Further explanation with examples included. Good Luck

1. CHAPTER 1:
THERMOCHEMISTRY
Pre – U Chemistry
Semester 2
THERMOCHEMISTRY

2. 1.1 Introduction
Energy is one of the most fundamental parts of our universe. We use
energy to do work. Energy lights our cities. Energy powers our vehicles,
trains, planes and rockets. Energy warms our homes, cooks our food,
plays our music, gives us pictures on television. Energy powers
machinery in factories and tractors on a farm.
According to the conservation of energy law, energy can be neither
created nor destroyed; it can only be converted from one form into
another
From the angle of chemistry, when a chemical reaction occur, energyFrom the angle of chemistry, when a chemical reaction occur, energy
changes occur generally in 2 ways, where it can be explained in terms
of kinetic energy and energetic energy. In this chapter, we focus more
on the study of energy changes, in the form of heat, which take place
during a chemical reaction occur, which is well known as
thermochemistry.
In order to understand thermochemistry, we must first understand what
is the difference between system and surrounding. System is the
specific part of substances that involved in chemical and physical
change, while surrounding is defined as the rest of the universe
outside the system.

3. There are generally 3 types of systems.
Open system Closed system Isolated system
An open system can
exchange mass and
energy, usually in the
form of heat with its
surroundings
closed system, which
allows the transfer of
energy (heat) but not
mass.
isolated system, which
does not allow the
transfer of either mass or
energy.

4. We shall focus more on a closed system throughout our lesson, with the
assumption that energy lost by system in a chemical reaction is the
same with the energy gained by surrounding. In thermochemistry
energy that were gained / lost by system were measured by heat energy.
In the laboratory, heat changes in physical and chemical processes are
measured with a calorimeter, a closed container designed specifically for
this purpose. Our discussion of calorimetry, the measurement of heat
changes, will depend on an understanding of specific heat and heat
capacity,
The specific heat capacity (c) of a substance is the amount of heatThe specific heat capacity (c) of a substance is the amount of heat
required to raise the temperature of one gram of the substance by one
degree Celsius. It has the units J g-1°C-1.
The heat capacity (C) of a substance is the amount of heat required to
raise the temperature of a given quantity of the substance by one degree
Celsius. Its units are J °C-1.
Specific heat is an intensive property whereas heat capacity is an
extensive property.
The relationship between the heat capacity and specific heat capacity
of a substance is C = c x m (mass)

5. 1.2 Enthalpy and Enthalpy Change
Measurement of energy transferred during chemical reaction is made
under control conditions. However, in a closed system, we assume that
there’s no changes in the volume of a system, hence no work is done
toward the heat change occur within the system. By that, we shall
deduce the energy transferred in a system is corresponding to the heat
transfer towards the surrounding. Heat transfer in this case is described
as enthalpy, H.
In a chemical reaction, where reactants products
The difference of energy changes occur on a chemical reaction is known
as enthalpy change, ∆H, as the difference between the enthalpies of the
products and the enthalpies of the reactants
H = [ΣΣΣΣ Hproduct – ΣΣΣΣ Hreactant].
Such enthalpy is also known as enthalpy change of reaction
Since the enthalpy changes is a quantitative value use to measure the
difference by the heat given off before and after a reaction, so it may be
a positive value or negative value

6. Enthalpy – heat content of the system
Enthalpy changes ; H ~ heat changes occur
during a chemical reaction.
H = [ΣΣΣΣ Hproduct – ΣΣΣΣ Hreactant]
Unit = kJ mol-1.
Σ Hproduct > Σ Hreactant Σ Hproduct < Σ Hreactant
H = positive (+ve) H = negative (–ve)
Endothermic exothermic

7. Process Endothermic Exothermic
Definition
Process of heat
absorbed by system
Process of heat
released by system
ΔH Positive Negative
Energy
profile

8. Process Endothermic Exothermic
θ Temperature decrease Temperature increase

9. 3 STEPS ON CALCULATING
ENTHALPY CHANGE
1 q = m c θθθθ
MVmass
2 mol =
3
∆H =
1000
MV
or
M
mass
R
mol
q

10. Question 1 : Solution
Equation : Zn (s) + H2SO4 (aq) ZnSO4 (aq) + H2 (g)
Step 1 : q = m c θθθθ @ q = (25.0) (4.18) (31.5 – 27.0)
q = 470.25 J
Step 2 : determine limitant
mol of Zn = mass / mol mol of H2SO4 = MV /1000mol of Zn = mass / mol mol of H2SO4 = MV /1000
= 6.00 / 65.3 = (0.100) (25.0) / 1000
= 0.0919 mol = 0.00250 mol (lim)
Step 3 : H = q / mol @ H = 470.25 / 0.00250
H = 188 100 J / mol @ – 188 kJ / mol

