Catalysts, Enzymes, Rates of reaction.

School Science Lessons
(topic17)
2024-08-29

Rates of reaction
Contents
17.1.0 Rates of reaction
17.2.0 Rates of reaction, clock reactions
17.3.0 Catalysis
17.4.0 Common ion effect
17.5.0 Chemical equilibrium
17.6.0 Gravimetric analysis
17.7.0 Hydrogen peroxide

17.1.0 Rates of reaction:
17.1.1 Catalysts and rate of reaction
17.1.2 Concentration and temperature, cobalt (II) chloride-6-water
17.1.3 Concentration of reactants, hydrochloric acid with magnesium
17.1.4 Concentration of reactants, hydrochloric acid with sodium thiosulfate
17.1.4 Concentration and rate of reaction, sodium thiosulfate with dilute hydrochloric acid
17.1.6 Dilute hydrochloric acid with zinc
17.1.7 Dilute hydrochloric acid with marble chips
17.1.8 Measure rates of reaction
17.1.9 Magnesium with hydrochloric acid
17.1.10 Potassium iodide with potassium iodate
17.1.11 Rate of reaction, k
17.2.0 Rates of reaction, clock reactions
17.1.13 Rates of reaction of aspirin
17.1.14 Size of particles and rate of reaction with balloons
17.1.15 Temperature and rate of reaction

17.2.0 Rates of reaction, clock reactions
17.2.1 Belousov-Zhabotinskii clock reaction, BZ reaction
17.2.2 Hydrogen peroxide clock reaction, Briggs-Rauscher oscillating reaction
17.2.3 Iodine clock reaction, hydrogen peroxide with potassium iodide
17.2.4 Iodate clock reaction
17.2.5 Old Nassau flag clock reaction
17.2.6 Persulfate-iodide clock reaction

17.3.0 Catalysis
17.3.1 Catalysis
17.3.2 Catalytic oxidation of methyl alcohol
17.3.3 Catalytic oxidation of ammonia solution, with oxygen gas
17.3.4 Bromine catalyses the oxidation of sulfur to sulfuric acid
17.3.5 Ethyl acetate with sodium hydroxide, autocatalytic hydrolysis
17.3.6 Heat potassium chlorate, manganese dioxide catalyst
17.3.7 Oxalic acid with potassium manganate (VII), autocatalysis
17.3.8 Oxidation of acetone vapour, copper catalyst
17.3.9 Potassium bromate with propanedioic acid, double autocatalytic reaction, oscillating reaction
17.3.10 Potassium iodide with hydrogen peroxide, reverse colour change
17.3.11 Sodium hypochlorite decomposition, cobalt sulfate catalyst
17.3.12 Sugar cube burning, combustible cube

17.4.0 Common ion effect
17.4.1 Common ion effect in ammonium chloride solution
17.4.2 Common ion effect to precipitate barium chloride from solution
17.4.3 Common ion effect to precipitate sodium chloride from solution
17.4.4 Common ion effect, sodium ethanoate and ethanoic acid

17.5.0 Chemical equilibrium
17.5.1 Chemical equilibrium
17.5.2 Effect of temperature on chemical equilibrium
17.5.3 Equilibrium between ICl and ICl3
17.5.4 Heat nitrogen tetroxide
17.5.5 Hydrolysis of antimony chloride
17.5.6 Hydrolysis of bismuth chloride
17.5.7 Law of mass action and reversible reactions

17.6.0 Gravimetric analysis
17.6.1 Weight of aluminium in aluminium sulfate
17.6.2 Weight of calcium in marble
17.6.3 Weight of iron in iron (II) ammonium sulfate
17.6.4 Weight of magnesium in magnesium sulfate
17.6.5 Weight of sulfate radical in sodium sulfate
17.6.6 Weight of tin in solder

17.7.0 Hydrogen peroxide
Hydrogen peroxide concentration 30% solution is not permitted in schools.
17.7.1 Hydrogen peroxide
17.7.2 Hydrogen peroxide as a reducing agent
17.7.3 Hydrogen peroxide as an oxidizing agent
17.7.4 Hydrogen peroxide bleaching action
17.2.2 Hydrogen peroxide clock reaction, Briggs-Rauscher oscillating reaction
17.7.6 Hydrogen peroxide concentration and storage
17.7.7 Hydrogen peroxide decomposition, with different catalysts
17.7.8 Hydrogen peroxide decomposition, with manganese (IV) oxide catalyst
17.7.9 Hydrogen peroxide on cut skin, (catalase enzyme)
17.7.10 Hydrogen peroxide on cut potato, (catalase enzyme)
17.7.11 Hydrogen peroxide on hair
17.7.12 Hydrogen peroxide concentration and enzyme activity
17.7.13 Hydrogen peroxide pH and enzyme activity
17.7.14 Hydrogen peroxide temperature and enzyme activity
17.7.15 Hydrogen peroxide with manganese (IV) oxide, height of suds
17.7.16 Hydrogen peroxide with catalase in raw beef liver
17.7.17 Hydrogen peroxide with catalase -.Disproportionation
17.2.3 Iodine clock reaction, hydrogen peroxide with potassium iodide
17.7.18 Hydrogen peroxide with potassium sodium tartrate, cobalt (II) chloride catalyst
17.7.19 Hydrogen peroxide with sodium thiosulfate, ammonium molybdate catalyst
17.7.20 Hydrogen peroxide with yeast, elephant's toothpaste reaction
17.7.21 Prepare molar volume of oxygen with hydrogen peroxide
17.3.10 Potassium iodide with hydrogen peroxide, reverse colour change
17.7.22 Prepare hydrogen peroxide solution
17.7.23 Tests for hydrogen peroxide

17.1.1 Catalysts and rate of reaction
See diagram 3.2.94: Catalysts and rate of reaction.
1. Put a cube of sugar in a glass or aerated water, e.g. fizzy lemonade.
The drink fizzes more, because the sugar cube provides more sites for the formation of carbon dioxide bubbles on the sharp corners of the sugar crystals.
The sugar is said to act as a physical catalyst.
2. Fill a burette with water and invert it in a container of water to measure the volume of a gas in the burette by downward displacement of water.
Use a flask or test-tube fitted with a one-hole stopper and bent delivery tube.
Add 2 mL of 20 volumes of hydrogen peroxide to 50 L of water in the flask.
Note the time then immediately add 1 g of manganese (IV) oxide to the flask, close the stopper and adjust the end of the delivery tube inside the burette.
Note the volume of oxygen gas in the burette at intervals of 15 seconds.
Use graph paper to plot the volume of oxygen gas produced every 15 seconds against the time of the reaction.
Manganese (IV) oxide is the best catalyst for this reaction.
MnO2 + 2H2O2 --> MnO2 + 2H2O + O2

17.1.2 Concentration and temperature, cobalt (II) chloride-6-water
Test the effect of concentration and temperature on the equilibrium position.
1. Dissolve 4 g cobalt (II) chloride-6-water in 40 mL water.
The solution contains Co(H2O)62+, it is pink.
2. Add concentrated hydrochloric acid until total volume is 100 mL, it is violet (between pink and blue).
3. Add more concentrated hydrochloric acid to the violet solution.
The solution contains CoCl42-, it is blue.
4. Add water to the violet solution, it turns pink.
Heat the solution, it turns blue.
5. Cool the solution with ice water, it turns pink.
6. Add sodium chloride to the pink solution, it turns blue.
Co(H2O)62+ (aq) + 4Cl- (aq) <--> CoCl42- (aq) + 6H2O (∆H +ve).

17.1.3 Concentration of reactants, hydrochloric acid with magnesium
Pour 50 mL 6 M hydrochloric acid solution into a 100 mL standard flask.
Make up the solution to the mark and mix it well.
This is Solution B, 3 M.
Pipette 50 mL Solution B into another 100 mL standard flask.
Make up this solution to the mark and mix it well.
This is Solution C, 1.5 M.
Pour the remaining solution into a 100 mL beaker.
Pipette 50 mL solution C into another 100 mL standard flask and make this solution up to the mark and mix it well.
Label this solution D, 0.75 M.
Pour the remaining solution into a 100 mL beaker.
Pipette 50 mL of solution D into a 100 mL beaker.
Clean the oxide coating from a 2 cm strip of magnesium ribbon.
Cut off four 0.5 cm pieces of magnesium ribbon.
Put one strip into solution A and record the time of the reaction until you have dissolved all of the magnesium.
Repeat with a new magnesium strip for each of solutions B, C, and D.
Graph the results with time as the vertical axis and concentration as the horizontal axis.
Note whether a linear relationship exists between acid concentration and the rate at which magnesium dissolves.
Mg + 2HCl --> MgCl2 + H2.

17.1.4 Concentration and rate of reaction, sodium thiosulfate with dilute hydrochloric acid
See diagram 3.2.92: Black cross no longer visible.
1. Dissolve 5 g sodium thiosulfate crystals in 500 mL water.
Add 5 mL hydrochloric acid to 50 mL the sodium thiosulfate solution.
2. Dissolve 20 g of sodium thiosulfate in 500 mL of water and put 50 mL of the solution in a container.
Place the container on a black cross marked on a sheet of paper.
Add 5 mL of dilute hydrochloric acid to the container and note the time.
Look down through the solution and note when the black cross is no longer visible.
Sulfur is produced during the reaction making the solution cloudy.
Repeat the experiment with 40 mL of sodium thiosulfate solution and 10 mL of water.
Add 5 mL of dilute hydrochloric acid.
The time when the black cross is no longer visible is greater.
Repeat the experiment with 30 mL of sodium thiosulfate solution and 20 mL of water.
The time when the black cross is no longer visible is still greater.
Repeat the experiment with 20 mL of sodium thiosulfate solution and 30 mL of water.
Use graph paper to plot the volume of the thiosulfate solution (concentration) against time taken for the reaction.
Na2S2O3 (aq) + 2HCl (aq) --> H2O (l) + SO2 (g) + S (s) [The S (s) causes the solution to become cloudy.]
(S2O3)2- (aq) + 2 H+ (aq) --> H2O (l) + SO2 (g) + S (colloidal)

17.1.5 Concentration of reactants, hydrochloric acid with sodium thiosulfate
See diagram 17.4.1: Observe the cloudiness in the solution
1. The reaction slowly forms sulfur that makes the solution cloudy.
Measure the rate of reaction by measuring the cloudiness in the solution.
Note the time until you can no longer observe a black cross on a piece of paper below the beaker.
Change the concentration of the sodium thiosulfate and keep the concentration of hydrochloric acid constant.
Dissolve 20 g sodium thiosulfate in 500 mL of water.
Pour 50 mL of this solution into a 100 mL beaker.
Put the beaker on a sheet of paper marked with a black cross.
Add 5 mL of 2 M acid and stir the acid into the solution.
Record the time when the cross is no longer visible through the precipitated sulfur in the solution.
2. Repeat the experiment with a smaller concentration of thiosulfate as follows:
Table 17.7.5

.	Sodium thiosulfate Solution	Deionized water
Beaker 1	30 mL	20 mL
Beaker 2	20 mL	30 mL
Beaker 3	10 mL	40 mL

Stir the solution, then add 5 mL of 2 M acid as before.
The time for the cross to become invisible is greater.
The reaction takes longer as the concentration decreases.
Draw a graph to plot concentration of the thiosulfate solution against time taken for the reaction.
Express concentration values as the volume of the original thiosulfate solution used, e.g. 50 mL, 30 mL, 20 mL and 10 mL.
Na2S2O3 (aq) + 2HCl (aq) --> H2O (l) + SO2 (g) + S (s).

