Halloween Reaction or Old Nassau Reaction

The Old Nassau or Halloween reaction is a clock reaction in which the color of a chemical solution changes from orange to black. Here's how you can do this reaction as a chemistry demonstration and a look at the chemical reactions that are involved.

Materials Needed

• Water
• Soluble starch
• Sodium metabisulphite (Na₂S₂O₅)
• Mercury(II) chloride
• Potassium iodate (KIO₃)

Read More

Briggs-Rauscher Oscillating Color Change Reaction

By Anne Marie Helmenstine, Ph.D.

Prepare the Solutions

• Solution A: Mix 4 g soluble starch in a couple of milliliters of water. Stir the starch paste into 500 ml boiling water. Allow the mixture to cool to room temperature. Add 13.7 g of sodium metabisulphite. Add water to make 1 liter of solution.
• Solution B: Dissolve 3 g mercury(II) chloride in water. Add water to make 1 liter of solution.
• Solution C: Dissolve 15 g potassium iodate in water. Add water to make 1 liter of solution.

Perform the Halloween Chemistry Demonstration

1. Mix 50 ml solution A with 50 ml of solution B.
2. Pour this mixture into 50 ml of solution C.

The color of the mixture will change to an opaque orange color after a few seconds as the mercury iodide precipitates. After another few seconds, the mixture will turn blue-black as the starch-iodine complex forms.

If you dilute the solutions by a factor of two then it takes longer for the color changes to occur. If you use a smaller volume of solution B the reaction will proceed more rapidly.

Chemical Reactions

1. Sodium metabisulfite and water react to form sodium hydrogen sulfite:
   Na₂S₂O₅ + H₂O → 2 NaHSO₃
2. Iodate(V) ions are reduced to iodide ions by the hydrogen sulfite ions:
   IO₃^- + 3 HSO₃^- → I^- + 3 SO₄^2- + 3 H⁺
3. When the concentration of iodide ions becomes sufficient for the solubility product of the HgI₂ to exceed 4.5 x 10^-29 mol³ dm^-9, then orange mercury(II) iodide precipitates until the Hg²⁺ ions are consumed (assuming an excess of I^- ions):
   Hg²⁺ + 2 I^- → HgI₂ (orange or yellow)
4. If I^- and IO₃^- ions remain, then an iodide-iodate reaction takes place:
   IO₃^- + 5 I^- + 6 H⁺ → 3 I₂ + 3 H₂O
5. The resulting statch-iodine complex is black to blue-black:
   I₂ + starch → a blue/black complex