GATE PSU Study Material Chemical Reaction Engineering

Chemical Engineering (GATE & PSUs)

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ChemicalReaction Engineering

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AppearedChemicalEngineering

27.52 18.34 24.77 15874GATE Syllabus: Chemical Engineering

Chemical Reaction Engineering: Theories of reaction rates; kinetics of homogeneous reactions,interpretation of kinetic data, single and multiple reactions in ideal reactors, non-ideal reactors; residencetime distribution, single parameter model; non-isothermal reactors; kinetics of heterogeneous catalyticreactions; diffusion effects in catalysis.

CONTENT

1. INTRODUCTION AND BASIC CONCEPT . 03-9

2. KINETICS OF REACTION . 10-17

3. DESIGN EQUATION FOR IDEAL REACTORS 18-31

4. INTERPRETATION OF BATCH REACTOR DATA . 32-44

5. DESIGN FOR SINGLE REACTION 45-66

6. DESIGN FOR MULTIPLE REACTIONS 67-82

7. HETEROGENEOUS REACTION 83-87

8. TEMPERATURE AND PRESSURE EFFECTS 88-97

9. BASIC ASPECTS OF NON IDEAL FLOW. 98-108

10. QUESTION BASED ON GATE PAPER .. 109-120

11. MODEL TEST PAPER-I WITH SOLUTION 121-133

12. PRACTICE SET-1, 2, 3 134-171

13. Interview Practice Bank 172-182



CHAPTER-1

INTRODUCTION AND BASIC CONCEPT

1. Every industrial chemical process is designed to produce economically a designed product from a varietyof starting material through a succession of treatment steps.

Figure: Show typical chemical process:

2. Raw material undergoes a number of physical steps to put them in the form in which they can be reactedchemically. Then material passes through the reactor. The products of reaction must then undergo furthertreatment.

3. In chemical reaction engineering subject we are concerned with chemical treatment step of a process. CREcombines the study of chemical kinetics (reaction rates and reaction mechanisms) with the reactors in which thereaction occurs.

A chemical reaction takes place by decomposition, combination or change in configuration of molecules.

4. We define rate of reaction in terms of component. If the rate of change in number of moles of this

component due to reaction is Nd idt

then

1 N Moles of formedV (Volume of fluid) (Time)i

d i ir

dt

i.e., the number of moles of component i formed per unit time per unit volume (mol/dm3s). If species i is areactant, the numerical value of ri is a negative number (e.g. ri = -3mol/dm3s or ri = 3mol/dm3s)If species i is a product, then ri= will be a positive number.

1 ii

dNr

V dt

For constant volume system

( )i ii iN dCd d

r Cdt V dt dt



Ci is the concentration of species i in reaction mixture.

Q.1. For a homogeneous reaction system, where Ci is the concentration of i at time t, Ni is the no. of moles ofi at time t, V is the reaction volume at time t and t is the reaction timeThe rate of reaction for species i is defined as

(a.) idCdt

(b.) idCdt

(c.)1 idNV dt

(d.) 1 idNV dt

Ans. (c)5. (i) Chemical reaction may be classified depending upon no. of phases involved as homogeneous andheterogeneous reaction. Homogeneous reaction is one which takes place in only one phase. A heterogeneousreaction is one which involves presence of more than one phases.(ii) Reaction may be classified according to catalytic and non catalytic reaction.(iii) Reaction may be classified on the basis of molecularity of reaction.(iv) The reaction may be classified based upon the heating effect as exothermic and endothermic reaction.(v) Reaction may be classified as reversible and irreversible reaction.(vi) Reaction may be classified based upon the order of reaction.

6. Relative rates of reaction:For the reaction:aA bB cC dD The relation between the rate of participants A, B, C, D in chemical reaction with their stoichiometric coefficientis

CA B Drr r r

a b c d

Also, if we compare CA rra c

A ca

r rc

Rate of disappearance of A = ac

(Rate of formation of C)

Q.2.For an elementary reaction

2 3kA B C (a.) Rate of appearance of C is equal to rate of disappearance of A(b.) Rate of disappearance of A is equal to rate of disappearance of B.(c.) Rate of appearance of C is equal to rate of disappearance of B.(d.) Rate of appearance of C is 3 times rate of disappearance of A.

