Chemistry: Industrial Chemistry
Melissa Lau
1. Replacing Natural Products
Industrial chemistry processes have enables scientists to develop replacements for natural products
Discuss the issues associated with shrinking world resources with regard to one identified
natural product that is not a fossil fuel, identifying the replacement materials used and/or
current research in place to find a replacement for the named material
Key Issues that lead to replacement by synthetics
 Increasing demand for product
 Inability of natural sources to meet demand
 Depletion of natural resources
 Competition for natural resource from other use (fats for food and soup)
 Increasing prices for natural product
 Increasing availability/decreasing cost of starting materials for synthetics
 Decreasing price for synthetic product (quantities produced increased)
 Greater reliability of supply and stability of price for the synthetic product
 Need to synthesise will continue because of increasing population, level of consumption per person
and natural resources in used are being depleted rapidly.
Rubber
 Natural rubber (isoprene) – replaced with synthetic rubber
 Increased demand for rubber during WWII and disruption of supplies lead to the development of
synthetic rubbers

Styrene-butadiene rubber (SBR) was the 1st replacement rubber, had the most impact as is the
most common
o used to make tyres
o components are by-products of petroleum refining
 Butadiene – comes from steam cracking various feedstock (ethane, propane,
naphtha)
 Styrene – obtained from benzene (which is obtained by catalytic reforming)
o Vulcanised to improve properties (polymer chains loosely coiled and cross linked by
sulphur atoms)
o Fewer double bonds than natural rubber = less likely to deteriorate (less likely to react)


Now approx. 80% of rubber production is synthetic
Neoprene (polychloroprene) is the most commonly used synthetic alternative
o formed by addition polymerisation of 2-chloro-1,3-butadiene or chloroprene
o structurally similar to isoprene
o contains a chlorine atom instead of a methyl group  superior resistance to chemicals
as it’s less reactive
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Chemistry: Industrial Chemistry
Melissa Lau
2. Chemical Equilibrium
Many industrial processes involve the manipulation of equilibrium reactions
Explain the effect of changing the following factors on identified equilibrium reactions
Pressure and Volume
 decrease in volume (increase in pressure) = favouring direction that decreases the total
concentration of gaseous molecules = position of equilibrium will shift to side with lesser number
of gas molecules
 increase in volume (decrease in pressure) = favouring direction that increases the total conc. of
gaseous molecules = position of equilibrium will shift to side with greater number of gas
molecules
Concentration
 increasing conc. of a substance = favour the reaction that decreases the conc. of that substance
 decreasing conc. of a substance = favour reaction that increases the conc. of that substance
Temperature
 increasing temp = favouring the direction that ABSORBS heat (ENDOTHERMIC reaction)
 decreasing temp = favouring the direction that RELEASES heat (EXOTHERMIC reaction)
Interpret the equilibrium constant expression from the chemical equation of equilibrium
reactions
Qualitative Features of chemical equilibrium
 Some chemical reaction are reversible and come to equilibrium
 Dynamic equilibrium: rate of forward and reverse reactions equal
 Final equilibrium state is same regardless from which direction it is approached
 LTP summaries way in which equilibrium adjusts itself to minimise disturbance
 Reversible reactions can be driven to almost completion by use of LTP
Equilibrium Law
For a reversible reaction at equilibrium and a constant temperature, the ratio of the concentrations
of reactant and products has a constant value called the ‘equilibrium constant’ (K).
Equilibrium Constant (K)
K represents the relationship between the concentrations of reactants and products at equilibrium
For any chemical reaction aA +bB  cC + dD
Where A, B = reactants
K = [R]r[S]s
C, D = products
[P]s[Q]q
A, b, c, d, are number of each species involved (give mole ratio)
Characteristics of K
 Every reaction has it’s own K – a constant at a particular temperature
 Where a reaction involves solids or liquids, these terms are not included in the equilibrium
expression as their concentrations do not vary
 The size of K gives an indication of the relative proportions of reactants and products at
chemical equilibrium
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Chemistry: Industrial Chemistry
Melissa Lau
o

