1. Exothermic and endothermic reactions

2. Chemical Reactions usually involve a temperature change (heat is given out or taken in)

3. Law of conservation of energy Energy cannot be created or destroyed, but only changed from one form into another

4. Exothermic Reactions Examples include: –Burning reactions including the combustion of fuels. –Detonation of explosives. –Reaction of acids with metals. Thermit reaction Magnesium reacting with acid Exothermic reactions increase in temperature.

5. Magnesium + Hydrochloric acid Gets hot 25 o C 45 o C magnesium Hydrochloric acid Heat energy given out Exothermic Reactions

6. Reactants convert chemical energy to heat energy. The temperature rises. 25 o C 45 o C Exothermic Reactions

7. 45 o C Almost immediately the hot reaction products start to lose heat to the surroundings and eventually they return to room temperature. 25 o C Exothermic Reactions

8. Energy / kJ) Progress of reaction (time) Energy Level Diagram for an Exothermic Reaction reactants Reactants have more chemical energy. Some of this is lost as heat which spreads out into the room. products Products now have less chemical energy than reactants.

9. Energy / kJ) Progress of reaction (time) Energy Level Diagram for an Exothermic Reaction reactants products

10. Exothermic reactions give out energy. There is a temperature rise and  H is negative. Exothermic Reaction - Definition products Energy / kJ) Progress of reaction reactants  H is negative

11. Heat changes also happen when substances change state.

12. An exothermic reaction When hydrocarbons burn in oxygen they produce carbon dioxide and water vapour. The reaction also involves the loss of heat so it is an exothermic reaction.

13. Objectives for today.. Define endothermic reaction Draw an energy profile diagram for an endothermic reaction Bond energy definition Mandatory experiment Homework – Write up today’s experiment in experiment copy

14. Endothermic reaction An endothermic reaction is when heat is taken in in a reaction.

15. Endothermic Reactions Endothermic chemical reactions are relatively rare. A few reactions that give off gases are highly endothermic - get very cold.

16. Cools Heat energy taken in as the mixture returns back to room temp. Starts 25°C Cools to 5°C Ammonium nitrate Water Endothermic reactions cause a decrease in temperature. Returns to 25°C Endothermic Reactions

17. 25 o C The cold reaction products start to gain heat from the surroundings and eventually return to room temperature. 5 o C The reactants gain energy. 25 o C This comes from the substances used in the reaction and the reaction gets cold. Eventually heat is absorbed from the surroundings and the mixture returns to room temperature. Overall the chemicals have gained energy. Endothermic Reactions

18. Endothermic reactions take in energy. There is a temperature drop and  H is positive. Endothermic Reaction Definition  H=+ products Energy / kJ Progress of reaction reactants

19. Homework Revision CH 1,2,3 HL 2015 Q4a,b Q5a,c 2014 Q4b, Q10b OL 2015 Q4a 2013 Q5b 2011 Q4a, 2010 Q4a 2009 Q4b 2008 Q11a

20. Heat of reaction The heat of reaction is the heat change when the number of moles of reactants indicated in the balanced equation react completely. For an exothermic reaction: the heat of reaction is always negative e.g ∆ H = -34kJ For an endothermic reaction: the heat of reaction is always positive e.g ∆ H = +34kJ

22. Measuring the heat of reaction of hydrochloric acid and sodium hydroxide A polystyrene cup is used as it is an insulator of heat – it doesn’t let the heat escape ( has negligible heat capacity) The equation for the reaction is HCl + NaOH → NaCl + H 2 O

23. Other precautions to ensure an accurate result! 1.Make sure both solutions are at the same temperature before you start! 2.Wash the thermometer and dry it before switching solutions. 3.Stir the mixture slowly and make sure none of the mixture is splashed out of the cup.

