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Corresponding Solutions for Chemical Reaction Engineering

CHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)

CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)

CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)

CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (19)

CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (22)

CHAPTER 6 DESIGN FOR SINGLE REACTIONS (26)

CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (32)

CHAPTER 11 BASICS OF NON-IDEAL FLOW (34)

CHAPTER 18 SOLID CATALYZED REACTIONS (43)

Chapter 1 Overview of Chemical Reaction Engineering

1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 ???→

?microbes CO 2 + H 2O

A typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.

Figure P1.1

Solution:

)/(1017.2)/(75.183132/100010001)0200()(313200031

3200013433

33s m mol day m mol day mol

m L mg g L mg day day m day

day m Vdt

dN r A A ??=?=-???-?-

=--

1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within the beds, based on the oxygen used.

Waste water

32,000 m 3/day

Clean water 32,000 m 3/day

200 mg O 2

needed/liter

Zero O 2 needed

Solution:

380010)1420(m V =???=

)/(9000101089.05.01024033hr bed molc hr

kgc

kgcoal kgc hr coal t N c ?-=?-=???-=?? )/(25.111

9000

80011322hr m kmolO t N V r r c c O ?=-?-=??-

=-=

)/(12000412000

190002hr bed mol dt dO ?=+?= )/(17.4800

)

/(105.113422

s m mol hr bed mol dt dO V r O ?=??==-

Chapter 2 Kinetics of Homogeneous Reactions

2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction? Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.

2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the rates of

formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-

2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate

expression

-r A = 2 C0.5 AC B

What is the rate expression for this reaction if the stoichiometric equation is written as

A + 2

B = 2R + S

Solution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.

2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate is

given by

-r A =

06]

][[1760C E A + , mol/m 3·s

What are the units of the two constants? Solution: ]

[]6[]][][[][03

A A C E A k s m mol r +=?=

- 3/][]6[m mol C A ==∴

m mol m mol m mol s m mol k 1

)/)(/(/][3333=

??=

2.5 For the complex reaction wi th stoichiometry A + 3B → 2R + S and with second -order

rate expression

-r A = k 1[A][B]

are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .

Solution: R B A r r r 2

31=-=-

2.6 A certain reaction has a rate given by

-r A = 0.005 C2 A , mol/cm 3·min

If the concentration is to be expressed in mol/liter and time in hours, what would be

the value and units of the rate constant? Solution:

min

)()(3

??-=??

-cm mol

r hr L mol r A A 2

2443'

300005.0106610)(min

A A A A C C r r cm mol mol hr L r =??=??=-???=

-∴ A

A A A A C C cm

mol mol L C cm

mol

C L mol C 33'

3'10)()(=??=∴?=?

42'32'103)10(300300)(A

A A A C C C r --?=?==-∴ 4'103-?=∴k

2.7 For a gas reaction at 400 K the rate is reported as -

dp A

= 3.66 p2 A, atm/hr (a) What are the units of the rate constant?

(b) What is the value of the rate constant for this reaction if the rate equation is

expressed as

-r A = - dt

dN V A

1 = k C

2 A , mol/m 3·s

Solution:

(a) The unit of the rate constant is ]/1[hr atm ? (b) dt

dN V r A

A 1-

Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to

22)(66

.366.3)(1RT C RT

P RT dt dP RT dt dP VRT V r A A A A A ==-=-

22)66.3(A A kC C RT ==

So we can get that the value of

1.12040008205.066.366.3=??==RT k

2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.

How much faster the decomposition at 650℃ than at 500℃? Solution:

586.7)9231

1731()10/(314.8/300)11(3211212=-??=-==K

K K mol kJ mol kJ T T R E k k Ln r r Ln

7.19701

=∴

r r

2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I find

Running speed, m/hr 150 160 230 295 370 Temperature, ℃

What activation energy represents this change in bustliness? Solution:

E RT

E e

k e

ak t cons ion concentrat f let ion concentrat f e

k r -

-=??=?='

00tan )()(

T Lnk Lnr A 1'-

=∴ Suppose T

x Lnr y A 1,==, so ,R

slope -= intercept 'Lnk =

)/(1-?h m r A

150 160 230 295 370

A Lnr

-3.1780 -3.1135 -2.7506 -2.5017 -2.2752 C T o /

13 16 22 24 28 3101

-?T

3.4947

3.4584

3.3881

3.3653

3.3206

-y = -5147.9 x + 15.686

Also K R

slope 9.5147-=-

=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=??-=

Chapter 3 Interpretation of Batch Reactor Data

3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate of

reaction when C A = 10 mol/liter? Note: the order of reaction is not known.

Solution: Information is not enough, so we can’t answer this kind of question.

3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A is

converted in a 5-minute run. How much longer would it take to reach 75% conversion? Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:

A A kC r =-

We know that for 1st -order reaction, kt C C Ln

=, 11kt C C Ln

A Ao =, 22

kt C C

Ln A Ao = Ao A C C 5.01=, Ao A C C 25.02=

So 21

)24(1)(11212Ln k

Ln Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=

- equ(1) min 521

)(111===

Ln k