SOLUTIONS MANUAL for Elements of Chemical Reaction Engineering 6th Edition by H Scott Foger by welldoneassistant

Solutions Manual for

Elements of Chemical Reaction Engineering Sixth Edition

H. Scott Fogler Ame and Catherine Vennema, The Arthur F. Thurnau, Ann Arbor, Michigan

Table of Contents Screen Shot of Web Home Page ....................................................................................... i Screen Shot of Interactive Computer Games (ICGs) ........................................................ ii Screen Shot of Polymath Living Example Problems (LEPs) ........................................... iv Algorithm to Decode ICGs .............................................................................................. v Sample Course Syllabus ................................................................................................. ix Solutions to Chapter 1 – Mole Balances Questions Q1-1 through Q1-12 Problems P1-1 through P1-8 Solutions to Chapter 2 – Conversion and Reactor Sizing Questions Q2-1 through Q2-6 Problems P2-1 through P2-11 Solutions to Chapter 3 – Rate Laws Questions Q3-1 through Q3-7 Problems P3-1 through P3-16 Solutions to Chapter 4 – Stoichiometry Questions Q4-1 through Q4-9 Problems P4-1 through P4-13 Solutions to Chapter 5 – Isothermal Reactor Design: Conversion Questions Q5-1 through Q5-13 Problems P5-1 through P5-26 Solutions to Chapter 6 – Isothermal Reactor Design: Moles and Molar Flow Rates Questions Q6-1 through Q6-7 Problems P6-1 through P6-13 Solutions to Chapter 7 – Collection and Analysis of Rate Data Questions Q7-1 through Q7-7 Problems P7-1 through P7-12 Solutions to Chapter 8 – Multiple Reactions Questions Q8-1 through Q8-5 Problems P8-1 through P8-18 Solutions to Chapter 9 – Reaction Mechanisms, Pathways, Bioreactions and Bioreactors Questions Q9-1 through Q9-4 Problems P9-1 through P9-23 Solutions to Chapter 10 – Catalysis and Catalytic Reactors Questions Q10-1 through Q10-9 Problems P10-1 through P10-23

Solutions to Chapter 11 – Nonisothermal Reactor Design: The Steady-State Energy Balance and Adiabatic PFR Applications Questions Q11-1 through Q11-14 Problems P11-1 through P11-10 Solutions to Chapter 12 – Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange Questions Q12-1 through Q12-6 Problems P12-1 through P12-27 Solutions to Chapter 13 – Unsteady State Nonisothermal Reactor Design Questions Q13-1 through Q13-5 Problems P13-1 through P13-11 Solutions to Chapter 14 – Mass Transfer Limitations in Reacting Systems Questions Q14-1 through Q14-8 Problems P14-1 through P14-17 Solutions to Chapter 15 – Diffusion and Reaction Questions Q15-1 through Q15-8 Problems P15-1 through P15-17 Solutions to Chapter 16 – Residence Time Distributions of Chemical Reactors Questions Q16-1 through Q16-5 Problems P16-1 through P16-13 Solutions to Chapter 17 – Predicting Conversion Directly from the Residence Time Distribution Questions Q17-1 through Q17-4 Problems P17-1 through P17-18 Solutions to Chapter 18 – Models for Nonideal Reactors Questions Q18-1 through Q18-6 Problems P18-1 through P18-20

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 l = Always assigned AA = Always assign one from the group of alternates O = Often assigned

I = Infrequently assigned S = Seldom assigned G = Graduate level N = Never assigned

E.g., means problem l 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.

l 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. l Q1-2A (8-10 min) i>Clicker l 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). l 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. l 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. l 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. l 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: V=

10 * 0.9 v0 C A 0 X A v0 X A FA0 X A FA0 X A = = = = kC A kC A0 (1 - X A ) k (1 - X A ) 0.23 * (0.1) - rA,exit

=391.3 dm3

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 • • • •

Aà 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

7.150679

3 k

0.23

4 Ke

5 ra

-2.3

-0.1071251

6 rb

2.3

0.1071251

2.3

0.1071251

7 V

100.

8 v0

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 2 Ke = 3 3 ra = -k * (Ca-Cb/Ke) 4 rb = -ra 5 v0 = 10

P1-2 Given A = 2*1010 ft 2

TSTP = 491.69R

H = 2000 ft

V = 4*1013 ft 3

T = 534.7 ° R

PO = 1atm

R = 0.7302

atm ft 3 lbmol R

yA = 0.02

FS = CO in Santa Ana winds

C S = 2.04*10−10

FA = CO emission from autos

lbmol ft

v A = 3000

P1-2 (a) Total number of lb moles gas in the system: PV N= 0 RT N=

1atm×(4 ×1013 ft 3 ) = 1.025 x 1011 lb mol " 3% $ 0.73 atm. ft ' ×534.69R $ lbmol.R '& #

P1-2 (b) Molar flowrate of CO into L.A. Basin by cars.

FA = y AFT = y A ⋅v A CT 3

•%no.%of%cars STP

3000 ft 1lbmol × ×400000 cars hr car 359 ft 3 FA = 6.685 x 104 lb mol/hr FT =

(See appendix B)

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 ⋅C S = 1.673 x 1013 ft3/hr ×2.04 ×10−10 lbmol/ft3 = 3.412 x 103lbmol/hr 1-5

C = 4*105 cars ft 3 per car at STP hr

P1-2 (e) Rate of emission of CO by cars + Rate of CO in Wind - Rate of removal of CO = dC FA + FS − voCco = V co dt

dNCO dt

(V=constant, Nco = CcoV )

P1-2 (f) t = 0 , Cco = CcoO t

Cco

CcoO A

co ∫ dt = V ∫ F + F −v C S

o co

V "$ FA + FS −voCcoO %' t = ln vo $# FA + FS −voCco '& P1-2 (g) Time for concentration to reach 8 ppm. lbmol 2.04 lbmol CCO0 = 2.04 ×10−8 ×10−8 , CCO = 3 4 ft ft 3 From (f), % V " F + F −v .C t = ln$$ A S O CO0 '' vo # FA + FS −vO .CCO & " % 3 $ 6.7×104 lbmol + 3.4 ×103 lbmol −1.673×1013 ft ×2.04 ×10−8 lbmol ' $ hr hr hr 4 ft 3 ft 3 ' = ln$ ' 3 3 $ 13 ft 4 lbmol 3 lbmol 13 ft −8 lbmol ' 1.673×10 6.7×10 + 3.4 ×10 −1.673×10 × 0.51×10 ' hr $# hr hr hr ft 3 & t = 6.92 hr P1-2 (h) (1)

to = 0

tf = 72 hrs

C co = 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 + bsin# π & + Fs − voCco = V co dt " 6% 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

2 B

3.0E+04

3 C

2.0E-10

2.134E-08

1.877E-08

4 F

341.23

5 t

72.

6 V

4.0E+13

7 v0

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$ dC a + bsin# π & − voCco = V co dt " 6% Now solving this equation using POLYMATH we get plot between Cco vs t (2) tf = 48 hrs

Fs = 0

1-7

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)=48

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

2 B

3.0E+04

3 C

2.0E-10

1.921E-08

1.71E-08

4 F

341.23

5 t

48.

6 V

4.0E+13

7 v0

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

P1-2 (i) Changing a è Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of the sine function by adding to the baseline. 1-8

Changing b è The amplitude of ripples is directly proportional to ‘b’. As b decreases amplitude decreases and graph becomes smooth. Changing v0 è As the value of v0 is increased the graph changes to a “shifted sin-curve”. And as v0 is decreased graph changes to a smooth increasing curve.

P1-3 (a) Initial number of rabbits, x(0) = 500 Initial number of foxes, y(0) = 200 Number of days = 500 dx = k x − k2 xy …………………………….(1) dt 1 dy = k3 xy − k4 y …………………………….(2) dt Given,

k1 = 0.02day −1 k2 = 0.00004 / (day × foxes) k3 = 0.0004 / (day × rabbits) k4 = 0.04day −1 See the following polymath code: Polymath Code: d(x)/d(t) = (k1*x)-(k2*x*y) d(y)/d(t) = (k3*x*y)-(k4*y) k1 = 0.02 k2 = 0.00004 k3 = 0.0004 k4 = 0.04 t(0)=0 t(f)=500 x(0)=500 y(0)=200

Output: POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 k1

0.02

2 k2

4.0E-05

3 k3

0.0004

4 k4

0.04

5 t

500.

6 x

500.

2.962693

519.4002

4.219969

7 y

200.

1.128572

4099.517

117.6293

Differential equations 1 d(x)/d(t) = (k1*x)-(k2*x*y) 2 d(y)/d(t) = (k3*x*y)-(k4*y)

1-9

Explicit equations 1 k1 = 0.02 2 k2 = 0.00004 3 k3 = 0.0004 4 k4 = 0.04

When, tfinal = 800 and k3 = 0.00004 /(day ´ rabbits )

POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 k1

0.02

2 k2

4.0E-05

3 k3

4.0E-05

4 k4

0.04

5 t

800.

6 x

500.

377.9769

2086.088

1467.831

7 y

200.

114.6959

1341.876

143.6569

Differential equations 1 d(x)/d(t) = (k1*x)-(k2*x*y) 2 d(y)/d(t) = (k3*x*y)-(k4*y)

1-10

Explicit equations 1 k1 = 0.02 2 k2 = 0.00004 3 k3 = 0.00004 4 k4 = 0.04

Plotting rabbits vs. foxes

P1-3 (b) By increasing k4 and decreasing k2, foxes verses rabbits plot tends to become circular P1-3 (c) Below are the graphs when death rate is taken in to account

1-11

P1-3 (d) To solve the system of equation, we can use Polymath Nonlinear Equation Solver Polymath Code: f(x) = (x^3)*y-4*y^2+3*x-1 x(0) = 2 f(y) = 6*y^2-9*x*y-5 y(0) = 2

Polymath Output:

Calculated values of NLE variables Variable Value f(x) Initial Guess 1 x

2.385039 2.53E-11 2.

2 y

3.797028 1.72E-12 2.

Nonlinear equations 1 f(x) = (x^3)*y-4*y^2+3*x-1 = 0 2 f(y) = 6*y^2-9*x*y-5 = 0

P1-4 Individualized solution

P1-5

P1-6 (a) – rA = k with k = 0.05 mol/h dm3 CSTR: The general equation is

F −F V = A0 A −rA Here CA = 0.01CA0 , v0 = 10 dm3/min, FA = 5.0 mol/hr Also, we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3 Substituting the values in the above equation we get, C v −C v (0.5)10 − 0.01(0.5)10 V = A0 0 A 0 = k 0.05

à V = 99 dm3

1-12

PFR: The general equation is dFA

= rA = k , Now FA = CAv0 and FA0 = CA0v0 =>

dV Integrating the above equation, we get

∫ k

C A0

dC A =

dC Av0 dV

= −k

∫ dV => V = k0 (C A0 −C A ) 0

Hence V = 99 dm Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration. P1-6 (b) –rA = kCA with k = 0.0001 s-1 CSTR: We have already derived that

C v −C v v C (1− 0.01) V = A0 0 A 0 = 0 A0 −rA kC A k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1 àV=

(10dm3 / hr)(0.5mol / dm3 )(0.99) 3

−1

(0.36hr )(0.01*0.5mol / dm )

=> V = 2750 dm3

PFR: From above we already know that for a PFR dC Av0

dV Integrating

v0 k

= rA = −kC A

= − dV

∫ C

C A0

dC A

∫

C ln A0 =V k CA

Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1 Substituting the values in above equation we get V = 127.9 dm3 P1-6 (c) –rA = kCA2 with k = 300 dm3/mol.hr CSTR:

C v −C v v C (1− 0.01) V = A0 0 A 0 = 0 A0 −rA kC 2 A

1-13

Substituting all the values we get

(10dm3 / hr)(0.5mol / dm3 )(0.99) 3

(300dm / mol.hr)(0.01*0.5mol / dm )

