“Difficult” problems Marcel Maeder Department of Chemistry University of Newcastle Australia...

“Difficult” problems

Marcel Maeder

Department of Chemistry

University of Newcastle

Australia

[email protected]

A “simple” problem

What is the pH of a 1M solution of acetic acid?

3 3K

K

CH COO H CH COOH

A H AH

What is the pH of a 1M solution of acetic acid?

4.7[ ]10

[ ][ ]

[ ] [ ] 1

AHK

A H

AH A M

2 equations and 3 unknowns: [AH], [A], [H]

Approximation 1: Acetic acid is a weak acid, only a very small fraction dissociates and acetic acid is the only source of protons

2

[ ] 1[ ][ ] [ ]

[ ] 1

and

[ ] [ ]

1[ ]

AHK

A H H

H

H

K

AH M

A

Approximation 2: acetic acid is the only source of protons

2

2

[ ] 1 [ ] 1 [ ][ ][ ] [ ][ ] [ ]

a quadratic equation

[ ] [ ] 1 0

with s

[

ol

] [ ]

ution:

[ ]

A

AH A HK

A H A H

H

H

H

K H

H

But, there are two sources of protons, acetic acid and the autodissociation of water

2

[ ] [ ] [ ] [ ] [ ]

[ ] [ ][ ]

1 [ ][ ] 1 [ ] [ ]

a cubic eqation![ ][ ] [ ][ ] ([ ] )[ ]

[ ]

w

w

w

w

AH A

H O OH

KH A OH A

H

KA H

H

KH

AH A HK

KA H A H H H

H

H

H

Activities

In reality the law of mass action really applies to activities and not concentrations.

{ }{ }{ }

where { } is the activity of the component

{ } [ ]

is the activity coeefficient of A

A

A

AHK

A H

A A

A A

More on that on Thursday

The “simple” problem of calculating the pH of a 1M acetic acid involves the solution of a cubic equation. This is possible, but the equation for that solution is very long and nobody ever uses it.

Things are much more difficult for polyprotic acids.

We need a general solution!

nomenclature

xyz

x y zxX yY zZ X Y Z

[ ]

[ ] [ ] [ ]x y z

xyz x y z

X Y Z

X Y Z

[ ] [ ] [ ] [ ]x y zx y z xyzX Y Z X Y Z

X, Y, Z are components

XxYyZz are species

Example: Cu + ethylene diamine, in water

NN

Components: M (Cu)

L (en)

H

Species: M

L

H

LH

LH2

ML …

Notation Species

m l h Formation constant

M (Cu2+) 1 0 0 β100=1

L (en) 0 1 0 β010=1

H (H+) 0 0 1 β001=1

LH (enH+) 0 1 1 011[ ]

[ ][ ]LH

L H

LH2 (enH22+) 0 1 2 2012 2

[ ]

[ ][ ]

LH

L H

ML (Cu(en)2+) 1 1 0 110[ ]

[ ][ ]ML

M L

ML2 (Cu(en)22+) 1 2 0 2120 2

[ ]

[ ][ ]

ML

M L

ML3 (Cu(en)32+) 1 3 0 3130 3

[ ]

[ ][ ]

ML

M L

MLH (Cu(enH)3+) 1 1 1 111[ ]

[ ][ ][ ]MLH

M L H

MLH-1 (Cu(en)(OH)+) 1 1 -1 -111-1 -1

[ ]

[ ][ ][ ]

MLH

M L H

H-1 (OH-) 0 0 -1 00-1 [ ][ ] WOH H K

3 components

11 species

2 3 -1

2 2 3 -1

2 -1

[ ] [ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] 2[ ] 3[ ] [ ] [ ]

[ ] [ ] [ ] 2[ ] [ ] - [ ] - [ ]

tot

tot

tot

M M ML ML ML MLH MLH

L L LH LH ML ML ML MLH MLH

H H LH LH MLH MLH OH

The task is to compute all species concentrations, knowing the total concentrations of the components and the equilibrium constants

There are 11 unknowns (the species concentrations)

We need 11 equations!

3 equations for the total concentrations of the components:

111 1 1 1

111

. .

