chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案.doc

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 m3/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) +O2 CO2 + H2O
A typical entering feed has a BOD (biological oxygen demand) of 200 mg O2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.
Waste water
32,000 m3/day
Waste water
Treatment plant
Clean water
32,000 m3/day
200 mg O2
needed/liter
Mean residence
time =8 hr
Zero O2 needed

Figure P1.1
Solution:

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%H2), 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.

Solution:

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 2NO2 + 1/2 O2 = N2O5 , what is the relation between the rates of formation and disappearance of the three reaction components?
Solution:

2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate expression
-rA = 2 C0.5 ACB
What is the rate expression for this reaction if the stoichiometric equation is written as
A + 2B = 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
-rA = , mol/m3·s
What are the units of the two constants?
Solution:

2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and with second-order rate expression
-rA = k1[A][B]
are the reaction rates related as follows: rA= rB= rR? If the rates are not so related, then how are they related? Please account for the sings , + or - .
Solution:

2.6 A certain reaction has a rate given by
-rA = 0.005 C2 A , mol/cm3·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:

2.7 For a gas reaction at 400 K the rate is reported as
- = 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
-rA = - = k C2 A , mol/m3·s
Solution:
(a) The unit of the rate constant is
(b)
Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to

So we can get that the value of

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:

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, ℃
13
16
22
24
28
What activation energy represents this change in bustliness?
Solution:

Suppose ,
so intercept

150
160
230
295
370

-3.1780
-3.1135
-2.7506
-2.5017
-2.2752

13
16
22
24
28

3.4947
3.4584
3.3881
3.3653
3.3206

-y = -5147.9 x + 15.686
Also , intercept = 15.686 ,

Chapter 3 Interpretation of Batch Reactor Data

3.1 If -rA = - (dCA/dt) =0.2 mol/liter·sec when CA = 1 mol/liter, what is the rate of reaction when CA = 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:

We know that for 1st-order reaction, ,
,
,
So equ(1)
equ(2)
So

3.3 Repeat the previous problem for second-order kinetics.
Solution: We know that for 2nd-order reaction, ,
So we have two equations as follow:
, equ(1)
, equ(2)
So ,

3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a -order rate. What would be the fraction converted in a half-hour run?
Solution:
In a order reaction: ,
After integration, we can get: ,
So we have two equations as follow:
, equ(1)
, equ(2)
Combining these two equations, we can get:, but this means , which is impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted .

3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?
Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,

And

3.6 After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction.
Solution:
In 1st order reaction, , dissatisfied.
In 2nd order reaction, , satisfied.
According to the information, the reaction is a 2nd-order reaction.

3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike—into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable—at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise?
Solution:
, , ,
, ,
So we obtain ,
,
3.9 The first-order reversible liquid reaction
A R , CA0 = 0.5 mol/liter, CR0=0
takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction.
Solution: Liquid reaction, which belongs to constant volume system,
1st order reversible reaction, according to page56 eq. 53b, we obtain

, , so we obtain eq(1)
eq(1)
, , so we obtain eq(2)
,
eq(2)
Combining eq(1) and eq(2), we obtain

So the rate equation is

3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor its concentration drops from CA0 = 2.03 mol/liter to CAf = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.
Solution: It’s a irreversible second-order reaction system, according to page44 eq 12, we obtain
,
so
so the rate equation is
3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as follows:
Aucrose products
Starting with a sucrose concentration CA0 = 1.0 millimol/liter and an enzyme concentration CE0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):
CA, millimol/liter
0.84
0.68
0.53
0.38
0.27
0.16
0.09
0.04
0.018
0.006
0.0025
t,hr
1
2
3
4
5
6
7
8
9
10
11
Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or
-rA = where CM = Michaelis constant
If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method.
Solution: Solve the question by the integral method:
,
,

,mmol/L

1
0.84
1.0897
6.25
2
0.68
1.2052
6.25
3
0.53
1.3508
6.3830
4
0.38
1.5606
6.4516
5
0.27
1.7936
6.8493
6
0.16
2.1816
7.1428
7
0.09
2.6461
7.6923
8
0.04
3.3530
8.3333
9
0.018
4.0910
9.1650
10
0.006
5.1469
10.0604
11
0.0025
6.0065
11.0276
Suppose y=, x=, thus we obtain such straight line graph

, intercept=
So ,

3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:
A R, -rA =
If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 10 mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction..
Solution:

Rearranging and integrating, we obtain:

