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Unit 3
Chemical Reactions
Menu
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The Chemical Industry
Hess’s Law
Equilibrium
Acids and Bases
Redox Reactions
Nuclear Chemistry
The Chemical Industry
The Chemical Industry
Major contributor to quality of
life and economy.
The Chemical Industry
• Quality of life
• Fuels (eg petrol for cars)
• Plastics (Polythene etc)
• Agrochemicals
(Fertilisers, pesticides
etc)
• Alloys (Inc. Steel for
building)
• Chemicals (eg Cl2 for
water purification)
• Dyes (for clothing etc)
• Cosmetics and medicines
• Soaps and detergents
• Etc!!!!
The Chemical Industry
• Contributes to
National Economy
• Major employer of
people at all skill
levels
• Revenue from
taxation on fuels etc
• Revenue from sales
of product
• Revenue from
exports of products
The Chemical Industry
• Research
chemists identify
a chemical route
to make a new
product, using
available
reactants.
The Chemical Industry
• Feasibility study
produces small
amounts of
product – to see if
the process will
work
The Chemical Industry
• The process is now
scaled up to go into
full scale production.
• Process so far will
have taken months.
• Many problems will
have been
encountered and will
have to be resolved
before full scale
production
commences
The Chemical Industry
• Chemical plant is
built in a suitable
site
• Operators employed
• Early production will
allow monitoring of
cost, safety,
pollution risks, yield
and profitability
The Chemical Industry
Unreacted feedstocks
recycled
Feedstock
MIXER
REACTION
VESSEL
BY-PRODUCT
SEPARATOR
PRODUCT
The Chemical Industry Feedstocks
Fossil fuels
Metallic ores
Minerals
Coal, oil and natural
gas
Haematite to make
iron,
Bauxite to make
aluminium
Limestone needed in
Blast furnace
Air
Water
Supplies O2 and N2
Can be used as a
reactanct or as a
coolant or in heat
exchangers.
The Chemical Industry
• Can be Continuous process
• Or can be Batch Process
The Chemical Industry
• Continuous
Process
• Used by big industries
where large quantities of
product are required
• Requires small workforce
• Often automated /
computer controlled
• Quality of product
checked remotely
• Energy efficiency usually
good
• Plants expensive to build
• Plants not flexible
The Chemical Industry
• Batch Process
• Make substance which
are required in smaller
amounts
• Process looks more like
the initial reaction
• Overhaul of system
needed regularly – time
and energy lost if plant
has to be shut down
• Plant can be more
flexible
• Plant is usually less
expensive to build
initially
The Chemical Industry:
The Costs Involved
Cost
Capital costs
Example
Fixed costs
Repayment of loan
Wages, Council tax
(Stay the same regardless of
whether plant runs at full or half
capacity)
Variable costs
(Vary dependent on whether
plant is running at full or half
capacity.)
Building the plant
Road and rail links
(Usually needs a substantial bank
loan)
Cost of raw materials, and other
chemicals required
The Chemical Industry
Industries can be classed as:
• Labour intensive
• Capital intensive
The Chemical Industry
• Service industries
(Catering,
education,
healthcare), are
labour intensive
The Chemical Industry
• Chemical industry
tends to be more
Capital intensive
as a large
investment is
required to buy
equipment and
build plants
The Chemical Industry
• Expectations of
work safety and a
clean
environment
increase during
the twentieth
century
• H & S legislation
protects
workforce
The Chemical Industry
• Tradition is important – steel
making continues in areas where it
was set up even if raw materials
are no longer available locally
• Transport options are important
The Chemical Industry
Choice of a particular chemical route is
dependent upon:
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Cost of raw materials
Suitability of feedstocks
Yield of product
Option to recycle unreacted feedstock
Marketability of by products
Costs of getting rid of wastes, and safety
considerations for workforce and locals
• Prevention of pollution
The Chemical Industry
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Industry.
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Hess’s Law
Hess’s law
• Hess’s law states that the enthalpy
change for a chemical reaction is
independent of the route taken.
• This means that chemical
equations can be treated like
simultaneous equations.
