School Science Lessons
Topic 17 Catalysis, chemical equilibrium, enzymes and biological catalysts,
gravimetric analysis, rates of reaction
Please send comments to: J.Elfick@uq.edu.au Table of contents 17.3.0 Catalysis 17.5.0 Chemical equilibrium 17.4.0 Enzymes and biological catalysts 17.2.0 Factors affecting rates of reaction 17.6.0 Gravimetric analysis 17.1.0 Rates of reaction 17.1.1 Rates of reaction, clock reactions 17.3.0 Catalysis
17.3.0 Catalysis, catalysts 18.104.22.168 Catalytic oxidation
of ammonia forms nitrogen monoxide, with red-hot platinum wire 22.214.171.124 Catalytic oxidation
of ammonia, with chromium (III) oxide catalyst 17.3.13 Bromine catalyses the oxidation of
sulfur to sulfuric acid 17.3.3 Ethanedioic acid-2-water (oxalic acid)
with potassium manganate (VII), autocatalysis 17.3.15 Ethyl acetate with sodium hydroxide,
autocatalytic hydrolysis 17.3.11 Heat potassium chlorate, manganese
dioxide catalyst 126.96.36.199 Hydrogen peroxide decomposition,
with different catalysts 17.3.12 Methanol with platinum wire catalyst,
catalytic oxidation 17.3.10 Oxidation of acetone vapour, copper
catalyst 17.3.7 Potassium bromate with propanedioic acid,
double autocatalytic reaction, oscillating reaction 17.3.5 Potassium iodide with hydrogen peroxide,
reverse colour change 17.3.14 Sodium hypochlorite decomposition,
cobalt sulfate catalyst 17.3.4 Sugar with potassium chlorate, spontaneous
188.8.131.52 Hydrogen peroxide
decomposition, with different catalysts
184.108.40.206 Hydrogen peroxide decomposition, with different catalysts 220.127.116.11 Hydrogen peroxide decomposition,
with manganese (IV) oxide catalyst 18.104.22.168 Hydrogen peroxide on cut skin, (catalase) 22.214.171.124 Hydrogen peroxide on cut potato,
(catalase) 126.96.36.199 Hydrogen peroxide on hair 188.8.131.52 Hydrogen peroxide with catalase enzyme
in raw beef liver, (catalase) 17.1.5 Hydrogen peroxide with manganese (IV)
oxide, rate of reaction, height of suds 184.108.40.206 Hydrogen peroxide with potassium
sodium tartrate, cobalt (II) chloride catalyst, visible activated complex 220.127.116.11 Hydrogen peroxide with sodium thiosulfate,
ammonium molybdate catalyst 18.104.22.168 Hydrogen peroxide with yeast, elephant's
toothpaste reaction 9.3.12 Tests for zymase and catalase
in yeast 17.5.0 Chemical equilibrium 17.5.0 Chemical equilibrium, the law of chemical
equilibrium (law of mass action) 22.214.171.124 Common ion effect in ammonium chloride
solution 126.96.36.199 Common ion effect to precipitate
sodium chloride from solution 188.8.131.52 Common ion effect to precipitate
barium chloride from solution 184.108.40.206 Effect of temperature on chemical
equilibrium, thermal dissociation of ammonium chloride 220.127.116.11 Explanation of group analysis 18.104.22.168 Heat nitrogen tetroxide (dinitrogen
tetroxide, N2O4) 22.214.171.124 Hydrolysis of antimony chloride 126.96.36.199 Hydrolysis of bismuth chloride 188.8.131.52 Law of mass action and reversible
reactions, effect of alteration of concentration 17.4.0 Enzymes and biological catalysts 17.4.1 Breakdown starch to sugar 17.4.3 Bromelain enzyme from pineapples 17.4.2 Fermentation using yeast 184.108.40.206 Hydrogen peroxide with catalase enzyme
in raw beef liver 17.6.0 Gravimetric analysis 17.6.2 Weight of aluminium in aluminium sulfate 17.6.3 Weight of calcium in marble, calcium
carbonate 17.6.1 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 17.1.0 Rates of reaction: 3.7.2 Rates of reaction Hydrochloric
acid reactions 17.1.4 Balloons, dilute hydrochloric acid with
marble chips 3.94 Catalysts and rate of reaction 3.7.1 Concentration and rate of reaction 3.92 Concentration and rate of reaction,
sodium thiosulfate with hydrochloric acid 17.1.1 Count bubbles, dilute hydrochloric acid
