C9 THERMOCHEMISTRY

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 9
THERMOCHEMISTRY

9.1 : Concept of Enthalpy

9.2 : Calorimeter
9.3 : Hess’s Law

9.4 : Born-Haber Cycle

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CHAPTER 9
THERMOCHEMISTRY

9.1 : CONCEPT OF ENTHALPY

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9.1 : Concept of Enthalpy

Learning Outcomes

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

(a) explain endothermic and exothermic
reactions using the energy profile diagram.

(b) state standard conditions of reaction and
define the following terms
i. enthalpy
ii. standard enthalpy

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(c) define and write thermochemical equation for
each enthalpy below

i. Formation.
ii. Combustion
iii. Atomisation
iv. Neutralisation
iv. Neutralisation
vi. Solution (dissolution)

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Introduction

 Thermochemistry is 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 is 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 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
Ea = Activation energy
Reactants H = enthalpy change

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) 14

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 18

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) 20

• 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) 21

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 many kind 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

If the conditions are the standard
conditions of 1 atm and 298 K, the
enthalpy of reaction is called the
standard enthalpy of 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 26

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oHndof it[iHon2Os (l)] = -286 kJ mol-1 at standard

Thermochemical equation:

H2(g) + ½O2(g) → H2O(l) Hof = -286 kJ

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Exercise 2

Write the thermochemical equations based
on the following abbreviations:

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

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Answer

i) ΔHf° [SO2(g)] = -296.1 kJ mo-1
S(s) + O2(g) → SO2(g) ΔHf° = -296.1 kJ

ii) ΔHf° [C6H12O6(s)] = -1258 kJ mol-1

6C(s) + 6H2(g) + 3O2(g) → C6H12O2(g) ΔHf° = -1258 kJ

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

Example :

Hof O2(g)=0 Hof Cl2(g)=0

Hof Na(s)=0 Hof Br2(l)=0

• Enthalpies of formation can be used to

calculate the enthalpy of reaction, H.

H =  H o (products) -  H o (reactants)
f f

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Exercise 3

Given:
Hof [C6H12O6 (s)] = -1258 kJ mol-1
Hof [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 =∑ ΔHf°(product) - ∑ ΔHf°(reactant)
= [6 x ΔHf°CO2 + 6 x ΔHf°H2O] –
[6 x ΔHf°O2 + ΔHf°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
C(s) + O2(g) → CO2(g)
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Exercise 4

Write the thermochemical equations based
on the following abbreviations:

i)Hoc[Al(s)] = -835 kJ mol-1
ii)Hoc[C2H5OH(l)] = -1371 kJ mol-1

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Answer

i) ΔHc° [Al(s)] = -835 kJ mo-1

Al(s) + ¾O2(g) → ½Al2O3(s) ΔHc° = -835 kJ

ii) ΔHc° [C2H5OH(l)] = -1371 kJ mol-1

C2H5OH(s) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔHc° = -1371kJ

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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
½Cl2(g) → Cl(g) ΔHa = +123 kJ

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Enthalpy of Neutralization, ∆Hn

 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) ΔHn= -57.3 kJ

• 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
Cl-(g) → Cl-(aq) ΔHhyd = - 363 kJ

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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

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CHAPTER 9
THERMOCHEMISTRY

9.2 : CALORIMETRY

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LEARNING OUTCOME

(a) Define
i. heat capacity, C
ii. specific heat capacity, c

(b) Calculate the heat change in a
i. Constant-Pressure Calorimeter
(simple calorimeter)
ii. Constant-Volume Calorimeter
(Bomb 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 25oC to
60oC.
Answer:0.39 Jg-1°C-1

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