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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?
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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: • 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 • 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
Ni 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+…
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 Ni2+ Absorbance = ·Concentration (Beer-Lambert Law)
Color is Additive py Ni2+ BF4- 0.0997 M Ni(BF4)2 w/ 0.585 pyridine Abs = 0[Ni]2+ + 1[Nipy]2+ + 2[Nipy2]2+ + 3[Nipy3]2++ …
Ni 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+…
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
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++ …
Matrix Form of Beer-Lambert Law Absorbances (n x p) (n x m) Concentration (m x p) Molar Absorbtivity 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). Each row represents one of the m chemical species. Concentration (m x p) This smaller concentration matrix contains the absolute concentration of each chemical species in each of the solution mixtures.
Matrix Absorbance Equation Abs = C 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.
Absorbances (n x p) e (n x m) Concentration (m x p) With Residual Error! Abs = eC + R 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 (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 Structure m = 6 6th eigenvalue is relatively small, but still possibly significant.
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 From 0 to 142 equivalents of pyridine. Pure pyridine (12.4 Molar) is dripped into 0.10 M Ni(BF4)2 Solution
Sivvu • Inputs • raw absorbance data • composition of solutions for mass balance equations • chemical reactions for equilibrium equations
Sivvu • Inputs • raw absorbance data • composition of solutions for mass balance equations • chemical reactions for equilibrium equations • guesses for 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
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
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 = ?
What’s going on in 25th solution? 0.0997 M Ni(BF4)2 w/ 0.585 pyridine