This document discusses thermo chemistry and key concepts like enthalpy, calorimetry, and Hess's law. It begins by defining open, closed, and isolated systems and explains endothermic and exothermic reactions using energy profile diagrams. It then discusses enthalpy and defines standard enthalpy. Various types of enthalpies are defined like enthalpy of formation, combustion, atomization, and more. Calorimetry is explained as a method to measure heat changes in reactions using calorimeters. Hess's law is introduced as the principle that the enthalpy change of a reaction is the same whether it occurs in one step or multiple steps.

1. CHAPTER 8:CHAPTER 8:
THERMOCHEMISTRYTHERMOCHEMISTRY
8.1 System
8.2 Concept of enthalpy
8.3 Calorimetry
8.4 Hess’s Law
8.5 Born-Haber cycle
1

2. 8.1 SYSTEM8.1 SYSTEM
2

3. LEARNING OUTCOMELEARNING OUTCOME
1. Describe:
i. open system
ii. closed system
iii. isolated system.
2. Explain endothermic and exothermic reactions using the
complete energy profile diagrams.
3

4. What happens to the ice cream
and the hot coffee
after 10 minutes?
4

5. FIRST LAW OF
THERMODYNAMICS
 The law states that energy can be
tansformed (changed from one form to
another) but cannot be created or
destroyed.

6. There are generally 3 types of systems:
Open System Closed system Isolated system
An open system a
system that can exchange
mass and energy, usually
in the form of heat with
its surroundings
closed system, which
allows the exchange of
energy with its
surroundings
isolated system that
does not allow the
exchange of either mass
or energy with its
surrounding

7. TERMS DEFINITION
Heat energy transferred between two
bodies of different temperatures
System any specific part of the universe
Surroundings everything that lies outside the
system
IMPORTANT TERMS

8. ENERGY
 the ability to do work
 Units of energy
• SI Unit: Joule (J)
1 J = 1 kgm2
s-2
• Older unit: calorie (cal)
1 cal = 4.184 J
8

9.  A study of heat change in chemical
reactions.
 Two types of chemical reactions:-
 Exothermic
 Endothermic
9

10. • ΔH is (-ve)
• Enthalpy of products < Enthalpy of reactants
• Heat is released from the system to the surroundings.
• E.g: combustion, neutralization etc.
2H2(g) + O2(g) 2H2O(l) + heat
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
10

11. Consider the following reaction:-
A (g) + B (g) → C (g) ΔH = −ve
(reactants) (product)
Energy profile diagram for exothermic reaction
A(g) + B(g)
C(g)
enthalpy
Reaction
pathway
ΔH = -ve
Ea
11

12.  ΔH is (+ve)
 Enthalpy of products > enthalpy of reactants
 Heat is absorbed by the system from the surrounding
 E.g: ice melting
Heat + H2O(s) H2O(l)
heat + H2O(l)  H2O(g)
12

13. Energy profile of diagram endothermic reactions
Consider the following reaction:-
A (g) + B (g) → C (g) ΔH = +ve
(reactants) (product)
A(g) + B(g)
C(g)
enthalpy
Reaction pathway
ΔH = +ve
Ea
13

14. 8.2 CONCEPT OF ENTHALPY8.2 CONCEPT OF ENTHALPY
14

15. LEARNING OUTCOMELEARNING OUTCOME
1. State standard conditions of reaction and define the following terms :
i. enthalpy
ii. standard enthalpy
2. Define enthalpy of :
i. formation
ii. combustion
iii. atomisation
iv. neutralisation
v. hydration
vi. solution
3. Write thermochemical equation for each of the following enthalpies
15

16. The heat content or total energy in the system
Commonly measured through heat change.
Examples: system undergoes combustion or
ionization.
16

17.  Heat given off or absorbed during a reaction
at constant pressure
 Hproducts < Hreactants  Hproducts > Hreactants
9.1-15
17

18. Enthalpy of reaction, ∆H:
◦ The enthalpy change associated with a
chemical reaction.
( ΔHreaction = ΣΔHf product – ΣΔHf reactant )
Standard enthalpy, ∆Hº
◦ The enthalpy change for a particular reaction
that occurs at 298K and 1 atm (standard
state)
18