24. Results

25. Calculations – What is the heat change in our reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

26. How many moles were reacted in our experiment?

27. Question: What is the heat of reaction? Balanced equation: HCl + NaOH NaCl + H 2 O 1 1 0.1 moles of HCl released ________of heat 1 mole of HCl would release_________ of heat The heat of reaction = - J (or – kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

28. Q260. Calculations for the heat of reaction 261. Calculate the heat of reaction (using the formula ΔH = mcΔT) for the reaction between and nitric acid sodium hydroxide from the following experimental results: Volume of nitric acid = 100 cm 3 of 1.0 M Volume of sodium hydroxide = 100 cm 3 of 1.0 M Initial temperature of the solutions = 17.5 o C Final temperature of the solutions = 24.4 o C Specific heat capacity of the mixture = 4080 J Kg —1o C —1

29. Heat change= (.2kg) x (4080Jkg -1 K -1 ) x 6.9 o C Heat change = 5630J 1. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

30. 2. How many moles of Nitric acid were reacted?: 100cm 3 of a 1M solution of HNO 3 was reacted 1 x 100 = 0.1 moles 1000 Number of moles of HNO 3 reacted = 0.1moles

31. Question: What is the heat of reaction? Balanced equation: HNO 3 + NaOH NaNO 3 + H 2 O 1 1 0.1 moles of HNO 3 = 5630J of heat 1 mole of HNO 3 = 5630 X10 = 56300 The heat of reaction = - 56300J (or – 56.3kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

32. Example from notes

33. Heat change= (.4kg) x (4060Jkg -1 K -1 ) x 6.9 o C Heat change = 11205.6 J 1. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

34. 2. How many moles of Nitric acid were reacted?: (1) = 0.2 MOLES 1000 Number of moles of HNO 3 reacted = 0.2moles 200cm 3 of a 1M solution of HNO 3 was reacted

35. Question: What is the heat of reaction? Balanced equation: HNO 3 + NaOH NaNO 3 + H 2 O 1 1 0.2 moles of HNO 3 = 11205.6 J of heat 1 mole of HNO 3 = 11205.6 x 5 = 56028 kJ The heat of reaction = -56025kJ (or – 56.025kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

36. Q261. Calculations for the heat of reaction A student carried out an experiment to measure the heat of reaction (neutralisation) of nitric acid by sodium hydroxide in a container made of plastic of negligible heat capacity. She used 100 cm3 of 1.0 M nitric acid and 100 cm3 of 1.0 M sodium hydroxide. The initial temperature of the solutions was 15.6 o C and the final temperature of the solution was 22.4 o C. Given that specific heat capacity of the solution is 4080 J Kg —1 K —1, calculate the heat of reaction. (Assume that the density of the solution is 1 g cm —3 )

37. Heat change= (.2kg) x (4080Jkg -1 K -1 ) x 6.8 o C Heat change = 5548.8J 1. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

38. 2. How many moles of Nitric acid were reacted?: 100cm 3 of a 1M solution of HNO 3 was reacted 1000cm 3 of solution = 1 mole in it. 100cm 3 of solution = x moles (1) = x 10 Number of moles of HNO 3 reacted = 0.1moles

39. Question: What is the heat of reaction? Balanced equation: HNO 3 + NaOH NaNO 3 + H 2 O 1 1 0.1 moles of HNO 3 = 5548J of heat 1 mole of HNO 3 = 5548 x 10 = 55480kJ The heat of reaction = -55480kJ (or – 55.48kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

40. Heat change= (.5kg) x (4060Jkg -1 K -1 ) x 3.4 o C Heat change = 6902J Question from WS:. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

41. 2. How many moles of Hydrochloric acid were reacted?: 250cm 3 of a 0.5M solution of HCl was reacted 1000cm 3 of solution = 0.5 mole in it. 250cm 3 of solution = (0.5/ 1000) x 250 = 0.125moles Number of moles of HCl reacted = 0.125moles

42. Question: What is the heat of reaction? Balanced equation: HCl + NaOH NaCl + H 2 O 1 1 0.125 moles of HCl = 6902J of heat 1 mole of HNO 3 = 6902 x 8 = 55216 The heat of reaction = -55216J (or – 55.216kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

43. Q266 (d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction. [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.] (e) How many moles of hydrochloric acid were used in the experiment? Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.

44. Heat change= (.1kg) x (4.2kJkg -1 K -1 ) x 6.8 K Heat change = 2.856J 1. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

45. Q266 (d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction. [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.] (e) How many moles of hydrochloric acid were used in the experiment? Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.

46. How many moles of Hydrochloric acid were reacted?: 50cm 3 of a 1M solution of HCl was reacted 1000cm 3 of solution = 1 mole in it. 50cm 3 of solution = x moles (50)(1) = x(1000) (1) x (50) = x 1000 Number of moles of HCl reacted = 0.05moles

47. Q266 (d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction. [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.] (e) How many moles of hydrochloric acid were used in the experiment? Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.

