Hydrocarbons and Fuels

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Transcript Hydrocarbons and Fuels

Principles to production
This unit studies how the chemical industry
applies key principles of physical chemistry in
order to turn research ideas into profitable
products without harming the environment.
Getting the most from
costly reactants
Overview
Learn about how industrial
processes are designed to
maximise profit and minimise
environmental impact.
Principles to production
From previous work you should know and understand the following:
•The Haber Process for the manufacture of ammonia.
•The Ostwald Process for the catalytic oxidation of ammonia.
•The fractional distillation of crude oil.
• Rate of reactions
• Reacting masses/mole calculations involving mass conc and volume
• Redox
a) Factors influencing the
design of an industrial process
Learning intention
Learn about the variety of factors
which influence industrial process
design in order to ensure
maximum profit with minimal
environmental impact.
Manufacturing chemicals
There are many stages leading up to the manufacture of a
new chemical on an industrial scale
New Higher Chemistry E Allan J Harris
New Higher Chemistry E Allan J Harris
Principles of design
• Industrial processes are designed to
maximise profit and minimise the
effect on the environment
7 principles of design
process
•
•
•
•
•
•
•
Availability of feedstocks
Cost of the feedstock
Sustainability of the feedstock
Opportunities for recycling
Energy requirements
Marketability of by-products
Product yield (or atom economy)
Principles of design -Availability, cost and
sustainability of feedstocks
Raw materials - major raw materials are fossil
fuels, water, air, metal ores, and minerals. They
must be converted to feedstocks before being
passed into the reactor.
Feedstocks are simple chemicals or mixtures of
chemicals derived from raw materials used in
further manufacture.
What is the cost?
Is the feedstock sustainable?
Principles of design – Batch and
continous
A chemical plant produces the desired products.
The process used to manufacture the product may be
operated in batch or continuous sequences.
Principles of design –Batch versus Continuous
manufacturing in the Chemical Industry
Batch
Continuous
For
OK for up to 100
tonnes per annum.
More versatile.
Good for multi-step
reactions.
OK for over 1000 tonnes per
annum.
Good for fast single step
processes.
Easy to automate.
Against
Contamination of
product is more
likely.
At times, no product
is made.
Safety more of an
issue.
Capital cost is high.
Less flexible.
Need to run at full capacity
to make a profit.
Batch – Drug development
Batch – whisky production
Continuous process - plastics
Continuous - Blast furnace
Principles of design - Costs
Capital Costs
Research and
development
Plant
Construction
Buildings
Infrastructures
Fixed
Costs
Depreciation of
plant
Labour
Variable
Costs
Raw material
Energy bills
Land purchase or Overheads
rental
Sales expenses
Effluent waste
treatment
Principles of design – Considerations
Energy in or out
Feedstock
REACTION
Temp, pressure, catalyst
products
Separation
By-products
Recycle loop
Consideration has to be given to:
Operating conditions
Energy consumption, generating your own, conservation, use of
catalysts, recycling of heat, (heat exchangers)
Environmental issues
Safety
Can the by-products be sold?
Cost of waste disposal
Yield and atom economy
Fertiliser Industry
Haber Process
Ammonia is manufactured from N2 and H2. The nitrogen is available
from the raw material, air. (something which is available naturally).
The other feedstock for the manufacture of NH3 is hydrogen which is
usually produced from methane.
Natural
gas
Water
CH4 (g)
AIR
alkali
H2O(g)
Catalyst
heat
Stage 1
Catalyst
Stage 2
Catalyst
Stage 3
N2(g) +
H2 (g)
CO2 removed
Sulphuric Acid Industry
Sulphuric acid is manufactured by the Contact Process.
Waste gases
Sulphur
Air
S
Stage 2
Oxygen
O2(g)
burner
heat
Stage
98% acid
oleum
Stage 3
SO2(g)
Catalytic SO3(g) absorber
feedstock Converter
(In 90%
H2SO4)
Cat=V2O5
1
water
Mixture to
dilute the acid
H2SO4
Location
Location - Grangemouth
New Higher Chemistry E Allan J Harris
Location – Teeside plant chemical works
New Higher Chemistry E Allan J Harris
Safety
Stockline Factory Explosion:
Nine men were killed in May
2004 when the ICL Plastics
factory in Glasgow’s Maryhill
exploded.
Piper Alpha oil rig on July 6,
1988
266 men 59 survivors
“LOSSIECHEM BRINGS ECOLOGICAL
CHEMICAL PLANT TO MORAY”
Sea water can be electrolysed to produce 3
products, sodium hydroxide, chlorine and
hydrogen.
Scientists
and
engineers
at
Lossiechem have produced a pilot plant that
successfully splits sea water using electricity
produced from renewable resources. The
company and local planners must now decide
whether to build a full scale plant here, bringing
much needed jobs to the local economy.
How do the 7 principles of design
process apply to Lossiechem
•
•
•
•
•
•
•
Availability of feedstocks
Cost of the feedstock
Sustainability of the feedstock
Opportunities for recycling
Energy requirements
Marketability of by-products
Product yield (or atom economy)
Factors influencing the process
chosen
• Availability and sustainability of feedstocks
• Salty water is available in abundance in the Moray
Firth and supplies are unlimited
• Cost of feedstocks
• Salty water would have to be pumped to the plant
but the cost of transport is minimal once the
pipeline has been built. The water may have to be
treated and filtered to ensure no effluent is
present in the final product, sodium hydroxide.
Factors influencing the process
chosen
• Opportunities for recycling
• Unreacted sodium chloride solution can be
separated from the sodium hydroxide and
fed back into the plant.
• Cost of feedstocks
• Salty water would have to be pumped to
the plant but the cost of transport is
minimal once the pipeline has been built.
The water may have to be treated and
filtered to ensure no effluent is present in
the final product, sodium hydroxide.
Factors influencing the process
chosen
• Energy requirements
• Electricity could be obtained from
hydorelectric sources or wind power.
Perhaps a wave powered device could
be used in the Firth.
• Marketability of the by-products
• There seem to be no by -products,
only 3 marketable products
Factors influencing the process
chosen
• Product yield
• Atom economy is 100%. All the
available chemicals are turned into
products.
Environmental
considerations
• Minimising waste
• Avoiding use of production of toxic
substances
• Designing biodegradeable products
Environmental considerations
• Minimising waste
• The production of electricity
produces no carbon dioxide and the
process only produces waste when the
water is purified. This could be used
as a fertiliser, but research is
needed.
Environmental considerations
• Avoiding the production of toxic
substances
• Chlorine is toxic but it is a desirable
product as it can be marketed. All the
chlorine produced would be kept
under pressure in a storage depot or
in a tanker.
Environmental considerations
• Products that will biodegrade
• This is not relevant to this process
but it is hoped that chlorine would
only be supplied to plastics
manufacturers if they were producing
a biodegradeable plastic.
Case study -Acrylics for the future
Lucite International
Methyl methacrylate (MMA) is
the monomer used to make
perspex or acrylic polymers.
Lucite International have
developed a new greener
process.
READ THE ARTICLE – MAKE
NOTES ON THE SEVEN
PRINCIPLES AND THE 12
PRINCIPLES OF GREEN
CHEMISTRY
DomAquaree hotel berlin
1 million litre aquarium built of acrylic made
withh MMA
7 principles of design process
•
•
•
•
•
•
•
Availability of feedstocks
Cost of the feedstock
Sustainability of the feedstock
Opportunities for recycling
Energy requirements
Marketability of by-products
Product yield (or atom economy)
b) Calculation of the mass or
volume of products
Learning intention
Learn how the theoretical mass or
volume of product can be
calculated from the balanced
reaction equation.
From previous studies
You should be able to
• Write Formulae
• Calculate percentage composition
• Calculate empirical formulae
• Calculate the number of moles in a given mass
• Calculate the number of moles of solute dissolved in a solution
Reacting Masses
Accurately weigh a crucible
Add approx 1.2g of Mg ribbon and reweigh
Place the crucible and lid in a silica triangle.
Heat gently at first then more strongly. Lift the lid with tongs
from time to time to admit more oxygen, but not enough to
let out the magnesium oxide.
Reacting masses
When the reaction is complete, the magnesium will
not glow more brightly when the lid is raised
Allow the crucible to cool
Reweigh the crucible
Calculations
Using the balanced equation calculate the mass of MgO you
would expect to be formed?
2Mg + O2
→
2MgO
Reacting masses
What mass of zinc sulphate will be produced on adding 6.0g
zinc to excess sulphuric acid?
14.9g
Reacting masses
What mass of sodium carbonate will react completely with
100cm3 of nitric acid concentration 1 mol l-1?
5.3g
The Mole
A mole is that amount of substance which contains as
many ‘elementary particles’ as there are carbon atoms 12g of
carbon-12.
Fe
55.8 g
Cu
Sn
63.5g
118.7g
I2
(126.9 x 2)
= 253.8g
One mole of an element is the relative atomic mass expressed
in grams.
One mole of an compound is the formula mass expressed in
grams.
You can calculate the number of moles (n) in a substances
How many moles in 18g of carbon?
n = mass/GFM
n= 18/12
n= 1.5 moles
1 mole Carbon ↔ 12g
(18/12) x 1 ↔ 18g
= 1.5 moles
Avogadro's number
Avogadro’s constant, NA, is the number of ‘elementary particles’
in one mole of any substance.
It has the value of 6.02 x 1023
Things to understand about Avogadro's number
• It is a number, just as is "dozen"; you can think of
Avogadro's
number asnumber
the "chemist's
dozen".
An Avogadro's
of standard
soft drink cans would cover the
• It is asurface
huge number,
far greater
in magnitude
than we
of the earth
to a depth
of
can visualize
over 200 miles.
6.02 x 1023
billion
If we602
were
able trillion
to count atoms at the
602,000,000,000,000,000,00
rate of
10 million per second, it would
take 0,000
about 2 billion years to count the
atoms in one mole.
Amadeo Avogadro (1766-1856)
Avogadro’s hypothesis states
that “equal volumes of
different gases, under STP,
contain equal numbers of
molecules.”
Avogadro's number is one of
the fundamental constants of
chemistry.
Avogadro never knew
his own number; it was named
in his honor by a French scientist
in 1909. Its value
23
6.02
x 10 Loschmidt, an
was first estimated
by Josef
billion in
trillion
Austrian chemistry602
teacher,
1895.
602,000,000,000,000,000,00
0,000
Equimolar amounts of substances contain equal
numbers of elementary entities
Copper
Water
Sodium chloride
Formula
Cu
H2O
Na+Cl-
1mole
63.5g
18g
58.5g
Molecules
Formula units
6.02x1023
molecules
6.02x1023 formula
units
Elementary entities Atoms
1 mole contains
6.02x1023 atoms
The Mole and Avogadro’s constant
Worked example 1.
Calculate the number of atoms in 6 g of carbon.
1 mole
12g
6g



6.02 x 1023 Carbon atoms
6.02 x 1023 Carbon atoms
(6/12 ) x 6.02 x 1023
= 3.01 x 1023 Carbon atoms
The Mole and Avogadro’s constant
Worked example 2.
How many molecules are in 6g of water?
1 Mole of water
 Avogadro’s constant of molecules
18 g
 6.02 x 1023 molecules
6g

(6/18) x 6.02 x 1023
= 2 x 1023
molecules
The Mole and Avogadro’s constant
Worked example 3.
Calculate the number of atoms in 4 g of bromine.
1 mole

6.02 x 1023 Br2 molecules
160g

(2 x 6.02 x 1023 ) Br atoms
4g

(4/160 ) x 1.204 x 1024
= 3.01 x 1022 bromine atoms
The Mole and Avogadro’s constant
Worked example 4.
Calculate the number of sodium ions in 71g of sodium sulphate
(Na+)2SO421 mole

6.02 x 1023 Formula units
142g

(2 x 6.02 x 1023 ) Na+ ions
71 g

(71/142 ) x 1.204 x 1024
= 6.02 x 1023 sodium ions
Calculations for you to try.
1.
How many atoms are there in 0.01 g of carbon?
2. How many oxygen atoms are there in 2.2 g of carbon dioxide?
3. Calculate the number of sodium ions in 1.00g of sodium carbonate.
4. Calculate the number of molecules in 25.5g of ammonia.
5.
A sample of the gas dinitrogen tetroxide, N2O4, contained 2.408
x 1022 oxygen atoms. What mass of dinitrogen tetroxide was
present?
1.
How many atoms are there in 0.01 g of carbon?
5.02 x 1020 C atoms
2. How many oxygen atoms are there in 2.2 g of carbon dioxide?
6.02 x 1022
O atoms
3. Calculate the number of sodium ions in 1.00g of sodium carbonate.
1.14 x 1021 Na+ ions
4. Calculate the number of molecules in 25.5g of ammonia.
9.03 x 1023 molecules
5. A sample of the gas dinitrogen tetroxide, N2O4, contained 2.408
x 1022 oxygen atoms. What mass of dinitrogen tetroxide was
present?
0.92 g
1. The mass of 1 mol of sodium is 23 g.
What is the approximate mass of one sodium atom?
A 6 × 1023g
B 6 × 10–23g
C 3.8 × 10–23g
D 3.8 × 10–24g
C
2. In which of the following pairs do the gases
contain the same number of oxygen atoms?
A 1 mol of oxygen and 1 mol of carbon monoxide
B 1 mol of oxygen and 0.5 mol of carbon dioxide
C 0.5 mol of oxygen and 1 mol of carbon dioxide
D 1 mol of oxygen and 1 mol of carbon dioxide
D
3. The Avogadro Constant is the same as the number of
A molecules in 16 g of oxygen
B electrons in 1 g of hydrogen
C atoms in 24 g of carbon
D ions in 1 litre of sodium chloride solution,concentration 1
mol l–1.
B
Mole and gas volume
The molar volume of a gas is its volume per mole, litre mol-1.
It is the same for all gases at the same temperature and pressure.
The value, though, is temperature and pressure dependent.
The molar volume of all gases is approximately
24 litre mol-1 at 20oC and 22.4 litre mol-1 at 0oC.
Volume (l)
Molar
Volume (l)
x
n
Measuring Molar gas volume
Finding out the molar volume by experiment.
Method 1
Density = mass/volume
Or mass = density x volume
Higher Chemistry Eric Alan and John Harris
1. Weight the flask with air in it. Use the density of air and the volume of the flask to
calculate the mass of air in the flask and hence the weight of the empty flask.
Mass of flask + air =
Mass of air = density x volume = 1.29 x 10-3 x ........................... =
Mass of empty flask =
2. The flask is filled with the gas and reweighed. Calculate the mass of gas.
g
g
g
Measuring Molar gas volume
O2 Oxygen
Mass of empty flask (g)
Mass of flask + gas (g)
Mass of gas (g)
Volume of flask i.e. volume
of gas (l)
Mass of one mole (g)
Volume of 1 mole
N2 Nitrogen
CO2 Carbon dioxide
Higher Chemistry Eric Alan and John Harris
3. The volume of the flask can be measured by filling with water and emptying into a measuring
cylinder.
Example calculation
You have found that ...................... g of O2 occupies .................... Litres
.........................g ↔ ..................... Litres
32g ↔ .
Litres
Therefore 1 mole, 32g of O2 will occupy ........................ Litres
Measuring Molar gas volume
Finding out the molar volume by experiment.
Method 2
Measuring Molar gas volume
Finding out the molar volume by experiment.
Method 2
Calculate the molar gas volume
____ g magnesium has produced ____cm3 hydrogen
____ g ↔ ________ cm3
24 g
↔ _________
The volume of one mole of H2 gas is ________ Litres
• http://media.rsc.org/Classic%20Chem%20e
xperiments/CCE-68.pdf
Why?
It may seem surprising that the
molar volume is the same for all
gases, even at the same temperature
and pressure. In a gas the molecules
have much more kinetic energy and
are relatively far apart so the volume
of the gas does not depend on the
sizes of the particles.
In a gas the at room temperature and
pressure the molecules only occupy
about 0.1% of the volume of a gas.
The rest is empty space!
Calculations involving molar gas volume
At room temperature and pressure the molar gas volume is 24 litres mol-1.
Worked example 1. Calculate the volume of 0.025 moles of oxygen.
1 mole