19. There are many kind of enthalpies such as:
TYPES OF ENTHALPIES
19

20. The heat changed when 1 mole of a compound is formed
from its elements in their most stable state.
H2 (g) + ½ O2(g) → H2O (l) ∆Hf = −286 kJ
The standard enthalpy of formation of any element in its
most stable state form is zero.
E.g. :-
∆H f(O2 (g) ) = 0
∆H f (Cl2 (g)) = 0
K(s) + ½ Br2(l)  KBr(s) ∆H = ∆Hf
20

21. The heat released when 1 mole of substance is
burned completely in excess oxygen.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = ∆Hcomb
C4H10(l) +13/2 O2(g)  4CO2(g) + 5H2O(l) ∆H = ∆Hcomb
21

22. The heat absorbed when 1 mole of gaseous
atoms is formed from its element
∆Ha is always positive because it involves only
breaking of bonds
E.g:-
Na(s) → Na(g) ∆Ha = +109 kJ
½Cl2(g) → Cl(g) ∆Ha = +123 kJ
22

23. The heat released when 1 mole of water, H2O is
formed from the neutralization of acid and base.
E.g:-
HCl(aq)+ NaOH(aq)→ NaCl(aq) +H2O(l) ΔHn = −58 kJ
23

24. The heat released when 1 mole of gaseous ions is
hydrated in water.
E.g:-
Na+
(g) → Na+
(aq) ∆Hhyd = − 406 kJ
Cl-
(g) → Cl-
(aq) ∆Hhyd = −363 kJ
24

25. • The heat changed when 1 mole of a substance is
dissolves in water.
E.g:
KCl(s) → K+
(aq) + Cl−
(aq) ∆Hsoln = +690 kJ
Enthalpy of Solution, ∆Hsoln
25

26.  Standard enthalpy of reaction
9.1-35
 The enthalpy change of a reaction carried
out at standard states (1 atm, 25O
C)
oO
26

27. •Shows the enthalpy changes.
E.g : H2O(s) → H2O(l) ΔH = +6.01 kJ
• 1 mole of H2O(l) is formed from 1 mole of H2O(s)
at 0°C, ΔH is +6.01 kJ
• However, when 1 mole of H2O(s) is formed from 1
mole of H2O(l), the magnitude of ΔH remains the
same but with the opposite sign of it.
H2O(l) → H2O(s) ΔH = – 6.01 kJ
27

28. The combustion of a sample of aluminium produces 0.250 mol
of aluminium oxide and releases 419 kJ of heat at
standard conditions.
Al(s) + 3/4O2(g) 1/2Al2O3(s)
(i) Calculate the standard enthalpy of combustion of
aluminium.
(ii) Determine the enthalpy of formation of Al2O3 and
write its thermochemical equation.
ExampleExample
28

29. (i) Al(s) + 3/4O2(g) 1/2Al2O3(s)
0.250 mol Al2O3 released 419 kJ heat
∴ 0.500 mol Al2O3 released ?
= 0.500x419
0.250
= 838 kJ
ΔHo
for combustion of Al is -838 kJmol-1
Answer
29

30. (ii) ΔHo
f Al2O3 = 2 x ΔHo
c Al
= 2 x (-838)
= -1.68 x 103
kJmol-1
Thermochemical equation:
2Al(s) + 3/2 O2(g) Al2O3(s) ΔHo
= - 1.68 x 103
kJ
30

31. 8.3 CALORIMETRY8.3 CALORIMETRY
31

32. LEARNING OUTCOMELEARNING OUTCOME
1. Define
i. heat capacity , C
ii. specific heat capacity, c
2. Calculate heat of reaction in a calorimeter for two possible
conditions.
i. Heat of reaction = heat absorb by medium
ii. Heat of reaction = heat absorb by calorimeter + heat
absorb by medium
32

33. CALORIMETRYCALORIMETRY
 A method used in the laboratory to measure the
heat change of a reaction.
 Apparatus used is known as the calorimeter
 Two types of calorimeter:
i. Simple calorimeter (constant pressure)
ii. Bomb calorimeter (constant volume)
33

