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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
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.
Process A: 1) BrO3- + Br- + H+ ---->
HOBr2 + HOBr 2) HOBr2 + Br- + H+ ---->
2HOBr 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.
Process B: 4) BrO3- + HBrO2 + H+ ---->
2 BrO2 + H2O 5) BrO2 + Ce3+ + H+ ---->
HBrO2 + Ce4+ 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.
Process C: 7) Malonic+ Br2 ---->
BromoMalonic+ Br- + H+ 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