17.1.6 Dilute hydrochloric acid with zinc
See diagram 17.1.1: Count bubbles, dilute hydrochloric acid with zinc
1. Put a piece of granulated zinc in dilute hydrochloric acid in a test-tube.
Count the number of bubbles of hydrogen gas that reach the surface of the solution every 30 seconds.
Draw a graph by plotting the number of bubbles along the vertical axis and time along the horizontal axis.
The reaction starts quickly then becomes slower until it stops.
The comparative rates of reaction is measured by the size of the extended balloons. 2. See diagram 17.1.2: Volume of gas, dilute hydrochloric acid with zinc
Repeat the experiment with a flask connected to a gas syringe to measure how much hydrogen gas forms in the reaction.
Zn + 2HCl --> ZnCl2 + H2

17.1.7 Dilute hydrochloric acid with marble chips
See diagram 17.3.3: Gas burette
This experiment is used to investigate:
1. the effect of changing the surface area of marble chips and
2. changing the concentration of hydrochloric acid. on the rate of reaction between marble chips and dilute hydrochloric acid.
1. The smallest particles show the most vigorous reaction:
If two pieces of substance with the same weight are ground into coarse and fine sizes, the fine size pieces would have a greater total surface area.
Be Careful! Hydrogen gas forms.
Break marble chips or granulated zinc into four sizes with a hammer.
Put 2 g of each into four test-tubes:
Test-tube 1, Original size as control,
Test-tube 2, Rice grain size,
Test-tube 3, Half rice grain size,
Test-tube 4, Coarse powder.
Add the same volume of 2 M hydrochloric acid to each test-tube.
Compare the reactions.
Rate of reaction of Test-tube 4 > Test-tube 3 > Test-tube 2 > Test-tube 1.
Collect the gas in a burette inverted over water.
Compare how much gas forms in unit time for each size of marble chips or zinc.
Weigh the container and note the loss in mass every half minute while the reaction goes on.
CaCO3 (s) + 2HCl (aq) --> CaCl2 (aq) + CO2 (g) + H2O (l)

17.1.8 Measure rates of reaction
Rates of reactions for laboratory experiments should not be so fast that an explosion occurs.
Also, they should not be so slow that you cannot observe or measure any change in reasonable time.
To measure the speed of a reaction some change must be measured, e.g. time taken for colour change, volume of gas formed, weight of reactants used.
Chemical reactions occur only if the particles of reactants can collide with sufficient energy, the activation energy, Ea.
Reactions will become faster if the number of collisions increases and if the energy of collision increases.
You can increase the rate of reaction by increasing concentration of reactants, light energy, particle size, pressure, temperature, and with catalysts.

17.1.9 Magnesium with hydrochloric acid
A reversible reaction can proceed in either direction by altering the conditions of the reaction, e.g. 1. altering the relative concentrations,
active mass 2. altering the temperature 3. altering the pressure.
Measure 10 mL of concentrated hydrochloric acid into 4 beakers, then add the following:
Beaker A add nothing, Beaker B add 10 mL water, Beaker C add 30 mL water, Beaker D add 70 mL water.
Simultaneously add 6 cm of magnesium ribbon to each beaker.
Note how the rate of reaction in each beaker is proportional to the concentrations of the acid.

17.1.10 Potassium iodide with potassium iodate
1. Put 30 mL of solution A containing 5 g of potassium iodate per litre in a measuring cylinder and dilute to 200 mL with water.
Ttransfer the solution to a beaker and add drops of starch solution.
2. Put 30 mL of solution B containing 10 g of hydrated sodium sulfite per litre and 2.5 mL of 2M sulfuric acid in a measuring cylinder and dilute to 200 L
3. Pour solution B into the beaker containing the diluted solution A.
Note the starting time and record the time in seconds when a dark blue coloration appears.
4. Repeat the experiment with 5 mL, 7.5 mL, 10 mL, 12.5 mL and 15 mL of 2M sulfuric acid.
Plot a graph of volume of acid to reciprocal of time and note the straight line result.
Reaction 1. is a slow and reaction.
The presence of iodine as a dark blue coloration indicates the completion of reaction 1.
Reaction 1. IO3- + 3HSO3- --> I- + 3HSO4-
Reaction 2. is rapid, but does not occur until reaction 1. is complete
Reaction 2. 5I- + IO3- + 6H+ --> 3I2 (s) + 3H2O.

17.1.11 Rate of reaction, k
For many chemical reactions, but not all, increasing the concentrationof reactants increases the rate of reaction.
The rate constant, k, is the constant for a given reaction at a given temperature.
H2 (g) + I2 (g) --> 2HI (g)
rate = k [H2 (g)] [I2 (g)], where [H2(g)] = concentration of hydrogen gas
If x = any substance, [x] = concentration of x.
If in a chemical reaction, [x] is doubled and the rate of reaction remains constant, then the rate of reaction is independent of [x].
If in a chemical reaction, [x] is doubled and the rate of reaction doubles, then the rate of reaction = k[x].

17.1.13 Rates of reaction of aspirin
Be Careful! Aspirin is a drug so do not let the students take them!
Soluble aspirins dissolve in water with a well defined end point and constants interaction time.
Use these tablets to investigate the effects of temperature, stirring, crushing and varying the water volume on reaction time.
Investigate the temperature change required to halve the reaction time between one tablet and known quantity of water at room temperature.
Investigate the number of tablets or fractions of tablets required to exactly double the reaction time in a known quantity of water at room temperature.

17.1.14 Size of particles and rate of reaction with balloons
See diagram 3.2.91: Size of particles and rate of reaction.
1. Use a hammer to break natural marble chips into four sizes:
1.1 coarse powder, 1.2 half a rice grain, 1.3 rice grain, 1.4. marble chips.
Put 2 g of each size separately into four test-tubes.
Blow up four balloons several times to stretch them.
Put 5 mL of dilute hydrochloric acid into each of the four balloons.
Slip the mouths of the balloons over the tops of the test-tubes, but do not let any acid enter the test-tubes.
Tip the acid from each balloon into the attached test-tube.
Note which balloon is the fastest and the slowest to expand, because of the production of carbon dioxide.
The coarse powder produces carbon dioxide in the shortest time.
2. Repeat the experiment without balloons, but with four conical flasks on a sensitive top balance.
Add 5 mL of dilute hydrochloric acid to 2 g of coarse powder and note the loss in weight every 30 seconds.
Then continue the experiment with the other sizes of marble chips.

17.1.15 Temperature and rate of reaction
Increase in temperature of 10oC usually doubles the rate of reaction.
The rate increases, because collisions between particles are more frequent and with more energy.
Also, more collisions per second may occur.
1. Add enough potassium manganate (VII) solution to dilute sulfuric acid to make it pink.
Put in a nail and record the time for the solution to lose the pink colour at room temperature.
Repeat the experiment at 10oC above room temperature.
Repeat the experiment at 20oC above room temperature.
Draw a graph to show time taken to decolorize against temperature.
The rate of reaction may double for each 10oC rise in temperature.
2. Iron with sulfuric acid.
Put 1 g of steel wool in a test-tube and add 10 mL of dilute sulfuric acid.
Heat the mixture in a beaker of hot water until all the steel wool has dissolved.
Put a lighted paper into the test-tube to show that the gas given off is hydrogen gas.
Filter the solution while hot and leave to cool.
If crystals do not form on cooling, add alcohol to cause crystallization.
Pour off the liquid and dry the crystals between absorbent paper.
Observe the shape of the crystals of green salt iron sulfate.
iron(s) + sulfuric acid (aq) --> hydrogen (g) + iron sulfate (aq).
3. Put 1 cm depth of iron filings in a test-tube.
Just cover the iron filings with a dilute acid solution.
Warm the test-tube until frothing starts.
Hydrogen gas is colourless and odourless, but any impurities in the iron filings give a nasty smell.
Test for hydrogen gas:
Stop heating, place ther thumb over the end of the test-tube, count to 5, apply lighted paper to the end of the test-tube, the hydrogen gas explodes with a loud pop.
Never test more than one test-tube full of hydrogen gas!
4. Dissolve 20 g of sodium thiosulfate in 500 mL of water and put 10 mL of the solution in a container.
Add 40 mL of water to the container.
Note the room temperature and add 5 mL of dilute hydrochloric acid.
Look down through the solution and note when the black cross is no longer visible.
Repeat the experiment at 30oC, 40oC, 50oC and 60oC.
Use graph paper to plot the temperature of the sodium thiosulfate solution against time taken for the reaction.
The time when the black cross is no longer visible becomes less as the temperaturerises.

17.2.1 Belousov-Zhabotinskii clock reaction, BZ reaction
This reaction is famous as an example of non-equilibrium thermodynamics.
Solution A: Dissolve 38.41g of malonic acid in 600 mL of water.
Add water to 1 litre (malonic acid, propanedioic acid, CH2(COOH)2)
Solution B: Dissolve 32.26g of malonic acid in 600 mL of water.
Add 7.021g of potassium bromide.
Add water to 1 litre (0.31 M malonic acid + 0.059 M potassium bromide)
Solution C: Dissolve 10.412g of ammonium cerium (IV) nitrate in 400 mL of water.
Slowly add 150 mL of concentrated sulfuric acid.
Solution gets hot.
Leave to cool to room temperature.
Add water to 1 litre (0.019 M ammonium cerium (IV) nitrate + 2.7 M sulfuric acid)
Solution D: 0.025M ferroin sulfate (direct purchase required).
(Ferroin, ferroin sulfate, C36H24FeN6O4S, Ferroin indicator solution)
1. Clock reaction
In a 1 litre flask, add 300 mL of solution A to 300 mL of solution B.
When solution clears, add 300 mL of solution C.
When solution fades to yellow Cerium colour, add 3.6 mL of solution D.
The chemical clock reaction commences as the solution cycles through green --> purple --> red --> blue, and back, with an initial 60 seconds total cycle time.
2. Concentric circles
Add 2.5 mL of Solution A and 2.5 mL of solution B to a Petri dish.
Add 2.5 mL of solution C.
Add 1 mL of solution D.
Mix well, then after 1 minute groups of concentric rings appear from the outside of the Petri dish then move towards the centre.
During the ferroin-catalyzed oxidation of malonic acid by acid bromate, successive bands of oxidation advance through reduced regions.
The reaction is studied as an example of non-equilibrium phenomena in that patterns form when the solution is stimulated.

