1. Ch 28 - Chemical equilibrium

2. 2Mg + O2 2MgO
Magnesium + oxygen
Magnesium oxide
This reaction usually goes to completion, and can only go one direction.

3. The Haber process N2 + 3H2 2NH3
This reaction is reversible – it goes both directions!
The reactants react to form the product……
and at the same time
the product is decomposing to form the reactants.

4. Dynamic equilibrium
Dynamic – reaction is still occurring
Equilibrium –forward reaction and the backward reaction are going at the same speed.
The symbol  is used to mean dynamic equilibrium.
In a reaction at equilibrium the concentrations of each reactant and product are constant (overall they are not changing!)
like the situation where a man is walking the wrong way along a moving pavement or escalator.
Neither have stopped but the man could remain in the same place for ever!
The man stays in the same place!

5. Equilibrium constant Kc
The equilibrium constant tells us the position of the equilibrium of a reaction …
It tells us whether the reaction mixture is made up of mostly reactants or mostly products!

6. A large equilibrium constant Kc tells us that reaction lies on the right hand side – at equilibrium there is more products than reactants.
2SO2 + O SO3
Say at equilibrium this reaction has an equilibrium constant of 45.
This tells us that there is a higher concentration of products compared to reactants at equilibrium!

7. A small equilibrium constant Kc tells us that reaction lies on the left hand side – at equilibrium there is more reactants than products
N2O NO2
Say at equilibrium this reaction has an equilibrium constant of 0.05.
This tells us that there is a higher concentration of reactants compared to products at equilibrium!

8. The formula for Kc aA + bB cC + dD Kc = [C]c [D]d [A]a [B]b
In the reaction
aA + bB
cC + dD
Kc = [C]c [D]d
[A]a [B]b

9. Check your learning…
What does reversible mean in terms of a chemical reaction?
Describe what is happening in a reaction when it reaches chemical equilibrium
Why is chemical equilibrium described as dynamic?
Tell what a large value of Kc means
Tell what a small value of Kc means

10. Review of yesterday’s lesson..
What is a reversible reaction?
What does equilibrium mean in terms of a chemical reaction?
Why are reactions at equilibrium said to be dynamic?
What is Kc, and what does it tell you about a reaction?

11. Calculations involving Kc – EXAMPLE 1
In a reaction between sulfur dioxide and oxygen
2SO SO3
It is found that at equilibrium the concentrations of SO2,02 and SO3 were 0.07mol-1L, 0.035mol-1L and 0.03mol-1L. Calculate the value of the equilibrium constant Kc

13. Calculations involving Kc – EXAMPLE 2
Mix together 180g of ethanoic acid and 138g of ethanol. At equilibrium there is 46g of ethanol in the reaction. Calculate Kc for the reaction
C2H5OH
CH3COOC2H5 +
CH3COOH +
H20

14. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

15. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles 3
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

16. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles 3
Change in moles
Equilibrium amount in moles 1
Equilibrium concentration in moles/L

17. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles 3
Change in moles
Down 2
Up 2
Equilibrium amount in moles 1
Equilibrium concentration in moles/L

18. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles 3
Change in moles
Down 2
Up 2
Equilibrium amount in moles 1 2
Equilibrium concentration in moles/L

19. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles 3
Change in moles
Down 2
Up 2
Equilibrium amount in moles 1 2
Equilibrium concentration in moles/L

20. Kc = [CH3COOC2H5 ][H20]
[CH3COOH ][C2H5OH]
Kc = [2][2]
[1][1]
Kc = 4

21. Review of equilibrium.. What is a reversible reaction?
What does equilibrium mean in terms of a chemical reaction?
Why are reactions at equilibrium said to be dynamic?
What is Kc, and what does it tell you about a reaction?

22. 2007 Q10a
(i)
Kc = [NH3]2
[N2][H2]3

23. Equilibrium amount in molesH2
NH3
Starting amount
In moles 3 9
Change in moles
Down 1
Down 3
Up 2
Equilibrium amount in moles 2 6
Equilibrium concentration in moles/L

24. Kc = [NH3]2
[N2][H2]3
Kc = [2]2 [2][6]3
Kc = 1
108
Kc =

25. 2004 Q9e) Calculations involving Kc –
In an experiment 6 moles of nitrogen and 18 moles of hydrogen were mixed and allowed to come to equilibrium in a 5L sealed vessel at a certain temperature. At equilibrium 6 moles of ammonia were present. Find the value for Kc
N2 + 3H2 2NH3

26. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

27. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

28. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
UP 6
Equilibrium amount in moles
Equilibrium concentration in moles/L

29. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
DOWN 3
DOWN 9
UP 6
Equilibrium amount in moles
Equilibrium concentration in moles/L

30. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
DOWN 3
DOWN 9
UP 6
Equilibrium amount in moles
Equilibrium concentration in moles/L

31. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
DOWN 3
DOWN 9
UP 6
Equilibrium amount in moles 3 9
Equilibrium concentration in moles/L

32. Equilibrium amount in molesH2
NH3
Starting amount
In moles 6 18
Change in moles
DOWN 3
DOWN 9
UP 6
Equilibrium amount in moles 3 9
Equilibrium concentration in moles/L
3/5 = 0.6
9/5 = 1.8
6/5 = 1.2

33. Kc = [NH3]2
[N2][H2]3
Kc = [1.2]2
[0.6][1.8]3
Kc = 0.4 L2/mol2

34. Exam questions on this
2010 Q7c
B – Not last part
2008 Q7b

35. Calculations involving Kc – 2010 Q7c
In an experiment 208.5g of Phosphorous chloride were placed in a 100L flask. The reaction was allowed to reach equilibrium. The mass of Chlorine gas present was 53.25g. Calculate the equilibrium constant for the reaction at 500K.
PCl5 PCl3 +Cl2

36. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

37. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
Equilibrium amount in moles
0.75
Equilibrium concentration in moles/L

38. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
up 0.75
Equilibrium amount in moles
0.75
Equilibrium concentration in moles/L

39. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
Down 0.75
up 0.75
Equilibrium amount in moles
0.75
Equilibrium concentration in moles/L

40. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
Down 0.75
up 0.75
Equilibrium amount in moles
0.25
0.75
Equilibrium concentration in moles/L

41. Equilibrium amount in moles
PCl5
PCl3
Cl2
Starting amount
In moles 1
Change in moles
Down 0.75
up 0.75
Equilibrium amount in moles
0.25
0.75
Equilibrium concentration in moles/L
0.25/ 100=
0.0025
0.75/100=
0.0075

42. Kc = [PCl3] [Cl2]
[PCl5]
Kc = [0.0075] [0.0075]
[0.0025]
Kc =0.025 mol/L

43. 2008 Q7)
If 11 moles of hydrogen gas and 11 moles of iodine gas mixed at 764K. At equilibrium 17 moles of HI are present.
H2 + I2 2HI

44. I) Write the expression for Kc
Kc = [HI]2
[H2][I2]

45. Equilibrium amount in moles
H2 + I2 2HI H2 i2 HI
Starting amount
In moles 11
Change in moles
Down 8.5
up 17
Equilibrium amount in moles
11 – 8.5 =
2.5 17
Equilibrium concentration in moles/L

46. Kc = [HI]2
[H2][I2]
Kc = [17]2
[2.5][2.5]
Kc = 46.25

47. Finding equilibrium concentrations when Kc is given
2005 Q9c) The value of Kc is 4.0 at 373K. What mass of ethyl ethanoate would be present in the equilibrium mixture if 15g of ethanoic acid and 11.5g of ethanol were mixed and equilribrium established at this temperature?

48. Kc = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH]

49. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

50. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles
0.25
Change in moles
Down x
Up x
Equilibrium amount in moles x x
Equilibrium concentration in moles/L

51. 4 = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH] 4 = [x][x] [0.25-x ][0.25-x]
If you cannot take the square root of all sides, then you should find the quadratic equation and use
4 = [x] 2
[0.25-x ]2
2 = [x]
[0.25-x ]

52. Sub x back into your table to see equilibrium conc.
2 [0.25-x ] = [x]
0.5 – 2x = x
0.5 = 3x
X = 0.167
Sub x back into your table to see equilibrium conc.