[ ]

[ ] [ ] [ ]

[ ] [ ][ ][ ]

e g

MLH

M L H

MLH M L H

8 equations for the species concentrations

2 3 -1

2 2 3 -1

2 -1

[ ] [ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] 2[ ] 3[ ] [ ] [ ]

[ ] [ ] [ ] 2[ ] [ ] - [ ] - [ ]

tot

tot

tot

M M ML ML ML MLH MLH

L L LH LH ML ML ML MLH MLH

H H LH LH MLH MLH OH

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

m l htot mlh

m l htot mlh

m l htot mlh

M m M L H

L l M L H

H h M L H

Generalised equations:

c_spec =beta.*prod(repmat(c',1,nspec).^Model,1); %species conc

c_tot_calc=sum(Model.*repmat(c_spec,ncomp,1),2)'; %comp ctot calc

Allow very compact MATLAB code:

How can we write a MATLAB program that can resolve these systems of many unknowns with many

equations and that can deal with any model?

The Newton-Raphson algorithm

The Taylor series expansion

2 ( )1 1 1( ) ( ) '( ) ''( ) ... ( ) ...

1! 2! !n nf x x f x f x x f x x f x x

n

2 ( )1 1 1( ) ( ) '( ) ''( ) ... ( ) ...

1! 2! !n n

nf x x f x f x x f x x f x x

( ) ( ) '( )f x x f x f x x

( ) ( ) '( )f x x f x f x x

20 22 24 26 28 30 32 34

0x

f(x)

f’(x)

x+x

f(x+x)

The function to minimise is:

The difference between the known total concentrations and the calculated values:

t[ ] [ ] [ ] [ ]

( ) [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

x y ztot xyz

x y ztot xyz

x y ztot xyz

X x X Y Z

Y y X Y Z

Z z X Y Z

d d c

[ ] [ ] [ ]X Y Zc

( )( + ) = ( ) +

d c

d c c d c cc

= +

d(c+c) d(c) c ( )d cc

= +

d(c+c) d(c) c ( )d cc

1( )

= ( )

d cc d c

c

2 31

1 1 1

2 31

2 2 2

2 31

3 3 3

d ddc c c

d ddc c c

d ddc c c

dJ

c

The Jacobian J

1

1

2 3

1 1

2 31

2 2 2

2 31

3 3 3

d dc c

d ddc c

dc

c

d ddc c c

dJ

c

2 1

2

2

[ ] [ ] [ ] [ ]

[ ] [ ]

[ ] [ ] [ ]

[ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ]

[ ]

[ ]

x y ztot xyzX

x y zxyz

x y zxyz

x y zxyz

x y z

X x X Y ZdX X

x X Y Z

X

x X Y Z

x X Y Z

X

x X Y Z

X

2

2

2 [ ] [ ]

[ ] [ ]

[ ] [ ] [ ]( )[ ] [ ] [

[ ]

[ ]

]

[ ] [ ] [ ]

[ ] [ ] [ ]

x y z x y z

x y z x y z x y z

x y z

x y z

x y z x y z

xy X Y Z xz X Y Z

X X

xy X Y Z y X Y Z yz X Y Z

Y Y Y

xz X Y Z yz X Y Z z X Y Z

Z Z

Y Z

X

Z

x X

d cJ

c

Compare actual total conc. with computed ones

e.g.

Guess initial concentrations for the free component concentrations

[X], [Y], [Z]

Calculate the concentrations of all species

e.g. XxYyZz=xyz [X]x[Y]y[Z]z

Calculate J acobian

Calculate shift vector and add to component concentrations

Difference

>0

0Exit

Calculate the total concentration of the components

e.g. _[ ] [ ] [ ] [ ]x y ztot calc xyzX x X Y Z

_[ ] [ ]x tot tot calcd X X

Compare actual total conc. with computed ones

e.g.

Guess initial concentrations for the free component concentrations

[X], [Y], [Z]

Calculate the concentrations of all species

e.g. XxYyZz=xyz [X]x[Y]y[Z]z

Calculate J acobian

Calculate shift vector and add to component concentrations

Difference

>0

0Exit

Calculate the total concentration of the components

e.g. _[ ] [ ] [ ] [ ]x y ztot calc xyzX x X Y Z

_[ ] [ ]x tot tot calcd X X

• Kinetics at non-constant temperature

• Kinetics and equilibria, taking into account changes in ionic strength

• Kinetics at non-constant pH

Why?

