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Title: Chemical Reactions


1
Chemical Reactions UNIT 3
2
b) Hess's Law
3
Hesss Law
Hesss Law states that the enthalpy change for
a reaction is always the same no matter what
route was taken to get from reactants to
products or The total energy for the overall
reaction is the sum of the energies of all the
other steps This is useful If we want to know the
enthalpy change for a reaction which is
impractical or impossible to do directly
4
Enthalpy
In Unit 1 we learned to measure calculate the
enthalpies of combustion, solution and
neutralisation We did this indirectly by heating
water with the reaction and using the
equation Energy change (Eh) cmDT Then working
out the enthalpy change (energy change for one
mole) by dividing Eh by the number of moles used
in the reaction Enthalpy change (DH)
Eh moles
5
Prescribed Practical 1
In PPA 1 you are asked to confirm Hesss Law from
the following KOH(s)
Route 1 (DH1) HCL(aq)
KCl(aq)
Route 2(a) (DH2a) H2O(l)
Route 2(b) (DH2b) HCl(aq)
KOH(aq)
The first route (1) adds acid directly to solid
KOH pellets The second 2 first (a) adds water to
the pellets, and then (b) adds the acid to the
KOH solution Hesss law should mean that DH1DH2a
DH2b
6
Prescribed Practical 1
The temperature changes in each step are measured
and cmDT used to find the energy released
(temperature increased) The mass of the pellets
are measured and the moles of KOH used calculated
using Moles mass/formula mass
Typically the calculations yield results for
route 1 and route 2 which are about 3 or 4
kjmol-1 different Consistent with experimental
error
7
Enthalpy calculations using Hesss Law
The required equation can be built up from
equations given e.g. Given the
following 1 CO ½ O2
CO2 DH1-283kjmol-1
CuO DH2-155kjmol-1
2 Cu ½ O2
Find the enthalpy change for 3 CuO CO
CO2 Cu DH3 ?
Looking at whats in each of the equations The
equation 3 can be built up using 1 and the
reverse of 2
8
The reverse of 2 we will label 4 4 CuO
Cu ½ O2 DH4155kjmol-1
1 CO ½ O2
CO2 DH1-283kjmol-1
We combine the two equations and add the DH
values, careful to remember the signs Then cancel
those species which appear on both sides to give
the required equation CuO CO ½ O2
Cu ½ O2 CO2 DH3-283 155kjmol-1 DH3
-128kjmol-1
9
Examples to try
The thermite reaction is a displacement reaction
used to join sections of railway The equation for
the reaction is 2Al(s) Fe2O3(s)
2Fe(s) Al2O3(s)
Calculate the enthalpy for the reaction given
that 2Al(s) 1 ½ O2(g)
Al2O3(s) DH1 -1600kjmol-1
Fe2O3(s) DH2 -820kjmol-1
2Fe(s) 1 ½ O2(g)
10
Example solution
The required reaction can be built up from
reaction 1 and the reverse of reaction 2 2Al(s)
1 ½ O2(g)
Al2O3(s) DH1 -1600kjmol-1
Fe2O3(s)
2Fe(s) 1 ½ O2(g) DH2 820kjmol-1
Adding the reactions and cancelling species which
appear on both sides gives us 2Al(s) 1 ½
O2(g) Fe2O3(s)
Al2O3(s) 2Fe(s) 1 ½ O2(g) DH1 DH2 -1600
(820) -780kjmol-1
11
Enthalpy calculations using Hesss Law
Building up an enthalpy cycle First you write a
balanced equation for the required reaction and
then build a cycle with the steps you have to
connect your reactant to the product
required e.g. There are two forms of phosphorous,
red (4P) and yellow (P4). Given the enthalpies of
combustion of each calculate the enthalpy of
conversion from red to yellow Start with balanced
equation 4P(red)
P4(yellow)
12
Using the balanced equation as a starting point
make a cycle where you have 2 alternative routes
to P4O10 4P(red) P4(yellow)
DH1
-2390kjmol-1
4 x 727.5kjmol-1
Enthalpy of combustion DH3
Enthalpy of combustion DH2
P4O10
Now enter all the DH values taking care to make
sure the sign matches the direction of the arrow
4P to P4O10 is 4 x the enthalpy of
combustion Using Hesss Law we can
show DH1 DH3 DH2 or DH1 (4x 727.5)
(-2930) 20kjmol-1
13
Examples to try
Calculate the energy change when one mole of
hydrogen peroxide decomposes to from water and
oxygen given H2(g) O2(g)
H2O2(l) DH1 -189kjmol-1
H2(g) ½ O2(g)
H2O(l) DH2 -286kjmol-1
14
Example solution
The common product in these reactions is
H2O H2(g) ½ O2
H2O(l) ½ O2
DH2
O2
DH3
DH1
H2O2(l)
Using Hesss Law DH2 DH1 DH3 -286 -189 DH
3 DH3 -286 189 -97kjmol-1
15
c) Equilibrium
16
Dynamic Equilibrium
ACID ALCOHOL ESTER WATER
Dynamic - The reactions are still
occuring. Equilibrium- When the rate of the
forward reaction equals the rate of the reverse
reaction.
17
Shifting Equilibrium
Cl2(g) ICl(l) ICl3(s)
Brown liquid
Yellow / green gas
Yellow solid
18
Le Chateliers Principle
If a system at equilibrium is subjected to any
change, the system readjusts itself to counteract
the applied change.
19
Pressure Change
Scholar Fig 2.13
20
Temperature Change
Scholar Fig 2.12
21
Lowering temperature favours exothermic
reaction Increasing temperature favours
endothermic reaction
22
Concentration Change
Scholar Fig 2.15
23
d) Acids Bases
24
You need to know
  • Metal oxides produce alkalis, non metal oxides
    produce acids.
  • All acids contain H ions, alkalis contain OH-.
  • Neutral solutions and water have pH 7.
  • Reactions of acids with metal oxides, metal
    carbonates metals.