34. constant–pressure
calorimeter
(simple calorimeter)
34

35. constant–volume
calorimeter
(bomb calorimeter)
35

36. ConstantConstant–P–Pressure Calorimeterressure Calorimeter
The outer Styrofoam cup
insulate the reaction mixture
from the surroundings (it is
assumed that no heat is lost to
the surroundings)
Heat release by the reaction is
absorbed by solution and the
calorimeter
36

37. Constant–Volume Calorimeter
37

38. Heat capacity, C
Is the amount of heat required to raise the
temperature of a given quantity of the substance by
one degree Celsius (J°C−1
)
Specific heat capacity, c
Is the amount of heat required to raise the
temperature of one gram of the substance by one
degree Celsius (J g−1
°C−1
).
Important Terms in Calorimeter
38

39. Heat released
by a reaction
=
Heat absorbed
by surroundings
• Surroundings may refer to the:
i. Calorimeter itself or;
ii. The medium(e.g. medium) and calorimeter
qreaction= mcΔT or qreaction= CcΔT
Basic Principle in CalorimeterBasic Principle in Calorimeter
39

40. Heat released by
reaction
= Heat absorbed by
calorimeter+water
q = heat released by reaction
mw= mass of water
Cw= specific heat capacity for water
Cc = heat capacity for calorimeter
∆T = temperature change
q = Cc∆T + mwcw∆T
40

41. In an experiment, 0.100 g of H2 and excess of O2 were
compressed into a 1.00 L bomb and placed into a
calorimeter with heat capacity of 9.08 x 104
J0
C−1.
The initial temperature of the calorimeter was 25.0000
C
and finally it increased to 25.155 0
C.
Calculate the amount of heat released in the reaction
to form H2O, expressed in kJ per mole.
Example 1
41

42. Heat released = Heat absorbed by the calorimeter
q = C∆T
= (9.08 X 104
J/o
c) X (0.1550
C)
= 14074 J
= 14.074 kJ
H2(g) + ½O2(g) → H2O(c)
mole of H2 = 0.100
2.0
= 0.05 mol
Answer
42

43. no moles of H2O = no mole of H2
0.05 mol of H2O Ξ 14.074 kJ energy
∴ 1 mol H2O released Ξ ? kJ
= 14.074
0.05
= 281.48 kJ
Heat of reaction, ∆H = - 281 kJ mol−1
43

44. Calculate the amount of heat released in a reaction in an
aluminum calorimeter with a mass of 3087.0 g and
contains 1700.00 mL of water. The initial temperature of
the calorimeter is 25.0°C and it increased to 27.8°C.
Given: -
Specific heat capacity of aluminum = 0.553Jg-1
°C-1
Specific heat capacity of water = 4.18 Jg-1
°C-1
Water density = 1.0 g mL-1
ΔT = (27.8 -25.0 )°C = 2.8°C
Example 2
44

45. q = mwcwΔT + mcccΔT
= (1700.0 g)(4.18 Jg-1
°C-1
)(2.8 °C) +
(3087.0 g)(0.553 Jg-1
°C-1
)(2.8°C)
= 2.5 x 104
J
= 25.0 kJ
Answer
45

46. The heat of neutralization for the following reaction
is -56.2 kJmol-1
.
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
100.0 ml of 1.50 M HCl is mixed with 100.0 ml of
1.50 M NaOH in a calorimeter having a heat
capacity of 15.2 Jo
C-1.
The initial temperature of
HCl and NaOH solution are 23.2o
C. Calculate the
final temperature for this reaction.
(ρsolution=1.00gml-1
; csolution = 4.18 Jg-1o
C-1
)
Exercise 1
46

47. Answer
47

48. A calorimeter contains 400 ml of water at
25.0o
C. If 600 ml of water at 60.0o
C is added
To it, determine the final temperature.
Assume that the heat absorbed by
calorimeter is negligible
Exercise 2
48

49. Answer
49

50. 8.4 HESS’S LAW8.4 HESS’S LAW
50

51. LEARNING OUTCOMELEARNING OUTCOME
1. State Hess’s Law
2. Apply Hess’s Law to calculate enthalpy
changes using the algebraic method and
energy cycle method
51