53. Equilibrium amount in moles
CH3COOH
C2H5OH
CH3COOC2H5
H20
Starting amount
In moles
0.25
Change in moles
Down x
Up x
Equilibrium amount in moles
0.25 – =
0.083
0.167
Equilibrium concentration in moles/L

54. Mass of ethyl ethanoate at equilibrium...
0.167 moles of ethyl ethanoate present
0167 x RMM = mass present at equilibrium
0.167x 88g = g, the mass of ethyl ethanoate at equilibrium

55. Objectives Calculating the value of Kc for a reaction
Le Chatelier’s principle.
Effect (if any) on equilibrium position of concentration

56. Homework Equilibrium HL 2012 q11b
Revision – Water and sodium thiosulfate titrations
HL 2007 Q1
Not on OL syllabus – instead 2011 q6, 2010 q1

57. 2003 Q11)Ai)ii)
The value of Kc is 50 at 721K. If 2 moles of hydrogen iodide gas were introduced into a sealed vessel at this temperature calculate the amount of hydrogen iodide gas present when equilibrium is reached?
H2 + I2 2HI

58. I) Write the expression for Kc
Kc = [HI]2
[H2][I2]

59. Equilibrium amount in moles
H2 + I2 2HI H2 i2 HI
Starting amount
In moles 2
Change in moles
Equilibrium amount in moles
Equilibrium concentration in moles/L

60. Equilibrium amount in moles
H2 + I2 2HI H2 i2 HI
Starting amount
In moles 2
Change in moles
Up x
Up 0.5 x
down x
Equilibrium amount in moles
0.5x
2- x
Equilibrium concentration in moles/L

61. Now sub x back into the table
50 = [HI]2 [H2][I2]
50 = [2 - x]2
[0.5x][0.5x]
50 = [2 - x]2
[0.5x]2
= [2 - x]
[0.5x]
3.5355x = 2 – x
4.5355x = 2
X = 0.441
Now sub x back into the table

62. Equilibrium amount in moles
H2 + I2 2HI H2 i2 HI
Starting amount
In moles 2
Change in moles
Up x
Up 0.5 x
down x
Equilibrium amount in moles
.2205
1.559
Equilibrium concentration in moles/L
Answer

63. b

64. Objectives for today Le chatliers principle
Effect of concentration and temp on equilibrium constant

65. Le Chatlier’s principle
When a system at equilibrium is subjected to a stress the position of the equilibrium changes to minimise the effect of the stress.

66. A possible “stress” = Concentration changes
For the following reaction:
H2 + I HI
At equilibrium the forward and backward reaction are happening at the same rate
ITS DYNAMIC no overall change in concentrations happens, even though both reactions are still happening!!! I2 H2 HI

67. H2 + I HI H2 I2 HI
Increasing the conc. of reactant(s) will be a stress on the equilibrium
To oppose this stress and reestablish the balance...
the forward reaction will be favoured, and the concentration of HI produced increases H2 I2 HI

68. HI I2 H2
A new equilibrium balance is set up, and a higher yield of HI will now be formed.

69. At equilibrium.. I2 H2 HI
But what will happen if some HI is removed from the mixture?
A Stress HI H2 I2

70. HI HI To oppose this stress and reestablish the balance...
the forward reaction will be favoured, and the concentration of HI produced increases H2 I2 HI I2 H2
A new equilibrium balance is set up, and a higher yield of HI will now be formed.

71. Learning check
What effect would adding more HI into the reaction have?
What effect would removing the reactants have? I2 H2 HI

72. 2010 Q9b

73. Check your learning.. What does equilibrium mean?
Why are reactions at equilibrium said to be dynamic?
What does Kc tell you about a reaction?
Does temp effect the value of Kc?
Does a catalyst effect the value of Kc?
State Le Chatlier’s Principle

74. Objectives Le Chatelier’s principle.
Effect (if any) on equilibrium position of concentration, pressure, temperature and catalyst.
Industrial application of Le Chatelier’s principle in the catalytic oxidation of sulfur dioxide to sulfur trioxide and in the Haber process.
Mandatory experiment 8.1 Simple experiments to illustrate Le Chatelier’s principle

76. A possible “stress” 2 = Temperature changes
For the following reaction:
H2 + I HI
The forward reaction has of Kjmol-1 so it is exothermic. GIVES OUT HEAT
So we can think about it like this:
+ heat I2 H2 HI
+ heat

77. + heat H2 I2 HI H2 I2 HI
+ heat
Increase the temperature of the reaction = a stress

78. H2 I2 HI
THe backward reaction is favoured to oppose the stress
+ heat
+ heat H2 I2 HI
A new equilibrium is set up , with the conc. Of reactants higher than before

79. Learning check.. What would happen if the temp was decreased?
+ heat H2 I2 HI
+ heat
+ heat HI
decrease the temperature of the reaction = a stress I2 H2

80. + heat HI
The forward reaction is favoured to oppose the stress I2 H2 HI
+ heat
A new equilibrium is set up , with the conc. Of products higher than before H2 I2

