Solution Manual LevenSpiel

Prévia do material em texto

By Octave Levenspiel
PREFACE 
This solution manual is result of hard work done by students of B. Tech. 
Chemical Engineering (Batch of 2k13), Nirma University. This manual 
provides solution to unsolved numerical from book “Chemical Reaction 
Engineering” by Octave Levenspiel. We hope this will help you in solving 
your queries regarding numericals. For any corrections or suggestions 
mail us at solvecre2k13@gmail.com 
Credit goes to….
Aashish Abhijeet Akshada
Aman Ankit Arpit
Rahul Bhuvan Brijesh
Samarpita Harsh Nirav
Harshit Jay Prashant
Yash Shardul Yagnesh
Krish Deep Manish
Chirag Maulik Mehul
Dhruv Jigar Parthvi
Darshan Prasann Shivang
Rohan Vedant Vishal
Huma Mihir Prerak
Jay Rochesh Sarthak
Fenil Siddharth Ravi
Sunil Ruchir Parth
Ravindra Viren Sunny
Yash Yuga Nirmal
Vikas Hardik
 Chapters Solved: 
 Chapter-2 
 Chapter-3 
 Chapter-4 
 Chapter-5 
 Chapter-7 
 Chapter-8 
 Chapter-9 
 Chapter-10 
 Chapter-11 
 Chapter-12 
 Chapter-13 
 Chapter-14 
 Chapter-18 
 Chapter-23 
 Chapter-24 
 Chapter-25 
 Chapter-26 
 
 
 
 
 
 
 
 
 
 
 
Problem 2.1: 
A reaction has the stoichiometric equation A + B = 2R. What is the order of reaction and rate 
expression? 
Solution 2.1: 
We require experimental kinetics data to answer this question, because order of reaction need 
not to match the stoichiometry, most oftenly it doesn’t. 
If this reaction is elementary, the order of reaction is 2. And rate expression is, 
− 𝑟𝐴 = 𝐾 𝐶𝐴 𝐶𝐵 
 
Problem 2.2: 
Given the reaction 2NO2 + 
𝟏
𝟐
 O2 = N2O5, what is the relation between the rates of formation and 
disappearance of the three reaction components? 
Solution 2.2: 
The relation between the rates of formation and disappearance of the three reaction 
components is, 
−𝑟𝑁𝑂2
2
= 
−𝑟𝑂2
1/2
=
−𝑟𝑁2𝑂5
1
 
 
Problem 2.3: 
A reaction with stoichiometric equation 
𝟏
𝟐
 A + B = R + 
𝟏
𝟐
 S has the following rate expression, 
− 𝑟𝐴 = 2 𝐶𝐴
0.5 𝐶𝐵 
What is the rate expression for this reaction if the stoichiometric equation is written as A + 2B = 
2R + S? 
Solution 2.3: 
It does not affect how we write the stoichiometry equation. The rate of reaction remains 
unchanged, 
− 𝑟𝐴 = 2 𝐶𝐴
0.5 𝐶𝐵 
 
Problem 2.4: 
For the enzyme-substance 𝐴
𝑒𝑛𝑧𝑦𝑚𝑒
→ 𝑅, the rate of disappearance of any substrate is given by 
− 𝑟𝐴 = 
1760[A][𝐸0]
6+𝐶𝐴
,
𝑚𝑜𝑙
𝑚3 𝑠
 
What are the units of the two constants? 
Solution 2.4: 
Assume 𝐶𝐴 is in mol/m
3 
Constant 6 has to be the same unit as of 𝐶𝐴, then and then only summation is possible. 
 6 mol/m3 
1760
𝑚𝑜𝑙
𝑚3

𝑚𝑜𝑙
𝑚3
𝑚𝑜𝑙
𝑚3
 = 
𝑚𝑜𝑙
𝑚3 𝑠
 
 1760 𝑠−1 
 
Problem 2.5: 
For the complex reaction with stoichiometry A + 3B  2R + S and with second order rate 
expression 
 rA = k1[A][B] 
are the reaction rates related as follows: rA = rB = rR? If the rates are not so related, then how are 
they related? Please account for the signs, + or . 
Solution 2.5: 
𝑟𝑅 = − 2𝑟𝐴 −
2
3
 𝑟𝐵 
 𝑜𝑟 
− 𝑟𝐴 = −
1
3
𝑟𝐵 = 
1
2
𝑟𝑅 
 
Problem 2.6: 
A certain reaction has a rate given by 
𝑟𝐴 = 0.005𝐶𝐴
2 , 
𝑚𝑜𝑙
𝑐𝑚3.𝑚𝑖𝑛
 
If the concentration is to be expressed in mol/liter and time in hours, what would be the 
value and unit of the rate constant? 
Solution 2.6: 
Given rate expression: 
 
𝑟𝐴 = 0.005𝐶𝐴
2 
𝑚𝑜𝑙
𝑐𝑚3. 𝑚𝑖𝑛
 
 
Change the unit 
−𝑟
𝐴=0.005𝐶𝐴 
2 𝑚𝑜𝑙
(10−3)𝑙𝑖𝑡∗
1
60ℎ𝑟
 
−𝑟 𝐴 = 300𝐶𝐴 
2 
𝑚𝑜𝑙
 𝑙𝑖𝑡.ℎ𝑟
 
 
Value of rate constant =300 
Unit of rate constant= 
𝑚𝑜𝑙
 𝑙𝑖𝑡.ℎ𝑟
 
 
Problem 2.7: 
For a gas reaction at 400 K the rate is reported as 
−
𝑑𝑝𝐴
𝑑𝑡
= 3.66𝑝𝐴
2 ,
𝑎𝑡𝑚
ℎ𝑟
 
(a) What are the unit of the rate constant? 
(b)What is the value of the rate constant for this reaction if the rate equation is expressed as 
−𝑟
𝐴 = 
−1
𝑉 
𝑑𝑁𝐴
𝑑𝑡 =𝑘𝐶𝐴
2, 
𝑚𝑜𝑙
𝑚3.𝑠
 
 
Solution 2.7: 
(a) 
Here the rate equation is expressed in homogeneous & order of the reaction is second ,so the 
unit of the rate constant K can be expressed by 
 K=(𝑡𝑖𝑚𝑒)−1 (𝑐𝑜𝑛𝑠)1−2 
 K=3.66(𝑡𝑖𝑚𝑒)−1 (𝑐𝑜𝑛𝑠)−1 
 (b) 
here 
−𝑟
𝐴 = 
−1
𝑉 
𝑑𝑁𝐴
𝑑𝑡 =𝑘𝐶𝐴
2, 
𝑚𝑜𝑙
𝑚3.𝑠
 
 
-𝑟𝐴=k𝐶𝐴
2 
So, now using ideal gas low 
PV=nRT 
𝑃𝐴 =
𝑛𝐴
𝑉
𝑅𝑇 = 𝐶𝐴𝑅𝑇  
−𝑑𝑃𝐴
𝑑𝑡
= 3.66𝑃𝐴
2 
 −
𝑑𝐶𝐴𝑅𝑇
𝑑𝑡
= 3.66(𝐶𝐴𝑅𝑇)
2  
𝑑𝐶𝐴
𝑑𝑡
= 3.66𝐶𝐴
2𝑅𝑇 
 −𝑟𝐴 = 3.66(0.0820) ∗ 400 ∗ 𝐶𝐴
2(
𝑚𝑜𝑙
𝑙𝑖𝑡. ℎ𝑟
) 
 K=120.13 (ℎ𝑟−1) (
𝑚𝑜𝑙
𝑙𝑖𝑡
)-1 
 
Problem 2.8: 
The decomposition of nitrous oxide is found to be proceed as follows 
 
𝑁2O—›𝑁2 + 
1
2
𝑂2 ; −𝑟𝑁2𝑂= 
𝑘1[𝑁2𝑂]²
1+𝑘2[𝑁2𝑂]
 
What is the order of reaction WRT 𝑁2O, and overall? 
 
 
 
Solution 2.8: 
If, 𝑘2 concentration is low, then order of reaction is 2
nd order. 
If Its concentration is high, the order of this reaction is 1st order. 
Overall order of reaction = 2-1 = 1. 
Therefore it is of 1st order. 
 
Problem 2.9: 
The pyrolysis of ethane proceeds with an activation energy of about 300 
𝐾𝐽
𝑚𝑜𝑙
 How much 
faster is the decomposition at 650°C than at 500°C? 
 
Solution 2.9: 
 
E=300 𝑘𝐽 𝑚𝑜𝑙.⁄ 
 
Let. Kinetic express by= -𝑟𝐴=k𝑐𝐴
𝑛 
where 𝑘0𝑒
−𝑒 𝑅𝑇⁄ 
-𝑟𝐴∞𝑘0𝑒
−𝑒 𝑅𝑇⁄ for same can be 
 
−𝑟𝐴1
−𝑟𝐴2
=
𝑒−𝑒 𝑅𝑇⁄
𝑒−𝑒 𝑅𝑇⁄
 = 𝑒
−𝑒
𝑅⁄ (
1
𝑝𝑖
−
1
𝑇2
) = 𝑒
−𝑒
𝑅 [
𝑇1−𝑇2
𝑇1𝑇2
] 
 
−𝑟
𝐴650
𝑟𝐴500
 = exp [
300
2.314
(
650−500
650×500
)] 
= 7.58 Time faster 
 
 
Problem 2.10: 
A 1100-K n-nonane thermally cracks (breaks down into smaller molecules) 20 times as rapidly 
as at 1000K. Find the activation energy for this decomposition. 
Solution 2.10: 
𝑟1100𝐾=20𝑟1000𝐾 
𝑟1100
𝑟1000
=20 
1 Arrhenius theory 
 
K=𝑘0𝑒
−𝑒 𝑅𝑇⁄ 
 
In same concentration 𝑘1=𝑘2 
𝑟1100
𝑟1000
 = 
𝑘1𝑒
−𝐸
𝑅𝑇1100
𝑘2𝑒
−𝐸
𝑅𝑇1000
 = 20 
Log20 = 
−𝐸
𝑅
(
1
1100
−
1
1000
) 
E=
𝑙𝑜𝑔20∗𝑅
9.09∗.00001
 
E=1189.96 
𝐽
𝑚𝑜𝑙
 
 
Problem 2.11: 
In the mid nineteenth century the entomologist henri fibre noted that French ants busily 
bustled about their business on hot days but rather sluggish on cool days. Checking this 
result with Orezon ants, I find 
 
Running 
speed (m/hr) 
150 160 230 295 370 
Temperature 
C 
13 16 22 24 28 
What activation represents this change in bustliness? 
 
Solution 2.11: 
Speed lnS T (
1
𝑇
*10−3) 
150 5.01 13 3.4965 
160 5.07 16 3.4602 
230 5.44 22 3.3898 
295 5.69 24 3.3670 
370 5.91 28 3.3220 
Slope of graph = −
𝐸
𝑅
=-5420 
E=(5420)*(8.314) 
 =45062 J/mol 
 
 
Problem 2.12: 
The maximum allowable temperature for a reactor is 800K. At present our operating set 
point is 780K, The 20-K margin of safety to account for fluctuating feed, sluggish controls, 
etc. Now, with a more sophisticated control system we would be able to raise our set point 
to 792 K with the same margin of safety that we now have. By how much can the reaction 
4.9
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6
3.3 3.353.4 3.45 3.5 3.55
lnS vs. ((1/T)*(10^-3))
rate, hence, production rate, be raised by this change if the reaction taking place in the 
reactor has an activation energy of 175 KJ/mole? 
Solution 2.12: 
As we know according to Arrhenius Law 
(-ra) = ko T
oe-Ea/RT 
Here To = 1 
ln(r1/r2) = ln(k1/k2) = (Ea/R )[1/T1-1/T2] Here 
T1= 780K 
T2= 800 K 
Ea= 175000 J/mol 
R =8.314 
Put in formula we get 
r2 = .51 r1 -------- equation 1 
similarly 
for temperature 
T3= 792K 
T2= 800 K 
 We get 
r3 = .77 r1------------ equation 2 
Hence from Eq. 1 and 2 we get 
r3 = 1.509 times of r2 
 
Problem 2.13: 
Every may 22 I plant one watermelon seed. I water it, I fight slugs, I pray, I watch my beauty 
grow, and finally the day comes when the melon ripens, I than harvest and feast. Of course, 
some years are sad, like 1980, when blue jay flew off with the seed. Anyway, 6 summers 
were a pure joy and for this I have tabulated, the number of growing days vs. the mean day 
time temperature, during the growing season. Does the temperature affect the growth 
rate? If so, represent this by an activation energy. 
Year 1976 1977 1982 1984 1985 1988 
Growing 
days 
87 85 74 78 90 84 
Mean 
temp °C 
22.0 23.4 26.3 24.3 21.1 22.7 
 
 
Solution 2.13: 
Temperature 
(
1
𝑇
∗ 10−3) 
Days ln(rate)=ln(
1
𝑑𝑎𝑦𝑠
) 
22 3.390 87 -4.466 
23.4 3.374 85 -4.443 
26.3 3.341 74 -4.304 
24.3 3.364 78 -4.357 
21.1 3.400 90 -4.500 
22.7 3.382 84 -4.431 
Slope of graph = −
𝐸
𝑅
 = 3392 
E = 3392 * 8.314 = 28201 J/mol 
 
 
Example 2.14: 
On typical summer days, field crickets nibble, jump, and chirp now and then, But at a night, 
when great numbers congregate, chirping seems to become a serious business and tends to 
be in unison.in 1897, A.E. Dolbear reported that this social chirping rate was dependent on 
the temperature as given by 
(Numbers of chirps in 15 s) + 40 = (temperature °F) 
Assuming that the chirping rate is directly measured of the metabolic rate, find the 
activation energy in kJ/mol of these crickets in the temperature range 60-80°F. 
 
Solution 2.14: 
(Numbers of chirps in 15 s) + 40 = (temperature °F) 
(Numbers of chirps in 15 s) = (temperature °F)-40 
For 60 °F 
(Numbers of chirps in 15 s) = 20 
y = -3.3922x + 7.0323
R² = 0.9441
-4.55
-4.5
-4.45
-4.4
-4.35
-4.3
-4.25
3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.4 3.41
Chart Title
For 80 °F 
(Numbers of chirps in 15 s) = 40 
dN = k *(metabolism rate) 
after integration we get 
40 – 20 =k(80-60) 
k= 1 Numbers of chirps/°F 
(-ra = metabolism rate) = ko Toe-Ea/RT 
Here To = 1 
For 60 °F 
(-rA = metabolism rate) = 60 
For 80 °F 
(-rA = metabolism rate) = 80 
ln(r1/r2) = = (Ea/R )[1/T1-1/T2] 
ln(60/80) = (Ea/8.314 )[1/60-1/80] 
Ea = -69.04 KJ/mol 
 
Problem 2.15: 
On doubling the concentration of the reactant, the rate of reaction triples. Find the reaction 
order. 
For the stoichiometry A + B (products) find the reaction order with respect to A and B. 
 
Solution 2.15: 
−𝑟𝐴 = 𝑘𝐶𝐴
𝑛 
𝑟2
𝑟1
=
𝑘(𝐶𝐴)
𝑛
𝑘𝐶𝐴
𝑛 = 2
𝑛 
 n=1.585 
 
 
CHAPTER NO. : - 2 
 
UNSOLVED EXAMPLE: - 2.16 to 2.23 
 
 
 
 
 
Submitted by:- 
11BCH044 (SUREJA VISHAL) 
13BCH044 (NAGPARA KUMAR) 
13BCH156 (PATEL KISHAN) 
12BCH157 (SUVAGIYA SHARAD) 
 
 
 
 
 
 
 
 
 
2.16 
CA 4 1 1 
CB 1 1 8 
-rA 2 1 4 
 
Solution 
Guess that, 
–rA = K*CAp*CBq 
In this Equation, There are three unknown k,p and q 
From question we can make three equations 
2 = K (4)p (1)q --------------------------------(1) 
1 = K (1)p (1)q --------------------------------(2) 
1 = K (1)p (8)q --------------------------------(3) 
Now Equation (3) divided by Equation (2), 
 
𝟒
𝟏
=
𝐊 (𝟏)𝐩 (𝟖)𝐪
𝐊 (𝟏)𝐩 (𝟏)𝐪
 
4 = (8)q 
(2)2 = (2)3q 
So, 3q = 2 
q = 2/3 
Now Equation (1) divided by Equation (2), 
𝟐
𝟏
=
𝐊 (𝟒)𝐩 (𝟏)𝐪
𝐊 (𝟏)𝐩 (𝟏)𝐪
 
2 = (4) p 
2 = (2)2p 
So, 2p = 1 
p = ½ 
Put the values of p, q in Equation (2), 
1 = K (1)1/2 (1)2/3 
1 = K (1)1/3 
K = 1 
So, from value of p, q, and K we get rate equation, 
-rA = CA1/2 CB2/3 
2.17 
CA 2 2 3 
CB 125 64 64 
-rA 50 32 48 
 
Solution 
Guess that, 
–rA = K*CAp*CBq 
In this Equation, There are three unknown k,p and q 
From question we can make three equations 
50 = K (2)p (125)q --------------------------------(1) 
32 = K (2)p (64)q --------------------------------(2) 
48 = K (3)p (64)q --------------------------------(3) 
Now Equation (3) divided by Equation (2), 
 
𝟒𝟖
𝟑𝟐
=
𝐊 (𝟑)𝐩 (𝟔𝟒)𝐪
𝐊 (𝟐)𝐩 (𝟔𝟒)𝐪
 
3/2 = (3/2)p 
So, p = 1 
P = 1 
Now Equation (1) divided by Equation (2), 
𝟓𝟎
𝟑𝟐
=
𝐊 (𝟐)𝐩 (𝟏𝟐𝟓)𝐪
𝐊 (𝟐)𝐩 (𝟔𝟒)𝐪
 
(5/4)2 = (5/4) 3q 
So, 3q = 2 
q = 2/3 
Put the values of p, q in Equation (1), 
50 = K (2)1 (125)2/3 
50 = K (2) (5)2 
K = 1 
So, from value of p, q, and K we get rate equation, 
-rA = CA CB2/3 
EXAMPLE 2.18:- 
The decomposition of reactant A at 4000𝑐 for pressure between 1 and 10 atm follows a 
first order rate law. 
A) Show that a mechanism similar to azo-methane decomposition 
𝐀 + 𝐀 ↔ 𝐀∗ + 𝐀 
𝐀∗ → 𝐑+ 𝐒 
Is consistent with observed kinetics 
Different mechanisms can be proposed to explain first order kinetics. To claim that this 
method is correct in the face of other alternatives requires additional evidence. 
B) For this purpose, what further experiments would you suggest we run and what 
results we expect to find? 
 
Solution 
𝐀 + 𝐀
𝐊𝟏
→ 𝐀∗ + 𝐀 
𝐀∗ + 𝐀
𝐊𝟐
→ 𝐀 + 𝐀 
𝐀∗
𝐊𝟑
→ 𝐑 + 𝐒 
Rate of disappearance of A, 
−𝐫𝐀 = 𝐊𝟏[𝐀]
𝟐 − 𝐊𝟐[𝐀][𝐀
∗] 
And rate of formation of A, 
𝐫𝐀
∗ = 𝐊𝟏[𝐀]
𝟐 − 𝐊𝟐[𝐀][𝐀
∗] + 𝐊𝟑[𝐀
∗] 
Also the rate of formation of intermediate is zero therefore, 
𝐫𝐀
∗ = 𝟎 
𝐊𝟏[𝐀]
𝟐 = [𝐀∗](𝐊𝟐[𝐀] + 𝐊𝟑) 
[A∗] =
K1[A]
2
K2[A] + K3
 
So, 
−rA = K1[A]
2 − K2[
K1[A]
2
K2[A] + K3
] 
 
−𝐫𝐀 =
𝐊𝟑𝐊𝟏[𝐀]
𝐊𝟐
 
 
Hence it is a first order reaction 
Now at low enough concentration of CA the rate of decomposition of A will reduce to 
 
−𝐫𝐀 = 𝐊𝟏[𝐀]
𝟐 
That is it would be a second order reaction. 
 
EXAMPLE 2.19 
Show that the following scheme 
 𝐍𝟐𝐎𝟓 
𝐊𝟏
→ 𝐍𝐎𝟐 + 𝐍𝐎𝟑
∗ 
𝐍𝐎𝟐 + 𝐍𝐎𝟑
∗
𝐊𝟐
→ 𝐍𝟐𝐎𝟓 
𝐍𝐎𝟑
∗
𝐊𝟑
→ 𝐍𝐎∗ + 𝐎𝟐 
𝐍𝐎∗ + 𝐍𝐎𝟑
∗
𝐊𝟒
→ 𝟐𝐍𝐎𝟐 
Proposed by R.Ogg J.Chem.Phys., 15,337(1947) is consistent with and can explain the 
first order decomposition of[𝑁2𝑂5]. 
Solution 
Rate of formation of NO2, 
𝐫𝐍𝐎𝟐 = 𝐊𝟏[𝐍𝟐𝐎𝟓] − 𝐊𝟐[𝐍𝐎𝟐][𝐍𝐎𝟑
∗ ] −
𝐊𝟒
𝟐
[𝐍𝐎∗][𝐍𝐎𝟑
∗ ] 
Rate of disappearance ofN2O5, 
−𝐫𝐍𝟐𝐎𝟓 = −𝐊𝟏[𝐍𝟐𝐎𝟓] + 𝐊𝟐[𝐍𝐎𝟐][𝐍𝐎𝟑
∗ ] ---- (1) 
Rate of intermediate, 
𝐫𝐍𝐎𝟑
∗ = 𝐊𝟏[𝐍𝟐𝐎𝟓] − 𝐊𝟐[𝐍𝐎𝟐][𝐍𝐎𝟑
∗ ] − 𝐊𝟑[𝐍𝐎𝟑
∗ ] − 𝐊𝟒[𝐍𝐎𝟑
∗ ][𝐍𝐎∗] 
Also, 
𝐫𝐍𝐎𝟑
∗ = 𝟎 
Therefore; 
K1[N2O5] = K2[NO2][NO3
∗ ] + K3[NO3
∗ ] + K4[NO3
∗ ][NO∗] 
K1[N2O5] = [NO3
∗ ] (K2[NO2] + K3 + K4[NO
∗]) 
 
[𝐍𝐎𝟑
∗ ] =
𝐊𝟏[𝐍𝟐𝐎𝟓]
𝐊𝟐[𝐍𝐎𝟐] + 𝐊𝟑 + 𝐊𝟒[𝐍𝐎∗]
 
Rate of intermediate, 
Also 
𝐫𝐍𝐎
∗ = 𝟎 
And, 
rNO
∗ = K3[NO3
∗ ] − K4[NO3
∗ ][NO∗] 
[NO∗] =
K3[NO3
∗ ]
K4[NO3
∗ ]
 
[𝐍𝐎∗] =
𝐊𝟑
𝐊𝟒
 
Therefore,NO3
∗ =
K1[N2O5]
K2[NO2] + K3 + K4[
K3
K4
]
 
𝐍𝐎𝟑
∗ =
𝐊𝟏[𝐍𝟐𝐎𝟓]
𝐊𝟐[𝐍𝐎𝟐] + 𝟐𝐊𝟑
 
 
So, now put value in equation (1), 
−rN2O5 = −K1[N2O5] + K2[NO2][NO3
∗ ] 
−rN2O5 = −K1[N2O5] + K2[NO2](
K1[N2O5]
K2[NO2] + 2K3
) 
−𝐫𝐍𝟐𝐎𝟓 = [𝐍𝟐𝐎𝟓](𝐊𝟐[𝐍𝐎𝟐] (
𝐊𝟏
𝐊𝟐[𝐍𝐎𝟐] + 𝟐𝐊𝟑
) − 𝐊𝟏) 
−rN2O5 = [N2O5]((
K1K2[NO2]
K2[NO2] + 2K3
) − K1) 
 
−rN2O5 = K1[N2O5]((
K2[NO2]
K2[NO2] + 2K3
) − 1) 
−𝐫𝐍𝟐𝐎𝟓 = 𝐊𝟏[𝐍𝟐𝐎𝟓] (
𝐊𝟐[𝐍𝐎𝟐] − 𝐊𝟐[𝐍𝐎𝟐] − 𝟐𝐊𝟑
𝐊𝟐[𝐍𝐎𝟐] + 𝟐𝐊𝟑
) 
Assume K3 >>> K2, K2 ≅ 0 
−𝐫𝐍𝟐𝐎𝟓 = −𝐊𝟏[𝐍𝟐𝐎𝟓] 
Therefore this is a first order reaction 
 
EXAMPLE-2.20 
Experimental analysis shows that the homogeneous decomposition of ozone proceeds 
with a rate 
-r𝐎𝟑 = k [O3]2 [O2]-1 
 
[1] What is the overall order of reaction? 
[2] Suggest a two-step mechanism to explain this rate. 
 
Solution:- 
The homogeneous decomposition of ozone proceeds as, 
2 O3 2 O2 
And follows the rate law 
-r𝐎𝟑 = K [O3]2 [O2]-1 
The two step mechanism, consistent with the rate suggested is, 
Step 1:- Fast, at equilibrium 
O3 K1 O2 + O 
O3 K2 O2 + O 
STEP 2:- Slow 
 O + O3 K3 2 O2 
The step 2 is the slower, rate determining step and accordingly, the reaction rate is 
-r𝐎𝟑 = 
−𝐝 [𝐎𝟑]
𝐝𝐭
 = K3 [O3] [O] 
Thus step 1 is fast and reversible, for this equilibrium step, we have 
𝐊 = 
𝐊𝟏
𝐊𝟐
=
[𝐎𝟐] [𝐎]
[𝐎𝟑]
 
 
[𝐎] = 
𝐊 [𝐎𝟑]
[𝐎𝟐]
 
Putting value of [O] from equation (2) into equation (1), we get 
-rO3= 
𝐝 [𝐎𝟑]
𝐝𝐭
= K3 * K 
[𝐎𝟑 ]𝟐
[𝐎𝟐]
 
Let, K = K3 * K 
-rO2 = k [O3]2 [O2]-1 
Overall order of reaction is 2 – 1 = 1 
 
EXAMPLE 2.21:- 
Under the influence of oxidizing agents hypo phosphorus acid is transformed into 
phosphorus acid 
H3PO2 oxidizing agent H3PO3 
The kinetics of this transformation present the following features. At a low concentration 
of oxidizing agent. 
rH3PO3 = K [oxidizing agent] [H3PO2] 
At a high concentration of oxidizing agent 
rH3PO3 = k [H*] [H3PO2] 
To explain the observed kinetics it has been postulated that, with hydrogen ions as 
catalyst normal unreactive H3PO2 is transformed reversibly int an active from the nature 
of which is unknown, this intermediate then react with the oxidizing agent to give H3PO3 
show that this scheme does explain the observed kinetics. 
Solution :- 
Hypothesize 
H3PO4 + H K1 X* + H 
H3PO4 + H K2 X* + H 
 X* + Ox K3 H3PO4 
 X* + Ox K4 H3PO4 
 [ where X* is an unstable intermediate] 
rH3PO3 = k3 [X*] [Ox] – k4 [H3PO3] ………(1) 
r[X*] = k1 [H3PO2] [H+] – k2 [X*] [H+] - k3 [X*] [Ox] + k4 [H3PO3] =0 
 (steady state assumption) 
Thus, 
 [X*] = 
𝐊𝟏 [𝐇𝟑𝐏𝐎𝟐][𝐇
+] +𝐊𝟒 [𝐇𝟑𝐏𝐎𝟑]
𝐊𝟐 [𝐇+]+𝐊𝟑 [𝐎𝐱]
 ………(2) 
Assuming k4 = 0, replacing (2) in (1) then gives, 
rH3PO3 = 
𝐊𝟏𝐊𝟑 [𝐇𝟑𝐏𝐎𝟐][𝐇
+] [𝐎𝐱]
𝐊𝟐 [𝐇+]+𝐊𝟑 [𝐎𝐱]
 ……...(3) 
now when k3 [ox] >> k2 [H+]…i.e. high oxidized concentration equations (3) gives 
 rH3PO4 = K1 [H3PO2] [H+] ………(4) 
on the other hand wen k2 [H+] >> k3 [Ox]..i.e. low oxidized concentration equation (3) 
gives, 
rH3PO3 = 
𝐊𝟏 𝐊𝟑
𝐊𝟐
 [H3PO2] [Ox] ………(5) 
Equation (4) & (5) is fit in the hypohesized model is accepted. 
 
EXAMPLE 2.22:- 
Come up with (guess and then verify) a mechanism that is consist with the experimentally 
found rate equation for the following reaction 
2 A + B A2B 
With 
r [A2B] = K [A] [B] 
Solution 
It is the non-elementary reaction. So, assume reaction mechanism, 
 
𝐀 + 𝐁 ↔ 𝐀𝐁∗ 
𝐀𝐁∗ + 𝐀 ↔ 𝐀𝟐𝐁 
𝟐𝐀 + 𝐁 ↔ 𝐀𝟐𝐁 
 
 2A K1 A2* ……...(1) 
 2A K2 A2 ………(2) 
 A2* + B K3 A2B ………(3) 
 A2B K4 A2* + B ………(4) 
From reaction (3) & (4) rate of formation of r [A2B], 
r [A2B] = K3 [A2*] [B] – K4 [A2B] ……….(5) 
In reaction (4) A2B is dispersed. 
Rate of disappearance, 
𝐫𝐀 = 𝐊𝟏[𝐀]
𝟐 
Rate of formation of A, 
𝐫𝐀 = 
𝐊𝟏[𝐀]
𝟐
𝟐
 
(2 moles of A consume give 1 mole of A * so ½) 
𝐫[𝐀𝟐
∗ ] = 
𝐊𝟏[𝐀]
𝟐
𝟐
− 𝐊𝟐[𝐀𝟐
∗ ] − 𝐊𝟑[𝐀𝟐
∗ ][𝐁] + 𝐊𝟒[𝐀𝟐𝐁] 
𝐫[𝐀𝟐
∗ ] = 𝟎 
0 = 
K1[A]
2
2
− K2[A2
∗ ] − K3[A2
∗ ][B] + K4[A2B] 
K3[A2
∗ ][B] + K2[A2
∗ ] = K4[A2B] + 
K1[A]
2
2
 
 
[A2
∗ ] [K3[B] + K2] = K4[A2B] + 
K1[A]
2
2
 
[𝐀𝟐
∗ ] = 
𝐊𝟒[𝐀𝟐𝐁] + 
𝐊𝟏[𝐀]
𝟐
𝟐 
[𝐊𝟑[𝐁] + 𝐊𝟐]
 
Put value of [A2*] in equation (5), 
𝐫𝐀𝟐𝐁 = 𝐊𝟑(
𝐊𝟒[𝐀𝟐𝐁] + 
𝐊𝟏[𝐀]
𝟐
𝟐
[𝐊𝟑[𝐁] + 𝐊𝟐]
) [𝐁] − 𝐊𝟒[𝐀𝟐𝐁] 
rA2B = K3
K1[A]
2
2
[B] + K3K4[A2B] [B] − 
K4[A2B] K3[B]
[K3[B] + K2]
− K4K2[A2B] 
𝐫𝐀𝟐𝐁 =
𝐊𝟑
𝐊𝟏[𝐀]
𝟐
𝟐
[𝐁] + 𝐊𝟐𝐊𝟒[𝐀𝟐𝐁]
[𝐊𝟑[𝐁] + 𝐊𝟐]
 
 
If K4 is very small, K4 ≅ 0 
𝐫𝐀𝟐𝐁 =
𝐊𝟑
𝐊𝟏[𝐀]
𝟐
𝟐
[𝐁]
[𝐊𝟑[𝐁] + 𝐊𝟐]
 
If K2 is very small, K2 ≅ 0 
𝐫𝐀𝟐𝐁 =
𝐊𝟏𝐊𝟑[𝐀]
𝟐[𝐁]
𝐊𝟑[𝐁]
 
So, from above equation we can conclude that our first assumption is wrong so, 
mechanism also wrong. 
Now, we can assume for second order reaction 
𝐀 + 𝐁 ↔ 𝐀𝐁∗ 
𝐀𝐁∗ + 𝐀 ↔ 𝐀𝟐𝐁 
𝟐𝐀 + 𝐁 ↔ 𝐀𝟐𝐁 
2A + B K1 AB* ……...(1) 
 AB* K2 A + B ………(2) 
 A2* + B K3 A2B ………(3) 
 A2B K4 A2* + B ………(4) 
From reaction (3) and (4) rate of formation of r[A2B], 
𝐫𝐀𝐁∗ = 𝐊𝟏[𝐀][𝐁] − 𝐊𝟐[𝐀𝐁
∗] − 𝐊𝟑[𝐀𝐁
∗][𝐁] + 𝐊𝟒[𝐀𝟐𝐁] …(5) 
𝐫[𝐀𝐁∗] = 𝟎 
0 = K1[A][B] − K2[AB
∗] − K3[AB
∗][B] + K4[A2B] 
[𝐀𝐁∗] = 
𝐊𝟏[𝐀][𝐁] + 𝐊𝟒[𝐀𝟐𝐁] 
[𝐊𝟑[𝐀] + 𝐊𝟐]
 
Put [AB*] value in equation (5), 
𝐫𝐀𝟐𝐁 = 𝐊𝟑 (
𝐊𝟏[𝐀][𝐁] + 𝐊𝟒[𝐀𝟐𝐁] 
[𝐊𝟑[𝐀] + 𝐊𝟐]
) [𝐀] − 𝐊𝟒[𝐀𝟐𝐁] 
𝐫𝐀𝟐𝐁 = 𝐊𝟑𝐊𝟏[𝐀]
𝟐[𝐁] + 𝐊𝟑𝐊𝟒[𝐀𝟐𝐁] [𝐀] − 
𝐊𝟒 𝐊𝟑[𝐀][𝐀𝟐𝐁]
[𝐊𝟑[𝐀] + 𝐊𝟐]
− 𝐊𝟒𝐊𝟐[𝐀𝟐𝐁] 
If K4 is very small, K4 ≅ 0 
𝐫𝐀𝟐𝐁 =
𝐊𝟑𝐊𝟏[𝐀]
𝟐[𝐁]
[𝐊𝟑[𝐀] + 𝐊𝟐]
 
If K2 is very small, K2 ≅ 0 
𝐫𝐀𝟐𝐁 =
𝐊𝟏𝐊𝟑[𝐀]
𝟐[𝐁]
𝐊𝟑[𝐀]
 
𝐫𝐀𝟐𝐁 = 𝐊𝟏[𝐀][𝐁] 
EXAMPLE 2.23:- 
Mechanism for enzyme catalyzed reactions. To explain the kinetics of enzyme-substrate-
reactions, Michaelis and Menten came up with the following mechanism, which uses an 
equilibrium assumption 
𝐀 + 𝐄 
𝐊𝟏
→ 𝐗 
 𝐗 
𝐊𝟐
→ 𝐀 + 𝐄 with 𝐊 = 𝐗
𝐀𝐄
=
𝐊𝟏
𝐊𝟐
 
𝐗 
𝑲𝟑
→ 𝐑 + 𝐄 
𝑬𝟎 = 𝐄 + 𝑿 
And where [E0] represents the total enzyme and [E] represents the free unattached 
enzyme. 
G.E. Briggs and J.B.S Haldane, Bochem.J. 19,338 (1925), on the other hand employed a 
steady state assumption in place of the equilibrium assumption 
𝐀 + 𝐄 
𝑲𝟏
→ 𝐗 
𝐗 
𝑲𝟐
→ 𝐀 + 𝑬 
 𝑬𝟎 = 𝐄 + 𝑿 
What final rate from -ra in terms of [A], [E0], K1, K2 and K3 does 
1) The Michalels and Menten mechanism gives?2) The Briggs-Haldane mechanism give? 
Solution 
𝐀 + 𝐄 
𝑲𝟏
→ 𝐗 ------ (1) 
𝐗 
𝑲𝟐
→ 𝐀 + 𝑬 ------ (2) 
𝐗 
𝑲𝟑
→ 𝐑 + 𝐄 ------ (3) 
𝐄𝟎 = 𝐄 + 𝐗 ------ (4) 
M-M assume that the reverse reaction of (1) approach equilibrium quickly, or 
𝐊 = 
𝐗
𝐀𝐄
=
𝐊𝟏
𝐊𝟐
 
B-H assume that quickly 
𝐝𝐲
𝐝𝐭
= 𝟎 ------ (5) 
for Michacli’s – Menten 
From equation (3) 
𝐫𝐑 = 𝐊𝟑𝐗 
From equation (5) 
𝐗 = 
𝐊𝟏
𝐊𝟐
 𝐀𝐄 ------ (6) 
Eliminate E with (4) 
𝐗 = 
𝐊𝟏
𝐊𝟐
 𝐀(𝐄𝟎 − 𝐗)------ (7) 
Or 
 
𝐗 = 
𝐊𝟏
𝐊𝟐
 𝐀𝐄𝟎/ 𝟏 + 
𝐊𝟏
𝐊𝟐
 𝐀 ------ (8) 
Equation (8) in (7) gives 
 
𝐫𝐑 = 
𝐊𝟑𝐀𝐄𝟎
𝐊𝟐
𝐊𝟏
+ 𝐀
 
These equation gives essentially the same result. 
 
(a) For the Briggs - Haldane Mechanisms 
From equation (3), 
 
𝐫𝐑 = 𝐊𝟑𝐗 ----- (9) 
From equation (6) 
𝐝𝐲
𝐝𝐭
= 𝟎 = 𝐊𝟏𝐀𝐄 − (𝐊𝟐 + 𝐊𝟑 )𝐗 
 
Eliminate E with (A) 
 𝐊𝟏𝐀(𝐄𝟎 − 𝐗) − (𝐊𝟐 + 𝐊𝟑 )𝐗 = 𝟎 
Or 
 
𝐗 =
𝐊𝟏𝐀𝐄𝟎
𝐊𝟏𝐀 + 𝐊𝟐 + 𝐊𝟑
 
 
Equation (10) in (9) gives 
 
𝐫𝐑 = 
𝐊𝟑𝐀𝐄𝟎
𝐊𝟐+ 𝐊𝟑
𝐊𝟏
+𝐀
 ---------- (10) 
 
These equation give essentially the same result. 
𝐊𝟐+ 𝐊𝟑
𝐊𝟏
 (Michaclis const.) 
 
CRE-1 Term Paper 
Example 3.1-3.11 
Submitted By: 
Yagnesh Khambhadiya (13BCH025) 
Chirag Mangrola (13BCH032) 
Fenil Shah (13BCH052) 
Ravindra Thummar (13BCH058) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Since the reaction order, hence concentration dependency is not known we 
are not given enough information to find the rate of reaction at higher 
concentration. 
-ra = kCa 
0.2 = k * 1 
K= 0.2 sec
-1
 
-ra = kCa =0.2*10 = 2 mol/l.s 
 
 
For given first order reaction by equation, 
 (
 
 
) = Kt 
For Xa = 0.5 then t = 5 min. 
So, K = 0.1386 
Now for 75% conversion Xa= 0.75 & K = 0.1386 
 (
 
 
) = 0.1386 * t 
t = 10 min 
So, for 75% conversion time required is 10min 
 
 
 For the second order disappearance of single reactant gives, 
kt=
 
 
 * ( 
 
 ) 
Now, for 50% conversion 
T50 = 
 
 
 *( 
(
 
 
)
 (
 
 
)
 ) = 5 min. 
Now, for 75% conversion 
T75 = 
 
 
 * (
(
 
 
)
 (
 
 
)
 ) = 15 min. 
So, extra time needed is 10 minutes. 
 
