 + [H+]/K 2 + [w flK iK 2 J

Thusthe maximum rateis proportional to the active fraction and