3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in
Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9℃:
H2SO4 + (C2H5)2SO4 → 2C2H5SO4H
Initial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter. Find a rate equation for this reaction.
Table P3.20
t, min
C2H5SO4H,
mol/liter
t, min
C2H5SO4H,
mol/liter
0
0
180
4.11
41
1.18
194
4.31
48
1.38
212
4.45
55
1.63
267
4.86
75
2.24
318
5.15
96
2.75
368
5.32
127
3.31
379
5.35
146
3.76
410
5.42
162
3.81
∞
(5.80)
Solution: It’s a constant-volume system, so we can use XA solving the problem:
i) We postulate it is a 2nd order reversible reaction system
The rate equation is:
, ,
,
When ,
So ,

After integrating, we obtain
eq (1)
The calculating result is presented in following Table.
, min

0
0
5.5
0
0
0
41
1.18
4.91
0.1073
0.2163
-0.2275
48
1.38
4.81
0.1254
0.2587
-0.2717
55
1.63
4.685
0.1482
0.3145
-0.3299
75
2.24
4.38
0.2036
0.4668
-0.4881
96
2.75
4.125
0.25
0.6165
-0.6427
127
3.31
3.845
0.3009
0.8140
-0.8456
146
3.76
3.62
0.3418
1.0089
-1.0449
162
3.81
3.595
0.3464
1.0332
-1.0697
180
4.11
3.445
0.3736
1.1937
-1.2331
194
4.31
3.345
0.3918
1.3177
-1.3591
212
4.45
3.275
0.4045
1.4150
-1.4578
267
4.86
3.07
0.4418
1.7730
-1.8197
318
5.15
2.925
0.4682
2.1390
-2.1886
368
5.32
2.84
0.4836
2.4405
-2.4918
379
5.35
2.825
0.4864
2.5047
-2.5564
410
5.42
2.79
0.4927
2.6731
-2.7254

5.8
2.6
0.5273
—
—
Draw ~ t plot, we obtain a straight line:

,

When approach to equilibrium, ,
so
So the rate equation is

ii) We postulate it is a 1st order reversible reaction system, so the rate equation is

After rearranging and integrating, we obtain
eq (2)
Draw ~ t plot, we obtain another straight line:

,
So

So the rate equation is

We find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when XAe =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that XAe =0.5.(The data that we use just have XAe =0.5273 approached to 0.5, so it causes to this.)

3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and CA alone determines this rate:
CA,mol/liter
1
2
4
6
7
9
12
-rA, mol/liter·hr
0.06
0.1
0.25
1.0
2.0
1.0
0.5
We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from CA0 = 10 mol/liter to CAf = 2 mol/liter.
Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.
0
4
8
12
16
20
0
2
4
6
8
10
12
14
Ca
-1/Ra

3.31 The thermal decomposition of hydrogen iodide
2HI → H2 + I2
is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:
T,℃
508
427
393
356
283
k,cm3/mol·s
0.1059
0.00310
0.000588
80.9×10-6
0.942×10-6
Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds.
According to Arrhenius’ Law,
k = k0e-E/R T
transform it,
- In(k) = E/R·(1/T) －In(k0)
Drawing the figure of the relationship between k and T as follows:

From the figure, we get
slope = E/R = 7319.1 intercept = - In(k0) = -11.567
E = 60851 J/mol k0 = 105556 cm3/mol·s
From the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is
- rA = 105556e-60851/R T·CA2

Chapter 4 Introduction to Reactor Design
4.1 Given a gaseous feed, CA0 = 100, CB0 = 200, A +B→ R + S, XA = 0.8. Find XB,CA,CB.
Solution: Given a gaseous feed, , ,
, find , ,
,

4.2 Given a dilute aqueous feed, CA0 = CB0 =100, A +2B→ R + S, CA = 20. Find XA, XB, CB.
Solution: Given a dilute aqueous feed, ,
, , find , ,
Aqueous reaction system, so
When ,
When ,
So ,
,
, which is impossible.
So ,

4.3 Given a gaseous feed, CA0 =200, CB0 =100, A +B→ R, CA = 50. Find XA, XB, CB.
Solution: Given a gaseous feed, , ,
, .find , ,
,
, which is impossible.
So

4.4 Given a gaseous feed, CA0 = CB0 =100, A +2B→ R, CB = 20. Find XA, XB, CA.
Solution: Given a gaseous feed, ，, ,
Find , ,
,

,
,

4.6 Given a gaseous feed, T0 =1000 K, π0＝5atm, CA0=100, CB0=200, A +B→5R,T =400 K, π=4atm, CA =20. Find XA, XB, CB.
Solution: Given a gaseous feed, , , ,
, , , , find , , .
, ,
According to eq page 87,

4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn to