4 Major contributor to quality of life and economy. The Chemical IndustryMajor contributor to quality of life and economy.
5 The Chemical Industry Quality of life Fuels (eg petrol for cars) Plastics (Polythene etc)Agrochemicals (Fertilisers, pesticides etc)Alloys (Inc. Steel for building)Chemicals (eg Cl2 for water purification)Dyes (for clothing etc)Cosmetics and medicinesSoaps and detergentsEtc!!!!
6 The Chemical Industry Contributes to National Economy Major employer of people at all skill levelsRevenue from taxation on fuels etcRevenue from sales of productRevenue from exports of products
7 The Chemical IndustryResearch chemists identify a chemical route to make a new product, using available reactants.
8 The Chemical IndustryFeasibility study produces small amounts of product – to see if the process will work
9 The Chemical IndustryThe process is now scaled up to go into full scale production.Process so far will have taken months.Many problems will have been encountered and will have to be resolved before full scale production commences
10 The Chemical Industry Chemical plant is built in a suitable site Operators employedEarly production will allow monitoring of cost, safety, pollution risks, yield and profitability
11 The Chemical Industry Unreacted feedstocks recycled SEPARATOR MIXER REACTIONVESSELBY-PRODUCTPRODUCT
12 The Chemical Industry - Feedstocks Fossil fuelsCoal, oil and natural gasMetallic oresHaematite to make iron,Bauxite to make aluminiumMineralsLimestone needed in Blast furnaceAirSupplies O2 and N2WaterCan be used as a reactanct or as a coolant or in heat exchangers.Fossil fuelsMetallic ores
13 The Chemical Industry Can be Continuous process Or can be Batch Process
14 The Chemical Industry Continuous Process Used by big industries where large quantities of product are requiredRequires small workforceOften automated / computer controlledQuality of product checked remotelyEnergy efficiency usually goodPlants expensive to buildPlants not flexible
15 The Chemical Industry Batch Process Make substance which are required in smaller amountsProcess looks more like the initial reactionOverhaul of system needed regularly – time and energy lost if plant has to be shut downPlant can be more flexiblePlant is usually less expensive to build initially
16 The Chemical Industry: The Costs Involved ExampleCapital costsBuilding the plantRoad and rail links(Usually needs a substantial bank loan)Fixed costs(Stay the same regardless of whether plant runs at full or half capacity)Repayment of loanWages, Council taxVariable costs(Vary dependent on whether plant is running at full or half capacity.)Cost of raw materials, and other chemicals required
17 The Chemical Industry Industries can be classed as: Labour intensive Capital intensive
18 The Chemical IndustryService industries (Catering, education, healthcare), are labour intensive
19 The Chemical IndustryChemical industry tends to be more Capital intensive as a large investment is required to buy equipment and build plants
20 The Chemical IndustryExpectations of work safety and a clean environment increase during the twentieth centuryH & S legislation protects workforce
21 The Chemical IndustryTradition is important – steel making continues in areas where it was set up even if raw materials are no longer available locallyTransport options are important
22 The Chemical IndustryChoice of a particular chemical route is dependent upon:Cost of raw materialsSuitability of feedstocksYield of productOption to recycle unreacted feedstockMarketability of by productsCosts of getting rid of wastes, and safety considerations for workforce and localsPrevention of pollution
23 The Chemical Industry Click here to repeat The Chemical Industry. Click here to return to the MenuClick here to End.
25 Hess’s lawHess’s law states that the enthalpy change for a chemical reaction is independent of the route taken.This means that chemical equations can be treated like simultaneous equations.Enthalpy changes can be worked out using Hess’s law.
26 Hess’s lawCalculate the enthalpy change for the reaction: C(s) + 2H2(g) CH4(g) using the enthalpies of combustion of carbon, hydrogen and methane.
43 Dynamic EquilibriumReversible reactions reach a state of dynamic equilibriumThe rates of forward and reverse reactions are equal.At equilibrium, the concentrations of reactants and products remain constant, although not necessarily equal.
