TOPIC 2 THERMOCHEMISTRY 20222023

CHAPTER 2
THERMOCHEMISTRY

2.1 : Concept of Enthalpy

2.2 : Calorimetry
2.3 : Hess’s Law

2.4 : Born-Haber Cycle

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CHAPTER 2
THERMOCHEMISTRY

2.1 : CONCEPT OF ENTHALPY

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2.1 : Concept of Enthalpy

Learning Outcomes

At the end of this topic, students should be able to:

a) State standard conditions of reaction.
b) Define the following terms:

i. enthalpy; and
ii. standard enthalpy.

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2.1 : Concept of Enthalpy

Learning Outcomes

c) Define each of the standard enthalpy below:
i. formation;
ii. combustion;
iii. neutralisation;
iv. atomisation;
v. hydration;
vi. solution (dissolution).

d) Write the thermochemical equation for each of the enthalpy
in (c).

e) Illustrate the energy profile diagram based on the
thermochemical equations for (c)(i), (c)(ii) and (c)(iii).

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Introduction

 Thermochemistry is the study of heat change in
chemical reactions.

 In a chemical reaction, the reactants and the
products are considered as a system.

 The system is the specific part of the universe
that is of interest in the study.

 The surroundings are everything that lies
outside the system.

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 Open system is a system that can exchange
mass and energy with its surroundings.

 Closed system is a system that allows the
exchange of energy with its surroundings.

 Isolated system is a system that does not
allow the exchange of either mass or energy
with its surroundings.

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 When there is a temperature difference between a
system and its surroundings, heat will be
transferred from a region of higher temperature to
one of lower temperature; once the temperatures
become equal, heat flow stops.

 Heat, q is defined as the energy transferred
between two bodies of different temperatures.

 Energy is the ability to do work.
 SI unit of energy is kg m2 s-2 or Joule (J)
 Non-SI unit of energy is a calorie (Cal)

1 Cal = 4.184 J

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Concept of Enthalpy

⚫ Enthalpy, H is a quantity reflecting the energy
content of a system at constant pressure.

⚫ The absolute value of the enthalpy of a system
cannot be measured.

⚫ However, the enthalpy change, ∆H of a
system can be determined.

⚫ The enthalpy change is the amount of heat
released or absorbed when a chemical reaction
occurs at constant pressure.

∆H = ∑H(products) – ∑H(reactants)

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 The energy change in a chemical reaction can
be represented using energy profile diagram.

 Two types of chemical reactions:
(a) Exothermic reaction
(b) Endothermic reaction

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Exothermic Reactions

• The reactants have higher enthalpy
compared to the products.

∆H = ∑H(products) – ∑H(reactants)

• The enthalpy change of the reaction, ΔH is
negative.

• Energy is released from the system to the
surroundings.

Example:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ∆H = –2044 kJ

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Energy Profile Diagram for Exothermic Reactions

Energy Where:

Ea = Activation energy
H = enthalpy change

Reactants Ea

H = -ve

Products

Reaction progress

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Example

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ∆H = –2044 kJ
Energy

C3H8(g) + 5O2( g) Ea

H = -2044 kJ

3CO2(g) + 4H2O(g) 13

Reaction progress

Endothermic Reactions

• The reactants have lower enthalpy compared to
the products.

∆H = ∑H(products) – ∑H(reactants)

• The enthalpy change of the reaction, ΔH is
positive.

• Energy is absorbed from the surroundings to the
system.
Example:
2C(s) + H2 (g) → C2H2 (g) ΔH = +211 kJ

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Energy Profile Diagram for Endothermic Reactions

Energy Where:
Ea = Activation energy
H = enthalpy change

Ea

Products

H = +ve

Reactants

Reaction progress

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Example

2C(s) + H2(g) → C2H2(g) ΔH = +211 kJ

Energy

Ea C2H2(g)

2C(s) + H2(g) H = +211 kJ

Reaction progress 17

Enthalpy of Reaction

• Enthalpy of reaction, ΔH:
The enthalpy change associated with a
chemical reaction.

• Standard Enthalpy, ΔHo is the enthalpy
measured at standard condition.

• Standard conditions:
Temperature, T = 25oC (298 K)
Pressure, P = 1 atm
Concentration of solution = 1.0 M

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Thermochemical Equation

⚫ A thermochemical equation is a balanced
chemical equation that shows the enthalpy
changes of a chemical reaction.