25
The pH scale
  • A continuous scale from less than 0 to more than
    14. H can vary from 10 mol l-1 to 10-15 mol l-1.
  • The concentration of hydrogen ions in solution is
    measured in pH units.
  • pH stands for the negative logarithm (to base 10)
    of the hydrogen ion concentration.
  • pH -log10H(aq)

26
Calculating pH
H moles l-1 pH
101 100 10-1 10-2 10-3 10-7 10-14
-1
0
1
2
3
7
14
27
pH
  • For water and neutral solutions in water, H and
    OH- ions come from the partial dissociation of
    water molecules.
  • In water and neutral solutions there is an
    equilibrium between H(aq) and OH-(aq) ions and
    water molecules.

H2O(l) H(aq) OH-(aq)
28
Equilibrium
  • An equilibrium is reached with the concentrations
    of both H and OH- ions equal to 10-7 mol l-1 at
    25ºC.
  • H OH-
  • 10-7moles l-1 10-7moles l-1
  • so
  • HOH- 10-7moles l-1 x 10-7moles l-1
  • 10-14moles2 l-2

Ionic product of water
29
Calculations
  • We can now calculate pH, H and OH- using the
    relationship
  • H 10-14 moles l-1
  • OH-
  • Or
  • OH- 10-14 moles l-1
  • H

10-14
H
OH-
30
Example 1
  • What is the concentration of OH- ions in a
    solution containing 0.01 moles l-1 of H ions.
  • H 10-2 moles l-1
  • OH- 10-14 10-14 10-12 M
  • H 10-2

31
Strong acids
  • Strong acids ionise completely into ions.
  • HCl(aq) H(aq) Cl-(aq)
  • H2SO4(aq) 2H(aq) SO4-(aq)

Ions are separate in solution, no HCl left.
32
Weak acids
  • Partially ionise. An equilibrium is reached
    between the ions and the molecule.
  • When diluted the equilibrium will shift to the
    right. As dilution increases the degree of
    ionisation (dissociation) also increases.

CH3COOH(aq) CH3COO-(aq) H(aq)
33
Strong weak acids
  • Differ in pH, conductivity and reaction rates.
  • By comparing equimolar solutions we can find out
    the differences.

