lecture 3
DESCRIPTION
Lecture 3. Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 3 – Thursday 1/17/2013. Review of Lectures 1 and 2 Building Block 1 Mole Balances (Review) - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/1.jpg)
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in which they take place.
Lecture 3
![Page 2: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/2.jpg)
Lecture 3 – Thursday 1/17/2013Review of Lectures 1 and 2Building Block 1
Mole Balances (Review)Size CSTRs and PFRs given –rA= f(X)Conversion for Reactors in Series
Building Block 2Rate LawsReaction OrdersArrhenius EquationActivation EnergyEffect of Temperature
2
![Page 3: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/3.jpg)
Reactor Differential Algebraic Integral
The GMBE applied to the four major reactor types (and the general reaction AB)
V FA 0 FA
rA
CSTR
Vrdt
dNA
A 0
A
A
N
N A
A
VrdNtBatch
NA
t
dFA
dVrA
A
A
F
F A
A
drdFV
0
PFRFA
V
dFA
dW r A
A
A
F
F A
A
rdFW
0
PBRFA
W3
Reactor Mole Balances Summary
![Page 4: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/4.jpg)
fedA moles reactedA moles X
D ad C
ac B
ab A
ncalculatio of basis asA reactant limiting ChooseD d C c B b A a
4
![Page 5: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/5.jpg)
Reactor Differential Algebraic Integral
A
A
rXFV
0CSTR
AA rdVdXF 0
X
AA r
dXFV0
0PFR
VrdtdXN AA 0
0
0
X
AA Vr
dXNtBatch
X
t
AA rdWdXF 0
X
AA r
dXFW0
0PBR
X
W5
Reactor Mole Balances SummaryIn terms of Conversion
![Page 6: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/6.jpg)
Levenspiel Plots
6
![Page 7: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/7.jpg)
Reactors in Series
7
reactorfirst tofedA of molesipoint toup reactedA of molesX i
Only valid if there are no side streams
![Page 8: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/8.jpg)
Reactors in Series
8
![Page 9: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/9.jpg)
Building Block 2: Rate Laws
9
Power Law Model:
BAA CkCr
βαβα
OrderRection OverallBin order Ain order
![Page 10: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/10.jpg)
Building Block 2: Rate Laws
10
C3BA2
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
BAAA CCkr 2
![Page 11: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/11.jpg)
Building Block 2: Rate Laws
11
If 2AA kCr
›Second Order in A›Zero Order in B›Overall Second Order
2A+B3C
![Page 12: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/12.jpg)
Relative Rates of Reaction
12
dDcCbBaA
dr
cr
br
ar DCBA
DadC
acB
abA
![Page 13: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/13.jpg)
Relative Rates of Reaction
13
Given
Then 312CBA rrr
C3BA2
sdmmolrA
310
sdmmolrr A
35
2B
sdmmolrr AC
31523
![Page 14: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/14.jpg)
Reversible Elementary Reaction
14
e
CBAA
AA
CBAACABAAA
KCCCk
kkCCCkCkCCkr
32
3232
A+2B3C
kA
k-A
![Page 15: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/15.jpg)
Reversible Elementary Reaction
15
Reaction is: First Order in A
Second Order in B
Overall third Order
sdmmolesrA 3 3dm
molesCA
smoledm
dmmoledmmolesdmmole
CCrk
BA
A2
6
233
3
2
A+2B3C
kA
k-A
![Page 16: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/16.jpg)
16
![Page 17: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/17.jpg)
Algorithm
17
iA Cgr Step 1: Rate Law
XhCi Step 2: Stoichiometry
XfrA Step 3: Combine to get
How to find XfrA
![Page 18: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/18.jpg)
Arrhenius Equation
18
RTEAek
k is the specific reaction rate (constant) and is given by the Arrhenius Equation.where:
T k AT 0 k 0
A 1013k
T
![Page 19: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/19.jpg)
Arrhenius Equation
19
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)
![Page 20: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/20.jpg)
Reaction Coordinate
20
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:CABCBABCA ::::::
The reaction coordinate is:
![Page 21: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/21.jpg)
21
![Page 22: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/22.jpg)
Collision Theory
22
![Page 23: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/23.jpg)
Why is there an Activation Energy?
23
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 of the kinetic energy from molecular collisions to 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.
![Page 24: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/24.jpg)
Distribution of Velocities
24
We will use the Maxwell-Boltzmann Distribution of Molecular Velocities. For a species af mass m, the Maxwell distribution of velocities (relative velocities) is:
f(U,T)dU represents the fraction of velocities between U and (U+dU).
23 2
2 2, 42
BmU k T
B
mf U T dU e U dUk T
![Page 25: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/25.jpg)
Distribution of Velocities
25
A plot of the distribution function, f(U,T), is shown as a function of U:
Maxwell-Boltzmann Distribution of velocities.
T2>T1T2
T1
U
𝑓 (𝑈 ,𝑇 )
![Page 26: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/26.jpg)
Distribution of Velocities
26
Given
Let
f(E,T)dE represents the fraction of collisions that have energy between E and (E+dE)
23 2
2 2, 42
BmU k T
B
mf U T dU e U dUk T
2
21 mUE
1 23 2
2,2
B
Ek T
B
f E T dE E e dEk T
![Page 27: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/27.jpg)
One such distribution of energies is in the following figure:
f(E,T)dE=fraction of molecules with energies between E+dE
27
![Page 28: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/28.jpg)
End of Lecture 3
28
![Page 29: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/29.jpg)
Supplementary Material
29
BC AB
kJ/Molecule
rBC
kJ/Molecule
r0
Potentials (Morse or Lennard-Jones)
VABVBC
rAB
r0
![Page 30: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/30.jpg)
Supplementary Material
30
For a fixed AC distance as B moves away from C the distance of separation of B from C, rBC increases as N moves closer to A. As rBC increases rAB decreases and the AB energy first decreases then increases as the AB molecules become close. Likewise as B moves away from A and towards C similar energy relationships are found. E.g., as B moves towards C from A, the energy first decreases due to attraction then increases due to repulsion of the AB molecules as they come closer together. We now superimpose the potentials for AB and BC to form the following figure:
One can also view the reaction coordinate as variation of the BC distance for a fixed AC distance:
l
BCA l
CAB BCAB rr
l
CBA
![Page 31: Lecture 3](https://reader036.vdocument.in/reader036/viewer/2022062411/5681697c550346895de1861d/html5/thumbnails/31.jpg)
Supplementary Material
31
Energy
r
BC
ABS2S1
E*
E1PE2P ∆HRx=E2P-E1P
Ea
Reference