Teacher Notes
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Sudsy Kinetics![]() IntroductionTeach kinetics concepts in a fun and sudsy way! This demonstration provides an interesting twist on the traditional “Old Foamey” or “Elephant Toothpaste” reaction. Not only will your students be amazed at the sudsy eruption—they will learn kinetics concepts along the way! Concepts
MaterialsHydrogen peroxide, H2O2, 30%, 20 mL*
Hydrogen peroxide, H2O2, 10%, 20 mL* Hydrogen peroxide, H2O2, 3%, 20 mL* Alconox® cleaner, 3–4 g* Sodium iodide solution, NaI, 2 M, 4–5 mL* Tap water Graduated cylinders, 10-mL, 3 Large, plastic demonstration tray, several inches deep Graduated cylinders, 100-mL, 3 Lighter or matches Graduated cylinders or Erlenmeyer flasks, 500-mL, 2 Spoon or scoop Wood splint *Materials provided in kit. Safety PrecautionsHydrogen peroxide solution, 30%, is severely corrosive to the skin, eyes and respiratory tract: very strong oxidant. Dangerous fire and explosion risk. Do not heat this substance. Sodium iodide is slightly toxic by ingestion. Although the Alconox detergent is considered nonhazardous, do not ingest the material. Do not stand over the reaction; steam and oxygen are produced quickly. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron. Please review current Safety Data Sheets for additional safety, handling and disposal information. DisposalPlease consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. The foam and any solution left in the cylinder or on the plastic tray may be rinsed down the drain with excess water according to Flinn Suggested Disposal Method #26b. ProcedurePart 1. Effect of Concentration on the Rate of the Reaction
Part 2. Old Foamey—Observing a Reaction Intermediate and Products
Part 3. Comparing the Rate of the Reaction with its Stoichiometry
Student Worksheet PDFTeacher Tips
Sample DataPart 1. Data Table {11830_Data_Table_1}
Part 2. Data Table {11830_Data_Table_2}
Part 3. Data Table {11830_Data_Table_3}
Answers to Questions
Concentration is directly proportional to rate of reaction. The lower the concentration is, the lower the reaction rate is, and therefore the longer the reaction takes. Also, reactants with lower concentrations seem to yield less product.
a. What is responsible for the initial brown color of the foam? The free iodine produced by the oxidation of sodium iodide by the hydrogen peroxide results in an initial brown color in the foam. b. What is responsible for the wood splint re-igniting? Oxygen, a flammable gas released by the decomposition of hydrogen peroxide, caused the wood splint to re-ignite.
H2O2(aq) + 3I–(aq) → 2OH–(aq) + I3–(aq)
2H2O2(aq) → 2H2O(l) + O2(g) + Heat Energy
The moles of peroxide were the same in each cylinder, even though the concentrations were different due to one being diluted with water. Higher concentration results in a higher reaction rate, but a higher rate does not result in more product, which depends on its stoichiometry. DiscussionHydrogen peroxide decomposes to produce oxygen and water according to the decomposition reaction shown: 2H2O2(aq) → 2H2O(l) + O2(g) + Energy
The reaction is quite slow unless catalyzed by a substance such as iodide ions, manganese metal, manganese dioxide, ferric ions and many other substances such as yeast or even blood. A catalyst is a substance that, when added to a reaction mixture, participates in the reaction and speeds it up, but is not itself consumed in the reaction. The iodide ion is used as a catalyst in this demonstration. While the complete mechanism is not known, the observations of this reaction are consistent with the following reactions: Step 1: Hydrogen peroxide and iodide mix to cause spontaneous formation of the brown color from the I3– intermediate with very little foam. This reaction shows that the catalyst is indeed participating in the reaction. H2O2(aq) + 3I–(aq) → 2OH–(aq) + I3–(aq) H2O2(aq) + I3–(aq) + 2OH–(aq) → 2H2O(l) + 3I–(aq) + O2(g) Overall Reaction: Combining Steps 1 and 2 gives the overall reaction shown below. Notice that the enthalpy or heat of reaction (ΔHrxn) is a negative value, indicating that the reaction is exothermic and releases heat. The free energy of the reaction (ΔGrxn), which takes into account not only the enthalpy but also the entropy of the reaction, is also a negative value, indicating that the reaction takes place spontaneously. If the reaction occurs spontaneously, then why is a catalyst needed? The iodide catalyst causes the reaction to occur at a reasonable rate—without it, the reaction would occur, but so slowly that it would not be observable. 1.11 g/mL x 15 mL = 16.7 g of solution
Why all the foam? The volume of oxygen that is released from the reaction can be calculated using 15 mL (as in Part 3) of 30% hydrogen peroxide. Hydrogen peroxide solution (30%) has a specific gravity of 1.11 g/mL. Therefore, the mass of solution can be determined: 1.11 g/mL x 15 mL = 16.7 g of solution
Since only 30% of the total volume of solution is H2O2, then 30% of 16.7 g of solution = 5.01 g of H2O2
Since the molecular weight of H2O2 is 34.02 g/mol, the number of moles of H2O2 can be calculated: 5.01 g x 1 mole/34.02 g = 0.147 mol of H2O2 used
From the balanced equation, there is a 2 to 1 ratio of hydrogen peroxide to oxygen gas. Thus, the number of moles of O2 released can be determined 0.147 mol H2O2 x 1 mole O2/2 mol H2O2 = 0.0736 mol O2
This shows that the reaction should produce 0.0736 moles of oxygen gas. To determine the volume of this amount of gas, use the ideal gas law, PV = nRT, assuming a reaction temperature of the steaming foam of approximately 100 °C (or 373 K) and standard pressure (1 atm). Solving for volume, V = nRT/P, where n = 0.0736 mol, R = ideal gas constant = 0.0821 L•atm/mol•K, T = 373 K and P = 1 atm, the volume of oxygen gas can be calculated: V = (0.0736 mol) (0.0821 L•atm/mol•K) (373 K)/1 atm = 2.25 L O2
The volume of oxygen expected, then, is 2.25 liters. Discuss with your students how they could have predicted that the foam would overflow out of the 500-mL container using this calculation. Student Exercise: Calculate the volume of foam that will be produced by 3% hydrogen peroxide, which has a specific gravity of 1.01 g/mL. Compare this to the 30% peroxide. Will the foam rise up and overflow out of a 500-mL container? Using the same procedure and steps as shown above, the volume of oxygen expected from 15 mL of 3% H2O2 is approximately 205 mL, using 100 °C (373 K). This is a high estimate since the reaction temperature is most likely less than 100 °C. Perhaps have students repeat the experiment, measuring the temperature of the reaction and then performing the calculations. The students can conclude that the foam will not overflow out of a 500-mL container. Finally, perform the demonstration for the students with 3% hydrogen peroxide. Allow students to determine if experimental results support their calculated results. ReferencesSpecial thanks to Walter Rohr of Eastchester High School in Eastchester, NY, for bringing this demonstration to our attention. Recommended Products
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