Lecture 7
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Today’s lecture
 Block 1: Mole Balances
 Block 2: Rate Laws
 Block 3: Stoichiometry
 Block 4: Combine
 California Professional Engineers Exam
Exam is not curved, 75% or better to pass
Problem 4-12
2
3
General Guidelines for California
Problems
4
Some hints:
1. Group unknown parameters/values on the same side of the
equation
example: [unknowns] = [knowns]
2. Look for a Case 1 and a Case 2 (usually two data points) to
make intermediate calculations
3. Take ratios of Case 1 and Case 2 to cancel as many unknowns
as possible
4. Carry all symbols to the end of the manipulation before
evaluating, UNLESS THEY ARE ZERO
Gas Phase PFR
The irreversible elementary reaction takes place in the gas
phase in an isothermal tubular (plug-flow) reactor. Reactant A
and a diluent C are fed in equimolar ratio, and conversion of A
is 80%. If the molar feed rate of A is cut in half, what is the
conversion of A assuming that the feed rate of C is left
unchanged? Assume ideal behavior and that the reactor
temperature remains unchanged. [From California
Professional Engineering/Engineers Exam.]
FA01 ; FI0
FA02  0.5 FA01 ; FI0
Unknown:
5
2A B
V, k, T, P, CA0 , FA0 , FI0 , V0

X  0.8
X?
Gas Phase PFR
Will the conversion increase or decrease?
INCREASE: Slower Volumetric Rate (Reactants
spend more time in the reactor
DECREASE: Concentration of Reactant Diluted
6
Gas Phase PFR
Assumptions:
T  T0 , P  P0 , V1  V2 ,
k1  k 2 , P1  P2 , CT 0  CT 02
1) Mole Balance:
dX  rA

dV FA 0
2) Rate Law:
 rA  kC2A
3) Stoichiometry: (gas phase)   0 1  X; T  T0 ; P  P0
7
A½B
Gas Phase PFR
1
1
  1  
2
2
CA 0 1  X 
CA 
1  X 
4) Combine:
 1  X  

 rA  kC 
 1  X  
2
A0
2
A0
dX kC

dV
FA 0
8
2
FA 0
V
kC2A 0
 1  X  


 1  X  
2
 1  X  
0  1  X   dX
X
2
Gas Phase PFR
kC2A 0 V
X
2
2
 21  ln1  X    X  1  
FA 0
1 X
2
kC
A 01V
Case 1:
 2.9
FA 01
2
kC
Case 2:
A 02 V
 RHS
FA 02
Take ratio of Case 2 to Case 1
9
Gas Phase PFR
Most make this assumption
to keep pressure the same,
CT02=CT01


kCA2 02V
2
2 






RHS
CA 02 FA 01
y A 02 CT 02
1 
FA 02

 
 
 
 F 
2
kCA 01V
2.9 CA 01  FA 02  y A 01 CT 01   A 01 


F
 A 02 
FA 01
1
y A 01 
2
10
y A 02
1

3
FA 02 1

FA 01 2
Gas Phase PFR
1 2 
3  1  RHS
1  1 
    2.9
2  2 
8 
 2.9  RHS
9 
1
 2  y A 02 
6
8 
 1  1 
 1 2
 1 2 X 2
 2.9  2 1  ln1  X 2    X 2  1  
9 
6  6 
 6 
 6  1  X 2
11
Solve with computer, X=0.758
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
12
End of Lecture 7
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