Solutions for Chapter 3 - Rate Law and Stoichiometry

P3-1

Individualized solution.

P3-2 (a)

Example 3-1

0 .. 008

0 .. 007 - t - - - -

0 .. 006 - t - - - # - - - - j 0 .. 005

~ ~ 0 .. 004 + - - - F

.iOi

0 .. 003

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

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

o

310

. T .•....

"

... ,.-++ .... _ ... .,

315 320

T(K)

For E = 60kJ/moi

=

1.32 X 10¹⁶

expC-- 6~~OJ

⁾

E = 240 kj/mol 35E-22

3E-22 t - - - + - - - i

25E-22 t---~._---___j

g

2E-22 t - - - - \ - - - I

:;; 1 5E'22 t - - - \ - - - - -

1E-22 t - - - ' b - - - -

5E-23 t---"'c--~

325

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

330 335

For E1 = 240kJ/moi

kl = 1.32 X 10¹⁶

exp( - 2~~OOJ)

E=60 kj/mol

6000000 . . ... -_ •.•

5000000

4000000 t----~-=--

"'

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

2000000 t - - - -

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

000295 0003 0.00305 00031 000315 0.0032 000325 1fT (11K)

P3-2 (b)

Example 3-2 Yes, water is alJeady considered inert.

3-1

P3-2 (c)

Example 3-3

The solution to the example at a conversion of 20% would remain unchanged. For 90 % conversions of the caustic soda, the final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of caustic soda is possible.

b e d P3-2 (d)

Example 3-4

A + - B

-c ^{+ -} ^D

a a a

So, the minimum value of @B

=

b/a

= - - 113 = 0.33

P3- 2 (e)

Example 3-5

For the concentration of N2 to be constant, the volume of reactor must be constant. V = Vo.

Plot:

180

160

140

120

f"100

::!:-

...

⁸⁰

1 0.5(1-0.14x)2

-r_A (I-X)(0.54--0.5X)

1/(-ra) vs X

02 04 06

08 12

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

P3-2 (0

Example 3-6

For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor.

Therefore the reverse reaction decreases ..

Cro ⁼constant and inerts are varied.

N₂0₄*"72N0₂

A ~ 2B

C

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

CA,e

Stoichiometry: £

= Y

^A0

6 = Y

^AO

(2 -1) = Y

^AO

Constant volume Batch:

3-2

C - 2N

^AO

X -2C ^X

B - - AO

Vo

Plug flow reactor:

C = _~AO(l- ^X) = ^C Ao(l- ^X)

^and

C

= 2F

X = 2C

^AO

X

A v

(1 + ^eX) ^(l + EX) vo (l +

eX) (1

+

eX)

CAO

Y;o;0

⁼

YAo(O.07176)mol! dm

Combining: For constant volume batch:

C 2 4C² X2

K _c-

^-~-

_-

^Ao ^.

CAe

AO

(1 - X) 4C_AO

For flow reactor:

C 2 4C2 X2

K =_._~= Ao K C (1 - X e )(1

+

eX e)

CAe

AO

(1 - X)(1

+

eX)

See Polymath program P3··2-LpoL

POLYMA TH Results NLES Report (safenewt)

Nonlinear equations

[1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao»I\() .. 5 = 0

[2] f(Xef) = Xef - (kc*(1-Xef)*(1 +eps*Xef)/(4*Cao»A0 .. 5 = 0 Explicit equations

[ l ) yao

=

[2] kc = 0.1

[3] Cao

=

0 .. 07174*yao [4] eps

=

^yao

1 ...•.••...