44 Changing the Equilibrium Using a catalyst does not change the position of the equilibrium.A catalyst speeds up both the forward and back reactions equally and so the equilibrium is reached more quickly.
45 Changing the Equilibrium Changes in concentration, pressure and temperature can alter the position of equilibrium.Le Chatelier’s Principle states that when we act on an equilibrium the position of the equilibrium will move to reduce the effect of the change.
46 Concentration Consider the equilibrium: A + B C + D If we increase the concentration of A, we speed up the forward reaction.This results in more C and D being formed.
47 Br2(aq) + H2O(l) 2H+(aq) + Br-(aq) + BrO-(aq) ConcentrationConsider the equilibrium:Br2(aq) + H2O(l) 2H+(aq) + Br-(aq) + BrO-(aq)The solution is red-brown, due the Br2 molecules.If we add sodium bromide, increasing the concentration of Br-, we favour the RHS and so the equilibrium moves to the left.The red-brown colour will increase.
48 PressureRemember:1 mole of any gas has the same volume (under the same conditions of pressure and temperature).This means that the number of moles of has are the same as the volumes.
49 PressureIncreasing pressure means putting the same number of moles in a smaller space.This is the same as increasing concentration.To reduce this effect the equilibrium will shift so as to reduce the number of moles of gas.
50 PressureIncreasing pressure favours the side with the smaller volume of gas. Consider:N2O4(g) 2NO2(g)1 mole moles1 volume volumesIf we increase the pressure we favour the forward reaction, so more N2O4 is formed.
51 Temperature An equilibrium involves two opposite reactions. One of these processes must release energy (exothermic).The reverse process must take in energy (endothermic).
52 TemperatureFirst consider an exothermic reaction.
53 Exothermic ReactionThis is the distribution of molecular energy
54 These molecules have sufficient energy to react Activation Energy
63 Increasing temperature leads to a greater increase in the number of molecules with sufficientactivation energy
64 TemperatureThe percentage increase in the number of molecules with sufficient activation energy is much greater in the endothermic reaction, compared to the exothermic reaction.
65 Thus both the endothermic and exothermic processes are speeded up by increasing temperature. However an increase in temperature has a greater effect on the endothermic process.
66 Increasing temperature favours the endothermic side of the equilibrium Increasing temperature favours the endothermic side of the equilibrium. Consider:N2O4(g) 2NO2(g) DH= +58 kJIf we increase the temperature we favour the forward reaction so more NO2 is formed.
67 The Haber ProcessThe Haber process involves the preparation of ammonia from nitrogen and hydrogen.N H2 2NH3 DH = -88 kJWe shall look at the factors affecting this equilibrium.
68 The Haber Process N2 + 3H2 2NH3 DH = -88 kJ A catalyst of finely divided iron is used to increase the reaction speed and so shorten the time needed to reach the equilibrium.
69 The Haber Process N2 + 3H2 2NH3 DH = -88 kJ 1 mole 3 moles 2 moles 1 vol 3 vols 2 vols4 vols 2 volsSince the RHS has a lower volume of gas than the LHS, higher pressure will favour the production of ammonia.A reaction chamber to withstand the higher pressure will cost much more.
70 The Haber Process N2 + 3H2 2NH3 DH = -88 kJ Since the forward reaction is exothermic more ammonia will be produced at low temperatures.At low temperatures the reaction is very slow so the rate of production of ammonia is low.
71 The Haber Process N2 + 3H2 2NH3 DH = -88 kJ To ensure maximum conversion the unreacted gases are recycled through the reaction chamber after reaction.
