True or false: The goal of fitting kinetic Reactive...
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L6 L6 - - 1 1 ChE 400 - Reactive Process Engineering ChE ChE 400 400 - - Reactive Process Engineering Reactive Process Engineering Review of Fitting Kinetic Data • True or false: The goal of fitting kinetic data is to find the true rate expression. • • What are the two general methods used to fit kinetic data? –
Transcript of True or false: The goal of fitting kinetic Reactive...
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Review of Fitting Kinetic Data
• True or false: The goal of fitting kinetic data is to find the true rate expression.
•
• What are the two general methods used to fit kinetic data?–
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Advantages and Drawbacks
•
•
•
•
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CST Data as Obtained
0.0
5.0
10.0
15.0
20.0
25.0
30.0
0 50 100 150 200Time, min
Con
cent
ratio
n, m
g/L
Cin = 15 mg/LCin = 20 mg/LCin = 25 mg/L
Review of Fitting Kinetic Data
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Transformed CST Data
-5.0
-4.0
-3.0
-2.0
-1.0
0.00 20 40 60 80 100 120 140 160 180 200
time, min
ln{(
Cin
-C)/(
Cin
-C0)
}
Cin = 15 mg/LCin = 20 mg/LCin = 25 mg/L
Transformed CST Data with Trendline (all data)
ln{(Cin-C)/(Cin-C0)} = -0.025 tR2 = 0.999
-5.0
-4.0
-3.0
-2.0
-1.0
0.00 20 40 60 80 100 120 140 160 180 200
time, min
ln{(
Cin
-C)/(
Cin
-C0)
}
Cin = 15 mg/LCin = 20 mg/LCin = 25 mg/L
Review of Fitting Kinetic Data
What is the order of the reaction?
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Determining k0, EASo, now that we know k, how can we obtain k0 and EA??
Remember k?
ln k
1/T
ln k0
-E/R
Plotting ln k
What’s the slope?What’s the slope?What’s the slope?
What’s the intercept?What’s the intercept?What’s the intercept?
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Kinetics of Many Processes in Nature Kinetics of Many Processes in Nature
follow Arrhenius Relationshipfollow Arrhenius RelationshipSome Examples• Cricket chirping• Ant walking• Tumor growth• Diffusion in solids [D = Do exp (-ED/RT)]
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Ant walking: Does it follow Arrhenius’ Law?Ant walking: Does it follow Arrhenius’ Law?
M. I. M. I. MaselMasel, Chemical Kinetics and Catalysis, Chemical Kinetics and Catalysis
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The Ants are walking one by one…The Ants are walking one by one…
0
1
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8
0 5 10 15 20 25 30 35 40 45
Temperature (C)
Run
ning
Spe
ed (c
m/s
)
Raw data Raw data (speed (speed vsvs temperature)temperature)
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.15 3.20 3.25 3.30 3.35 3.40 3.45 3.50 3.55 3.60
1000/T (K-1)
ln (w
laki
ng s
peed
)
LinearizedLinearized datadata((lnln u u vsvs 1/T)1/T)
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Summary: Determining Rate Coefficients
1.1. set upset up
2.2. measuremeasure
3.3. setset--up up
4.4. linearizelinearize
5.5. use use
6.6. do this do this
7.7. linearizelinearize, ,
8.8. use linear use linear -- slopeslope-- interceptintercept
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Example: determining k
t (s) [NO2] (mol/l)
0 0.01
50 0.00787
100 0.00649
200 0.00481
300 0.0038
Nitrogen dioxide is introduced into an isothermal batch reactor at a temperature of 300°C. Its decomposition to nitrogen oxide and oxygen is followed over time and the data below is obtained. Is the reaction first order or second order? Determine the reaction rate coefficient.
2 NO2 <=> 2 NO + O2
order integrated rate law linear plot for
1
2
…so, let’s do it!……so, let’s do it!so, let’s do it!
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Example: 1st order kinetics
1st order fit
y = -0.0032x - 4.6679R2 = 0.9824
-6
-5.8
-5.6
-5.4
-5.2
-5
-4.8
-4.6
-4.4
-4.2
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0 50 100 150 200 250 300 350
t [s]
ln {[
NO
2 / (
mol
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Example: 2nd order kinetics
2nd order fit
y = 0.5431x + 99.833R2 = 1
100
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0 50 100 150 200 250 300 350
t [s]
1/[N
O2]
/ (m
ol/l)
perfect fit indicates
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Determining Reaction OrderGreat, so if I know the reaction order n, I can determine k. What if I do not know n?