 
-ra = k 
 
Cao 
 
 
 = k 
 
∫
 
 
 
 
 = k 
 
2[- 
Case I : Xa= 0.75 when t = 10 min 
2[- 
k’ = 0.1 
Case II : Xa= ? when t = 30 min 
2[- 
Xa=0.75 
 
 
Since the fractional disappearance is independent of initial concentration we have 
first order rate equation, 
 (
 
 
) = Kt 
Where, Ca = Monomer concentration 
We also can find the rate constant, thus replacing values. 
Case I: Cao= 0.04 mol/l 
 
 
 
 = k*34 
So, k= 0.00657 . 
Case II:. Cao= 0.8 mol/l 
 
 
 
 = k*34 
k= 0.00657 . 
Hence the rate of disappearance of monomer is given by, 
-ra = (0.00657 ) * C 
 
 
Cao = 1 
 
 
 
t = 8min then conversion 80% 
t = 18 mins then conversion 90% 
Assume the reaction is second order 
-ra = k* 
 
 
 
 
= k t 
 
 
= k* *8 
k=0.5 (mol/l)
-1
(min)
-1 
 
 
= k* *18 
k=0.5 (mol/l)
-1
(min)
-1 
From the same value of k in both case, the reaction is second order 
-ra = k* 
 
 
-ra = 0.5* 
 
 
 
 
 
Magoos batting habits and his losses can be described by, 
 (
 
 
) = kA 
Where, A = money in hand 
Now, t=0 then A0 = 180 
 t=2hrs then A= 135 
K= 
 
 
 * (
 
 
) = 
 
 
 * (
 
 
) = 0.144 
After raise we have, 
 t= 0 then A0=? 
 t= 3hrs then A= 135 
For unchanged habits, 
 (
 
 
) = 0.144*3 
 A0= 208 
Hence his raise is 208-180 = 28 
 
 
 
 
 
Equimolor quantities of H2 and NO are used… 
Cao= Cbo 
Co = Cao + Cbo 
Co = 2Cao 
Ca α Pa 
Pao = Po /2 
Cao= 
 
 
 
Cao= 
 
 
 
Po= 200 mmHg= 0.263 atm 
Cao = 
 
 
 =5.414 * 10
-3 
ln Cao = -5.22 
t1/2 = 265 sec 
ln t1/2 = 5.58 
 
P(mm Hg) 200 240 280 320 360 
P(atm) 0.263 0.316 0.368 0.421 0.474 
Cao* 10^-3 5.414 6.5 7.57 8.67 9.76 
ln Cao -5.22 -5.04 -4.88 -4.75 -4.63 
t1/2 265 186 115 104 67 
ln t1/2 5.58 5.23 4.74 4.64 4.2 
 
 
Slope = (n-1)=2.275 
n = 3.275 
 
y = 2.2756x - 6.2815 
R² = 0.9786 
0
1
2
3
4
5
6
4.6 4.8 5 5.2 5.4
ln
 t
1
/2
 
-ln Ca0 
 
For a first order reversible reaction, 
A R 
Where, = Reversible reaction 
 K1 = Forward reaction rate const. 
 K2= Backward reaction rate const. 
Ca0 = 0.05
 
 
 
Cr0 = 0 
M= 0/0.5=0 
Xae = 0.67 
The integral conversion in a batch reactor is given by, 
-ln(1- 
 
 
 ) = k1 * t *
 
 
 
 
Replacing values we then find, 
-ln(1- 
 
 
 
 
 
 
 
 
 ) = k1 * t *
 
 
…..(1) 
k1= 0.0577 
Now from thermodynamics we know that, 
K = 
 
 
 = 
 
 
 =
 
 
 
So, K1= (0.667/0.332) K2…… (2) 
From equation (1) & (2) 
K1= 0.057750 
K2= 0.028875 
 
Thus, rate expression for the disappearance of A. 
-ra = 0.057750 Ca - 0.028875 Cr 
 
 
For the second order disappearance of single reactant gives, 
k*t=
 
 
 * ( 
 
 ) ….. (1) 
Here it is given that Ca0 = 2.03 
 
 
 & Ca = 1.97 
 
 
 
So, Xa = 
 
 
 by putting values of Ca & Cao ,Xa will be 0.029 
So, putting value of Xa and Cao in equation (1) 
K (1) = 
 
 
 * ( 
 
 ) 
K = 0.0147 
So, for second order rate expression is –ra = 0.0147( ) 
 
From the table of data, 
 
 
 Reaction is second order. 
 
 
k = 1 * 10
-5
 (mol/l)
-1
(min)
-1 
 
 
 
 
 
 
t = 5*60 =300 min 
Cao = 500 mol/m
3
 
 
 
 = 0.003 + 
 
 
 
Ca = 200 mol/m
3 
Xa = 
 
 
 = 
 
 
 = 0.6 
y = 1E-05x + 0.001 
R² = 1 
0
0.001
0.002
0.003
0.004
0.005
0.006
0 200 400 600
Ca t 1/ca 
1000 0 0.001 
500 100 0.002 
333 200 0.003003 
250 300 0.004 
200 400 0.005 
Ex - 3.12 
Aqueous A at a concentration C,, = 1 mollliter is introduced into a batch reactor where it reacts 
away to form product R according to stoichiometry A R. The concentration of A in the reactor 
is monitored at various 
times, as shown below: 
 
T (min) 0 100 200 300 400 
Ca mol/m
3
 1000 500 333 250 200 
 
Find the rate for the reaction 
Ans: For batch reactor t = -cao∫
 
 
 
 
 , Cao = 100 mol/m
3
 
 
 
 = ln 
 
 
 
 
 
 slope = 
 
 
 
 0.0045 = 
 
 
 
 0.0045 = 
 
 
 
 K = 4.5 min
-1 
 ln(cao/ca) t 
0 0 
.693 100 
1.099 200 
1.386 300 
1.609 400 
Ex – 3.13 
Betahundert Bash by likes to play the gaming tables for relaxation. He does not expect to win, 
and he doesn't, so he picks games inwhich losses area given small fraction of the money bet. He 
plays steadily without a break, and the sizes of his bets are proportional to the money he has. If at 
"galloping dominoes" it takes him 4 hours to lose half of his money and it takes him 2 hours to 
lose half of his money at "chuk-a-luck," how long can he play both games simultaneously if he 
starts with $1000 and quits when he has $10 left, which is just enough for a quick nip and carfare 
home? 
Ans: 
 Money 
 
→ Game 1 Money 
 
→ Game2 
 Game 1: Game 2: 
 K1 =0 .693 / (t1/2 )1 K1 = 0.693 / (t1/2 )2 
 =0 .693 / 4 = 0.693 / 2 
 = 0.17325 hour
-1
 = 0.3465 hour
-1
 
 
(- rate of reaction) = (K1 + K2).CA 
 
 
 = M e
-(k1+k2) 
M = M0 e
-(k1+k2)t
 
10 = 1000 e
-(0.3465 + 0.17325)t 
0.01 = e-(0.5197
t)
 
 
 
 
 
 
t = 8.82 hour 
Ex - 3.14 
For the elementary reactions in series 
A 
 
 R 
 
 S , K1 = K2 at t = 0 
CA0 = CA , CRO = CSO = 0 
find the maximum concentration of R and when it is reached. 
Ans : 
-ra = K1Ca +ra = K1Ca – K2CR +rS = 
K2CR 
 
 
 = K1Ca 
 
 
 = K1Cao e
-K
1
t 
- K2Cr 
Ca = Ca0e
-Kt 
 
 
 +K2Cr =K1Ca0 e
-K
1
t
 
 
CR = ∫ K1 CA0e
-k
1
t 
e
-k
2
t
. dt + c 
 
CRe
K
2
t
= K1CA0 t + c 
 at t = 0, 
CR = K1CA0t . e
-kt 
CR = 0, 
 C = 0, 
 
 
 
 
k1cA0e
-kt
+ k1cA0e
-kt
 .t. (-k) = 0 
 
 k1 – k1
2
.t = 0 
 
 t = ⁄ 
 (cr)max= k1cA0 . 
 
 
 . e
(-1/k1) . k1
 
 Cr = 
 
 ⁄ 
 
 
Ex – 3.15 
At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as 
follows: 
 Sucrose → product 
Starting with a sucrose concentration CAo = 1.0 millimol/liter and an enzyme concentration 
CEO = 0.01 millimollliter, the following kinetic data are obtained in a batch reactor 
(concentrations calculated from optical 
rotation measurements): 
 
CA .84 .68 .53 .38 .27 .16 .09 .04 .018 .006 .0025 
t 1 2 3 4 5 6 7 8 9 10 11 
 
Determine whether these data can be reasonably fitted by a kinetic equation of the Michaelis-
Menten type, or 
 
-rA = 
 
 
 
 
If the fit is reasonable, evaluate the constants K3 and Cm. Solve by the integral method. 
 
Ans: 
 
 
-rA = 
 
 
 
 = 
 
 
 
 ∫
 
 
 
 
= ∫ 
 
 
 
 -(CA – CA0) – CM ln 
 
 
 = K3 . t. CE0 
 T = 
 
 
 . (
 
 
 – CA) + 
 
 
 
 Compare with , Y = mx + c 
 m = 
 
 
 
 c = 
 
 
 
 
x = ln [ 
 
 
 – CA] 
 
Y=time 
-.665 1 
-.294 2 
0.104 3 
0.587 4 
1.039 5 
1.672 6 
2.317 7 
3.178 8 
3.999 9 
5.109 10 
 
 
Slop = 1.5511 
Intercept = 2.885 
C = 2.885 = 
 
 
 
K3 = 
 
 
 
K3 = 35.02 
 
M = slope = 1.5511 = 
 
 
 
Cm = 1.5511 * 35.02 * 0.01 
Cm = 0.5431 
 
 
 
 
 
Ex - 16. 
At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as 
follows: 
 Sucrose → product 
Starting with a sucrose concentration Cao = 1.0 millimol/liter and an enzyme concentration CEO 
= 0.01 millimollliter, the following kinetic data are obtained in a batch reactor (concentrations 
calculated from optical rotation measurements): 
CA .84 .68 .53 .38 .27 .16 .09 .04 .018 .006 .0025 
t 1 2 3 4 5 6 7 8 9 10 11 
 
Determine whether these data can be reasonably fitted by a kinetic equation of the Michaelis-
Menten type, or 
-rA = 
 
 
 
If the fit is reasonable, evaluate the constants K3 and Cm. Solve by the diffrential method. 
Ans: 
-rA = 
 
 
 
 
 
 = 
 
 
 
 
 
 = 
 
 
 + 
 
 
 
Plot CA v/s t & take slope at different CA 
CA 
 
 
 
 
 
 
 
 
0.38 0.139 2.63 7.2 
0.225 0.097 4.44 10.3 
0.1 0.052 10 19.23 
0.075 0.042 12.82 23.8 
 
 
 
 
Slope = 1.623 
Intercept = 3.0035 
 = 
 
 
 
 = 
 
 
 
 K = 33.33 
 
Slope = 1.623 = 
 
 
 
M = 1.623 * 33.33*0.01 
M = 0.54 
 
 
 
 
 
Ex - 17 
Enzyme E catalyzes the transformation of reactant A to product R as follows: 
 
A 
 
-rA = 
 
 
 
 
If we introduce enzyme (CEO = 0.001 mollliter) and reactant (CAo = 10mollliter) into a batch 
reactor and let the reaction proceed, find the time needed for the concentration of reactant to drop 
to 0.025 mollliter. Note that the concentration of enzyme remains unchanged during the reaction. 
 
 
Ans : 
 
CEO = 0.001 M 
CA0 = 10 M 
For batch reactor, t = -CAO∫
 
 
 
 = -CAO ∫
 
 
 
 
Ceo = .001M 
 
 = -CAO ∫
 
 
 
 
 + 
 
 
 
 = - CAO [
 
 
 * ln CA + 
 
 
 8 CA ]
 
 = -10 [ -3.678 - 7.305 ] 
 
 
Ex - 18 
An ampoule of radioactive Kr-89 (half life = 76 minutes) is set aside for a day. What does this do 
to the activity of the ampoule? Note that radioactive decay is a first-order process. 
Ans : 
t1/2 = 76 min. 
 k = 
 
 
 = 9.1184 * 10
-3 
 t = 1440min
-1 
 CA = CAO e 
–(.00911*1440) 
 = CAO . 1.98*10
-6 
 
% concentration = ( CAO - 
 
 ⁄ ) * 100 
 = (
 
 
 )* 100 
 = 99.99 % 
 
Activity reduced to 99.99 % 
 
 
Ex – 19 
Find the conversion after 1 hour in a batch reactor for 
A→R , -ra = √ 
 
 , CA0 = 1M 
Ans: For batch reactor, 
 t = -CAO∫CAO = 1M 
 t = 109.87 min. 
1 = -1 ∫
 
 
 
 
 
 -1 = 2 [√ 
 
] 
 -0.5 = √ - 1 
 CA = 0.25 
 XA = 1 - 
 
 ⁄ 
 = 1 - ⁄ 
 
 
 
Ex - 20. 
M. Hellin and J. C. Jungers, Bull. soc. chim. France, 386 (1957), present the data in Table P3.20 
on the reaction of sulfuric acid with diethylsulfate in aqueous solution at 22.9 degree Celsius: 
 
H2SO4 + (C2H5)2SO4 → 2C2H5SO4H 
 
Initial concentrations of H2S04 and (C2H5)2SO4 are each 5.5 mollliter. Find a rate equation for 
this reaction. 
 
 
t, (min.) C2H5SO4H,(M) t, (min.) C2H5SO4H,(M) 
0 0 180 4.11 
41 1.18 194 4.31 
48 1.38 212 4.45 
55 .63 267 4.86 
75 2.2 318 5.15 
96 2.75 368 5.32 
127 3.31 379 5.35 
146 3.76 410 5.42 
162 3.81 ∞ 5.8 
 
 
XA = 0.75 
 Ans: 
H2SO4 + (C2H5)2SO4 → 2 C2H5SO4 
A + B → 2R 
 CAO = CBO = 5.5 M 
 -ra = K CA CB 
 = K CAO . CBO. (1-XA). (1-XB) ( CAO.XA = CBO. XB ) 
 = K CAO. (1-XA).(1-MXA) 
 = K. (1-XA)
2
 
 
 
 
 
 = K. (1-XA)
2
 
 
 
 = (1 – Xa)
2 
.K 
∫
 
 
 = K ∫
 
 
 
 
 
 
 
 = 
 
 
 
 
 
 
 
 
 
 
 
Slope = .0027 
 
 
 
 = .0027 
 K = 5.5 * 0.0027 
 
 Xa 
 
 
 
 
0.107 0.119 
0.125 0.142 
0.148 0.173 
0.204 0.256 
 0.25 0.33 
0.346 0.529 
0.392 0.644 
K = 0.01485 min 
-ra = 0.01485 CA. CB 
 
 
 
 
 
Ex – 21 
A small reaction bomb fitted with a sensitive pressure-measuring device is flushed out and then 
filled with pure reactant A at 1-atm pressure. The operation is carried out at 25
0
C, a temperature 
low enough that the reaction does not proceed to any appreciable extent. The temperature is then 
raised as rapidly as possible to 100°C by plunging the bomb into boiling water, and the readings 
in Table are obtained. The stoichiometry of the reaction is 2A → B, and after leaving the bomb 
in the bath over the weekend the contents are analyzed for A; none can be found. Find a rate 
equation in units of moles, liters, and minutes which will satisfactorily fit the data. 
 
T,(min) ∏ (atm) T,(min) ∏ (atm) 
1 1.14 7 0.85 
2 1.04 8 0.832 
3 0.982 9 0.815 
4 0.94 10 0.8 
5 0.905 15 0.754 
6 0.87 20 0.728 
 
Ans: 
 
Constant volume reaction 
 V1 = V2 
P2 = (
 
 ⁄ ). P1 
 = ( ⁄ ). 1 
 = 1.25 atm 
PA = PA0 – ( ⁄ )(P – P0) 
 = 1.25 – ⁄ ). (1.14 – 1.25 ) 
 = 2P -2P0 + P0 
 = 2P – P0 
 
 CA1 = 
 
 ⁄ 
 = 0.041 
 
t 0 1 2 3 4 5 6 7 8 9 10 15 20 
CA 0.041 0.034 0.027 0.023 0.021 0.018 0.016 0.015 0.013 0.012 0.0114 0.0089 0.0067 
 
 
Assume 2
nd 
order reaction 
 K = { [
 
 
] – [
 
 
]}* ⁄ ) 
K1 = [
 
 
 - 
 
 
]*
 
 
 = 5.02 M
-1 
min.
-1
 
K2 = [
 
 
 - 
 
 
]*
 
 
 = 6.32 M
-1 
min.
-1 
K3 = [
 
 
 - 
 
 
]*
 
 
 = 6.36 M
-1 
min.
-1
 
K4 = [
 
 
 - 
 
 
]*
 
 
 = 5.81 M
-1 
min.
-1 
K5 = [
 
 
 - 
 
 
]*
 
 
 = 6.23 M
-1 
min.
-1 
K6 = [
 
 
 - 
 
 
]*
 
 
 = 6.35 M
-1 
min.
-1 
There is not much variation in the value of K. 
Avg. K = 
 
 
 
K = 6.067 
 
 
 . min
-1
 
Rate equation - rA = 6.067 CA
2
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ex. 4.1 
Gaseous Feed, 
A+B→R+S 
 CA0=100 , CB0=200, XA =0.8 
For XA=0,VI=100A+200B+0R+0S=300, 
For XA=1, VF=0A+100B+100R+100S=300, So εa=0 
CA= CA0(1- XA) = 100(1-0.8)=20 …ans(1) 
XBCB0= bCA0XA 
XB= CA0XA/ CB0=(100/200)(0.8)=0.4 …ans(2) 
CB= CB0(1- XB)=200(1-0.4)=120 …ans(3) 
 
Ex. 4.2 
Dilute aqeous feed, So εa=0 
A+2B→R+S 
CA0=100 , CB0=100,CA=20 
CA= CA0(1- XA) , So XA =1-(20/100)=0.8 …ans(1) 
XBCB0= bCA0XA, 
XB= CA0XA/ CB0=2(100/100)(0.8)=1.6(impossible) …ans(2 & 3 ) 
 
Ex. 4.3 
Gaseous Feed, 
A+B→R 
 CA0=200 , CB0=100, CA =50 
For XA=0 ,VI=200A+100B+0R=300, So εa=( VF- VI)/ VI =(100-300)/300 = -0.667 
For XA=1 , VF=0A-100B+200R=100, 
XA=(CA0-CA)/(CA0+ εaXA) = (200-50)/(200-0.667*50) = 0.9 
XBCB0= bCA0XA 
XB= CA0XA/ CB0=(200/100)(0.9)=1.8 (impossible)…ans(2) 
CB= CB0(1- XB)=200(1-1.8)= impossible …ans(3) 
 
Ex. 4.4 
Gaseous Feed, 
A+2B→R 
 CA0=100 , CB0=100, CA =20 
For XA=0, VI=100A+100B+0R=200, So εa=( VF- VI)/ VI =(0-200)/200 = -1 
For XA=1, VF=0A-100B+100R=0, 
XA=(CA0-CA)/(CA0+ εaXA) = (100-20)/(100-2*50) = 1 
XBCB0= bCA0XA 
XB= CA0XA/ CB0=(100/100)(1)=2 (impossible)…ans(2) 
CB= CB0(1- XB)=100(1-2)= impossible …ans(3) 
 
Ex. 4.5 
Gaseous Feed, 
A+B→2R 
 CA0=100 , CB0=200, T0=400 K , T= 300 K, π0=4 atm , π=3 atm , CA =20 
For XA=0, VI=100A+200B+0R=300, So εa=( VF- VI)/ VI = 0 
For XA=1, VF=0A+100B+200R=300, 
XA=(CA0-CA)(Tπ0/T0π)/(CA0+ εaXA) (Tπ0/T0π) 
= (100-20)(300*4/400/3)/(100+0*20)( 300*4/400/3) = 0.8 
XBCB0= bCA0XA 
XB= CA0XA/ CB0=(100/200) = (0.8) …ans(2) 
CB= CB0(1- XB)=200(1-0.4) = 120 …ans(3) 
 
Ex. 4.6 
Gaseous Feed, 
A+B→5R 
 CA0=100 , CB0=200, T0=1000 K , T= 400 K, π0=5 atm , π=4 atm , CA =20 
For XA=0, VI=100A+200B+0R=300, So εa=( VF- VI)/ VI = 2 
For XA=1, VF=0A+100B+500R=600, 
XA=(CA0-CA)(Tπ0/T0π)/(CA0+ εaXA) (Tπ0/T0π) 
= (100-20)(400*5/1000/4)/(100+2*20)( 400*5/1000/4) = 0.75 …ans(1) 
XBCB0= bCA0XA 
CB/CA0= ((CB0/ CA0)-b XA/a )/(Tπ0/T0π)=((200/100)-0.75)/( 400*5/1000/4) , 
So CB= 100 …ans(2) 
 XBCB0= bCA0XA 
XB=0.75 …ans(3) 
 
Ex. 4.7 
1 lit/min → →28 lit/min 
A→31R 
X=Popcorn produced 
(1-X)=disappearance of A 
So, outlet = 31X+(1-X)1=28 lit/min 
X=27/30=0.9 
Ans : 90% popcorn popped. 
 
 
 
 
 
 
 
 
Ex. 5.1 
2A→R+2S 
XA=0.9 
Space velocity(s) = 1/min 
So, space time(τ)=1/s 
 =1 min …ans(1) 
Here, εa=(3-2)/2 = 0.5 VF = V0(1+ εa XA) 
 =V0(1+0.5*0.9)=1.45 V0 
So. Holding time, ŧ = V/Vf 
 = V/V0*1.45 
 =0.69 min 
But in the plug flow reactor since the gas reacts progressively as it passes through the reactor, 
expands correspondly , therefore holding time = 1 min. …ans (2) 
 
Ex 5.2 
Liquid phase reaction , XA=0.7 , time = 13 min , first order reaction 
For first order reaction, -rA=k CA 
-ln (1- XA) = kt , so, k=0.0926 / min 
Case -1(for mixed flow reactor), 
Here τm= CA0*XA/(-rA) 
 = CA0*XA/(k*CA0(1- XA)) 
 =(0.7/(0.926*0.3)) 
 τm= 25.2 min … ans(1) 
Case -2(for plug flow reactor), 
Same as batch reactor, τp = τ = 13 min …ans(2) 
 
 
 
Ex. 5.3 
Aqueous monomer A , V = 2 litre , ν = 4 lit/min , CA0 = 1 mol/lit , CA = 0.01 mol/lit 
 A +A → R + A → S + A → T + … 
Here, τ = V/ν = 2/4 = 0.5 min 
For reaction rate of A, 
 -rA = (CA0 – CA) / τ 
 =(1-0.01)/0.5 = 1.98 mol/(lit*min) …ans(1) 
For W, CW = 0.0002 mol/lit 
-rW = (-CW0 +CW) / τ 
= (0.0002-0)/0.5 = 0.0004 mol/(lit*min) …ans(2). 
 
Ex 5.4 
-rA=k CA
1.5 
XA=0.7 
For mixed flow reactor , τ = V/ V0 
 = XA/ (-rA) 
 = XA*CA0/(k*CA0
1.5*(1- XA)
1.5) 
From above equation , k = XA/(( CA0
0.5*(1- XA)
1.5)*( V/ V0)) …equ (1) 
Putting XA=0.7 
 So, k = 4.260/( CA0
0.5(V/ V0)) 
Now if volume is doubles so V’ = 2V 
So , τ = V’/ V0 
 = XA*CA0/(k*CA0
1.5*(1- XA)
1.5) 
 8.52 = XA/(1- XA)
1.5 … from equ (1) 
Take square both side , 
72.6 = XA
2/(1- XA)
3 
So, 72.6 – 72.6 XA3 + 216 XA2 - 217 XA =0 
Solving by numerical method we get XA=0.7947 … ans 
 
 
Ex. 5.5 
A +B → R 
 -rA=200 CA*CB 
CA0= 100 , CB0 = 200 , ν = 400 lit/min , XA=0.999 
For second order reaction , (different reactants) 
k τ *CA0*(M-1)= ln((M- XA)/(M* (1- XA)) 
here M=2, so substituting values, 
V/ν =ln((2-0.999)/2(1-0.999)) / (200*0.10*1) 
= 0.31 
So V=0.31 * 400 = 124 litre …ans. 
 
Ex. 5.6 
A↔R (reversible reaction) 
-rA=0.04 CA*0.01 CB 
so k1 = 0.04 , k2 = 0.01, M= CR0/ CA0 = 0 
 k1/ k2 = XAe/(1- XAe) from there XAe = 0.8 …ans (1) 
For first order plug flow reactor (reversible reation), 
V/(ν* CA0) = ∫ 𝑑𝑋𝐴
XAf
0
/(-rA) 
-rA=0.04 CA0(1- XA) - 0.01 CA0(M + XA) 
So by integrating term we get, 
-ln (1 - XA/ XAe ) 
= k1* V/ν*(M + 1) / (M + XAe ) 
Substituting all values,we get 
-ln (1 - XA/ XAe ) = 0.64 
XA/ XAe = 0.48 
 XA = 0.38 …ans 
 
 
 
 
Ex 5.7 
Half time for this reaction (τ1/2) = 5.2 days 
Mean resident time (τ) = 30 days 
So k = 0.693/ τ1/2 =0.693/5.2 = 0.1333 /day 
For mixed flow reactor, CA/ CA0 = 1 / ( 1 + k τ) = 0.2 
So fraction of activity = 1 - CA/ CA0 = 0.8 
Ans = 80% 
 
Ex. 5.8 
A↔R (reversible reaction) 
-rA=0.04 CA*0.01 CB 
so k1 = 0.04 , k2 = 0.01, M= CR0/ CA0 = 0 
 k1/ k2 = XAe/(1- XAe) from there XAe = 0.8 …ans (1) 
For first order mixed flow reactor (reversible reation), 
V/(ν* CA0) = XA /(-rA) 
-rA=0.04 CA0(1- XA) - 0.01 CA0(M + XA) 
 
V/ν = 0.10 XA / (0.04 CA0(1- XA) - 0.01 CA0(M + XA)) 
Substituting all values,we get 
XA= 0.8-0.8 XA -0.01 XA 
 XA = 0.40 …ans 
Ex. 5.9 
A + E(enzyme) → R 
-rA=0.1 CA/(1+0.5 CA) 
Here, ν = 25 lit/min , CA0 = 2 mol/ litr , XA= 0.95 
Now, V/(ν) = ∫ 𝑑𝐶𝐴
CAf
CA0
/(-rA) 
Substituting -rA=0.1 CA/(1+0.5 CA) and CA= CA0 (1 - XA) 
So, V/(ν) = [-10(ln (0.10/2) - 5(0.10-2)] = 39.452 
So, V = 986.25 lit …ans. 
 
Ex 5.10 
A → 2.5(all products) 
-rA=10 CA , plug flow reactor 
Here, ν = 100 lit/min , CA0 = 2 mol/ litr , V= 22 L 
 
From the stoichiometry εa = (2.5-1)/1 = 1.5 
If we write equation for plug flow reaction, 
(V/ν)* CA0 = ∫ 𝑑𝑋𝐴
XAf
0
/(-rA) 
CA = CA0(1- XA)/(1+ εa XA) 
So (V/ν)* CA0 * k = - (1+ εa) ln(1- XA) -1.5 XA 
4.4 = -2.5 ln (1- XA) - 1.5 XA 
XA = 0.733 …ans. 
 
 
 
 
 
 
 
 
 
 
 
1 
 
Group 12 
13bch024, 13bch033, 13bch048, 13bch059 
 
Solution 5.11: 
Solution- 
 
A → R −rA = 0.1CA/ (1 + 0.5CA) 
 
According to the performance equation of mixed flow reactor, 
τm
CA0
 = 
XA
−rA
 - (1) 
Now substituting the value of XA from the equation, 
 CA = CAO(1 − XA) - (2) 
We get, 
 τm = 
CAO −CA
0.1CA
(1 + 0.5CA)
 - (3) 
Now, conversion is 95 % so, XA= 0.95 
Therefore, substituting the values in the equation, we get 
 τm = 
(2−0.1)(1+0.5(0.1))
0.1(0.1)
 = 199.5 min 
 
Therefore, 
V = τmvo = 199.5(25) L = 4987.5 L = 5 m
3 
 V = 5 𝐦𝟑 
 
 
 
 
 
 
 
2 
 
 
 
 
 
Problem 5.12: 
Solution- 
 
A + B → R −rA= 200CACB (mol)/(litre).(min) 
 
Now, 
CA = 0.1 mol/L 
CB = 0.4 mol/L 
vo = 400 L/min 
Hence, from the equation: CA = CAO(1 − XA) , we get XA = 0.99 
 τm = 
XA.CA0
−rA
 , 
 
CA = CAO(1 − XA) 
 CB = CAO(MB − (b/a)XA) Here, MB = 200/100 = 2 
Now, b = a = 1 
 
−rA = 200(0.1)
2(1 − XA)(2 − XA) 
 = 0.0202 
τm = 
XA.CA0 
−rA
 = τm = 
(0.99)(0.1)
0.0202
 = 49.9 min 
 
Therefore, 
V = τmvo = 49.9(400) L = 19960 L = 19.96 m
3 
 V = 19.96 𝐦𝟑 
 
 
3 
 
 
 
 
Problem 5.13 
Solution- 
 
4PH3 → P4 (g) + 6H2 −rPH3 = 10CPH3 h
−1; k = 10 
XA = 0.75 
P = 11.4 atm 
FAO = 10 mol/hr 
PH3 = (2/3) of feed Inerts = (1/3) of feed 
 
As it is a gaseous reaction we need to find ε𝐴 Hence, 
ε𝐴 = (1 + 6 - 4)(2) / 4(3) = 0.5 
The reaction is of a first order hence, the integral law can be written as 
τp = 
1
10
[(1 + ε𝐴)(-ln(1-XA) - ε𝐴 XA] 
τp = 0.17 
 
CAO = 
𝑃𝐴𝑂
𝑅𝑇𝑜
 = 
11.4(2/3)
0.082(649+273)
 = 0.1 mol/L 
v0 = 
𝐹𝐴0
𝐶𝐴0
 = 10/0.1 = 100 L/h 
Therefore, 
V = τpvo = 0.17(100) L = 17 L 
V = 17 L 
 
 
 
 
4 
 
 
 
 
Problem 5.14 
Solution- 
 
3A → R −rA = 54 mmol/ L.min = 0.054 mmol/ L.min 
CAO = 0.6 mol/L 
CA = 0.33 mol/L 
FAO = 0.54 mol/min 
 
Now, 
ε𝐴 = 2/3 
 
Hence, 
CA = 
𝐶𝐴0(1− XA)
(1+ ε𝐴XA)
 
330 = 660
(1−XA)
(1+(
2
3
)XA)
 therefore, XA = 0.75 
Therefore, 
τp = ∫
𝑑𝑋𝐴
−𝑟𝐴
0.75
0
 = 
𝐶𝐴0XA
−𝑟𝐴
 = 9.17 min 
 
V = τpvo = τp
FAO
CAO
 = 9.17(540/660) = 7.5 L 
 
 
 
 
 
 
5 
 
 
 
 
 
Problem 5.15 
Solution- 
 
2 A → R −𝑟𝐴 = 0.05 CA
2
 mol/L s 
Now, 
τm = 
XA.CA0
−rA
 
ε𝐴 = 0.5 
From the equation, 
CA = 
𝐶𝐴0(1− XA)
(1+ ε𝐴XA)
 
330 = 660
(1−XA)
(1− 0.5XA)
 , therefore on solving we get XA = 0.66 
 
Now on expanding the rate expression we get, 
−𝑟𝐴 = 0.05𝐶𝐴0
2(
(1−XA)
(1− 0.5XA)
)2 
Hence from, 
τm = 
XA.CA0
−rA
 = 54.442 minTherefore, 
Vo = 
𝑉
τm
 = 2/54.442 = 0.036 L/min 
 
 
 
 
6 
 
 
 
 
 
Problem 5.16 
Solution- 
 
A → 3 R −𝑟𝐴 = 0.6 CA 𝑚𝑖𝑛
−1 
The feed contains of 50% A and 50% inerts 
vo = 180 litre/min 
CA0 = 0.3 mol/litre 
V = 1000 L = 𝑚3 
Now, 
τm = 
XA.CA0
−rA
 = 
V
v0
 
ε𝐴 = 2(0.5) = 1, 
 
Hence the rate expression can be expanded as, 
−𝑟𝐴 = 0.6 CA
(1−XA)
(1+ εAXA)
 = 0.6 CA
(1−XA)
(1+ XA)
, 
τm = 
XA.CA0
−rA
 = 
V
v0
 = 
XA(1+ XA)
CA0(1− XA)
 
we obtain, 
3𝑋𝐴
2 + 13𝑋𝐴 − 10 = 0 
 
XA = 
−13± √169+40(3)
2(3)
 = 0.67 
 
 
 
 
7 
 
 
 
 
 
Problem 5.17 
Solution- 
 
2O3 → 3O2 −𝑟𝑜𝑧𝑜𝑛𝑒 = kCozone
2
 , k = 0.05 litre/ (mol.s) 
 
The rate expression shows that it is a second order reaction so, the integral expression is 
defined as, 
𝑘𝜏𝑝𝐶𝐴0 = 2𝜀𝐴(1 + 𝜀𝐴) ln(1 − 𝑋𝐴) + 𝜀𝐴
2𝑋𝐴 + (𝜀𝐴 + 1)
2 𝑋𝐴
1 − 𝑋𝐴
⁄ 
 
Now, 
𝐶𝐴0= 
𝑃𝐴0
𝑅𝑇0
 = 1.5(0.2)/(0.082)(95 + 273) = 0.01 mol/L 
𝜀𝐴= (3-2)/2 x (0.2) = 0.1 
 
Hence, putting in the equation we get 
𝜏𝑝 =
𝑉
𝑣0
=
1
0.05(0.01)
{2[0.1(1.1)𝑙𝑛0.5 + 0.12(0.5) + 1.12(1))} 
𝜏𝑝 = 2125.02 s 
 
V = 2125.02 (1 L/s) = 2125 L = 2.125 m3 
V = 2.125 𝐦𝟑 
 
 
 
 
 
 
8 
 
 
 
 
 
Problem 5.18 
Solution- 
 
2A → R −𝑟𝐴 = 0.05CA
2
, mol/litre.s 
 
FA0 = 1 mol/litre 
V = 2 L 
𝑣0 = 0.5 litre/min 
 
Therefore, 
𝜏𝑝 =
𝑉
𝑣0
 = 
2 𝐿
0.5 𝐿/𝑚𝑖𝑛
 = 4(60) s = 240 s 
The integral rate expression for the 2nd order reaction is given by, 
𝑘𝜏𝑝𝐶𝐴0 = 2𝜀𝐴(1 + 𝜀𝐴) ln(1 − 𝑋𝐴) + 𝜀𝐴
2𝑋𝐴 + (𝜀𝐴 + 1)
2 𝑋𝐴
1 − 𝑋𝐴
⁄ 
 Now, 
𝜀𝐴 = 0 
Therefore, the equation will reduce to 
𝑘𝜏𝑝𝐶𝐴0 =
𝑋𝐴
1 − 𝑋𝐴
 
So, substituting the values we get, 
𝑘𝜏𝑝𝐶𝐴0 =
𝑋𝐴
1 − 𝑋𝐴
 
𝑋𝐴 = 
𝑘𝜏𝑝𝐶𝐴0
1+𝑘𝜏𝑝𝐶𝐴0
 = 
0,05(240)(1)
1+(0.05)(240)(1)
 = 0.92 
 
 
 
9 
 
 
 
 
 
Problem 5.19 
Solution 
 
P = 3 atm A → 3R 
T = 30°𝐶 
𝐶𝐴𝑂 = 0.12 mol/litre 
 
𝑉𝑂 𝐶𝐴 
0.06 30 
0.48 60 
1.5 80 
8.1 105 
 
Now, 
τm = 
XA.CA0
−rA
 = 𝑉
𝑣0
 – (1) 
Similarly, 
𝜀𝐴 =
(3−1)1
1
 =2 
Now, from the equation 
𝐶𝐴 = 
𝐶𝐴0(1−𝑋𝐴)
(1+𝜀𝐴𝑋𝐴)
 
We get different values of 𝑋𝐴 and from that we calculate different values of –𝑟𝐴 
Hence, we obtain the following data 
 
 
 
 
XA −rA 
0.5 3.6 
10 
 
0.25 14.4 
0.143 25.74 
0.045 44.18 
 
 
Hence, from the rate expression 
−rA = 𝑘𝐶𝐴
𝑛 
Taking log on both sides 
ln(−rA) = lnk + lnCA 
 
Now plotting ln(−rA) on Y-axis and lnCA on X-axis 
We get, 
 
 
Hence, slope = n = (3.78-1.2809)/(4.653-3.4011) = 2 
Similarly, k = 250 
 
Hence, final rate expression will be 
−rA = 250𝐶𝐴
2 
 
 
 
 
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.5 1 1.5 2 2.5 3 3.5 4
Chart Title
11 
 
 
 
 
Problem 5.20 
Solution- 
 
A → R 
CA0 = 0.1 mol/litre 
V = 1 m3 
 
𝑣 𝐶𝐴 
1 4 
16 20 
24 50 
 
For mixed flow reactor, performance equation is written as 
𝜏𝑚 =
𝐶𝐴0 − 𝐶𝐴
−𝑟𝐴
 
Therefore, 
−𝑟𝐴 =
𝐶𝐴0 − 𝐶𝐴
𝜏𝑚
 
−𝑟𝐴 =
(100 − 𝐶𝐴)𝑣0
𝑉
 
 
So, 
−rA CA ln(−rA) ln(CA) 
96 4 4.5643 1.386 
480 20 6.173 2.995 
1200 50 7.09007 3.91 
 
−rA = 𝑘𝐶𝐴
𝑛 
Now taking log on both sides, 
ln(−rA) = lnk + lnCA 
 
12 
 
 
Therefore, slope = n= 1 
Similarly the value of k can be evaluated by k = 
𝐶𝐴
−𝑟𝐴
 = 
50
1200
= 0.0417 𝑚𝑖𝑛−1 
So, 
−𝑟𝐴 = 0.0417𝐶𝐴 mmol/L.min 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 2 4 6 8
Chart Title
13 
 
Problem 5.21 
Solution- 
 A → R 
CA0 = 1.3 mol/litre 
𝐶𝐴 = 0.3 mol/litre 
𝐶𝐴 −𝑟𝐴 
0.1 0.1 
0.2 0.3 
0.3 0.5 
0.4 0.6 
0.5 0.5 
0.6 0.25 
0.7 0.1 
0.8 0.06 
1.0 0.05 
1.3 0.045 
2.0 0.042 
 
 It is a batch reactor so, the performance equation of the reactor is, 
∫
𝑑𝐶𝐴
−𝑟𝐴
𝐶𝐴0
𝐶𝐴
 = ∫
𝑑𝐶𝐴
−𝑟𝐴
1.3
0.3
 = 
Therefore, we need to find the area under the curve of −𝑟𝐴 and CA from 0.3 to 1.3 
 
Hence, in this plot the area under the curve from 0.3 to 1.3 we can calculate the residence 
time 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250 300 350 400 450
14 
 
t = ∫
𝑑𝐶𝐴
−𝑟𝐴
1.3
0.3
 = 12.6 min 
 
 
Problem 5.22 
Solution- 
Given, 
𝑋𝐴 = 0.8 
𝐹𝐴0 = 1000 mol/hr 
𝐶𝐴0 = 1.5 mol/litre 
From, 𝐶𝐴 = 𝐶𝐴0(1 − 𝑋𝐴) we get CA = 0.3 mol/litre 
Now, the performance equation of the plug flow reactor is 
𝜏𝑝 = ∫
𝑑𝐶𝐴
−𝑟𝐴
𝐶𝐴0
𝐶𝐴
 
Similarly from the example 5.21 we obtain 
∫
𝑑𝐶𝐴
−𝑟𝐴
1.3
0.3
 = 12.8 min 
Now, the time can be found by finding out the area between 1.3 to 1.5 
 
So, the area between 1.3 to 1.5 we obtain the area = 4.5 min 
Therefore, 
The total area = 12.8 +4.5 min 
 = 17.3 min 
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250 300 350 400 450
15 
 
Problem 5.23 
Solution- 
𝑋𝐴 = 0.75 
𝐹𝐴0 = 1000 mol/hr 
𝐶𝐴0 = 1.2 mol/litre 
(a) The performance equation of mixed flow reactor is 
𝛕𝐦 = 
𝐗𝐀.𝐂𝐀𝟎
−𝐫𝐀
 
Now from the equation, 𝐶𝐴 = 𝐶𝐴0(1 − 𝑋𝐴) so, substituting the values we get 𝐶𝐴 = 0.3 mol/L 
and −𝑟𝐴 = 0.5 mol/L 
 Therefore, 
𝑉
FA0
=
(𝐶𝐴0 − 𝐶𝐴)
−𝑟𝐴
 
V = 1000(1.2-0.3)/1.2(0.5) (60) = 1500 L 
 
(b) 
𝐹𝐴0 = 2000 mol/hr 
Hence, when the feed rate is doubled the volume also doubles. 
Therefore, V = 50 ltr 
 
 
(c) 
𝐶𝐴0 = 2.4 mol/litre 
𝐶𝐴 = 0.3 mol/litre 
Therefore putting it in the performance equation we get, 
𝑉
FA0
=
(𝐶𝐴0 − 𝐶𝐴)
−𝑟𝐴
 
V = 1000(2.4-0.3) /(2.4)(0.5)(60) = 29.167 ltr 
 
 
 
 
16 
 
Problem 5.24 
Solution- 
A → 5 R 
V = 0.1 litre 
CA0 = 100 mmol/litre = 0.1 mol/litre 
𝐹𝐴0 mmol/ltr 𝐶𝐴 mmol/ltr 
300 16 
1000 30 
3000 50 
5000 60 
εA = 4 
τm = 
XA.CA0
−rA
=
𝑉
𝑣0
 
−rA = 
v0XA.CA0
V
 = 10FA0XA 
Therefore, 
𝑋𝐴 =
1−
𝐶𝐴
𝐶𝐴0
1+𝜀𝐴
𝐶𝐴
𝐶𝐴0
 , from this we generate the following data 
FA0 CA XA −rA 
300 16 0.512 1536.6 
1000 30 0.318 318.8 
3000 50 0.167 5000 
5000 60 0.118 5882.4 
 
−rA = 𝑘𝐶𝐴
𝑛 
Now taking log on both sides, 
ln(−rA) = lnk + nlnCA 
 
Slope = 1; k = 0.01 
−𝐫𝐀 = 𝟎. 𝟎𝟏𝐂𝐀 
0
2
4
6
8
10
0 1 2 3 4 5
Chart Title
17 
 
Problem 5.25 
Solution- 
XA = 0.75; It is a batch reactor 
CA0 = 0.8 mol/litre 
CA in feed stream CA in exit stream Holding time 
2 0.65 300 
2 0.92 240 
2 1 250 
1 0.56 110 
1 0.37 360 
0.48 0.42 24 
0.48 0.28 200 
0.48 0.20 560 
 
Hence, holding time = residence time = 𝜏 
𝜏 = − ∫
𝑑𝐶𝐴
−𝑟𝐴
𝐶𝐴
𝐶𝐴0
 ; 
So, we generate the following data, by using the correlation of 𝜏 =
𝐶𝐴0−𝐶𝐴
−𝑟𝐴
 
CA 1/(−rA) 
0.65 222 
0.92 222 
1 250 
0.56 250 
0.37 572 
0.42 400 
0.28 1000 
0.2 2000 
 
On plotting the graph of 1/(−rA) vs CA 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250
18 
 
Therefore, area under the curve will give the residence time 
 𝝉 = 300 sProblem 5.26 
Solution- 
If it is a mixed flow reactor 
τm = 
XA.CA0
−rA
=
𝑉
𝑣0
 
Following the same procedure and plotting the same graph of 1/(−rA) vs CA we find the area 
of the complete rectangle because it is a MFR so, 
 
 
𝜏 = 900 s 
 
 
 
 
 
 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250
19 
 
Problem 5.27 
Solution- 
(a) The conversion drops from 80 to 75 % so, it is quite possible that Imbibit fell into the vat 
containing alcohol. The reason for the drop in conversion is due to decrease in the volume of 
liquid available for the reaction. 
 