• In traditional measurements temperature, ionic strength, pH are kept constant using buffers, excess inert salt, thermostatting

• All these external means are only required to simplify the computations. If it is possible to take changes in these parameters into account, buffers etc are no more required

Kinetics at non-constant temperature

Example: 1 3

2

k k

k A B C

1 2

1 2 3

3

[ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ]

A k A k B

B k A k B k B

C k B

Differential equations cannot be solved explicitly → numerical integration: Euler, Runge-Kutta, MATLAB ODE solvers

This function computes the derivatives of the concentrations vs. time (c_dot) at time t for a given set of concentrations c.

The rest is done by the ODE-solver

1 2

1 2 3

3

[ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ]

A k A k B

B k A k B k B

C k B

Translation into MATLAB language, the function ode_AeqBtoC.m is used by the MATLAB ode-solvers

function c_dot=ode_AeqBtoC(t,c,flag,k)

% A <-> B

% B --> C

c_dot(1,1)=-k(1)*c(1)+k(2)*c(2); % A_dot

c_dot(2,1)= k(1)*c(1)-k(2)*c(2)-k(3)*c(2); % B_dot

c_dot(3,1)= k(3)*c(2); % C_dot

ode_AeqBtoC.m

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

conc

.

Fast first reaction and slow second reaction

Difficult to observe both in one measurement

c0=[1;0;0]; % initial conc of A, Cat, B and C

k=[.2;.1;.01]; % rate constants k1 and k2

times=[0:1:300]';

[times,C] = ode45('ode_AeqBtoC',times,c0,[],k); % call ode-solver

figure(1); plot(times,C) % plotting C vs t

xlabel('time');ylabel('conc.');

AeqBtoC.m

Arrhenius equation

Rate constants are temperature dependent, only constant at constant temperature.

Arrhenius equation describes rate constant k as a function of the temperature (Eyring equation could be used as well)

-1 -1

A = pre-exponential factor

Ea = activation energy

R = gas constant ( = 8.314 J K mol )

T = temperature in K

Ea-RTk =Ae

Arrhenius plotEa-RT

Ea 1ln(k)=ln(A)- ×

R T

k =Ae

ln(k)

1/T

slope: -Ea/R

A function is required that computes the derivatives of the concentrations vs. time, at one particular time t

At that time the rate constants are computed using the Eyring parameters, based on the temperature T at that time

T has to be interpolated from the temperatures temp measured at the times times

function c_dot=ode_AeqBtoC_T(t,c,flag,k,temp,A,Ea,times)

% A <-> B

% B --> C

R=8.314; % gas constant J K-1 mol-1

T=lolipop(times,temp,t,2,5); % interpolation to comp T at particular time t

k=A.*exp(-Ea./(R*(T+273))); % rate constants at T

c_dot(1,1)=-k(1)*c(1)+k(2)*c(2); % A_dot

c_dot(2,1)= k(1)*c(1)-k(2)*c(2)-k(3)*c(2); % B_dot

c_dot(3,1)= k(3)*c(2); % C_dot

ode_AeqBtoC_T.m

A =[12.13 1.77 36.79]; % pre-exp factor for each rate constant

Ea=[1e4 7e3 2e4]; % activation energies

temp=20+0.2*times; % temperatures at measurement times

[times,C] = ode45('ode_AeqBtoC_T',times,c0,[],k,temp,A,Ea,times); % ode-solver



AeqBtoC.m … contiuned

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

conc

.

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

conc

.at 20°C at 20-80°C

Similar to temperature program in GC or non-isocratic mobile phase in HPLC

Modeling - Fitting

• The core of the fitting algorithm is the modeling of the measurement and in most cases this is the computation of the matrix C

• This is exactly what we have done so far for non-isothermal kinetics• The parameters used to model the concentration profiles can, at least

potentially, also be fitted to appropriate data• In this case non-isothermal measurements allows the determination

of the activation parameters of all rate constants• Much more convenient than classical method of repeating the

measurement at different temperatures with subsequent Arrhenius analysis

Newton-Gauss Algorithmii) Calculation the residuals and ssq according to model and estimated parameters

Calculation of residual matrix

R = A – C (C\A)

400 420 440 460 480 500 520 540 560 580 600-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

Wavelength (nm)