11. Question 2 : Solution
Equation : Na2SO4 + Ba(NO3)2 2 NaNO3 + BaSO4
Step 1 : q = m c θθθθ
@ q = (20.0 + 30.0) (4.18) (34.0 – 30.0)
q = 836 J
Step 2 : determine limitant
mol Na2SO4 = MV /1000 mol Ba(NO3)2 = MV /1000mol Na2SO4 = MV /1000 mol Ba(NO3)2 = MV /1000
= (0.500) (20.0) / 1000 = (0.300) (30.0) / 1000
= 0.010 mol = 0.0090 mol (lim)
Step 3 : H = q / mol @ H = 836 / 0.0090
H = 92889 J / mol – 92.9 kJ / mol

12. Question 3 : Solution
Equation : 2 KI + Pb(NO3)2 2 KNO3 + PbI2
Step 1 : q = m c θθθθ
@ q = (20 + 30) (4.18) (34 – 29)
q = 1045 J
Step 2 : determine limitant
mol KI = MV /1000 mol Pb(NO ) = MV /1000mol KI = MV /1000 mol Pb(NO3)2 = MV /1000
= (0.18) (30) / 1000 = (0.15) (20) / 1000
= 0.0054 mol = 0.0030 mol
*Since 2 mol of KI ≡ 1 mol of Pb(NO3)2 ;
KI is limitant mol of reaction = 0.0027 mol
Step 3 : H = q / mol @ H = 1045 / 0.0027 mol
H = 387037 @ = – 390 kJ / mol

13. Question 4 : Solution
NaCl + AgNO3 NaNO3 + AgCl
Step 3 : From H and mol ; find q
Step 2 :
mol of AgNO3 = MV / 1000 mol of NaCl = MV / 1000
= (1.00)(10.0) / 1000 = (0.800)(15.0) / 1000
= 0.010 mol (lim) = 0.012 mol= 0.010 mol (lim) = 0.012 mol
so q = H x mol @ q = (– 63400) (0.010)
q = 634 J
Step 1 : q = m c θθθθ @ θθθθ = q / mc
θθθθ = 634 / [(4.18) (10.0 + 15.0)
θθθθ = 6.07 oC

14. 1.2.2 Standard condition for calculating enthalpy changes
The standard conditions of temperature and pressure for
thermochemical measurement are 298 K and 1 atm. Any
enthalpy changes measured under these conditions is
described as standard enthalpy of reaction, and the symbol
is written as H∅∅∅∅
In this Chapter, there are a total of 9 standard enthalpy
change of reaction that we shall learned through.change of reaction that we shall learned through.
There are 3 basic rules applied when thermochemical
equations were used for calculations.

15. The total amount of energy released or absorbed is directly
proportional to the number of moles of the reactant used. For
example, in the combustion of methane :
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) H∅ = – 890 kJ mol-1
If there’s 2 mole of methane, CH4 are combusted
2 CH4 (g) + 4 O2 (g) 2 CO2 (g) + 4 H2O (l) H =
The enthalpy change for the reverse reaction is equal in
magnitude but opposite in sign to the enthalpy change for themagnitude but opposite in sign to the enthalpy change for the
forward reaction.
Na+ (g) + Cl– (g) NaCl (s) H∅ = – 770 kJ mol-1
If the reaction is reversed
NaCl (s) Na+ (g) + Cl– (g) H∅ =
The value of H∅∅∅∅ for a reaction is the same whether it occurs in
one step or a series of steps. This shall be further discussed on the
coming sub-topic about Hess' Law

16. Enthalpy change of formation, H∅
f
Energy changes occur when 1 mol of substance is
formed from its individual elements under standard
condition.
E.g. : H2 (g) + ½ O2 (g) H2O (l)
Note the following important things
The physical states of the substance involved isThe physical states of the substance involved is
stated accordingly under standard condition
The Hf
∅ of water is not written as H2O (g) as it
is not in gas under standard condition.
2 H2 (g) + O2 (g) 2 H2O (l) is not consider as
standard as substance formed is not 1 mole.
H∅
f of pure element is = 0 kJ / mol

17. Examples
CO2 : C (s) + O2 (g) CO2 (g)
MgCO3 : Mg (s) + C (s) + 3/2 O2 (g) MgCO3 (s)
NH3 : ½ N2 (g) + 3/2 H2 (g) NH3 (g)
NaCl : Na (s) + ½ Cl2 (g) NaCl (s)
C H O : 6 C (s) + 6 H (g) + 3 O (g) C H O (s)C6H12O6 : 6 C (s) + 6 H2 (g) + 3 O2 (g) C6H12O6 (s)
SO3 : 1/8 S8 (s) + 3/2 O2 (g) SO3 (g)
CH3COOH :2C(s) + 2H2(g) + O2(g) CH3COOH (l)
Al2O3 : 2 Al (s) + 3/2 O2 (g) Al2O3 (s)

18. H∅
f and Stability of Compound
Product formed via exothermic process are more
stable than product formed via endothermic process
Example : compare H∅
f of sodium halide
NaI < NaBr < NaCl < NaF
H∅
f more exothermic
Reaction between Na and X2 more vigorous
Stability of compound formed increase
Hrxn of a reaction can be calculated using H∅
f
Hrxn = Σ H∅
f (products) – Σ H∅
f (reactants)