with granulated zinc 17.2.0 Factors affecting rate of reaction 17.1.3 Gas burette, dilute hydrochloric acid
with marble chips (calcium carbonate) 17.1.5 Hydrogen peroxide with manganese (IV)
oxide, rate of reaction, height of suds 17.1.0 Measure rates of reaction 220.127.116.11 Rate of reaction depends on concentration 3.91 Size of particles and rate of
reaction 3.93 Temperature and rate of a reaction 17.1.2 Volume of gas, dilute hydrochloric acid
17.1.1 Rates of reaction,
clock reactions 17.1.11 Belousov-Zhabotinskii clock reaction,
17.1.8 Hydrogen peroxide clock reaction, Briggs-Rauscher oscillating
reaction, hydrogen peroxide, potassium iodate 17.1.6 Iodine clock reaction, hydrogen peroxide,
potassium iodide 17.1.9 Iodate clock reaction, potassium iodate,
sodium metabisulfite (sodium bisulfite) 17.1.7 Old Nassau flag clock reaction, (orange
and black) sodium metabisulfite, mercury (II) iodide 17.1.10 Persulfate-iodide clock reaction 18.104.22.168 Reactions of sodium
thiosulfate, (See 4. Concentrated hydrochloric acid with sodium thiosulfate
solutions) 22.214.171.124 Rate of reaction
depends on concentration 126.96.36.199 Magnesium with hydrochloric acid 188.8.131.52 Potassium iodide with potassium iodate
17.2.0 Factors affecting rates
of reaction 184.108.40.206 Concentration of reactants, hydrochloric
acid with magnesium 17.2.2 Concentration of reactants, hydrochloric
acid with sodium thiosulfate (hypo) 17.2.1 Particle size, dilute hydrochloric acid
with marble chips 17.2.4 Rates of reaction of aspirin 17.2.3 Temperature and rate of reaction, hydrogen,
iron with sulfuric acid 17.1.0 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 you must measure some change, e.g. time taken for colour change,
pH change, precipitate appears, reagents disappear, temperature change,
volume of gas that forms 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. 17.1.1 Count bubbles, dilute hydrochloric acid
with zinc See diagram 17.1.1: Count bubbles, dilute
hydrochloric acid with zinc
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. 17.1.2 Volume of gas, dilute hydrochloric acid
with zinc 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. 17.1.3 Gas burette, dilute hydrochloric acid with
marble chips See diagram 17.3.3: Gas burette | See: Saturation vapour pressure over water
Add zinc or marble chips to dilute hydrochloric acid. 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. 17.1.4 Balloons, dilute hydrochloric acid with
marble chips See diagram 3.2.91: Balloons, dilute hydrochloric
acid with marble chips
Add marble chips to dilute hydrochloric acid. Show the relative production
of hydrogen gas by attaching a previously stretched balloon to each test-tube. 17.1.5 Hydrogen peroxide with manganese (IV) oxide
(manganese dioxide), height of suds
Manganese (IV) oxide does not take part in the chemical reaction.
Its function is to provide an increased surface area for the reactants.
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 (MnO2,
manganese dioxide) 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.
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. 17.1.6 Iodine clock reaction, hydrogen peroxide,
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 ethanoic
acid (glacial acetic acid, CH3CO2H), 4.1 g sodium
ethanoate (sodium acetate, CH3CO2Na), 50 g potassium
iodide (KI), and 9.4 g sodium thiosulfate-5-water (Na2S2O3.5H2O).
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. 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. 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
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 to combine with all the
thiosulfate. 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.