48. Question: What is the heat of reaction? Balanced equation: HCl + NaOH NaCl + H 2 O 1 1 0.05 moles of HCl = 2.856kJ of heat 1 mole of HNO 3 = 2.856 /20 = 57.12 J of heat The heat of reaction = -57.12kJ The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

49. Q267 (f) Calculate the number of moles of acid neutralised in this experiment. In an experiment to measure the heat of reaction for the reaction between sodium hydroxide with hydrochloric acid, a student added 50 cm3 of 1.0 M HCl solution to the same volume of 1.0 M NaOH solution in a polystyrene foam cup.Taking the total heat capacity of the reaction mixture used in this experiment as 420 J K–1, calculate the heat released in the experiment if a temperature rise of 6.7 ºC was recorded. Hence calculate the heat of reaction for NaOH + HCl → NaCl + H2O

50. How many moles of Hydrochloric acid were reacted?: 50cm 3 of a 1M solution of HCl was reacted 1000cm 3 of solution = 1 mole in it. 50cm 3 of solution = (( 1/1000) x 50) = 0.05 moles Number of moles of HCl reacted = 0.05moles

51. Q267 (f) Calculate the number of moles of acid neutralised in this experiment. In an experiment to measure the heat of reaction for the reaction between sodium hydroxide with hydrochloric acid, a student added 50 cm3 of 1.0 M HCl solution to the same volume of 1.0 M NaOH solution in a polystyrene foam cup.Taking the total heat capacity of the reaction mixture used in this experiment as 420 J K–1, calculate the heat released in the experiment if a temperature rise of 6.7 ºC was recorded. Hence calculate the heat of reaction for NaOH + HCl → NaCl + H2O

52. Heat change= (.1kg) x (420Jkg -1 K -1 ) x 6.7 o C Heat change = 281.4J 1. What is the heat change in the reaction: Heat change = mc ∆ T Mass in kilograms Specific heat capacity Temperature rise

53. Question: What is the heat of reaction? Balanced equation: HCl + NaOH NaCl + H 2 O 1 1 0.05 moles of HCl = 281.4J of heat 1 mole of HCl = (281.4 x 20) = 5628 J of heat The heat of reaction = -5628J ( - 5.628kJ) The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

54. 2009 Q3, (b) which of the substances was identified by adding silver nitrate solution? KCl ( has chloride ions!), result = white precipitate soluble in ammonia solution (c)what as other reagent added with conc. Sulfuric acid that resulted in a brown ring forming? Iron (II)sulfate solution salt which gave positive result =KNO3 (d)Test the samples for the phosphate anion = add ammonium molbdate, add a few drops of nitric acid. A yellow precipitate indicates a positive result

55. (e) Write a balanced equation for one of the equations which gave a white ppt with the BaCl 2 solution... Must be the salts with sulfite Na 2 SO 3.7H20 or sulfate ions Na 2 SO 4.7H 2 O What was observed when student added dilute HCl? White ppt dissolves in the sulfite Na 2 SO 3.7H20, But doesn't in the sulfate Na 2 SO 4.7H 2 O (f)Suggest a way to confirm the presence of the last salt - Hydrogen carbonate salt NaHCO 3 remains. Add dilute HCl, carbon dioxide gas should be released( limewater will go cloudy), add magnesium sulfate to a fresh sample. No white precipitate should be formed

56. Bond energy

57. Breaking chemical bonds Breaking chemical bonds requires energy – is an endothermic process. Heat taken in Energy needed to overcome the bonding between the atoms Energy in chemicals Energy needed

58. Activation Energy. –Before new bonds can be formed we need to break some existing chemical bonds. activation energy (E a E act ) –This requires an energy input –known as the activation energy (E a or E act ) –Activation energy is the energy input into a reaction to allow chemical bonds to be broken and the reaction to be initiated

59. Making chemical bonds Energy will be given out in an exothermic process when bonds are formed. Heat given out Energy given out as bonds form between atoms Energy in chemicals Energy given out

60. Bond energy This is the energy needed to break 1 mole of covalent bonds or The energy released when 1 mole of covalent bonds are made

61. Changes to chemical bonds Again some existing bonds are broken (endothermic) Energy taken in as old bonds break And new bonds are formed (exothermic) Energy in chemicals reactants products Energy given out as new bonds form HH Overall exothermic – in this case

