Solutions manual Elements of Chemical Reaction Engineering, 7th Edition, by Scott Fogler (COMPLETE)

Solutions Manual for
Elements of Chemical
Reaction Engineering
Seventh Edition



H. Scott Fogler
Ame and Catherine Vennema Professor of Chemical Engineering and
the Arthur F. Thurnau Professor
Bryan R. Goldsmith
Associate Professor of Chemical Engineering

Eranda Nikolla
Professor of Chemical Engineering

Nirala Singh
Associate Professor of Chemical Engineering

University of Michigan, Ann Arbor

May 2025




Hoboken, New Jersey

, Synopsis for Chapter 1 – Mole Balances
Mole balances are the first building block of the chemical reaction engineering algorithm.
General: The goal of these problems are to reinforce the definitions and provide an understanding of the
mole balances of the different types of reactors. It lays the foundation for step 1 of the algorithm in
Chapter 5.

Key to Nomenclature
I = Infrequently assigned
AA = Always assign one from the group of S = Seldom assigned
alternates G = Graduate level
O = Often assigned N = Never assigned
E.g., means problem P1-3B will be assigned every time I teach the course, problem AA P1-8 means that
this problem or one of the other problems with the prefix AA is always assigned for this chapter, Problem
l P1-2 will be infrequently assigned, Problem O P1-6B will often be assigned, and Problem S P3-16B is
seldom assigned.

Alternates: In problems that have a dot in conjunction with AA means that one of the problems, either
the problem with a dot or any one of the alternates are always assigned.

Time: Approximate time in minutes it would take a B student to solve the problem.



 Q1-1A (9 seconds) Questions Before Reading (QBR).
(a) John Falconer at the University of Colorado gives workshops on Teaching in which he points out
that students have a better comprehension if they ask themselves a question before reading
the text. The first question of each chapter, Q1, is just such a question.
(b) The students are asked, at a minimum read through the Questions to help put the chapter and
their studies in perspective.
(c) I encourage using the i>Clicker questions.
 Q1-2A (8-10 min) i>Clicker
 Q1-5A (5-75 min) through Q1-12A. To get a “feel” of the resources available, the students should
spend a total of about 50-75 minutes on these questions.


Computer Simulations and Experiments (5-15 minutes per simulation)
These problems are interactive and are a minor paradigm shift in the way we use homework problems.
Here the students are asked to explore the reaction and the reactor in which they occur to get an
intuitive feel and understanding of the reactor system. This procedure is called Inquiry Based Learning
(IBL).
 P1-1A (10-15 min) Good introduction to the use of Wolfram and Python.


Problems
I P1-2B (60 min) Problem reinforces wide range of applications of CRE and problem is given in the
web module which can be accessed from the Web Home Page (www.umich.edu/~elements). Many
students like this straight forward problem because they see how CRE principles can be applied to
S1-1

, an everyday example. It is often assigned as an in-class problem where parts (a) through (f) are
printed out from the web and given to the students in class. Part (g) is usually omitted.
 P1-3B (45 min) I always assign this problem so that the students will learn how to use
Polymath/MATLAB, Wolfram and Python before needing it for chemical reaction engineering
problems. Most problems will use either Polymath or MATLAB to solve the end of chapter problems.
 P1-4A (30 min) The Interactive Computer Games (ICGs) have been found to be a great motivation
for this material. This ICG will help student AIChE chapters prepare for the Jeopardy Competition at
the Annual AIChE Meeting.
 P1-5A (10 min) Old Exam Question (OEQ) to reinforce the convention and stoichiometry in mole
balances.
O P1-6B (30 min) A hint of things to come on sizing reactors. Fairly straight forward problem to make
a calculation. Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the
student an idea of things to come in terms of sizing reactors in chapter 4.
I P1-7A (30 min) Helps develop critical thinking and analysis.
AA P1-8A (20 min) Puzzle problem to identify errors in the solution. Many students especially those
who enjoy working Sudoku or crossword puzzles enjoy working these types of problems.