Solution:Relative Ratesof reaction1 2 3

CA B rr r

Ar = rate of disappearance of A



Br = rate of disappearance of BrC = Rate of appearance of C

3( )C Ar r Option (D) is correct

7. Many variable may affect the rate of a chemical reaction. In homogeneous system the temperature,pressure and composition affect the rate of reaction.In heterogeneous system factors that affect the rate of reaction are rate of mass transfer i.e., heat transfernature of catalyst if present.Heterogeneous reactions involve more than one phase.

Rate of reaction is expressed in measure other than volume such as reaction surface area or catalyst weight.'

ir Number of moles of i formed per unit time per unit mass of catalyst (mol/s.g catalyst08. Rate lawAn algebraic equation that relates the reaction rate and concentration

A PRODUCTS

-r ( ) ( , )A n A B

A A A B

A Bk T f C C

or

r k C C

kA Rate constant/ specific reaction rateOrder of reaction (n) = + For the reaction where all the stoichiometric coefficients are 1 e.g.

A + B C + DA B C Dk k k k k

Imp. Rate Laws Are determined by experimental observation.

Q.3.Rate expression of reaction2 30.02 / minA Ar C mol cm

If concentration is to be expressed in mol/l and time in hours, the value of rate constant would be(a.) 0.12hr-1 (mol/L)-1 (b.) 0.012hr-1(mol/L)-1(c.) 12hr-1(mol/L)-1 (d.) 1.2hr-1(mol/L)-1

Solution: 1L = 10cm3

1hr = 60min



22

2

2 1 1

0.021 1

10 60

0.02 600

12

A A

A A

A A

molr C

hr

molr C

L hrr C mol L hr

1 1

2 2 2

1 1

1 1

12

12 ( / )

A

A

Clearly kr molL hr

unity of kC mol L

unit of k mol L hrk hr mol L

Option C is correct

Q.4.The rate of reaction for a gas phase reaction is expressed as22.56 /A A

dp P atm hrdt

The Units of rate constant are

(a.) 1 1( )hr atm (b.) 1 3( )hr atm (c.) 1hr atm (d.) 1 1( / )hr mol L

Solution: 1 12 2/ /A

A

dp dt atm hrk atm hrp atm

TYPES OF REACTIONS9. (i) A homogeneous reaction is one which takes place in only one phase; i.e., all the reacting material andproducts and catalyst (if any) will be present in a similar phases.

Example: 2 21NO( ) O ( ) NO ( )2

g g g A hetrogeneous reaction is one which involves the presence of more than one phase.

Example: 2 5V O ( )2 2 31SO ( ) O ( ) SO ( )2

sg g g (ii) Catalytic reaction are those reaction which involves use of catalyst to enhance the rate of reaction.

Example: Ni2 4 2 2 6HeatEthylene EthaneC H H C H

Non catalytic reactions are those reaction which does not involve use of catalyst.

Example: 2 21NO O NO2

(iii) The reaction may be classified based upon the molecularity of a reaction i.e., based upon the number ofmolecules that takes part in the reaction as unimolecular, bimolecular and trimolecular reaction.



(a) Unimolecular Reaction:

Example:2 2 Decomposition

2 42 2

H C CH| | 2C HH C CH

(b) Bimolecular Reaction:Decomposition

2 22HI( ) I ( ) H ( )g g g (c) Trimolecular Reaction:

Oxidation2 22NO O 2NO

(iv) The reaction may be classified as exothermic and endothermic reaction.Exothermic reaction is one in which heat is evolved.

Example: Cu2 3CO 2H CH OH Heat Endothermic reaction is one in which heat is absorbed

2 3Al O2 5 2 4 2Dehydration

ethyl alcoholC H OH C H H O Heat

(v) The reaction may be classified as reversible and irreversible reaction. Reversible reaction are those inwhich forward and reverse reaction takes place simultaneously.