0.025

24 litres
0.025/
1
x 24 = 0.6 litres
Worked example 2. Calculate the number of moles in 72 litres of
hydrogen.
1 mole
72/
24
x1
= 3 moles

24 litres

72 litres
Calculations involving molar gas volume
What is the mass of steam in 180 cm3 of the gas, when the molar
volume is 24 litres mol-1?
0.135 g
Calculations involving mass and volume
Worked example 1.The equation below shows the reaction between
calcium carbonate and hydrochloric acid.
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g)
+ H2O(l)
20g of calcium carbonate reacts with excess hydrochloric acid.
Calculate the volume of carbon dioxide gas formed. (Take the molar
volume to be 23.0 litre mol-1)
Write the
balanced equation
Show mole ratio
Change moles into
required units
Use
proportion
CaCO3(s) + 2HCl(aq)  CaCl2 (aq) + CO2(g)
+ H2O(l)
1 mol

1mol
100 g

23.0 litres
20 g

20/
100
x 23.0 litre
4.6 litres
Calculations involving density and molar volume
The molar volume is the volume occupied by one mole of a gas.
Worked example 1. In an experiment the density of carbon dioxide was
measured and found to be 1.85 g l-1.
Calculate the molar volume of carbon dioxide.
So 1 mole,
1.85 g

44 g

1 litre
44/
1.85
x 1 = 23.78 litres
Worked example 2. A gas has a molar volume of 24 litres and a
density of 1.25 g l -1.
Calculate the mass of 1 mole of the gas.
So 1 mole,
1 litre

24 litre

1.25g
24/
1
x 1.25 = 30 g
Calculations for you to try.
1.
Calculate the number of moles in 0.36 litres of argon (molar gas
volume = 24 litres mol-1 ).
0.015moles
2. Calculate the volume of 0.04 moles of CO2. (molar gas volume = 24
litres mol-1 ).
0.96L
3. Under certain conditions oxygen has a density of 1.44 g l-1. Calculate
the molar volume of oxygen under these conditions.
22.22 litres
4. A gas has a density of 2.74 g l-1 and a molar volume of 23.4 litre mol-1.
Calculate the molecular mass of the gas.
64.1 g
Gas Volume Calculations
Calculate the volume of nitrogen dioxide gas produced when 100cm3 of
nitrogen is sparked in excess oxygen
N2
(g)
+ 2O2
1mole
1vol
100cm3
(g)
 2NO2
(g)
2moles
2vol
200cm3
200cm3 of NO2
Gas Volume Calculations
Combustion of methane
What volume of C02, is produced if 100 cm3 of O2 is used to
completely to burn some CH4 gas?
Balanced equation
CH4 (g) + 2O2
(g)
 CO2 (g) + 2 H20 (l)
50cm3 of CO2
Gas Volume Calculations
50 cm3 of C0, is burned with 20cm3 of oxygen.
a) Which gas is in excess
b) What is the volume and composition of the resulting gas
mixture
Balanced equation
CO (g) + ½ O2
(g)
 CO2 (g)
CO in excess
10cm3 CO + 40cm3 CO2 = 50cm3 of gas
Gas Volume Calculations
Combustion of propane
10 cm3 of propane gas is mixed with 75cm3 of O2 and the mixture
exploded.
a) Calculate the volume and the composition of the resulting gas
mixture.
Balanced equation
C3H8 (g) + 5O2
(g)
 3CO2 (g) + 4H20
(l)
25cm3 of O2 + 30cm3 of CO2 = 55cm3 of gas
b) What will be the change in volume when the resulting gas
mixture is shaken with dilute sodium hydroxide solution.
Volume will decrease by 30cm3 (CO2) removed
c) Calculations concerning
reactions which involve solutions
Learning intention
Learn how to calculate quantities
of reactant or product for reactions
involving solutions, using the
concentration.
Calculations involving concentration
n=CxV
1000
n=number of moles
C=concentration mol l-1
V=volume in Litres
Standard solutions
Standard solutions are solutions of
known concentrations. They are made
in volumetric glassware and can be
made from a solid or from another
solution.
Making a standard solution
What mass of solid should be dissolved in 100cm3
of water to prepare a 0.1 mol l-1 solution of
copper(II) sulphate?
n=CxV
1000
n = 0.1 x 100
1000
n= 0.01 moles
1 mole

0.01 moles

123.5 g
=(0.01/1)x123.5
=1.235g
Making a standard solution
weigh out the solid.
Making a standard solution
Next, dissolve the solid in a
small amount of solvent.
Making a standard solution
Finally, fill the solvent to the
line.
Aqueous solutions
• Most solutions are aqueous i.e. the
solvent is water
• When preparing aqueous solutions
deionised water (or distilled water) is
added why?
Calculations involving concentration
How many moles are in 100cm3 of sodium hydroxide
concentration 0.1 mol l-1?
0.01 moles
Neutralisation Reactions/titrations
What volume hydrochloric acid 1.0 mol l-1 is needed to neutralise
50 cm3 of potassium hydroxide solution concentration 0.25 mol l-1?
12.5cm3
Calculations for you to try.
1.
2.
Calculate the concentration of potassium
hydroxide (KOH ) if 14.8 cm3 is required to
neutralise 20cm3 of 0.1 mol/l nitric acid (HNO3).
0.135 mol/l
Calculate the volume of 0.15 mol/l sulphuric
acid(H2SO4) if it is neutralised by 25 cm3 of
0.25mol/l sodium hydroxide (NaOH).
20.83 cm3
d) Excess
Learning intention
Learn how to calculate how much
of a particular reactant is in
excess
from
the
balanced
equation.
Excess
You can use the relative numbers of moles of substances, as shown in
balanced equations, to calculate the amounts of reactants needed or
the amounts of products produced.
A limiting reactant is the substance that is fully used up and
thereby limits the possible extent of the reaction. Other reactants
are said to be in excess.
Calculations involving excess
As soon as one of the reactant in a chemical reaction is used up the
reaction stops. Any other reactant which is left over is said to be ‘in
excess’. The reactant which is used up determines the mass of product
formed.
You can use the relative numbers of moles of substances, as shown in
balanced equations, to calculate the amounts of reactants needed or
the amounts of products produced.
A limiting reactant is the substance that is fully used up and
thereby limits the possible extent of the reaction. Other reactants
are said to be in excess.
Calculations involving excess
Worked example. Which reactant is in excess when 10g of calcium
carbonate reacts with 100cm3 of 1 mol l-1 hydrochloric acid?
CaCO3 is in excess.
Calculations involving excess
Worked example 2. 1.2g of magnesium was added to 100cm3 2 mol
l-1 hydrochloric acid. Calculate the reagent in excess?
Acid in excess.
Graphs and Rates of Reaction
e.g. Zn + 2HCl  ZnCl2 + H2
Zn in excess
2mol l-1 HCl 20oC
2mol l-1 HCl 40oC
Faster, but same amount of
gas produced
Vol
H2
cm 3
1 mol l-1 HCl 20oC
Half the gas produced
time /s
HCl is limiting reagent
Calculations for you to try.
1.
The graph below was obtained when 1.0g of powdered zinc was
added to excess hydrochloric acid 1.0 mol l-1, copy the graph and
sketch a line to show what you would expect if the reaction was
repeated using
a) 2.0 mol l-1 HCl and 1.0g Zn
b) 1.0 mol l-1 HCl and 0.75g Zn
Vol
H2
cm 3
time /s
2. a) Calculate which reagent is in excess when 3.27g of zinc is reacted
with 100cm3 of 2.0 mol l-1 hydrochloric acid.
b) What mass of hydrogen gas will be produced?
0.1 g
Calculations involving excess
Examples. For each of the following reactions calculate which
reagent is in excess?
a) 4.86g magnesium added to 250cm3 2 mol l-1 hydrochloric acid
HCl
b) 2.7g aluminium added to 200cm3 1 mol l-1 hydrochloric acid
Al
c) 2.43g magnesium added to 200cm3 1 mol l-1 sulphuric acid
H2SO4
d) 3.27g zinc added to 100cm3 0.2 mol l-1 hydrochloric acid.
Zn
Calculations involving excess and molar gas volume
An experiment was carried out to measure the concentration
of hypochlotite ions (ClO-) in a sample of bleach. In this
experiment the bleach sample is reacted with excess hydrogen
peroxide.
H2O2(aq) + ClO-(aq) H2O(l) + Cl-(aq) + O2(g)
By measuring the volume of oxygen given off, the
concentration of the bleach can be calculated.
80cm3 of oxygen was produced from 5.0cm3 of bleach.
Calculate the concentration of the hypochlorite ions in the
bleach (Take the molar gas volume to be 24 Litres mol-1).
Calculations for you to try.
1. What mass of calcium oxide is formed when 0.4 g of calcium reacts
with 0.05 mole of oxygen?
2Ca +
O2

2CaO
0.56g
2. What mass of hydrogen is formed when 3.27g of zinc is reacted with
25cm3 of 2 mol l-1 hydrochloric acid?
Zn
+ 2HCl