=> V = 660 dm3

PFR: dC Av0

dV Integrating

∫

C A0

=> V =

= rA = −kC 2A V v 1 1 ) =V = − dV => 0 ( − 2 k C A C A0 C

dC A A

∫ 0

10dm3 / hr

1 1 − ) = 6.6 dm3 3 0.01C C 300dm / mol.hr A0 A0 (

P1-6 (d) CA = 0.001CA0

NA0

∫ N −r V

Constant Volume V=V0

C A0 dC A

∫C

−rA

Zero order: .999C Ao 1 t = "#C A0 − 0.001C A0 $% = = 9.99h k 0.05 First order: ! 1 $ 1 !C $ 1 t = ln## A0 && = ln# & = 69078s = 19.19h k " C A % 0.0001 " .001 %

Second order: 1" 1 1 % 1 " 1 1 % '= t= $ − − $ ' = 6.66h k $# C A C A0 '& 300 # 0.5⋅0.001 0.5 &

P1-7 Enrico Fermi Problem P1-7 (a) Population of Chicago = 4,000,000 Size of Households = 4 Number of Households = 1,000,000 Fraction of Households that own a piano = 1/5 Number of Pianos = 200,000 Number of Tunes/year per Piano = 1 Number of Tunes Needed Per Year = 200,000 Tunes per day = 2 1-14

Tunes per year per tuner =

250!days 2 × = 500/yr/tuner yr day

200,000!tunes 1 × = 400 Tuners yr 500!tunes / yr / tuner

P1-7(b) Assume that each student eats 2 slices of pizza per week. Also, assume that it is a 14” pizza, with 8 pieces. Hence, the area of 1 slice of pizza = 19.242 inch2 = 0.012414 m2 Thus, a population of 20000, over a span of 4 months, eats 20000 * 2 slices * 4 months * 4 weeks/month = 640000 slices of pizza, with a total area of 640000 * 0.012414 m2 = 7945 m2 of pizza in the fall semester. P1-7(c) Assume you drink 1L/day Assume you live 75 years*365days/year = 27375 days 1L/day*27375 days = 27375 L drank in life Bathtub dimensions: 1m*0.7m*0.5m = 0.35m3 = 350L/tub Bathtubs drunk = 27375L*1tub/350L = 78 tubs P1-7(d) Jean Valjean, Les Misérables.

P1-8 Mole Balance: F − "F V"=" A0 A −rA Rate Law :

−rA = kC 2A Combine: F − "F V"=" A0 A kC 2A

FA0 = v0C A = 3 FA = v0C A = 3

dm3 2molA 6molA . = s s dm3

dm3 0.1molA 0.3molA . = s s dm3

mol 3 s V"=" = 19000 1900dm dm3 3 dm mol 2 (0.03 )(0.1 ) mol.s dm3 The incorrect part is in step 6, where the initial concentration has been used instead of the exit concentration. (6 − 0.3)

1-15

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1-16

Synopsis for Chapter 2 – Conversion and Reactor Sizing General: The overall goal of these problems is to help the student realize that if they have –rA = f(X) they can “design” or size a large number of reaction systems. It sets the stage for the algorithm developed in Chapter 5. See Chapter 1 Synopsis for nomenclature guide to problem assignments. Questions l Q2-1A (4 seconds) Questions Before Reading (QBR). O Q2-2A (20 min) Secondly, I also encourage going through the i>Clicker questions AFTER the students have completed all the reading and homework associated with this chapter. l Q2-4A (20-25 min) Firstly, I encourage students to take the Solomon/Felder Inventory of Learning Styles test (https://www.engr.ncsu.edu/stem-resources/legacy-site/learning-styles/) and then use Appendix I.2 to see how they can best use the text and interactive web materials. O Q2-5A (15 min) If a student did not visit the University of Colorado’s LearnChemE site in Chapter 1, I recommend they view one or two screencasts now. I Q2-6A (7 min) NFPA. Now is also the time to visit the tutorials of the Safety Website (http://umich.edu/~safeche/) to become acquainted with the wealth of safety resources available on the safety website. Interactive Computer Games (ICG) l P2-1A (20-25 min) Because this interactive game has so many choices of reactions to maximize the conversion, the time to play the game is a little longer than other ICGs. Problems O P2-2A (45 min) Helps the student explore the example problems in this chapter. Parts (d) and (e) take a little longer than the other parts. O P2-3B (35 min) Reinforces use of the Levenspiel plots. AA P2-4B (40 min) Requires the student to construct a Levenspiel plot. Alternative to problems P2-5B, P2-7B, and P2-10C. AA P2-5B (30 min) This problem is a reasonably challenging trial and error problem that reinforces Levenspiel’s plots and reactor staging. O P2-6B (45 min) Novel application of Levenspiel plots from an article in CEE by Professor Alice Gast formerly at Massachusetts Institute of Technology, now President of Imperial College, London. AA P2-7B (30 min) Straight forward problem alternative to problems P2-4B and P2-10C. The answer gives ridiculously large reactor volume. The point is to encourage the student to question their numerical answers. Alternative to P2-4B, P2-5B and P2-10C. I P2-8A (30 min) Helps the students get a feel of real reactor sizes. l P2-9D (2 min) Great motivating problem. Students from all universities around the world remember this problem long after the course is over. S2-1

AA P2-10C (45 min) Alternative problem to P2-4B, P2-5B, and P2-7B. l P2-11B (45 min) This problem is a departure from the other problems in this chapter because it is a batch reactor.

S2-2

Solutions for Chapter 2 – Conversion and Reactor Sizing Q2-1 Individualized solution.

Q2-2 Individualized solution.

Q2-3 Individualized solution.

Q2-4 Individualized solution.

Q2-5 Individualized solution.

Q2-6 Individualized solution.

P2-1 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual.

P2-2 (a) For a batch reactor, 𝑑𝑋 𝑡 = 𝑁%& ' & −𝑟% 𝑉 400 dm3= 400/1000 m3 =0.4 m3 For 10% conversion, 5& 𝑑𝑋 𝑡 = (100/0.4) ' & −𝑟% The area under the curve of 1/-rA vs X needs to be found out till X=0.1 Using trapezoidal rule, Area= 0.5*(2.22+2.7)*0.1= 0.246 Thus, time= 250*0.246 = 61.5 s To find BR times for X=0.5 and X=0.8, apply Simpson’s rule and find time in similar way as described above P2-2 (b) Example 2-1 through 2-3 For Example, 2-1 If flow rate FAO is cut in half. v1 = v/2 , F1= FAO/2 and CAO will remain same. Therefore, volume of CSTR in example 2-1,

2-1

If the flow rate is doubled, F2 = 2FAO and CAO will remain same, Volume of CSTR in example 2-1, V2 = F2X/-rA = 12.8 m3 For Example, 2-2, If flow rate is cut to half, V= 0.5*2.165 m3 =1.083 m3 If flow rate is doubled, V=2*2.165= 4.33 m3 For Example, 2-3, If flow rate is cut to half, VCSTR= 0.5*6.4 m3 =3.2 m3 VPFR= 0.5*2.165 m3 =1.08 m3 If flow rate is doubled, VCSTR= 2*6.4 m3 =12.8 m3 VPFR=2*2.165= 4.33 m3 In a 4.5 m3 PFR, We need to find the value of X at which area under the curve of (Fao/-rA) and X is 4.5 From Figure E2-2.1 it can be seen that 80% conversion is achieved in 2.165 m3. Since data is only available till X=0.8, it is expected that with 4.5 m3, conversion achieved would be near to 100 % In a 4.5 m3 CSTR, 𝐹%& ∗ 𝑋 4.5 = −𝑟% So, we need to find a value of (FA0/-rA) and X which when multiplied gives 4.5 We can see that it is achieved at approximately 74 % conversion at which (FA0/-rA) is 6.1 P2-2 (c) 𝐹%& ∗ 𝑋 𝑉= −𝑟% So, we need to find a value of (FA0/-rA) and X which when multiplied gives 2.5 We can see that it is achieved at approximately 63.5 % conversion at which (FA0/-rA) is 3.9 P2-2 (d) Conversion achieved in a 2.4 m3 CSTR is 62.5 %. Now we add a 1 m3 PFR volume to it. We need to find X at which area under the curve of (Fao/-rA) and X is 1. The starting point is X=0.625. So, by trial and error, using trapezoidal rule, we find at 80 % conversion, below equation is satisfied 1= 0.5 (8+3.7)*(0.8-0.625) So, Conversion achieved is 80 %.

2-2

P2-2 (e) Example 2-4

x Now, FAO = 0.4/2 = 0.2 mol/s, Table: Divide each term

in Table 2-3 by 2.

X [FAO/-rA](m3)

0.1 0.545

0 0.445

Reactor 1 V1 = 0.82m3 V = (FAO/-rA)X

0.2 0.665

0.4 1.025

0.6 1.77

0.7 2.53

0.8 4

Reactor 2 V2 = 3.2 m3

!F $ 0.82 = ## A0 && X1 " −rA %X

( )

"F % 3.2 = $$ A0 '' X2 # −rA &X

( )

By trial and error, we get: X1 = 0.546 and X2 = 0.8 Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673 P2-2 (f) Example 2-5 (1) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR.

Conversion X1 = 0.20 X2 = 0.60 X3 = 0.65

Original Reactor Volumes V1 = 0.188 (CSTR) V2 = 0.38 (PFR) V3 = 0.10 (CSTR)

For PFR, X1 = 0.2

Using trapezoidal rule, XO = 0, X1 = 0.2 2-3

Worst Arrangement V1 = 0.23 (PFR) V2 = 0.53 (CSTR) V3 = 0.10 (CSTR)

1 (𝑋5 − 𝑋& )(𝑓(𝑋& ) + 𝑓(𝑋5 )) 2 −𝑟% = 0.2(1.28+0.94)/2 = 0.222 m3

𝑉5 =

For CSTR,

F V2 = A0 X2 − X1 = 1.32 0.6 − 0.2 = 0.53-m3 −rA

At X2 = 0.6 A0 = 1.32 m ,

(

−rA

)

(

)

For 2nd CSTR,

F V3 = A0 X3 − X2 = 2)m3 0.65− 0.6 = 0.1)m3 −rA

At X3 = 0.65, A0 = 2 m3 ,

(

−rA

)

(2) For first CSTR, At X=0.2,

FA0 = 0.94 m3 ; V1 = 0.94&m3 (0.2) = 0.188&m3 - rA

(

)

From previous example; V1 (volume of first CSTR) = 0.188 m3 Also, the next reactor is PFR, Its volume is calculated as follows

For next CSTR, X3 = 0.65,

, V3 =

(3) Now the sequence of the reactors remain unchanged. But all reactors have same volume. First CSTR remains unchanged Vcstr = .1 = (FA0/-rA )*X1 ð X1 = .088 Now For PFR:

By estimation using the Levenspiel plot X2 = .183

2-4

(

)

For CSTR, VCSTR2 = => X3 = .316 P2-2 (g) Residence time, 𝜏 is given as 2 sec = Since, 𝜏 = > ?