2 5 3 3 2 5 2C H OH CH COOH CH COOC H H OH

Irreversible reaction are those which can proceed only in one direction.

6 6 3 6 5 2 2C H HNO C H NO H O (vi) The reaction may be classified as first order, second order and third order reaction.

First order reaction:

Example: Decomposition2 5 2 21N O NO + O2

Second order reaction: (Saponification of ester)Example: 3 2 5 3 2 5CH COOC H NaOH CH COONa C H OH Third order reaction:

Example: 2 2 22 NO H N O H O Single and Multiple Reaction:(i) When a single stoichiometric equation and single rate equation are chosen to represent the progress of thereaction, we have a single reaction. When more than one stoichiometric equation is chosen to observedchanges then more than one kinetic expression is needed to follow the changes in the concentration of all thereaction component and we have multiple reaction.



(ii) Multiple reaction may be classified asSeries Reaction: A B S

Parallel Reaction:and

A BC S

Side by side

Complicated Reaction:A B C C B D

Here reaction proceeds in parallel with respect to B but in series with respect to A, C and D.

Elementary and Non Elementary Reaction:Consider a single reaction with stoichiometric equation A B C According to rate law,Rate of disappearance of A is given by

A A BK C Cr Such reaction in which the rate equation corresponds to a stoichiometric equation are called elementaryreaction.When there is no direct corresponding rate and stoichiometry, then the reaction is called non-elementaryreaction.The best example of a non-elementary reaction is that between hydrogen and bromine.

2 2H Br 2HBr Which has rate expression

12

1 2 2HBr

22

K [H ][Br ][HBr]K + [Br ]

r

10. Molecularity and Order of ReactionThe molecularity of an elementary reaction is the number of molecules involved in the reaction and this hasbeen found to have the values of one, two or occasionally three.Molecularity refers only to an elementary reaction.Consider a reaction

A B C D Pa b c d then rate expression

A A B C DK C C C Ca b c d

r a + b + c + d = n

A

S

B

CompetitiveType



where a, b, .. d are not necessarily related to the stoichiometric coefficient.We call the power to which the concentration are raised the order of reaction. Hence the reaction is

ath order with respect to Abth order with respect to Bnth order overall

Since the order refers to the empirically found rate expression, it may not to be an integer and can have afractional value.But, the molecularity of a reaction must be an integer because it refers to mechanism of reaction and appliesonly to elementary reaction.

11. Rate constant (K)The rate constant of a chemical reaction is a measure of the rate of reaction when all the reactants are at unitconcentration.So,for any reaction say A B

A AK Cr If we assume that the concentration are equal to unity then

A Kr Where K is rate constant. The rate constant K depends on the temperature. It increasees with temperature. Kis also called specific rate constant.The units of K depends upon the units of time and concentration.

Suppose : A AK Cn

r

Then we know units of A( ) mol/l. secr

and AmolClit .

1

1

molsec.

mollit .K (sec)lit .mol

lit .

n

n

1 1K (time) (concentration) n Where n is the order of reaction.

Q.5.For a gas reaction at T(K), the rate is given by2

' /A Adp k p atm hrdt

If the rate equation is expressed as



21 ( / )AA AdN

r kC mol L hrV dt

The rate constant k is given by(a.) k = krT (b.) k = k (c.) k = k (RT)2 (d.) k = k/RT

Solution:

2

2 2

2 2

'

( ) ' ( )

' ( )

A AA A

A A

AA

dp pk p cdt RTd

c RT k c RTdt

dcRT k c RTdt

2

2

2 2

' ( )

' ( )' ( )

'( )

AA

A A

A A

dc k c RTdt

r k c RT

kc k c RTk k RT

Option (A) is correct

12. For reversible reactions: FORWARDBACKWARD

A B C

Equilibrium constant (kC) = AA

kk

kA Rate constant of forward reactionk

-A Rate constant of backward reaction

During the course of a chemical reaction, the rate of a reaction decreases as the reaction proceeds

Q.6.The irreversible reaction is the special case of reversible reaction ifConcentration of reactant at equilibrium condition is ___________Solution: Zero

Conceptual Numerical Answer Practice Set

:::::::::::::::::::::::::SAMPLE PART::::::::::::::::::::::::::::::



CHAPTER-2

KINETICS OF REACTION

1. In homogeneous system, the possible factors that affect the rate of reaction are temperature, compositionand pressure.Hence for homogeneous system we can write rate of reaction as a function of concentration and temperature.