Small value of K (<10-30) = small concentration of products therefore reactants are
favoured
o Very small value = reaction hardly occurs
o Large value of K (>1030) = small concentration of reactants therefore products are
favoured.
o Very large value = reaction goes to completion
Size of K indicates the relative strength of acids and the solubility of substances
Reaction Quotient [not in syllabus]
 Right hand side of expression called the reaction quotient (Q)
 A reaction is at equilibrium if Q is equal to the equilibrium constant L.
 If reaction not at equilibriums Q different from K and chemical reaction occurs until Q equals K
(reached equilibrium)
o Q<K reaction goes from left to right until Q=K
o Q>K reaction goes from right to left until Q=K
o Q=K reaction is at equilibrium
 Unit for equilibrium constants varies. If until not given, concentrations must always be in mol/L
Identify that temperature is the only factor that changes the value of K for a given equation.
K is dependant on temperature
Temperature
Endothermic
Exothermic
Increase
K increases
K decreases
Decrease
K decreases
K increases
Identify data, plan and perform a first-hand investigation to model an equilibrium reaction
Modelling an Equilibrium Reaction
AIM: to model an equilibrium reaction
EQUIPMENT: 2 100mL measuring cylinders labelled A & B, 10mL and 5mL graduated pipettes
METHOD:
1. Fill A with 100mL of water
2. Invert the 10mL pipette in A and allow water to rise. Seal tip with finer and transfer trapped
water to B. – waterA  waterB (forward reaction)
3. Invert 5mL pipette into B and allow water to rise. Seal tip with finger and transfer trapped
water to A – waterB  waterA (reverse reaction)
4. Record volumes of water in A and B.
5. Repeat until water volumes remain constant for 5 transfer cycles.
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Chemistry: Industrial Chemistry
Melissa Lau
Choose equipment and perform a first – hand investigation to gather information and
qualitatively analyse an equilibrium reaction
Qualitative Analysis of Temperature on Equilibrium
The effect of temperature on the equilibrium:
2NO2(g)  N2O4(g)
ΔH = -57kJ
AIM: To determine the effect of temperature on an equilibrium reaction by observing colour change.
EQUIPMENT: thermometer, three tubes of NO2 gas, cold water, hot water, white paper
HYPOTHESIS: The tube of NO2 gas places in the hot water will darken and the tube in cold water
will lighten.
METHOD
1. Record air temperature
2. Place one of the tubes in a test tube rack. This is the control
3. Record the temperature of the heated water and place the second tube in it
4. Record the temperature of the cooled water and place the third tube in it
5. Leaves the tubes in the water for five minutes
6. Hold up the tubes against a piece of white paper and compare the colours of the gases.
RESULTS:
The tube in the heated water darkened in colour
The tube in the cooled water lightened in colour
CONCLUSION:
The effects of temperature on equilibrium can be observed through the colour change.
DISCUSSION:
The tube in the heated water darkened because the reaction favoured the direction that absorbed
heat (according to L.C.P) which in the above equilibrium reaction is the direction that results in a
production of the brownish yellow NO2 gas. The tube in the cooled water lightened because the
reaction was forced in the direction of the exothermic reaction, resulting in the production of the
colourless N2O4 which lightened the colour of the gas in the tube.
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Chemistry: Industrial Chemistry
Melissa Lau
3. Sulfuric Acid
Sulfuric acid is one of the most important industrial chemicals
Outline three uses of sulfuric acid in industry
Sulfuric acid is produced on a greater scale than any other manufacturing chemical. It is the cheapest
strong acid to produce.
Fertiliser Production
 superphosphate fertiliser: make natural rock phosphate (insoluble calcium phosphate) soluble
for plants
 ammonium sulfate: neutralising ammonia with sulfuric acid
o 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
Cleaning Iron
 very corrosive  clean iron or steel (grease, dirt, oxide layer) before galvanising or
electroplating  plating metal binds better
Dehydrating Agent
 drying agent in manufacture of explosives, dyes and detergents
 brings about condensation reactions  polymers, esters
Lead-acid car batteries – electrolyte
 Sulfuric acid can form insoluble lead sulfate with lead ions  keeps concentration of lead ions
low  battery can deliver more constant voltage.
Describe the processes used to extract sulphur from mineral deposits, identifying the
properties of sulphur which allow its extraction and analysing potential environmental issues
that may be associated with its extraction
Frasch Process – used to extract sulfur from mineral deposits
1. Super heated steam (165°C) pumped
into deposit, melting the sulfur (low MP
119°C)  forms emulsion
2. Compressed air pumped down inner
pipe forcing up molten sulfur to surface
(low density)
3. On cooling, sold sulfur separates from
liquid water (sulfur is insoluble)
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Chemistry: Industrial Chemistry
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Environmental Issues
 Sulfur is easily oxidised to sulfur dioxide or reduced to hydrogen sulphide, both of which are
serious air pollutants at low concentrations,
o Care to reduce release of these into atmosphere
 Superheated water may have dissolved impurities in deposit
o Water must be reused and not discharged into environment
 Unlike other mining ventures, the mine can’t be back-filled with processed tailings
Other extraction process
 Some can be obtained from natural gas and crude oil
 Some can contain up to 25% hydrogen sulfide (especially oil/gas from Canada and France)
 Hydrogen sulfide removed, producing elemental sulfur
Outline the steps and conditions necessary for the industrial production of H2SO4 from its raw
materials
Gather, process and present information from secondary sources to describe the steps and
chemistry involved in the industrial production of H2SO4 and use available evidence to analyse
the process to predict ways in which the output of sulfuric acid can be maximised
Contact Process
1. Oxidation of Sulfur  sulfur dioxide (combustion furnace)
S + O2 (g) → SO2 (g)

Δ = -297 kJ/mol
Rapid exothermic reaction  goes to completion
2. Oxidation of Sulfur dioxide  sulfur trioxide (catalytic converter)
 Reversible reaction: Controlled conditions for max yield
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Chemistry: Industrial Chemistry
2SO2 (g) + O2 (g)  2SO3 (g)
Melissa Lau
Δ = -99 kJ/mol
3. Sulfur Trioxide  Sulfuric Acid (Absorption Tower)
 The next step would be to react SO3 with water forming sulfuric acid:
SO3 (g) + H2O (l) → H2SO4 (l)


