Industrial Chemistry

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Industrial Chemistry

• Hesss law, Fertiliser, Sulphuric Acid,
  Petrochemical, Pharmaceutical and Chemical
  Industries

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Index
Hesss Law and its experimental verification
Hesss Law calculations
Industrial Chemistry
Fertiliser Industry and Haber process
Sulphuric acid industry
Petrochemical industry and natural gas
Pharmaceutical industry
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Hesss Law and calculations
Hesss law states that enthalpy change is
independent of the route taken
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Verification of Hesss Law
?H enthalpy change
The conversion of solid NaOH to sodium chloride
solution can be achieved by two possible routes.
One is a direct,single-step process, (adding HCl
(aq) directly to the solid NaOH) and secondly a
two-step process (dissolve the solid NaOH in
water first, then add the HCl(aq)) All steps are
exothermic.
If Hesss Law applies, the enthalpy change for
route 1 must be the same for the overall change
for route 2.
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Experimental Confirmation of Hesss Law
2.50g of KOH added to a dry, insulated beaker.
Before adding the acid, its temperature is
recorded. The final temperature rise 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 temperature rise. Use
the equation below to calculate ?H2 and ?H 3
?H 2
Knowing the specific heat capacity for water, it
is then possible to calculate the Enthaply
change for this reaction.
?H 3
?H 2 ?H 3 ?H 1 will verify Hesss law
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Combining Equations
?H c C -394 kJ mol 1
?H c H -286 kJ mol 1
Hesss law can be used to calculate enthalpy
changes that cannot be directly measured by
experiment.
Route 1
Route 1 cannot be carried out in a lab, as
Carbon and Hydrogen will not combine directly.
The products of combustion act as a stepping
stone which enables a link with carbon and
hydrogen (the reactants) with propane (the
product)
Route 2a involves the combustion of both carbon
and hydrogen
and
3C (s) 3O2 (g) ? 3CO2 (g)
3H2 (s) 1.5 O2 (g) ? 3H2O (g)
Route 2b involves the reverse combustion of
propane
3CO2 (g) 3H2O(l) ? 3C2H6 (g) 41/2O2 (g)
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?H1
3
3
3
3
?H 1
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Alternative approach to calculate the ?Hf of
propane
C(graphite) O2 (g) ? CO2(g) ?Ho298
-394 kJmol-1 H2(g) ½O2(g) ? H2O(g)
?Ho298 -286 kJmol-1 C3H6(g) 4½O2(g) ? 3H2O(g)
3CO2(g)?Ho298 -2058.5 kJmol-1
3C(graphite) 3H2 (g) ? C3H6(g)
?Hf ?
Re-write the equations so that the reactants and
products are on the same side of the arrow as
the equation you are interested in. Multiply
each equation so that there are the same number
of moles of each constituent also.
3C(graphite) 3O2 (g) ? 3CO2(g)
?Hc 3 x -394
3H2(g) 1½O2(g) ? 3H2O(g)
?Hc 3 x -286
3H2O(g) 3CO2(g) ? C3H6(g) 4½O2(g)
?Hc 2058.5
Equation has been reversed (enthalpy now has
opposite sign)
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Now add the equations and also the corresponding
enthalpy values
3C(graphite) 3H2(g) ? C3H6(g)
?Hf (3 x -394) (3 x -286) (2058.5)
?Hf 18.5 kJ mol-1
Time to go!
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3
The products of combustion act as a stepping
stone which enables a link with benzene and
hydrogen (the reactants) with hexane (the product)
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Alternative approach to problem 2
C6H12(l) 9O2 (g) ? 6H2O(g) 6CO2(g)
?Ho298 -3924 kJmol-1 H2(g) ½O2(g) ?
H2O(g) ?Ho298 -286 kJmol-1C6H6(g) 7½O2(g)
? 3H2O(g) 6CO2(g) ?Ho298 -3273 kJmol-1
C6H6(l) 3H2(g) ? C6H12(l)
?Hf ?
Re-write the equations so that the reactants and
products are on the same side of the arrow as
the equation you are interested in. Multiply
each equation so that there are the same number
of moles of each constituent also.
C6H6(l) 7½O2 (g) ? 6CO2(g) 3H2O(g)
?Hc -3273
3H2(g) 1½O2(g) ? 3H2O(g)
?Hc 3 x -286
6H2O(g) 6CO2(g) ? C6H12(g) 9O2(g)
?Hc 3924
Equation has been reversed (enthalpy now has
opposite sign)
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Now add the equations and also the corresponding
enthalpy values
C6H6(l) 3H2(g) ? C6H12(l)
?Hf -3273 (3 x -286) 3924
?Hf -207 kJ mol-1
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3. Use the enthalpy changes of combustion shown
in the table to work out the enthalpy change of
formation of ethyne, C2H2.
Substance C(graphite) H2(g) C2H2(g)
?Ho(combustion) -395 kJmol-1 -286 kJmol-1 -1299 kJmol-1
Second method
Required equation, 2C(graphite) H2(g) ?
C2H2(g) ?Hf ?
? 2C(graphite) 2O2(g) ? 2CO2(g) ?Hc
2 x -395 kJ mol-1
? H2(g) ½O2(g) ? H2O(g) ?Hc -286
kJ mol-1
C2H2(g) 2½O2(g) ? 2CO2(g) H2O(g) ?Hc
-1299 kJ mol-1
? 2CO2(g) H2O(g) ? C2H2(g) 2½O2(g)
?Hc 1299 kJ mol-1
Adding bulleted equations gives us
2C(graphite) H2(g) ? C2H2(g)
?Hf (2 x -395) (-286) 1299
?Hf 223 kJ mol-1
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4. Using the following standard enthalpy changes
of formation, ?Hof / kJmol-1 CO2(g), -394
H2O(g), -286 C2H5OH(l), -278 calculate the
standard enthalpy of combustion of ethanol i.e.
the enthalpy change for the reaction C2H5OH(l)
3O2(g) ? 2CO2(g) 3H2O(l)
C(graphite) 3H2(g) ½O2(g) ? C2H5OH(l) ?Hf
-278
? 2C(graphite) 2O2(g) ? 2CO2(g) ?Hf 2 x
-394
? 3H2(g) 1½O2(g) ? 3H2O (g) ?Hf 3 x
-286
? C2H5OH(l) ? C(graphite) 3H2(g) ½O2(g) ?Hf
278
Add bulleted equations
C2H5OH(l) 3O2(g) ? 2CO2(g) 3H2O(g)
solve equation for ?Hc
?Hc 278 (2 x -394) (3 x -286)
?Hc -1368 kJ mol-1
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Industrial Chemistry
The UK chemical industry is the nations 4th
largest manufacturing industry and the 5th
largest in the world. The 3 largest sections are
(a) food, drink and tobacco, (b) mechanical
engineering and (c) paper, printing and
publishing
All chemical plants require a source of raw
materials, which can either be non-living eg
minerals, or living, eg plants and
micro-organisms. (collectively known as biomass).
A chemical plant produces the desired products.
The process used to manufacture the product may
be operated in continuous or batch sequences.
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Economic aspects
REACTION Temp, pressure, catalyst
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Choices to be made
1. Cost, availability of feedstocks 2. Yield of
the reaction 3. Can un-reacted materials be
recycled 4. Can by-products be sold 5. Cost of
waste disposal 6. Energy consumption, generating
your own, conservation, use of catalysts,
recycling, (heat exchangers), 7. Environmental
issues
Value added, eg the value of the products from
crude oil
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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
hydrogen, like nitrogen, a feedstock for the
manufacture of NH3. Hydrogen is usually produced
from methane.
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Haber Process
Stage 1
CH4 (g) H2O (g) ? CO (g) 3H2 (g)
?H1 210 kJ
Stage 2
4N2 (g) O2 (g) 2H2(g) ? 2H2O (g) 4N2
(g)
?H2 -484 kJ
?H3 -41 kJ
Stage 3
CO (g) H2O (g) ? CO2 (g) H2 (g)
In order to achieve a ratio of 3x hydrogen to
nitrogen, stage 1 and 3 need to be 3.5x greater
than stage 2.
Combining the three stages
3.5 CH4 (g) 4N2 (g) O2(g) 5H2O (g) ? 4N2
(g) 12H2 (g) 3.5 CO2
?H1 (210 x 3.5) kJ
?H2 -(484) kJ
?H3 -(41 x 3.5) kJ
(?H1 ?H2 ?H3 ) ?Htotal -41 kJ
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Haber process
Reaction Conditions
Low temperature shifts the equilibrium to the
right, but means a slow reaction rate. Fe
catalyst improves this. A high pressure favours
also shifts the equilibrium to the right
because this is the side with fewer gas molecules.
Temperatures around 500oC and pressures of over
150 atmospheres give a yield of ammonia of about
15.
Product removal In practice, equilibrium is not
reached as unreacted gases are recycled and the
ammonia gas is liquefied.
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Sulphuric Acid Industry
Sulphuric acid is manufactured by the Contact
Process.
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Suphuric Acid
The raw materials for the manufacture of H2SO4
are H2O, O2 from air and S or a compound
containing sulphur.

