1. Exothermic and endothermic reactions

2. (heat is given out or taken in)
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.
Exothermic reactions increase in temperature.
Magnesium reacting with acid
Thermit reaction

5. Exothermic Reactions Magnesium + Hydrochloric acid Heat energy given
out
Gets hot
Hydrochloric
acid

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

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

8. Energy Level Diagram for an
Exothermic Reaction
Energy / kJ)
Progress of reaction (time)
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. Exothermic Reaction - Definition
Exothermic reactions give out energy.
There is a temperature rise and H is negative.
products
Energy / kJ)
Progress of reaction
reactants
H is negative

10. Heat changes also happen when substances change state.

11. 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.

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

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

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

15. Endothermic Reactions
The cold reaction products start to gain heat from the surroundings and eventually return to room temperature.
25o C
5o C
25o C
The reactants gain energy.
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.

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

17. 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

19. The equation for the reaction is
Measuring the heat of reaction of hydrochloric acid and sodium hydroxide
The equation for the reaction is
HCl + NaOH → NaCl + H2O
A polystyrene cup
is used as it is an insulator of heat – it doesn’t let the heat escape ( has negligible heat capacity)

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

21. Results

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

23. 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 cm3 of 1.0 M
Volume of sodium hydroxide = 100 cm3 of 1.0 M
Initial temperature of the solutions = 17.5 oC
Final temperature of the solutions = 24.4 oC
Specific heat capacity of the mixture = 4080 J Kg—1oC—1

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

25. 2. How many moles of Nitric acid were reacted?:
100cm3 of a 1M solution of HNO3 was reacted
x = 0.1 moles
1000
Number of moles of HNO3 reacted = 0.1moles

26. Question: What is the heat of reaction?
Balanced equation:
HNO3 + NaOH NaNO3 + H2O
0.1 moles of HNO3 = 5630J of heat
1 mole of HNO = X10 = 56300
The heat of reaction = J (or – 56.3kJ)
The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

27. 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 oC and the final temperature of the solution was 22.4 oC.
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)

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

29. 2. How many moles of Nitric acid were reacted?:
100cm3 of a 1M solution of HNO3 was reacted
1000cm3 of solution = 1 mole in it.
100cm3 of solution = x moles
(1) = x 10
Number of moles of HNO3 reacted = 0.1moles

30. Question: What is the heat of reaction?
Balanced equation:
HNO3 + NaOH NaNO3 + H2O
0.1 moles of HNO3 = 5548J of heat
1 mole of HNO3 = 5548 x 10 = 55480kJ
The heat of reaction = kJ (or – 55.48kJ)
The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

31. Q262. Calculations for the heat of reaction
A student mixed 250 cm3 of 0.5 M HCl with an equal volume of 0.5 M NaOH in a plastic container. The original temperature of both solutions was 14.8 oC and the final temperature was 18.2 oC. Calculate the heat of reaction (neutralisation) of hydrochloric acid and sodium hydroxide.
Assume that the density of the final solution is 1 g cm—3 and its specific heat capacity is 4060 J Kg—1 K—1.

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

33. 2. How many moles of Hydrochloric acid were reacted?:
250cm3 of a 0.5M solution of HCl was reacted
1000cm3 of solution = 0.5 mole in it.
250cm3 of solution = (0.5/ 1000) x 250 = 0.125moles
Number of moles of HCl reacted = 0.125moles

34. Question: What is the heat of reaction?
Balanced equation:
HCl + NaOH NaCl + H2O
0.125 moles of HCl = 6902J of heat
1 mole of HNO = 6902 x 8 = 55216
The heat of reaction = J (or – kJ)
The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

35. 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.

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

37. 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.

38. How many moles of Hydrochloric acid were reacted?:
50cm3 of a 1M solution of HCl was reacted
1000cm3 of solution = 1 mole in it.
50cm3 of solution = x moles
(50)(1) = x(1000)
(1) x (50) = x
1000
Number of moles of HCl reacted = 0.05moles

39. 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.

40. Question: What is the heat of reaction?
Balanced equation:
HCl + NaOH NaCl + H2O
0.05 moles of HCl = kJ of heat
1 mole of HNO = /20 = J of heat
The heat of reaction = kJ
The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

41. 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

42. How many moles of Hydrochloric acid were reacted?:
50cm3 of a 1M solution of HCl was reacted
1000cm3 of solution = 1 mole in it.
50cm3 of solution = (( 1/1000) x 50) = 0.05 moles
Number of moles of HCl reacted = 0.05moles

43. 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

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

45. Question: What is the heat of reaction?
Balanced equation:
HCl + NaOH NaCl + H2O
0.05 moles of HCl = J of heat
1 mole of HCl = (281.4 x 20) = 5628 J of heat
The heat of reaction = -5628J ( kJ)
The negative sign is because the reaction is exothermic ( heat is given out – the temperature went up!)