72 The Haber Process N2 + 3H2 2NH3 DH = -88 kJ To achieve the most profitable production of ammonia the following conditions are used:iron powder as catalyst250 atmospheres pressuretemperature of 500oC - 600oC
73 The Haber Process Unreacted N2 + H2 recycled SEPARATOR REACTION MIXER ChamberFe catalystNH3
74 Equilibrium Click here to repeat Equilibrium. Click here to return to the MenuClick here to End.
91 H2O(l) H+(aq) + OH-(aq) If we look at water, pH=7[H+] = [OH-] = 10-7 mol/l[H+] x [OH-] = mol2/l2This is due to the equilibrium in water:H2O(l) H+(aq) + OH-(aq)For any solution
92 Thus we can find [H+] for any solution. What is [H+] of a solution with pH 10?[OH-] = 10-4 mol/l[H+] x 10-4 = mol2/l2Thus [H+] = 10-10
93 Strong and Weak AcidsA strong acid is one which completely dissociates in solution:HCl(aq) H+(aq) + Cl-(aq)A weak acid is one which partially dissociates in solution:CH3CO2H(aq)H+(aq) + CH3CO2-(aq)
94 We can compare eqimolar solutions of strong and weak acids e. g We can compare eqimolar solutions of strong and weak acids e.g. 0.1 mol/l hydrochloric acid and 0.1 mol/l ethanoic acid.We compare pH, conductivity, reaction rates and stoichiomery.
95 Test100 ml 0.1 mol/l HCl100 ml 0.1 mol/l CH3CO2HpH13ConductivityVery highLowRate of reactionFastSlowStoichiomeryReacts with 0.4 g NaOH
96 The differences between the properties of strong and weak acids are caused by the fact that weak acids contain many fewer H+ ions than strong acids.Both acids can produce the same number of H+ ions, its just that weak acids do so more slowly.
98 Strong and Weak BasesA strong base is one which completely dissociates in solution:NaOH(aq) Na+(aq) + OH-(aq)A weak base is one which partially dissociates in solution:NH4OH(aq)NH4+(aq) + OH-(aq)
99 We can compare eqimolar solutions of strong and weak bases e. g We can compare eqimolar solutions of strong and weak bases e.g. 0.1 mol/l sodium hydroxide and 0.1 mol/l ammonium hydroxide.When we compare pH, conductivity, reaction rates and stoichiomery we find similar results to the comparison of weak and strong acids.
100 NH3(g) + H2O(l) NH4OH(aq) NH4OH(aq) NH4+(aq) + OH-(aq) Weak BasesA solution of ammonia is a weak base.NH3(g) + H2O(l) NH4OH(aq)NH4OH(aq) NH4+(aq) + OH-(aq)
101 Acids + BasesA strong acid and a strong base produce a salt which is neutral.A strong acid and a weak base produce a salt which is acidic.A weak acid and a strong base produce a salt which is basic.
105 Na2CO3(aq) 2Na+(aq) + CO32-(aq) Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present.
106 Na2CO3(aq) 2Na+(aq) + CO32-(aq) Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present.H2O(l) H+(aq) + OH-(aq)
107 Na2CO3(aq) 2Na+(aq) + CO32-(aq) The ions set up an equilibrium. Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present.H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium.
108 Basic Salts Sodium carbonate is completely ionised. Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present.H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium.2H+(aq) + CO32-(aq) H2CO3(aq)
109 Basic Salts This removes of H+(aq) from water. H2O(l) H+(aq) + OH-(aq)
110 Basic Salts This removes of H+(aq) from water. H2O(l) H+(aq) + OH-(aq)The OH-(aq) left behind make the resulting solution basic.
122 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) RedoxAn oxidation reaction is one where electrons are lost.Zn(s) Zn2+(aq) + 2eA reduction reaction is one where electrons are gained.Cu2+(aq) + 2e Cu(s)A redox reaction is one in which both oxidation and reduction are occurring.Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
123 Redox An oxidising agent is a substance which accepts electrons. This means that an oxidising agent must itself be reduced.
124 Redox A reducing agent is a substance which donates electrons. This means that a reducing agent must itself be oxidised.
125 RedoxWe should be able to recognise oxidising and reducing agents from the reaction equation.5Fe MnO H+ 5Fe Mn H2OFe2+ is oxidised to Fe3+ so MnO4– acts as an oxidising agent.
126 Writing Ion-Electron Equations. Simple equations can be obtained from the data booklet.More complex equations are written using the following routine.
