1. IGCSE CHEMISTRY SECTION 4 LESSON 2

2. Content The iGCSE Chemistry course Section 1 Principles of Chemistry Section 2 Chemistry of the Elements Section 3 Organic Chemistry Section 4 Physical Chemistry Section 5 Chemistry in Society

3. Content Section 4 Physical Chemistry a)Acids, alkalis and salts b)Energetics c)Rates of reaction d)Equilibria

4. Lesson 2 b) Energetics 4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic 4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation in which heat energy changes can be calculated from measured temperature changes 4.12 calculate molar enthalpy change from heat energy change 4.13 understand the use of ΔH to represent enthalpy change for exothermic and endothermic reactions 4.14 represent exothermic and endothermic reactions on a simple energy level diagram 4.15 understand that the breaking of bonds is endothermic and that the making of bonds is exothermic 4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.

5. Exothermic and endothermic reactions

6. In a chemical reaction there is usually a temperature change.

7. Exothermic and endothermic reactions In a chemical reaction there is usually a temperature change. In an EXOTHERMIC reaction, the temperature goes UP. Heat energy is given OUT.

8. Exothermic and endothermic reactions In a chemical reaction there is usually a temperature change. In an EXOTHERMIC reaction, the temperature goes UP. Heat energy is given OUT. In an ENDOTHERMIC reaction, the temperature goes DOWN. Heat energy is taken IN.

9. Exothermic and endothermic reactions Examples of EXOTHERMIC reactions: Combustion is a common example of an exothermic reaction. Methane + oxygen  Carbon dioxide + water + Heat energy Carbon + oxygen  Carbon dioxide + heat energy (coal)

10. Exothermic and endothermic reactions Examples of EXOTHERMIC reactions: Neutralisation is another example of an exothermic reaction. Acid + Alkali  Salt + Water + Heat energy

11. Exothermic and endothermic reactions Examples of EXOTHERMIC reactions: Neutralisation is another example of an exothermic reaction. Acid + Alkali  Salt + Water + Heat energy The symbol to show a change in the amount of heat energy is ∆H (pronounced delta H)

12. Exothermic and endothermic reactions For an EXOTHERMIC reaction, ∆H is negative. This means that heat has been lost from the reaction. Increasing energy reactants products Loss of energy Time

13. Exothermic and endothermic reactions Examples of ENDOTHERMIC reactions: Endothermic reactions tend to be less common. Dissolving ammonium nitrate crystals in water is an endothermic reaction. Ammonium + Water  ammonium nitrate – Heat nitrate solution energy Temperature of reactants = 20 o C Temperature of products = 13 o C

14. Exothermic and endothermic reactions For an ENDOTHERMIC reaction, ∆H is positive. This means that heat has been gained in the reaction. Increasing energy reactants products Gain in energy Time

15. Exothermic and endothermic reactions Let’s just recap

16. Exothermic and endothermic reactions Let’s just recap EXOTHERMIC

17. Exothermic and endothermic reactions Let’s just recap EXOTHERMIC Heat is given out

18. Exothermic and endothermic reactions Let’s just recap EXOTHERMIC Heat is given out ∆H is negative

19. Exothermic and endothermic reactions Let’s just recap ENDOTHERMIC

20. Exothermic and endothermic reactions Let’s just recap ENDOTHERMIC Heat is taken in

21. Exothermic and endothermic reactions Let’s just recap ENDOTHERMIC Heat is taken in ∆H is positive

22. Exothermic and endothermic reactions Another word for the heat energy content of a compound is ENTHALPY

23. Exothermic and endothermic reactions You need to know that the unit of energy is the JOULE (J).

24. Exothermic and endothermic reactions You need to know that the unit of energy is the JOULE (J). 1000 joules = 1 kilojoule (kJ)

25. Calculating enthalpy changes

26. The enthalpy change of a reaction is the enthalpy of the products minus the enthalpy of the reactants.

27. Calculating enthalpy changes The enthalpy change of a reaction is the enthalpy of the products minus the enthalpy of the reactants. Eg. Enthalpy of combustion of a fuel

28. Calculating enthalpy changes Eg. Enthalpy of combustion of a fuel The amount of energy released or absorbed in a reaction can be measured using this formula: heat energy = mass of x rise in x specific heat produced water temperature capacity of water (joules) (grams) ( o C) J/kg/ o C

29. Calculating enthalpy changes thermometer Draught shield Insulating card water Copper calorimeter Spirit burner Measurements: Temperature of water before and after heating Mass of water used Mass of fuel used

30. Calculating enthalpy changes EXAMPLE: the temperature of 100g of water was increased by 32.6 o C by heating with 0.46 grams of ethanol. Assuming no heat losses from the apparatus, what was the enthalpy of combustion of ethanol, C 2 H 5 OH?