19. Example 5
CO (g) + ½ O2 (g) CO2 (g)
Hrxn = Σ H∅
f (products) – Σ H∅
f (reactants)
= (Hf CO2) – Hf (CO + O2)
= (– 393 kJ / mol) – ( – 110 kJ / mol + 0)
= – 283 kJ
Example 6Example 6
2 FeCl2 (s) + Cl2 2 FeCl3 (s)
Hrxn = Σ H∅
f (products) – Σ H∅
f (reactants)
= (2 x Hf FeCl3) – Hf (2 FeCl2 + Cl2)
=(2 x –405 kJ / mol) – ( 2 x –341 kJ / mol + 0)
= – 128 kJ

20. *Extra Note – Cyclohexene, C6H10 contain one carbon – carbon
double bond, C=C. When cyclohexene undergoes hydrogenation, the
enthalpy change is –120 kJ / mol.
If a benzene ring (which has 3 C=C), react with hydrogen :
supposedly the enthalpy of hydrogenation must be 3 x -120 = -360 kJsupposedly the enthalpy of hydrogenation must be 3 x -120 = -360 kJ
/ mol. However, when experiment involving hydrogenation is carried
out, the ∆H of benzene is – 208 kJ / mol, indicating that benzene
molecule does not contain three double bonds in its structure.
1 mol of benzene is 152 kJ / mol more stable than 1 mole of
cyclohexene.
The more stable the structure, the less heat given out during a
reaction. The real structure of benzene is a resonance hybrid
between the structure above

21. 1.4 Enthalpy change of combustion, H∅
c
Energy liberated occur when 1 mol of substance is
burned with excess air (oxygen) under standard
condition.
E.g. : CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)
Note a few things in the thermochemical equation
above :above :
The standard combustion of substance must be 1
mole of the reactant burned. The mole of
oxygen used must be balanced accordingly.
Oxygen is always combust in excess
For H∅
c is always exothermic. MAKE SURE
THE ‘ – ‘ MUST BE PLACED.

22. Examples
C : C (s) + O2 (g) CO2 (g) [= H∅
fof CO2]
H2 : H2 (g) + ½ O2 (g) H2O (l) [= H∅
fof H2O]
C2H5COOH : C2H5COOH (l) + 7/2 O2 (g)
3 CO2 (g) + 3 H2O (l)
C2H5OH : C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l)
Mg : Mg (s) + ½ O (g) MgO (s)Mg : Mg (s) + ½ O2 (g) MgO (s)
P : P4 (s) + 5 O2 (g) P4O10 (s)
Al : Al (s) + 3/4 O2 (g) ½ Al2O3 (s)
C6H12O6 : C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l)

23. Calorimeter ~ instrument used to measure the heat
transferred during a chemical reaction.
Simple calorimeter :

24. Advantages :
Simple to be prepared and set-up
Disadvantages :
The experimental value is always lesser than the actual
H∅
c because of the following reason
Heat is easily lost to surrounding
Combustion of the sample is incompleteCombustion of the sample is incomplete
Combustion is not done under standard condition

25. Example 7
Step 1 : q = m c θθθθ
@ q = (150) (4.18) (71.0 – 27.8)
q = 27.1 kJ
Step 2 : calculate the mol of pentane burned
mol = mass / RMM = 4.30 / 72mol = mass / RMM = 4.30 / 72
= 0.0597 mol
Step 3 : H = q / mol @
H = 27.1 kJ / 0.0597 mol
H = – 454 kJ / mol

26. Bomb Calorimeter
Bomb calorimeter consist of a
thick stainless steel pressure
vessel called “bomb”
“Bomb” is then sealed after
weighted sample is placed.
A volume of water is added
to ensure the surface is covered
Pure oxygen is pumped into the
valve until 25 atm. initial temperature
is recorded. Temperature of water is
taken from time until it reached maximum temperature. The difference of
temperature is taken as θ.
Then, benzoic acid (C6H5COOH) is used to calibrate the instrument to
determine the heat capacity of the instrument.
Heat capacity ~ heat required to raise the temperature of the whole
apparatus by 1 K
)(
)(
,
θchangeetemperatur
qchangeenthalpy
CcapacityHeat =

27. Steps of calculating H∅
c using bomb calorimeter
Calibration Sample
(using benzoic acid) (burned sample)
∆H = q
mol
mol = mass
RMM
mol = mass
RMM
C = q / θθθθ
mol = mass
RMM
q = ∆H x mol
q = C θθθθ
q = ∆H x mol

28. Example 10 :
mol = 0.625
122
= 5.12 x 10-3mol
q = -3230 x 5.12 x 10-3
ΔH = 16.2 kJ
0.0123
=-1310 kJ/ mol
mol = 0.712
q = 10.5 x 1.54
= 16.2 kJ
q = -3230 x 5.12 x 10-3
= 16.5 kJ
C =16.5 / 1.58
= 10.5 kJ / K
mol = 0.712
58
= 0.0123 mol

29. 1.5– Hess Law
~ stated that the heat absorbed or liberated during a
chemical reaction, is independent of route by which the
chemical changes occur.
Consider the following equation : A + B C + D required 2
steps A + B Z Z C + D
A + B
HM
Z
HN
C + D