(b) The revelation couldn’t be made because Watson asked for some dill (for smoking) but he 
didn’t have any idea that smoking near a vat of ethanol isn’t a very good idea. 
 
 
 
Problem 5.28 
Solution- 
2A→ R + S 
The PFR is operating at 100℃ and 1 atm 
FA0 = 100 mol/hr 
XA = 0.95 
The feed contains 20% inerts and 80% A 
So, 
−rA = kCA
n 
The reaction is of 1st order 
So, the integral form of equation in terms of conversion is obtained which is 
𝑘𝑡 = −ln (1 − 𝑋𝐴), which can be also written as 𝑋𝐴 = 1 −
𝐶𝐴
𝐶𝐴0
 – (1) 
Now, 
CA= 
𝑝𝐴
𝑅𝑇
 
So, replacing the term in the equation (1) we get, 
𝑘𝑡 = 𝑙𝑛
𝑝𝐴0
𝑝𝐴
 
Hence, we now generate a data 
 
 
20 
 
T ln (𝑝𝐴0/𝑝𝐴) 
0 1 
20 1.25 
40 1.47 
60 1.78 
80 2.22 
100 2.702 
140 4 
200 7.14 
260 12.5 
330 25 
420 50 
 
 
 
Therefore, slope = k = 0.01116 s−1 
The reaction is 1st order reaction hence, the rate expression is given by 
−rA = 0.01116CA 
We can also write the equation like 
kτp = k
V
v0
= −ln (1 − XA) Therefore, V = −
𝑣0
𝑘
ln (1 − 𝑋𝐴) 
𝑣𝑜 =
𝐹𝐴0
𝐶𝐴0
=
𝐹𝐴0𝑇0𝑅
𝑝𝐴0
=
100(373)(0.082)
0.8
= 3823.25
𝐿
ℎ
=
1.06𝐿
𝑠
 
 
V = 1.06ln (1-0.95)/0.01116 = 284.54 L = 2.84 m3 
 
 
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50 100 150 200 250 300 350 400 450
Chart Title
21 
 
Problem 5.29 
Solution- 
 
Now, for the same problem if we take a MFR then, the performance equation is written as 
Plotting Pa v τ 
 
Slope = rate = 0.000375 
τm = 
(pA0−pA)
−rA
 = (0.8-0.04)/ (0.000375) = 2026.67 s 
𝑣𝑜 =
𝐹𝐴0
𝐶𝐴0
=
𝐹𝐴0𝑇0𝑅
𝑝𝐴0
=
100(373)(0.082)
0.8
= 3823.25
𝐿
ℎ
=
1.06𝐿
𝑠
 
V = 2026.67(1.06) = 2.148 L = 2.15 m3 
 
 
 
 
 
 
 
 
 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250 300 350 400 450
22 
 
Problem 5.30 
Solution- 
A → R 
CA0 = 0.8 mol/litre 
FA0 = CA0(V) = 0.8 mol/s 
0.8 mol/litre 
XA = 0.75 
V = 1 litre/s 
 
(a) The reactor is a plug flow reactor 
The performance equation is 
V
FA0
=
τ
CA0
= ∫
dXA
−rA
XA
0
 
It can be modified as 
𝑉
𝐹𝐴0
= 𝜏 = − ∫
𝑑𝐶𝐴
−𝑟𝐴
𝐶𝐴
𝐶𝐴0
 
Hence, now from the equation, 𝜏 =
𝐶𝐴0−𝐶𝐴
−𝑟𝐴
 we can generate the following table 
 
 
τ CA0 CA −rA 
50 2 1 0.02 
16 1.2 0.8 0.025 
60 2 0.65 0.0225 
22 1 0.56 0.02 
4.8 0.48 0.42 0.0125 
72 1 0.37 0.00875 
40 0.48 0.28 0.005 
1122 0.48 0.2 0.0025 
 
Hence, on plotting the graph of 1/( −rA) vs CA 
23 
 
 
So, for PFR we need to find the area under the curve. 
Therefore, Area = τ = 70 s 
So, putting in the performance equation we get, 
V= (70)0.8/0.8 = 70 ltr 
 
(2) The reactor is MFR 
For MFR, we need to find the rectangular area which is 
Area = τ = 320 s 
V = 320(0.8)/ (0.8) = 320 
Therefore, volume required = 320 ltr 
 
 
 
0
50
100
150
200
250
300
350
400
450
0 0.2 0.4 0.6 0.8 1 1.2
 
 
 
 
 
Solution:- 
 
Case 1:- n1=1, n2=2, n3=3 
Use MFR with some particular concentration of A. 
Case 2) n1=2, n2=3, n3=1 
Use PFR with low XA 
Case 3) n1=3, n2=1, n3=2 
Use MFR with high XA 
 
 
7.2
 
 
Solution:- 
A + B → R 
A → S 
Low CA, high CB 
 
 
 
 
7.3
 
 
Solution:- 
A + B → R 
2A → S 
2B → T 
Use low CA and low CB 
Slowly add A & B 
Don’t add excess of A or B. 
 
 
 
 
7.4 
 
 
 
A + B → R 
A → S 
First add B completely then add A slowly & steadily. 
 
 
 
 
 
7.5 
 
 
A + B → R 
2A → S 
Keep CA low , CB high. 
 
 
 
 
 
 
Solution:- 
 
FOR TANK-1 
𝜓𝑅/𝐴 =
0.4 − 0
1 − 0.4
= 
0.4
0.6
= 0.67 
𝜏1 =
𝐶𝑅 − 𝐶𝑅0
𝑘1 ∗ 𝐶𝐴
 
2.5 =
0.4
𝑘1 ∗ 0.4
 
 K1 = 0.4 
𝜏1 =
𝐶𝑆 − 𝐶𝑆0
𝑘2 ∗ 𝐶𝐴
 
2.5 =
0.2
𝑘1 ∗ 0.4
 
 
 
 K2 = 0.2 
 
For tank – 2 
𝜏2 =
𝐶𝐴1 − 𝐶𝐴2
−𝑟𝐴
 
5 =
0.4 − 𝐶𝐴2
(𝐾1 + 𝐾2)𝐶𝐴2
 
3 𝐶𝐴2 = 0.4 − 𝐶𝐴2 
𝐶𝐴2 = 0.1
𝑚𝑜𝑙
ℎ𝑟
 
 𝐶𝑅2 = 𝐶𝑅1 + 𝜓𝑅
𝐴
(𝐶𝐴1 − 𝐶𝐴2) 
 = 0.4 +
𝐾1
𝐾1+𝐾2
(0.4 − 0.1) 
 = 0.6 𝑚𝑜𝑙/ℎ𝑟 
𝐶𝑆2 = 𝐶𝑆1 + 𝜑𝑆
𝐴
(𝐶𝐴1 − 𝐶𝐴2) 
 = 0.2 +
𝐾1
𝐾1+𝐾2
(0.3) 
 = 0.3 mol/hr 
 
 
 
Solution:- 
FOR TANK-1 
𝜓𝑅
𝐴
=
𝑁𝑅1 − 𝑁𝑅0
𝑁𝐴0 − 𝑁𝐴1
 
=
0.2 − 0
1 − 0.7
 
= 0.2857 
𝜏 =
𝐶𝐴0 ∗ 𝑋𝐴
−𝑟𝑅
 
=
𝐶𝑅 − 𝐶𝑅0
𝑘1 ∗ 𝐶𝐴
2 
 2.5 =
0.2−0
𝐾1∗0.42
 
 K1= 0.5 
 
𝜏 =
𝐶𝑆 − 𝐶𝑆0
𝑘2 ∗ 𝐶𝐴
 
 
 
2.5 =
0.7 − 0.3
𝐾2 ∗ 0.4
 
𝐾2 = 0.4 
FOR TANK -2 
𝜏2 =
𝐶𝐴1 − 𝐶𝐴2
−𝑟𝐴
 
10 =
0.4 − 𝐶𝐴2
𝐾1𝐶𝐴2
2 + 𝐾2𝐶𝐴2
 
5𝐶𝐴2
2 + 4𝐶𝐴2 = 0.4 − 𝐶𝐴2 
5𝐶𝐴2
2 + 5𝐶𝐴2 − 0.4 = 0 
𝐶𝐴2 = 0.074 
 
𝐶𝑅2 = 𝐶𝑅1 + 𝜓𝑅
𝐴
(𝐶𝐴1 − 𝐶𝐴2) 
= 0.2 +
𝐾1𝐶𝐴2
𝐾1𝐶𝐴2
(0.4 − 0.074) 
 = 0.2 + 0.5∗0.074(0.0326)
0.5∗0.074+0.4
 = 0.227 
 
𝐶𝑆2 = 𝐶𝑆1 + 𝜓𝑆
𝐴⁄
(𝐶𝐴1 − 𝐶𝐴2) 
 = 0.7 +
𝐾2
𝐾1𝐶𝐴2+𝐾2
(0.4 − 0.074) 
 𝐶𝑆2 = 0.998 
 
 
 
 
Solution::--- 
CR = CR0 + ψ R/A (CA0-CA) 
 = 0 + 
0.4∗4
0.4∗4+2
(40 − 4) 
 = 1.6*36/3.6 
 = 16 mol/L 
CS = 36-16 = 20 mol/L 
τ = 
𝐶𝐴0−𝐶𝐴
−𝑟𝐴
 
τ = 
40−4
0.4(4)2+2(4)
 
τ = 36/14.4 
τ = 2.5 sec 
 
 
 
 
 Solution::--- 
𝑑𝐶𝑅
𝑑𝑡
= 𝑟𝑅 = 𝑘1𝐶𝐴
2 = 0.4𝐶𝐴
2 
𝑑𝐶𝑆
𝑑𝑡
= 𝑟𝑆 = 𝑘2𝐶𝐴 = 0.4𝐶𝐴 
𝜓𝑅
𝐴⁄
 =
𝑟𝑅
−𝑟𝐴
 
 =
𝐾1𝐶𝐴
2
𝐾1𝐶𝐴
2 + 𝐾2𝐶𝐴
 
 =
0.4𝐶𝐴
0.4𝐶𝐴 + 2
 
𝐶𝑅 = ∫ 𝜓
𝐶𝐴0
𝐶𝐴
𝑑𝐶𝐴 
= ∫ (
0.4𝐶𝐴
0.4𝐶𝐴 + 2
)
40
4
𝑑𝐶𝐴 
 
 
= ∫ (1 −
2
0.4𝐶𝐴 + 2
)
40
4
𝑑𝐶𝐴 
= (40 − 4) − [
2
0.4
 ln (0.4𝐶𝐴 + 2)]4
40
 
= 36 − 5 𝑙𝑛
18
3.6
 
𝐶𝑅 = 27.953 
𝐶𝑆 = 36 − 𝐶𝑅 = 8.047 
𝜏 = ∫
𝑑𝐶𝐴
−𝑟𝐴
40
4
 
 =∫
𝑑𝐶𝐴
4𝐶𝐴
2 + 2𝐶𝐴
 
40
4
 
 = 2.5 ∫
𝑑𝐶𝐴
𝐶𝐴
2 + 5𝐶𝐴 + (
5
2)
2 − (
5
2)
2
40
4
 
 = 2.5 ∫
𝑑𝐶𝐴
(𝐶𝐴 +
𝜋
2)
2 − (
5
2)
2
40
4
 
 = 0.346 sec 
 
 
 
 
 
 
Solution::--- 
 𝑟𝑅 = 0.4 ∗ 𝐶𝐴
2 
𝑟𝑆 = 2 ∗ 𝐶𝐴 
𝜓𝑆
𝐴⁄
=
𝑑𝐶𝑆
−𝑑𝐶𝐴
= 
2𝐶𝐴
2𝐶𝐴 + 0.4𝐶𝐴
2 = 
1
1 + 0.2𝐶𝐴
= 
10
10 + 2𝐶𝐴
 
 
Ca ψ 
0 1 
1 0.833333 
10 0.333333 
20 0.2 
40 0.111111 
 
 
 
𝐴 = (
10
10 + 2𝑓𝐴
) (40 − 𝑐𝐴𝑓) = 
400 − 10𝑐𝐴𝑓
10 + 2𝑐𝐴𝑓
 
𝑑𝐴
𝑑𝐶𝐴
= 10 + 2𝐶𝐴𝑓(−10) − (400 − 10𝐶𝐴𝑓)(2) 
0 = −100 − 20𝐶𝐴𝑓 − 800 + 20𝐶𝐴𝑓 
−100 − 20𝐶𝐴𝑓 = 800 − 20𝐶𝐴𝑓 
𝐶𝐴𝑓 = 0 
XA = 100% 
𝜏
40
=
1
2(0) + 0.4(0)
 
𝜏 = ∞ 
 
0
0.2
0.4
0.6
0.8
1
1.2
-10 0 10 20 30 40 50
ψ
 
Ca
ψ 
 
 
 
 
 
 SOLUTION::--- 
𝜓𝑅
𝐴⁄
 = 
𝐾1𝐶𝐴
2
𝐾1𝐶𝐴
2 + 𝐾2𝐶𝐴
=
0.4𝐶𝐴
0.4𝐶𝐴 + 2
=
𝐶𝐴
𝐶𝐴 + 5
 
𝑑𝜓𝑅
𝐴⁄
𝑑𝐶𝐴
= (𝐶𝐴 + 5) ∗ 1 − 𝐶𝐴 ∗ 1 
 
Ca ψ 
0 0 
1 0.166667 
10 0.666667 
20 0.8 
40 0.888889 
 
 
 
 
𝐴 =
40𝐶𝐴𝑓 − 𝐶𝐴𝑓
2
𝐶𝐴𝑓 + 5
 
𝑑𝐴
𝑑𝐶𝐴𝑓
= (𝐶𝐴𝑓 + 5)(40 − 2𝐶𝐴𝑓) − (
40𝐶𝐴𝑓 − 𝐶𝐴𝑓
2
(𝐶𝐴𝑓 + 5)2
) 
40𝐶𝐴𝑓−2𝐶𝐴𝑓
2 + 200−10𝐶𝐴𝑓−40𝐶𝐴𝑓 + 𝐶𝐴𝑓
2 = 0 
𝐶𝐴𝑓
2 + 10𝐶𝐴𝑓 − 200 = 0 
𝐶𝐴𝑓 =
−10 + 30
2
= 10 
𝑋𝐴 = 1 − (
10
40
) 
𝜏
40
=
0.75
0.4𝐶𝐴
2 + 2𝐶𝐴
 
𝜏 =
30
0.4 ∗ (10)2 + 2 ∗ (10)
 
 τ = 
30
40+20
 = 0.5 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-5 0 5 10 15 20 25 30 35 40 45
ψ 
CRE-I SPECIAL ASSIGNMENT 
QUESTION: [7.12.] 
Reactant A in a liquid either isomerizes or dimerizes as follows: 
𝐴 → 𝑅𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑟𝑅 = 𝑘1𝐶𝐴 
𝐴 + 𝐴 → 𝑆𝑢𝑛𝑤𝑎𝑛𝑡𝑒𝑑 𝑟𝑆 = 𝑘2𝐶𝐴
2 
(a) Write 𝜑 (
𝑅
𝐴
) and 𝜑 [
𝑅
(𝑅+𝑆)
]. 
With a feed stream of concentration CA0, find CR,max which can be 
formed: 
(b) In a plug flow reactor. 
(c) In a mixed flow reactor. 
A quantity of A of initial concentration CA0 = 1 mol/liter is dumped 
into a batch reactor and is reacted to completion. 
(d) If Cs = 0.18 mol/liter in the resultant mixture, what does this tell 
of the kinetics of the reaction? 
ANSWER: 
(a) 𝜑 [
𝑅
(𝑅+𝑆)
] = 
𝑟𝑅
𝑟𝑅+ 𝑟𝑆
 = 
𝑘1𝐶𝐴
𝑘1𝐶𝐴 + 𝑘2𝐶𝐴
2 
 
 𝜑 (
𝑅
𝐴
) = 
𝑟𝑅
−𝑟𝐴
 = 
𝑘1𝐶𝐴
𝑘1𝐶𝐴 + 2𝑘2𝐶𝐴
2 
 
(b) 
We get CR,max when we have CAF = 0. 
So, CR,max = ∫ 𝜑𝑑𝐶𝐴
𝐶𝐴0
0
 = ∫
1
1 + 
2𝑘2
𝑘1
𝐶𝐴 
𝑑𝐶𝐴
𝐶𝐴0
0
 = 
𝑘1
2𝑘2
 {ln( 1 +
2𝑘2
𝑘1
𝐶𝐴)}
𝐶𝐴0
0
 
 CR,max = 
𝑘1
2𝑘2
 {ln( 1 +
2𝑘2
𝑘1
𝐶𝐴0) − ln 1} = 
𝑘1
2𝑘2
 {ln( 1 +
2𝑘2
𝑘1
𝐶𝐴0)} 
(c) 
 CRm = 𝜑𝑓(𝐶𝐴0 − 𝐶𝐴) 
CRm, max = 1(𝐶𝐴0 − 0) = 𝐶𝐴0 
(d) 
CS = 0.18 gives CR = CA0 – CA – CS = 1 – 0 – 0.18 = 0.82 
Also, 
CR can be calculated from the following equation: 
CR,max = 
𝑘1
2𝑘2
 {ln( 1 +
2𝑘2
𝑘1
𝐶𝐴0)} 
Let’s say, K = k1/k2 and assume K = 4 and K = 5. 
So, CR calculated from the above equation is 0.84 and 0.81 
respectively. 
K 4 5 
CR calculated 0.84 0.81 
Hence, for different values of k1 and k2, CR,max is different. It may be 
more or less than the CR value. 
 
QUESTION: [7.13.] 
In a reactive environment, chemical A decomposes as follows: 
 𝐴 → 𝑅, 𝑟𝑅 = 𝐶𝐴
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟𝑠
 
 𝐴 → 𝑆, 𝑟𝑆 = 1
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
 
For a feed stream 𝐶𝐴0 = 4
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟𝑠
 what size ratio of two mixed flow 
reactors will maximize the production rate of R? Also give the 
composition of A and R leaving these two reactors. 
 
ANSWER: 
 
𝜑 = 
𝑟𝒓
−𝑟𝐴
= 
𝐶𝐴
1 + 𝐶𝐴
 
𝐶𝐴 → 0, 𝜑 → 0 𝑎𝑛𝑑 𝐶𝐴 → ∞, 𝜑 → 1 
𝐶𝑅 = 𝜑∆𝐶𝐴 = (
𝐶𝐴1
1 + 𝐶𝐴1
) (4 − 𝐶𝐴1)+ = (
𝐶𝐴2
1 + 𝐶𝐴2
) (𝐶𝐴1 − 𝐶𝐴2) 
 
CA1 and CA2 are not known, but there is a fixed CA1 and CA2 value that 
maximizes CR and is what makes dCR/dCA1 = 0 
𝑑𝐶𝑅
𝑑𝐶𝐴1
= 0 = 
(1 + 𝐶𝐴1)(4 − 2𝐶𝐴1) − (4𝐶𝐴1 − 𝐶𝐴1
2 )(1)
(1 + 𝐶𝐴1)2
+ 
𝐶𝐴2
1 + 𝐶𝐴2
(1) 
 
So, 
4 − 2𝐶𝐴1 − 𝐶𝐴1
2
(1 + 𝐶𝐴1)2
= −
𝐶𝐴2
1 + 𝐶𝐴2
 
If CA2 = 0.5 mol, 
4 − 2𝐶𝐴1 − 𝐶𝐴1
2
(1 + 𝐶𝐴1)2
= −
0.5
1 + 0.5
 
13 − 4𝐶𝐴1 − 2𝐶𝐴1
2 = 0 
𝐶𝐴1 = 
4 ± √42 − 4(13)(−2)
2
2(−2)
= 1.7386
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
 
𝐶𝑅2 =
1.7386
1 + 1.7386
(4 − 1.7386) +
0.5
1 + 0.5
(1.7386 − 0.5)
= 1.8485 
 
 
 
 
 
 
 
 
 
 
QUESTION: 
Consider the parallel decomposition of A of different orders as 
follows: 
𝐴 → 𝑅, 𝑟𝑅 = 1 
 
𝐴 → 𝑆, 𝑟𝑆 = 2𝐶𝐴 
 
𝐴 → 𝑇, 𝑟𝑇 = 𝐶𝐴
2 
Determine the maximum concentration of desired product 
obtainable in (a) plug flow, (b) mixed flow. 
[7.14.]: R is the desired product and𝐶𝐴0 = 2. 
 ANS: 
𝜑𝑅 = 
1
 1 + 2𝐶𝐴+ 𝐶𝐴
2 
(a) 
𝐶𝑅,𝑚𝑎𝑥 = ∫
𝑑𝐶𝐴
1 + 2𝐶𝐴 + 𝐶𝐴
2
2
0
= ∫
𝑑𝐶𝐴
(𝐶𝐴 + 1)2
2
0
= ( 
(𝐶𝐴 + 1)
−1
−1
)
0
2
= −
1
1 + 2
+ 
1
1
= 
2
3
 
(b) 
𝜑𝑅 = 
1
1 + 2𝐶𝐴 + 𝐶𝐴
2 
When, 𝐶𝐴 = 0; 𝐶𝑅 = 𝐶𝑅,𝑚𝑎𝑥 
So, 𝐶𝑅𝑚, 𝑚𝑎𝑥 = 𝜑𝐶𝐴 = 0 (2 − 0) = 1(2) = 2 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟 
[7.15.]: S is the desired product and𝐶𝐴0 = 4. 
ANS: 
𝜑𝑅 = 
2𝐶𝐴
1 + 2𝐶𝐴 + 𝐶𝐴
2 = 
(a) 
When, 𝐶𝐴 = 0; 𝐶𝑆 = 𝐶𝑆𝑃, 𝑚𝑎𝑥 
 𝐶𝑆𝑃, 𝑚𝑎𝑥 = ∫ 
𝑑𝐶𝐴
1
2𝐶𝐴
+ 1+ 
𝐶𝐴
2
= 2 ∫
𝐶𝐴𝑑𝐶𝐴
(𝐶𝐴+ 1)2
4
0
 
4
0
 
∫
𝑥𝑑𝑥
(𝑎 + 𝑏𝑥)2
= 
1
𝑏2
[𝑙𝑛(𝑎 + 𝑏𝑥) + 
𝑎
𝑎 + 𝑏𝑥
] 
So, 
𝐶𝑆𝑃, 𝑚𝑎𝑥 = 2 {
1
1
[𝑙𝑛(1 + 𝐶𝐴) + 
1
1 + 𝐶𝐴
]}
0
4
 
𝐶𝑆𝑃, 𝑚𝑎𝑥 = 2 {
1
1
[𝑙𝑛(1 + 4) +
1
5
− 𝑙𝑛(1 + 0) − 1]} = 1.6188
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
 
(b) 
𝐶𝑆𝑃, 𝑚𝑎𝑥 = 𝜑𝐶𝐴𝑓(𝐶𝐴0 − 𝐶𝐴𝑓) 
𝐶𝑆𝑚 = 
2𝐶𝐴
2𝐶𝐴 + 1 + 𝐶𝐴
2 (4 − 𝐶𝐴) = 2 {
𝐶𝐴(4 − 𝐶𝐴)
2𝐶𝐴 + 1 + 𝐶𝐴
2} 
𝑑𝐶𝑆𝑚
𝑑𝐶𝐴
= 2 {
(2𝐶𝐴 + 1 + 𝐶𝐴
2)[𝐶𝐴(−1) + 4 − 𝐶𝐴] − (4𝐶𝐴 − 𝐶𝐴
2)(2 + 2𝐶𝐴)
(2𝐶𝐴 + 1 + 𝐶𝐴
2)2
}
= 0 
So, 
2{(2𝐶𝐴 + 1 + 𝐶𝐴
2)(2 − 𝐶𝐴) − (4𝐶𝐴 − 𝐶𝐴
2)(1 + 𝐶𝐴) } = 0 
3𝐶𝐴
2 + 𝐶𝐴 − 2 = 0 
Therefore, 
𝐶𝐴 =
−1 ± √1 − 4(3)(−2)
2
2(3)
= 
2
3
 
𝐶𝑆𝑚, 𝑚𝑎𝑥 = 
2 (
2
3)
2 (
2
3) + (
2
3)
2
+ 1
(4 − 
2
3
) = 1.6
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
 
 
[7.16.]: T is the desired product and𝐶𝐴0 = 5. 
ANS: 
𝜑𝑇 = 
𝐶𝐴
2
1 + 2𝐶𝐴 + 𝐶𝐴
2 = 
1
1 + 
2
𝐶𝐴
+ 
1
𝐶𝐴
2 
 
(a) 
 CTP is maximum when CAf = 0. 
𝐶𝑇𝑃, 𝑚𝑎𝑥 = ∫ 
𝑑𝐶𝐴
2
𝐶𝐴
+ 1 + 
1
𝐶𝐴
2
= ∫
𝐶𝐴
2𝑑𝐶𝐴
(𝐶𝐴 + 1)2
5
0
 
5
0
 
∫
𝑥2𝑑𝑥
(𝑎 + 𝑏𝑥)2
= 
1
𝑏3
[𝑎 + 𝑏𝑥 − 2𝑎 𝑙𝑛(𝑎 + 𝑏𝑥) − 
𝑎2
𝑎 + 𝑏𝑥
] 
So, 
𝐶𝑇𝑃, 𝑚𝑎𝑥 = {1 + 𝐶𝐴 − 2𝑙𝑛(1 + 𝐶𝐴) − 
1
1 + 𝐶𝐴
}
0
5
= 6 − 2𝑙𝑛6 − 
1
6
− [1 − 2𝑙𝑛1 − 1] = 2.2498
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
 
(b) 
𝐶𝑇𝑚 = 
𝐶𝐴
2
1 + 2𝐶𝐴 +𝐶𝐴
2 (5 − 𝐶𝐴) 
So, 
𝑑𝐶𝑇𝑚
𝑑𝐶𝐴
= 
(1 + 2𝐶𝐴 + 𝐶𝐴
2)[(5 − 𝐶𝐴)(2𝐶𝐴) + 𝐶𝐴
2(−1)] − 𝐶𝐴
2(5 − 𝐶𝐴)(2𝐶𝐴 + 2)
(1 + 2𝐶𝐴 + 𝐶𝐴
2)2
= 0 
So, 
𝐶𝐴{(𝐶𝐴 + 1)
2[−𝐶𝐴 + 2(5 − 𝐶𝐴)] − 𝐶𝐴(5 − 𝐶𝐴)(2𝐶𝐴 + 2)} = 0 
(𝐶𝐴 + 1){(𝐶𝐴 + 1)(− 𝐶𝐴 + 10 − 2𝐶𝐴) − (5𝐶𝐴 − 𝐶𝐴
2)(2)} = 0 
(𝐶𝐴 + 1)(10 − 3𝐶𝐴) − 2(5𝐶𝐴 − 𝐶𝐴
2) = 0 
𝐶𝐴
2 + 3𝐶𝐴 − 10 = 0 
𝐶𝐴 = 
−3 ± √9 − 4(1)(−10)
2
2
= 2 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟 
𝐶𝑇𝑚, 𝑚𝑎𝑥 = 
22
22 + 2(2) + 1
= 
8
9
 = 0.89 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟 
 
 
 
QUESTION: 
Under ultraviolet radiation, reactant A of CA0 = 10 kmol/m3 in a 
process stream (𝜗 = 1 𝑚3/min) decomposes as follows: 
𝐴 → 𝑅, 𝑟𝑅 = 16𝐶𝐴
0.5 
 
𝐴 → 𝑆, 𝑟𝑆 = 12𝐶𝐴 
 
𝐴 → 𝑇, 𝑟𝑇 = 𝐶𝐴
2 
We wish to design a reactor setup for a specific duty. Sketch the 
scheme selected, and calculate the fraction of feed transformed into 
desired product as well as the volume of reactor needed. 
[7.17.]: Product R is the desired material. 
 ANS: 
The reaction of the desired product is the lowest order, so it is better 
to use a mixed reactor with high conversion. 
 
CRm, max is obtained when CAf = 0. 
𝐶𝑅𝑚, 𝑚𝑎𝑥 = 𝜑(𝐶𝐴0 − 𝐶𝐴𝑓) = 1(10 − 0) = 10 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟 
𝜑𝑅 = 
16𝐶𝐴
0.5
 16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2 
𝐶𝑅𝑚 = 𝜑𝑅(𝐶𝐴0 − 𝐶𝐴) 
𝜏𝑚 = 
𝐶𝐴0 − 𝐶𝐴
 16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2 
𝑉 = 𝜏𝑚(𝜗0) 
 
Select a high conversion and make the calculations for each of them. 
XA CA 𝜏 V 𝜑 CR 
0.98 0.2 1.0130 1.0130 0.7370 5.8960 
0.99 0.1 1.5790 1.5790 0.8070 7.9894 
0.995 0.05 2.3803 2.3803 0.8558 8.5159 
 
As we can see above, going from XA = 0.99 to 0.995, ΔCR = 0.5265 
mol / L and to achieve this, ΔV = 0.8013 m3 (almost 1 m3) is required, 
then XA = 0.995 could be selected. 
 
[7.18.]: Product S is the desired material. 
 ANS: 
The desired reaction is the middle order, so there corresponds an 
intermediate concentration, which makes maximum performance. 
 
𝜑𝑆 = 
12𝐶𝐴
 16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2 
(A) 
If you cannot recirculate unreacted A, then use a mixed reactor until 
the concentration giving 𝜑máx and thereafter a piston. 
(B) 
If the A can be recirculated unreacted economically, then use a mixed 
reactor with concentration giving 𝜑máx. 
 
(A) 
𝑑𝜑𝑆
𝑑𝐶𝐴
= 
12(16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2) − 12𝐶𝐴(12 + 8𝐶𝐴
−0.5 + 2𝐶𝐴)
(16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2)
2
= 0 
So, 
18𝐶𝐴
0.5 − 𝐶𝐴
2 = 0 
 
𝐶𝐴 = 4
𝑘𝑚𝑜𝑙
𝑚3
 
𝐶𝑆𝑚 = 0.5(10 − 4) = 3
𝑘𝑚𝑜𝑙
𝑚3
 
𝜏𝑚 = 
10 − 4
16(4)0.5 + 12(4) + 42
= 0.0625 𝑚3 = 62.5 𝑙𝑖𝑡𝑒𝑟 
 
𝐶𝐴 4 3 2 1 0.6 0.4 0.11 0.02 
𝜑𝑆 0.5 0.4951 0.4740 0.4138 0.3608 0.2501 0.1988 0.0959 
 
Suppose, XA = 0.998, implies CA = 0.02 
𝐶𝑆𝑃 = ∫ 𝜑𝑑𝐶𝐴
4
0.02
= 
∆𝐶𝐴
2
[𝜑0 + 𝜑𝑓 + 2 ∑ 𝜑𝑖
𝑓−1
𝑖=1
] 
𝐶𝑆𝑃 = 
1
2
[0.4138 + 0.5 + 2(0.4740 + 0.4951)]
+
0.4
2
[0.4138 + 0.2501 + 2(0.3608)]
+ 
0.09
2
[0.2501 + 0.0959 + 2(0.1988)] 
𝐶𝑆𝑃 = 1.7367 𝑚𝑜𝑙/𝑚
3 
𝐶𝑆, 𝑡𝑜𝑡𝑎𝑙 = 3 + 1.7367 = 4.7367 𝑚𝑜𝑙/𝑚
3 
 
𝜏𝑃 = ∫
𝑑𝐶𝐴
−𝑟𝐴
 ≈ 
∆𝐶𝐴
2
𝐶𝐴𝑓
𝐶𝐴0
{(−𝑟𝐴)0
−1 + (−𝑟𝐴)𝑓
−1 + 2 ∑(−𝑟𝐴)𝑖
−1
𝑓−1
𝑖=1
} 
𝐶𝐴 4 3 2 1 0.6 0.2 0.11 0.02 
−𝑟𝐴 96 72.71 50.62 29 19.95 9.60 6.64 2.50 
 
So, 
𝜏𝑃 = 
1
2
[
1
29
+ 
1
96
+ 2 (
1
72.71
+ 
1
50.62
)]
+ 
0.4
2
[
1
29
+ 
1
9.60
+ 2 (
1
19.95
)]
+ 
0.09
2
[
1
9.6
+ 
1
2.5
+ 2 (
1
6.64
)] = 0.1399 𝑚𝑖𝑛 
 
(B) 
 
At point D, balance the flow: 
𝜗0(𝑅 + 1)(4) = 0 + 𝜗0𝑅(10) which gives R = 2/3. 
Now, 
𝑉𝑚
𝜗0(𝑅+1)
= 
10−4
96
 which gives 𝑉𝑚 = 0.104 𝑚
3 = 104 𝑙𝑖𝑡𝑒𝑟𝑠 
 
[7.19.]: Product T is the desired material. 
 ANS: 
The reaction occurs where T is of the highest order. So we should use 
a flow reactor. 
𝜑𝑆 = 
𝐶𝐴
2
 16𝐶𝐴
0.5 + 12𝐶𝐴 + 𝐶𝐴
2 
𝐶𝑇𝑃 = ∫ 𝜑𝑑𝐶𝐴
𝐶𝐴𝑓
𝐶𝐴0
≈ 
∆𝐶𝐴
2
[𝜑0 + 𝜑𝑓 + 2 ∑ 𝜑𝑖
𝑓−1
𝑖=1
] 
𝜏𝑃 = ∫
𝑑𝐶𝐴
−𝑟𝐴
 ≈ 
∆𝐶𝐴
2
𝐶𝐴𝑓
𝐶𝐴0
{(−𝑟𝐴)0
−1 + (−𝑟𝐴)𝑓
−1 + 2 ∑(−𝑟𝐴)𝑖
−1
𝑓−1
𝑖=1
} 
Most T is formed when CAf = 0, but that requires 𝜏 = ∞, so let’s 
choose XA = 0.998. 
𝐶𝐴 𝜑 −𝑟𝐴 
0.02 0.0959 2.5031 
0.11 0.1988 6.6387 
0.2 0.2501 9.5954 
0.6 0.3601 0.3608 
1 0.0345 29.0000 
2 0.0790 50.6274 
3 0.1238 72.7128 
4 0.1667 96.0000 
5 0.2070 120.7771 
6 0.2446 147.1918 
7 0.2795 175.3320 
8 0.3118 205.2548 
9 0.3418 237.0000 
10 0.3696 270.5964 
 
So, 
𝐶𝑇𝑃 = 
1
2
 {[0.0345 + 0.3696
+ 2(0.079 + 0.1238 + ⋯ + 0.3118 + 0.3418)]
+ (1 − 0.05)[0.0345 + 0.0598]} 
𝐶𝑇𝑃 = 1.9729
𝑘𝑚𝑜𝑙
𝑚3
 
𝜏𝑃 = ∫
𝑑𝐶𝐴
−𝑟𝐴
10
4
+ ∫
𝑑𝐶𝐴
−𝑟𝐴
4𝑓
0.02
 
𝜏𝑃 = 
1
2
 {[270.6−1 + 96−1
+ 2(120.8−1 + 147.2−1 + 175.3−1 + 205.3−1
+ 237−1) + 0.1399]} = 0.0369 + 0.1399 
𝜏𝑃 = 0.1768 𝑚𝑖𝑛 
This gives, V = 177 litres 
 
 
SOLUTION 7.20 
 
Since we know that fractional yield is defined by, 
 Φm=
𝐶𝑅
𝐶𝐴0−𝐶𝐴
=
𝐶𝑅
100−𝐶𝐴
 
CA CR Φm 
90 7 0.7 
80 13 0.65 
70 18 0.6 
60 22 0.55 
50 25 0.5 
40 27 0.45 
30 28 0.4 
20 28 0.35 
10 27 0.3 
0 25 0.25 
 
Given, 
CA0=100 
CAf=20 
CRp=∫ Φm
𝐶𝐴𝑂
𝐶𝐴𝑓
dCA 
 =
ΔCA
2
(Φ0 + Φf) 
 =
(100−20)
2
*(0.75+0.25) 
 =40 mol/ lit 
 
 
 
 
 
 
 
SOLUTION 7.21 
 
CRm = Φm (ΔCA) =0.35 (100 – 20) = 28 mol/lit 
 
SOLUTION 7.22 
y = mx + b 
Φm = 0.25 + (0.4/80) CA 
CR = Φm (100 – CA) 
=(0.25 + 0.005 CA)(100 – CA) 
CR= 25+0.25CA-0.005CA2 
𝑑𝐶𝑅
𝑑𝐶𝐴
=0 (for maximum CR) 
 =0.25-0.005(2)CA 
Therefore, CA=25 mol/lit 
 
Now putting this value in equation, we get 
CR= 28.125 mol/lit 
 
 
 
 
 
 
 
 
 
SOLUTION 7.23 
 
Given, 
A + B → R + T rR = 50 CA 
A + B → S + U rS = 100 CB 
Given that CA0+CB0=60mol/m3 
And they both are equimolar, therefore CA0=30 mol/m3 = CB0 
Also given, FA0=FB0=300 mol/hr 
and, XA=0.9 
Now we know, 
CA0=
𝐹𝐴0
𝑣0
 
or, vo=
𝐹𝐴0
𝐶𝐴0
=
300
30
=10 m3/hr 
Now, rate of disappearance of A= rate of formation of R and S 
Therefore, -rA=rR+rS 
 =50CA+100CB 
 =150CA→ (1) 
Now, 
Performance equation curve for MFR is given by: 
τ
𝐶𝐴0
=
𝑋𝐴
−𝑟𝐴
=
𝑉
𝐹𝐴0
 
Therefore, 
τ=
𝑋𝐴∗𝐶𝐴0
150𝐶𝐴
=
𝑋𝐴∗𝐶𝐴0
150𝐶𝐴0(1−𝑋𝐴)
=
0.9
150∗0.1
=0.06hr 
Now, τ=
𝑉
𝑣0
 
or, V=Volume of reactor 
 = 0.06*10= 0.6 m3= 600 litres. 
SOLUTION 7.24 
 
Performance equation for a PFR or Plug Flow Reactor is given by: 
τ
𝐶𝐴0
=
𝑉
𝐹𝐴0
=
∫ 𝑑𝑋𝐴
−𝑟𝐴
 
 
Here, we know 
-rA=150CA 
 =150*CA0*(1-XA) 
Therefore, 
τ
𝐶𝐴0
=
∫ 𝑑𝑋𝐴
150 ∗ 𝐶𝐴0 ∗ (1 − 𝑋𝐴)
 
or, τ = −
ln (1−XA)
150
 
 = -
ln 0.1
150
=0.0153hr 
 
Therefore, V= τ ∗ (v0) 
 = 0.0153*10=0.15 m3= 150 litres. 
 