• Enthalpy changes can be worked
out using Hess’s law.
Hess’s law
• Calculate the enthalpy change for
the reaction:
C(s) + 2H2(g)  CH4(g)
using the enthalpies of combustion
of carbon, hydrogen and methane.
Hess’s law
First write the target equation.
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
Then write the given equations.
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
H2 + ½ O2  H2O DH= -286 kJ
CH4+ 2O2 CO2 + 2H2O DH=-891 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
Build up the target equation from
the given equations.
If we multiply we must also
multiply DH.
If we reverse an equation we
reverse the sign of DH.
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
H2 + ½ O2  H2O DH= -286 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CH4 + 2O2  CO2+2H2O DH=-891 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CO2+2H2O  CH4 + 2O2 DH=+891 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CO2+2H2O  CH4 + 2O2 DH=+891 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
We can add all the equations,
striking out species that will
appear in equal numbers on both
sides.
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CO2+2H2O  CH4 + 2O2 DH=+891 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CO2+ 2H2O  CH4 +2O2 DH=+891 kJ
C + 2H2  CH4 DH=(-394 –
572 + 891)kJ = -75 kJ
Hess’s law
C(s) + 2H2(g)  CH4(g) DH=?
C + O2  CO2 DH= -394 kJ
2H2 + O2  2H2O DH= -572 kJ
CO2+ 2H2O  CH4 +2O2 DH=+891 kJ
C + 2H2  CH4 DH=-75 kJ
Hess’s Law
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Equilibrium
Dynamic Equilibrium
• Reversible reactions reach a state
of dynamic equilibrium
• The rates of forward and reverse
reactions are equal.
• At equilibrium, the concentrations
of reactants and products remain
constant, although not necessarily
equal.
Changing the Equilibrium
• Using a catalyst does not change
the position of the equilibrium.
• A catalyst speeds up both the
forward and back reactions equally
and so the equilibrium is reached
more quickly.
Changing the Equilibrium
• Changes in concentration, pressure
and temperature can alter the
position of equilibrium.
• Le Chatelier’s Principle states that
when we act on an equilibrium the
position of the equilibrium will
move to reduce the effect of the
change.
Concentration
• Consider the equilibrium:
A+BC+D
If we increase the concentration of
A, we speed up the forward
reaction.
This results in more C and D being
formed.
Concentration
• Consider the equilibrium:
Br2(aq) + H2O(l)  2H+(aq) + Br-(aq) + BrO-(aq)
The solution is red-brown, due the Br2
molecules.
If we add sodium bromide, increasing
the concentration of Br-, we favour the
RHS and so the equilibrium moves to
the left.
The red-brown colour will increase.
Pressure
• Remember:
• 1 mole of any gas has the same
volume (under the same conditions
of pressure and temperature).
• This means that the number of
moles of has are the same as the
volumes.
Pressure
• Increasing pressure means putting
the same number of moles in a
smaller space.
• This is the same as increasing
concentration.
• To reduce this effect the
equilibrium will shift so as to
reduce the number of moles of gas.
Pressure
• Increasing pressure favours the side
with the smaller volume of gas.
Consider:
N2O4(g)  2NO2(g)
1 mole
2 moles
1 volume
2 volumes
• If we increase the pressure we favour
the forward reaction, so more N2O4 is
formed.
Temperature
• An equilibrium involves two
opposite reactions.
• One of these processes must
release energy (exothermic).
• The reverse process must take in
energy (endothermic).
Temperature
• First consider an exothermic
reaction.
Exothermic Reaction
This is the
distribution
of
molecular
energy
Activation Energy
These molecules
have sufficient
energy to react
Now increase the
molecular energy
by heating
Now these
can react
Increasing temperature leads to a small increase
in the number of molecules with sufficient
activation energy
Temperature
• Now consider an endothermic
reaction.
Endothermic Reaction
This is the
distribution
of
molecular
energy
Activation Energy
These molecules
have sufficient
energy to react
Now increase the
molecular energy
by heating
Now these
can react.
Increasing temperature leads to a greater increase
in the number of molecules with sufficient
activation energy
Temperature
• The percentage increase in the
number of molecules with
sufficient activation energy is
much greater in the endothermic
reaction, compared to the
exothermic reaction.
