THERMOCHEMISTRY/THERMODYNAMICS.

Thermochemistry is the study of energy

changes accompanying chemical reactions.
During chemical reactions, energy is
absorbed to break the bonds between
atoms and energy is given out when
new bonds are formed.
Thermochemistry investigates the;
position of equilibrium in the reaction,
feasibility of the reaction and,
how far the reaction can go.
TYPES OF CHEMICAL REACTIONS
Chemical reactions are grouped according
to whether heat is gained/ lost during the
reaction.

Endothermic reactions
• Are reactions which proceed with

absorption of heat from the

surrounding.
• It is denoted with a positive

enthalpy change
• Products are at a higher energy

level than the reactants

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Energy level diagram for endothermic
reactions
Potential Energy

-1
( kJmol ) H2 Products

∆ H = + ve

H1 Reactants

Reaction coordinate

Examples of endothermic reactions

include;

• Photosynthesis
• Dissolution of Ammonium nitrate in

water

Exothermic reactions
• Are reactions which proceed
with evolution of heat to the
surrounding
• Denoted with a negative
enthalpy change
• Products are at a lower energy
level than the reactants

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 Energy level diagram for an exothermic
reaction
Potential energy

(kJmol-1

H 1 reactants

∆H = negative

H2 products

Examples include Reaction coordinate

• Dissolution of Sulphuric Acid in

water
• Combustion
• Thermit reaction i.e. reduction
of Fe2O3 by aluminium
ENTHALPY (H)
Is the energy or heat content of a
substance.

Enthalpy change (∆H)

Is the quantity of heat absorbed or given
out at constant pressure.

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Enthalpy change = Enthalpy of products –
enthalpy of reactant

∆H = H2− H1
Where H2 = enthalpy of products
H1 = enthalpy of reactants

Standard enthalpy change (∆HΘ)

Is the heat change when molar quantities of
reactants in their standard states react to
give products.
NB: Standard state of the substance is the
pure substance in the specified state (solid,
liquid or gas) at 1 atmosphere and 298K.
The values of enthalpy change depend on;
✓ Temperature at which reaction is
taking place
✓ Pressure of the gaseous reactants and
products
✓ Physical state of reactants (solid, liquid
or gas)
✓ Amount of reactants and products

TYPES OF ENTHALPY CHANGES

Standard enthalpy change of formation
(∆𝐇𝐟𝛉)

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Is the enthalpy change that occurs when
one mole of a substance is formed from
its elements in their normal physical
state at standard temperature and
pressure (STP)
e.g. Al 1676 KJmol-1
Enthalpy of formation of substances are
always negative and cannot be measured
experimentally but are determined
indirectly using the Born Haber cycle.
Standard enthalpy of formation of
elements at standard conditions is
equal to zero.
Standard enthalpy of formation is related
to the stability of substances i.e.
the more negative the value of standard
enthalpy change of formation, the more
stable the compound will be,
and the positive value of ∆Hfθ indicates
that the compound is likely to be thermally
unstable or may not form at all.
Standard enthalpy change of combustion
∆𝐇𝐜𝛉)
Is the enthalpy change that occurs when
one mole of a substance is completely

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burnt in O2 under standard conditions of
temperature and pressure (STP) e.g. heat
of combustion of Carbon.

C(s) + O2 (g) ⟶ CO2 (g) ∆Hcθ = − 393KJmol−1

Experimental determination of
enthalpy of combustion of
ethanol Setup:

Thermometer
Stirrer
Calorimeter

water

Heat shield

Burner

Ethanol

Procedure:

•A known amount of cold water is

weighed into a calorimeter.

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 • The initial temperature of the cold
water is determined, T0OC
• Some ethanol is added to a burner

and the mass of the burner plus the

ethanol before burning is
determined, wog.
• The wick is lit and the flame kept

steady under water.

• The water is constantly stirred until

a reasonable rise in temperature is

noted.
• The flame is extinguished and the

highest temperature T1OC is

determined.
Results

• Mass of water = mg
• Initial temperature of water = T0 C
O

• Final temp. of water = T1 C

O

• Mass of burner ethanol before

burning = wog
• Mass of burner ethanol after
burning = w1g
• SHC of water = C Jg C
-1 -1

Treatment of results:

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 Assumption:

✓ Neglecting heat gained by calorimeter;

heat evolved by burning ethanol is
equal to heat absorbed by water.
✓ Temperature rise, T = (T1 T0) 0C
✓ Mass of ethanol burnt = (wo w1)g
✓ Heat evolved = heat absorbed by
water RMM of ethanol = 46
If (wo w1) g of ethanol evolves mc T J
46g will evolve −
Heat of combustion = − J

LAWS OF THERMOCHEMISTRY
There are 2 laws of thermochemistry both
of which are based on the principle of
conservation of energy.