 
 
 
 
 
 
 
 
SOLUTION 7.25 
 
Given, 
CBeverywhere=3 mol/m3 
 Φ(R/A)=
50𝐶𝐴
50𝐶𝐴+100𝐶𝐵
 
 
Therefore, 
CRF= ∫ Φ(R/A)dCA
𝐶𝐴0
𝐶𝐴𝐹
 
 =∫
𝐶𝐴
𝐶𝐴+2𝐶𝐵30
3
dCA 
 Given CB=3 
CRF=18.68 
 
 Similarly, CSF=∫ Φ (
S
A
) dCA
𝐶𝐴0
𝐶𝐴𝐹
 
 =∫
2𝐶𝐵
𝐶𝐴+2𝐶𝐵
30
3
dCA 
or, CSF=8.32 
 
Finally, 
V=(FA0/CA0)∫
𝑑𝐶𝐴
−𝑟𝐴
𝐶𝐴𝐹
𝐶𝐴0
= 
300
30
∗ ∫
𝑑𝐶𝐴
50𝐶𝐴+100(3)
300
30
 
 = 0.2773 m3= 277.3 litres. 
 
 
 
 
 
SOLUTION 7.26 
 
φ=
𝑑𝐶𝑅
−𝑑𝐶𝐴
 
 
 
CA CR dCA Φ 
90 1 10 0 
80 4 10 0.3 
70 9 10 0.5 
60 16 10 0.7 
50 25 10 0.9 
40 35 10 1 
30 45 10 1 
20 55 10 1 
10 64 10 0.9 
0 71 10 0.7 
 
 
 
 
 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 20 40 60 80 100
Φ
Φ
a) For MFR, 
 
CR=φ10(CA0-CAf) 
 =0.9(100-10) 
 =0.9*90 
 =81 mol/lit 
 
 
b) For PFR, 
φNmixed=
𝜑1(𝐶𝐴0−𝐶𝐴1)+𝜑2(𝐶𝐴1−𝐶𝐴2)+⋯
(𝐶𝐴0−𝐶𝐴𝑁)
 
 =
10
100
(0.3+0.7+0.9+0.5+3+0.7+0.9) 
 = 0.1*7= 0.7 
 
Therefore, 
CR= 𝜑(𝐶𝐴0 − 𝐶𝐴𝑓) 
 = 0.7*(100-10)= 0.7*90=63 mol/lit 
 
 
c) Given, 
 
CA0=70 
Therefore, 
 For MFR, 
 
CR= φ10(CA0-CAf) 
 =0.9(70-10) 
 =0.9*60 
 = 54 mol/lit 
 
d) For PFR, 
 
φNmixed=
𝜑1(𝐶𝐴0−𝐶𝐴1)+𝜑2(𝐶𝐴1−𝐶𝐴2)+⋯
(𝐶𝐴0−𝐶𝐴𝑁)
 
 = 
10
70
(0.7+0.9+3+0.7+0.9+0.5) 
 = 0.142*6.7=0.9571 
 
CR= φNmixed(CA0-CAf) 
 
 =0.9571*(70-10)=57.426 mol/lit 
 
 
SOLUTION 7.28 
 
Given, 
A→ R rR=1 
A→ S rS=2CA 
A→ T rT=CA2 
 
 
 
Now, given flow rate of the reactor = 100 lit/sec 
i.e. v0=100 
 
Now –rA = rR+rS+rT 
 = 1+2CA+CA2 
 = (1+CA)2 
 
Now, taking the first case i.e. MFR from 2-1 
 
Therefore, here, CA0=2; CAf=1 
 
For MFR, 
τm=
𝐶𝐴0−𝐶𝐴𝑓
−𝑟𝐴
=
2−1
4
= 0.25 
 
therefore, V1= τ*(v0) 
 =0.25*100= 25 litres 
 
 
Now taking the second case for PFR, 
Here CA1=1; CA2=0 
 
Therefore, 
τp=∫
𝑑𝐶𝐴
−𝑟𝐴
0
1
=∫
𝑑𝐶𝐴
(1+𝐶𝐴)2
0
1
 
 =-[
1
1+𝐶𝐴
]= 1-
1
2
= 0.5 
 
Therefore, 
 We get, V2= 0.5*(100) 
 = 50 litres 
 
Hence, total volume required = V1+V2 
 = 25+50=75 litres 
 
 
SOLUTION 7.27 
 
Let B be the no of british ships and F be the no of French Ships and if French wins 
dF/dt=-KF…..(1) 
dB/dt=-KB….(2) 
Div 1 by 2 
dF/dB=F/B 
Integrating we get, 
∫ 𝐹𝑑𝐹 = ∫ 𝐵𝑑𝐵
𝐵
𝐵0
𝐹
𝐹0
 
F2-F02=B2-B02 
Since british is losing B=0 
F2=332-272=360 
F=19 ships(approx) 
 
PFR 
PFR 
PFR 
PFR 
Sketch the best contacting patterns for both continuous and non-continuous 
operations. 
Solution 8.1 Continuous 
 
 
 
 
 
 
 
A 
B 
B 
A 
A 
B 
A 
B 
A 
A 
A 
B 
(Drop by drop) 
B B 
B 
(Slowly) 
Instantaneously Instantaneously 
A 
(a) 
(b) 
(c) 
(d) 
(a) (b) (c) (d) 
8.1 (a) r1=k1CACB2 
 r2=k2CRCB 
(b) r1=k1CACB 
 r2=k2CRCB2 
(c) r1=k1CACB 
 r2=k2CR2CB 
(d) r1=k1CA2CB 
 r2=k2CRCB 
Non-Continuous 
PFR 
Batch 
 
Solution 8.2 
a) PFR b) MFR 
CA0 = 1 mol/L CR0 = CS0 = 0 
𝐶𝑅,𝑚𝑎𝑥
𝐶𝐴0
 = 
1
[(𝑘2 𝑘1)⁄
1/2+1]
2 
𝐶𝑅,𝑚𝑎𝑥
𝐶𝐴0
 = e-1 CRmax = 
1
[(0.1 0.1⁄ )1/2+1]
2 
CR,max = 
1
𝑒
 = 0.3679 mol/L CRmax = 0.25 mol/L 
tmax = k-1 = 
1
0.1
 = 10 min tmax = 
1
√𝑘1𝑘2
 
recidance time = 
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟
𝑣𝑜𝑙𝑢𝑚𝑎𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒
 tmax = 10 min 
10 = 
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟
1000
60⁄
 Volume of reactor = 166.67 L 
Volume of reactor = 166.67 L 
Here the volume to get maximum of R is 166.67 L but the maximum 
concentration of R in PFR is 0.3679 mol/L and in MFR 0.25 mol/L. So 
maximum concentration of R can be obtained from PFR. 
 
 
Solution8.3 
Assuming parallel reaction A 
CA0 =100 mol/L CR0=CS0=0 
Run CA CR CS 
𝑪𝑹
𝑪𝒔
= 
𝒌𝟏
𝒌𝟐
 
1 75 15 10 1.5 
2 25 45 30 1.5 
τ = 
𝐶𝑅−𝐶𝑅0
𝑘1𝐶𝐴
 ---(1) 
τ = 
𝐶𝑆−𝐶𝑆0
𝑘2𝐶𝐴
 ---(2) 
Residence time will be equal from both equations 
From eq. (1) and (2) 
𝐶𝑅
𝑘1
= 
𝐶𝑆
𝑘2
&
𝐶𝑅
𝐶𝑠
= 
𝑘1
𝑘2
 
For both the runs the value of 
𝑘1
𝑘2
 is same which is 1.5 
Therefore, the assumption of parallel reaction is correct. 
 
Solution8.4 
CA0 =100 mol/L 𝐶𝑅0=CS0=0 
Assuming parallel reaction taking place 
 
Run CA CR CS 
𝑪𝑹
𝑪𝒔
= 
𝒌𝟏
𝒌𝟐
 
1 75 15 10 1.976 
2 25 45 30 0.666 
 
τ = 
𝐶𝑅−𝐶𝑅0
𝑘1𝐶𝐴
---(1) 
τ = 
𝐶𝑆−𝐶𝑆0
𝑘2𝐶𝐴
---(2) 
Residence time will be equal from both equations 
From eq. (1) and (2) 
𝐶𝑅
𝑘1
= 
𝐶𝑆
𝑘2
 
𝐶𝑅
𝐶𝑠
= 
𝑘1
𝑘2
 
For both the runs the value of 
𝑘1
𝑘2
 is coming different 
Therefore, the assumption of parallel reaction is not correct. 
Now, assuming series reaction taking place 
Run CA CR CS 
𝑪𝑹
𝑪𝒔
= 
𝒌𝟏
𝒌𝟐
 CA/CA0 CR/CA0 CS/CA0 
𝒌𝟐
𝒌𝟏
 
1 50 33.3 16.7 1.976 0.5 0.33 0.167 0.8 
2 25 30 45 0.666 0.25 0.3 0.45 0.8 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution8.5 
Assuming parallel reaction 
Run CA CR CS 
𝑪𝑹
𝑪𝑺
= 
𝒌𝟏
𝒌𝟐
 
1 50 40 10 4 
2 20 40 40 1 
 
τ = 
𝐶𝑅−𝐶𝑅0
𝑘1𝐶𝐴
 ---(1) 
 
τ = 
𝐶𝑆−𝐶𝑆0
𝑘2𝐶𝐴
 ---(2) 
Residence time will be equal from both equations 
From eq. (1) and (2) 
𝐶𝑅
𝑘1
= 
𝐶𝑆
𝑘2
 
𝐶𝑅
𝐶𝑠
= 
𝑘1
𝑘2
 
For both the runs the value of 
𝑘1
𝑘2
 is coming different 
Therefore, the assumption of parallel reaction is not correct. 
Now, assuming series reaction 
Run CA CR CS 
𝑪𝑹
𝑪𝑺
= 
𝒌𝟏
𝒌𝟐
 CA/CA0 CR/CA0 CS/CA0 
𝒌𝟐
𝒌𝟏
 
1 50 40 10 4 0.5 0.4 0.1 0.3 
2 20 40 40 1 0.2 0.4 0.4 0.3 
 
 
 
Solution 8.6 
As grinding will progress as, 
large particles → appropriate particles→ small particles 
denoting this process as, 
A → R → S 
a) Basis: 100 particles (10 A, 32 R and 58 S) 
Too small particles so you have to reduce the residence time, 
increasing the feed flow to make an estimate, assume that a series of 
first reaction order may represent milling process. 
XA = 0.9 and CR/CA0 = 0.32 is that k2/k1≈0.7 
 
If 
𝑘2
𝑘1
≈ 0.7 ⇒CRmax / CA0 = 0.5 and XA= 0.75 and will get 25% 
of very large particles, 50% of particles of appropriate size and 25% 
of very small particles. 
 
b) Multi-stage is better, as single stage grinder will act as single MFR 
and multi-stage grinder will act as MFR in series which will act as 
PFR and in PFR we get more uniform product. So using multistage 
grinder is best option. 
 
 
 
 
 
Solution 8.7 
a) A0 = 1 B0=3 
B consumed is 2.2 so ∆B=0.8 
S=0.2 
𝑘2
𝑘1
= 0.85 
For S=0.6 and
𝑘2
𝑘1
= 0.85 from graph R=0.3 
A+B→ R &R+B→ S 
To produce 0.6 mol of S, B required is 0.6 and R required is 0.6 
The final mol of R is 0.3 so total mol of R must be produced is 0.9 
and to produce it 0.9 bole of A and B required 
Therefore ∆B=0.6+0.9 = 1.5 
The final concentration of A, B, R and S is 0.1, 1.5, 0.3 and 0.9 
respectively. 
 
 
 
 
 
 
 
 
 
b) 
 
Only S is found so as the R is formed it is directlyconverted to S 
and it indicates that k2>> k1 which cannot be determined form the graph 
shown above. 
 
 
 
 
c) 
 
∆B
𝐴
 = 1 and S=0.25 from graph 
𝑘2
𝑘1
= 0.4 
 
 
Solution 8.8 
a) Here, 
𝑘2
𝑘1
 = 0.8 
From graph concentration of diethyl-aniline is 0.2mol and 
mono-ethyl-aniline is 0.4mol 
Water will be formed in second reaction is 0.2mol 
For producing 0.2 mol of di-ethyl-aniline 0.2 mol of mono-
ethyl-aniline will be consumed so total mono-ethyl-aniline produced 
in first reaction is 0.6 mol and 0.6 mol of water will be formed 
Total mol of water formed is 0.8 
0.6 mol of ethanol in first and 0.2 mol ethanol in second reaction will 
be consumed so total ethanol consumed is 0.8 mol 
0.6 mol of aniline will be consumed in first reaction 
 
 
Final composition of mixture from batch reactor 
Aniline Ethanol 
Mono-ethyl-
aniline 
Di-ethyl-
aniline 
Water 
0.4 0.2 0.4 0.2 0.8 
 
 
 
 
 
b) 
 
For 70% conversion of ethanol value of ∆B = 1.4 
∆B
𝐴
 = 1.4 
For this data final moles of mono-ethyl-aniline and di-ethyl-aniline are 
0.2 mol and 0.6 mol respectively. 
Ratio of mono to di-ethyl-aniline is 0.333 
 
 
c) 
 
From graph for equimolar feed in plug flow reactor for highest 
concentration of mono-ethyl-aniline the conversion of A is 67% 
From graph ∆B = 0.92 
Therefore, overall conversion of B is 92% considering both reactions 
 
 
 
Solution 8.9 
Considering aniline (A), ethanol (B), mono-ethyl-aniline (R) and 
di-ethyl-aniline (S) 
 
A0 = 1 mol, B0 = 2 mol 
∆B
𝐴
= 0.56 
After reaction conversion of A is 40% so 60%will remain unconverted 
A = 0.6 
B reacted in first reaction is 0.4 mol 
The remaining final mixture R is 3 parts and S is 2 parts to produce 2 
parts of S, 2 parts of R must react in second reaction so total R formed in 
first reaction is 5 parts which is 0.4 mol. 
2 parts reacted so remaining moles of R is 0.4-(
2
5
 * 0.4) = 0.24 
S formed is 0.16 mol 
The value from graph 
𝑘2
𝑘1
= 1 
The final composition in outlet of MFR 
Aniline Ethanol 
Mono-ethyl-
aniline 
Di-ethyl-
aniline 
Water 
0.6 0.44 0.24 0.16 0.56 
 
 
 
 
Solution 8.10 
 
pH k1 k2 k1/k2 
2 11 3 3.666667 
3 8 4 2 
4 7 5 1.4 
5 8 6 1.333333 
6 11 7 1.571429 
 
a) Here, the value of 
𝑘1
𝑘2
is highest at pH=2 so the pH which should be 
maintained should be slight higher than 2 and should be maintained 
throughout the reactor to get maximum of R as the k1 will be higher 
compare to k2 then R will be more formed. MFR should be used to 
maintain pH throughout the reactor as 100% mixing and uniform pH 
will be there in MFR. 
b) As, the pH required to be constant no PFR is used and in MFR no 
stages with different pH should be kept. 
CRE-I Term paper 
Sums 8.11-8.21 
Group members: 13BCH012, 13BCH013, 13BCH037, 13BCH045, 13BCH055 
Problem 8.11 
As S is desired it requires both k1 as high as k2 
pH k1 k2 
2 3 11 
3 4 8 
4 5 7 
5 6 8 
6 7 11 
When pH=6, k1=7(maximum value) and k2=11(maximum value), so it is best to work with pH=6 
and keep it constant. 
 
Problem 8.12 
(a)R is in series with A and S, so it is best to use the plug flow reactor. 
(b) As in parallel reactions CRis the area under the curve Ф vs CA. So, let’s take Ф = f (CA) 
 Ф = −
𝑑𝐶𝐴 
𝑑𝐶𝑅
 = (k1* CA – k4*CR) / k12*CA , where k12= k1+k2 
It is not possible to separate variables and integrate because Ф is also a function of CR; but it is a 
linear differential equation of first order with integral factor. 
𝑑𝐶𝑅 
𝑑𝐶𝐴
 – (k4*CR)/ (k12*CA) =-
𝑘1 
𝑘12
 
Comparing this equation with, 
𝑑𝑦
𝑑𝑥
 + P(x) y = Q(x) 
y = CR , x = CA , P(x) = - 
𝑘4 
𝑘12∗𝐶𝐴
 , Q(x) = -
𝑘1 
𝑘12
 
y = е− 𝑃 𝑥 𝑑𝑥 𝑄(𝑥)е 𝑃 𝑥 𝑑𝑥 𝑑𝑥 
 𝑃 𝑥 𝑑𝑥 = − (
𝑘4 
𝑘12∗𝐶𝐴
)dCA = ln𝐶𝐴
𝑘4/𝑘12 
е− 𝑃 𝑥 𝑑𝑥 = CA
 
е 𝑃 𝑥 𝑑𝑥 = 𝐶𝐴
−𝑘4/𝑘12
 
 𝑄 𝑥 е 𝑃 𝑥 𝑑𝑥𝑑𝑥 = -
𝑘1 
𝑘12
 𝐶𝐴
𝑘4/𝑘12 dCA = (
−𝑘1 
𝑘12
) ∗
𝐶𝐴
(1−
𝑘4
𝑘12
)
 
(1− 
𝑘4 
𝑘12
)
 
Put all these values in main equation, 
CR= 𝐶𝐴
−𝑘4/𝑘12 {(
−𝑘1 
𝑘12
)*( 𝐶𝐴
(1−
𝑘4
𝑘12
)
 / (1- 
𝑘4 
𝑘12
)) + constant} 
When CA = CA0 then CR = 0 and constant = (
𝑘1 
𝑘12
) ∗ 
𝐶𝐴0
(1−
𝑘4
𝑘12
)
 
(1− 
𝑘4 
𝑘12
)
 
CR= 𝐶𝐴
𝑘4/𝑘12 (
−𝑘1 
𝑘12
) ∗
𝐶𝐴
(1−
𝑘4
𝑘12
)
 
(1− 
𝑘4 
𝑘12
)
 + (
𝑘1 
𝑘12
) ∗ 
𝐶𝐴0
(1−
𝑘4
𝑘12
)
 
(1− 
𝑘4 
𝑘12
)
 
Dividing the whole equation for CA0 and removing the common factor constant ratio within the 
key and performing the indicated multiplication, 
𝐶𝑅 
𝐶𝐴0
 = 
𝑘1 
𝑘12
1−
𝑘4
𝑘12
 
𝐶𝐴0𝐶𝐴0
−
𝑘4 
𝑘12 𝐶𝐴
𝑘4
𝑘12
𝐶𝐴0
− 
𝐶𝐴
(1−
𝑘4
𝑘12
+
𝑘4
𝑘12
)
𝐶𝐴0
 
𝐶𝑅 
𝐶𝐴0
= 
𝑘1
𝑘12 − 𝑘4
 
𝐶𝐴
𝐶𝐴0
 
𝑘4
𝑘12
− 
𝐶𝐴
𝐶𝐴0
 
Note that you could have obtained from the equation 8.48 (p. 195) by k34 = k3+ k4= k4 (because 
k3= 0) 
𝐶𝑅 
𝐶𝐴0
= 
𝑘1
𝑘12
 
𝑘12
𝑘4
 
𝑘4
𝑘4−𝑘12 =
3
4
 
4
0.2
 
0.2
0.2−4
 = 0.865 
CR = 2(0.865) = 1.73mol 
 
Problem no. 8.13 
(a) 
 
k1=40, k2=10, k3=0.1, k4= 0.2 
CS = 0.2CA0 
k12 = k1 + k2 = 50; k34 = k3 + k4 = 0.1 + 0.2 = 0.3 
𝐶𝑆 
𝐶𝐴0
=
𝑘1𝑘3
𝑘34−𝑘12
(
е1−𝑘34 𝑡
𝑘34
− 
е−𝑘12 𝑡
𝑘12
exp⁡(−k12t)
k12
) + 
𝑘1𝑘3
𝑘34−𝑘12
 + 
𝐶𝑅0𝑘3
𝐶𝐴0𝑘34
 (1 – е−𝑘34𝑡) + 
𝐶𝑆0
𝐶𝐴0
 
We have, CA0 + CR0 + CT0 + CU0 = CA + CR + CT + CU 
Or CA0 = CA + CR + CT + CU 
Therefore, 
0.2𝐶𝐴0
𝐶𝐴0
 =
(40)(0.1)
0.3−50
(
е−0.3𝑡
0.3
 - 
е−50𝑡
50
) + 
40∗0.1
50∗0.3
 +0 +0 
→ 0.2 = 
−0.0804
15
(50e
-0.3t
 – 0.3e-50t) + 0.2667 
Or 12.44 = 50e
-0.3t
 – 0.3e-50t 
→ ln12044 = ln50 – 0.3t – ln0.3 + 50t 
→2.52 = 3.91 + 49.7t + 1.2 
So, t = -0.0521 minutes. 
Negative value of time is not possible. Thus there is some error in the data. 
(b) 
 
The above rate constants show that the first reaction is a slow reaction and the second reaction is 
a fast reaction. 
CA0 + CR0 + CT0 + CU0 = CA + CR + CT + CU 
𝐶𝑆 
𝐶𝐴0
=
𝑘1𝑘3
𝑘34−𝑘12
(
е1−𝑘34 𝑡
𝑘34
− 
е−𝑘12 𝑡
𝑘12
) + 
𝑘1𝑘3
𝑘34∗𝑘12
 
Let CA0 = 100 mol/lit 
So, CS = 0.2*100 = 20 mol/lit 
→
0.2𝐶𝐴0
𝐶𝐴0
 =
(0.2)(10)
30−0.03
(
е−30𝑡
30
 - 
е−0.03𝑡
0.03
) + 
0.02∗10
0.03∗30
 
Or 0.02 = 6.67x10
-3(
0.03е−30𝑡−30е−0.03𝑡
30∗0.03
) + 0.222 
Therefore, 0.2 – 0.222 = 7.414*0.03*10-3e-30t – 70417*30*e-0.03t*10-3 
0.0222 = -2.224x10
-4
e
-30t
 + 0.2224xe
-0.03t 
 -8.411 – 30t – 1.4961 + 0.03t = -3.867 
 -8.417 – 1.4961 + 3.8167 = 29.97t 
Thus t = 
−6.0904
29.97
 = - 0.2032 minutes 
Which is negative so the data given is incorrect. 
 
Problem 8.14 
 
(a) Arguably k3> k4 and k5> k6. we cannot conclude anything about k1 and k2 because although 
the branch of R there are fewer moles than by the branch of S may be that k1> k2 and k3 and k4 to 
be small and has accumulation of R may also be that k1<k2 and k1<k3 and k4 so that all R formed 
pass T and U. 
(b)If the reaction was now complete and is T, U, V and W only by the branch above 5 moles of T 
and 1mole of U formed, so it was 6 moles of R transformed to U and T, whereas below the 
branch 9 mol of V and 3 mol of W were formed, meaning that there were 12 mol ofS. In this 
case can be concluded that, 
k1<k2 
Formation rate of R = 
𝑑𝐶𝑅 
𝑑𝑡
 = k1CACB 
Formation rate of S = 
𝑑𝐶𝑆 
𝑑𝑡
 = k2CACB 
dCR = 
𝑘1
𝑘2
 dCs ═>
𝑘1
𝑘2
 = 
𝐶𝑅
𝐶𝑆
 = 
6
12
 ═> k2=2k1 
Formation rate of T = 
𝑑𝐶𝑇 
𝑑𝑡
 = k3CACB 
Formation rate of U = 
𝑑𝐶𝑈 
𝑑𝑡
 = k4CACB 
dCT = 
𝑘3
𝑘4
 dCU ═>
𝑘3
𝑘4
 = 
𝐶𝑇
𝐶𝑈
 = 
5
1
 ═> k2=5k1 valid for (a) 
Formation rate of V = 
𝑑𝐶𝑉 
𝑑𝑡
 = k5CACB 
Formation rate of W = 
𝑑𝐶𝑊 
𝑑𝑡
 = k6CACB 
dCV = 
𝑘5
𝑘6
 dCW ═>
𝑘5
𝑘6
 = 
𝐶𝑉
𝐶𝑊
 = 
9
3
 ═> k2=3k1 valid for (a) 
 
Problem 8.15 
 
For series reaction, the overall rate constant is given by, 
1
𝑘13
 = 
1
𝑘1
 + 
1
𝑘3
 = 
1
0.21
 + 
1
4.2
 = 5 
So, k13= 0.2 
Overall rate constant is given by,K=k13+k2= 0.2 + 0.2 = 0.4 
Since S is desired product, if we use mixed flow reactor, there will be more mixing and hence 
intermediate formation will be less. So we can use plug flow reactor. 
By using equation, 
𝐶𝑆𝑚𝑎𝑥
𝐶𝐴0
 =
𝑘123
𝑘4
𝑘4
 𝑘4−𝑘123 =
0.4
0.004
0.004
0.004−0.4 = 0.9545 
CSmax= 0.9545 CA0 
ԏ = 
ln⁡(
𝑘2
𝑘1
)
𝑘2−𝑘1
 = 
ln⁡(
0.004
0.4
)
0.004−0.4
 = 8.128 [therefore k1 = k123 = 0.4] 
As we know equation, 
CA = CA0е−𝑘1𝑡
 
But CA0 = 100 е
- (0.4*8.128) 
 =100*0.03872 
 =3.872 mol/L 
XA= 
𝐶𝐴0−𝐶𝐴
𝐶𝐴0
 = 
100−3.872
100
 = 96.12% 
Problem no. 8.16 
The initial gravel handling rate of shovel is 10 ton/min. It is also mentioned that the shovel’s 
gravel handling rate is directly proportional to the size of the pile still to be moved. So it shows 
that the removal of the sand using shovel can be compared to the first order reaction. 
The conveyor on the other hand works at the uniform rate of 5 ton/min. This shows that the 
working of the conveyor is independent of the amount of sand in the hopper. So it can be 
considered as the zero order reaction. 
So for shovel, 
10 ton/min=k1 * 20000……… (1) 
k1= 10/20000= 5 * 10
-4 
For Conveyor,
 
5 ton/min= k2 
(a) 
𝐶𝑏𝑖𝑛
𝐶𝐴0
= 1-K(1-ln K) 
K= 
𝑘2
𝐶𝐴0𝑘1
 = 
5
20000∗5∗10−4
 = 0.5 
Cbin= 20000 * (1-0.5(1-ln 0.5)) = 3068.52 tonnes 
 
(b) tRmax= ln(1/K)/k1= ln(1/0.5)/(5*10
-4
)= 1386.29 min 
(c) Input= Output 
k1 *CA0= k2 
5*10
-4
 *CA0= 5 
CA0= 10000 tonne 
 
(d) When Cbin=0, 
𝑘2𝑡
 𝐶𝐴0
 + е−𝑘1𝑡 = 1 
By using trial and error method, 
t=0.0132min 
 
 
Problem no. 8.17 
A → R → S 
Where, A = Garbage still to be collected 
 A0 = 1140 tonnes (initial garbage load) 
 R = Garbage in bin 
 S = Garbage in incinerator 
Since the rate of collection is proportional; to amount of garbage still to be collected – 
Therefore, A → R is a First order reaction. 
Since the conveyor is operated at uniform rate to the incinerator – 
R → S is a zero order reaction. 
Therefore, 
−𝑑𝐴
𝑑𝑡
 = k1A 
At t = 0, 
−𝑑𝐴𝑜
𝑑𝑡
 = 6 tons/min 
k1= 
−𝑑𝐴
𝑑𝑡
A
 = 
6
1140
 = 
1
240
 min
-1
 
A → R … (n1 = 1, k1 = (1/240) min
-1
) 
R → S … (n2 = 0, k2 = 1 ton/min) 
(a) For 95% of A collected – 
𝑑𝑅
𝑑𝑡
 = k1A – k2 
From the given equation: 
𝐶𝐴
𝐶𝐴0
 = е−𝑘1𝑡 
A0 = 1140 tonnes 
A (at 95%) = 0.05*1440 = 72 tonnes 
k1 = 
1
240
 
𝐴
𝐴𝑜
 = 
72
1140
 = e
-t/240 
Therefore, 
1
20
 = e
-t/240 
Or, ln (
1
20
) = 
−𝑡
240
 
So, t = 718 min = 12 hrs. 
 
(b) For maximum collection in bin, 
𝑅𝑚𝑎𝑥
𝐴𝑜
 = 1-k (1 - lnk) 
Thus, k = 
(𝑘2/𝐴𝑜)
𝑘1
 = 
(1/1440 )
(1/240)
 = 0.1667 
Therefore, Rmax = 1140[1- (0.1667) (1 – ln (0.1667))] = 769.9 ton 
(c) tmax =
1
𝑘1
ln (
1
𝑘
) = 240 ln (
1
0.1667
) = 430 min 
(d) Bin empties when R = 0 
Therefore, R = 0 = A0 (1 – e
-k1*t
) – k2t 
Implies, 0 = 1440(1 – e-t/240) – t 
Or, e
-t/240
 + 
𝑡
1440
 = 1 
By using Goal Seek method, 
t = 1436 min 
Problem no. 8.18 
Suppose Upper Slobbovians = U and Lower Slobbovians = L 
Here it is given that, 
−𝑑𝑈
𝑑𝑡
 = L and 
−𝑑𝐿
𝑑𝑡
 = U 
Last week, after the encounter of 10U and 3L, 8 U survived 
It means that 2 U died in this encounter and all the 3 L died. 
So, when 1U dies 1.5 L die 
(a) Upper Slobbovians are better fighters. And one Upper Slobbovian is as good as 1.5 
Lower Slobbovians. 
(b) When 10 U and 10 L encountered 
1.5L → 1U 
10L →
1∗10
1.5
 = 6.6 ≈ 7U 
So here we can conclude that, after this encounter 3 U lived and 7 U died. Whereas all the 10 L 
were killed. 
 
 
Problem no. 8.19- 
Since chemical X is added slowly and it dissolves quickly, this shows that it is a zero order 
reaction. 
Since Y decomposes slowly it shows that it is a 1
st order
 reaction. 
Thus it can be said that, the overall reaction is zero order followed by 1
st 
order. 
X→Y … (n=1) 
Y→Z … (n=2) 
 
CX0 = 100 mol/m
3 
Now, 
−𝑑𝐶𝑋
𝑑𝑡
 = k1 = 
100 
0.5
 = 200 mol/ (m
3.hr) … as n=0 
−𝑑𝐶𝑌
𝑑𝑡
 = k2CY 
Overall rate constant can be given by ‘k’ where – 
K = 
𝑘2∗𝐶0
𝑘1
 = 
1.5∗100
200
 = 0.75 
(a) At half hour Ymax 
𝐶𝑌𝑚𝑎𝑥
𝐶𝑋0
= 
1
𝑘
(1– e-k) = 
1
0.75
(1 – e-0.75) = 0.7035 
 Or, CYmax = 100(0.7035) = 70.35 mol/m
3 
 Thus, tmax = 
𝐶𝑋0
𝑘1
 = 
100
200
 = 0.5 hours 
(b) After 1 hour, CX0 = 0 
𝐶𝑌
𝐶𝑋0
 = 
1
𝑘
(е𝑘1−𝑘1𝑡 - е−𝑘2𝑡) = 
1
0.75
(e
-0.75-1.5
-e
-1.5
) = 0.3323 
 CY = 100(0.3323) = 33.23 mol/m
3 
 CZ = CX0 – CX - CY = 100 – 0 – 33.23 = 66.7 mol/m
3 
 
Problem no. 8.20 
(1) 
𝑑𝐶𝑋
𝑑𝑡
 ≠ 0 and 
𝑑𝐶𝑅
𝑑𝑡
 ≠ 0 
CA = CA0 - kt 
CA - CA0 = kt 
From initial slope of CX Vs t and CR Vs t, which is not zero we can say that A gives X and R in a 
Parallel reaction. 
A→X 
A→R 
(2) But when A is completely exhausted, still CX decreases and CR and thus X decomposes to 
give the final product R. 
A→X→R 
Therefore A→X, A→R, and X→R 
 
We know, CA0 = CX + CA + CR 
Therefore CX = 100 – CA – CR 
t(min) 0 0.1 2.5 5 7.5 10 20 ∞ 
CA(mol/m
3
) 100 95.8 35 12 4 1.5 - 0 
CR(mol/m
3
) 0 1.4 26 41 52 60 80 100 
CX(mol/m
3
) 0 2.8 39 47 44 38.5 20 0 
 
CA = CA0e
-kt 
Or lnCA = ln CA0 – kt 
 
 
Thus from the graph of lnCA vs t we get a straight line with intercept lnCA0. Thus it follows 1
st 
order kinetics. 
We know, ln
𝐶𝐴0
𝐶𝐴
 = (k1 + k2)*t 
And k1 + k2 = k12 
Therefore, k12 = 
ln 95.8 −ln⁡(12)
0.1−5
 
Or k12 = 0.42 min
-1
 
(
𝑑𝐶𝑅
𝑑𝑡
)t=0 = 
1.4
0.1
 = k2CA0 
Thus k2 = 
1.4
 0.1 ∗(100)
 = 0.14 min
-1
 
(
𝑑𝐶𝑋
𝑑𝑡
)t=0 = 
2.8
0.1
 = k1CA0 
So k1 = 
2.8
 0.1 ∗(100)
 = 0.28 min
-1
 
Therefore, X is the intermediate product and has a maximum at 5 minutes. 
→ for CR0 = CX0 = 0, 
𝐶𝑋𝑚𝑎𝑥
𝐶𝐴0
 = 
𝑘1
𝑘2
*
𝑘12
𝑘34
𝑘34
𝑘34− 𝑘12 
Here K34 = k3 
Thus 
𝐶𝑋𝑚𝑎𝑥
𝐶𝐴0
 = 
47
100
 = 0.47 
Or 0.47 = 
0.28
0.42
∗
0.42
𝑘3
𝑘3
𝑘3− 0.42 
By using seek method we get, k3 = 0.067 
To verify whether the method is correct, 
Recalculating CR, 
𝐶𝑅
𝐶𝐴0
 = 
𝑘1𝑘3
𝑘3−𝑘12
 (
е−𝑘3𝑡
𝑘3
 - 
е−𝑘12 𝑡
𝑘12
) + 
𝑘1
𝑘2
 + 
𝑘2
𝑘12
(1 – е−𝑘12 𝑡) 
= 
0.28(0.067)
0.067−0.42
(
е−0.067∗5
0.067
−
е−0.42∗5
0.42
) + 
0.28
0.42+ 
0.14
0.42
(1-е−0.42∗5) 
→ 
𝐶𝑅
𝐶𝐴0
 = 0.4087 
Therefore CR = 0.4087x100 = 40.87 or 41 mol/m
3
 
At CXmax, i.e. t = 5 minutes 
CR = 41 mol/m
3
 
 
Problem no. 8.21 
 
k1 =6 hr
-1
 
k2 = 3hr
-1
 
k3 = 1hr
-1
 
CA0 = 1 mol/litre 
𝐶𝑅𝑚𝑎𝑥
𝐶𝐴0
 = 
𝑘1
𝑘12
∗
𝑘12
𝑘34
𝑘34
𝑘34− 𝑘12 
Since k1 = k12 for one case, 
𝐶𝑅𝑚𝑎𝑥
𝐶𝐴0
 = 
6
6
(
6
4
)
(4/6)
 
= 
3
2
2
3
 
Therefore, CRmax = 
4
9
 = 0.44 mol/litre 
tmax = 
ln⁡(4/6)
4−6
 = 0.202 hours. 
 