52. Hess’s Law states that when reactants are converted to
products, the change in enthalpy is the same whether the
reaction takes place in one step or in the series of steps.
The enthalpy change depends only on the initial and final
states of the reactants and products but is independent of
the path taken.
A B
C
∆Ho
1
∆H0
3
∆Ho
2
∆Ho
1 = ∆Ho
2 + ∆Ho
3
HESS’S LAWHESS’S LAW
52

53. 53

54. QUESTION :
Given the following enthalpies of reaction,
C(s) + O2(g) CO2(g) ∆H = - 393 k
H2(g) + 1/2O2 H2O(g) ∆H= - 286
kJ
C2H6(g) + 7/2O2(g) 2CO2(g)+ 3H2O(g) ∆H = -1560k
Calculate the enthalpy formation for , C2H6(g)
Algebraic MethodAlgebraic Method
54

55. C(s) + O2(g) CO2(g) ∆H = - 393 kJ
H2(g) + 1/2O2 (g) H2O(g) ∆H= - 286 kJ
C2H6(g) + 7/2O2(g) 2CO2(g)+ 3H2O(g) ∆H = -1560kJ
Step 1
☻ List all the thermochemical equations involved
Algebraic MethodAlgebraic Method
55

56. )(62
?
)(2
3
)(
2
g
HC
f
H
g
H
s
C  →
=∆
+
Step 2
☻ Write the enthalpy of formation reaction for C2H6
56

57. -84kJ/mol
3
H
2
H
1
H
f
H
84kJ-
)(62
?
)(2
3
)(
2
_______________________________________________________________
1560kJ
3
H
)(22
7
)(62)(2
3
2(g)
2CO(iii)
-286kJ3
2
H
)(2
3
)(22
3
)(2
33)(
kJ-3932
1
H
)(2
2
)(2
2
)(
22)(
=
∆+∆+∆=∆
 →
=∆
+
+=∆+→+
×=∆→+×
×=∆→+×
g
HC
f
H
g
H
s
C
g
O
g
HC
g
OHreverse
g
OH
g
O
g
Hii
g
CO
g
O
S
Ci
Step 3
☻Add the given reactions so that the result is the desired
reaction.
57

58. Given the following enthalpies of reaction,
C(s) + 2F2(g) CF4 ∆H= -680 kJ
H2(g) + F2(g) 2HF(g) ∆H= -537 kJ
C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g) ∆H= -2490 kJ
Calculate the enthalpy change for the reaction
between carbon and hydrogen to form ethane, C2H4(g)
2C(s) + 2H2(g) C2H4(g)
58

59. 2C(s) + 4F2(g) 2CF4 ∆H= 2x(-680 kJ)
H2(g) + F2(g) 2HF(g) ∆H= 2x(-537 kJ)
2CF4(g) + 4HF(g) C2H4(g) + 6F2(g) ∆H= +2490 kJ
___________________________________________
2C(s) + 2H2(g) C2H4(g)
∆ H = +2490 + 2(-680) + 2(-537)
= +56 kJ/mol
59

60. Given :
H2(g) + F2(g) 2HF(g) ΔH = -573 kJ
C(s) + 2F2(g) CF4(g) ΔH = -680 kJ
2C(s) + 2H2(g) C2H4(g) ΔH = +52.3 kJ
Using algebraic method, calculate the enthalpy change, ΔH for
the following reaction :
C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)
60

61. 61

62. 2C (s) + 3H2 (g) C2H6 (g)
2CO2 (g) + 3H2O(g)
2O2(g) 3/2 O2(g)H
H
H
H
7/2 O2(g)
O
O
O
O
f
1
2
3
= 2(-393)
k
= 3(-286)
= - (-1560)
Draw the energy cycle and apply Hess’s Law to
calculate the unknown value.
Energy Cycle MethodEnergy Cycle Method
kJ
kJ
kJ
62