Properties 0.1 M HCl 0.1M CH3COOH
H pH Conductivity Rate of reaction with Mg Rate of reaction with CaCO3 0.1 mol l-1 1 High Fast Fast 0.0013 mol l-1 2.88 Low Slow Slow
34
Neutralising weak acids
CH3COOH CH3COO- H
  • As the reaction proceeds H ions are used up.
  • This forces the equilibrium to the right, the
    ionisation step.
  • Eventually all of the molecules dissociate.
  • Therefore it requires the same amount of base to
    neutralise a certain volume of either 0.1M HCl or
    0.1M CH3COOH

35
Strong weak alkalis
  • Strong bases fully dissociate.
  • Weak bases only partially dissociate.

Properties 0.1M NaOH 0.1M NH3(aq)
OH- pH Conductivity 0.1mol l-1 13 High 0.1mol l-1 11.12 Low
36
pH of salt solutions
  • In SG you learned that and acid an alkali react
    to form a neutral salt.
  • This is only true if using a strong acid a
    strong alkali or a weak acid and a weak alkali.

37
Salts
Acid Alkali pH of salt if soluble
Strong Strong Weak Strong Weak Strong Neutral Acid Alkali
38
Soaps
  • Soaps are salts of strong alkalis and weak acids.
  • Usually KOH or NaOH is used at the alkali.
  • A long chain alkanoic acid (fatty acid) is used,
  • This makes soaps alkaline in nature.

39
e) Redox Reactions
40
Oxidation
is
Loss
Reduction
is
Gain
41
Displacement
  • When a metal that is more likely to lose
    electrons (more reactive) pushes another metal
    out of a compound.
  • Magnesium displaces copper from a solution of Cu
    (II) ions.

Mg(s) Mg2(aq) 2e-
Oxidation
Reduction
Cu2(aq) Mg(s) Cu(s) Mg2(aq)
REDOX
42
Zinc displaces silver from a solution of Ag(I)
ions
Zn(s) Zn2(aq) 2e-
Oxidation
( x2)
Ag e- Ag(s)
2
2
2
Reduction
2Ag (aq) Zn(s) 2Ag(s) Zn2(aq)
Redox
43
Oxidising / reducing agents.
  • An oxidising reagent is one that gains electrons
    and as a result is reduced itself.
  • A reduction reagent is one that loses electrons
    and as a result is oxidised itself.

44
WritingIon-Electron Equations
45
Balance all elements except oxygen and water
46
If hydrogen is unbalanced add H to balance.
If oxygen is unbalanced add H20 to balance.
47
2
7
Check charges on each side.
48
2
2
Add appropriate number of electrons to the side
which is short of negative charges
49
Electrolysis
  • Qualitative
  • Only interested in what is produced.
  • Quantitative
  • Measuring the volume of products.

50
Electrolysis
  • Reduction occurs at the cathode (-ve
    electrode)
  • Oxidation occurs at the anode (ve electrode)
  • As with redox reactions the total number of
    electrons lost must equal the total number of
    electrons gained.

51
Quantitative electrolysis
  • The quantity of electricity passed through the
    electrolyte is measured in coulombs (C).
  • This value can be linked to quantity of product
    as 1 mole of any substance is discharged during
    electrolysis by
  • n x 96500C
  • Where n is the number of electrons in the ion
    electron equation!

52
Q Charge (coulombs) I current (amps) t
time (seconds)
Q
t
I
53
Molten sodium chloride is electrolysed for 32
minutes and 10 seconds using 10A. Calculate the
mass of product at each electrode
Quantity of electricity used, Q It
10 x 1930
19300C
a) -ve electrode product is Na
Na e- Na
1 mol
1 mol
96500C
23g
1C 23/96500
19300C 23/96500 x 19300
4.6g of Na
54
b) ve elctrode product is Cl2
2Cl- Cl2 2e-
1 mol
2 mol
71g 2 x 96500
193000C
193000C 71g
1C 71/193000
19300C 71/193000 x 19300
7.1g of Cl2
55
f) Nuclear Chemistry
56
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57
What you need to know
  • Atoms consist of Protons, neutron electrons.