ZnCl2
+ H2
0.05 g
e) Reversible reactions
Learning intention
Learn how reversible reactions
may reach equilibrium instead of
completely converting reactants to
products. Find out how dynamic
equilibrium is defined in terms of
reaction rates and concentrations
of reactant and product.
Dynamic equilibrium
Reactants
Products
Chemical reactions which take place in both directions are
called reversible reactions.
The following is an example of a reversible reaction hydrogen and iodine reacting to form hydrogen iodide.
H2(g) + I2(g)⇄ 2HI(g)
The equilibrium can be arrived at from different starting
points.
The position of an equilibrium does not depend on the
starting position.
Dynamic equilibrium
A reversible reaction attains a state of equilibrium
when the rate of the forward reaction is equal to the
rate of the reverse reaction.
In the equilibrium mixture both the forward and
reverse reactions are taking place.
Graph 1
Reaction
rate
Forward reaction
equilibrium
Backward reaction
Time
Dynamic equilibrium
Since the rates of the forward and reverse reactions
are equal, the concentration of reactants and products
remain constant, though not necessarily equal.
The system is said to be at dynamic equilibrium.
Graph 2
Concentration
Forward (reactants)
Backward (products)
Time
Position of equilibrium
At equilibrium, the concentration of
the products and the reactants will
remain constant
The concentration of reactants will
probably not equal the concentration
of the products.
Position of equilibrium
concentration
products
reactants
time
Equilibrium
At equilibrium the concentration of products
and reactants remains the same.
Equilibrium
A reversible reaction can reach
equilibrium in a closed system.
N2
+ 3H2
⇄
2NH3
A reaction reaches equilibrium when
the rate of the forward reaction
equals the rate of the reverse
reaction.
Direction
• The equilibrium position will be the same
whether we start with only the products or
only the reactants
• Iodine dissolves in both cyclohexane and
water/KI.
• The experiment shows one boiling tube set
up with 100% iodine in cyclohexane and one
with 100% iodine in water/KI.
Iodine equilibrium
A
B
cyclohexane
Iodine/KI solution
Iodine/cyclohexane
KI solution
Final equilibrium is the same
• A
B
f) Altering Equilibrium Position
Learning intention
Learn how chemists alter the
position of equilibrium to increase
product yield, by changing factors
such as concentration, pressure,
temperature and also how the use
of a catalyst can ensure
equilibrium is reached more
quickly.
Le Chatelier's Principle
“For a system in equilibrium, alteration of one of the
factors (pressure, temperature or concentration) will
cause the position of equilibrium to shift to reduce the
effects of the imposed conditions”.
Le Chateliers Principle
An equilibrium will move to undo any
change imposed upon it.
If the forward reaction is favoured we say
the equilibrium has moved to the right.
If the reverse reaction is favoured we say
the equilibrium has moved to the left
Shifting the equilibrium position
The proportion of products to reactants in an equilibrium mixture is
described as the equilibrium position.
A+B⇄C+D
If the conversion of A and B into C and D is small the position of
equilibrium lies to the left, or to the side of the reactants.
If the equilibrium mixture is largely composed of C and D, the
position of equilibrium lies to the right, or to the side of the
products.
Temperature may alter the
position of equilibrium
Temperature may alter the
position of equilibrium
Heating a reversible reaction at equilibrium shifts the
reaction in the direction of the ENDOTHERMIC
REACTION
Cooling a reversible reaction at equilibrium shifts the
reaction in the direction of the EXOTHERMIC
REACTION
The equilibrium will move to undo any change imposed
upon it.
Exothermic reaction
A + B ⇄ C + D + Energy -ΔH
Forward reaction is exothermic
Increasing temperature shifts equilibrium to the left.
Conc product
Low temp
faster
High temp - favours reactants
so less product formed.
Endothermic reaction
Energy + A + B ⇄ C + D +ΔH
Forward reaction is endothermic
Increasing temperature shifts equilibrium to the right.
Conc product
High temp- favours products so
more product formed.
Low temp
faster
Temperature may alter the position of equilibrium
N 2 O4
dinitrogen tetraoxide
(colourless)
⇄
2NO2 Δ +ve
nitrogen dioxide
(dark brown)
Higher Chemistry Eric Alan and John Harris
Temperature may alter the position of equilibrium
N2O4
dinitrogen tetraoxide
(colourless)
⇄
2NO2 Δ +ve
nitrogen dioxide
(dark brown)
Increasing the temperature will cause the equilibrium to move to
lower the temperature. The forward reaction takes in energy so
the equilibrium moves to the right producing more NO2 and less
N2O4. So the colour becomes darker.
Decreasing the temperature will cause the equilibrium to move to
raise the temperature. The reverse reaction gives out energy so
the equilibrium moves to the left producing more N2O4 and less
NO2. So the colour becomes lighter.
Temperature may alter the position of equilibrium
A mixture of cobalt chloride and conc HCl sets up the following equilibrium:
Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve
Heat the equilibrium mixture
turns blue
Cool the equilibrium mixture
turns pink
SSERC
Temperature may alter the position of equilibrium
A mixture of cobalt chloride and conc HCl sets up the following equilibrium:
Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve
If the temperature is increased the equilibrium will favour
the forward reaction because that will lower the
temperature. The equilibrium move to the right, therefore
the solution becomes blue in colour.
If the temperature is reduced the equilibrium will favour
the reverse, exothermic reaction because that will
increase the temperature. The equilibrium move to the
left, therefore the solution becomes pinker in colour.
Concentration may alter the
position of equilibrium
Concentration may alter the
position of equilibrium
Consider the following reaction at equilibrium
A+B⇄C+D
An increase in concentration of A or B will speed up the
forward reaction, thus increasing the concentration of
C and D.
A similar effect can be achieved by reducing the
concentration of C or D.
Concentration may alter the position of equilibrium
A mixture of cobalt chloride and conc HCl sets up the following equilibrium:
Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve
Adding extra Cl- ions forces the equilibrium to try to remove
these. The forward reaction is favoured because this uses up
chloride ions.
The equilibrium has moved to the right so the solution becomes
blue in colour.
Concentration may alter the position of equilibrium
Add 10 cm3 iron (III) chloride to a test tube. Iron (III) ions are yellow. Add potassium
thiocyanate solution until the solution goes orange. Red coloured iron thiocyanate ions
form. The equilibrium position now lies in the middle, roughly equal amounts of both
coloured ions are present.
Fe 3+
yellow
+ CNS -
⇄
[FeCNS]2+
red
Concentration may alter the position of equilibrium
Fe 3+
yellow
Higher Chemistry Eric Alan and John Harris
+ CNS -
⇄
[FeCNS]2+
red
In B, the Fe 3+ ions are added. The
equilibrium moves to use them up,
favouring the forward reaction.
The equilibrium moves right and the solution
becomes more red
Concentration may alter the position of equilibrium
Fe 3+
yellow
Higher Chemistry Eric Alan and John Harris
+ CNS -
⇄
[FeCNS]2+
red
In C, CNS- ions are added. The
equilibrium moves to remove these,
favouring the forward reaction.
The equilibrium moves right and the
solution becomes more red
Concentration may alter the position of equilibrium
Fe 3+
yellow
Higher Chemistry Eric Alan and John Harris
+ CNS -
⇄
[FeCNS]2+
red
In D, the Fe 3+ ions are removed. The
equilibrium moves to replace this,
favouring the reverse reaction.
The equilibrium moves left and the solution
becomes more yellow
Concentration may alter the position of equilibrium
ICl + Cl2 ⇄
brown liquid
ICl3
yellow solid
Concentration may alter the position of equilibrium
ICl + Cl2 ⇄
brown liquid
ICl3
yellow solid
Increasing the concentration of a chemical will cause the
equilibrium to move to use up the chemical.
Increasing the concentration of chlorine will cause the equilibrium
to move to use up the chlorine. The forward reaction uses up the
chlorine so the equilibrium moves to the right producing more
yellow solid and less brown liquid.
Decreasing the concentration of a chemical will cause the
equilibrium to move to form the chemical.
Decreasing the concentration of chlorine will cause the equilibrium
to move to form the chlorine. The reverse reaction produces
chlorine so the equilibrium moves to the left producing more brown
liquid and less yellow solid.
Pressure may alter the
position of equilibrium
Pressure may alter the
position of an equilibrium
The pressure exerted by a gas is caused by the freely
moving molecules bombarding the walls of the
container.
An increase in the number of molecules will result in an
increase in pressure, assuming the size of the container
is kept constant.
An increase in pressure will cause the equilibrium to
counteract this effect i.e. it will reduce the pressure.
By favouring the side with less gas molecules.
N2O2 (g) ⇄ 2NO2 (g)
colourless
1mole
Higher Chemistry Eric Alan and John Harris
brown
2moles
Pressure may alter the
position of an equilibrium
N2O4
⇄
colourless
1 mole, so fewer particles
Lowers the pressure
2NO2
brown
2 moles
Increasing the pressure will cause the equilibrium to
move to decrease the pressure. The equilibrium
will move to reduce the number of gas particles.
The equilibrium moves to the left producing more
N2O4 and less NO2 so the colour lightens.
Pressure may alter the
position of an equilibrium
N2O4
⇄
colourless
1 mole, so fewer particles
2NO2
brown
2 moles
increase the pressure
Decreasing the pressure will cause the equilibrium to
move to increase the pressure. The equilibrium will
move to increase the number of gas particles.
The equilibrium moves to the right producing more
NO2 and less N2O4 so the colour darkens.
Soda water
• Fizzy drinks contain carbon dioxide in solution
under pressure.
• In this experiment, methyl red indicator is used.
• It is red below pH 4.2 and yellow above pH 6.3.
• Boil some soda water containing a little methyl red.
This will expel the carbon dioxide, which is less
soluble at high temperatures. What happens to the
carbon dioxide and what pH change is seen?
Explanation
• Heating the soda water expels the
carbon dioxide, which is less soluble
at high temperatures,
• The colour change of the indicator
from red to yellow shows the pH
rising as the carbon dioxide, which
makes an acidic solution, is removed.
Soda water contains carbon dioxide
that has been dissolved in it under
pressure
• The equilibria involved are:
• CO2(g)
⇄
CO2(aq)
• CO2(aq) + H2O(l) ⇄ H2CO3(aq)
(carbonic acid)
(1)
(2)
• H2CO3(aq) ⇄ H+(aq) + HCO3(aq)
(3)
(hydrogencarbonate ions)
• HCO3(aq) ⇄ H+(aq) + CO3 (aq)
carbonate ions
(4)
Effect of pressure
• Pour a few cm3 of soda water into the beaker and
add a few drops of methyl red
• Draw about 4 cm3 of this solution into the syringe.
• Use a finger to cover the nozzle, pull the plunger
out to the 20 cm3 mark and lock it with the nail.
Shake the syringe
Effect of pressure
• Hold the syringe vertically with the nozzle
pointing upwards, remove your finger and
the nail, and push in the plunger to expel
the gas but not the solution.
• Stopper the syringe again and repeat the
above cycle. More CO2 bubbles will be seen
and the indicator will turn more towards a
yellow colour.
• Several more cycles can be repeated until
the indicator becomes yellow.
Observation
• What is happening to the gas? Why?
• What is happening to the pH?
• Why?
Temperature and energy
Observation
• What is happening to the gas?
• If the pressure is reduced, the
equilibrium in (1) tries to increase the
pressure, so it moves to the left
producing more gas. Gas is seen
coming out of solution
• What is happening to the pH? It
increases
• Why?
Observation
• Why?
• If the equilibrium in (1) moves left, it
decreases the amount of CO2(aq) in
(2) which moves left.
• This in turn reduces the reactants in
(3) and (4), which move to the left,
reducing the concentration of H+(aq)
ions, thereby increasing the pH
• http://media.rsc.org/Classic%20Chem%20D
emos/CCD-39.pdf
Effect of chemicals not present
in the equilibrium
Dissolving chlorine in water produces the
hypochlorite ion, ClO-, which has a bleaching
effect.
Cl2 +
H 2O
⇄
2H +
+ ClO-
+ Cl-
Effect of neutralisation
Cl2 +
H2O
⇄
2H +
+ ClO-
+ Cl-
Adding an alkali will remove hydrogen
ions from the equilibrium which will
move to the right to replace them
The bleaching effect will be increased.
Effect of precipitation
Cl2 +
H2O
⇄
2H +
+ ClO-
+ Cl-
Adding silver nitrate will remove chloride
ions from the equilibrium as the precipitate
silver nitrate is formed.
The equilibrium will move to the right to
replace them so the bleaching effect will
be increased.
Effect of acid
Cl2 +
H2O
⇄
2H +
+ ClO-
+ Cl-
Adding an acid causes the equilibrium to
move to use up H+ ions. The equilibrium
moves to the left producing more toxic Cl2
This can be fatal and accidents caused by
mixing bleach and acid are not unusual.
Catalysts
• Catalysts increase the rate at which
an equilibrium is formed but do not
effect the equilibrium position.
• The rate of the forward AND
reverse reactions are speeded up
equally.
7 principles of design
process
•
•
•
•
•
•
•
Availability of feedstocks
Cost of the feedstock
Sustainability of the feedstock
Opportunities for recycling
Energy requirements
Marketability of by-products
Product yield (or atom economy)
The Haber Process
• How does the Haber process illustrate the
7 principles of design for an industrial
process?
• How is the equilibrium manipulated to move
as far to the right as possible? Give 3
ways.
• How is the effect of temperature on rate
and on yield balanced?
Haber process
The Haber Process combines nitrogen from the air with hydrogen
derived mainly from natural gas (methane) into ammonia. The reaction is
reversible and the production of ammonia is exothermic.
http://www.rmtech.net/uses_
of_ammonia.htm
The Haber Process and equilibrium
• N2 + 3H2 ⇄
2 NH3
- ΔH
• In the reactor, ammonia is removed
after each cycle of the gases through
the reactor. What effect would that
have on the equilibrium position?
The Haber Process and equilibrium
• N2 + 3H2 ⇄
2 NH3
- ΔH
• Increasing the temperature would speed
up the rate of reaction, but what would
happen to the amount of ammonia
produced in the equilibrium?
• The reacting gases are pumped around
the reactor at a suitable ‘flow rate’
The Haber Process and equilibrium
You need to shift the position of the equilibrium as far as
possible to the right in order to produce the maximum
possible amount of ammonia in the equilibrium mixture.
The forward reaction (the production of ammonia) is
exothermic.
According to Le Chatelier's Principle, this will be favoured
if you lower the temperature. The system will respond by
moving the position of equilibrium to counteract this - in
other words by producing more heat.
In order to get as much ammonia as possible in the
equilibrium mixture, you need as low a temperature as
possible. However, 400 - 450°C isn't a low temperature!
The Haber Process and equilibrium
Tonnes of
ammonia
200 oC
400 oC
time
If the flow rate is adjusted so that the gases spend
x time in the reactor, more ammonia is produced per
day at the higher temperature, despite the lower
yield at equilibrium
The Haber Process and equilibrium
The lower the temperature you use, the slower the reaction
becomes. A manufacturer is trying to produce as much
ammonia as possible per day. It makes no sense to try to
achieve an equilibrium mixture which contains a very high
proportion of ammonia if it takes several years for the
reaction to reach that equilibrium.
You need the gases to reach equilibrium within the very short
time that they will be in contact with the catalyst in the
reactor.
The compromise
400 - 450°C is a compromise temperature producing a
reasonably high proportion of ammonia in the equilibrium
mixture (even if it is only 15%), but in a very short time.
The Haber Process and equilibrium
• N2 + 3H2
⇄
2 NH3
- ΔH
• What would be the effect of increasing
the pressure?
• What might be the drawbacks, both
technical and economic of a high
pressure?
The Haber Process and equilibrium
N2 + 3H2
⇄
2 NH3
- ΔH
Notice that there are 4 molecules on the left-hand side of
the equation, but only 2 on the right.
According to Le Chatelier's Principle, if you increase the
pressure the system will respond by favouring the reaction
which produces fewer molecules. That will cause the
pressure to fall again.
In order to get as much ammonia as possible in the
equilibrium mixture, you need as high a pressure as possible.
200 atmospheres is a high pressure, but not amazingly high.
The Haber Process and equilibrium
Increasing the pressure brings the molecules
closer together. In this particular instance, it will
increase their chances of hitting and sticking to
the surface of the catalyst where they can react.
The higher the pressure the better in terms of
the rate of a gas reaction.
The Haber Process and equilibrium
Very high pressures are very expensive to produce on two counts.
You have to build extremely strong pipes and containment vessels
to withstand the very high pressure. That increases your capital
costs when the plant is built.
High pressures cost a lot to produce and maintain. That means that
the running costs of your plant are very high.
The Haber Process and equilibrium
The compromise
200 atmospheres is a compromise pressure chosen on economic
grounds. If the pressure used is too high, the cost of generating it
exceeds the price you can get for the extra ammonia produced.
The Haber Process
• The catalyst is iron, which is cheap.
• What is the effect a catalyst on a
reversible reaction?
The Haber Process and equilibrium
In the absence of a catalyst the reaction is
so slow that virtually no reaction happens in
any sensible time. The catalyst ensures
that the reaction is fast enough for a
dynamic equilibrium to be set up within the
very short time that the gases are actually
in the reactor.
.
The Haber Process and equilibrium
The Haber Process and equilibrium
When the gases leave the reactor they are
hot and at a very high pressure. Ammonia is
easily liquefied under pressure as long as it
isn't too hot, and so the temperature of
the mixture is lowered enough for the
ammonia to turn to a liquid. The nitrogen
and hydrogen remain as gases even under
these high pressures, and can be recycled
The Haber Process and equilibrium
• http://www.freezeray.com/flashFiles/theHa
berProcess.htm
In which of the following reactions would an
increase in pressure cause the equilibrium
position to move to the left?
A CO(g) + H2O(g) →CO2(g) + H2(g)
B CH4(g) + H2O(g) → CO(g) + 3H2(g)
C Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
D N2(g) + 3H2(g) → 2NH3(g)
B
If ammonia is added to a solution containing
copper(II) ions an equilibrium is set up.
Cu2+(aq) + 2OH–(aq) + 4NH3(aq) → Cu(NH3)4(OH)2(aq)
(deep blue)
If acid is added to this equilibrium system
A the intensity of the deep blue colour will increase
B the equilibrium position will move to the right
C the concentration of Cu2+(aq) ions will increase
D the equilibrium position will not be affected.
C
D
g) Percentage Yield and Atom
Economy
Learning intention
Learn how the efficiency of a
chemical
reaction
can
be
measured in terms of percentage
yield and atom economy.
Making only as much as we need
PERCENTAGE YIELD
and ATOM ECONOMY
Most of the substances we use every day are made
from RAW MATERIALS, often through complex
chemical reactions.
Reactants
(raw
materials)
Chemical reactions
Products
Pottington
Braunton
The chemical industry is a multi billion pound international industry
producing millions of products vital to our civilisation and well being.
Chemical Engineers play a crucial role and are much in demand –
there are many opportunities and high levels of pay!
Chemical Engineers are much concerned with:
% YIELD and ATOM ECONOMY..
What is
green chemistry?
• The sustainable design of chemical products and
chemical processes.
• It minimises the use and generation of chemical
substances that are hazardous to human health or the
environment.
Green chemistry
principles
• Better to prevent waste than to treat it or clean it up.
• Chemical processes should aim to incorporate all
reactants in the final product.
• Chemical processes should aim to use and generate
substances with minimal toxicity to human health and
the environment.
The green
chemical industry
• Modern chemists design reactions with the highest
possible atom economy in order to minimise environmental
impact.
• Chemists achieve this by reducing raw material and
energy consumption.
Percentage yield
Actual yield
% yield =
x 100
Theoretical yield