Iodine clock reaction, hydrogen peroxide, potassium iodide
Solution A Prepare a solution of potassium iodide, sodium thiosulfate
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
hydrogen peroxide + iodide ion + hydrogen ion --> triiodide ion
I3- (aq) + 2S2O32-
(aq) --> 3I- (aq) + S4O62-
(aq) 17.1.7 Old Nassau flag clock reaction 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
IO3- (aq) + 3HSO3- (aq)
--> I- (aq) + 3SO42- (aq) + 3H+
3. With excess iodide ions, mercury (II) iodide, HgI2,
forms as an orange precipitate
Hg2+ (aq) + 2I- (aq) --> HgI2
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
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". 17.1.8 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 malonic acid (propanedioic acid, CH2(COOH)2)
+ 3.4 g manganese sulfate monohydrate, (MnSO4.H2O),
to 800 mL deionized water + 4 g of freshly made soluble starch, stir
to dissolve then 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
Hydrogen peroxide + iodate --> iodine + oxygen (auto-catalytic)
IO3- + 2H2O2 + CH2(COOH)2
+ H+ --> ICH(COOH)2 + 2O2 + 3H2O
17.1.9 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
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
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 reduces iodine back to
iodine + bisulfite ion + water --> iodide ion + bisulfate ion +
When no more bisulfate ion exists because it is all used up reducing
the iodine, the remaining excess iodine forms blue-black colour with starch.
Use sodium persulfate (sodium peroxodisulfate) Na2S2O8,
or potassium persulfate (potassium peroxydisulfate or KPS) K2S2O8,
or ammonium persulfate (ammonium peroxydisulfate), (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)
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. 17.1.11 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 ferrion sulfate (direct purchase required)
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 nonequilibrium
phenomena in that patterns form when the solution is stimulated. 17.2.1 Particle size, dilute
hydrochloric acid with marble chips (calcium carbonate)
The smallest particles show the most vigorous reaction because if
you grind two pieces of substance with the same weight into coarse and
fine sizes, the fine size pieces would have total surface area greater than
the coarse pieces.
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
Rate of reaction of Test-tube 4 > Test-tube 3 > Test-tube 2
> Test-tube 1. 17.2.2 Concentration of reactants, hydrochloric
acid with sodium thiosulfate See diagram 17.4.1: Measure 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:
Sodium thiosulfate Solution
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)
220.127.116.11 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 17.2.3 Temperature and rate of reaction, hydrogen,
iron with sulfuric acid
1. 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. 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. To tests for hydrogen gas, remove from
heating, place your thumb over the end of the test-tube, count to five,
apply a lighted paper to the end of the test-tube, the hydrogen gas explodes
with a loud pop sound. Never test more than one test-tube full of hydrogen
gas! 17.2.4 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. 17.3.0 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. 18.104.22.168 Hydrogen peroxide decomposition, with
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 bubbles. 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 commercial
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”
we say that sulfuric acid is a negative catalyst for the decomposition
of hydrogen peroxide. Sodium hydroxide and ammonia are “positive” catalysts. 22.214.171.124 Hydrogen peroxide with yeast, elephant's
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. 126.96.36.199 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
2. Put 2 cm of 20 vols hydrogen peroxide (20 vol. solution) in two
test-tubes, Test-tube 1 & Test-tube 2.
Test-tube 1. 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
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
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
altered during the reactions. Iron chloride and manganese dioxide
are catalysts or "chemical accelerators". 188.8.131.52 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.
Put 3% hydrogen peroxide in a clear plastic container and record its
1. 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
2. Repeat the experiment with the same size piece of liver 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
2H2O2 (l) --> H2O (g) + O2
184.108.40.206 Hydrogen peroxide with potassium sodium
tartrate, cobalt (II) chloride catalyst, visible activated complex 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
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
C4H4O62- (aq) + 5H2O2
(aq) --> 4CO2 (g) + 2OH- (aq) + 6H2O
(l) 220.127.116.11 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) 18.104.22.168 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.
22.214.171.124 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.
126.96.36.199 Hydrogen peroxide on hair
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. 17.3.3 Ethanedioic acid-2-water (oxalic acid) with
potassium manganate (VII), autocatalysis
Solution A: 6 g of ethanedioic acid-2-water (oxalic acid) in 300 mL
Solution B: 100 mL of 0.001 M potassium manganate (VII) solution,
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) 17.3.4 Sugar with potassium chlorate, spontaneous
combustion BE CAREFUL! Do this experiment
in a fume cupboard or in the open, behind a glass screen.