3 𝑚C 𝑣& = = 1.5 = 75 𝑚 C /𝑚𝑖𝑛 2 𝑠 P2-3

X FAO/-rA (m3)

0 0.89

0.1 1.08

0.2 1.33

0.4 2.05

0.6 3.54

0.7 5.06

0.8 8.0

V = 1 m3 P2-3 (a) Two CSTRs in series For first CSTR, V = (FAo/-rAX1) X So, we have to find a value of (FAo/-rAX1) and X in such a way that on multiplication it gives 1 So at (FAo/-rAX1) = 2.3, X= 0.435 => X1 = 0.435 For second CSTR, V = (FAo/-rAX2) (X2 – X1) Again, by trial and error, we have to find a condition which satisfy 1=(FAo/-rAX2)*( X2-0.435) => X2 = 0.66 P2-3 (b) Two PFRs in series

2-5

For X1, we need to find area under the curve such that area =1. We can use trapezoidal rule to estimate the conversion taking two points at a time to get X1 = 0.565

X2 = 0.775

P2-3 (c) Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 V = (0.5FAO/-rAX1) X1 2=(FAo/-rAX1) X Solving in similar manner as part a, we get, Xout = 0.585 P2-3 (d) Two PFRs in parallel with the feed equally divided between the two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 So, similar to part b, the area under the curve must be 2 in this case i.e. V=2

Xout = 0.775 P2-3 (e) A PFR followed by a CSTR, XPFR = 0.565 (using part(b)) V = (FAo/-rA-XCSTR) (XCSTR – XPFR) Now, 1=(FAo/-rA-XCSTR) (XCSTR – 0.565) By looking at graph we get, XCSTR = 0.732 P2-3 (f) A CSTR followed by a PFR, XCSTR = 0.435 (using part(a))

1= Area under the curve (with starting point as 0.435) By calculating area, we get XPFR = 0.71 P2-3 (g) A 1 m3 PFR followed by two 0.5 m3 CSTRs, For PFR, XPFR = 0.565 (using part(b)) CSTR1: V = (FAo/-rA-XCSTR) (XCSTR – XPFR) = 0.5 m3 XCSTR = 0.673 CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 – XCSTR1) = 0.5 m3 XCSTR2 = 0.75 Two PFR in parallel or series was a better option as exit conversion was higher P2-3 (h) A CSTR and a PFR are in parallel with flow equally divided 2-6

Since the flow is divided equally between the two reactors, the overall conversion is the average of the CSTR conversion (part C) and the PFR conversion (part D) Xo = (0.585 + 0.775) / 2 = 0.68

P2-4 Exothermic reaction: A → B + C X 0 0.20 0.40 0.45 0.50 0.60 0.80 0.90

r(mol/dm3.min) 1 1.67 5 5 5 5 1.25 0.91

1/-r(dm3.min/mol) 1 0.6 0.2 0.2 0.2 0.2 0.8 1.1

P2-4 (a) For a batch reactor, To solve this problem, first plot 1/–rA vs. X from the chart above. - I-

t=NA0∫& NA0=400 moles V=400 dm3 t=(400 /400)*(area under the curve) For X=0.4, area under curve is =0.4*0.2+0.5*0.4*0.8=0.24 dm3.min/mol, thus t= 0.24 mins

JKL =

Similarly, For X=0.8, area under curve is =0.4*0.2+0.5*0.4*0.8+0.2*0.2+0.2*0.2+0.5*0.2*0.6= 0.38 dm3.min/mol, thus t= 0.38 mins P2-4 (b) Use mole balance as given below. CSTR: Mole balance: PFR:

=>VCSTR = 24 dm3

Mole balance: = 300(area under the curve) VPFR = 72 dm3 P2-4 (c) For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant over this conversion range. 2-7

P2-4 (d) VCSTR = 105 dm3 Mole balance:

Use trial and error to find maximum conversion. At X = 0.70, 1/–rA = 0.5, and X/–rA = 0.35 dm3.min/mol Maximum conversion = 0.70 P2-4 (e) From part (a) we know that X1 = 0.40. Use trial and error to find X2. Mole balance: Rearranging, we get

At X2 = 0.64, Conversion = 0.64 P2-4 (f) From part (a), we know that X1 = 0.40. Use trial and error to find X2. Mole balance: At X2 = 0.908, V = 300 x (area under the curve) => V = 300(0.24) = 72dm3 Conversion = 0.908. 2-8

P2-4 (g)

P2-5 We must first find a CSTR up to X = 0.2 to minimize the volume, and hence the cost of reactor. We can either use a CSTR or a PFR for X = 0.2 to X = 0.6 since the rate is independent of X in that range. A PFR for X > 0.6 is preferable since it will reduce the volume required and the cost. Let’s calculate the volumes for the different cases. i) X = 0 to X = 0.2 Volume for CSTR = 20*0.2 = 4 dm3 Volume for PFR = 0.5*(50+20)*0.2 = 7 dm3 ii) X = 0.2 to X = 0.6 Volume for CSTR = Volume for PFR = 0.4*20 = 8 dm3 iii) X > 0.6 Volume for PFR < Volume for CSTR; for the same conversion. Hence a PFR would minimize the cost. First, we use the 4 dm3 CSTR worth $2000. Conversion at this stage = 0.2 After this, we have $8000, and we see that the PFR minimizes the cost (So, we can have more conversion with the same money) Hence, we can use a 12 dm3 PFR followed by a 4 dm3 PFR, totally worth $8000, which will give the highest conversion. The first 8 dm3 of PFR would give conversion up to 0.6, and the rest (16 – 8) dm3 = 8 dm3 would give the conversion of:

Therefore, X = 0.8 Final conversion = 0.8

P2-6 (a) Individualized Solution P2-6 (b) 1) In order to find the age of the baby hippo, we need to know the volume of the stomach. The metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of what the mother eats then Fao (baby) = ½ Fao (mother). The Levenspiel Plot is shown:

2-9

Autocatalytic Reaction

Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the baby’s stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.

P2-6 (b) 2) If Vmax and mao are both one half of the mother’s then

and since then

will be identical for both the baby and mother. Assuming that like the stomach the intestine volume is proportional to age then the volume of the intestine would be 0.75 m3 and the final conversion would be 0.40 P2-6 (c) Vstomach = 0.2 m3 From the web module we see that if a polynomial is fit to the autocatalytic reaction we get: 2-10

= 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499 And since Vstomach = X, solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3 Xstomach = .067 For the intestine, the Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178. Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.

P2-6 (d) PFR→ CSTR PFR: Outlet conversion of PFR = 0.111

CSTR: We must solve V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499) X=0.42 Since the hippo gets a conversion over 30% it will survive.

2-11

P2-7

2A + B → 2C P2-7 (a) PFR volume necessary to achieve 50% conversion Mole Balance

Volume = Geometric area under the curve of (FA0/-rA) vs X)

V = 150000 m3 P2-7 (b) CSTR Volume to achieve 50% conversion Mole Balance

V = 50000m3 P2-7 (c) Volume of second CSTR added in series to achieve 80% conversion

V2 = 150000m3 P2-7 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion VPFR = 90000m3 P2-7 (e) For CSTR, V = 60000 m3 (CSTR) Mole Balance

X = 0.463 For PFR, V = 60000 m3 (PFR) 2-12

Mole balance

X = 0.134

P2-7 (f) Real rates would not give that shape. The reactor volumes are absurdly large.

P2-8 Problem 2-8 involves estimating the volume of three reactors from a picture. The door on the side of the building was used as a reference. It was assumed to be 8 ft high. The following estimates were made: CSTR h = 56ft d = 9 ft V = πr2h = π(4.5 ft)2(56 ft) = 3562 ft3 = 100,865 L PFR Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(11) = 3036 ft D = 1 ft V = πr2h = π(0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L Answers will vary slightly for each individual.

P2-9 No solution necessary.

P2-10 (a) The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in series and containing equal amounts of catalyst can be calculated from the following figure.

The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR. This figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR. 2-13

P2-10 (b) Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining the area of the shaded region in the figure below.

The area of the rectangle is approximately 23.2 kg of catalyst. P2-10 (c) The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of the shaded rectangle shown in the figure below.

The area of the rectangle is approximately 7.6 kg of catalyst. P2-10 (d) The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the shaded region in the figure below.

The necessary catalyst weight is approximately 22 kg. 2-14

P2-10 (e) The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating the area of the shaded region in the graph below.

The necessary catalyst weight is approximately 13 kg. P2-10 (f)

P2-11 X 0 0.20 0.40 0.60 0.80

-r(mol/dm3.min) 4 3 2.2 1.5 1

1/-r(dm3.min/mol) 0.25 0.33 0.454 0.666 1

To solve this problem, first plot 1/–rA vs. X from the chart above

2-15

For calculating the area from the above graph, approximate the above curve as polynomial of degree 2 and its equation can be obtained from trendline option in excel. Then do the definite integral from limits you wish.

We know that for a batch reactor, from mole balance we get the following equation:

The volume term can be taken out from integral and then Nao/V can be substituted as Cao which is equal to 2. P2-11 (a) The time for 10% conversion can be found out by evaluating the integral from 0 to 0.1 t = 0.051min for conversion of 10% t = 0.34min for conversion of 50% t = 0.783min for conversion of 80% P2-11 (b) Plot the rates for the above time which you have got and get –r vs t plot

2-16

Synopsis for Chapter 3 – Rate Laws General: The second building block is the rate law. The purpose of these problems is to help students understand how to write a rate law and how activation energy affects the rate of reaction. See Chapter 1 Synopsis for nomenclature guide to problem assignments. Questions l Q3-1A (20 seconds) Questions Before Reading (QBR). l Q3-2A (15-18 min) The i>Clicker questions are an excellent review for the student chapter Jeopardy competition held at the Annual AIChE Meeting in the Fall. It is recommended that the student read through the questions and then spend 1 minute (2 minutes at most) thinking about the answer to the question. O Q3-3C – Q3-6A (10-15 min each) I Q3-7B (5 min) Not fearing for my own safety from attack from the crickets, I collected this data from my yard and put it on the audio in the Summary Notes. The students should compare it with Sheldon Cooper’s analysis in the Big Bang Theory and on government websites. Computer Simulations and Experiments l P3-1B (a) (10 min maximum) Wolfram and Python explore change in the energy distribution of the collisions with temperature. O P3-1B (b) (8 min) This stochastic simulation was developed by Kaushik Nagaraj of IIT Bombay. Students should change the initial concentrations and run at least one simulation. Problems O P3-2B (a) – (b) (10 min) Lead the student to explore the example problems. I P3-2B (c) (75 min) Requires additional reading on collision theory on PFR on the web and is considered advanced material or graduate material. AA P3-3B (15 min) Uses Figure 3-4 and LEP 3-1B to study collision frequencies at different temperatures. O P3-4B (15 min) Helps the student understand reaction coordinate plots. AA P3-5B (30 min) Expands on P3-3B and P3-4B. AA P3-6A (30 min) Further discussion of the activation energy. I P3-7A (30 min) Picture of cricket and the table may encourage the student to work this problem without it being assigned. When making the Arrhenius plot for each of the insects, one finds they all have virtually the same activation energy. AA P3-8B (15 min) Real industrial example of competing effects between concentration and temperature. l P3-9B (20 min) Coupled with YouTube video “Black Widow,” this is a great motivating problem for the students (https://www.youtube.com/watch?v=CE7b9teZwqs). AA P3-10B (35 min) This problem is not too straight forward to derive this rule of thumb analytically. S1-1

AA P3-11C (15 min) Very short, if the student has read the text before starting the problem. Straight forward calculation to find the activation energy. AA P3-12B (6-10 min) Straight forward problem. 5-10 minutes if the student has read the text before starting to write the rate law. l P3-13A (10 min) Straight forward problem. 5-10 minutes if the student has read the text before starting to write the rate law. AA P3-14A (20 min) Somewhat tricky problem to find the rate law. AA P3-15B (10 min) Find the rate law for reversible reactions. S P3-16B (10 min) Open-ended problem to guess how to incorporate light into rate law.

S1-2

Solutions for Chapter 3 – Rate Laws Useful Links: 1. Click on the link given below to download Wolfram/python codes for Chapter 3: http://www.umich.edu/~elements/5e/03chap/live.html 2. Click on the link given below to view Wolfram tutorial (for running Wolfram codes): http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf

Q3-1 Individualized solution.

Q3-2 Individualized solution.

Q3-3 (a) Individualized solution. (b) Individualized solution. (c) kB = 50 (dm3/mol)2/s and kC = 50 (dm3/mol)2/s (d) Individualized solution. (e) Individualized solution.

Q3-4 For ethylene oxide, the following GHS symbols are applicable: 1. Health hazard 2. Gas cylinder 3. Exclamation mark 4. Flame 5. Skull and crossbones

Q3-5 Individualized solution.