30. Example : In the reaction of formation of SO3, it is a 2 steps reaction.
Step 1 : 1/8 S8 (s) + O2 (g) SO2 (g) H1 = – 297 kJ / mol
Step 2 : SO2 (g) + ½ O2 (g) SO3 (g) H2 = – 99 kJ
Overall: 1/8 S8 (s) + 3/2 O2 (g) SO3 (g) Hf
∅ =
Energy / kJ
1/8 S8 (s) + 3/2 O2 (g)
– 396 kJ / mol
1/8 S8 (s) + 3/2 O2 (g)
SO2 (g) + ½ O2 (g)
SO3 (g)

31. Using Hess’s Law, the energy required to form intermediate can
also be determined.
Example : In the reaction of processing ammonia, the equation is
N2 (g) + 3 H2 (g) 2 NH3 (g) H = – 92.2 kJ
The 2 steps involve in the process of forming ammonia
Step 1 : N2 (g) + 2 H2 (g) N2H4 (g) H1 = x kJ/mol
Step 2 : N2H4 (g) + H2 (g) 2 NH3 (g) H2 = – 187 kJ / mol
Since H required is N + 2 H N HSince Hrxn required is N2 + 2 H2 N2H4
While : N2 (g) + 3 H2 (g) 2 NH3 (g) H = – 92.2 kJ
Eq.2 is reversed 2 NH3 (g) N2H4 (g) + H2 (g) H2 = + 187 kJ
N2 (g) + 2 H2 (g) N2H4
∆Hf = + 94.8 kJ / mol

32. Energy / kJ
N2H4 (g) + H2 (g)
N2 (g) + 3 H2 (g)N2 (g) + 3 H2 (g)
2 NH3 (g)

33. Example 9 :
Find H2 (g) + O2 (g) H2O2 (l)
H2 (g) + ½ O2 (g) H2O (l) Hf
∅ = - 286 kJ/mol (1)
H2O2 (g) H2O (l) + ½ O2 (g) H = - 188 kJ/mol (2)
Reverse equation (2) => equation (3)
H2O (l) + ½ O2 (g) H2O2 (g) H = +188 kJ (3)
H2 (g) + ½ O2 (g) H2O (l) Hf
∅ = - 286 kJ/mol (1)
So, when equation (1) + (3)
H2 (g) + O2 (g) H2O2 (l) H = - 98 kJ/mol

34. Energy / kJ
H2 (s) + O2 (g)
H2O2 (l)H2O2 (l)
H2O (g) + ½ O2 (g)

35. Example 10 :
From these data,
S(rhombic) + O2(g) → SO2(g) Hrxn = - 296.06 kJ/mol
S(monoclinic) + O2(g) → SO2(g) Hrxn = - 296.36 kJ/mol
Calculate the enthalpy change for the transformation
S(rhombic) → S(monoclinic)
(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)
Since the equation required is
S(rhombic) → S(monoclinic)
Make sure S(rhombic) is at the left while S(monoclinic) is at the right
By reversing eq (2) and compare to eq (1)
S(rhombic) + O2 (g) → SO2(g) Hrxn = - 296.06 kJ/mol
SO2(g) → S(monoclinic) + O2 (g) Hrxn = + 296.36 kJ/mol
-----------------------------------------------------------------------------------
S(rhombic) → S(monoclinic) Hrxn = + 0.30 kJ / mol

36. Energy / kJ
sulphur (rhombic) + O2 (g)
sulphur (monoclinic) + O2 (g)
SO2 (g)

37. 1.5.3 Relationship between Hc
∅ and Hf
∅ using Hess Law
For example, in determining the Hf
∅ of butane, C4H10. Given the
Hc
∅ for C4H10, C and H2 are – 2 877 kJ / mol ; – 393 kJ / mol
and -296 kJ / mol respectively.
Solution : C (s) + O2 (g) CO2 (g) Hc
∅ = - 393 kJ / mol .. (1)
H2 (g) + ½ O2 (g) H2O (l) Hc
∅ = - 286 kJ / mol .. (2)
C4H10 (l) + 13/2 O2 (g) 4 CO2 (g) + 5 H2O (l)
Hc
∅ = - 2877 kJ /mol .. (3)
4 C (s) + 5 H2 (g) C4H10 (l) Hf
∅ = ? kJ / mol . (4)
Since the equation of formation require 4 C (s) and 5 H (g), soSince the equation of formation require 4 C (s) and 5 H2 (g), so
the overall equation for (1) and (2) are multiply by 4 and 5
respectively, where as in equation (3) are reversed.
4 C (s) + 4 O2 (g) 4 CO2 (g) Hc
∅ = – 1572 kJ
5 H2 (g) + 5/2 O2 (g) 5 H2O (l) Hc
∅ = – 1430 kJ
4 CO2 (g) + 5 H2O (l) C4H10 (l) + 13/2 O2 (g)
Hc
∅ = + 2877 kJ
4 C (s) + 5 H2 (g) C4H10 (l) Hf
∅ = – 125 kJ / mol

38. Energy / kJ
4 C (s) + 5 H2 (g) + 13/2 O2 (g)
C4H10 (l) + 13/2 O2 (g)
4 CO2 (g) + 5 H2 (g) + 5/2 O2 (g)
4 CO2 (g) + 5 H2O (l)