31. Calculating enthalpy changes heat energy = mass of x rise in x specific heat produced water temperature capacity of water (joules) (grams) ( o C) J/kg/ o C heat energy = 100 x 32.6 x 4.2 produced (joules) heat energy = 13692 (for 0.46g of ethanol) produced (joules)

32. Calculating enthalpy changes heat energy = mass of x rise in x specific heat produced water temperature capacity of water (joules) (grams) ( o C) J/kg/ o C heat energy = 100 x 32.6 x 4.2 produced (joules) heat energy = 13692 (for 0.46g of ethanol) produced (joules) So what is the heat energy produced per mole of ethanol?

33. Calculating enthalpy changes heat energy = mass of x rise in x specific heat produced water temperature capacity of water (joules) (grams) ( o C) J/kg/ o C heat energy = 100 x 32.6 x 4.2 produced (joules) heat energy = 13692 (for 0.46g of ethanol) produced (joules) So what is the heat energy produced per mole of ethanol? Another word for the heat energy content of a compound is ENTHALPY Remember that a mole of a substance is the relative molecular mass in grams

34. Calculating enthalpy changes heat energy = 13692 (for 0.46g of ethanol) produced (joules) Ethanol C 2 H 5 OH = 24 + 5 + 16 + 1 = 46

35. Calculating enthalpy changes heat energy = 13692 (for 0.46g of ethanol) produced (joules) Ethanol C 2 H 5 OH = 24 + 5 + 16 + 1 = 46 So the mass of one mole of ethanol is 46g

36. Calculating enthalpy changes heat energy = 13692 (for 0.46g of ethanol) produced (joules) Ethanol C 2 H 5 OH = 24 + 5 + 16 + 1 = 46 So the mass of one mole of ethanol is 46g The energy released from the combustion of one mole of ethanol = 13692 x 46/0.46 = 1369200 J = 1369.2 kJ

37. Calculating enthalpy changes Burning ethanol is an EXOTHERMIC reaction, so heat energy is lost (∆H is –ve) The molar enthalpy of combustion of ethanol is therefore: = -1369.2 kJ/mol

38. Calculating enthalpy changes Use a simple calorimeter to measure the enthalpy change in reactions such as dissolving, neutralisation and displacement.

39. Calculating enthalpy changes thermometer Insulating lid Chemical mixture (eg. acid + alkali) Polystyrene cup (insulated)

40. Calculating enthalpy changes Enthalpy of neutralisation = the heat produced when an acid and alkali react to produce one mole of water.

41. Calculating enthalpy changes Enthalpy of neutralisation = the heat produced when an acid and alkali react to produce one mole of water. eg. 50cm 3 of 2M hydrochloric acid were added to 50cm 3 of 2M sodium hydroxide in a polystyrene cup. The temperature rose by 13.7 o C.

42. Calculating enthalpy changes eg. 50cm 3 of 2M hydrochloric acid were added to 50cm 3 of 2M sodium hydroxide in a polystyrene cup. The temperature rose by 13.7 o C. If there were no heat losses, and a final volume of 100cm 3, what was the enthalpy of neutralisation?

43. Calculating enthalpy changes HCl + NaOH  NaCl + H 2 O

44. Calculating enthalpy changes HCl + NaOH  NaCl + H 2 O 1 1 1 1

45. Calculating enthalpy changes HCl + NaOH  NaCl + H 2 O 1 1 1 1 Number of moles in 50cm3 of 2M HCl = 50 x 2 = 0.1 1000

46. Calculating enthalpy changes HCl + NaOH  NaCl + H 2 O 1 1 1 1 Number of moles in 50cm3 of 2M HCl = 50 x 2 = 0.1 1000 From the equation, 0.1 moles of HCl will produce 0.1 moles of H 2 O

47. Calculating enthalpy changes HCl + NaOH  NaCl + H 2 O 1 1 1 1 Number of moles in 50cm3 of 2M HCl = 50 x 2 = 0.1 1000 From the equation, 0.1 moles of HCl will produce 0.1 moles of H 2 O Heat released in producing 0.1M = 100 x 13.7 x 4.2 = 5754J of water

48. Calculating enthalpy changes Heat released in producing 0.1M = 100 x 13.7 x 4.2 = 5754J of water Enthalpy of = 5754 x 1/0.1 J/mol neutralisation

49. Calculating enthalpy changes Heat released in producing 0.1M = 100 x 13.7 x 4.2 = 5754J of water Enthalpy of = 5754 x 1/0.1 J/mol neutralisation = - 57540 J/mol = - 57.54 kJ/mol (remember: because this is an exothermic reaction, ∆H is negative)

50. Making and breaking bonds

51. Why are there energy (temperature) changes in chemical reactions?

52. Making and breaking bonds Why are there energy (temperature) change sin chemical reactions? Because chemical reactions involve the making and breaking of bonds!