17.2.2 Hydrogen peroxide clock reaction, Briggs-Rauscher oscillating reaction
1. Solution A.
Add 43 g potassium iodate (KIO3) + 5 mL sulfuric acid (H2SO4), to 800 mL deionized water, stir to dissolve then dilute to 1 litre.
2. Solution B.
Add 15.6 g of malonic acid + 3.4 g manganese sulfate monohydrate, to 800 mL deionized water + 4 g of soluble starch, stir to dissolve, dilute to 1 litre.
3. Solution C.
Dilute 400 mL 30% hydrogen peroxide (H2O2) to 1 litre.
4. Mix thoroughly 300 mL of solutions A. and B., then add 300 mL of solution C.
The 3 mixed colourless solutions oscillate through colourless --> amber--> blue --> colourless, for about 5 minutes, then stays blue-black.
Hydrogen peroxide + iodate --> iodine + oxygen (auto-catalytic) (blue)
IO3- + 2H2O2 + CH2(COOH)2 + H+ --> ICH(COOH)2 + 2O2 + 3H2O.

17.2.3 Iodine clock reaction, hydrogen peroxide with potassium iodide
First version
1. Solution A: Add drops of water to 0.2 g soluble starch, pour this paste into a beaker containing 1 cc boiling water and stir.
Pour this solution into a 1 litre beaker and dilute to 800 mL.
Add 30 mL glacial acetic acid + 4.1 g sodium acetate + 50 g potassium iodide, + 9.4 g sodium thiosulfate-5-water.
Dissolve all solutes by stirring and when cooled to room temperature make up solution to 1 litre.
The solution is slightly cloudy.
Solution B: Dilute 500 mL 20 vols hydrogen peroxide to 1 litre.
Put 100 mL of each solution in separate beakers.
Pour the contents of one beaker into another and stir quickly or use a magnetic stirrer.
After about 20 seconds the colourless mixture suddenly turns dark blue.
Repeat the experiment at room temperature, but note the time taken for the colour change after mixing.
2. Repeat the experiment at 10oC above room temperature.
Note the time taken for the colour change after mixing, about double the time compared to 1. 3.
3. Repeat the experiment at 10oC below room temperature.
Note the time taken for the colour change after mixing, about half the time compared to 1. 4.
4. Repeat the experiment at room temperature using half the concentration of solution B, i.e. 50 mL solution B + 50 mL water.
Note the time taken for the colour change after mixing, about double the time compared to (1.), because the reaction rate is halved.
The amount of hydrogen peroxide has been halved.
5. Repeat the experiment at room temperature using half the concentration of solution A, i.e. 50 mL solution A + 50 mL water.
Note the time taken for the colour change after mixing, the same as for (1.
). The reaction rate has been halved, but the amount of thiosulfate also has been halved, so you only need to make half the amount of iodine.
If the sodium thiosulfate was not in solution A, but in a solution C, the result of 5. would be the same as in 4.
H2O2 (aq) + 2I- (aq) + 2H+ (aq) --> I2 (aq) + 2H2O (l).
Hydrogen peroxide reacts with iodide ions to form iodine.
The ethanoic acrid with sodium ethanoate buffers the reaction I2 (aq) + 2S2O32- (aq) --> 2I- (aq) + S4O62- (aq) +
The iodine reacts with thiosulfate to form tetrathionate ions and iodide ions return to the solution.
As soon as all the thiosulfate is converted to tetrathionate ions the remaining iodine reacts with the starch solution to form a blue black colour.
Second version
Iodine clock reaction, hydrogen peroxide, potassium iodide.
Solution A Prepare a solution of potassium iodide, sodium thiosulfate and starch..
Solution B Add sulfuric acid to hydrogen peroxide solution..
Add solution A to solution B..
H2O2 (aq) + 3I- (aq) + 2H+ (aq) --> I3- (aq) + 2H2O (aq) slow reaction.
hydrogen peroxide + iodide ion + hydrogen ion --> triiodide ion + water.
I3- (aq) + 2S2O32- (aq) --> 3I- (aq) + S4O62- (aq).

17.2.4 Iodate clock reaction
Concentration and temperature affects rate of reaction, potassium iodate, sodium metabisulfite (sodium bisulfite)
Solution A.
Dissolve 4.3 g potassium iodate, KIO3, in on litre of water
Solution B.
Prepare a starch solution by dissolving 5 g of starch in 800 mL hot water, boil, then leave to cool.
Add 0.2 g sodium metabisulfite. Add 5 cc of M sulfuric acid.
Dilute to 1 litre.
1. Note the time for colour change, colourless to dark blue, by adding 50 mL of solution A in beaker A to so 50 mL of solution B in beaker B.
Pour the solution back and forth, B --> A --> B --> A.
The reaction may take 5 to 6 minutes.
The reaction is designed to be a slow colour change reaction, a clock reaction.
2. Repeat the experiment with half concentration of solution A.
3. Repeat the experiment with solution A warmed in a water bath to 35oC.
Increase of concentration will increase the rate of reaction, because of increased chance of particles reacting.
Increased temperature will increase the rate of reaction, because of increase of speed of particles increases chance of particles reacting.
IO3-(aq) + 3HSO3- (aq) --> I- (aq) + 3HSO4-(aq)
iodate ion + bisulfite ion --> iodide ion + bisulfate ion
IO3- (aq) + 5I- (aq) + 6H+ (aq) --> 3I2 + 3H2O (l) iodate ion oxidizes the iodide ion to iodine
iodate ion + iodide ion + hydrogen ion --> iodine + water
However, the iodine is reduced immediately back to iodide by the bisulfite:
I2 (aq) + HSO3- (aq) + H2O (l) --> 2I- (aq) + HSO4-(aq) + 2H+ (aq) bisulfite
The bisulfite reduces iodine back to iodide ion.
iodine + bisulfite ion + water --> iodide ion + bisulfate ion + hydrogen ion
When no more bisulfate ion exists, because it is all used up reducing the iodine, the remaining excess iodine forms a blue-black colour with starch.

17.2.5 Old Nassau flag clock reaction
The "Old Nassau flag" was orange and black.
This experiment should NOT be done in schools, because it uses mercuric (II) chloride, HgCl2!
1. Sodium metabisulfite reacts with water to form sodium hydrogen sulfite, colourless reaction
Na2S2O5 (aq) + H2O (l) --> 2NaHSO3 (aq).
2. Hydrogen sulfite ions reduce iodate (V) ions to iodide ions, colourless reaction
IO3- (aq) + 3HSO3- (aq) --> I- (aq) + 3SO42- (aq) + 3H+ (aq).
3. With excess iodide ions, mercury (II) iodide, HgI2, forms as an orange precipitate
Hg2+ (aq) + 2I- (aq) --> HgI2 (s).
4. When all the mercury is used up reacting with iodine, the remaining iodate and iodide ions react to form iodine and a blue-black iodine starch complex forms.
IO3- (aq) + 5I- (aq) + 6H+ (aq) --> 3I2 (aq) + 3H2O (l)
The orange precipitate and the black iodine complex are the colours of Princeton University, sometimes referred to as "Old Nassau".

17.2.6 Persulfate-iodide clock reaction
Use sodium persulfate Na2S2O8, or potassium persulfate K2S2O8, or ammonium persulfate (NH4)2S2O8.
The reaction mixture remains colourless for several minutes after the reactants are mixed.
1. Solution A.
Prepare 10 mL of 0.1 M potassium iodide solution + 5 mL of 0.01 M sodium thiosulfate solution.
Solution B.
Prepare 10 mL of 0.1 M ammonium persulfate solution + 1 drop of starch solution
Pour solution B into solution A.
Pour the solution back and forth, A --> B --> A --> B
Record the time when the starch-I2 complex forms.
Record the temperature of the solution.
2. Solution A.
Prepare 10 mL of 0.1 M potassium iodide solution + 5 mL of 0.01 M sodium thiosulfate solution
Solution B.
Prepare 5 mL of 0.1 M ammonium persulfate solution + 1 drop of starch solution.
Add 5 mL of water to keep the volume constant.
Pour solution B into solution A.
Pour the solution back and forth, A --> B --> A --> B
Record the time when the starch-I2 complex forms.
Record the temperature of the solution
S2O82- (aq) + 2I- (aq) --> 2SO42- (aq) + I2 (aq)
Slow reaction, but the iodine formed is soon changed to iodine in the next reaction
I2 (aq) + 2S2O32- (aq) --> 2I- + S4O6- (aq)
Fast reaction
I- (aq) + starch --> blue-black complex
When all the thiosulfate ion is used up, the concentration of I2 increases and forms a blue complex with starch.
The rate of reaction is the rate of consumption of the S2O82- ion.

17.3.1 Catalysis
Catalysts increase the rate of reactions without themselves being chemically changed.
A catalyst can change the rate of a chemical reaction without itself being permanently changed.
They provide an alternative pathway for the reactions and so decrease the activation energy needed.
Substances that slow the rate of reactions are called inhibitors.

17.3.2 Catalytic oxidation of methyl alcohol
Be Careful!
Have ready a piece of cardboard or glass to put over the beaker if the methyl alcohol ignites!
Make a platinum spiral by winding platinum wire around a glass rod and leave a length of wire above the spiral.
Put 1 cm of methanol (methyl alcohol) in a small beaker and warm it gently with an electric heater.
Do NOT use a Bunsen burner!
Heat the spiral strongly with the electric heater and transfer the glowing spiral to the beaker.
Hold the length of wire above the spiral, so that the spiral is just above the methyl alcohol.
The spiral continues to glow and you can notice the smell of formaldehyde given off from the reaction
2CH3OH + O2 --> 2HCHO + 2H2O.

17.3.3 Catalytic oxidation of ammonia solution, with oxygen gas
Make a platinum spiral by winding platinum wire around a glass rod and leave a length of wire above the spiral.
Put 1 cm of 880 ammonia solution in a small beaker and pass oxygen gas from an oxygen cylinder through the mixture.
Observe brown fumes of nitrogen dioxide or white fumes of ammonium nitrate and ammonium nitrite.
4NH3 + 7O2 --> 4NO2 + 6H2O

17.3.4 Bromine catalyses the oxidation of sulfur to sulfuric acid
Do the experiment in a fume cupboard.
Put 1 cc of flowers of sulfur into "evaporating basin 1" and "evaporating basin 2".
Add 5 mL of concentrated nitric acid to each evaporating basin.
Be Careful!
Add one drop of bromine to "evaporating basin 1" only.
Be Careful!
Warm each evaporating basin for 2 minutes.
Pour the solutions into "test-tube 1" and "test-tube 2".
Add hydrochloric acid, then barium chloride solution, to each test-tube.
A precipitate of barium sulfate forms only in "test-tube 1".
In this catalysis, intermediate compounds form, which are more readily decomposed.
2S + Br2 --> S2Br2 sulfur monobromide
2S2Br2 + 2H2O --> SO2 + 4HBr + 3S
SO2 + 2HNO3 --> H2SO4 + 2NO2
2HBr + 2HNO3 --> 2H2O + 2NO2 + Br2.

