no data left behind

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 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

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.

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 G = -187.52 kJ/mol

• N2 + O2 2NO G = 125.02 kJ/mol

• 2H2 + O2 2H2O 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



[Nipy5]2+ + py [Nipy6]2+…

Ni

Pyridine and Nickel

Ni2+ and BF4-

in methanol

equivalents of pyridine

added: 1 2 3 4 5

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+

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

What we know is not very much

[Ni]2+ + py [Nipy]2+ G1

[Nipy]2+ + py [Nipy2]2+ G2




[Nipy5]2+ + py [Nipy6]2+…

Ni

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

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

5

8

y

x

13

32

2x + 3y = 8

3x – y = 5

Matrix Multiplication

C = AB(n x p) (n x m) (m x p)

Matrix Algebra


22

2

2

210

Nipy

Nipy

Ni

Abs

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

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.

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.

Component Concentrations


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.

Matrix Absorbance Equation

component

CAbs

Absorbances(n x p)

(n x m)


So the problem is essentially factoring a matrix.

…Or solving np equations for mn + mp unknowns.

Abs = C

With Residual Error!

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.

Abs = C + R

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

(16 x 461) matrix of random numbers

All 16 eigenvalues contribute about the same to the structure of the matrix

Non-random Data

50 data curves of Ni2+

solution with zero to 142 equivalents of pyridine.

How many additive factors exist in this data?

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

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

-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

Sivvu


equations• chemical reactions for equilibrium equations

Inputs

Sivvu


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


Process• calculates concentrations from G’s

• solves for wavelength dependent colors

Sivvu



• solves for wavelength dependent colors• calculates root mean square of residuals

Sivvu



• solves for wavelength dependent colors• Calculates root mean square of residuals• Searches for G’s that minimize rms residual

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

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

So what color is [Nipy]+2?

Ni++

NiPy++

NiPy2++

NiPy3++

NiPy4++

Ni++

NiPy++

NiPy2++

NiPy3++

NiPy4++

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

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.

is Uv-vis in reverse.S IV V U

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

no data left behind

Documents