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Transcript
```Chapter
1
Chapter
Solid State
"A true friend is one soul in two bodies." -Aristotle
Introduction :
The physical state of matter is the interplay of intermolecular forces of attraction like dipoledipole interactions, dipole-induced dipole interactions, London forces, hydrogen bonding etc.
SYLLABUS
1.1 CLASSIFICATION OF SOLIDS
(a) Characteristic properties of solids
(d) Distinguish between crystal lattice and unit
cell
(b) Classification of Solids
(c) Properties of crystalline solids
(e) Types of unit cells
(f) Seven types of unit cells or Fourteen Bravais
(d) Properties of amorphous solids
(e) Anisotropy
lattice
(g) Number of lattice points in one unit cell of
(f) Isotropy
(g) Isomorphous
the crystal structure
(h) Number of atoms in one unit cell of the
(h) Polymorphous
(i) Glass
crystal structure
(i) Solved examples
(j) Distinguish between Crystalline solids &
Amorphous Solids
(k) Classification of crystalline solids
(l) Solid ice is lighter than water
(m) Ionic solids are hard and brittle
(n) Allotropic forms of carbon
Theoretical MCQs
1.3 PACKING IN SOLIDS
(a) Coordination number
Theoretical MCQs
(b) Linear packing in one dimension OR Close
packing in one dimension (Stage - I)
(c) Close packing in two dimensions. (Stage II) OR AAAA type and ABAB type of two
1.2 UNIT CELL & BRAVAIS LATTICES
(a) Crystal lattice
dimensional arrangement. OR Square close
packing and hexagonal close packing in
(b) Lattice point
(c) Unit cell
two dimension
(d) AAAA type of three dimensional packing :
2
(stage III) or Explain simple cubic structure
(SCC)
(e) ABAB type of three dimensional packing
OR Hexagonal close packing in three
dimension
(f) ABCABC type of three dimensional packing
1.5 IMPERFECTIONS OR DEFECTS IN
CRYSTAL STRUCTURE
(a) Defects in crystal structure
(b) Types of defects in the solids or crystal
structures
(c) Point defects
or cubic close packed structure (ccp) in
three dimension
Theoretical MCQs
(g) Distinguish between hexagonal close
packing and cubic close packing.
(h) Distinguish between between tetrahedral
void and octahedral void.
(i) Coordination numbers in the following :
(j) Define
(k) Calculation of packing efficiency in
crystals
(l) Obtain a relation for the density of the crystal
1.6 ELECTRICAL
AND
MAGNETIC
PROPERTIES OF CRYSTALLINE SOLID
(a) Electrical properties of solids
(b) Band theory or origin of electrical properites
in solids
(c) Semiconductors and their types
(d) Uses of semiconductors
(e) Origin of magnetic properties in solids
(f) Diamagnetism
(g) Paramagnetism
Theoretical MCQs
1.4 PACKING EFFICIENCY AND DENSITY OF
UNIT CELL
(h) Ferromagnetism
(i) Antiferromagnetism
(j) Ferrimagnetism
(k) Gouy’s method for determining magnetic
(a) Packing in voids of ionic solids
(b) Radius ratio for ionic compounds
properties of the substances
(c) Packing of ions in cubic closed packed
Theoretical MCQs
(ccp) or hexagonal closed packed (hcp)
structures
(d) Structure of NaCl unit cell (rock salt
structure)
Theoretical MCQs
Unique Solutions ®
HOURS BEFORE EXAM
NUMERICALS WITH SOLUTIONS
NUMERICALS FOR PRACTICE
HIGHER ORDER THINKING SKILLS
S.Y.J.C. Science - Chemistry - Part I
3
1.1
CLASSIFICATION OF SOLIDS :
Concept Explanation :
(a) Characteristic properties of solids :
1. Solid of fixed composition has fixed mass, volume, shape and density.
2. Solid state is generally denser than liquid or gaseous state of the same matter, except ice
(solid H2O) which is lighter than liquid water.
3. Most of the solids are hard, incompressible and rigid. Some of the solids like K, Na, P are
soft. The intermolecular distances are shortest hence, intermolecular forces of attraction
are stronger compared to liquid and gaseous state and hence, solids cannot be compressed
and they are rigid.
4. All solids have characteristic melting points which depend on the extent of intermolecular
forces present in solid state. Stronger the intermolecular forces of attraction, higher is the
melting point.
5. The constituent particles of solids like molecules, atoms or ions have fixed stationary positions
in solid and can only oscillate about their equilibrium or mean positions. Hence, they have
fixed shape and cannot be poured like liquids.
(b) Classification of solids :
Depending on the presence or absence of orderly arrangement of the constituent particles,
the solids are classified into two types :
1. Crystalline solids : A homogeneous solid in which the constituent particles like atoms,
ions or molecules are arranged in a definite repeating pattern throughout the solid is called
crystalline solid. For example, diamond, NaCl, K2SO4, etc.
2. Amorphous solids : A substance which appears like solid but does not have perfectly
ordered crystalline structure of constituent particles is called amorphous solid.
For example : tar, glass, plastics, rubber, butter, etc.
(c) Properties of crystalline solids :
1. A homogeneous solid in which the constituent particles like atoms, ions or molecules are
arranged in a definite repeating pattern throughout the solid is called crystalline solid.
2. A crystalline solid consists of a large number of small crystals called unit cells, each having
a definite characteristic geometrical shape and properties of the crystalline solid.
3. The intermolecular forces of attraction are at maximum.
4. The forces in the crystalline solid may involve ionic bonds, covalent bonds, hydrogen bonds
and van der Waals forces.
5. All pure crystalline solids have sharp melting point.
6. The physical properties like refractive index, electrical conductance etc. of crystalline solids
change with change in directions. The ability of crystalline solids to change values of physical
properties when measured in different directions is called anisotropy.
Chapter - 1 Solid State
4
7.
8.
They have definite enthalpy of fusion and it depends upon arrangement of particles.
When cut with a sharp edged tool, they split into two pieces and new surfaces are plain
and smooth.
(d) Properties of amorphous solids :
1. A substance which appears like solid but does not have perfectly ordered crystalline structure
of constituent particles is called amorphous solid.
2. An amorphous solid does not have regular arrangement of constituent particles.
3. Amorphous solids do not have sharp melting points. With increase in temperature they
gradually soften, become less viscous and melt over a range of temperature.
4. Amorphous solids are called supercooled liquids or pseudo solids.
5. Physical properties do not change with change in directions hence, amorphous solids are
isotropic in nature.
6. Amorphous solids behave like fluids and very slowly float under gravity.
7. They do not have definite enthalpy of fusion.
8. When cut with a sharp edged tool, they split into pieces with irregular surfaces.
(e) Anisotropy :
Definition: The physical properties like refractive
index, electrical conductance etc of crystalline solids
change with change in directions. The ability of
crystalline solids to change values of physical properties
when measured in different directions is called
anisotropy.
Explanation : This property is due to different
arrangement of constituents in different directions.
Example: Refractive Index, electrical conductance,
dielectric constant etc.
(Different arrangements of constituent particles about different directions, AB, CD and EF.)
(f) Isotropy :
Definition: The ability of amorphous solids to exhibit identical physical properties even
though measured in different directions is called isotropy.
(g) Isomorphous :
Definition : When two or more crystalline substances have the same crystal structure,
they are said to be isomorphous and the phenomenon is called as isomorphism. For example,
Cr2O3 and Fe2O3, NaF and MgO.
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(h) Polymorphous :
Definition : A single substance which crystallizes in two or more forms under different
conditions is called polymorphous and this phenomenon is called as polymorphism. For
example, carbon exists as diamond and graphite, or sulphur exists as rhombic and monoclinic
forms.
(i) Glass
1. Glass is an optically transparent material produced by fusing together silicon dioxide (SiO2),
sodium oxide (Na2O), boron oxide (B2O3) and a trace amount of transition metal oxides
to impart colour to the glass.
Types of Glass :
Quartz glass
: Only silicon dioxide, SiO2.
Pyrex glass
: 60-80 % SiO2, 10-25 % B2O3 and remaining amount of Al2O3.
Soda lime glass : 75 % SiO2, 15% Na2O and 10% CaO.
Red glass
: trace amount of gold and copper.
Yellow glass
: UO 2.
Blue glass
: CoO or CuO.
Green glass
: Fe2O3 or CuO.
(j) Distinguish between Crystalline solids and Amorphous solids.
Crystalline Solids
1.
2.
3.
4.
5.
6.
7.
8.
Amorphous Solids
These solids have definite characteristic
shape.
They have sharp melting point.
1.
These solids have irregular shape.
2.
When cut with a sharp edged tool, they
split into two pieces and new surfaces are
plain and smooth.
They have definite enthalpy of fusion.
They are anisotropic.
They are true solids.
3.
They do not have definite melting point.
They gradually softens over a range of
temperature.
When cut with a sharp edged tool, they
cut into two pieces with irregular surfaces.
They have regular arrangement of the
constituent particles. They are said to
exhibit long range order.
Example : Copper, Silver, Iron, Common
salt, Zinc sulphide etc.
7.
4.
5.
6.
8.
They do not have definite enthalpy of fusion.
They are isotropic.
They are pseudo-solids and super-cooled
liquids.
They do not have regular arrangement of
constituent particles. They may have short
range order.
Example : Glass, Rubber, Plastic etc.
Chapter - 1 Solid State
6
(k) Classification of crystalline solids :
Crystalline solids are classified into four main types as follows :
1. Molecular Solids : These are crystalline substances in which the constituent particles are
molecules. The molecules are held together by dispersion forces or London forces, dipoledipole forces or hydrogen bonds. These are further subdivided into :
(a) Polar molecular solids : In such solids, molecules are formed by polar covalent
bonds. For example HCl, NH3(s), solid SO2 etc. In these, molecules are held by
relatively stronger dipole-dipole forces. These solids are soft non-conductor of electricity,
have low melting and boiling points and exists in gaseous state under normal conditions
of temperature and pressure. In polar molecules, there is separation of positive and
negative charges hence, the molecule arrange themselves in such a way that opposite
charges of neighbouring molecules are brought together.
