Industrial Chemistry

Hess’s law, Fertiliser, Sulphuric Acid, Petrochemical,

Pharmaceutical and Chemical Industries

IndexHess’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 calculationsHess’s law states that

“enthalpy change is independent of the route

taken”

Verification of Hess’s Law

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) directlyto 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 = enthalpy change

H 1 = H 2 H 3+

NaOH (s) NaCl (aq)Route 1H 1

NaOH (s)

NaOH (aq)

+ H2O (l)

NaCl (aq)

+ HCl aq)

Route 2

H 2 H 3

Experimental Confirmation of Hess’s Law

2.50g of KOH added to a dry, insulatedbeaker. Before adding the acid, its temperatureis recorded. The final temperature riseafter adding the acid is also recorded.

Knowing the specific heat capacity for water, it is then possible to calculate the Enthaply change for this reaction.

1. 2.50g of KOH added to a dry, insulatedbeaker. 2. Before adding the water, its temperatureis recorded. The final temperature riseafter adding the water is also recorded. 3. Now add the acid, again, recording the finaltemperature rise. Use the equation below to calculate H2 and H 3

H 2

H 2 + H 3 = H 1 will verify Hess’s law

Route 1 H 1

50 ml 1mol l-1 HCl

Route 2 H 2 + H 3

50 ml H2O 50 ml HCl

then

H 1 = c m TH = c m T

H 3

Combining EquationsHess’s law can be used to calculate enthalpy changes that cannot be directly measured by experiment.

Route 1 Route 1 cannot be carried outin a lab, as Carbon and Hydrogen will not combine directly.

The products of combustion act as a stepping stone which enables a link withcarbon and hydrogen (the reactants) with propane (the product)

Route 2a involves the combustion of both carbon and hydrogen

3C (s) + 3O2 (g) 3CO2 (g)

H c C= -394 kJ mol –1 H c H = -286 kJ mol –1

Route 2b involves the reverse combustion of propane

3CO2 (g) + 3H2O(l) 3C2H6 (g) + 41/2O2 (g)

3H2 (s) + 1.5 O2 (g) 3H2O (g)and

3C (s) + 3H2 (g) C3H6 (g)

3CO2 (g) + 3H2O(g)

Route 2a Route 2b

H 1

3C (s) + 3H2 (g) C3H6 (g)Route 1

3CO2 (g) + 3H2O(g)

Route H 2a Route H 2b

H 1 = H 2a H 2b+

H1

Route 2a H c C= -394 kJ mol –1 H c H = -286 kJ mol –1

H 2a = -2040 kJ mol –1

H 2a= -( x 394) = -1182 kJ mol -1 -( x 286) = -858 kJ mol -1+3 3

3 3

Route 2b

H 2b= + 2058.5 kJ mol –1 (note the reverse sign)

H c Propane = -2056 kJ mol –1

= -2040 kJ mol -1 +(+ 2058.5 kJ mol –1) = 18.5 kJ mol -1

Alternative approach to calculate the ΔHf of propane

3C(graphite) + 3H2 (g) C3H6(g)

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

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 2Calculate the enthalpy change for the reaction:

The products of combustion act as a stepping stone which enables a link withbenzene and hydrogen (the reactants) with hexane (the product)

C6H6(l) + 3H2 (g) C6H12(l) Route 1

6H2O(g) + 6CO2(g)

Route 2a Route 2b

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 c hexane = -3924 kJ mol –1

H 1 = H 2a + H 2b = ( -3273 + ( x - 286)) + 3924 = 207 kJ mol -1 3

3

Alternative approach to problem 2

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) ΔHc = 3 x -286

6H2O(g) + 6CO2(g) C6H12(g) + 9O2(g) ΔHc = +3924

Equation has been reversed; (enthalpy now has opposite sign)

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

Now add the equations and also the corresponding enthalpy values

C6H6(l) + 3H2(g) C6H12(l)

ΔHf = -3273 + (3 x -286) + 3924

ΔHf = -207 kJ mol-1

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

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) ΔHc = +1299 kJ mol-1

ΔHf = (2 x -395) + (-286) + 1299

ΔHf = +223 kJ mol-1

Adding “bulleted” equations gives us 2C(graphite) + H2(g) C2H2(g)

4. Using the following standard enthalpy changes of formation, ΔHo

f / 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)

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) ΔHc = ?

Alternative method

C(graphite) + 3H2(g) + ½O2(g) C2H5OH(l) ΔHf = -278

● 2C(graphite) + 2O2(g) 2CO2(g) ΔHf = 2 x -394

● 3H2(g) + 1½O2(g) 3H2O (g) ΔHf = 3 x -286

● C2H5OH(l) C(graphite) + 3H2(g) + ½O2(g) ΔHf = +278

Add bulleted equationsC2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

solve equation for ΔHc

ΔHc = 278 + (2 x -394) + (3 x -286)

ΔHc = -1368 kJ mol-1

Industrial ChemistryThe UK chemical industry is the nation’s 4th largest manufacturingindustry 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

A chemical plant produces the desired products. The process usedto manufacture the product may be operated in continuous orbatch sequences.

