complete modul sem 1 2021 2022

SCI 1054 CHEMISTRY SEMESTER 1
b) Pi bond

A covalent bond resulting from the formation of a molecular orbital by side-to-side overlap
of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the
atoms, denoted by the symbol π. It occurs in molecules with double or triple bonds

Here's another illustration showing how the side-to-side overlapping occurs:

Figure 4.7 : Overlapping of two p orbitals

Example: N2

N :     
1s 2s 2p
valence electrons

p p
zπ
z
π
p

y

p p
xσ
x

p

y

NN

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Example 2:

+ p p
zπ
Nitrogen monoxide ion, NO z
+
π p
The Lewis structure is :N ≡ O:
There is a σ bond and two π bonds. y

7N :      p σ p
1s 2s 2p
valence electrons x x

8O+ :      p N  O+
1s 2s 2p
valence electrons y

4.4.1. Hybridization

Hybridization is defined as the concept of mixing two atomic orbitals with the same energy
levels to give a degenerated new type of orbitals. This intermixing is based on quantum
mechanics. The atomic orbitals of the same energy level can only take part in hybridization and
both full filled and half-filled orbitals can also take part in this process provided they have equal
energy.
During the process of hybridization, the atomic orbitals of similar energy are mixed together
such as the mixing of two „s‟ orbitals or two „p‟ orbital‟s or mixing of an „s‟ orbital with a „p‟
orbital or „s‟ orbital with a „d‟ orbital.

4.4.1.1 Type Hybridization

Based on the types of orbitals involved in mixing, the hybridization can be classified as sp, sp2,
sp3, sp3d, sp3d2, sp3d3. Let us now discuss the various types of hybridization along with their
examples.

a) Formation of hybrid orbital: sp

sp hybridization is observed when one s and one p orbital in the same main shell of an atom
mix to form two new equivalent orbitals. The new orbitals formed are called sp hybridized
orbitals. It forms linear molecules with an angle of 180°

 This type of hybridization involves the mixing of one „s‟ orbital and one „p‟ orbital of
equal energy to give a new hybrid orbital known as an sp hybridized orbital.

 sp hybridization is also called diagonal hybridization.
 The shape of sp hybrid orbital will form a linear and the angle is 180o

Examples of sp hybridization:
 All compounds of beryllium like BeF2, BeH2, BeCl2
 All compounds of carbon-containing triple Bond like C2H2.

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Figure 4.8 : sp hybridization

i. Beryllium chloride ( BeCl2)

 The beryllium chloride (BeCl2) molecule is predicted to be linear by VSEPR .
 The orbital diagram for the valence electrons in Be is:

Ground state Be: 
2s 2p

 In ground state, Be atom does not form covalent bond with Cl atom because
electrons are paired in the 2s orbital in Be atom.

 Electron from 2s orbital is promoted to a 2p orbital,

Excited state Be: 
2s 2p

 Mixing 2s orbital and 2p orbital to form two equivalent sp orbitals.

Hybrid state Be:  
sp

 These two hybrid orbitals lie on the same line, angle between them is 180o.
 Each of BeCl bond is then formed by the overlapping of a sp hybrid orbitals atom Be

and p orbital atom Cl resulting BeCl2 molecule with a linear geometry.

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sp hybrid orbital of BeCl2

ii. Formation of ethyne molecule (C2H2)

The acetylene molecule contains a carbon-carbon triple bond. The molecule is linear.
Hybridization of C atom in C2H2

Ground State :
2s 2p

Excited State:
(Promotion of e)

2s 2p

sp–hybridized state :
sp unhybridized

 Promoting a 2s orbital electron into a 2p orbital result in the following excited state.
 Mixing 2s orbital with one 2p orbital generates two sp hybrid orbitals and that the two

2p orbitals remains unchanged.

Figure 4.9 : sp hybridization

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b) Formation of hybrid orbital : sp2

sp2 hybridization is observed when one s and two p orbitals of the same shell of an atom
mix to form 3 equivalent orbitals. The new orbitals formed are called sp2 hybrid orbitals.
 sp2 hybridization is also called trigonal hybridization.
 It involves mixing of one „s‟ orbital and two „p‟ orbitals of equal energy to give a new

hybrid orbital known as sp2.
 A mixture of s and p orbital formed in trigonal symmetry and is maintained at 1200.

 All the three hybrid orbitals remain in one plane and make an angle of 120° with one
another. Each of the hybrid orbitals formed has 33.33% s character and 66.66% „p‟

character.
 The molecules in which the central atom is linked to 3 atoms and is sp2 hybridized have

a triangular planar shape.

Examples of sp2 hybridization
 All the compounds of Boron i.e. BF3, BH3
 All the compounds of carbon containing a carbon-carbon double bond, Ethene (C2H4)

i. Hybridization of B atom in BF3

Ground State :

Excited State : 2s 2p
(Promotion of e) 2s 2p

sp2 –hybridized state :

sp2

 Boron atom has one unpaired electron and can form only one bond with other orbital.

 This will form a very unstable BF molecule.
 The electron in 2s orbital is promoted to an unoccupied 2p orbital.
 The mixing the 2s orbital with the two 2p orbitals form three sp2 hybrid orbitals.
 Overlap of boron sp2 hybrid orbitals and 2p orbital flourine atom to form sigma ( )

bond in each of the BF bond.
 The BF3 molecule is planar with all the FBF angle equal to 120o .

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Orbital overlap of BF3

ii. Hybridization of carbon atom in ethene molecule
 In ethene molecule contains a carbon-carbon double bond and has planar geometry.

Ground

Excited State: 2s 2p
(Promotion of e) 2s 2p
sp2–hybridized state :

sp2
 Promoting a 2s electron into a 2p orbital
 Mixing the 2s orbital and two 2p orbitals generates three sp2 hybrid orbital. One of 2p

orbital remains unchanged.

c) Formation of hybrid orbital : sp3
When one „s‟ orbital and 3 „p‟ orbitals belonging to the same shell of an atom mix together
to form four new equivalent orbitals, the type of hybridization is called a tetrahedral
hybridization or sp3. The new orbitals formed are called sp3 hybrid orbitals.

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 These are directed towards the four corners of a regular tetrahedron and make an angle

of 109°28‟ with one another.
 The angle between the sp3 hybrid orbitals is 109.280
 Each sp3 hybrid orbital has 25% s character and 75% p character.
 Example of sp3 hybridization: ethane (C2H6), methane.