(b) Non-polar molecular solids : Crystalline solids in which the constituent particles are
either atoms like those of noble gases or non-polar molecules. e.g. solid helium, solid
argon, solid CO2, solid hydrogen etc. In these, the constituent particles are held by
weak-weak dispersion or London forces. These are also soft solids having very low
melting point and are non-conductor of electricity. These are liquid or gaseous state
at room temperature and pressure and are volatile.
(c) Hydrogen bonded molecular solids : The molecule of such solids contain hydrogen
bonds between them. For example, consider solid water (ice). The negative end of one
molecule (Oδ-) attracts the positive end of another molecule (Hδ+) forming hydrogen
bond. These solids contain hydrogen bond between H and F, O or N. These exists as
liquids or even gases at normal conditions of temperature and pressure and solidify on
cooling. These solids are non-conductor of electricity and have higher melting and
boiling points compared to non-polar and polar molecular solids.
2. Ionic Solids : These solids consist of positively and negatively charged ions arranged in
definite manner throughout the solid. The constituent ions are held together by strong
electrostatic force of attraction (coulombic forces). The charges on ions and their arrangement
is such that they balance each other and molecule is electrically neutral.
The actual arrangement of ions in ionic solids depends upon :
(i) size of cation and anion, (ii) the charge on the ions, and (iii) the ease with which anions
can be polarized.
The three dimensional arrangement of cations and anions is such that the opposite forces
are completely compensated or neutralized. Hence, these solids are hard and brittle and
possess higher melting point. As the +ve and –ve charges compensated, there is no free
electrons in the solid hence, are non-conductor of electricity. However, when fused (molten)
or dissolved in aqueous solution, the ions get separated and they conduct electricity. The
common examples of ionic solids are NaCl, KCl, KNO3, LiF, Na2SO4 etc.
Unique Solutions ®
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3. Metallic solids : In metallic solids, the constituent particles are metal ions i.e. positive ions
(kernels) immersed in a sea of electrons. The solids are formed by the atoms of same
element which contribute one or more electrons towards sea of mobile electrons.
Since, the electrostatic force of attraction between the positively charged kernels and free
mobile electrons is very strong, metallic bond is a strong bond. The free mobile electrons
are delocalized and are responsible for electrical and thermal conductivity. In metallic solids,
atoms are arranged one over the other in the form of spheres. Under the influence of stress,
these layers slips one over the other and since, electrons are delocalized, they quickly adjust
the new environment without breaking the solid. i.e., they are malleable and ductile.
4. Covalent Solids : Covalent or network solids are those in which the constituents particles
are atoms which are held together by covalent bonds. Depending on the nature of covalent
bond they are hard or brittle, giant molecules and have high melting points. Depending upon
the availability of mobile electrons, they may be either good conductor or insulators. e.g.,
diamond, graphite and fullerene which are allotropic modification of carbon. The other
example of covalent solids are carborundum (silicon carbide), quartz (SiO2), boron nitride
(BN) etc.
(l) Solid ice is lighter than water
1. Ice has hexagonal three dimensional crystal structure formed due to intermolecular hydrogen
bonding which leave almost half the space vacant.
2. The structures of liquid water and solid ice are almost identical.
3. On melting of ice, some hydrogen bonds are broken and vacant spaces are occupied by
water molecules, and they are accommodated more closely.
4. In the solid ice, vacant space is more than in liquid water. Therefore, the density of ice is
less than water.
5. Hence, ice is lighter than water and ice floats on the surface of water.
The structure of ice
Chapter - 1 Solid State
8
(m) Ionic solids are hard and brittle :
1. In ionic crystalline solids, constituent particles are
positively charged cations and negatively charged
anions placed at alternate lattice points.
2. The ions are held by strong coulombic electrostatic
forces of attraction compensating opposite forces.
Hence, they are hard.
3. Since there are no free electrons, they are not malleable and on applying a shearing force,
ionic crystals break into small units. Hence they are brittle.
(n) Allotropic forms of carbon :
Carbon has three allotropic modifications :
1. Diamond
2. Graphite
3. Fullerene
1. Diamond : In diamond, each carbon is sp3 hybridised and is bonded to four other carbon
atom by single covalent bond. Each carbon atom lies at the centre of regular tetrahedron
and the other four carbon atoms are present at the corners of tetrahedron. Hence, there
is three dimensional network of strong covalent bonds.
This makes diamond extremely hard crystal with very high melting point (3843K).
As all the valence electrons of carbon are strongly held in carbon-carbon bonds, it is poor
conductor of electricity.
2. Graphite : In graphite, each carbon atom is sp2 hybridised and is bonded to three other
carbon atoms by single covalent bond while the fourth electron form π-bond with other
carbon atom. Thus, it forms hexagonal rings in two dimension. These array of rings form
layers. The layers are separated by a distance of 340pm. The large distance between these
layers indicate that only weak Van der Waal’s forces exists which hold these layers together,
is responsible for the soft nature of graphite. The electrons forming π-bonds are delocalized
and are relatively free to move under the influence of electric field. Hence, it is good
conductor of electricity.
3. Fullerene : In 1985, new allotrope of carbon was
discovered by passing high power laser on carbon and found
to contain sixty carbon atoms i.e., C60. It had soccer-ball
shape i.e. hollow sphere. Like graphite, each carbon is sp2
hybridised and the delocalized molecular orbits are spread
over the complete structure of Fullerene.
Fullerene react with transition metal to form a catalyst. With
The structure of C60,
potassium, it reacts to form K35C60 which is super-conductor.
Buckminster - fullerene
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Type of Solid Constituent
Particles
Bonding/
Attractive
Forces
Examples
Physical
Nature
Electrical Melting
conductivity Point
1. Molecular solids Molecules
(a) Non polar
(b) Polar
(c) Hydrogen
bonded
Dispersion or
London forces
Dipole-dipole
interactions
Hydrogen
bonding
Soft
Ar, CCl4
H2, I2, CO2
HCl, SO2
Soft
Insulator
Very low
Insulator
Low
H2O (ice)
Insulator
Low
2. Ionic solids
Ions
Coulombic or
electrostatic
NaCl, MgO, Hard but Insulators
ZnS, CaF2 brittle
in solid
state but
conductors
in molten
state and
in aqueous
solutions
3. Metallic solids
Positive
Metallic
ions in a
bonding
sea of
delocalised
electrons
Fe, Cu, Ag, Hard but
Mg
malleable
and
ductile
Conductors Fairly
high
in solid
state as
well as in
molten
state
4. Covalent or
network solids
Atoms
SiO2
(quartz),
SiC, C
(diamond),
AIN,
C(graphite)
Hard
Insulators
Soft
Conductor
(exception)
Covalent
bonding
Chapter - 1 Solid State
Hard
High
Very
high
10
•
*1.
Ans.
Give characteristics of solid state?
The characteristics of solid states are :
(a) They have definite three dimensional arrangement of constituent (atoms / ions / molecules).
(b) They have definite shape, size and volume.
(c) They have definite melting points and boiling points.
(d) They have long range order.
*2.
Ans.
Give the classification of solids.
On the basis of orderly arrangement of constituent particles (atoms, ions or molecules), solids
are classified into two type.
(a) Crystalline solids
(b) Amorphous solids
*3.
Classify the following solids into different types :
(a) Plastic
(b)
(d)
(c) S8 molecule
(e) Tetra phosphorus decoxide
(f)
(g) Brass
(h)
(i) Graphite
(j)
(k) NaCl
(l)
P4 molecule
Iodine molecule
Ammonium phosphate
Rubidium
Diamond
Silicon
Ans.
Types of solids
Non Polar moleculear solids Amorphous
Plastic
I2
Metallic
Brass
Rubidium
Covalent
Graphite
Diamond
Silicon
P4 molecule
S8 molecule
Tetra
Phosphorous
Decoxide
*4.
Ans.
Explain crystalline solids and amorphous solids.
Refer 1.1 (b)
5.
Ans.
Define : (a) Aniosotropy (b) Isotropy (c) Isotropy (d) Polymorphous
Refer 1.1 (e), (f), (g), (h)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
Ionic
Ammonium
phosphate
NaCl
11
*6.
Ans.
*7.
Ans.
*8.
Ans.
*9.
Ans.
What is glass? Distinguish between crystalline solids and amorphous solids? Give examples.
Glass is an optically transparent material produced by fusing together silicon dioxide (SiO2),
sodium oxide (Na2O), boron oxide (B2O3) and a trace amount of transition metal oxides to
impart colour to the glass.
Ans. Refer 1.1 (j)
Explain : (a) Molecular solids
(b) Hydrogen bonded molecular solids
(c) Ionic solids
(d) Metallic solids
(e) Covalent solids
(f) Coordination number
Refer 1.1 (k)
(f) Coordination number : The number of nearest neighbours of any constituent particles
is called its coordination number.
Write a note on : (a) Diamond (b) Graphite (c) Fullerene.
(a) Diamond : In diamond, each carbon is sp3 hybridised and is bonded to four other carbon
atom by single covalent bond. Each carbon atom lies at the centre of regular tetrahedron
and the other four carbon atoms are present at the corners of tetrahedron. Hence, there
is three dimensional network of strong covalent bonds.
This makes diamond extremely hard crystal with very high melting point (3843K). As all
the valence electrons of carbon are strongly held in carbon-carbon bonds, it is poor conductor
of electricity.
(b) Graphite : In graphite, each carbon atom is sp2 hybridised and is bonded to three other
carbon atoms by single covalent bond while the fourth electron form ð-bond with other
carbon atom. Thus, it forms hexagonal rings in two dimension. These array of rings form
layers. The layers are separated by a distance of 340pm. The large distance between these
layers indicate that only weak Van der Waal’s forces exists which hold these layers together,
is responsible for the soft nature of graphite. The electrons forming ð-bonds are delocalized
and are relatively free to move under the influence of electric field. Hence, it is good
conductor of electricity.
(c) Fullerene : In 1985, new allotrope of carbon was discovered by passing high power laser
on carbon and found to contain sixty carbon atoms i.e., C60. It had soccer-ball shape i.e.
hollow sphere. Like graphite, each carbon is sp2 hybridised and the delocalized molecular
orbits are spread over the complete structure of Fullerene. Fullerene react with transition
metal to form a catalyst. With potassium, it reacts to form K35C60 which is super-conductor.
[Refer fig. 1.1 (n) - 3]
Explain why : (a) Ionic solids are hard and brittle, (b) Solid ice is lighter than water.
(a) [Refer 1.1 (l)]
(b) [Refer 1.1 (m)]
Chapter - 1 Solid State
12
10.
Ans.
Classify the following solids as ionic, covalent, metallic, molecular or amorphous solid.
(a) Tetraphosphorous decoxide
(b) Graphite
(c) SiC
(d) Rubidium
(f) Ammonium phosphate
(e) I 2
(g) Brass
(h) Si
(j) Solid CO2
(i) P 4
(k) LiBr
Metallic
: Rubidium
Ionic
: LiBr, (NH4)2SO4
Covalent or Network : Graphite, SiC, Si
Molecular : P4O10, I2, Solid CO2
Amorphous : Nil
11.
Ans.
Write a feature which will distinguish a metalic solid from ionic solid.
Metallic solids are good conductor of heat and electricity while ionic solids are insulator in solid
state but are conductor in molten and aqueous solution.
12.
Ans.
Write a distinguishing feature of metallic solid.
Metallic solids are good conductor of heat and electricity.
13.
Ans.
What is difference between glass and quartz while both are made from SiO4 tetrahedral?
Glass is amorphous while Quartz is crystalline solid.
14.
Ans.
Some of very old glass objects appear slightly milky instead of being transparent. Why?
This is because of some crystallisation in that region.
15.
Ans.
Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely
high temperature. What type of solid is it?
Network or Covalent solid.
16.
Ans.
Ionic solid conduct electricity in molten state but not in solid state. Explain.
In Ionic solids, the electrostatic force of attraction between oppositely charged ions are strong.
Hence, ions are not available for conductance. However, in molten or in aqueous solution, ions
are available for conductance.
17.
Ans.
What type of solids are electrical conductors, malleable?
Metallic.
19.
Ans.
Classify the following as amorphous or crystalline solid.
(a) Naphthalene (b) Teflon
(c) Benzoic acid
(d) Potassium nitrate
(e) Cellophane
(f) Fibre glass
(g) Copper
(h) Zinc sulphide
Amorphous solids : (b) Teflon (e) Cellophane (f) Fibre glass
Crystalline solids : (a) Naphthalene (c) Benzoic acid (d) Potassium nitrate (g) Copper
(h) Zinc sulphide.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
13
Multiple Choice Questions :
•
1.
Theoretical MCQs :
The solids in which the magnitude of physical properties changes with the direction of measurement
are .......
a) amorphous
b) crystalline
c) colloidal
d) metalloid
In solids the constituents particles may be .......
a) atoms
b) ions
c) molecules
d) any one of the above three
A single substance that exists in two or more forms is called .......
a) polymorphous
b) amorphous
c) isomorphous
d) mono-morphous
The force that hold kernels together in the crystal is called .......
a) ionic bond
b) covalent bond
c) metallic bond d) hydrogen bond
Graphite is a .......
a) molecular crystal
b) covalent crystal c) ionic crystal
d) metallic crystal
Diamond is a .......
a) molecular crystal
b) covalent crystal c) ionic crystal
d) metallic crystal
The major binding force of diamond is .......
a) ionic bond
b) covalent bond
c) dipole-dipole attraction
d) induced dipole-dipole attraction
The major binding force in silicon is .......
a) ionic bond
b) covalent bond
c) dipole-dipole attraction
d) induced dipole-dipole attraction
*2.
*3.
4.
*5.
*6.
*7.
*8.
•
9.
10.
*11.
12.
The solids which are brittle and hard are .......
a) ionic solid
b) molecular solid
c) covalent solid
d) metallic solid
Pyrex glass is obtained by fusing together .......
b) SiO2, Na2CO3 and CaCO3
a) SiO2, MgO and CuO
c) SiO2, B2O3 and Al2O3 d) SiO2 and Na2SiO3
The major binding force in graphite is .......
a) ionic bond
b) covalent bond c) hydrogen bond d) London force
In which of the following pairs both the solids belong to same type?
a) solid CO2, ZnS
b) CaF2, Ca
c) graphite, ice
d) SiC, AlN
Chapter - 1 Solid State
14
1.2
UNIT CELL AND BRAVAIS LATTICES :
Concept Explanation :
(a) Crystal lattice:
Definition : A regular arrangement of the constituent particles (atoms, ions or molecules)
of a crystalline solid having similar environment in three dimensional space is called crystal
lattice or space lattice.
(b) Lattice point :
Definition : A position occupied by a crystal constituent particle like an atom, ion or a
molecule in the crystal lattice is called lattice point or lattice site.
(c) Unit cell :
1. Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice)
which when repeated in different directions produces the crystal line solid (lattice).
2. The crystal is considered to consist of an infinite number of unit cells.
3. The unit cell possesses all the characteristics of the crystalline solid.
4. A unit cell is characterised by following parameters :
(a) Edges or edge lengths : The intersection of two
faces of crystal lattice is called as edge. The three
edges denoted by a, b and c represent the dimensions
(lengths) of the unit cell along three axes. These edges
may or may not be mutually perpendicular.
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(b) Angles between the edges (or planes) : There are three angles between the edges
of the unit cell represented as α, β and γ.
(i) The angle α is between edges b and c.
(ii) The angle β is between edges a and c.
(iii) The angle γ is between edges a and b.
The crystal is defined with the help of these parameters of its unit cell.
(d) Distinguish between crystal lattice and unit cell :
Crystal lattice
1.
2.
3.
4.
5.
Unit cell
It is a regular arrangement of constituent
particles of a crystalline solids in three
dimensional space.
It is made up of a large number of unit
cells.
1.
Crystal lattice is defined in terms of
properties of unit cell.
Crystal lattice can be prepared, handled
and studied experimentally.
3.
It is a macroscopic system.
5.
2.
4.
It is the smallest repeating structural unit
which when repeated in space in all
dimensions generates a crystal lattice.
It is one fundamental unit of crystal lattice
which possesses all the properties of the
crystal.
Unit cell describes the fundamental
properties of the crystal lattice.
Unit cell is a hypothetical object which
is the smallest possible imaginary part of
the crystal lattice and cannot be handled
and studied individually.
It is a microscopic molecular size system.
(e) Types of unit cells :
Basically unit cells are of two types as follows :
1. Primitive unit cells : The unit cells in which the constituent particles like atoms, ions or
molecules are present only at the corners of the unit cell are called primitive unit cells or
simple unit cells.
2. Non-primitive or centered unit cells : The unit cells in which the constituent particles
are present at corners as well as at some other positions in the unit cell are called nonprimitive or centred unit cells. The centred unit cells are of three types :
(a) Body centred unit cell : A unit cell in which the constituent particles are present at
the corners as well as at its body centre is called body centred unit cell.
(b) Face centred unit cell : A unit cell in which the constituent particles are present at
the corners as well as at the centre of each face is called face centred unit cell or cubic
close packed (CCP) unit cell.
(c) End centred unit cell : A unit cell in which the constituent particles are present at the
corners as well as at the centres of two opposite faces is called end centred unit cell.
Chapter - 1 Solid State
16
(f) Seven types of unit cells Or Fourteen Bravais lattices
1. Simple or primitive : It has points at all
the corners of the unit cell i.e. it consist
of one atom at all the corners of cube
and has all edges (sides) equal i.e. a = b
= c and α = β = γ = 90º.
2. Body centred cube : In this type of crystal, in
addition to the lattice point present at the corners,
one atom (lattice point) is present at the body
centre of the cube. The edge lengths are a = b
= c and α = β = γ = 90º; e.g. Na, Rb, Fe, Ti, W,
U, Zr, etc.
3. Face centred cube : In this type of crystal, there
are points at all the corners as well as at the centre
of each face. The edge lengths are a = b = c and
α = β = γ = 90º; e.g. Cu, Al, Ni, etc.
Unique Solutions ®
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17
(a) Tetragonal lattice :
It is of two types. In this type of unit cell, two edges
are same while the third edge is different
(i) Simple or primitive tetragonal : The
lattice points are present at the corners of
tetragon. The edge lengths are a = b ≠ c
and α = β = γ = 90º; e.g. SnO2
(ii) Body centred tetragonal : In this type
of unit cell, eight points are present at the
corners of tetragon and one point is present
at the centre. The edge lengths are a = b
≠ c and α = β = γ = 90º; e.g. TiO2, CaSO4
etc.
(b) Orthorhombic lattice : In orthorhombic unit cell, all edges are different i.e. a ≠ b
≠ c and all angles are same i.e. α = β = γ = 90º. It is further classified as :
(a) simple or primitive (b) body centred orthorhombic (c) face centred and (d)
end-centred orthorhombic unit cell.