Amol[Temperature dependent term, Concentration dependent term]l. s

r f

2. According to rate law rate of reaction is directly proportional to the reactant concentration. Rate law is alsocalled rate equation, kinetic expression.

3. Consider a irreversible reactionA B ....... R S ..........a b r s

According to rate law we can write

A A B A BK (C , C ) K C C ..........a br f Where K is rate constant a and b are the power of concentration of reactant A, B .. on which rate of reactiondepends.4. Kinetics Model for Non-Elementary Reactions:To explain the kinetics of non elementary reaction we assume that a sequence of elementary reaction isactually occurring but we can not observe the intermediate formed because they are only present in veryminute quantity. Hence we observe only initial reactant and final products that appears to be a single reaction.

For example: 2 2A D 2AD If the kinetic of reaction shows that reaction is non-elementary then we can postulate a series of elementarysteps.

2A 2A*

2A* D AD D*

A* D* AD where * sign shows unobserved intermediate.

Types of Intermediate:(i) Free radicals:Free atoms that contain one or more unpaired electrons are called free radicals.



Example: 3 2 5 3CH , C H , I , H , CCl

They are highly reactive and unstable.(ii) Ions and Polar Substance: Electrically charged atoms, molecules or fragments of molecules are calledions.

Example: 3N , N , OH , Ia

(iii) Molecules: Consider a consecutive reactionA B C

If the intermediate B is highly reactive its mean life time will be very small and its concentration in thereacting mixture can become too small to measure.In this condition B may not be observed and can be considered to be a reactive intermediate.

(iv) Transition Complex: Unstable forms of molecule or unstable association of molecules which can eitherdecompose to give product or by further collision return to molecules in the normal state. Such unstableforms are called transition complex.Reaction scheme involving these four kinds of intermediate can be of two types:(a) Non chain reaction:

Reactants (Intermediates)*(intermediates)* Products

(b) Chain Reaction:Reactant (Intermediate)* [Initiation](Intermediate)* Reactant (Intermediate)* Product [Propagation](Intermediate)* Product [Termination]

Testing Kinetics Models:Now we will see that how to test correspondence between experiment and a proposed mechanism.In matching the predicted rate expression with the experimentally found, we rely on the following rule:(i) If any component: Takes part in more than one reaction then its net rate of change is equal to sum of therates of change of components i in each of the elementary reaction in which it is participated.

, neti ir r = All elementary reaction participated.

(ii) As the intermediates are highly reactive they have very short life time and they are present in very smallquantities, hence their net rate of formation is taken as zero.

*A 0r This is called the steady state approximation.

5. Temperature Dependent Term of a Rate Equation



For many reactions and particularly elementary reactions, the rate expression can be written as a product of atemperature dependent term and a concentration dependent term or

1 2(temperature) (concentration)ir f f

2K (concentration)ir fFor such reactions the temperature dependent term, the reaction rate constant has been found in practically allcases.

(a) Temperature dependency from Arrhenius law:From Arrhenius law rate constant K is given by:

ERTK Ko e

Where Ko is called the frequency or pre-exponential factor.