Cre Term Paper 
Submitted by – 13bch005, 13bch023,13bch041,13bch042 
 
Problem 9.1 
For the reacting system of example 9.4 
 A) What τ is required for the 60 % conversion of reagent using the optimal progression of 
temperature in a flow reactor in piston? 
B)Find the outlet temperature of the reactor. Use k10=33*10^6 , k20= 1.8*10^18 
Use any information you need from the example 9.4 
Solution 
A) 
The treaty system in the example 9.4 A↔ R with -rA = k1 CA - K2 CR, where k1 =exp 
(17,34 - 48900/RT) and k2 = exp (42.04 - 124200/RT) and C0= 4 mol/L. The chart shown in 
the example, to not be squared, makes the collection of data from it is very vague and that is 
why we are going to develop the necessary data, without using the chart. 
If you want to find the optimal profile is considered that in the same 
 
With the previous equations and data taken from the example you can assess the temperature 
of the optimal profile for each XA and then see how varies - r with XA along the optimal 
profile 
XTo 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 
Tópt (K) 368 386,47 373,54 365,43 359,02 353,35 347,84 342,04 335,22 
-rto 15.54 12.29 9.04 5.89 3.79 2.43 1.49 0.84 0.40 
1/-RTO 0.06 0.08 0.11 0.17 0.26 0.41 0.67 1.19 2.5 
With these values are estimated the volume of flow reactor in piston, using area under the 
curve 
τ= CA0 
𝑑𝑋𝐴
−𝑟𝐴
𝑋𝑎
0 = 4
 
𝑑𝑋𝐴
−𝑟𝐴
0.6
0 = 0.558 min
 
 
 
B) 
In the table above we see that if X = 0.6 The temperature in the optimal profile is 347,84 K = 
74,84°C 
 
Problem 9.2 (p. 238) 
You want to convert the concentrated aqueous solution of previous example (C0 = 4 mol/L; 
F0 = 1000 mol/min) up to 70 % with the smaller size of reactor for thorough mixing. Make 
an outline of recommended system, indicating the temperature of the input and output current 
and the required space weather 
Solution 
In the table that appears in the problem 9.1 we take the temperature and the speed of the 
optimal profile X = 0.7 
T = 342.04 - 273 = 69°C, -rA = 0.84 mol/l min 
𝜏m= CA0XA/-rA= 4(0.7)/.84 = 3.33 min 
XA= Cp’(T-T0)/-∆Hr , T0= T+( XA∆HR)/CP= 20◦C 
 
Problem 9.3 (p. 238) 
With regard to the flow reactor in piston which operates on the optimal profile 
The example 9.4 (C0 = 4 mol/L; F0 = 1000 mol/min; X = 0.8; Tmin = 5°C; Tmax = 95°C) 
and power and current product to 25°C, much heat or 
Cooling will be required 
(A)to the supply current 
(B)In the reactor itself 
(C)For the output current 
Solution 
In the table that appears in the problem 9.1 The temperature of the optimal profile for XA = 
0.8 is 335,22 K = 62,22°C 
25°C 95°C 
 
62.2°C 25°C 
 
 
Q3 
 
Q1 
 
 
Q2 
 
Q1= cp’(Tsat—Texit)=250*4187*70◦C= 73272,5J/mol A= 73272,5 *1000= 73272.5kJ/min 
 
Q3= cp’(Tsat—Texit )=250 *4187*(25-62.2)*1000=-38939.1kJ/min 
 
Q2= cp’(Tsat—Texit )=250*4180*(62.2-95)*1000=34333.4kJ/min 
 
 
 
 
 
 
 
 
 
 
 
 
 
There is that to provide the power 73272.5 kJ/min, while the output current there is that 
extract 38939,1 J/min. There are also removed from the reactor 34333.4 kJ/min. In total there 
are that remove 73272.5 kJ/min. I suggest the following 
 
 
 
 
 
Problem 9.4 (p. 238) 
It plans to carry out the reaction of the example 9.4 (C0 = 4 mol/L; F0 = 1000 mol/min) in a 
flow reactor in piston which is maintained at 40°C to XTO = 90%. Find the required volume 
Solution 
Constant density system 
 
 
k1τ /XA = ln(XA/(X*-XA)) 
From example 9.4 
k1=exp (17.34-48900/RT)min
-1
 
 K= exp( 75300/RT-24.7) R=8.314 J/mol K 
T = 40 + 273 = 313 
k313 = 0.2343 min-1 , 
K313 = 69.1405 
X AE = K/ (K+1) = 0.98 
 
τ p = 
0.98 
Ln 
0.98 
=10.48 min 
 0.2343 0.98 -0.9 
 
 
 
Problem 9.5 (p. 238) 
Rework the example 9.4 replacing C0 with 1 mol/L/h. 
Example 9.4. Using the optimal progression of temperature in a flow reactor in piston to the 
reaction of the previous examples. Tmax = 95 °C 
 (a)Calculate the time space and the volume required for the 80 % conversion of 1000 mol 
to/min with C0 = 4 mol/L 
B)Plot the temperature and conversion profile along the reactor 
Solution 
Constant density system because it is Liquid 
-rA = k1 CA0 (1 - X) - k2 CA0 XA = CA0[k1 (1 – XA) - k2 XA] 
(-rA)1 = [k1 (1 – XA) - k2 XA] 
(-rA)4 = 4 [k1 (1 – XA) - k2 XA] = 4 (-rA)1 
0.8 dXA 
 
 
0.8 dXA 0.8 dXA 
 
∫ 
 
= ∫ 
 
 = 4 ∫ 
 
 
 
(-rA )1 
 
 
 
(-rA )4 (-r A)4 
 
 
 
0 
 
 
 
1 
 
0 
 
0 
 
 
 
 
 
4 
 
0.8 
dXA 
 
∫ 
 
 
=0.405 taken from the example 9.4 ( p. 230) (-rA )4 
 
 
0 
 
 
 
0.8 
dXA 
 
=4(0.405) 
 
∫ 
 
 
 
(-rA ) 1 
 
 
 
 
 
 
 
 
0.8 
dX A 
 
0.8 
 
τ p = CA0 
 
∫ 
 
=1 ∫ dXA 
 
 
=(1)(4)(0.405)=1.63 min (-rA )1 (-rA) 
 
 
 
0 
 
 
 
 
 
 
 
 
 
v0=FA0/CA0 
=1000L/min 
 
V= τv0=1620L 
As they are reactions of first order the change in the value of C0 did not affect the value of τ. 
The volume if it is affected because F0 remained constant and therefore v0 increased 4 
times,causing the volume is 4 times more. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Volume 
(L) 
 
 
 
 
Problem 9.6 (p. 238) 
Rework the example 9.5 replacing CA0 with 1 mol/L 
Example 9.5. The concentrated solution of the previous examples (C0 = 4 mol/L; F0 = 1000 
mol/min is going to be 80% converted into a complete mix reactor 
(A)What size reactor is required? 
 (B) What should be the transfer of heat if the power is at 25°C and the output current must be 
at the same temperature? 
Solution 
A) in the table of the problem 9.1 appears which the speed in reported the optimal profile X = 
0.8 is 0.4 mol/l min; but for CA0 = 4 mol/l and that is needed is the corresponding to C0 = 1 
mol/L 
-rA)1=-rA)4/4 =0.4/4= 0.1 mol/Lmin 
τ=CA0XA/-rA=1(0.8)/0.1= 8 min, v0=1000L/min 
V = τv0= 8(1000)= 8000 L 
 
Case1 
Q =cp(T-T0’)+ HXA 
0=250*4187*(62,22-T0’)+(-18000)(4187)(0.8) 
T0’= 4.62◦C 
Q1= 250(4187)(4.62-25)=21332.765J/mol A 
Q2=250(4187)(25-62.22)=-38960.035J/mol A 
Case 2 
Q1' = 250(4,187)(62,22 -25)+(18000)(4,187)(0.8)= -21332,765 J / molA 
It can be seen thatboth forms of exchange of heat are equivalent 
Problem 9.7 (p. 238) 
Rework the example 9.6; but with C0 = 1 mol/l instead of C0 = 4 mol/L and recital FA0 = 
1000 mol to/min 
Example 9.6. Find the size of the flow reactor in piston required to convert up to 80 % the 
1000 mol to/min with C0 = 4 mol/L, which is used in the example 9.5 
Solution 
In the example 9.6 appears that 
 
0.8 
 
dXA 
 
Τ4mol / L = 4 ∫ 
 
= 8.66 min 
 
(-rA )4mol/L 
 
 
 
 
 
 
 
 
 
0.8 
 
 dX A 
 
Τ1mol / L = (1)∫ 
 
(-rA )1mol/L 
 
 
0 
 
 
 
(-rA )4mol / L/4 (-rA )4mol / L = 
 
4 (-rA )1mol / L 
 
(-rA )1mol / L = 
 
 
0.8 
 
dX A 
 
Τ1mol / L = (1)∫ 
 
(-rA )1mol/L 
 
 
0 
 
 
 
 
 
(-rA )4mol / L 
(-rA )4mol / L = 
 
4 (-rA )1mol / L 
 
(-rA )1mol / L = 
 
 4 
 
 
0.8 
 
 dX A 
 
0.8 
 dX A 
 
Τ1mol / L = (1)∫ 
 
= 4 ∫ 
= 8.66 min 
 
 
(-rA )4mol / L 
 
(-rA) 
 
0 
 
0 
 
 4mol / L 
 
V1mol/L= τ1mol/L(v0)= 8.66(1000/1)=8660L=8.66m
3 
As is noted for a reaction of the first order, the τ is not dependent on CA0; but the volume yes 
because FA0 remains constant, that is to say that v0 varied. 
 
Problem 9.8 (p. 238) 
Solution 
(-rA)4mol/L = 4 (-rA)1 mol/L 
 
1 
 
= 
1 
 
1 
 
(-rA ) 
4mol / L 
4 
 
(-rA 
 
) 
 
 
 
 
 
1mol / L 
 
1 
 
= 4 
 
1 
 
(-rA ) 
 
(-rA ) 
4mol / L 
 
1mol / L 
 
 
 
The scale of the fig. E9.7 (p. 234) is multiplied by 4 if it reduces the C0 from 4 to 1 mol/L 
Area under the curve of 1/-Rto vs Xto the example 9.7 with 4 mol/L CA0 = 1.2 Area under 
the curve of 1/-Rto vs X with 1 mol/L CA0 = 1.2 (4) = 4.8=τ/C0 
Τ =4.8 (1) = 4.8 min (the same τ the example) 
V = τ v0 = 4.8 (1000) = 4800 L (4 times larger than that of the example) 
Space weather is not affected by the variation of the concentration; but if F0 remains constant 
volume v0 Varies 
 
Problem 9.9 (p. 238) 
You want to carry out the reaction of the example 9.4 in a reactor complete mix up to 95 % 
conversion of a power supply with CA0 = 10 mol/L and a volumetric flow of 100 L/min What 
size reactor is required? 
Solution 
T= 
𝐸2−𝐸1
𝑅
𝑙𝑛
𝑘02𝐸2
𝑘01𝐸1
+𝑙𝑛
𝑋𝐴
1−𝑋𝐴
=
124200 −48900
8.314
𝑒42.04(124200 )
𝑒17.34(48900 )
+𝑙𝑛
.95
1−0.95
= 316.94 K 
-rA=10 𝑒𝑥𝑝 17.34−
48900
𝑅𝑇
 1− 𝑋𝑎 − 𝑒𝑥𝑝 42.04−
124200
𝑅𝑇
 (𝑋𝑎) 
-rA= 0.0897L/mol min 
Τm = CA0 X A /-rA = 10(0.95)/.0897=105.91min 
V =τmv0 =10590,85 L ≈10.6 m3 
 
Problem 9.10 (p. 239) 
As E1 < E2, low temperature at the beginning of the reaction is good to produce more of R 
As E3 > E4, increasing the temperature when the reaction has already advanced induces to 
produce S which is the desired product. As E3 < E5 also E3 < E6,lowering the temperature in 
the final stages of the reaction while deduce the disintegration of S hence we can maximize 
concentration of S. 
 
Problem 9.11 (p. 239) 
If we analyze the E / R is concluded that in the first stage reaction (decomposition of A) the 
temperature should be in the high and then in low end. Let the values of the kinetic constants 
in the 2 extreme temperatures 
 
T(° C) K1 K2 K1/K2 K3 K4 K3/K4 
10 0.62 7.27 0.085 1.51 10
-6 
3.84 10
-7 
4.00 
90 66.31 163.82 .4 1.71 10
-3 
4.40 10
-3 
0.39 
 
An analysis of the values of the constants is concluded what we already knew and more; 
We know that the profile should be decreasing because the first reaction must occur in the 
decomposition of A and R is favored with high temperatures (k1 / k2 = 0.4 at 90 ° C), after 
lowering the temperature must 
because the formation of S is favored by low temperatures (k3 / k4 = 4.00 at 10 ° C). Add to 
this, we did not know that k1 and k2 >> K3 and K4, then A is exhausted practically without 
R being reacted yet. It can therefore assumed that in the first tank only decomposition of A 
occurs and in the second of R. Thus the first tank is kept at 90 ° C and second at 10 ° C 
 
𝜑𝑟 =
𝑟𝑟
−𝑟𝑎
=
𝑘1
𝑘1 + 𝑘2
=
66.31
163.82 + 66.31
= 0.288 
𝜑𝑠 =
𝑟𝑠
−𝑟𝑟
=
𝑘3
𝑘3 + 𝑘4
=
1.51 ∗ 10−6
1.51 ∗ 10−6 + 3.84 ∗ 10−7
= 0.8 
𝜑 
𝑆
𝐴
 = 𝜑𝑅 ∗ 𝜑𝑆 = 0.288 ∗ 0.8 = 0.2304 
 
 
 
Problem 9.12 (p. 239) 
The reversible reaction in the gaseous phase to ​↔ ​R is going to be carried out in a reactor                                 
of thorough mixing. If operated at 300 K the required volume of the reactor is 100 L for a                                     
60 % conversion. What should be the volume of the reactor for the same power and the                                 
same conversion; but operating at 400 K. 
Data:  To Pure  K​1 ​= 10​3​exp (­2416/T)  C​p​' ​= 0 
  
K = 10 to 300 K  H​r ​= ­8000 cal/mol to 300 K 
Solution 
With the data to 300 K you can calculate v​0 ​and v​0 ​the required volume to 400 K. It                                 
should be noted that v​0 ​varies by varying the temperature and that H​R ​is constant because                           
C​p​' ​= 0 
 
 =   =    =   = tm v0
V m
−rA
C XA0 A C XA0 A
K C (1− X )− K C X1 A0 A 2 40 A
XA
K (1− X )− K X1 A 2 4
 
  =   V 0 V m )( XA
K (1− X )−K X1 A 2 A  
 = 0.318K1(300k) min
−1  
 =   =   = 0.0318K2(300k) K
K1
10
0.318 min−1  
 =    = 18.2 L/minV 0 0.6
100[0.318(1−0.6)−0.0318(0.6)]  
 =    exp  = 10exp  = 4.485K400k K300k − ( )][ R
Hr 1
400 −
1
300 − ( )][ 8314
8000 1
400 −
1
300  
 = 2.382K1(400k) min
−1  
  =   = 0.531K2(400k) 4.485
2.382 min−1  
 =   = 18.02  = 24.03 L/minV 0(400k) ( )V 0(300k) 300
400 )(300
400  
V =   = 24.03  = 22.73L( )V 0
XA
K (1− X )−K X1 A 2 A
][ 0.62.382(1−0.6)−0.531(0.6)   
Example 10.1 
Given the two reaction 
 
 
→ -r1=k1CACB 
 
 
→ -r2=k2CRCB 
Where R is the desired product and is to be maximized. Rate the four schemes shown in 
Fig.P10.1-either “good” or “not so good,” Please, no complicated calculations, just reason it out. 
 
Answer: 
 
The distribution of products (R / S) is determined by the ratio of k1 / k2 because reactions are of 
the same order with respect to A and B, so if I want more R, as in a simple reaction, which XA 
require is greater. This is achieved by working with maximum speeds. Assuming isothermal 
operation, -rA grows when the concentration is high, so considering all of the above 
(D) The CA and CB better because high ( Very good) 
(A) is eqully good because here CA is in excess quantity. so formation of R is really high 
compare to formation of S. 
(C) Not so good. beacuse CB is in excess quantity. So rate of formation is S is very high compare 
to rate of formation of R. 
(B) The worse because both concentrations are low. 
 
Example 10.2 
Repeat Problem 1 with just one change 
-r2 = k2CRCB
2 
Answer: 
 
It is a typical reaction series and parallel performance tells us that precisely. We must analyze 
separately the components in series and componentsin parallel. 
A → R → S are in series so it should not be high. 
The reagent parallel, B, has a lower order than the desired reaction unwanted, so B should be 
maintained at low concentrations and also we take difference is high so A has as high as 
possible. 
(A) The best Design because high CA and low CB. 
(B) and (D) Intermediate because in (B) CA and CB and lower (D) CA and CB high, 
only meet one requirement both 
(C) The worse Design because CA low and CB high means difference is less so Desired quantity 
is R which is less form because Formation R it instantaneously react with B (It is in excess 
quantity ) and form S. Which is undesired for us. 
 
Example: 10.3 
Repeat Problem 1 with just one change 
-r2 = k2CR
2CB 
Answer: 
 
It is a parallel reaction and performance series tells us just that. We must analyze separately the 
components in series and components in parallel. 
A → R → S are in series so CA should be high 
The reagent parallel, B, has the same order that the desired reaction the unwanted, so B does not 
influence the distribution of products. 
(A) and (d) are the best because high 
(B) and (C) worse (not so good) design because rate is mainly dependent on CR and also CA is in 
denominator so A is low necessary but after it decreases concentration of R. and formation of S 
is high in (C). 
 
Example: 10.4 
For the reaction 
 → -r1 = k1CACB 
 → -r2 = k2CRCB
2 
Where R is the desired product, which of the following ways of running a batch reactor is 
favourable, which is not ? See Fig. P10.4 
 
 
Answer: 
(A) better because high CA and CB low 
(B) Intermediate because CA = CB low. considering the third alternative where the contents of 
the two beakers are rapidly mixed together, the reaction being slow enough so that it does not 
proceed to any appreciable extent before the mixture becomes uniform. During the first few 
reaction increments R finds itself competing with a large excess of A for B and hence it is at a 
disadvantage. Carrying through this line of reasoning, we find the same type of distribution curve 
as for the mixture in which B is added slowly to A. 
 
(C) The CA worse because low CA and high CB, is not suitable. For the first alternative pour A a 
little at a time into the beaker containing B, stirring thoroughly and making sure that all the A is 
used up and that the reaction stops before the next bit is added. With each addition a bit of R is 
produced in the beaker. But this R finds itself in an excess of B so it will react further to form S. 
The result is that at no time during the slow addition will A and R be present in any appreciable 
amount. The mixture becomes progressively richer in S and poorer in B. This continues until the 
beaker contains only S. 
Example: 10.5 
The Oxydation of Xylene. The violent oxidation of xylene simply produces CO2 and H2O; 
however, when oxidation is gentle and carefully controlled, it can also produce useful quantities 
of valuable phthalic anhydride as shown in Fig.P10.5.Also,because of the danger of 
explosion,the fraction of xylene in the reacting mixture must be kept below 1%.Naturally,the 
problem in this process is to obtain a favourable product distribution. 
(a) In a plug flow reactor what values of the three activation energies would require that we 
operate at the maximum allowable temperature? 
(b) Under what circumstances should the plug flow reactor have a falling temperature 
progression? 
 
Answer: 
a) If E1> E2 and E1> E3 
The most desired is the activation energy and will be favored by high temperatures, so it must 
work to the maximum allowable temperature. 
b) If E1> E3 and E1 <E2 
At first you must work with high temperatures to encourage 1 against the reaction 3 and then the 
temperature should drop to not favor the profile step 2. It is then to be used is decreasing. 
 
Example 10.6 
The Trambouze Reactions - Reactions in parallel. Given the set of elementary reactions with a feed 
of CA0 =1moll/liter and V = 100 literslmin we wish to maximize the fractional yield, not the 
production of S, in a reactor arrangement of your choice. 
 
 
a) Do you judge that arrangement of Fig.P10.6 is the best set up? 
b) If not,suggest a better scheme. Sketch your scheme and calculate the volume of the 
reactors you plan to use. 
 
 
Answer: 
 
A → R rR = k0 k0 = 0.025 mol/L min 
A → S rS = k1CA k1 =0.2 min
-1 
A → T rS = k2CA k2 = 0.4 L/mol min 
The order of the desired reaction determines how they should be the concentrations, whether 
high or low. In this case 
Order A→R < Order A → S < Order A → T 
 
The order of reaction is the desired intermediate. CA favor high A → R and CA low to A → T. It 
is obvious that there must be an intermediate concentration A → S. To find that CA makes 
performance maximum must be raised that dφ (S / A) / dCA = 0 
 
Φ (
 
 
) = 
 
 
 
 
 
 
 
 
 = 
( ) 
 
 = 0 
 
0.025 + 0.2 CA + 0.4 CA
2 + 0.2 CA – 0.8 CA
2 = 0 
 
The only possible solution is CA = 0.25 mol / L and the concentration value making the most 
appropriate performance 
 
It is best to work with a mix Flow reactor that throughout the reactor instant performance equals 
0.5 which is its maximum value. 
 
b) 
 
 
CS = Φ (
 
 
)(CA0- CA) = 0.5(1-0.25) = 0.375 
 
Example 10.7 
For the set of elementary reaction of problem 10.6, with a feed of CA0 = 1 mol/liter and v = 100 
liters/min we now wish to maximize the production rate of intermediate S (not the fractionl 
yield) in a reactor arrangement of your choice. Sketch your chosen scheme and determine CS max 
obtainable. 
 
Answer: 
Given the shape of the curve φ vs CA there to work with a reactor will complete mixture CA= 1 to 
0.25 mol / L, which will have the maximum performance and with this a series of plug flow 
reactor that CA = 0.25 go from to 0 mol / L, to take advantage as much as possible high yields. 
The final concentration is 0 because if ΔCA ↑ Δ CR↑ and CR wants maximum 
Cs max = CS m + CS P 
Cs max = 0.5(1-0.25) = 0.375 mol/liter 
Cs p = ∫ 
 
 
= ∫
 
 
 
 
 
Dividing numerator and denominator by 0.4 
Cs p = 0.5∫
 
 
 
 
 = 0.5∫
 
 
 
 
 
 
Cs p = 0.5 {
 
 
[ln(0.25 + CA) + 
 
 
]0.250 } 
 = 
 
 
 
 
 
Example 10.8 
 
Automobile Antifreeze Ethylene glycol and diethylene glycol are usedas automobile antifreeze, 
and are produced by the reactions of ethylene oxide with water, as follows: 
 
A mole of either glycol in water is as effective as the other in reducing the freezing point of 
water; however, on a molar basis the diethylene glycol is twice as expensive as the ethylene 
glycol. So we want to maximize ethylene glycol, and minimize the diethylene glycol in the 
mixture. 
 
One of the country's largest suppliers produced millions of kilograms of antifreeze annually in 
reactors shown in Fig. P10.8a. One of the company's engineers suggested that they replace their 
reactors with one of the type shown in Fig. P10.8b. What do you think of this suggestion? 
 
 
Answer: 
H2O+ ethylene → Ethylene Oxide 
Ethylene+ ethylene → Diethylene Oxide 
If H2O = A, Y = ethylene oxide, ethylene glycol =R and diethylene = S 
The reaction can be expressed as 
 
A + B → R (reaction 1) 
R + B → S (reaction 2) 
Analyzing the components in series A → R → S is the most convenient plug flow reactor and the 
reactor of Figure (b) may be considered as diameter / length such by their relationship. 
 
As for the addition of B, the component parallel can not conclude nothing because the reaction 
order of this component is not known in the desired and undesired reaction. If the desired out the 
higher order, then the addition of B, which raises the concentration of this component is 
appropriate. 
 
We select PFR because length/diameter ratio is high in PFR and also undesired quantity 
(diaethyelene glycol ) is less. 
 
Example 10.9 
The Homoeneous catalytic Reaction. Consider the elementary reaction 
A + B 
 
→ 2B, -rA = kCACB with k = 0.4 liter/mol min 
For the following feed and reactor space time 
Flow rate v = 100 liters/min 
Feed compositin {
 
 
 
Space time τ = 1 min 
We want to maximize the concentration of B in the product stream. Our clever computer [see 
problem 8, Chem. Eng. Sci., 45, 595-614 (1990)] gives the design of Fig. P10.9 as its best try. 
 
 
 
Do you think that this is the best way to run this reaction? If not, suggest a better scheme. Do not 
bother to calculate reactor size, recycle rate, etc. Just indicate a better scheme. 
 
Answer: 
As this is a simple reaction system criteria used is As the production efficiency, the more 
efficient is the system having higher speeds. So to know what is more suitable reactor is need to 
know how the conversion varies with –rA 
 
-rA = kCACB (Elementary system) 
CA = CA0(1-XA) 
CB =CB0-CA0XA+2CA0XA (for stoichiometry) 
CB = CA0(M + XA) ,where M = 
 
 
 = 
 
 
 = 1.22 
-rA = k CA0
2(1-XA)(M + XA) 
 
If, XA↑, (1-XA)↓ so you can increase or decrease, according to the relative weight of the factors 
One way to know how varies –rA with XA is to find the derivative of the function. 
 
 
 
 = kCA0
2[(1-XA)(1) + (M + XA)(-1)] = kCA0
2(1-M-2XA) < 0 
 
If we consider that M = 1.22 and analyze the derivative we see that it is negative over the entire 
range of conversions, indicating that the function is decreasing for any value of XA, i.e, the speed 
decreases to increase conversion. 
Another way to know how to vary the speed conversion, less accurate, it is to evaluate the 
function in the interval. 
-rA = 0.4 (0.45)
2 (1 – XA) (1.22 + XA) 
-rA = 0.081 (1 – XA) (1.22 + XA) 
XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 
-rA* 
10-2 
9.9 9.6 9.2 8.6 7.8 7.0 5.9 4.7 3.3 1.7 
 
 
 
The higher speed is XA = 0 (high concentrations) so that the reactor It is best piston without 
recycling, because the recycle low profile concentrations occurring in the reactor and thus lower 
the speed occurring in the reactor. 
 
Compare both PFR and MFR 
 
Here Plug flow is best design. but here recycling is not required if recycle then it is against the 
assumption of PFR. 
 
Example 10.10 
To Color Cola Drinks. When viscous corn syrup is heated it caramelizes (turns a deep dark 
brown). However, if it is heated a bit too long it transforms into carbon 
 
 
The caramelized liquid is sent by railroad tank cars to the cola syrup formulators, who then test 
the solution for quality. If it is too light in color-penalty; if it has too many carbon particles per 
unit volume, then the whole tank car is rejected. There is thus a delicate balance between 
underreacting and overreacting. 
At present a batch of corn syrup is heated at 154°C in a vat for a precise time. Then it is rapidly 
discharged and cooled, the vat is thoroughly cleaned (very labor intensive), and then is 
recharged. 
The company wants to reduce costs and replace this costly labor intensive batch operation with a 
continuous flow system. Of course it will be a tubular reactor (rule 2). What do you think of their 
idea? Comment. please, as you sit and sip your Coke or Pepsi. 
Answer: 
The substitution is theoretically substantiated that over piston take place the same story of 
concentrations, and therefore speeds, which occur with time in the batch, and It was further 
added advantage of continuous operation. 
However, in this system is formed inevitably solid, which adhere to the reactor wall to some 
extent (which is why the cleaning batch reactor was a daunting) task is that if the liquor is 
viscous and heated through the walls will create a temperature gradient and coal to be formed 
will adhere to the pipe walls preventing proper operation. or that think about stopping and clean 
the pipe. 
10.11 A-R-S 
 T U Cao = 6 CRo=0.6 CTo=0 
The feed has CR/CT=∞ 
a) So to maximise CR/CT do not react at all, the design of figure is not 
good. 
b) To minimise CR/CT , run to completion. All the R will disappear. 
Next reaction 1 is second order 
Reaction 2 is first order keep CA low 
So, use a large MFR you will end up with more CT 
 
 
10.12 Cao = 90 mol/m3, Cbo = 10 mol/m3 
 A+B  B+B -rA=kCaCb 
If we want minimum volume, which will be efficient in terms of production and 
with maximum speed, we have to know how rate varies with speed 
-rA = k CA CB 
CA = CA0 (1 – XA) = CA0 (1-XA) 
CB = CB0 – CA0XA + 2CA0XA = CA0 (M + XA) where M = CA0/CB0 
-rA = k CA02 (1- XA) (M + XA) 
))(1(
)( 2 XaMXakCao
dXa
rad





 
 
 
M = 1/9 = 0.11, so the derivative is positive for low values XA and 
negative for high values and maximum -rA the derivative is 0, so 
XA = (1-M)/2 = 0.44 
 
Speed will be : (-rA/k) = 8100 (1 – XA) (1/9 + XA) 
 
XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 
-rA/k 900 1600 2100 2400 2500 2400 2100 1600 900 810 
 
 
 
Use a MFR 
 
 
 
 
10.13 A+B B +B 
 -rA=KCACB CAf = 9 CBf= 91 
 CAo= 90 CBo=10 
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0 0.2 0.4 0.6 0.8 1
Y-Values
Y-Values
 
Use a MFR followed by a PFR 
 
10.14 
𝐴 + 𝐵 → 𝐵 + 𝐵, -rA = kCACB 
CA0 = 90 mol/m
3 CAf = 72 mol/m
3 
CB0 = 10 mol/m
3 CBf = 28 mol/m
3 
XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 
-rA/k 900 1600 2100 2400 2500 2400 2100 1600 900 
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0 0.2 0.4 0.6 0.8 1
Y-Values
Y-Values
 
As curve k/-rA vs XA downward slope in the range XA = 0 to 0.2, it is best to 
work with a mixing reactor complete, which will have the higher speed range 
and therefore will require the least τ. And for the 20% conversion it will give 
the lower volume as compared the other reactors. 
 
10.15 
 Since E1>E2 use a high temperature or T = 90
0C 
 At 900C, k1 = 30e
-20000/8.314(363) = 0.0397 min-1 
 k2 = 1.9e
-15000/8.314(363) = 0.0132 min-1 
 Therefore, k2/k1 = 0.3322 
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0 0.2 0.4 0.6 0.8 1
Y-Values
Y-Values
 
 Therefore, Form the above fig, CRmax/CA0 = 0.58 
 topt = [ln(k2/k1)]/[k2-k1] = (ln 0.3322)/(0.0132-0.0397) 
 = 4.16 min-1 
10.16The following system of elementary reactions have 
 
C6H6 + Cl2 → C6H5Cl + HCl k1 = 0,412 L/kmol h 
C6H5Cl + Cl2 → C6H4Cl2 + HCl k2 = 0,055 L/kmol h 
 
A = C6H6 
B = Cl2 
C = C6H5Cl 
D = C6H4Cl2 
A + B → R 
R + B → S 
 
It is best to use a plug flow reactor. You should not use the first recycle reactor. 
It should work at low conversions of benzene. For example, 
XA =0.4, 
with k2 / k1 = 0.1335, CR / CA0 = 0.3855, CS / CA0 = 0.0145 and CR / CS = 26.68. 
 
 
 
 
10. 17- Let A = C3H6 B=O2 R=C3H4O 
 Then A+B  R K2/K1=0.1 
 A+4.5B S K3/K1=0.25 
 AR S 
From the Denbigh reaction 
CRmax/CAo = K1/K2(K12/K3)^(K3/K3-K12) 
=10/11*(11/25) ^(2.5/2.5-11) 
=0.588 
10.18 
 
It is a series parallel system , so selection is based on analyzing their series and 
parallel components as separate components . Parallel components are A → R 
and A → T and if we see the orders of the reactions, lower order is desired , 
therefore taking into account parallel components concentration must be kept 
low during the course of the reaction . 
Series components are A → R → S and R is the desired as having into account 
the components in series the concentration of A must remain as high as possible 
during the course of the reaction . 
There is a contradiction and the final decision depends on the relative weight of 
reactions . 
Considering the kinetic constants 
T (K) k1 k2 k3 k1/k3 k1/k2 
360 0.1306 0.0560 0.00006 2176 2.33 
396 0.9700 0.1530 0.00120 808.3 6.3 
Analysis of this table we see that between 360 and 396 K ,k3 <<k1 and k1 and 
k2 are almost similar. 
 
The reaction A → T ½ can be neglected because it occurs slowly compared to 
the other . if so then the plug flow reactor is the most convenient . 
 
b ) Working with a piston 396 K 
12
2
2
1 kk
k
k
k
Cao
Cr 







 
Cr=0.707 mol/L 
s
kk
k
k
252.2
12
1
2
ln














 
Xa=1-exp(k1τ)=0.888 
 
 
 
 
QUES: 10.19 Phthalic Anhydride from Naphthalene. 
The accepted mechanism for the highly exothermic solid catalysed oxidation of 
naphthalene to produce 
Phthalic anhydride is 
 
 R 
 
A S T 
 
 
Where 
k1 = k2 = 2 * 10
13 exp (-159000/RT) [hr-1] 
k3 = 8.15 * 10
17 exp (-209000/RT) [hr-1] 
k4 = 2.1 * 10
5 exp (-83600/RT) [hr-1] 
 
And where 
A = naphthalene (reactant) 
R = naphtha Quinone (postulated intermediate) 
S = phthalic anhydride (desired product) 
T = CO2 + H2O (waste products) 
 
and the Arrhenius activation energy is given in units of J/mol. This reaction is to 
be run somewhere between 900 K and 1200 K. 
A local optimum reactor setup discovered by the computer [see example 1, 
Chem. Eng. Sci., 49, 1037-1051 (1994)] is shown in Fig. P10.19. 
 
(a) Do you like this design? Could you do better? If so, how? 
(b) If you could keep the whole of your reactors at whatever temperature and τ 
value desired, and if recycle is allowed, how much phthalic anhydride could be 
produced per mole of naphthalene reacted? 
 
Suggestion: Why not determine the values of k1, k2, k3, and k4 for both 
extremes of temperature, look at the values, and then proceed with the solution? 
 
 
SOLUTION: 
 Given, 
k1 = k2 = 2 * 10
13 exp (-159000/RT) [hr-1] 
k3 = 8.15 * 10
17 exp (-209000/RT) [hr-1] 
k4 = 2.1 * 10
5 exp (-83600/RT) [hr-1] 
 
 
 R 
 
A S T 
 
Since k4 value is much smaller than the other values of k we use the highest 
allowable temperature of 1200K. 
 
k1 = 2.4 * 10
6 hr-1 
k2 = 2.4 * 10
6 hr-1 
k3 = 650 * 10
6 hr-1 
k4 = 48.2 hr
-1 
 
Now, 
Since k3 value is much bigger than k2 value we can approximate the reaction as 
follows. 
 
 
A R T 
 
 Now we find CRmax 
 
CRmax
CA0
= (
𝐾5
𝐾6
)
𝐾6
𝐾6−𝐾5
 = (
4.8∗106
48.2
)
48.2
48.2−(4.8∗106)
 
 = 0.99988 
 
tplug = 
ln(
𝑘2
𝑘1
)
𝑘2−𝑘1
 = 2.4 * 10-6 hr = 0.0086 sec = time of residence 
With such a small time and high conversion practically we can use any type of 
reactor but we are using a straight plug flow reactor with no recycle which is the 
best. 
Now we find CR for this residence time 
𝐶𝑅
𝐶𝐴𝑜
=
𝑘5
𝑘5−𝑘6
[𝑒−𝑘1𝑡 − 𝑒−𝑘2𝑡] = (-1) [0 – 0.9762] = 0.9762 = CR 
 
 
10.20. Professor Turton dislikes using reactors in parallel, and he cringed when he saw my 
recommended "best" design for Example 10.1. He much prefers 
using reactors in series, and so for that example he suggests using the design of Figure El0.la, 
but without any recycle of fluid. Determine the fractional yield of S, Φ (S/A), obtainable with 
Turton's design, and see if it matches that found in Example 10.1 
 
SOLUTION: 
Here, 
 
τ 1 = 
𝑉
25
=
𝐶𝐴0−𝐶𝐴1
(−𝑟𝐴)1
 
 
 
τ 2 = 
𝑉
50
=
𝐶𝐴1′−𝐶𝐴2
(−𝑟𝐴)2
 where CA1’ = 
1
2
 𝐶𝐴1 +
1
2
 
 
 
τ 3 = 
𝑉
75
=
𝐶𝐴2′−𝐶𝐴3
(−𝑟𝐴)3
 where CA2’ = 
2
3
 𝐶𝐴2 +
1
3
 
4.8 * 106 48.2 
 
 
τ 4 = 
𝑉
100
=
𝐶𝐴3′−0.25
(−𝑟𝐴)25
 where CA3’ = 
3
4
 𝐶𝐴3 +
1
4
 
 
 
It is a set of 4 equations and 4 unknowns , so it is determined ; but must be solved by trial and 
error and once you know the CR concentrations determine the output of each reactor. It is 
clear that the procedure is quite cumbersome and graphical method for solving the material 
balance of the reactor mixture can help to the solution. According to the method the operating 
point reactor located where the curve ( -rA ) vs CA is cut with balance materials in the plane ( 
-rA ) -CA is a straight line passing through the reactor inlet concentration and whose slope -1 / 
τi . 
 
 
(-rA) = 0.025 + 0.2 CA + 0.4 (CA)
2 
 
 
 
 
 
CA 0.6 0.55 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 
(- 0.625 0.256 0.225 0.196 0.169 0.144 0.121 0.100 0.081 0.064 0.049 0.036 
rA) 
 
 
 
 
 
 
 
 
 
 
 
 
The steps followed in solving the problem are: 
1. With these values the curve ( -rA ) is obtained vs CA 
2. It is assumed CA1 and τ1 is calculated by equation 1 
3. τ1 is calculated V 
4. V is calculated τ2 , and τ1 τ3 
5. CA ! '= ½ CA1 + 1/2, and τ2 operating point is calculated 
reactor 2 
6. CA2 '= 2/3 +1/3 CA2 and τ3 the operating point is calculated 
reactor 3 
7. CA3 '=¾ CA3 +1/4 and τ4 operating point is calculated 
reactor 4 
8. If CA4 = 0.25, the assumed value of CA1 is correct, if not return to 
Step 2 
 
 
Suppose CA1 = 0.3 
 
 
 
V = 
 
(1 −0.3)(25) 
 
=144.6281 L 0.025 + 0.2(0.3)+ 0.4(0.32 ) 
τ2 = 
144.6281 
= 2.8925 min 
 
⇒ 
1 
 = 0.3457 
 
 
50 τ2 
 
τ3 = 144.6281 =1.9283 min ⇒ 1 = 0.5186 min −1 
 
75 
 
 
τ3 
 
τ2 = 
144.6281 
=1.4462 min ⇒ 
1 
 
= 0.6914 
min −1 
 
 
100τ2 
 
 
 
C′A1 = 02,3 + 12 = 0.65 mol / L 
 
C′A2 = 2(03,3) + 13 = 0.5333 mol / L 
 
C′A3 = 3(04,3) + 14 = 0.475 mol / L 
 
Suppose CA1 = 0.25 
 
 
V = 
 
(1 −0.25)(25) 
 
=187.5 L 0.025 + 0.2(0.25)+ 0.4(0.252 ) 
τ2 = 
187.5 
 
= 3.75 min 
 
⇒ 
1 
 
= 0.2667 
min −1 
 
 
50 
 
τ2 
 
τ3 = 187.5 = 2.5 min ⇒ 1 = 0.4 min −1 
 
75 
 
 
τ3 
 
τ2 = 
187.5 
 
=1.875 min 
 
⇒ 
1 
 
= 0.5333 
 
 
100 
 
τ2 
 
 
C′A1 = 
0.25 
 
+ 
1 
= 0.625 mol / L 
 
2 2 
 
 C′A2 = 2(0.25) + 1 = 0.5 mol / L 
 
3 
 
 
3 
 
C′A3 = 3(0.25) + 1 = 0.4375 mol / L 
 
 
4 
 
4 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
CRE 1 Unsolved examples chapter 11 
11.1. A pulse input to a vessel gives the results shown in Fig. P1l.l. 
(a) Check the material balance with the tracer curve to see whether the results are consistent. 
(b) If the result is consistent, determine t, V and sketch the E curve. 
Ans. Area Under the curve=Cpulse×t= 0.05×5= 0.25 mol min. /lit. 
a. M= 1 mol at t=0 (given); ν= 4 lit. / min. 
Therefore, 
𝑀
ν
= ¼ = 0.25 mol min. / lit. 
Since Area Under the curve equals 
𝑀
ν
 the results are consistent. 
 
b. Taverage = 
(Ʃ ti Ci)
Ʃ Ci
= 
(5×0.05) 
0.05
= 5 min. 
Therefore, Taverage= 
V
ν
 
V = Taverage × ν = 5 × 4 = 20 liters 
E = (Cpulse × ν)/ M = (0.05 × 4)/ 1 = 0.2 
 
 
 
 
 
 
 
 
11.2. Repeat Problem P1l.l with one change: The tracer curve is now as shown in Fig. P11.2. 
 
Ans. From the previous sum 
𝑀
ν
 = ¼ = 0.25 mol min. / lit. 
 Area Under the curve=Cpulse×t= 0.05×6= 0.3 mol min. /lit. 
 Therefore, 
𝑀
ν
 ≠ Cpulse×t 
 The tracer comes out late hence the experiment is not done correctly. 
 
 
11.3. A pulse input to a vessel gives the results shown in Fig. P11.3. 
(a) Are the results consistent? (Check the material balance with the experimental tracer curve.) 
(b) If the results are consistent, determine the amount of tracer introduced 
M, 
Ans. ν= 4 cm3/s V= 60 cm3 
In Pulse Input taverage= 
V
ν
= 60/4 = 15 s 
From the graph (assuming the value of tip to be C) 
Area = 
1
2
[(3 × 𝐶) + (6 × 𝐶)] = 9 ×
𝐶
2
 
taverage= 
Ʃ𝑡𝐶∆𝑡
𝐴𝑟𝑒𝑎
 = 
[(16×0)+(19×𝐶)+(25×0)]
9𝐶
2
 = 
38
9
 
Since taverage is not the same the result if not consistent 
 
 
11.4. A step experiment is made on a reactor. The results are shown in Fig. P11.4. 
(a) Is the material balance consistent with the tracer curve? 
(b) If so, determine the vessel volume V, 7 
Ans. Since Value of Cmax is not given this problem cannot be solved 
 
 
11.6. A pipeline (10 cm I.D., 19.1 m long) simultaneously transports gas and liquid from here to 
there. The volumetric flow rate of gas and liquid are 60 000 cm3/s and 300 cm3/s, respectively. 
Pulse tracer tests on the fluids flowing through the pipe give results as shown in Fig. P11.6. What 
fraction of the pipe is occupied by gas and what fraction by liquid? 
 
 
 
 
 Ans. V= 𝜋/4*D2H 
= 𝜋/4*102*10-4*19.1 [Given D=10cm and H= 19.1m] 
∴ VT= 0.15 
VG=60000 cm
3/s 
tRG = 2s 
VG = 60000 * 10
-6 * 2 m3 
= 0.12 m3 
VL = 300 cm
3/s 
tRL = 100s 
∴ VL = 300 * 10-6 * 100 m3 
= 0.03 m3 
∴ % Volume occupied by gas =0.12/0.15 * 100 =80% 
% Volume occupied by liquid =0.03/0.15 * 100=20% 
A liquid macro fluid reacts according to A + R as it flows through a vessel. Find the conversion 
of A for the flow patterns of Figs. P11.7 to P1l.ll and kinetics as shown. 
 