• Historical method for evaluating reaction
efficiency.
• Measures the proportion of the desired product
obtained compared to the theoretical maximum.
• Gives no indication of the quantity of waste
produced.
% YIELD is the amount of product you actually make
as a % of the amount you should theoretically make
REACTANTS
PRODUCT
%
YIELD
ABOUT
75%
+
SHOULD
make this
much
ACTUALLY
make this
much
Old fashioned example: Cement from limestone
Limekiln
PERCENTAGE YIELD 1
LIMESTONE (calcium carbonate) is used to make
QUICKLIME (calcium oxide) for cement making
CaCO3
RFM:
100
CaO
56
+
CO2
RAM
Ca 40
O 16
C 12
44
So, THEORETICALLY, 100 tonnes of limestone should
produce 56 tonnes of quicklime.
Why? –
next
BUT the ACTUAL YIELD is only 48 tonnes
slide
So..the PERCENTAGE YIELD is only 48 x 100 = 87.5%
56
PERCENTAGE YIELD 2
Very few chemical reactions have a yield of
100% because:
• The raw materials (eg limestone) may not be pure
• Some of the products may be left behind in the
apparatus
• The reaction may not have completely finished
• Some reactants may give some unexpected products
•Some reactions involve an equilibrium
Careful planning and
design of the equipment
and reaction conditions
can help keep % yield high
Atom economy
• In an ideal reaction, all reactant atoms end up
within the useful product molecule. No waste is
produced!
• Inefficient, wasteful reactions have low atom
economy.
• Efficient processes have high atom economy and
are important for sustainable development. They
conserve natural resources and create less waste.
Atom economy
Mass of desired product(s)
% atom economy =
x 100
Total mass of reactants
• A measure of the proportion of reactant included in
the final useful product.
• A reaction may have a high percentage yield but a
low percentage atom economy, or vice versa.
High atom economy
All reactant atoms included in the desired
product.
Low atom economy
Some reactant atoms not included in the
desired product.
Atom Economy
Compare these two industrial reactions
2Mg
+
O2
→
2MgO
What do you notice about
each one?
USEFUL PRODUCT
(antacids, fire resistant
coatings, electrical
insulators)
Think raw materials, useful
products, waste products
CaCO3
→
USEFUL PRODUCT
(cement, glass,
agriculture etc)
CaO
+
CO2
ATOM ECONOMY is the mass of the product you
want as a % of the mass of all the products you make
REACTANTS
PRODUCTS
+
Stuff you
want
Stuff you
also get
but don’t
want
ATOM
ECONOMY
about 50%
CALCULATING ATOM ECONOMY
Often, chemical reactions produce unwanted products along
with the product you want.
ATOM ECONOMY is the mass of product you want as a
% of the mass of all the reactants
Waste product
Useful product
CaCO3
RFM:
100
ATOM
ECONOMY
→ CaO
56
+
CO2
44
= mass useful product X 100%
mass of all reactants
Atom Economy = 56 / (56 + 44) = 56 / 100 = 56 %
ATOM ECONOMY is the mass of product you want as a
% of the mass of all the reactants
2Mg + O2
RFM: 48
→
32
2MgO
80
Atom Economy = 80 / 80 x 100% = 100 %
Fe2O3 + 3CO
160
RAM
Mg 24
Fe 56
C 12
O 16
84
(obviously)
→ 2Fe + 3CO2
112
132
Atom Economy = 112 / 244 x 100% = 45.9 %
Find the atom economy for these 2 methods of extracting
copper:
2. Heat copper
1. Heat copper
sulphide with
oxide with carbon
oxygen
RAM Cu 64,
O 16,
2CuO + C → 2Cu + CO2
2(80)
12
Mass of
copper =
128
128
44
Mass of all
reactants
160 + 12 = 172
ATOM
128 X 100
=
ECONOMY
172
= 74.4 %
C 12,
S 32
CuS + O2 → Cu + SO2
96
32
64
64
Mass of
Mass of all
copper = 64 reactants
= 96 + 32 = 128
ATOM
64 X 100
=
ECONOMY
128
= 50 %
Real example:
Paracetamol
The non-prescription analgesic
market (paracetamol, aspirin,
ibuprofen) is worth about £21
billion annually.
Maximising % yield and atom
economy in the reactions at left is
vital to save money and conserve
energy and resources
SUMMARY
% YIELD is the
amount of product you
actually make as a % of
the amount you should
theoretically make
ATOM ECONOMY is the
mass of the product you
want as a % of the mass of
all the products you make
%
YIELD
ABOUT
75%
SHOULD
make this
much
ACTUALLY
make this
much
Stuff you
also get but
don’t want
Stuff you
want
ATOM
ECONOMY
about 50%
Raw materials are scarce and expensive and so must
be carefully conserved. Also, chemical processes
need to produce as little waste as possible, minimise
costs, energy use and pollution.
CHEMICAL ENGINEERS must plan to maximise:
PERCENTAGE YIELD
by…
Using the most EFFICIENT
REACTION CONDITIONS
& APPARATUS
to…
Reduce energy use, costs
and conserve raw materials
ATOM ECONOMY
by…
Choosing the most
EFFICIENT REACTION
to make the product
to…
Reduce waste and pollution
Example 1
What is the percentage atom economy for the following reaction for
making hydrogen by reacting coal with steam?
C(s) +
2H2O(g) → CO2(g)
+ 2H2(g)
12 g
2(2 + 16) g
[12 + (2 × 16)] g
12 g
36 g
44 g
Total mass of reactants
= 12 + 36 = 48 g
2(2 × 1) g
4g
Mass of desired product
=4g
Example 1 (contd)
% atom economy = mass of desired product × 100
total mass of reactants
=
4 × 100
48
= 8.3%
This reaction route has a very low atom economy and
is an inefficient method of producing hydrogen.
Example 2
Calculate the percentage atom economy for the reaction below.
CH3
H3C
C
CH3
C6H12
Total mass of reactants
= [(6 × 12) + (12 × 1)]
= 84 g
acid
H3C
CH CH2
CH3
C
H3C
C6H12
Mass of desired product
= [(6 × 12) + (12 × 1)]
= 84 g
C
CH3
Example 2 (contd)
% atom economy = mass of desired product × 100
total mass of reactants
=
84 × 100
84
= 100%
This reaction route has a very high atom economy as
all reactant atoms are incorporated into the desired
product.
Example 3
Hydrazine (N2H4) is used for rocket fuel. Calculate the atom economy for
hydrazine production.
2NH3
+
NaOCl
N2H4
NH3
2 mol
NaOCl
1 mol
N2H4
1 mol
NaCl
1 mol
H2O
1 mol
34 g
74.5 g
32 g
58 g
18 g
+
Total mass of reactants
Mass of desired product
= 34 + 74.5 = 108.5 g
= 32 g
NaCl
+
H2O
Example 3 (contd)
% atom economy = mass of desired product × 100
total mass of reactants
=
32 × 100
108.5
= 30%
This reaction route has an atom economy of 30%. The
remaining 70% is waste product (NaCl and H2O).
Catalysts
• Have a crucial role in improving atom economy.
• Allow the development of new reactions requiring fewer starting
materials and producing fewer waste products.
• Can be recovered and re-used.
• Allow reactions to run at lower temperatures, cutting energy
requirements.
Methanol and atom economy
Methanol
• Methanol is a potential
fuel of the future
• It can be made from
methane which in turn
could be made by
anaerobic
fermentation of waste
organic material
Laptop with methanol fuel cell
green technology
Methanol
• It is added to other fuels, it is the fuel for
drag car racing and it is being widely used in
fuel cells to power vehicles as it has not been
easy to set up the infrastructure needed for
hydrogen based fuel cells
Carbon neutral?
• If methane is combusted in a limited supply of
oxygen, carbon monoxide forms
•
CH4 + ½ O2
→
CO + 2H2
• The products are then passed over a catalyst at
50-100 atmospheres and 250oC
• CO + 2H2
→
CH3 OH
Carbon neutral?
CO + 2H2
28
4
→
CH3 OH
32
Atom economy = mass of desired products x 100
Total mass of reactants
= 32
16 + 16
x 100 = 100%
Carbon neutral?
All the carbon in the methane is converted into
carbon in the methanol – it is said to be carbon
neutral
The carbon dioxide released when methanol burns
contains the same amount of carbon that was
reclaimed from the CO
The amount of carbon dioxide released when the
methanol burns is the same as the amount the
methane would have released if it was burned
completely. No additional carbon dioxide results
from the process
Carbon neutral?
• What were the conditions needed for
step 1? How were they achieved?
Limited supply of oxygen
• What were the conditions needed for
step 2? How would these have been
achieved?
High pressure and temperature – ie
energy costs (where will this energy
come from?
• In your opinion, how true is the claim
that the process is carbon neutral?
Green Chemistry
• Look at the 12 principles of Green
Chemistry
• How many of them apply to this
process? The catalysts are zinc
oxide, aluminium oxide or copper.
% yield
Percentage yield = actual yield/theoretical yield x 100%
Preparation of zinc sulphate
• Zinc sulphate is used as a paste,
mixed with zinc oxide to treat acne.
• Aim to produce zinc sulphate and
calculate the % yield
Preparation of zinc sulphate
• It is an astringent
– it closes the
pores of the skin
to keep out
bacteria
• This product
contains peroxide,
zinc sulphate, tea
tree oil and sage
Preparation of zinc sulphate
• It can also be used to prevent
sunburn!
What to do
• Add a weighing boat to the balance and zero it
• Add between 2.2 – 2.5 g zinc oxide
• Record the accurate mass
• Measure out 25cm3 1.0 mol l-1 H2SO4
• Heat the acid to about 50 oC
• Add the zinc oxide, stirring constantly
What to do next
• Cool the reaction mixture
• Weigh an evaporating dish and record the mass accurately
• Weigh a filter paper, fold the filter paper, place in a funnel and
filter the zinc sulphate solution into the dish
• Heat to nearly dry, then leave to fully evaporate and cool.
• Reweigh the dish and product
Calculation % Yield
Write the balanced equation
Which reactant is in excess and which one
limits the amount of product formed?
How much product is expected (using the
limiting reactant in the calculation)
Calculate the yield = actual mass
expected mass
x 100
Preparation of sodium
citrate % yield
Sodium citrate is a food additive
(E331) e.g. in pepsi
What to do
• Weigh a beaker accurately
• Add approx 2g citric acid to the beaker
and reweigh
• Add 25cm3 1.0 mol l-1 NaOH and stir until
all the crystals have dissolved
• Evaporate the beaker almost to dryness
then leave to dry completely
• Reweigh the beaker
For the calculation
3 mol NaOH reacts with 1 mol citric acid
Na
+ 3NaOH ↔
Na
Na
+ 3H2O
The calculation
1. 3NaOH + C6H8O6 → CH5O7Na3 + 3H2O
2. Work out the theoretical yield.
3. Use the actual yield from your results and
the theoretical yield to calculate the
percentage yield.
Chemical Energy
Overview
Study the role of enthalpy in
influencing the design of a
chemical reaction, and use Hess's
Law
to
calculate
reaction
enthalpies.
a) Enthalpy
Learning intention
Learn how industrial chemists
predict and control the quantity of
heat taken in/ given out during
chemical
reactions
via
an
understanding of chemical energy
(enthalpy).
Learn how to
calculate the enthalpy change for
a specific chemical reaction.
Enthalpy Chemical energy
Enthalpy (H) is a measure of the energy stored in a
chemical. In the previous sections we have seen potential
energy diagrams during exothermic and endothermic
reactions. It is important to recognise we cannot measure
the enthalphy of a chemical, however we can measure the
enthalphy change in a reaction.
Enthalpy changes and industrial processes
Seveso: Italian dioxin crisis
• On midday of July 10,
1976 an explosion
occurred in a TCP
(2,4,5-trichlorophenol)
reactor in the ICMESA
chemical company in
Meda, Italy.
Enthalpy changes and industrial processes
Seveso: Italian dioxin crisis
• A toxic cloud escaped into the atmosphere containing high
concentrations of TCDD, a highly toxic form of dioxin.
• Downwind from the factory the dioxin cloud polluted a densely
populated area of six kilometres long and one kilometre wide,
immediately killing many animals.
• A neighbouring municipality that was highly affected is called Seveso.
The accident was named after this village. The dioxin cloud affected a
total of 11 communities.
•
Read more: http://www.lenntech.com/environmentaldisasters.htm#3._Seveso:_Italys_dioxin_crisis#ixzz15SH4fvbR
Enthalpy changes and industrial processes
Bhopal Union carbide plant
Enthalpy changes and industrial processes
December 3,1984: Bhopal,India.
• Shortly after midnight, a poisonous gas cloud escaped from the Union
Carbide India Limited (UCIL) pesticide factory. The cloud contained
15 metric tons of methyl isocyanate (MIC), covering an area of more
than 30 square miles.
• The gas leak killed at least 4.000 local residents instantly and caused
health problems such as oedema for at least 50.000 to perhaps 500.000
people.
•
This event is now known as the worst industrial environmental
disaster to ever have occurred.
• Read more: http://www.lenntech.com/environmentaldisasters.htm#ixzz15SEf7zzG
Enthalpy changes and industrial processes
Thermite and ice ?
• http://www.mhhe.com/physsci/chemistry/es
sentialchemistry/flash/activa2.swf
b) Enthalpies of combustion
Learning intention
Learn the definition of enthalpy of
combustion, which can be directly
measured using a calorimeter.
Enthalpy of combustion
• The enthalpy of combustion is the heat energy
given out when 1 mole of fuel burns completely in
oxygen.
• The enthalpy of combustion of methane can be
represented by the equation
• CH4(g) + O2 (g)
CO2(g) + H2 O(l)
Enthalpy of combustion
The heat energy released when alcohols burn can be measured
The enthalpy of combustion of a substance is the amount of energy
given out when one mole of a substance burns in excess oxygen.
Specific heat capacity
Calculating the energy change during a chemical reaction in water.
E = c. m.  T
c
m
T
=specific heat capacity
=mass in Kg
=temperature change
The mass of water can be calculated by using the fact that 1 ml = 1 g.
The value for c is usually taken as 4.18 kJ kg
–1 oC-1
Measuring the enthalpy of combustion of alcohols
Weigh a filled alcohol burner
Measure 50 cm3 water into a copper calorimeter
Take temperature of the water
Light the burner and use it to heat the water to approx 60oC
Stir the water and take the highest temperature reached
Reweigh the burner and remaining fuel
Enthalpy of combustion
Procedure
1. Weigh the spirit burner (already containing ethanol) with its cap on and
record its mass. (The cap should be kept on to cut down the loss of
ethanol through evaporation)
2. Using the measuring cylinder, measure out 100 cm3 of water into the
copper can.
3. Set up the apparatus as directed by your teacher/lecturer.
4. Measure and record the temperature of the water.
5.Remove the cap from the spirit burner and immediately light the burner.
6.Slowly and continuously stir the water with the thermometer. When the
temperature has risen by about 10 °C, recap the spirit burner and measure
and record the maximum temperature of the water.
7. Reweigh the spirit burner and record its mass.
Specific heat capacity
Calculation
(a)The heat energy gained by the water (Eh) can be calculated
using the formula:
Calculating the energy change during a chemical reaction in water.
E = c. m.  T
c
m
T
=specific heat capacity
=mass in Kg
=temperature change
The mass of water can be calculated by using the fact that 1 ml = 1 g.
The value for c is usually taken as 4.18 kJ kg –1 oC-1
CALCULATION
Suppose 0.25 g of ethanol had been burned and the temperature of the water had risen by
12.5 °C.
The heat energy gained by the water (Eh) is calculated using the formula:
Eh = c m T
Eh = 4.18 x 0.10 x 12.5
= 5.225 kJ
We assume that the heat energy released by the burning ethanol is gained only by the water.
The heat energy released on burning 0.25 g of ethanol = 5.225 kJ
Ethanol: CH3CH2OH
Mass of 1 mole = 2(12) + 6(1) + 16 = 46 g
We can now calculate the heat energy released on burning 1 mole of ethanol.
0.25g
46g
5.225kJ
46
0.25
x
5.225
= 961 kJ
The enthalpy of combustion of ethanol = - 961 kJ mol-1
(A negative sign is used because combustion is an exothermic reaction)
Measuring the enthalpy of combustion of alcohols
The heat energy gained by the water (Eh) is calculated
using the formula:
Eh
= c m ∆T
Eh
=
x
x
=
kJ
Measuring the enthalpy of combustion of alcohols
We assume that the heat energy released by the
burning alcohol is gained only by the water.
The heat energy released on burning ……….. g of ……………anol
So one mole .............. g of ................anol
………….. kJ
....................kJ
The enthalpy of combustion of …………anol = -………….. kJ mol-1
(A negative sign is used because combustion is an
exothermic reaction)
Sources of inaccuracy
• Heat loss to surroundings
• Ignore heat rise of calorimeter
• Incomplete combustion
• Possible loss of fuel by evaporation
from wick
Calorimetry
To eliminate these inaccuracies a bomb
calorimeter is used
The burning fuel (or food) is supplied with
oxygen to encourage complete
combustion
The combustion chamber is entirely
surrounded so there is no heat loss to
the surroundings
Commercial ‘bomb’
calorimeters
The calorimeter is heated electrically.
Energy required to heat the entire
apparatus by 1 0C is calculated.
Enthalpy of combustion
Worked example 1.
0.19 g of methanol, CH3OH, is burned and the heat energy given out
increased the temperature of 100g of water from 22oC to 32oC.
Calculate the enthalpy of combustion of methanol.
–704 kJ mol-1
Enthalpy of combustion
Worked example 1.
0.19 g of methanol, CH3OH, is burned and the heat energy given out
increased the temperature of 100g of water from 22oC to 32oC.
Calculate the enthalpy of combustion of methanol.
–704 kJ mol-1
Worked example 2.
0.22g of propane was used to heat 200cm3 of water at 20oC. Use the
enthalpy of combustion of propane in the data book to calculate the
final temperature of the water.
33.3oC
Calculations for you to try.
1.
0.25g of ethanol, C2H5OH, was burned and the heat given out raised
the temperature of 500 cm3 of water from 20.1oC to 23.4oC.
Calculate the enthalpy of combustion of ethanol
2. 0.01 moles of methane was burned and the energy given out raised
the temperature of 200cm3 of water from 18oC to 28.6oC. Calculate
the enthalpy of combustion of methane.
3. 0.1g of methanol, CH3OH, was burned and the heat given out used to
raise the temperature of 100 cm3 of water at 21oC.
Use the enthalpy of combustion of methanol in the data booklet to
calculate the final temperature of the water.
4. 0.2g of methane, CH4, was burned and the heat given out used to
raise the temperature of 250 cm3 of water
Use the enthalpy of combustion of methane in the data booklet
to calculate the temperature rise of the water.
1.
0.25g of ethanol, C2H5OH, was burned and the heat given out raised
the temperature of 500 cm3 of water from 20.1oC to 23.4oC.
Calculate the enthalpy of combustion of ethanol
-1269 kJ mol-1
2. 0.01 moles of methane was burned and the energy given out raised
the temperature of 200cm3 of water from 18oC to 28.6oC. Calculate the
enthalpy of combustion of methane.
-886.2 kJ mol-1
3. 0.1g of methanol, CH3OH, was burned and the heat given out used to
raise the temperature of 100 cm3 of water at 21oC.
Use the enthalpy of combustion of methanol in the data booklet to
calculate the final temperature of the water.
26.4oC
4. 0.2g of methane, CH4, was burned and the heat given out used to
raise the temperature of 250 cm3 of water
Use the enthalpy of combustion of methane in the data booklet
to calculate the temperature rise of the water.
10.66oC
Enthalpy of solution
The enthalpy of solution of a substance is the energy change when one
mole of a substance dissolves in water.
Higher Chemistry Eric Alan and John Harris
Enthalpy of solution
This is the energy change when 1 mole of
solute dissolves in water
The enthalpy of solution for NaOH can be
represented by the equation
NaOH(s)
Na+(aq)
+ OH-(aq)
Measuring enthalpy of solution
•
Add 20ml water to 2 plastic beakers and take the temperature of both
•
Accurately record the mass of approximately 2g of the 2 solids
•
Add NaOH to one and record the final temperature reached
•
Add ammonium nitrate to the other and record the final temperature
mass (g)
sodium hydroxide
ammonium nitrate
start temp
( oC)
final temp
( oC)
change in temp
( oC)
Measuring the enthalpy of solution
The heat energy gained by the water (Eh) is calculated
using the formula:
Eh
= c m ∆T
Eh
=
x
x
=
kJ
Measuring the enthalpy of solution
The heat energy released on dissolving ……….. g of NaOH
So one mole .............. g of NaOH
………….. kJ
.............kJ
The enthalpy of solution of NaOH is -………….. kJ mol-1
(A negative sign is used because energy was released
the temperature increased i.e. exothermic reaction)
Measuring the enthalpy of solution
The heat energy released on dissolving ……….. g of NH4NO3
So one mole .............. g of NH4NO3
………….. kJ
.............kJ
The enthalpy of solution of NH4NO3 is +………….. kJ mol-1
(A positive sign is used because energy was taken in
the temperature decreased i.e. endothermic
reaction)
Data book values
NH4 NO3
+25.69kJ mol -1
NaOH
-44.51 kJ mol
-1
Comment on sources of inaccuracy
Enthalpy of solution
The enthalpy of solution of a substance is the energy change when one
mole of a substance dissolves in water.
Worked example 1. 5g of ammonium chloride, NH4Cl, is completely
dissolved in 100cm3 of water. The water temperature falls from 21oC to
17.7oC. Calculate the enthalpy of solution of ammonium chloride.
14.77 kJ mol-1
Calculations for you to try.
1.
2.
8g of ammonium nitrate, NH4NO3, is dissolved in 200cm3 of water.
The temperature of the water falls from 20oC to 17.1oC. Calculate
the enthalphy of solution.
