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Transcript of No Data Left Behind Modeling Colorful Compounds in Chemical Equilibria Mike DeVries D. Kwabena...
![Page 1: No Data Left Behind Modeling Colorful Compounds in Chemical Equilibria Mike DeVries D. Kwabena Bediako Prof. Douglas A. Vander Griend.](https://reader031.fdocuments.net/reader031/viewer/2022032723/56649cff5503460f949cfeb8/html5/thumbnails/1.jpg)
No Data Left Behind
Modeling Colorful Compounds in Chemical Equilibria
Mike DeVries
D. Kwabena Bediako
Prof. Douglas A. Vander Griend
![Page 2: No Data Left Behind Modeling Colorful Compounds in Chemical Equilibria Mike DeVries D. Kwabena Bediako Prof. Douglas A. Vander Griend.](https://reader031.fdocuments.net/reader031/viewer/2022032723/56649cff5503460f949cfeb8/html5/thumbnails/2.jpg)
Outline
• What is chemical equilibrium?
• What makes color data good or not-so-good?
• How does matrix algebra work again?
• What is Sivvu and how does it work?
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What is Chemical Equilibrium?
When chemicals react, they ultimately form a balance between products and reactants such that the ratio is a constant:
[products]/[reactants] = Kequilibrium
Log(products) – Log(reactants) = LogKeq = -G/RT
G is called free energy.
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Example
Seal .04 mole of NO2 in a 1 liter container.
2NO2 N2O4G = -5.40 kJ/mol (@ 25°C)
[N2O4]/[NO2]2 = Keq = exp(-G/RT) = 8.97
Let x amount of NO2 that reacts.
(0.5x)/(.04-x)2 = 8.97
x = .01304, or 32.6% reacts
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Demonstration
A B
1. Secretly choose A or B and submit choice.2. As directed, choose A or B again:
If last time you submitted A, now submit B. If last time you submitted B,
Submit B again if coin flip shows tails Submit A if coin flip shows heads
3. Repeat #2 as directed.
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Multiple Equilibria
Solving one equilibrium equation can be tricky.
Solving simultaneous equilibria requires a computer.
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Exhaust Example
Calculate the equilibrium amounts of CO2, N2, H2O, CO, O2, NO, and H2 after burning 1 mole of C3H8 in air.
4 mass balance equations, 1 for each element:• Carbon = 3, Hydrogen = 8, Nitrogen = 40 and Oxygen = 10
3 equilibrium equations:• 2CO + O2 2CO2 G = -187.52 kJ/mol
• N2 + O2 2NO G = 125.02 kJ/mol
• 2H2 + O2 2H2O G = -247.86 kJ/mol
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Exhaust Example
4 mass balance equations:• Carbon = 3• Hydrogen = 8• Nitrogen = 40• Oxygen = 10
3 equilibrium equations:• [CO2]2/[CO]2/[O2] = 41874
• [NO]2/[N2]/[O2] = 0.001075
• [H2O]2/[H2]2/[O2] = 1134096
Species Amount
CO2 2.923
N2 19.99
H2O 3.980
CO 0.07667
O2 0.03471
NO 0.02732
H2 0.02006
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Metal Complexation Reactions
[Ni]2+ + py [Nipy]2+ G1
[Nipy]2+ + py [Nipy2]2+ G2
[Nipy2]2+ + py [Nipy3]2+ G3
[Nipy3]2+ + py [Nipy4]2+ G4
[Nipy4]2+ + py [Nipy5]2+ G5
[Nipy5]2+ + py [Nipy6]2+…
Ni
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Pyridine and Nickel
Ni2+ and BF4-
in methanol
equivalents of pyridine
added: 1 2 3 4 5
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UV-visible Spectroscopy
Absorbance = · Concentration
(Beer-Lambert Law)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
400 500 600 700 800 900 1000
Wavelength(nm)
Abs
orba
nce
0.1046 M Ni(BF4)2 in methanol
Ni2+
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Color is Additive
Ni 2+
BF 4-
py
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …
0.0997 M Ni(BF4)2 w/ 0.585 pyridine
0
0.2
0.4
0.6
0.8
1
1.2
400 500 600 700 800 900 1000
Wavelength(nm)
Abs
orba
nce
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What we know is not very much
[Ni]2+ + py [Nipy]2+ G1
[Nipy]2+ + py [Nipy2]2+ G2
[Nipy2]2+ + py [Nipy3]2+ G3
[Nipy3]2+ + py [Nipy4]2+ G4
[Nipy4]2+ + py [Nipy5]2+ G5
[Nipy5]2+ + py [Nipy6]2+…
Ni
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The Problem
We don’t know the wavelength-dependent colors or the equilibrium constants!
We can’t measure the independent color (absorptivities) because all the compounds are present together.
We don’t know the amounts of the compounds because they have equilibrated.
Almost all the data is composite.
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The Solution
It is possible, using advanced mathematical computations, to isolate information about pure species without chemically isolating them.
How? Generate more composite data by making more mixtures with differing amounts of reactants.
Model all the data according to chemical equilibria and the Beer-Lambert law for combining absorbances.
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Why it Works
Each data point corresponds to a single equation. For each point on the same curve, the [concentrations] are
the same. For each point at the same wavelength, the molar
absorptivities, n, are the same. With enough solution mixtures then, there will be more
equations than unknowns. This is known as an overexpressed mathematical system
which can theoretically be solved with error analysis.