39. Example 11 :
Given Hc
∅ of C2H2 and C6H6 are – 1300 kJ / mol and – 3270
kJ/mol respectively. Find 3 C2H2 (g) C6H6 (l)
C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l) Hc
∅ = –1300 (1)
C6H6 (l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) Hc
∅ = –3270 (2)
Multiply equation (1) by 3
Reverse equation (2)Reverse equation (2)
3C2H2 (g) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l) Hc
∅ = –3900
6 CO2(g) + 3 H2O(l) C6H6 (l) + 15/2 O2 (g) Hc
∅ = +3270
3 C2H2 C6H6 (l) H∅ = – 630 kJ

40. Energy / kJ
3 C2H2 (g) + 15 / 2 O2 (g)
C6H6 (l) + 15/2 O2 (g)
3 H2O (g) + 6 CO2 (g)

41. Example 12 :
C3H6 (g) + H2 (g) C3H8 (g) H∅ = –124 kJ/mol
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Hc
∅ = –2222 kJ/mol
H2 (g) + ½ O2(g) H2O (l) Hc
∅ = – 286 kJ / mol
Find for C3H6(g) + 9/2 O2(g) 3 CO2(g) + 3 H2O(l) Hc
∅ = ?
Reverse equation (3)
H2O (l) H2 (g) + ½ O2(g) Hc
∅ = + 286 kJ / molH2O (l) H2 (g) + ½ O2(g) Hc = + 286 kJ / mol
C3H6 (g) + H2 (g) C3H8 (g) H∅ = –124 kJ/mol
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Hc
∅ = –2222 kJ/mol
C3H6(g) + 9/2 O2(g) 3 CO2(g) + 3 H2O(l) Hc
∅= –2060 kJ/mol

42. Energy / kJ
C3H6 (g) + H2 (g) + 5 O2 (g)
C3H8 (g) + 5 O2 (g)
C3H6 (g) + 9/2 O2 (g) + H2O (l)C3H6 (g) + 9/2 O2 (g) + H2O (l)
3 CO2 (g) + 4 H2O (l)

43. 1.6 Enthalpy change of Neutralisation H ∅
neut
~ amount of energy liberated when 1 mol of hydrogen ion from
acid react with 1 mol of hydroxide ion from alkali to form 1
mole of water under standard condition.
Equation : H+ (aq) + OH– (aq) H2O (l)
H∅
neut for strong acid and strong base under standard
condition is – 57.3 kJ / mol.
The value of H∅
neut will be different with weak acid / base isThe value of H∅
neut will be different with weak acid / base is
used or if the acid used is a polyproctic acid
In laboratory, H∅
neut can be determine using simple cup
calorimeter (MPM Experiment 6)
The ways of calculating H∅
neut is still the same as we
learned previously.

44. Example
Equation : HCl (aq) + NH3 (aq) NH4Cl (aq)
Step 1 : q = m c θθθθ @ q = (25.0 + 30.0) (4.18) (31.3 – 27.6)
q = 850.63 J
Step 2 : determine limitant
mol of HCl = MV / 1000 mol of NH3 = MV /1000
= (1.00)(25.0)/1000 = (0.800)(30.0) / 1000= (1.00)(25.0)/1000 = (0.800)(30.0) / 1000
= 0.025 mol = 0.024 mol (lim)
Step 3 : H = q / mol @ H = 850.63 / 0.0240
H = – 35443 J / mol @ – 35.4 kJ / mol

45. H∅
neut for weak acid or weak alkali reaction.
If H∅
neut is ≠ 57.3 kJ / mol depend on :
the example above, it can be tell that, the H∅
neut for
weak acid and strong alkali is ≠ - 57.3 kJ / mol. This is
due to, some heat is absorbed by CH3COO-H to break the
O-H to form hydrogen ion. Therefore, it is less exothermic
than the expected value.
Basicity of an acid : HCl H+ + Cl– [monoproctic acid]
H2SO4 2 H+ + SO4
2- [diproctic acid]
H3PO4 3 H+ + PO4
3- [triproctic acid]H3PO4 3 H + PO4 [triproctic acid]
For example, when NaOH (aq) react with H2SO4 (aq)
Stage 1 :H2SO4 (aq) + NaOH (aq) NaHSO4 (aq) + H2O (l)
H∅
neut = –61.95 kJ / mol
Stage 2 : NaHSO4 (aq) + NaOH (aq) Na2SO4 (aq) + H2O (l)
H∅
neut = –70.90 kJ / mol
Overall : 2 NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2 H2O (l)
H∅
neut = [-61.95 + (-70.90)]= - 132.85 kJ

46. Reaction involving HF :
HF (aq) + NaOH (aq) NaF (aq) + H2O (l)
H∅
neut= –102.4 kJ / mol
The reaction become more exothermic than expected
despite that HF is consider as a weak acid. When HF is
dissolve in water, H-F dissociate in water to form H+ and F-.
The enthalpy of hydration, H∅
hyd of the fluoride ion is very
exothermic, making the overall process to be much
exothermicexothermic
F- (g) + water F- (aq) H∅
hyd = – 63.4 kJ/mol

