Rate Law & Stoichiometry Tables - Reactor Engineering Course Block 3

CH3: Rate Law & Stoichiometry

RE3

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Chemical Reaction Engineering Methodology


CH3: Elements of Chemical Reaction EngineeringH. Scott Fogler (4th Edition)

Chemical Reaction Engineering Methodology



Content

• Section 1: Rate Laws

– Introduction to Rate Laws

– Reaction Order

– Reaction Rate Constant k

• Section 2: Stoichiometry

– Batch System

– Continuous Flow Systems

• Liquid Phase (constant volume)

• Gas Phase (change in volume)


Section 1

Rate Laws


Solve this problem!

• Reaction data:

AB

k = 0.23 min-1

• Size it for a PFR and a CSTR @ 80% Conversion.

• It follows an elementary rate of reaction law


Solve this problem!

• Reaction data:

AB

k = 0.23 min-1



Something is “missing”…


Solve this problem!

• Reaction data:

AB

k = 0.23 min-1



Something is “missing”…Where is our Rate of Reaction vs. Conversion Data!?


Solve this problem!


Why study Rate Laws?

• We’ve seen before how to calculate volumes of reactors with our equations

• These equations depend on having a function of rate of reaction vs. conversion

• This is very rare actually

• What happen if you don’t have it?– We can apply theoretical concepts

– Or go to the lab and do experimental data


Why study Rate Laws?

• Once we get this data… we can continue as we done in Chapter 2

• Sizing is now done with the same equations

• We will find out that…

– Rate of reaction depend on concentration

– Concentration depend on flows/Conversion

– Our Mathematical processes will change!


Introduction to Rate Law

• Basic Definitions

– Homogeneous reaction: Only one phase reaction

– Heterogeneous reaction: More than one phase reaction

– Irreversible reaction: Only happens to one direction.

• A B and not B A

– Reversible reaction: May happen on both directions• A B and BA


Relative Rates of Reaction

• From aA+bB cC+dD

• We get for the limiting reactant:

– A + b/a ·B c/a ·C + d/a ·D

• Interesting that…

– If 1 mol of A reacts, then c/a moles of C appears

– If 1 mol of A reacts, then b/a moles of B disappears

• Therefore, there is a relationship between rates of reaction/production


Relative Rates of Reaction

• Actually, the relationship is as follows

• So for 2NO+O2 2NO2

• So if there are 4 gmol/s of NO2 reacting…

rNO/-2 = rO2/-1 = rNO2/+2

rNO/-2 = rNO2/+2 rNO2= (+2/-2)·rNO

rNO2= (-1)·4 = -4 gmol of NO2 per second


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Rate of Reaction Law

• For many reactions may be written

A + B Products

-rA=k·f(CA,CB)

• Where k is a function of Temperature k(T)

• Concentrations of reactants

• No Product Concentration involved

• This is a Rate Law


Power Models

• Many times you can get:

A + B Products

-rA=k·CAaCB

b

• The rate law is based on:

– Concentration of the Species being reacted

– Raised to a certain power


Reaction Order

• Refers to the “powers” being raised in the equation

• Overall Order of Reaction = a+b+c…• Example:

A + B Products

-rA=k-rA=k·CA

-rA=k·CACB

-rA=k·CACB2

Order = 0Order = 1Order = 1+1 =2Order = 1+2 =3

NOTE: k values vary





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Elementary Rate Law

• This is a very special case of the Rate Law of Power Models

• If a reaction such as:

aA + bB Products

• Follows the next law:

-rA=k·CAaCB

b

• Then this is an Elementary Rate Law


Elementary Rate Law

• Examples:

AB

A+2B 3C

½A + B + C ¼ D + 5F

• What would be the Elementary Rate laws?


Elementary Rate Law

• Examples:

AB

A+2B 3C

½A + B + C ¼ D + 5F

-rA=k·CA

-rA=k·CACB2

-rA=k·CA1/2CBCC





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Elementary Rate Law

• Once again… this is done by observation

• You can’t propose an elementary rate law for areaction if you haven’t prove it in the lab!

• You can check in data bases for commonreactions and know if they follow anelementary rate law


Elementary Rate Law

• A reaction that does not follows an Elementary Rate Law is defined as a Non-Elementary Rate Law


Non-Elementary Rate Law

• Many reactions do not follow the Elementary Rate Law

• This is just an overview, it is not typically included in a Reactor Engineering Course

• We discuss this in Reaction Mechanisms CH7


Non-Elementary Rate Law:Equilibrium Reactions

• The rate law of every species 0

• This is due to Equilibrium

• Lets suppose that A B

• But the same quantity of BA

• Probably you are familiar with this equation


Reaction Rate Constant “k”

• We’ve analyzed reaction rates

– They depend on Concentrations

– Some times they have raised powers

– But there is also a constant in the rate law!