When 0.1 mol of a compound dissolves in 100cm3 of water the
temperature of the water rises from 19oC to 22.4oC . Calculate
the enthalpy of solution of the compound.
3. The enthalpy of solution of potassium chloride, KCl, is + 16.75kJ mol-1.
What will be the temperature change when 14.9g of potassium
chloride is dissolved in 150cm3 of water?
1.
8g of ammonium nitrate, NH4NO3, is dissolved in 200cm3 of water.
The temperature of the water falls from 20oC to 17.1oC.
24.2 kJ mol-1
2.
When 0.1 mol of a compound dissolves in 100cm3 of water the
temperature of the water rises from 19oC to 22.4oC . Calculate
the enthalpy of solution of the compound.
-14.2 kJ mol-1
3. The enthalpy of solution of potassium chloride, KCl, is + 16.75kJ mol-1.
What will be the temperature change when 14.9g of potassium
chloride is dissolved in 150cm3 of water?
-5.34 oC
Enthalpy of neutralisation
The enthalpy of neutralisation of a substance is the amount of energy
given out when one mole of water is formed in a neutralisation reaction.
Enthalpy of neutralisation
Worked example 1. 100cm3 of 1 mol l -1 hydrochloric acid, HCl, was
mixed with 100 cm3 of 1 mol -1 sodium hydroxide, NaOH, and the
temperature rose by 6.2oC.
-51.8 kJ mol-1.
Calculations for you to try.
1.
400 cm3 of 0.5 mol l-1 hydrochloric acid. HCl, was reacted with 400
cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by
6.4oC . Calculate the enthalpy of neutralisation.
2.
250 cm3 of 0.5 mol l-1 sulphuric acid. H2SO4, was reacted with 500
cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by
2.1oC. Calculate the enthalpy of neutralisation.
3.
100cm3 of 0.5 mol l-1 NaOH is neutralised by 100cm3 of 0.5 mol l-1
HCl. Given that the enthalpy of neutralisation is 57.3 kJ mol-1,
calculate the temperature rise.
1.
400 cm3 of 0.5 mol l-1 hydrochloric acid. HCl, was reacted with 400
cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by
6.4oC . Calculate the enthalpy of neutralisation.
2. 250 cm3 of 0.5 mol l-1 sulphuric acid. H2SO4, was reacted with 500
cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by
2.1oC. Calculate the enthalpy of neutralisation.
-26.32 kJ mol-1.
3.
100cm3 of 0.5 mol l-1 NaOH is neutralised by 100cm3 of 0.5 mol l-1
HCl. Given that the enthalpy of neutralisation is 57.3 kJ mol-1,
calculate the temperature rise.
3.4oC
c) Hess’s law
Learning intention
Learn how to calculate enthalpy
changes for a reaction by
application of Hess’s Law.
Hess law
From previous work you should know and understand the
following:
• Enthalphy (H) is a measure of the energy stored in a
chemical.
•The enthalphy change in a chemical reaction
ΔH = Hproducts – Hreactants
•If a reaction is exothermic ΔH is negative; if
endothermic ΔH is positive.
Hess’s Law and calculations
Hess’s law states that
“the enthalpy change for a
reaction depends only on the
initial and final states of the
reaction and is independent of
the route by which the reaction
may occur.”
Hess law can be shown diagrammatically in different
ways
e.g. carbon can burn to produce carbon dioxide directly by route X
or it can form the intermediate CO by route Y, then the CO can
burn to produce CO2 by route Z.
C + O2
Enthalpy
H = Y
CO + ½O2
According to Hess law
X=Y+Z
H = X
CO2
H = Z
Hess law can be shown diagrammatically in different
ways
e.g. carbon can burn to produce carbon dioxide directly by route X
or it can form the intermediate CO by route Y, then the CO can
burn to produce CO2 by route Z.
+ O2
H = X
C
CO2
+ ½O2
H = Y
+ ½O2
CO
H = Z
According to Hess law
X=Y+Z
A more useful way is to set out the equations like
simultaneous equations in maths.
e.g. carbon can burn to produce carbon dioxide directly by route X
or it can form the intermediate CO by route Y, then the CO can
burn to produce CO2 by route Z.
C
+
O2
 CO2
C + ½O2  CO
CO + ½O2  CO2
C
+ O2
 CO2
H = X
H = Y
H = Z
H = X = Y + Z
Verification of Hess’s Law
H = enthalpy change
KOH (s)
H 2
+ H2O(l)
Route 1
H 1
Route 2
KOH (aq)
KCl (aq)
+ HCl(aq)
H 3
The conversion of solid KOH to KCl solution can be achieved by two
possible routes. Route 1 is a single-step process, (adding HCl (aq) directly
to the solid KOH) and Route 2 is a two-step process (dissolve the solid KOH in
water, then adding the HCl(aq)) All steps are exothermic.
If Hess’s Law applies, the enthalpy change for route 1 must be the same as the
overall change for route 2.
H 1
= H 2 +
H 3
Verification of Hess’s Law
Route 2 H 2 + H 3
Route 1 H 1
50 ml 1mol l-1 HCl
50 ml H2O
50 ml HCl
then
2.50g of KOH added to a dry, insulated
beaker.
Before adding the acid, its temperature
is recorded. The final temperature
after adding the acid is also recorded.
1. 2.50g of KOH added to a dry, insulated
beaker.
2. Before adding the water, its temperature
is recorded. The final temperature rise
after adding the water is also recorded.
3. Now add the acid, again, recording the final
Knowing the specific heat capacity for
water, it is then possible to calculate the temperature.
Use the equation below to calculate
Enthaply change for this reaction.
H 1 = cm T
H2 and H 3
H = cm T
H 1 = H 2 + H 3 will verify Hess’s Law
Hess’s Law Calculations
Previously we have used experimental data to work out
ΔH combustion, solution, neutralisation etc.
To do this we have carried out the reaction.
However sometimes we want to work out the enthalpy
change for a reaction which cannot be measured
directly because it does not usually occur e.g.
C(s) + 2H2(g) → CH4 (g)
C(s) + 2H2(g) → CH4 (g)
We can work out the enthalpy change for this reaction
using Hess law from the enthalpy of combustion of C(s)
H2(g) and CH4(g)
Because the reaction is for forming one mole of compound
from its elements in their standard state this is called
enthalpy of formation.
Enthalpy of formation
• The enthalpy of formation is the energy needed
to make one mole of a compound from its
elements in their standard state.
• This cannot be measured directly for many
compounds
Step 1
Write desired
“key equation”
C(s) + 2H2(g) → CH4 (g)
Step 2
C(s) + O2(g)
→ CO2(g)
ΔH = -394
Write out the H2(g) + ½ O2(g) → H2O(l)
ΔH = -286
equations for CH (g) + 2O (g) → CO (g) + 2H O(l) ΔH = -891
4
2
2
2
enthalphy of
Step
3
combustion
C(s) + O2(g)
→ CO2(g)
ΔH = -394
C(s) H2(g)
Compare
with
CH4(g) and the 2H2(g) + O2(g) → 2H2O(l)
desired
ΔH = (2x-286)
values from
equation
and
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH = +891
databook.
rearange.
Step 4
Cancel and add
C(s) + 2H2(g) → CH4 (g)
ΔH = -75kJmol-1
Hess’s Law Calculations
Calculate the enthalpy formation propene
3C (s) + 3H2 (g)
Route 2a
Route 1
C3H6 (g)
Route 2b
3CO2 (g) + 3H2O (l)
Route 1 cannot be
carried out
in a lab, as carbon and
hydrogen
will not combine
directly.
The enthalpy of combustion reactions can act as a stepping stone which
enables a link with carbon and hydrogen (the reactants) with propene
Route
2a involves the combustion of both carbon and hydrogen
(the product)
3C (s) + 3O2 (g)  3CO2 (g)
and
3H2 (g) + 1½O2 (g)  3H2O (l)
Route 2b involves the reverse combustion of propene
3CO2 (g) + 3H2O(l)  C3H6 (g) + 4½O2 (g)
H1
3C (s) + 3H2 (g)
Route 2a
Route 1
C3H6 (g)
Route 2b
3CO2 (g) + 3H2O(l)
H 1 = H 2a + H 2b
H comb hydrogen = -394 kJ mol
–1
H comb carbon = -286 kJ mol –1
Route 2a
H 2a = (3 x -394) + (3 x -286)
H 2a = -2040 kJ
Route 2b H comb Propene = -2058.5 kJ mol –1
H 2b = + 2058.5 kJ (note the reverse sign)
H 1 = -2040 kJ + (+2058.5) = + 18.5 kJ mol
-1
Examples to try
The following equation shows the formation of carbon
monoxide
C(s) + ½ O2(g) → CO(g)
Use the enthalpies of combustion of carbon and carbon
monoxide to calculate the enthalphy change for this
reaction
ΔH = -111 kJmol-1
Examples to try
The following equation shows the formation of ethanol
from carbon, hydrogen and oxygen
2C(s) + 3H2(g) + ½ O2(g) → C2H5OH(l)
Use the enthalpies of combustion of carbon, hydrogen
and ethanol to calculate the enthalphy change for this
reaction
ΔH = -279 kJmol-1
Note: Oxygen is one of the elements present in ethanol but it is not
involved in deriving the required key equation. The calculation is based
on enthalpies of combustion. Oxygen gas supports combustion; it does
not itself have an enthalpy of combustion.
Given the equations
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) ΔH = a J mol–1
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) ΔH = b J mol–1
Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s) ΔH = c J mol–1
then, according to Hess’s Law
Ac=a–b
Bc=a+b
Cc=b–a
D c = – b – a.
d) Bond enthalpies
Learning intention
The molar bond enthalpy is the
energy required to break one mole
of bonds. The enthalpy change of
a gas phase reaction can be
estimated as the sum of bondmaking
and
bond-breaking
enthalpies.
Breaking bonds
If we want to break a covalent bond between two
atoms, we need
to
to overcome the
put energy
in
attractive force.
C
C
© Nueyer
Bond breaking is an endothermic process.
H-H bond enthalpy
The bond enthalpy is the energy required to
break one mole of H-H bonds
The product is 2 H atoms in the gaseous
state.
This is equal and opposite to the energy given
out when 2 H atoms in the gaseous state
form 1 mole of H-H bonds.
Diatomic molecules
For a diatomic molecule, the bond
enthalpy values can be measured
directly
H-Cl
O=O
Br-Br
Cl-Cl
Mean bond enthalpies
Mean bond enthalpies have to calculated as
the value of a C-H bond enthalpy will vary
according to the environment it is in.
Energy used to break a bond (endothermic)
will equal that released when the bond is
made (exothermic)
Using bond enthalpies
∆H can be calculated from bond enthalpies using the
equation
H =  H bonds broken +  H bonds made
Making bonds
The opposite is true if we want to make new bonds.
Energy is released when new chemical bonds are
formed.
Bond making is an exothermic process.
Breaking or making the same chemical bond will
require the same energy to be put in or released.
H H
2H
2H
H H
∆H = 432 kJ
∆H = –432 kJ
Bond enthalpy values
The values for bond enthalpies are found on page 9
of the data book.
For some bonds, the mean bond enthalpy is quoted.
This is to give an average value to work from since
the precise enthalpy value for a bond may be
different in different molecules.
C
For example, the energy needed to break a C
bond in ethane (C2H6) will be different to the energy
needed to break a C
C bond in decane (C10H22)
Gaseous state
The bond enthalpies quoted in the data book are the
energies required to break 1 mole of a particular
bond between a pair of atoms in the gaseous state.
We can use these bond enthalpies to approximately
calculate the enthalpy change for a given reaction.
Example 1
What is the enthalpy change when hydrogen is added
to ethyne to produce ethane?
C2H2 (g) + 2H2 (g)
C2H6 (g)
To answer this we must look at what types of bonds
must be broken in the reactants and formed in the
products.
Potential energy
C2H2 (g) + 2H2 (g)
In this reaction, we must
first break all the bonds
inside the reactant
molecules.
This will require energy to
be put in.
C2H6 (g)
Next, new bonds must be
formed between the atoms
in the product molecule.
This releases energy.
Reaction pathway
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H
2 2 (g)
H C
C
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
H
To answer this question we can follow these
steps.
Step One: Draw the full structural formulae of all
the molecules from the equation. This will show
exactly what bonds are involved.
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)
H C
C
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
Bond Breaking
1xC
2 xC
2 xH
C
H
H
Step Two:
Make a list of all the bonds being broken in the
reactants
H
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)
H C
C
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
H
Bond Breaking
1 x C C = 835
2 x C H = 2 x 414 = 828
2 x H H = 2 x 432= 864
Step Three:
Fill in the values for the bond enthalpies from Page 9
of the data book.
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)
H C
C
Bond Breaking
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
H
Bond Making
1 x C C = 835
1xC
2 x C H = 2 x 414 = 828 6 x C
2 x H H = 2 x 432= 864
Step Four:
C = 346
H = 6 x 414 = 2484
Repeat this process for the Bond Making steps.
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)
H C
C
Bond Breaking
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
H
Bond Making
1 x C C = 835
1 x C C = 346
2 x C H = 2 x 414 = 828 6 x C H = 6 x 414 = 2484
2 x H H = 2 x 432= 864
Total
in = 2527 kJ
Total given out = –2830 kJ
Step put
Five:
Remember:
Remember:
Calculate the total energy put in breaking bonds and
Bond breaking is an
Bond making is an
total energy given out making new bonds.
endothermic process
exothermic process
What is the enthalpy change when hydrogen is added
to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)
H C
C
Bond Breaking
H
+
H H
H H
C2H6 (g)
H
H
H C
C
H
H
H
Bond Making
1 x C C = 835
1 x C C = 346
2 x C H = 2 x 414 = 828 6 x C H = 6 x 414 = 2484
2 x H H = 2 x 432= 864
Total put in = 2527 kJ
Total given out = –2830 kJ
∆H = 2527 + (–2380)
Step Six:
= 2527 – 2380
Calculate the enthalpy
change
for-1the reaction.
= – 303
kJ mol
Enthalpy of formation
Bond enthalpies are also useful when calculating the
enthalpy change in making 1 mole of a compound from
its elements in their standard states.
For example, the enthalpy of formation of
methane can be represented by:
C (s) + 2H2 (g)
CH4 (g)
The data book quotes the value for this reaction as
∆Hformation = –75 kJ mol–1
Enthalpy of formation
To calculate the enthalpy of formation for methane
using bond enthalpies, we would need to change 1
mole of solid carbon atoms into 1 mole of gaseous
carbon atoms.
This change, C (s)
C (g), is known as the
enthalpy of sublimation of carbon and has a
value of 715 kJ mol–1 (see page 9 of the data
book).
Example 2
H H H
C
(s)
Calculate theHenthalpy
H H Hof formation of propane.
C (s) +
H C C C H
H H H H
C (s)
H H
3C (s) + 4H2 (g)
C3H
H8 (g)
Example 2
Calculate the enthalpy of formation of propane.
3C (s) + 4H2 (g)
C3H8 (g)
H H H
C (s)
H H H H
C (s) +
H C C C H
H H H H
C (s)
H H H
Bond Breaking
3 x C(s) C(g) = 3 x 715 = 2145
4 xH
H = 4 x 432= 1728
Total put in = 3873 kJ
Example 2
Calculate the enthalpy of formation of propane.
3C (s) + 4H2 (g)
C3H8 (g)
H H H
C (s)
H H H H
C (s) +
H C C C H
H H H H
C (s)
H H H
Bond Breaking
3 x C(s) C(g) = 3 x 715 = 2145
4 xH
H = 4 x 432= 1728
Total put in = 3873 kJ
Example 2
Calculate the enthalpy of formation of propane.
3C (s) + 4H2 (g)
C3H8 (g)
H H H
C (s)
H H H H
C (s) +
H C C C H
H H H H
C (s)
H H H
Bond Breaking
Bond Making
3 x C(s) C(g) = 2145
2 x C C = 2 x 346= 692
4 xH
H = 1728
Total put in = 3873 kJ
8 xC
H = 8 x 414 = 3312
Total given out = –4004 kJ
Example 2
Calculate the enthalpy of formation of propane.
3C (s) + 4H2 (g)
C3H8 (g)
H H H
C (s)
H H H H
C (s) +
H C C C H
H H H H
C (s)
H H H
Bond Breaking
Bond Making
3 x C(s) C(g) = 2145
2 x C C = 2 x 346= 692
4 xH
H = 1728
Total put in = 3873 kJ
8 xC
H = 8 x 414 = 3312
Total given out = –4004 kJ
∆H = 3873 + (–4004)
= 3873 – 4004
= – 131 kJ mol-1
Oxidising and reducing
agents
Overview
In this section, study how
elements, molecules and group
ions can act as oxidising or
reducing agents and learn to write
ion-electron and redox equations.
Elements as oxidising and
reducing agents
Learning intention
Learn how the electronegativity of
an element influences its ability to
gain or lose electron(s) and
therefore to act as an oxidising or
reducing agent.
Molecules and group ions can act
as oxidising and reducing agents
Learning intention
Learn about examples of molecular
and ionic compounds found in the
electrochemical series which can
behave as oxidising or reducing
agents.
Redox Equations
Redox reactions include reactions which involve the loss
or gain of electrons.
The reactant giving away (donating) electrons is called
the reducing agent (which is oxidised)
The reactant taking (accepting) electrons is called
the oxidising agent (which is reduced)
Both oxidation and reduction happen simultaneously,
however each is considered separately using
ion-electron equations.
O.I.L. R.I.G. Oxidation is loss, reduction is gain of electrons
Displacement reactions
Some metals are more reactive than others. In
this experiment, a strip of metal is added
to a solution of another. If the metal is more
reactive than the metal in solution, this
metal displaces (pushes out) the less reactive
metal from the solution.
-
X
√
√
√
-
√
√
X
X
-
X
X
X
√
-
What is the order of reactivity of these metals (from most to
least)?
Magnesium, zinc, lead, copper
Write a symbol equation for the reaction of magnesium and zinc
nitrate.
Mg(s) + Zn2+(NO3-)2(aq) → Mg2+(NO3-)2(aq) + Zn(s)
Rewrite the equation omitting the spectator ions.
Mg(s) + Zn2+(aq) + 2NO3-(aq) → Mg2+(aq) + 2NO3-(aq) + Zn(s)
Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)
Write a half equation for the reducing agent in the above
reaction
Mg(s) → Mg2+(aq) + 2e-
Reactions of halogens
The Group 7 elements are called the halogens. This
experiment involves some reactions of the halogens.
reaction
(colour change)
Bleached (fast)
Bleached (slow)
Bleached (very
slow)
No reaction
No reaction
No reaction
reaction
(colour change)
reaction
(colour change)
Which halogen solution is the strongest bleaching agent?
Chlorine
Which halogen is the most reactive.
Chlorine
Write a symbol equation for the reaction of chlorine and
potassium bromide
Cl2(aq) + 2K+Br-(aq) → 2K+Cl-(aq) + Br2(aq)
Cl2(aq) + 2Br-(aq) → 2Cl-(aq) + Br2(aq)
Write a half equation for the oxidising agent in the above
reaction
Cl2(aq) + 2e- → 2Cl- (aq)
v
1A
Row
1
H
2A
3
Li
4
Be
11 12
Na Mg
e.g.
e.g.
½O2 +
Note that, in general,
½Cl2 + e- 
Mg 
2 1
5 6 7 8 9 10 2
B C N O F Ne
13 14 15 16 17 18 3
Al Si P S Cl Ar
O2Cl-
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Sc Ti
V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4
19 20
K Ca
37 38
Rb Sr
He
5A 6A 7A
2e-3A 4A
Mg2+
+
2e-
39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6
Cs Ba La Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn
Al  Al3+ + 3e87 88 89 90 91 92 93 94 95 97 98 99 100 101 102 103 104 105 106 107
Fr Ra Ac Th Pa U Np Pu Am Bk Cl Es Fm Md No Lr Unq UnpUnhUns
109
Une
7
• Metals on the LHS of the Periodic Table ionise by
electron loss and are called reducing agents
• Non-metals on the RHS of the Periodic Table ionise
by electron gain and are called oxidising agents
Oxidising and reducing agents
The elements with high eletronegativities tend to form
ions by gaining electrons (reduction) and so act as
oxidising agents
v
1A
Row
1
H
2A
3
Li
4
Be
11 12
Na Mg
He
5A 6A 7A
2e-3A 4A
½O2 +
Note that, in general,
½Cl2 + e- 
5 6 7 8 9 10 2
B C N O F Ne
13 14 15 16 17 18 3
Al Si P S Cl Ar
Mg 
O2Cl-
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Sc Ti
V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4
19 20
K Ca
37 38
Rb Sr
2 1
Mg2+ +
2e-
39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6
Cs Ba La Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn
Al  Al3+ + 3e87 88 89 90 91 92 93 94 95 97 98 99 100 101 102 103 104 105 106 107
Fr Ra Ac Th Pa U Np Pu Am Bk Cl Es Fm Md No Lr Unq UnpUnhUns
109
Une
7
The elements with low eletronegativities tend to form ions
by losing electrons (oxidation) and so act as reducing agents
Redox and the
Electrochemical Series
Eo/V Oxidising agents
Li+(aq) + e  Li(s)
Na+(aq) + e  Na(s)
Mg2+(aq) + 2e  Mg(s)
Pb2+(aq) + 2e  Pb(s)
Increasing powerful reducing agent
(write the reaction backwards)
2H+(aq) +2e  H2(g)
Cu2+(aq) + 2e  Cu(s)
Ag+(aq) + e  Ag(s)
Hydrogen reference
Increasing powerful oxidising agent
(write the reaction as it appears)
Considering the two ion-equations,
Mg 2+ (aq ) + 2e-  Mg (s)
and Ag + (aq) + e-  Ag ,
Mg, being higher up the electrochemical series, would act as the reducing
agent. (i.e. the ion-electron equation would be written backwards).
While Ag would be written as it appears in the electrochemical series.
Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s)
Ion-electron and redox equations
Learning intention
Learn to represent oxidation and
reduction reactions by ion-electron
equations, and to combine these
to write balanced redox equations.
Cells and Redox
A metal higher
in the series
A metal lower
in the series
Ion
bridge
Ions of
metal
higher in
ECS
Ions of metal
lower in ECS
Metal atoms will be oxidised.
Metal ions in solution will be reduced,
Metal atoms are the reducing agent. Metal ions are the oxidising agent.
e.g. Zn  Zn
2+
+ 2e-
Overall redox equation
E.g. Cu 2+ + 2e-  Cu
Zn(s) + Cu 2+ (aq)
Zn 2+ (aq) + Cu (s)
Cells and Redox
magnesium(s) + silver nitrate(aq)  magnesium nitrate(aq) + silver(s).
The reducing agent in this reaction is the Mg as it will
donate electrons to the silver ions .
The oxidising agent is the Ag+ ions as they accept electrons
from the Mg
Oxidation: Mg(s)  Mg2+(aq) + 2 e-
Reduction:
Mg
2Ag+(aq) + 2e-  2Ag(s)
(s)
Half equations
or ion-equations
+ 2Ag+ (aq)  Mg2+ (aq) + 2Ag (s)
Redox equation, electrons cancel out
Writing REDOX equations
Consider the reaction between sodium and water:
Na(s) + H2O(l)

