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Higher chemistry unit 3

Higher chemistry unit 3. Chemical reactions. Unit 3. The chemical industry Hess’s law Equilibrium Acids & bases Redox reactions Nuclear chemistry. The chemical industry. You need to know Haber process Oswald process Fractional distillation of crude oil Cracking

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Higher chemistry unit 3

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  1. Higher chemistry unit 3 Chemical reactions

  2. Unit 3 • The chemical industry • Hess’s law • Equilibrium • Acids & bases • Redox reactions • Nuclear chemistry

  3. The chemical industry • You need to know • Haber process • Oswald process • Fractional distillation of crude oil • Cracking • Addition polymerisation • Condensation polymers • Manufacture of fuels

  4. Cosmetics & toiletries Disinfectant Polishes & cleaners Veterinary medicines Inks, dyes & pigments Paper industry Metal treatment Synthetic fibres UK chemical industry • Major global contributor, 6th largest manufacturer of chemicals. Plasitics Paints Detergents Explosives Adhesives Pharmaceuticals Aerosols Fertilisers Herbicides Pesticides

  5. Chemical industry • Contributes to both the quality of our life and to the national economy. • Over 150,000 Scot's work in the chemical industry.

  6. Manufacturing a new chemical • First there must be a demand. • In the rainy season demand for fungicides quadruples. • R&D (research & development) • Chemists find the most suitable route to making the desired chemical. Considering costs, HASAWA, timings etc.

  7. Manufacturing a new chemical 3. Laboratory Process • Chemists refine the process on a small scale to make it viable. 4. Pilot plant • A medium scale investigation of product quality. Engineering requirements assessed. 5. Production plant • Full scale plant design, construction and start up! Processing is constantly reviewed in the laboratory & on plant

  8. The Contact process (sulphuric acid)

  9. Step 1 Burning sulphur S(l) + O2(g) SO2(g) • Molten sulphur is sprayed into a furnace and burned in a blast of dry air. • The resulting mixture of 10% SO2 and 10% O2 is cooled from 1000°C to 400°C

  10. Step 2 Sulphur dioxide conversion • In the converter SO2 combines with oxygen in the presence of vanadium (V) oxide (catalyst) producing SO3 • 99.5% conversion to SO3 desirable. SO2(g) + ½ O2(g) SO3(g)

  11. Step 3 Sulphur trioxide absorption SO3 + H2O(l) H2SO4 • SO3 is absorbed in 98% sulphuric acid at 100-150°C. • Water alone is not used as the dissolving process is so quick a mist of sulphuric acid would be produced causing major environmental problems.

  12. Raw materials & feedstock • A feedstock is a reactant that other chemicals can be extracted or synthesised from. • Feedstock's are derived from raw materials. • Main raw materials include: • Fossil fuels • Ores • Air & water • Organic materials

  13. Batch processing • Reactants are added to the reactor. • Reaction is started and process is carefully monitored testing throughout. • At the end of the reaction the reaction vessel is emptied and product passed on to separation, purification and formulation stages. • Reaction vessel is cleaned down and the process starts again.

  14. Reactants flow into the reactor at one end and products flow out the other end. Continuous process

  15. Pro’s & cons

  16. Economic aspects

  17. The UK’s chemical industry is capital rather than labour intensive

  18. Energy • Strive to use as little energy as possible. • Use an exotherm to heat a reactor elsewhere in plant. • Waste heat can be used to generate electricity. • Sell waste heat to local housing schemes.

  19. Safety • All steps in a process must be risk assessed. • Any risks must be fully investigated and alternative routes researched. • PPE (personal protective equipment) is a last resort. • H&S is the biggest obstacle to overcome in designing a process.

  20. Location of industry • Large flat area. • Water close at hand. • Good transport links (rail/sea/road) • Skilled workforce at hand.

  21. Factors that influence route to product • Availability of feedstock's • Yield of products attainable • Opportunities for recycling • Marketing of products & by-products • Health & safety • Competition Now complete page 134 & 135

  22. Unit 3 • The chemical industry • Hess’s law • Equilibrium • Acids & bases • Redox reactions • Nuclear chemistry

  23. You need to remember… • The enthalpy change in a reaction is the difference in potential energy between reactants & products. • H and measured in kJ mol-1 • Exothermic H = -ve • Endothermic H = +ve

  24. Hess’s law Enthalpy change for a chemical reaction is independent of route taken.

  25. Hess’s law • The enthalpy change of a chemical reaction depends only on the chemical nature and physical states of the reactants & products. It is independent or any intermediate steps.

  26. Formation of 1 mole of methane from its elements Hx C(s) + 2H2(g) CH4(g) + O2 + O2 -2O2(g) H1 H2 H3 Reverse of combustion of methane 2H2O(l) + CO2(g) C(s) + O2(g) CO2(g) 2H2(g)+O2(g) 2H2O(l) CO2+ 2H2O CH4+ 2O2 = 2 x -286 = -572kJ = -394kJ H3 = +891kJ H2 H1

  27. According to Hess’s law Hx = H1 + H2 + H3 = (-394) + (-572) + 891 = - 75kJ mol-1 • Now compare with the enthalpy of formation of methane from the databook.

