1 / 161

Unit 3

Unit 3. Chemical Reactions. Menu. The Chemical Industry Hess’s Law Equilibrium Acids and Bases Redox Reactions Nuclear Chemistry. The Chemical Industry. The Chemical Industry. Major contributor to quality of life and economy. Quality of life. Fuels (eg petrol for cars)

reed
Download Presentation

Unit 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Unit 3 Chemical Reactions

  2. Menu • The Chemical Industry • Hess’s Law • Equilibrium • Acids and Bases • Redox Reactions • Nuclear Chemistry

  3. The Chemical Industry

  4. The Chemical Industry Major contributor to quality of life and economy.

  5. 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 medicines Soaps and detergents Etc!!!! The Chemical Industry

  6. Contributes to National Economy Major employer of people at all skill levels Revenue from taxation on fuels etc Revenue from sales of product Revenue from exports of products The Chemical Industry

  7. The Chemical Industry • Research chemists identify a chemical route to make a new product, using available reactants.

  8. The Chemical Industry • Feasibility study produces small amounts of product – to see if the process will work

  9. The Chemical Industry • The 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 employed • Early production will allow monitoring of cost, safety, pollution risks, yield and profitability

  11. The Chemical Industry Unreacted feedstocks recycled SEPARATOR MIXER Feedstock REACTION VESSEL BY-PRODUCT PRODUCT

  12. The Chemical Industry - Feedstocks

  13. The Chemical Industry • Can be Continuous process • Or can be Batch Process

  14. Continuous Process Used by big industries where large quantities of product are required Requires small workforce Often automated / computer controlled Quality of product checked remotely Energy efficiency usually good Plants expensive to build Plants not flexible The Chemical Industry

  15. Batch Process Make substance which are required in smaller amounts Process looks more like the initial reaction Overhaul of system needed regularly – time and energy lost if plant has to be shut down Plant can be more flexible Plant is usually less expensive to build initially The Chemical Industry

  16. The Chemical Industry: The Costs Involved

  17. The Chemical Industry Industries can be classed as: • Labour intensive • Capital intensive

  18. The Chemical Industry • Service industries (Catering, education, healthcare), are labour intensive

  19. The Chemical Industry • Chemical industry tends to be more Capital intensive as a large investment is required to buy equipment and build plants

  20. The Chemical Industry • Expectations of work safety and a clean environment increase during the twentieth century • H & S legislation protects workforce

  21. The Chemical Industry • Tradition is important – steel making continues in areas where it was set up even if raw materials are no longer available locally • Transport options are important

  22. The Chemical Industry Choice of a particular chemical route is dependent upon: • Cost of raw materials • Suitability of feedstocks • Yield of product • Option to recycle unreacted feedstock • Marketability of by products • Costs of getting rid of wastes, and safety considerations for workforce and locals • Prevention of pollution

  23. The Chemical Industry Click here to repeat The Chemical Industry. Click here to return to the Menu Click here to End.

  24. Hess’s Law

  25. Hess’s law • Hess’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 law • Calculate the enthalpy change for the reaction: C(s) + 2H2(g)  CH4(g) using the enthalpies of combustion of carbon, hydrogen and methane.

  27. Hess’s law First write the target equation.

  28. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? Then write the given equations.

  29. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ H2 + ½ O2  H2O DH= -286 kJ CH4+ 2O2 CO2 + 2H2O DH=-891 kJ

  30. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? Build up the target equation from the given equations. If we multiply we must also multiply DH. If we reverse an equation we reverse the sign of DH.

  31. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ

  32. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ H2 + ½ O2  H2O DH= -286 kJ

  33. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ

  34. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CH4 + 2O2  CO2+2H2O DH=-891 kJ

  35. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CO2+2H2O  CH4 + 2O2DH=+891 kJ

  36. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CO2+2H2O  CH4 + 2O2DH=+891 kJ

  37. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? We can add all the equations, striking out species that will appear in equal numbers on both sides.

  38. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CO2+2H2O  CH4 + 2O2DH=+891 kJ

  39. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CO2+ 2H2O  CH4 +2O2DH=+891 kJ C + 2H2 CH4DH=(-394 –572 + 891)kJ = -75 kJ

  40. Hess’s law C(s) + 2H2(g)  CH4(g) DH=? C + O2  CO2DH= -394 kJ 2H2 + O2  2H2O DH= -572 kJ CO2+ 2H2O  CH4 +2O2DH=+891 kJ C + 2H2 CH4DH=-75 kJ

  41. Hess’s Law Click here to repeat Hess’s Law. Click here to return to the Menu Click here to End.

  42. Equilibrium

  43. Dynamic Equilibrium • Reversible reactions reach a state of dynamic equilibrium • The 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. Concentration • Consider 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. Pressure • Remember: • 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. Pressure • Increasing 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. Pressure • Increasing pressure favours the side with the smaller volume of gas. Consider: N2O4(g)  2NO2(g) 1 mole 2 moles 1 volume 2 volumes • If we increase the pressure we favour the forward reaction, so more N2O4 is formed.

More Related