Q3-6 NFPA diamond colors and GHS symbols that can be categorized into that color: 1. Blue: Health hazard, exclamation mark, corrosion, skull and cross bones 2. Red: Flame, flame over circle 3. Yellow: Explosive bomb 4. White: Environment, exclamation mark, gas cylinder

Q3-7 (a) From P3-7, 52150

ln(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑟𝑝𝑠/𝑚𝑖𝑛𝑢𝑡𝑒) = 179 − 𝑇 In the recording, approximately 20 chirps were heard in 10 seconds i.e, 120 chirps/min This gives T = 300 K

3-1

(b) The correlation given in the website is, 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑟𝑝𝑠 + 37 = 𝑐𝑇 15

Where T is in Fahrenheit and c is a proportionality constant Arrhenius equation: 𝑘 = 𝐴 𝑒 −𝐸𝑇 𝑘 = 𝐴(1 − 𝐸𝑇) (approximation of the exponent term)

P3-1 (a) (i) For a low value of E, irrespective of value of T the fraction of molecules is almost equal to 1 (can be seen by below graphs)

For activation energy as high as 10 Kcal/mol, the fraction of molecules with energy greater than activation energy increases as temperature increases but even at high values of T=5000K , still the fraction approximates to be 0.5-0.6

, Fraction of molecules with energy greater than Activation Energy is, 0.5667

3-2

(ii)

2550 K (can be seen by varying Wolfram parameters)

(iii)

Individualized solution

P3-1 (b) (i) The equilibrium concentration changes, but the equilibrium conversion of A remains the same in all the three cases (50%). The time required to attain equilibrium decreases with increasing particle number. 1. We can conclude that conversion of A does not depend on no of A or B molecules present initially provided number of molecules of A is less than or equal to number of molecules of B. 2. Equilibrium concentration and equilibrium time will for sure vary depending on the no of A and B molecules present initially (ii) The trajectories remain similar, but the time taken to attain equilibrium changes, as the rate constants are lower. (iii) The forward/reverse reaction goes to completion. (iv) This is the nature of a stochastic simulation. The number of molecules in the simulation are very less, as compared to the large number in the deterministic model (number of molecules are of the order of the Avogadro number). The fluctuations reduce as you increase the number of molecules, and the stochastic model is identical to a deterministic model when there are infinite molecules. (v) The fluctuations in the trajectories reduce. (vi) No, equilibrium only means that the forward reaction rate is equal to the reverse reaction rate. It does not mean that the reactions stop occurring. This can be seen from the fact that there is always a small fluctuation about the equilibrium concentration. (vii) 1. By this whole experiment one can conclude that equilibrium conversion of A is independent of no of reacting molecules present initially provided number of molecules of A is less than or equal to number of molecules of B. 2. We can conclude that by increasing the no of molecules, one moves to a more realistic model. 3. Both forward and backward rate constants affect the time to attain equilibrium.

3-3

P3-2 (a)

𝑙𝑛𝑘 = 𝑙𝑛𝐴 −

𝐸 1 ( ) 𝑅 𝑇

From the graph of ln k vs 1/T above, we get: 𝑙𝑛𝐴 = 28.817, Thus, 𝐴 = 3.27 ∗ 1012 𝑠 −1 ,

𝐸 = −11299 𝐾 𝑅 𝐸 = 93.94 𝑘𝐽/𝑚𝑜𝑙

𝑘 = 3.27 ∗ 1012 𝑒𝑥𝑝 (− P3-2 (b) Example 3-1 For, E = 60kJ/mol

11299 −1 )𝑠 𝑇

For, E = 240kJ/mol

T (K)

k (1/sec)

1/T

ln(k)

T (K)

k (1/sec)

1/T

ln(k)

310

1023100

0.003226

13.83918

310

4.78E-25

0.003226

-56.0003

315

1480488

0.003175

14.2087

315

2.1E-24

0.003175

-54.5222

320

2117757

0.003125

14.56667

320

8.77E-24

0.003125

-53.0903

325

2996152

0.003077

14.91363

325

3.51E-23

0.003077

-51.7025

330

4194548

0.00303

15.25008

330

1.35E-22

0.00303

-50.3567

335

5813595

0.002985

15.57648

335

4.98E-22

0.002985

-49.0511

3-4

(1) For lower activation energy, we observe that the increase in rate constant is gradual. The slope of the plot: ln(k) vs 1/T is steeper for higher activation energy. (2) From the plots we can see that higher the activation energy smaller the rate constant at a given temperature. Therefore, higher activation energy will lead to slower reaction rates. The molecules will have to collide with greater energies for the reaction to occur. Activation energy could be a measure of the affinity molecules have to react with each other. P3-2 (c) Solution will not be given

P3-3 (a) Refer to Fig 3-4 The fraction of molecular collisions having energies less than or equal to 2.5 Kcal is given by the area under the curve, f(E,T)dE from EA = 0 to 2.5 Kcal. From wolfram code, answer is At 300 K, Area= 1-0.0388 =96.1 % At 500 K, Area =1-0.17 =83 % P3-3(b) The fraction of molecular collisions having energies between 3 and 4 Kcal is given by the area under the curve f(E,T) from EA = 3 to 4 Kcal. At 300 K, Area = 0.018-0.00383 = 1.4 % At 500 K, Area = 0.11-0.045 = 6.5 % P3-3 (c) The fraction of molecular collisions having energies greater than the activation energy E A= 25 Kcal is given by the area under the curve f(E,T) from E A =25 to 50 Kcal. At 300 K, Area = 0 % At 500 K, Area = 6.97E-9 % 3-5

P3-4 (a)

Considering the molecules to be linear harmonic oscillator:

P3-4 (b,c) Use Figure 3-1(b) to know the position of RAB = 2.8 A° in Figure 3-1(a)

P3-5 (a) Use the following equation to find out the fraction of molecules with energy greater than EA = 25 kcal, 1

𝐹(𝐸 > 𝐸𝐴 , 𝑇) =

2 𝜋

𝐸𝐴 2 𝐸 ) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 ) 1( 𝑅𝑇 2

3-6

(1)

Substituting 𝐸𝐴 = 25 kcal/mol and T = 300 K, 500 K, 800 K and 1200 K 𝑘𝑐𝑎𝑙

𝐹 (25 𝑚𝑜𝑙 , 300) = 5.844× 10-18 𝑘𝑐𝑎𝑙

𝐹 (25 𝑚𝑜𝑙 , 500) = 7.835× 10-11 𝑘𝑐𝑎𝑙

𝐹 (25 𝑚𝑜𝑙 , 800) = 7.303× 10-7 𝑘𝑐𝑎𝑙

𝐹 (25 𝑚𝑜𝑙 , 1200) = 1.09× 10-4 P3-5 (b) We need to find the ratio of the quantity, 𝐹(25 𝑘𝑐𝑎𝑙/𝑚𝑜𝑙, 𝑇) − 𝐹(15 𝑘𝑐𝑎𝑙/𝑚𝑜𝑙, 𝑇) at 300 K and 1200 K The ratio of the faction of energies between 15 and 25 kcal at 300 K to that for the same energy range (15–25 kcal) at 1200 K =

𝑘𝑐𝑎𝑙

𝐹(25 𝑚𝑜𝑙 ,300𝐾)−𝐹(15 𝑚𝑜𝑙 ,300𝐾) 𝐹(25 𝑚𝑜𝑙 ,1200𝐾)−𝐹(15 𝑚𝑜𝑙 ,1200𝐾)

= 1.467× 10-8

P3-5 (c) The fraction of collisions with energy greater than E A is given by the equation, 𝐹(𝐸 > 𝐸𝐴 , 𝑇) = This equation is plotted as a function of

1 2

𝐸

𝐴 ) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 ) 1( 𝑅𝑇

𝜋2

𝐸𝐴 𝑅𝑇

𝐸

for 𝐴 > 3 using MATLAB 𝑅𝑇

At 700 K, the fraction of collisions with energy greater than 25 kcal/mol is given by: 𝐸𝐴 25000 = 2×700 = 17.86 𝑅𝑇

Substituting in (1) gives, 𝑘𝑐𝑎𝑙

𝐹 (𝐸 > 25 𝑚𝑜𝑙 , 700𝐾) = 8.354 × 10-8

3-7

(1)

P3-5 (d) The fraction of collisions with energy greater than E A is given by the equation, 1

𝐹(𝐸 > 𝐸𝐴 , 𝑇) =

2 𝜋

𝐸𝐴 2 𝐸 ) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 ) 1( 𝑅𝑇 2

(1)

At 500 K, the fraction of collisions with energy greater than 15 kcal/mol is given by: 𝐸𝐴 15000 = 2×500 = 15 𝑅𝑇

Substituting in (1) gives, 𝑘𝑐𝑎𝑙

𝐹 (𝐸 > 15 𝑚𝑜𝑙 , 500𝐾) = 9.713 × 10-5 P3-5 (e) The plots were generated using MATLAB

As can be seen from the plots, as the activation energy increases the fraction of molecules with energy greater than the activation energy at a given temperature decreases.

P3-6 (a) The fraction of collisions having energy between E = 3 and E = 5 is the area under the graph between those two boundaries For Figure P3-6 (a), Area =2*0.5(sum of parallel sides)*height = 2*0.5*(0.1875 + 0.25)*1 = 0.4375 For Figure P3-6 (b), Area=0.5*(0.2+0.15)*2=0.35

3-8

P3-6 (b) For Figure P3-6 (a), Fraction of collision having energy grater than 5 kcal/mol= 0.5*3*0.1875=0.281 For Figure P3-6 (b) Fraction of collision having energy greater than 5 kcal/mol= 0.5*3*0.15=0.225 P3-6 (c) For both the figures 8

∫ 𝑓(𝐸, 𝑇)𝑑𝐸 = 1 0

P3-6 (d) Since f(E,T) = 0 for E > 8 kcal, the fraction with energies greater than 8 kcal = 0 P3-6 (e) The fraction with energies greater than EA ,8 kcal is zero P3-6 (f) At 400K, activation energy will be higher than the value at 300 K. So, X axis will have higher nos. Since total area under curve has to remain constant to 1, so y axis value has to be lower. The graph may look like this

P3-7 (a) Note: This problem can have many solutions as data fitting can be done in many ways. Using Arrhenius Equation For Fire flies: T(in K) 1/T 294 0.003401 298 0.003356 303 0.003300

Flashes/min 9 12.16 16.2

ln(flashes/min) 2.197 2.498 2.785

Plotting ln (flashes/min) vs. 1/T, We get a straight line. For Crickets: T(in K) 1/T x103

chirps/min

ln(chirps/min)

287.2 293.3 300

80 126 200

4.382 4.836 5.298

3.482 3.409 3.333

Plotting ln (chirps/min) Vs 1/T, we get a straight line. ➔Both, Fireflies and Crickets data follow the Arrhenius Model. ln y = A + B/T , and have the similar activation energy. 3-9

P3-7 (b) For Honeybee: T(in K) 1/T x103 298 3.356 303 3.300 308 3.247

V(cm/s) 0.7 1.8 3

ln(V) -0.357 0.588 1.098

Plotting ln (V) vs. 1/T, almost straight line. ln (V) = 44.6 – 1.33E4/T At T = 40oC (313K) V = 6.4cm/s At T = -5oC (268K) V = 0.005cm/s (But bee would not be alive at this temperature) P3-7 (c) For Ants: T(in K) 283

1/T x103 3.53

V(cm/s) 0.5

ln(V) -0.69

293 303

3.41 3.30

2 3.4

0.69 1.22

311

3.21

6.5

1.87

Plotting ln (V) vs. 1/T, We get almost a straight line. So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same. Insect Cricket Firefly Ant Honeybee

Activation Energy 52150 54800 95570 141800

P3-7 (d) There is a limit to temperature for which data for any one of the insects can be extrapolate. Data which would be helpful is the maximum and the minimum temperature that these insects can endure before death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless. P3-7 (e) 1) The rate at which the beetle can push a ball of dung is directly proportional to its rate constant, therefore –rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant −𝐸𝑎

𝑘 = 𝐴𝑒𝑥𝑝 ( 𝑅𝑇 ) From the data given -rA 6.5 13 18

T(K) 300 310 313

1/T 0.003333 0.003226 0.003195

ln k 1.871802 2.564949 2.890372

3-10

Refer to P3-8 (similar procedure) Therefore, A = 1.299X1011 E = 59195.68 J/mol k = 1.299X1011 exp(-7120/T) Now at T = 41.5 C = 314.5 K k = 19.12 cm/s Therefore, beetle can push dung at 19.12 cm/s at 41.5 C P3-7 (e) 2) Individualized solution

P3-8 (a) There are two competing effects that bring about the maximum in the corrosion rate: Temperature and HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the column and consequently the rate of corrosion should increase. However, the HCN concentrations (and the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion rate somewhere around the middle of the column. P3-8 (b) The component ‘Corrosion’ can be applied to this problem

P3-9 Antidote did not dissolve from glass at low temperatures.