11.7) 
 
 
Ans. ∫ 𝐸. 𝑑𝑡
∞
0
 =1 
∴ 2*E=1 
∴E= 0.5 
= C/C0 = ∫ (
C
C0
) . Edt 
∞
0
 
 (C/C0) for a batch material is given by Eq
n 
Here n = 0.5 and k= 2 mol0.5/litre0.5 * min 
(n-1) kCA0
nt = (CA/CA0)
1-n -1 
∴ -1/2 * 2 * 1 * t = (CA/CA0)1/2 -1 
∴ (CA/CA0) = (1-t)2 
∴ C/C0 = ∫ (1 − 𝑡)
2
0
2 . (½). dt 
= -1/2* [(1-t)3/3]
2
0
 
= -1/2*[-1/3-1/3] 
= -1/2 * -2/3 
= 1/3 
= XA= 1-1/3 = 2/3 
11.8) 
 
Ans. C/C0 =∫ (C/C0)
∞
0
. E.dt 
∫ 𝐸. 𝑑𝑡
∞
0
 = 1 
∴ 0.5= 0.5 * 0.5 * E 
∴ E = 2 
Here n = 2, k=2 litre/mol * min and CA0= 2 mol/litre 
(n-1) kCA0
nt = (CA/CA0)
1-n -1 
1 * 2 * 22 * t = (CA/CA0)
-1 -1 
1 + 8t = (CA/CA0)
-1 
= (CA/CA0) = 1/1+8t 
=2∫ 1/1 + 8𝑡
0.5
0
 * dt 
=1/4 [ln (1+8t)]
0.5
0
 
=1/4 [ln (5)] 
= XA = 0.6 
11.9 
CA0 = 6 mol/Lit 
 -rA= k 
k = 3 mol/lit min 
 
Ans. The Performance Equation is 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅
= ∫
𝐶𝐴
𝐶𝐴0
𝐸𝑑𝑡
∞
0
 
For t< CA0 / k = 6/3 = 2min; we have CA / CA0 = 1 – ( kt / CA0 ) 
For t> 2 min; we have CA / CA0= 0 
Replacing it in the above equation, 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
= ∫(1 −
𝑘𝑡
𝐶𝐴0
)𝐸𝑑𝑡
2
0
 
Putting E =1/3, 
 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
=
1
3
∫(1 − 3 ∗
𝑡
6
)𝑑𝑡
2
0
 
 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
=
1
6
∫( 2 − 𝑡 )𝑑𝑡
2
0
 
 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
=
1
3
 
 
So the conversion is XA = 1 – CA / CA0 = 2/3 = 0.666 = 66.6 % 
 
11.10 
CA0 = 4 mol /lit 
 -rA= k 
k = 1mol/lit min 
 
Ans. Here Dirac Delta function is used. It is Zero everywhere except one point and that point is 
at Infinite value. The Equation to be used here is: 
CA
CA0
̅̅ ̅̅ ̅
= ∫ f(t) δ (t − t0) dt
∞
0
 
We know that, ∫ 𝑓(𝑡) 𝛿 (𝑡 − 𝑡0) 𝑑𝑡
∞
0
= 𝑓(𝑡0) 
For t< CA0 / k = 4/1 = 4 min; CA / CA0 = 1 – ( kt / CA0 ) 
For t> 4min ;CA / CA0 = 0 
Here f (t) = 1 – (kt / CA0) 
So putting it in the above equation and by Dirac Delta function, 
 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
= ∫(1 −
𝑘𝑡
𝐶𝐴0
)𝛿 (𝑡 − 3) 𝑑𝑡
4
0
 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
= 1 −
𝑘𝑡
𝐶𝐴0
| 𝑎𝑡 𝑡 = 3 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅ ̅̅
= 0.25 
So XA = 0.75 = 75% 
 
11.11 
CA0 = 0.1 mol /lit 
-rA= k 
k = 0.03mol/lit min 
 
 
Ans. The Performance Equation is :- 
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅
= ∫
𝐶𝐴
𝐶𝐴0
𝐸𝑑𝑡
∞
0
 
For t< CA0 / k = 0.1/0.03 = 3.33 min; CA / CA0 = 1 – ( kt / CA0 ) 
For t> 3.33 min ;CA / CA0 = 0 
Here Nothing is leaving the Reactor after t = 3.33 min. so everything has been reacted .so
𝐶𝐴
𝐶𝐴0
̅̅ ̅̅
= 0 
and XA = 1. 
11.12-11.14. Hydrogen sulfide is removed from coal gas by contact with a moving bed of iron 
oxide particles which convert to the sulfide as follows: 
Fe203 -+ FeS 
 
In our reactor the fraction of oxide converted in any particle is determined by its residence time t 
and the time needed for complete conversion of 
 
 
 
11.12 
The reaction occurring here is Fe2O3 → FeS. 
1-X = (1-t/𝜏 )3 where t<1 and 𝜏 = 1 hr 
X = 1 where t ≥ 1 hr 
 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 −
t
𝜏
)
31
0
 Edt 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 − t )3
1
0
 Edt 
Here E=1 [ From figure Area = E*t so we have 1=E*1 which results to E=1 ] 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 − 𝑡3 − 3𝑡 + 3𝑡2
1
0
 ) dt 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = 0.25 
�̅� = 0.75 = 75 % 
Find the conversion of iron oxideto sulfide if the RTD of solids in the contactor is approximated by the 
curve 
 
Ans. 
For solid particles, 
1 − X = 
CA
CA0
̅̅ ̅̅ ̅
= ∫ f(t) δ (t − t0) dt
∞
0
 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 −
t
𝜏
)
31
0
 E dt , where E= 𝛿 (𝑡 − 𝑡0) 
We know that, ∫ 𝑓(𝑡) 𝛿 (𝑡 − 𝑡0) 𝑑𝑡
∞
0
= 𝑓(𝑡0) 
1 − 𝑋 = ( 1 − 𝑡 )3|𝑎𝑡 𝑡 = 0.5 
1 − 𝑋 = 0.125 
 
Ans: X= 0.875 
 
 
 
 
 
 
Ans. 
For solid particles, 
1 − X = 
CA
CA0
̅̅ ̅̅ ̅
= ∫ f(t) δ (t − t0) dt
∞
0
 
 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 −
t
𝜏
)
31.5
0.5
 E dt 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 − t )3
1.5
0.5
 E dt 
Here E=1 [ From figure Area = E*t so we have 1=E*1 which results to E=1 ] 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = ∫ ( 1 − 𝑡3 − 3𝑡 + 3𝑡2
1.5
0.5
 ) dt 
1 − 𝑋̅̅ ̅̅ ̅̅ ̅ = 0 
Ans: �̅� = 1 = 100 % 
 
Q 11.15 
Cold solids flow continuously into a fluidized bed where they disperse rapidly enough so that 
they can be taken as well mixed. They then heat up, they devolatilize slowly, and they leave. 
Devolatilization releases gaseous A which then decomposes by first-order kinetics as it passes 
through the bed. When the gas leaves the bed decomposition of gaseous A stops. From the 
following information determine the fraction of gaseous A which has decomposed. Data: Since 
this is a large-particle fluidized bed containing cloudless bubbles, assume plug flow of gas 
through the unit. Also assume that the volume of gases released by the solids is small compared 
to the volume of carrier gas passing through the bed. 
Mean residence time in the bed: 6 = 15 min, 6 = 2 s for carrier gas For the reaction: 
A →products, -rA = kcA, k = 1 s-l 
 
Ans. 
Here A is released uniformly. 
 
From the curve area under curve must be 1, therefore E=0.5 
The Performance Equation is 
𝑪𝑨
𝑪𝑨𝟎
̅̅ ̅̅
= ∫
𝑪𝑨
𝑪𝑨𝟎
𝑬𝒅𝒕
∞
𝟎
 
For first order reaction, 
CA
CA0
= e−kt 
 
CA
CA0
̅̅ ̅̅
= ∫ e−ktEdt
∞
0
 
1 2 
CA
CA0
= [e−t]0
2 
CA
CA0
= 0.4323 
Ans: X=1 - 
𝑪𝑨
𝑪𝑨𝟎
= 0.5677 
 
Q 11.16 
Reactant A (CAo = 64 mol/m
3) flows through a plug flow reactor (r = 50 s), and reacts away as 
follows: 
 
Determine the conversion of A if the stream is: 
(a) a microfluid, 
(b) a macrofluid. 
A 11.16 
(a) For Microfluid: 
 
 𝜏
𝐶𝐴0
̅̅ ̅̅ ̅̅
=
∫ (𝑑𝐶𝐴)
𝐶𝐴
𝐶𝐴𝑜
𝑘𝐶𝐴
1.5 𝐶𝐴𝑜
 
 
 𝜏 = 
[𝐶𝐴
−1.5+1]𝐶𝐴0
𝐶𝐴
−𝑘[−1.5+1]
 
50*0.005*0.5 = 
1
𝐶𝐴
0.5 - 
1
𝐶𝐴𝑜
0.5 
CA = 16, Since CAO= 64 
X = 
𝐶𝐴𝑂− 𝐶𝐴
𝐶𝐴𝑂
 = 0.75 Ans: X = 0.75 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
CRE-2 Group No-11 Chapter –13 
(13BCH054, 055, 056, 057) 
 
 
 
 
 
 
 
 
13.4 is not done as per the talk with you sir 
 
13.5 
 
 
Solving it with the help of dispersion model – 
Using equation 8; 
σ2 = 2(DL/u3) 
 Or; σ2 is proportional to L 
Therefore; 
σc2 - σb2 = σb2 - σa2 
Or σc2-1024 = 1024-256 
Thus; 
σc2=1792 
Hence Width = kσc = 42.3 meters. 
 
 
 
 
 
13.7 
For finding σ2 for this flow – 
Re = (dup/ µ) = (0.225)(1.1)/(0.9x10-6) = 3.2x105 
Now since (D/udt) = 0.22 
From Eqation 8 – (D/uL)=(D/udt)(dt/L) = 0.22(0.255/1000000) = 5.61x10-8 
(σ2/t2) = (σ0)2 = 2(D/uL) = 2(561x10-8) = 11.22x10-8 
Therefore; σ = (11.22x10-8)0.5x(100000/1.1) = 11022x10-8 
So the width at 1σ = (304.5)(1.1) = 335m 
From 5/95 to 95/5 Figure 12 shows that this includes – 
(1.65σx2) of width 
Therefore; the 5/95 to 95/5 % width is – (335m)(1.655x2) = 1105m. 
 
 
 
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Chapter 14: Tanks in Series Model 
14.1 From experiment 
T tmean C 
0-2 1 2 
2-4 3 10 
4-6 5 8 
6-8 7 4 
8-10 9 2 
10-12 11 0 
 
The last observation was interpolated linearly. 
 
taverage = 
Ʃ𝑡𝑡𝑡𝑡
Ʃ𝑡𝑡
 = 1(2)+ 3(10)+ 5(8)+ 7(4)+ 9(2)+ 11(0)
2+10+8+4+2
 = 118
26
 = 4.538 
σ2 = Ʃ𝑡𝑡
2𝑡𝑡
Ʃ𝑡𝑡
− � t(avg)�2 = 12(2)+ 32(10)+ 52(8)+ 72(4)+ 92(2)+ 112(0)
2+10+8+4+2
 - (4.538)2 = 4.4038 
1
𝑁𝑁
 = σ2θ = 
𝜎𝜎2
�𝑡𝑡(𝑎𝑎𝑎𝑎𝑎𝑎)�2 = 4.40384.5382 = 0.2138 
N= 4.68 tanks 
 
14.2 
For N= 10 
CMAX= 100 mmol ; σ2= 1 min. 
a. EθMAX ∝ CMAX 
0
2
4
6
8
10
12
1 3 5
C vs T
C vs T
EθMAX = 
𝑁𝑁 (𝑁𝑁−1)(𝑁𝑁−1) (𝑁𝑁−1)! e-(N-1) 
EθMAX (for 10 tanks) = 1.317 
EθMAX (for 20 tanks) = 1.822 
𝐸𝐸𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) 
𝐸𝐸𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) = 𝑡𝑡𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)𝑡𝑡𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) 
1.317
1.822 = 100𝑥𝑥 
 X = 138.3 mmol 
The maximum concentration of leaving tracer is 138.3 mmol 
b. σ2 ∝ N 
𝜎𝜎2 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡𝑡𝑡) 
𝜎𝜎2 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡𝑡𝑡) = 1020 
1
𝑥𝑥
 = 10
20
 
Therefore x= 2 
The spread is √22 = 1.414 min. 
c. The relative spread increases with increase in number of tanks. (ans) 
 
14.3 
The above problem is similar to the mixed flow reactor 
V= 1.25 × 109 bills; ν = 109 bills/year 
Taverage = 1.25 × 109/ 109 = 1.25 years 
E(t) = 𝐹𝐹𝑓𝑓𝑎𝑎𝐹𝐹𝑡𝑡𝐹𝐹𝑓𝑓𝑡𝑡 𝑓𝑓𝑓𝑓 𝑡𝑡𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑓𝑓 𝑓𝑓𝑓𝑓 𝑛𝑛𝐹𝐹𝑏𝑏𝑏𝑏𝑡𝑡
𝑌𝑌𝑛𝑛𝑎𝑎𝑓𝑓
 = 𝑛𝑛
𝑛𝑛0
 
Where b is number of number of bills and b0 is total number of bills 
a. F(t) = 1- et/t(average) = 1- et/1.25 
b. The number of bills in circulation which are older than 21 years is by the following formula: 
b = ∫ 𝑏𝑏𝑓𝑓 𝐸𝐸(𝑡𝑡)𝑑𝑑𝑡𝑡∞21 where b0= 1.25 × 109 
E(t) = ∫𝐹𝐹(𝑡𝑡)𝑑𝑑𝑡𝑡 = ∫(1 − exp � 𝑡𝑡
𝑡𝑡(𝑎𝑎𝑎𝑎𝑛𝑛𝑓𝑓𝑎𝑎𝑎𝑎𝑛𝑛)�) 𝑑𝑑𝑡𝑡 
Therefore E(t) = 1
𝑡𝑡𝑡𝑡𝑎𝑎𝑎𝑎𝑓𝑓𝑡𝑡𝑎𝑎𝑎𝑎
 𝑒𝑒𝑡𝑡/𝑡𝑡(𝑎𝑎𝑎𝑎𝑛𝑛𝑓𝑓𝑎𝑎𝑎𝑎𝑛𝑛) = 0.8e-0.8t 
Therefore b = ∫ 0.8 exp (−0.8𝑡𝑡)𝑑𝑑𝑡𝑡∞21 = 63.2 bills ≈ 63 bills 
 
14.4 
V=1.25*10^15 bills 
v=10^9 bills 
T=(V/v)= 1.25*10^6 
(a) 
F(t)=𝑒𝑒−( t1.25∗10^6) 
t=0 ; F(t)=0 
t=∞ ; F(t)=1 
(b) 
b=∫ b0E(t)∞10 
Where b0 = 1.25 ∗ 10^15 
E(t)=∫ F(t)dt = ∫ 1 − exp �t
T
�dt 
b=∫ (1.25 ∗ 1015)( 1
1.25∗10^6∞10 exp� −t1.25∗10^6� 
 =1.009*10^9 bills 
 =109 bills 
 
14.5 
 kT=lncA0
cA
= ln 1000
1
= 6.9078 
For the small deviation from plug flow , by the tanks in series model first calculate ỽ2 from the tracer 
curve. 
α=8-12=4 
ỽ2 = α24 = 23 
T=10 
1N = ỽ∅2 = ỽ2T2 = �23�102 = 0.67 ∗ 10−2 
N=150 tanks 
kT=6.9078/150=0.0461 
For tank in series 
cA0
cA
= 1(1+kT)N = 1(1+0.0461)150 
=1/863 cA0cA = 0.00116 cA = 1.16 
 
 
 
 
 
 
14.6 
fully suspended 
solid particle leaves 
 
 v=1 m^3/min 
 
 1m^3 slurry 
 
 
Conversion complete after 2 minutes in the reactor. 
So there is no any dead zone is present because 1 minutes for each reactor. 
 
First approximate each pulse by plug flow pattern. 
𝐴𝐴2
𝐴𝐴1
=1
3
= 𝑅𝑅
𝑅𝑅+1
 
R=0.5 
𝑉𝑉𝑉𝑉1(𝑅𝑅+1)𝑎𝑎 = 1; 
Putting value R; 
𝑉𝑉𝑉𝑉1(0.5+1)1 = 1; 
𝑉𝑉𝑉𝑉1 = 3
2
; 
𝑉𝑉𝑉𝑉1(𝑅𝑅+1)𝑎𝑎 + 𝑉𝑉𝑉𝑉2𝑅𝑅𝑎𝑎 = 3; 
Putting values and find Vp2; 
3/2(0.5+1)1 + 𝑉𝑉𝑉𝑉2(0.5)1 = 3; 
Vp2=1; 
Vactual =(3/2)+1=2.5; 
Vdead= 5-2.5= 2.5; 
Now considering that the pulse output has width, 
∆𝜃𝜃
𝜃𝜃𝑛𝑛𝑎𝑎𝑥𝑥
= 2
√(𝑁𝑁1−1); 
2/3
1
= 2
√(𝑁𝑁1−1); 
N1=10 tanks; 
N2=(2/3)*N1; 
N2=6.67 tanks; 
Tavg=∑𝑡𝑡𝐹𝐹
∑ 𝐹𝐹
; 
Tavg=2.49=2.5; 
𝑉𝑉1+𝑉𝑉2
𝑎𝑎
= 1.5 + 1.0 = 2.5; 
14.8 
 Find out number of tanks. 
 Tavg = ∑𝑡𝑡𝐹𝐹𝑡𝑡𝑡𝑡
∑𝐹𝐹𝑡𝑡𝑡𝑡
; 
 =320
40
; 
 =8 min 
𝜎𝜎2 = ∑ 𝑡𝑡𝑡𝑡
∑ 𝑡𝑡
− �
∑ 𝑡𝑡𝑡𝑡𝑑𝑑𝑡𝑡
∑ 𝑡𝑡𝑑𝑑𝑡𝑡
�
2 ; 
 =�250+490+810+1210
40
� − �
320
40
�
2 ; 
 =5; 1
𝑁𝑁
= 𝜎𝜎2(𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)2 ; 
N=64
5
; 
N=12.8 
N=13. 
14.9 For N tanks in series. 
Finding out number of tanks 
N=1+4(𝜃𝜃𝑛𝑛𝑎𝑎𝑥𝑥
∆𝜃𝜃
)^2; 
N1peak=104 tanks; 
N2peak=101 tanks; 
N3peak=101 tanks; 
N4peak=97 tanks; 
So that average is 100 tanks. 
Model 
 
14.10 
(a) Area under curve c vs t curve 
� cidti∞
0
= � ci∆ti 
 =3000 
Mean resident time =∑citi∆ti
∑ ci∆ti
= 34.66 
(b) E curve 
C Time E=C/q 
 
CA/CAo 
 
35 0 0 0 
38 10 0.00333 0.0202 
40 20 0.00666 0.0245 
40 30 0.0133 0.0223 
39 40 0.0166 0.0183 
38 50 0.022 0.0136 
35 60 0.0233 0.0099 
 
 
 
 
 
 
(c) Variance=∑citi
2∆ti
∑ ci∆ti
− Ti2 
 
 =22.45 
(d) How many tanks in series in this vessel equipment to? 
 1N = ỽ∅2 = ỽ2T2 = (22.45)^234.662 
 
N=2.38 tanks 
 
 
 
Example:-14.11 
A reactor has flow characteristics given by the nonnormalized C curve in Table P14.11, and by the 
shape of this curve we feel that the dispersion or tanks-in-series models should satisfactorily 
represent flow in the reactor. 
(a) Find the conversion expected in this reactor, assuming that the dispersion model holds. 
(b) Find the number of tanks in series which will represent the reactor and the conversion expected, 
assuming that the tanks-in-series model holds. 
Time Concentration 
1 0 
2 9 
3 57 
4 81 
5 90 
6 90 
7 86 
8 77 
10 67 
15 47 
20 32 
30 15 
41 7 
52 3 
70 0 
 
(c) Find the conversion by direct use of the tracer curve. 
(d) Comment on the difference in these results, and state which one you think is the most reliable. 
 
Solution:- 
To find out non ideal characteristics of reactor determine proper D/uL 
To use for dispersion model, or the proper N value to use for tank in series model. 
 
This is done by two way :- matching tracer concentration curve with dispersion model or with tank in 
series model , or by calculating ỽ^2 fromthat D/uL or N. 
Lets us use the latter procedure. So first calculate T and ỽ^2 from the table of data with the help of 
given equ. 𝑇𝑇 = ∫ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡∞0
∫ 𝑡𝑡𝑡𝑡𝑡𝑡
∞
0
= ∑𝑡𝑡𝐹𝐹𝑡𝑡𝐹𝐹∆𝑡𝑡𝐹𝐹
∑𝑡𝑡𝐹𝐹∆𝑡𝑡𝐹𝐹
 
ỽ2 = ∫ (𝑡𝑡 − 𝑇𝑇)2𝑡𝑡𝑡𝑡𝑡𝑡∞0
∫ 𝐶𝐶𝑑𝑑𝑡𝑡
∞
0
= ∫ 𝑡𝑡2𝐶𝐶𝑑𝑑𝑡𝑡∞0
∫ 𝐶𝐶𝑑𝑑𝑡𝑡
∞
0
− 𝑇𝑇2 
ƩC=213 T=2149/213=10.09min 
Ʃtc=2149 ỽ2 = �37685
213
� − (10.09)2 = 75.11 
Ʃt^2c=37685 ỽ∅2 = ỽ2𝑇𝑇2 = 75.116(10.09)^2 = 0.7378 
Next determine the behavior of ideal PFR 
k=o.456min−1 kT=4.6 
so for PFR x𝑎𝑎 = 1 − 𝑒𝑒kT = 1 − e4.6 = 0.99 
(a) Use the dispersion model 
 
Here related ỽ∅2 with D/uL so, 
ỽ∅
2 = 0.7378 = 2 � DuL� − 2( DuL)2[1 − 𝑒𝑒−uLD ] 
Solve by trial and error.this gives D/uL=1 
 
 
 
v/vp constant kt=4.6 and D/uL=1 
 
 
 xa=0.89 
 
From fig x(disp)=0.89 
 
(b) Use tank in series model 
 
N= 1
ỽ∅
2= 10.7385 = 1.35 Tanks 
 
From fig 6.5we can find the X(tank)=83% 
 
(c) Use the tracer data directly 
 
From c
c0
= ∑( c
c0
)E∆t 
To find out E curve make the area under C curve unity, 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
E= c
area
= c
∑ci∆t
 
 Now we know that 
c
c0
= ∑� c
c0
�E∆t 
∑ c =213 
∑ e =0.1911 
 
X=0.88(from curve) 
 
(d) Which answer is most reliable 
Naturally the direct use of tracer gives most reliable answer. In this given model RTD 
come from dispersion model D/uL=1. Thus wind accepted that ans to part (a) and (c) 
should agree. 
 
 
 
 
 
 
 
 
 
 
 
 
Time Concentration 
𝑒𝑒−0.465(citi213) 
1 0 0 
2 9 0.0506 
3 57 0.0456 
4 81 0.0158 
5 90 0.0050 
6 90 0.0015 
7 86 0.0005 
8 77 0.0002 
10 67 0.0001 
15 47 16*10^-6 
20 32 5*10^-6 
30 15 1*10^-6 
41 7 0 
52 3 0 
70 0 0 
 
 
 
 
 
 
 
 
Solid Catalysed Reactions 
 
Solution 18.1 
We cannot tell just by looking at the reactor and the catalyst bed height whether the reactor is integral or 
differential. It isn’t the depth or the packing or the conversion level that counts but the assumptions used when 
the data from the reactor are analysed. If we assume it as a constant rate of reaction throughout the reactor, then 
it is a differential reactor but if we assume a changing rate of reaction with changing position, then it becomes a 
integral reactor. It is just the matter of how we are looking at the rate. 
 
Solution 18.2 
Since the volume is made 3 times the original volume but the temperature, feed flow rate, the amount of the 
catalyst remains the same, according to the performance equation of the mixed flow reactor, 
𝑊
𝐹𝐴0
= 
𝑋𝐴
−𝑟𝐴
 
Thus there will be no change in the conversion. 
 
Solution 18.3 
a) We have been given data for the weight of the catalyst, molar flow rate and the conversion. Since it is a 
packed bed reactor, the performance equation is as follows. 
W
FAO 
= ∫
dXA
−rA
XA
0
 
Writing it in the differential form, 
𝑑𝑊
𝐹𝐴0
= 
𝑑𝑋𝐴
−𝑟𝐴
 
Thus making the subject as –rA, 
−𝑟𝐴 = 
𝑑𝑋𝐴
𝑑 (
𝑊
𝐹𝐴
)
 
 
Thus we see that the slop of the cure at the xA= 0.4 gives us the rate of the equation. 
Thus taking the slope of the curve at the point Xa= 0.4 
−𝑟𝐴 = 
0.482 − 0.23
0.8 − 0.1
= 0.36 
𝑘𝑔𝑚𝑜𝑙𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑
𝑘𝑔 𝑐𝑎𝑡 . ℎ𝑟
 
 
 
 
 
 
 
 
 
 
 
 
b) For finding the amount of catalyst required if the molar flow is 400 kmol/hr and the conversion required is 
Xa=0.4, 
From the curve we can seethat for Xa= 0.4, the ration of W/FAO is 0.575. 
Therefore, W= 0.575 X 400 = 230 kg Cat. 
c) For completely mixed flow reactor, 
W
FAO 
= 
XA
−rA at 0.4
 
W
FAO 
= 
XA
−rA at 0.4
= 
0.4
0.36
 
 
W = 400 X (0.4/0.36) = 444 kg catalyst 
Solution 18.4 
A  R n=1/2 th order reaction with infinite (very large) recycle hence R = ∞ 
 
W = 3 kg of catalyst 
CA0 =2 mol/m
3 , CA = CAout = 0.5 mol/m
3 
V = 1 m3/hr 
A = 1-1/1 = 0 
 
XA = (CA0 − CA)/(CA0 + A ∗ CA) 
XA = 
2−0.5
2+00.5
 
XA=0.75 
 
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
X
A
(W/FAO)
FA0 = molar flow rate of gas A 
FA0 = 2 mol/m
3  1 m3/hr 
 = 2 mol/hr 
 
 
W
FA0
 = 
XAout
−rAout
 (for MFR) ______________ (1) 
Here, -rAout = k × CAout1/2 _________________(2) 
 
 k = 
FA0∗XAout
W∗CAout
 _________________________(3) 
 = 
20.75
30.5
 
 = 1 m3/kg of catalyst hr 
Put the value of k in equation (2) to get value of -rAout 
 
-rAout = 1 m3/kg of catalyst hr  0.51/2 mol/m3 
-rAout = 0.707 mol / kg of catalyst hr 
 
Hence, -rA (mol / kg of catalyst hr ) = (1 m3/kg of catalyst hr) (CA mol / m3)1/2 
 
SOLUTION 18.5 
Very large recycle, A→3R, n=1and feed contains 50% A and 50% inerts. 
Here Ɛ=1, FA0=2 mol/hr, W=3 kg, CAout = 0.5 mol/m3, CAo=2 mol/m3 
With large recycle,use the performance equation of a mixed flow reactor, 
𝑊
𝐹𝐴0
 = 
𝑋𝐴𝑜𝑢𝑡
−𝑟𝐴𝑜𝑢𝑡
 , -rAout=k’CAout … first order decomposition of A 
∴
𝑊
𝐹𝐴0
 = (
1−
𝐶𝐴𝑜𝑢𝑡
𝐶𝐴𝑜
1+
𝐶𝐴𝑜𝑢𝑡
𝐶𝐴𝑜
) × 
1
𝑘′𝐶𝐴𝑜𝑢𝑡
 
∴ k’= 0.8 m3/(h∙ kg ∙cat) 
Therefore, -rA’ = -
1
𝑊
 
𝑑𝑁𝐴
𝑑𝑡
 = 0.8 m3/(h∙ kg ∙cat) ∙ CA mol/m3 
SOLUTION 18.6 
No recycle, A→3R, n=2 and feed contains 25% A and 75% inerts, 4 volumes 
ƐA=(6-4)/4 =0.5 
CA = CAo(
1−𝑋𝐴
1+0.5XA
) 
With No recycle, it becomes a plug flow reactor. 
So, 
𝑊
𝐹𝐴0
 =∫
𝑑𝑋𝐴
−𝑟𝐴′
𝑋𝐴
0
 
-rA’= k’CA2 
𝑊
𝐹𝐴0
 =∫
𝑑𝑋𝐴
(
1−𝑋𝐴
1+0.5XA
)
2
𝑋𝐴
0
 
= ∫
(1+0.5𝑋𝐴)2𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 
=∫
(1+𝑋𝐴+0.25𝑋𝐴2)𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 
=∫
𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 + ∫
(𝑋𝐴)𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 + 0.25 ∫
𝑋𝐴2 𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 
I = ∫
𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 = 
𝑋𝐴
(1−𝑋𝐴)
 
II = ∫
(𝑋𝐴)𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 = 
𝑋𝐴
(1−𝑋𝐴)
 + ln(1-XA) 
III = 0.25 ∫
𝑋𝐴2 𝑑𝑋𝐴
(1−𝑋𝐴)2
𝑋𝐴
0
 = 0.25 
𝑋𝐴
(1−𝑋𝐴)
 + 0.50 ln (1-XA) + 0.25XA 
I+II+III = 2.25 
𝑋𝐴
(1−𝑋𝐴)
 + 1.50 ln (1-XA) + 0.25XA 
𝑊
𝐹𝐴0
= 
1
𝑘′𝐶𝐴𝑜
 [2.25 
𝑋𝐴
(1−𝑋𝐴)
 + 1.50 ln (1 − XA) + 0.25XA] 
XA = 0.67 
k’= [2.25 
𝑋𝐴
(1−𝑋𝐴)
 + 1.50 ln (1 − XA) + 0.25XA] ×
𝐹𝐴0
𝐶𝐴02×𝑊
 
k’= 0.512 m6/(mol.h.kg cat) 
-rA’ = -
1
𝑊
 
𝑑𝑁𝐴
𝑑𝑡
 = 0.512 m6/(mol.h∙ kg ∙cat) ∙ (CA mol/m3)^2 
 
Solution 18.7 
Mixed Flow Behaviour 
A → R 
∈ a =
1−1
1
 = 0 
CAO =10 
mol
m3
 ; W=4 gm 
Performance Equation 
W
FAO 
= 
XA
−rA
 
For No. 1: 
vo = 3 
m3
h
 , XA = 0.2 
FAO = Vo * CAO 
 = 3 * 10 
 = 30 
mol
h
 
-rA = XA * 
FAO
W
 
 = 0.2 * 30/4 
 = 1.5 
mol
h g cat
 
CA = CAO * (1-XA) since ∈ a = 0 
 = 10 (0.8) 
CA = 8 
mol
m3
 
 
For No. 2: 
Vo = 2 
m3
h
 XA = 0.3 
FAO = Vo * CAO 
 = 2 * 10 
 = 20 
mol
h
 
-rA = XA * 
FAO
W
 
 = 0.3 * 20/4 
 = 1.5 
mol
h g cat
 
CA = CAO * (1-XA) since ∈ a = 0 
 = 10 (0.7) 
CA = 7 
mol
m3
 
 
 
 
 
vo 3 2 1.2 
FAO 30 20 12 
XA 0.2 0.3 0.5 
CA 8 7 5 
-rA 1.5 1.5 1.5 
So the reaction is independent of CA (0 
th Order) 
-rA = K CA 
n 
-rA = 1.5 
mol
h g cat
 
Solution 18.8 
 Plug Flow 
CAO = 60 
mol
m3
 vO = 3 
lt
min
 = 3 * 10-3 
m3
min
 
FAO = vO * CAO 
 = 60 * 3 * 10
-3 
 = 0.18 
mol
min
 
W
FAO 
= ∫
dXA
−rA
XA
0
 
A → R , ∈ a =
1−1
1
 = 0 
CA = CAO * (1-XA) 
XA = 1- 
CA
CAO
 
XA1 = 1- 
30
60
 = 0.5 
W1 
FAO 
 = 
0.5
0.18
 = 2.78 
XA2 = 1- 
20
60
 = 0.67 
W2 
FAO 
 = 
1
0.18
 = 5.56 
XA3 = 1- 
10
60
 = 0.833 
W3 
FAO 
 = 
2.5
0.18
 = 13.89 
Guessing the rate as 
First order 
-rA = k * CA
 
 = k * CAO * (1-XA) 
W
FAO 
= ∫
dXA
−rA
XA
0
 
 = ∫
dXA
k ∗ CAO ∗ (1 − XA)
XA
0
 
 
 = 
− ln(1− XA)
k∗ CAO
 
W
FAO 
= 
1
𝑘∗𝐶𝐴𝑂 
 * ln ( 
1
1−𝑋𝐴
 ) 
y = mx 
 1 2 3 
W
FAO 
 
2.78 5.56 13.89 
ln ( 
1
1−𝑋𝐴
 ) 0.693 1.098 1.79 
 
 
 
W
FAO 
 vs ln ( 
1
1−𝑋𝐴
 ) is not coming straight line. So our assumption of First order is wrong. 
 
Guessing Second order: 
-rA = k * CA
2 
 = k * CAO
2
 * (1-XA)
2 
W
FAO 
= ∫
dXA
−rA
XA
0
 
 = ∫
dXA
k ∗ 𝐶𝐴𝑂
2 ∗ (1 − 𝑋𝐴)2
XA
0
 
W
FAO 
= 
1
𝑘∗ 𝐶𝐴𝑂
2 * 
𝑋𝐴
1− 𝑋𝐴
 
 1 2 3 
W
FAO 
 
2.78 5.56 13.89 
𝑋𝐴
1 − 𝑋𝐴
 
1 2.03 4.998 
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 2 4 6 8 10 12 14 16
ln
 (
 1
/(
1
-X
A
 )
 )
W/(FAO)
First order
 
 
W
FAO 
 vs 
𝑋𝐴
1−𝑋𝐴
 is coming straight line. So our assumption of Second order is correct. 
Slope = 
1
𝑘∗ 𝐶𝐴𝑂
2 = 2.773 
 
1
𝑘∗ 602
 = 2.773 
 k = 10-4 
m6
mol min g cat
 
-rA = 10
-4 * CA
2 
Solution 18.9 
We have reaction A→R, CAO= 10mol/m3 
 
 
Ca XA FA0 1/-rA 
8 0.2 0.14 3.571429 
6 0.4 0.42 0.595238 
4 0.6 1.67 0.0998 
2 0.8 2.5 0.05 
1 0.9 1.25 0.088889 
 
0
1
2
3
4
5
6
0 2 4 6 8 10 12 14 16
X
A
/(
1
-X
A
 )
W/(FAO)
Second order
 
For packed bed reactor with CAO= 8 mol/m3 and XA=0.75, CA= 2 mol/m
3. 
Also the area under the curve for 1/-rA vs CA from CA0= 8 mol/m
3 to CA= 2 mol/m
3 is 5.08. 
According to the performance equation of the Packed bed reactor, 
𝑊
𝐹𝐴𝑂
= ∫
𝑑 𝑋𝐴
−𝑟𝐴
 
We get, W= 1000 X 5.08/8 = 615 kg of catalyst is required. 
 