Particle Mass Charge Location Symbol
Proton Neutron Electron 1 1 1/2000 1 0 -1 Nucleus Nucleus Orbitals 11P or 11H 10n 0-1e
58
Radioactivity
  • Radioactive decay involves a change in the
    nucleus of an atom.
  • Unstable nuclei (radioisotopes) transform
    themselves into more stable nuclei by releasing
    energy.

59
Name Symbol Charge Mass Stopped by Distance in Air
Alpha a or 42He 2 4 Paper A few cm
Beta ß or 0-1e -1 0 Thin metal foil A few m
Gamma ? 0 0 Great thickness or concrete / thick lead sheet A few km
60
Radioactivity
  • Is the result of an unstable nuclei rearranging
    to form a stable nuclei with the release of
    energy.
  • Radiation is connected solely with the nucleus.
  • The stability of a nucleus depends on its proton
    to neutron ratio.
  • For small stable nuclei, the protons neutrons
    are equal.

61
Too many neutrons
  • When an atom has more neutrons than protons it
    can stabilise itself by changing a neutron into a
    proton and an electron. (ie ß particle is
    projected)
  • 10n 11p 0-1e

62
  • Neutron, proton ratio is reduced therefore
    nucleus is more stable.
  • Example Carbon-14 emits a ß particle to form a
    stable isotope of nitrogen.
  • 146C 147N 0-1e

63
Too few neutrons
  • When isotopes with low atomic number have too few
    neutrons, electron capture might take place.
  • An electron is removed from the first energy
    layer and combined with a proton to form a
    neutron.
  • Electrons rearrange to fill first layer.
  • 11p 0-1e 10n

64
  • Argon 37 undergoes electron capture to form an
    isotope of chlorine.
  • 3718Ar 0-1e 3717Cl
  • Beyond atomic number 83 almost all isotopes are
    unstable. They gain stability if they decrease in
    mass. Emission of alpha particles reduces the
    mass.

65
Radioactive equations
  • Loss of a particle 42He
  • Equivalent to losing 2 protons. (drop 2 atomic
    numbers and by a mass of 4 amu)
  • 23290Th 22888Ra 42He

Must add up to 232
66
Loss of ß particle
  • 22888Ra 22889Ac 0-1e
  • 21684Po 21685At 0-1e
  • Frequently the daughter nuclei is also
    radioactive, therefore a chain of disintegrations
    occurs until a stable nucleus is achieved.

67
Half life
  • The time taken for activity or mass of a
    radioisotope to half.
  • This half life is independent of chemical or
    physical state, mass of sample being
    investigated, temperature, pressure,
    concentration or the presence of a catalyst.

68
Half life
  • The half life is the time taken for the
    radioactive activity to decrease by a half.
  • Abbreviated to t1/2

69
Constancy of half-life
  • 1g Radium
  • 1g Radium oxide
  • 1g Radium sulphate
  • Half life will be the same.
  • Different number of radium atoms so the intensity
    will differ.

70
Example
A count rate from a sample of 75Se was found to
fall about 1.5 of its original value in 720
days. Calculate the half-life of 75Se.
of Isotope 100 50 25 12.5 6.25
3.125 1.56 No. of Half-lives
1
2
3
4
5
6
6 Half-lives needed to reduce count to 1.5. 720
days 6 Half lives 1 Half-life 720/6
120 days
71
Applications Medical
  • Radioactive isotopes are used in the treatment of
    cancer.
  • ? emitter 6027Co is used in treatment of deep
    seated tumours.
  • ß emitter 3215P is much less penetrating, which
    makes it useful as a treatment for skin cancer.

72
Applications Industrial
  • A source of highly penetrating radiation e.g.
    6027Co or 19277Ir can be used to examine castings
    welds for imperfections.
  • Modern smoke alarms contain americium-241 an a
    emitter, even tiny levels of smoke will interfere
    with the normal level of radiation given out
    triggering the alarm.

73
Applications Energy production
  • Nuclear fission is when we bombard nuclei with
    slow moving neutrons to release energy.