P3-10 (a) If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1. Then with k = Ae-E/RT in general,

and

3-11

, or

Therefore:

which can be approximated by

. Consequently, for this doubling rate rule of thumb to be

valid, the temperature at which the doubling will take place must be related to the activation energy by this relationship. P3-10 (b) Individualized solution

P3-11 From the given data -rA(dm3/mol.s)

T(K)

1/T (𝐾 −1 ) 0.003333

ln (k) -8.00637

0.002

300

k= -rA/ (4*1.5) 0.00033333

0.046

320

0.00766667

0.003125

-4.87087

0.72

340

0.12

0.002941

-2.12026

8.33

360

1.38833333

0.002778

0.328104

Plotting ln(k) vs (1/T), we have a straight line:

𝐸𝑎

Since,𝑙𝑛𝑘 = 𝑙𝑛𝐴 − 𝑅𝑇 therefore ln A = 41.99 and E/R = 14999 P3-11 (a) Activation energy (E),

 E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol 3-12

P3-11 (b) Frequency Factor (A), ln A = 41.99 𝑑𝑚6 1

⇒ 𝐴 = 1.72 ∗ 1018 𝑚𝑜𝑙2 . 𝑠 P3-11 (c)

𝑘 = 1.72 ∗ 1018 𝑒𝑥𝑝 (

−14999 𝑇

)

(1)

Given T0 = 300K 𝑑𝑚6 1 Therefore, putting T = 300K in (1), we get 𝑘(𝑇0 ) = 3.33 ∗ 10−4 𝑚𝑜𝑙2 . 𝑠 Hence,

𝐸 1 1 𝑑𝑚6 1 𝑘 = 𝑘(𝑇0 )𝑒𝑥𝑝 ( ( − )) . 𝑅 300 𝑇 𝑚𝑜𝑙 2 𝑠

P3-12 (a) –ra=kCACB P3-12 (b) –ra=kPAPB P3-12 (c) –ra=kCA P3-12 (d) –ra=k

P3-13 (a) C2H6 → C2H4 + H2

Rate law: -rA =

P3-13 (b) C2H4 + 1/2O2 → C2H4O

Rate law: -rA =

, k = [ s-1 ] , k = [ dm1.5/s.mol0.5 ]

P3-13 (c) (CH3)3COOC(CH3)3 ↔ C2H6 + 2CH3COCH3 A

↔ B

Rate law: -rA = k[CA – CBCC2/KC] , k = [ s-1 ] P3-13 (d) n-C4H10 ↔ I- C4H10

Rate law: -rA = k[

–

/Kc], k = [ s-1 ]

P3-13 (e) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH A

↔

Rate law: -rA = k [CACB – CCCD/KC], k = [ dm3/mol.s ] P3-13 (f) 2CH3NH2 ↔ (CH3)2NH + NH3

Rate law: -r’A = kP2CH3NH2, k = [ mol/kg cat.s. Pa2]

P3-13 (g) (CH3CO)2O + H2O → 2CH3COOH A+

B →

Rate law: -rA = kCACB, k = [ dm3/mol.s ]

3-13

P3-14 (a) (1) –rA = kACACB2 (2) –rA= kACB (3) –rA= kA (4) –rA= kACA/CB P3-14 (b) (1) H2 + Br2 → 2HBr

(2) H2 + I2 → 2HI

P3-15 (a)

Rate law: -rHBr =

Rate law: -rA =

A→B A→B –rA = kCA

We need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (B → A) is rArev = k–1CB = then:

From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so:

Solving for KC gives: P3-15 (b)

If we assume the rate law for the reverse reaction 2D → A + 2B is r–A = k–1 CD then:

From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so: 3-14

Solving for KC gives:

P3-15 (c)

If we assume the rate law for the reverse reaction 𝐶 + 𝐷 → 𝐴 + 𝐵is

then:

From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so:

Solving for KC gives:

P3-16 Assuming the reactions to be elementary: 2 Anthracene →Dimer

Similarly, for the second reaction: Norbornadiene → Quadricyclane

where β is greater than 1.0 and is related to the intensity of the light.

3-15

Page intentionally blank

3-16

Synopsis for Chapter 4 – Stoichiometry General: This chapter adds our third building block, stoichiometry. The goal is to express concentrations in terms of either molar flow rates or conversion and then to write the rate laws in terms of these variables for both liquid phase and gas phase reactions. Questions Most often (O) I require the student to think through O Q4-3A, O Q4-4A and O Q4-5A. l Q4-1A (20 seconds) Questions Before Reading (QBR). Q4-2A (10-12 min) I usually assign i>Clicker questions. l Q4-6A (2 min) Queen is the answer. Q4-8A (15 min) As usual, if a student is having difficulty with stoichiometry the LearnChemE screencasts are encouraged. Q4-9A (12 min) Ties in safety to CRE. Computer Simulations and Experiments l P4-1A (5-10 min) Use Wolfram for part (a) and part (b). Part (vi) is always assigned for Example 4-5 that compared the equilibrium conversion for batch and flow systems. Interactive Computer Games (ICG) l P4-2A (15 min) This ICG is excellent preparation for the Student Chapter Jeopardy Competition at the Annual AIChE meeting. Problems O P4-3A (25 min) Straight forward problem on chemical equilibrium. AA P4-4B (20 min) Very easy problem on stoichiometry, but parts (c) and (d) may surprise you. l P4-5A (20-45 min) I always assign two or three parts (a) through (d). Gives practice at writing the rate law as a function of conversion and help reinforce the CRE algorithm. O P4-6A (30 min) this problem will be continued in an Example Problem in Chapter 13. AA P4-7B (30 min) Old Exam Question (OEQ). Similar to Problem P4-4B. l P4-8B (30 min) Old Exam Question (OEQ). Good problem! Students need to choose their basis of calculation carefully. AA P4-9B (45 min) Old Exam Question (OEQ). This problem is a good problem as it relates to an industrial process and an industrial reaction. AA P4-10B (30 min) Old Exam Question (OEQ). Good problem on equilibrium calculation. I usually assign one or two problem P4-10B to P4-12B on equilibrium conversion. AA P4-11B (15 min) Another good equilibrium calculation. AA P4-12B (10 min) Equilibrium conversion calculation. S4-1

AA P4-13C (50 min) Fairly challenging problem. Requires thinking, not just formula plugging.

S4-2

Solutions for Chapter 4 – Stoichiometry Useful Links: 1. Click on the link given below to download Wolfram/python codes for Ch-4 http://umich.edu/~elements/5e/04chap/live.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

Q4-1 Individualized solution

Q4-2 Individualized solution

Q4-3 Individualized solution

Q4-4 For part (a): CD=CA0*(X/3)=3*(0.2/3)=0.2 mol/dm3 CB=3*0.133=0.4 mol/dm3 CD and CB decreases when initial concentration is lower, and A becomes the limiting reactant.

Q4-5 Individualized solution

Q4-6 Individualized solution

Q4-7 Individualized solution

Q4-8 Individualized solution

Q4-9 Individualized solution

P4-1 (a) Example 4-4 (i) By increasing thetaB, the rate profile is affected only for the initial levels of conversion. Increase in thetaB makes the rate to fall down. This is because increase in thetaB causes the partial pressure of O2 to rise, hence decreasing the rate of SO2 oxidation. 4-1

(ii)

Kp has the least effect, and KSO3 has the greatest effect (can be seen by varying Wolfram parameters)

Increased the value of Kp to 30 times (from 64 to 1967) but still the shape of graph remains almost same

Increased the value of KSO3 to just 10 times and the see how the graph completely changed (iii)

1. By varying the sliders, one can say that Kp and epsilon have almost a negligible effect on the rate vs conversion profile. The greatest effect the rate profile is of K SO3. 2. By increasing thetaB, the rate profile is affected only for the initial levels of conversion. 3. Increasing PSO2 and k increases rate for a given conversion while opposite is the case for thetaB, KO2 and KSO3.

4-2

(iv) Polymath: Polymath code: d(X)/d(t) = 1 Fao = 3 PSO20 = 4.1 epsilon = -0.14 PSO2 = PSO20* (1-X)/(1+epsilon*X) PSO3 = PSO20* X/(1+epsilon* X) thetaB = 0.54 PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X) k = 9.7 KO2 = 38.5 KSO3 = 42.5 KP = 930 ra = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 yaxis = Fao/(-ra) X(0)=0 t(0) = 0 t(f) = 0.4

Polymath Output: POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

epsilon

-0.14

Fao

9.7

KO2

38.5

930.

KSO3

42.5

PO2

2.214

1.476695

2.214

1.476695

PSO2

4.1

2.605932

4.1

2.605932

PSO20

4.1

10 PSO3

1.737288

11 ra

-0.5651727

-0.0045241

12 t

0.4

13 thetaB

0.54

14 X

0.4

15 yaxis

5.308112

663.1111

Differential equations 1 d(X)/d(t) = 1 Explicit equations 1

Fao = 3

PSO20 = 4.1

epsilon = -0.14

PSO2 = PSO20* (1-X)/(1+epsilon*X)

PSO3 = PSO20* X/(1+epsilon* X)

thetaB = 0.54

PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)

k = 9.7

4-3

KO2 = 38.5

10 KSO3 = 42.5 11 KP = 930 12 ra = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 13 yaxis = Fao/(-ra)

At X = 0.4 FA0/–rA’ = 663.11 W = (FA0/–rA’)X at X = 0.4 Hence, W = 265.24 g (v) This program is to find Xe: Polymath code: f(X) = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 PSO20 = 4.1 epsilon = -0.14 PSO2 = PSO20* (1-X)/(1+epsilon*X) thetaB = 0.54 PSO3 = PSO20* X/(1+epsilon* X) k = 9.7 KO2 = 38.5 KSO3 = 42.5 KP = 930 PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X) X(min) = 0.5 X(max) = 1

Polymath Output: POLYMATH Report Nonlinear Equation

Calculated values of NLE variables Variable Value 1 X

f(x)

Initial Guess

0.997577 -1.251E-09 0.5 ( 0 < X < 1. )

Variable Value 1

epsilon

-0.14

9.7

KO2

38.5

930.

KSO3

42.5

PO2

0.1963961

PSO2

0.0115472

PSO20

4.1

PSO3

4.754015

10 thetaB

0.54

Nonlinear equations 1 f(X) = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 = 0

4-4

Explicit equations 1

PSO20 = 4.1

epsilon = -0.14

PSO2 = PSO20* (1-X)/(1+epsilon*X)

thetaB = 0.54

PSO3 = PSO20* X/(1+epsilon* X)

k = 9.7

KO2 = 38.5

KSO3 = 42.5

KP = 930

10 PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)

Hence, Xe = 0.9976 Now we will use Xe in the following Polymath Program Polymath code: d(X1)/d(t) = 1 d(X2)/d(t) = 0.99*Xe/0.3 Fao = 1000 PSO20 = 4.1 epsilon = -0.14 PSO21 = PSO20* (1-X1)/(1+epsilon*X1) PSO31 = PSO20* X1/(1+epsilon* X1) thetaB = 0.54 PO21 = PSO20* (thetaB-X1/2)/(1+epsilon* X1) k = 9.7 KO2 = 38.5 KSO3 = 42.5 KP = 930 ra1 = -k *(PSO21* Sqrt(PO21)-PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2 yaxis1 = Fao/(-ra1) Xe = 0.9976 PSO22 = PSO20* (1-X2)/(1+epsilon*X2) PSO32 = PSO20* X2/(1+epsilon* X2) PO22 = PSO20* (thetaB-X2/2)/(1+epsilon* X2) ra2 = -k *(PSO22* Sqrt(PO22)-PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2 yaxis2 = Fao/(-ra2) X1(0)=0 X2(0)=0 t(0) = 0 t(f) = 0.3

Polymath Output: Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

epsilon

-0.14

Fao

1000.