• We need to analyze that constant

-rA=k·CAaCB

b



• It is actually not a constant, it depends

– Temperature

– Type of reaction

– Reactants

– Catalyst

• Since we will see Isothermal Reactors, we can, for the moment, take it as a constant



• This is Arrhenius Equation

KA(T): The Rate Constant @ that TemperatureA: Pre-exponential Factor / Frequency Factore: the mathematical operation “exponential”E: Activation energy of that specific reactionR: The ideal gas constantT: Absolute Temperature



• This is Arrhenius Equation

As Temperature increases k increasesThis is due to the Kinetic theory

More temperature more collisionsMore collisions more reactions




• Collision theory…

• Please check your Chemistry notes!


A: Pre-exponential Factor / Frequency Factor

• Its an empirical relationship between Temperature and Rate Coefficient

• Depends on how often molecules collide when all concentrations are 1 mol/L

• Depends on whether the molecules are properly oriented when they collide.


A: Pre-exponential Factor / Frequency Factor

• The units are identical to those of the rate constant and will vary depending on the order of the reaction.

• A is the total number of collisions* per second and

*(leading to a reaction or not)


E: Activation energy of that specific reaction

• The minimum energy that must be input to a chemical system with potential reactants to cause a chemical reaction

• The activation energy of a reaction is usually denoted by Ea and is measured in KJ/mol

• Catalysts are used to lower this energy

– Improves speed (time of reaction)











• What if we get a constant at one temperature…

• It is still possible to relate to another temperature

• We will use this equation


Activation Energy: Exercise




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• Show the values of temperatures for:

– 2·k(25ºC) T1

– 3·K(25ºC) T2

– 10·k(25ºC) T3

• Given k(25ªC) = 0.025 and E/R = 2.53





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Section 2

Stoichiometry


Introduction

• Express Concentration in terms of Conversion

aA + bB cC + dD

&

A + b/a·B c/a·C + d/a·D


The “Problem”

• Now, we may have a rate of reaction as follows:

-rA=k·CACB

• How do we relate CA and CB to terms of Conversion of A?

• We need stoichiometric relationships!


Why Stoichiometry?

• Will help us relate all quantities

– rate of reactions

– concentrations

– flows

– conversions

• In terms of A and not other species

– (B,C,D, etc.)


Stoichiometry for Batch Reactors

• t = 0 will be our “initial” condition

• t = t will be our Final condition, or “any time” condition

aA + bB cC + dD



Stoichiometry for Batch ReactorsAll is based on Species “A”aA + bB cC + dD






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This equation is the actual change in moles due to the reaction



This equation is the actual change in moles due to the reaction

A +2B C

3A + C 2D

A+ ½·B C +2D

A + B C + D

0+1/1 -2/1 -1 = -2

2/3 -1/3 -1 = -2/3

1+2 -1/2 -1 = ½

1+1-1-1 = 0





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• Terms of Concentration





• All is expressed in terms of A

– Initial Concentration of A

– Conversion of A

– Stoichiometric Values based on A ratios



• These Equations are valid only when the Volume is constant

• Examples:– Liquid phase reactions– Isobaric and Isothermal Reactions of gases (no change

in moles)

• These Equations does not apply in– Gaseous phase reactions with change in volume

• Changes in pressure change in volume• Changes in temperature change in volume• Isobaric & Isothermal reactions but change in moles

change in volume




Exercise 3-2

CH3: Elements of Chemical Reaction Engineering

H. Scott Fogler (4th Edition)


Exercise 3-2

• Liquid phase, it is easier (no need to mess with gases!

• Suppose it is a Batch Reactor -> Volume





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Exercise 3-2This table is the “answer”





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Exercise 3-3CH3: Elements of Chemical Reaction

EngineeringH. Scott Fogler (4th Edition)


Exercise 3-3





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Continuous Flow SystemsStoichiometry Table

• Now we analyze continuous flow reactors!



• Same story, now for continuous flow















• Turning those Flows into Concentrations (more useful)



• Defining this concept:





• These Equations are valid only when the Volumetric flow is constant

• Examples:

– Liquid phase reactions

– Isobaric and Isothermal Reactions of gases (no change in moles)



• These Equations does not apply in

–Gaseous phase reactions with change in volume/volumetric flow rates

• Changes in pressure change in volume

• Changes in temperature change in volume

• Isobaric & Isothermal reactions but change in moles change in volume


Gas Reaction: Change in MolesReactor Stoichiometry

• The past equations are not valid when

– There is a change in Volume/Volumetric Flow

– This happen when there is:

• Change in Temperature

• Change in Pressure

• Change in Moles (A+B C)

• We analyze now the Change in Moles



• Change in Pressure (Pressure Drop) is seen in other Chapter

• Change in Temperature (Non-Isothermal Design) is seen in CH8


Stoichiometry for Gas Phase Reactors



• Isothermal Process (Same Temperature operation

• Isobaric Process (No drop or change in Pressure)