3. Monoclinic lattice : In monoclinic
unit cell, all edges are different, a
≠ b ≠ c and α = β = 90º but γ ≠ 90º
i.e. two of the angles are same but
the third angle is different. It is also
of two types :
(a) primitive monoclinic and
(b) end-centred monoclinic
Chapter - 1 Solid State
18
4. Triclinic lattice : In this, all edges as well
as angles are different i.e. a ≠ b ≠ c and α
≠ β ≠ γ ≠ 90º
5. Hexagonal lattice : This type of unit cell
has two same edge length i.e.
a = b but the third edge has different length.
Also two angles are same i.e.
α = β = 90º but γ = 120º.
6. Rhombohedral lattice : In this type of unit
cell, all edges have same length i.e. a = b =
c. Two axial angles are of 90º but the third
angle is more than 90º i.e. α = β ≠ γ
CRYSTAL SYSTEMS AND BRAVAIS LATTICES
Crystal System
Cubic
Tetragonal
Orthorhombic
Monoclinic
Triclinic
Hexagonal
Rhombohedral
Unique Solutions ®
Types of Lattices
Simple, Body centred, Face centred
Simple, Body cent red
Simple, Body centred, Face centred, End centred
Simple, End centred
Simple
Simple
Simple
S.Y.J.C. Science - Chemistry - Part I
19
Types of unit cells their edge lengths, angles and examples.
No.
Crystal
system
Type
Edge
length
Angle
Example
1. Cubic
Simple/primitive a = b = c α = β = γ = 90º
2. Cubic
Body centred
a = b = c α = β = γ = 90º
Fe, Rb, Na, Ti, W, U, Zr
3. Cubic
Face centred
a = b = c α = β = γ = 90º
Cu, Al, Ni, Au, Ag, Pt
4. Tetragonal
Primitive
a = b ≠ c α = β = γ = 90º
SnO2
5. Tetragonal
Body centred
a = b ≠ c α = β = γ = 90º
TiO2, CaSO4
6. Orthorhombic Primitive
a ≠ b ≠ c α = β = γ = 90º
Rhombic sulphur
7. Orthorhombic Body centred
a ≠ b ≠ c α = β = γ = 90º
KNO 3
8. Orthorhombic Face centred
a ≠ b ≠ c α = β = γ = 90º
BaSO 4
9. Orthorhombic End centred
a ≠ b ≠ c α = β = γ = 90º
MgSO4; 7H2O
10. Monoclinic
Primitive
a ≠ b ≠ c α = β = 90º, γ ≠ 90º Monoclinic sulphur
11. Monoclinic
End centred
a ≠ b ≠ c α = β = 90º, γ ≠ 90º Na2SO4; 10H2O
12. Triclinic
Primitive
a ≠ b ≠ c α ≠ β ≠ γ ≠ 90º
13. Hexagonal
Primitive
a = b ≠ c α = β =90º, γ = 120º ZnO, BeO, CoS, SnS
14. Rhombohedral Primitive
a = b = c α = β = γ ≠ 90º
Polonium
K2Cr2O7, H3BO3
Calcite, NaNO3, FeCO3
(g) Number of lattice points in one unit cell of the crystal structures.
1. Simple cubic
2. Face centred cubic
3. Body centred cubic
4. Face centred tetragonal.
Lattice points in one unit cell represent the positions of atoms, ions or molecules in the unit cell.
1. Simple cubic unit cell : In this primitive unit cell, the lattice points are at 8 comers of the
unit cell. Hence there are 8 lattice points.
2. Face centred cubic unit cell : In this unit cell, the lattice points are at 8 corners and 6 face
centres.
(In cubic close packing unit cell, the lattice points are also at edge centres and body centre.)
3. Body centred cubic unit cell : In this, the lattice points are at 8 corners and one at body
centre.
4. Face centred tetragonal cubic unit cell : In this, the lattice points are at 8 corners and 6 face
centres.
Chapter - 1 Solid State
20
(h)
1.
2.
3.
4.
Number of atoms in one unit cell of the crystal structures :
Simple cubic
Body centred cubic
Face centred cubic
Hexagonal unit cell
1.
Number of atoms in simple cubic (scc) crystal :
In simple or primitive cubic unit cell, there are 8 atoms at 8
corners. Each corner contributes l/8th atom to the unit cell.
∴ Number of atoms present in the unit cell =
1
×8=1
8
Hence, the volume of the unit cell is equal to the volume of one
atom.
2. Number of atoms in body centred cubic (bcc) crystal :
In this unit cell, there are 8 atoms at 8 corners and one additional
atom at the body centre. Each corner contributes l/8th atom, to
the unit cell, hence, due to 8 corners.
Number of atoms = 8 ×
1
= 1 atom.
8
An atom at the body centre wholly belongs to the unit cell.
∴ Total number of atoms present in bcc unit cell = 1 + 1= 2.
Hence the volume of unit cell is equal to the volume of two atoms.
3. Number of atoms in face centred cubic (fcc cubic crystal) :
In this unit cell, there are 8 atoms at 8 corners and 6 atoms at
6 face centres. Each corner contributes l/8th atom to the unit
cell, hence, due to 8 corners,
Number of atoms =
1
×8=1
8
Each face centre contributes half of the atom to the unit cell, hence, due to 6 face centres,
∴ Number of atoms =
1
×6=3
2
Total number of atoms present in fee unit cell = 1+ 3 = 4.
Hence, the volume of the unit cell is equal to the volume of four atoms.
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S.Y.J.C. Science - Chemistry - Part I
21
4. Hexagonal unit cell : In this unit cell, 12 atoms are present at 12 corners and 2 atoms
are present at the centres of two opposite hexagonal faces.
Each corner, contributes l/6th atom to the unit cell since it is common for 6 unit cells, hence,
due to 12 corners.
1
× 12 = 2
Number of atoms =
6
Each face centre contributes half atom to the unit cell, hence, due to 2 face centres,
Number of atoms = 1/2 × 2 = 1.
Total number of atoms in hexagonal unit cell = 2+1=3.
(i) Solved Examples :
1.
Sol.
∴
2.
Sol.
∴
∴
∴
∴
3.
Sol.
A compound formed by element A and B, crystallises in the cubic arrangement in which A atoms
are at the corners of a cube and B atoms are at the face centres. What is the formula of
compound?
1
The number of A atoms in the unit cell of cube = 8 × = 1
8
B atoms are at the centre of face and each face is shared by two cubes. The number of B atoms
1
are = 6 × = 3.
2
The formula of compound is AB3
If three elements P, Q and R crystallises in a cubic solid lattice with P atoms at the corners,
Q atoms at the cube centres and R atoms at the centre of edges, then write the formula of the
compound.
As P atoms are present at the corners of cube
1
no. of P atoms in unit cell = 8 × = 1
8
Q atoms are present at the cube centres
no. of Q atoms in unit cell = 1
R atoms are present at the edges. There are 12 edges and atom at each edge is shared by four
atoms.
1
no. of R atoms in unit cell = 12 × = 3
4
Formula of compound is PQR3
Sodium crystallises in bcc unit cell. Calculate the approximate number of unit cell in 9.2g of
sodium (atomic mass of Na = 23)
9.2 × 6.022 × 10 23
= 2.4088 × 1023 atoms
No. of atoms present in 9.2g of sodium =
23
Chapter - 1 Solid State
22
A bcc unit cell contains 2 atoms
2.4088 × 10 23
= 1.2044 × 1023
2
∴
Number of unit cells =
4.
Silver crystallises with fcc unit cell. Each side of the unit cell has a length of 400 pm. Calculate
If a is the edge length of unit cell, then fcc diagonal = 2 a
fcc diagonal is also equals to 4r
4r = 2 a
1.414 × 400
2a
r=
=
= 141.4 pm
4
4
Sol.
∴
5.
Sol.
•
1.
Ans.
2.
Ans.
3.
Ans.
4.
Ans.
5.
Ans.
Gold crystallises in fcc unit cell (atomic radius = 0.144 nm). What is the length of the side of
the cell?
a
For fcc unit cell, radius of atom, r =
2 2
a = r ⋅ 2 2 = 0.144 × 2 × 1.414 = 0.407 nm
What is a primitive cell?
A primitive cell has particles (atoms) at all the corners of the unit cell.
Name the crystal system in which all the three axes are of equal length which are inclined at
the same angle but the angle is not equal to 90º.
Rhombohedral.
What is meant by the term coordination number? What is the coordination number of atoms in
(a) a cubic close packing and (b) body centred cubic close packing.
It is number of nearest neighbours with which a given sphere is in contact.
(a) 12 (b) 8
How many atoms can be assigned to it’s unit cell, if an element forms :
(a) body centred cubic cell and (b) face centred cubic cell?
(a) 2 (b) 4
If a is the edge length of a body centred cubic structure and r is the radius of the atom, then
how are these two related?
For a body centred cubic structure, edge length and radius is related as : r =
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
3
a.
4
23
6.
Ans.
How are unit cell and space lattice related?
Space lattice is obtained by repeating the unit cell in three dimensions. The spatial arrangement,
stoichiometry and density of unit cell and space lattice are equivalent.
7.
Ans.
What is the total number of atoms present per unit cell in a face centred cubic structure?
The total number of atoms present is 4.
8.
Ans.
∴
9.
Ans.
∴
∴
10.
Ans.
∴
11.
Ans.
If three elements X, Y and Z crystallises in a cubic solid with X atoms at the corners, Y atoms
at the cube center and Z atoms at the faces of cube, then write the formula of the compound.
1
Atom X per unit cell = 8 × = 1
8
Atom Y per unit cell = 1
1
Atom Z per unit cell = 6 × = 3
2
Formula of the compound = XYZ3
In a solid AB ‘A’ atoms have cubic close packing arrangement and B atoms occupy all the
octanedral sites. If all the face centred atoms are along one of the axis are removed, then what
will be the resultant stoichiometry of the compound?
In cubic close packing structure, there are 8A at the corners of the cube and 6A atoms on the
face centres. If all the face centred atoms along one of the axis are removed, it means removal
of 2A atoms. Therefore, only 4A atoms will be left on faces.