9.7

KO2

38.5

930.

KSO3

42.5

PO21

2.214

1.669102

2.214

1.669102

PO22

2.214

0.2197559

2.214

0.2197559

4-5

PSO20

4.1

10 PSO21

4.1

2.995825

4.1

2.995825

11 PSO22

4.1

0.0588832

4.1

0.0588832

12 PSO31

1.283925

13 PSO32

4.698973

14 ra1

-0.5651727

-0.0092831

15 ra2

-0.5651727

-5.276E-06

16 t

0.3

17 thetaB

0.54

18 X1

0.3

19 X2

0.987624

20 Xe

0.9976

21 yaxis1

1769.371

1.077E+05

22 yaxis2

1769.371

1.895E+08

Differential equations 1 d(X1)/d(t) = 1 2 d(X2)/d(t) = 0.99*Xe/0.3 Explicit equations 1

Fao = 1000

PSO20 = 4.1

epsilon = -0.14

PSO21 = PSO20* (1-X1)/(1+epsilon*X1)

PSO31 = PSO20* X1/(1+epsilon* X1)

thetaB = 0.54

PO21 = PSO20* (thetaB-X1/2)/(1+epsilon* X1)

k = 9.7

KO2 = 38.5

10 KSO3 = 42.5 11 KP = 930 12 ra1 = -k *(PSO21* Sqrt(PO21)-PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2 13 yaxis1 = Fao/(-ra1) 14 Xe = 0.9976 15 PSO22 = PSO20* (1-X2)/(1+epsilon*X2) 16 PSO32 = PSO20* X2/(1+epsilon* X2) 17 PO22 = PSO20* (thetaB-X2/2)/(1+epsilon* X2) 18 ra2 = -k *(PSO22* Sqrt(PO22)-PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2 19 yaxis2 = Fao/(-ra2)

At X = 0.3 FA0/–rA’ = 107,700 W = (FA0/–rA’)X at X = 0.3 Hence, W = 32,310 g = 32.31 kg

4-6

At X = 0.99Xe FA0/–rA’ = 1.895 x 108 W = (FA0/–rA’)X at X = 0.99Xe Hence, W = 1.872 x 108 g = 1.872 x 105 kg P4-1 (b) Example 4-5 (i) Area under the figure E4-5.1 multiplied by the mole flow rate will be the volume of PFR required. (ii) Area under the figure E4-5.1 multiplied by the initial concentration will be the time required for the reaction to occur in a batch reactor. (iii) Kc = 0.4 and CA0 = 0.09 (Red and blue curves are farthest apart)

(iv)

When Kc is minimum and CA0 is maximum (can be seen from below graph). Both red and blue curves merge together.

4-7

(v)

(v-vi) 1) We can see that when Kc is set to minimum and CT0 at maximum, the line becomes parallel to X-axis taking value equal to 1 for all values of mole fraction. Under these conditions both of the conversions obtained take their minimum values. 2) Another case when Kc is at its maximum and CT0 at minimum, the ratio of conversion is still equal to 1 but this time both of the conversions are individually equal to 1 for almost all values of mole fraction. 3) The above results can easily be justified as when you have a high initial concentration but a low rate constant, no matter which reactor we choose, conversion would be very low but in the other extreme case the conversion is very high irrespective of the reactor we choose.

P4-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual.

P4-3 2A  B CA0 = 4 mol/dm3 Xe = 0.6 P4-3 (a) Kc = CB/ CA2 ε = –1/2 CA = CA0(1–X)/(1+ εX) = 4*0.4/0.7 = 2.286 mol/dm3 CB = 4*0.3/0.7 = 1.714 mol/dm3 Therefore Kc = 0.328 dm3/mol P4-3 (b) For a liquid phase reaction, CA = 1.6 mol/dm3 CB = 1.2 mol/dm3 Thus, Kc = 0.469 dm3/mol 4-8

P4-3 (c) CA = CA0(1–X)/(1+ εX) CB = CA0(0.5*X)/(1+ εX) –rA = k(CA^2– CB/Kc) –rA = k((CA0(1–X)/(1+εX)) ^2 – (CA0(0.5*X)/(1+ εX) )/Kc) –rA = 2((2.0(1–X)/(1– 0.5X)) ^2 – (2.0(0.5X)/(1–0.5X) )/0.5) P4-3 (d) For a constant volume reaction, ε = 0 Therefore, –rA = k((CA0(1–X)) ^2 – CA0(0.5*X)/Kc)

P4-4

1 1 B→ C 2 2 1  1  1 1   = y A0 =    − − 1 = − 2  2  2 2  A+

P4-4 (a)

C B0 = C A0 = 0.1 P4-4 (b)

C A = C A0



mol dm 3

(1− X) = 0.1 (1− 0.25) = (0.1)(0.75) = 0.086 mol ( ) 1  0.875 dm 3 (1+ X) 1− X  

2 

X  0.25      ( 0.1)( 0.125) = 0.0143 mol 2 2  CC = C A0   = ( 0.1)  = 0.875 dm3  1  (1 +  X ) 1 − X    2 



P4-4 (c)

 1   1  1 − X  C A 0 1 − X  2   2  = C = 0.1 mol C B = C A0 =  = A0 (1 + X ) dm 3  1  1 − X   2  P4-4 (d) C B = 0.1



mol dm 3

P4-4 (e) rA rC = , rA = −2rC = −4 mol dm 3 min −4 2 4-9



P4-4 (f) 2A + B  C ε = δyA0 = –0.5 CA0 = 2 mol/dm3 Irreversible rate of reaction

–rA = kCA2CB 3 −𝑟𝐴 = 𝑘𝐶𝐴0 (

1−𝑋 2 ) 1 − 0.5𝑋

P4-4 (g) –rA = 7.11 mol/dm3s

P4-5 (a) Liquid phase reaction, O

CH2–OH

CH2 – CH2 + H2O → CH2–OH A + B → C CA0 = 16.13mol/dm3 Stoichiometric Table: Species Ethylene oxide Water Glycol

CB0 = 55.5 mol/dm3 Symbol A B C

Initial CA0=16.13 mol/dm3

Change –CA0X

CB0= 55.5 mol/dm3,  B = 3.441

–CA0X

Remaining CA= CA0(1–X) = (1–X) mol/dm3 CB = CA0(  B –X)

CA0X

=(3.441–X) mol/dm3 CC = CA0X mol/dm3

Rate law:

–rA = kCACB

Therefore,

2 –rA = k C AO (1–X) (  B –X) = k (16.13)2(1–X) (3.441–X)

At 300K

 CSTR =

E = 12500 cal/mol, k = 0.1dm3/mol.s

X = 0.9,

(16.13)( 0.9 ) C AO X = 2.196sec = 2 −rA ( 0.1)(16.13) (1 − 0.9 )( 3.441 − 0.9 )

and, V = τ *vo = 2.196 sec X 200 liters/sec = 439.2 liters At 350K, k2 = k exp((E/R)(1/T–1/T2))= 0.1exp((12500/1.987)(1/300–1/350)) = 1.99 dm3/mol.s Therefore,

 CSTR =

(16.13)( 0.9 ) C AO X = 0.110sec , = 2 −rA (19.99 )(16.13) (1 − 0.9 )( 3.441 − 0.9 )

and, V = τ* vo = 0.110 X 200 liters = 22 liters 4-10

P4-5 (b) Isothermal, isobaric gas-phase pyrolysis, C2H6 C2H4 + H2 A → B + C Stoichiometric table: Species symbol Entering C2H6 A FA0 C2H4 B 0 H2 C 0 FTO=FA0

Change –FA0X +FA0X +FA0X

Leaving FA=FA0(1–X) FB=FA0X FC=FA0X FT=FA0(1+X)

 = yao  = 1(1+1–1) = 1 v = vo(1+  X) => v = vo(1+X) P CA0 = yA0 CTO = yA0 RT =

(1)(6atm )

 m3 atm  0.082   (1100 K ) K .kmol  

= 0.067 kmol/m3 = 0.067 mol/dm3

CA =

(1 − X ) mol/dm3 FA FAO (1 − X ) = = C AO v vO (1 + X ) (1 + X )

CB =

F (X ) FB X = AO = C AO mol/dm3 v vO (1 + X ) 1 + X ( )

CC =

FC F (X ) X = AO = C AO mol/dm3 v vO (1 + X ) (1 + X )

Rate law: –rA = kCA= kCA0

(1 − X ) =0.067 k (1 − X ) (1 + X ) (1 + X )

If the reaction is carried out in a constant volume batch reactor, =>(  = 0) CA = CA0(1–X) mol/dm3 CB = CA0 X mol/dm3 CC = CA0 X mol/dm3 P4-5 (c) Isothermal, isobaric, catalytic gas phase oxidation,

1 O2 → C2H4O 2 1 + B → C 2

C2H4 + A

–rA’ = kApApB0.5 pA = yA0P = yA0CRT = CART pB = yB0P = yB0CRT = CBRT Therefore, rA’ = kACACB0.5(RT)1.5 4-11

(1) For a fluidized batch reactor, V is constant CA = CA0(1–X) CB = CA0(θB – b/a X) θB = 0.5 (stoichiometric feed) b/a = 0.5 Therefore, CB = CA0(1–X)/2 Therefore, –rA’ = kACA01.5(1–X)1.5(RT)1.5/20.5 (2) Stoichiometric table: Species Symbol Entering Change Leaving C2H4 A FA0 –FA0X FA=FA0(1–X) O2 B FB0 1 FB=FA0(  B –X/2) – FA0X

C2H4O

1 FBO 2 FAO 1 B = = = FAO FAO 2

y AO =

+ FA0X

FC=FA0X

FAO FAO 2 = = FTO FAO + FBO 3

(6atm ) P 2 mol = = 0.092 3 3 RT 3  dm atm.dm   0.082  (533K ) mol.K   F (1 − X ) C AO (1 − X ) 0.092 (1 − X ) F C A = A = AO = = v vO (1 +  X ) (1 − 0.33 X ) (1 − 0.33 X )

C AO = y AO CTO = y AO

X  FAO   B −  F 2  0.046 (1 − X )  CB = B = = v vO (1 +  X ) (1 − 0.33 X ) CC =

0.092 ( X ) FC FAO X = = v vO (1 +  X ) (1 − 0.33 X )

If the reaction follow elementary rate law Rate law: –rA’ = kACACB0.5(RT)1.5, T = 533 K

 0.092 (1 − X )  0.046 (1 − X )  −rA = k     (1 − 0.33 X )   (1 − 0.33 X ) 

P4-5 (d) Isothermal, isobaric, catalytic gas-phase reaction in a PBR C6H6 + 2H2 → C6H10 A + 2B → C Given:

 0 = 50 dm 3 / min 4-12

0.5

(RT)1.5

Stoichiometric Table: Species Symbol C6H6 A H2 B C6H10 C

Entering FA0 FB0=2FA0 0

FB 0 2 FA0 = =2 FA 0 FA 0

FA 0 FA 0 1 = = FT 0 FA0 + FB 0 3

B =

C A 0 = CT 0 y A 0 = CA =

CB =

CC =

 



y A0 =

Change –FA0X –2FA0X FA0X

Leaving FA=FA0(1–X) FB=FA0(θB–2X) Fc=FA0X

1 2  = y A0 = (1 − 2 − 1) = − 3 3

P 1 6atm mol 1 = 0 . 055  =   3 RT  3  0.0821 dmmolatm 3 dm 3  K * ( 443K )  

FA0 (1 − X ) (1 − X ) = C A0  0 (1 + X ) (1 − 23 X )

FA0 ( B − 2 X ) (2 − 2 X ) (1 − X ) = C A0 = 2 * C A 0  0 (1 + X ) (1 − 23 X ) (1 − 23 X )

FA 0 X X = C A0  0 (1 + X ) (1 − 23 X )

Rate Law: NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or concentration units are used. See below for an example.

