Rate Law & Stoichiometry Tables - Reactor Engineering Course Block 3

CH3: Rate Law & Stoichiometry
RE3
Chemical Engineering Guy
www. Chemical Engineering Guy .com

Chemical Reaction Engineering Methodology
CH3: Elements of Chemical Reaction Engineering
H. Scott Fogler (4th Edition)

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:
AB
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:
AB
k = 0.23 min-1
Something is “missing”…

Solve this problem!
• Reaction data:
AB
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 BA

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·CA
aCB
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·CACB
2
Order = 0
Order = 1
Order = 1+1 =2
Order = 1+2 =3
NOTE: k values vary

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·CA
aCB
b
• Then this is an Elementary Rate Law

Elementary Rate Law
• Examples:
AB
A+2B  3C
½A + B + C  ¼ D + 5F
• What would be the Elementary Rate laws?

Elementary Rate Law
• Examples:
AB
A+2B  3C
½A + B + C  ¼ D + 5F
-rA=k·CA
-rA=k·CACB
2
-rA=k·CA
1/2CBCC

Elementary Rate Law
• Once again… this is done by observation
• You can’t propose an elementary rate law for a
reaction if you haven’t prove it in the lab!
• You can check in data bases for common
reactions and know if they follow an
elementary 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 BA
• 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·CA
aCB
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 Temperature
A: Pre-exponential Factor / Frequency Factor
e: the mathematical operation “exponential”
E: Activation energy of that specific reaction
R: The ideal gas constant
T: Absolute Temperature

• This is Arrhenius Equation
As Temperature increases  k increases
This is due to the Kinetic theory
More temperature  more collisions
More collisions  more reactions

• Collision theory…
• Please check your Chemistry notes!

• 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.

• 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)

• 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

• 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

More Problems of this Section: Rate Laws?
• Need more Problems? Check out the course!
– www.ChemicalEngineeringGuy.com
• Courses
–Reactor Engineering
»Solved Problems Section
• CH3 – Rate Laws and Stoichiometry

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

All is based on Species “A”aA + bB  cC + dD

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

• 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

Exercise 3-2
• Liquid phase, it is easier (no need to mess
with gases!
• Suppose it is a Batch Reactor -> Volume

Exercise 3-2
This table is the “answer”

Exercise 3-3
Engineering

Exercise 3-3

Continuous Flow Systems
Stoichiometry Table
• Now we analyze continuous flow reactors!

Stoichiometry Table
• Same story, now for continuous flow

Stoichiometry Table

Stoichiometry Table
• Turning those Flows into Concentrations (more useful)

Stoichiometry Table
• Defining this concept:

Stoichiometry Table
• 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)

Stoichiometry Table
• 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 Moles
Reactor 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

Batch 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)

Continuous 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!

• 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 Conversion

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

Flow Rate in
terms of Xa

Master Equation!
Everything in terms of
Conversion of A!
Flow Rate in terms of Xa
Volumetric Flow in terms of Xa

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

Visual Summary
Engineering

Visual Summary
Set the Equation in terms of A  CH3: Elements of Chemical
Reaction Engineering

Visual Summary
CH3: Elements of Chemical

Exercises of Stoichiometry Tables for
Engineering

Exercises of Stoichiometry Tables for

More Problems for Space-Time and Spatial-Velocity?
• Need more Problems? Check out the course!
– www.ChemicalEngineeringGuy.com
• Courses
–Reactor Engineering
»Solved Problems Section
• CH3 – Rate Law & Stoichiometry

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
H. Scott Fogler (1st Edition)
Chemical Reactor
Analysis and Design
Fundamentals
J.B. Rawlings and J.G.
Ekerdt (1st Edition)
Elements of Chemical

Bibliography
Elements of Chemical Reaction Engineering
We’ve seen  CH3

Rate Law & Stoichiometry Tables - Reactor Engineering Course Block 3

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Rate Law & Stoichiometry Tables - Reactor Engineering Course Block 3