E is called activation energy of the reaction.R = gas constantT = absolute temperature, KAt the same concentration, but at two different temperature, Arrhenius law indicate that

2 2

1 1 1 2

K E 1 1ln lnK R T T

r

r

1

2 1 2

1 1ln t Et R T T

Unit of Ko = unit of KKo is not a function of temperatureKo (also known as frequency factor) does not affect the temperature sensitivity of reaction

Also 21 1 2( ) 1 1ln xRHK T

K T R T T

Here K(T2) and K (T1) are equilibrium constant at T1 & T2 respectively & HRX is the heat of reaction

Also 2(ln )d EK

dT RT

If temperature is low change in K w.r.t T is high and vice versa.Q.1.From Arrhenius law, a plot of ln k versus 1/T gives a straight line with a slope of (-E/R). The units of E/R

are

(a.) K/cal (b.) cal/K (c.) cal (d.) K

Solution:1

1 1E cal mol KR cal mol K

Option (d) is correct



Q.2.From the chemical reaction kA B an increase in temperature from 295K to 305K causes the rate ofreaction to double. The energy of activation for this reaction must be(a.) 12389Kcal/mol (b.) 12.38Kcal/mol(c.) 1.2Kcal/mol (d.) 1238Kcal/mol

Solution:1 1

2

1 2 2 1

1

1

1.9871 1ln2

2 1 1ln1.987 295 305

r R cal mol KEr R T T r r

r Er

On solving E = 12389.47 cal/molOr E = 12.38Kcal/molOption (b) is correct

Figure: Sketch showing temperature dependency of the reaction rate.Activation Energy and Temperature Dependency:The temperature dependency of reaction is determined by the activation energy and temperature level of thereaction as shown in figure. Conclusion from figure are as follows:

(i) From the Arrhenius law, the value of the frequency factor Ko does not affect the temperature sensitivity.

(ii) Any given reaction is much more temperature sensitive at a low temperature than at a high temperature.(iii) Reactions with high activation energy are very temperature sensitive, reaction with low activation energyare relatively temperature insensitive.

(iv) From Arrhenius law a plot of ln K vs 1T

given a straight line with large slope for large E and small slope

for small E.



Activation Energy:The excess energy of the reactants required to dissociate into products.

In Exothermic reaction, reactants have more energy than productsReaction

Reaction

Energy of product - Energy of Reactantve (for exothermic reactions)

HH

(b) Temperature Dependency from the Collision Theory:Collision theory is based upon the concept that before molecule react, there must be collision between thereactant molecules and only those collision in which the colliding molecules have a certain minimum amountof energy are effective. Further this concept leads to the rate expression based upon the frequency ofmolecular collisions and the fraction collisions that are effective.From collision theory rate constant K is given by

1 E2 RTK T e

2 21 1 2 1

1 1 1ln ln2

k TEk R T T T

(c) Temperature Dependency from Transition State Theory:The fundamental postulate of the theory is that (i) The reacting molecules must form unstable intermediatecalled activated complex before being converted (i.e., there is formation of activated complex) to product. (ii)There exists an equilibrium between the activated complex and reactants at all the time.According to transition state theory rate constant is given by:

ERTK T e

or 2 21 1 2 1

1 1ln lnk TEk R T T T

This equation describe the temperature dependency from Transition State Theory.

6. Difference between Collision Theory and Transition State Theory:



Collision Theory Transition State Theory1. Collision theory does not agree more closely withexperiments.

1. Transition state theory agrees more closely withexperiments.

2. Collision theory is based on Kinetic theory ofgases.

2. Transition state theory is based on statisticalmechanics.

3. Collision theory view that decomposition ofactivated complex is very rapid.

3. Transition state theory view that formation ofactivated complex is very rapid.

4. Collision theory view that formation of activatedcomplex is slow and rate controlling.

4. Transition state theory view that decompositionof activated complex is slow and rate controlling.

5.1 E2 RTK T e

5.E

RTK T e

SOLVED EXAMPLE

1. The decomposition of 2 5N O is postulated to occur by the following mechanism.

1

2

K *2 5 2 3K

N O NO NO3K*

3 2NO NO* O 4K*

3 2NO* NO 2 NO

using steady state approximation derive an expression for rate of decomposition of 2 5N O .