1. “Try” different reaction orders (i.e. search for the best fit of the data to the integrated rate laws)Works only for reasonably simple reaction orders (n = -1, 0, 1, 2)
t (min) CA (mol/L)0 2.005 1.609 1.3515 1.1022 0.8730 0.7040 0.5360 0.35
0.0
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time (min)
CA
(mol
/L)
Example: The following is measured data from some reaction A -> P
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Reaction Order: Example cont’d
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0 10 20 30 40 50 60
time (min)
ln(C
A/C
A,0) (
dim
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(1) Assume 1st order, linearize, and replot
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Reaction Order: Example cont’d(1) Assume 2nd order, linearize, and replot
0.0
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2.5
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time (min)
[1/C
A - 1
/CA
,0] (
L/m
ol)
ktCC AA
=−0,
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1/CA0.5 - 1/CA,0
0.5 = 0.0165tR2 = 0.9998
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time (min)
1/C
A0.5 -
1/C
A,0
0.5
Reaction Order: Example cont’d(1) Assume order 1.5, linearize, and replot
r = k CA1.5 into BR design equation:
with constant volume assumption:
…and integration yields:
What is k??
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Determining Reaction OrderObviously, trial-and-error fits will not help us except for very simple cases. Hence we need a generally applicable way…
2. Initial Rates (differential) Method:Measure reaction rate at several different initial concentrations.
From r = k [A]n => ln r = n ln[A] + ln kHence, the logarithmic plot of initial reaction rate r0 vs [A]0 yields the reaction order n as the slope.
•••
ln [A]0
ln r0
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Reaction Order, cont’dIn practice, one rarely deals with simple reactions with only one reactantA -> products. Therefore, the reaction order needs to be determined foreach reactant separately. This is done by the…
Isolation Method:Isolation Method:The concentrations of all but one reactant are kept constant during an experiment, thus isolating the effect of this reactant on the reaction rate.
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Summary: Determining Reaction Orders
•• Integral + Trial and ErrorIntegral + Trial and Error:
•• Differential MethodDifferential Method
•• Isolation MethodIsolation Method
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Designing a Kinetic Experiment
1. Select
2. set up
3. measure
4. determine
5. measure
6. determine
That’s it! You now know how to design That’s it! You now know how to design an experiment to determine all relevant an experiment to determine all relevant kinetic data for any reaction system:kinetic data for any reaction system:
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A Glimpse of Theory…
As we saw above, kinetic experiments are difficult and lengthy
procedures (and hence also expensive…), so one would really LOVE
to replace experiments by theoretical calculationsreplace experiments by theoretical calculations.
While we are still far away from being able to do this reliably,
theoretical chemistry has made great progress in past decades
and is becoming an increasingly important tool in chemical kinetics
(even in industrial application!).
The following will give just a very brief glimpse of some of the
very basics of those calculations.
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A (very brief) Theory of Reaction Rates
All chemical reactions occur by collisions between moleculescollisions between molecules (or by collision of molecules with surfaces – that’s catalysis, and we’ll talk about that a bit later…).One very simple theory of reactions builds on this observation, and is hence called
‘collision theory’‘collision theory’.
Consider the simple bimolecular reaction:
In the framework of the collision theory, the rate of this reaction is described by:r = r =
You can’t react if you don’t collide!You can’t react if you don’t collide!
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Probability of Successful Collision
Boltzmann-Distributionp(E) = exp{-E/RT}p(Ep(E) = exp{) = exp{--E/RT}E/RT}
fraction of successful collisions(= probability distribution)
The probability of a ‘successful’ collisionprobability of a ‘successful’ collision can be estimated based on the fact that a bond needs to be broken during a reactive event. This requires a certain amount of energy, E. Since the energy distribution (for thermal energy) is governed by a BoltzmannBoltzmanndistributiondistribution, the probability that a molecule at temperature T has acquired the
necessary energy is p(Ep(E) = exp{) = exp{--E/RT}E/RT}.