1st Law:

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 States that, “the amount of heat energy
required to decompose a compound into its
elements is equal in magnitude but
opposite in sign to the amount of heat
energy given out when the compound is
formed.”

e.g. NaCl (s) (aq) Na (aq) + Cl (aq)

H=
H=

2nd Law (Hess’ law of heat summation)

States that, “enthalpy change of a
chemical reaction at constant
temperature and pressure is the same
irrespective of the number of stages
passed but will depend on the initial and
final stages of the reaction. Consider
conversion of A to B.

Applications of Hess’s law

• Used to determine heats of reaction

which cannot be determined
experimentally from given

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 thermochemical data e.g. determining
the enthalpy of formation of a
compound by using enthalpy of
combustion of elements that make up a
compound. When carrying out
calculations involving Hess‟s law;
✓ All enthalpy changes must have correct
signs.
✓ Reversing the reaction also changes
the sign of the enthalpy change
✓ Multiplying or dividing thermochemical
equation by certain number also
multiplies or divides the enthalpy
change by the same number.
✓ States of the reactants and products
must always be specified.

Examples:
1. Calculate the enthalpy change of
formation of Carbon monoxide given
that the enthalpy of combustion of
carbon is 393kJmol-1, and the
enthalpy of combustion of carbon
monoxide is 286kJmol-1

Solution:

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Calculate the enthalpy change of formation of
methane from the following thermochemical data.
Enthalpy of combustion of Carbon =
393kJmol-1
Enthalpy of combustion of hydrogen =
286kJmol-1
Enthalpy of combustion of methane =
890kJmol-1

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Activity:
3.Calculate the enthalpy of reaction;
2C(s) 2H2 (g) C2H4 (g)
Given that enthalpy of combustion of carbon =
393kJmol-1
Enthalpy of combustion of hydrogen =
286kJmol-1

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 Enthalpy of combustion of ethane = 393kJmol-
1

4.Calculate the heat of formation of

ethanol given that the heat of
combustion of liquid ethanol is
649.15kJmol-1 while the enthalpy of
formation of CO2 gas and water are
186.7kJmol-1 and 135.1kJmol-1
respectively.
Calculate the heat change of the
5.

reaction C2H2 H2O CH3CHO from

the following data.
Molar heat of combustion of C2H2 =
1260kJmol-1
Molar heat combustion of CH3CHO =
1160kJmol-1
6. Calculate the enthalpy of

combustion of Benzene given that

the enthalpy of combustion of
Carbon is 393kJmol-1, enthalpy of
formation of Benzene is +49kJmol-1,
and enthalpy of formation of
water is 285kJmol-1.

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 CALCULATION OF ENTHALPY OF REACTION
FROM;
a) Enthalpy of formation:
Enthalpy of a reaction is the
difference between the total
enthalpy of formation of products
and the total enthalpies of
formation of reactants. Consider the
reaction;

Example:
1. The chemical equation for the
combustion of methane is
CH4 (g) + 2O2 (g) CO2(g) +
2H2O(l)
Calculate the standard enthalpy
change of combustion of methane
using the values of standard enthalpy
change for methane, CO2 and water
given below.
74.4kJmol-1
kJmol-1

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 O) =
285.8 kJmol-1

NB. Enthalpy of formation of elements in their

standard states is 0, giving
Activity
1. Calculate the heat change in the
reaction;

Given the following standard heats of

formation in kJmol-1
NH 46.1kJmol-1
H 285.8kJmol-1
NO2 (g) 33.2kJmol-1

b) Bond enthalpy (bond energy)

• Bond enthalpy is the standard
enthalpy change that occurs when
one mole of a covalent bond is
formed from its constituent
gaseous atoms.
• Therefore it is referred to as the
bond energy of formation ,

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 denoted by a negative value of
enthalpy.
It can also be defined as the heat
change that occurs when one
mole of a covalent bond is broken
down into its constituent gaseous
atoms
Therefore it is referred to as the bond
dissociation energy , denoted by
positive value of enthalpy e.g.
H2 (g) 2H(g) H = + 431 kJmol-1
H=
• The standard enthalpy of a reaction
is the difference between the sum
of average standard bond
enthalpies of products and sum of
the average standard bond
enthalpies of reactants.

θ
∆H = Bonds broken – bonds formed
R

Example:
1. Calculate the enthalpy change for the
following reaction; N2(g) 3H2(g)
2NH3(g)

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 Bond Mean of
bond
enthalpy
N N 945
H H 436
N H 391

2. The equation of formation of a nitrogen

monoxide gas.

