HSC Chemistry Option: Industrial Chemistry

HSC Chemistry Option: Industrial ChemistryReplacing natural productsChemical equilibriumSulfuric acid productionSodium hydroxide productionSaponificationThe Solvay Process: Na2CO3 production1. Replacing natural productsVarious industrial products have replaced naturally occurring chemicalsThese may have been depleted or are no longer economically viable to produceNatural fertiliser: GuanoRich in nitrates and phosphates formed from sun-baked bird and bat droppingsFormed over thousands of years on hot, dry islands of Peru and Nauru (ie. Non-renewable)Guano deposits have been rapidly depleted as intensive farming practices developedSynthetic fertilisers, why?Developed to cope with a growing population and shrinking supplies of natural fertilisersProduction driven by development of important industrial advances in the early 20th centuryHaber Process (manufacture of NH3)Ostwald Process (manufacture of HNO3)Synthetic replacement: NH4NO3Combining the products of the Haber & Ostwald Process:NH3(g) + HNO3(aq) NH4NO3(aq)Good replacement as it is rich in nitratesAlternative replacementSuperphosphate mixture of hydrated calcium sulfate and calcium dihydrogen phosphateCa3(PO4)2 + 2H2SO4 + 4H2O 2CaSO4.2H2O + Ca(H2PO4)2Made from crushed phosphate rocks treated with sulfuric acidWidely used to improve soil fertility and replace mineralsNatural rubberNatural addition polymer obtained from rubber trees typically grown in South East AsiaAlso known as polyisoprenePure natural rubber is not very usefulSoft when heated, inelastic and susceptible to chemical attack due to double bondsBut S-S cross-links can be introduced in a process known as vulcanisationThis made rubber harder with greater elasticityVulcanised rubber

Synthetic rubberStyrene-butadiene rubber (SBR)Made from by-products of petroleum refiningCan be vulcanised and is less susceptible to attackWhy develop SBR?Escalating demand for rubber to make car tyresReduced supply from South East Asia during WWIIWhat factors may lead to a natural product being replaced by a synthetic one?Problems with natural productUnable to keep up with demandCompetition from another useUnpredictable, unreliable or unavailable supplyDepletion or rising cost of natural resourcesAdvantages of synthetic productGreater availability or lower cost of raw materialsGreater reliability of supply and stability of priceEstablished industrial processes availableSuperior properties compared to natural product2. Chemical equilibriumThe equilibrium constant is known as KConsider aA + bB cC + dDForward rate: Rf Kf[A]a[B]bReverse rate: Rb Kb[C]c[D]dAt equlibrium, Rf = RbKf[A]a[B]b = Kb[C]c[D]dK = Kf/Kb = [C]c[D]d/[A]a[B]b (conceptual only!)The equilibrum constant (K)For a reversible reaction, at equilibrium and at a constant temperature, the ratio of product to reactant concentrations expressed as

has a constant value K

Features of KThe magnitude of K can tell us about the relative equilibrium positionIf K has a large value (>1000), equilibrium lies well the right (ie. Products favoured)If K has a small value ( K then the reaction proceeds from right to left until Q = K (towards reactants)Examples: Write an expression for K2NO(g) + O2(g) 2NO2(g)2Pb(NO3)2(s) 2PbO(s) + 4NO2(g) + O2(g)CuO(s) + SO3(g) CuSO4(s)Answers: Write an expression for KK = [NO2]2/[NO]2[O2]K = [NO2]4[O2]K = 1/[SO3]

A word of cautionNote that K uses the coefficients and direction of the equilibrium reaction as written!2NO(g) + O2(g) 2NO2(g)K = [NO2]2/[NO]2[O2]2NO2(g) 2NO(g) + O2(g)K = [NO]2[O2]/[NO2]2NO(g) + O2(g) NO2(g)K = [NO2]/[NO][O2]More examplesConsider N2(g) + 3H2(g) 2NH3(g). At equilibrium, [N2] = 0.10mol/L, [H2] = 0.3mol/L & [NH3] = 0.05mol/L. Find the value of K. [answer = 0.93]At a given temperature, 0.50mol of CF4 and 0.40mol of CO2 were placed in a 10.0L container and allowed to react to form COF2. At equilibrium, 0.15mol of CF4 remained. Calculate K. [answer = 65.3]MICEM = mole ratioI = initial concentrationC = change in concentrationE = equilibrium concentrationExample 2 CF4(g) + CO2(g) 2COF2(g)M112I0.5mol/10L = 0.05mol/L0.4mol/10L = 0.04mol/L0CEPut in the initial concentrations, remembering to divide moles by volume!Example 2 CF4(g) + CO2(g) 2COF2(g)M112I0.5mol/10L = 0.05mol/L0.4mol/10L = 0.04mol/L0CE0.15mol/10L = 0.015mol/LFill in the equilibrium concentrations that you are givenExample 2 CF4(g) + CO2(g) 2COF2(g)M112I0.5mol/10L = 0.05mol/L0.4mol/10L = 0.04mol/L0C-0.035mol/LE0.15mol/10L = 0.015mol/LFind the change in concentration for one (or more) set of values you have.Example 2 CF4(g) + CO2(g) 2COF2(g)M112I0.5mol/10L = 0.05mol/L0.4mol/10L = 0.04mol/L0C-0.035mol/L-0.035mol/L2 x +0.035mol/L = +0.07mol/LE0.15mol/10L = 0.015mol/LChanges in concentration are STOICHIOMETRIC ie. They follow the mole ratio. Fill in the other changes of concentration, using a negative symbol if the substance isused up or a positive symbol if the substance is created.Example 2 CF4(g) + CO2(g) 2COF2(g)M112I0.5mol/10L = 0.05mol/L0.4mol/10L = 0.04mol/L0C-0.035mol/L-0.035mol/L2 x +0.035mol/L = +0.070mol/LE0.15mol/10L = 0.015mol/L0.005mol/L0.070mol/LFill in the equilibrium concentrations using simple maths in the I and C column.You can now calculate K remember K is ONLY calculated using equilibrium concentrations!More examples3. A 1L container is filled with 0.50mol of HI at 448C. The value of the equilibrium constant, Kc, for the reaction H2(g) + I2(g) 2HI(g) at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the vessel at equilibrium?