• Thus both the endothermic and
exothermic processes are speeded
up by increasing temperature.
• However an increase in
temperature has a greater effect
on the endothermic process.
• Increasing temperature favours
the endothermic side of the
equilibrium. Consider:
N2O4(g)  2NO2(g) DH= +58 kJ
• If we increase the temperature we
favour the forward reaction so
more NO2 is formed.
The Haber Process
• The Haber process involves the
preparation of ammonia from
nitrogen and hydrogen.
N2 + 3H2  2NH3 DH = -88 kJ
• We shall look at the factors
affecting this equilibrium.
The Haber Process
N2 + 3H2  2NH3 DH = -88 kJ
A catalyst of finely divided iron is
used to increase the reaction speed
and so shorten the time needed to
reach the equilibrium.
The Haber Process
N2 + 3H2  2NH3 DH = -88 kJ
1 mole 3 moles
2 moles
1 vol
3 vols
2 vols
4 vols
2 vols
Since the RHS has a lower volume of
gas than the LHS, higher pressure will
favour the production of ammonia.
A reaction chamber to withstand the
higher pressure will cost much more.
The Haber Process
N2 + 3H2  2NH3 DH = -88 kJ
Since the forward reaction is
exothermic more ammonia will be
produced at low temperatures.
At low temperatures the reaction is
very slow so the rate of production
of ammonia is low.
The Haber Process
N2 + 3H2  2NH3 DH = -88 kJ
To ensure maximum conversion
the unreacted gases are recycled
through the reaction chamber after
reaction.
The Haber Process
N2 + 3H2  2NH3 DH = -88 kJ
To achieve the most profitable
production of ammonia the
following conditions are used:
iron powder as catalyst
250 atmospheres pressure
temperature of 500oC - 600oC
The Haber Process
Unreacted N2 + H2
recycled
N2 + H2
MIXER
REACTION
Chamber
Fe catalyst
SEPARATOR
NH3
Equilibrium
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Acids and Bases
pH
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pH is a scale of acidity.
It can be measured using:
pH paper
Universal Indicator solution
A pH meter.
• We carry out an
experiment where
we progressively
dilute acid.
Tube 1
10 ml 0.1 mol/l
hydrochloric acid
Transfer
1 ml of
acid from
Tube 1
Tube 2
Add 9 ml
of water
to Tube 2
0.01 mol/l
hydrochloric
acid
Tube 2
• Repeat this process five more
times so you have a series test
tubes.
1
2
3
4
5
6
7
• Concentrations are:
1
[H+]
2
10-1 10-2
3
4
10-3 10-4
5
10-5
6
10-6
7
10-7
• Add Universal Indicator:
1
[H+]
pH
2
10-1 10-2
1
2
3
4
10-3 10-4
3
4
5
10-5
5
6
7
10-6
10-7
6
7
• Look for a relationship between
the concentration of acid and the
pH.
• If [H+] = 10-x
• pH = x
• We repeat the
experiment but
this time we
progressively
dilute alkali.
Tube 1
10 ml 0.1 mol/l
sodium hydroxide
• You now have five test tubes,
numbered as below.
13
12
11
10
9
8
7
• Concentrations are:
13
[OH-]
12
10-1 10-2
11
10
10-3 10-4
9
8
7
10-5
10-6
10-7
• Add Universal Indicator:
13
12
[OH-] 10-1 10-2
pH
13
12
11
10
10-3 10-4
11
10
9
8
7
10-5
10-6
10-7
9
8
7
• Look for a relationship between
the concentration of alkali and the
pH.
• If [OH-] = 10-y
• pH = 14-y
[H+]
mol/l
10-1
10-2
10-3
10-4
10-5
10-6
10-7
pH
1
2
3
4
5
6
7
[OH-]
mol/l
10-1
10-2
10-3
10-4
10-5
10-6
10-7
pH
13
12
11
10
9
8
7
• If we look at water, pH=7
• [H+] = [OH-] = 10-7 mol/l
[H+] x [OH-] = 10-14 mol2/l2
This is due to the equilibrium in water:
H2O(l)  H+(aq) + OH-(aq)
• For any solution
[H+] x [OH-] = 10-14 mol2/l2
• Thus we can find [H+] for any
solution.
