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Industrial Chemistry
Hess’s law, Fertiliser, Sulphuric Acid,
Petrochemical, Pharmaceutical and
Chemical Industries
Index
Hess’s Law and its experimental verification
Hess’s Law calculations
Industrial Chemistry
Fertiliser Industry and Haber process
Sulphuric acid industry
Petrochemical industry and natural gas
Pharmaceutical industry
Hess’s Law and calculations
Hess’s law states that
“enthalpy change is independent
of the route taken”
Verification of Hess’s Law
H = enthalpy change
Route 1
NaOH (s)
Route 2
NaOH (s)
H 2
H 1
NaCl (aq)
NaCl (aq)
+ H2O (l)
NaOH (aq)
+ HCl aq) H 3
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 Hess’s Law applies, the enthalpy change for route 1 must be the same for the
overall change for route 2.
H 1
= H 2 +
H 3
Experimental Confirmation of Hess’s Law
Route 1 H 1
50 ml 1mol l-1 HCl
Route 2 H 2 + H 3
50 ml HCl
50 ml H2O
then
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. H 2
3. Now add the acid, again, recording the final
Knowing the specific heat capacity for
H 3
water, it is then possible to calculate the temperature rise.
Use the equation below to calculate H2 and H 3
Enthaply change for this reaction.
H 1 = c m T
H = c m
H 2 + H
3
T
= H 1 will verify Hess’s law
H c C= -394 kJ mol –1
H c H = -286 kJ mol –1
Combining Equations
Hess’s law can be used to calculate enthalpy changes that cannot be
directly measured by experiment.
3C (s) + 3H2 (g)
Route 1
Route 2a
C3H6 (g)
Route 2b
Route 1 cannot be carried out
in a lab, as Carbon and Hydrogen
will not combine directly.
3CO2 (g) + 3H2O(g)
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
3C (s) + 3O2 (g)  3CO2 (g)
and
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)
H1
3C (s) + 3
3H2 (g)
3
Route 1
Route H 2a
C3H6 (g)
Route H 2b
3CO2 (g) + 3H2O(g)
H 1 = H 2a +
Route 2a
H c C= -394 kJ mol –1
H 2a = -(3x 394) = -1182 kJ mol
-1
+
H 2b
H c H = -286 kJ mol –1
-(3 x 286) = -858 kJ mol -1
H 2a = -2040 kJ mol –1
Route 2b H c Propane = -2056 kJ mol –1
H 2b = + 2058.5 kJ mol
H 1 =
–1
(note the reverse sign)
-2040 kJ mol -1
+
(+ 2058.5 kJ mol –1)
= 18.5 kJ mol
-1
Alternative approach to calculate the ΔHf of propane
C(graphite) + O2 (g)  CO2(g)
ΔHo298 = -394 kJmol-1
H2(g) + ½O2(g)  H2O(g)
ΔHo298 = -286 kJmol-1
C3H6(g) + 4½O2(g)  3H2O(g) + 3CO2(g)ΔHo298 = -2058.5 kJmol-1
3C(graphite) + 3H2 (g)  C3H6(g)
ΔHf = ?
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) ΔHc = 3 x -394
3H2(g) + 1½O2(g)  3H2O(g)
ΔHc = 3 x -286
3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(g)
ΔHc = +2058.5
Equation has been reversed; (enthalpy now has opposite sign)
3C(graphite) + 3O2 (g)  3CO2(g) ΔHc = 3 x -394
3H2(g) + 1½O2(g)  3H2O(g)
ΔHc = 3 x -286
3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(g)
ΔHc = +2058.5
Now add the equations and also the corresponding enthalpy values
3C(graphite) + 3H2(g)  C3H6(g)
ΔHf = (3 x -394) + (3 x -286) + (+2058.5)
ΔHf = +18.5 kJ mol-1
Time
to go!
Example 2
Calculate the enthalpy change for the reaction:
3 2 (g)
C6H6(l) + 3H
Route 1

C6H12(l)
Route 2b
Route 2a
6H2O(g) + 6CO2(g)
The products of combustion act as a stepping stone which enables a link with
benzene and hydrogen (the reactants) with hexane (the product)
Route 2a involves the combustion of both benzene and hydrogen
C6H6(g) + 7½O2(g)  3H2O(g) + 6CO2(g)
H c benzene = -3273 kJ mol –1
H2(g) + ½O2(g)  H2O(g)
H c hydrogen = -286 kJ mol –1
Route 2b involves the reverse combustion of hexane
6CO2 (g) + 6H2O(l) => C6H12 (g) + 71/2O2 (g)
H 1 = H 2a + H 2b
H c hexane = -3924 kJ mol –1
= ( -3273 + (3x - 286)) + 3924 = 207 kJ mol
-1
Alternative approach to problem 2
C6H12(l) + 9O2 (g)  6H2O(g) + 6CO2(g) ΔHo298 = -3924 kJmol-1
H2(g) + ½O2(g)  H2O(g)
ΔHo298 = -286 kJmol-1
C6H6(g) + 7½O2(g)  3H2O(g) + 6CO2(g) ΔHo298 = -3273 kJmol-1
C6H6(l) +
3H2(g)  C6H12(l)
ΔHf = ?