17.3.5 Ethyl acetate with sodium hydroxide, autocatalytic hydrolysis
1. Make the following solutions:
Flask A: 100 mL of 0.5 M sulfuric acid,
Flask B: 0.5 M hydrochloric acid,
Flask C: 0.5 M acetic acid.
Leave the 3 flasks in a thermostat at 25oC so that all the contents are at the same temperature.
Titrate 2 mL of each acid in Flask A, Flask B and Flask C separately against 0.1 M sodium hydroxide, with phenolphthalein indicator, and record the results.
2. Add 5 mL of ethyl acetate solution to Flask A, Flask B and Flask C then leave them in a thermostat for 15 minutes.
Titrate 2 mL of the Flask A, Flask B and Flask C solutions (+ ethyl acetate) against 0.1 M sodium hydroxide solution, with phenolphthalein indicator.
3. Repeat the titrations every 15 minutes for 2 hours and tabulate the results.
Increased titration with time occurs, because of the formation of acetic acid by hydrolysis of ethyl acetate.
The catalytic action of the hydrogen ions in the original acid is increased by the hydrogen ions from the acetic acid produced by the hydrolysis.
The rate of hydrolysis in the presence of a mineral acids in Flask A and Flask B is higher than in Flask C where the hydrogen ion concentration is lower.
This reaction is an example of autocatalysis.
CH3COOC2H5 + H2O --> CH3COOH + C2H5OH.

17.3.6 Heat potassium chlorate, manganese dioxide catalyst
Be careful!
In some school systems this experiment is not allowed, because potassium chlorate may explode!
* Mix 0.5 g of manganese dioxide with 2 g of potassium chlorate and put the mixture in "ignition tube 1".
* Put 0.5 g of manganese dioxide in "ignition tube 2".
* Put 2g of potassium chlorate in "ignition tube 3".
Insert the ignition tubes vertically and close together in a sand tray and place a safety screen between you and the sand tray.
Slowly heat the sand tray.
Use a glowing splint to tests for oxygen at the openings of the ignition tubes.
Oxygen appears first from Ignition tube 1 and later from the other ignition tubes.
When the reaction is complete, wash the contents into a beaker.
Stir the contents to dissolve all the potassium chloride and any remaining potassium chlorate.
Filter the mixture, dry the residue on the filter paper and weigh the manganese dioxide residue to show that there is no loss in weight.

17.3.7 Oxalic acid with potassium manganate (VII), autocatalysis
Solution A: 6 g of ethanedioic acid-2-water (oxalic acid) in 300 mL of water.
Solution B: 100 mL of 0.001 M potassium manganate (VII) solution (potassium permanganate).
Pour 150 mL of Solution A into two beakers, then add 5 mL of concentrated sulfuric acid to each beaker.
Add 50 mL of the potassium manganate (VII) solution to each beaker.
Add a small crystal of manganese (II) chloride, MnCl2, to one beaker then stir the solutions in both beakers.
The solution containing the crystal of manganese (II) chloride starts to lose colour and becomes colourless in about a minute.
The other solution does not change in colour for two or three minutes but, when sufficient Mn2+ ions are present, it starts to become colourless.
The Mn2+ autocatalyses the solution as follows.
2MnO4- (aq) + 5C2O42+ (aq) + 16H3O+ (aq) --> 2Mn2+ (aq) + 10CO2 (g) + 24H2O (l).

17.3.8 Oxidation of acetone vapour, copper catalyst
1. Heat a copper wire coil or copper coil to red heat in a Bunsen burner flame then hang it just above a very thin layer of acetone in a beaker.
Note the shimmering colours of the copper surface caused by the heat being maintained in the copper coin from the heat of the chemical reactions on its surface.
Also note the colours of black copper (II) oxide< red copper (I) oxide (red) and pink copper metal.
You can warm the beaker in hot water to produce sufficient acetone vapour.
Be careful!
Do NOT smell the oxidation products, because they contain ketones that may be a health hazard.
The red heat is maintained as long as some acetone remains, because it keeps warm through the heat of the exothermic reaction.
The reaction is safe except that at the top of the beaker where air dilutes the vapour a flame may occur.
If this happens, move the hot copper quickly out and cover the beaker to extinguish the flame.
2. Drop the red-hot copper into a beaker with a 2 mm thin layer of acetone at the bottom.
A sizzling sound caused by the cooling of the copper ends in a loud crescendo when liquid acetone contacts the copper to increase the rate of cooling.
Be Careful!
3. Put 1 cm of acetone (propanone) in a beaker without a pouring lip.
Attach one end of a copper wire to a strip of copper or copper coin.
Wind the other of the copper wire around the stem of a glass stirring rod.
Place the stirring rod across the beaker and wind it around so that when the strip of copper hangs down it is still 1 cm above the surface of the acetone.
Put a cover over the beaker and place it in a fume cupboard.
Wear goggles.
Light a Bunsen burner well away from the acetone in the beaker.
Heat the copper until it is red hot.
Turn off the Bunsen burner, remove the cover over the beaker and quickly place the glass stirring rod over the beaker.
This action allows the heated strip of copper to be glowing still red hot over the acetone.
The strip of copper produces a flickering gold flame as it catalyses the oxidation of acetone in the acetone vapour to carbon dioxide.
Turn off the lights to see the flickering better.
Heating the copper just supplies enough heat to initiate the reaction that is highly exothermic to keep the coin at a high temperature after the reaction starts.
The reaction continue until all the acetone evaporates or the coin is lifted out of the beaker.
CH3COCH3 (g) + 3/2 O2 (g) --> CH3CHO (g) + CO2 (g) + H2O (g)
acetone + oxygen --> acetaldehyde + carbon dioxide + water.

17.3.9 Potassium bromate with propanedioic acid, double autocatalytic reaction, oscillating reaction
In this reaction, bromine ions form to give a red colour, but some intermediate product also forms to react with bromine ions to give a colourless solution.
Use a clean beaker washed in deionized water.
Add 75 mL concentrated sulfuric acid to 750 mL deionized water (NOT tap water!).
Be Careful!
Leave the hot acid solution long enough to cool to room temperature slowly.
Stir the cooled solution fast enough to form a vortex.
Add 9 g propanedioic acid HOOC.CH2.COOH, malonic acid.
Add 8 g potassium bromate (V), KBrO3.
Add 1.8 g manganese (II) sulfate, MnSO4.H2O.
Observe a red colour that oscillates from red to colourless, with increasing time between oscillations.
3HOOC.CH2.COOH (aq) + 4BrO3- --> 4Br- (aq) + 9CO2 (g) + 6H2O (l).

17.3.10 Potassium iodide with hydrogen peroxide, reverse colour change
Dissolve potassium iodide crystals in water.
Add drops of starch solution and dilute hydrochloric acid.
The solution is colourless.
Add drops of dilute hydrogen peroxide solution.
The solution turns blue black.
Iodide ions are oxidized to iodine that gives starch a blue black colour.
2KI (aq) + H2O2 (l) --> 2KOH (aq) + I2 (g)
Add drops of dilute sodium thiosulfate solution.
The solution turns colourless.
The sodium thiosulfate reduces the iodine back to iodide ions that are colourless.
I2 (g) + Na2S2O3 (aq) --> NaI (aq) + Na2S2O5 (aq)
Wait until the blue black colour returns.
Add drops of sodium thiosulfate solution and it disappears again.
The first reaction is still going slowly.
The second reaction is much slower.

17.3.11 Sodium hypochlorite decomposition, cobalt sulfate catalyst
Warm a test-tube half full of sodium hypochlorite solution and observe that no decomposition occurs.
Add a two drops of cobalt sulfate solution and tests for oxygen with a glowing splint.
2NaOCl --> 2NaCl + O2 (g).

17.3.12 Burning sugar cube, combustible cube
1. Use a match to set fire to a sugar cube.
It will not burn.
Dip the sugar cube in ash.
The sugar cube will light much more easily.
Carbon in the ash gets hotter and raises the temperature of the surrounding sugar crystals to the ignition temperature to burn the sugar.
When the first crystals are burning, they give out enough heat for other sugar crystals nearby to continue the burning.
The ash acts as a catalyst causing the sugar to ignite, but not ignite itself.
The ash remains after the sugar cube is totally changed to carbon dioxide and water.
2. Fasten a sugar cube in a wire loop.
Rub one corner of the cube in fine ash.
Hold the sugar cube at an angle over a dish.
Heat the corner with a Bunsen burner, also held at an angle.
The corner starts to burn with a blue flame.
Black, sticky liquid drops into the dish and small rings of smoke form.
Drop the wire loop into the sink and turn on the tap.

17.4.1 Common ion effect in ammonium chloride solution
On addition of ammonium chloride to a solution of ammonium hydroxide, the concentration of OH- decreases.
The increased concentration of ammonium ion from ammonium chloride reduces the (OH-) concentration in ammonia solution,
NH3 (aq) ("ammonium hydroxide") solution.
Add a drop of phenolphthalein to a dilute solution of ammonium hydroxide.
The solution goes pink.
Add solid ammonium chloride drop by drop until the colour disappears.
NH4Cl produces NH4+ ions, which increases the speed of the back reaction.
NH4OH <--> NH4+ + OH-
Add NH4Cl (NH4)2SO4 to increase the concentration of NH4+1, and the number of collisions per second between NH4+1 and OH-1.
The equilibrium shifts to the left, and the concentration of the OH-1 decreases.
The ion NH4+1, is common to both the ammonium hydroxide solution and the added ammonium chloride salt, so it is called a "common ion".

17.4.2 Common ion effect, precipitate barium chloride from solution
Add 3 drops of concentrated hydrochloric acid to a saturated solution of barium chloride.
Barium chloride precipitates as white crystals.
Increase in the concentration of the chloride ion favours the backward reaction with subsequent precipitation of barium chloride.
BaCl2 <--> Ba2+ + 2Cl-.

17.4.3 Common ion effect to precipitate sodium chloride from solution
Add a 3 drops of concentrated hydrochloric acid to a saturated solution of table salt.
Sodium chloride precipitates as white crystals.
Increase in the concentration of the chloride ion favours the backward reaction with subsequent precipitation of common salt.
Increase in the concentration of the chloride ion favours the backward reaction with subsequent precipitation of common salt.
NaCl <--> Na+ + Cl-
HCl --> H+ + Cl-.

17.4.4 Common ion effect, sodium ethanoate and ethanoic acid
A common ion occurs in solution and in another substance added to the solution.
The common ion effect refers to the change in concentration of solute ions on addition of a substance to a solution of another substance.
Hydrochloric acid is a completely dissociated strong acid.
Ethanoic acid is a less dissociated weak acid.
Make three solutions and add universal indicator to each:
Solution A: 100 mL 2 M hydrochloric acid, then add universal indicator until red.
Solution B: 100 mL 2 M ethanoic acid, then add universal indicator until orange.
Solution C: Add 13.5 g sodium ethanoate-3-water to 80 mL 2 M ethanoic acid.
Add more ethanoic acid to 100 mL, then add universal indicator until yellow.
Put 1 g calcium carbonate powder + 1 mL liquid detergent into 3 measuring cylinders (A1, B1, and C1), containing 100 mL water.
Simultaneously, put solution A into measuring cylinder A1, put solution B into measuring cylinder B1, and put solution C into measuring cylinder C1.
Observe the measuring cylinders:
A1 has the fastest froth produced, height h.
B1 has the second fastest froth produced, height h.
C1 has the slowest froth produced, height h/2.
In this mixture of sodium ethanoate and ethanoic acid, the equilibrium moves left, decreasing the concentration of hydrogen ions.
CH3CO2H (aq) <--> CH3CO2- (aq) + H+ (aq).

