0

The first order reactions A → B ↔ C occurs within a tank. The rate constants are k1=0.1 hr-1

,

k2=0.2 hr-1

and k3=0.1 hr-1

for A → B, B →C and C → B, respectively. The initial concentrations

of species are CA0=5 M and CB0=CC0=0.

a) Derive the set of differential equations describing the change in concentration three

species in the tank.

1

b) Solve the system of equations and plot the change in concentration of each species over

time (in the same figure) by using MATLAB. Select an appropriate time interval for the

integration.

2

MATLAB Code;

%CHE 386 - HW 2-question 1-b

%Pınar ALP, Gizem BAYRAM, Semra TUNCAY

close all

clear all

format compact

format long

clc

tic

% time interval for the integration can be taken as an input

Tfinal = input('Please give an end time [hr]: ');

k1=1;

k2=2;

k3=1;

Ca0=5;

Cb0=0;

Cc0=0;

options = odeset('RelTol',1e-10,'AbsTol',[1e-10 1e-10 1e-15]);

[T, C]= ode23(@(t,C) concfuntttt(t,C,k1,k2,k3) ,[0 Tfinal],[Ca0 Cb0 Cc0],options);

%calculating the max Cb and the time where Cb is max

maxCb=max(C(:,2));

j=1;

while abs(max(C(:,2))-C(j,2))>=0.005

j=j+1;

end

tmaxCbhr=T(j); %hour

tmaxCbmin=T(j)*60;%minute

%calculating the time of balance

h=1;

while C(h,1)>=0.005

if h< length(T)

h=h+1;

timeofbalance=T(h);%hr

else

timeofbalance=Tfinal;

disp('Concentrations have not reached to balance yet')

break

end

3

end

maxi=max(C);

maxim=max(maxi);

if timeofbalance==Tfinal

figure

plot(T,C(:,1),'-',T,C(:,2),'-.',T,C(:,3),'--')

title('Concentration vs Time')

grid on

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalance)])

ylim([0 round(maxim+0.5)])

legend('Ca(t)','Cb(t)','Cc(t)')

xlabel('Time [hr]')

ylabel('Concentration [M]')

else

figure

plot(T,C(:,1),'-',T,C(:,2),'-.',T,C(:,3),'--')

title('Concentration vs Time')

grid on

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalance+0.5)])

ylim([0 round(maxim+0.5)])

legend('Ca(t)','Cb(t)','Cc(t)')

xlabel('Time [hr]')

ylabel('Concentration [M]')

end

toc

The function (concfuntttt) that is used above is;

function dC=concfuntttt(t,C,k1,k2,k3)

dC=zeros(3,1);

dC(1)=-k1*C(1);

dC(2)=k1*C(1)-k2*C(2)+k3*C(3);

dC(3)=k2*C(2)-k3*C(3);

end

4

c) The rate constants for different reaction conditions, R1, R2 and R3, are given below.

Imagine that you will perform the experiments in different conditions with given initial

concentrations. Your aim is to obtain product B as much as possible. Before performing

the experiments, you want to model the system and simulate them in MATLAB. Run your

MATLAB code written in question b for the six experiments. Select an appropriate time

interval for the integration in each simulation. You will obtain six figures and comment

them. Write your comments below each figure, and answer below questions. Which

experiment condition (reaction condition and initial concentration) do you prefer for

achieving your aim? Why? How much time do you need for one batch?

5

For the 1st experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

1 R1 0.1 0.2 0.1 5 0 0

The graph drawn by using MATLAB is;

In the 1st experiment, reaction reaches to balance in about 7 hours and concentration of B reaches

to 1.67 M at the end of the reaction. k2 is higher than k3 which means conversion of B to C is

higher than conversion of C to B. At the end concentration of C is higher than B, thus this

experiment is more favorable for high concentration of C.

