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Kjemisk reaksjonsteknikk
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 FA0 FA GA
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Reactor Mole Balance Summary
The GMBE applied to the four major reactor types (and the general reaction AB) Reactor Differential Algebraic Integral Batch NA t CSTR PFR FA V PBR FA W 3 3
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Reactor Mole Balances in terms of conversion
FA=FA0-FA0X=FA0(1-X) Reactor Differential Algebraic Integral Batch X t CSTR X PFR PBR X W 4 4
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Reactor Mole Balances in terms of residence time and conversion
FA=CAΦV , CA=CA0(1-X), τ=V/ ΦV X t Batch reactor X CSTR X τ PFR 5 5
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Two types of problems Levenspiel Plots PFR CSTR
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 ….. x 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 9
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Reactors in Series Exothermic reaction in an adiabatic reactor
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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 13
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Chemical Engineers have Simple Solutions !!!
"Everyone has Problems - but Chemists have Solutions" Chemical Engineers have Simple Solutions !!!
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Example : Ammonia decomposition
2NH3=3H2+N 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) FAr,s (ml/min) XNH Can we determine the reactor order? (n=0,1,2 ? ) 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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2A+B3C If Second Order in A Zero Order in B Overall Second Order 18
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Reversible Reaction Elementary
This equation is thermodynamically consistent. 19
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Reversible Reaction kA A+2B 3C k-A Reaction is: First Order in A
Second Order in B Overall third Order 20
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Arrhenius Equation k is the specific reaction rate (constant) and is given by the Arrhenius Equation. where: k T Svante August Arrhenius was a Swedish scientist, received the Nobel Prize for Chemistry in 1903 Arrhenius found the temperature dependence of reaction rate, which is the most important equation in catalysis. 22
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Arrhenius Equation where: E = Activation energy (cal/mol)
R = Gas constant (cal/mol*K) T = Temperature (K) A = Frequency factor (same units as rate constant k) (units of A, and k, depend on overall reaction order) 23
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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 24
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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). The molecules need energy to disort or stretch their bonds in order to break them and thus form new bonds As the reacting molecules come close together they must overcome both stearic and electron repulsion forces in order to react. 25
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Collision probability
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f(E,T)dE=fraction of molecules with energies between E+dE
One such distribution of energies is in the following figure: 29
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Rate expression – Gas phase reaction
A+B = 2C
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Rate expression – Catalysed reaction
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