NaOH(aq)
+
½H2(g)
Consider how the ions are formed in this reaction
Na(s)
H2O(l) +
Na(s)

Na+(aq) +
e-
e-
OH-(aq) + ½H2(g)
 Na+(aq)+
e-
A sodium atom loses
an electron

H2O(l) + e-  OH-(aq) + ½H2(g)
and, we could say that a water
molecule must be accepting
the electron
Na(s) 
Na+(aq) +
e-
H2O(l) + e-  OH-(aq) + ½H2(g)
OIL
RIG
These are called ion-electron equations
(or ionic half equations).
Na(s) 
Na+(aq) +
e-
Electrons cancel!
H2O(l) + e-  OH-(aq) + ½H2(g)
Reduction and oxidation occur simultaneously. Adding
the two equations together gives us the overall
equation for a reaction.
Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
For each of the following reactions, combine the oxidation and reduction
step to form a balanced equation.
a) Cu(s)  Cu2+ + 2eAg+(aq) + e-  Ag(s)
b) MnO4- + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+ + e-
c) Ce4+(aq) + e-  Ce3+(aq)
2Br-(aq)  Br2(aq) + 2e-
Balancing Redox equations
Most redox reaction you will come across will occur in
neutral or acidic conditions.
1. Make sure there are the same number of atoms of each
element being oxidised or reduce on each side of the
half equation.
2. If there are any oxygen atoms present, balance them by
adding water molecules to the other side of the half-equation.
3. If there are any hydrogen atoms present, balance them
by adding hydrogen ions on the other side of the
half-equation.
4. Make sure the half-reactions have the same overall
charge on each side by adding electrons.
For basic solutions H atoms are balanced using H2O and then the same
number of OH- ions to the opposite side to balance the oxygen atoms
1. Write down what you know….
sulphur dioxide is oxidised to sulphate ions
SO2(g)

