No Data Left Behind
Modeling Colorful Compounds in
Chemical Equilibria
Mike DeVries
D. Kwabena Bediako
Prof. Douglas A. Vander Griend
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?
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.
Example
Seal .04 mole of NO2 in a 1 liter container.
2NO2  N2O4
G = -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
Demonstration
AB
Secretly choose A or B and submit choice.
As directed, choose A or B again:
1.
2.


If last time you submitted A, now submit B.
If last time you submitted B,


3.
Submit B again if coin flip shows tails
Submit A if coin flip shows heads
Repeat #2 as directed.
Multiple Equilibria

Solving one equilibrium equation can be
tricky.

Solving simultaneous equilibria requires a
computer.
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
N2 + O2  2NO
2H2 + O2  2H2O
G = -187.52 kJ/mol
G = 125.02 kJ/mol
G = -247.86 kJ/mol
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
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+
Ni
G4
[Nipy4]2+ + py  [Nipy5]2+
G5
[Nipy5]2+ + py  [Nipy6]2+…
Pyridine and Nickel
Ni2+ and
BF4- in
methanol
equivalents
of pyridine
added:
1
2
3
4
5
UV-visible Spectroscopy
0.1046 M
Ni(BF4)2 in
methanol
0.9
0.8
Ni2+
Absorbance
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
400
500
600
700
800
900
1000
Wavelength(nm)
Absorbance = ·Concentration
(Beer-Lambert Law)
Color is Additive
py
1.2
Absorbance
1
0.8
0.6
0.4
0.2
0.0997 M
Ni(BF4)2 w/
0.585 pyridine
0
400
500
600
700
800
900
Wavelength(nm)
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …
1000
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+
Ni
G4
[Nipy4]2+ + py  [Nipy5]2+
G5
[Nipy5]2+ + py  [Nipy6]2+…
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.
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.
Why it Works
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …





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.
Matrix Algebra Refresher
2x + 3y = 8
3x – y = 5
2 3   x  8
3  1  y   5

   
Matrix Multiplication
C = AB
(n x p)
(n x m)
(m x p)
Matrix Algebra
Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2+ + …

Abs    0
1
 Ni2 

2 

 2   Nipy  
Nipy  2 
2



Matrix Form of Beer-Lambert Law
Abs p    C
n
n
m
m
p
m
 C
Absorbances
(n x p)
Abs  

component
Molar
Absorbtivity
Abs  
(n x m)
Concentration
 C
(m x p)

component
m
n  # of wavelengths
m  # of chemical species
p  # of mixture solutions
Measured Absorbances
Every column is a UV-vis curve.
Every row is a wavelength
Absorbances
(n x p)
n = number of wavelengths
p = number of solution mixtures
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.
Molar Absorptivities
Every column represents one of the m chemical species.
Every row is a wavelength

(n x m)
This smaller matrix contains all of the molar
absorbtivity values for each pure chemical
species in the mixtures at every wavelength.
Component Concentrations
Every column corresponds to one of the p solution mixtures (UVvis curve).
Concentration
(m x p)
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.
Matrix Absorbance Equation
Abs = C
Concentration
(m x p)
 C
Absorbances
(n x p)
Abs  

component

(n x m)
So the problem is
essentially factoring a
matrix.
…Or solving np
equations for mn + mp
unknowns.
With Residual Error!
Abs = C + R
Concentration
(m x p)
Absorbances
(n x p)

(n x m)
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.
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.

Random Matrix Structure
Eigenvalue Significance
1.4
1.2
1
0.8
(16 x 461) matrix of random numbers
0.6
0.4
All 16 eigenvalues contribute about the same to the structure of the matrix
0.2
0
1
2
3
4
5
6
7
8
9
Component
10
11
12
13
14
15
16
Non-random Data
50 data curves of Ni2+
solution with zero to 142
equivalents of pyridine.
How many additive factors exist in this data?
-0.15
-0.2
400
500
600
700
800
Wavelength (nm)
900
1000
Data Matrix Structure
Relative Significance
7
absorbance data
random numbers
6
5
m=6
4
3
6th eigenvalue is relatively
small, but still possibly
significant.
2
1
0
10
20
30
Factor Ratios
40
50
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.
Sivvu

Inputs
•
raw absorbance data
Absorbance Data
n = 305 wavelengths
p = 50 solution mixtures
Sivvu

Inputs
•
•
raw absorbance data
composition of solutions for mass balance
equations
Composition of Solutions
1
Ni++
Log(Concentration)
0.5
Py
0
From 0 to 142
equivalents of
pyridine.
-0.5
-1
-1.5
Pure pyridine (12.4 Molar) is dripped into 0.10 M Ni(BF4)2 Solution
-2
1
4
7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55
Solution number
Sivvu

Inputs
•
•
•
raw absorbance data
composition of solutions for mass balance
equations
chemical reactions for equilibrium equations
Inputs
Sivvu

Inputs
•
•
•
•
raw absorbance data
composition of solutions for mass balance
equations
chemical reactions for equilibrium equations
guesses for G’s
Inputs
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
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
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
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
Structure of Residual
Residual Factors in Measured Data
0.3
1
2
3
4
5
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
400
500
600
700
800
900
1000
Ni
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 = ?
So what color is [Nipy]+2?
Ni++
NiPy++
NiPy2++
NiPy3++
NiPy4++
Ni++
NiPy++
NiPy2++
NiPy3++
NiPy4++
What’s going on in 25th
solution?
1.2
0.0997 M
Ni(BF4)2 w/
0.585 pyridine
Asorbance
1
0.8
0.6
0.4
0.2
0
400
500
600
700
800
Wavelength (nm)
900
1000
Summary
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.

SIVVU
is Uv-vis in reverse.
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