1
1
No. of A atoms per unit cell = 8 × + 4 × = 3
8
2
No. of B atoms in ccp structure are 12 at edge corner and one at body centre. Therefore, no.
1
of B atoms per unit cell = 12 × + 1 = 4
4
stoichiometry of compound = A3B4
Sodium crystallizes in bcc unit cell. Calculate the approximate number of unit cells in 9.2g of
sodium (At mass = 23)
(CBSE sample paper 2011)
23
6.022 × 10 × 9.2
= 2.4088 × 1023 atoms
No. of atoms in 9.2 g of sodium
=
23
A bcc unit cell contains 2 atoms,
2.4088 × 10 23
= 1.2044 × 1023
No. of unit cells present
=
2
Explain : Coordination number.
Coordination number can be 4, 6, 8 or 12. Higher coordination number indicate that sphere are
compactly arranged, space between them is less and are denser. For example in primitive or
simple cubic unit cell, every sphere is in contact with four sphere i.e. a single sphere is surrounded
Chapter - 1 Solid State
24
by total number of six spheres. Thus, it’s coordination number is six. Similarly, the coordination
number of body centred cube is eight while coordination number of face centred cube is twelve.
*12.
Ans.
*13.
Ans.
*14.
Ans.
What is unit cell? Explain Bravais Lattice.
Unit cell is smallest part (position) which when repeated in different direction generates the
entire lattice. Bravais has proved that crystals do not have simple lattice i.e. having lattice point
only at the corners. They can be arranged in maximum of fourteen types. The arrangement is
called Bravais Lattice.
Explain with the help of a diagram.
(a) Seven types of unit cell.
(c) Two types of tetragonal unit cells.
(e) Two types of monoclinic unit cell.
(g) Primitive hexagonal unit cell.
Refer 1.2 (f)
(b) Three types of cubic cells.
(d) Four types of orthorhombic unit cells.
(f) Triclinic unit cell
Give the number of lattice points in one unit cell of the crystal structures.
(a) Simple cubic.
(b) Face centred cubic.
(c) Body centred cubic
(d) Face centred tetragonal.
(a) No. of lattice point in simple cube : In simple cube, the atoms are present at the corners
of cube. Atoms present at the corner contribute 1/8 to each cube because it is spared by
8 cube.
∴ The no. of atoms present in each unit cell (lattice point)
1
= 8 corners atoms × atoms per unit cell
8
= 1 atom
(b) Face centred cube : It has points at all the corners as well as at the centre of each of six
faces which is shared by two unit cells. Therefore, contribution of each of atom at the face
per unit cell is 1/2.
Thus, No. of atoms present at corners per unit cell
1
= 8 corners atoms × atoms per unit cell
8
= 1 atom
No. of atoms present at faces per unit cell
1
atoms per unit cell
= 6 atoms at the faces ×
8
= 3 atoms
∴ Total no. of atoms per unit cell = 1 + 3 = 4 atoms.
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(c) Body centred cube : It has points at all the corners as well as body centre of the cube. The
points (atoms) present at the corner is shared by 8 other cube. Thus, the contribution of
each corner atoms is only 1/8th.
Thus, the no. of atoms present at the corner per unit cell
1
= 8 corners atoms × atoms per unit cell
8
= 1+1
= 2 atoms
15.
Ans.
•
1.
2.
3.
4.
5.
•
6.
7.
Distinguish between Crystal Lattice and Unit Cell.
Ref. 1.2 (d)
Multiple Choice Questions :
Theoretical MCQs :
The crystal system with unit cell dimension, a = b ≠ c and α = β = 90º and γ = 120º has .......
structure.
a) rhombohedrals
b) hexagonal
c) cubic
d) tetragonal
In the face centred unit cell, the points are present at the .......
a) corners and face centres of the unit cell b) corners of the unit cell
c) face centres of the unit cell
d) corners and centres of the unit cell
The number of atoms in body centred cube is .......
a) 1
b) 2
c) 4
d) 8
The number of atoms present in a face-centred cube is .......
a) 1
b) 2
c) 3
d) 4
The most unsymmetrical and symmetrical systems are respectively .......
a) tetragonal, cubic
b) triclinic, cubic
c) rhombohedral, hexagonal
d) orthorhombic, cubic
The relation a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90º belong to the crystal system .......
a) triclinic
b) monoclinic
c) rhombic
d) cubic
If the edge length in a crystal is a = b ≠ c and the angle between them is α = β = γ = 90º, what
type of crystal lattice is?
a) monoclinic
b) cubic
c) orthorhombic
d) tetragonal
Chapter - 1 Solid State
26
1.3
PACKING IN SOLIDS :
Concept Explanation :
(a) Coordination number :
The coordination number of constituent particle of the crystal lattice is the number of
particles surrounding a single particle in the crystal lattice.
More the coordination number more tightly the particles are packed in the crystal lattice.
(b) Linear packing in one dimension OR close packing in one dimension:
(Stage - I)
The constituent particles of the
crystalline
solid
are
represented by spheres of
equal size.
In one dimensional arrangement, the spheres are arranged in a tow, touching each other.
Each sphere is in contact with two of its neighbours. Thus the coordination number is 2.
(c) Close packing in two dimensions. (Stage - II)
OR
AAAA type and ABAB type of two dimensional arrangement.
OR
Square close packing and hexagonal close packing in two dimension
Two dimensional close packing crystal structure can be generated by placing one dimensional
linear crystal structure over another to form multiple layers. This staking of linear rows
may be taking place in two different ways giving two different two dimensional structures
as follows :
1. AAAA type two dimensional close packing or square close packing :
(a) Different one dimensional
identical rows are placed one
over the other.
(b) In this arrangement each
sphere in a row is placed
over another sphere of
another row.
(c) In this, spheres have
horizontal as well as vertical
alignment. All the rows of
spheres are identical in
planar structure.
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(d) All crests of spheres as well as all the depressions formed by the arrangement are
aligned.
(e) If the first row is labelled as A type, then second and all subsequent rows are also
identical, hence are of A type. Therefore, this planar two dimensional close packing
is called AAAA…… type packing.
(f) In this arrangement, each sphere is in contact with four other spheres around it. Hence,
the coordination number of each sphere is four and the packing is called two dimensional
or planar square close packing.
(g) In this, packing efficiency is 52.4 %. In this comparatively empty space is more. The
shape of void is square.
2. ABAB type two dimensional packing or hexagonal close packing :
(a) In this arrangement, the
crests of the spheres of
one row are placed into
the depressions or
troughs
formed
spheres of next row.
(b) This gives closest type
packing of spheres and
reduces the unoccupied
or
void
space
considerably.
(c) In this arrangement, the positions of spheres of neighbouring rows do not match each
other since crests of spheres of one row are in contact with depressions or troughs
of next row.
(d) If one row of spheres is labelled as A then the next row will be B, third row will again
be A, fourth row B and so on. Hence this planar or two dimensional close packing is
called ABAB... type packing.
(e) In this arrangement, each sphere is in contact with or touching six other spheres around
it shown by a hexagon obtained by joining the centres of six spheres.
(f) Hence, the coordination number of each sphere is six and the packing is called two
dimensional or planar hexagonal close packing.
(g) In this, the packing efficiency is 60.4% which is more than linear close packing.
(h) The size of hole is very small and has triangular shape.
Chapter - 1 Solid State
28
(d) AAAA type of three dimensional packing: (stage - III) OR Explain simple
cubic structure (SCC) :
All the real structures of crystalline solids are three dimensional structures obtained by
staking one above the other two dimensional AAAA type square close packed crystalline
structures or two dimensional ABAB type hexagonal close packed structures.
1. Three dimensional close packing from two dimensional AAAA type square close
packed layers :
(a) In this arrangement, AAAA
type of two dimensional crystal
layers are placed over the
other such that all the spheres
of the successive layers are
exactly above the spheres of
the lower layers.
(b) All spheres of different layers
are perfectly aligned
horizontally and vertically
forming unit cells having
primitive or simple cubic
structure.
(c) Since all the layers are identical and if each layer
is labelled as layer A, then whole three dimensional
crystal lattice will be of AAAA... type.
(d) Each sphere is in contact with six surrounded
spheres, four being in the same plane, one in the
above layer and one in the below layer. Hence the
coordination number of each sphere is six.
(e) This arrangement is called AAAA type or Simple
cubic structure.
(f) In this arrangement the packing efficiency is 52.4% and empty or void space is 47.6%.
(e) ABAB type of three dimensional packing OR Hexagonal close packing
in three dimension :
1. Placing second layer over the first layer :
(a) Consider a two dimensional hexagonal close packed layer 'A' with triangular voids 'a'
and 'b'.
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(b) Place a similar layer above it such that the spheres of the second layer are placed in
the depressions of the first layer.
(c) Since the spheres are aligned differently let us call the second layer as 'B'.
(d) It is observed that not all the triangular voids of the first layer are covered. In this
arrangement, if triangular voids of the type 'b' are occupied then 'a' type are unoccupied,
i.e., only half of the voids of first layer are occupied.
(e) When a triangular void of the first layer is covered by a sphere of the second layer
(or vice versa), tetrahedral void is formed.
(f) The remaining half of the triangular voids of the first layer are covered by the triangular
voids of the second layer to form a octahedral void.
Chapter - 1 Solid State
30
2. Placing third layer over the second layer :
(a) Tetrahedral voids of the second layer may
be covered by the spheres of the third layer.
(b) In this case, the spheres of the third layer
are exactly aligned with those of the first layer.
(c) Thus the pattern of spheres is repeated in
alternate layers.
(d) This pattern is called as ABAB and this
structure is called hexagonal close packed
(hcp).
(e) In this, packing efficiency is 74%. The
coordination number of each sphere is 12.