⚫ The thermochemical equation should be
included:

a) the physical state of products and reactants.
b) the enthalpy change, ΔH of a reaction.

Example: H = -890 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 19

• When a thermochemical equation is reversed,
the value of H is reversed in sign.

Example:

CH4(g) + 2O2(g) → CO2(g)+ 2H2O(l) H = -890 kJ

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) H = +890 kJ

• When both sides of the equation is multiplied
by a factor n, the value of H for the new
equation must change by the same factor n.

Example: H = -890 kJ
ΔH = -1780 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
2CH4(g) + 4O2(g) → 2CO2(g) + 4H2O(l) 20

Exercise 1

Given the thermochemical equation:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H = -890 kJ

Calculate the heat released when 10.0 g of CH4
is combusted.

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Answer

Mole of CH4 = 10.0 g
16 g/mol

= 0.625 mol

1 mol CH4 - 890 kJ
0.625 mol CH4 - 556.25 kJ

Heat released = 556.25 kJ

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Types of Enthalpies

 There are various types of enthalpies, such as :
▪ Enthalpy of formation
▪ Enthalpy of combustion
▪ Enthalpy of atomisation
▪ Enthalpy of neutralisation
▪ Enthalpy of hydration
▪ Enthalpy of solution (dissolution)

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KEEP IN MIND

When the conditions are 1.0 atm and
298 K, the enthalpy of the reaction is
referred to as the standard enthalpy of
the reaction.

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Enthalpy of Formation, ∆Hf

 The heat change when one mole of a
compound is formed from its elements in
their stable states under a stated temperature
and pressure.

Example:
Hf [H2O(g)] = -242 kJmol-1 at 100oC and 1
atm

Thermochemical equation:

H2(g) + ½O2(g) → H2O(g) ∆Hf = -242 kJ mol-125

Standard Enthalpy of Formation, ∆Hf°

 The standard enthalpy of formation, ∆Hfo
is the heat change when one mole of a
compound is formed from its elements in
their stable states at 1 atm and 298 K.

Example:

coHndfoit[iHon2Os (l)] = -286 kJ mol-1 at standard

Thermochemical equation: Hfo= -286 kJ mol-1

H2(g) + ½O2(g) → H2O(l)

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Exercise 2

Write the thermochemical equations based
on the following abbreviations:

i) Hfo [SO2(g)] = -296.1 kJ mol-1
ii) Hof [C6H12O6(s)] = -1258 kJ mol-1

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Answer

i) Δ [SO2(g)] = -296.1 kJ mol-1
S(s) + O2(g) → SO2(g) Δ = -296.1 kJ mol-1
ii) Δ [C6H12O6(s)] = -1258 kJ mol-1

6C(s) + 6H2(g) + 3O2(g) → C6H12O2(g) Δ = -1258 kJ mol-1

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 The standard enthalpy of formation of any
element in its most stable state form is ZERO.

Example :

H o O2(g)=0 H o Cl2(g)=0
f f

Hof Na(s)=0 H o Br2(l)=0
f

• Enthalpies of formation can be used to

calculate the enthalpy of reaction, H.

H =  Hof (products) -  Hfo (reactants)

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Exercise 3

Given:
Hfo[C6H12O6 (s)] = -1258 kJ mol-1
Hfo[CO2 (g)] = -392 kJ mol-1
Hof [H2O(l)] = -286 kJ mol-1

Using the data given above, calculate the
enthalpy of the following reaction at standard
condition:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

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Answer

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

ΔH =∑ Δ (product) - ∑ Δ (reactant)
= [6 x Δ CO2 + 6 x Δ H2O] –
[6 x Δ O2 + Δ C6H12O6]
= [6 x (-392 kJ) + 6 x (-286 kJ) ] –
[6 x (0) + (-1258 kJ) ]
= -2810 kJ

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Enthalpy of Combustion, ∆HC

Enthalpy of combustion is the heat released
when one mole of substance is
completely burned in excess oxygen at the
stated conditions of temperature and
pressure.
The value is always negative because
combustion is an exothermic process.

Example: ∆Hc = -393 kJ mol-1
C(s) + O2(g) → CO2(g)
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Exercise 4

Write the thermochemical equations based
on the following abbreviations:

i)H o [Al(s)] = -835 kJ mol-1
c

ii)Hco[C2H5OH(l)] = -1371 kJ mol-1

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Answer

i) Δ [Al(s)] = -835 kJ mo-1

Al(s) + ¾O2(g) → ½Al2O3(s) Δ = -835 kJ mol-1

ii) Δ [C2H5OH(l)] = -1371 kJ mol-1

C2H5OH(s) + 3O2(g) → 2CO2(g) + 3H2O(g) Δ =-1371 kJ mol-1

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Enthalpy of Atomisation, ∆Ha

 Enthalpy of atomisation is the heat
absorbed when one mole of gaseous atoms
is formed from its element at the stated
conditions.