Abs

orba

nce

400 420 440 460 480 500 520 540 560 580 600-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Wavelength (nm)

Abs

orba

nce

400 420 440 460 480 500 520 540 560 580 6000

0.2

0.4

0.6

0.8

1

1.2

1.4Simulated Spectra

Wavelength (nm)

Abs

orba

nce

--- =

ssq = SS r2I,j = 1.66

Kinetics, taking into account changes in ionic strength

1

2

k

k A B C

1 2

1 2

[ ] [ ] [ ][ ] [ ]

[ ] [ ][ ] [ ]

A B k A B k C

C k A B k C

1 2

1 2

[ ] [ ] { }{ } { }

[ ] { }{ } { }

A B k A B k C

C k A B k C

{A} = activity of species A

Kinetics, taking into account changes in ionic strength

Activity coefficients

For Ionic compounds, activity coefficients can be approximated in dilute solutions as (Debye-Hückel):

With A parameter depending on dielectric constant of solvent, in water A~0.51

zi charge of species

Ionic strength of solution, computed as:

{ } [ ]AA A

2 log

1

A z

21[ ]

2 i ii

C z

1

2

k

k A B C

1 2

1 2

[ ] [ ] { }{ } { }

[ ] { }{ } { }

A B k A B k C

C k A B k C

2 log ...

1A

AA z

2 2 21([ ] [ ] [ ] )

2 A B CA z B z C z

As the concentrations change during the reaction, the ionic strength and all activity coefficients change as well.

Example

1

2

2 34 4( )

k

kSO Fe Fe SO

1M Na2SO41M Fe(ClO4)3 1M Na+

0.5 M SO42-

0.5 M Fe3+

1.5 M ClO4-

+ →

1M Na+

0.5 M SO42-

0.5 M Fe3+

1.5 M ClO4-

2 2 3 24 4

2 2

1([ ] [ ]2 [ ]3 [ ])

21

(1 0.5 2 0.5 3 1.5)21

(1 2 4.5 1.5)24.5

Na SO Fe ClO

2

24

2

3

0.5 1 4.5

1 4.5

0.5 2 4.5

1 4.5

0.5 3 4.541 4.5

10 0.46

10 0.044

10 8.7 10

Na

SO

Fe

Note: Debye-Hückel equation not valid at such high concentrations !

1

2

k3+ 2- +4 4k

Fe +SO Fe(SO )

3 2 3 24 1 4 2 4

3 24 1 4 2 4

[ ] [ ] { }{ } { ( ) }

[ ( ) ] { }{ } { ( ) }

Fe SO k Fe SO k Fe SO

Fe SO k Fe SO k Fe SO

4 4

3 2 3 24 1 4 2 4

3 21 4 2 4

3 21 4 2 4

4

[ ] [ ] [ ] [ ] [ ( ) ]

[ ][ ] 0.044 8.7 10 0.4 [ ( ) ]

[ ][ ] [ ( ) ]

6

Fe SO FeSOFe SO k Fe SO k Fe SO

k Fe SO k Fe SO

k Fe SO k Fe SO

-53.8×10 0.46

The kinetics and the equilibrium position are dramatically affected by the very small activity coefficients !

function c_dot=ode_AplusBeqC_I(t,c,flag,k,c_I,charges,mode,A)

% A + B <--> C including activities

[gamma,mu] = act_coeffs_I([c;c_I],charges,mode,A); % act coeff and ionic strength

act=c.*gamma(1:length(c)); % activities instead of conc.

% c_dot(1,1)=- k(1)*c(1)*c(2)+k(2)*c(3); % A_dot ignoring activities

c_dot(1,1)=-k(1)*act(1)*act(2)+k(2)*act(3); % A_dot

c_dot(2,1)= c_dot(1,1); % B_dot

c_dot(3,1)= -c_dot(1,1); % C_dot

c_dot=ode_AplusBeqC_I.m

function [gamma,mu] = act_coeffs (conc,charges,mode,A)

% calculating activity coefficients

mu = sum(1/2*(conc.*(charges.^2)));

if mode==1

gamma = ones(size(conc));

elseif mode==2

log_gamma = (-A*(charges.^2)*(mu^0.5))/(1+(mu^0.5)); % defining log g

gamma = 10.^log_gamma; % defining activity coefficient, gamma

end

2 log

1

A z

21[ ]

2 i ii

C z

0 50 100 150 200 250 3000

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

time

conc

.