17.5.1 Chemical Equilibrium
Chemical equilibrium / dynamic equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction.
In other words, there is no net change in the concentration of reactants and products.
1. Equilibrium exists only in a closed system.
No reactants are put in and no reactants are taken out.
A closed system cannot exchange matter with its surroundings, but it may exchange energy with its surroundings.
2. Limestone (calcium carbonate) is heated in a furnace to form quicklime (calcium oxide) and carbon dioxide.
The reverse reaction cannot occur, because the carbon dioxide is sucked out of the furnace.
Also, the calcium oxide is steadily and replaced by limestone.
This is a steady state system.
If the furnace were closed, at high temperature both decomposition and formation of calcium carbonate would occur.
When these process occur at the same rate then equilibrium exists.
CaCO3 (s) --> CaO (s) + CO2 (g).
3. Add different concentrations of iron (II) nitrate solution to potassium thiocyanate solution.
At equilibrium, the concentration of the SCN2+solution has a blue colour.
It is a property of this equilibrium reaction that does not change.
If sodium thiocyanate NaSCN or iron (III) nitrate Fe(NO3)3 is added to the equilibrium mixture it changes colour.
Fe3+ (aq) + SCN - (aq) <--> FeSCN2+(aq).
4. The following equation shows that both forward and reverse reactions are going on.
It does not show the position of equilibrium.
The position at equilibrium shows whether there are more reactants or more products at equilibrium.
It can shift if reactants or products are added or removed.
N2O4 (g) <--> 2NO2 (g)
In the above reaction, if, at equilibrium, N2O4 is added, the system moves to the right, i.e. some N2O4 changes to NO2 until equilibrium is reached.
In the above reaction, if, at equilibrium, N2O4 is removed, the system moves to the left, i.e. some NO2 changes to N2O4 until equilibrium is reached.
Position at equilibrium can shift if the temperature changes, but it will not shift, because solid is added or removed from a system.
Position at equilibrium does not change if a catalyst is added to the reaction.
However, but if the system is not at equilibrium, it will reach equilibrium quicker, because of the influence of the catalyst.
5. Le Chatelier's principle states that if the conditions of a system at equilibrium are altered, changes will occur in the system against the change in conditions.
When a system in equilibrium is subjected to a change in conditions, it adjusts itself so as to oppose that change.
Henri Le Chatelier (1850 - 1936).
If a stress is applied to a system in dynamic equilibrium, the system behaves in a way that relieves the stress.
6. The law of chemical equilibrium (law of mass action)
For reaction aA + bB <--> eE + fF (lower case = number of species, upper case = different types of species, e.g. 3 species of oxygen gas = 3O2) at equilibrium:
[E]e × [F]f / [A]a × [B]b= K, where K is the
equilibrium constant of that reaction at that temperature.
If K is large, at equilibrium, the concentration of products is much greater than the concentration of the reactants.
If K is small, at equilibrium, the concentration of products is much smaller than the concentration of the reactants.

17.5.2 Effect of temperature on chemical equilibrium
Heat 1 mL of ammonium chloride in a dry test-tube damp red litmus paper fixed inside the mouth.
The ammonium chloride sublimes and condenses on the side of the test-tube, leaving a clear space where the test-tube is hot.
The damp litmus paper turns blue.
The clear space contains the colourless gases ammonia and hydrogen chloride formed by the decomposition of the ammonium chloride.
Ammonia is lighter than hydrogen chloride so it diffuses faster and reaches the red litmus paper first.
Recombination to form ammonium chloride occurs in the cooler part of the test-tube
NH4Cl <--> HCl + NH3 [hot ">, cold <"].

17.5.3 Equilibrium between ICl and ICl3
See diagram 17.4.2: Equilibrium reaction between ICI and IC3
Put 0.1 g iodine in the U-tube.
With the 3-way tap in the CB position, turn on the filter pump to pass air bubbles through the sodium hydroxide solution.
With the 3-way tap in the AB position, turn on the filter pump and let drops of concentrated hydrochloric to fall on 10 g potassium permanganate.
As the chlorine passes over the iodine, first iodine monochloride, ICI, forms as a brown liquid.
then iodine trichloride, ICI3, forms as a yellow solid.
Turn the 3-way tap in the CB position, remove stopper 1 from the U-tube to draw in air.
The yellow solid turns into a brown liquid.
Replace stopper 1 in the U-tube and turn the 3-way tap to the AB position.
The brown liquid turns into a yellow solid.
Increasing concentration of chlorine moves equilibrium to the right, obeying Le Chatelier's principle.
With the 3-way tap in the AC position to get rid of the chlorine, tighten the screw clips.
When you dip the U-tube in water just below boiling point, the yellow solid turns into a brown liquid.
When you dip the U-tube in ice water, the brown liquid turns into a yellow solid.
Increase of temperature moves the equilibrium to the left.
Decrease of temperature moves the equilibrium to the right.
Dispose of the chemicals by using different sinks.
Pour the sodium hydroxide solution into a laboratory sink and wash it down with plenty of water to follow.
Wash the contents of the flask generating the chlorine down the sink in the fume cupboard with plenty of water to follow.
ICl (l) + Cl2 (g) --> ICl3 (s).

17.5.4 Heat nitrogen tetroxide
Dinitrogen tetroxide, N2O4
See diagram 17.5.6.2: Heat nitrogen tetroxide
Heat a test-tube containing the lead nitrate and also heat the centre of the long horizontal delivery tube.
The colour of the gas in the hot part of the delivery tube is a darker brown than in the cooler part of the test-tube, because more brown NO2 molecules are there.
The gas phase dissociates to form nitrogen dioxide.
N2O4 (pale yellow) <--> 2NO2 (brown) (pale yellow <-- cooling | heating --> brown).

17.5.5 Hydrolysis of antimony chloride
Repeat the above experiment using antimony chloride in the place of bismuth chloride.
The reactions and the explanations are similar to those for bismuth.
SbCl3 + H2O <--> SbOCl (s) (antimony oxychloride) + 2HCl.

17.5.6 Hydrolysis of bismuth chloride
Put some bismuth chloride in a test-tube, add 1 mL of water and observe white bismuth oxychloride forming.
Equilibrium was reached when certain concentrations of bismuth oxychloride and hydrochloric acid form.
BiCl3 + H2O <--> BiOCl (s) + 2HCl
[BiOCl] [HCl]2 / [BiCl3] [H2O] = k (Equilibrium constant)
Add more drops of water and observe the white bismuth oxychloride reappear.
Add drops of concentrated hydrochloric acid until the white precipitate disappears.
This occurs because some bismuth oxychloride and some hydrochloric acid reacted to form more bismuth chloride and water.
By these changes, the value of the expression assumed the original mathematical value of K.

17.5.7 Law of mass action and reversible reactions
Effect of alteration of concentration
Reaction: A + B --> C + D
A chemical reaction may stop although some of the reacting substances remain.
If A and B are the reacting substances and C and D are the resulting substances, an equilibrium occurs with some A and B remaining unchanged and some C and D formed.
Initially A and B react at a rate that depends on their concentrations.
A change in the concentration of either A or B produces a change in the rate of the reaction.
The rate of the forward action is proportional to the product of the concentrations of A and B, so rate of reaction of A and B is proportional to (Concentration of A) × (Concentration of B) =
k1 (Concentration of A) × (Concentration of B).
However, as soon as A and B react, their concentrations decrease, so the rate of reaction continuously decreases.
The reaction between A and B will have formed some C and D, and the concentrations of A B will increase and in turn react to form A and B with the rate of reaction proportional to the product of
their concentrations.
Rate of reaction of C and D is proportional to (Concentration of C) × (Concentrationof D) = k2 (Concentration of C) × (Concentration of D).
When there is apparently no further action, an equilibrium is reached with the rate of reaction of A and B forming C and D, equal to the rate of reaction of C and D forming A and B.
At equilibrium, k1 (Concentration of A) × (Concentration of B) = k2 (Concentration of C) × (Concentration of D).
k1 / k2 = the equilibrium constant, k.
At equilibrium, the product of the concentrations of C and D, divided by the product of the concentrations of A and B has a definite value.
So if at equilibrium the concentration of A is increased by adding of more of it, the concentrations of B, C, and D will assume new values, but the value of the expression (K) will remain unchanged.
This will involve the combination of some A and B to form more C and D, i.e., the previous equilibrium concentration of B will be decreased and those of C and D will be increased.

17.6.1 Weight of aluminium in aluminium sulfate.
Dissolve 5 g of aluminium sulfate in 50 mL of demineralized water.
Add drops of dilute sulfuric acid and heat to boiling.
Add excess of ammonium hydroxide and heat to boiling.
Filter, leave to dry, ignite the filter paper in a crucible, leave to cool, then weigh.
Al2SO4 + NH4OH --> Al(OH)3 + NH4SO4
AL(OH)3 --> Al2O3 + H2O
Molecular weight of Al2O3 = 102.
Weight of aluminium sulfate = wa g,
Weight of aluminium oxide = xb g.
So weight of aluminium in oxide = (54 / 102) × wb g.
Percentage weight of aluminium in aluminium sulfate = [(54 × wb) / 102] / wa × 100%.

17.6.2 Weight of calcium in marble
Stir 1.5 g of marble powder, calcium carbonate, into 10 mL of demineralized water then add 5 mL of concentrated hydrochloric acid.
Heat the solution until the marble dissolves, then add water to make the volume 50 mL.
Add ammonium solution until the solution is alkaline by litmus test, then heat to boiling.
Add 3 g of crushed ammonium oxalate, stir, and again heat to boiling.
Leave the solution to form a precipitate.
Wash the liquid and precipitate into a filter paper.
Wash the precipitate with water until the filtrate contains no more chloride ions as tested with silver nitrate solution.
Dry the precipitate of calcium oxalate, ignite the filter paper in a furnace, leave to cool, then weigh the final residue of calcium oxide.
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
CaCl2 + (NH4)2C2O4 --> CaC2O4 + NH4Cl
CaC2O4 + "O" --> CaO + 2CO2
(The "O"-->is from air)
Molecular weight of CaO = 56
Weight of calcium carbonate = wa g.
Weight of calcium oxide = wb g.
So weight of calcium in calcium oxide = (40 / 56) X wb g.
Percentage weight of calcium in marble = (40 wb / 56 wa) X 100%.

17.6.3 Weight of iron in iron (II) ammonium sulfate.
Dissolve 2 g of iron (II) ammonium sulfate in 50 mL of demineralized water.
Heat to boiling then add 2 mL of concentrated nitric acid and continue boiling for two minutes.
Leave to cool.
Add ammonium hydroxide and stir until precipitation is complete.
Test with litmus paper.
Heat again to boiling and filter.
Use a wash bottle to wash any iron (III) hydroxide from the beaker and stirring rod.
Leave the filter paper and filtrate to dry.
Weigh a crucible.
Transfer the dry filter paper and precipitate to the crucible.
Ignite the filter paper, leave to cool and weigh.
Heat the crucible until it becomes red, leave to cool and weigh until again until the weight remains constant.
2Fe2O3 --> 4Fe + 3O2
Molecular weight of Fe2O3 = 160
Molecular weight of Fe = 56
So 2 Fe, 2 × 56 is equivalent to 1 Fe2O3, 160
If weight of iron (II) ammonium sulfate = 1.96 g
If weight of iron (III) oxide = 0.40 g
So weight of iron in iron (III) oxide = 112 × 0.40 / 160 = 0.28 g
Percentage weight of iron in iron (II) ammonium sulfate = (0.28 / 1.96) × 100 = 14.29%.