62. The formation of nitrogen (IV) oxide (formula NO 2 ) from reaction of nitrogen with oxygen in car engines has a  H value of +33.2kJ per mol of nitrogen oxide. 1.Write a word equation for the reaction. 2.Write a chemical equation for the reaction. 3.Is  H positive or negative? 4.Is the reaction exothermic or endothermic? 5.Draw an simple energy diagram for the reaction (showing bond breaking and forming.) 6.Which involves the biggest energy change: bond breaking or bond forming? Activity

63. 1.Nitrogen + oxygen  nitrogen(IV)oxide 2.N 2 + 2O 2  2NO 2. 3.  H positive (+33.2kJ/mol). 4.The reaction is endothermic. 5.Energy diagram 6.Bond breaking involves the biggest energy change. Answer

64. Energy / kJ) Progress of reaction reactants products  H= - Activation Energy and Exothermic Reactions E a = + Activation energy

65. Energy / kJ) Progress of reaction reactants products  H=+ E a = + Activation energy Activation Energy and Endothermic Reactions

66. Copy the energy diagram and use it to help you explain why garages can store petrol safely but always have notices about not smoking near the petrol pumps. Energy / kJ) Progress of reaction Petrol + oxygen Carbon dioxide + water ActivationE nergy Activity

67. Check your learning What is an exothermic reaction? What is an endothermic reaction? Define heat of reaction

68. This is an exothermic reaction Bond Forming Bond Breaking Progress of reaction Energy in chemicals O O O O H C H H H O O O O C H HHH O C O O O H H H H HH Burning Methane

69. Energy / kJ) Progress of reaction reactants products  H= - Activation Energy and Exothermic Reactions E a = + Activation energy

70. Energy / kJ) Progress of reaction reactants products  H=+ E a = + Activation energy Activation Energy and Endothermic Reactions

71. The reaction is exothermic but requires the Activation energy to be provided before the reaction can get underway. This is necessary to break some of the bonds in the oxygen or petrol before new bonds can start forming. Energy / kJ) Progress of reaction Petrol + oxygen Carbon dioxide + water ActivationE nergy Answer

72. Learning objectives Definitions: Heat of combustion, Heat of Formation, Hesses Law Doing calculations incvolving Hesses Law

73. Homework Thermochemistry Ol 2007 Q3 Revision Ch 4,5 HL 2007 Q4(f), 2004 Q4j, 2003 Q5(c) OL 2007 Q4H, Q5a,c, Q11a

74. Heat of combustion The heat change that occurs when 1 mole of substance is burned in excess oxygen Uses of heat generated from burning substances Transportation Generating electricity

76. A bomb calorimeter Use : measuring the energy contents of fuels or foods

77. Kilogram Calorific value Amount of heat generated when 1 kilogram of fuel when it is completely burned. FuelsGross calorific value/ MJ kg −1 Ethanol30 General purpose coal (5–10% water)32–42 Peat (20% water)16 Diesel fuel46 Gas oil46 Heavy fuel oil43 Kerosine47 Petrol44.8–46.9 Wood (15% water)16 Natural gas`54 Hydrogen141.9

78. Heat of formation The heat change when 1 mole of a substance is formed from its elements in their standard states C (S) + 2H 2(g) = CH 4(g) ∆H = -74.9kJmol -1 H 2(g) + S (S) + 2O 2(g) = H 2 SO 4(g) ∆H = -811kJmol -1 O 2(g) = O 2(g) ∆H = 0kJmol -1 The heat of formation of any element is zero

79. Check your learning.. Define each of the following… Heat of reaction Heat of combustion Kilogram Calorific Value Heat of formation Hesses Law

80. Hess’s Law The heat change for a reaction is the same whether it takes place in one step or in a series of steps This rule can be used for calculations using Hess’s Law ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

81. Elements in standard states ∆H r = ∑ ∆H f products - ∑ ∆H f reactants ReactantsProducts ∑ ∆H f reactants ∑ ∆H f products ∆H r

82. An example What is the heat of reaction of SO 2 + 1/2O 2 SO 3 ? The heat formation of SO 3 and SO 2 are -395 kJmol -1 and -297kJmol -1 respectively ∆H r = ∑ ∆H f products - ∑ ∆H f reactants ∑ ∆H f products = 1 ( -395) = - 395 kJmol -1 ∑ ∆H f reactants = 1 ( - 297 ) + 1/2 (0) = -297 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = (-395) – (-297) = - 98 kJmol -1 heat of reaction for the given equation