Reaction is exothermic so SO3 not directly absorbed in water  produce fog of
H2SO4  doesn’t work properly
Instead, SO3 gas is dissolved in conc. sulfuric acid to form oleum mixture
SO3 (g) + H2SO4 (l) → H2S2O7 (l)

Oleum reacts with water to form sulfuric acid:
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
Describe the reaction conditions necessary for the production of SO2 and SO3
Possible Ways to increase the yield include:
 Adding excess oxygen - however this isn’t very cost effective
 High pressure (Mole ratio is 3:2) – plants that can withstand the high temperatures are
extremely expensive and not worth the small increase in yield that it would produce
 Low temperature – since the reaction is exothermic, however as temperature increases the rate
of reaction becomes excruciatingly slow. A temperature of 400ºC produce yields around 99%. In
order to deal with the slow reaction rate, a catalyst is used: Vanadium (V) oxide, V2O5
Final reaction conditions used are:
 Pressure little above atmospheric
 Small excess of oxygen
 Catalyst of Vanadium (V) oxide supported on silica
 Temperatures of catalyst beds of 550ºC (for high rates) and 400ºC (for high conversion)
Apply the relationship between rates of reaction and equilibrium conditions to the production of
SO2 and SO3



Maximum yield is achieved from shifting equilibrium position forward
o Involves excess of O2, high pressure, low temperature (exothermic)
Compromise between equilibrium conditions, reaction rate and economic factors  O2 and high
pressure is not cost effective for small increase in yield
Temperature  temp of 400°C produces 99% yield but reaction rates are too slow.
o V205 allows faster reaction rate at moderate temperatures.
o SO2 is first passed over catalyst beds at high temp for quick reaction and then reduced
temp for increase in yield.
Describe, using examples, the reactions of sulfuric acid acting as: an oxidising agent, a
dehydrating agent
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Chemistry: Industrial Chemistry
Melissa Lau
Oxidising Agent:
 Dilute sulfuric acid can oxidise reactive metals,  produce hydrogen gas. The hydrogen ion
acts as the oxidant.
Zn(s)+ 2H+  Zn 2+ +H2(g)

Concentrated sulfuric acid will oxidise metals such as aluminium, which have a resistant oxide
coating. Sulfur dioxide is released and the oxidant is the hydrogen sulfate ion.
Sn(s) +2H+ + 2HSO4-  Sn2+ +SO2(g) +2H2O(l) + SO42-
Dehydrating Agent:
 Concentrated sulfuric acid rapidly absorbs moisture from the air.
 Use to produce dry air for the combustion of sulfur in the contact process
 Concentrated sulfuric acid is also used to dehydrate ethanol to form ethene and water.
 dehydration of sugar to carbon or ethanol
C2H5OH  C2H6 + H2O
Perform first-hand investigations to observe the reactions of sulfuric acid acting as: an
oxidising agent, a dehydrating agent
Oxidising Agent:
METHOD:
1.
Place zinc in test tube.
2.
Add a few drops of dilute sulfuric acid
RESULTS: hydrogen gas given off.
Zn(s) + H2SO4 (aq)  ZnSO4 (s) +H2(g)
Dehydrating Agent:
METHOD:
1.
Put several spoons of sugar (sucrose) in beaker. Put beaker in fume cupboard.
2.
Add 10mL conc. Sulfuric acid.
RESULTS: The sugar turns brown and then black as carbon forms. The heat released causes the
water and acid to begin to vaporize.
C12H22O11 (s)  12C (s) + 11H2O (l)
Safety Issues:
 Wear safety glasses to avoid eye damage.
 Extremely corrosive  protective clothing, careful handling
 Use fume cupboard
Describe and explain the exothermic nature of sulfuric acid ionisation
 Diluting conc. sulfuric acid involves the ionisation of H2SO4 into hydronium and HSO4- ions.
H2SO4 (l) + H2O (l) → H3O+ (aq) + HSO4- (aq)
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Chemistry: Industrial Chemistry
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
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Melissa Lau
o Involves breaking O-H bond  endothermic (requires energy)
o But formation of hydronium ion  exothermic. Releases far greater amount of energy
 Therefore overall reaction is exothermic (–95kJ/mol)
Serious risk of vaporization of water and splashing of acid
To avoid this add acid slowly to water.
The same process occurs in the second ionisation step, but because the H+ has to leave the
HSO4– ion, which already has a negative charge, it is a weak acid so the reaction occurs to a
lesser extent
Identify and describe safety precautions that must be taken when using and diluting
concentrated sulfuric acid






Add acid to water as heat produced by hydration can cause acid to boil and spit.
Dilution should take place in fume cupboard.
Wear protective clothing and covered shoes to prevent skin contact if spills occur
Have baking soda (sodium hydrogen carbonate) and water available in case of spills (direct use
of strong base neutralises by releases a lot of heat and solution could boil)
Store conc. Sulfuric acid away from water and reducing agents
Acid should be stored in glass bottles no larger than one litre for easy/safe handling
Use available evidence to relate the properties of sulfuric acid to safety precautions necessary
for its transport and storage