1. The reaction of the concentrated sulfuric acid with the sugar released
heat. The heat then activated the release of oxygen from the potassium
chlorate. The oxygen released by the potassium chlorate further oxidized
the sugar. This further oxidation released so much heat that the sugar bursts
into flames. Mix sugar or powdered sugar (castor sugar) with an equal amount
of potassium chlorate crystals in an evaporating dish. Push a dent in the
top of the heap of powder. Add one drop of concentrated sulfuric acid.
A spontaneous combustion occurs. BE CAREFUL!
2KClO3 --> 2KCl + 3O2
2. Repeat the experiment with potassium nitrate. The reaction is slower
because while potassium chlorate loses all its oxygen, potassium nitrate
loses only one third of it oxygen.
2KNO3 --> 2KNO2 + O2 17.3.5 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
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. 17.3.7 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 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, and 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 so that the heated strip of copper is 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 once
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 17.3.11 Heat potassium chlorate,
manganese dioxide catalyst
Be careful! In some school systems this experiment
is not allowed because potassium chlorate may explode.
1.1 Mix 0.5 g of manganese dioxide with 2 g of potassium chlorate
and put the mixture in "ignition tube 1".
1.2 Put 0.5 g of manganese dioxide in "ignition tube 2".
1.3 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. 17.3.12 Methanol with platinum
wire catalyst, catalytic oxidation
Be careful! Have ready a piece of cardboard or glass to put over the
beaker if the methyl alcohol ignites.
1. 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, holding 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
2CH3OH + O2 --> 2HCHO + 2H2O
2. Repeat the experiment with 880 ammonia solution instead of methanol
and pass oxygen gas from an oxygen cylinder through the mixture. Observe
brown fumes of nitrogen dioxide or white fumes of ammonium nitrate and
4NH3 + 7O2 --> 4NO2 + 6H2O 17.3.13 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.
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
2HBr + 2HNO3 –> 2H2O + 2NO2 +
Br2 17.3.14 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) 17.3.15 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
3. Repeat the titrations every 15 minutes for 2 hours and tabulate
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
CH3COOC2H5 + H2O -->
CH3COOH + C2H5OH
17.4.1 Breakdown starch to
sugar See 9.130: Hydrolysis of starch by
salivary amylase (ptyalin) See 9.142: Tests for starch, Fehling's
tests for starch
Salivary amylase enzyme breaks down starch into the reducing sugars
(+) glucose and maltose. Reducing sugars do not react with iodine solution
and starch does not react with Fehling's solution. The sugars reduce copper
(II) in Fehling's solution to brick-red copper (I) oxide.
Pour 10 mL of dilute starch solution into a test-tube. Add 1 mL of
saliva and stir. After 2 minutes use a dropper put 2 drops of the solution
on a white tile. To tests for starch, add iodine solution and note the intensity
of the blue black colour. To tests for reducing sugars, add Fehling's
No. 1 and No. 2 solutions and heat. Repeat the experiment every 2 minutes
with clean droppers. Note the decreasing intensity of the blue colour
that shows that starch is being used up. Repeat the experiment. Put 3
drops of the reaction
mixture in a test-tube. Add 3 mL of Fehling's solution. Heat the mixture.
Note the intensity of the brick-red colour increasing with time. 17.4.2 Fermentation using yeast See 3.38: Carbon dioxide and fermentation
During fermentation, enzymes breakdown carbohydrates and other organic
molecules in the absence of oxygen. 17.4.3 Bromelain enzyme from
Add pineapple juice to milk. The milk protein begins to coagulate
and degrade as it reacts with the bromelain. Also, pineapple juice will
also remove the gelatine emulsion surface on black and white photographic
film! 17.5.0 Chemical Equilibrium
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. If different concentrations of iron (II) nitrate solution are added
to potassium thiocyanate solution, at equilibrium the concentration of
the FeSCN2+ solution has a certain 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+
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 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 towards
counteracting 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). 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. 17.5.1 Concentration and temperature, cobalt (II)
Test the effect of concentration and temperature on the equilibrium
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.
The solution 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) 17.5.2 Common ion effect, sodium ethanoate and
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
Solution B: 100 mL 2 M ethanoic acid, then add universal indicator
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
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
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) 17.5.3 Equilibrium between ICl and ICl3 See diagram 17.5.3: Equilibrium reaction
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) 188.8.131.52 Magnesium with
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. 184.108.40.206 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, then transfer
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 mL with water.
