CHEMICAL REACTION ENGINEERING 1

RPKPS

Main reference:Fogler, H.S., 2006, ”Elements of Chemical Reaction Engineering”, 4th Ed., Pearson Education, Inc., Prentice Hall Professional Technical Reference, New Jersey.

Community:elisa.ugm.ac.idTeknik Reaksi Kimia 1_TK_Budhijanto

1.MOLE BALANCESObjectives:After completing Chapter 1, you will be able to:

Define the rate of chemical reaction. Apply the mole balance equations to a batch

reactor, CSTR, and PFR. Describe photos of real reactors.

Chemical kinetics:the study of chemical reaction rates and reaction mechanisms.

Reactor:an equipment in which reactions occur

Chemical Reaction Engineering (CRE):

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combines the study of chemical kinetics with the reactors.

How is a chemical engineer different from other engineers?It is primarily a knowledge of chemical kinetics and reactor design that distinguishes the chemical engineer from other engineers.

Why do we have to study CRE?The selection of a reaction system that operates in the safest and most efficient manner can be the key to the economic success or failure of a chemical plant. For example, if a reaction system produces a large amount of undesirable product, subsequent purification and separation of the desired product could make the entire process economically unfeasibleSpecific example: production of maleic anhydride (raw material for various industries, e.g. polyester resins, paint, etc)Main reaction:

Side reactions:

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Reaction condition: 2 – 5 bar, 400 - 450°C, vanadium catalyst

Chemical Identity:

A chemical species is said to have reacted when it has lost its chemical identity.

The identity of a chemical species is determined by the kind, number, and configuration of that species' atoms.

Three ways a chemical species can lose its chemical identity:

1. Decomposition

2. Combination

3. Isomerization 

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Reaction Rate:

the rate at which a species looses its chemical identity per unit volume.

It can be expressed as the rate of disappearance of a reactant or as the rate of appearance of a product. Consider species A:

A B

rA = the rate of formation of species A per unit volume−rA = the rate of a disappearance of species A per unit volumerB = the rate of formation of species B per unit volume

If B is being created at a rate of 0.2 moles per decimeter cubed per second, ie, the rate of formation of B is,

rB = 0.2 mole/(dm3∙s)

Then A is disappearing at the same rate:−rA = 0.2 mole/(dm3∙s)

The rate of formation of A isrA = −0.2 mole/(dm3∙s)

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For a catalytic reaction, we refer to −rA’, which is the rate of disappearance of species A on a per mass of catalyst basis.

NOTE: dCA/dt is not the rate of reaction

Example: Is sodium hydroxide reacting?

Sodium hydroxide and ethyl acetate are continuously fed to a rapidly stirred tank in which they react to form sodium acetate and ethanol:

NaOH + CH3COOC2H5 CH3COONa + C2H5OH

The product stream, containing sodium acetate and ethanol, together with the unreacted sodium hydroxide and ethyl acetate, is continuously withdrawn from the tank at a rate equal to the total feed rate. The contents of the tank in which this reaction is taking place may be considered to be perfectly mixed. Because the system is operated at steady state, if we were to withdraw liquid samples at some location in the tank at various times and analyze them chemically, we

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would find that the concentrations of the individual species in the different samples were identical. Because the species concentrations are constant and therefore do not change with time,

dC A

dt=0

where A is NaOH. If we defined

r A=dC A

dt

thenr A=0

which is incorrect because C2H5OH and CH3COONa are being formed from NaOH and CH3COOC2H5 at a finite rate. Consequently, the rate of reaction as defined by

r A=dC A

dt

cannot apply to a flow system and is incorrrect if it is defined in this manner.

rA is an algebraic law!

r A=dC A

dt

is simply a mole balance that is only valid for a constant volume batch system.

Consider species j:

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rj is the rate of formation of species j per unit volume [e.g. mol/(dm3∙s)]

rj is a function of concentration, temperature, pressure, and the type of catalyst (if any)

rj is independent of the type of reaction system (batch, plug flow, etc.)

rj is an algebraic equation, not a differential equation

We use an algebraic equation to relate the rate of reaction, −rA, to the concentration of reacting species (e.g. CA) and to the temperature (T) at which the reaction occurs. Example:

A product−r A=k (T )C A

−r A=k (T )C A2

−r A=k 1 (T )C A

1+k2 (T )CA

Which one is the correct one?It must be determined from experimental observation.