Figure 4.10 : Formation of sp3 hybridization

i. Hybridization of C atom in CH4

Ground State :

Excited State: 2s 2p
(Promotion of e) 2s 2p

sp3–hybridized state :
sp3

 The carbon atom has two unpaired electrons (one in each of the two 2p orbitals).

 It can form only two  bonds with hydrogen in its ground state.
 Four C –H bonds in methane, one electron from the 2s orbital has to be excited to

unoccupied 2p orbital.
 In the exited state there are four unpaired electrons in carbon atom. The 2s orbital and

2p orbitals mix to form sp3 hybrid orbitals.
 The four sp3 hybrid orbitals of carbon overlap with s orbital of hydrogen to form  bonds

in methane molecule.

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ii. Hybridization of nitrogen atom in ammonia ( NH3 ).
Ground state :

2s 2p

sp3–hybridized state :

sp3
 The nitrogen atom has three unpaired electrons. The electron will not be promoted from

2s orbital to 2p orbital.
 However, 2s and 2p orbitals hybridize to form sp3 orbitals.
 Overlapping of sp3 orbitals with s orbital of hydrogen atoms to form sigma bonds.
 Three of the four hybridize orbitals form covalent N-H bonds
 The fourth hybrid orbital accommodates the lone pair electrons in nitrogen

iii. Hybridization of oxygen atom in water (H2O)
Ground state :

2s 2p

sp3–hybridized state :

sp3
 The H2O molecule has a V-shaped geometry based on VSEPR.
 The 2s and 2p orbitals hybridized to form sp3 hybrid orbitals.

d) Formation of hybrid orbital : sp3d

sp3d hybridization involves the mixing of 3p orbitals and 1d orbital to form 5 sp3d hybridized
orbitals of equal energy. They have trigonal bipyramidal geometry.

 The mixture of s, p and d orbital forms trigonal bipyramidal symmetry.
 Three hybrid orbitals lie in the horizontal plane inclined at an angle of 120° to each

other known as the equatorial orbitals.
 The remaining two orbitals lie in the vertical plane at 90 degrees plane of the equatorial

orbitals known as axial orbitals.
 Example:Hybridization in Phosphorus pentachloride (PCl5)

i. Hybridization in Phosphorus pentachloride (PCl5)
 The phosphorus pentahydride molecule is predicted to be trigonal bipyramidal geometry

base on VSEPR.
 The orbital diagram for the ground state valence electrons in P is:

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

Ground state :

3s 3p 3d

 Promoting a 3s electron into a 3d orbital result in the following exited state:
Excited state :

3s 3p 3d

 Mixing the one 3s , three 3p and one 3d orbitals generates five sp3d hybrid orbitals.:
sp3d–hybridized state :

sp3d

Figure 4.11 :

Hybridization in Phosphorus pentachloride (PCl5)

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e) Formation of hybrid orbital : sp3d2

 sp3d2 hybridization has 1s, 3p and 2d orbitals, that undergo intermixing to form 6
identical sp3d2 hybrid orbitals.

 These 6 orbitals are directed towards the corners of an octahedron.

 They are inclined at an angle of 90 degrees to one another.

i. Hybridization in sulphur hexaflouride (SF6)

 The sulphur hexaflouride (SF6) molecule has octahedral geometry base on VSEPR.
 The ground state electron configuration of S is:

Ground state :

3s 3p 3d

 Promoting a 3s and 3p electrons into a 3d orbital result in the following excited state
Excited state :

3s 3p 3d

 Mixing the 3s, three 3p and two 3d orbitals generate six sp3d2 hybrid orbitals.
sp3d2–hybridized state :

sp3d2

FIGURE 4.12 : Hybridization in Sulphur Hexaflouride (SF6)
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4.5 Intermolecular Forces

Intermolecular forces are generally much weaker than covalent bonds. Thus, when a molecular
substance changes states the atoms within the molecule are unchanged. The temperature at
which a liquid boils reflects the kinetic energy needed to overcome the attractive intermolecular
forces (likewise, the temperature at which a solid melts). Thus, the strength of the
intermolecular forces determines the physical properties of the substance

4.5.1 Van der Waals forces
There are three types of van der waals forces
i) dipole-dipole forces
ii) induce dipole-dipole forces
iii) London dispersion forces.

The intermolecular forces increase in strength according to the following:
London dispersion < dipole-dipole < H-bonding < covalent bond

(a) Dipole-Dipole Forces
A dipole-dipole forces exists between neutral polar molecules. Polar molecules attract one
another when the partial positive charge on one molecule is near the partial negative charge on
the other molecule.
The polar molecules must be in close proximity for the dipole-dipole forces to be significant.
Dipole-dipole forces increase with an increase in the polarity of the molecule.

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Effect of boiling point
For polar molecules with similar molecular mass, the boiling points increase with increasing
dipole moments.
Relationship between boiling point and dipole moments:

Substance Molecular Dipole moment, Boiling Point
Mass (amu) u (D) (°K)
Propane 0.1 231
Dimethyl ether 44 248
Methyl chloride
Acetaldehyde 46 1.3 249
Acetonitrile 294
50 2.0
355
44 2.7

41 3.9

Table 5.2

For comparable molar mass, boiling point increases as the net dipole
moment increases

(b) Induce dipole-dipole forces
When an oxygen gas comes close to a polar H2O molecule, the electrons can shift to one side of
nucleus to produce a very small dipole moment that lasts for only an instant

Other example is HCl and Argon atom

δ+ H----Cl δ- δ+ Ar δ-

By distorting the distribution of electrons around the argon atom, the polar HCl molecule
induces a small dipole moment on this atom, which creates a weak induced dipole-dipole force
of attraction between the HCl molecule and the Ar atom. This is very weak, with a bond energy
about 1 kJ/mol.

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(c) London Dispersion Forces
Nonpolar molecules would not seem to have any basis for attractive interactions. However,
gases of nonpolar molecules can be liquefied indicating that if the kinetic energy is reduced,
some type of attractive force can predominate. Fritz London (1930) suggested that the motion
of electrons within an atom or non-polar molecule can result in a transient dipole moment.