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
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 220.127.116.11 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)
× (Concentration of 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
So 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. 18.104.22.168 Hydrolysis of bismuth
Put some bismuth chloride in a test-tube, add 1 mL of water and observe
white bismuth oxychloride forming because 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
Add drops of concentrated hydrochloric acid until the white precipitate
disappears 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.
22.214.171.124 Hydrolysis of antimony
Repeat the above experiment using antimony chloride in the place of
bismuth chloride. The reactions and the explanations are similar to those
SbCl3 + H2O <--> SbOCl (s) (antimony oxychloride)
+ 2HCl 126.96.36.199 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- 188.8.131.52 Common ion effect
to precipitate barium chloride from solution
Add a 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- 184.108.40.206 Effect of temperature
on chemical equilibrium, thermal dissociation of ammonium chloride
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
220.127.116.11 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-
Adding NH4Cl, (NH4)2SO4,
or any soluble ammonium salt, increases the concentration of NH4+1,
and increases the number of collisions per second between NH4+1
and OH-1. So 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". 18.104.22.168 Heat nitrogen tetroxide
(dinitrogen tetroxide, N2O4) See diagram 22.214.171.124: Heat nitrogen
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) 126.96.36.199 Explanation of group
A saturated solution can remain in equilibrium with undissolved molecules
of the solute.
Two equilibria exist: 1. an equilibrium between the undissolved solute
and dissolved molecules, and 2. an equilibrium between dissolved molecules
and ions formed by dissociation
(XY) <--> XY <--> X+ + Y-
Equilibrium 1.: (XY) <--> XY
Equilibrium 2.: XY <--> X+ + Y-
(XY) = undissolved molecules
XY = dissolved but unionized molecules
X+ + Y- = ions
The tendency of the solid to pass into solution depends on its active
mass, solution pressure. If the temperature remains constant, the active
mass remains constant because, by the law of mass action, concentration
dissolved molecules / concentration undissolved molecules = the constant,
K. The concentration of dissolved molecules is also a constant. Also,
by the law of mass action, if [concentration X+] [concentration
Y-] / [concentration dissolved molecules] = a constant, in a
saturated solution, the product of the concentrations of the ions is a constant.
So if large concentrations of ions are brought together into the same solution,
ions of X+ and Y- will precipitate out of solution
as solid molecules until the concentrations of the remaining ions in solution
have such values that the product of their concentrations equals the specific
constant, the solubility product. Group I
Lead, silver and mercury (I) are precipitated as chlorides by chloride
ions from hydrochloric acid. The concentrations of silver and chloride
ions that can remain in solution are small. So when a solution of a silver
salt containing silver ions is mixed with hydrochloric acid, most of the
silver and chloride ions form molecular silver chloride and leave the solution
as a solid phase until the
remaining ions attain equilibrium.
(concentration silver ions) × (concentration chloride ions) = 1
× 10-10, the solubility product.
The solubility product has a constant value, so adding excess chloride
ions reduces the concentration of the silver ions to a negligible quantity.
Only silver chloride, lead chloride and mercury (I) chloride have low
solubility products. So the ions of other metals remain in solution in the
presence of high concentrations of chloride ions. Group II
Assume hydrogen sulfide is ionized
H2S <--> 2H+ + S2-
By the law of mass action (concentration H+)2
× (concentration S2-) / (concentration unionized H2S)
= the constant, 1.1 × 10-22.
In a neutral solution, the concentration of sulfide ions is low because
hydrogen sulfide is a weak electrolyte. The concentration of hydrogen
ions is also low.
In the acid solution used for Group II, the concentration of hydrogen
ions is increased by the presence of the strong acid. So to maintain
the value of the solubility product constant, the concentration of the
sulfide ion is reduced below its already small value in neutral solution.
However, the amount of sulfide ions is enough to allow the solubility
products of the sulfides of mercury (II), lead, copper and bismuth to be
exceeded. Also, cadmium sulfide may precipitate if the acid is not too
concentrated. So in Group II, all the sulfides of mercury (II), lead, copper,
bismuth and cadmium precipitate.