− rA ' = kPA PB

mol mol = * atm * atm 2 3 kgcat  min kgcat  min atm Notice that if you use concentrations in this rate law, the units will not work out.

PA = y A0 P = ( y A0 C ) * RT = C A RT − rA = kPA PB = kC A C B ( RT ) 3 = 4kC A0 2

(1 − X ) 3 ( RT ) 3 3 2 (1 − 3 X )

Design Equation for a fluidized CSTR:

W= W=

FA0 X − rA '

FA0 X (1 − 23 X ) 3 4kC A0 (1 − X ) 3 ( RT ) 3 3

 0 X (1 − 23 X ) 3 4kC A0 (1 − X ) 3 ( RT ) 3 2

Evaluating the constants:

k = 53

mol at 300 K kgcat  min atm 3 4-13

At 170°C (443K),

J   80000   EA  1   1 mol mol  1 − 1  = 1663000  = 53 exp  k 443 = k 300 exp   −  J kgcat  min atm  300 K 443K   R  T300 T443   8.314 mol  K   Plugging in all the constants into the design equation: X = 0.8 3 2 50 dm min  0.8  (1 − 3 0.8) W= = 5.25  10 −7 kgcat 3 3 mol mol 2 dm3 atm 4  1663000 kgatmin ( 0 . 055 ) ( 1 − 0 . 8 ) ( 0 . 0821  443 K ) mol  K atm3 dm3 3

At 270°C (543K),

J   80000   EA  1   1 mol mol  1 − 1  = 90790000  = 53 exp  k 543 = k 300 exp   −  J kgcat  min atm  300 K 543K   R  T300 T543   8.314 mol  K   Plugging in all the constants into the design equation: X = 0.8 3 2 50 dm min  0.8  (1 − 3 0.8) W= = 5.22 10 −9 kgcat 3 3 mol mol 2 dm 3 atm 4  9079000 kgatmin atm3 (0.055 dm3 ) (1 − 0.8) (0.0821 molK  543K ) 3

P4-6 (a) Let A = ONCB B = NH3

C = Nibroanaline D = Ammonium Chloride

A + 2B ⎯⎯ → C+D -rA = kC ACB P4-6 (b) Species A B C D P4-6 (c) For batch system, CA=NA/V

Entering FA0 FB0 = ΘBFA0 =6.6/1.8 FA0 0 0

–rA = kNANB/V2

4-14

Change –FA0X –2 FA0X FA0X FA0X

Leaving FA0(1–X) FB= FA0(ΘB – 2X) FC=FA0X FD=FA0X

P4-6 (d)

-rA = kC ACB FA =

N A N A N A0 F F = = (1 − X ) = C A0 (1 − X ), C A = A = A = C A0 (1 − X ) V V0 V0 v v0

FB =

N B N B N A0 F = = ( B − 2 X ) = C A0 ( B − 2 X ), CB = B = C A0 ( B − 2 X ) V V0 V0 v0

−rA = kC A2 0 (1 − X )( B − 2 X )

B =

CB 0 6.6 = = 3.67 C A0 1.8

C A0 = 1.8

kmol m3

−rA = k (1.8 ) (1 − X )(3.67 − 2 X ) 2

P4-6 (e) 1) At X = 0 and T = 188°C = 461 K 2

− rA0 = kC A2 0  B = 0.0017 −rA0 = 0.0202

m3 kmol  kmol  1.8 3  3.67 = 0.0202 3 kmol min  m  m min

kmol m3 min

2) At X = 0 and T = 250C = 298K

 E  1 1  k = k O exp  −    R  TO T   cal   11273   m mol  1 − 1   k = 0.0017 exp   cal  461 298   kmol. min  1 . 987   mol.K   m3 = 2.03  10 −6 kmol. min 3

m3 kmol   kmol   −rA = 2.03 10 1.8 3   3.67 3  kmol min  m   m  −6

–rA0 = kCA0aCB0 = 2.41 X 10–5 kmol/m3min 3)

 E  1 1  k = k0 exp   −    R  T0 T  

4-15

  cal 11273     m 1 1 mol k = 0.0017 exp  −    kmol min 1.987 cal  461 K 561 K     mol K 3

k = 0.0152

m3 kmol min

−rA0 = kC A0 2CB 0 2

m3 kmol   kmol   −rA = 0.0152 1.8 3   3.67 3  kmol min  m   m  −rA = 0.1807

kmol m3 min

P4-6 (f) rA = kCA02(1–X)(θB–2X) At X = 0.90 and T = 188C = 461K 1) at T = 188 C = 461 K

  kmol  m3 1.8 3  (1 − 0.9 )(3.67 − 2(0.9 )) − rA =  0.0017 kmol. min  m   kmol = 0.00103 3 m min 2

2) at X = 0.90 and T = 25C = 298K

  kmol  m3 −6 1.8 3  (1 − 0.9 )(3.67 − 2(0.9 )) − rA =  2.03  10 kmol. min  m   kmol = 1.23  10 −6 3 m min 2

3) at X = 0.90 and T = 288C = 561K

  kmol  m3 1.8 3  (1 − 0.9 )(3.67 − 2(0.9 )) − rA =  0.0152 kmol. min  m   kmol = 0.0092 3 m min 2

P4-6 (g) FA0 = 2 mol/min 1) For CSTR at 25oC –rA = 1.23  10

V = =

FAO (1 − X ) − rA X =0.9

−6

kmol m 3 min

2mol / min 0.1 = 162.60m 3 mol 1.23  10 −3 3 m min 4-16

2) At 288oC, –rA = 0.0092

V = =

FAO (1 − X ) − rA X =0.9

kmol m 3 min

2mol / min 0.1 = 21.739m 3 mol 0.0092 3 m min

P4-7 (a) It is given that the feed is equimolar, so CB0 = 2 mol/dm3 P4-7 (b) For every mole of B, two moles of A are required. Therefore, A is the limiting reagent. P4-7 (c) From the stoichiometry of the reaction, we can write (NA0 – NA)/2 = (NB0 – NB) As NA0 = NB0 and conversion of A = 0.25 we get, NA = (3/4) NA0 NB = (7/8) NA0 NC = (1/8) NA0 Let the initial volume be V0 , CA0 = NA0 /V0 As the reaction is in gas phase, volume of the reaction will change which will affect the concentrations. NT= (7/4) NA0 V= (7/4) V0 Therefore, CB= NB /V = (½) CA0 = 1 mol/dm3 P4-7 (d) For the given reaction, 1

𝛿 = 2 − 2 − 1 = −1 𝑦𝐴0 = 0.5 𝜀 = 𝑦𝐴0 ∗ 𝛿 = −0.5 Therefore, concentrations of each species is given by, CA = CA0 (1–X)/(1+ εX) CB = CA0 (1–X/2)/(1+ εX) CC = CA0 X/(2(1+ εX)) Rate expression is given by, –rA = kf CA 2 CB – kb CC –rA = kf (CA 2 CB – CC /Kc) Substituting the equations derived for concentrations, –rA = kf CA0 ((CA0 (1–X)/(1+ εX))2 (1–X/2)/(1+ εX) – X/(2(1+ εX))/Kc) – (1) Substitute the given values to obtain the final expression for rate. P4-7 (e) equilibrium conversion is given by, rA = 0 (CA0 (1–X)/(1+ εX))2 (1–X/2)/(1+ εX) – X/(2(1+ εX))/Kc = 0 2 Kc CA02 (1 –X3/2+2X2 –5X/2) = X (1+ 2εX+ε2X2) CA0 = 2.0 (units) ε = –1/2 Kc= 0.5 4-17

Substituting gives a cubic equation, –2.25 X3 +9 X2 –11 X +4 =0 Whose roots are given by, X = 2, 0.67 and 1.33 Clearly, the only practical root is Xe = 0.67 which is the equilibrium conversion. P4-7 (f) (1) Substitute X = 0 in (1) –rA=2*2(2*2*(1)*(1)–0)= 16 mol/dm3.s (2) Substitute X= 0.5 in (1) to get –rA =4.45 mol/dm3.s (3) Substitute X=0.99*0.67 =0.663 in (1) to get –rA =0.099 mol/dm3.s

P4-8 (a) Isothermal gas phase reaction.

1 3 N 2 + H 2 → NH 3 2 2 Making H2 as the basis of calculation:

1 2 H 2 + N 2 → NH 3 3 3 1 2 A+ B → C 3 3 Stoichiometric table: Species Symbol Initial change H2 A FA0 –FA0X N2 B FB0=  B FA0 –FA0X/3 NH3

+2FA0X/3 FC=(2/3)FA0X

P4-8 (b)

2 1 

 

Leaving FA=FA0(1–X) FB=FA0(  B –X/3)

 =  − − 1 = − 3 3 3 1  2   16.4 ( atm ) C AO = 0.5 = 0.2 mol/dm3  atm.dm3   0.082  (500 K ) mol.K  

 = y AO = 0.5   −  = − 3 3

C AO (1 − X ) 0.2 (1 − 0.6 ) = = 0.1mol / dm3 0.6   (1 +  X ) 1 −  3   2 C ( X ) 2 0.2 ( 0.6 ) CNH3 = CC =  AO =  = 0.1mol / dm3 3 (1 +  X ) 3  0.6  1 −  3  

CH 2 = C A =

4-18

P4-8 (c) kN2 = 40 dm3/mol.s (1) For Flow system: 3

−rN2 = k N2 CN2  2 CH 2  2 1

 X   2   2 1 −   2 3    (1 − X )     = 40 (C AO )    1 − X    1 − X        3     3   3

 2  (1− X )   −rN 2 = 1.6 1− X    3  

(2) For batch system, constant volume. 

P4-9 (a)

2A + 9B

2C + 4D + 4E

A+ B

C + 2D + 2E

We are going to rework the class problem for the case of 3.5% naphthalene and 96.7% air and for a pressure of 10 atm and a temperature of 500 K. A feed under these conditions’ naphthalene is the limiting reactant.

4-19

Stoichiometric Table Species

Symbol

Change

Out

Naphthalene

FA0

–FA0X

FA = FA0 (1 – X)

Oxygen

FB0 = B FA0

–

9 2

FA0X

FB = FA0 (B –

9 2

Phthalic Anhydride

FC = 0

+FA0X

FC = FA0 X

Carbon Dioxide

FD = 0

+2 FA0X

FD = 2FA0 X

Water

FE = 0

+2 FA0X

FE = 2FA0 X

Nitrogen

FI = I FA0

–––

FI = I FA0 FT = FT0 +  FA0X

IT0

 = [1 + 2 + 2 – P4-9 (b) Species

9 2

– 1] = –

Symbol

Change

Out

NA0

–NA0X

NA = NA0 (1 – X)

NB0 = B NA0

Naphthalene Oxygen

–

9 2

NA0X

NB = NA0 (B –

9 2

Phthalic Anhydride

NC = 0

+NA0X

NC = NA0 X

Carbon Dioxide

ND = 0

+2 NA0X

ND = 2NA0 X

Water

NE = 0

+2 NA0X

NE = 2NA0 X

Nitrogen

NI = I NA0

–––

NI = I NA0

IT0

For isothermal constant volume batch 𝑁

𝑁

∗𝐹

(0.965)∗(0.21)

𝑉 = 𝑉0, 𝐶𝐴 = 𝑉𝐴, 𝐶𝐵 = 𝑉𝐵 , Isothermal T = T0 Now, 𝐹

𝑦

𝐴0

𝐹

𝑦 ∗𝐹

𝐴0

𝑇0

𝜃𝐼 = 𝐹 𝐼0 = 𝑦 𝐼0 ∗𝐹𝑇0 =

0.035 (0.965)∗(0.79)

𝑇0

0.035

=5.79

=21.78

For constant Volume, (1) Concentration

9 2

𝐶𝐵 = 𝐶𝐴0 (𝜃𝐵 − 𝑋) 𝐶𝐷 = 𝐶𝐴0 (0 + 2𝑋) (2) Total pressure 𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵 + 𝑃𝐶 + 𝑃𝐷 + 𝑃𝐸 + 𝑃𝐼

