1. Higher Chemistry Increasing the Amount of Product
NEW LEARNING
Equilibrium and factors affecting equilibrium.
Hess’ Law
REVISION
Moles, mass and concentration.
Calculations from equations.
Reversible Reactions

2. Starter Questions
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3. Starter Questions
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4. Lesson 1: Reversible Reactions
Today we will learn to
Explain how reversible reactions reach equilibrium.
We will do this by
Experimenting with some reactions and drawing graphs to represent equilibria.
We will have succeeded if
We can explain what equilibrium is using some examples.

5. Reversible Reactions
Many reactions are reversible, this is where some of the products formed in the reaction turn back into the original reactants again.
Activity 3.3: Reversible Reactions:
Carry out one of the three suggested experiments in pairs and write down your observations. Be prepared to explain what happened to your classmates.
Copper sulfate
Ammonium Chloride
Cobalt Chloride

6. Equilibrium
In a reversible reaction, if the conditions are not altered a balance point will be reached. At this point the reaction is said to be in equilibrium. At equilibrium the reactants change into products at exactly the same rate at which the products change back into reactants.
At equilibrium the concentrations of the reactants and products remain constant but not necessarily equal.
products
reactants
Rate of change of reactants
Rate of change of products =

7. Equilibrium Products of an Equilibrium Reaction with Time
At the start of the reaction there will be 100% reactants. However, as the reaction proceeds the reactants will be used up and the products will be formed. Eventually the reaction will reach equilibrium.
At equilibrium, the concentration of reactants and products are constant.

8. Starter Questions

9. Starter Questions
What would be the effect on the amount of sodium benzoate produced if this was an equilibrium?

10. Lesson 2: Factors Affecting Equilibrium
Today we will learn to
Explain how concentration, then pressure and temperature can affect equilibrium.
We will do this by
Experimenting with the different species on equilibria and observing the effects.
We will have succeeded if
We can explain ho concentration affects yield in an equilibrium.

11. Equilibrium
The graph below illustrates that at equilibrium the rates of the forward and reverse reaction must be equal.

12. Position of Equilibrium
If, once an equilibrium has been established the reaction conditions are changed then the position of equilibrium will be altered.
If a new equilibrium is established where there is an increase in products we say the equilibrium has shifted to the right. If the new equilibrium results in an increased amount of reactants then we say the equilibrium has shifted to the left.
Changing reaction conditions such as concentration, temperature and pressure will cause a shift in equilibrium position.

13. Position of Equilibrium
To increase the yield of ammonia or of sulphur trioxide, we need to pick conditions which shift the equilibria to the right.

14. Position of Equilibrium

15. Position of Equilibrium

16. Definition
Choose three new words you have learnt in this topic and write dictionary definitions.

17. Starter Questions
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18. Starter Questions
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19. Starter Questions
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20. Lesson 3: Factors Affecting Equilibrium
Today we will learn to
Explain how pressure and temperature can affect equilibrium.
We will do this by
Experimenting with the NO2 equilibrium.
We will have succeeded if
We can explain how temperature and pressure affect different equilibria.

21. Effect of Pressure on Equilibrium.

22. Effect of Pressure on Equilibrium.

23. Effect of Pressure on Equilibrium.
In the gaseous state molecules have high energies and are fast moving. Pressure is the result of gas molecules bombarding the walls of the vessel in which the gas is contained. The greater the number of molecules in a given volume the greater the pressure.
The pressure on the right hand side is greater than the pressure on the left hand side because there are more molecules.
NOTE:
1. Pressure only affects the equilibrium of a system that involves gases.
2. A pressure change will alter equilibrium only if there are different numbers of moles of gases on each side.

24. Effect of Temperature on Equilibrium.

25. Effect of Temperature on Equilibrium.
Increasing the temperature favours the endothermic reaction.
Decreasing the temperature favours the exothermic reaction.

26. Effect of Pressure on Equilibrium.
A(g) + B(g) ⇌ C(g) + D(g)
Changing pressure will not affect this equilibrium since there are the same number of moles of gas on each side of the equation.
2A(g) + B(g) ⇌ C(g) + D(g)
Changing pressure will affect this equilibrium since there are different numbers of moles of gases on each side of the equation.
Increasing the pressure will favour the side with the lowest number of moles in the gas state.
Decreasing the pressure will favour the side with the highest number of moles in the gas state.