Lead sulfide 4 × 10-28, Copper sulfide 8 × 10-45,
Mercury (II) sulfide 4 × 10-54,
Cadmium sulfide 3.6 × 10-29, Manganese sulfide 1.4 ×
10-15, Zinc sulfide 1.2 × 10-24
The concentration of sulfide ion in acid solution is not enough to
allow the higher solubility products of the sulfides of manganese, zinc,
cobalt or nickel to be reached with any possible concentration of the metal
ion, so these sulfides do not precipitate. They precipitate later in Group
IV, where the precipitating agent is the highly ionized salt, ammonium
sulfide, and the concentration of the sulfide ion from it is high. Group III
The precipitating agent is ammonia solution.
NH4OH <--> NH4+ + OH-
By the law of mass action:
(concentration NH4+) × (concentration OH-)
/ (concentration unionized NH4OH) = a constant.
Ammonia solution is a weak base, so does not ionize much and the value
of the constant is only about 2 × 10-5. Most of the ammonia
solution will be dissolved but not ionized. The small hydroxyl ion concentration
in a solution of ammonia solution that is also fairly concentrated with
respect to ammonium chloride is still large enough to cause a precipitation
of the hydroxides of ferric iron, chromium and aluminium, but not great
enough to precipitate the hydroxides of zinc, manganese, cobalt and nickel.
Manganese hydroxide may precipitate slightly if the concentration of ammonium
chloride is not sufficiently great. Group IV
In Group II the presence of hydrogen ions from the added acid reduces
the concentration of sulfide ions, but this reduced value was enough
to allow the solubility products of the metallic sulfides in the group
to be exceeded.
In Group IV, hydrogen sulfide is added to a solution made alkaline
with ammonia solution and so contains excess of hydroxyl ions.
By the law of mass action: (concentration H+)2
× (concentration S2-) / (concentration unionized H2S)
= a constant.
The ionic product of water [H+] × [OH-1] =
(10-14). The hydroxyl ions from the ammonia solution lower the
concentration of hydrogen ions causing an increased concentration of S2-
ion. In Group IV, the metal sulfides not already precipitated in Group
II because of their high solubility products, precipitate. So the ionic
concentration of the sulfide ion is controlled by variation of the
concentration of the ions it is associated with, i.e. hydrogen ions. Group V
The metals still remaining in solution include barium, strontium,
calcium, magnesium, sodium and potassium. Barium, strontium and calcium
are precipitated as carbonates by the addition of CO32-
ions in alkaline solution because of the low values of the solubility
[X2+] × [CO32-] = K, between 10-8
The solubility product of magnesium carbonate is low, 10-5,
and in neutral solution would be precipitated, but its precipitation is
prevented in this group by the ammonium ions, mainly from the ammonium
chloride added before Group III. With the large NH4+
ion concentration from this source, the concentration of CO32-
ions is reduced below the concentration to precipitate magnesium as a
/ [NH4)2CO3] = K 17.6.1 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
2Fe2O3 –> 4Fe + 3O2
Molecular weight of Fe2O3 = 160
Molecular weight of Fe = 56
So 2 Fe, 2 × 56 is equivalent to 1 Fe2O3,
Let weight of iron (II) ammonium sulfate = 1.96 g
Let 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%. 17.6.2 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
AL(OH)3 –> Al2O3 + H2O
Molecular weight of Al2O3 = 102.
Let weight of aluminium sulfate = wa g,
Let 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%. 17.6.3 Weight of calcium in
marble, calcium carbonate.
Stir 1.5 g of marble powder 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
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 +
CaCl2 + (NH4)2C2O4
—> CaC2O4 + NH4Cl
CaC2O4 + "O" –> CaO + 2CO2
(The “O” is from air)
Molecular weight of CaO = 56
Let weight of calcium carbonate = wa g.
Let weight of calcium oxide = wb g.
So weight of calcium in calcium oxide = (40 / 56) × wb g.
Percentage weight of calcium in marble = (40 wb / 56 wa) × 100%. 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
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
Molecular weight of Mg2P2O7, magnesium pyrophosphate
Let weight of magnesium sulfate = aw g.
Let 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 ×
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
Let weight of sodium sulfate crystals = aw g.
Let 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 +
Molecular weight of SnO2 = 151
Let weight of solder = aw g
Let 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.