0 50 100 150 200 250 3000

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

time

conc

.

0.01M Na2SO4

0.01M Fe(ClO4)3

k1=10k2=0.01

without with ionic strengthKinetics is much slower but also equilibrium position very different !

% AplusBeqC

% A + B <--> C

c0=[.01;.01;0]; % initial conc of Fe, SO4 and Fe(SO4)

k=[10;.01]; % rate constants k1 and k2

times=[0:1:300]';

[times,C] = ode45('ode_AplusBeqC',times,c0,[],k); % call ode-solver



c_I=[2*c0(1); 3*c0(2)]; % Na ClO4

charges=[3; -2; 1; 1; -1]; % Fe SO4 Fe(SO4) Na ClO4

mode=2;

A=0.51;

[times,C] = ode45('ode_AplusBeqC_I',times,c0,[],k,c_I,charges,mode,A);, % ode-solver



AplusBeqC.m

Equilibrium calculations are more complex:

In kinetics, using an ODE solver, the only requirement is to compute the derivatives of the concentrations vs. time for a given set of concentrations. Thus, it is possible to calculate the activity coefficients at that moment and incorporate into the derivatives.

In equilibria, the activity coefficients depend on the concentrations and the concentrations depend on the activities. An internal iterative process is required.

Titration of phosphoric acid:1

2

3

3 24 4

34 2 4

34 3 4

2

3

PO H HPO

PO H H PO

PO H H PO

24

24

2 24 4

1 3 34 4

{ } [ ]

{ }{ } [ ][ ]HPO

HPO H

HPO HPO

PO H PO H

3 2 2 24 4

1([ ] [ ] 3 +[ ] 2 + )

2H PO HPO

2 4

0.5 1

110H H PO

Iterative approach:

Guess -valuesCalculate concentrationsCalculate new -values

Newton-Raphson algorithm

• The core is the computation of the difference d between actual and computed total concentrations c, and

• The derivatives of these with respect to the total concentrations

dc

OLD

c_spec =beta.*prod(repmat(c',1,nspec).^Model,1); % species conc

c_tot_calc=sum(Model.*repmat(c_spec,ncomp,1),2)'; % comp ctot calc

d =c_tot-c_tot_calc; % diff actual and calc total conc

NEW, with activity coefficients

it=1;

while it<99

act_comp = c.*gamma(1:ncomp);

act_spec = beta.*prod(repmat(act_comp',1,nspec).^Model,1); % species activities

c_spec = act_spec./gamma(1:nspec); % species activities

[gamma_new,mu] = act_coeffs_I([c_spec c_tot_I],charges,mode,A); % calc gamma for next it

if all(abs(gamma-gamma_new)<1e-3)

break

end

gamma=gamma_new;

it=it+1;

end

c_tot_calc = sum(Model.*repmat(c_spec,ncomp,1),2)'; % comp tot

d = c_tot-c_tot_calc; % diff actual and calc total conc

3 22 4 2 4{ } { }{ } H PO PO H

34

3 34 4{ } [ ]

{ } [ ]

PO

H

PO PO

H H

2 42 4 2 4[ ] { }/

H POH PO H PO

22 4

1([ ] 1 )

2H PO

2 4

0.5 1

110 H PO

constant ?no

yes

guess

act_comp = c.*gamma(1:ncomp);

act_spec = beta.*prod(repmat(act_comp',1,nspec).^Model,1);

c_spec = act_spec./gamma(1:nspec);

[gamma_new,mu]= act_coeffs_I([c_spec c_tot_I],charges,mode,A);

if all(abs(gamma-gamma_new)<1e-3)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.0350

2

4

6

8

10

12

vol. added (L)

pH

2 3 4 5 6 7 8 9 10 11 120

0.005

0.01

0.015

0.02

0.025

10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 11

0

1

2

3

4

5

6

7

8

9

x 10-3

10 ml of 0.03M H3PO4 (pK 11,7,3)

titrated with 35ml of 0.01M NaOH

HPO42- PO4

3-

[HPO42-] [PO4

3-] at pH = pK

Kinetics at non-constant pH

example:cyclam + Cu2+ → Cu(cyclam)2+

N N

NN

Cu

NH HN

HNNH

2++ Cu

1

2

3

4

K+ +

K+ + 2+2

K2+ + 3+2 3

K3+ + 4+3 4

L + H LH

LH + H LH

LH + H LH

LH + H LH

NH HN

HNNH

Cyclam has 4 secondary amines that are protonated depending on the pH of the solution