A model to Explain London Dispersion Forces:

Helium atoms (2 electrons)

The average distribution of electrons around each nucleus is spherically symmetrical. The atoms
are non-polar and posses no dipole moment The distribution of electrons around an individual
atom, at a given instant in time, may not be perfectly symmetrical. Both electrons may be on
one side of the nucleus. The atom would have an apparent dipole moment at that instant in
time (i.e. a transient dipole).

A close neighboring atom would be influenced by this apparent dipole - the electrons of the
neighboring atom would move away from the negative region of the dipole. Due to electron
repulsion, a temporary dipole on one atom can induce a similar dipole on a neighboring atom
This will cause the neighboring atoms to be attracted to one another. This interaction is called
the London dispersion force (or just dispersion force) and only significant when the atoms are
close together.

4.5.1.2 Strength of van der Waals

The strength of van der Waals forced is influenced by 2 factors:
a) Molecular size (molecular mass)

- Larger and heavier atoms and molecules exhibit stronger force.
- The larger the atom, the more electrons it has and the more easily the valence electrons

are displaced from their average position. Thus they are less tightly held and can more
easily form the temporary dipole that produce the attraction.
b) Molecular shape
- Molecules with large surface area allow greater contact between molecules, thus increase
van der Waals force than molecules with more compact shape.
- For example, n-butane have higher boiling point than 2-methylpropane although both
compounds have the same molecular formula, C4H10.

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CH CH CH CH CH CHCH
3223 33
n-butane CH
(long and linear shape) 3

BIGGER surface area 2-methypropane
(Spherical shape)
LESS surface area

c) Polarity
The intermolecular forces increase in strength according to the following:

London dispersion < dipole-dipole < H-bonding < covalent bond

4.5.2 Hydrogen bonding
A hydrogen atom in a polar bond (e.g. H-F, H-O or H-N) can experience an attractive force with
the other electronegative atom from molecule or ion which has an unshared pair of electrons
(usually an F, O or N atom on another molecule). Hydrogen bonds are considered to be dipole-
dipole type interactions.
A bond between hydrogen and an electronegative atom such as F, O or N is quite polar:

The hydrogen atom has no inner core of electrons because all valence electron has been
consumed to make covalent bond. H atom will carry partial positive charge (∂+) and O will carry
partial negative charge (∂-).

This positive charge is attracted to the negative charge of an electronegative atom in a nearby
molecule. Because the hydrogen atom in a polar bond is electron-deficient on one side (i.e. the
side opposite from the covalent polar bond) this side of the hydrogen atom can get quite close
to a neighboring electronegative atom (with a partial negative charge) and interact strongly
with it (remember, the closer it can get, the stronger the electrostatic attraction). This
attraction is known as hydrogen bonding.

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Figure 4.12 : Hydrogen bonding among water and ammonia molecules.
Hydrogen bonds are still weaker than typical covalent bonds. But they are stronger than dipole-
dipole and or dispersion forces.
The hydrogen bond increase in strength according to the following:

H-F > H-O > H-N

Hydrogen bonding is one type of intermolecular forces,that exist
BETWEEN two or more molecules, NOT in the same molecule

4.5.2.1 The Effect Of Hydrogen Bonding
Hydrogen bonding effects:

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① Boiling point
As we refer to figure below,

Figure 5.5 Boiling point of some hydrogen compounds.

Based on group 4A:
 The boiling points of hydrides of group 14 elements (CH4,SiH4, GeH4 and SnH4)
displays normal behavior. CH4 < SiH4 < GeH4 < SnH4
 The type of intermolecular forces exist between the molecules are only van der
Waals
 Hydrogen bonding is not present.
 The boiling point increases regularly when
 the relative molecular mass increases.
 because the van der Waals forces increase as the molecular size
increases.

For hydrides of group 5A, 6A and 7A elements
 hydrogen bonding are not present except NH3, H2O & HF
 The increase in the boiling points for the hydrides of each periodic group is due
to the increase in van der Waals forces as the molecular size increases.
 The strength of H-bonding is proportional to the polarity of the bond.②
 Thus, H-F > H2O > NH3

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However, boiling point increases in such order

H2O > HF > NH3
 Even though fluorine is more electronegative than oxygen, H2O has higher

boiling point than HF because there are 4 H-bonding per H2O molecule compared
to only 2 per HF molecule.

 HF has higher boiling point than NH3 because of stronger H-bonding of HF.

Tips for checking boiling point:
1. Type of intermolecular forces

London dispersion < dipole-dipole < H-bonding < covalent
bond
2. Molar mass
② Solubility
Most organic compounds are insoluble in water except amines (e.g: C2H5NH2), an alcohols (e.g:
C2H5OH) and carboxylic acid (e.g: CH3COOH)
Amines are soluble in water because the -NH2 group in amines can form H-bonding with the
water molecules. Alcohol and carboxylic acids are soluble in water because the -OH groups in
the compounds can form hydrogen bonding with the water molecules.

However, NOT all organic compounds that contain -NH2 group or –OH groups are soluble in
water. As the relative molecular mass increases, the non-polar hydrocarbon portion becomes
larger.

Why Is Water More Dense Than Ice?
③ Density
H2O is unusual in its ability to form an extensive hydrogen bonding network. When cooled to a
solid the water molecules arrange themselves to form tetrahedral arrangement in such a way to
maximize the amount of hydrogen bonding between them.

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The leaves a relatively large amount of space between them and gives rise to an “open”
structure. This arrangement of molecules has greater volume (is less dense) than liquid water,
thus water expands when frozen.
The three-dimensional structure of the covalent bonds and the hydrogen bonding in ice is
shown in the figure.

Figure 4.13: Hydrogen bonding (dotted lines) between water molecules in ice.
This “open” structure of the ice accounts for the fact that is less dense than water at 0oC. When
ice melts, some of the hydrogen bonds are broken.This allows the water molecules to be more
compactly arranged, resulting in a 9% decrease in volume. Water has a higher density than ice.

4.5.4 Metallic Bond
Metallic bonding is the electrostatic attraction between a network of positive ions and the
„sea‟ of delocalized electrons.
The structures and properties of crystals such as melting points, boiling points, density and
hardness are determined by the kinds of forces that hold the particles together.
In a metallic bond, metal atoms can be imagined as an array of positive ions immersed in a sea
of delocalized valence electrons. A metallic bond can be defined as the electrostatic force
between the positively charged metallic ions and the „sea‟ of electrons .The great cohesive force
resulting from delocalization is responsible for a metal‟s strength. The mobility of the delocalized
electrons makes metal good conductors of heat and electricity.