SO42-(aq)
2. Balance the oxygen atoms by adding water
SO2(g) + 2H2O(l)
 SO42-(aq)
3. Balance the hydrogen atoms by adding hydrogen ions
SO2(g) + 2H2O(l)  SO42-(aq) +
4H+(aq)
4. Balance the charges by adding electrons
SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq) + 2echarge is zero
4 - and 4 + equals zero
Write a balanced ion-electron equation for each of the
following reactions
MnO4IO3-
 Mn2+
 I2
Pb2+
 PbO2
ClOMn2+
Sb
 Cl MnO2
 SbO+
VO2+
 V3+
HOBr
 Br2
C2 H5OH
 CH3CHO
MnO4-
+ 8 H+ + 5e -  Mn2+
2 IO3- + 12 H+ + 10e -  I2
Pb2+
ClO-
+ 2H2O
+ 2 H+
 PbO2
+ 2e -  Cl-
+ 4H2O
+ 6H2O
+ 4 H+
+ H2O
+ 2e -
Mn2+
Sb
+ 2H2O
+ H2O
 MnO2
+ 4 H+
+ 2e -
 SbO+
+ 2 H+
+ 3e -
VO2+
2 HOBr
+ 2 H+ + e -  V3+
+ H2O
+ 2 H+ +2 e -  Br2
+ 2 H2O
C2 H5OH
 CH3CHO
+ 2 H+
+2 e -
Redox Reactions
From a reaction you should be able to
identify
a) spectator ions if present
b) half equations for
reduction/oxidation
c) the reducing and oxidising agents
Mg(s)
+
2 H+(aq) → Mg 2+(aq)
No spectator ions shown
+
H2(g)
Mg(s)
+
2 H+(aq) → Mg 2+(aq)
+
H2(g)
Mg(s)
+
2 H+(aq) → Mg 2+(aq)
+
H2(g)
Mg(s)
→ Mg 2+(aq)
2 H+(aq)
→
H2(g)
Mg(s)
2 H+(aq)
→ Mg 2+(aq)
→
H2(g)
Mg(s)
2 H+(aq)
→ Mg 2+(aq)
+
2e-
→
+
H2(g)
2e-
Mg(s)
→ Mg 2+(aq)
+
2e-
oxidation
oxidising agent H+(aq)
2 H+(aq)
+
2e-
→
H2(g)
reduction
reducing agent Mg(s)
2I -(aq)
+
2 Fe3+(aq)
→ I2(aq)
No spectator ions shown
+
2 Fe2+(aq)
2I -(aq)
+
2 Fe3+(aq)
→ I2(aq)
+
2 Fe2+(aq)
2I -(aq)
2 Fe3+(aq)
→ I2(aq)
+
2e-
→
+
2 Fe2+(aq)
2e-
2I -(aq)
→ I2(aq)
+
2e-
oxidation
oxidising agent Fe3+(aq)
2 Fe3+(aq)
+
2e-
→
2 Fe2+(aq)
reduction
reducing agent I-(aq)
2NaBr(aq)
+
Cl2 (g)
→
2 NaCl(aq)
Show charges on ions
+
Br2(aq)
2Na+Br -(aq) + Cl2 (g)
→
2 Na+Cl -(aq)
Omit spectator ions
Two Na+ ions
+
Br2(aq)
2Br -(aq)
+
Cl2 (g)
→
2 Cl -(aq)
+
Br2(aq)
2Br -(aq)
+
Cl2 (g)
→ 2 Cl -(aq)
+
Br2(aq)
2Br -(aq)
Cl2 (g)
→ Br2(aq)
→
2 Cl -(aq)
2Br -(aq)
Cl2 (g)
→ Br2(aq)
+
2e-
→
+
2 Cl -(aq)
2e-
2Br -(aq)
→ Br2(aq)
oxidising agent Cl 2 (aq)
Cl2 (g)
+
2e-
→
+
2e-
oxidation
2 Cl -(aq)
reduction
reducing agent Br - (aq)
2AgNO3 (aq)
+
Cu(s)
→ 2 Ag (s)
Show charges on ions
+
Cu(NO3)2 (aq)
2Ag+NO3- (aq) +
Cu(s)
→ 2 Ag (s)
Omit spectator ions
Two NO3- ions
+
Cu2+(NO3-)2 (aq)
2Ag+(aq)
+
Cu(s)
→ 2 Ag (s)
+
Cu2+ (aq)
2Ag+(aq)
+
Cu(s)
→ 2 Ag (s)
+
Cu2+ (aq)
Cu(s)
2Ag+(aq)
→ Cu2+ (aq)
→
2 Ag (s)
Cu(s)
2Ag+(aq)
→ Cu2+ (aq)
+
2e-
→
+
2 Ag (s)
2e-
Cu(s)
→ Cu2+ (aq)
+
2e-
oxidation
oxidising agent Ag+ (aq)
2Ag+(aq)
+
2e-
→
2 Ag (s)
reduction
reducing agent Cu (s)
K2SO4 + Br2 + H2O K2SO4 + 2HBr
Show charges on ions
(K+)2SO32- + Br2 + H2O (K+)2SO42- + 2H+Br -
Omit spectator ions
Two K+ ions
SO32- + Br2 + H2O  SO42- + 2H+Br -
SO32-(aq) + Br2(l)
+
2→
SO
H2O(l)
4 (aq)
+
2H+ Br - (aq)
SO32-(aq)
Br2(l)
+
H2O(l)
→
SO42-(aq)
→
+
2H+ (aq)
2 Br - (aq)
SO32-(aq)
Br2(l)
+
H2O(l)
+
2e-
→
SO42-(aq)
→
+
2H+ (aq)
2 Br - (aq)
+
2e-
SO32-(aq)
+
H2O(l)
→
SO42-(aq)
+
2H+ (aq)
+
oxidation
oxidising agent Br2 (l)
Br2(l)
+
2e-
→
2 Br - (aq)
reduction
reducing agent SO32-(aq)
2e-
2Cr 3+ + 7H2O + 3Cl2 Cr2O72- + 14H+ + 6Cl-
No spectator ions shown
2Cr 3+ + 7H2O + 3Cl2 Cr2O72- + 14H+ + 6Cl-
2Cr 3+ + 7H2O  Cr2O72- + 14H+
3Cl2

6Cl-
2Cr 3+ + 7H2O  Cr2O72- + 14H+ + 6e oxidation
oxidising agent Cl2 (l)
3Cl2
+6e-

6Clreduction
reducing agent Cr3+(aq)
5CrCl2 + KMnO4+ 8HCl

5CrCl3 + KCl +MnCl2 + 4H2O
Show charges on ions
5Cr2+(Cl-)2 + K+MnO4-+ 8H+Cl -

5Cr3+(Cl-)3 + K+Cl- +Mn2+(Cl-)2 + 4H2O
Omit spectator ions
One K+ ion and eighteen Cl- ions
5Cr2+ + MnO4- + 8H+

5Cr3+ + Mn2+ + 4H2O
5Cr2+ + MnO4- + 8H+

5Cr3+ + Mn2+ + 4H2O
5Cr2+
5Cr3+

+ 5eoxidation
oxidising agent MnO4- / H+
-
MnO4 +
8H+
+
5e-
 Mn2+ + 4H2O
reduction
reducing agent Cr2+
Molecules and group ions can act
as oxidising and reducing agents
Learning intention
Learn about examples of molecular
and ionic compounds found in the
electrochemical series which can
behave as oxidising or reducing
agents.
Molecules and group ions can act as
oxidising and reducing agents
Compounds can also act as oxidising or reducing agents.
The electrochemical series contain a number of ions and molecules.
Oxidising and reducing agents can be selected using an electrochemical series.
1. Permaganate MnO4- is a strong
oxidising agent
Oxidising agents such as potassium permaganate (KMnO4) are
used in explosives and fire works
MnO4-(aq) + 8H+ +5e- → Mn2+(aq) + 4H2O(l)
Mn(VII)
Mn(II)
The potassium permanganate oxidises the glycerol to carbon
dioxide and water (hence the steam) and is itself reduced.
2. Dichromate and permanganate in
acidic solutions are strong oxidising
agents.
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
Orange
green
MnO4-(aq) + 8H+ + 5e- → Mn2+(aq) + 4H2O(l)
Purple
colourless
Both half equations contain H+ ions
showing they are strong oxidising
agents in the presence of acid
3. An Oxidation and Reduction
reaction – Blue Bottle experiment
A conical flask contains a colourless solution. When shaken, a blue colour
forms. After a few seconds, the blue colour fades and the solution
again becomes colourless.
The process can be repeated. It is an oxidation followed by a reduction
process
What to do
1. Put some water in the conical flask. Put in the stopper. Shake
vigorously to check for leaks. If there are none, pour the
water away and proceed.
2. Put 100 cm3 of potassium hydroxide solution into a conical
flask.
3. Add 3.3 g dextrose.
4. Add 3–4 drops of methylene blue indicator.
5. Put a stopper on the flask.
6. Shake vigorously.
7. When the solution clears, repeat the process.
8. It is necessary periodically to remove the stopper.
• Methylene blue is reduced by the
alkaline dextrose solution to produce
a colourless solution.
• When the solution is shaken, it is
oxidised by the oxygen in the flask to
produce the blue dye.
• The stopper must be taken off to
allow more oxygen in.
5. Hydrogen peroxide is a strong
oxidising agent
Oxidising agents such as hydrogen peroxide are used in many
bleaching products
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
Elephant Toothpaste Chemistry
The overall equation for this reaction
is:
2 H2O2(aq) --> 2 H2O(l) + O2(g)
The catalyst is potassium iodide. Some of the iodide ions are
converted to I2 molecules by the hydrogen peroxide
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
2I-(aq) → I2(s) + 2eH2O2(aq) + 2H+(aq) + 2I-(aq) → I2(s) + 2H2O(l)
Everyday uses for strong
oxidising agents
Learning intention
Learn how strong oxidising agents
are put to use in a variety of
everyday substances including
bleach and fungicides.
Everyday uses for strong oxidising
agents
• Oxidising agents are widely employed
because of the effectiveness with which
they can kill fungi and bacteria, and can
inactivate viruses.
• The oxidation process is also an effective
means of breaking down coloured
compounds making oxidising agents ideal
for use as “bleach” for clothes and hair.
Oxidising agents as bleach
• Oxidising agents are widely used in
everyday life because of the
effectiveness with which they can kill
fungi, and bacteria and can inactivate
viruses. The oxidising agents include
bleaches and chemicals such as
potassium permanganate, KMnO4.
Potassium permaganate
Potassium permanganate solution
will react with any organic matter
in a pond including algae,
bacteria, fish, particulate and
dissolved organic, and organic
bottom sediments. It has been
used in fish ponds to treat
common fish pathogens such as
gill parasites and external
bacterial and fungal infections.
Potassium permaganate
• Athlete's foot, another fungal
infection in humans can be treated
with very dilute potassium
permanganate solution.
Hydrogen peroxide
The oxidation process is also an effective means of
breaking down coloured compounds making oxidising
agents ideal for use as "bleach" for clothes and hair.
Hydrogen peroxide, H2O2(aq) can be used to bleach
hair and it behaves here as an oxidising agent. In the
reaction, coloured pigments in the hair, related to the
pigment melanin are oxidised to colourless substances.
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
Chlorine is a strong oxidising agent
Teacher demo
Chlorine gas was bubbled through
tomato juice what did you observe?
How does Chlorine
in swimming pools
work?
Chlorine kills bacteria though a fairly simple chemical
reaction. The chlorine solution you pour into the water
breaks down into many different chemicals, including
hypochlorous acid (HOCl) The hypochlorus acid is a strong
oxidising agent and kills microorganisms by removing
electrons from the bacteria destroying the cell walls and
enzymes and structures inside the cell, rendering them
harmless.
Sodium hypochlorite
Sodium hypochlorite has a wide application
as a bleaching agent for textiles and
textile laundering. It also acts as a
powerful disinfectant. In solution the
hypochlorite ions form hypochlorous acid,
HClO. This is the oxidising agent.
2HClO-(aq) + 2H+(aq) + 2e– → Cl2(g) + 2H2O(l)
Household bleach is, in general, a solution
containing 4-6% sodium hypochlorite
Experiment
4 drops of yellow food colouring (E102) or
4 drops of blue food colouring (E 124) are
dissolved in 40 cm3 of water.
A solution containing 4 drops of household
bleach in 20 cm3 of water is added to the
solution and the mixture stirred.
The hypochlorite oxidises the colourings
taking the solution through a number of
colour changes.
Luminol Chemilumenescence
• Luminol can be oxidised by a solution
of bleach to an aminophthalate ion
which is produced in an excited state
and emits light as it drops back into a
ground state.
A bleach which is a reducing
agent
Aran sweaters used to be
bleached using sulphur dioxide gas
on the wet sweaters. The reaction
involves the dissolving of the gas
to make a solution containing
hydrogen sulphite. This produces
some sulphite ions.
SO2(g) + H2O(l) → H2SO3 (aq)
SO32-(aq) + H2O(l) → SO42-(aq) + 2H+ + 2e- Oxidation
This is not an example of a bleach which is an
oxidising agent. As the reaction is oxidation,
SO2(g) in water and SO32-(aq) are examples of
a bleach which is a reducing agent.
They are described as anti-oxidants and are
used to sterilise glass bottles in wine-making
and are present in some foods. If you see E220
amongst the products of a food stuff, this
represents sulphur dioxide.
Experiment
Add a flower petal to a test tube SO2
• What did you observe?
• Soundbite molecules: Potassium permanganate
This Royal Society of Chemistry article looks at
the uses of potassium permanganate, a compound
used by chemists as a strong oxidising agent.
• Chemical functional definitions - Bleach systems
This Procter and Gamble article looks at what
bleach is, why it is used in cleaning products
and how it improves the laundry process.
The following reactions take place when nitric
acid is added to zinc.
NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l)
Zn(s) → Zn2+(aq) + 2e–
How many moles of NO3– (aq) are reduced by
one mole of zinc?
A 2/3
B 1
C 3/2
A
D 2
In which reaction is hydrogen gas acting as an
oxidising agent?
A H2 + CuO → H2O + Cu
B H2 + C2H4 → C2H6
C H2 + Cl2 → 2HCl
D H2 + 2Na → 2NaH
D
Iodide ions can be oxidised using acidified potassium
permanganate solution. The equations are:
2I–(aq) → I2(aq) + 2e–
MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
How many moles of iodide ions are oxidised
by one mole of permanganate ions?
A 1.0
B 2.0
C 2.5
D 5.0
D
Chemical analysis as part
of Quality Control
Overview
Learn how analytical chemistry
techniques
such
as
chromatography and volumetric
analysis can be used to control
the quality of products produced in
an industrial process.
a) Chromatography can be used to
check the composition and purity
of reactants and products
Learning intention
Learn how differences in size and
polarity of molecules can be used
to separate mixtures of reactants
and products by chromatography
to produce a chromatogram.
Chemical Analysis
Chromatography
Chromatography is one technique
widely used for analysis
Chromatography derives from the
Greek word ‘Chroma’ meaning colour
and ‘graphia’ to mean writing.
Chromatography
Separation depends on the
equilibrium between 2 phases
– a mobile phase
(shoppers)
– a stationary phase ( shops)
Paper chromatography
Stationary phase: paper
Mobile phase : solvent
Substances that are more soluble in the
solvent travel further
Substances that have more of an affinity
for the paper (stationary phase) travel less
far
To Affinity… and beyond!
• Lets go shopping..
• We are all on a train going to Glasgow Central.
• When we get off the train we all have a different
‘affinity’ to different shops.
• Some will go to Costa, some to the Tie Rack and
some to WH Smith.
• It is how we, as the mobile phase, separate out on
the stationary phase, Glasgow Central.
Chromatography
Indigo ink is a mixture of dyes (liquids)
As the solvent
travels up the paper
the different liquids
travel different
Indigo ink contains
distances
yellow and green
dyes – the mixture
of inks has been
seperated
Solvent
Calculating Rf.
Paper
• http://chemsite.lsrhs.net/FlashMedia/html/p
aperChrom.html
• Stationary phase – solid, liquid or gas
supported on a solid i.e. uniform absorbent
paper.
• Mobile phase – liquid or gas.
• Common use – separating ball point pen
inks.
Thin layer chromatography
• Stationary phase: glass or Al supports thin
layer of cellulose or silica
• Mobile phase : solvent
• TLC quicker and prevents spots spreading
• Fluorescent dye can be incorporated to
enable spots to be located by UV
Thin Layer
• http://www.chemguide.co.uk/analysis/chro
matography/thinlayer.html
• Stationary phase – silica/alumina backed
metal/plastic/glass (can include UV plates)
• Mobile phase – solvent or a mixture of
solvents.
• Common use – cannabis analysis and amino
acid separation.
Column chromatography
• The column is packed with inert material
• Substances take a different time to travel
through the column according to their
relative affinity for the solvent and the
column material
• Each substance has a specific retention
time that can be used to identify it.
Column
• http://chemsite.lsrhs.net/FlashMedia/html/c
olumnChrom.html
• Stationary phase – usually silica.
• Mobile phase – eluent and sample.
• Common use – separating food dyes.
Gas Liquid chromatography
• Widely used in many industries
• Locally used by Diageo to test every mash
produced by a distillery and every batch of alcohol
produced
• Ensures the distillery is converting as much sugar
to alcohol as possible
• Checks for impurities in the whisky
Gas-Liquid Chromatography
• Stationary: High bp liquid held on an inert
powdered support material, packed into a
coiled column in an oven!
• Mobile phase : inert gas eg Helium or
nitrogen
• Heat vaporises sample, carried by gas
through hot column
Gas Liquid chromatography
GLC cont
• Rate of travel through column depends on relative
affinity for gas or inert material
• As substance emerged, analysed by detector,
producing a graph
• Identification is by ‘retention time’
• Height of peak indicates amount of substance
Gas column
• Stationary phase- the
column which contains a
liquid stationary phase.
• Mobile phase – flow of
inert gas. Usually Ne,
Ar, N2
• Common use – drink
driving – breath test
analysis.
GC Chromatograms
Retention Time
• The retention time of
a solute is taken as the
elapsed time between
the time of injection
of a solute and the
time of elution of the
peak maximum of that
solute.
Identification
• The Retention times of known
compounds can been measured and
recorded in a data base
• The Retention times of substances in
the sample can then be compared to
the data base and identified
Chromatography is used to
identify drugs in urine
samples taken from athletes.
The mixture of liquids in the
urine is injected, samples of
known drugs have already
been analysed.
Steroid A takes 18mins 50
sec to pass through the
chromatography machine.
Has the athlete tested below
been using Steroid A?
Chemical analysis
• Analaysis is very important in industry for
quality control purposes
• Checking for contaminants
• Checking the level of product e.g alcohol in
whisky, aspirin in tablets etc
Analysis
Analysis is widely carried out for a range of
applications. A few examples are:
1. soil analysis
2. samples of blood, urine etc in health
3. Water purification analysis
4. swimming pools etc
5. environmental health
Variety is the spice of life!
•
•
•
•
•
•
•
•
Column Chromatography
Paper Chromatography
Ion Exchange
Thin Layer
Liquid Chromatography
Gas Chromatography
High Performance Liquid Chromatography
Gel Permeation
Case Studies
http://media.rsc.org/Chemistry%20at%20the
%20races/RSC%20Horseracing%20Part%2
05.pdf
(neigh chance!!)
• http://media.rsc.org/Classic%20Chem%20D
emos/CCD-24.pdf
b) Volumetric titration
Learning intention
Learn how solution concentrations
can be accurately determined in
volumetric titrations.
Redox titrations
Redox Titrations
Titration is a technique for measuring the concentration
of a solution. A solution of known concentration is used to
work out the unknown concentration of another solution.
Redox titrations involve solutions of reducing and
oxidising agents.
At equivalence-point of a redox titration precisely enough
electrons have been removed to oxidise all of the reducing
agent.
Vitamin C - Redox titration
Youtube video
Vitamin C (ascorbic acid) is an important component of our diet. In
its absence the protein, collagen, cannot form fibres properly and
this results in skin lesions and blood vessel fragility.
Although vitamin C occurs naturally in many fruits and vegetables
many people take vitamin C tablets to supplement their intake.
Vitamin C can undergo a redox reaction with iodine in which the
vitamin C is oxidised and the iodine molecules are reduced
OH H
H
C C
HO
C H H
C
O
C
HO
C
O
I2(aq) + 2e
OH
OH H
H
C C
O
C H H
C
O
C
O
C
O
2I-(aq)
OH
+ 2H+(aq) + 2e
Redox Titrations, Vitamin C
I2 (aq) + 2e-  2I - (aq)
reduction
C6H8O6 
oxidation
C6H6O6 + 2H+ (aq) + 2e-
I2 (aq) + C6H8O6
Blue/Black (in the

