Chemical Engineering Chapter 5 Solutions For Collection And Analysis Rate Data Individualized Solution Individualized Solution

5-1

Solutions for Chapter 5 – Collection and Analysis of

Rate Data

P5-1 (a) Individualized solution

P5-1 (b) Individualized solution

P5-1 (c) Individualized solution

P5-1 (d) Individualized solution

P5-1 (e) Individualized solution

P5-1 (f) Individualized solution

P5-1 (g) Individualized solution

P5-1 (h) Example 5-1

The graphical method requires estimations of the area under and above curves on a plot as well as in

reading the intersection of lines on the plot. This can lead to small inaccuracies in each data point.

P5-1 (i) Example 5-2

Assuming zero order reaction:

Rate law:

‘

A

dC k

dt

!=

CA = CAO – k’t

Ist order reaction

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 50 100 150 200 250 300

t(min.)

ln(1/CA)

zero order reaction

0

0.01

0.02

0.03

0.04

0.05

0.06

0 50 100 150 200 250 300

t(min.)

CA(mol/dm3)

5-2

Assuming first order reaction:

Rate law:

‘

A

dC k C

dt

!=

Or, ln(1/CA) = k’t + 3

t(min.)

0

50

100

150

200

250

300

CA(mol/dm3)

0.05

0.38

0.0306

0.0256

0.0222

0.0195

0.0174

Since none of these plots are straight lines, its is not Ist order reaction or second order reaction.

P5-1 (j) Example 5-3 Because when α is set equal to 2, the best value of k must be found.

P5-1 (k) Example 5-4

CHCl,0 (mol/dm3)

1.0

4.0

2.0

0.1

0.5

‘‘ 2 7

,0 ( / . ) 10

HCl

r mol cm s! “

1.2

2.0

1.36

0.36

0.74

See Polymath program P5-1-k.pol.

POLYMATH Results

Nonlinear regression (L-M)

Model: r = k*(Ca^alfa)

Variable Ini guess Value 95% confidence

k 0.1 1.0672503 0.0898063

R^2 = 0.9812838

R^2adj = 0.9750451

Rmsd = 0.0341709

P5-1 (l) Example 5-5

rate law:

4 2

CH CO H

r kP P

! “

=

Regressing the data

r’(gmolCH4/gcat.min)

PCO (atm)

PH2 (atm)

5.2e-3

1

30e-3

4.08

1

5.25e-3

1

4

5-3

POLYMATH Results

Nonlinear regression (L-M)

Model: r = k*(PCO^alfa)*(PH2^beta)

Variable Ini guess Value 95% confidence

k 0.1 0.0060979 6.449E-04

alfa 1 1.1381455 0.0850634

R^2 = 0.9869709

R^2adj = 0.9782849

Rmsd = 4.176E-04

Therefore order of reaction = 1.14

Again regressing the above data putting

1

!

=

POLYMATH Results

Nonlinear regression (L-M)

Model: r = k*(PCO^0.14)*(PH2)

Variable Ini guess Value 95% confidence

R^2 = -0.8194508

R^2adj = -0.8194508

Rmsd = 0.0049354

P5-2 Solution is in the decoding algorithm given with the modules.

P5-3 Individualized solution

P5-4 (a)

The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor.

HbO2  Hb + O2

Rate law: -rA=k

n

A

C

Mole balance:

( )

1n

n

AO AO

dX

F kC X

dV =!

c

dV A dz=

, where AC is the tube cross sectional area

( )

1

n

Ao c

Ao

kC A

dX X

dz F

=!

therefore,

( )

ln ln ln 1

,

n

Ao

dX a n X

dz

where a

F

! ” = + #

$ %

& ‘

=

Electrode Position

1

2

3

4

5

6

7

Position (cm)

0

5

10

15

20

25

30

(1-XA)

1.0000

0.9807

0.9618

0.9432

0.9252

0.9075

0.8900

X

–

0.0193

0.0189

0.0186

0.018

0.0177

0.0175

X/ z (cm-1)

–

0.00386

0.00378

0.00372

0.00360

0.00354

0.00350

A histogram plot of X/ z vs. z is then produced. The values of dX/dz are evaluated using equal-area

graphical differentiation:

Using the values obtained above, a plot of ln(dXA/dz) vs. ln(1-XA) is produced and a line

is fit to the data

( )

ln 5.5477 1.0577 ln 1

dX X

dz

! ” =#+#

$ %

& ‘

n = 1

ln(a) = -5.5

( )

ln ln ln 1

,

n

Ao c

Ao

dX a n X

dz

kC A

where a

F

! ” = + #

$ %

& ‘

=

5-5

( )

6 3

6

3 1

6 3 2

2.3 10 /

45.7 10 /

45.7 10 / 4.1 10 4.1

2.3 10 / 0.0196

Ao

AO

n

AO C

C mol cm

F moles s

F a moles s

k cm s

C A moles cm cm

!