S1-2

, Solutions for Chapter 1 – Mole Balances

Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-1
http://umich.edu/~elements/5e/01chap/obj.html#/

2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf

3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf


Q1-1 Individualized solution.


Q1-2 Individualized solution.


Q1-3
For CSTR:
F X F X v0CA0 X A v0 X A = 10 * 0.9 =391.3 dm3
V= A0 A = A0 A = =
 rA,exit kCA kCA0 (1 X A ) k (1 X A ) 0.23* (0.1)

Q1-4 Individualized solution.


Q1-5 Individualized solution


Q1-6 Individualized solution


Q1-7 (a)
The assumptions made in deriving the design equation of a batch reactor are:
- Closed system: no streams carrying mass enter or leave the system
- Well mixed, no spatial variation in system properties
- Constant Volume or constant pressure

Q1-7 (b)
The assumptions made in deriving the design equation of CSTR, are:
- Steady state
- No spatial variation in concentration, temperature, or reaction rate throughout the vessel

Q1-7 (c)
The assumptions made in deriving the design equation of PFR are:
- Steady state
- No radial variation in properties of the system
1-1

,Q1-7 (d)
The assumptions made in deriving the design equation of PBR are:
- Steady state
- No radial variation in properties of the system

Q1-7 (e)
For a reaction
A➔ B
 -rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=]
moles/ (dm3.s).
 -rA’ is the rate of disappearance of species A per unit mass (or area) of catalyst [=] moles/
(time. mass of catalyst).
 rA’ is the rate of formation (generation) of species A per unit mass (or area) of catalyst [=]
moles/ (time. mass catalyst).
 -rA is an intensive property, that is, it is a function of concentration, temperature, pressure,
and the type of catalyst (if any), and is defined at any point (location) within the system. It is
independent of amount. On the other hand, an extensive property is obtained by summing
up the properties of individual subsystems within the total system; in this sense, -rA is
independent of the ‘extent’ of the system.


Q1-8 Individualized solution.


Q1-9 Individualized solution.


Q1-10 Individualized solution.


Q1-11 Individualized solution.


Q1-12 Individualized solution.


P1-1 (a) Example 1-3




1-2

,The above graph represents intial Ca and Cb profiles for k=0.23 and v0 = 10.
(i) With an increase in k (lets take k =0.35) for same volume and v0 , Ca decreases and Cb increases




Now lets make k=0.23(initial value) and make an increase in v0 (change from 10 to 15) for same volume.
We notice that now Ca increases and Cb decreases.
All of these graphs of concentration profiles are taken from Wolfram player by shifting the sliders.




1-3

,We can observe that varying rate constant has more effect on concentration profiles as compared to
varying volumetric flow rate.
(ii) CA decreases and CB increases with an increase in k and Ke, and a decrease in v0 for the same
volume.
(iii) Individualized solution
(iv) See the following polymath code:
Polymath Code:
d(Ca)/d(V) = ra / v0
d(Cb)/d(V) = rb / v0
k = 0.23
Ke=3
ra = -k * (Ca-Cb/Ke)
rb = -ra
v0 = 10
V(0)=0
V(f)=100
Ca(0)=10
Cb(0)=0

Output:




POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca 10. 2.849321 10. 2.849321
2 Cb 0 0 7.150679 7.150679
3 k 0.23 0.23 0.23 0.23
4 Ke 3. 3. 3. 3.
5 ra -2.3 -2.3 -0.1071251 -0.1071251
6 rb 2.3 0.1071251 2.3 0.1071251
7 V 0 0 100. 100.
8 v0 10. 10. 10. 10.