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Figure 4.14 : Representation of metal ions in a metallic crystal.

As the name implies, metallic bonding is confined to metal and near metals, many of which are
to be found in Group 1 , 2 and 13 of the Periodic Table. The electrons flow through the metal
(they form a current under the influence of an applied electric field) as if they were like gas
particles that could travel freely in the volume of a container. The key to metallic bonding is
that certain electrons are delocalized. These delocalized electrons are not bound to individual
atoms and they can therefore serve to bind large numbers of metal atoms together.
Metals have low electron attracting powers and thus low ionization energies. As a result, the
atoms in the metallic lattice release their outer valence electrons to form a cluster of positive
ions held together by oppositely charged clouds of delocalized valence electrons.

4.5.3.1 Physical properties of metals

Metals have the following properties:
1. Malleable
2. Good electrical conductor
3. Good thermal/heat conductor

4.5.3.2 The change of melting points and boiling points

A melting point of a metal indicates the strength of its metallic bonding in its structure. Due to
the fact that the mobility of electrons in their valence orbital is responsible for the formation of
metallic bonding, this strength (or melting point) depends mainly on the number of valence
electrons. Generally, the strength of metallic bonding is directly proportional to the number of
valence electron per atom divided by the radius of a metal.

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The strength of metallic no. of valence electron per atom
bonding 

Radius of a rate

So, the bonding will be weaker in sodium (one valence electron) as compared to magnesium
(two valence electrons) and aluminum (three valence electrons).

Element Na Mg Al Li Be B
Melting point (o C) 98 650 660 180 1280 2300
Boiling point (o C) 892 1107 2450 1330 2480 3930
No. of valence electrons 1 2 312 3

Table 4.2 : The melting points and the boiling points for elements in Group 1, 2, 13

It does not matter how slowly you go as long as you do not stop.
_Confucius

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CHAPTER 5 CHEMICAL
EQUILIBRIUM

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5.0 CHEMICAL EQUILIBRIUM

5.1 The concept of equilibrium

5.1.1 Reversible reaction

Reversible reaction can go in both directions. As soon as some product molecules are formed,
the reverse process begins to take place and reactant are formed from product.

 The reversible sign:

Sulfur dioxide and sulfur trioxide exist in equilibrium as indicated in the following equation:

+2SO2(g) O2 (g) 2SO3 (g)

The net concentration of SO2, O2 and SO3 do not change. However, SO2 and O2 combining to
form SO3 at the same time SO3 decomposed to form SO2 and O2.

Non-reversible reactions proceed in only one direction .The reaction proceeds toward the
formation of product. In the following reaction, the formation of KCIO3 from the mixture of KCI
and O2 does not take place.

2KClO 3 (g) +2KCl (s) 3O2(g)

YOU HAVE BRAINS IN
YOUR HEAD.

YOU HAVE FEET IN YOUR
SHOES.

YOU CAN STEER
YOURSELF ANY
DIRECTION YOU
CHOOSE.

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5.1.2 State of equilibrium
A system reached an equilibrium when a forward and a reverse reaction proceed at equal
rates, and the concentrations of reactants and products remain constant over time. The system
is said to be in dynamic equilibrium.

Concentration C and D (products)

A and B (reactants)
ta Time

A+B C+D

Figure 5.1 Concentration versus time graph for the reversible reaction.

The concentration of C and D increase with increasing of time. In contrast, the concentration of
A and B decrease with increasing of time. Up to a point, (ta) both reactions has no further
change in the concentration of products or reactant.

Rate

A+ B C+D
Time
Time

Figure 5.2 Rate versus time graph for the reversible reaction.
In reversible reaction, the rate of forward reaction equaTlsimtoe rate of reverse reaction (Figure
5.2)

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5.1.3 Law of chemical equilibrium

The equilibrium concentration of reactants and products are related by law of mass action/ law
of chemical equilibrium.

Consider general reaction bB cC + dD

aA +

At equilibrium, [ C ]c. [ D ]d = K
[ A ]a. [ B ]b

where : K = equilibrium constant, [ ] = concentration

Example 5.1.1:
The following equilibrium concentrations are measured at 483K; [CO]=1.03M, [H2]=0.322M and
[CH3OH]=1.56M . Determine the equilibrium constant.

+CO(g) 2H2(g) CH3OH(g)

Solution 5.1.1: K = [CH3OH]
[CO][H2]2

= 1.56
1.03 (0.322)2

= 14.5

5.2 EQUILIBRIUM CONSTANTS

5.2.1 The equilibrium constant expression
The concentrations in the equilibrium constant expression are usually given in moles per liter
(M), and so the symbol K is often written with a subscript c for “concentration” as in Kc.

+aA(g) bB(g) +cC(g) dD(g)

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Kc = [ C ]c [ D ]d
[ A ]a [ B ]b

When the reactants and products in a chemical equation are gaseous, we can formulate the
equilibrium expression in terms of partial pressures instead of molar concentrations.

aA(g) + bB(g) +cC(g) dD(g)
Kp =
(PC)c(PD)d
(PA)a(PB )b

where : PA = partial pressure of gas A, etc.
Kp = equilibrium constant for the reaction in term of pressure of reactants.

Important!
Kc = equilibrium concentration (for solution)
Kp = equilibrium partial pressure (for gaseous compound)

5.2.2 The relationship between Kp and Kc
For reactions involving gases, Kp and Kc are not necessarily equal. For general equation:

+aA(g) bB(g) +cC(g) dD(g)

Kp = [PC]c[PD]d and Kc = [D]d[C]c
[PA]a[PB]b [A]a[B]b

Using n for moles and V for liters and assuming ideal gas behavior,

PV = nRT

To obtain the concentration of gas X, in a mixture:

[X] = nx = Px
V RT

Where Px is its partial pressure. From this it follows that;
Px =[X]RT

Substituting this relationship into the expression for Kp,

Kp = PDdPCc = [D]d(RT)d[C]c(RT)c
PAaPBb [A]a(RT)a[B]b(RT)b

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This can be rearranged to give
Kp = [D]d[C]c (RT)(d+c) – ( a+b)
[A]a[B]b

or Kp = Kc(RT)n

n =(number of moles gaseous products) - (number of moles of
gaseous reactants)

Example 5.2.1:

Calculate Kc for the following reaction at STP

CaCO 3(s) +CaO(s) CO2(g)

Kp= 2.1 x 10-4atm-1

Solution 5.2.1:
Kp = Kc(RT)n

So, Kc = Kp
(RT)n

= 2.1 x 10-4
0.08206 x 273.15

= 9.3696 x 10-6

Example 5.2.2:
The equlibrium constant Kp for the decomposition of phosphorus pentachloride to phosphorus
trichloride and chlorine gas is found to be 1.05 at 250C. If the equilibrium partial pressures of
PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial
pressure of Cl2 at 250C.