! !

!

=“

“

= = “=

” “

Hence rate law is,

3

4.1

A A

mol

r C

dm s

!=

P5-4 (b)

First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)

See Polymath program P5-4-b.pol.

POLYMATH Results

Polynomial Regression Report

Variable Value 95% confidence

a0 2.918E-14 0

a2 -6.14E-05 0

a3 7.767E-06 0

a6 -1.6E-10 0

General

Order of polynomial = 6

Statistics

R^2 = 1

R^2adj = 0

Next we differentiate our expression of X(z) to find dX/dz and knowing that

( )

ln ln ln 1

,

n

Ao c

Ao

dX a n X

dz

kC A

where a

F

! ” = + #

$ %

& ‘

=

Linear regression of

ln dX

dz

! “

# $

% &

as a function of

( )

ln 1 X!

gives us similar vaules of slope and intercept

as in the finite differences.

POLYMATH Results

Linear Regression Report

5-6

Variable Value 95% confidence

a0 -5.531947 0.0241574

General

Regression including free parameter

Number of observations = 7

Statistics

R^2 = 0.9482059

R^2adj = 0.9378471

Rmsd = 0.0044015

( )

3 1

6 3 2

45.7 10 / 3.96 10 4.0

2.3 10 / 0.0196

AO

n

AO C

F a moles s

k cm s

C A moles cm cm

! !

!

“

= = “=

” “

Hence rate law is,

1.28

3

4.0

A A

mol

r C

dm s

!=

P5-5 (a)

Liquid phase irreversible reaction:

A  B + C ; CAO = 2 mole/dm3

AO A

A

C C kC

!

“

#=

ln ln ln

AO A

A

C C k C

!

“

#

$ % = +

& ‘

( )

Space time (

!

)min.

CA(mol/dm3)

ln(CA)

ln((CAO–CA)/

!

)

15

1.5

0.40546511

-3.4011974

38

1.25

0.22314355

-3.9252682

300

0.75

-0.28768207

-5.4806389

By using linear regression in polymath:

See Polymath program P5-5-a.pol.

POLYMATH Results

Linear Regression Report

Model: y = a0 + a1*lnCa

ln ln ln

AO A

A

C C k C

!

“

#

$ % = +

& ‘

Variable Value 95% confidence

a0 -4.6080579 0.0162119

R^2 = 0.9999443

R^2adj = 0.9999258

5-7

Hence,

3slope

!

=“

ln(k) = intercept = -4.6

P5-5 (b) Individualized solution

P5-5 (c) Individualized solution

P5-6 (a)

Constant voume batch reactor: A B +C

Mole balance:

A

dC kC

dt

!

“=

Integrating with initial condition when t = 0 and CA = CAO for

1.0

!

“

mol/dm3.

t (min.)

CA (mol/dm3)

0

2

5

1.6

9

1.35

22

0.87

40

0.53

See Polymath program P5-6-a.pol.

POLYMATH Results

Nonlinear regression (L-M)

Model: t = (1/k)*((2^(1-alfa))-(Ca^(1-alfa)))/(1-alfa)

Variable Ini guess Value 95% confidence

k 0.1 0.0329798 3.628E-04

alfa 2 1.5151242 0.0433727

R^2 = 0.9997773

R^2adj = 0.9997327

Rmsd = 0.1007934

Hence , rate law is

1.5 3

0.03 / .

A

dC C mol dm s

dt

!=

P5-6 (b) Individualized solution

5-8

P5-6 (c) Individualized solution

P5-6 (d) Individualized solution

P5-7 (a)

Liquid phase reaction of methanol and triphenyl in a batch rector.

CH3OH + (C6H5)3CCl  (C6H5)3COCH3 + HCl

t (h)

CA(mol/dm3)

0

0.1

2

0.0735

10

0.0357

n

B

m

AA CkCr =!