83. Example from notes – HL The combustion of liquid benzene is described by the following equation 2C 6 H 6 + 15O 2 12CO 2 + 6H 2 0 Given that the heats of formation of carbon dioxide gas, liquid water and liquid benzene are -394,-286 and 49kJmol -1 respectively, calculate the heat of combustion of liquid benzene ∑ ∆H f products = 12 ( -394) + 6 (-286) = -6444kJmol -1 ∑ ∆H f reactants = 2 ( 49) + 14 (0) = 98 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = -6444 – = - 6542kJmol -1 heat of reaction for the given equation ( for 2 moles of benzene) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of benzene is ( - 6542/ 2) = - 3271kJmol -1 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

84. Try now… Hl 2006 Q6c, 2008 Q6b

85. Q272. Using Hess’s Law 2006 Q6c The combustion of cyclohexane is described by the following equation C 6 H 12 + 9O 2 6CO 2 + 6H 2 0 Given that the heats of formation of cyclohexane, carbon dioxide gas, liquid water are -156, -394,-286 kJmol -1 respectively, calculate the heat of combustion of cyclohexane ∑ ∆H f products = 6( -394) + 6 (-286) = -4080kJmol -1 ∑ ∆H f reactants = 1( -156) + 9(0) = -156 kJmol -1 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = -4080 – (-156) = -3924 kJmol -1 heat of reaction for the given equation ( for 1 mole of cyclohexane) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of cyclohexane is - 3924 kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

86. Q274. Using Hess’s Law 2008 q6b I.Write a balanced equation for the combustion of ethanol C2H5OH. II.Given that the heats of formation of ethanol, carbon dioxide and water are -278, -394,-286 kJmol -1 respectively, calculate the heat of combustion of ethanol. i. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 0 ii) ∑ ∆H f products = 2( -394) + 3(-286) = -1646kJmol -1 ∑ ∆H f reactants = 1( -278) + 3(0) = -278 kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants ∆H r = -1646 – (-278) = -1368 kJmol -1 heat of reaction for the given equation ( for 1 mole of ethanol) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of ethanol is - 1368 kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

87. 2012 q4c (c) DEFINE: average mass of atom(s) of element // relative to (based on) 1 / 12 the mass of a carbon-12 atom (2 × 3)

88. 2011 q4a

89. 2011 Q5a,b

91. 2011 Q5c

92. Q271. Using Hess’s Law The combustion of butane is described by the following equation 2C 4 H 10 + 13O 2 8CO 2 + 10H 2 0 Given that the heats of formation of butane, carbon dioxide gas, liquid water are -125,-394 and -286kJmol -1 respectively, calculate the heat of combustion of butane ∑ ∆H f products = 8( -394) + 10 (-286) = -6012kJmol -1 ∑ ∆H f reactants = 2 ( -125) + 13 (0) = -250kJmol -1 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = - 6012 – (-250) = - 5762kJmol -1 heat of reaction for the given equation ( for 2 moles of butane) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of butane is (5762 / 2) = - 2881kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

93. Q273. Using Hess’s Law Propane may be used in gas cyclinders for cooking appliances Propane burns according to the following equatioon: C 3 H 8 + 5O 2 3CO 2 + 4H 2 0 (i) Given that the heats of formation of propane, carbon dioxide gas, liquid water are -104, -394,-286 kJmol -1 respectively, calculate the heat of combustion of propane ∑ ∆H p = 3 ( -394) + 4 (-286) = -2326 kJmol -1 ∑ ∆H r = 1( -104) + 5(0) = -104 ∆H r = ∑ ∆H p - ∑ ∆H r ∆H r = - 2326 – (-104) = - 2222kJmol -1 heat of reaction for the given equation ( for 1 mole of propane) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of propane is - 2222kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

94. If 500kJ of energy are needed to boil a kettle of water what mass of propane gas must be burned to generate this mass of heat? Give your answer to the nearest gram x moles of propane = 500kJ of heat released when combusted 1 mole of propane = 2222Kj of heat released when combusted 500 x1 = x 2222 0.225022502 = x 0.225 = number of moles of propane needed to release 500 kJ of heat when combusted Find number of moles needed Find the number of grams needed