???The issue with this question is that you do not have any changes in concentration values to go off!Assign one of the changes in concentration with a molar ratio of 1 as x mol/LSolution to Example 3 H2(g) + I2(g) 2HI(g)M112I 0mol/L0mol/L0.50mol/LC+ x mol/L+ xmol/L-2x mol/LEx mol/Lx mol/L0.5-2x mol/LK = [HI]2/[H2][I2] = 50.5(0.50 - 2x)2/x2 = 50.5(0.50 - 2x)2 = 50.5x20.25 - 2x + 4x2 = 50.5x246.5x2 + 2x - 0.25 = 0 x = [-2 (22 - 4(-0.25 x 46.5)]/[2 x 46.5]x = 0.055

[H2]eq = 0.055mol/L[I2]eq = 0.055mol/L[HI]eq = 0.5 2(0.055) = 0.390mol/LMore examples4. At 448C the equilibrium constant Kc, for the reaction H2(g) + I2(g) 2HI(g) is 50.5. Predict the direction in which the reaction will proceed to reach equilibrium at 448C if we start with 2.0x10-2mol of HI, 1.0x10-2mol of H2 and 3.0x10-2mol of I2 in a 2L container.

[answer: Q < K]3. Sulfuric acid productionSulfuric acidThe worlds most produced chemicalProduction of fertilisers eg. Superphosphate and ammonium sulfateSteel processing prior to galvanisingElectrolyte in lead acid batteriesDetergents eg. Alkyl benzene sulfonates (anionic)Production of TiO2 from ilmenite (FeTiO3) as a pigment for paints, plastics, paper and sunscreenMedications for sulfur-based drugsRaw material: SulfurSulfur is a raw material for H2SO4 productionAlthough SO2 may also be used if availableMost sulfur exists on Earth as sulfide ores or part of crude oil and natural gasSome sedimentary deposits of elemental sulfur can be found in Louisiana and Texas (USA), Mexico and PolandThe Frasch ProcessSuperheated water is pumped down the outer pipeThis melts the sulfur forming a water-sulfur emulsionCompressed air is blown down the central pipe to force the emulsion up to the surfaceWhen cooled, solid sulfur separates from water, giving 99.5% SProperties of sulfurSeveral properties of sulfur are critical to the success of the Frasch ProcessLow MP (113C): easily melted by the superheated water (160-170C/1.5MPa)Low density (2.07g/cm3): emulsion can be lifted to the surface by compressed airLow solubility in water and inert: sulfur is easily separated from water when cooledEnvironmental issuesSulfur is readily oxidised to SO2 or reduced to H2S both are air pollutantsUsed water must be cooled and may contain dissolved impurities in the depositShould not discharge this back into natural waterGround subsidences are possible due to the difficulty of back-filling underground cavernsAlternative raw materialsSulfur is produced when H2S is removed from natural gas and crude oilTo make H2SO4, sulfur is converted to SO2SO2 may be obtained directly as a by-product of smelting sulfide ores2ZnS(s) + 2O2(g) 3ZnO(s) + 2SO2(g)The Contact ProcessInvolves 3 major steps:S(l) + O2(g) SO2(g) H = -297kJ/mol2SO2(g) + O2(g) 2SO3(g) H = -99kJ/molSO3(g) + H2O(l) H2SO4(l) H = -133kJ/molHowever its the second equation that is considered the actual contact process itselfThe Contact Process2SO2(g) + O2(g) 2SO3(g) H = -99kJ/molNote that this is an equilibrium reactionHow could we optimise the process?Equilibrium yieldRate of reaction or kineticsEconomicsWhat conditions favour greater yield of SO3?Lower temperaturesHigher pressuresRemoval of productHigher reactant concentrationsWhat conditions favour greater reaction rate?Higher temperaturesHigher pressuresUse of a catalystWhat conditions favour low cost?Lower temperaturesLower pressures (also for safety)Unreacted gases and heat released are recycled backCompromise conditions550C at 1st catalyst bed, 400C at later bedsModerately high temperature initially gives about 70% conversion relatively quicklyReduces temperatures then help to increase yieldPressure slightly above 100kPaHigher pressures favour yield but do not increase it significantly so this avoids excess costCompromise conditionsV2O5 catalyst on porous silicaIncreases rate to compensate for lower temperaturesAbout 5:1 air:SO2 ratio (ie. 1:1 SO2:O2)Slight excess of oxygen to favour increased yieldRemoval of SO3 product into concentrated H2SO4Overview of the whole Contact Process1. Combustion of sulfur to sulfur dioxideS(l) + O2(g) SO2(g) H = -297kJ/molMolten S and excess dry air is reacted in furnanceThe air is dried by passage through H2SO4A heat exchanger lowers temperature of output gas from 1000C to about 400CHeat is recycled or used to generate electricity