For table 2 data: CAO



CBO =>

‘m

A A

r k C!=

where

‘n

BO

k kC=

Using eqn 5-21,

(1 ) (1 ) (1 ) (1 )

(0.01)

1 1

‘ (1 ) ‘ (1 )

m m m m

AO A A

C C C

t

k m k m

! ! ! !

!!

= =

! !

POLYMATH Results

Nonlinear regression (L-M)

Model: t = (1/k)*((0.1^(1-m))-(Ca^(1-m)))/(1-m)

Variable Ini guess Value 95% confidence

k 1 1.815656 0.0109025

Nonlinear regression settings

Precision

R^2 = 1

R^2adj = 0.9999999

Rmsd = 3.268E-04

Therefore, m = 2

For first set of data, equal molar feed => CA = CB

Hence, rate law becomes

2 (2 )n n

A A B A

r kC C kC +

!= =

t (h)

CA(mol/dm3)

0

1.0

5-9

1.389

0.816

2.78

0.707

16.66

0.37

(1 (2 )) (1 (2 )) ( 1 ) ( 1 )

(0.1)

1 1

(1 (2 )) ( 1 ))

n n n n

AO A A

C C C

t

k n k n

!+!+! ! ! !

!!

= =

!+! !

POLYMATH Results

Nonlinear regression (L-M)

Model: t = (1^(-1-n)-Ca^(-1-n))/(k*(-1-n))

Variable Ini guess Value 95% confidence

n 3 0.8319298 0.0913065

Nonlinear regression settings

Precision

R^2 = 0.9999078

R^2adj = 0.9998848

Rmsd = 0.0233151

P5-7 (b) Individualized solution

P5-8 (a)

At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed

there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.

P5-8 (b)

Constant volume reactor at T = 504°C = 777 K

Data for the decomposition of dimethylether in a gas phase:

Time

0

390

777

1195

3155

!

5-10

01

A

y=

3 1 2

!

=“=

02

A

y

! “

= =

( )

0

0 0

1

P

V V X V

P

!

” #

=$=

% &

‘ (

because the volume is constant.

( )

01P P X

!

= +

at t = ∞, X = XAF = 1

0

1A

A

N

dN dX r

V dt V dt

=!=

Assume

A A

r kC!=

(i.e. 1st order)

( )

01

A A

C C X=!

(V is constant)

Then:

( )

0 0 1

A A

dX

C kC X

dt =!

and

0

P P

X

P

!

“

=

Therefore:

0

1dX dP

dt P dt

!

=

[ ]

( )

0

0 0 0

11 1

P P

dP k

k P P

P dt P P

!

! ! !

” #

$

=$= + $

% &

‘ (

or

[ ]

( )

0

1

dP k P P

dt

!

= + “

[ ]

000

1

P t

P

dP kdt

P P

!

=

+“

# #

Integrating gives:

( )

0 0

2624

ln ln ln

1 3 936

P P kt

P P P P P

!

“ # “ # ” #

= = =

$ % $ % $ %

+& & &

‘ (

$ % ‘ (

‘ (

Therefore, if a plot of ln

624

936 P!

versus time is linear, the reaction is first order. From the figure below,

Therefore the rate law is:

5-11

y = 0.00048x – 0.02907

-0.4

0

0.4

0.8

1.2

1.6

0 1000 2000 3000 4000

P5-8 (c) Individualized solution

P5-8 (d) The rate constant would increase with an increase in temperature. This would result in the

pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true

fro colder temperatures.

P5-9

Photochemical decay of bromine in bright sunlight:

P5-9 (a)

Mole balance: constant V

A

A A

dC r kC

dt

!

= = “

Differentiation

T (min)

10

20

30

40

50

60

Δt (min)

10

5-12

After plotting and differentiating by equal area

-dCA/dt

0.082

0.061

0.042

0.030

0.0215

0.014

ln(-dCA/dt)

-2.501

-2.797

-3.170

-3.507

-3.840

-4.269

ln CA

0.896

0.554

0.207

-0.128

-0.478

-0.821

P5-9 (b)

A

A B

dN Vr F

dt = =

0.0344 0.0344

min min

A

ppm mg

rl

=!=!

at CA = 1 ppm

5-13

( ) min 1 3.7851 1

25000 0.0344 60 0.426

min 1000 1 453.6

B

mg g l lbs lbs

F gal l hr mg gal g hr

! “ ! “! “! “

! “

= =

# $ # $# $# $

# $

% &

% & % &% &% &

P5-9 (c) Individualized solution

P5-10 (a)

Gas phase decomposition

A B +2C

Determine the reaction order and specific reaction rate for the reaction

Assume the rate law as:

n

A

AkC

dt

dC !=

Integrating:

!