6

For the 2nd

experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

2 R1 0.1 0.2 0.1 5 0 5

The graph drawn by using MATLAB is;

This experiment reaches to balance about same as experiment 1 in about 7 hours. Concentration

of B reaches to 3.3 M higher than the 1st experiment since although it has the same k values it has

extra 5M of C at the beginning. k2 is higher than k3 which means conversion of B to C is higher

than conversion of C to B. At the end concentration of C is higher than B, thus this experiment is

favorable for high concentration of C.

7

For the 3rd

experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

3 R2 0.2 0.3 0.3 5 0 0

The graph drawn by using MATLAB is;

This experiment is reached to balance in about 3.5 hours shorter than first two experiments.

Concentration of B reaches to about 2.5M. It has higher concentration of B than 1st experiment

since they both have same start up conditions with 5M of A and 0 M of C at the beginning,

however k values are higher in the 3rd

experiment. Although it has higher k values than 2nd

experiment, 2nd

experiment reaches to higher concentration of B since it has 5M of both A and C

at the beginning whereas 3rd

experiment has only A. k2 and k3 is equal to each other which means

the rate of conversion of C to B and B to C is equal to each other. This is not the best conditions

for higher concentration of B.

8

For the 4th

experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

4 R2 0.2 0.3 0.3 5 0 5

The graph drawn by using MATLAB is;

This experiment is reached to balance in about 3.5 hours than first two experiments almost same

with the 3rd

experiment. The concentration of B reaches to 5M. This experiment reaches to balance

since k1 is higher than it is in first two experiments and consumed faster however almost same

with the 3rd

experiment since both of them have the same k values. Moreover k2 and k3 values are

higher as well. k2 and k3 is equal to each other which means the rate of conversion of C to B and B

to C is equal to each other. Since both k values are higher in this experiment than first two

concentration of B is higher at the end. The concentration of B is higher than experiment 3 since at

the beginning of the experiment there are 5M of both A and C, as they both converted to B.

9

For the 5th

experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

5 R3 0.3 0.4 0.6 5 0 0

The graph drawn by using MATLAB is;

The reaction is reached to balance earlier than experiments 1, 2, 3 and 4 since k values are higher

than others. Moreover, as it can be seen from the graph, concentration of B is reached to 3M. It

should be noted that conversion of C to B is faster and more than conversion of B to C since k3 is

higher than k2 which increases the concentration of B makes it relatively higher. The

concentration of B is a little bit higher in the 2nd

experiment since there are 5M of both A and B

at the beginning whereas there is only 5M of A in the beginning of the experiment 5 however

really close since k values are higher in the 5th

experiment whereas the concentration of B is lot

higher in the 4th

experiment since there are 5M of both A and B at the beginning whereas there is

only 5M of A in the beginning of the experiment 5 and k values are relatively closer. At the end

when the reaction reached to balance A is all used up, concentration of B is higher than C as

expected. About 2 hours needed for this batch to reach equilibrium.

10

For the 6th

experiment the conditions are;

Experiment # Conditions k1, hr-1

k2, hr-1

k3, hr-1

CA0, M CB0, M CC0, M

6 R3 0.3 0.4 0.6 5 0 5

The graph drawn by using MATLAB is;

The reaction is reached to balance earlier than experiments 1, 2, 3 and 4 since k values are higher

than others. It is almost same with experiment 5 as the k values are the same. Moreover, as it can

be seen from the graph, concentration of B is reached to 6M, which is the highest of them all

although it has the same k vales as experiment 5. This result makes sense since at the beginning

of the reaction there are 5M of both A and C, as both of them is converted to B. Also conversion

of C to B is faster and more than conversion of B to C since k3 is higher than k2 which increases

the concentration of B makes it higher than others. At the end when the reaction reached to

balance A is all used up, concentration of B is higher than C as expected. About 2 hours needed

for this batch to reach equilibrium. This experiment has the best conditions for the higher

concentration of B.

11

The graph above shows 6 experiments on the same graph. Different colors indicates different

experiments such as black ones are the concentrations of experiment 6, green ones are the

concentrations of experiment 1 etc. Different shapes indicate the concentration of different

materials. As it can clearly be seen from the graphs experiments 5 and 6 used up A and reached

the equilibrium faster than experiments 3 and 4 which are faster than experiments 1 and 2. In

these experiments the aim is to produce most of the B. Straight lines that increases from 0

indicates concentration material B since there is no B in the beginning of the any experiment.