Concentrated sulfuric acid is largely unionised and so can be stored in iron or steel containers
When transporting concentrated sulfuric acid in metals containers, contamination with water
must be avoided else the acid will react vigorously with the container
Dilute sulfuric acid contains many hydrogen ions which can damage metals such as iron or
steel, so it must be stored in glass or plastic
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Chemistry: Industrial Chemistry
Melissa Lau
4. Sodium Hydroxide
The industrial production of sodium hydroxide requires the use of electrolysis
Explain the difference between galvanic cells and electrolytic cells in terms of energy
requirements
Galvanic (Voltaic) Cells
Chemical energy  electrical energy
Two half cells with separate electrolytes and salt bridge
Chemical reaction is spontaneous
Eo total is positive
Anode = negative terminal (oxidation)
Cathode = positive terminal
Oxidation at anode, releasing electrons into circuit 
becomes more positive and attracts negative ions from
solution
Electrolytic Cells
Electrical energy  chemical energy
Electrodes usually in the same electrolyte
Chemical reaction is forced by applying a
voltage
Eo total is negative
Anode = positive electrode (oxidation)
Cathode = negative electrode
Electrons flow from negative terminal to positive terminal
Electrons flow from negative battery terminal
to negative cathode
Uses: batteries
Uses: extract Al from Al2O3, electroplating,
purifying copper
Outline the steps in the industrial production of sodium hydroxide from sodium chloride
solution and describe the reaction in terms of net ionic and full formulaic equations
Electrolysis of molten sodium chloride
 Sodium metal & chloride gas produced by the electrolysis of molten NaCl in the Down’s Cell
 In the molten state, Na+ and Cl- ions can move through the melt.
Sodium metal is produced at the circular iron cathode:
Na+(l) +e-  Na(l)
Chlorine is evolved at the graphite anode:
2Cl-(l)  Cl2(g) + 2e10 - 26
Chemistry: Industrial Chemistry
Melissa Lau
The overall reaction is:
2Na+(l) + Cl-(l)  2Na(l) + Cl2(g)




products kept separate by iron screen between electrodes (otherwise they would react and
revert to NaCl
Molten sodium less dense than electrolyte  rises to surface  tapped off
Cl gas collected and dried  compressed and liquefied  storage and transport
Molten NaCl electrolyte contains some calcium chloride  lower MP of NaCl from 801°C to
600°C.
Identify data, plan and perform a first hand investigation to identify the products of the
electrolysis of sodium chloride
Electrolysis of Sodium Chloride
AIM: to determine the products of the electrolysis of sodium chloride
METHOD:
1. Put wad of cotton in bottom of a U-tube.
2. Fill tube with NaCl solution. Add a few drops of phenolphthalein.
3. Connect electrodes to power pack.
RESULTS: bubbles form on both sides.
 At positive cathode  bleach like odour detected
 At negative anode  indicator turns purple (detects OH ions). A burning match test would
confirm presence of H
CONCLUSION: The products of the electrolysis of aqueous NaCl produced hydrogen, chlorine and
hydroxide ions.
Cathode: 2H2O (l) + 2e-  H2 (g) +2OHAnode: 2Cl-  Cl2 (g) + 2e-
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Chemistry: Industrial Chemistry
Melissa Lau
Analyse information from secondary sources to predict and explain the different products of the
electrolysis of aqueous and molten sodium chloride
Electrolysis of molten salts is simpler than electrolysis of aqueous solutions because there are no
competing electrode reactions (only one is possible in each case)
Conditions used
Cathode Product
Anode Product
Molten
Conc., aqueous, inert electrodes
(membrane & diaphragm
processes)
Dilute, aqueous, inert electrodes
Conc., aqueous, Hg electrodes
(Mercury process)
Na
H2 [Water has high reduction
potential compared to
sodium]
H2
Cl2
Cl2 [High conc. of Cl ions in
solution  Cl produced rather
than oxygen/
O2 [ 2H2O  O2 + 4H+ + 4e- ]
Na
Cl2
Distinguish between the three electrolysis methods used to extract sodium hydroxide (mercury
process, diaphragm process, membrane process) by describing each process and analysing
the technical and environmental difficulties involved in each process.
Electrolysis of aqueous sodium chloride solution
 Electrolysis of dilute NaCl produces hydrogen at cathode and oxygen at anode (water
molecules reduced and oxidised more easily than sodium and chloride ions)
 Electrolysis of concentrated NaCl
o Cl ions oxidised at anode.
o Products depend on conc. of electrolyte solution  Higher conc. of chloride ion, high
discharge of chlorine
 Electrolysis of conc. NaCl = basis of huge chlor-alkali industry: hydrogen, chlorine and sodium
hydroxide manufactured from relatively cheap sodium chloride
Overall reaction:
2NaCl(aq) + 2H2O(l)  2NaOH(aq) + Cl2(g) + H2(g)
Ionic equation:
2Na+(aq) + 2Cl-(aq) + 2H2O(l)  2Na+(aq) + 2OH-(aq) +Cl2(g) + H2(g)
Net ionic equation:
2Cl-(aq) + 2H2O(l)  2OH-(aq) + Cl2(g) + H2 (g)
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Chemistry: Industrial Chemistry
Melissa Lau
Mercury Process
Component /
Chemical
Function
Equation
Mercury (flows)
Cathode. Combines with Na to
form amalgam
2Na+(aq) +2e- + Hg(l)  2Na(Hg)
Platinised titanium
plates
Support anode reaction
2Cl-(aq)  Cl2(g) + 2e-
Decomposer
Na-Hg amalgam sprayed into
water  Na reacts vigorously
2Na(Hg) + 2H2O(l)  2NaOH(aq) + H2(g)+ 2Hg(l)