4C_AO

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

<: 0 .. 8 +---~---...,-.mIIIJIIl''---i

.~

^Qi ^0.7

~;~~-~:i - -;-;:;~~~~_=--

§

0 .. 6 ____

o _{E 0 .. 5}

:::J

~ 04

'3 0 .. 3 1 - - - -

Jr

0 .. 2 + - - - 1 0 .. 1 - 1 - - - 1

o ... ···1· .. · ... " .. ," ... 1 .... · ... , ... · .. T .... · ... · .. T"· ... '1.'''... .. ... " .. ' ... "1' ... " ... , •• 'r~ ... ".' .. ' " .. · .. ·T·~ .. ' .. · .. · " .... n~ .. ·T

o

0 .. 1 0 .. 2 0.3 04 0 .. 5 0 .. 6 0 .. 7 0 .. 8 0.,9 y inerts

3-3

-Batchl -FIOW.J

Y_inert Yao 0 0.1 0.2 0.3 0.4

-

0.5 0.6 0.7 0.8 0.9 0.95 0.956

P3-2 (g)

No solution will be given

P3-2 (h)

1 1

A+-B-7-C

2 2

Xeb

1 0.44

0.9 0.458

0.8 0.4777

0.7 0.5

0.6 0.525

0.5 0.556

0.4 0.5944

0.3 0.6435

0.2 0.71

0.1 0.8112

0.05 0.887

0.044 0.893

Rate law: - r

A

⁼

kACA 2C

1 dm

and

kA = 25- - - s ( mol J

~=.~.=

rc -1 -112 112

1 (dm ³ ^J2

kc =k

=12.5- - s mol P3-2 (i)

A+3B

^-7

2C

Rate law: -

r

= k

C

B at low temperatures.

At equilibrium,

C

_Ce

K

= .. ---- '

C I12C

^3/2

A,e B,e

Xef

0.508 0.5217 0.537 0.5547 0.576 0.601 0.633 0.6743 0.732 0.8212 0.89 0.896

At "Iuil;b,;um, -, A = 0 , so we can sugge" th,t -

r

= k

^{A (}

C

^A^1/2

C

^B^{112 -}

~ ~ J

But at t

=

_{0, Cc}

=

So the rate law is not valid at t = 0 ..

3-4

Next guess:

P3-3

Solution is in the decoding algorithm available separately from the author.

P3-4 (a)

Note: This problem can have many solutions as data fitting can be done in many ways.

Using Arrhenius Equation For Fire flies:

T(in 11T --- r;:::-_.

Flashes/ In(flash

K) min es/min)

294 0 .. 00340 9 1

298 0.00335 12.16 6

303 0.00330 16.2

-

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

2.197 2.498 2 .. 785

See Polymath program P3-4-firetlics.pol.

For Crickets:

T(in K) 11T chrips/ In(chirPsi x10³ min min)

~2 ^3.482 ⁸⁰ ^{4 .. 382}

293 .. 3 3.409 126 4.836

300 3.333 200 5.298

Plotting In(chups/min) Vs lIT, we get a straight line ..

-+

Both, Fireflies and Crickets data follow the Arrhenius Model.

In y = A + BIT , and have the same activation energy.

See Polymath program P3-4-crickcts.pol.

3-5 2.66

2.52

2.38

2.24

210L---~·----~----~--~

3.30E-:3 .332E-3 3.34E1

h

3.36E-3 3.38E-3 3.40E-3

5.3

5.1

·~.5

4..3 '---~--~--~--

3.HE-3 1.36E-3 3.39Eih 1.42E-3 3.45E-3 3.48E-.3

P3-4 (b)

F H or oney! ee: b T(in K) 1rr x10³

298 3.356

303 3.,300

rsoa-

^3,,247

V(cm/s) In(V) 0.7 -0.357 1.8 0 .. 588

3 1.098

Plotting In(V) Vs liT, almost straight lme.

In(V) = 44.,6 - 1.33E4/T

At T = 40°C(313K) V = 6Acmls

At T = -5°C(268K) V = 0,005cmls(But bee would not be alive at this temperatme)

See Polymath program P3-4-bces.pol.

P3-4 (c)

For ants'

T(in K) 1rr x1Q" V(cm/s) In(V) - , - - -

283 3,53 0,,5 -0,,69

293 3,41 2 0 .. 69

303-

_3,,30 _3,4 _1,22

---,,-,---'-

311 3,,21 6,,5 1.,87

~-- ' - - - - ' - - - "

Plotting In(V) Vs liT, almost straight line,

- -

See Polymath program P3-4-ants.pol.