47. 1.6 Standard Enthalpy Change of Atomisation, H∅
atom
~ energy absorbed when 1 mole of gaseous atoms are formed
from its element under standard condition.
Eq : A (s) A (g) H∅
atom = + ve kJ/mol
Example :
Mg (s) Mg (g) ¼ P4 (s) P (g)
½ Cl2 (g) Cl (g) 1/8 S (s) S (g)
CH4 (g) C (g) + 4 H (g) PBr3 (s) P (g) + 3 Br (g)
Since the reaction required the substance involve to become
gaseous atom, so the process involved an endothermic
process.
For a solid, the H∅
atom involves 2 processes. For example, in
sodium, Na, to become a gaseous sodium, the solid metal
undergoes melting process before vapourising to gas.
Energy required to change 1 mol of solid to liquid is named as
enthalpy change of fusion, while the energy required to change 1
mol of liquid to gas to called as enthalpy change of vapourisation,
according to the following equation

48. Na (s) → Na (l) ∆Hfusion
Na (l) → Na (g) ∆Hvapourisation
Since noble gas exist naturally as monoatom gas the Enthalpy
Change of Atomisation for noble gas 0
As for the Bond enthalpy, it is the energy required to break the bond
between 2 covalently bond atoms.
For example, the bonding enthalpy of chlorine gas
Na (s) Na (g) H∅
atom
For example, the bonding enthalpy of chlorine gas
Cl – Cl (g) 2 Cl (g) H∅
BE = + 242 kJ / mol
Compare to the , H∅
atom of chlorine atom ;
½ Cl2 (g) Cl (g) H∅
atom = + 121 kJ / mol

49. 1.7 Ionisation energy, H∅
IE
~ energy absorbed when 1 mole of electron is removed from a
gaseous atom under standard condition.
Eq : A (g) A+ (g) + e- H∅
IE = + ve kJ/mol
The process is always endothermic as heat is absorbed to
free one mole of electron from an atom (to overcome the
electrostatic forces of attraction between the nucleus and
outermost electron)outermost electron)
Generally, when goes down to Group, ionisation energy
decrease, while across the Period, ionisation energy
increase. These trend shall be further discussed in Chapter 3
It is believed that, when enormous amount of energies is
supplied, electrons in an atom can be removed completely
from an atom. The energies required to consecutively
remove the electrons from an atom is called as successive
ionisation energies

50. The total ionisation energy is the sum of all the successive
ionisation of the element involve. Example
1st IE of Al : Al (g) Al+ (g) + e– H∅
IE = + 577 kJ / mol
2nd IE of Al : Al+ (g) Al2+(g) + e– H∅
IE = + 1820 kJ / mol
3rd IE of Al : Al2+ (g) Al3+(g) + e– H∅
IE = + 2740 kJ / mol
Overall : Al (g) Al3+ (g) + 3e– HIE = + 5137 kJ
The information of the 1st until the 4th ionisation energy of
elements can be obtained through Data Booklet suppliedelements can be obtained through Data Booklet supplied
during examination

51. 1.8 Electron Affinity H∅
EA
~ energy liberated when 1 mole of electron is received from gaseous
atom under standard condition.
Eq : O (g) + e- O– (g) H∅
EA = – X kJ / mol
For 1st Electron Affinity, the process is always exothermic, since
upon receive an electron, the energy carries by the electron is
released upon combining with the gaseous atom.
The trend of of 1st electron affinity is the same as in Ionisation
energy, where 1st electron affinity decrease when going downenergy, where 1st electron affinity decrease when going down
to group, whereas the 1st electron affinity increase when going
across Period.
However, unlike 2nd ionisation energy, after an atom received an
electron an form negative charged ion, upon receiving the second
electron, a repulsion forces is felt between the anion and electron
receive, due to the mutual charge between both substance.
Hence, for second electron affinity, heat is absorbed
(endothermic) by the anion to overcome the repulsion forces
between the anion and electron.

52. When forming O2– from O– (2nd EA), electron is received by
negative ion.
repulsion forces formed between anionrepulsion forces formed between anionrepulsion forces formed between anionrepulsion forces formed between anion
and electron received. Heat is absorbedand electron received. Heat is absorbedand electron received. Heat is absorbedand electron received. Heat is absorbed
to overcome the forces of repulsion.to overcome the forces of repulsion.to overcome the forces of repulsion.to overcome the forces of repulsion.
1st EA : O (g) + e– O– (g) H∅
EA= – 142 kJ / mol
2nd EA : O– (g) + e– O2– (g) H∅
EA = + 844 kJ / mol
e-
2nd EA : O– (g) + e– O2– (g) H∅
EA = + 844 kJ / mol
Overall : O (g) + 2e– O2– (g) HEA = + 702 kJ