H2 collected and stored  valuable by-product
Mercury does not react  recycled and reused
Advantages
 High quality NaOH and Cl
Disadvantages
 Mercury is toxic heavy metal  health and environmental issues
Escape into waterways through leakages, vaporisation or when electrolytic cells cleaned
In food chain  poisoning (muscle wasting, paralysis, death)
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Chemistry: Industrial Chemistry
Melissa Lau
Diaphragm process (Nelson Diaphragm Cell)
Cell Component
Reaction
Anode = unreactive metal eg titanium, carbon
2Cl-(aq)  Cl2(g) + 2e-
Cathode = Asbestos lined steel mesh prevents
products mixing
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
1. Steam passed over cathode steel mesh and condenses
2. Condensed steam washed out Na ions and OH ions to bottom  hot NaOH solution at bottom
3. Solid NaOH crystallised out.
Disadvantages: asbestos is known carcinogen
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Chemistry: Industrial Chemistry
Melissa Lau
Membrane process


Anode and cathode separated by water impermeable membrane (synthetic polymers eg Teflon)
Membrane only allows Na+ to pass.
Cell Component
Reaction
Anode = unreactive metal eg titanium
2Cl-(aq)  Cl2(g) + 2e-
Cathode = unreactive metal eg steel
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
Advantages
 Does not use asbestos or mercury  cheaper and safer.
 Virtually pure NaOH produced (no oxidant in spent brine)
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Chemistry: Industrial Chemistry
Melissa Lau
5. Saponification
Saponification is an important organic process
Describe saponification as the conversion in basic solution of fats and oils to glycerol and salt
of fatty acids.
Saponification – fats and oils react with conc. NaOH (or KOH) to form glycerol and the salt of the fatty
acid (soap)
Fat/oil + base  glycerol + soap
Fat/oil + conc. NaOH or KOH  glycerol + sodium/potassium salt of fatty acid
Eg.
Glyceryl tristearate
+ sodium hydroxide 
sodium stearate (soap)
+ glycerol
Glycerol is an alkanol with 3 hydroxy groups and the formula CH2OHCHOHCH2OH. Its systematic
name is 1, 2, 3-propanetriol.
Esters are carbon compounds with the general formula RCOOR' where R and R' are alkyl groups.
Esters can be made by the reaction of an alkanol and an alkanoic acid.
Alkanol + alkanoic acid
ester + water
Fats and oils are esters made from glycerol (1, 2, 3-propanetriol) and long chain fatty acids such as
stearic acid (CH3(CH2)16COOH). Different acids combined with glycerol produce different fats and oils.
Gather process and present information from secondary sources to identify a range of fats and
oils used for soap making