(each sphere has 6 neighboring spheres in
its own layer, 3 spheres in the layer above
and 3 spheres in the layer below it.).
(f) ABCABC type of three dimensional packing OR Cubic close packed
structure (ccp) in three dimension :
1. Placing second layer over the first layer:
(a) Consider a two dimensional hexagonal close packed layer 'A' with triangular voids 'a'
and 'b'.
(b) Place a similar layer above it such that the spheres of the second layer are placed in
the depressions of the first layer.
(c) Since the spheres are aligned differently let us call the second layer as 'B'.
(d) It is observed that not all the triangular voids of the first layer are covered. In this
arrangement, if triangular voids of the type 'b' are occupied then 'a' type are unoccupied,
i.e., only half of the voids of first layer are occupied.
(e) When a triangular void of the first layer is covered by a sphere of the second layer
(or vice versa), a tetrahedral void is formed.
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31
(f)
The remaining half of the triangular voids of the first layer are covered by the triangular
voids of the second layer to form a octahedral void.
2. Placing third layer over the second layer :
(a) The third layer may be placed
in such a way that its spheres
cover the octahedral voids.
(b) In this case, the spheres of the
third layer are not aligned with
those of either the first or the
second layer.
(c) Let us call the third layer as
C.
(d) This pattern is called as
ABCABC and this structure is
called Cubic close packed
(ccp).
(e) This arrangement is also called
face-centered cubic structure.
(f) In this, packing efficiency is
Close cubic packing (ccp)
74%. The coordination number
of each sphere is 12.
(g) Distinguish between hexagonal close packing and cubic close packing.
Hexagonal close packing
1.
In this three dimensional building of unit cell,
spheres of third layer are placed on
tetrahedral voids of the second layer.
Cubic close packing
1.
2.
2.
3.
4.
5.
In this, the spheres of third layer lie directly
above the spheres of first layer.
In this first and third layers are identical.
In this first and fourth layers are different.
The arrangement of packing is ABAB type.
3.
4.
5.
In this three dimensional building of unit cell,
the third layer is placed over octahedral voids
created by the first two layers.
In this, the spheres of third layer do not lie
above the spheres of first layer.
In this first and third layers are different.
In this first and fourth layers are identical.
The arrangement of packing is ABCABC
type.
Chapter - 1 Solid State
32
(h) Distinguish between tetrahedral void and octahedral void.
Tetrahedral void
1.
2.
3.
4.
5.
6.
Octahedral void
This vacant space or void is surrounded
by four atomic spheres.
The atom in tetrahedral void is in contact
with four atoms placed at four corners of
a tetrahedron.
This void is formed when a triangular
void made by three coplanar spheres, is
in contact with fourth sphere above or
below it.
The coordination number of a tetrahedral
void is 4.
There are two tetrahedral voids per sphere
or atom.
The tetrahedral voids are present on the
body diagonals of the unit cell.
(i)
1.
2.
3.
1.
2.
3.
4.
5.
6.
This vacant space or void is surrounded
by six atomic spheres.
The atom in octahedral void is in contact
with six atoms placed at six corners of an
octahedron.
Octahedral void is formed by two sets of
equilateral triangles pointing in opposite
directions with six spheres.
The coordination number of an octahedral
void is 6.
There is one octahedral void per sphere
or atom.
The octahedral voids are present on the
edges and 1 at the body centre of the
unit cell.
Coordination numbers in the following :
At the corner of scc
Body centred cubic (bcc) structure
Face centred cubic (fcc) structure
1. At the corner of scc : If we consider an atom at the
comer of simple cubic structure (scc) structure then it
is in contact with six identical equidistant spheres,
namely three from three corners of the unit cells (1, 2,
3) and three of neighbouring unit cells, (4, 5, 6). Hence
coordination number of the sphere is 6.
2. Body centred cubic (bcc) structure : In this unit
cell, 8 eight atoms are present at eight corners while
one additional atom is at the body centre. The atom at
the body centre is in contact with eight identical and
equidistant atoms. Hence its coordination number
is 8.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
33
3. Face centred cubic (fcc) structure : In this unit cell, eight atoms are present at eight
corners and six atoms are present at six face centres. Each sphere in the structure is in
contact with surrounding twelve equidistant neighbouring spheres, six in the same layer,
three in the layer above and three in the layer below. Hence the coordination number of
an atom or sphere is 12.
(j) Definition :
1. Packing fraction : The fraction of the total space in unit cell occupied by the constituent
particles is called packing fraction.
Packing fraction =
Volume occupied by all constituent particles
Volume of the unit cell
2. Packing efficiency : It is the percentage of total space in the unit cell occupied by the
constituent particles like atoms, ions or molecules of the crystal.
Packing efficient
=
Volume occupied by all constituent particles
× 100
Volume of the unit cell
= Packing fraction × 100
3. Void space in the crystal : A vacant space or a space not occupied by the constituent
particles in the unit cell is called a void space. It is given by,
Percentage of void space = 100 packing efficiency
OR
Fraction of void space
= 1 Packing fraction.
(k)
1.
2.
3.
Calculation of packing efficiency in following crystals :
Simple cubic (scc) structure
Body centred cubic (bcc) structure
Face centred cubic (fcc) structure OR Cubic close packing (ccp) structure.
1. Packing efficiency in simple cubic (scc) crystal : In this primitive
unit cell, 8 atoms of element are present at 8 corners. Two atoms
are in contact along the edge AB of length a of the unit cell.
If r is the radius of the atom, then AB = a = 2r
a
∴ r=
2
Since each corner atom contributes l/8th of the atom.
Chapter - 1 Solid State
34
Total number of atoms in unit cell =
1
× 8 = 1 atom
8
Volume of unit cell = a3 = (2r)3 = 8r3
4 3
Volume of 1 atom = πr
3
Volume occupied by 1 atom
Packing fraction =
Volume of unit cell
4/3πr 3
=
8r 3
π
=
6
= 0.524
∴ Packing efficiency = 0.524 × 100 = 52.4 %
Percentage of void space = 100 – 52.4 = 47.6 %
2. Body centred cubic (bcc) structure : In this unit cell, there are 8 atoms at 8 corners
and one atom at the body centre.
1
∴ Total number of atoms = 8 × + 1 = 1 + 1 = 2
8
The atoms are in contact along the body diagonal BF. Let a be the edge length and r the
Consider a triangle BCE.
BE2 = BC2 + CE2 = a2 + a2 = 2a2
Consider triangle BEF.
F
r
BF2 = BE2 + EF2
= 2a2 + a2
A
r D
= 3a2
a
r
3a
BF =
E
From figure, BF = 4r
3a
∴ 4r
=
3
∴ R
=
a
4
r
B
3
Volume of unit cell = a3 =
Volume of 1 atom =
4 3
πr
3
Unique Solutions ®
⎛ 4
⎞
64 3
× r⎟
⎜
r
=
3 3
⎜ 3
⎟
⎝
⎠
a
bcc structure
S.Y.J.C. Science - Chemistry - Part I
C
35
Packing fraction =
=
=
=
Volume occupied by 2 atom
Volume of unit cell
4
2 × πr 3
3
64 3
r
3 3
3π
8
3 × 3.142
8
= 0.68
∴ Packing efficiency = 0.68 × 100 = 68 %
∴ % of void space = 100 – 68 = 32 %
3. Face centred cubic (fcc) structure : In this unit cell, there are 8 atoms at 8 corners and
6 atoms at 6 face centres.
1
Number of atoms due to 8 corners = 8 × = 1
8
1
×6 = 3
Number of atoms due to 6 face centres =
2
∴ Total number of atoms in the unit cell = 1+ 3 = 4
The atoms are in contact along the face diagonal BD.
Let a be the edge length and r, the radius of an atom.
Consider a triangle BCD.
BD2 = BC2 + CD2
= a2 + a2 = 2a2
∴
BD = 2a
From figure, BD = 4r
∴
4r =
∴
r =
r =
Hence,
2a
a
2
a =
2 2
4
a
2 2
a = 2 2
Volume of unit cell = a3 = ( 2 2r ) 3 = 16 2r 3
Chapter - 1 Solid State
36
Volume of 1 atom =
∴ Packing fraction =
4 3
πr
3
Volume occupied by 4 atoms
Volume of unit cell
= 4
=
=
⎛4 3⎞
⎜⎝ πr ⎟⎠
3
16 2r 3
16/3πr 3
16 2r 3
π
3 2
3.142
=
3 × 1.414
= 0.74
∴ Packing efficiency = 0.74 ×100 = 74 %
∴ % of void space = 100 – 74 = 26 %
(l) Obtain a relation for the density of the crystal :
Consider a cubic unit cell of edge length a. Then the volume of the unit cell is = V = a ×
a × a = a3. Let M be the atomic mass of the constituting element or metal. If NA is the
M
Avogadro number, then the mass of one atom will be N .
A
If z is the number of atoms present in the unit cell, then the mass of unit cell will be,
M
Mass of unit cell = Mass of z atoms = z = N
A
Then the density of the unit cell is,
Density of unit cell
∴
=
d
=
d
=
Unique Solutions ®
Mass of unit cell
Volume of the unit cell
z×
M
NA
a3
z×M
a3 × NA
S.Y.J.C. Science - Chemistry - Part I
37
1. What is coordination number ?
Ans : Refer 1.3 (a)
2. Explain linear packing in one dimension. OR Close packing in one dimension.
Ans : Refer 1.3 (b)
3.
Explain the following :
Planar packing arrangement of spheres.
OR
Close packing in two dimensions.
OR
AAAA type and ABAB type of two dimensional arrangement. OR
Square close packing and hexagonal close packing in two dimension
Ans : Refer 1.3 (c)
4. Explain AAAA type of three dimensional packing. Or Explain simple cubic structure (SCC).
Ans : Refer 1.3 (d)
5. Explain ABAB type of three dimensional packing or Explain hexagonal close packing structure.
Ans : Refer 1.3 (e)
6.
