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Linear Equations in Chemistry
Example1: Adsorbtion Spectroscopy
Some examples of typical problems
λ
A brief review …
Determinants
Quantum mechanics
Need to find the concentration of each species
from the total adsorbtion A(λ)
Beer-Lambert
To Determine the Concentrations
The Beer-Lambert Law: in a multicomponent
system the total adsorbtion is the sum of the
contributions from each component.
Say a 4 component system;
Atotal (λ ) = lε 1c1 + lε 2 c2 + lε 3c3 + lε 4 c4
l – thickness of the sample
εi – adsorbtion of species I
ci – concentration of species I
(known)
(known)
(unknown)
IR Data for a 4 Component System
pxylene
mxylene
oxylene
ethylAtotal
benzene
λ
εl
εl
εl
εl
12.5
1.502
0.0514
0
0.0408
0.1013
13.0
0.0261
1.1516
0
0.0820
0.09943
13.4
0.0342
0.0355
2.532
0.2933
0.2194
14.3
0.0340
0.0684
0
0.3470
0.03396
How do we solve for the concentrations ?
How about for a 20 component system ?
In a 4 component system:
Measure at 4 different wavelengths
Set up 4 simultaneous equations and solve !
Atotal (λ1 ) = lε 1 (λ1 )c1 + lε 2 (λ1 )c2 + lε 3 (λ1 )c3 + lε 4 (λ1 )c4
Atotal (λ2 ) = lε 1 (λ2 )c1 + lε 2 (λ2 )c2 + lε 3 (λ2 )c3 + lε 4 (λ2 )c4
Atotal (λ3 ) = lε 1 (λ3 )c1 + lε 2 (λ3 )c2 + lε 3 (λ3 )c3 + lε 4 (λ3 )c4
Atotal (λ4 ) = lε 1 (λ4 )c1 + lε 2 (λ4 )c2 + lε 3 (λ4 )c3 + lε 4 (λ4 )c4
Example 2: Chemical Kinetics
The Arrhenius equation takes the linear form
ln(k ) = ln( A) −
Ea
RT
Measure at 2 different temperatures to get 2
simultaneous equations to solve for A and Ea
Ea
RT1
E
ln(k 2 ) = ln( A) − a
RT2
ln(k1 ) = ln( A) −
1
Other Examples
Balancing complex chemical equations
Mass spectroscopic analysis of mixtures
Reminder: Solution by Inspection
1
x + 8y = 9
2
2x + y = 5
x2
2 x + 32 y = 36
2x + y = 5
Subtract
Regression (fitting)
– Michaelis-Menten [S](k) fitting
– All model fitting
Quantum Mechanics – Huckel Theory
Reminder: Graphical Interpretation
Solving 2 linear equations is simply finding
where two lines cross
y =1
x=2
31 y = 31
OK – but for multiple equations this process rapidly
becomes tedious & eventually intractable
The General Problem (2 variables)
Find x and y given;
a11 x + a12 y = b1
a21 x + a22 y = b2
y1
x1
Solving n equations is where n lines cross in
an n-dimensional space !
Where a11, a12, a21,a22 and b1, b2 are known
constants
The General Solution (2 variables)
x=
b1a22 − b2 a12
a11a22 − a12 a21
y=
b2 a11 − b1a21
a11a22 − a12 a21
Note: the denominators are the same and can
be written as a determinant;
a11
a21
a12
≡ a11a22 − a12 a21
a22
Determinants (2x2)
(top left * bottom right) - (top right * bottom left)
a11
a21
a12
≡ a11a22 − a12 a21
a22
2
General Solution in Determinants (2x2)
b1 b2
a
a22
x = 12
a11 a12
a21 a22
b1 b2
a
a
y = 21 11
a11 a12
a21 a22
For example
1
x + 8y = 9
2
2x + y = 5
Determinants are the practical method for
solving linear problems with many unknowns
Determinants ….
And it is…
a11
-
a22
a32
a11
a21
a31
a11
The Rules …. !
a12
a22
a32
+
a13
a23 =
a33
Take each element in turn and multiply
it by the determinant of the 2x2 matrix
left after crossing out its row and column
a23
a21
+ a13
a33
a31
y=
− 15 1
b2 a11 − b1a21
2 =1
=
a11a22 − a12 a21 − 15 1
2
a12
a22
a32
a22
a32
a13
a23 =
a33
a23
a21
− a12
a33
a31
a23
a21
+ a13
a33
a31
a22
a32
For example..
Associate a sign with each of the
elements of the first row
a23
a21
− a12
a33
a31
b1a22 − b2 a12
− 31
=
=2
a11a22 − a12 a21 − 15 1
2
A 3x3 determinant can be computed in terms
of 2x2 determinants as follows…
There is a simple prescription for computing
them …..
+
x=
Computing Determinants (3x3 case)
Allow you to solve large sets of linear
equations and …
a11
a21
a31
1
a12 = 8 b1 = 9
2
a21 = 2 a22 = 1 b2 = 5
a11 =
a22
a32
2 1
= 2 × 4 − 1× 3 = 8 − 3 = 5
3 4
1 −1 2
0 3
0 = 1× (3 × −2) − −1(0) + 2(0 − 3 × 2)
2 −2 −2
= −6 + 0 − 12 = −18
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Terminology
Writing in Terms of Cofactors
The signed 2x2 determinants associated with
each element of the first row are called
cofactors. The cofactor of a11 is A11 …
a11
a21
a31
a12
a22
a32
a13
a23
a33
A11 =
a22
a32
a11
a21
a31
a12
a22
a32
a13
a23 = a11 A11 + a12 A12 + a13 A13
a33
a23
a33
Note that the minus sign is part of A12
The General Case
The 4x4 Case
The determinant of any array can be
computed in this way.
a11
a21
a31
a41
For example;
A 4x4 determinant can be expanded in the in
terms of 3x3 cofactors, which in turn get
expanded in terms of 2x2 cofactors…
Computers are very good at this !
Summary
Systems of linear equations turn up all the
time in chemistry.
With,
a12
a22
a32
a42
a13
a23
a33
a43
a14
a24
= a11 A11 + a12 A12 + a13 A13 + a14 A14
a34
a44
a22
a23
a24
A11 = + a32
a33
a34
a42
a43
a44
etc.
Quantum Mechanics
The interactions between atoms are governed
by the quantum mechanical behaviour of the
electrons.
These problems can be solved using
determinants.
There is a simple (but involved!) method for
computing determinants of arbitrary size.
Lets look in the simplest way at the reason for
the formation of a chemical bond…
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