74
Nuclear fusion
  • When small, light atomic nuclei are combined to
    form a larger, heavier nucleus large amounts of
    energy are released.
  • This is the process behind the energy given out
    by the sun and other stars. The hydrogen bomb
    also uses nuclear fusion.

75
Nuclear power generation
Advantages Disadvantages
No greenhouse gas No acid rain Fewer mining deaths with Uranium than coal Visual impact Disastrous accident Background radiation Disposal Capital cost greater Production of plutonium linked to nuclear weapons
76
a) Chemical Industry
77
The chemical industry
  • You need to know
  • Haber process
  • Oswald process
  • Fractional distillation of crude oil
  • Cracking
  • Addition polymerisation
  • Condensation polymers
  • Manufacture of fuels

78
UK chemical industry
  • Major global contributor, 6th largest
    manufacturer of chemicals.

Plasitics Paints Detergents Explosives Adhesives
Pharmaceuticals Aerosols Fertilisers Herbicides Pe
sticides
79
Chemical industry
  • Contributes to both the quality of our life and
    to the national economy.
  • Over 150,000 Scot's work in the chemical industry.

80
Manufacturing a new chemical
  • First there must be a demand.
  • In the rainy season demand for fungicides
    quadruples.
  • RD (research development)
  • Chemists find the most suitable route to making
    the desired chemical. Considering costs, HASAWA
    (health safety at work act), timings etc.

81
Manufacturing a new chemical
  • 3. Laboratory Process
  • Chemists refine the process on a small scale to
    make it viable.
  • 4. Pilot plant
  • A medium scale investigation of product quality.
    Engineering requirements assessed.
  • 5. Production plant
  • Full scale plant design, construction and start
    up!

Processing is constantly reviewed in the
laboratory on plant
82
Raw materials feedstock
  • A feedstock is a reactant that other chemicals
    can be extracted or synthesised from.
  • Feedstock's are derived from raw materials.
  • Main raw materials include
  • Fossil fuels
  • Ores
  • Air water
  • Organic materials

83
Batch processing
  • Reactants are added to the reactor.
  • Reaction is started and process is carefully
    monitored testing throughout.
  • At the end of the reaction the reaction vessel is
    emptied and product passed on to separation,
    purification and formulation stages.
  • Reaction vessel is cleaned down and the process
    starts again.

84
Continuous process
  • Reactants flow into the reactor at one end and
    products flow out the other end.

85
Pros cons
Batch Continuous
Small scale plant cheaper to build More flexible so can be used for other processes More appropriate for multistep process Can use reactants in any state Large quantities (gt1000 tonnes per year) Only economic when operating at capacity No time wasted emptying cleaning vessel Smaller workforce required
86
Economics!
  • Capital costs building of the plant.
  • Fixed costs salaries, loans.
  • Variable costs plant out put, cost of raw
    materials, getting rid of waste.

87
Economic aspects
Capital costs Fixed costs Variable costs
RD Plant construction Buildings Infrastructures Depreciation of plant Labour Land purchase or rental Sales expenses Raw materials Energy Overheads Effluent treatment/ disposal
88
UK chemical plant is capital intensive -
expensive to build and run and provide relatively
small employment. (They can recycle heat from
exothermic reactions.) ( Labour intensive
industry- provide high employment)
89
Energy
  • Strive to use as little energy as possible.
  • Use an exotherm to heat a reactor elsewhere in
    plant.
  • Waste heat can be used to generate electricity.
  • Sell waste heat to local housing schemes.

90
Safety
  • All steps in a process must be risk assessed.
  • Any risks must be fully investigated and
    alternative routes researched.
  • PPE (personal protective equipment) is a last
    resort.
  • HS is the biggest obstacle to overcome in
    designing a process.

91
Location of industry
  • Large flat area.
  • Water close at hand.
  • Good transport links (rail/sea/road).
  • Skilled workforce at hand.

92
Factors that influence route to product
  • Availability of feedstock's
  • Yield of products attainable
  • Opportunities for recycling
  • Marketing of products by-products
  • Health safety
  • Competition
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