• Sources of sulphur
• SO2 from smelting of ores, eg ZnS. The SO2 is
  converted into sulphuric
• acid rather than released into the atmosphere.
• CaSO4, the mineral anhydrite, is roasted with
  coke (C) and SiO2 (sand)
• S deposits in the ground.
• S can be extracted from oil and natural gas.

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Sulphuric Acid
Cost considerations
Capital costs The cost of building the plant
and all the associated costs of all buildings
Variable costs The cost that changes throughout
the year and is dependant of how much product is
sold. Buying raw materials, treating waste and
despatching the product
Fixed costs The cost of the staff, local rates,
advertising and utility bills.
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Petrochemical Industry
Grangemouth is one of the UKs major oil
refineries and petrochemical plants. The crude
oil is processed to increase its market value.
Oil refining is a continuous process. The crude
oil is processed to increase its market value.
The fractions produced have many uses and heavier
fractions are further processed by processes such
as cracking which produces key feedstock for the
plastic industry.
Refinery gas, eg propane and butane bottled gas
Petrol, which is further purified and blended
Naphtha, feedstock for the plastic industry
Kerosine, aviation fuel
Diesel,
Fuel oil, eg ships, oil-fired power stations,
industrial heating
residue, lubricating oil, waxes, bitumen
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Natural gas
The market value of Natural Gas is increased by
desulphurisation and separating it into its
constituent parts. Natural gas becomes a liquid
at below -161oC. Fractional distillation is then
used to separate out the constituents of natural
gases in a continuous process.
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Pharmaceutical Industry
Drugs alter the biochemical processes in our
bodies, for example, changing the way we feel
and behave. Drugs which lead to an improvement in
health are called medicines.
Once a new drug is discovered, it will be
patented, the licence lasting 20 years. Many
years of trials may be needed before the drug
even becomes commercially available. The
Government is also involved in this process,
providing the necessary licensing for the new
drug. The Chemical Industry earns 1000 million
pounds a year in invisible earning for
licensing fees for patented chemicals and
processes.
Once the necessary licensing has been granted a
pilot plant will be built for small scale
production to allow for product evaluation. Full
scale production is then implemented, where
safety, environmental and energy saving factors
have to be considered.