Fats are solids at room temperature while oils are liquids.
The only difference between fats and oils and ordinary esters is that in fats and oils the one
alcohol molecule is combined with three acid molecules.
The acids involved are usually long-chain ones such as saturated palmitic and stearic acids and
unsaturated ones such as oleic, linoleic and linolenic acids.
Carboxylic acids obtained from fats and oils are called fatty acids
Animal fats: tallow (cattle fat), lard (pig fat), whale oil, fish oil.
Vegetable oils: olive oil, palm oil, coconut oil and soybean oil.
Describe the conditions under which saponification can be performed in the school laboratory
and compare these with industrial preparation of soap
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Chemistry: Industrial Chemistry
Melissa Lau
Laboratory production of soap
 Relatively simple
 Heat mixture of relatively pure oil/fat with excess NaOH solution
 Soap filtered off, dried
 No attempt to remove glycerol.
Industrial manufacture of soaps
 Made on huge scale in enormous vats (Kettle Method)
 Starting oils and fats often not pure
 NaOH expensive – use is minimised
 Glycerol removed from aqueous residue by distillation as it is a valuable by-product (use in
pharmaceuticals, toiletries, paints, confectionary, etc)
o Soap-glycerol mixture with brine  glycerol dissolves in brine  separated
 Remaining salt from brine is reused
 Raw soap carefully washed and dried to meet national standards
 Blended with additives (eg perfumes, stabilisers)
Perform a first hand investigation to carry out saponification and test the product
Saponification
AIM: to produce soap
METHOD:
1. Combine 25mL water and 25mL ethanol in 250mL beaker. (ethanol is dehydrating agent)
2. Add 12.0g NaOH
3. Add mixture to a 500mL beaker containing 10g copha
4. Heat on hot plate, stirring for 1h
5. Filter. Allow soap to dry
SAFETY:
 NaOH is a strong base: gloves & goggles, handle with care, wash with water if spilt, if on skin
flush with water for 15 min
 Ethanol is flammable: use hot plate rather than naked flame, cross ventilation
RESULTS: White solid formed (soap)
Testing the Soap
pH: 9.5 (slightly alkali since we didn’t neutralise the soap with dilute acid)
Lathering ability:
1. Put equal amounts of our soap and a commercial soap into 2 test tubes with equal amounts
of water. Shake vigorously
2. Measure height of suds.
Result: commercial soap suds twice as high
Account for the cleaning action of soap be describing its structure
Soaps dissolve in water:
H 20
RCOO  Na  (s) 
 RCOO  (aq)  Na  (aq)
The resulting solution:
 sodium or potassium cation (no function in cleaning process)
 anion from fatty acid = surfactant
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Chemistry: Industrial Chemistry
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Surfactants (wetting agents) = Changes surface tension of water, allowing it to spread out and wet
surfaces evenly rather than forming droplets.
Structure of surfactants
Large ions consisting of negatively charged group and non-polar hydrocarbon chain.
 Negatively charged carboxylate (alkanoate) group at end forms hydrogen bonds with water 
hydrophilic end (‘water loving’)  surfactant ion dissolves readily in water
 Non polar hydrocarbon chain  hydrophobic (‘water hating’)  mixes readily and forms
dispersion forces with non-polar dirt, grease or oil.
How Soaps Work
1. Surfactant dissolves in water
2. Surfactant ions orientate themselves in grease and water (hydrocarbon chain forms dispersion
forces with grease, polar ‘heads’ directed towards water)
3. Agitation begins to separate grease from surface
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Chemistry: Industrial Chemistry
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Explain the soap, water and oil together form an emulsion with the soap acting as an emulsifier
Emulsion = mixture of two liquids that are dispersed and suspended in one another. Neither liquid will
dissolve in the other.
1.
2.
3.
4.
Add small amount of soap to oil water mixture
Hydrocarbon (non-polar) ends of surfactant form dispersion forces with oil, surrounding them
Polar ends of surfactant stick out and repel other polar ends from other oil droplets
Oil droplets can’t join up with each other  remain dispersed in liquid
Perform a first hand investigation to demonstrate the effect of soap as an emulsifier
Soap as an Emulsifier
AIM: To investigate soap as an emulsifier
METHOD:
1. Test Tube 1: Add 10mL water and 2mL oil to test tube. Shake vigorously for 10s. Leave to
rest
2. Test Tube 2: Add 10mL soap solution and 2mL oil to test tube. Shake vigorously for 10s.
Leave to rest.
3. Compare observations
RESULTS:
 Test tube 1: separate oil layer on top of water
 Test Tube 2: milky white mixture. No distinct layers
Perform a first hand investigation to gather information and describe the properties of a names
emulsion and relate these properties to its uses
Making an emulsion
AIM: to investigate a simple emulsion - a vinaigrette
METHOD:
4. Combine 2mL vinegar + pinch of mustard powder in test tube. Shake.
5. Add 4mL olive oil in 1mL portions. Shake after each addition
6. Shake vigorously 10 times. Leave to rest. Observe using hand lens
RESULT: droplets (water i.e. vinegar) suspended in oil.
CONCLUSION: salad dressing is a water-in-oil emulsion. Properties (taste, evenness, colour and
smell) are important for its use
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Chemistry: Industrial Chemistry
Emulsion
Milk
Mayonnaise
Cosmetic Creams
Paints
Melissa Lau
Contents
Fat droplets in water
Proteins are natural emulsifiers. Additional emulsifiers added to keep fat
suspended
Oil water and vinegar
Egg added to prevent separation
Oil and water
Other chemicals for perfume and colour
Pigments, solvents, polymers
Distinguish between soaps and synthetic detergents in terms of: the structure of the molecule,
chemical composition, effect in hard water
Detergents = cleaning agents that contain surfactants
Soap
Detergent
Made From
Chemical
Composition
Fatty acids in animal & vegetable oils
Na or K salts of long chain (alkanoic)
fatty acids
Hydrocarbon chain from petroleum
Usually hydrocarbons with sulfate or
sulfonate end
Similar
Structure
Head – ionic or polar
Tail – long non polar hydrocarbon
ANIONIC
Saponification: heating fats/oils (esters)
with NaOH or KOH. Precipitation with
NaCl
ANIONIC, CATIONIC or NON-IONIC
Alkanol from petroleum reacted with
H2SO4 sulfonic acid --< react with
NaOH  sodium sulfonate
Effect in hard water
Doesn’t lather well
Soap anions form  ppt with cations
(Ca2+ and Mg2+ in hard water forms
scum)
Lathers in hard water (doesn’t ppt with
mineral salts)
Biodegradability
Biodegradable
Straight chain hydrocarbon –
biodegradable
Branched chain – non biodegradable
Phosphates
No phosphates
Can be mixed with phosphates that
pollute
Cost
Solubility
Cheaper to make
Not very soluble
More expensive
Soluble in water
Other
Deteriorates with age
Doesn’t deteriorate with age
Very stable
Manufacture
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Chemistry: Industrial Chemistry
Melissa Lau
Distinguish between anionic, cationic and non-ionic synthetic detergents in terms of: chemical
composition uses
Anionic
Cationic
Non-ionic
Negative head (eg sulfate(SO42–)
or sulfonate(SO3–))
Long hydrocarbon end from
petroleum – benzene ring  alkyl
benzene sulfonates/ sulfates
Highly sudsing
general cleaning, particularly in
laundry and dishwashing
detergents
Positive head (usually ammonium
compounds)
Hydrophilic polar end with many
oxygen atoms  form H bonds