27. Factors Affecting Equilibrium

28. Factors Affecting Equilibrium
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -ve
Sulfuric acid is a very important chemical and is used in a great deal of chemical processes. It is manufactured by dissolving sulfur trioxide in water. In order to increase the yield of the sulfur trioxide in the above reaction you should use an excess of by using large quantities of air. The reaction should be carried out at a temperature and using a pressure.

29. ConsolidationTask
Complete Quick Test 8
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30. Quick Test Answers 1a. Favours reactants b. Favours products
2a. Low temperature & high pressure b. Low temperatures mean a very slow rate of reaction and high pressure are expensive due to thick pipes and compressors to produce high pressure.
3a. Go darker b. Go lighter
4a. Favours reactants b. Favours products
c. Favours products d. Favours reactants
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31. Helpful Tips
Write 5 top tips or golden rules about the topic for students taking the lesson next year.

32. Starter Task
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33. Starter Task 4a. Favours reactants b. Favours products
c. Favours products d. Favours reactants
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34. Starter Task
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35. Starter Task
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36. Starter Task
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37. Starter Task
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38. Lesson 4: Hess’ Law Today we will learn to
Calculate the enthalpy change of a reaction which it is not possible to measure.
We will do this by
Learning Hess’ law and then proving it by experiment.
We will have succeeded if
We can use Hess’ Law for different reactions..

39. Hess’ Law Hess’s Law—When you can’t measure enthalpy values
The enthalpy of formation of methane from carbon and hydrogen is impossible to measure. The reason for this is because if you reacted carbon with hydrogen you would produce not just methane but a whole range of other hydrocarbons too. However it is possible to work out the theoretical enthalpy change by experimentation using Hess’s Law. Hess’s Law states that the enthalpy change for a given reaction will be the same no matter which route is taken.
In the reaction A à B an alternative route could be to go from A to B via an intermediate compound C.
A à C à B.

40. Hess’ Law ΔH1 = ΔH2 + ΔH3 This leads to the following relationship:
This means that if we can measure the enthalpy changes for the reaction going from A àC and C à B then the enthalpy change for the unknown A àB can be calculated.

43. Confirming Hess’ Law
According to Hess's Law the overall enthalpy change involved in converting solid potassium hydroxide into potassium chloride solution will be the same no matter whether the direct or indirect route is taken.

44. Confirming Hess’ Law Route and Step Initial Temperature (oC)
Highest Temperature (oC)
Temperature Change (oC)
Enthalpy Change (kJmol-1)
Total Enthalpy Change (kJmol-1)
Direct Route (1, ΔH1)
Indirect Route (2A, ΔH2)
Indirect Route (2B, ΔH3)

45. Calculation
Route 1
Suppose 1.25 g of potassium hydroxide had been added to 25 cm3 of hydrochloric acid and the temperature
of the reaction mixture had risen by 23.5 °C.
Eh = c m ΔT
Eh = 4.18 x x 23.5
= kJ
potassium hydroxide: KOH
Mass of 1 mole = = 56 g
1.25 g releases kJ
56 g releases (2.456 x 56) / 1.25
= 110 kJ
Hence ΔH1 = kJ mol-1 (A negative sign is used because the reaction is exothermic)

46. Calculation
Route 2A
Suppose 1.18 g of potassium hydroxide had been added to 25 cm3 of water and the temperature of the reaction mixture had risen by 10 °C.
Eh = c m ΔT
Eh = 4.18 x x 10
= kJ
potassium hydroxide: KOH
Mass of 1 mole = = 56 g
1.18 g releases kJ
56 g releases (1.045 x 56) / 1.18
= 49.6 kJ
Hence ΔH2= kJ mol-1 (A negative sign is used because the reaction is exothermic)

47. Calculation
Route 2B
Suppose the temperature of the reaction mixture had risen by 5.5oC when the hydrochloric acid was added. NB The total volume is now 50cm3
Eh = c m ΔT
Eh = 4.18 x x 5.5
= kJ
From part 2A, we know that 1.18 g of KOH was initially added.
1.18 g releases kJ when it reacts with the HCl
56 g releases (1.150 x 56) / 1.18
= 54.6kJ
Hence ΔH3= kJ mol-1 (A negative sign is used because the reaction is exothermic)
Overall enthalpy change Route 2 = ΔH2 - ΔH3 = – 54.6 = -104kJmol-1

48. K U I As a result of the lesson today I: Know… Understand…
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K U I
As a result of the lesson today I:
Know…
Understand…
Can use the information in the following other situations….