17.6.4 Weight of magnesium in magnesium sulfate
Dissolve 2 g of magnesium sulfate in 50 mL of demineralized water.
Add 10 mL of ammonium chloride solution and then ammonium hydroxide until alkaline after stirring.
If magnesium hydroxide precipitates, add more ammonium chloride.
Heat and add disodium hydrogen phosphate (disodium phosphate) solution in excess, stir, and leave to settle.
Filter then wash the precipitate with ammonium hydroxide to remove ammonium chloride.
Test for the presence of chloride with silver nitrate solution acidified with nitric acid.
Heat the solid gradually then strongly, leave to cool, then weigh.
MgSO4 + NH4OH + Na2HPO4 --> Mg(NH4)PO4 + Na2SO4 + H2O
magnesium sulfate + ammonium hydroxide + disodium hydrogen phosphate --> magnesium ammonium phosphate + sodium sulfate + water
2Mg(NH4)PO4 --> Mg2P2O7 + 2NH3 + H2O
magnesium ammonium phosphate --> magnesium pyrophosphate + ammonia + water
Molecular weight of Mg2P2O7, magnesium pyrophosphate = 222
Weight of magnesium sulfate = aw g.
Weight of magnesium pyrophosphate = bw g.
So weight of magnesium in the pyrophosphate = (48 /222) × bw g.
Percentage magnesium in magnesium sulfate [(48 bw /222)] / aw × 100%.

17.6.5 Weight of sulfate radical in sodium sulfate
Dissolve 3 g of sodium sulfate crystals in 50 mL of demineralized water.
Add 5 mL dilute hydrochloric acid and 5 mL of ammonium chloride solution, then heat to boiling.
Add excess barium chloride solution to assist precipitation of barium sulfate, then and heat to boiling.
Leave to settle and then decant the liquid into filter paper.
Wash the precipitate onto a filter paper then dry in an oven.
Add a drops of concentrated nitric acid then heat the crucible until it becomes red.
Leave to cool then weigh.
Molecular weight of BaSO4 = 233
Weight of sodium sulfate crystals = aw g.
Weight of barium sulfate = bw g.
So weight of sulfate in barium sulfate = (96 / 233) × bw g.
Percentage weight of sulfate in sodium sulfate = [(96 bw / 233) / aw] × 100%.

17.6.6 Weight of tin in solder.
Make finely divided solder with a rough file.
Add 10 mL of concentrated nitric acid to 1 g of finely-divided solder in an evaporating basin.
Heat gently in a fume cupboard.
When the reaction stops, add two drops of nitric acid and heat gently again, until no more nitrogen dioxide forms.
Dilute the contents to 50 mL and filter.
Wash the precipitate with dilute nitric acid.
Dry the precipitate, ignite the filter paper, leave to cool, then weigh the final product.
The tin was first oxidized to metastannic acid, hydrated tin (IV) oxide, formula H2Sn5O11 or SnO2.xH2O.
It is used as an opacifying colour in ceramics and as an abrasive in the glass industry.
5Sn + 20HNO3 --> H2Sn5O11 + 20NO2 + 9H2O
H2Sn5O11 --> 5SnO2 + H2O
Molecular weight of SnO2 = 151
Weight of solder = aw g
Weight of tin (IV) oxide = bw g
So weight of tin in tin (IV) oxide = 119 bw / 151 g
Percentage of tin in solder = (119 bw /151 aw) × 100.

17.7.1 Hydrogen peroxide
Hydrogen peroxide > 30% W/W causes painful white blisters on the skin that heal slowly.
Hydrogen peroxide, 8% at a beauty supply store, for preparing "peroxide blondes".
Hydrogen peroxide, 6% W / V, 20 volume, mouthwash: 1 part + 9 parts water, irritant, disinfectant.
Hydrogen peroxide, 3%, 3 g per 100 mL, mouthwash 60 mL in equal amount of water, disinfectant.
Low cost: from pharmacies and drugstores, 3% (10 vols), and 6% (20 vols), hydrogen peroxide, from beautician and hairdressers' supply stores (for blonding hair), 12% (40 vols), hydrogen peroxide, from some pool supplies stores, higher concentrations ("Liquid Oxygen").
Hydrogen peroxide, H2O2, "peroxide", peroxide (of hydrogen), bleach, antiseptic, wastewater treatment for BOD (biological oxygen demand).
Contact with the eyes can cause serious long-term damage.
Corrosive solution can cause skin burns.
Decomposition may cause oxygen pressure build-up.
Tightly sealed bottles of hydrogen peroxide may explode due to a build up of pressure of oxygen gas.
Forms potentially explosive compounds with ketones (acetone), alcohols, (ethanol), esters, glycerine (glycerol), aniline, triethylamine and sodium carbonate.
Do not mix hydrogen peroxide with powdered Mg, Zn or Al.
A 3% solution is unstable.
It decomposes slowly in storage to yield water and oxygen.
It can decompose rapidly in the presence of catalysts.
It is difficult to transport, because of regulations governing is transport by air, sea and land.
It is sold as a colourless solution for disinfecting and for bleaching hair, silk, and wool.
It readily parts with one of its two oxygen atoms and forms water.
These oxygen atoms cause the disinfecting and bleaching actions.
The strengths of the solutions usually sold are marked as "10 volumes" and "20 volumes", that refer to the volume of oxygen given by 1 cc of the liquid.
So "20 volumes" means that 1 cc of the solution yields 20 cc of oxygen.
If possible, buy a small bottle of 20 volume solution.
It is usually kept in dark glass bottles, because light hastens the decomposition into water and oxygen.
Hydrogen peroxide solution, 30 % (w/w) in H2O, contains stabilizer MSDS: Harmful if swallowed.
Risk of serious damage to the eyes.
Heating may cause an explosion.
Contact with combustible material may cause fire.
Harmful by inhalation and if swallowed.
Causes severe burns.
Use hydrogen peroxide to remove blood stains.

17.7.2 Hydrogen peroxide as a reducing agent.
H2O2 - 2e- --> 2H + + O2 (electron loss)
1. Add drops of hydrogen peroxide solution to 5 mL of potassium permanganate solution acidified by dilute sulfuric acid.
Note the gas given off and the potassium permanganate loses its colour.
Keep the gas given off in the test-tube and test for oxygen with a lighted splint.
In this reaction, hydrogen peroxide acting as a reducing agent loses electrons and the permanganate ion is reduced by gain of electrons.
5H2O2 - 10e- --> 10H + + 5O2 (g)
2MnO4- + 16H + + 10e- --> 2Mn2+ + 8H2O
2MnO4- + 6H + + 5H2O2 --> 2Mn2+ + 8H2O + 5O2 (g).
2. Add a slight excess of sodium hydroxide solution to 5 mL of silver nitrate solution.
Note the brown precipitate of silver oxide.
Pour off the supernatant liquid.
Add drops of hydrogen peroxide to the silver oxide.
Oxygen is given off as the silver oxide is reduced to black metallic silver.
Hydrogen peroxide acts as a reducing agent and loses electrons.
Silver oxide is reduced and gains electrons.
H2O2 - 2e- --> 2H ++ O2 (g)
Ag2O + 2H+ + 2e- --> 2Ag (s) + H2O
H2O2 + Ag2O --> 2Ag (s) + H2O + O2
Add more drops of hydrogen peroxide and show that the finely divided silver acts as a catalyst.
3. Add 1 cc of lead dioxide to 5 mL of dilute nitric acid, Add drops of hydrogen peroxide.
Note the oxygen given off and how lead dioxide slowly dissolves.
Test the solution for lead ions by adding a drop of potassium chromate solution.
In this reaction hydrogen peroxide acts as a reducing agent and lead dioxide is reduced by gain of electrons.
PbO2 + 2H + + 2e- --> PbO + H2O
Lead monoxide and nitric acid then form lead nitrate solution.

17.7.3 Hydrogen peroxide as an oxidizing agent
Hydrogen peroxide turns an iodide solution brown, forming iodine and perhaps precipitating black crystals of iodine.
H2O2 + 2H + + 2e- --> 2H2O (electron gain)
1. Add hydrogen sulfide solution to 5 mL of lead acetate solution.
Note the precipitate of black lead sulfide.
Pour off the liquid.
Add 5 mL of hydrogen peroxide and shake the solution.
Add more hydrogen peroxide if needed to oxidize the lead sulfide to white lead sulfate.
Hydrogen peroxide acts as an oxidizing agent and accepts electrons.
Lead sulfide is oxidized by loss of electrons.
4H2O2 + 8H + + 8e- --> 8H2O
PbS + 4H2O - 8e- --> PbSO4 + 8H+
PbS + 4H2O2 --> PbSO4 + 4H2O.
2. Add drops of hydrogen peroxide solution to 2 cm of potassium iodide solution in a test-tube.
2H+ + 2I- + H2O --> 2H2O + I2
2I- - 2e- --> I2 (iodide ion is oxidized)
2H+ + H2O2 + 2e- --> 2H2O (H2O2 is reduced).
3. Add drops of potassium iodide solution to 20 vols (6%) hydrogen peroxide solution.
Then add the same number of drops of dilute sulfuric acid.
Heat gently.
Note any colour change.
Add drops of starch solution.
A blue black colour suggested oxidation of 2I- to I2.
H2O2 (aq) + 2H+ (aq) + 2e-- --> 2H2O (l)
2I- (aq) --> I2 (s) + 2e-
H2O2 (aq) + 2H+ (aq) + 2I- (aq) --> I2 (s) + 2H2O (l)
Or
I2 (s) + I- (aq) --> I3- (aq)
H2O2 (aq) + 2H+ (aq) + 3I- (aq) --> I3- (aq) + 2H2O (l).
4. Add drops of hydrogen peroxide solution to 5 mL of potassium iodide solution acidified by dilute sulfuric acid.
Test the brown colour of iodine with a drop of starch solution.
Hydrogen peroxide as oxidizing agent accepts electrons.
Iodide ions is oxidized by loss of electrons.
H2O2 + 2H+ + 2e- --> 2H2O
2I- - 2e- --> I2 (s)
H2O2 + 2H+ + 2I- --> 2H2O+ I2 (s).

17.7.4 Hydrogen peroxide bleaching action
Put a piece of red or blue litmus paper and a strand of dark-coloured hair into a test-tube and shake drops of hydrogen peroxide in the test-tube.
Both the paper and hair are bleached.