Calculate the enthalpy change of

formation of nitrogen monoxide using the
given bond energies.

Bond Mean of
bond
enthalpy
N N 946kJmol-1
H H 496 kJmol-1
N H 630 kJmol-1
ENTHALPY OF ATOMISATION AND AVERAGE
BOND ENTHALPY
a) Enthalpy of

Atomisation/sublimation/vapourisation
• Is the enthalpy change that
occurs when one mole of

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 gaseous atoms is formed from
the elements in their normal
physical states at standard
conditions.
• Enthalpy change is always
positive. E.g.
Na

C(s) C(g)
For a compound, the standard enthalpy of
atomization is the heat required to
convert one mole of a compound in its
natural state and under standard
conditions into free gaseous atoms.
For any di-atomic molecule, enthalpy of
atomization is HALF the bond energy.
Average bond enthalpy/bond energy term
✓ Is the average value of the standard
enthalpy changes required to break a
particular covalent bond in a full range
of molecules in which that bond may
be found.
✓ Calculations based on average
standard bond enthalpies give
approximate results since true
standard bond enthalpies vary from
compound to compound.

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Examples:

1. Calculate the C – C bond energy in

ethane given that
C(s) C(g) 713kJmol-1
H +
436 kJmol-1

84kJmol-1
Solution
Using the energy level diagram.

From Hess‟ law of constant heat

summation
(a) The standard enthalpy change of formation of
SiCl4 is 610kJmol-1. The standard enthalpy change
of atomization of Si and Cl are +338 and +
122kJmol-1 respectively. Use these values to
construct a Born Haber cycle for formation of SiCl4
from its elements and indicate the energy changes
involved.
(b) Calculate the average bond energy of
the Si-Cl bond.
Solution

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 a)

2. Calculate the enthalpy of formation of

chloromethane using the equation.
C (graphite) CH3Cl (g)
given the following thermochemical data;
Bond energy of C – Cl = 336kJmol-1
Dissociation energy of H H = 433kJmol-1
Dissociation energy of Cl Cl = 248kJmol-1
Bond energy of C H = 413kJmol-1
Enthalpy of vapourisation of graphite =
715kJmol-1

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BORN – HABER CYCLE

✓ Is the enthalpy cycle used to calculate

the standard enthalpy changes which
occur when an ionic compound is
formed from its elements.
✓ It is an application of Hess‟ law to ionic
compounds.
Enthalpy changes involved in Born Haber
cycle include
Standard enthalpy of atomization

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Is the enthalpy change when one mole of
gaseous atoms is formed from an element
in standard state.
Standard enthalpy of ionization ( H )
Is the enthalpy change accompanying the
removal of one mole of electrons from an
atom in gaseous phase e.g. for Sodium;
Na 498kJmol-1
• When an electron is removed from

a singly positively charged ion in the

gaseous phase, enthalpy change is
described as 2nd ionization enthalpy
• i.e. Na

electron affinity/Electron gain enthalpy

Is the standard enthalpy change
accompanying the addition of one electron
to an atom in the gas phase. i.e.
X

the 2nd electron affinity in positive

because work must be done to add the
2nd electron against repulsive forces
exerted by the first electrons.

LATTICE ENERGY

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Is the enthalpy change when one mole of
an ionic compound is formed from its
gaseous ions under standard conditions i.e.
M = negative)
OR: Is the enthalpy absorbed when one
mole of an ionic compound is decomposed
into its gaseous ions.
MX = positive)
Consider the formation of one mole of MX
solid where M is a metal and X is a gas
with a diatomic molecule.

Equation of formation of MX
M(s) MX(s)
Born Haber cycle showing energy changes
involved:

∆H = ∆H + ∆H + ∆H + EA + ∆H
𝐟 𝐬𝐮𝐛 𝐈𝐄 𝐚𝐭 𝐋
Example:

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1. (a) Draw a Born Haber for the
formation of solid Rubidium chloride from
its elements (b) Calculate the electron
affinity of chlorine atom. Use the
following data;
Lattice energy of Rubidium chloride = −665kJmol-1
Dissociation energy of chlorine gas molecule =
226kJmol-1
Heat of atomization of Rubidium metal= 84kJmol-1
Standard heat of formation of solid Rubidium
chloride is – 439kJmol-1
I.E. of Rubidium atom = 397kJmol-1

Activity
Draw a Born Haber cycle for the reaction of
Lithium(Solid) and F(g) to form LiF(s) and
indicate clearly whether the change is
exo/endothermic.