4-20

1 2

NT = NT0 +  NA0X  = [1 + 2 + 2 –

𝜃𝐵 = 𝐹𝐵0 = 𝑦𝐵0 ∗𝐹𝑇0 =

9 2

– 1] = –

1 2

Where 𝑃𝐴 = 𝐶𝐴0 (1 − 𝑋) ∗ 𝑅𝑇

9 2

𝑃𝐵 = 𝐶𝐴0 (𝜃𝐵 − 𝑋) ∗ 𝑅𝑇 𝑃𝐶 = 𝐶𝐴0 (0 + 𝑋) ∗ 𝑅𝑇 𝑃𝐷 = 𝐶𝐴0 (0 + 2𝑋) ∗ 𝑅𝑇 𝑃𝐸 = 𝐶𝐴0 (0 + 2𝑋) ∗ 𝑅𝑇 𝑃𝐼 = 𝐶𝐴0 (𝜃𝐼 ) ∗ 𝑅𝑇 Where T=500 K and R =0.082 𝑃𝑜 (0.035)(10 𝑎𝑡𝑚) 𝐶𝐴0 = 𝑦𝐴0 = = 0.0085 𝑚𝑜𝑙/𝑑𝑚3 𝑑𝑚3 𝐾 𝑅𝑇 (0.082 ) (500 𝐾) 𝑎𝑡𝑚. 𝑚𝑜𝑙 𝑃𝑇 = 𝐶𝐴0 (1 + 𝜃𝐵 + 𝜃𝐼 − 0.5 ∗ 𝑋)𝑅𝑇=0.0085*(1+5.79+21.78–0.5X)*0.082*500 𝑃𝑇 =0.36(28.57–0.5*X) (3) Rate law −𝑟𝐴 = 𝑘𝐴 𝐶𝐴2 𝐶𝐵 −𝑟𝐴 = 𝑘𝐴 ∗ (𝐶𝐴0 ∗ (1 − 𝑋))2 ∗ 𝐶𝐴0 ∗ (𝜃𝐵 − 9⁄2 𝑋) 9 3 (1 −𝑟𝐴 = 𝑘𝐴 𝐶𝐴0 − 𝑋)2 (5.79 − 𝑋) 2 (4) Time to achieve 90 % Conversion 0.9 𝑑𝑋

t=(NA0/V)*∫0

−𝑟𝐴 0.9 1 𝑑𝑋 57.5 t=𝐶 2 ∗𝑘 ∫0 9 == 2 ∗0.01= 921 days 2 (0.0085) (1−𝑋) ∗(5.79− 𝑋) 𝐴0 𝐴 2

P4-9 (c) Rate Law: −𝑟𝐴 = 𝑘𝐴 𝐶𝐴2 𝐶𝐵 Gas Phase Flow System:

𝑣 = 𝑣0 (1 + 𝜀𝑋)

𝑃0 𝑇 𝑃 𝑇0

𝐹𝐴 𝐹𝐴0 (1 − 𝑋) 𝑃0 𝑇 = 𝑣 𝑣0 (1 + 𝜀𝑋) 𝑃 𝑇0 (1 − 𝑋) 𝑃0 𝑇 𝐶𝐴 = 𝐶𝐴0 (1 + 𝜀𝑋) 𝑃 𝑇0 𝑃𝑜 (0.035)(10 𝑎𝑡𝑚) 𝐶𝐴0 = 𝑦𝐴0 = = 0.0085 𝑚𝑜𝑙/𝑑𝑚3 𝑑𝑚3 𝐾 𝑅𝑇 (0.082 ) (500 𝐾) 𝑎𝑡𝑚. 𝑚𝑜𝑙 𝐶𝐴 =

9 2 𝑋) 𝑃0 𝑇 𝐶𝐵 = 𝐶𝐴0 (1 + 𝜀𝑋) 𝑃 𝑇0 (𝜃𝐵 −

𝜖 = 𝑦𝐴0 ∗ 𝛿= (0.035)*(–0.5)=–0.0175 9

(5.79− 𝑋)

2 𝐶𝐵 = 0.0085 (1−0.0175 𝑋)

For isothermal flow system with no pressure drop 9 3 (1 𝑘𝐴 𝐶𝐴0 − 𝑋)2 (5.79 − 2 𝑋) −𝑟𝐴 = (1 − 0.0175𝑋)3 4-21

P4-9 (d) For CSTR, 𝐹

∗𝑋

𝐴0 V= −𝑟

𝐴

At X=0.98, –rA=3.6e–12 V=10*0.98/(3.6e–12)=2.72*1012 dm3

For PFR, 0.98 𝑑𝑋

V= (FA0)*∫0

−𝑟𝐴

𝐹𝐴0

0.98 (1−0.0175∗𝑋)3 𝑑𝑋 10∗29.76 V= *∫ = 9 0.01∗(0.0085)^3 𝑘∗CA0 3 0 (1−𝑋)2 ∗(5.79− 𝑋) 2

V=4.85*1010 dm3 P4-9 (e) Individualized solution

P4-10 P4-10 (a) P0 = PA0 = CA0RT PA = PA0(1–Xe)/(1 + ɛXe) PB = PA0Xe/2/(1 + ɛXe) CA = PA / RT CB = PB / RT Kc = CB/CA2 CA0 = 4 mol/dm3 Xe = 0.6 1

ɛ =𝑦𝐴0 ∗ 𝛿 = 1 ∗ (2 − 1) = −1/2 Substituting the values of CA0, Xe and ɛ, we get CA = 4*(1–0.6)/(1–0.5*0.6) =2.28 mol/dm3 CB = 4*(0.6/2)/(1–0.5*0.6)=1.71 mol/dm3 Kc = CB/CA2 = 0.329 dm3/mol P4-10 (b) Kc = CB/CA2 CA = CA0(1–Xe) CB = CA0Xe CA0 = 4 mol/dm3 Xe = 0.6 Substituting the values of CA0 and Xe, we get CA = 1.6 mol/dm3 CB = 1.2 mol/dm3 Kc = CB/CA2 = 0.46875 dm3/mol

P4-11 Rate law: –rA=kA*(CA–Cc^3/Kc)

4-22

P4-11 (a) Constant Volume: At equilibrium, –rA=0, X=Xe, therefore, 𝐶𝑐3 𝐾𝑐 = 𝐶𝐴 For constant volume, 𝐶𝑐 = 3 ∗ 𝐶𝐴0 ∗ 𝑋 𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋) 𝐶𝐴0 = 𝑃𝐴0 /(𝑅𝑇) =10/(0.082*400)=0.3 mol/dm3 Now, Kc is 2 27 ∗ 𝐶𝐴0 ∗ 𝑋3 𝐾𝑐 = (1 − 𝑋) Using the value of Kc=0.25, CA0=0.3, X= Xe= 0.4 P4-11 (b) Constant Pressure: 𝐶𝑐 = 3 ∗ 𝐶𝐴0 ∗ 𝑋/(1+𝜖𝑋) 𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)/(1+𝜖𝑋) 𝜖 = 𝑦𝐴0 ∗ 𝛿 𝑦𝐴0 = 1, 𝛿 = 3−1 = 2 Thus, 𝜖=2 Now, Kc is 2 27 ∗ 𝐶𝐴0 ∗ 𝑋3 𝐾𝑐 = (1 − 𝑋) ∗ (1 + 2𝑋)2 Solving for X, X=Xe= 0.585

P4-12 Rate law: –rA=kA*(CA^3–Cc/Kc) P4-12 (a) Constant Volume: At equilibrium, –rA=0, X=Xe, therefore, 𝐶𝑐 𝐾𝑐 = 3 𝐶𝐴

4-23

For constant volume, 1 𝐶𝑐 = ( ) ∗ 𝐶𝐴0 ∗ 𝑋 3 𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋) 𝐶𝐴0 = 𝑃𝐴0 /(𝑅𝑇) =10/(0.082*400)=0.3 mol/dm3 Now, Kc is 1 (3) ∗ 0.3 ∗ 𝑋 𝐾𝑐 = 0.33 (1 − 𝑋)3 Using the value of Kc=2.5, X= Xe= 0.27 So, equilibrium conversion is lower in P4-12B compared to P4-11B Constant Pressure: 𝐶𝑐 = (1/3) ∗ 𝐶𝐴0 ∗ 𝑋/(1+𝜖𝑋) 𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)/(1+𝜖𝑋) 𝜖 = 𝑦𝐴0 ∗ 𝛿 𝑦𝐴0 = 1, 1 𝛿 = − 1 = −2/3 3 Thus, 𝜖 = −2/3 Now, Kc is 1 (3) ∗ 0.3 ∗ 𝑋/(1 − 2/3𝑋) 𝐾𝑐 = 0.33 (1 − 𝑋)3 /(1 − 2/3𝑋)3 Solving for X, X=Xe= 0.136

P4-13

Given: Gas phase reaction A + B → 8C in a batch reactor fitted with a piston such that V = 0.1P0

( ft ) 3

k = 1.0

lb mol 2 sec

−rA = kC A2 CB NA0 = NB0 at t = 0 V0 = 0.15 ft3 T = 140°C = 600°R = Constant

4-24

P4-13 (a)

y A0 =

N A0 = 0.5 N A0 + N B 0

 = 8 −1−1 = 6  = y A0 = 3 V0 P0 (1 +  X ) and T = 1 , P0 = 10V0 , and P = 10V T  T0 P   T0 

Now V =

Therefore

10V02 (1 +  X ) 10V

V 2 = V02 (1 +  X )

N B = N A0  B − X 

N A = N A0 1 − X 

kN A2 N B kN A0 1 − X  − rA = kC CB = = 3 V3 V03 (1 +  X )2 Therefore 3

N B0 =1 N A0

y P  N A0 =  A0 0  V0  RT 

2 A

 y P  1 − X  −rA = k  A0 0  3  RT  (1 +  X )2 3

1 − X  lb mol −rA = 5.03*10 3 3 (1 + 3 X )2 ft sec 3

−9

P4-13 (b)

V 2 = V02 (1 +  X ) 0.22 = 0.152 (1 +  X ) X = 0.259 − rA = 8.63*10−10

B =

lb mol ft 3 sec

4-25

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4-26

Synopsis for Chapter 5 – Isothermal Reactor Design: Conversion General: This chapter combines all the building blocks to solve isothermal reaction engineering problems. Questions l Q5-1A (19 seconds) Questions Before Reading (QBR). I Q5-3A (10-60 min) Gives Guidelines and Ground Rules for formulating an original problem on Chemical Reaction Engineering. O Q5-4A (15-30 min) Takes 30 minutes if you cook the spaghetti and is an excellent motivational problem in which a one or more students have actually brought cooked spaghetti to their instructor. I Q5-9A (5 min) Based on the disparity in the numbers in that one would stop after making ethylene and sell that instead? O Q5-10A (5 min) This video is a fun and inventive video on the letters, “CSTR”. In the song the letters “YMCA” are replaced by “CSTR”. Computer Simulations and Experiments l P5-1B (a) – (f) (10-15 min per simulation) Uses Wolfram and ask the students to spend between 10 and 15 minutes max on the simulation and writing conclusions. Also, go to the Chapter Summary Notes in the left hand column in the Chapter 5 page of the Website there is a number of self-tests that will help the student get a feel on how they are doing. Interactive Computer Games (ICG) l P5-2B (25 min) World Famous Game. This game is based on a real murder case. However, in the game, the murderer and the victim are chosen randomly each time a student signs on. One of the most popular problems in the book. Problems l P5-3A (8 min) Multiple choice, short and sweet. Always assigned. l P5-4B (8 min) Multiple choice, short and sweet. Most always assigned. AA P5-5A (30 min) Excellent problem similar to a California Exam Problem in that it requires an intermediate calculation to find k. I P5-6C (40 min) Straight forward calculation and plot that can be obtained using Polymath. l P5-7B (25 min) Old Exam Question (OEQ). Excellent problem requiring an intermediate calculation similar to California Exam Questions. l P5-8B (25 min) Old Exam Question (OEQ). California Professional Engineers Registration Exam Problem. The problems are necessary to show the student the level of understanding they are expected to achieve if they want to become a registered professional engineer.

S5-1