Solving ODEs for chemical rate equations
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Oliver Wilson
on 15 Nov 2022
Commented: Star Strider
on 15 Nov 2022
I am trying to model a chemical equation of the form rate = k [a]^1 *[b]^-1. I have managed witht the help of some you tube resources to model a an equation for one reactant but am unsure of how to include the second. Here is the code i have got so far and any help would be appreciated.
clear all
close all
clc
K = 0.06;
timespan = [0 30]';
A0 = 0.5;
first = @(t,A) -K*A
[t,A_calc] = ode45(first,timespan,A0)
plot(t,A_calc, 'o','Linewidth',1,'Markersize',10)
2 Comments
Accepted Answer
Star Strider
on 15 Nov 2022
Here is prototype code for integrating the solution for a different kinetic parameter estimation problem with four rate equations —
t=[0.1
0.2
0.4
0.6
0.8
1
1.5
2
3
4
5
6];
c=[0.902 0.06997 0.02463 0.00218
0.8072 0.1353 0.0482 0.008192
0.6757 0.2123 0.0864 0.0289
0.5569 0.2789 0.1063 0.06233
0.4297 0.3292 0.1476 0.09756
0.3774 0.3457 0.1485 0.1255
0.2149 0.3486 0.1821 0.2526
0.141 0.3254 0.194 0.3401
0.04921 0.2445 0.1742 0.5277
0.0178 0.1728 0.1732 0.6323
0.006431 0.1091 0.1137 0.7702
0.002595 0.08301 0.08224 0.835];
theta0 = rand(10,1);
[theta,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@kinetics,theta0,t,c,zeros(size(theta0)));
fprintf(1,'\tRate Constants:\n')
for k1 = 1:numel(theta)
fprintf(1, '\t\tTheta(%2d) = %8.5f\n', k1, theta(k1))
end
tv = linspace(min(t), max(t));
Cfit = kinetics(theta, tv);
figure
hd = plot(t, c, 'p');
for k1 = 1:size(c,2)
CV(k1,:) = hd(k1).Color;
hd(k1).MarkerFaceColor = CV(k1,:);
end
hold on
hlp = plot(tv, Cfit);
for k1 = 1:size(c,2)
hlp(k1).Color = CV(k1,:);
end
hold off
grid
xlabel('Time')
ylabel('Concentration')
legend(hlp, compose('C_%d',1:size(c,2)), 'Location','best')
function C=kinetics(theta,t)
c0=theta(end-3:end);
[T,Cv]=ode45(@DifEq,t,c0);
%
function dC=DifEq(t,c)
dcdt=zeros(4,1);
dcdt(1)=-theta(1).*c(1)-theta(2).*c(1);
dcdt(2)= theta(1).*c(1)+theta(4).*c(3)-theta(3).*c(2)-theta(5).*c(2);
dcdt(3)= theta(2).*c(1)+theta(3).*c(2)-theta(4).*c(3)+theta(6).*c(4);
dcdt(4)= theta(5).*c(2)-theta(6).*c(4);
dC=dcdt;
end
C=Cv;
end
Each rate equation needs to be coded separately (here in the ‘DifEq’ function), then the integration can proceed.
.
2 Comments
Star Strider
on 15 Nov 2022
As always, my pleasure!
I would code that as:
dcdt(1) = K(1)*C(2)/C(1);
where ‘K’ is a vector of parameters to be estimated (‘theta’ in the code I posted), ‘C(1)’ is the concentration of ‘a’ and ‘C(2)’ is the concentration of ‘b’. Using the -1 exponent to indicate division is inefficient. Just do the division directly.
Code the other equations similarly.
I keep track of the vector elements that represent concentrations in a comment line, for example:
% C(1) = a, C(2) = b, C(3) = c
and if necessary, do the same thing for the parameters:
% K(1) = K_a, K(2) = K_b
in order to not lose track of them. That also makes it easier to display them correctly later.
.
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