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using the data below;
Enthalpy of atomization of Fluorine
molecule = 150kJmol-1
Enthalpy of sublimation
of Lithium metal =
155kJmol-1 Ionization
energy of Lithium atom
= 518kJmol-1 Standard
enthalpy of formation of
solid LiF =?
Lattice enthalpy = 1030 kJmol-1
Electron affinity of F = 351 kJmol-1

1. Calculate the lattice energy of

CaCl2 from the following
thermochemical data; Standard
enthalpy of atomization of CaCl2 =
795kJmol-1
Standard enthalpy of atomization of
Chlorine =

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FACTORS AFFECTING LATTICE ENERGY
Ionic charge
The larger the ionic charge, the bigger the
lattice energy
because of the strong electrostatic forces
of attraction, between the oppositely
charged ions, e.g. the lattice energy of
magnesium chloride is larger than that of
Sodium Chloride because the greater
charge of Mg2+ strongly attracts the
chloride ions.
Ionic size/radius:
Lattice energy is inversely proportional to
the ionic size i.e. large cat ions have less
attraction for anions because of the
reduced effective nuclear charge. (Large
distance of approach) e.g. NaCl has a
bigger lattice energy than KCl because the
K has a bigger radius than Na .
Consider the halides of Na and their values
of lattice energy.
NaF 795kJmol 1
NaCl 761kJmol 1
NaBr
742kJmol 1 NaI
699kJmol 1

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ENTHALPY OF SOLUTION
Is the enthalpy change that occurs when
one mole of crystal ionic compound is
dissolved in a specified number of moles of
water at standard conditions.
MX(s)
The enthalpy change accompanying the
dissolution of an ionic salt in water is either
positive or negative
The dissolution of an ionic compound
involves two energy terms;
i. Separating the ionic lattice into oppositely
charged gaseous ions ie
The energy required is the lattice
dissociation and is an endothermic
process ii. Hydration/ solvation of the
oppositely charged gaseous ions and it‟s an
exothermic process.ie

M M

Definition
Enthalpy of Hydration/ Solvation is the
heat evolved when 1 mole of gaseous ions
is completely surrounded by water
molecules i.e.

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 = lattice dissociation
solvation (Hydration)
Born Haber cycle showing the
relationship between the enthalpy of
solution, lattice dissociation energy and
hydration energy of salt MX

MX(s) MX (aq)
lattice
∆

dissociation
H hydration enthalpy

M (g) X

OR
M (g) X-(g)

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 Lattice dissociation enthalpy
∆ H hydration

MX(s)

∆ H solution

M +( aq) + X –( aq

Factors affecting enthalpy of solution

• Lattice energy
• Hydration energy

When the
✓ lattice dissociation is greater than

the hydration, enthalpy of solution is

endothermic,
✓ the hydration energy is greater

than lattice dissociation enthalpy,

the enthalpy of solution is
exothermic.
For an ionic compound to dissolve in water,
the lattice dissociation energy should be
overcome by the hydration enthalpy so
that an overall negative enthalpy of
solution is obtained i.e. compounds that
dissolve exothermically are more

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soluble than those which dissolve
endothermically.

Example:
The enthalpies of hydration and lattice
energy of some salts are given below;

Salt Enthalpy of Lattice

hydration energy
(kJmol-1) (kJmol-1)
KCl − 692 + 718
LiCl − 883 + 862

a) Calculate the heats of solution of K

and Li Chloride. State whether
dissolution is endothermic or
exothermic process in each case.
b) Which of the following two salts will
be more soluble in water at a given
temperature. Give a reason for your
answer.
c) How would you expect the
solubility of these two salts to change
with temperature?

Solution

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 a) ∆H = ∆H hydration + ∆H lattice
energy soln
⟹ ∆H soln (KCl) = − 692 + 718
= + 26kJmol-1
∆Hsoln (LiCl) = − 883 + 862
= − 21kJmol-1
KCl, dissolution is endothermic because
enthalpy of solution is positive
LiCl dissolution is exothermic because
enthalpy of solution is negative
b) LiCl, because enthalpy of solution

is negative
c) KCl, solubility increases with

increase in temperature because it

requires more energy to dissolve
(enthalpy of solution is positive)
LiCl, solubility decreases with increase in
temperature because enthalpy of solution
is exothermic.

2 (a) Define the term molar enthalpy of hydration

(b) The table below shows the enthalpies of
Mg2+ and Cl−

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 Ions Enthalpy of
hydration
(kJmol 1)
Mg2 1891
381
State whether the values of enthalpies of
hydration given in the table above is
positive or negative. Give a reason for your
answer
Calculate the enthalpy of hydration of

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