53. 1.9 Lattice Energy, H∅
LE
~ energy liberated when 1 mole of solid crystal lattice is
formed from oppositely charged gaseous ions under
standard condition.
Eq : M+ (g) + X– (g) MX (s) H∅
LE = –X kJ/mol
LE – always negative (exothermic) : heat is released when
ionic bond is formed.
Examples of writing thermochemical equation :Examples of writing thermochemical equation :
NaF : Na+ (g) + F- (g) NaF (s)
MgO : Mg2+ (g) + O2- (g) MgO (s)
CaCl2 : Ca2+ (g) + 2 Cl- (g) CaCl2 (s)
K2O : 2 K+ (g) + O2- (g) K2O (s)
Al2O3 : 2 Al3+ (g) + 3 O2- (g) Al2O3 (s)
AlN : Al3+ (g) + N3- (g) AlN (s)

54. Factors influencing Lattice Energy –
i) charge of ion ii) inter-ionic distance
Charge of ions (Zn+ . Zn–) Inter-ionic distance (r+ + r–)
Greater the charge ; greater
the forces of attraction ;
greater the value of Lattice
Energy (more exothermic)
Smaller the distance, greater
the attraction forces
between ions, greater the
lattice energy
−+
−+
+
•
∝
rr
ZZ
energyLattice
nn

55. Compound
Total
charge
∑ Ionic
radius
Compound
Total
charge
Ionic
radius
NaF 1 0.231 NaCl 1 0.276
KBr 1 0.328 KCl 1 0.314
CaO 4 0.239 MgO 4 0.205
Al O 6 0.190 K O 2 0.273
The trend of lattice energy of these 8 compounds are
KBr < KCl < NaCl < NaF < K2O < CaO < MgO < Al2O3
Lattice energy increase
Al2O3 6 0.190 K2O 2 0.273

56. 1.10 Born Haber Cycle
Lattice energy cannot be determined experimentally. They can
only be obtained by applying Hess’s Law in an energy cycle called
Born-Haber Cycle, which is a cycle of reactions used for
calculating the lattice energies of ionic crystalline solids.
There are basically 5 types of Born Haber Cycle which is mostly
tested all times.
i) A+B- ii) A2+B2
- iii) A2
+B2- iv) A2+B2- v) A2
3+B3
2-
To build the Born Haber cycle, students must be able to writeTo build the Born Haber cycle, students must be able to write
Hf
∅∅∅∅ of the compound and H∅∅∅∅
LE.
Here, we are going to build the Born Haber cycle using the 5
examples aboveSodium chloride, NaCl
Calcium chloride, CaCl2
Potassium oxide, K2O
Magnesium oxide, MgO
Chromium (III) oxide, Cr2O3

57. ∆Hatom of Na
Na (g) + ½ Cl2 (g)
∆HIE of Na
Na+ (g) + ½ Cl2 (g) + e-
∆Hatom of Cl
Na+ (g) + Cl (g) + e-
∆HEA of Cl
Na+ (g) + Cl- (g)
∆HLEof NaCl
Na (s) + ½ Cl2 (g)
NaCl (s)
∆Hf of NaCl
∆Hatom of Na ∆HLEof NaCl
H∅
f = H∅
LE + [ H∅
atom Na + H∅
atom Cl + H∅
1st IE Na + H∅
1st EA Cl]
H∅
LE = (-411) – [(+108) + (+121) + (+494) + (-364)]
= – 770 kJ/mol

58. ∆H of Ca
Ca (g) + Cl2 (g)
∆H1st IE of Ca +
∆H2nd IE of Ca
Ca2+ (g) + Cl2 (g) + 2 e-
2 x ∆Hatom of Cl
Ca2+ (g) + 2 Cl (g) + 2 e-
2 x ∆HEA of Cl
Ca2+ (g) + 2 Cl- (g)
∆H of CaCl
Ca (s) + Cl2 (g)
CaCl2 (s)
∆Hf of CaCl2
∆Hatom of Ca ∆HLEof CaCl2
H∅
f = H∅
LE + [ H∅
atom Ca + 2 H∅
atom Cl + H∅
1st IE Ca +
H∅
2nd IE Ca + 2 H∅
1st EA Cl]
H∅
LE = (-795) – [(+132) + 2(+121) +(+590) +(1150) + 2(-364)]
= – 2181 kJ/mol

59. 2 X ∆H of K
2 K (g) + ½ O2 (g)
2 X ∆H1st IE of K
2 K+ (g) + ½ O2 (g) + 2 e-
∆Hatom of O
2 K+ (g) + O (g) + 2 e-
∆HLEof K2O
∆H1st EA +
∆H2nd EA
2 K+ (g) + O2- (g)
2 K (s) + ½ O2 (g)
K2O (s)
∆Hf of K2O
2 X ∆Hatom of K
H∅
f = H∅
LE + [2 H∅
atom K + ½ H∅
BE O + 2 H∅
1st IE K +
H∅
1st EA O + H∅
2nd EA O]
H∅
LE = (-362) – [2(+129) + ½(+498) + 2(418) + (-141)+(+844)]
= – 2408 kJ/mol