.the best condition for the material B is provided by 6th

experiment. The last experiment reaches

balance statement faster and produces most of B. This result makes sense since there are5 M of

both A and C at the beginning of the reaction as both of them is converted to B. Moreover, k3 is

higher than k2, which indicates that C is converted to B faster than B is converted to C. The 6th

experiment which is favorable for the aim which is to get most of B, needs about 2 hours to

reach to balance.

12

Appendix

MATLAB code for question c;

%CHE 386 - HW 2-question 1-c

%Pınar ALP, Gizem BAYRAM, Semra TUNCAY

close all

clear all

format compact

format long

clc

tic

% time interval for the integration can be taken as an input

Tfinal = input('Please give an end time [hr]: ');

%Experimet 1

k1=1;

k2=2;

k3=1;

Ca0=5;

Cb0=0;

Cc0=0;

[ T, C,maxCb, tmaxCbhr,tmaxCbmin,timeofbalance,maxim] = experiment( k1,k2,k3,Ca0, Cb0,

Cc0,Tfinal);

%Experimet 2

k1=1;

k2=2;

k3=1;

Ca0=5;

Cb0=0;

Cc0=5;

[ T2, C2,maxCb2, tmaxCbhr2,tmaxCbmin2,timeofbalance2,maxim2] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal);

%Experimet 3

k1=2;

k2=3;

k3=3;

Ca0=5;

Cb0=0;

Cc0=0;

[ T3, C3,maxCb3, tmaxCbhr3,tmaxCbmin3,timeofbalance3,maxim3] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal);

%Experimet 4

k1=2;

k2=3;

13

k3=3;

Ca0=5;

Cb0=0;

Cc0=5;

[ T4, C4,maxCb4, tmaxCbhr4,tmaxCbmin4,timeofbalance4,maxim4] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal);

%Experimet 5

k1=3;

k2=4;

k3=6;

Ca0=5;

Cb0=0;

Cc0=0;

[ T5, C5,maxCb5, tmaxCbhr5,tmaxCbmin5,timeofbalance5,maxim5] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal);

%Experimet 6

k1=3;

k2=4;

k3=6;

Ca0=5;

Cb0=0;

Cc0=5;

[ T6, C6,maxCb6, tmaxCbhr6,tmaxCbmin6,timeofbalance6,maxim6] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal);

timeofbalancee=[timeofbalance timeofbalance2 timeofbalance3 timeofbalance4 timeofbalance5

timeofbalance6];

timeofbalancem=max(timeofbalancee);

figure(7)

if timeofbalancem==Tfinal

plot(T,C(:,1),'g.',T,C(:,2),'g-',T,C(:,3),'g-.',...

T2,C2(:,1),'m.',T2,C2(:,2),'m-',T2,C2(:,3),'m-.',...

T3,C3(:,1),'c.',T3,C3(:,2),'c-',T3,C3(:,3),'c-.',...

T4,C4(:,1),'b.',T4,C4(:,2),'b-',T4,C4(:,3),'b-.',...

T5,C5(:,1),'r.',T5,C5(:,2),'r-',T5,C5(:,3),'r-.',...

T6,C6(:,1),'k.',T6,C6(:,2),'k-',T6,C6(:,3),'k-.')

title('Concentration vs Time')

grid on

maxime=[maxim maxim2 maxim3 maxim4 maxim5 maxim6];

maximem=max(maxime);

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalancem) ])

ylim([0 round(maximem+0.5)])

legend('Ca(t)ex1','Cb(t)ex1','Cc(t)ex1',...

14

'Ca(t)ex2','Cb(t)ex2','Cc(t)ex2',...

'Ca(t)ex3','Cb(t)ex3','Cc(t)ex3',...

'Ca(t)ex4','Cb(t)ex4','Cc(t)ex4',...

'Ca(t)ex5','Cb(t)ex5','Cc(t)ex5',...

