Kjemisk reaksjonsteknikk - PowerPoint PPT Presentation

Title: Kjemisk reaksjonsteknikk


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• Lecture 2
• Kjemisk reaksjonsteknikk

• Review of Lecture 1 and 2 (chapter 1 and 2)
• Rate Laws
• Reaction Orders
• Arrhenius Equation
• Activation Energy
• Effect of Temperature

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General Mole Balance

• General Mole Balance on System Volume V

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Reactor Mole Balance Summary
The GMBE applied to the four major reactor types
(and the general reaction A?B)
Reactor Differential Algebraic Integral




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Reactor Mole Balances in terms of conversion
FAFA0-FA0XFA0(1-X)
Reactor Differential Algebraic Integral




X
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Reactor Mole Balances in terms of residence time
and conversion
FACAFV , CACA0(1-X), tV/ FV
Batch reactor
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CSTR
PFR
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Two types of problems

• Design problem Design a reactor to achieve
  certain conversion
• Operating problem in a existing reactor to find
  conversion or the residence time to reach the
  certain conversion

Levenspiel Plots
PFR
CSTR
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Reactors in Series
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PFR vs CSTR in series
..
Is the PFR always better than the CSTR in terms
of reactor size to achieve a identical
conversion?
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Reactors in Series
Only valid if there are no side streams
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Reactors in Series
Exothermic reaction in an adiabatic reactor
How can we minimize the reaction size?
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Basic Definitions

• Homogeneous reactions involve only one phase
• Heterogeneous reactions involve more than one
  phase, and reactions occur at interfaces of two
  phases
• Irreversible reactions occur at only one
  direction
• Reversible reactions occur at both directions,
  depending one the approach to equilibrium
• Molecularity of the reaction is the number of the
  atoms, ions or molecules involved in a reaction
  step. Unimolecular, bimolecular and termolecular
  refer to reactions involving one, two and three
  atoms or molecules in one reaction step,
  respectively
• Elementary reaction involves only one bond
  breaking or formation
• Non-elementary reaction could involve multi
  elementary reaction steps

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Kinetics
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Kinetics - Power Law Model
A reactor follows an elementary rate law if the
reaction orders just happens to agree with the
stoichiometric coefficients for the reaction as
written. e.g. If the above reaction follows an
elementary rate law 2nd order in A, 1st order in
B, overall third order
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                                                                  "Everyone has Problems -but Chemists have Solutions"

Chemical Engineers have Simple Solutions !!!
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Example Ammonia decomposition

• 2NH33H2N2 11 kJ/mol
• The kinetic study was performed in a fixed bed
  reactor (6 mm diameter) on Fe/Al2O3 catalysts at
  atmospheric pressure and total flow of 100 ml/min
  with Ar as the balance. 100 mg catalysts mixed
  with 1 g SiC were loaded in the reactor.
• FNH3.s (ml/min)
  20 40 80
• FAr,s (ml/min)
  80 60 20
• XNH3
  0.050 0.051 0.050
• Can we determine the reactor order? (n0,1,2 ? )
• How can we reduce the conversion from 5.0 to
  2.5

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Relative Rates of Reaction
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Relative Rates of Reaction
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2AB?3C

• Second Order in A
• Zero Order in B
• Overall Second Order

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Reversible Reaction
This equation is thermodynamically consistent.
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Reversible Reaction
Reaction is First Order in A Second Order
in B Overall third Order
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Arrhenius Equation
k is the specific reaction rate (constant) and is
given by the Arrhenius Equation. where
Svante August Arrhenius was a Swedish scientist,
received the Nobel Prize for Chemistry in 1903
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Arrhenius Equation
where E Activation energy (cal/mol) R Gas
constant (cal/molK) T Temperature (K) A
Frequency factor (same units as rate constant
k) (units of A, and k, depend on overall reaction
order)
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Reaction Coordinate
The activation energy can be thought of as a
barrier to the reaction. One way to view the
barrier to a reaction is through the reaction
coordinates. These coordinates denote the energy
of the system as a function of progress along the
reaction path. For the reaction
The reaction coordinate is
Transition state theory
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Why is there an Activation Energy?
We see that for the reaction to occur, the
reactants must overcome an energy barrier or
activation energy EA. The energy to overcome
their barrier comes from the transfer to the
kinetic energy from molecular collisions and
internal energy (e.g. Vibrational Energy).

1. The molecules need energy to disort or stretch
   their bonds in order to break them and thus form
   new bonds
2. As the reacting molecules come close together
   they must overcome both stearic and electron
   repulsion forces in order to react.

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Collision probability
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One such distribution of energies is in the
following figure
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Rate expression Gas phase reaction
AB 2C
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Rate expression Catalysed reaction