Solution 18.10 
For Mixed Flow reactor, the concentration inside the reactor will be same as that of the leaving stream. Given 
that the initial reactant concentration is 10 mol/m3 and the conversion is 90 %, the outlet concentration of the 
stream will be 1 mol/m3. Thus from the above tabulated data, the rate constant can be found out. 
𝑘 =
−𝑟𝐴
𝐶𝐴
 
 𝑇ℎ𝑢𝑠 𝑘 =
11.25
1
= 11.25 m3/mol. Min 
Thus the amount of the catalyst required is given by 
𝑊 = 
𝐹𝐴𝑂
𝑘
𝑋
𝑋𝐴
𝐶𝐴
 
𝑊 = 1000 𝑋 0.9/(11.25 𝑋 1) 
𝑊 = 80 𝑘𝑔 
0
0.5
1
1.5
2
2.5
3
3.5
4
0 1 2 3 4 5 6 7 8 9
1
/-
rA
Ca
 
 
 
Term Paper 
Of Cre-II, 
Chapter No:18 
Solid Catalysed Reactions 
Sum No: 18.11-18.20 
 Submitted By: 
 
 Samarpita Chakraborty 
(13BCH013) 
 Harsh Dodia 
(13BCH014)Nirav Donga 
(13BCH015) 
Harshit Goyal 
(13BCH018) 
 
 
 
 
18.11) 
 
 XB=0.8 
 
-rA=
50𝐶𝐴
1+.02𝐶𝐴
 Mol/kg.hr 
For Plug Flow 
T’= 
𝑊
𝑣
= ∫ 𝑑𝑋𝐴/−𝑟′𝐴
𝑋𝐴𝑓
0
=1/k’ ∫
1+.02𝐶𝑎
50𝐶𝑎
𝑑𝐶𝑎
100
20
 
K’T’= [ln (Cao/Ca) + k(Cao-Ca)] 
 
1000
50
[ln 
100
20
 + .02 (100-20)] 
 𝑊 = 64.19 𝐾𝑔𝑐𝑎𝑙 
18.12) 
 
 
 XB=0.8 
 
-rA= 8 CA
2
 Mol/kg.hr 
For Plug Flow 
T’= 
𝑊
𝑣
= ∫ 𝑑𝑋𝐴/−𝑟′𝑎
𝑋𝐴𝑓
0
=1/k’ ∫ d𝐶𝑎
100
20
/−𝑟′𝑎 
T’= 1/8 
∫ d𝐶𝑎
100
20
8𝐶𝑎2
 
V=1000 m3/hr 
Cao=100 mol/m
3 
V=1000m3/hr 
CS.=100mol/m
3 
 
 
 
 -1/8 [1/Ca0-1/Ca] 
 -1/8 [1/100 -1/20] 
 T’= W/V= 0.005 
 𝑊 = 0.005 𝑥 1000 = 5 𝑘𝑔 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 
 
18.13) 
 
 
 
 Xa= ? 
-ra’=.6 CACB 
For CAo=0.01 CBo 
We can assume that CB= Constant 
 -rA’ = CACB = .6 CA. 10 = 6 CA 
 T’= 
𝑤
𝑣
 = 
4
10
= .4 kg.hr/m
3
 
 For First order 
 
CA
𝐶𝐴𝑜
 = 𝑒−𝑘′𝑇′= 𝑒−6 .4= .0907 
 XA=90.9% 
 XB= .909% 
 
18.14) 
 
 
 
 
v=10 m
3
/hr 
Cao=0.1 mol/m
3
 and Cbo= 10 mol/m
3
 
W= 4 kg 
Mol.wt of feed= 0.255 kg/kmol 
Mol wt. of product= 0.070 kg/kmol 
Feed enters at 630 ⁰C and 1 atm (vaporized) 
Density of liquid feed =869 kg/m
3
 
Denisty of catalyst particles = 950 kg/m
3
 
Half the feed is cracked at rate 60 m
3
 liq/ m
3 
reactor. Hr 
 
 
 
 
 
 
 
For Plug flow 
𝜏/Cao = V/W∫
𝑑𝑋𝑎
−𝑟𝑎′
 
Where V= volume of catalyst and W= weight of catalyst 
We know, Cao = Pao/RT 
 = 1 atm/0.082 x (630 + 273) 
 = 0.01350 mol/litre 
 = 13.50 mol / m
3
 
For Plug flow and 1
st
 order kinetics, 
k’’’ 𝜏’’’= ln[Cao/Ca] 
k’’’ (1/60)= ln (13.5/7.75) 
k’’’ = 41.5888 hr-1 
We know, 
k’’’ 𝜏’’’= k’ 𝜏’ [𝜏’ is 𝜏′′′x (Density of the catalyst particles-bulk Density of packed bed) - as in 
solid catalyzed reactions volume of catalysts can be taken as the void volume of reactor.] 
(41.588)(1/60) =k’ ((950-700)/60) 
k’= 0.166 m3 liquid/kg catalyst. Hr 
18.15) 
 8atm 700⁰C 
A 3R 
ε=2 
 
 
 
 
𝑊
𝐹𝑎𝑜
=
𝑋𝑎
−𝑟𝑎′
 
𝐶𝑎
𝐶𝑎𝑜
=
1−𝑋𝑎
1+2𝑋𝑎
 
 
Ca= PA/RT= 
8
.08206∗973
=0.1mol/lit 
 
Ca Ca/Cao Xa=(1-
Ca/Cao)/(1+2Ca/Cao) 
Fao=vCao -ra 
=Xa 
Fao/W 
0.08 0.8 0.076923077 10 0.769 
0.05 0.5 0.25 2.2 0.55 
0.02 0.2 0.571428571 0.4 0.228 
0.01 0.1 0.75 0.1 0.075 
0.005 0.005 0.863636364 0.05 0.0482 
 
 
-ra vs Ca 
 
The Trendline of the curve gives us the relation: 
-rA’= 9.9132CA 
y = 9.9132x 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 0.02 0.04 0.06 0.08 0.1
 
 
 
 
18.16) 
 
For 1
st
 order solid catalyzed reaction, without any complex piping conversion is XA = 0.8 and 
thus 
 k’ 𝜏’ =ln [CAO/CA] 
 k’ 𝜏’ = -ln [1-Xa] 
 k’ 𝜏’ = -ln[1-0.8] 
 k’ 𝜏’= 1.6094 (Damkohler number for 1st order reactions) 
at points 2, 3, 4 the conversion is same at steady state. 
𝜏’/CA0 =∫
𝑑𝑋𝑎
−𝑟𝑎
 = ∫
𝑑𝑋𝑎
𝑘′ 𝐶𝑎
 
 k’ 𝜏′/CA0 = ∫
𝑑𝑋𝑎
𝐶𝑎𝑜(1−𝑋𝑎)
𝑋𝑎1
𝑋𝑎2
 
 k’ 𝜏′= - ln [(1-Xa1)/(1-Xa2)] 
 1.6094 = - ln [(1-Xa1)/(1-Xa2)]……………………………………………… 1 
 
Taking mole balance, 
CA0v0 + CA1v1 = CA3v3 
And CA2= CA3= CA4 
 CA0v0 + CA1v1 = CA3v3 
 
 
 
 (100)(100) + CA1(100) = CA3(200)………………………………………………2 
Solving equation 1 and 2, 
1.6094 = - ln [(1-Xa1)/(1-Xa2)] 
 𝑒1.6094 =[(1-Xa2)/(1-Xa1)] 
 5(1 − 𝑋𝐴1)= 1 − XA2 
 5CA1/CA0 = CA2/CA0 
 CA2/CA1 = CA3/CA1 = CA4/CA1 = 5 
Therefore solving equation 2 we get, 
CA4 = 11.11 x 5 = 55.55 
Thus, XA4 =1-(55.55/100) = 0.444 
 
18.17) (a) 
 
Cao=2mol/lit Caout=0.5mol/lit 
V0=1lit/hr A R R=∞ 
 Second order 
 - ra´=k´Ca
2 
T’= 
𝑊𝐶𝑎𝑜
𝐹𝑎𝑜
=
𝑊
𝑣0
= 
𝐶𝑎0−𝐶𝑎
− ra´
 
k´=
𝑉0
𝑊
 .
𝐶𝑎0−𝐶𝑎
𝐶𝑎∗𝐶𝑎
 
 =
1
3
 . 
2−0.5
0.5∗0.5
= 2lit2/mol.gm.hr 
 
(b) For packed bed (assume plug flow) 
 
V=1lit/hr 
W=3gm 
 
 
 
 
 
 
 
 
 Xa= 0.8 
For plug flow 
T’=
𝑊
𝑣0
= ∫
𝑑𝐶𝑎
𝑘𝐶𝑎.𝐶𝑎
𝐶𝑎𝑜
𝐶𝑎
=
1
𝑘
. [ 1
𝐶𝑎
−
1
𝐶𝑎𝑜
] 
W=
1000
2
. [ 1
0.2
−
1
1
]=2000gm=2kg 
 
(c) Add inert solid 
This does not change the rate equation or performance equation 
Therefore, adding inert make the reactor bigger but does not change the weight of the catalyst 
needed for the process 
18.18) 
 A  R in a packed bed reactor. 
 Cao 
mol/m3 
Ca 
mol/m3 
Vo 
(lit/hr) Fao Xa (-)ra' 1/(-)ra' 
10 1 5 0.05 0.9 0.045 22.22222 
W kg cat 2 20 0.2 0.8 0.16 6.25 
1 3 65 0.65 0.7 0.455 2.197802 
 
6 133 1.33 0.4 0.532 1.879699 
 
9 540 5.4 0.1 0.54 1.851852 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
V0=1000lit/hr 
Ca0=1mol/lit 
W= ? 
0
5
10
15
20
25
0 0.2 0.4 0.6 0.8 1
1
/-
ra
' 
Xa 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
∫
𝑑𝐶𝑎
−𝑟𝑎
=14 
( Area under the curve for PFR i.e with no recycle of exit fluid for 
 Xa= 0.75) 
 
 
(Ca0-Ca)/-ra = 36 
(Area under the curve for CSTR i.e with very high recycle and for 
 Xa= 0.75) 
 
Hence, 
Part 1: For PFR, 
W = FAO/CAO ∫
𝑑𝐶𝑎
−𝑟𝑎
= (1000 𝑥 14)/8 = 1750 𝑘𝑔 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 
Part 2 : For CSTR, 
W= FAO/CAO x (CA0-CA)/-ra 
 = (1000/8) x 36 
 = 4500 kg catalyst 
 
0
5
10
15
20
25
0 2 4 6 8 10
1
/-
ra
' 
Ca 
 
 
 
18.19) 
 1atm 609K 
2A R 
V=100 cm
3 
 
 
This batch circulation system is a constant volume system, so έa=0 
Ca= PA0/RT= 
101325
8.314∗609
=20mol/m
3 
PA= PA0 + 
𝑎
∆𝑛
(𝜋𝑜 − 𝜋) 
 
=1+
2
−1
(1 − 𝜋) 
 
=2 𝜋 − 1……………………(1) 
 
t 𝜋 PA Ca lnCa 1/Ca 
0 1 1 20 3 0.05 
4 0.75 0.5 10 2.3 0.1 
8 o.67 0.33 6.67 1.9 0.135 
6 0.60 0,2 4 1.38 0.25 
36 0.55 0.1 2 0.693 0.5 
 
 
 
 
 
 
 
 
 
 
 
Slope of 1/Ca vs t is 0.0125. 
Therefore rate equation is ra=0.0125Ca
2 
 
 
 
 
 
18.20) 
 Mw= µT
2∈= L2(-rA
’’’
/CA)obs/De 
Mw= .15 
-rA
’’’
/CA= .88 
De= 2*10
-6 
.15= L
2
*.88/2*10
-6 
L= 5.83* 10
-4
 m 
 = 58.3mm 
L= R/3 (Sphere) 
L= D/6 
D= 350 mm 
18.21. A reaction A→R is to take place on a porous catalyst pellet (d = 6mm, de = 10m3/m cat. 
s). How much is the rate slowed by pore diffusional resistance if the concentration of reactant 
bathing the particle is100 mol/m3 and the diffusion-free kinetics are given by 
 
 -ra = 0.1ca2 𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚3.𝑐𝑐𝑐𝑐𝑚𝑚.𝑠𝑠 
 
ans. 
 
For nth order reaction MT = 
𝑑𝑑𝑝𝑝6
 �(𝑛𝑛+1)∗𝐾𝐾′′𝐶𝐶𝑎𝑎𝑛𝑛−1
2∗𝐷𝐷𝑒𝑒
 
 
 
 = 
6∗10−3
6
 �(2+1)∗0.1∗1002−1
2∗10−6
 
 = 3.873 
 
 From the Graph value of E = 0.23 
 
 
18.22 In the absence of pore diffusion resistance a particular first-order gas- phase reaction 
proceeds as reported below. 
 
What size of spherical catalyst pellets ( g e 10^-3= cm3/cm cat as) would ensure that pore 
resistance effects do not intrude to slow the rate of re- action? 
Given P= 1atm 
CAo =P/RT 
CAo = 1/(0.08206*673) 
CAo = 1.8*10^-5 mol/cm^3 
XAe = CAo- CA/ CAo 
XAe = (1.8-1)10^-5/1.8*10^-5 
XAe = 0.444 Mt = L(K’’’/De*XAe)^(1/2) MT = 0.4 By Solving above equation We get L=0.4cm For Spherical particle R =3L R= 0.0795 cm R = 0.8mm 
 
18.23. The first-order decomposition of A is run in an experimental mixed flow reactor. Find the 
role played by pore diffusion in these runs in effect determine whether the runs were made under 
diffusion-free, strong resistance or intermediate conditions. 
 
dp w Cao v xa 
3 1 100 9 0.4 
12 4 300 8 0.6 
 
A → 𝑅𝑅 
ans. 
 
No dp w CAO v xa CA K’ 
1 3 1 100 9 0.4 60 6 
2 12 4 300 8 0.6 120 3 
 
For Mixed Flow 
(1) (2) 
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 
 = 100(1 – 0.4) = 300(1 – 0.6) 
 = 60 = 120 
For No diffusion 
𝐾𝐾2
𝐾𝐾1
 =1 
For Strong resistance 
 𝐾𝐾2
𝐾𝐾1
= 𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
 = 1
4
 
From the given data : 
𝜏𝜏′= 𝑤𝑤
𝑣𝑣
 = 𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊′𝐶𝐶𝐴𝐴
 
𝐾𝐾1
′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)
𝑊𝑊′∗𝐶𝐶𝐴𝐴
 𝐾𝐾2′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)𝑊𝑊′∗𝐶𝐶𝐴𝐴 
 = (100−60)∗9
1∗60
 = (300−120)∗8
4∗120
 
 = 6 = 3 
 
𝐾𝐾2
𝐾𝐾1
= 36 = 12 
 
So that, diffusion take place at intermediate condition 
 
18.24 The first-order decomposition of A is run in an experimental mixed flow reactor. Find the 
role played by pore diffusion in these runs in effect determine whether the runs were made under 
diffusion-free, strong resistance or intermediate conditions. 
 
dp w Cao v xa 
4 1 300 60 0.8 
8 3 100 160 0.6 
A → 𝑅𝑅 
 
ans 
No dp w CA0 V xa CA K’ 
1 4 1 300 60 0.8 60 240 
2 8 3 100 160 0.6 40 80 
 
For Mixed Flow 
(1) (2) 
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 
 = 300(1 – 0.8) = 100(1 – 0.6) 
 = 60 = 40 
For No diffusion 
𝐾𝐾2
𝐾𝐾1
 =1 
For Strong diffusion 
 𝐾𝐾2
𝐾𝐾1
= 𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
 = 1
2
 
From the given data : 
𝜏𝜏′= 𝑤𝑤
𝑣𝑣
 = 𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊′𝐶𝐶𝐴𝐴
 
𝐾𝐾1
′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)
𝑊𝑊′∗𝐶𝐶𝐴𝐴
 𝐾𝐾2′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)𝑊𝑊′∗𝐶𝐶𝐴𝐴 
 = (300−60)∗60
1∗60
 = 
(100−40)∗160
3∗40
 
 = 240 = 80 
 
𝐾𝐾2
𝐾𝐾1
= 80240 = 13 
So that, diffusion take place at strong resistance condition 
 
18.25 The first-order decomposition of A is run in an experimental mixed flow reactor. Find the 
role played by pore diffusion in these runs in effect determine whether the runs were made under 
diffusion-free, strong resistance or intermediate conditions. 
 
 
dp w Cao v xa 
2 4 75 10 0.2 
1 6 100 5 0.6 
 
A → 𝑅𝑅 
 
ans 
No dp w CA0 V xa CA K’ 
1 2 4 75 10 0.2 60 
2 1 6 100 5 0.6 40 
 
For Mixed Flow 
(1) (2) 
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0(1 - 𝑋𝑋𝐴𝐴) 
 = 75(1 – 0.2) = 100(1 – 0.6) 
 = 60 = 40 
For No diffusion 
𝐾𝐾2
𝐾𝐾1
 =1 
For Strong diffusion 
 𝐾𝐾2
𝐾𝐾1
= 𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
 = 1
2
 
From the given data : 
𝜏𝜏′= 𝑤𝑤
𝑣𝑣
 = 𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊′𝐶𝐶𝐴𝐴
 
𝐾𝐾1
′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)
𝑊𝑊′∗𝐶𝐶𝐴𝐴
 𝐾𝐾2′ = 𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴)𝑊𝑊′∗𝐶𝐶𝐴𝐴 
 = (75−60)∗10
4∗60
 = 
(100−40)∗5
6∗40
 
 = 0.625 = 1.25 
 
 
 
𝐾𝐾2
𝐾𝐾1
= 1.250.625 = 2 
So that, diffusion take place at strong pore diffusion 
 
Solution 18.26 
 
dp CA -rA’ T,K 
1 20 1 480 
2 40 1 480 
2 40 3 500 
 
A→ R; CA0 = 50 
Now, dp2 / dp1 = k1/k2 = k0e-E/RT1/ k0e-E/RT2 = 2 
Similarly, 
dp3 / dp2 = k2/k3 = k0e-E/RT2/ k0e-E/RT3 = 1 
dp3 / dp1 = k1 /k3 = e-E/RT1/ e-E/RT3 = 2 
On solving we get, 
E = -69139.224 J = -69.13 kJ 
 
Solution 18.27 
 
Quantity of 
Catalyst 
Pellet Diameter Flow rate of 
given feed 
Recycle Rate Measured 
reaction rate 
1 1 1 High 4 
4 1 4 Higher still 4 
1 1 1 Higher still 3 
4 2 4 High 3 
 
All these runs were made at the same W/FAo. We are also told that the recycle rate is high 
enough to have mixed flow throughout. Thus changing the recycle rate from run 1 to run 2, or 
from run 3 to run 4 with no change in rate shows that film resistance doesn’t influence the rate. 
Next, if particles were non-porous then the smaller size (runs 1 & 2) should have double the rate 
of the larger size (runs 3 & 4). This is not what is measured hence the results are not consistent 
with the guess of non-porous particles. (1) If in the regime of no pore diffusion resistance the 
rates should be same for large and small pellets. (2) If in the regime of strong pore diffusion the 
rate should halve from going 4 to 2 in going to the larger particle. 
Our results are half-way between these two extremes, hence we must be operating under 
conditions where pore diffusion is just beginning to influence the rate. Hence we can conclude 
that, porous particles, no film resistance, pore resistance just beginning to intrude and effect the 
rate. 
 
Solution 18.28 
dp W/FA0 CR,max/CA0 
4 1 0.5 
8 2 0.5 
 
A→ R → S 
Under the best possible condition we may accept CR, max/CA0 to be 0.5 and to obtain that we may 
use the plug flow pattern and catalyst size to be small. 
 
 
 
Solution 18.29 
For mixed flow the reactor performance equation is: W/FA0 = XA/-rA’ and since the data show 
that the rate is inversely proportional to pellet size shows that pore diffusion controls in both runs 
shows: 
k2 /k1 = 0.5 and by eliminatingpore diffusion in backmix flow we can get – CR/CA0 = 0.44 
and by eliminating pore diffusion and using plug flow we can get - CR/CA0 = 0.63 
So, use smaller pellets in a packed and we should be able to achieve as high as CR/CA0 = 0.63 
 
 
Solution 18.30 
CA0 = 10 mol/m3 
A packed bed catalytic reactor is used. 
(a) It should be operated in the strong diffusion regime. 
(b) We should use a plug flow. 
 
 
 
 Solution: First order catalytic reaction, A → R (Strong pore diffusion resistance) 
 -ln 𝐶𝐶𝐴𝐴
𝐶𝐶𝐴𝐴𝐴𝐴
= 𝑘𝑘𝑘𝑘 
 -ln(1-𝑋𝑋𝑎𝑎) = kt 
 1 − 𝑋𝑋𝑎𝑎 = 𝑒𝑒−𝑘𝑘𝑘𝑘 
 In the strong pore diffusion resistance, Double the pellet size rate 
constant is same 
 K1 = K2 = K 
 𝑋𝑋𝑎𝑎1 = 0.632 , 𝑘𝑘1 = ? 
 1 − 𝑋𝑋𝑎𝑎1 = 𝑒𝑒−𝑘𝑘1𝑘𝑘1 
 1 – 0.632 = 𝑒𝑒−𝑘𝑘𝑘𝑘1 
 Kt1 = 1 
 For 18 mm pellets, t2= 0.5t1 so Kt2= 0.5. 
 1 − 𝑋𝑋𝑎𝑎2 = 𝑒𝑒−𝑘𝑘2𝑘𝑘2 
 1 – Xa2 = 𝑒𝑒−0.5 
 𝑋𝑋𝑎𝑎2= 0.393 
 
 
 
Solution: 
P= 1atm, T=336 ºC, Xa= 0.8, De= 2*10-6 𝑚𝑚
3
𝑚𝑚 𝑐𝑐𝑎𝑎𝑘𝑘.𝑠𝑠 
L=1.2∗10−2
2∗3
= 0.002𝑚𝑚 
Cao= 𝑃𝑃𝑎𝑎𝐴𝐴
𝑅𝑅𝑅𝑅
= 101325
8.314∗(336+273) = 20 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚3 
𝜏𝜏
𝐶𝐶𝑎𝑎𝑚𝑚
= �𝑑𝑑𝑋𝑋𝑎𝑎
−𝑟𝑟𝑎𝑎
 
𝜏𝜏
𝐶𝐶𝑎𝑎𝑚𝑚
= 20000.01 �𝑋𝑋𝑎𝑎 ∗ (1 +∈𝑎𝑎 𝑋𝑋𝑎𝑎)2𝐶𝐶𝑎𝑎𝑚𝑚2 ∗ (1 − 𝑋𝑋𝑎𝑎)2 
𝜏𝜏 = 120 ∗ 𝐶𝐶𝑎𝑎𝑚𝑚 � 𝑑𝑑𝑋𝑋𝑎𝑎(1 − 𝑋𝑋𝑎𝑎)2 = 140 � 11 − 𝑋𝑋𝑎𝑎 − 1� = 140 � 11 − 0.8 − 1� = 0.1 
τ = 
𝑊𝑊
𝐹𝐹𝐴𝐴𝐴𝐴
→ 𝑊𝑊 = 0.1 ∗ 1 = 0.1𝑚𝑚3 = 200 𝑘𝑘𝑘𝑘 
 
 
 
 
 
Solution: 
 P= 1atm, T= 336 ºC, v= 4cm3/s, Xa= 0.8 when 10 gm of 1.2 mm catalyst 
 Design a fluidized bed of 1-mm particles (assume mixed flow of gas) and 
packed bed of 1.5 cm particles which need minimum catalyst? 
Cao= 𝑃𝑃𝑎𝑎𝐴𝐴
𝑅𝑅𝑅𝑅
= 101325
8.314∗(336+273) = 20 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚3 k= −𝑟𝑟𝑎𝑎𝐶𝐶𝑎𝑎 = 12.820∗(1−0.8) = 3.2 𝑠𝑠𝑒𝑒𝑠𝑠−1 
𝜏𝜏
𝐶𝐶𝑎𝑎𝑚𝑚
= 𝑋𝑋𝑎𝑎
−𝑟𝑟𝑎𝑎
 
τ = 
𝑊𝑊
𝐹𝐹𝐴𝐴𝐴𝐴
= 0.01
4∗10−6
= 2500 𝑠𝑠𝑒𝑒𝑠𝑠 
-ra = 20∗0.8
2500
= 6.4 ∗ 10−3 𝑚𝑚𝑚𝑚𝑚𝑚
𝑘𝑘𝑘𝑘.𝑠𝑠 = 12.8 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚3.𝑠𝑠 
Mw= 
𝐿𝐿2∗(−𝑟𝑟𝑎𝑎)
𝐶𝐶𝑎𝑎∗𝐷𝐷𝑒𝑒
= 12.8∗(0.0002)2
20∗0.2∗10−6 = 0.128 
For Packed bed: 
L= 2.5*10-3 m 
MT = L�𝐾𝐾
𝐷𝐷
= 2.5*10-3� 3.2
10−6
 = 4.472 → ↋ = 𝑘𝑘𝑎𝑎𝑡𝑡ℎ𝑀𝑀𝑇𝑇
𝑀𝑀𝑇𝑇
= tanh(4.472)
4.472 = 0.2236 
 𝑊𝑊
𝑉𝑉
= 1
𝑘𝑘 ∗∈
𝑙𝑙𝑙𝑙
𝐶𝐶𝑎𝑎𝑚𝑚
𝐶𝐶𝑎𝑎
= 4500 𝑘𝑘𝑘𝑘
𝑚𝑚3. 𝑠𝑠 
 
For Fluidized Bed: 
L= 1.67*10-4 m 
MT = L�𝐾𝐾
𝐷𝐷
= 1.67*10-4� 3.2
10−6
 = 0.2987 → ↋ = 𝑘𝑘𝑎𝑎𝑡𝑡ℎ𝑀𝑀𝑇𝑇
𝑀𝑀𝑇𝑇
= tanh(0.298)
0.298 = 0.98 
𝑊𝑊
𝑉𝑉
= 1
𝑘𝑘 ∗∈
𝐶𝐶𝑎𝑎𝑚𝑚𝑋𝑋𝑎𝑎
𝐶𝐶𝑎𝑎(1 − 𝑋𝑋𝑎𝑎) = 2500 𝑘𝑘𝑘𝑘𝑚𝑚3. 𝑠𝑠 
Fluidized bed is better because it’s required less amount of catalyst. It needs 
only 55.55% of the packed bed catalyst. 
 
 
 
Problem 18.34 & 18.35 
In aqueous solution, and in contact with the right catalyst, reactant A is 
converted to product R by the elementary reaction AR. Find the mass of 
catalyst needed in a packed bed reactor for 90% conversion of 104 mol A/hr of 
feed having CAo= 103 mol/m3. For this reaction 
34. k"' = 8 x 10-4 m3/m3 bed. s 
35. k"' = 2 m3/m3 bed. s 
Additional data: 
Diameter of porous catalyst pellets= 6 mm 
Effective diffusion coefficient of A in the pellet = 4 X 104 m3/m cat. s 
Void age of packed bed = 0.5 
Bulk density of packed bed = 2000 kg/m3 of bed 
Solution 35: 
 k’’’= 2 ( 𝐦𝐦𝟑𝟑
𝐦𝐦𝟑𝟑 bed sec) * 2 ( 𝐦𝐦𝟑𝟑 bed1 m3 catalyst) 
 = 4 ( m3
m3 cat sec) 
 MT = L √k′′′
D
 
 = 6 ∗ 10^ −3 
6
√
4
4 ∗10^−8 
 = 10 
 ε = 1
MT 
 = .1 
 τ’’’= 𝐶𝐶𝐴𝐴0∗𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
𝐹𝐹𝐴𝐴0
 
 = ∫ 𝑑𝑑𝐶𝐶𝐴𝐴
𝑘𝑘′′′∗ 𝐶𝐶𝐴𝐴 ∗ ε 
 = 1
𝑘𝑘′′′∗ ε ln 𝐶𝐶𝐴𝐴0𝐶𝐶𝐴𝐴 
vcat = 𝐹𝐹𝐴𝐴0
𝐶𝐶𝐴𝐴0∗ 𝑘𝑘′′′∗ 𝜀𝜀 ln 𝐶𝐶𝐴𝐴0𝐶𝐶𝐴𝐴 
 = 10^4
3600∗1000∗4∗.1 ln 101 
 = 0.016 
 W = Vcat * Pcat 
 = Vcat *( Pbulk
voidage
) 
 = 0.016 * 2000
0.5 
 = 64 kg 
Solution 34: 
 k’’’= 8 * 10^-4 ( m3
m3 bed sec) * 2( m3 bed1 m3 catalyst) 
 = 16 * 10^-4 ( m3
m3 cat sec) 
MT = L √k′′′
D
 
 = 6 ∗ 10^ −3 
6
√
16∗10^−4
4 ∗10^−8 
 = 0.2 
ε =1 
 τ’’’= 𝐶𝐶𝐴𝐴0∗𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
𝐹𝐹𝐴𝐴0
 
 = ∫ 𝑑𝑑𝐶𝐶𝐴𝐴
𝑘𝑘′′′∗ 𝐶𝐶𝐴𝐴 ∗ ε 
 = 1
𝑘𝑘′′′∗ ε ln 𝐶𝐶𝐴𝐴0𝐶𝐶𝐴𝐴 
Vcat = 𝐹𝐹𝐴𝐴0
𝐶𝐶𝐴𝐴0∗ 𝑘𝑘′′′∗ 𝜀𝜀 ln 𝐶𝐶𝐴𝐴0𝐶𝐶𝐴𝐴 
 = 10^4
3600∗1000 ∗ 16∗10^−4∗1 ln 101 
 = 4 
W = Vcat * Pcat 
 = Vcat *( Pbulk
voidage
) 
 = 4 * 2000
0.5 
 = 16000 kg 
Problem 18.36 
A first-order catalytic reaction A(l) → R(l) is run in a long, narrow vertical 
reactor with up flow of liquid through a fluidized bed of catalyst particles. 
Conversion is 95% at the start of operations when the catalyst particles are 5 
mm in diameter. The catalyst is friable and slowly wears away, particles 
shrink and the fine powder produced washes out of the reactor. After a few 
months each of the 5-mm spheres has shrunk to 3-mm spheres. What should 
be the conversion at this time? Assume plug flow of liquid. 
(a) Particles are porous and allow easy access for reactants (no resistance to 
pore diffusion). 
(b) Particles are porous and at all sizes provide a strong resistance to pore 
diffusion. 
Solution 18.36: 
a) –𝑟𝑟𝐴𝐴0
𝑟𝑟𝐴𝐴′
 = 𝑘𝑘𝐶𝐶𝐴𝐴
𝑘𝑘𝐶𝐶𝐴𝐴′
 =1 = 𝐶𝐶𝐴𝐴0(1−𝑋𝑋𝐴𝐴)
𝐶𝐶𝐴𝐴0(1−𝑋𝑋𝐴𝐴′) 
 
1 – XA = 1 – XA’ 
1 – XA’ = 1 – 0.05 
XA’ = 0.95 
 
b) rA= k CA 
 
 𝑟𝑟𝐴𝐴1
𝑟𝑟𝐴𝐴2
 = 𝑘𝑘 𝐶𝐶𝐴𝐴1
𝑘𝑘𝐶𝐶𝐴𝐴2
 = 𝑅𝑅2
𝑅𝑅1
 
 𝐶𝐶𝐴𝐴0(1−𝑋𝑋𝐴𝐴)
𝐶𝐶𝐴𝐴0(1−𝑋𝑋𝐴𝐴′) = 35 
 1−.95
1−𝑋𝑋𝐴𝐴′
 = 0.6 
 1 – XA’ = 0.0833 
 XA’ = 0.916 
 
Soln: Denoting catalyst palate without grains as 1 and with grains as 2. Therefore, in 
strong pore diffusion regime (as grains are having strong pore diffusion); 
𝑀𝑀𝑇𝑇2
𝑀𝑀𝑇𝑇1
= 𝜀𝜀1
𝜀𝜀2
= 𝑅𝑅2
𝑅𝑅1
= 𝟎𝟎.𝟎𝟎𝟎𝟎 = 𝑟𝑟′𝐴𝐴1
𝑟𝑟′𝐴𝐴2
= 𝑊𝑊2
𝑊𝑊1
= 𝑉𝑉2 ∗ .75
𝑉𝑉1
 
• Reason for last 3 terms; 
𝑟𝑟′𝐴𝐴 = 𝑋𝑋𝐴𝐴 ∗ 𝐹𝐹𝐴𝐴𝑊𝑊 
But, 
𝐹𝐹𝐴𝐴 = 𝐶𝐶𝐴𝐴 ∗ 𝑣𝑣𝐴𝐴 = 𝑁𝑁𝐴𝐴 ∗ 𝑣𝑣𝐴𝐴𝑉𝑉 
• As between grains there is 25% of void age and there exist free pore diffusion 
hence in only 75% of total volume there will be strong pore resistance. 
𝑊𝑊2
𝑊𝑊1
= 0.01 
• Hence 1% of earlier catalyst weight is required to obtain same 
results. 
𝑉𝑉2 ∗ .75
𝑉𝑉1
= 0.01 
𝑉𝑉2
𝑉𝑉1
= 0.0133 
• Therefore 1.33% of earlier volume is required. 
 
 
 
 
Soln: As catalyst is still in strong pore diffusion regime even after 
impregnating only outer surface hence formulas used in previous 
numerical are applicable. Denoting throughout impregnated 
catalyst as 1 and catalyst impregnated only at outer surface as 2. 
𝑀𝑀𝑇𝑇2
𝑀𝑀𝑇𝑇1
= 6.36 = 1.05 
𝜀𝜀2
𝜀𝜀1
= 11.05 = 0.9524 = 𝑊𝑊2𝑊𝑊1 
𝑊𝑊2 = 0.9524 ∗ 𝑊𝑊1 
 
• Assuming W kg of Pt was used earlier then amount of Pt saved by 
impregnating only outer surface will be; = 𝑊𝑊1 −𝑊𝑊2 = 0.0476 ∗ 𝑊𝑊 
• Means 4.76% kg of earlier amount of Pt used can be saved. 
 
Soln: 
𝐶𝐶𝐴𝐴 = 𝑝𝑝𝐴𝐴𝑅𝑅 ∗ 𝑇𝑇 
𝑋𝑋𝐴𝐴 = 0.5 = 1 − 𝑝𝑝𝐴𝐴0𝑝𝑝𝐴𝐴 
• Therefore, pA at 50% conversion is 361.9≅362mm-Hg. 
 
• From given data we can obtain above curve and 50% conversion 
(362 mm-Hg) is achieved at distance 39.97 m. 
• Now each 1m length of tube consists 1cm3 of bulk catalyst. Hence 
till distance 39.97 cm3 of bulk catalyst will be there. 
y = -0.0006x3 + 0.1517x2 - 15.055x + 759.7
0
100
200
300
400
500
600
700
800
0 10 20 30 40 50 60 70 80 90
P A
 (i
n 
m
m
-H
g)
Distance (in m)
PA v/s Distance
Now, 
𝑉𝑉𝑟𝑟
𝐹𝐹𝐴𝐴0
= 𝑉𝑉𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 
𝐶𝐶𝐴𝐴0 ∗ 𝑣𝑣0
 
• From ideal gas law, 
PV=nRT 
 
𝐶𝐶𝐴𝐴0 = 𝑛𝑛𝑉𝑉 = 𝑃𝑃𝐴𝐴0𝑅𝑅 ∗ 𝑇𝑇 = 122400 𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚3 
 
Therefore, 
𝑉𝑉𝑟𝑟100 ∗ 1033600 𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠 = 39.97 𝑐𝑐𝑚𝑚
3122400 𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚3 ∗ 10 𝑐𝑐𝑚𝑚3𝑠𝑠 
 
Vr = 2.48*106 cm3 
Vr = 2.48 m3 
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Group no: 7: 13bch036, 13bch037, 13bch038, 13bch040 
Cre II- term paper II 
Chapter 24: Fluid-Fluid reactors design 
 
24.1 
 
 
 
 
 
 
 
 
 
 
24.2 
A = agitated tank HA=0, K=0 
MH = 
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙
 = 
√3.6∗10−6∗0∗𝐶𝑏
0.114
 = 0 
𝐸𝑖 = 1 +
𝐶𝑏 ∗ 𝐻𝑎
𝑃𝐴1
= 1 
𝐸𝑖 ≥ 1 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑖 = 𝐸 = 1 
𝐺𝑎𝑠 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
1
𝐾𝑎𝑔 ∗ 𝑎
=
1
0.36
= 2.778 
𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑟𝑒𝑖𝑠𝑖𝑡𝑎𝑛𝑐𝑒 =
𝐻𝑎
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
= 0 
𝐿𝑖𝑞𝑢𝑖𝑑 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
𝐻𝑎
𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝐸
= 0 
−𝑟𝐴 =
𝑃𝐴
1
𝑘𝑎𝑔 ∗ 𝑎
=
1000
2.778
= 360 
Fg (Yin - Yout) = -ra*Vr = Fg*(PAin - PAout)/Π 
(i) Volume: 
𝑉𝑟 = 
90000
360
(1000 − 100)
105
= 22.5 𝑚3 
(ii) 100% gas film resistance 
 
24.3 
Straight mass transfer, counter current, HA= 18, PA2 = 100 
By material balance between points 1 & 2, we can find CA1 
90000
105
(1000 − 100) = 
900000
55556
(𝐶𝐴1 − 0) 
Or CA1 = 50 mol/m3 
At equilibrium, PA*= HA*CA or PA1* = 18*50= 900 Pa 
Now from the PA vs CA diagram we can see that the operating and equilibrium lines are 
parallel. Thus the rate of transfer is the same everywhere. Hence, 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎 + 
𝐻𝐴
𝑘𝑎𝑙 ∗ 𝑎
(𝑃𝐴 − 𝑃𝐴 ∗) = 
1
1
0.36 +
18
72
(1000 − 900) 
−𝑟𝐴′′′′ = 33 𝑚𝑜𝑙/(𝑚3 ∗ ℎ𝑟) 
(i) Volume 
𝑉𝑟 =
𝐹𝑔
𝛱
ʃ 
𝑑𝑝𝑎
−𝑟𝑎′′′′
= 
90000
105
(1000 − 100)
33
= 24.5 𝑚3 
(ii) 100% gas film resistance 
 
24.4 
Tower 
(KAg*a*PA) top = 0.36*100 = 36 
(KAg*a*PA) bottom = 0.36*1000 = 360 
(KAl*a*CB) top = 5.556*72 = 400 
(KAl*a*CB) bottom = 72*555556 = 4000 
CBin = 55.55 mol/m3 
CBout = 10% of Cbin = 5.55 mol/m3 
 (KAl*a*CB) bottom > (KAg*a*PA) bottom 
(KAl*a*CB) top > (KAg*a*PA) top 
Therefore gas film resistance is controlling one. 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎
𝑃𝐴 
 
(i) Volume 
𝑉𝑟 =
𝐹𝑔
𝛱
ʃ 
𝑑𝑋𝑎
−𝑟𝑎
= 
𝐹𝑔
𝑘𝑎𝑔 ∗ 𝑎 ∗ 𝛱
∫
𝑑𝑝𝑎
𝑝𝑎
= 
90000
0.36
𝑙𝑛
10
1
=
103
102
5.75 𝑚3 
(ii) 100% gas film resistance 
 
24.5 
By doing material balance, 
Fg (YA1- YA2) = Fl (XA1- XA2) 
90000 (
0.01
0.99
− 
0.001
0.999
) = 900000(0.001 − 𝑋B1) 
CB1 = 5 mol/m3 
We need to calculate Ei and MH to find out the rate 
𝐸𝑖 = 1 + 
𝐷𝑏 ∗ 𝐶𝑏
𝑏 ∗ 𝐷𝑎 ∗ 𝐶𝑎
= 1 +
𝐶𝑏
𝐶𝑎
= 1 +
𝐶𝑏 ∗ 𝐻𝑎
𝑃𝑎𝑖
= 1 +
105 ∗ 𝐶𝑏
𝑃𝑎𝑖
 
MH = 
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙2
 = ∞ 
Since 𝐸𝑖 ≪ 𝑀H , we have E=Ei 
At top point 2, 
Assuming that all the resistance lies in the gas film and liquid film resistance is negligible so 
taking PA2*= PA2 
𝐸𝑖 = 1 + 
55.56 ∗ 105
100
= 5.556 ∗ 104 
Hence E = 5.556*104. Now rate equation, 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎 + 
𝐻𝐴
𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝑒 +
𝐻𝐴
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
𝑃𝐴 
 
 = 
1
1
0.36 + 
105
72(5.556 ∗ 104)
+ 
105
∞
𝑃𝐴 = 0.36 ∗ 𝑃𝐴 
We can see that gas film resistance is not negligible. 
At bottom (point 1), 
 
𝐸𝑖 = 1 + 
5 ∗ 105
1000
= 501 
MH = 
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙2
 = ∞ 
Since 𝐸𝑖 ≪ 𝑀h, we have E=Ei so rate equation is, 
−𝑟𝐴′′′′ = 
1
1
0.36 + 
105
72(501)
+ 
105
∞
𝑃𝐴 = 0.18 ∗ 𝑃𝐴 = 
1
1
0.36 + 
1
0.36 + 0
𝑃𝐴 = 0.18 ∗ 𝑃𝐴 
This shows that gas and liquid film resistance is 50-50% 
 
Trying 2nd time, 
Guessing PAi = 500, and repeating the procedure, we get Ei = 1001 and from the rate 
expression 1/3rd of the resistance is in the gas film. Thus our guess was wrong. 
Trying 3rd time, guessing PAi = 500 or 10% resistance in the gas film. Then Ei = 5001 
−𝑟𝐴′′′′ = 
1
1
0.36 + 
1
3.6 + 0
𝑃𝐴 = 0.32 ∗ 𝑃𝐴 
So −𝑟𝐴′′′′mean = (0.36+0.33)/2 = 0.345 PA 
Hence, 
(i) Volume 
𝑉𝑟 =
𝐹𝑔
𝛱
ʃ 
𝑑𝑝𝑎
−𝑟𝑎′′′′
= 
90000
105
∫
𝑑𝑝𝑎
0.345 ∗ 𝑝𝑎
= 
90000
34500
𝑙𝑛
10
1
=
103
102
6 𝑚3 
 
(ii) Gas film resistance is 91% of total, and 9% is from liquid film. 
 
