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CHME 312, Reaction Engineering, 2011 Spring Final Exam,

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CHME 312, Reaction Engineering, 2011 Spring

Final Exam,

Open Text

(H. Scott Fogler, Elements of Chemical Reaction Engineering, 4th Ed.)

Note: For partial credit, please write your answer clearly and legibly.

For better credit, do all algebra and substitute digits, and check final figures carefully.

1. Ethane is decomposed with about 300 kJ/mol of activation energy. How fast the rate of decomposition is at 600°C comparing with that of 500°C. (15)

Sol)

- 

 

 

RT

A E k exp

210











 

 

 







 



 



 

 



 

 

K 1 273 600

1 273

500 1 K

mol 3144 J . 8

J 10 exp 300

1 exp 1

3

2 1 1

2

T T R E k

k

2. A converted to R, S, and T with different type of reaction order (a), (b), (c). Suppose that each reaction is elementary, S is desired product, and A is supplied at CA0, determine whether you use CSTR or PFR, and determine whether you maintain exit conversion at high or low to obtain a maximum selectivity. (45)









T

(Desired) S

R A

3 2 1

(a) n1 = 1, n2 = 2, n3 = 3 (b) n1 = 2, n2 = 3, n3 = 1

(c) n1 = 3, n2 = 1, n3 = 2, where n1, n2, n3 are reaction order of (1), (2), and (3), respectively.

(2)

2/5 Sol)

- rRk1CAn1,rSk2CAn2 , rTk1CAn3

- 1 3

2

n A 3 n A 1

n A 2 T

R S (D/U)

C k C k

C k r

r S r

 

 

(a) n1 = 1, n2 = 2, n3 = 3

3 A 3 1 A 1

2 A 2 n

A 3 n A 1

n A 2

(D/U) 1 3

2

C k C k

C k C

k C k

C S k

 

  ☜ Maximum value, CSTR

(b) n1 = 2, n2 = 3, n3 = 1

2 A 3 1 A 1

3 A 2 n

A 3 n A 1

n A 2

(D/U) 1 3

2

C k C k

C k C

k C k

C S k

 

  ☜ Monotonically increasing, PFR, low XA

(c) n1 = 3, n2 = 1, n3 = 2

2 A 3 3 A 1

1 A 2 n

A 3 n A 1

n A 2

(D/U) 1 3

2

C k C k

C k C

k C k

C S k

 

  ☜ Maximum value, CSTR, high XA

3. (a) Derive the expression of design equation (20) and (b) determine the reaction order for the gas-phase decomposition of A by applying integral method. (20)

A → B + 2C

This reaction was carried out in the laboratory in an isothermal batch system in which the total pressure was recorded at various times during the reaction and given in table. Only pure A was initially present in the reaction vessel.

Time(min) 0.0 2.5 5.0 10.0 15.0 20.0

Total Pressure (mmHg) 7.5 10.5 12.5 15.8 17.9 19.4

Sol)

1. Postulate a rate law.

-rA = kCαA

2. A combination of the mole balance on a constant-volume batch reactor and the rate law

(3)

3/5

A A

A r kC

dt

dC  

 (1)

Rewrite the design equation in terms of the measured variable

For the case of a constant-volume batch reactor, we recall Equations (3-26) and (3-38):

) 1 ) (

1 (

A0 A0

A C X

V X

C N   

(2)

0 0

0 (1 )

T X T P

V P

V  

 

 

For isothermal operation and constant volume (V = V0, P = P0) )

1 ( ) 1 (

0 0

0 0

0

P P P

P y P P

X

A

   (3)

) 1 (

0 0

P P P

X

A

 (4)

By equation (2), (4)

 

RT P P

C PA0 ( 0)/

A

 

Pure A initially, yA01and thereforePA0P0, δ = 1 + 2 – 1 = 2

   

RT P P RT

P P P RT

P P C P

2 3 2 / ) (

/ )

( 0 A0 0 A0

A0 A

 

 

  

(5)

By equation (1), (5)



 

 

RT

P k P

dT dP

RT 2

3 2

1 A0

Let k'k

2RT

1, then k'

3PT0 PT

dT

dP   (a)

(4)

4/5 (b)

Guess 0th order, k' dTdP  Integrating PT = PT0 + kt This plot gives the right graph Not linear!

Try 1st order

T0 T

' 3P P

dT k

dP  

Integrating,

t P k

P

P '

T T0

T0

3

ln 2 

 

This plot gives the right graph Linear!

Just in case, try 2nd order

⇒ Compare the linearity

T0 T

2

' 3P P

dT k

dP  

Integrating,

t P k

P dP

T

T

P P

T '

2 T

0(3 T0 ) 

,

or kt

P P P

' T0 T

T0 2

1 3

1  

This plot gives the right graph Not linear!

(5)

5/5 4. If adsorption constants of CO and NO are KCO = 1 atm-1 and KNO = 3 atm-1, what is the ratio

of sites containing CO to those containing NO when the conversion is 30% and 99%?

What fraction of sites is occupied by NO and by CO. The feed is equal molar in CO and NO with PO = 2 atm.

Sol)

0.33

3 1 K K C

sites C NO to CO

NO CO S

NO S

CO   

- Fraction of CO sites.

- For X=0.3

at 20% conversion 15% of the sites are occupied by CO and 9.4% are vacant.

For X=.99

at 99% conversion 1.8% of the sites are occupied by CO and 89% are vacant.

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