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Solution 5.2.2:
+PCl3(g) Cl2(g)
P Cl5(g)

Kp = PPCI3PCI2
PPCI5

PCl 2 = (1.05)(0.875)
(0.463)

= 1.98 atm

5.2.3 Homogeneous equilibrium

In homogeneous equilibrium, all substances are in the same phase. For example:

+3H2(g) N2(g) 2NH3(g)

Kc = [NH3]2
[N2] [H2]3

5.2.4 Problems involving Equilibrium Constant (Application of ICE Table)

The approach in solving equilibrium constant problems could be summarized as follows:
i) Express the equilibrium concentration of all species in terms of the initial concentrations

and a single unknown x, which represents the change in concentration.
ii) Write the equilibrium constant expression in terms of the equilibrium concentrations.

Knowing the value of the equilibrium constant, solve x
iii) Having solved for x, calculate the equilibrium concentrations all species.

TRUST ME..I’M A CHEMISTRY STUDENT!

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Example 5.2.3:

The equal amount of hydrogen and iodine are injected into a 1.50 L reaction flask at a fixed
temperature.

H2(g) + I2(g) 2HI(g)

At equilibrium, analysis shows that the flask contains 1.80 mole of H2, 1.80 mole of I2, and
0.520 mole of HI. Calculate Kc.
Solution 5.2.3:

Convert the amounts(mole) to concentration (mole/L), using the flask volume of 1.50 L

[ H2 ] = 1.80 /1.50 = 1.20 M
[ I2 ] = 1.80 /1.50 = 1.20 M
[ HI ] = 0.347 /1.50= 0.347 M

Kc = [ HI ]2
[ H2 ][ I2 ]

= ( 0.347 )2
(1.20)(1.20)

= 8.36 x 10-2

Example 5.2.4:
At equilibrium, 0.200 mole of hydrogen halide is injected into a 2.00L reaction flask at fixed
temperature:

2HI(g) H2(g) + I2(g)
If [ HI ]=0.078 M at equilibrium, calculate Kc.

Solution 5.2.4:

[ HI ] = 0.200 mol/2.00L = 0.100M

Initial 2HI(g) H2(g) I2(g)
Change 0.100 0 0
Equilibrium +x +x
concentration(M) -2x x x
0.100-2x

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Solving for x, using the known [ HI ] at equilibrium:
[ HI ] = 0.100 M – 2x = 0.078M
x = 0.011 M

Therefore, the equilibrium concentrations are:
[ H2 ] = [ I2] = 0.011 M and [ HI ] = 0.078 M
Kc = [ H2 ][ I2 ]
[ HI ]2
= ( 0.011 )( 0.011 )
(0.078)2
= 0.020

Example 5.2.5:

At 440C, the equilibrium constant Kc for below reaction has a value of 49.5.

H2(g) + I2(g) 2HI(g)

If 0.200 mole of H2 and 0.200 mole of I2 are placed into a 10.0 L vessel and permitted to react
at this temperature, what will be the concentration of each substance at equilibrium?

Solution 5.2.5 :

Kc = [ HI ]2
[ H2 ][ I2 ]

= 49.5
[ H2 ] = 0.200 /10.0 = 0.0200 M
[ I2 ] = 0.200 /10.0 = 0.0200 M
[ HI ] = 0.0 M

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Initial H2(g) I2(g) 2HI(g)
Change 0.0200 0.0200 0
Equilibrium
concentration(M) -x -x +2x

0.0200-x 0.0200-x 0 + 2x

( 2x)2 = 49.5
(0.0200-x)(0.0200-x)

2x = 7.04
0.0200 –x

x = 0.0156

The equilibrium concentrations are :

[ H2 ] = 0.0200 – 0.0156 = 0.0044 M, [ I2 ]= 0.0200 – 0.0156 = 0.0044 M
[ HI ] = 2 (0.0156) = 0.0312 M

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5.3 The Le Châtelier’s Principle

Le Châtelier’ principle: If a system at equilibrium is subjected to a change of pressure,
temperature or the number of moles of a component, there will be a tendency for a net
reaction in the direction that reduces the effect of this change.

5.3.1 Factors affecting an equilibrium position:

A. Change in temperature
Temperature affects the value of Kc and Kp. The effect of temperature depends on whether the
forward reaction is exothermic / endothermic.

i. Exothermic reaction

Exothermic reactions give out/ release heat to the surrounding. Eg:

N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH= -92kJ

Since the forward reaction is exothermic, the reverse reaction is endothermic.

ΔH= -92kJ the (-) sign means energy has been released from the reaction

The equation may be rewritten by writing heat as one of the products of the reaction.
N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH = +92 kJ

Case 1:
Increasing the temperatures of the system in equilibrium will shift the equilibrium position to
the left so as to absorb the added heat.

N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH = -92kJ

Case 2:
Decreasing the temperatures of the system – equilibrium will shift to the right.

N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH= +92 kJ

Thus the values of Kc and Kp decrease with increase in temperature.

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If the system is subjected to certain change, direction of reaction must proceed to
the way to REDUCE it! LE CHÂTELIER’S PRINCIPLE

ii. Endothermic reaction

Endothermic reactions absorb heat from the surrounding. Eg:

N2O4 (g) 2 NO2 (g) ΔH = +57 kJ

Since the forward reaction is endothermic, the reverse reaction is exothermic.

The equation may be rewritten by writing heat as one of the products of the reaction.
N2O4 (g) + 57 kJ 2 NO2 (g)

Case 1:
Increasing the temperatures of the equilibrium system shift the equilibrium position to the right
so that the added heat can be absorb.

Case 2:
Decreasing the temperatures – shift the equilibrium to the left.
Thus, the values of Kc / Kp increase with increasing the temperature.

B. Change in concentration
Case 1:
Increasing the concentration of a reactant shifts the equilibrium towards product
formation (to the right) because the rate of the forward reaction is increased.