• What is [H+] of a solution with pH
10?
[OH-] = 10-4 mol/l
[H+] x 10-4 = 10-14 mol2/l2
Thus [H+] = 10-10
Strong and Weak Acids
• A strong acid is one which
completely dissociates in solution:
HCl(aq)  H+(aq) + Cl-(aq)
• A weak acid is one which partially
dissociates in solution:
CH3CO2H(aq)H+(aq) + CH3CO2(aq)
• We can compare eqimolar
solutions of strong and weak acids
e.g. 0.1 mol/l hydrochloric acid and
0.1 mol/l ethanoic acid.
• We compare pH, conductivity,
reaction rates and stoichiomery.
Test
100 ml 0.1
mol/l HCl
pH
1
100 ml 0.1
mol/l
CH3CO2H
3
Conductivity Very high
Low
Rate of
Fast
reaction
Stoichiomery Reacts with
0.4 g NaOH
Slow
Reacts with
0.4 g NaOH
• The differences between the
properties of strong and weak
acids are caused by the fact that
weak acids contain many fewer H+
ions than strong acids.
• Both acids can produce the same
number of H+ ions, its just that
weak acids do so more slowly.
Weak Acids
• Solutions of ethanoic acid, carbon
dioxide and sulphur dioxide are
weak acids.
CH3CO2H(aq)H+(aq) + CH3CO2(aq)
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3(aq)  2H+(aq) + CO32-(aq)
SO2(g) + H2O(l)  H2SO3(aq)
H2SO3(aq)  2H+(aq) + SO32-(aq)
Strong and Weak Bases
• A strong base is one which
completely dissociates in solution:
NaOH(aq)  Na+(aq) + OH-(aq)
• A weak base is one which partially
dissociates in solution:
NH4OH(aq)NH4+(aq) + OH-(aq)
• We can compare eqimolar solutions of
strong and weak bases e.g. 0.1 mol/l
sodium hydroxide and 0.1 mol/l
ammonium hydroxide.
• When we compare pH, conductivity,
reaction rates and stoichiomery we find
similar results to the comparison of
weak and strong acids.
Weak Bases
• A solution of ammonia is a weak
base.
NH3(g) + H2O(l)  NH4OH(aq)
NH4OH(aq) NH4+(aq) + OH-(aq)
Acids + Bases
• A strong acid and a strong base
produce a salt which is neutral.
• A strong acid and a weak base
produce a salt which is acidic.
• A weak acid and a strong base
produce a salt which is basic.
Basic Salts
Basic Salts
• Sodium carbonate is completely
ionised.
Basic Salts
• Sodium carbonate is completely
ionised.
Na2CO3(aq)  2Na+(aq) + CO32-(aq)
Basic Salts
• Sodium carbonate is completely
ionised.
Na2CO3(aq)  2Na+(aq) + CO32-(aq)
Water is also present.
Basic Salts
• Sodium carbonate is completely
ionised.
Na2CO3(aq)  2Na+(aq) + CO32-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
Basic Salts
• Sodium carbonate is completely
ionised.
Na2CO3(aq)  2Na+(aq) + CO32-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
The ions set up an equilibrium.
Basic Salts
• Sodium carbonate is completely
ionised.
Na2CO3(aq)  2Na+(aq) + CO32-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
The ions set up an equilibrium.
2H+(aq) + CO32-(aq)  H2CO3(aq)
Basic Salts
• This removes of H+(aq) from
water.
H2O(l)  H+(aq) + OH-(aq)
Basic Salts
• This removes of H+(aq) from
water.
H2O(l)  H+(aq) + OH-(aq)
The OH-(aq) left behind make the
resulting solution basic.
Acid Salts
Acid Salts
• Ammonium chloride is completely
ionised.
Acid Salts
• Ammonium chloride is completely
ionised.
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Acid Salts
• Ammonium chloride is completely
ionised.
