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)
reaction orderCHAPTER 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()(31320003
1320001343333s m mol day m mol day mol g 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 =⨯⨯⨯=
)
/(9000101089.05.0102403
3hr bed molc hr kgc
kgcoal kgc hr coal
t N c
⋅-=⨯-=⨯⨯⨯-=∆∆
)/(25.1119000800113
22hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=
)/(12000412000
190002
hr bed mol dt dO ⋅=+⨯=
)/(17.4800)
/(105.113422s 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:  Be
cause 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  = A
06]][[1760C E A + ,  mol/m 3·s      What are the units of the two constants?
Solution:  ]
[]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴
s
m mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=
2.5  For the complex reaction with 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
131=-=-
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  22443'300005.0106610)(min
A 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)()(=⋅⋅=∴⨯=⨯
2
'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
-dt
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 C2 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

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