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …
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Matrix Algebra Refresher
5
8
y
x
13
32
2x + 3y = 8
3x – y = 5
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Matrix Multiplication
C = AB(n x p) (n x m) (m x p)
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Matrix Algebra
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …
22
2
2
210
Nipy
Nipy
Ni
Abs
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Matrix Form of Beer-Lambert Law
component
CAbs
component
mCAbs
Absorbances(n x p)
(n x m)
Concentration(m x p)
m
mp
nm
n CAbs p
Mol
arA
bsor
btiv
ity
n # of wavelengths
m # of chemical species
p # of mixture solutions
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Measured Absorbances
Absorbances(n x p)
n = number of wavelengths
p = number of solution mixtures
Every column is a UV-vis curve.
Every row is a wavelength
So there are a total of np absorbance data points, each associated with a distinct equation.
The total absorbance at any particular point is the sum of the absorbances of all the chemical species in solution according to Beer-Lambert Law.
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Molar Absorptivities
(n x m)
Every column represents one of the m chemical species.
Every row is a wavelength
This smaller matrix contains all of the molar absorbtivity values for each pure chemical species in the mixtures at every wavelength.
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Component Concentrations
Concentration(m x p)
Every column corresponds to one of the p solution mixtures (UVvis curve).
Each row represents one of the m chemical species.
This smaller concentration matrix contains the absolute concentration of each chemical species in each of the solution mixtures.
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Matrix Absorbance Equation
component
CAbs
Absorbances(n x p)
(n x m)
Concentration(m x p)
So the problem is essentially factoring a matrix.
…Or solving np equations for mn + mp unknowns.
Abs = C
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With Residual Error!
Absorbances(n x p)
(n x m)
Concentration(m x p)
Residual(n x p)
Given data matrix of absorbances, find the absorptivity and concentration matrices that result in the smallest possible values in the residual matrix.
Abs = C + R
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Factor Analysis
What is m?• How many pure chemical species?• How many mathematically distinct (orthogonal)
components are needed to additively build the entire data matrix?
The eigenvalues of a matrix depict the additive structure of the matrix.
Requires computers. Matlab is very nice for this.
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Random Matrix Structure
(16 x 461) matrix of random numbers
All 16 eigenvalues contribute about the same to the structure of the matrix
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Non-random Data
50 data curves of Ni2+
solution with zero to 142 equivalents of pyridine.
How many additive factors exist in this data?
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Data Matrix Structure400 500 600 700 800 900 1000
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15Initial Factors in Measured Data
Wavelength (nm)
Abs
orba
nce
0 10 20 30 40 501
2
3
4
5
6
7
Factor Ratios
Rel
ativ
e S
igni
fican
ce
absorbance data
random numbers
6th eigenvalue is relatively small, but still possibly significant.
m = 6
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Equilibrium-Restricted Factor Analysis
Factoring a big matrix into 2 smaller ones does not necessarily give a positive or unique answer.
We also the force the concentration values to adhere to equilibrium relationships.
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Sivvu
Inputs• raw absorbance data
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Absorbance Data
n = 305 wavelengths
p = 50 solution mixtures
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance
equations
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Composition of Solutions
-2
-1.5
-1
-0.5
0
0.5
1
1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55
Solution number
Lo
g(C
on
cen
trati
on
)
Ni++
Py
From 0 to 142 equivalents of pyridine.
Pure pyridine (12.4 Molar) is dripped into 0.10 M Ni(BF4)2 Solution
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance
equations• chemical reactions for equilibrium equations
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Inputs
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance
equations• chemical reactions for equilibrium equations• guesses for G’s
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Inputs
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance equations• chemical reactions for equilibrium equations• guesses for G’s
Process• Calculates concentrations from G’s
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance equations• chemical reactions for equilibrium equations• guesses for G’s
Process• calculates concentrations from G’s
• solves for wavelength dependent colors
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance equations• chemical reactions for equilibrium equations• guesses for G’s
Process• calculates concentrations from G’s
• solves for wavelength dependent colors• calculates root mean square of residuals
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Sivvu
Inputs• raw absorbance data• composition of solutions for mass balance equations• chemical reactions for equilibrium equations• guesses for G’s
Process• calculates concentrations from G’s
• solves for wavelength dependent colors• Calculates root mean square of residuals• Searches for G’s that minimize rms residual
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Structure of Residual
400 500 600 700 800 900 1000-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3Residual Factors in Measured Data
1
2
34
5
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Now we know a lot!
[Ni]2+ + py [Nipy]2+ G1 = -6.76(2) kJ/mol
[Nipy]2+ + py [Nipy2]2+ G2 = -3.52(2) kJ/mol
[Nipy2]2+ + py [Nipy3]2+ G3 = 0.64(3) kJ/mol
[Nipy3]2+ + py [Nipy4]2+ G4 = 5.8(5) kJ/mol
[Nipy4]2+ + py [Nipy5]2+ G5 = ?
[Nipy5]2+ + py [Nipy6]2+ G6 = ?
Ni
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So what color is [Nipy]+2?
Ni++
NiPy++
NiPy2++
NiPy3++
NiPy4++
Ni++
NiPy++
NiPy2++
NiPy3++
NiPy4++
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What’s going on in 25th solution?
0
0.2
0.4
0.6
0.8
1
1.2
400 500 600 700 800 900 1000
Wavelength (nm)
As
orb
an
ce
0.0997 M Ni(BF4)2 w/ 0.585 pyridine
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Summary
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Conclusions
Factor Analysis extracts much useful information about complex systems from easy experiments.
Forcing the concentrations to satisfy chemical equilibria greatly enhances stability and sensibleness.
is Uv-vis in reverse.S IV V U
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Acknowledgements American Chemical Society Petroleum Research
Fund
Research Corporation Cottrell College Science Award
Pleotint L.L.C.
Calvin College Research Fellowship
Jack and Lois Kuiper Mathematics Fellowship