Equilibrium Direction

Product

Reactant

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In order to achieve equilibrium, reaction must proceed to the right in order to reduce the
disturbance. (Disturbance = increase concentration of reactant).
Case 2:
Increasing the concentration of a product shifts the equilibrium towards the reactant
formation (to the left) because the rate of the reverse reaction is speeded up.

Equilibrium direction

Reactant Product

In order to achieve equilibrium, reaction must proceed to the left in order to reduce the
disturbance. (Disturbance = increase concentration of product)
Case 3:
Decreasing the concentration of a reactant (by removal or by compounding it with
something else or by precipitation) shifts equilibrium to the left (toward the reactants
formation) because the forwards reaction is slowed down.

Case 4:
Decreasing the concentration of a product shifts equilibrium to the right (towards the
products formation) because the reverse reaction is slowed down and the forwards reaction is
favored.

Example 5.3.1:
Synthesis of HI
H2(g) +
I2(g) 2HI(g)

Increasing the concentration of H2 to a reaction mixture will disturb the equilibrium of the
system. The system will reduce the disturbance by eliminating some of the added H2. The
additional H2 reacts with some of the I2 to form more HI. As a result, the position of equilibrium
shifts to the right.

C. Change in pressure (volume)
Change in pressure is significant only on equilibrium systems with gaseous components.
Pressure changes can occur in three ways:

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 Adding or removing gaseous reactant or product from equilibrium mixture
 Changing the volume of equilibrium system
Adding an inert gas (does not take part in the reaction) at constant volume

For the reaction involving gaseous components, Le Chatelier‟s principle predicts the following:

 If the total pressure on the equilibrium is increased, reaction will proceed to the
direction that decreases the number of moles of gas.

 If the total pressure on the equilibrium is decreased, reaction will proceed to the
direction that increases the number of moles of gas.

Case 1 : Adding or removing gaseous reactant or product from equilibrium mixture

The effect in changes the concentration on equilibrium system is similar like we have discussed
in part A.

Case 2: Changing the volume of the equilibrium system.

Note 1

Decrease in volume of equilibrium system is equivalent to an increase in pressure. As the
volume of the system is reduced, equilibrium direction will shift to the direction with fewer
moles of gases.

Note 2

Increase in volume of equilibrium system is equivalent to a decrease in pressure. As the
volume of the system is increased, equilibrium direction will shift to the direction with more
moles of gases.

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N2O4(g) 2NO2(g) N2O4(g) 2NO2(g) N2O4(g) 2NO2(g)

To the right

To the left

Disturbance : increase in volume

When the volume of equilibrium system is increased, the pressure of the system will decreased.
The equilibrium will shift to the right producing more NO2. So, [N2O4] will decrease and
[NO2] will increase.

N2O4(g) 2NO2(g)
1 mole 2moles

Disturbance : decrease in volume
When the volume of equilibrium system is decreased, the pressure of the system will increased.
The equilibrium will shift to the left producing more N2O4. So, [N2O4] will increase and [NO2] will
decrease.

N2O4(g) 2NO2(g)
1 mole 2 moles

Case 3: Adding an inert gas (does not take part in the reaction) at constant
volume

Adding an inert gas at constant volume has no effect on the position of the equilibrium.
Addition of inert gas at constant volume increases the total number of mole of gaseous
molecules and hence the total pressure. But the partial pressure of each gas in the equilibrium
mixture remains unchanged. So all the reactant and product concentration remain the same.

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Example 5.3.2:
Increased pressure will cause more collisions to take place between gaseous molecules
(pressure is a measure of the number of particles per unit volume). In the below reaction, the
increase in pressure:

 will favour the reaction involving the most number of gas molecules.
 favours reverse reaction more than the forward reaction.
 speeded up the reverse reaction and shifts the equilibrium to the left.

N2O4(g) 2NO2(g)

If the pressure is decreased (by increasing volume), the equilibrium has shifted to the right
because the reverse reaction has slowed down. In the below reaction, the decrease in pressure:

 will favour the reaction involving the fewer number of gas molecules
 will proceed to the site that producing more number of moles to attain equilibrium
 favours forward reaction more than the reverse reaction
 speed up the forward reaction and shifts the equilibrium to the right.

N2O4(g) 2NO2(g)

Note: Pressure changes only matter if there is a different number of moles of gaseous
molecules on each side. Pressure changes are irrelevant if there are no gas molecules in the
reaction, e.g.

H2(g) + I2(g) 2HI(g)
2 moles of gas 2 moles of gas
Here, there is no change in equilibrium position.

Example 5.3.3:

N2(g) +3H2(g) 2NH3(g)
4 moles of gas 2 moles of gas

In this reaction the number of gas molecules decreases as the product is formed; four gaseous
reactant molecules produce two gaseous product molecules.
This means that the pressure in the system can be decreased if the position of equilibrium shifts
to the right.

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Therefore, decreasing the volume of a mixture of gases that are in chemical equilibrium shifts
the position of equilibrium in the direction of the fewest number of gas molecules.
The Effects of catalyst on equilibrium

A catalyst provides an easier path for the forward and reverse reaction. A catalyst will not shift
an equilibrium position because both rates are equally increased. The equilibrium is achieved
faster.

SIMPLIFY!
4 factors that affecting an equilibrium position:

Change in Change in Change in Effect of
temperature concentration pressure catalyst

Case study

Example 5.3.4 : Predicting the effects of a change in temperature on the equilibrium
position

How does an increase in temperature affect the equilibrium concentration of the underlined
substance and Kc for the following reactions?

a) CaO(s) + H2O(l) Ca(OH)2(aq) Ho = -87 kJ
b) SO2(g) S(s) + O2(g) Ho = 297 kJ

Solution 5.3.4:
Increasing the temperature adds heat to the equilibrium system. System is already in „hot‟
condition as the reaction releases heat (exothermic). So the system shifts to the direction to
absorb the heat, that is favours the endothermic reaction. Equilibrium direction shift to the left.
Kc and [Ca(OH)2] will decrease.

CaO(s) + H2O(l) Ca(OH)2(aq) Ho = -87 kJ

*Adding heat shifts the system to the left : [Ca(OH)2] and Kc will decrease

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Increasing the temperature adds heat to the equilibrium system. System is in “cold” condition
as the reaction absorbs heat (endothermic). So the system shifts to the direction to absorb
heat, that is favours the endothermic reaction. Equilibrium direction shift to the right. [SO2] will
decrease and Kc will increase.