24.6 
Agitated Tank, HA = 105 , k = 2.6*107 
Fg (YOUT – YIN) = Fl (XIN – XOUT) 
90000 (
1000
105
− 
100
105
) = 900 (
5556
55556
− 
𝐶𝑏𝑜𝑢𝑡
55556
) 
CBout = 555.96 mol/m3 
MH = 
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙2
 = 
√3.68∗10−6∗2.6∗107∗555.960.722
 = 300.46 
 
PAi = 100 KPa 
𝐸𝑖 = 1 +
𝐶𝑏 ∗ 𝐻𝑎
𝑃𝑎𝑖
𝐸𝑖 = 1 +
500 ∗ 105
100
= 5 ∗ 105 
𝐸𝑖 ≥ 𝑀H, MH = E = 300.46 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎 + 
𝐻𝐴
𝑘𝑎𝑙 ∗ 𝑎 +
𝐻𝐴
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
𝑃𝐴 
 
−𝑟𝐴′′′′ =
1
1
0.72 + 
105
144 ∗ 300 +
105
2.6 ∗ 105 ∗ 555.56 ∗ 0.9
𝑃𝐴 = 0.710 ∗ 𝑃𝐴 
 
(i) Volume 
𝑉𝑟 =
𝐹𝑔
𝛱
ʃ 
𝑑𝑝𝑎
−𝑟𝑎′′′′
= 
90000
105
(
1000 − 100
100 ∗ 0.719
) = 11.26 𝑚3 
 
(𝑖𝑖)𝑇ℎ𝑢𝑠 𝑔𝑎𝑠 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
1
0.72
(
1
0.719)
= 99% 
 
 
24.7 
Fg (YOUT – YIN) = Fl (XIN – XOUT) 
90000 (
0.01
0.99
−
0.001
0.999
) =
900000
55556
(55.56 − 𝐶𝑏𝑜𝑢𝑡) 
CB = 5 mol/m3 
At exit conditions, MH = 
√3.68∗10−6∗2.6∗105∗500
0.722
 = 30 
𝐸𝑖 = 1 +
105 ∗ 500
1000
> 𝑀𝐻 
Therefore, Ei = MH =30 
−𝑟𝐴′′′′ =
𝑃𝐴
1
0.72 + 
105
144 ∗ 40 +
105
2.6 ∗ 105 ∗ 500 ∗ 0.9
 
−𝑟𝐴′′′′ = 0.041*PA 
Therefore, 
𝑉𝑟 = 
𝐹𝑔
𝛱
∗
∆𝑃
−𝑟𝐴′′′′
= 
90000(1000 − 100)
105(0.041) ∗ 100
= 197.6 𝑚3 
(i) Volume = 197.6 m3 
 (ii) Liquid film dominates (94.3%) 
 
 
24.8 
HA= 1000, k=2.6*103, Tower 
𝐹𝑔
𝛱
(𝑃𝐴1 − 𝑃𝐴2) =
𝐹𝑙
𝑏 ∗ 𝐶𝑡
(𝐶𝑏2 − 𝐶𝑏1) 
90000
105
(1000 − 900) =
900000
55556
(𝐶𝑏2 − 𝐶𝑏1) 
CBout = 5.556 mol/m3 
 
At top, MH = 
√3.68∗10−6∗2.6∗103∗55.56
0.72
 = 1 
𝐸𝑖 = 1 +
55.56 ∗ 105
100
= 55561 > 𝑀𝐻 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎 + 
𝐻𝐴
𝑘𝑎𝑙 ∗ 𝑎 +
𝐻𝐴
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
𝑃𝐴 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 ∗ 100 +
105
2.6 ∗ 1000 ∗ 55.56 ∗ 0.9
 
 
−𝑟𝐴′′′′ = 0.0718 
1/−𝑟𝐴′′′′ = 13.92 
 
At bottom, MH = 
√3.68∗10−6∗2.6∗103∗55.56
0.722
 = 0.3167 
But MH can’t be less than 1. Therefore it is equal to 1. 
𝐸𝑖 = 1 +
105 ∗ 5.556
1000
= 55561 > 𝑀𝐻 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 +
105
2.6 ∗ 1000 ∗ 5.556 ∗ 0.9
 
 
−𝑟𝐴′′′′ = 0.714 
1/−𝑟𝐴′′′′ = 1.4 
 
At the middle, at PA = 3000 Pa, CA = 44.44 mol/m3 
MH = 
√3.68∗10−6∗2.6∗103∗44.44
0.722
 = 0.895 
𝐸𝑖 = 1 +
105 ∗ 44.44
1000
> 𝑀𝐻 
So E = MH = 1 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 +
105
2.6 ∗ 1000 ∗ 44.44 ∗ 0.9
 
−𝑟𝐴′′′′ = 0.359 
1/−𝑟𝐴′′′′ = 2.58 
𝑉𝑟 = 
𝐹𝑔
𝛱
∫
𝑑𝑝𝑎
−𝑟𝐴′′
103
102
 
 
𝑉𝑟 = 2685.6 m3 
 
24.9 
Fg (Yin-Yout) = Fl ( XBOUT – XBIN) 
90000 (
0.01
0.99
−
0.001
0.999
) =
900000
55556
(55.56 − 𝐶𝑏𝑜𝑢𝑡) 
CB = 5 mol/m3 
At top, MH = 
√3.68∗10−6∗2.6∗107∗55.56
0.722
 = 100 
𝐸𝑖 = 1 +
105 ∗ 55.56
1000
> 𝑀𝐻 
Therefore, Ei = MH =100 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 ∗ 100 +
105
2.6 ∗ 107 ∗ 55.56 ∗ 0.9
 
−𝑟𝐴′′′′ = 6 
 
At bottom, MH = 
√3.68∗10−6∗2.6∗107∗5
0.722
 = 30 
𝐸𝑖 = 1 +
105 ∗ 5
1000
> 𝑀𝐻 
Therefore, Ei = MH =30 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 ∗ 30 +
105
2.6 ∗ 1000 ∗ 5 ∗ 0.9
 
−𝑟𝐴′′′′ = 20.4 
 
In the middle, at PA = 547 Pa, CA = 30.56 mol/m3 
MH = 
√3.68∗10−6∗2.6∗107∗30.56
0.722
 = 74 
𝐸𝑖 = 1 +
105 ∗ 30.56
1000
> 𝑀𝐻 
So E = MH = 74 
−𝑟𝐴′′′′ =
100
1
0.36 + 
105
0.72 ∗ 74 +
105
2.6 ∗ 107 ∗ 30.56 ∗ 0.9
 
−𝑟𝐴′′′′ = 25.4 
1/−𝑟𝐴′′′′ = 0.0393 
𝑉𝑟 = 
𝐹𝑔
𝛱
∫
𝑑𝑝𝑎
−𝑟𝐴′′
103
102
 
(i) Volume 
𝑉𝑟 = (9000/105)*66= 59.4 m3 
(ii) liquid film dominates (84%) 
 
24.10 
HA= 1000, k=2.6*105 , Tower 
𝐹𝑔
𝛱
(𝑃𝐴1 − 𝑃𝐴2) =
𝐹𝑙
𝑏 ∗ 𝐶𝑡
(𝐶𝑏2 − 𝐶𝑏1) 
90000
105
(1000 − 900) =
900000
55556
(𝐶𝑏2 − 𝐶𝑏1) 
CBout = 5.556 mol/m3 
 
At top, MH = 
√3.63∗10−6∗2.6∗105∗55.56
0.72
 = 10 
𝐸𝑖 = 1 +
55.56 ∗ 103
100
= 556.6 > 𝑀𝐻 
Therefore, Ei = MH =10 
−𝑟𝐴′′′′ =
100
1
0.36 + 
103
0.72 ∗ 10 +
103
2.6 ∗ 105 ∗ 55.56 ∗ 0.9
 
 
−𝑟𝐴′′′′ = 23.95 mol/m3*hr 
1/−𝑟𝐴′′′′ = 0.0417 
 
At bottom, MH = 
√3.68∗10−6∗2.6∗105∗5.56
0.722
 = 3.16 
But MH can’t be less than 1. Therefore it is equal to 1. 
𝐸𝑖 = 1 +
103 ∗ 5.556
1000
= 6.55 > 𝑀𝐻 
Therefore, Ei = MH =3.16 
−𝑟𝐴′′′′ =
100
1
0.36 + 
103
0.72 ∗ 3.16 +
105
2.6 ∗ 105 ∗ 5.556 ∗ 0.9
 
 
−𝑟𝐴′′′′ = 137.93 
1/−𝑟𝐴′′′′ = 0.0073 
 
At the middle, at PA = 300 Pa, CA = 44.44 mol/m3 
MH = 
√3.68∗10−6∗2.6∗107∗44.44
0.722
 = 8.95 
𝐸𝑖 = 1 +
105 ∗ 44.44
1000
> 𝑀𝐻 
So E = MH = 8.95 
−𝑟𝐴′′′′ =
300
1
0.36 + 
105
0.72 ∗ 8.95 +
105
2.6 ∗ 105 ∗ 44.44 ∗ 0.9
 
−𝑟𝐴′′′′ = 69.136 
1/−𝑟𝐴′′′′ = 0.0144 
𝑉𝑟 = 
𝐹𝑔
𝛱
∫
𝑑𝑝𝑎
−𝑟𝐴′′
103
102
 
 
𝑉𝑟 = 0.9*10.8 = 9.72 m3 
 
24.11 
𝐶𝐴 =
𝑃𝐴𝑖𝑛
𝐻𝐴
= 
101325
3500
= 28.95 𝑚𝑜𝑙/𝑚3 
CBin = 300 mol/m3, since CB is greater than CA we have excess of B, so CB can be taken as 300 
mol/m3 
 
MH = 
√1.4∗10−9∗0.433∗300∗120
0.25
 = 2.05 
𝐸𝑖 = 1 +
300 ∗ 3500
101325
= 11.36 > 𝑀𝐻 
So E = MH = 2.05 
−𝑟𝐴′′′′ =
101325
0 + 
3500
0.025 ∗ 2.05 +
3500
0.433 ∗ 300 ∗ 0.08
 
−𝑟𝐴′′′′ = 1.476 
1/−𝑟𝐴′′′′ = 0.677 
𝑉𝑟 = 𝐹𝐴0 ∫
𝑑𝑋𝐴
−𝑟𝐴′′
𝑋𝐴
0
 
FA0 = V0*CA0 = 0.0363*18.95 = 1.05 m3/hr 
𝑋𝐴 = 
𝑉𝑟
𝐹𝐴0
(−𝑟𝐴′′′′) = 
0.6042
1.05
∗ 1.48 = 85% 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 
 
24.12 
Cl2 + 2NaOH  P 
K= 2.6*109 km3/mol.hr, HA = 105 Pa m3/mol.hr 
Acs = 55 m3 
Fg = 100, Fl = 250 mol/s*m2 
D = 1.5*109 m2/s 
CB (IN) = 10 % * 250 = 25 mol/m3 
𝐹𝑔
𝛱
(𝑃𝐴1 − 𝑃𝐴2) =
𝐹𝑙
𝑏 ∗ 𝐶𝑡
(𝐶𝑏2 − 𝐶𝑏1) 
100
105
(2.36 ∗ 0.99) =
250
2 ∗ 2736
(25 − 𝐶𝐵2) 
CB2 = 24.9 m3 / mol 
MH = 
√1,5∗10−9∗2.6∗109∗24.9
0.45
 = 0.218 
𝐸𝑖 = 1 +
24.5 ∗ 105
0
= ∞ > 𝑀𝐻 
But MH can’t be less than 1. Therefore E is equal to 1. 
When PA = 1000, 
−𝑟𝐴′′′′ =
𝑃𝐴
1
133 + 
105
133 ∗ 1 +
105
2.6 ∗ 109 ∗ 24.9 ∗ 0.9
= 1.329 ∗ 10−3𝑃𝐴 
 
𝐻 = 
𝐹𝑔
𝐴𝑐𝑠 ∗ 𝜋
∫
𝑑𝑃𝑎
0.1329 ∗ 𝑃𝑎
= 
100
55 ∗ 1
∗
𝑙𝑛10
0.1329
= 3.14 𝑚
103
102
 
 
24.13 
For agitated tank, 
K= 2.6*105, HA = 105 
At the beginning, CB = 555.6 mol/m3 
MH = 
√3.6∗10−6∗2.6∗105∗555.6
1.44
 = 15.84 
𝐸𝑖 = 1 +
555.6 ∗ 105
1000
= 55561 > 𝑀𝐻 
So E = MH = 15.84 
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
𝐹𝑔 + 
1
𝐾𝑎𝑔 ∗ 𝑎 +
𝐻𝑎
𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 +
𝐻𝑎
𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
 
 
−𝑟𝐴′′′′ =
1000
105 ∗ 1.62
9000 ∗ 0.9 + 
1
0.72 +
105
144 ∗ 15.84 +
105
2.6 ∗ 105 ∗ 555.6 ∗ 0.9
 
−𝑟𝐴′′′′ = 15.33 mol/m3*hr 
1/−𝑟𝐴′′′′ = 0.0652 
 
At the end of the run, MH = 
√3.68∗10−6∗2.6∗105∗55.6
1.44
 = 5.01 
𝐸𝑖 = 1 +
105 ∗ 55.56
1000
= 5561 > 𝑀𝐻 
Therefore, Ei = MH =5.01 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
105 ∗ 144 ∗ 5.01
+ 0
 
 
−𝑟𝐴′′′′ = 6.25 mol/m3.hr 
1/−𝑟𝐴′′′′ = 0.16 
 
At some intermediate run, CA = 138.9 mol/m3 
MH = 15.84/2 = 7.92 
𝐸𝑖 = 1 +
105 ∗ 138.9
1000
= 13890 > 𝑀𝐻 
So E = MH = 7.92 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
144 ∗ 7.92 + 0
 
−𝑟𝐴′′′′ = 9.168 mol/m3.hr 
1/−𝑟𝐴′′′′ = 0.109 
 
The length of the time that we have to bubble gas through the liquid is 
𝑡 = 
𝐹𝑙
𝑏
∫
𝑑𝐶𝑏
−𝑟𝐴′′
103
102
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 51.38 = 46.2 ℎ𝑟 
The minimum time needed if all the A reacts with B, 
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
= 
1.62(555.6 − 55.56)
9000(1000/99000)
= 8.91 ℎ𝑟 
Therefore, the efficiency of the use of A = 8.91/46.24 = 19.26% 
 
 
24.14 
For agitated tank, 
K= 2.6*109, HA = 105 
At thebeginning, CB = 555.6 mol/m3 
MH = 
√3.6∗10−6∗2.6∗109∗555.6
1.44
 = 1583.6 
𝐸𝑖 = 1 +
555.6 ∗ 105
1000
= 55561 > 𝑀𝐻 
So E = MH = 1583.6 
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
𝐹𝑔 + 
1
𝐾𝑎𝑔 ∗ 𝑎 +
𝐻𝑎
𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 +
𝐻𝑎
𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
 
 
−𝑟𝐴′′′′ =
1000
105 ∗ 1.62
9000 ∗ 0.9 + 
1
0.72 +
105
144 ∗ 1583.6 +
105
2.6 ∗ 109 ∗ 555.6 ∗ 0.9
 
−𝑟𝐴′′′′ = 45.81 mol/m3*hr 
1/−𝑟𝐴′′′′ = 0.0218 
 
At the end of the run, MH = 
√3.68∗10−6∗2.6∗109∗55.6
1.44
 = 506.5 
𝐸𝑖 = 1 +
105 ∗ 55.56
1000
= 5561 > 𝑀𝐻 
Therefore, Ei = MH =506.5 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
144 ∗ 506.5 + 0
 
 
−𝑟𝐴′′′′ = 43.93 mol/m3.hr 
1/−𝑟𝐴′′′′ = 0.0227 
 
At some intermediate run, CA = 138.9 mol/m3 
MH = 1583.6/2 = 791.8 
𝐸𝑖 = 1 +
105 ∗ 138.9
1000
= 13890 > 𝑀𝐻 
So E = MH = 395.9 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
144 ∗ 791.8 + 0
 
−𝑟𝐴′′′′ = 44.9 mol/m3.hr 
1/−𝑟𝐴′′′′ = 0.022 
 
The length of the time that we have to bubble gas through the liquid is 
𝑡 = 
𝐹𝑙
𝑏
∫
𝑑𝐶𝑏
−𝑟𝐴′′
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 10.987 = 9.88 ℎ𝑟 
The minimum time needed if all the A reacts with B, 
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
= 
1.62(555.6 − 55.56)
9000(1000/99000)
= 8.91 ℎ𝑟 
Therefore, the efficiency of the use of A = 8.91/9.88 = 90.1% 
 
24.15 
For agitated tank, 
K= 2.6*103, HA = 105 
At the beginning, CB = 555.6 mol/m3 
MH = 
√3.6∗10−6∗2.6∗103∗555.6
1.44
 = 1.584 
𝐸𝑖 = 1 +
555.6 ∗ 105
1000
= 55561 > 𝑀𝐻 
So E = MH = 1.58 
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
𝐹𝑔 + 
1
𝐾𝑎𝑔 ∗ 𝑎 +
𝐻𝑎
𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 +
𝐻𝑎
𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
 
 
−𝑟𝐴′′′′ =
1000
105 ∗ 1.62
9000 ∗ 0.9 + 
1
0.72 +
105
144 ∗ 1.584 +
105
2.6 ∗ 105 ∗ 555.6 ∗ 0.9
 
−𝑟𝐴′′′′ = 2.175 mol/m3*hr 
1/−𝑟𝐴′′′′ = 0.459 
 
At the end of the run, MH = 
√3.68∗10−6∗2.6∗103∗55.6
1.44
 = 0.5 
𝐸𝑖 = 1 +
105 ∗ 55.56
1000
= 5561 > 𝑀𝐻 
Therefore, Ei = MH =0.5 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
105 ∗ 144 ∗ 0.5
+ 0
 
 
−𝑟𝐴′′′′ = 0.7096 mol/m3.hr 
1/−𝑟𝐴′′′′ = 1.41 
 
At some intermediate run, CA = 555.6/4= 138.9 mol/m3 
MH = 15.84/4 = 0.792 
𝐸𝑖 = 1 +
105 ∗ 138.9
1000
= 13890 > 𝑀𝐻 
So E = MH = 1 
−𝑟𝐴′′′′ =
1000
20 + 1.3889 +
105
144 ∗ 1 + 0
 
−𝑟𝐴′′′′ = 1.66 mol/m3.hr 
1/−𝑟𝐴′′′′ = 0.6 
 
The length of the time that we have to bubble gas through the liquid is 
𝑡 = 
𝐹𝑙
𝑏
∫
𝑑𝐶𝑏
−𝑟𝐴′′
103
102
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 350 = 315 ℎ𝑟 
The minimum time needed if all the A reacts with B, 
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
= 
1.62(555.6 − 55.56)
9000(1000/99000)
= 8.91 ℎ𝑟 
Therefore, the efficiency of the use of A = 8.91/315 = 2.82% 
 
24.16 
K= 2.6*1011 m3/mol.hr, HA = 103 Pa/m3*mol 
MH = 
√1.5∗10−9∗2.6∗1011∗24.9
45
 = 2.18 
Ei = ∞, therefore, Ei =MH = 2.18 
−𝑟𝐴′′′′ =
1
1
𝑘𝑎𝑔 ∗ 𝑎 + 
𝐻𝐴
𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝐸 +
𝐻𝐴
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
𝑃𝐴 
 = 
1
1
133 + 
103
133 + 
103
2.6 ∗ 1011 ∗ 24.09 ∗ 0.9
𝑃𝐴 = 0.1328 ∗ 𝑃𝐴 
 
𝐻 = 
𝐹𝑔
𝐴𝑐𝑠 ∗ 𝜋
∫
𝑑𝑃𝑎
0.1328 ∗ 𝑃𝑎
= 
100
55 ∗ 1
∗
𝑙𝑛10
0.1328
= 31.524 𝑚
103
102
 
 
Chapter 25 : Fluid-Particle Reactions:Kinetics 
Roll Number : Rohan Patel (13BCH041) 
 Shivang Patel (13BCH042) 
 Vedant Patel (13BCH043) 
 Vishal Patel (13BCH045) 
 
Q 25.1 A batch of solids of uniform size is treated by gas in a uniform environment. Solid is 
converted to give a nonflaking product according to the shrinking-core model. Conversion is 
about 7/8 for a reaction time of 1 h, conversion is complete in two hours. What mechanism is 
rate controlling? 
A 25.1 
Assume spherical particles 
τ=2 hr t= 1 hr, Xb=0.875 
Film Diffusion 
 
τ
 
 
 
 Xb 
Film Diffusion not dominating. 
Ash Diffusion 
 
 
 
 
τ
 
Ash Diffusion is dominating. 
Reaction 
 
 
 0.5=
 
τ
 
Reaction control is also dominating 
Hence, Ash Diffusion and Reaction controls are dominating. 
 
Q 25.2 In a shady spot at the end of Brown Street in Lewisburg, Pennsylvania, stands a Civil 
War memorial-a brass general, a brass cannon which persistent undergraduate legend insists may 
still fire some day, and a stack of iron cannonballs. At the time this memorial was set up, 1868, 
the cannonballs were 30 inches in circumference. Today due to weathering, rusting, and the 
once-a-decade steel wire scrubbing by the DCW, the cannonballs are only 29.75 in. in 
circumference. Approximately, when will they disappear completely? 
A 25.2 
R=4.78 inch 
rc=4.74 inch t=148 years 
As it is a reaction controlling mechanism. 
 
 
 
 
Now, 
 
τ
 
 
 
 
Hence the time in another 17,538 years. 
 
Q 25.3 Calculate the time needed to burn to completion particles of graphite (R, = 5 mm, p, = 2.2 
gm/cm
3
, k" = 20 cm/sec) in an 8% oxygen stream. For the high gas velocity used assume that 
film diffusion does not offer any resistance to transfer and reaction. Reaction temperature = 
900°C. 
A 25.3 
As it is a shrinking core particle ash film diffusion will not participate in the reaction. Also it is 
given that film diffusion is negligible. Hence only reaction control is applicable. 
R=5 mm 
ρb=2.2 gm/cm
3
=0.1833mol/cm
3 
k’’=20 cm/sec 
 
 
 
 
 
 
 
b=1 
 ρ
 
 
 
 
 
Hence 1.53 hours are needed for complete combustion of graphite particles. 
 
Q 25.4 Spherical particles of zinc blende of size R = 1 mm are roasted in an 8% oxygen stream 
at 900°C and 1 atm. The stoichiometry of the reaction is Assuming that reaction proceeds by the 
shrinking-core model calculate the time needed for complete conversion of a particle and the 
relative resistance of ash layer diffusion during this operation. 
Data 
Density of solid, p, = 4.13 gm/cm
3
 = 0.0425 mol/cm
3
 
Reaction rate constant, k" = 2 cm/sec 
For gases in the ZnO layer, De = 0.08 cm
2
/sec 
Note that film resistance can safely be neglected as long as a growing ash layer is present. 
A 25.4 
Fro the reaction, b= 2/3 
τtotal = τash + τreaction 
For ash layer control, 
CAG = Pa / RT = 0.08*101325/(8.314*1173) = 0.83118*10
-2
 mol/L 
τash = 
 
 
 = 1598.02 sec 
For Reaction control, 
τreaction = 
 
 
 
 = 3834.90sec 
τtotal = 1598.02 + 3834.90 = 5432.94sec 
 = 90.54min 
 
Relative Resistance of ash layer = 
 
 
 
 = 29.41 % 
 
Q On doubling the particle size from R to 2R the time for complete conversion triples. What is 
the contribution of ash diffusion to the overall resistance for particles of size, 
 
Q 25.5 R? 
A 25.5 
Let 1 refer to particle size of R 
 2 refer to particle size of 2R 
Then, τ2 = 3τ1……. (1) 
τ1 = τ1ash + τ1reaction…….(2) 
τ2 = τ2ash + τ2reaction……(3) 
Put τ2ash = 4τ1ash…..(4) 
 τ2reaction = 2τ1reaction……(5) 
So, from equation (1),(2),(3),(4) and (5), 
3τ1 = 4τ1ash + 2τ1reaction….(6) 
So from equation (2) and (6), 
τ1ash = τ1reaction 
therefore % contribution of ash diffusuion for R = 50% 
 
Q 25.6 R? 
A 25.6 
Let 1 refer to particle size of R 
 2 refer to particle size of 2R 
Then, τ2 = 3τ1……. (1) 
τ1 = τ1ash + τ1reaction…….(2) 
τ2 = τ2ash + τ2reaction……(3)Put τ2ash = 4τ1ash…..(4) 
 τ2reaction = 2τ1reaction……(5) 
So, from equation (1),(2),(3),(4) and (5), 
3τ1 = 4τ1ash + 2τ1reaction….(6) 
So from equation (2) and (6), 
τ1ash = τ1reaction 
put values from equation (4) and (5), 
 τ2ash /4 = τ2reaction/2 
so, τ2reaction = τ2ash /2 
therefore % contribution of ash diffusion for R = 66.66% 
 
Q Spherical solid particles containing B are roasted isothermally in an oven with gas of constant 
composition. Solids are converted to a firm nonflaking product according to the SCM as follows: 
 
 
From the following conversion data (by chemical analysis) or core size data (by slicing and 
measuring) determine the rate controlling mechanism for the transformation of solid. 
 
Q 25.7 
Dp (mm) Xb t (min) 
1 1 4 
1.5 1 6 
A 25.7 
For finding out the corresponding rate control mechanism, considering the constant size 
spherical particles if it is: 
Film diffusion controlling : The value of R/τ will remain constant for both cases. 
Reaction controlling: The value of R/τ will remain constant for both cases. 
Ash diffusion control : The value of R
2/τ will remain constant for both cases. 
Substituting the value of τ from both cases: 
Dp 
(mm) 
Xb t 
(min) 
Film Diffusion 
control 
Ash film diffusion control Reaction Control 
 
 
 
 
 
 
 
 
 
 
 
 
1 1 4 0.25 0.0625 0.125 
1.5 1 6 0.1667 0.0938 0.125 
 
From the above table it implies that the rate controlling mechanism for the transformation of 
solid is Reaction Control. 
 
Q 25.8 
Dp (mm) Xb t (sec) 
1 0.3 2 
1 0.75 5 
 
A 25.8 
For finding out the corresponding rate control mechanism, considering the constant size 
spherical particles if it is: 
Film diffusion controlling : The value of R/τ will remain constant for both cases. 
Reaction controlling: The value of R/τ will remain constant for both cases. 
Ash diffusion control : The value of R
2/τ will remain constant for both cases. 
Substituting the value of τ from both cases: 
Dp 
(mm) 
Xb t (sec) Film Diffusion 
control 
Ash film diffusion control Reaction Control 
 
 
 
 
 
 
 
 
 
 
 
 
 
1 0.3 2 0.15 0.00436 0.0280 
1 0.75 5 .015 0.01547 0.037 
From the above table it implies that the rate controlling mechanism for the transformation of 
solid is Film diffusion control. 
 
Q 25.9 
Dp (mm) Xb t (sec) 
1 1 200 
1.5 1 450 
 
A 25.9 
For finding out the corresponding rate control mechanism, considering the constant size 
spherical particles if it is: 
Film diffusion controlling : The value of R/τ will remain constant for both cases. 
Reaction controlling: The value of R/τ will remain constant for both cases. 
Ash diffusion control : The value of R
2/τ will remain constant for both cases. 
Substituting the value of τ from both cases: 
Dp 
(mm) 
Xb t 
(sec) 
Film Diffusion 
control 
Ash film diffusion control Reaction Control 
 
 
 
 
 
 
 
 
 
 
 
 
1 1 200 0.005 0.00125 0.0025 
1 1 450 0.0022 0.00125 0.00167 
 
From the above table it implies that the rate controlling mechanism for the transformation of 
solid is Ash film diffusion control. 
 
 
Q 25.10 
Dp (mm) Xb t (sec) 
2 0.875 1 
1 1 1 
A 25.10 
For finding out the corresponding rate control mechanism, considering the constant size 
spherical particles if it is: 
Film diffusion controlling : The value of R/τ will remain constant for both cases. 
Reaction controlling: The value of R/τ will remain constant for both cases. 
Ash diffusion control : The value of R
2/τ will remain constant for both cases. 
Substituting the value of τ from both cases: 
Dp 
(mm) 
Xb t (sec) Film Diffusion 
control 
Ash film diffusion control Reaction Control 
 
 
 
 
 
 
 
 
 
 
 
 
 
2 0.875 1 0.875 0.5 0.5 
1 1 1 0.5 0.25 0.5 
From the above table it implies that the rate controlling mechanism for the transformation of 
solid is Reaction control. 
 
Q 25.11 Uniform-sized spherical particles UO, are reduced to UO, in a uniform 
environment with the following results: 
 
If reaction follows the SCM, find the controlling mechanism and a rate 
equation to represent this reduction. 
A 25.11 
Method : 1 
For 
Gas Film Controlling : 
 
τ 
 = 
Reaction Controlling : 
 
τ 
 = 
 
Ash Film Controlling : 
 
τ 
 = 
 
 
Now, Find the value of R.H.S of all equation... 
t, hr 
 
 
 
0.18 0.45 0.1807 0.0861 
0.347 0.68 0.316 0.2365 
0.453 0.8 0.4125 0.374 
0.567 0.95 0.6316 0.6928 
0.733 0.98 0.7286 0.819 
 
Then, Plot R.H.S. vs time(hr) and whichever is nearer to diagonal line it will be controlling 
mechanism. 
 
The assumption of reaction control gives a better straight line fit than does the assumption for 
gas film and ash film. 
Method 2 : 
Find the value of τ for any 2 or 3 time for all three mechanism and in whichever mechanism all τ 
are same then it will be controlling mechanism. 
Time,hr 0.18 0.453 0.733 
Xb 0.45 0.8 0.98 
Gas Film Controlling 0.4 0.56625 0.7479 
Reaction Controlling 0.996 1.09 1.004 
Ash Film Controlling 2.09 1.211 0.894 
From the above table it implies that the rate controlling mechanism for the transformation of 
solid is Reaction control. 
 
Q 25.12 A large stockpile of coal is burning. Every part of its surface is in flames. In a 24-hr 
period the linear size of the pile, as measured by its silhouette against the horizon, seems to 
decrease by about 5%. 
(a) How should the burning mass decrease in size? 
(b) When should the fire burn itself out? 
(c) State the assumptions on which your estimation is based. 
A 25.12 
(a). As it is written that surface is in flames, the burning mass should decrease in size by 
Shrinking Core Model (SCM) 
0 
0.2 
0.4 
0.6 
0.8 
1 
0 0.2 0.4 0.6 0.8 1 
R
.H
.S
. o
f 
e
q
u
at
io
n
s 
time (hr) 
Gas Film Controlling 
Reaction Controlling 
Ash Film Controlling 
(b). The coal fire will burn out when entire coal will be covered by the ash layer and it will no 
longer be able to diffuse oxygen through the ash layer and the fire should burn itself out. 
(c). The outer surface of the solid particle and the deeper layers do not take part in the reaction 
until all the outer layer has transformed into solid or gaseous product. The reaction zone then 
moves inward(into solid), constantly reducing the size of core of unreacted solid and leaving 
behind completely converted solid(solid product) and inert material (inert constituent of the solid 
reactant). And we refer to converted solid and inert material as ash. At any time the solid 
comprises of a core surrounded by a envelope. The envelope consist of a solid product and inert 
material. Ash layer may contribute to resistance. 
CHAPTER 26 
FLUID-PARTICLE REACTORS: DESIGN 
 
SOLUTION OF UNSOLVED EXAMPLES 
 
PREPARED BY 
CHIRAG MANGAROLA (13BCH032) 
MAULIK SHAH (13BCH033) 
MEHUL MENTA (13BCH034) 
DHRUV MEWADA (13BCH035) 
26.1 
Solution 
Given: Shrinking core model with ash diffusion as controlling mechanism 
Conversion = 80% 
To find conversion when the size of reactor is doubled 
Feed rate, flow pattern, gas environment are considered constant 
For plug flow, SCM, Ash diffusion 
𝑡𝑝
𝜏
= 1 − 3 1 − 𝑋𝐵 
2
3 + 2 1 − 𝑋𝐵 
 = 1 − 3 1 − 0.8 
2
3 + 2 1− 0.8 
 = 0.374 
On doubling the size of reactor 𝑡𝑝
′ = 2𝑡𝑝 
2𝑡𝑝
𝜏
= 1 − 3 1 − 𝑋𝐵
′ 
2
3 + 2 1 − 𝑋𝐵
′ 
So, 𝑋𝐵
′ = 96.5% 
 
26.2 
Solution 
For MFR same gas environment and same feed rate so no change in conversion. 
 
 
 
26.3 
Solution 
Reactor type: tubular 
Hard solid product (SCM/Reaction control) 
𝜏 = 4hr for 4mm particles 
To find: residence time for 100% conversion of solids 
For plug flow SCM/reaction control 
𝜏 = 
𝜌𝐵 𝑅
𝑘′′ 𝑏 𝐶𝐴𝑔 
 
This shows that 𝜏 α R 
𝜏2𝑚𝑚 = 2𝑕𝑜𝑢𝑟𝑠 
𝜏1𝑚𝑚 = 2𝑕𝑜𝑢𝑟𝑠 
For 100% conversion residence time should be more than 4hours. 
 
 
26.4 
Solution 
We need to find conversion of 1 mm, 2mm & 4mm particle sizes in 15 min of time. 
 
 For 4mm particle 
 
 
15
4 ∗ 60
= 1 − 1 − 𝑥𝑏 
1
3 
 
 1 − 𝑥𝑏 = 0.824 
𝑥𝑏 = 0.176 
 
 For 2 mm particle 
 
15
2 ∗ 60
= 1 − 1 − 𝑥𝑏 
1
3 
 
 1 − 𝑥𝑏 = 0.669 
𝑥𝑏 = 0.331 
 
 For 1mm particle 
 
15
1 ∗ 60
= 1 − 1 − 𝑥𝑏 
1
3 
 
 1 − 𝑥𝑏 = 0.423 
 
𝑥𝑏 = 0.577 
 
 Over all conversion 
 
 1 − 𝑥𝑏 = 0.5 ∗ 0.176 + 0.3 ∗ 0.331 + 0.2 ∗ 0.577 
𝑥𝑏 = 0.6973 
 
 
Solution 
Given: SCM with reaction control 
Conversion 60% 
To find: conversion when reactor is made twice as large keeping amount of solids and gas 
environment constant 
W1=W2 (Amount of solids) 
Concentration CA01=CA02 
𝑊
𝐹𝐴0
= 
𝑡 
𝐶𝐴0
 
Here same gas environment so FA01=FA02 
Therefore, 𝑡 1 = 𝑡 2 
Conversion is same 
Size of reactor does not play role but the weight of solids passed play a role in changing 
conversion in this case. 
 
 
𝑡 =
𝑤
𝐹0
= 100𝑠𝑒𝑐 
 
1 − 𝑥𝑏 =
1
4
∗
𝜏
𝑡 
−
1
20
 
𝜏
𝑡 
 
2
+
1
120
∗ 
𝜏
𝑡 
 
3
 
 
𝜏
𝑡 
= 1.246 
 
𝜏 = 124.65 
 
1 − 𝑥𝑏 = 0.02 
0.02 =
1
4
∗
𝜏
𝑡 
−
1
20
 
𝜏
𝑡 
 
2
+
1
120
∗ 
𝜏
𝑡 
 
3
 
 
𝑡 =
124.6
0.081
= 1538.27 
 
𝑡 › 𝜏 because of presence of non idealities. 
 
𝑤 = 𝑡 ∗ 𝐹0 
 
𝑤 = 15382.7 𝑔 
 
Information from Example 26.3 Modification 
 30% of 50μm radius particles (τ=5min) 
 40% of 100μm radius particles (τ=10min) 
 30% of 200μm radius particles (τ=20min) 
 Reactor type: Fluidized bed reactor (MFR) 
 Unchanging particle (SCM) with reaction control 
 Feed rate = 1kg solids/min 
 Solids in bed = 10kg 
 Mean residence time = 10min 
 Ash film control 
 τ(R=100μm) =10min 
 
 
Solution 
For ash controlling mechanism τ α R2 
So, τ(R=50μm) = 
τ R=100μm ∗(502)
1002
 = 
10∗(502)
1002
 = 2.5min 
And, τ(R=200μm) = 
τ R=100μm ∗(2002)
1002
 = 
10∗(2002)
1002
 = 40min 
For ash diffusion 
1-𝑋 𝐵 = 
1
5
τ(Ri )
𝑡 
− 
19
420
 
τ(Ri )
𝑡 
 
2
 
𝐹(𝑅𝑖)
𝐹
 
 = 0.3 
1
5
∗ 
2.5
10
 − 
19
420
∗ 
2.5
10
 
2
 + 0.4 
1
5
∗ 
10
10
 − 
19
420
∗ 
10
10
 
2
 + 
 0.3 
1
5
∗ 
40
10
 − 
19
420
∗ 
40
10
 
2
 
 = 0.0989 
Conversion = 0.901 = 90.1% 
 
 
 
 
 
Information from Example 26.4 Modification 
 A(gas) + B(solid)  R(gas) + S(solid) 
 τ = 1hr (with environment of CA0 gas concentration) 
 Gas and solid both in mixed flow 
 CA0 = CB0 
 𝑋 𝐵 = 0.9 
 FB0 = 1 ton/hr 
 Stoichiometric feed rate of A and B 
 Twice gas to solid 
Stoichiometric ratio 
 
 
 
 
 
 
Solution 
Material balance 
2(CA0 – CA) FA0 = (CB0 – CB) FB0 
2(CA0 – CA) = 0.9 CA0 
Therefore, CA = 0.55 CA0 
The solid will be in the environment of the gas of concentration 0.55CA0 
τ α 
1
𝐶𝐴0
 from equation τ = 
𝜌𝐵 𝑅
𝑏 𝑘 𝐶𝐴0
 
So τ = 
1
0.55
 = 1.818h 
For reaction control; 
 1 − 𝑋 𝐵 = 
1
4
τ
𝑡 
− 
1
20
 
τ
𝑡 
 
2
= 0.1 
On solving this equation 
τ
𝑡 
= 0.435 
 𝑡 = 
τ
0.435
= 
1.818
0.435
= 4.18 𝑕𝑟 
W = 𝑡 FB0 = 4.18 tons 
Weight of bed required is 4.18tons. 
 
 
 
 
 
 
 
 
Information from Example 26.4 Modification 
 A(gas) + B(solid)  R(gas) + S(solid) 
 τ = 1hr (with environment of CA0 gas concentration) 
 Gas and solid both in mixed flow 
 CA0 = CB0 
 𝑋 𝐵 = 0.9 
 FB0 = 1 ton/hr 
 Stoichiometric feed rate of A and B 
 Gas is in plug flow 
 
 
 
Solution 
CAf = 0.1CA0 
Taking logarithmic mean for the gas 
𝐶𝐴 = 
𝐶𝐴0−𝐶𝐴𝑓
𝑙𝑛
𝐶𝐴0
𝐶𝐴𝑓
 = 
𝐶𝐴0−0.1𝐶𝐴0
𝑙𝑛
𝐶𝐴0
0.1𝐶𝐴0
= 0.39𝐶𝐴0 
The solid will be in the environment of the gas of concentration 0.55CA0 
τ α 
1
𝐶𝐴0
 from equation τ = 
𝜌𝐵 𝑅
𝑏 𝑘 𝐶𝐴0
 
So τ = 
1
0.39
 = 2.564h 
For reaction control; 
 1 − 𝑋 𝐵 = 
1
4
τ
𝑡 
− 
1
20
 
τ
𝑡 
 
2
= 0.1 
On solving this equation 
τ
𝑡 
= 0.435 
 𝑡 = 
τ
0.435
= 
2.564
0.435
= 5.89 𝑕𝑟 
W = 𝑡 FB0 = 5.89 tons 
Weight of bed required is 5.89tons. 
 
 
Reaction step is rate controlling (SCM) 
Assuming that Fibers are cylinder in shape with same particle size and the useful product is 
solid (unchanging particle size) 
Solution 
1-XB = 1 −
t
τ
 
2
 
1-𝑋 𝐵 = 1 − 𝑋𝐵 𝐸 𝑑𝑡
𝜏
0
 
 = 1 −
t
τ
 
2
𝑒
−𝑡
𝑡 𝑑𝑡
𝜏
0
 
 = 1 −
2𝑡
𝜏
+ 
𝑡
𝜏
 
2
 𝑒
−𝑡
𝑡 𝑑𝑡
𝜏
0
 
 = 1 + 
𝑡
𝜏
 
2
−
2𝑡𝑡 
𝜏2
− 
2𝑡 2
𝜏2
+ 
2𝑡
𝜏
+ 
𝑡 
𝜏
 𝑒
−𝑡
𝑡 
0
𝜏
 
 = −𝑒−𝑇 + 𝑒−𝑇 −
2𝑒−𝑇
𝑇
−
2𝑒−𝑇
𝑇2
+ 2𝑒−𝑇 + 
𝑒−𝑇
𝑇
+ 1 −
1
𝑇
−
2
𝑇2
 [where T = 
𝜏
𝑡 
 ] 
 1-𝑋 𝐵= −
𝑒−𝑇
𝑇
+
2(1−𝑒−𝑇)
𝑇2
+ 2𝑒−𝑇 + 1 −
1
𝑇
 
 
 
 
 
 
 
Solution 
26.11 23.12 
For SCM/Reaction Control, 
t/τ=1-(1-XB)
1/3 
so from above equation τ=1 hr(60min) 
1-𝑋B= (1 − 𝑋
𝜏
0 B
)Edt 
1-𝑋B= (1 − 𝑡/𝜏)
𝜏
0
3
*(60/90)dt 
=(2/3)* (1 − 𝑡)
1
0
3
*dt 
=(2/3)*(1/4) 
 =0.167 
𝑋B=83.3% 
For SCM/Reaction Control, 
t/τ=1-(1-XB)
1/3 
so from above equation τ=1 hr(60min) 
1-𝑋B= (1 − 𝑋
𝜏
0 B
)Edt 
1-𝑋B= (1 − 𝑡/𝜏)
0.75
0.25
3
*(60/30)dt 
 =2* (1 − 𝑡)
0.75
0.25
3
*dt 
 =2*0.078 
 =0.156 
𝑋B=0.8437 
26.13 26.14 
For SCM/Reaction Control, 
t/τ=1-(1-XB)
1/3 
so from above equation τ=1 hr(60min) 
1-𝑋B= (1 − 𝑋
𝜏
0 B
)Edt 
1-𝑋B= (1 − 𝑡/𝜏)
1
0.5
3
*(60/60)*dt 
 = (1 − 𝑡)
1
0.5
3
*dt 
 =0.0156 
𝑋B=0.9843 
For SCM/Reaction Control, 
t/τ=1-(1-XB)
1/3 
so from above equation τ=1 hr(60min) 
1-𝑋B = (1 − 𝑋
𝜏
0 B
)Edt 
1-𝑋B = (1 − 𝑡/𝜏)
𝜏
0
*1*dt 
 =1∗ (1 − 𝑡)
0.75
0
*dt 
 =1*0.4687 
 =0.4687 
𝑋B=0.5313 
 
 
 
 
 
 
 
 
 
Compiled by 
Manish Makwana 
makwana.manish357@outlook.com 
	Cover page
	PREFACE
	Credits
	Index
	2.1 to 2.15
	2.16 to 2.23
	3.1 to 3.11
	3.12 to 3.21
	4.1 to 4.7
	5.1 to 5.10
	5.11 to 5.30
	7.1 to 7.11
	7.12 to 7.19
	7.20 to 7.28
	8.1 to 8.10
	8.11 to 8.21
	9.1 to 9.11
	9.12
	10.1 to 10.10
	10.11 to 10.20
	11.1 to 11.16
	12.1 to 12.12
	13.1 to 13.7
	13.9 to 13.13
	14.1 to 14.11
	18.1 to 18.10
	18.11 to 18.20
	18.21 to 18.30
	18.31 to 18.40
	13BCH025,027,028,030
	Solution_cre2assignment
	CRE term paper
	New Doc
	23.1 to 23.9
	24.1 to 24.16
	25.1 to 25.12
	26.1 to 26.10
	Compiled by