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solution manual Business Research Methods Schindler 14th edition solution manual Case Studies in Finance Managing for Corporate Value Creation Bruner Eades Schill 8th edition solution manual Chemical Engineering Design, Principles, Practice and Economics

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Solution Manual Contemporary Engineering Economics Park 7th edition - Updated 2024 Complete Solution Manual With Answers

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  • October 21, 2024
  • 19
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

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  • solution manual contemporary engineering economics
  • solution manual contemporary engineering

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  • Bioprocess Engineering Basic
  • Bioprocess Engineering Basic
  • Bioprocess Engineering Basic
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Chapter 3

Problem 3.1

S E k2
k1
 ES1 k3
k4
(ES) 2 
k5
P  E
rate  k5  ES2

a.) Equilibrium approach
* r values are the same as k values all are rate constants

r1 S E   r2  ES1 (1)
r3  ES1  r4  ES2 (2)
 E0    E    ES1   ES2 (3)
r
from (2)  ES1  4  ES2 (4)
r3
r2 r4  ES2
from (1) + (4)  E   (5)
r1r3 S

Substitute (4) and (5) into (3):

 r2 r4  r1 r4  S  r1 r3  S 
 E 0      ES2
 r1 r3  S 
r1 r3  E 0  S
  ES2 
r2 r4  r1 r4  S  r1 r3  S
r5 E 0 S r2 r
rate  where Km1  , Km 2  4
Km 2  Km1  S  S r1 r3

b.) Quasi–steady–state approach

d  ES1
 r1  E  S  r2  ES1  r4  ES 2  r3  ES 1  0 (6)
dt
d  ES2
 r3  ES1  r4  ES2  r5  ES2  0 (7)
dt
r4  r5
from (7)  ES1   ES2 (8)
r3

r r r r r r 
from (6) + (8):  E    2 4 2 5 3 5   ES2 (9)
 r1 r3  S 

substitute (8) and (9) into (3):



1

,  r2 r4  r2 r5  r3 r5  r1 r4 S   r1 r5 S   r1 r3 S  
 E 0      ES2
 r1 r3  S  
 r1 r3  E 0  S 
 ES2   
 r2 r4  r2 r5  r3 r5  r1 r4 S   r1 r5 S   r1 r3 S  
r5 E0 S
 rate 
 Km2  r5 r3  Km1  S  r5 r1  S
r r
where Km1  2 and Km 2  4
r1 r3

Problem 3.2
* r values are the same as k values all are rate constants
E S r1
r1
 ES r2
r2
EP
d  ES
 r1  E S  r2  E  P   r1  ES  r2  ES  0 (1)
dt
d P
rate   r2  ES  r2  E  P  (2)
dt
 E0    E    ES ,  E    E0    ES (3)

r1  r2
from (1):  E    ES (4)
r1 S  r2  P 

 r  S  r2  P   r1  r2 
Combine (3) + (4):  E 0    1   ES
 r1  S   r2  P  
E 0  r1  S  r2  P  
  ES  (5)
r1  S  r2  P   r1  r2

Substitute (3) and (5) into (2):

r1 r2 E 0  S  r2 r2 E 0  P 
rate   r  2  E 0   ES   P 
r1  r2  r1  S  r2  P 

Substituting (ES) in terms of E0:

r1 r2 E  S   r1 r2 E 0  P 
rate 
r1  r2  r1  S  r2  P 

LetVS = r2 E0 and VP = r-1 E0

r1 Vs  S  r2 VP  P 
rate 
r1  r2  r1  S  r2  P 


2

, Divide top and bottom by (r-1 + r2) and let
1 r 1 r
 1 and  2
K m r1  r2
1
K P r1  r2


 rate 
V s K1m  S   VP K P  P
S P
1 1 
K m KP

Problem 3.3

k 1  k 2
a) Km   4.5 105 M
k1

b) E 0  106 M
S  103 M
V S
V  m1
Km  S
but Vm  r2 E0
 V  9.58 104 Msec1


Problem 3.4.

At low substrate concentrations So< 150 mg/l substrate inhibition is negligible.
V0 = Vm0 So / (Km + So) or 1/V0 = 1/ Vm + Km/Vm (1/S0)
For S0 < 150 mg/l Plot 1/V0 versus 1/S0
Slope = Km/Vm0 = 13.8 y-intercept = 1/ Vm = 0.023
Then, Vm = 43.5 mg/l-h and Km = 600 mg/l

At high substrate concentrations above 150 mg/l , substrate inhibition is significant.
V0 = Vm0/ (1+ S0/Ksı) or 1/V0 = 1/Vm0 + S0/Vm Ksı
For S0 > 150 mg/l plot 1/V0 versus S0
Slope = 1/ Vm0Ksı = 2.59x 10-3 Then, Ksı = 8.9 mg/l

Low Ksı indicates severe substrate inhibition.

Problem 3.5.

Plot 1/V versus 1/S at different inhibitor concentrations
Since the lines intercept at the same point on y-axis inhibition is competitive (Constant Vm,
increased Km).
For I = 0 , No inhibitor : From the intercept on y axis , 1/V m = 0.2 and Vm = 5 mM/h
And from the intercept on X-axis, - 1/Km = -1.2 and Km = 0.83 mM
From 1/V versus 1/S plot for I = 1.3 mM and S 0 = 0.50 mM V = 1.3 mM/h
Then, V = Vm S/ (Km(1+I/Ksı)) + S , 1.3 = 5(0.5)/ (0.83(1+1.3/Kı) + 0.5 )
Then Kı = 1.82 mM

3

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