53. Making and breaking bonds C H H HH

54. C H H HH Breaking chemical bonds during a reaction

55. Making and breaking bonds When breaking a chemical bond, energy is put IN

56. Making and breaking bonds When breaking a chemical bond, energy is put IN If energy is put in, then this must be ENDOTHERMIC

57. Making and breaking bonds When breaking a chemical bond, energy is put IN If energy is put in, then this must be ENDOTHERMIC When making a chemical bond, energy is given OUT

58. Making and breaking bonds When breaking a chemical bond, energy is put IN If energy is put in, then this must be ENDOTHERMIC When making a chemical bond, energy is given OUT If energy is given out, then this must be EXOTHERMIC

59. Making and breaking bonds When breaking a chemical bond, energy is put IN If energy is put in, then this must be ENDOTHERMIC When making a chemical bond, energy is given OUT If energy is given out, then this must be EXOTHERMIC BREAKING bonds is ENDOTHERMIC, MAKING bonds is EXOTHERMIC

60. Making and breaking bonds So what happens in an endothermic reaction? REACTANTS  PRODUCTS Energy In Energy out More energy has to be put IN to break the old bonds than is released when the new bonds are formed.

61. Making and breaking bonds So what happens in an exothermic reaction? REACTANTS  PRODUCTS Energy in Energy out More energy is given OUT when new bonds are formed than is taken in when the old bonds are broken.

62. Making and breaking bonds Each type of bond (eg. C-C) has a specific bond energy – it is the same whether the bond is being made or broken.

63. Making and breaking bonds Each type of bond (eg. C-C) has a specific bond energy – it is the same whether the bond is being made or broken. BONDBOND ENERGY kJ/mol C – H435 C = O803 O = O497 O – H464 YOU DON’T NEED TO LEARN THESE!

64. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O

65. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O In methane there are four C-H bonds. Each one has a bond energy of 435 kJ/mol. So to break methane apart requires 4 x 435 kJ/mol = 1740 kJ/mol

66. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H

67. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H)

68. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H) 4 x 435 + 2 x 497 2 x 803 + 4 x 464

69. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H) 4 x 435 + 2 x 497 2 x 803 + 4 x 464 = 1740 + 994 1606 + 1856

70. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H) 4 x 435 + 2 x 497 2 x 803 + 4 x 464 = 1740 + 994 1606 + 1856 = 2734 kJ/mol = 3462 kJ/mol

71. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H) 4 x 435 + 2 x 497 2 x 803 + 4 x 464 = 1740 + 994 1606 + 1856 = 2734 kJ/mol = 3462 kJ/mol 2734 - 3462 = - 728 kJ/mol

72. Making and breaking bonds Methane + Oxygen  Carbon Dioxide + Water CH 4 2O 2 CO 2 2H 2 O O=O O=C=O H-O-H O=O H-O-H 4 x (C-H) 2 x (O=O) 2 x (C=O) 4 x (O-H) 4 x 435 + 2 x 497 2 x 803 + 4 x 464 = 1740 + 994 1606 + 1856 = 2734 kJ/mol = 3462 kJ/mol 2734 - 3462 = - 728 kJ/mol The negative sign shows that this is an EXOTHERMIC reaction and heat energy is given OUT

73. Making and breaking bonds We can show the energy changes in the form of an energy level diagram.

74. Making and breaking bonds We can show the energy changes in the form of an energy level diagram. Progress of reaction Energy (kJ/mol)

75. Making and breaking bonds We can show the energy changes in the form of an energy level diagram. Progress of reaction Energy (kJ/mol) 2734 kJ/mol

76. Making and breaking bonds We can show the energy changes in the form of an energy level diagram. Progress of reaction Energy (kJ/mol) 2734 kJ/mol3462 kJ/mol

77. Making and breaking bonds We can show the energy changes in the form of an energy level diagram. Progress of reaction Energy (kJ/mol) 2734 kJ/mol3462 kJ/mol -728 kJ/mol

78. End of Section 4 Lesson 2 In this lesson we have covered: Exothermic and endothermic reactions Calculating enthalpy changes Making and breaking bonds