Problem 1.1.: The Convention for Rates of Reaction Consider the reaction

A + 2B 3Cin which the rate of disappearance of A is 5 moles of A per dm3 per second at the start of the reaction.At the start of the reaction (a)   What is −rA?

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(b)   What is the rate of formation of B? (c)   What is the rate of formation of C? (d)   What is the rate of disappearance of C? (e)   What is the rate of formation of A, rA? (f)    What is –rB?

General Mole Balance Equation

A mole balance on species j at any instant in time, t:

F j0−F j+G j=F j0−F j+∫❑

V

r jdV=d N j

dt

WhereNj represents the number of moles of species j in the system at time t.V is the reaction volume.

Batch Reactors

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Assumption: the reaction is perfectly mixed.d N j

dt=r jV

Ort=∫

N j0

N j d N j

r jV

There are two types of batch reactor for gas phase reactions.

1.Constant volume batch reactorsdC j

dt=r j

2.Constant pressure batch reactorsr j=

1V

d N j

dt= 1V

d (C jV )dt

=dC j

dt+C j

VdVdt

=dC j

dt+C j

d ( lnV )dt

Continuous Stirred Tank Reactor (CSTR = Vat = Backmix Reactor)

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F j0=F j=0

∫❑

V

r jdV=d N j

dt

Assumption: the reaction is perfectly mixed.Thus:1. ∫

❑

V

r jdV=r jV

2. The conditions in the exit stream (e.g. concentration, temperature) are identical to those in the tank.

F j0−F j+r jV=d N j

dt

At steady state:d N j

dt=0

We get:F j0−F j+r jV=0

OrV=

F j0−F j−r j

Because:F j=C j v

Wherev = the volumetric flow rate (volume/time)

V=C j0 v0−C j v

−r j

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Fj0 Fj,exit

∆V

V

Fj│V) Fj│V+∆V

Tubular Reactor

Assumption: plug flow no radial variation in reaction rate the reactor is referred to as a Plug Flow Reactor (PFR).

The differential volume, ΔV, is sufficiently small such that there are no spatial variations in reaction rate within its volume.At steady state:

F j|V+∫ΔVr jdV−F j|V +ΔV=0

F j|V+r j ΔV−F j|V +ΔV=0

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F j|V +ΔV−F j|VΔV

=r jlimit ∆V→0:

dF jdV

=r jOr

V=∫F j 0

F j dF jr j

Summary:General Mole Balance Equation:

F j 0+∫Vr jdV−F j=

dN j

dt

Reactor Mole Balance RemarkBatch dN j

dt=r jV

Perfectly mixed

CSTRV=

F j 0−F j−r j

Perfectly mixed; Steady State

PFR dF jdV

=r jSteady State

Akhir Kuliah 1Problem 1.2.: How Large Is It? Consider the liquid phase cis-trans isomerization of 2-butene.

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which we will write symbolically asA → B

The first order (−r A=k C A ¿ reaction is carried out isothermally in a tubular reactor in which the volumetric flow rate, v, is constant = 10 L/min. The exiting concentration of A is 10% of its entering concentration. k = 0,23/min.

a. Sketch the concentration profile inside the PFR

b.Determine the reactor volume

Problem 1.3.:The same as Problem 2b, but instead of using a PFR, the reaction occured in an isothermal CSTR.

Problem 1.4.:A 200 L constant-volume batch reactor is pressurized to 20 atm with a mixture of 75% A and 25% inert. The gas-phase reaction is carried out isothermally at 227 C.

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V = 200 LP = 20 atmT = 227°C

a. Assuming that the ideal gas law is valid, how many moles of A are in the reactor initially? What is the initial concentration of A?

b. If the reaction is first order:

Calculate the time necessary to consume 99% of A.

c. If the reaction is second order:

Calculate the time to consume 80% of A. Also calculate the pressure in the reactor at this time if the temperature is 127°C.

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