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The Belousov-Zhabotinskii Reaction (classic)
Solutions:
| Solution A: |
0.23M KBrO3 ( potassium bromate) |
| Solution B: |
0.31M ( malonic acid) & 0.059M KBr (potassium bromide) |
| Solution C: |
0.019M Ce(NH4)2(NO3)6 (
ammonium cerium(IV) nitrate) & 2.7M H2SO4 (
sulfuric acid) |
| Solution D: |
0.025M ferroin sulfate (1,10-phenanthroline iron(II) sulfate) |
Demonstration taken from "Chemical Demonstrations
A Handbook for Teachers of Chemistry"
by Bassam Z. Shakhashiri, Vol. 2 (1992) University of Wisconsin Press
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Outline of Operations and Observations
- In a 1 liter flask, add 300 mL solution A to 300ml solution B.
Solution becomes amber.
- Wait until the solution clears, add 300 mL solution C, mixture
turns orange with the smell of bromine.
- Wait until the orange color fades
to the yellow cerium solution color and
add 3.6 mL solution D.
- Color changes from gorgeous green to
pompous purple to rusty red
to blink of blue (if you blink you miss
the blue) and finally back to gorgeous
green. Cycle takes ~1 min, with the cycle
lengthening with time.
- If you pour some of the mixture as a thin layer into a Petri dish, beautiful
patterns, with blue concentric circles,
called targets, on a red background will
develop.
| Overall Reaction: |
|
3 CH2(CO2H)2 + 4
BrO3- |
-----> |
4 Br- + 9 CO2 +
6 H2O |
The reaction can be split into three overall processes.
Process A occurs when the bromide ion concentration rises above a certain
critical level.
Process B involves radical and one-electron transfers. The dominant process
is determined by the bromide ion concentration.
Process C regenerates the bromide ion and reduces the catalyst back to its
lower oxidation state.
In process A, bromate ions oxidize bromide ions to produce bromine.
As the concentration of bromide ions decreases, so does the rate of
eqn 1. The bromate ions then compete for reaction with the hypobromous acid
(HBrO2) and switch the system to process B.
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Process A: |
| |
1) |
BrO3- + Br-
+ H+ |
---->
|
HOBr2 + HOBr |
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2) |
HOBr2 + Br- +
H+ |
---->
|
2HOBr |
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3) |
HOBr + Br- +
H+ |
---->
|
Br2 + H2O |
In the process B part of the reaction, cerium oxidizes from oxidation
state (III) to oxidation state (IV). This gives the color change from red
to blue. Because two BrO2
radicals are produced in eqn 4, and each reacts rapidly to form an HBrO2
molecule, this part of the reaction constitutes an autocatalytic cycle. The
autocatalysis causes the rate of this process to increase very quickly once
it has switched on, so red changes rapidly to blue. The growth in the concentration
of HBrO2 is limited by eqn 6. The switch between processes A and
B will occur when the rates of steps 1 and 3 are approximately equal.
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Process B: |
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4) |
BrO3- + HBrO2
+ H+ |
---->
|
2 BrO2 + H2O |
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5) |
BrO2 + Ce3+ +
H+ |
---->
|
HBrO2 + Ce4+ |
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6) |
2 HBrO2 |
---->
|
HOBr + BrO3- +
H+ |
In process C, the bromide ion is regenerated and reduce the catalyst
back to its lower oxidation state. While this part of the reaction is not
well understood, the following representation can be used. Malonic acid reacts
with bromine to give bromomalonic acid. If this is then oxidized by the cerium
(IV), bromide and cerium (III) will be produced. The oxidized form of the
catalyst can also react directly with malonic acid, so there may be fewer
than one bromide ion per cerium (III) ion produced. Process C sees the blue
change to red and resets the chemical
clock for the next oscillation.
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Process C: |
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7) |
Malonic+ Br2 |
---->
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BromoMalonic+ Br- +
H+ |
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8) |
2Ce(IV) + Malonic+
BromoMalonic+ H+ |
---->
|
Br- + 2Ce(III)
and other products. |
where is known as the stoichiometric factor.
In the simplest computer analyses, is assumed to be a constant, say =1.
In modelling used to match complex oscillations, we attempt to allow (the
number of bromide ions produced as two cerium ions are reduced) to be a function
of the instantaneous concentrations of other species, such as HOBr.
The potential of the Ce(III)-Ce(IV) couple, at 25oC,
is given by the Nernst equation:
E = Eo - 0.059 log [Ce(III)]/[Ce(IV)]
Hazards
Because sulfuric acid is a strong acid and a powerful dehydrating agent,
it can cause burns. Spills on skin should be washed with copious amounts
of water and treated medically. Spills on inantimate surfaces should
be neutralized with sodium bicarbonate and rinsed clean.
Malonic acid is a strong irritant to skin, eyes and mucous membranes.
Bromates are strong oxidizing agents. Mixtures of bromates with finely
divided organic materials, metals, carbon or other combustible materials are
easily ignited, sometimes explosively. Ingestion of potassium bromate
can cause vomiting, diarrhea and renal injury.
Personal Protective Equipment
Use gloves (disposable nitrile or PVC gloves will work), a lab coat and
eye protection. Do not directly inhale the minor amount of bromine generated.
Protect the class with a splash shield.
Disposal
At the end of the color change demonstration, discuss the importance of
protecting the environment and responsibly husbanding wastes. Carefully
add sodium bicarbonate to the solution in a container large enough to accomidate
the bubbles created. Once enough sodium bicarbonate has been added for
the pH to be about 7, dispose in the santary sewer system, as there is nothing
hazardous. Of course, comply with all applicable regulations
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