'Ca(t)ex6','Cb(t)ex6','Cc(t)ex6')

xlabel('Time [hr]')

ylabel('Concentration [M]')

else

plot(T,C(:,1),'g.',T,C(:,2),'g-',T,C(:,3),'g-.',...

T2,C2(:,1),'m.',T2,C2(:,2),'m-',T2,C2(:,3),'m-.',...

T3,C3(:,1),'c.',T3,C3(:,2),'c-',T3,C3(:,3),'c-.',...

T4,C4(:,1),'b.',T4,C4(:,2),'b-',T4,C4(:,3),'b-.',...

T5,C5(:,1),'r.',T5,C5(:,2),'r-',T5,C5(:,3),'r-.',...

T6,C6(:,1),'k.',T6,C6(:,2),'k-',T6,C6(:,3),'k-.')

title('Concentration vs Time')

grid on

timeofbalancee=[timeofbalance timeofbalance2 timeofbalance3 timeofbalance4 timeofbalance5

timeofbalance6];

timeofbalancem=max(timeofbalancee);

maxime=[maxim maxim2 maxim3 maxim4 maxim5 maxim6];

maximem=max(maxime);

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalancem+0.5) ])

ylim([0 round(maximem+0.5)])

legend('Ca(t)ex1','Cb(t)ex1','Cc(t)ex1',...

'Ca(t)ex2','Cb(t)ex2','Cc(t)ex2',...

'Ca(t)ex3','Cb(t)ex3','Cc(t)ex3',...

'Ca(t)ex4','Cb(t)ex4','Cc(t)ex4',...

'Ca(t)ex5','Cb(t)ex5','Cc(t)ex5',...

'Ca(t)ex6','Cb(t)ex6','Cc(t)ex6')

xlabel('Time [hr]')

ylabel('Concentration [M]')

end

toc

The “experiment“ function that is used above is;

function [ T, C,maxCb, tmaxCbhr,tmaxCbmin,timeofbalance, maxim] = experiment(

k1,k2,k3,Ca0, Cb0, Cc0,Tfinal)

options = odeset('RelTol',1e-10,'AbsTol',[1e-10 1e-10 1e-15]);

[T, C]= ode23(@(t,C) concfuntttt(t,C,k1,k2,k3) ,[0 Tfinal],[Ca0 Cb0 Cc0],options);

%calculating the max Cb and the time where Cb is max

15

maxCb=max(C(:,2));

j=1;

while abs(max(C(:,2))-C(j,2))>=0.005

j=j+1;

end

tmaxCbhr=T(j); %hour

tmaxCbmin=T(j)*60;%minute

maxi=max(C);

maxim=max(maxi);

h=1;

while C(h,1)>=0.005

if h< length(T)

h=h+1;

timeofbalance=T(h);%hr

else

disp('Concentrations have not reached to balance yet')

break

end

end

if timeofbalance==Tfinal

figure

plot(T,C(:,1),'-',T,C(:,2),'-.',T,C(:,3),'--')

title('Concentration vs Time')

grid on

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalance)])

ylim([0 round(maxim+0.5)])

legend('Ca(t)','Cb(t)','Cc(t)')

xlabel('Time [hr]')

ylabel('Concentration [M]')

else

figure

plot(T,C(:,1),'-',T,C(:,2),'-.',T,C(:,3),'--')

title('Concentration vs Time')

grid on

%no matter what you enter as Tfinal the graph shows the appropriate time interval

%where concentrations reached balance and easy to be seen on the graph

xlim([0 round(timeofbalance+0.5)])

ylim([0 round(maxim+0.5)])

legend('Ca(t)','Cb(t)','Cc(t)')

xlabel('Time [hr]')

ylabel('Concentration [M]')

16

end

end

The function (concfuntttt) that is used in the experiment function is;

function dC=concfuntttt(t,C,k1,k2,k3)

dC=zeros(3,1);

dC(1)=-k1*C(1);

dC(2)=k1*C(1)-k2*C(2)+k3*C(3);

dC(3)=k2*C(2)-k3*C(3);

end