#

$

&‘

=‘‘ 1

1

11

1

n

t

1

0.025

4.1

1.410987

-3.6888795

3

0.01

9.8

2.2823824

-4.6051702

5

0.075

1.3

0.26236426

-2.5902672

See Polymath program P5-10-a.pol.

POLYMATH Results

Linear Regression Report

Model: lnt = a0 + a1*lnCa0

Variable Value 95% confidence

a0 -2.3528748 0.1831062

a1 -1.0128699 0.0492329

Regression including free parameter

Number of observations = 5

Statistics

R^2 = 0.9993004

R^2adj = 0.9990673

5-14

POLYMATH Results

Nonlinear regression (L-M)

Precision

R^2 = 0.9986943

R^2adj = 0.9982591

Rmsd = 0.0531391

5-15

P5-10 (b)

We know,

!

“

#

$

%

&

‘

=‘

‘

1

0

2/1

11

)1(

)12(

(

A

C

t

k

Solving for k at 110° C

( )

(2 1)

2 1 1

‘ 20

2(2 1) 0.025 .min

lt

kgmol

!

!” #

= =

$ %

!& ‘

From these values,

2

1

1 2

20

ln 8.314 ln

. 10.5

1 1

373 383

kJ

R

kmol K

E

K K

T T

! “

# $

% &

= = =

! ” ! “

‘

‘# $

# $ % &

% &

76.5 kJ/mol

P5-11

The values of k1 and k2 may depend on your initial guess. Look for the lowest s2. You could try

“rO3=kCBu

COZ

O3 + wall  loss of O3 k1

O3 + alkene  products k2

Rate law:

3

O

r!

=

3

O

dC

dt

= k1 + k2

Bu

Oz

C

Using polymath nonlinear regression we can find the values of k1 and k2

Run ozone rate Ozone concentration Butene concentration

# (mol/s.dm3) (mol/s.dm3) (mol/s.dm3)

Ozra CO3 Cbu

1 1.5e-7 0.01 1e-12

3 3.5e-7 0.015 1e-10

5 8.8e-7 0.001 1e-08

POLYMATH Results

Nonlinear regression (L-M)

Model: Ozra = k1+k2*Cbu/CO3

Variable Ini guess Value 95% confidence

5-16

Nonlinear regression settings

Max # iterations = 300

Precision

R^2 = 0.7572693

R^2adj = 0.6965866

Rmsd = 4.531E-08

Rate law:

( ) ( )

3

7 3

3.5 10 0.05 / .

Bu

O

C

r mol dm s

C

!

!=“+

P5-12

Given: Plot of percent

decomposition of NO2 vs V/FA0

2

% Decomposition of NO

X= 100

Assume that

n

A A

r kC!=

X has a linear relationship with

0A

V

F

as

shown in the figure.

Therefore the reaction is zero order.

P5-13

2 2 6 2

6 2SiO HF H SiF H O+!+

C S

S 2

S

A ñ ä

N = moles of SiO =

MW

AC = cross-sectional area

ρS = silicon dioxide density

F

V

N = moles of HF=

100 MW

w

!

w = weight percentage of HF in solution

5-17

MWF = molecular weight of HF = 20.0

Assume the rate law is

S F

r kC

!

“=

Mole balance:

S

dN r V

dt =

100

C S

S F

Ad wV

k V

MW dt V MW

!

“#

$ %

&=‘ (

) *

100

S

C S F

kMW

dVw

dt A MW

!

” #

#

$ %

&=‘ (

) *

dw

dt

!

“#

$=

where

100

S

C S F

kMW V

A MW

!

“

#“

$ %

=& ‘

( )

ln ln ln

dw

dt

!” #

$ %

&= +

‘ (

) *

ln d

dt

!

” #

$

% &

‘ (

-16.629

-15.425

-14.32

-13.816

-13.479

ln w

2.079

2.996

3.497

3.689

3.871

where

d

dt

!

” #

$

% &

‘ (

is in

m

min

From linear regression between

ln d

dt

!

” #

$

% &

and ln w we have:

1.775

100

S

C S F

kMW V

A MW

!

“!

# $

=% &

‘ (

( )( )( )( )

6 2

10 *10 10 2 sides 1000 wafers 0.2

C

A m m m

!

= =