63. ∆Ho
f = 2(∆Ho
1) + 3(∆Ho
2) +
∆Ho
3
= -786-858+1560
= -84 kJ mol-1
63

64. Given:
S(s) + O2(g) SO2(g) ΔH = −287 kJ
SO2(g) + ½O2(g) SO3(g)ΔH = −92 kJ
Calculate ΔHf for the following reaction
using energy cycle method:
S(s) + 3/2O2(g) → SO3(g)
EXERCISE
64

65. 65

66. 8.58.5 Born-Haber CycleBorn-Haber Cycle
66

67. LEARNING OUTCOMELEARNING OUTCOME
1. Define
i. lattice energy
ii. electron affinity
1. Explain the following effects on the magnitude of lattice
energy
i. ionic charge
ii. ionic radii
3. Explain the dissolution process of ionic solids
4. Construct Born-Haber cycle for simple ionic solids using
energy cycle diagram and energy level diagram
5. Calculate enthalpy changes using Born-Haber cycle
67

68.  Lattice energy formation is the energy
released when one mole of a solid (ionic
compound) is formed from its gaseous ions
Na+
(g) + Cl-
(g) → NaCl(s) ∆Hlattice = -771 kJ
(lattice formation)
Lattice Energy, ∆Hlattice
68

69.  Lattice energy dissociation is the energy
required to completely separate one mole of a
solid (ionic compound) into its gaseous ions
NaCl(s) → Na+
(g) + Cl-
(g) ∆Hlattice = +771 kJ
(lattice dissociation)
Lattice Energy, ∆Hlattice
69

70. The magnitude of lattice energy increases as:-
۩ the ionic charges increase
☞ ions attract each other more strongly
۩ the ionic radii decrease
☞ they get closer together
70
Effect on the magnitude of
lattice energy

71. E.g.
∆H for MgO is more negative than ∆H for Na2O
because Mg2+
is smaller in size and has bigger
charge than Na+
∴ ∆Hºlattice (MgO) > ∆Hºlattice (Na2O)
71

72. Electron AffinityElectron Affinity
Definition Electron Affinity
The amount of energy change to added 1 mole of electron into 1 mole of gaseous
atoms or ions in their ground state.
These reactions usually exothermic (release energy) because when an electron is
added to a neutral atom, it will experience an attraction of nucleus and release an
amount of energy.
Atom(g) + e–
 ion–
(g) ∆E = EA1 = -ve
72

73.  However, affinity does not always release energy. In some cases, affinity requires
energy.
Example: Formation of oxide, O2-
;
EA1=-ve
O-
(g)+ e-
O2-
(g) EA2=+ve
O(g)+ e-
O-
(g)
O2-
* so the values of EA are generally negative.
* the higher (more negative) the EA, the more easily it
accepts an electron.
73

74. KEEP IN MINDKEEP IN MIND
First Ionisation Energy
- Is the minimum energy required to remove an
electron from a neutral gaseous atom in its ground
state.
X(g) X+
(g) + e-
Second Ionisation Energy
- Is the energy required to remove an electron from a
gaseous positive ion in its ground state.
X+
(g) X2+
(g) + e-
74

75.  Dissolution is the process by which a solid or liquid forms a solution in a
solvent.
 Occur when an ionic solid dissolve in water
 Water molecules are polar
 Most ionic crystals are soluble in water
 Ions in the solid crystal can be separated from each other and
converted to the gaseous ions (∆Hlattice)
 The attraction forces between gaseous ions and polar water molecules
cause the ions to be surrounded by water molecules (∆Hhyd)
∆Hsoln = ∆Hlattice dissociation+ ∆Hhyd
Dissolution Process of Ionic Solid
NaCl(s) Na+
(aq) + Cl-
(aq)
75

76. Lattice Energy
dissociation Heat of Hydration
Heat of solution
Na+
(g) and Cl-
(g)
NaCl(s) Na+
(aq) and Cl-
(aq)
76

77. Based on the data given below:
ΔHo
hydration Na+
= -390 kJ mol-1
ΔHo
hydration Cl-
= -380 kJ mol-1
ΔHo
solution NaCl = +6 kJ mol-1
(i) Construct energy cycle diagram to represent the
dissolution of NaCl
(ii) Calculate the lattice energy of NaCl
77