with water
Do not ionise in water
Low sudsing
hair conditioner, fabric softener
(positive end attaches to negative
charges in fabric, reduce static)
disinfectants and antiseptics
(ammonium ions disrupt cell walls
of some bacteria),
low foam applications:
dishwasher powders, paints,
pesticides, cosmetics
Not used in dishwashers – glass
has negative surface, attracts
positive heads, leaving tails to
make glass slippery
Solve problems and use available evidence to discuss, using examples, the environmental
impacts of the use of soaps and detergents
Environmental concerns:
Soap
Biodegradable – usually straight carbon chains
No phosphates
Detergents
Straight chain  biodegradable
Branched chain  Non biodegradable
May contain phosphates that pollute. Cause
eutrophication of waterways (algal blooms)
In many countries there are restrictions on the amount of phosphate used in washing detergents. In
Australia there is a voluntary codes in operation.
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Chemistry: Industrial Chemistry
Melissa Lau
6. The Solvay process
The Solvay process has been in use since the 1980s
Identify the raw materials used in the Solvay Process and name the products
CaCO3(s) + 2NaCl (aq)
Na2CO3(aq) + CaCl2(aq)
Reactants = calcium carbonate (limestone), sodium chloride, ammonia
Products = sodium carbonate, calcium chloride (ammonia recovered again to be reused)
Under normal conditions, CaCO3 doesn’t react with NaCl  use of ammonia (which is later recovered)
Describe the uses of sodium carbonate
Glass Making
 more than half of NaCO3 produced is used for glass making
 Melt sodium carbonate, calcium carbonate and sand (silicon dioxide)
Water Treatment
 Used as water softener (ppt with Mg and Ca) industrially and domestically (in some washing
powders as washing soda).
 Also used to make zeolite builders used to replace phosphate builders in washing powders and
liquids
Making Soaps and Detergents
 Cheaper alternative to sodium hydroxide
Paper making
 NaCO3 used to make sodium hydrogen sulfite (used in one method of paper making)
Common Base
 Used in many chemical factories. Cheaper than NaOH
Manufacture of sodium hydrogen carbonate
 aka sodium bicarbonate, bicarbonate of soda
 Used as baking soda and in fire extinguishers
 NaHCO2 in from Solvay process in contaminated with ammonia  heat to remove ammnica
 Na2CO3 + CO2 is a more efficient way of getting it
Removing sulfur dioxide from flue (waste) gases in power stations
 More common now, esp overseas where power station have to meet stringent emission controls
SO2 + Na2CO3  CO2 + Na2SO3 (sodium sulfite)
Petrol Refining
Identify, given a flow chart, the sequence of steps used in the Solvay process and describe the
chemistry involved in: brine purification, hydrogen carbonate formation, formation of sodium
carbonate, ammonia recovery
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Chemistry: Industrial Chemistry
Melissa Lau
STEP 1: Brine purification (30% NaCl solution required for Solvay process)
Brine = conc. NaCl solution.
1. Source:
o Brine wells – salt water stored deep underground
o Underground rock salt deposits - pump water in to dissolve
o Solar evaporation – from sea water
2. Impurities (normally Ca, Mg, some Fe) need to be removed to avoid precipitating with sodium
hydrogen carbonate to form side products
o
NaCO3 to ppt calcium:
Ca2+(aq) + CO32–(aq) → CaCO3(s)
o
OH to ppt Mg (MgCO3 is slightly soluble  does not ppt with carbonate at low conc.)
Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s)
o
Hydroxide also precipitates iron and other heavy metals, if they are present:
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
3. Flocculent added to coagulate ppt
4. Ppt skimmed off the brine
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Chemistry: Industrial Chemistry
Melissa Lau
STEP 2: Sodium Hydrogen Carbonate Formation
1. Heat calcium carbonate in kiln to form Calcium oxide and carbon dioxide. CaO removed,
used later in ammonia recovery
CaCO3(s) → CaO(s) + CO2( g)
Coke (carbon) in kiln produces more carbon dioxide when heated, providing heat to decompose
calcium carbonate
C(s) + O2(g) → CO2(g)
2. Ammonia dissolved in purified brine  ammonical brine.
3. Carbon dioxide dissolved in ammonical brine  NaHCO3 ppt forms. Carried out at low
temperature (decreases solubility of NaHCO3)
NaCl(aq) + NH3( aq) + CO2( g) + H2O(l ) → NaHCO3(s) + NH4Cl(aq) [neutral]
CO2( g) + NH3( aq) + H2O(l ) → HCO3-(s) + NH4+(aq)
[net ionic]
4. Mixture is filtered
o
Filtrate (ammonium chloride)  to ammonia recovery
o
Ppt (NaHCO3)  washed and dried
STEP 3: Sodium Carbonate Formation
1. Heat NaHCO3 to 300°C.
2NaHCO3(s) → Na2CO3(s) + CO2( g) + H2O( g)
2. Remove NaCO3. Reuse CO2
STEP 4: Ammonia Recovery
1. The calcium oxide is converted to a calcium hydroxide solution in the lime dissolver:
CaO(s) + H2O(l ) → Ca(OH)2(aq)
2. This calcium hydroxide solution is mixed with the filtrate from Step 3 (containing NH4Cl) and the
mixture heated to regenerate ammonia which is re-used in Step 1:
Ca(OH)2(aq) + 2NH4Cl(aq) → CaCl2(aq) + 2H2O(l ) + 2NH3( g)
Overall Reaction:
2NaCl(aq) + CaCO3(s) → Na2CO3(s) + CaCl2(aq)
Perform a first hand investigation to assess risk factors and then carry out a chemical step
involved in the Solvay process, identifying and difficulties associated with the laboratory
modelling of the step
Process information to solve problems and quantitatively analyse the relative quantities of
reactants and products in each step of the process
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Chemistry: Industrial Chemistry
Melissa Lau
Laboratory Modelling of the Solvay Process
AIM: To model the Solvay process by performing a series of simple chemical reactions
METHOD:
1. Heating calcium carbonate  CaO and CO2
a. Place marble chip (CaCO3) on wire gauze. Heat
strongly for about 5 minutes. Record observations
b. Put chip in test tube. Add 5mL water and 5 drops
universal indicator. Shake. Record observations.
c. Repeat step b using untreated marble chip. Record
observations.
2. Sodium hydrogen carbonate formation:
a. Transfer 50mL cold ammonical brine into conical
flask. Add dry ice (CO2). NaHCO3 precipitates.
b. Filter reaction mixture. Dry the NaHCO3.
3. Sodium Carbonate formation
a. Transfer dried NaHCO3 into side arm test tube.
Immerse delivery tube into limewater.
b. Heat NaHCO3 strongly to convert it to Na2CO3
c. Record observations
RESULTS
1. Treated: pH 10 (purple). Untreated: pH7.5 (green)
 CaO ionises in water, forms Ca(OH)2 (basic)
2. Bubbles violently. Tiny white particles form close to CO2 source. Not enough ppt to do next step
(used fresh sample)
3. Limewater turns cloudy  presence of CO2 gas
DIFFICULTIES
 Can’t collect CO2 from heating marble chip (could use Kipps Generator)
 Amount of CO2 produced is too small to ppt out the NaHCO3
 Ppt is very fine and in small amounts  hard to filter
 Use dry ice pellets: ready source of CO2. Cold temp lowers solubility of NaHCO3, assisting precipitation
reaction.
SAFETY
Ammonia (in ammonical brine): volatile, toxic fumes  cross ventilation, fum cupboard, use conical flask
(narrow neck holds in gas)
Discuss environmental issues associated with the Solvay process and explain how these issues
are addressed.
Solvay process produces less pollution than previous methods: reactions take place in a tower and byproducts (ammonia, CaO and CO2) are re-used
Calcium chloride
 Calcium chloride is difficult to dispose of.
 Some uses: drying agent, concrete additive, melt road ice.
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Chemistry: Industrial Chemistry