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Water is also present.
Acid Salts
• Ammonium chloride is completely
ionised.
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
Acid Salts
• Ammonium chloride is completely
ionised.
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
The ions set up an equilibrium.
Acid Salts
• Ammonium chloride is completely
ionised.
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Water is also present.
H2O(l)  H+(aq) + OH-(aq)
The ions set up an equilibrium.
NH4+(aq) + OH-(aq)  NH4 OH(aq)
Acid Salts
• This removes of OH-(aq) from
water.
H2O(l)  H+(aq) + OH-(aq)
Acid Salts
• This removes of OH-(aq) from
water.
H2O(l)  H+(aq) + OH-(aq)
The H+(aq) left behind make the
resulting solution acidic.
Acids and Bases
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Bases.
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Redox Reactions
Redox
• An oxidation reaction is one where
electrons are lost.
Zn(s)  Zn2+(aq) + 2e
• A reduction reaction is one where
electrons are gained.
Cu2+(aq) + 2e  Cu(s)
• A redox reaction is one in which both
oxidation and reduction are occurring.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Redox
• An oxidising agent is a substance
which accepts electrons.
• This means that an oxidising agent
must itself be reduced.
Redox
• A reducing agent is a substance
which donates electrons.
• This means that a reducing agent
must itself be oxidised.
Redox
• We should be able to recognise
oxidising and reducing agents from
the reaction equation.
• 5Fe2+ + MnO4- + 8H+ 
5Fe3+ + Mn 2+ + 4H2O
• Fe2+ is oxidised to Fe3+ so MnO4–
acts as an oxidising agent.
Writing Ion-Electron
Equations.
• Simple equations can be obtained
from the data booklet.
• More complex equations are
written using the following
routine.
Writing Ion-Electron
Equations.
• Write the reactants and products.
2IO3-  I2
• Add H2O to the side with less oxygen.
2IO3-  I2 + 6H2O
• Add H+ to the other side.
2IO3- + 12H+  I2 + 6H2O
• Balance charge by adding electrons.
2IO3- + 12H+ + 10e-  I2 + 6H2O
Combining Oxidation and
Reduction Equations.
• Combining the ion-electron half
equations produces the overall
reaction equation.
• This must be done so that the
number of electrons on opposie
sides are equal, and so cancel each
other out.
Combining Oxidation and
Reduction Equations.
• Oxidation
Combining Oxidation and
Reduction Equations.
• Oxidation
2IO3- + 12H+ + 10e-  I2 + 6H2O
Combining Oxidation and
Reduction Equations.
• Oxidation
2IO3- + 12H+ + 10e-  I2 + 6H2O
• Reduction
Combining Oxidation and
Reduction Equations.
• Oxidation
2IO3- + 12H+ + 10e-  I2 + 6H2O
• Reduction
SO32- + H2O  SO42- +2H+ + 2e-
Combining Oxidation and
Reduction Equations.
• Oxidation
2IO3- + 12H+ + 10e  I2 + 6H2O
• Reduction multiplied by 5
5SO32- + 5H2O  5SO42- +10H++ 10e
Combining Oxidation and
Reduction Equations.
• Add the equations
2IO3- + 12H+ + 10e  I2 + 6H2O
5SO32- + 5H2O  5SO42- +10H++ 10e
2IO3- + 2H+ + 5SO32- 
I2 + H2O + 5SO42We now can extract the mole
relationship – 2 moles iodate react with
5 moles of sulphite
Redox Titrations.
• These can be carried out to
calculate concentration.
• Many use permanganate or
starch/iodine reactions which are
self-indicating – the colour change
of the reaction tells you when the
end point is reached.
Redox Titrations.
• It was found that 12.5 ml of of 0.1
mol/l acidified potassium
dichromate was required to oxidise
the alcohol in a sample of 1 ml of
wine.
• Calculate the mass of alcohol in 1
ml of wine.
Redox Titrations.
• Equations:
Cr2O72-+14H+ + 6e 2Cr 3+ + 7H2O
C2H5OH + H2O CH3COOH
+4H++4e
Mole Relationship
2 moles dichromate react with 3
moles ethanol
1 mole dichromate react with 1.5
moles ethanol
Redox Titrations.