Explain ABCABC type of three dimensional packing or Explain Cubic close packed structure
(ccp).
Ans : Refer 1.3 (f)
*7. Distinguish between hexagonal close packing and cubic close packing.
Ans : Refer 1.3 (g)
*8. Distinguish between tetrahedral void and octahedral void.
Ans : Refer 1.3 (h)
9.
Find the coordination numbers in the following :
(a) At the corner of scc
(b) Body centred cubic (bcc) structure
(c) Face centred cubic (fcc) structure.
Ans : Refer 1.3 (i)
10. What is packing fraction ?
Ans : Refer 1.3 (j) (1)
11.
Ans:
What is packing efficiency ?
Refer 1.3 (j) (2)
12. What is a void space in the crystal ?
Ans : Refer 1.3 (j) (3)
Chapter - 1 Solid State
38
*13.
Calculate packing efficiency in the following crystals :
(a) Simple cubic (scc) structure
(b) Body centred cubic (bcc) structure
(c) Face centred cubic (fcc) structure OR Cubic close packing (ccp) structure.
Ans : Refer 1.3 (k)
14. Obtain a relation for the density of the crystal.
Ans : Refer 1.3 (l)
15.
Write the packing fraction, packing efficiency and void space for the following crystal structures:
(1) SCC (2) BCC (3) FCC (4) HCP
Ans.
Crystal structure
Packing fraction
Packing efficiency
Void space
SCC
BCC
FCC
HCP
0.524
0.68
0.74
0.74
52.4%
68%
74%
74%
47.6 %
32%
26%
26%
(1)
(2)
(3)
(4)
*16.
Ans.
•
1.
2.
•
*3.
Explain with the help of neat diagrams AAAA and ABAB and ABCABC type of three dimensional
packings OR Explain the packing and voids in ionic solids.
Refer 1.3
Multiple Choice Questions :
Theoretical MCQs :
Which type of crystal structure is obtained by ABAB arrangement of spheres?
a) tetragonal
b) hexagonal
c) octahedral
d) cubic
ABC ABC ....... closest packing of spheres in solid gives
a) cubic close packing
b) hexagonal close packing
c) tetragonal close packing
d) octahedral close packing
The ratio of close packed atoms to tetrahedral holes in cubic packing is .......
a) 1 : 1
*4.
b)
1:2
c) 2 : 1
d)
1:3
The ratio of close packed atoms to octahedral holes in hexagonal close packing is .......
a) 1 : 1
b)
Unique Solutions ®
1:2
c) 2 : 1
d)
S.Y.J.C. Science - Chemistry - Part I
1:3
39
1.4
PACKING EFFICIENCY AND DENSITY OF UNIT CELL :
Concept Explanation :
(a) Packing in voids of ionic solids :
1. Ionic solids are formed from cations and anions.
2. The number of oppositely charged ions are appropriately
adjusted such that the charges are balanced and the
compound formed is electrically neutral.
3. Cations and anions are always at alternate positions for
more stability.
4. Anions occupy the lattice points while the relatively smaller cations occupy either tetrahedral
or octahedral voids.
5. Generally, if cation is relatively smaller than anions, then the cations occupy tetrahedral
voids while bigger cations occupy octahedral voids.
6. In case, the size of cation is very large then, the packing of anions is adjusted by separating
them and larger cations are accommodated in larger cubic holes.
(b) Radius ratio rule for ionic compounds :
1. The ratio of radius of cation (r+) to the radius of anion (r-) is known as radius ratio of the
ionic solid and represented as
r+
2.
r–
The radius ratio rule is useful in predicting the crystalline structures of ionic solids.
3.
The structure of an ionic compound depends upon stoichiometry of it and the sizes of ions.
4.
If r+ and r− are the radii of cation and anion respectively, then r+/r− represents the radius ratio.
5.
Greater the radius ratio, greater is the coordination number of cation.
Coordination number
Crystalline structure
Example
0.155 – 0.225
3
Plane triangular
B 2O 3
0.225 – 0.414
4
Tetrahedral
ZnS
0.414 – 0.732
6
Octahedral
NaCl
0.732 – 1
8
Cubic
CsCl
Chapter - 1 Solid State
40
(c) Packing of ions in cubic closed packed (ccp) or hexagonal closed
packed (hcp) structures :
1. In case of cubic closed packed (ccp) or hexagonal closed packed (hcp) structures, there
are two tetrahedral holes per anion and one octahedral hole per anion in the array.
2. In ionic compounds with formula AB, if the cation has a smaller size it occupies tetrahedral
voids. However, since the number of tetrahedral voids are double the number of anions,
only half of the tetrahedral voids are occupied and the other half are vacant. This type of
structure is known as Zinc blende structure. Eg: ZnS, CuCl etc.
3. In ionic compounds with formula AB, if the cation has a bigger size it occupies octahedral
voids. However, since the number of octahedral voids are equal to the number of anions,
all the octahedral voids are occupied. This type of structure is known as rock salt structure.
Eg: NaCl, AgBr etc.
4. Consider the ionic crystals (A2B) of Li2O, Li2S, Na2O, Na2S, etc. in which size of cation
is less than the size of anion (r+ < r−). There are total four anions, while eight cations are
packed in eight tetrahedral voids. This type of structure is called Antifluorite structure
5. In ionic compounds with formula AB2, the cation has a bigger size it forms ccp structure.
Since the number of anions are twice the number of cations they occupy all the tetrahedral
voids. This type of structure is known as Fluorite structure. Eg: BaCl2, SrF2 etc
(d) Structure of NaCl unit cell (rock salt structure) :
1. NaCl unit cell has face centred cubic (fcc) structure.
2. Na+ ions exist in octahedral voids at 12 edge centres and at body centre.
Since edge centre atom contributes 1/4 th atom, the total number of Na+ ions will be,
1
× 12 + 1 (body centre) = 4
4
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
41
3.
4.
5.
6.
Cl− ions exist at 8 corners and 6 face centres. Hence, the total number of Cl− ions will
1
1
be, × 8 + × 6 = 4
8
2
Thus, unit cell contains 4Na+ and 4Cl− ions, or 4NaCl molecules.
The coordination number of Na+ and Cl% is 6.
The edge length of NaCl unit cell : a = 2(rc + ra)
(KCl crystal has the same structure like NaCl.)
•
*1.
Ans.
Explain the packing in voids of ionic solids.
Refer 1.4 (a)
2.
Ans.
Refer 1.4 (b)
*3.
Ans.
Explain radius ratio rule for ionic compounds.
Refer 1.4 (b)
4.
Ans.
Explain the packing of ions in cubic closed packed (ccp) or hexagonal closed packed (hcp)
structures.
Refer 1.4 (c)
5.
Ans.
Explain the structure of NaCl unit cell. (rock salt structure)
Refer 1.4 (d)
•
1.
2.
3.
4.
Multiple Choice Questions :
Theoretical MCQs :
The number of nearest neighbours around each atom is a face centred cube is .......
a) 4
b) 6
c) 8
d) 12
In a body centred cube, the number of nearest neighbouring atoms are .......
a) 2
b) 4
c) 6
d) 8
+ –
In ionic compound has radius ration (r /r ) in the range of 0.225 to 0.414, what is the structure
and coordination number of compound?
a) cubic and 8
b) octahedral and 6
c) tetrahedral and 4
d) trigonal and 3
The inter ionic distance for cesium chloride crystal is .......
a)
2
3
a
b)
3
a
2
c)
Chapter - 1 Solid State
3a
d)
2a
3
42
*5.
*6.
7.
8.
The packing efficiency for a body centred cubic structure is .......
a) 0.42
b) 0.53
c) 0.68
d) 0.82
An ionic crystal lattice has r+/r– radius ratio of 0.524. its coordination number is .......
a) 2
b) 4
c) 6
d) 8
In hcp mode of staking, a sphere has coordination number .......
a) 4
b) 6
d) 8
d) 12
If edge of bcc crystal of an element is ‘a’ cm. M is the atomic mass and NA, is the Avogadro’s
number, then the density of crystal is .......
a)
9.
•
10.
11.
*12.
*13.
*14.
15.
4M
a 3 NA
b)
2N A
Ma 3
c)
M
a 3N A
d)
Ma 3
1N A
In which of the following structures, the anion has maximum coordination number?
a) NaCl
b) ZnS
c) CaF2
d) Na2O
In the closest packing of atoms .......
a) the size of tetrahedral void is greater than that of octahedral voids.
b) the size of tetrahedral void is smaller than that of octahedral voids.
c) the size of tetrahedral and octahedral voids are same.
d) the size of tetrahedral void is greater than that of octahedral voids.
Compound AxBy crystallises in hexagonal cubic close packing lattice. The element A occupy
2/3 of tetrahedral voids. Calculate x and y .......
a) x = 1, y = 2
b) x = 2, y = 3
c) x = 3, y = 4
d) x = 4, y = 3
The number of octahedral sites per sphere in fcc structure is .......
a) 1
b) 2
c) 3
d) 4
The number of tetrahedral sites per sphere in ccp structure is, .......
a) 1
b) 2
c) 3
d) 4
An ionic compound AxBy occurs in fcc type crystal structure with B ion at the centre of each
face and A ion occupying corners of the cube. Give the formula AxBy.
a) AB 3
b) AB4
c) A3B
d) A4B
The packing fraction for a body centred cube is .......
a) 0.42
b) 0.54
c) 0.68
d) 0.74
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
43
1.5
IMPERFECTIONS OR DEFECTS IN CRYSTAL STRUCTURE :
Concept Explanation :
(a) Defects in crystal structure :
1. Any deviation from perfectly ordered arrangement of atoms, ions or molecules in the
crystal lattice is called a defect.