127 Writing Ion-Electron Equations. Write the reactants and products.2IO3- I2Add H2O to the side with less oxygen.2IO3- I2 + 6H2OAdd H+ to the other side.2IO H+ I2 + 6H2OBalance charge by adding electrons.2IO H+ + 10e- I2 + 6H2O
128 Combining Oxidation and Reduction Equations. Combining the ion-electron half equations produces the overall reaction equation.This must be done so that the number of electrons on opposie sides are equal, and so cancel each other out.
133 Combining Oxidation and Reduction Equations. 2IO H+ + 10e I2 + 6H2OReduction multiplied by 55SO H2O 5SO H++ 10e
134 Combining Oxidation and Reduction Equations. Add the equations2IO H+ + 10e I2 + 6H2O5SO H2O 5SO H++ 10e2IO H+ + 5SO32- I2 + H2O + 5SO42-We now can extract the mole relationship – 2 moles iodate react with 5 moles of sulphite
135 Redox Titrations. These can be carried out to calculate concentration. Many use permanganate or starch/iodine reactions which are self-indicating – the colour change of the reaction tells you when the end point is reached.
136 Redox Titrations.It was found that 12.5 ml of of 0.1 mol/l acidified potassium dichromate was required to oxidise the alcohol in a sample of 1 ml of wine.Calculate the mass of alcohol in 1 ml of wine.
138 Redox Titrations.12.5 ml of 0.1 mol/l dichromate contain x0.1 moles dichromate.1.25x10-3 molesMoles of alcohol = 1.25x10-3 x1.5= 1.875x10-3Mass of alcohol = 46 x 1.875x10-3 g= g
139 ElectrolysisElectrolysis takes place when electricity is passed through an ionic liquid.Chemical reaction take place at the electrodes – reduction at the negative electrode and oxidation at the positive electrode.
140 ElectrolysisThe electrode reactions can be represented by ion electron equations.In the electrolysis of nickel(II) chloride the reactions are:+ electrode 2Cl- Cl e- electrode Ni e Ni
141 Electrolysis + electrode 2Cl- Cl2 + 2e - electrode Ni2+ + 2e Ni In both of these ion electron equations one mole of product is produced by two moles of electrons.
142 The FaradayTo find the value for one mole of electrons multiply Avogadro’s number by the charge on the electron (1.6x10-19 coulombs)One mole of electrons is called a Faraday and is 96,500 coulombs.
143 The Faraday Using the value for the Faraday and the equation: Charge = Current x Time (Coulombs) (Amps) (Seconds) we can carry out many calculations.
144 Redox Reactions Click here to repeat Redox Reactions. Click here to return to the MenuClick here to End.
148 Stable nuclei All stable nuclei fit in a narrow band Some nuclei are unstable because they need too much energy to hold them together.Thus they split apart, sending out some small particles.
149 Radioactive decay Particle Symbol Nature Stopped by alpha a Sheet of paperbetabFew cm of aluminiumgammagradiationMany cms of lead
150 a decay a decay takes place when the nucleus ejects a helium nucleus. This causes a change in the nucleus.
151 b decay b decay takes place when the nucleus ejects an electron. This causes a change in the nucleus.
152 g decay g decay takes place when the nucleus loses energy. This is the extra energy which is no longer needed to hold the nucleus together.
153 Nuclear EquationsWhen we write a nuclear equation the sum of the mass numbers and atomic numbers on each side must be equal.
154 Half lifeHalf life is the time which it takes for the radioactivity to half.For any radioactive substance this time is constant.
155 Half lifeThe decay of individual nuclei within a sample is random and is does not depend of chemical or physical state of the element.Half lives of individual elements may vary from seconds to thousands of years.
156 Half lifeCalculations involving half life usually involve precise fractions e.g.3H is a b-emitting isotope with a half life of 12.3 years. How long will it take for the radioactivity of a sample to drop to 1/8 of its original value?
157 Half life3H is a b-emitting isotope with a half life of 12.3 years. How long will it take for the radioactivity of a sample to drop to 1/8 of its original value?Time y 24.6y 36.9yFraction ½ ¼ 1/8
158 Half lifeFor examples where the numbers are more complex the quantity of radioactive material against time is best estimated from a graph.