12.5 ml of 0.1 mol/l dichromate contain
0.0125x0.1 moles dichromate.
1.25x10-3 moles
Moles of alcohol = 1.25x10-3 x1.5
= 1.875x10-3
Mass of alcohol = 46 x 1.875x10-3 g
= 0.08625 g
Electrolysis
• Electrolysis takes place when
electricity is passed through an
ionic liquid.
• Chemical reaction take place at the
electrodes – reduction at the
negative electrode and oxidation at
the positive electrode.
Electrolysis
• The electrode reactions can be
represented by ion electron
equations.
• In the electrolysis of nickel(II)
chloride the reactions are:
+ electrode 2Cl-  Cl2 + 2e
- electrode Ni2+ + 2e  Ni
Electrolysis
+ electrode 2Cl-  Cl2 + 2e
- electrode Ni2+ + 2e  Ni
• In both of these ion electron
equations one mole of product is
produced by two moles of
electrons.
The Faraday
• To find the value for one mole of
electrons multiply Avogadro’s
number by the charge on the
electron (1.6x10-19 coulombs)
• One mole of electrons is called a
Faraday and is 96,500 coulombs.
The Faraday
• Using the value for the Faraday
and the equation:
• Charge =
Current x Time
(Coulombs) (Amps) (Seconds)
we can carry out many
calculations.
Redox Reactions
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Reactions.
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Nuclear Chemistry
Stable nuclei
• Nuclei contain protons and
neutrons.
• Energy is needed to hold these
particles together.
• We can plot the number of protons
against the number of neutrons.
Stable nuclei
Number of protons v number of neutrons
neutrons
150
100
50
0
0
20
40
protons
60
80
100
Stable nuclei
• All stable nuclei fit in a narrow
band
• Some nuclei are unstable because
they need too much energy to hold
them together.
• Thus they split apart, sending out
some small particles.
Radioactive decay
Particle
Symbol
Nature
alpha
a
beta
b
0
1
gamma
g
radiation
4
2
He
e
Stopped by
Sheet of paper
Few cm of
aluminium
Many cms of
lead
a decay
 a decay takes place when the
nucleus ejects a helium nucleus.
• This causes a change in the
nucleus.
M
M 4
4
A
A 2
2
X  
Y  He
U   Th  He
239
92
235
90
4
2
b decay
 b decay takes place when the
nucleus ejects an electron.
• This causes a change in the
nucleus.
P   Q e
C   N  e
M
A
14
6
M
A1
14
7
0
1
0
1
g decay
 g decay takes place when the
nucleus loses energy.
• This is the extra energy which is no
longer needed to hold the nucleus
together.
Nuclear Equations
• When we write a nuclear equation
the sum of the mass numbers and
atomic numbers on each side must
be equal.
239
94
239
94
100
1
Pu  01n  137
52Te  42 Mo 30 n
Pu  n   Te Mo 3 n
1
0
137
52
100
42
1
0
Half life
• Half life is the time which it takes
for the radioactivity to half.
• For any radioactive substance this
time is constant.
Half life
• The decay of individual nuclei
within a sample is random and is
does not depend of chemical or
physical state of the element.
• Half lives of individual elements
may vary from seconds to
thousands of years.
Half life
• Calculations involving half life
usually involve precise fractions
e.g.
• 3H is a b-emitting isotope with a
half life of 12.3 years. How long
will it take for the radioactivity of a
sample to drop to 1/8 of its
original value?
Half life
• 3H is a b-emitting isotope with a
half life of 12.3 years. How long
will it take for the radioactivity of a
sample to drop to 1/8 of its
original value?
• Time
12.3y
24.6y
36.9y
• Fraction
½
¼
1/8
Half life
• For examples where the numbers
are more complex the quantity of
radioactive material against time is
best estimated from a graph.
Half life
Activity
120
100
80
60
Series1
40
20
0
1
2
3
4
Time
5
6
7
8
9 10 11 12
The Nuclear Chemistry
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Chemistry.
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The End
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