C6H6O6 + 2H+ + 2I- (aq)
colourless
presence of starch)
Iodine, those concentration is known (in the burette)
acts as an oxidising agent.
Vitamin C, the unknown concentration (in the conical flask)
is a reducing agent.
Starch is added to show when the end-point is reached.
Procedure
1. Add a vitamin C tablet to the beaker.
2. Add some deionised water (approximately 50 cm3) to the beaker
and stir the mixture until the tablet has dissolved.
3. Carefully add the resulting solution to the 250 cm3 standard flask.
Rinse out the beaker several times with water and add the washings
to the flask.
4. Add water to the standard flask to bring the volume of the
solution up to the graduation mark on the neck.
5. Stopper the flask and invert it several times to make sure the
solution is thoroughly mixed.
RedoxTitration
What to do:
Carefully fill the burette
with iodine solution.
Carefully pipette exactly
25 ml of vitamin C solution
into the conical flask. Add a
Couple drops of starch solution.
Add the iodine until a
permanent blue/black colour appears
in the conical flask.
A rough titration is done
first to give a rough
equivalence-point (endpoint), then repeated
more accurately to give
concordant results (+or0.1cm3).
61. After rinsing the pipette with a little of the vitamin C solution, pipette 25 cm3
of it into the conical flask. Add a few drops of starch solution to the vitamin C
solution in the conical flask.
82 .After rinsing the burette with a little iodine solution, fill the burette with the
iodine solution.
3. Note the initial burette reading. Since the solution has a dark colour, it is
difficult to see the bottom of the meniscus. Take the burette reading from the
top of the meniscus.
4. Add the iodine solution slowly from the burette whilst gently swirling the
solution in the conical flask. Initially you will see a blue/black colour as the iodine
reacts with the starch but this will rapidly disappear as the iodine reacts with the
vitamin C.
15. Near the end-point of the titration the colour disappears more slowly. At this
point add the iodine solution drop by drop until the solution just turns a blue/black
colour and remains so. This is the end-point of the titration i.e. all the vitamin C has
reacted. Note the final burette reading. Repeat the titrations until concordant
results are obtained.
.
Calculation
(a)Knowing the average volume and concentration of the
iodine solution used in the redox titration, the number of
moles of iodine can be calculated.
(b)With the result from step (a) and the balanced equation
for the redox reaction, we can work out the number of
moles of vitamin C in 25 cm3 of the vitamin C solution.
This can be scaled up to find the number of moles of
vitamin C in 250 cm3 of the vitamin C solution.
(c)Your final answer in step (b) will, of course, be equal to
the number of moles of vitamin C in the tablet. Using this
result and the mass of one mole of vitamin C (176 g) we
can finally work out the mass of vitamin C in the tablet.
Redox titrations
From standard grade you should be familiar with
volumetric titration questions
The same procedures can be used with a balanced
redox equation
Volumetric titrations
H2SO4 + 2NaOH → Na2SO4 + 2H2O
20 cm3 of sodium hydroxide solution was neutralised by 15cm3 of
0.1mol l-1 sulphuric acid.
Calculate the concentration of the sodium hydroxide solution.
0.15 mol l-1
Volumetric titrations
What volume of potassium hydroxide, KOH, concentration 2 mol l-1 is
required to neutralise 50cm3 of H2SO4, concentration 0.5 mol l-1 ?
0.05 Litres
Volumetric titrations
Examples for you to try
1) What volume of sulphuric acid, concentration 2 mol l-1 is
required to neutralise 25cm3 of KOH, concentration 4
mol l-1?
3
25 cm
2) If 50cm3 of KOH solution is neutralised by 17.8cm3 of
H2SO4 (2 mol l-1), what is the concentration of the
alkali?
1.42mol l-1
3) What volume of HCl, concentration 1.0 mol l-1 is required
to neutralise 100cm3 of NaOH solution concentration
50cm3
0.5moll-1?
Redox titrations
The same procedures can be used with a balanced redox equation
Permanganate ions react with hydrogen peroxide in acidic solution.
2MnO4- + 6H+ + 5H2O2
→
2Mn2+ + 8H2O + 5O2
25cm3 of hydrogen peroxide solution reacted with 16cm3 of permanganate
solution, concentration 0.1 mol l-1.
Calculate the concentration of the hydrogen peroxide solution.
0.16 mol l-1
Redox titrations
Try the following example
1. The overall equation for the reaction of I2 with SO32- ions is:
I2 + SO32- + H2O → 2I- + SO42- + 2H+
Calculate the volume of iodine solution (concentration 0·5mol 1-1)
needed to completely react with 50cm3 of sodium sulphite solution
of concentration 0·2mol 1-1.
20cm3
2. The overall equation for the reaction of Fe3+ ions with I- ions is:
2Fe3+ + 2I- → I2 + 2Fe2+
Calculate the volume of iodide solution (concentration 0·2mol 1-1)
needed to completely react with 25cm3 of iron (III) nitrate
solution of concentration 0·1mol 1-1.
12.5cm3
Redox titrations
Try the following example
a) Write the equation for the overall reaction between
acidified dichromate ions and bromide ions.
2Cr2O72- + 14H+ + 6Br- → 2Cr3+ + 7H2O + 3Br2
b) Calculate the volume of bromide solution, concentration
0.5 mol l-1 that would react completely with 50cm3 of 0.2
mol l-1 solution of acidified dichromate.
120cm3
Redox titrations
Try the following example
2Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
SO32- + H2O → SO42- + 2H+ + 2e-
a) Write the ion-electron equation for the
i) reduction of dichromate ions
ii) oxidation of sulphite ions
b) Write the equation for the overall reaction between
dichromate ions and sulphite ions.
2Cr2O72- + 14H+ + 3SO32- + 3H2O → 2Cr3+ + 7H2O + 3SO42-+ 6H+
c) Calculate the volume of dichromate solution 0.5 mol l-1
that would react completely with 30cm3 of 0.25 mol l-1
solution of sulphite ions
5cm3
Redox titrations
A 50·0cm3 sample of contaminated water containing chromate ions was
titrated and found to require 27·4 cm3 of 0·0200 mol l–1 iron(II) sulphate
solution to reach the end-point.
The redox equation for the reaction is:
3Fe2+(aq) + CrO42–(aq) + 8H+(aq) → 3Fe3+(aq) + Cr3+(aq) + 4H2O(l)
Calculate the chromate ion concentration, in mol l–1, present in the sample
of water.
Show your working clearly.
Concentration of CrO42- = 0·00365 (mol l-1 )
Redox titrations
MnO4- + 8H++ 5Fe2+
→ Mn2+ + 4H2O + 5Fe3+
8.25g of an iron (II) salt was dissolved in 250 cm3 of pure water. 25.0 cm3 aliquots were
pipetted from this stock solution and titrated with a standard solution of 0.02 moll-1
acidified potassium permaganate solution.
The titration values obtained were 24.95 cm3, 23.80 cm3 and 23.82 cm3.
(i) Calculate the average titration volume.
(ii) What is meant by a standard solution?
(iii) Calculate the number of moles of potasssium permaganate which reacted with the
25.0cm3 aliquot.
(iv) Calculate the number of moles of iron (II) in the 25cm3 aliquot.
(v) Calculate the number of moles of iron in the original sample.
(vi) Calculate the total mass of iron in the original sample of the iron(II) salt.
(vii) calculate the % iron in the salt.
(viii) Why is the potassium permanganate solution acidified?
Redox Titrations
Titration is a technique for measuring the concentration
of a solution. A solution of known concentration is used to
work out the unknown concentration of another solution.
Redox titrations involve solutions of reducing and
oxidising agents.
At equivalence-point of a redox titration precisely enough
electrons have been removed to oxidise all of the reducing
agent.
Redox Titrations
Aim : To estimate the iron(II) content of an iron tablet, a
tablet is first dissolved in distilled water. This solution
is then titrated against previously standardised
potassium manganate(VII) solution. The reaction is
represented by the equation:
5 Fe2+ (aq) + 8H+ (aq) + MnO4- (aq)
purple
 5 Fe3+ (aq)
+ Mn2+ (aq) + 4H2O(l)
colourless
Redox Titrations
5 Fe2+ (aq) + 8H+ (aq) + MnO4- (aq)
 5 Fe3+ (aq)
purple
+ Mn2+ (aq) + 4H2O(l)
colourless
1. Find the mass of one iron tablets.
2. Crush the tablet in a mortar and pestle. Transfer all the
ground material to a beaker where it is dissolved in
about 100 cm3 of distilled water.
3. All of this solution (including washings) is transferred to
a 250 cm3 volumetric flask and the solution made up to
the mark with deionised water. The volumetric flask
should be stoppered and inverted several times. This is
the solution containing iron(II) ions.
RedoxTitration
What to do:
Carefully fill the burette
with potassium permanganate 0.001moll-1
Carefully pipette exactly
25 ml of iron (II) sulphate
into the conical flask.
Add the permanganate until a
permanent purple colour appears
in the conical flask.
A rough titration is done
first to give a rough
equivalence-point (endpoint), then repeated
more accurately to give
concordant results.
Questions relating to the
experiment
• In this experiment why is dilute sulfuric acid
added to the permaganate solution?
• Why are burette readings taken from the top of the
meniscus?
• How is the end-point of the titration detected?
• Why is a rough titration carried out?
• Why is more than one titration carried out
subsequently?
The chemistry of fizzy sweets
A gelatine sweet incorporates soluble flavourings
and sweeteners
A common ingredient is citric acid which gives the
sharp, citrus taste.
A role of the analytical chemist is to check that the
gum contains the correct amount of citric acid
confirming that the plant is working efficiently.
Jelly sweet analysis
• How do chemists check what it contains?
The reaction
C6H8O7 + 3NaOH
C6H5O7Na3+3H2O
Citric acid is neutralised by sodium hydroxide
Phenolphthalein indicator is colourless in acid but
changes to pink once the acid is used up and the
alkali is in excess
What to do
• Accurately weigh out approximately 2g of jelly
sweet.
• Add to 75cm3 warm water and stir until the jelly
and the soluble citric acid are dissolved.
• Transfer this to a 100 cm3 volumetric flask,
together with the washings and make up to the
mark.
What to do
• Fill a burette with 0.001 mol l-1 NaOH
• Pipette 25cm3 of the citric acid into a
conical flask and add a few drops of
phenolphthalein indicator
• Titrate against the NaOH until a pink
colour is seen. Repeat until concordant
results are obtained.
Calculation
Calculate the percentage mass of citric acid in the sweet.
Citric acid : NaOH
1 mol : 3 mol
Redox Titrations, Vitamin C
I2 (aq) + 2e-  2I - (aq)
reduction
C6H8O6 
oxidation
C6H6O6 + 2H+ (aq) + 2e-
I2 (aq) + C6H8O6
Blue/Black (in the

C6H6O6 + 2H+ + 2I- (aq)
colourless
presence of starch)
Iodine, those concentration is known (in the burette)
acts as an oxidising agent.
Vitamin C, the unknown concentration (in the conical flask)
is a reducing agent.
Starch is added to show when the end-point is reached.