All chemical plants require a source of raw materials, which caneither be non-living eg minerals, or living, eg plants and micro-organisms. (collectively known as biomass).

Batch Continuous

For OK for up to 100 tonnes per annum Versatile Good for multi-step reactions

OK for over 1000 tonnes per annum Good for f ast single step processes Easy to automate

Against Contamination At times, no product is made. Safety

Capital cost Less flexible Need to run at full capacity

Economic aspects

REACTIONTemp, pressure, catalyst

Recycle loop

Operating conditions

Costs, capital, fixed and variable

Use of energy

Location of the Chemical industry

Safety and the environment

Consideration has to be given to

Feedstockpreparation

Energy in or out

Separationproducts

Co-products

Choices to be made1. Cost, availability of feedstocks2. Yield of the reaction3. Can un-reacted materials be recycled4. Can by-products be sold5. Cost of waste disposal6. 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

Crudeoil

naphtha propene polypropene carpeting

£’s per tonne1x£’s per tonne 3x£’s per tonne 8x£’s per tonne 20x£’s per tonne

Fertiliser IndustryHaber process

Ammonia is manufactured from N2 and H2. The nitrogen is available fromthe 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.

Naturalgas

WaterCatalyst

CH4 (g)

H2O(g)

heat

Stage 1

AIR

Catalyst

Stage 2 Stage 3

CatalystN2(g) + H2 (g)

alkali

CO2 removed

Haber ProcessStage 1

Stage 2

Stage 3

CH4 (g) + H2O (g) CO (g) + 3H2 (g)

4N2 (g) + O2 (g) + 2H2(g) 2H2O (g) + 4N2 (g)

ΔH1 = +210 kJ

ΔH2 = -484 kJ

ΔH3 = -41 kJCO (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.

ΔH1 = (+210 x 3.5) kJ ΔH2 = -(484) kJ ΔH3 = -(41 x 3.5) kJ

(ΔH1 + ΔH2 + ΔH3 ) ΔHtotal = -41 kJ

3.5 CH4 (g) + 4N2 (g) + O2(g) + 5H2O (g) 4N2 (g) + 12H2 (g) + 3.5 CO2

Combining the three stages

Haber processReaction Conditions

Low temperature shifts the equilibrium to the right, but meansa slow reaction rate. Fe catalyst improves this.A high pressure favours also shifts the equilibrium to the right becausethis is the side with fewer gas molecules.

Temperatures around 500oC and pressures of over 150 atmospheresgive 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.

N2 (g) + 3H2 (g) 2NH3 (g)

ΔHf = -92 kJ

Sulphuric Acid IndustrySulphuric acid is manufactured by the Contact Process.

Sulphur

Airburner

S

O2(g)

heatStage 1

CatalyticConverterCat=V2O5

SO2(g)

SO3(g)

Stage 2

feedstock

water

absorber

mixerH2SO4

98% acidWaste gases

Stage 3

Suphuric AcidThe raw materials for the manufacture of H2SO4 are H2O, O2 from airand S or a compound containing sulphur.

Sources of sulphura. 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) ΔH = -299 kJ

Stage 2 2SO2 (g) + O2 (g) 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 AcidCost considerations

Capital costs: The cost of building the plant and all the associatedcosts of all buildings

Variable costs: The cost that changes throughout the year andis 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 IndustryGrangemouth is one of the UK’s major oil refineries and petrochemicalplants. The crude oil is processed to increase its market value.

Oil refining is a continuous process. The crude oil is processed toincrease its market value. The fractions produced have many usesand heavier fractions are further processed by processes such ascracking 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 andseparating it into its constituent parts. Natural gas becomes a liquid at below -161oC. Fractional distillation is then used to separateout the constituents of natural gases in a continuous process.

Naturalgases

sulphur

methane

ethanepropane

butane

Gas gridCracker (ethene)

LPG

petrol

Pharmaceutical Industry

Drugs alter the biochemical processes in our bodies, for example, changing the way we feel and behave. Drugs which lead to animprovement in health are called medicines.

Once a new drug is discovered, it will be patented, the licence lasting20 years. Many years of trials may be needed before the drug evenbecomes commercially available. The Government is also involved inthis process, providing the necessary licensing for the new drug.The Chemical Industry earns £1000 million pounds a year in ‘invisibleearning’ 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, environmentaland energy saving factors have to be considered.