17.7.6 Hydrogen peroxide concentration and storage
Hydrogen peroxide 50% w / w solution
Hydrogen peroxide, 100 volume, vols, 30% w / w solution
Hydrogen peroxide, 8-20%, Hydrogen peroxide, 40 volume, vols, 12% w/v solution
Hydrogen peroxide, 5-8%, Hydrogen peroxide, 20 volume, vols, 6% w /v solution (hair and teeth bleach, antiseptic)
Hydrogen peroxide, < 5%, Hydrogen peroxide, 10 volume, vols, 3% w/v solution
Hydrogen peroxide may be sold in pharmacies in two strengths as follows:
1. 3% aqueous solution w / v, 3 g / mL, 10 volume, 10 vols (10% volume)
2. 6% aqueous solution w / v, 6 g / mL, 20 volume = 20 vols (20% volume)
The strength of an aqueous solution is represented by the volumes of oxygen that 100 cm3 of the liquid solution will give on decomposition.
So a 20 volume concentration means that when 1 volume of hydrogen peroxide solution is decomposed, it produces 20 volumes of oxygen.
As a hydrogen peroxide 20 volume solution contains 6% H2O2, 60g of H2O2 in 1000g of solution.
Atomic mass of H2O2 (H atomic mass 1.008, oxygen atomic mass 16) = 34.016, so the concentration of a 20 volume solution is 60 / 34.016 = 1.76 m.
The smaller concentration, 3% w / v H2O2, is less stable and decomposes faster at room temperatures, so the actual concentration is probably less than 3%.
Protect hydrogen peroxide solution from light and store in a cool place.
Keep it in a brown glass bottle closed with a glass stopper, paraffined cork or plastic screw cap.
Airlines may not be allowed to carry hydrogen peroxide as freight.
Hydrogen peroxide test strips are available that contain a peroxide reagent to detect production of hydrogen peroxide by certain bacteria, e.g. Streptococcus pneumoniae.
Hydrogen peroxide is used as an antiseptic where catalase enzyme in blood catalyses the decomposition of hydrogen peroxide to water and oxygen.

17.7.7 Hydrogen peroxide decomposition
1. Some substances can increase the rate of reaction for the decomposition of hydrogen peroxide.
Put two small equal amounts of hydrogen peroxide solution gently into four test-tubes.
Note the oxygen bubbles.
2H2O2 (l) --> 2H2O (l) + O2 (g).
Test for oxygen with a glowing splint.
Add the following substances:
Test-tube 1, Nothing added (control),
Test-tube 2, Manganese (IV) oxide,
Test-tube 3, Iron (III) chloride,
Test-tube 4, Copper (II) sulfate.
Note the bubbles.
Tests for oxygen with a glowing splint.
2. Decomposition of hydrogen peroxide by heat.
Heat drops of hydrogen peroxide in a test-tube.
When effervescence begins, put a glowing wood spill into the test-tube.
The spill relights, showing that oxygen is being produced.
Observe the rate at which hydrogen peroxide decomposes by the addition of acid or alkali.
Put 2 cm of hydrogen peroxide into each of two test-tubes.
To test-tube 1, add drops of a solution of sodium hydroxide or strong ammonia solution.
To test-tube 2, add sulfuric acid or citric acid.
Put both test-tubes into a beaker of hot water for one minute.
Test both test-tubes for oxygen.
Oxygen forms from test-tube 1 containing the alkali, but very little from the other test-tube.
Alkalis speed up the formation of oxygen, while acids slow it down.
For this reason acid is often added to containers of hydrogen peroxide to help it to keep its oxygen.
A catalyst is a substance that helps a chemical action without being changed itself.
A substance that hinders instead of helping a chemical action is called a "negative" catalyst.
So we say that sulfuric acid is a negative catalyst for the decomposition of hydrogen peroxide.
Sodium hydroxide and ammonia are "positive" catalysts.

17.7.8 Hydrogen peroxide decomposition, with manganese (IV) oxide catalyst
1. Fill a test-tube with hydrogen peroxide to a depth of about 1 cm and add a little of manganese dioxide.
Tests for oxygen with a glowing splint.
A rapid evolution of oxygen occurs and the manganese dioxide is not lost.
2H2O2 (l) --> 2H2O (l) + O2 (g).
2. Put 2 cm of 20 vols hydrogen peroxide (20 vols solution) in two test-tubes, Test-tube 1 and Test-tube 2.
Add 5 drops of sodium hydroxide solution.
Test-tube 2. Add 5 drops of dilute sulfuric acid solution.
Immerse both test-tubes in a beaker half full of hot water.
Use a glowing splint to tests for oxygen in the mouth of the test-tubes.
Oxygen is found in Test-tube 1, but not in Test-tube 2.
3. Set up a conical flask fitted with a one hole stopper and delivery tube that leads into a beaker of water.
Invert a closed burette full of water over the end of the delivery tube.
Pour 50 mL of water in the flask and add 2 mL of 20 vols (6%) hydrogen peroxide solution.
Add 1 g manganese (IV) oxide and immediately insert the stopper with the delivery tube into the flask.
Note how much oxygen forms every 15 seconds.
Plot on a graph how much oxygen forms every 15 seconds against the time of the reaction.
4. Repeat the experiment by adding more hydrogen peroxide solution to the same test-tube.
The manganese (IV) oxide is not "used up", because more oxygen forms.
Repeat the experiment with 1 g copper (II) oxide.
Add 2 mL of 20 vols (6%) hydrogen peroxide solution.
5. Repeat the experiment with 1 g zinc oxide.
Add 2 mL of 20 vols (6%) hydrogen peroxide solution.
Plot a graph for each experiment.
Manganese (IV) oxide is the better catalyst in these reactions.
Warm some hydrogen peroxide solution gently in a test-tube and hold a glowing splinter of wood in the mouth of the test-tube.
6. Place two small equal amounts of hydrogen peroxide in separate test-tubes.
Add some iron chloride to one test-tube and a little manganese dioxide to the other.
Apply the glowing splinter test to any gas given off.
Hydrogen peroxide decomposes to give off oxygen when heated, but will decompose without heating when iron chloride and manganese dioxide are added to it.
The black colour of manganese dioxide and the brown colour of iron chloride remains after the reactions, so these chemicals may not have been altered during the reactions.
Iron chloride and manganese dioxide are catalysts or "chemical accelerators".

17.7.9 Hydrogen peroxide on cut skin
When you pour hydrogen peroxide onto a cut in your skin it bubbles, because the catalase enzyme in your bodily fluids catalyzes the decomposition of hydrogen peroxide.

17.7.10 Hydrogen peroxide on cut potato
See diagram 9.161: Hydrogen peroxide on cut potato
Cut a potato and immediately add hydrogen peroxide.
Observe the bubbles forming, because of the catalase from the cut potato cells.

17.7.11 Hydrogen peroxide on hair
1. Add hydrogen peroxide to some cut hair, not hair on your head.
Observe the colour change.
Hydrogen peroxide is used to whiten hair to produce "peroxide blondes".
However, do not do this yourself without getting expert advice on blonding your own hair.
The bleach used by hairdressers contains 8-10% hydrogen peroxide in a thick gel.
2. Baking soda opens up the cuticle to allow the hydrogen peroxide to get into the hair cortex more easily and it break down the melanin in hair.
To lighten hair with hydrogen peroxide and baking soda, mix the two ingredients into a paste, and spread it evenly throughout the hair.
Leave for 15 minutes and then wash the hair

17.7.12 Hydrogen peroxide concentration and enzyme activity
1. Prepare capsicum extract from capsicum, diced, pureed in a blender then filtered through a microfibre cloth.
2. Label 3 large test tubes containoing 5 mL of 0%, 3% and 6% hydrogen peroxide solution, and add 2 mL of distilled water in each test-tube.
3. Use a disposable pipette to add 1 mL of capsicum extract to each test-tube and immediately start timing with a stopwatch.
4. After 30 seconds, record the height of the oxygen foam, (in cm). in each test-tube.
Continue recording height every 30 seconds for 2 minutes.
5. Compare the heights of foam in the three test-tubes.

17.7.13 Hydrogen peroxide pH and enzyme activity
1. Prepare capsicum extract from capsicum, diced, pureed in a blender then filtered through a microfibre cloth.
2. Label 3 large test tubes with test-tube 1, test-tube 2 and test-tube 3.
. Place 2 mL of 0.05M hydrochloric acid in test-tube 1, 2 mL distilled water in test-tube 2, and 2 mL 0.05M sodium hydroxide in test-tube 3.
Add 2 mL 6% 6% hydrogen Peroxide into each tube.
3. Use a disposable pipette to add 1 mL of capsicum extract to each test-tube and immediately start timing with a stopwatch.
4. After 30 seconds, record the height of the oxygen foam, (in cm). in each test-tube.
Continue recording height every 30 seconds for 2 minutes.
5. Compare the heights of foam in the three test-tubes.

17.7.14 Hydrogen peroxide temperature and enzyme activity
1. Prepare capsicum extract from capsicum, diced, pureed in a blender then filtered through a microfibre cloth.
2. Measure 2 mL of 6% hydrogen peroxide solution and pour it into 3 large test-tubes labelled 'hot', 'room temperature' and 'cold'.
. Place one test-tube in a hot water bath, one test-tube in a cold water bath and leave one test-tubeat room temperature for 5 minutes.
. Record temperatures in the three test-tubes.
. Remove the test-tubes from water baths and place them in test tube rack when their temperatures have stabilised.
3. Use a disposable pipette to add 1 mL of capsicum extract to each test-tube and immediately start timing with a stopwatch.
4. After 30 seconds, record the height of the oxygen foam, (in cm). in each test-tube.
Continue recording height every 30 seconds for 2 minutes.
5. Compare the heights of foam in the three test-tubes.
The volume of foam approximates the quantity of oxygen released by the enzyme catalase in the capsicum extract.
2H2O2 --> 2H2O + O2.
...catalase.

17.7.15 Hydrogen peroxide with manganese (IV) oxide, height of suds
Manganese (IV) oxide (manganese dioxide, MnO2), does not take part in the chemical reaction.
Its function is to provide an increased surface area for the reactants.
1. Pour 10 mL of hydrogen peroxide, 5 mL of detergent and 5 mL of glycerine into two identical measuring cylinders: Measuring cylinder 1. and Measuring cylinder 2.
Stir both solutions.
Measuring cylinder 1. Add crystals of manganese (IV) oxide and stir again.
A mass of bubbles arises to form suds.
The glycerine increases the surface tension of the liquid to delay the collapsing of the bubbles.
The hydrogen peroxide decomposes into water and the oxygen that forms the bubbles.
The suds are higher in the measuring cylinder containing the manganese (IV) oxide, because it acts as a catalyst.
Measuring cylinder 2. Control without manganese (IV) oxide.
2. Repeat the experiment with dust or dirt or ashes instead of manganese (IV) oxide.
The heights of the detergent suds are similar to the heights using manganese (IV) oxide.
This observation shows that manganese (IV) oxide does not take part in the chemical reaction.
Its function is to provide an increased surface area for the reactants.