81. 3H2(g) + N2 (g)  2NH3 (g) H=-92kJ/mol
Is the forward reaction exothermic or endothermic?
Will heating the mixture give an equilibrium mixture with more or less ammonia?
exothermic
less

82. Another possible stress = Pressure
This applies to gas reactions only
Here the rule depends upon the number of gas molecules on each side of the equation
The more gas molecules present the higher the pressure
2NO2(g)  N2O4 (g)

83. Changing the pressure = a stress
Get less gas molecules in forward direction
2NO2(g)  N2O4 (g)
Get more gas molecules in backward direction
Increasing the pressure will cause the reaction to favour the forward direction- give more N2O4
Decreasing pressure will cause the reaction to favour the backward direction - gives more NO2 at equilibrium..
The higher the pressure the more the reaction moves in the direction with less gas molecules.

84. 3H2(g) + N2 (g)  2NH3 (g) H=-92kJ/mol
Are there more gas molecules of reactant or product?
Will raising the pressure give an equilibrium mixture with more or less ammonia?
reactant
more

85. A possible stress = adding a catalyst
A catalyst changes the rate of a chemical reaction without getting used up itself.
If you add a catalyst to a reaction mixture at equilibrium the rate of the forward reaction and backward reaction will be increased/decreased to the same extent.
IT WILL HAVE NO EFFECT!

86. 2005

87. 2005

88. 2010 q8

89. 2010

91. Objectives Le Chatelier’s principle.
Effect (if any) on equilibrium position of concentration, pressure, temperature and catalyst.

92. Now try… HL 2008 q7 (a)(c)– part b should already be done
Hl 2007a (iii) – other parts already done

93. 2008 Q7

95. HL 2007 q10a

96. Le Chatelier’s Principle
Le Chatelier’s Principle
CoCl42-
H2O
Co(H2O)62+ +
4Cl- +
H= +
H= -
Exothermic (Heat is given out)
Endothermic ( Heat is taken in)

97. CrO42- Cr2O72- Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the system adjusts to relieve the stress
CrO42-
Cr2O72- +
H2O
2H+ +

98. Fe(CNS)2- FeCl3 Le Chateliers Principle
If a stress is applied to a system at equilibrium, the system adjusts to relieve the stress
Fe(CNS)2-
FeCl3 +
CNS- +
3Cl-

99. Le Chatliers’ Principle
Industrial applications: 1.The Haber Process
For the manufacture of ammonia in the chemical industry the following reaction is used:
N2 + 3H NH3
The reaction is reversible, so it is impossible to convert all the reactants to products!

100. The Haber Process N2 + 3H2 2NH3 1) Using High pressures
Using Le Chatliers principle we can choose the best conditions to favour the forward reaction over the reverse reaction
This will allow us to make as much ammonia as possible!
N2 + 3H NH3
1) Using High pressures
This will cause the forward reaction to be favoured as there are less gas molecules on the product side!

101. The Haber process 2) Using Low temperatures
This will favour the forward reaction (because the forward reaction is exothermic, and heat is given out to counteract the drop in temperature.)
N2 + 3H NH3

102. The Haber Process
/use this animation if you wish to show the process step by step.

103. The Haber Process
Use this animation if you wish to show the process as a whole

104. Compromising with these process
Using very high pressures can cause safety issues.
Using very low temperatures can slow down a reaction.
So we can not always use the conditions that would be best for product yield 
– there is usually a compromise involved in this processes!

105. Le Chatliers’ Principle
Industrial applications: 2.The Contact Process
For the manufacture of sulfuric acid in the chemical industry the following reaction is used:
2SO2 + 2O SO3
The reaction is reversible, so it is impossible to convert all the reactants to products!

106. The Contact Process 2SO2 + 2O2 2SO3 1) Using High pressures
Using Le Chatliers principle we can choose the best conditions to favour the forward reaction over the reverse reaction
2SO2 + 2O SO3
1) Using High pressures
This will cause the forward reaction to be favoured as there are less gas molecules on the product side!

107. The Contact process 2) Using Low temperatures
This will favour the forward reaction (because the forward reaction is exothermic, and heat is given out to counteract the drop in temperature.)
2SO2 + 2O SO3

108. Compromising with these process
Using very high pressures can cause safety issues.
Using very low temperatures can slow down a reaction.
So we can not always use the conditions that would be best for product yield 
– there is usually a compromise involved in these processes!

109. Try these now...
HL 2009 Q11a
HL 2004 Q9a,b,c,d