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(Simple) Collision Theory
The collision cross section (assuming hard spheres) is
The collision rate can be calculated from the The collision rate can be calculated from the kinetic theory of gaseskinetic theory of gases::
The frequency of collisions of a molecule of A with B is given by:
ZZABAB ==
Therein, the average thermal velocity u is given by <u> =
The ‘reduced molecular mass’ is given by: μ =
The The collision cross sectioncollision cross section for two for two molecules can be regarded asmolecules can be regarded as
σAB =
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ZAB =
(Simple) Collision Theory
Or, with concentrations:
We can hence write for the reaction rate:
k0 = kk00 = =
Compare this to the usual rate law for bimolecular (elementary) reactions:r = k0 exp{-EA/RT} CA CB
We can therefore calculate the pre-exponential factor for a 2nd order reaction from:
ZAB =
Putting this all together, we obtain for the collision ratecollision rate:
r =
ZAB =
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kk00 = {8 RT/(= {8 RT/(π μπ μ)})}1/2 1/2 ππ ddABAB22
(Simple) Collision Theory, cont’d
This value gives in effect the maximal possible rate coefficientmaximal possible rate coefficient for bimolecular reactions. In reality, the rate coefficients are typically significantly lower than this, due to steric as well as energetic considerations.
ExampleExample: .OH + H. -> .O. + H2
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Example 2:
This value gives in effect the maximal possible rate coefficientmaximal possible rate coefficient
kk00 = {8 RT/(= {8 RT/(π μπ μ)})}1/2 1/2 ππ ddABAB22
(Simple) Collision Theory, cont’d
for bimolecular reactions. In reality, the rate coefficients are typically significantly lower than this, due to steric as well as energetic considerations.
Steric and energetic factors can reduce the rate coefficient by many orders of magnitude. Collision theory can hence only give an upper limit to k!(Comparison with “similar” reactions with known steric factors occasionally allows us to estimate the steric factor and hence to calculate k0 from collision theory.
=> rcorrected = P Z exp{-E/RT} )
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ExampleEstimate the rate coefficient for bimolecular gas phase reactions. Assume room temperature.
k0 = {8 RT/(π μ)}1/2 π dAB2kk00 = {8 RT/= {8 RT/(π μ(π μ)})}1/2 1/2 ππ ddABAB22
μ =dAB ≈
kk00 ==
Dimensions:
[k0] =
-> to obtain the usual dimensions for a 2nd order rate constant, we need to multiplywith Avogadro’s number NA= 6.02 * 1023 (atoms)/mol
R = 8.3 J/mol K, T = 300 KValues:
Let’s assume simple molecules: NLet’s assume simple molecules: N2 2 and Oand O22
Compare with experimental valuesCompare with experimental values::K + BrK + Br2 2 --> > KBrKBr + Br+ Br kk0,exp0,exp≈≈CHCH33 + CH+ CH3 3 --> C> C22HH66 kk0,exp0,exp≈≈HH22 + C+ C22HH4 4 --> C> C22HH66 kk0,exp0,exp≈≈ kk00 = k= k0,theo 0,theo × P ; × P ; P ≈P ≈
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Collision Theory: “Summary”
frequencyof
collision
frequencyof
collision
fraction ofmolecules
facing each other correctly
fraction ofmolecules
facing each other correctly
fraction ofmolecules
with sufficientkinetic energyto break bonds
fraction ofmolecules
with sufficientkinetic energyto break bonds
k = x x
Which of these terms does Collision Theory capture (more or less)correctly, and which does it not?
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Collision Theory: Unimolecular RctnsCollision theory does offer interesting insights into unimolecular reactions: It is experimentally observed that unimolecular reactions show a surprising decrease in the measured reaction coefficient towards very low pressures (so-called “fall-off” regime). How can that be explained?
In reality, unimolecular reactions require intermolecular collisions!
A -> B , r = k CA
is in reality the sum reaction for a reaction sequence!A + A <=> A* + A , r1 =
A* -> B , r2=
Since A* is a highly reactive intermediate, let’s assume it’s concentration is essentially constant:
dCA*/dt =
CA* =
Hence: rB = r2 =
High pressure limit:High pressure limit:
Low pressure limit:Low pressure limit:
