Untangling Complex Systems, page 46
k 4
−
+
2 →
BrO3 + HOBr + H
Ce4+ + CH (
)
5
k
−
2 COO
OH →
f Br
2
At 293.15 K, the values of the rate constants are:
k
3
−
9
1
−
1 = 2 mol dm s
k
6
2
−
6
1
−
2 = 10 mol
dm s
k
2
−
6
1
−
3 = 10 mol
dm s
k
−1
3 −1
4 = 2000 mo
ol dm s
1
3
1
k 5 1 mol− dm s−
=
Write the differential equations for the three variables X = [HBrO ],
2
Y = [Br−], and
Z = [M( n+1)+] and solve them by numerical integration, fixing the value of [H+] to 0.8 M,
of [CH (COOH) ] to 0.02 M, of the coefficient
− the values
2
2
f to 0.6, and assigning to [BrO3 ]
of (1) 0.18 M, (2) 0.15 M, (3) 0.12 M, (4) 0.1 M, and (5) 0.08 M, respectively. Determine
the relationship between the period of the oscillations and [BrO−3], i.e., the exponential n
of the relation T
n
∝
−
[BrO3 ] . If you choose MATLAB to solve the differential equations by
numerical integration, use stiff solvers, like “ode15s” and “ode23s,” because the changes
of the concentrations of the intermediates are substantial and show spikes.
8.4. The activation energies (Pellitero et al. 2013) determined at 293.15 K, for the five steps of
the Oregonator (see exercise 8.3) are E
54 kJ/mol, E
25 kJ/mol, E
60 kJ/mol,
act,1 =
act,2 =
act,3 =
E
64 kJ/mol, and E
70 kJ/mol, respectively. The values of the kinetics con-
act,4 =
act,5 =
stants of the five reactions of the Oregonator are reported in exercise 8.3. Calculate the
period (∆ T) of the oscillations at T 293.15 K,
303.15 K, and
313.15 K, and for
1 =
T 2 =
T 3 =
[BrO−3] = .
0 18M, [CH (COOH) ]
2
2 = 0.1 M, [H+] = 0.8 M, and f = 0.6. Finally, determine
the activation energy for the Arrhenius-type relationship between ∆ T−1 and T.
8.5. The goal of this exercise is to observe the spontaneous formation of chemiluminescent
oscillations with the Orbán reaction. Go to the laboratory, wear a white coat, gloves, safety
glasses and prepare the following solutions using deionized water as the solvent:
• Solution A: H O 1 M. You may buy H O in a drugstore and find that its concentra-
2
2
2
2
tion is expressed in volumes. The concentration in volume is the ratio between the
volume of gaseous O (measured in standard conditions, i.e., at 273.15 K and 1 atm)
2
224
Untangling Complex Systems
that evolves for complete decomposition of the H O contained in a given volume and
2
2
the volume of the solution. For example, if we buy a 36-volume solution of H O , the
2
2
grams of H O we have in 1 L are x = (36 22 4
. )68 =
2
2
109 g according to the following
decomposition reaction of H O considered in standard conditions:
2
2
2H2O2( →
+
aq
2H
)
2O( liq
O
)
2( gas)
68 g
22.44 L
The concentration of 109 g of H O in 1 L of solution can also be expressed as either
2
2
10.9% or [(109 / 34
1
g
g mol− )/1L] = .
3 2 M. For the preparation of H O 1 M, take
2
2
15.6 mL of H O 3.2 M and add water until you have 50 mL of total volume.
2
2
• Solution B: KSCN 0.15 M. Dissolve 0.729 g of KSCN (MW = 97.17 g·mol−1) in 50 mL
of deionized water.
• Solution C: luminol 3.7·10−3 M in NaOH 0.1 N. First, prepare a solution of NaOH 0.1 N
by dissolving 2 g of NaOH in 500 mL of deionized water. Then, dissolve 0.0328 g of
luminol (C H N O ) in 50 mL of the NaOH 0.1 N solution.
8
7
3
2
• Solution D: CuSO 6·10−4 M. Dissolve 9.6·10−3g of CuSO in 100 mL of deionized
4
4
water.
Insert a magnetic stir bar into a small dry beaker and place the beaker on a stirrer. Add the
volumes that are listed in Table 8.6 for five distinct experiments.
After stirring the reactive solution for a few minutes, take 3 mL of it and put it in a
fluorometric cuvette. The solution within the cuvette must be maintained under stirring.
Record the chemiluminescent oscillations by using a fluorimeter (note that you do not
have to switch on the lamp of the fluorimeter to record the chemiluminescent signals).
Fix the wavelength of emission at 436 nm. Choose large widths of the slits and do not
change them in the five experiments to compare the intensity of the chemiluminescent
signals.
Find out the best conditions to observe strong chemiluminescent signals and fast
oscillations.
8.6. Write the system of three differential equations for the mechanism [8.25] and solve it
numerically for the following values of the parameters: k
1 A = 0.5; k 2 = 0.0028; k 3 = 0.0084;
k
0 = 0.05 (representing the flow rate). The constant concentrations introduced in the open
system are X
, , ]
in = 4; Y in = 1; Z in = 9 and the initial conditions are [ X 0 Y 0 Z 0 = [10, 10, 10].
You can use “ode45” to solve the system of differential equations in MATLAB. Compare
the results of this exercise with those depicted in Figure 8.6.
TABLE 8.6
Volumes of the Reactive Solutions A, B, C, D, H2O2 3.2 M, and NaOH 0.1 M for Observing
the Orbán Reaction in Five Distinct Experiments
V (H2O2 3.2 M)
V (sol B)
V (sol C)
V (NaOH 0.1 M)
V (sol D)
V (sol. A) (mL)
(mL)
(mL)
(mL)
(mL)
(mL)
Exp. 1
1
0
1
1
0
2
Exp. 2
1
0
1
2
0
1
Exp. 3
1
0
2
1
0
1
Exp. 4
1
0
1
1
1
1
Exp. 5
0
1
1
1
0
2
The Emergence of Temporal Order in a Chemical Laboratory
225
8.7. It has been demonstrated (Fujii and Rondelez 2013) that the following abstract model can
account for the mechanism illustrated in Figure 8.9:
v
X
X
1
→
2 X
v = ⋅(
)⋅ ⋅
1
k 1 pol A 1+ b⋅ A⋅ X
X + Y
v 2
→
Y
2
v =
⋅(
)⋅ ⋅
2
k 2 pol X Y
v
X
X
3
→
waste
v =
⋅(
)⋅
3
k 3 exo
Y
1+ K
v
Y
Y
4
→
waste
v =
⋅(
)⋅
4
k 4 exo
Y
1+ K
Note that prey growth, prey and predator degradations obey Michaelis-Menten kinetics.
Write the system of two differential equations for prey and predator and solve
it by numerical integration and the following values for the parameters (Padirac
et al. 2013): k
3
−
2
−
1
−
1 = 3×10
nM min ; pol = 1 7
. nM; A = 140 nM; b = × −
−
6 10 5
2
nM ;
k
3
2
1
= × −
−
−
2
1
1
= −
−
−
2
4 10 nM min ;
k 3 10 nM min ;
exo = 25 nM;
K = 34 nM;
k
3
1
1
= × −
−
−
4
4 10 nM min .
8.8. The fundamental elementary steps of a pH oscillator are listed as follows.
H + Red
k 1
A →
HRedA
HRed
k−1
A
→
H + RedA
Ox + HRed
k 2
A + H →
Y
HRed
k 3
A + Y →
3 H
k 4
Y →
P
5
Red
k
B + H →
Q
Write the differential equations for the four variables H, RedA, HRedA, and Y assuming that the concentrations of Ox and RedB are fixed. Solve numerically the differential equations.
In analogy to the real situation, assume that only RedA and H have nonzero inflow, while
all the species are allowed to flow out the reactor. Use the following values for the rate
constants and the concentrations of the reagents: k
1 = (5 × 1010 M−1s−1); k−1 = (3 × 103 s−1);
k
(the inverse of the
2 = (3.077 × 106 M−2s−1); k 3 = (106 M−1); k 4 = (11 s−1); k 5 = (2.5 M−1s−1); k 0
residence time) = (10−3 s−1); [ Ox] = 6.5 × 10−2 M; [ Red ]
]
B = 2.0 × 10−2 M; [ RedA 0 = 6 × 10−2 M;
[ H] 0= 2 × 10−2 M. All these values refer to the bromate-sulfite-ferrocyanide reaction (Luo
and Epstein 1991).
8.9. Verify that in a biochemical multi-gene circuit similar to the repressilator, if the number
of repressors is odd, we can observe oscillations, whereas if the number of repressors is
even, the circuit converges to a fixed point. Test this rule of thumb for circuits with 2,
3, 4 and 5 repressors linked in a ring structure. Use the simple model of transcriptional
regulation (see equation [8.39]) and the following values for the parameters: α =
0
0 1
. ,
α =100, K = 1, n = 2 5
. , kdm = 1, kdp = 5. As initial conditions fix the amount of the mRNA
226
Untangling Complex Systems
and protein of the first repressor to 10, and the amount of the mRNAs and proteins of
all the other repressors to 5.
8.10. An oscillatory dynamic involving a synthetic biochemical circuit containing intercel-
lular communication and affecting the abundance of a population of bacteria has been
recently presented (Balagaddé et al. 2005). Many bacteria communicate and coordinate
gene expression through quorum sensing. There are bacteria, like Escherichia coli, which
produce and release N-acyl-homoserine lactone ( A) as a chemical signal molecule. As the
cell density N increases, the concentration of the signal molecule also increases. The signal
molecule A, in turn, modulates the expression of a killer gene R that produces a killer pro-
tein E. The killer protein reduces the cell density N. When the population of the bacteria
becomes again low, A also becomes low, and the activity of the killer gene becomes neg-
ligible. The population enjoys a renewed exponential growth. The delayed negative effect
played by E on N may induce oscillations in the cell density. The ordinary differential
equations (ODEs) proposed to describe the dynamics of such system are
dN
N
= k 1 N 1−
k
− d NE − kf N
dt
K 1
dA = k
(
)
A N − dA + k f
A
dt
dR = kRA− dRR
dt
dE = kER− dEE
dt
In such equations, kf is the flow rate.
1. Solve the system of ODEs numerically by using the values for the parameters listed in
Table 8.7.
2. Transform the differential equations in elementary steps and discuss to which “ primary
oscillator” may belong such synthetic biochemical circuit.
TABLE 8.7
Values of the Parameters Regarding the
Abundance of a Population of Bacteria
Influenced by Intercellular Communication
Parameter
Value
k1
0.7 hr−1
kf
0.1 hr−1
K1
3 × 109 mL−1
kd
0.004 nM−1 hr−1
kA
4 × 10−7 nM mL hr−1
dA
0.1 hr−1
kR
1.0 hr−1
dR
0.7 hr−1
kE
1.0 hr−1
dE
0.7 hr−1
N(t = 0)
1 × 107 mL−1
The Emergence of Temporal Order in a Chemical Laboratory
227
8.9 SOLUTIONS TO THE EXERCISES
8.1. The differential equation describing how [ X] changes over time is:
d [ X ]
= k (
]) ′ (
])[ ]2
0 [
]
r CA − [ X
+ kr CA −[ X X − k X
dt
If we divide both terms of the equation by C and we indicate with x the ratio ([ X ]/ C ), we A
A
obtain:
dx
= (1− x)( k
′ 2 2 )
r + kr CAx
− k 0 x
dt
The steady-state solutions are the intersection points between the straight line k 0 x and
the cubic curve − k′ 2 3
2 2
r CAx + k′ r CAx − kr x + kr. For the numerical values given in the text
of the exercise, i.e., k
−1
2
1
r = 1 t , ′ r =
− −
k
100 M t , CA = 1 M (note that the autocatalytic pro-
cess speeds up the rate of the transformation A → X of two orders of magnitude), we find
that when k 0 is small or large (for instance 15 t−1 and 35 t−1, respectively), the straight line intersects the cubic curve only at one point. Graphically, we see that such steady states are
stable (see plots a and c of Figure 8.16). On the other hand, when k 0 is medium, the straight line intersects the cubic curve at three points. Two of them are stable whereas the third (at
the intermediate value of x) is unstable (see plot b of Figure 8.16).
8.2. For the preparation of the stock solutions, we weight the salts and we dissolve them in spe-
cific amounts of water. By performing such operations, we make systematic and perhaps
also random errors. The systematic errors depend on the limited accuracy of our balance,
our pipettes, and our flasks. For instance, we weight 11.7023 g of KBrO by using a techni-
3
cal balance whose inherent uncertainty is ± 0.0001 g. Let us suppose that the execution
of the operation is perfect, and we do not introduce any random error. The final value of
the KBrO mass will be
3
m = (11.7023 ± 0.0002) g. The next step is to solubilize the salt
in deionized water. If we use a flask of volume V = (200.00 ± 0.15) mL and we avoid any
random error, like, for instance, the parallax error (see Appendix D), the concentration of
KBrO solution will be:
3
KBrO
3 = 0.3503 mol/L. Its uncertainty is estimated by using
0
s
s
s
te
te
te
Ra
Ra
Ra
0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6
0.0
0.2
0.4
0.6
(a)
x
(b)
x
(c)
x