Solution: We know thatRate of decomposition is given by

*2 51 2 5 2 2 3

[N O ] K [N O ] K [NO ][NO ]ddt

(a)By steady state hypothesis.

* *

3 3 4 3[NO*] K [NO ] K [NO*][NO ] 0d

dt (i)

*

* * *31 2 5 2 2 3 3 3 4 3

[NO ] K [N O ] K [NO ][NO ] K [NO ] K [NO*][NO ]ddt

(ii)From equation (i)

*

3 3 3*

4 3 4

K [NO ] K[NO*]K [NO ] K (iii)

Put equation (iii) in equation (ii)* * *

1 2 5 2 2 3 3 3 3 3K [N O ] K [NO ][NO ] K [NO ] K [NO ]



* 1 2 532 2 3

K [N O ][NO ]K [NO ] 2K (iv)

Put equation (iv) in equation (a)We get,

2 5 1 2 51 2 5 2 2

2 2 3

[N O ] K [N O ]K [N O ] K [NO ]K [NO ] 2K

ddt

2 21 2 5

2 2 3

K [NO ]K [N O ] 1K [NO ] 2K

2. At 500 K the rate of a bimolecular reaction is ten times the rate at 400 K. Find the activation energy for thisreaction.(a) From Arrhenius law (b) From Collision theorySolution: (a) We have from Arrhenius law

Eln( ) ln KRT o

r

2

1 2 1

E 1 1lnR T T

r

r

2

1 1 2

E 1 1lnR T T

r

r

We have, 2 110r r

1T 400 K

2T 500 KR = 1.987 cal/mol K

1

1

10 E 1 1ln1.987 400 500

r

r

E = 9150 cal/mol K

(b) From Collision theory we know that1 E2 RTK T e

1 E2 RTK Tor e

12E 1ln ln T ln K

R T or



12

1 11

E 1ln ln T ln KR T o

r

12

2 22

E 1ln ln T ln KR T o

r

12

2 21

1 1 2 21

TE 1 1ln lnR T T T

r

r

1T = 400 K, 2T = 500 K

2 110r r

12

11

1 2

10 E 1 1 (500)ln ln1.987 400 500 (400)

r

r

E = 8707 cal/mol Ans.

3. On doubling the concentration reactant rate of reaction triples. Find the reaction order.

Solution: Let A AK Cn

r

Suppose at concentration1A

C rate of reaction is1 1A A

K Cnr

At concentration2A

C rate of reaction is2 2A A

K Cnr

If2 1A A

C 2C we have

2 1A A3 ( )r r

2 2

1 1

A A

A A

K CK C

n

n

r

r

2 2

1 1

A A

A A

CC

n

n

r

r

1

1 1

AA1

A A

(2C )3( )(C )

n

n

r

r

3 2n

ln 3 = n ln 2 1.58 1.6n

4. A certain reaction has a rate given by 2 3A A0.005 C mol/cm minr



If the concentration is expressed in mole/lit and time in hours. What would be the value and units of rateconstants.

Solution: 2 3A A0.005 C , mol/cm minr 23

2A A 3 3

cm mol mol0.005 C ,mol . min cm cm min

r

K =3cm0.005

mol . min

We know that 1 liter = 31000 cm

3 11cm1000

l 60 min = 1h

1 min = 1 h60

K =

33

1 lit(0.005 cm )1000 cm

1 hmol . min

60 min

K = 4 lit3 10mol. h

Value of K = 43 10

Units of K = litmol h

5. The pyrolysis of ethane proceeds with an activation energy of about 75000 cal. How much faster is thedecomposition at 650C than at 500C ?

Solution: 2 6C H PE = 75000 cal/mol

Let 1K be the rate constant at 1T and 2K be the rate constant at 2T

1T 500 C 773K

2T 650 C 923K

2

1 1 2

K E 1 1lnK R T T

R = 1.987 cal/mol K



2

1

K 75000 1 1lnK 1.987 773 923

2

1

Kln 7.951K

2

1

K 2838K

2 1K 2838 KAt 650C, the decomposition is faster by a factor 2838 than at 500C.

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