46. Bond energy

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

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

49. Energy given out as bonds form between atoms
Making chemical bonds
Energy will be given out in an exothermic process when bonds are formed.
Heat given out
when they have bonded there is more friction within the newly bonded substance thus heat is made and given off, thus energy is released
Energy in chemicals
Energy given out
Energy given out as bonds form between atoms

50. 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

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

52. Activity
The formation of nitrogen (IV) oxide (formula NO2) from reaction of nitrogen with oxygen in car engines has a H value of +33.2kJ per mol of nitrogen oxide.
Write a word equation for the reaction.
Write a chemical equation for the reaction.
Is H positive or negative?
Is the reaction exothermic or endothermic?
Draw an simple energy diagram for the reaction (showing bond breaking and forming.)
Which involves the biggest energy change: bond breaking or bond forming?

53. Nitrogen + oxygen  nitrogen(IV)oxide N2 + 2O2  2NO2.
Answer
Nitrogen + oxygen  nitrogen(IV)oxide
N2 + 2O2  2NO2.
H positive (+33.2kJ/mol).
The reaction is endothermic.
Energy diagram
Bond breaking involves the biggest energy change.

54. Activation Energy and Exothermic Reactions
reactants
H= -
Energy / kJ)
products
Progress of reaction

55. Activation Energy and Endothermic Reactions
products
H=+
Energy / kJ)
reactants
Progress of reaction

56. Activity
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.
ActivationEnergy
Petrol + oxygen
Energy / kJ)
Carbon dioxide + water
Progress of reaction

57. Answer
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.
ActivationEnergy
Petrol + oxygen
Energy / kJ)
Carbon dioxide + water
Progress of reaction

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

59. 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

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

62. Kilogram Calorific value
Amount of heat generated when 1 kilogram of fuel when it is completely burned.
Fuels Gross calorific value/ MJ kg−1
Ethanol
General purpose coal (5–10% water) 32–42
Peat (20% water) 16
Diesel fuel 46
Gas oil
Heavy fuel oil 43
Kerosine
Petrol –46.9
Wood (15% water) 16
Natural gas `54
Hydrogen

63. Heat of formation
The heat change when 1 mole of a substance is formed from its elements in their standard states
C(S) + 2H2(g) = CH4(g) ∆H = -74.9kJmol-1
H2(g) + S(S) +2O2(g)= H2SO4(g) ∆H = -811kJmol-1
O2(g)= O2(g) ∆H = 0kJmol-1
The heat of formation of any element is zero

64. 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
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

65. An example ∑ ∆Hf products = 1 ( -395) = - 395 kJmol-1
What is the heat of reaction of SO2 + 1/2O2 SO3? The heat formation of SO3 and SO2 are -395 kJmol-1 and -297kJmol-1 respectively
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
∑ ∆Hf products = 1 ( -395) = kJmol-1
∑ ∆Hf reactants = 1 ( ) + 1/2 (0) = -297
∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = (-395) – (-297) = - 98 kJmol-1 heat of reaction for the given equation

66. Q270. Using Hess’s Law
The combustion of liquid benzene is described by the following equation
2C6H6 + 15O2 12CO2+ 6H20
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
∑ ∆Hf products = 12 ( -394) + 6 (-286) = -6444kJmol-1
∑ ∆Hf reactants = 2 ( 49) + 14 (0) = 98
∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = – = kJmol-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 ( / 2) = kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
∆Hr = ∑ ∆Hp - ∑ ∆Hr

67. Q271. Using Hess’s Law
The combustion of butane is described by the following equation
2C4H O2 8CO2+ 10H20
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
∑ ∆Hf products = 8( -394) + 10 (-286) = -6012kJmol-1
∑ ∆Hf reactants = 2 ( -125) + 13 (0) = -250kJmol-1
∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = – (-250) = kJmol-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) = kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

68. Q272. Using Hess’s Law
The combustion of cyclohexane is described by the following equation
C6H12 + 9O2 6CO2+ 6H20
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
∑ ∆Hf products = 6( -394) + 6 (-286) = -4080kJmol-1
∑ ∆Hf reactants = 1( -156) + 9(0) = -156 kJmol-1
∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = – (-156) = 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 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