17.7.16 Hydrogen peroxide with catalase in raw beef liver
Hydrogen peroxide with catalase enzyme in raw beef liver.
Hydrogen peroxide, acting as an oxidizing agent, it is toxic to cells, so it is a useful disinfecting agent that disrupts the metabolism of bacteria.
Our body cells contain an enzyme called catalase that accelerates the conversion of toxic hydrogen peroxide to water and oxygen gas.
It is a reactive oxygen metabolic by-product that regulates some oxidative stress-related states related to asthma, inflammatory arthritis, atherosclerosis, diabetic vasculopathy, osteoporosis, and some
neurodegenerative diseases.
Be Careful!
This reaction can be violent and the steam formed may be hot!
Use safety glasses and nitrile chemical-resistant gloves.
1. Put 3% hydrogen peroxide in a clear plastic container and record its temperature.
Put a 6 mm piece of liver in the hydrogen peroxide.
The mixture starts to bubble and foam.
Note the height of the foam.
Record the temperature each minute for 5 minutes.
The temperature immediately rises, levels, then decreases.
The hydrogen peroxide is decomposed, oxygen gas is given off, bubbles form and heat energy is given off.
Put a glowing splint in the test-tube of liver and hydrogen peroxide.
The splint flames in the oxygen.

17.7.17 Hydrogen peroxide with catalase -.Disproportionation
This type of reaction where something is oxidized and the same species is also reduced is called a disproportionation reaction.
It happens where an element has a number of possible oxidation states, (oxidation numbers) both higher and lower.
So some atoms/ions of that element lose electrons while others gain electrons.
This reaction is an example of the type of redox reaction called disproportionation, where the atom is simultaneously reduced and oxidized to form two different molecules.
2H2O2 (l) --> 2H2O (g) + O2 (g).
2. Repeat the experiment with the same size piece of liver, as in 17.7.16, but very finely chopped.
The height of the foam is greater, because the catalase has more access to the hydrogen peroxide.
3. Repeat the experiment with very finely chopped liver in water, heated to 50oC, or boiled for 3 minutes.
No bubbling occurs when hydrogen peroxide is added, because you have destroyed the enzyme catalase in the beef liver, i.e. it is "denatured".
4. Repeat the experiment with a piece of muscle tissue, "meat".
Little reaction occurs, because muscle has little catalase.
5. Repeat the experiment with a piece of potato.
No reaction occurs.
6. Add 2 cm of hydrogen peroxide to 2 cm of 0.1 M acetic acid solution into a test-tube.
Use litmus paper to find the approximate pH for the solution.
Put a 6 mm piece of liver in the test-tube.
No reaction, because the catalase is denatured by the acid.
7. Add 2 cm of hydrogen peroxide to 2 cm of 0.1 M ammonium hydroxide solution.
Use litmus paper to find the approximate pH for the solution.
No reaction, because the catalase is denatured by the alkali.
8. Put 6 g manganese dioxide into 100 mL 10% hydrogen peroxide.
Observe similar bubbling and relights a glowing wooden splint to show the presence of oxygen.
2H2O2 (l) --> 2H2O (g) + O2 (g).

17.7.18 Hydrogen peroxide with potassium sodium tartrate, cobalt (II) chloride catalyst
See diagram 16.3.7: Potassium sodium tartrate
Use a 250 mL beaker.
Add 20 mL 6% hydrogen peroxide solution to a solution of 5 g potassium sodium tartrate-4-water (Rochelle salt) in 60 mL water.
Heat to 75oC.
Stir and observe that gases are formed and a pink colour forms.
Add 5 mL of solution containing 0.2 g cobalt (II) chloride-6-water.
Observe frothing and pink colour.
Be Careful!
Test gases with limewater.
Carbon dioxide is in gases produced by the reaction.
Test gases with a glowing splint, not extinguished, because some oxygen from hydrogen peroxide.
Later, observe frothing stops and pink colour returns.
Equation in 2 parts:
C4H4O62- (aq) + 3H2O2 (aq) --> 2HCOO- (aq) + 2CO2 (g) + 4H2O (l)
Co2+ --> Co3+
Pink cobalt ions oxidized to a green activated complex with tartrate
2HCOO- (aq) + 2H2O (aq) --> 2OH- (aq) + 2CO2 + 2H2O
Co3+ --> Co2+
Basic equation: Green activated complex with tartrate reduced to pink cobalt ions
C4H4O62- (aq) + 5H2O2 (aq) --> 4CO2 (g) + 2OH- (aq) + 6H2O (l).

17.7.19 Hydrogen peroxide with sodium thiosulfate, ammonium molybdate catalyst
Dissolve together in deionized water: 8.7 g sodium thiosulfate-5-water, 3.8 g sodium ethanoate-3-water (sodium acetate tri-hydrate) or 2.3 g anhydrous sodium ethanoate, and 0.5 g sodium hydroxide.
The sodium ethanoate buffers the sodium hydroxide.
Make up the solution to 1 litre and add universal indicator until each solution is blue.
Pour 225 mL of the solution into 3 test-tubes labelled as follows:
1. "With catalyst", 2. "No catalyst", 3. "Control".
Dissolve 14 mL of 20 vols hydrogen peroxide in deionized water.
Make up to 40 mL and divide into two 20 mL portions.
Add 0.08 g ammonium molybdate to test-tube 1. "With catalyst", and shake to dissolve.
Add the 20 mL portions to test-tubes: 1. "With catalyst" and 2. "No catalyst".
Observe colour changes after 5 minutes:
1. "With catalyst": changes from blue to green to yellow to orange to orange red.
2. "No catalyst": Same colour changes, but slower.
3. "Control": No colour change.
Na2S2O3 (aq) + 4H2O2 (aq) --> Na2SO4 (aq) + H2SO4 (aq) + 3H2O (l).

17.7.20 Hydrogen peroxide with yeast, elephant's toothpaste reaction
1. Put 125 mL of 6% hydrogen peroxide into a plastic drink bottle.
Add dishwashing liquid and swirl to mix.
Mix one sachet of dry yeast with 100 mL of warm water then pour this mixture into the plastic drink bottle.
The mixture rapidly produces foam, which squirts out of the plastic drink bottle like tooth paste.
The foam is warm, because of the exothermic reaction.
Yeast produces catalase enzyme.
Adding yeast to hydrogen peroxide catalyses the decomposition of hydrogen peroxide into water, oxygen gas and heat
2H2O2 --> O2 + 2 H2O.
2. Mix 4 parts of hydrogen peroxide with 2 parts of liquid dish soap and a few drops of food colouring, not cochineal.
Add this mixture to a plastic soda bottle and place it in the sink.
Mix one packet of active yeast with warm water, and leave it for 5 minutes.
Use a funnel to pour the yeast mixture into the plastic soda bottle.
Observe the foam produced by the reaction.

17.7.21 Prepare molar volume of oxygen with hydrogen peroxide
See diagram: 13.1.6: Molar volume of oxygen
12.5.1: Saturation Vapour Pressure over water, (SVP)
Put 15 mL of 3% w/w (3 g H2O2 /100 g solution) hydrogen peroxide solution into flask A.
Put 0.05 g of yeast in a small test-tube then lower the test-tube into flask A.
Weigh flask A and its contents, W1.
Attach Plastic tube 1 and Plastic tube 2 to flask B only.
Put water into flask B leaving a space in the neck of the flask.
Add water to a beaker until it is one third full.
Siphon water into the beaker by blowing into the open end of Rubber tube1 or by using a pipette bulb.
Raise and lower the beaker to remove any air bubbles from Plastic tube 2.
Adjust the height of the beaker so that the levels of the water in the beaker and in flask B are the same.
Connect Plastic tube one to flask A.
Raise the beaker to check for leaks in the apparatus.
Again, adjust the height of the beaker so that the levels of the water in the beaker and in flask B are the same.
Close the pinch clamp.
Replace the glass tube in the beaker and open the pinch clamp to allow some water to flow into the beaker.
With the pinch clamp still open, tip flask A so that the yeast falls into the hydrogen peroxide solution.
Swill flask A until the reaction is completed when the water level in the beaker does not change.
Again, adjust the height of the beaker so that the levels of the water in the beaker and in flask B are the same.
Close the pinch clamp.
Remove the stopper in flask A, insert a thermometer and note the temperature of the gas inside, T1.
Repeat this measurement with flask B, T2.
Disconnect Plastic tube 1 from flask A and again weigh flask A and its contents, W2.
Measure the oxygen produced, by measuring the volume or weight water in the beaker, V.
Find the vapour pressure of water at that temperature from the Table of saturated vapour pressure over water, Psvp.
Note the room temperature.
Note the barometric pressure from a barometer or ask the weather bureau or local airport.
Calculation 1.
Calculate the volume at STP of 32 g, one mole, of oxygen gas.
(W2 - W)1 = Loss in weight
VO2 = volume of water in the beaker = volume of oxygen collected
Tf = average temperature in the flasks = (T1 + T2) / 2
Patm = atmospheric pressure = pressure of oxygen in the flask, PO2 + saturation vapour pressure of water at that temperature, Psvp.
So PO2 = (Patm - Psvp)
VO2 = volume of oxygen in the flask
Tstp = temperature at STP (Standard Temperature and Pressure) = 0oC, 273 K
Pstp = pressure at STP = 760 mm Hg = 101325 Pa
Vstp = volume at STP.
P1V1 / T1 = P2V2 / T2 (Boyle's law and Charles's law)
(P1 × V1) / T1 = (P2 × V2) / T2
(PO2 × VO2) / Tf = (Pstp × Vstp) / Tstp
So Vstp = VO2 [(PO2 / Pstp) × (Tstp / Tf)]
Relative molecular mass of oxygen = 32 g
So number of moles of oxygen = (W2-W1) / 32
So the molar volume of oxygen at stp = Vstp / number of moles of oxygen = litres / mole
A mole of an ideal gas occupies 22.4 litres at STP.
Calculation 2.
If barometric pressure = 1016 kPa, average temperature = 20oC, loss in weight of flask = 0.2 g, volume of oxygen collected at average temperature = 140 mL
Pressure of oxygen in apparatus = (barometric pressure - SVP at 20oC) = (1016 - 2.3) = 1013.7 kPa
Vstp = 140 [(1013.66 / 101.325) × (293 / 273)] = 503.17 mL
Number of moles = 0.2 / 32 = 0.00625 moles
Molar volume = 140 / 0.00625 = 22400 = 22.4 litres = 22.4 L / mole

17.7.22 Prepare hydrogen peroxide solution
Dilute 5 mL of a syrup of phosphoric acid with its own volume of water.
(Remember: Acid to water!) Slowly add barium peroxide while cooling the test-tube under the tap.
Filter the solution to remove the barium phosphate.
The hydrogen peroxide solution may contain excess phosphoric acid and barium ions.
2H3PO4 + 3BaO --> Ba3(PO4)2 (s) + 3H2O2.

17.7.23 Tests for hydrogen peroxide
Acidify 2 cm of potassium dichromate solution with dilute sulfuric acid.
Cover the solution with 2 cm of ether.
Add a drop of diluted hydrogen peroxide.
The blue colour in the ether layer may come from perchromic acid (HCrO5).
This reaction is a test for hydrogen peroxide.
Ionization reaction, Ka = 2.4 × 10-12
H2O2 + H2O <--> H3O+ + HO2-.