SO2(g) S(s) + O2(g) Ho = 297 kJ

*Adding heat shifts the system to the right : [SO2] will decrease and Kc will increase.

Example 5.3.5: Predicting the effects of a change in concentration on the
equilibrium position

To improve air quality and obtain a useful product, sulfur is often removed from coal and
natural gas by treating the fuel contaminant, hydrogen sulfide with O2.

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

What happens to

a) [H2O] if O2 is added
b) [O2] if H2S is removed

Solution 5.3.5:

Write the reaction quotient to see how Qc is affected by each disturbance, relative to Kc. This
affect tells us the direction in which the reaction proceeds for the system to retain equilibrium
and how each concentration changes.
Writing the reaction quotient: Qc = [H2O]2
[H2S]2[O2]

a) When O2 is added, the denominator of Qc increases, so Qc < Kc. The reaction proceeds
to the right until Qc = Kc again, so [H2O] increases.

b) When H2S is removed, the denominator of Qc decreases, so Qc>Kc. As the reaction
proceeds to the left to reform H2S, more O2 is produced as well, so [O2] increases.

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Example 5.3.6 : Predicting the effects of a change in volume (pressure) on the
equilibrium position.
How would you change the volume of each of the following reactions to increase the yield of
the products?

a) S(s) + 3F2(g) SF6(g)
b) Cl2(g) + I2(g) 2ICl(g)
Solution 5.3.6:

Whenever gases are present, a change in volume causes a change in concentration. For
reactions in which the number of moles of gas changes, if the volume decreases (pressure
increases), the equilibrium position shifts to relieve the pressure by reducing the number of
moles of gas. A volume increase (pressure decrease) has the opposite effect.

a) With 3 mole of gas on the left and only 1 mole on the right, we decrease the volume
(increase the pressure) to form more SF6.

b) The number of moles of gas is the same at both sides of the equation, so a change in
volume (pressure) will have no effect on the yield of ICI.

Example 5.3.7 :

Consider the following reaction at equilibrium

H2 (g) + CO2 (g) H2O (g) + CO (g) ∆H = +ve

Using the Le Chatelier‟s Principle, explain how the equilibrium would shift (if any) and how it will effect
the concentration of H2 if;

a) CO2 is added.
b) H2O is added.
c) catalyst is added.
d) the temperature is increased.
e) the volume of the container is decreased.
Solution 5.3.7
a) system will shift to the right. [H2] decrease.
b) system will shift to the left. [H2] increase.
c) no net change. [H2] remain unchanged.
d) system will shift to the right. [H2] decrease.
e) no net change. [H2] remain unchanged.

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Example 5.3.8 :

The value of Kc for the following reaction at 25C is 9.0 x 10-22.

C(s) + CO2 (g) 2CO(g)

Describe completely what will happen if 2.0 mol of CO and 1.0 mol of CO2 are mixed in a 500
mL container. Determine in which direction will the equilibrium proceed.

Solution 5.3.8

[ CO] = 2 mol/ 0.5L = 1M
[CO2] = 1 mol/ 0.5L = 0.5M
Qc = [CO]2/ [CO2]

= (1M)2/0.5M
= 0.5 M3

Qc > Kc,
thus undergo reverse reaction until eq is achieved

Example 5.3.9 :

Consider the following equilibrium:
4NH3 (g) + 3O2 (g) 2N2 (g) + 6H2O (g) ΔH = -1531 kJ

Predict the effect on the equilibrium position when the following changes are made to
the system. Explain.
a) Some of the ammonia is removed from the system.
b) The pressure of the system is increased.
c) The temperature of the system is raised.
d) Helium is added at a constant volume.
e) Helium is added at a constant pressure.

Solution 5.3.9

a) The equilibrium position shifts backward in order to increase the amount of NH3
b) The equilibrium position shifts backward, to the direction that reduces the number of

moles of gases.
c) The equilibrium position shifts backward, to the direction that absorbs heat.
d) The addition of He increases the total pressure but not the partial pressures. Thus, there

is no shift in the equilibrium position.
e) The addition of He at a constant pressure causes the volume of container to increase,

thus the partial pressures decrease. The equilibrium position shifts forward, to the
direction with the larger number of moles of gas.

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CHAPTER THERMOCHEMISTRY
6

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6.0 THERMOCHEMISTRY
Thermochemistry is the study of heat change in chemical reactions. Almost all chemical
reactions absorb or produce (release) energy (in the form of heat).
6.1 Enthalpy (H)

Enthalpy, H refers to the total value of energy of a system.
a) Enthalpy of reaction, ΔH

The absolute heat energy (enthalpy) of a substance cannot be determined. However, the
difference between the enthalpy of a system before and after the reaction can be measured.
Enthalpy of reaction is the difference between the enthalpy of the products and the enthalpy of
the reactants.

ΔH = Σ∆H products – Σ∆H reactants

For a general reaction: A+B→C+D

ΔH = Σ∆H products – Σ∆H reactants

So, ΔH = (HC + HD) – (HA + HB)
The enthalpy change of a reaction is the heat energy absorbed or released in a chemical
reaction, for the number of moles of reactants shown in the balanced equation.

(i) Exothermic reactions
Exothermic reaction is accompanied by a rise in temperature of the surrounding (as heat given
out to the surrounding)

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SURROUNDING
SYSTEM

ENERGY

Heat, H is negative
Figure 6.1 Exothermic reaction

 Enthalpy of products < Enthalpy of reactants
 ΔH is negative
 This means energy is released from the system to its surroundings

(system  surrounding)

Figure 6.2 Energy profile diagram for exothermic reaction
(ii) Endothermic Reactions
Endothermic reaction is accompanied by a drop in the temperature of the surrounding (as
heat absorbed from the surrounding)

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SYSTEM SURROUNDING
ENERGY

Heat, H is positive
Figure 6.3 Endothermic reaction
 Enthalpy of products > enthalpy of reactants
 ΔH is positive
 Energy is absorbed by the system from its surrounding
(surrounding  system)

Figure 6.4 Energy profile diagram for endothermic reaction
6.2 Thermochemical equation
The thermochemical equation shows the enthalpy changes as well as the mass relationship.
The stoichiometry coefficients refer to the number of moles of the substance.
The values of ΔH depends on the number of moles of substances involved and the direction of
the reaction.
Example:

H2O(s)  H2O(l) ΔH = +6.01 kJ

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When 1 mole of liquid water is formed from 1 mole of ice at 0° C the enthalpy change is 6.01
kJ. The forward reaction is endothermic reaction,

For reverse reaction, the magnitude of ΔH for the equation remains the same but its sign
changes. Therefore, when 1 mole of ice formed from 1 mole of liquid water, the reaction will
become exothermic.

