MODIFICATION INSTRUCTIONS
In this
chapter instructions are given in order to create different arithmetical data
and schemes for the exercises, as well as for the creation of new exercises with
the desirable prescriptions.
The arithmetical
data and schemes should totally agree with the theory types, whether they have
been mentioned or not. This is what the scientific ethics impose after all. For
this purpose a number of different calculative programs may be used. One of
these programs is Excel, by Microsoft, which is installed in most computers. For
the use of this program the following specific instructions are given.
EXAMPLE OF HOW TO USE THE PROGRAM EXCEL BY MICROSOFT
First order reactions
In order to find the shape
of the curve in first order reactions e.g. A
P, we work out the following familiar type
C = C_{0}e^{kt}.
In
a work sheet of Excel we set at the positions F_{1 }and F_{2},
the prices 0.8 and 0.3465 respectively (we have a satisfactory approach if we
give the price 0.0346 at position F_{2}). These prices correspond to the
parameters C =0.8 and k=0.03465 of the type C = C_{0 }e^{kt }.
At the positions from A1 to A17
(more or less may be chosen), we set the prices from 0, 10 up to 160 which
correspond to time in seconds. At the position B1 we
keyboard the function =$F$1*EXP($F$2*A1) which
is the insertion of the type C = 0.8e^{0,03465t } in the
Excel program, then we press "enter". We repeat the procedure at the positions B_{2
}to B_{17}. The program has the capability of automatic repetition.
The columns A and B result as shown in the following table. The next step is to
give directives in order to design the diagram.
A 
B 
C 
D 
E 
F 
G 
H 
I 
G 

1 
0 
0.8 
0.8 

2 
10 
0.565727 
0.03465 

3 
20 
0.400059 


4 
30 
0.282905 

5 
40 
0.200059 

6 
50 
0.141473 

7 
60 
0.100044 

8 
70 
0.070747 

9 
80 
0.050029 

10 
90 
0.035379 

11 
100 
0.025018 

12 
110 
0.017692 

13 
120 
0.012511 

14 
130 
0.008847 

15 
140 
0.006256 

16 
150 
0.004424 

17 
160 
0.003129 
The
curve that arises corresponds to a first order reaction, this is obvious from
the time that the it takes for the concentration to decrease to onehalf its
initial value. Every 2 seconds, C decreases onehalf its initial value, as the
following type predicts, t_{1/2 }=
0.693 / k for first order reactions.
By changing the prices at the
positions F_{1} and F_{2} we create curves with desirable
prescriptions (in the Excel program).
If we want to create the
above curve only, we keyboard at the position B1 the function with the
simplified form: =0.8*EXP(0.03465*A1) and
continue in a similar way.
For the creation of nicer schemes with even more
choices, we can use another program, Illustrator by Adobe, having as a guide the
curve that we created in Excel, since Illustrator is not a calculative program.
In order to achieve this, we transfer Excel’s scheme in Illustrator using the
directions copy and paste, we work out the scheme in order to isolate the curve,
which then we transfer to the appropriate lattice axis. We can also give prices
that we took from Excel program. As an example we refer the scheme
of exercise 10 as well as in the chapter “Data
of simple reactions with diagrams”.
Second order reactions
For the shape of a second order reaction
curve e.g. 2AP, we work out the familiar type C = C_{0} / (1+ktC_{0})._{
} In a n Excel’s
work sheet we set at the positions F_{1} and F_{2}, the prices
0.4 and 0.125 respectively. These prices correspond to the parameters C_{0}
= 0.4 and k = 0.125 of the type C = C_{0}
/ (1+ktC_{0}).
We remind that independently of the number of the reactants
the shape of the curve is the same.
At
the positions from A_{1} to A_{18} (more or less positions can
be chosen), we set the prices from 0 to 17 which correspond to time in seconds.
At the position B1 we keyboard the function =$F$1/(1+$F$2*A1*$F$1)
which is the insertion of the type C = 0.4 /
(1+0.125t0.4)
in the Excel program, then we press "enter". We repeat
the procedure at the positions B2 to B17. Thus, the columns A and B result in
the following table. Afterwards we give directives in order to design the
diagram.
A 
B 
C 
D 
E 
F 
G 
H 
I 
G 