Example:

Na(s) → Na(g) ΔHa = +109 kJ mol-1
½Cl2(g) → Cl(g) ΔHa = +123 kJ mol-1

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Enthalpy of Neutralization, ∆Hneutralization

 Enthalpy of neutralization is the heat released
when one mole of water is formed from the
neutralization of acid and base at the stated
conditions.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O(l) ΔHneutralization= -57.3 kJ mol-1

• The enthalpy of neutralization of strong acid-

strong base is almost constant  -57.3 kJ mol-1

• The enthalpy of neutralization which involves

weak acid or weak base is less than -57.3 kJ
mol-1 due to partial dissociation in water.

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Enthalpy of Hydration, ∆Hhyd

 Enthalpy of hydration is the heat released
when one mole of gaseous ions is hydrated
in water.
Example:
Na+(g) → Na+(aq) ΔHhyd = - 406 kJ mol-1
Cl-(g) → Cl-(aq) ΔHhyd = - 363 kJ mol-1

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Enthalpy of Solution, ∆Hsoln

• Enthalpy of solution is the heat change when

one mole of a substance dissolves in
solvent(water) to form an infinitely dilute
solution.

• Also known as enthalpy of dissolution

Example:

KCl(s) → K+(aq) + Cl-(aq) ΔHsoln=+690 kJ mol-1

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CHAPTER 2
THERMOCHEMISTRY

2.2 : CALORIMETRY

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LEARNING OUTCOMES

a) Define:
i. heat capacity; and
ii. specific heat capacity.

b) Calculate the heat change in a constant-
volume calorimeter (bomb calorimeter).

c) Calculate the heat change in a constant-
pressure calorimeter.

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Calorimetry

 Calorimetry is the experimental technique to
determine the heat change of a reaction.

 The apparatus used is known as a calorimeter.
 A calorimeter consist of an insulated vessel in

which a chemical reaction is allowed to take
place and the heat released or absorbed is
measured.
 Examples of calorimeter
◦ Simple calorimeter
◦ Bomb calorimeter

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 The quantity of heat, q absorbed by an object
is proportional to its temperature change, ∆T.
q α ∆T
q = C ∆T
ΔT = Tfinal – Tinitial

 Every object has its own capacity for
absorbing heat

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Heat Capacity, C

• 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.

Heat capacity, C = q
ΔT

• Unit : J°C-1

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Specific Heat Capacity, c

• 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.

Specific heat capacity, c = q
m x ΔT

Unit : Jg-1°C-1

Heat, q = mc∆T

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Exercise 1

Calculate the specific heat capacity of copper
given that 204.75 J of energy raises the
temperature of 15 g of copper from 25°C to
60°C.
Answer:0.39 Jg-1°C-1

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Answer

c = q ΔT
mx

= 204.75 J
15 g x (60°C - 25°C)

= 0.39 Jg-1°C-1

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Heat Change, q

• The equation for calculating the heat
change, q is given by :

q = mc∆T or q = C∆T

q = heat released by substance
m = mass of substance
c = specific heat capacity
C = heat capacity
∆T = temperature change

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 Heat is a form of energy which flows from a
hot body to a colder body when the two are
placed in thermal contact.

 The heat released is equal the heat absorbed.
qreleased + qabsorbed = 0
- qreleased = qabsorbed

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Exercise 2

Calculate the amount of heat needed to
increase the temperature of 250 g of water
from 20°C to 56°C.
(specific heat capacity of water = 4.18 Jg-1°C-1)

Aqns=wemrcΔT 49
= 250 g x 4.18 Jg-1°C-1 x (56-20) °C

= 37620 J
= 37.62 kJ

Simple calorimeter:
(Constant-pressure Calorimeter)

 The outer Styrofoam cup insulate
the reaction mixture from the
surroundings (it is assumed that
no heat is lost to the
surroundings)

 It is suitable for measuring
enthalpy change for reactions
that take place in aqueous
solution.

 Heat release by the reaction(qrxn)
is absorbed by solution(qs)
and the calorimeter(qcal). 50