So activity of bees, ants, crickets and fireflies follow Arrhenius modeL So activity increases with an

20 , - - - ,

1.4

o .. s

0.2

-0.4

-1.0 L-_ _ ^~_ _ ^~_ _ ^~__ ^~_ _ __' 3.25E-3 ,'.2'7E-3 3.29E

lr

HIE-, 3.3.3E-3 3.36E-3 20 r - - - ,

1.4

o.S

0.2

-10L--~---

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

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

---:-c-

--

Insect Activation Energy Cricket 52150

Firefly 54800 Ant 95570 --- Honeybee 141800

P3-4 ^(d)

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

P3-5

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

3-6

HCN-H2S04 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column, These two opposing factors results in the maximum of the cOIrosion rate somewhere around the middle of the column.,

P3-6

Antidote did not dissolve from glass at low temperatures.

P3-7 (a)

If a reaction rate doubles for an increase in lOoC, at T = T 1 let k = kl and at T = T 2 = T 1+ 10, let k = k2 = 2k_{l •}Then with k = Ae-^EIRTin general,

kl

= Ae-E1R1i and

k2

= Ae-E'RT2 , or

k2 _ -~(-) ^E

- - e

kl R

Therefore:

In (k2 J ^{(T. (T.} ⁺ ¹⁰⁾⁾

k I l l

(In 2)(7; (7; +10))

E

R--

R·---,-· "

(7;-7;) 10

T. (T. + 10) = lOE

1 1

RIn2

!OED ⁵ which can be approximated by T

= ----

RIn2

P3-7 (b)

Equation 3-18 is

k =

^{Ae RT}

E( 1 I)

Dividing gives

~2,

⁼^e^-IiTz""-"1;" , or

3-7

[1.99 cal ] [

273

K][

373

E = mol K

In(·050)

7960

cal

lOOK

.001

mol

A

k e

^RT^,=

10-

min

^-I

exp

. 1

(

1.99_Cal

](273K)

molK

P3-7 (c)

Individualized solution

P3-8

= 2lO0min-

When the components inside air bag are ignited, following reactions take place, 2NaN3 ^--72Na + 3N2 ... ·.. . , ' , ... ' .. ' ,,(1)

lONa + 2KN03 --> K20 + 5Na20 + _{N2 ,} , ,(2) K20 + Na20 + _Si02^--7alkaline silicate glass" ",.,.,(3)

5x rxn(l) + rxn(2) + rxn(3) = rxn(4)

NaN₃+ 0.2KN0₃+ _01Si02-7 04Na20 + _1.6N2+ complex/l0., ... , ,(4) Stoichiometric table:

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

Species Symbol Initial Change Final

---

NaN_{3 _} A NA -NAX NA(1-X)

KN03 B NA~B -O.2XNA NA( ^{BB -}02X) Si02 C

- -

NABe ··OIXNA NA( Be -01X)

Na20 D 0 O.4XNA O.4XNA

N2 E 0 1.6XNA 1.6XNA

Given weight of NaN₃= 150g

Therefore, no., of moles of NaN3 = 2 .. 3

Mwt of NaN₃= 65

1 moles of NaN₃requires 02 mole of KN0₃

=>

Moles ofB, KN0₃

=

02(2..3)

=

0.46 moles Mwt ofKN0₃

=

lOLl Therefore, grams ofKN0₃required = 046 x lOLl = 46..5 g

1 moles ofNaN₃requires 01 mole of Si02,

Moles of C, Si02

=

0.1(2.3)

=

023 moles Mwt of Si0₂

=

60,08 Therefore, grams of Si02 required = 0,23 x 60,,08 = 13,8 g

Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards:

• Store cars in cool, dry, ventilated areas"

• Avoid Physical damage of the bag in cal.

• It is stable under ordinary conditions of storage" Decomposes explosively upon heating (over 221 0 For 105⁰C), shock, concussion, or friction,

3-8

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

P3-9 (a)

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

dt

linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and therefore halve the cooking time ..

P3-9 (b)

When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only be 100°C.

When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than double that of boiling water.