H2O(l)  H2O(s) ΔH = 6.01 kJ

6.4 Standard Conditions in Thermochemical Measurements

The enthalpy change of a chemical reaction depends on:
 Pressure
 Temperature
 Concentration of solutions
 Physical state of the substances
 Type of allotropes

The standard conditions used in thermochemistry measurement are:
 Pressure is fixed at 1 atm or 101 kPa
 Temperature is fixed at 25C or 298 K
 The concentration of solutions is at 1.0 mol dm-3
 The most stable allotrope at 298 K and 1 atm is used.

The enthalpy change measured under these conditions is called the standard enthalpy
change of reaction, and is represented by ΔH or ΔH(298).

6.5 Types of Enthalpy

Standard enthalpy change (H) is measured under standard condition.
Standard condition: Pressure 1 atm, temperature 25C.

Enthalpy change (H) is measure under non-standard condition.

A) Standard Enthalpy of Formation (Hf)

 The heat change when 1 mole of a compound is formed from its elements in their stable
or standard states at a constant pressure of 1 atm and fixed temperature of 298 K.

H2 (g) + ½ O2(g) → H2O (l) H  = 296 kJ mol1
f

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 Every different compound will has their respective Hf value. But the standard enthalpy
of formation of any element in its most stable state form is ZERO.

Hf (O2 ) = 0 H  (Cl2) = 0
f

 The value of standard enthalpy of formation of a substance can be either positive or
negative.

2 C (s) + H2 (g) → C2H2 (g) H  = +237 kJ mol1
f

C(graphite) + O2(g) → CO2(g) H  = 394 kJ mol1
f

i) Finding Hf of substance

 The standard enthalpy of formation of some substances cannot be directly measured
from experiments.

 For example, carbon does not combine directly with hydrogen, under any conditions, to
produce ethyne, C2H2. The standard enthalpy of formation of ethyne is calculated from
other known enthalpy values using Hess’ Law.

 In the case below, given the standard enthalpy of combustion for CO2, H2O and C2H2.
From these values, we can find standard enthalpy of formation Hf for ethyne.

C(graphite) + O2(g) → CO2(g) Hc = 394 kJ
H2 (g) + ½ O2(g) → H2O (l) Hc = 296 kJ
C2H2 (g) + 5/2 O2 (g) → 2 CO2 (g) + H2O (l) Hc = 1 311 kJ

A thermochemical cycle is then constructed. C2H2 (g)
2 C (graphite) + H2 (g)

(-394 x 2) -296 5/2 O2
-1311

By Hess‟ Law: 2 CO2x(g+) (-1+311)H=2O2(l()-394) + (-286)
x = + 237 kJmol-1

Note : Algebraic method is the other alternative method in finding Hf of ethyne.

ii) Finding standard enthalpy change of reaction by using Hf value

 The standard enthalpy of formation can be used to calculate the enthalpy change of a
reaction.

 The enthalpy change of a reaction  ΔHf (product) -  ΔHf (reactants)

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For a general reactions:

A+B → C+D

ΔHreaction =  ΔHf ( C + D) -  ΔHf (A + B)

Example: 6.6.1

The ΔHf for the following reaction is – 566 kJ.
2 CO (g) + O2 (g) → 2 CO2 (g)

Given that ΔHf of CO2 = - 394 kJ mol-1, calculate the ΔHf of CO.
Solutions:

Lets the standard enthalpy of formation of CO = x kJ mol-1

2 CO (g) + O2 (g) → 2 CO2 (g)
ΔHºf : 2x 0 (- 394 x 2)

 - 566 = (- 394 x 2 ) – (2x + 0)
x = - 111 kJ mol-1

B) Standard Enthalpy of Combustion (Hc)

 the heat released (evolved) when 1 mole of substance is burnt completely in
excess oxygen at a standard reference state of 1 atm and 25°C.

C(s) + O2(g) → CO2(g) Hc = 393 kJ mol1
C3H8(g) + 5 O2(g) → 3 CO2 (g) + 4 H2O (l) Hc =  2220 kJ mol1

 Hc value is always negative.

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6.3 Calorimetry
In the laboratory, heat changes in physical and chemical processes are measured with
calorimeter.

The equation for calculating the heat change (q) is given by:
q = m c ΔT

Where, q = heat change
m = mass of the sample
ΔT = temperature change

Calorimeter is an apparatus/device used in measuring the heat of chemical reactions or
physical changes as well as heat capacity.
There are 2 types of calorimeter that are commonly used:

 Simple calorimeter or coffee-cup calorimeter (used in neutralization reaction)
 Bomb calorimeter (used in combustion reaction)

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A) Simple Calorimeter

Figure 6.5 Simple calorimeter (coffee-cup)
 By using a simple calorimeter, heat is assumed to be absorb only by the water.
 Simple calorimeter can be used to determine enthalpy of neutralization.
 Heat released by reactions = heat absorbed by water

q = mw cw ΔT

Where, mw = mass of water
cw = specific heat of water
T = temperature change

Example 6.3.1
50.0 mL of 2.0 M HCl was reacted with 50.0 mL of 2.0 M NaOH in a coffee cup calorimeter. The
temperature of the solution increased from 22.0oC to 35.6oC. Determine the enthalpy of
neutralisation.
(Specific heat capacity of solution= 4.184 Jg-1 C-1, d=1.00 g/mL)
Solution:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
ΔT = (35.6 - 22)°C = 13.6°C

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Heat released = heat absorbed by water
q = mwcwΔT
= (100 g)(4.18 Jg-1 °C-1)(13.6°C)
= 5684.8 J
= 5.6848 kJ

So, q = - 5.6848 kJ (heat RELEASED)

n of HCl @ NaOH 50 x 2

= 1000

= 0.1 mol

Enthalphy of neutralization, H q
= n
= - 5.6848 kJ

0.1 mol

= - 56. 4 kJ/mol

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