49. Lesson 5: Hess’ Law Today we will learn to
Apply Hess’ law to exam style questions.
We will do this by
Working through exam worked examples then trying some ourselves.
We will have succeeded if
We can use Hess’ Law on unseen questions

50. Starter Task
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51. Starter Task
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52. Starter Task
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53. Using Hess’ Law Worked Example 1
Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon, hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1560kJmol-1 respectively.
The first thing that you must do is write a balanced equation for the equation that you are required to calculate the enthalpy change for. This is called the Target Equation.
TE: 2C(s) + 3H2(g)  C2H6(g) ∆H = ?
Next you need to construct balanced equations for each of the substances in your target equation, using the information given in the question.
(1) C(s) + O2(g)  CO2(g) ∆H = -394kJ
(2) H2(g) + ½O2(g)  H2O(l) ∆H = -286kJ
(3) C2H6(g) ½O2(g)  2CO2(g) H2O(l) ∆H = -1560kJ

54. Using Hess’ Law 2C(s) + 3H2(g)  C2H6(g) ∆H = -76kJmoll-1
Worked Example 1
Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon, hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1560kJmol-1 respectively.
These equations can now be rearranged to give the target equation. Note that whatever change you make to the equation you must also make to the enthalpy value!
2 x (1) 2C(s) + 2O2(g)  2CO2(g) ∆H = -778kJ
3 x (2) 3H2(g) ½O2(g)  3H2O(l) ∆H = -858kJ
reverse (3) 2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
Once everything that appears on both the reactant side and the product side have been cancelled you should be left with your target equation!
2 x (1) 2C(s) + 2O2(g)  2CO2(g) ∆H = -778kJ
3 x (2) 3H2(g) ½O2(g)  3H2O(l) ∆H = -858kJ
reverse (3) 2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
2C(s) + 3H2(g)  C2H6(g) ∆H = -76kJmoll-1

55. Using Hess’ Law Worked example 2
Use the enthalpies of combustion in your data booklet to find the enthalpy change for the reaction between ethyne and hydrogen to produce ethane.
TE: C2H2 + 2H2  C2H6 ∆H = ?
(1) C2H ½O2  2CO H20 ∆H = -1300kJ
(2) H ½O2  H20 ∆H = -286kJ
(3) C2H ½O  2CO H2O ∆H = -1560kJ
(1) C2H ½O2  2CO H20 ∆H = -1300kJ
2 x (2) 2H O2  2H20 ∆H = -572kJ
reverse (3)2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
C2H2 + 2H2  C2H ∆H= -312kJmoll-1

56. ConsolidationTask
Complete Quick Test 9
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57. Quick Test Answers -129 kJmol-1 2. -312 kJmol-1 3. +20 kJmol-1
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58. Explain what you have learnt today and how you have learnt it
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What? How?
Explain what you have learnt today and how you have learnt it ?

59. Starter Task
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60. Starter Task
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61. Starter Task
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62. Lesson 6: Bond Enthalpies
Today we will learn to
Use bond enthalpies to calculate reaction enthalpies.
We will do this by
Learning about energy changes when bonds are made or broken and doing some calculations.
We will have succeeded if
We can apply bond enthalpies to unknown enthalpies of reaction.

63. Bond Enthalpies
In order to break a bond between two atoms energy must be supplied, if the bond reforms then the same amount of energy is released. This energy is called the bond enthalpy.
E.G H—H  2H ΔH = +432 kJmol-1 Bond Breaking – Endothermic
2H  H—H ΔH = -432 kJmol-1 Bond Making - Exothermic
The enthalpy of a reaction can be calculated using the bond enthalpies, by adding up the total number of bonds broken and then taking away from this the number of new bonds formed.

64. Bond Enthalpies -

65. Bond Enthalpies -

66. ConsolidationTask
Complete Quick Test 10
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