P3-9 (C)

No solution will be given

P3-10 (a)

1) C2H6 ~ C2H4 + H2 Rate law: -fA =

kC

_C2H6

2) C2H4 + ₁₁₂₀₂-+ C2H40 Rate law: -fA =

kC

_C2H4C~:2 3) (CH3)3COOCCCH3)3 ~ C2H6 + _2CH3COCH3

A ~ B + 2C

Rate law: -fA = k[CA - CBCc²IKcJ 4) n-C4HIO +-+ 1-C4HlO Rate law: orA = k[

C

_n^{C H}₄ ₁₀

--C

_iC

IKe]

4HlO

5) CH3COOC2Hs + _C4H90H +-+ CH3COOC4H9 + C₂HsOH

A + B +-+ C + D

P3-10 (b)

2A + B - - C (1)

(2)

(3) (4)

P3-10 (C)

orA = kCACB2

orA = kC_B orA = k orA = kCAC_B-¹

(1) C₂H₆ ~ C2H4 + H2

Rate law: -fA = k[CACB - CcCnlKcJ

3-9

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

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

P3-11 (a)

Liquid phase reaction,

o

/"-

^CH2-·-OH

^I

CH2 - CH2 + H2^0·~CHT-OH A +B--~ C CAO = 1 Ibmol/ft3

CBO = 3.47 lbmol/fe

Stoichiometric Table·

Species Ethylene oxide

Water

f-- Glycol

' - - - -

Rate law:

Therefore, At 300K

Symbol Initial

A CAo=1Ibmol/ft3

--

CBO= 347lbmol/ft1,

----

BB=347

C 0

----

- - - - _ ..

_ - - - - -

-rA = kCACB

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

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

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

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

= X Ibmol/ft3

0.1

3 3

k = O . .ldm ImoLs

=--x35.315ft Ilbmol.s = 0.0035ft Ilbmol.s 1000

CAOX (1)(0.9)

-1

'rCSTR

= - - - =

- - 2 - - - -

=

1000 . .56 s -r_A

(0.0035)(1) (1- 0.9)(3.47 - 0.9)

At 350K,

k2 = k exp«E/R)(lIT-lIT 2»= O.0035exp«125001l.987)(11T-IIT2»= 0.071 dm3/moLs Therefore,

CAOX (1)(0.9)

^{_ - 1}

'rCSTR = - - = 2 - 49.3s

-rA

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

P3-11 (b)

Isothermal, isobaric gas-phase pyrolysis, CZH6 C2H4 + H2

A ~ B + C

Stoichiometric table:

3-10

....

-

e

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

C2H4 B

H2 C

=> v = vo(l+X) CAO = YAO CTQ = YAO -

RT

0 +FAOX FB=FAoX

0 +FAoX Fc=FAOX

FTO=FAO F~FAO(1+X)

(I)(6atm)

= = 0..067 kmol/m³= 0..067 mol/dm³

(0.082

m3atm J(1100K)

K.kmol

CA = FA

=

^FAO(l-

^X) = ^{C (1- X)}

^mol/dm³

v v

⁰

(l + X)

(1 + X)

FB FAO(X) X 3

C_B=- = -

=C

_{AO - - -} mol/dm

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

Fe FAO(X) X 3

Cc = -

= =

CAO . - - - mol/dm

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

Rate law:

(I-X) (I-X)

=kCAO - - - =o..o.67k---

(I+X) (I+X)

-fA = kCA

If the reaction is carried out in a constant volume batch reactor, =>(

e

= 0.)

CA = CAO(1-X) mol/dm³ CB = CAO X mol/dm³ Cc = CAOX mol/dm³

P3-11 (C)

Isothermal, isobaric, catalytic gas phase oxidation, C2H4 + -1 O2 ~ C2~0

A + -B

1

~ C

2

Stoichiometric table:

Species C2~

O

C2H4

O

Symbol A B C

---

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

0. _+FAOX _Fc=FAoX

3-11

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

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

FA FAo(1-X) CAO (1-X) 0.092(1-X)

C - - - --~---.:..

A -

V - Vo

(1 ⁺ eX) - (1-0.33X) - (1-0.33X) FB FAo(eB-~) 0.046(1-X)

C - - - - --'---'-

B-

V -

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

e V

vo(1+eX) (1-0.33X)

If the reaction follow elementary rate law

_ {0.092(1- X)}{0.046(1- X)}O.5

=>

-r

-k .