Melissa Lau
Discharge into rivers  increase calcium and chloride ion concentrations  affect ecosystem
At Osborne, it has been discharged into the ocean for many years.
Solid waste
 Waste = unburnt calcium carbonate, sand and clays from the kiln
 Osborne Plant (1997) pumped 200 tonnes per day into adjacent river  huge sludge deposits.
 Not toxic but unsightly, nuisance and blocks shipping channels.
 Research new uses for waste (eg fertiliser, landfill, brick manufacture)
Ammonia
 There is often some loss of ammonia to atmosphere  significant air pollutant
 Addressed with careful monitoring and good design.
Dust control
 improved truck loading facilities & keeping vehicles on the asphalt roadways
 Upgrading of dust suppression systems in the plant
 using a wetting solution to suppress dust in open areas
 using bag filters to reduce dust in the bicarbonate plant
 Installation of dust scrubbing systems.
Noise suppression - Reduced by enclosure of noisy areas, using silencers to dampen noise and
community monitoring to identify sources of noise.
Heat - process is exothermic  waste water must be cooled before it is returned to rivers or ocean.
Use available evidence to determine the criteria used to locate a chemical industry using the
Solvay process as an example
Penrice is the only manufacturer of soda ash and sodium bicarbonate in Australia and in 2004 supplied
74% of the soda ash and 88% of sodium bicarbonate used by Australian industry.
Factors that influence location of
chemical industry
Proximity to supply of raw materials
Proximity to Market
Availability of transport for raw
materials and product
Solvay Process – Penrice (Osborne SA)
Osborne is on 35km strip of low-lying land along Gulf of St
Vincent
Salt – coastal location allows easy access to sea water which is
pumped into shallow ponds, purified and crystallised
Limestone – quarry in the Barossa Valley sends trainload each
day.
Supplies Australian region
Raw: Limestone transported by train
Product: 48000 tonnes of sodium bicarbonate and
325000tonnes sodium carbonate transported by road, rail & sea
each year.
Availability of amenities for workers
and family (eg housing, schools,
shops)
Osborne is western suburb of Adelaide
Facilities for waste disposal
Until recently discharged into Port River.
Other uses now being found (eg land fill)
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