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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.

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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

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General Guidelines for California Problems

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

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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.]

F A 01 ; F I 0 X  0 .

8 2

A

B

F A 02  0 .

5 F A 01 ; F I 0 X  ?

Unknown: V ,  k , T , P , C A 0 , F A 0 , F I 0 , V 0

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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

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Gas Phase PFR

Assumptions: T  T 0 , P  P 0 , V 1  V 2 , k 1  k 2 , P 1  P 2 , C T 0  C T 02 1) Mole Balance: dX dV   r A F A 0 2) Rate Law:  r A  kC 2 A 3) Stoichiometry: (gas phase)    0  1   X  ; T  T 0 ; P  P 0 A  ½ B

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Gas Phase PFR

  1 2  1   1 2 C A  C A  1 0   1   X X   4) Combine:  r A  kC 2 A 0     1 1   X  X     2 dX dV V   kC 2 A 0 F A 0     1 1   X  X     2 F A 0 kC 2 A 0 0  X     1 1    X X     2 dX

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Gas Phase PFR

kC 2 A 0 V F A 0  2   1     1  X    2 X   1    2 X 1  X

Case 1:

kC 2 A 01 V  2 .

9 F A 01

Case 2:

kC 2 A 02 V  RHS F A 02 Take ratio of Case 2 to Case 1

 10

Gas Phase PFR

Most make this assumption to keep pressure the same, C T02 =C T01

kC

2

A

02

V F A

02

kC

2

A

01

V F A

01 

RHS

2.9

  

C A

02

C A

01   2  

F A

01

F A

02    

y A

02

y A

01

C T

02

C T

01   2     1

F A

01

F A

02    

y A

01  1 2

y A

02  1 3

F A

02

F A

01  1 2   

   11

Gas Phase PFR

 1    3 1 2     2     1 1 2     

RHS

2.9

 8 9     

RHS

 2 

y A

02    1 6  8 9      2    1 6    1  1 6    

X

2    1 6   2

X

2   1  1  2 6  1 

X

2

X

2 Solve with computer, X=0.758

12 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance

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End of Lecture 7