95. 0.225 moles of propane, how many grams? (44)(0.225) = 9.901 Answer :9.901 g of propane would generate 5000Kj of heat If 500kJ of energy are needed to boil a kettle of water what mass of propane gas must be burned to generate this mass of heat? Give your answer to the nearest gram Find number of moles needed Find the number of grams needed X RMM

96. Q275. Using Hess’s Law I.Write a balanced equation for the combustion of methanol CH 3 OH. II.Given that the heats of formation of ethanol, carbon dioxide and water are -239, -394,-286 kJmol -1 respectively, calculate the heat of combustion of methanol. i. CH 3 OH + 1½O 2 1CO 2 + 2H 2 0 ii) ∑ ∆H f products = 1( -394) + 2(-286) = -966 kJmol -1 ∑ ∆H f reactants = 1( -239) + 1.5(0) = -239 kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants ∆H r = -966 – (-239) = - 727 kJmol -1 heat of reaction for the given equation ( for 1 mole of methanol) Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of methanol is - 727 kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants

97. example from notes - Using Hess’s Law The combustion of methane is described by the following balanced equation CH 4 + 2O 2 CO 2 + 2H 2 0 ∆H = -890.4kJmol-1 Given that the heats of formation of carbon dioxide gas, liquid water are -394,- 286 kJmol -1 respectively, calculate the heat of formation of methane ∆H r = -890.4kJmol -1 ∑ ∆H f products = 1( -394) + 2 (-286) = -966kJmol -1 ∑ ∆H f reactants = 1( x ) + 2(0) = x kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants -890.4kJmol -1 = -966kJmol -1 – (x kJmol -1 ) - 75.6kJmol -1 = x kJmol -1 Heat of formation of methane is - 75.6kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants Let x = heat of formation of methane

98. Q277) Using Hess’s Law The combustion of ethyne is described by the following balanced equation C 2 H 2 + 2½O 2 2CO 2 + H 2 0 ∆H = -1299kJmol-1 Given that the heats of formation of carbon dioxide gas, liquid water are -394,- 286 kJmol -1 respectively, calculate the heat of formation of ethyne ∆H r = -1299kJmol -1 ∑ ∆H f products = 2( -394) + 1(-286) = -1074kJmol -1 ∑ ∆H f reactants = 1( x ) + 2½ (0) = x kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants -1299kJmol -1 = -1074kJmol -1 – (x kJmol -1 ) - 225kJmol -1 =- x kJmol -1 Heat of formation of ethyne is 225kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants Let x = heat of formation of ethyne

99. Q278) Using Hess’s Law The combustion of propane is described by the following balanced equation C 3 H 8 + 5O 2 3CO 2 + 4H 2 0 ∆H = -2222kJmol-1 Given that the heats of formation of carbon dioxide gas, liquid water are -394,- 286 kJmol -1 respectively, calculate the heat of formation of propane ∆H r = -2222kJmol -1 ∑ ∆H f products = 3( -394) + 4(-286) = -2326kJmol -1 ∑ ∆H f reactants = 1( x ) + 5 (0) = x kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants -2222kJmol -1 = -2326kJmol -1 – (x kJmol -1 ) - 104kJmol -1 = x kJmol -1 Heat of formation of propane is - 104kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants Let x = heat of formation of propane

100. Q279) Using Hess’s Law Write a balanced equation for the combustion of ethanol i. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 0∆H = -1300kJmol-1 Given that the heats of formation of carbon dioxide gas, liquid water are -394, -286 kJmol -1 respectively, calculate the heat of formation of ethanol ∆H r = -1300kJmol -1 ∑ ∆H f products = 2( -394) + 3 (-286) = -1646kJmol -1 ∑ ∆H f reactants = 1( x ) + 3(0) = x kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants -1300kJmol -1 = -1646kJmol -1 – (x kJmol -1 ) - 346kJmol -1 = x kJmol -1 Heat of formation of ethanol is - 346kJmol -1 ∆H r = ∑ ∆H f products - ∑ ∆H f reactants Let x = heat of formation of ethanol

101. Homework Thermochemistry HL 2015 Q6c, 2009 Q6e OL 2015 q10b Revision – ch 6,7 Hl 2015 Q11a Ol 2015 Q4c, q5 Test next Monday part 1 ch 1-7, part 2 thermochemistry ch 24