78. (i) Na+
(g) + Cl-
(g) NaCl(s)
Na+
(aq) + Cl-
(aq)
(ii) = ΔHo
hydrationNa+
+ ΔHo
hydration Cl-
+
(-ΔHo
solutionNaCl)
= -390 + (-380) – 6
ΔHo
lattice
ΔHo
=
-6 kJ
ΔHo
=
-380 kJ
ΔHo
=
-390 kJ
Lattice
energy
78

79. Based on the data given below:
(i) Construct an energy cycle diagram to represent the
dissolution of LiCl
(ii) Calculate the lattice energy of LiCl
79
enthalpy hydration of
Li+
-510 kJ/mol
enthalpy hydration of
Cl-
-413 kJ/mol
enthalpy of solution of
LiCl
-77kJ/mol

80. 80

81.  Energy cycle for ionic compounds
 Connects enthalpy of formation with lattice
energy
Born-Haber Cycle
81

82. 82

83. Energy Cycle
Diagram

84. Find the lattice energy from the following
data:
Enthalpy of atomisation of potassium : +90 kJ mol−1
First ionisation energy of potassium : +418kJ mol−1
Atomisation energy of chlorine : +121kJ mol−1
Electron affinity of chlorine : –364 kJ mol−1
Enthalpy of formation of potassium chloride : –436 kJ mol−1
84

85. K(s) + 1/2Cl2(g) KCl(s)
K(g) Cl(g)
K+
(g) + Cl-
(g)
ΔHf= -436 kJ
ΔHa=+ 90 kJ
IE= +418 kJ
ΔHa= +121 kJ
EA = -364 kJ
ΔHf = ΔHa(K)+ IE + ΔHa(Cl) + EA + hΔ latt
-436 = 90+418+121-364+ HΔ latt
∴ HΔ latt = -701 kJ/mol
ΔHlattice
85

86. Construct a Born-Haber cycle for the formation of magnesium
fluoride, MgF2 by using the data below :
Based on the Born-Haber cycle, determine the lattice energy of
magnesium fluoride.
Enthalpy atomisation of Mg +148 kJ/mol
Enthalpy atomisation of F +159 kJ/mol
First ionization of Mg +738 kJ/mol
Second ionization of Mg +1450 kJ/mol
Electron affinity of F -328 kJ/mol
Enthalpy formation of MgF2 -1123 kJ/mol
86

87. 87

88. Energy level diagram of Born-Energy level diagram of Born-
Haber cycleHaber cycle
In the Born-Haber cycle energy diagram,
by convention, positive values are
denoted as going upwards, negative
values as going downwards.
88

89. Example :Example :
 Given;
i. Enthalpy of formation NaCl = - 411 kJmol-1
ii. Enthalpy of sublimation of Na = + 108 kJmol-1
iii. First ionization energy of Na = + 500 kJmol-1
iv. Enthalpy of atomization of Cl = + 122 kJmol-1
v. Electron affinity of Cl = - 364 kJmol-1
vi. Lattice energy of NaCl = ?
Consider the enthalpy changes in the formation of sodium
chloride.
89

90. A Born-Haber energy cycle diagram for NaCl
Na(s) + ½ Cl2(g)
Na(g) + ½ Cl2(g)
NaCl(s)
energy
E=0
Na(g) + Cl(g)
Na+
(g) + e + Cl(g)
Na+
(g) + Cl-
(g)
∆HaNa
∆HaCl
Ionisation
Energy of Na
Electron Affinity of Cl
Lattice energy
∆Hf NaCl
-ve
+ve
90

91. Calculation:Calculation:
[ ]
( )[ ]
kJ777H
kJ364kJ122kJ500kJ108kJ411H
EAHIEHHH
HEAHIEHH
lattice
lattice
)Cl(aS
0
flattice
lattice)Cl(aS
0
f
−=∆
−++++−−=∆
+∆++∆−∆=∆
∆++∆++∆=∆
From Hess’s Law:
∆Hf NaCl = ∆HaNa + ∆HaCl +IENa +
EACl + Lattice Energy
91

92. END OF
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