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) ΔHc = -3273
3H2(g) + 1½O2(g)  3H2O(g)
6H2O(g) + 6CO2(g)  C6H12(g) + 9O2(g)
ΔHc = 3 x -286
ΔHc = +3924
Equation has been reversed; (enthalpy now has opposite sign)
C6H6(l) + 7½O2 (g)  6CO2(g) + 3H2O(g)
ΔHc = -3273
3H2(g) + 1½O2(g)  3H2O(g)
ΔHc = 3 x -286
6H2O(g) + 6CO2(g)  C6H12(l) + 9O2(g)
ΔHc = +3924
Now add the equations and also the corresponding enthalpy values
C6H6(l) + 3H2(g)  C6H12(l)
ΔHf = -3273 + (3 x -286) +
ΔHf = -207 kJ mol-1
3924
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)
ΔHo(combustion)
-395 kJmol-1
-286 kJmol-1
-1299 kJmol-1
“Second method”
“Required” equation, 2C(graphite) + H2(g)  C2H2(g)
ΔHf = ?
● 2C(graphite) + 2O2(g)  2CO2(g)
ΔHc = 2 x -395 kJ mol-1
● H2(g) + ½O2(g)  H2O(g)
ΔHc = -286 kJ mol-1
C2H2(g) + 2½O2(g)  2CO2(g) + H2O(g)
ΔHc = -1299 kJ mol-1
● 2CO2(g) + H2O(g)
 C2H2(g) + 2½O2(g)
Adding “bulleted” equations gives us
ΔHc = +1299 kJ mol-1
2C(graphite) + H2(g)  C2H2(g)
ΔHf = (2 x -395) + (-286) + 1299
ΔHf =
+223 kJ mol-1
4. Using the following standard enthalpy changes of formation,
ΔHof / 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)
Alternative method
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
ΔHc = ?
C(graphite) + 3H2(g) + ½O2(g)  C2H5OH(l)
ΔHf = -278
● 2C(graphite) + 2O2(g)  2CO2(g)
ΔHf = 2 x -394
● 3H2(g) +
ΔHf = 3 x -286
1½O2(g)
 3H2O (g)
● C2H5OH(l)  C(graphite) + 3H2(g) + ½O2(g)
Add bulleted equations
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
solve equation for ΔHc
ΔHc = 278 + (2 x -394) + (3 x -286)
ΔHc = -1368 kJ mol-1
ΔHf = +278
Industrial Chemistry
The UK chemical industry is the nation’s 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.
Batch
For
OK for up to 100
tonnes per annum
Versatile
Good for multistep reactions
Against Contamination
At times, no
product is made.
Safety
Continuous
OK for over 1000
tonnes per annum
Good for fast
single step
processes
Easy to automate
Capital cost
Less flexible
Need to run at full
capacity
Economic aspects
Energy in or out
Feedstock
preparation
REACTION
Temp, pressure, catalyst
products
Separation
Recycle loop
Consideration has to be given to
Operating conditions
Costs, capital, fixed and variable
Use of energy
Location of the Chemical industry
Safety and the environment
Co-products
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
Crude
oil
£’s per tonne
naphtha
1x£’s per tonne
propene
3x£’s per tonne
polypropene
carpeting
8x£’s per tonne
20x£’s per tonne
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.
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
Haber Process
Stage 1
CH4 (g) + H2O (g)  CO (g) + 3H2 (g)
Stage 2
4N2 (g) + O2 (g) + 2H2(g)  2H2O (g) + 4N2 (g) ΔH2 = -484 kJ
Stage 3
CO (g) + H2O (g)  CO2 (g) + H2 (g)
ΔH1 = +210 kJ
ΔH3 = -41 kJ
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
ΔH1 = (+210 x 3.5) kJ
ΔH2 = -(484) kJ
(ΔH1 + ΔH2 + ΔH3 ) ΔHtotal = -41 kJ
ΔH3 = -(41 x 3.5) kJ
Haber process
Reaction Conditions
N2 (g) + 3H2 (g)
ΔHf = -92 kJ
2NH3 (g)
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.
Sulphuric Acid Industry
Sulphuric acid is manufactured by the Contact Process.
Waste gases
Sulphur
Air
98% acid
S
O2(g)
burner
heat
Stage 1
SO2(g)
feedstock
Stage 3
Stage 2
Catalytic
Converter
Cat=V2O5
SO3(g)
absorber
water
mixer
H2SO4
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
a. SO2 from smelting of ores, eg ZnS. The SO2 is converted into sulphuric
acid rather than released into the atmosphere.
b. CaSO4, the mineral anhydrite, is roasted with coke (C) and SiO2 (sand)
c. S deposits in the ground.
d. S can be extracted from oil and natural gas.
Stage 1
S (l) + O2 (g)  SO2 (g)
Stage 2
2SO2 (g) + O2 (g)
ΔH = -299 kJ
2SO3 (g)
ΔH = -98 kJ
The catalyst does not function below 400 0C, a 99% yield is obtained.
Stage 3
SO3 (g) + H2O (l)  H2SO4 (l)
ΔH = -130 kJ
The acid produced is absorbed in 98% H2SO4,. If dissolved in water
too much heat is created and gases are lost to the atmosphere.
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.
Petrochemical Industry
Grangemouth is one of the UK’s 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
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.
methane
ethane
propane
Natural
gases
Gas grid
Cracker (ethene)
LPG
butane
sulphur
petrol
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.