• The change in moles is due to the nature of the reaction

– A+B C+D (no change 2 moles and 2 moles)

– A C + D (1 moles vs. 2 moles)

– 1/2A+2B C (2.5 moles vs. 1 mole)



• Batch Reactors

– Batch Reactor

• Continuous Flow Reactors

– CSTR

– PFR

– PBR


Gas Reaction: Change in MolesBatch Reactor Stoichiometry

• We apply:

– Ideal gas law

• PV = nRT

– Z-Compressibility Chart if real gas is needed

• PV = ZnRT

• Our study is based on Real Gas

– if ideal, just assume Z0 and Z = 1



• Initial Condition P0V0=Z0NT0RT0

• Final Condition PV=NTRT

• Divide Final/Initial Conditions

P/P0·V/V0=Z/Z0·NT/NT0·R/R·T/T0

(P/P0)·(V/V0)=(Z/Z0)·(NT/NT0)·(T/T0)



• NT = NT0 + δNA0XA

• Important note on δ

– δ is by definition, the change in moles!

– If δ = 0 … then you have no mole change!

– You can treat this as a “liquid-phase” reaction



• If NT = NT0 + δNA0XA

• (NT/NT0) = (NT0 + δNA0XA)/NT0



• Substituting– (P/P0)·(V/V0)=(Z/Z0)·(NT/NT0)·(T/T0) transforms into:

V = V0·(P0/P)·(T/T0)·(Z/Z0)·(1+εX)

• If ideal V = V0·(P0/P)·(T/T0)·(1+εX)

• If no -ΔP V = V0·(T/T0)·(1+εX)

• If Isothermal Design:

V = V0·(1+εX)



• Just be sure that this correction of volume is when you have MOLE CHANGE!

• NOTE Batch reactors don’t change in volume (they are fixed vessels)


Gas Reaction: Change in MolesContinuous Flow Reactor Stoichiometry

• The same applies for continuous flow reactors

• If there is any change in moles due to

– Temperature

– Pressure

– The nature of the reaction itself

– Change in volumetric flow rates

• You will not be able to apply those equations



Volumetric Flow in terms of Molar Flow

Flow is Conversion dependent!





If Xa = 0%, 25% or 80% … the flows of Fa, Fb, Fc, Fd CHANGE

therefore Ft changes too!

FT is Conversion dependent!





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• Lets turn FT to terms of Conversion

FT in terms of Xa

Volumetric Flow in terms of Molar Flow

Volumetric Flow in terms of Conversion







Volumetric Flow in terms of Molar Flows

We want everything dependent of Conversion of A!



We need Flow Rate of j in terms of Conversion… we already have that! Go back on tables to see it!



Flow Rate in terms of Xa



• The “nu” Concept

• If it is a Reactant Negative sign for reacting

• If it is a Product Positive sign for being produced








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Master Equation! Everything in terms of

Conversion of A!


Volumetric Flow in terms of Xa



If ideal, isothermal, no pressure changes, and Initial Concentration


Visual Summary







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Visual SummarySet the Equation in terms of A CH3: Elements of Chemical

Reaction EngineeringH. Scott Fogler (4th Edition)





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Visual Summary







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Visual Summary







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Exercises of Stoichiometry Tables for Continuous Flow Systems








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• CH3 – Rate Law & Stoichiometry


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End of Block RE3

• We’re done studying rate laws and its stoichiometry

• Rate Laws are important to solve our problems, since our design equations need them

• Now you know that you don’t need experimental data, you could suppose a rate law (of course knowing it is suitable)


End of Block RE3

• Once with a rate law, it’s easy to size the reactor

• You know how to calculate the Order of a Reaction

• You know how to apply an Elementary Rate Law to a reaction

• You understand the importance of the “k” constant in the rate law


End of Block RE3

• We’ve seen the Arrhenius equation and how to account for each term

• We jumped to the “mathematics”. Given that our rate of reactions are based on ONLY one reactant, we need to base all the functions on that reactant

• Stoichiometry comes handy here!


End of Block RE3

• You now know that there is a difference between liquid-phase, gas-phase and constant volume/voluemtric flow rates

• Worst case scenario change of moles in a gaseous reaction

• Isothermal, Isobaric is supposed in these chapter

• Isothermal and Isobaric effects are seen in further chapters


End of Block RE3

• You may now continue with CH4 which is the design of Isothermal Reactors!

• Congratulations!


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Text Book & Reference

Essentials of Chemical Reaction EngineeringH. Scott Fogler (1st Edition)

Chemical Reactor Analysis and Design FundamentalsJ.B. Rawlings and J.G.

Ekerdt (1st Edition)

Elements of Chemical Reaction EngineeringH. Scott Fogler (4th Edition)


Bibliography

Elements of Chemical Reaction EngineeringH. Scott Fogler (4th Edition)


We’ve seen CH3