69. Q273. Using Hess’s Law ∑ ∆Hp = 3 ( -394) + 4 (-286) = -2326 kJmol-1
Propane may be used in gas cyclinders for cooking appliances
Propane burns according to the following equatioon:
C3H8 + 5O2 3CO2+ 4H20
(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
∑ ∆Hp = 3 ( -394) + 4 (-286) = kJmol-1
∑ ∆Hr = 1( -104) + 5(0) = -104
∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = – (-104) = kJmol-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 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

70. 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 moles of propane = 500kJ of heat released when combusted
1 mole of propane = 2222Kj of heat released when combusted
500 x1 = x
2222
= x
0.225 = number of moles of propane needed to release 500 kJ of heat when combusted

71. 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
0.225 moles of propane , how many grams? (44)(0.225) = Answer :9.901 g of propane would generate 5000Kj of heat

72. Q274. Using Hess’s Law i. C2H5OH + 3O2 2CO2+ 3H20 ii)
Write a balanced equation for the combustion of ethanol C2H5OH.
Given that the heats of formation of ethanol, carbon dioxide and water are , -394,-286 kJmol-1 respectively, calculate the heat of combustion of ethanol.
i. C2H5OH + 3O2 2CO2+ 3H20
ii)
∑ ∆Hf products = 2( -394) + 3(-286) = -1646kJmol-1
∑ ∆Hf reactants = 1( -278) + 3(0) = -278 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
∆Hr = – (-278) = 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 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

73. Q275. Using Hess’s Law i. CH3OH + 1½O2 1CO2+ 2H20 ii)
Write a balanced equation for the combustion of methanol CH3OH.
Given that the heats of formation of ethanol, carbon dioxide and water are , -394,-286 kJmol-1 respectively, calculate the heat of combustion of methanol.
i. CH3OH + 1½O2 1CO2+ 2H20
ii)
∑ ∆Hf products = 1( -394) + 2(-286) = -966 kJmol-1
∑ ∆Hf reactants = 1( -239) + 1.5(0) = -239 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
∆Hr = – (-239) = 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 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants

74. Q276b) Using Hess’s Law ∆Hr = -890.4kJmol-1
The combustion of methane is described by the following balanced equation
CH4 + 2O2 CO2+ 2H20 ∆H = kJmol-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
∆Hr = kJmol-1
∑ ∆Hf products = 1( -394) + 2 (-286) = -966kJmol-1
∑ ∆Hf reactants = 1( x ) + 2(0) = x kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
-890.4kJmol-1 = -966kJmol-1 – (x kJmol-1 )
- 75.6kJmol-1 = x kJmol-1
Heat of formation of methane is kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of methane

75. Q277) Using Hess’s Law ∆Hr = -1299kJmol-1
The combustion of ethyne is described by the following balanced equation
C2H2 + 2½O2 2CO2+ H20 ∆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
∆Hr = -1299kJmol-1
∑ ∆Hf products = 2( -394) + 1(-286) = -1074kJmol-1
∑ ∆Hf reactants = 1( x ) + 2½ (0) = x kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
-1299kJmol-1 = kJmol-1 – (x kJmol-1 )
- 225kJmol-1 =- x kJmol-1
Heat of formation of ethyne is 225kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of ethyne

76. Q278) Using Hess’s Law ∆Hr = -2222kJmol-1
The combustion of propane is described by the following balanced equation
C3H8 + 5O2 3CO2+ 4H20 ∆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
∆Hr = -2222kJmol-1
∑ ∆Hf products = 3( -394) + 4(-286) = -2326kJmol-1
∑ ∆Hf reactants = 1( x ) + 5 (0) = x kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
-2222kJmol-1 = kJmol-1 – (x kJmol-1 )
- 104kJmol-1 = x kJmol-1
Heat of formation of propane is kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of propane

77. Q279) Using Hess’s Law ∆Hr = -1300kJmol-1
Write a balanced equation for the combustion of ethanol
i. C2H5OH + 3O2 2CO2+ 3H20 ∆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
∆Hr = -1300kJmol-1
∑ ∆Hf products = 2( -394) + 3 (-286) = -1646kJmol-1
∑ ∆Hf reactants = 1( x ) + 3(0) = x kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
-1300kJmol-1 = kJmol-1 – (x kJmol-1 )
- 346kJmol-1 = x kJmol-1
Heat of formation of ethanol is kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of ethanol