4 protonation equilibria and 4 protonation constants define all concentrations [L], [LH+], …

L

L

LH

H

L

3

H

2

LH

4

k2+ 2+ 2+ +2

k2+ 3+ 2+ +3

k2+ 2+

k2+ + 2+ +

k2+ 4+ 2+4

Cu + L ML only at very high pH

Cu + LH ML + H only at very hig

Cu

h pH

Cu + LH ML

+ LH ML +2H

Cu + LH ML +3H

++4H only at very low pH

N N

NN

Cu

2+LH2k

2++ Cu

Each of the five differently protonated forms of cyclam reacts with Cu2+ with a particular rate constant.

E.g.

NH2+ HN

+H2NNH

LH2

LH3

k2+ 2+ 2+ +2

k2+ 3+ 2+ +3

Cu + LH ML +2H

Cu + LH ML +3H

c_dot(1,1)=-k(1)*c(1)*c(2)-k(2)*c(1)*c(3); % M_dot

c_dot(2,1)= c_dot(1,1); % L_dot

• At any time we need to know [Cu], [LH2] and [LH3].

• [Cu] and [L]tot are treated in the usual way by the ode-solver

• The total concentration of L, [L]tot=[L]+[LH+]+…+[LH44+], is

defined by the kinetics

• [LH2] and [LH3] are computed, knowing [L]tot , [H]tot and the equilibrium constants

function c_dot=ode_CuCyclam (t,c,flag,k,c_0,equil_model,beta)

% kinetics

% M + LH > ML + H

% M + LH2> ML + 2H

% equilibria

% L,LH,LH2,LH3,LH4,OH

% c =[M Ltot]

% c_0 =[M_0, Ltot_0, H_tot]

c_tot=[c(2),c_0(3)]; %[Ltot,Htot], the conc of free protons is constant

c_comp_guess = [1e-10 1e-10]; % initial guess for [L],[H]

L_eq=NewtonRaphson(equil_model, beta, c_tot, c_comp_guess,i); % [L] [H] [LH] [LH2] [LH3] [LH4] [OH]

c=[c(1) L_eq(3) L_eq(4)]; % [M], [LH], [LH2]

c_dot(1,1)=-k(1)*c(1)*c(2)-k(2)*c(1)*c(3); % M_dot

c_dot(2,1)= c_dot(1,1); % L_dot

ode_CuCyclam.m

%Main_CuCyclam

c_0=[0.004 0.005 0.015]; % initial concs Ltot,M,H

k=[1050000 0.135]; % rate constants

times=[1:100:10000]';

spec_names = {'L' 'H' 'LH' 'LH2' 'LH3' 'LH4' 'OH'};

equil_model = [ 1 0 1 1 1 1 0; ... % L

0 1 1 2 3 4 -1] ; % H

log_beta = [ 0 0 11.59 22.21 23.82 26.24 -14];

beta =10.^log_beta;

[times,C] = ode45('ode_CuCyclam',times,c_0(1:2),[],k,c_0,equil_model,beta); % C=(Ltot M)

c_comp_guess = [1e-10 1e-10]; % computing L_eq from Ltot and H

for i=1:length(times)

L_eq(i,:)=NewtonRaphson(equil_model, beta, [C(i,1) c_0(3)], c_comp_guess,i); % [L] [H] [LH] [LH2] [LH3] [LH4] [OH]

c_comp_guess = L_eq(i,1:2); % using calc conc as estimates for next iteration

end

plot(times,C,'.',times,L_eq);

Main_CuCyclam.m

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000

0.002

0.004

0.006

0.008

0.01

0.012

0.014

[M] [L]tot

[H][LH2]

[LH3]

[LH4]

c_0=[0.004 0.005 0.015]; % initial concs Ltot,M,H

“Difficult” problems Marcel Maeder Department of Chemistry University of Newcastle Australia...

Documents

Transcript of “Difficult” problems Marcel Maeder Department of Chemistry University of Newcastle Australia...