A (1-0.33X) (1-0.33X) P3-11 (d)

Isothermal, isobaric, catalytic gas phase reaction in a PBR C6H6 + 2H2 ~ C6HlO

A + 2B ~ C Stoichiometric table:

Species Symbol Entering Change

Benzene A FAO -FAOX

H2 B FBO=2FAO -2FAOX

- -

--

C6HlO C 0 FAOX

1 2

e=y , AO J=-(1-2-1)=-,- 3 3

- - - - Leaving

FA=FAO(1-X) FB=F AO(

BB -

2X) Fc=FAOX -,-

P (1) ^6atm ⁽ 1 )

CAO = CroYAO = RT - ^{= (} 3) - ⁼ 0.055mol / dm 3 0.082 atm.dm (443.2K) 3

mol.K

_ FA _ FAo(1-X) _ CAo (1-X) _ 0.055(1-X)

C_A- - - - - - - -

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

3-12

C = FB = F _AO (BB

-2X)

=

^O.l1(I-X)

v

o (1+£X) (1-~X)

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

vo (1+£X) (1-~X) (1-~X)

If the reaction follow elementary rate law.

Rate law:

·-r

A

^'=

kCAC;

(1 X)3

-r

A

^'=

0.0007k---

₃

(1- ^~ X)

For a fluidized CSTR:

W = FAoX.

-r ' _A

W = FAoX _

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

(1-~X ⁾

k=53 -

mol

" - - at300K

kgcat

min

atm

k

_jexp(E(l.. __

!.J)

⁼

53exp(80~9.Q_(_I_""_-I-_)J=

^1663000

^mol

R

T

8.314 300 443

kgcatminatm

FAO = CAO

*

Vo Vo = 5 dm³/min

atX= 0.8

W = 0.0024 kg of catalyst

3-13

P3-12

A + -B 1 -7 C

2

Stoichiometric table for the given problem will be as follows A ssummg gas pl ase h

Species Symbol Entering Change

C2H4 A FAO -FAOX

O2 B FBo = eBFAO -112 FAOX

-

I FI = elFAO ^---_.

C2H4O C 0 FAOX

1 2 F

^AO

1

8

_B = - - - = -

e =_FIO F = 0.79 F

I FAD' 10

0.21

^BD

F

_AO

2 Y

=

-.:iQ..

F = 0.30,

F

_TO

Dalam dokumen Soludons Manual for Elements of Chemical Reaction Engineering (Halaman 69-82)

Solutions for Chapter 3 - Rate Law and Stoichiometry

P3-1

P3-2 (a)

o

"

=

expC-- 6~~OJ

g

exp( - 2~~OOJ)

"'

P3-2 (b)

P3-2 (c)

b e d P3-2 (d)

A + - B

-c + - D

a a a

=

= - - 113 = 0.33

P3- 2 (e)

...

1 0.5(1-0.14x)2

P3-2 (0

C

= Y

6 = Y

(2 -1) = Y

C - 2N

X -2C X

Vo

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

C

= 2F

X = 2C

X

(1 + eX) (l + EX) vo (l +

+

CAO

Y;o;0

YAo(O.07176)mol! dm

K c-

-

CAe

AO

+

CAe

AO

+

See Polymath program P3··2-LpoL

=

=

=

.~

~;~~-~:i -~~ -;-;:;~~~~~~_=--

§

Jr

o

-

P3-2 (g)

P3-2 (h)

2 2

A

kACA 2C

1 dm

kA = 25- - - s ( mol J

rc -1 -112 112

1 (dm 3 J2

kc =k

=12.5- - s mol P3-2 (i)

A+3B

2C

r

= k

C

C

C

K

C I12C

r

= k

C

-c ^{+ -} ^D

X -2C ^X

C = _~AO(l- ^X) = ^C Ao(l- ^X)

(1 + ^eX) ^(l + EX) vo (l +

K _c-

_-

~;~~-~:i - -;-;:;~~~~_=--

1 (dm ³ ^J2

P3-4 ^(d)

k2 _ -~(-) ^E

In (k2 J ^{(T. (T.} ⁺ ¹⁰⁾⁾