Overview of the whole Contact Process2. SO2 catalytically converted to SO32SO2(g) + O2(g) 2SO3(g) H = -99kJ/molSO2 and excess O2 pass into a converter where they contact porous V2O5 catalystGases pass several times over catalyst beds at lower temperaturesUnreacted gases are recycled, giving 99.7% conversion with 0.3% SO2 released to the atmosphereOverview of the whole Contact Process3. SO3 dissolved in conc. H2SO4 to form oleumSO3(g) + H2SO4(l) H2S2O7(l)Avoids dissolving SO3 in water directlyThis is very exothermic and would produce a mist of H2SO4 droplets that would be difficult to collect4. Oleum hydrated to form conc. H2SO4 (98%)H2S2O7(l) + H2O(l) 2H2SO4(l)You produce twice as much H2SO4 as you put inProperties of sulfuric acidColourless and oily, non-volatile liquidHighly corrosive with strong affinity for waterStrong acid: H2SO4 ionises completely in waterConcentrated 98% H2SO4Not much free water present, so virtually all H2SO4 molecules with very few H3O+ or HSO4-Most water molecules exist as H2SO4.H2OAs water is added (dilution), H2SO4 molecules are ionises in a highly exothermic reactionH2SO4(l) + H2O(l) HSO4-(aq) + H3O+(aq) H = -90kJ/molH2SO4: Dehydrating agentConc H2SO4 can readily remove water from various compoundsHydrated to anhydrous copper sulfateCuSO4.5H2O(l) CuSO4(s) + 5H2O(l)Dehydration of ethanolC2H5OH(l) C2H4(g) + H2O(l)Sucrose to carbonC12H22O11 12C(s) + 11H2O(l)H2SO4: Oxidising agentH2SO4 is a moderately strong oxidant takes electrons from other speciesOxidisation of a reactive metal (up to Pb, with dilute H2SO4)Mg(s) Mg2+(aq) + 2e-2H+(aq) + 2e- H2(g)Mg(s) + 2H+(aq) Mg2+(aq) + 2e-H2SO4: Oxidising agentOxidation of a less reactive metal (conc H2SO4)Cu(s) Cu2+(aq) + 2e-H2SO4(l) + 2H+(aq) +2e- SO2(g) + 2H2O(l)Sulfur within H2SO4 being reducedCu(s) + H2SO4(l) + 2H+(aq) Cu2+(aq) + SO2(g) + 2H2O(l)Add sulfate to both sides making equation more accurat as there isnt H+ to begin with bus is needed there to balance the redox reactionCu(s) + 2H2SO4(l) CuSO4(aq) + 2H2O(l) + SO2(g)H2SO4: Oxidising agentOxidation of iodide (conc H2SO4)2I-(aq) I2(aq) + 2e-H2SO4(l) + 2H+(aq) +2e- SO2(g) + 2H2O(l)2I-(aq) + H2SO4(l) + 2H+(aq) I2(aq) SO2(g) + 2H2O(l)Conc H2SO4: Safety precationsAlways wear gloves, safety goggles and a lab coatWhen diluting, add a small volume of acid to a large volume of waterUse a glass rod to avoid splashingNever add water to conc acid otherwise boiling/splitting may occurIf spilt, wash off quickily with tap water or isolate and use a base like NaHCO3 to neutralise bench spillWhy so much safety?Conc H2SO4 is highly corrosive, strong dehydrating agent and oxidising agentWill attack many materials and living tissue

Storage and transportConc H2SO4 has very few free H3O+ ionsWill not attack iron/steel as a passivating layer is formed so transport in steel drums/tankersBut care must be taken to avoid moistureDilute H2SO4 has plenty of free H3O+ ionsStore in glass/plastic containers not iron/steelDifficult and costly to transportFor regular use, store in 1-2.5L glass bottlesShould be well sealedUse in plastic trays to avoid bench/shelf contactStore in cool, ventilated places well away from any metal, bases and water