Differential equations
1 d(Ca)/d(V) = ra / v0
2 d(Cb)/d(V) = rb / v0

1-4

, Explicit equations
1 k = 0.23
Ke = 3
ra = -k * (Ca-Cb/Ke)
rb = -ra
v0 = 10



P1-2
Given
A  2*1010 ft2 TSTP  491.69R H  2000 ft

V  4 *1013 ft3 T = 534.7  R PO = 1atm
3
atm ft 10 lbmol
R  0.7302 CS  2.04 *10 C = 4*105 cars
lbmol R 3
yA = 0.02 ft
ft3
FS = CO in Santa Ana winds FA = CO emission from autos vA 3000 per car at STP
hr

P1-2 (a)
Total number of lb moles gas in the system:
PV
N 0
RT
1atm(4 1013 ft3)
N=   = 1.025 x 1011 lb mol
atm. ft3 
0.73  534.69R
 
lbmol.R 


P1-2 (b)
Molar flowrate of CO into L.A. Basin by cars.
• no. of cars
FA  yAFT  yA vA CT
STP
3000 ft3 1lbmol
F
T   400000 cars (See appendix B)
hrcar 359 ft3
FA = 6.685 x 104 lb mol/hr

P1-2 (c)
Wind speed through corridor is U = 15mph
W = 20 miles
The volumetric flowrate in the corridor is
vO = U.W.H = (15x5280)(20x5280)(2000) ft3/hr = 1.673 x 1013 ft3/hr

P1-2 (d)
Molar flowrate of CO into basin from Santa Ana wind.
FS : v0 CS

= 1.673 x 1013 ft3/hr 2.04 1010 lbmol/ft3
= 3.412 x 103lbmol/hr
1-5

, P1-2 (e)
dNCO
Rate of emission of CO by cars + Rate of CO in Wind - Rate of removal of CO =
dt
dCco
FA  FS  voCco  V (V=constant, Nco  CcoV )
dt

P1-2 (f)
t = 0 , Cco  CcoO
t Cco
dC co
 dt  V  F FS  voCco

0 CcoO A 
 F F v C 
V A S o coO 

t  ln
 
vo  FA  FS  v C 
o co

P1-2 (g)
Time for concentration to reach 8 ppm.
lbmol 2.04 lbmol
CCO0  2.04 108 , C CO  108
ft3 4 ft3
From (f),
V  FA  FS  vO.CCO0 
t  ln 
v F  F  v .C 
o  A S O CO 
 
8 lbmol 
 3
4 lbmol 3 lbmol 13 ft
 6.7 10  3.4 10 1.67310  2.04 10 
4 ft3  hr hr hr ft3 

lbmol  
ln
ft 3  lbmol lbmol ft 3
1.6731013 6.7 104  3.4 103 1.6731013  0.51108 
hr  hr hr 
 hr ft3 
t = 6.92 hr

P1-2 (h)
(1) to = 0 tf = 72 hrs
Cco = 2.00E-10 lbmol/ft3 a = 3.50E+04 lbmol/hr
vo = 1.67E+12 ft3 /hr b = 3.00E+04 lbmol/hr
Fs = 341.23 lbmol/hr V = 4.0E+13 ft3
 t dC
a  bsi n   F  v C  V co
6 s o co dt
 
Now solving this equation using POLYMATH we get plot between Cco vs. t

See the following polymath code:




1-6

, Polymath Code:
v0 = 1.67*10^12
A= 35000
B = 30000
F = 341.23
V = 4*10^13
d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V
C(0)=2.0e-10
t(0)=0
t(f)=72

Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A 3.5E+04 3.5E+04 3.5E+04 3.5E+04
2 B 3.0E+04 3.0E+04 3.0E+04 3.0E+04
3 C 2.0E-10 2.0E-10 2.134E-08 1.877E-08
4 F 341.23 341.23 341.23 341.23
5 t 0 0 72. 72.
6 V 4.0E+13 4.0E+13 4.0E+13 4.0E+13
7 v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12

Differential equations
1 d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V

Explicit equations
1 v0 = 1.67*10^12
2 A = 35000
3 B = 30000
4 F = 341.23
5 V = 4*10^13




 t 
a  dCco
(2) tf = 48 hrs Fs = 0 bsin   voCco  V
 6  dt
Now solving this equation using POLYMATH we get plot between Cco vs t


1-7