60. Mg (g) + ½ O2 (g)
∆H1st IE of Mg +
∆H2nd IE of Mg
Mg2+ (g) + ½ O2 (g) + 2 e-
∆Hatom of O
Mg2+ (g) + O (g) + 2 e-
∆HLEof MgO
∆H1st EA +
∆H2nd EA of O
Mg2+ (g) + O2- (g)
Mg (s) + ½ O2 (g)
MgO (s)
∆Hf of MgO
∆Hatom of Mg
H∅
f = H∅
LE + [ H∅
atom Mg + ½ H∅
BE O + H∅
1st IE Mg +
H∅
2nd IEMg + H∅
1st EA O + H∅
2nd EA O]
H∅
LE = (-612) – [(+146) + ½(+498)+(736) + (1450) + (-141)+(+844)]
= – 3896 kJ/mol

61. 2 Cr (g) + 3/2 O2 (g)
2 x (∆H1st IE of Cr +
∆H2nd IE of Cr +
∆H3rd IE of Cr)
2 Cr3+ (g) + 3/2 O2 (g) + 6 e-
3 x ∆Hatom of O
2 Cr3+ (g) + 3 O (g) + 6 e-
∆HLEof Cr2O3
3 x (∆H1st EA O +
∆H2nd EA of O)
2 Cr3+ (g) + 3 O2- (g)
2 Cr (s) + 3/2 O2 (g)
Cr2O3 (s)
∆Hf of Cr2O3
2 x ∆Hatom of Cr
2 Cr (g) + 3/2 O2 (g)
H∅
LE = – 16408 kJ/mol

62. 1.12 Enthalpy Change of Hydration, hhyd
In terms of Thermochemistry, the solubility of ionic compound in
water depend on 2 factors
The enthalpy change of hydration
Lattice energy of the salt involved
Standard enthalpy change of hydration, H∅
hyd is
..
...under standard condition.
~ energy liberated when one mole of gaseous ion is hydrated by
water. ...under standard condition.
Equation :
Intermolecular forces occur during hydration of ions are ion-dipole
forces, which were stronger than hydrogen bonding. Diagram
below shows the ion-dipole forces between a positively and
negatively charged ion with water respectively.
water.
Mn+ (g) + water Mn+ (aq) Hhyd = – x kJ/mol
Qn- (g) + water Qn- (aq) Hhyd = – x kJ/mol

64. Since the intermolecular forces between ion and water is
strong, the H∅
hyd is always exothermic. Similar to lattice
energy, the magnitude of H∅
hyd depends on 2 factors :
Charge of ion - Greater the charge of ion, stronger the
attraction between the water and ion, the more exothermic it
is enthalpy change of hydration of ions
Size of ion - Smaller the size of ion, stronger the attraction
between the ions and water, the more exothermic it is thebetween the ions and water, the more exothermic it is the
enthalpy change of hydration of ions

65. Qn-
δ-
Mn+
δ+
Hhyd for cation increase Hhyd for anion increase
Na+ < Mg2+ < Al3+ l– < Br– < Cl–

66. 1.12 Enthalpy change of solution, Hsoln
Energy change when 1 mole of solute is dissolved in a large
excess water to form an infinite dilute solution.
For ionic substance : MX (s) + water M+ (aq) + X- (aq)
Some covalent subs : C6H12O6 (s) + water C6H12O6 (aq)
Hsoln is determined by Hhyd and HLE
HLE : M+ (g) + X- (g) MX (s) [reverse]
H : M+ (g) + X- (g) + water M+ (aq) + X- (aq)Hhyd : M+ (g) + X- (g) + water M+ (aq) + X- (aq)
– HLE : MX (s) M+ (g) + X- (g)
MX (s) + water M+ (aq) + X- (aq)
As a conclusion, Hsoln = Hhyd + (– HLE)
If Hsoln = - ve, then the salt is soluble in water
If Hsoln = + ve, then the salt is insoluble in water

67. In the solubility of Group 2 sulphate
both lattice energy and enthalpy change of hydration are
proportional to of the ions. Hence, when going down to Group 2
sulphate, both of these energies . As the size of
metal ion ..
However, the rate decrease in lattice energy is
than the rate of decrease in H∅
hyd
This is because the size of sulphate ion is much larger than the
size of metal ions, so even though the size of cation increases,
decrease
increase
Less significant
size of metal ions, so even though the size of cation increases,
the increase of (r+ + r-) is very small. This makes the lattice energy
changes become less significant when goes down to Group 2.
While in H∅
hyd it depend on both cation and anion. Since the
H∅
hyd for anion is constant, so the H∅
hyd is mainly depend on
the size of cation. When goes down to Group 2, the metal ion size
. , making H∅
hyd become
exothermic. So, the . of the heat become more
significant thus causing the rate of H∅
hyd is greater than lattice
energy.
increase less
decrease

68. Group 2 sulphate Be SO4 Mg SO4 Ca SO4 Sr SO4 Ba SO4
∆Hsolution (kJ / mol) -95.3 -91.2 + 17.8 + 18.70 +19.4
Solubility
(g / 100mL)
41.0 36.4 0.21 0.010 0.00025
∆Hhydration
∆Hlattice energy
BeSO4 Mg SO4 CaSO4 Sr SO4 BaSO4

69. Example : Solubility of Group 2 sulphate :
Sr2+
Ba2+
SO42-Be2+
Mg2+
Ca2+