1 
0 
0.4 
0.4 

2 
10 
0.266667 
0.125 

3 
20 
0.2 

4 
30 
0.16 

5 
40 
0.133333 

6 
50 
0.114286 

7 
60 
0.1 

8 
70 
0.088889 

9 
80 
0.08 

10 
90 
0.072727 

11 
100 
0.066667 

12 
110 
0.061538 

13 
120 
0.057143 

14 
130 
0.053333 

15 
140 
0.05 

16 
150 
0.047059 

17 
160 
0.044444 
The
resulting curve seems to correspond in a second order reaction, as we can see
from the time it takes to decrease to one half its initial value. This
first decrease takes place after 20 seconds. The second decrease, once again to
one half its previous value, takes place 40 seconds later from the first, which
is 20+40=60 seconds since initiation and the fourth takes place 140 seconds
since initiation. These prices are predicted from the half life time which has
the type
t_{1/2 }= 1 / kC_{0 }for second order reactions.
1 /0.125^{.}0.4 = 20
1 / 0.125^{.}0.2 = 40.
For the creation of schemes
in a specialized schemes program we follow the procedure that was mentioned in
the case above (first order reactions). As an example we refer to the
solution scheme of
exercise 21. as well as in the chapter “Data of
simple reactions with diagrams”.
Third order reactions
In order to predict the shape of the curve of a third order reaction e.g. 3AP,
we work
out the familiar type C = C_{0} / (1+2ktC_{0}^{2})^{0.5
}
In a n Excel’s work sheet we set in the positions F1 and F2, the prices 0.4
and 9.375 respectively. These prices correspond to the parameters C_{0 }=
0.4 and k = 9.375 of the type C = C_{0} / (1+2ktC_{0}^{2})^{0.5
}
We remind that independently of the number of the reactants the shape of the
curve is the same.
At the positions A1 to A19, we set the prices from 0 to 19 which correspond to time in seconds. At the position B1 we keyboard the function
=$F$1/(1+2*$F$2*A1*$F$1^2)^0.5, which
is the insertion of the type
C = 0.4 / (1+2^{.}9.375t0.4^{2})^{0.5}
in Excel program and then we press "enter". We repeat the
procedure at the positions B2 to B19. Therefore, the columns A and B result in
the following table. Finally we give directives in order to design the diagram.
A 
B 
C 
D 
E 
F 
G 
H 
I 
J 

1 
0 
0.4 
0.4 

2 
0.5 
0.252982 
9.375 

3 
1 
0.2 


4 
1.5 
0.170561 

5 
2 
0.151186 

6 
2.5 
0.137199 

7 
3 
0.126491 

8 
3.5 
0.117954 

9 
4 
0.11094 

10 
4.5 
0.105045 

11 
5 
0.1 

12 
5.5 
0.095618 

13 
6 
0.091766 

14 
6.5 
0.088345 

15 
7 
0.08528 

16 
7.5 
0.082514 

17 
8 
0.08 

18 
8.5 
0.077703 

19  9 
0.075593 
Once again from the time that it takes for the concentration to decrease to one half its initial value we can ascertain that the curve corresponds in a third order reaction. The first decrease of the concentration to one half its initial value takes place 1 minute later and the second 4 minutes later from the first, which is 1+4=5 minutes since initiation. These prices are predicted from the half life time which has the type
t_{1/2 }= 1,5 / kC_{0}^{2}_{ }for
third order reactions 1.5 /9.375^{.}0.4^{2} = 1
1.5 / 9.375^{.}0.2^{2} = 4
As an example we refer to the chapter “Data of simple reactions with diagrams”.