  28. Formation of 1 mole of propan-1-ol from its elements Hx 3C(s) + 4H2(g) + ½ O2(g) C3H7OH(g) + 3O2 + 2O2 -5O2(g) H1 H2 H3 Reverse of combustion of propanol 4H2O(l) + 3CO2(g) 3C(s) + 3O2(g) 3CO2(g) 4H2(g)+2O2(g) 4H2O(l) 3CO2+ 4H2O C3H7OH+ 5O2 = 4(-286) = -1144kJ = 3(-394kJ) H3 = +2020kJ H2 H1 = -1182kJ

  29. According to Hess’s law Hx = H1 + H2 + H3 = (-1182) + (-1144) + 2020 = - 306kJ mol-1 • Now compare with the enthalpy of formation of propan-1-ol from the databook.

  30. Unit 3 • The chemical industry • Hess’s law • Equilibrium • Acids & bases • Redox reactions • Nuclear chemistry

  31. You need to know… • Metal oxides produce alkalis, non metal oxides produce acids. • All acids contain H+ ions, alkalis contain OH-. • Neutral solutions and water have pH 7. • Reactions of acids with metal oxides, metal carbonates & metals.

  32. The pH scale • A continuous scale from less than 0 to more than 14. • The concentration of hydrogen ions in solution is measured in pH units. • pH stands for the negative logarithm (to base 10) of the hydrogen ion concentration. • pH = -log10[H+(aq)] Square brackets mean concentration Of what is inside them

  33. The pH scale • M represents moles l-1. • Square brackets [] represent the concentration in moles l-1.

  34. Calculating pH pH = - log10[H+] 0 0 -1 1 -2 2 3 -3 -7 7 -14 14

  35. pH • For water and neutral solutions in water, H+ and OH- ions come from the partial dissociation of water molecules. • In water and neutral solutions there is an equilibrium between H+(aq) and OH-(aq) ions and water molecules. H2O(l) H+(aq) + OH-(aq)

  36. Equilibrium • An equilibrium is reached with the concentrations of both H+ and OH- ions equal to 10-7 M (1/107 mol l-1) at 25ºC. • [H+] = [OH-] 10-7moles l-1 = 10-7moles l-1 so [H+][OH-] = 10-7moles l-1 x 10-7moles l-1 = 10-14moles2 l-2 Ionic product of water

  37. Calculations • We can now calculate pH, [H+] and [OH-] using the relationship… [H+] = 10-14 moles l-1 [OH-] Or [OH-] = 10-14moles l-1 [H+] 10-14 [H+] [OH-]

  38. Example 1 • What is the concentration of OH- ions in a solution containing 0.01 moles l-1 of H+ ions. [H+] = 10-2 moles l-1 [OH-] = 10-14 10-14 = 10-12 M [H+] 10-2

  39. Your turn!!! • What is the concentration of H+ ions in a solution containing 0.1 moles l-1 of OH- ions? • What is the pH of a solution in question 2. 10-13 moles l-1 pH = 13

  40. By the end of the period… • Complete questions on page 151 of your text book. • Complete pages 1-4 (stop @ strong & weak) in worksheets. • If finished, have a go at your equilibrium homework due tomorrow!!!

  41. Strong acids • Strong acids dissociate completely into ions. HCl(aq) H+(aq) + Cl-(aq) H2SO4(aq) 2H+(aq) + SO4-(aq) Ions are separate in solution, no HCl left.

  42. Weak acids… • Partially dissociate. An equilibrium is reached between the ions and the undissociated molecule. • When diluted the equilibrium will shift to the right. As dilution increases the degree of ionisation (dissociation) also increases. CH3COOH(aq) CH3COO-(aq) + H+(aq)

  43. Strong & weak acids • Differ in pH, conductivity and reaction rates. • By comparing equimolar solutions we can find out the differences.

  44. Neutralising weak acids CH3COOH CH3COO- + H+ • As the reaction proceeds H+ ions are used up. • This forces the equilibrium to the right, the ionisation step. • Eventually all of the molecules dissociate. • Therefore it requires the same amount of base to neutralise a certain volume of either 0.1M HCl or 0.1M CH3COOH

  45. Stoichiometry (mole ratio of reactants) • So the stoichiometry of the neutralisation reactions are the same, NaOH + HCl NaCl + H2O NaOH + CH3COOH CH3COONa + H2O 1 mol 1 mol 1 mol 1 mol

  46. Carbonic acid (weak) • Carbonic acid formation CO2(aq) + H2O H2CO3(aq) O HO C OH

  47. Sulphurous acid (weak) SO2 + H2O HSO3− + H+ O HO S OH

  48. Strong & weak alkalis • Strong bases fully dissociate. • Weak bases only partially dissociate. • The stoichiometry of strong and weak alkalis is the same.

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