2. Crystalline solids are formed by regular repetition of large number of unit cells in all directions.
3. There is a short range as well as a long range order in the arrangement of constituents
particles in solids.
4. The real crystals are never perfect. They always contain some imperfection in the crystal
formation.
5. The crystalline solids are formed by cooling a substance in the liquid state.
6. If the crystallisation takes place faster, then the orderly arrangement of constituent particles
cannot take place which produces defect in solids.
7. The defects can be minimized by carrying out crystallisation at the slowest rate, so that
orderly arrangement of particles can take place.
8. The crystal defects change the original physical and chemical properties of crystalline solids.
(b) Types of defects in the solids or crystal structures :
The defects in crystalline solids are of two types viz.,
1. Point defect 2. Line defect.
1. Point defects are further classified as :
(a) Vacancy defect or Schottky defect
(b) Interstitial defect or Frenkel defect
(c) Impurity defect :
This is further classified as :
(i)
Substitution impurity defect
(ii) Interstitial impurity defect.
2. Line defects are further classified as :
(a) Edge dislocation
(b) Screw dislocation.
(c) Point defects :
1. Point defect : The defect is due to a fault in the arrangement of a point i.e. a constituent
particle, e.g. an atom or an ion or a molecule in the crystalline solid.
2. The point defects are classified as follows :
(a) Vacancy defect or Schottky defect :
(i) Sometimes during crystallisation, some of the places of the constituent particles
remain unoccupied and the defect generated is called Vacancy defect.
Chapter - 1 Solid State
44
(ii) In case of ionic solids, the vacancies
are produced due to absence of cations
and anions in stoichiometric proportions
from their positions. This defect is
called Schottky Defect.
(iii) In these ionic solids, electrical neutrality
is maintained but the actual density is
less than the expected value.
(iv) These defects are observed in the
solids with cations and anions of almost
equal size. E.g. NaCl, KCl, CsCl, etc.
(b) Interstitial defect or Frenkel defect :
(i) When cation or anion from ionic solid
leaves a regular lattice site and
occupies a place between the regular
lattice points or interstitial points. This
defect is called Frenkel or interstitial
defect.
(ii) Frenkel defect arises in case of ionic
solids with relatively smaller cationic
size which can fit into interstitial space
and difference in ionic radii is large.
(iii) This defect is observed in AgCl solid
by Ag+ ions, in ZnS solid by Zn2+ ions,
etc.
(iv) Frenkel defect does not alter the density of the crystalline solid.
(c) Impurity defect :
(i) This defect arises when a cation from its regular site in ionic crystal lattice is
replaced by different cations.
(ii) If the impurity cation is substituted in the place of regular cation, then it is called
substitution impurity defect.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
45
(iii) If the impurity of cation is present in the interstitial positions then it is called
interstitial impurity defect.
(iv) The interstitial impurity defect changes the properties of the original crystalline
solid.
(v) Alloys are formed by substitution defects. E.g. Brass, in which Cu metal is
substituted by zinc metal. Stainless steel involves an interstitial impurity defect by
carbon atoms.
Non-Stoichiometric defects:
1. The defects discussed so far do not disturb the stoichiometry of the crystalline substance.
2. However, in some solids the constituent elements are in non-stoichiometric ratio due to
defects.
3. These defects are of two types :
(a) Metal excess defect
(b) Metal deficiency defect.
(a) Metal excess defect :
1. Metal excess defect due to anionic vacancies:
(i) Alkali metal halides like NaCl and KCl show
this type of defect.
(ii) When NaCl crystals are heated in an
atmosphere of Sodium vapour, the Na atoms
are deposited on the surface of the crystal.
(iii) The Cl– ions diffuse to the surface of the
crystal and combine with Na atoms to give
NaCl.
(iv) This happens by loss of electron by Na atoms
to form Na+ ions.
(v) The released electrons diffuse into the crystal and occupy anionic sites.
(vi) As a result the crystal has excess of sodium.
(vii) The anionic sites occupied by unpaired electrons are called F-centres.
(viii) They impart yellow colour to the crystals of NaCl.
(ix) The colour results by excitation of these electrons when they absorb energy from
the visible light.
2. Metal excess defect due to presence of extra cations at interstitial sites :
(i) ZnO is white in colour at room temperature.
(ii) On heating it loses oxygen and turns yellow.
1
→ Zn 2+ + O 2 + 2e –
(iii) ZnO ⎯⎯
2
Chapter - 1 Solid State
46
(iv) Now there is excess of Zn2+ in the crystal which moves to interstitial sites and
electrons to the neighbouring interstitial sites.
(b) Metal deficiency defect :
(i) There are many solids which are difficult to prepare in the stoichiometric
composition and contain less amount of metal.
(ii) Eg : in FeO the actual composition varies from Fe0.93O to Fe0.96O. In these
crystals some Fe+2 ions are missing and the loss of positive charge is made up
by the presence of Fe+3.
•
*1.
Ans.
Explain interstitial defects and impurity defects?
Refer 1.5 (c) 2(b, c)
*2.
Ans.
*3.
Ans.
Explain the terms : (a) Schottky defect, (b) Frenkel defect
(a) Refer 1.5 (c) (ii)
(b) Refer 1.5 (b) (i)
What are point defects?
The defects which is due to a fault produced in the arrangement of a point i.e. a constituent
particle like atom, ion or molecule in a crystalline solid.
4.
Ans.
What is meant by defects in crystal structure or solid?
Refer 1.5 (a)
5.
Ans.
Why do defects in the crystalline solids arise ?
Refer 1.5 (a)
6.
Ans.
Mention the types of defects in the solids or crystal structures.
Refer 1.5 (b)
7.
Ans.
Explain point defects.
Refer 1.5 (c)
8.
Write the type of defect in the following : (a) KC1 (b) AgCl (c) Brass (d) Stainless steel
(e) ZnS (f) Bronze.
Ans.
(a)
(b)
(c)
(d)
(e)
(f)
Substance
Type of defect
KCl
AgCl
Brass
Stainless steel
ZnS
Bronze
Schottky defect or vacancy defect
Frenkel defect or interstitial defect
Substitution impurity defect
Interstitial impurity defect
Schottky defect or vacancy defect
Substitution impurity defect
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
47
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
In crystalline solid few of the cations moved from their positions into the interstitial .......
position. The defect is called as, .......
a) interstitial defect
b) Frenkel defect
c) Schottky defect d) line defect
Due to Frenkel defect the density of ionic solid .......
a) increases
b) decreases
c) remains same d) fluctuates
*2.
1.6
Electrical and Magnetic properties of crystalline solid :
Concept Explanation :
Solids exhibit an interesting range of variation electrical conductivity over 27 orders of magnitude
ranging from 10–20 to 107 Ohm–1m–1. Based on their conductivity, solids are classified into :
(i) Conductors (ii) Insulators and (iii) Semi-conductors
(a) Electrical propertiesof solids :
1. Conductors : The solid which allow the current to flow through it is called conductor.
They have conductivity in the range of 104 to 107 Ohm–1m-1. e.g. metals.
2. Insulators : The solids which do not allow passage of electric current through them are
called insulators. They have very low conductivity ranging between 10–20 to 10–10
Ohm–1 m–1. e.g. wood, rubber, sulphur, phosphorus etc.
3. Semi-conductors : The solids whose conductivities are in between that of conductors and
insulators are called semiconductors. Their conductivities are in the range of 10–6 to 104
Ohm–1 m–1 and the conductivity is mainly due to presence of an impurity or some defects.
In most of the solids, the conductivity is through the movement of electron under the influence
of electric field while in ionic solid, it is by ions. In solids (metals), thus, conductivity strongly
depends on the number of valence electron available for conductance. The electrical
conductivity of conductors, insulators and semi conductors can be explained on the basis
of band theory.
(b) Band theory OR Origin of electrical properties in solids :
1. Metals are good conductors of heat and electricity. In order to explain the conducting
properties of metals a band theory was developed.
2. The theory assumes that the atomic orbitals of the metal atoms overlap to form molecular
orbitals which are spread all over the crystal structure.
3. With increase in number of atoms participating in crystal formation, the number of molecular
orbitals containing electrons is greatly increased.
Chapter - 1 Solid State
48
4.
5.
6.
7.
8.
9
As the number of molecular orbitals increases, the energy difference between adjacent
molecular orbitals decreases and when it becomes very less, the orbitals merge into one
another forming continuous bands which extent over the entire crystal.
The interaction between these large number of orbitals leads to formation of bonding and
anti bonding molecular orbitals.
All the molecular orbitals are very close to each other and cannot be distinguished from
one another and therefore, these orbitals are collectively called a band.
There are two types of bands of molecular orbitals as
(a) Valence band : The atomic orbitals with filled electrons from the inner shells form
valence bands, where there are no free mobile electrons since they are involved in
bonding.
(b) Conduction band : Atomic orbitals which are partially filled or empty on overlapping
form closely placed molecular orbitals giving conduction bands where electrons are
delocalized and can conduct, heat and electricity.
In metallic crystals, the valence bands and conduction bands overlap or are very close and
energy difference between them is very less, so that electrons from valence bands can
easily be excited to conduction bands.
A metal crystal is considered as an electron sea model and viewed as a three dimensional
array of metal cations immersed in a sea of mobile electrons. Consider magnesium metal,
2
2
6
2
12Mg having electronic configuration 1s , 2s , 2p , 3s , 3pº, 3dº. In metallic crystal it exists
as Mg2+. The atomic orbitals 1s2, 2s2, 2p6, after overlapping form valence bands while
empty orbitals, 3sº, 3pº, 3dº form conduction bands.
In insulators, energy difference between valence band and conduction band is very high,
hence high energy is required to excite the electrons from valence band to conduction band.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
49
10. If the energy difference
between valence band
and conduction band is
moderate, then the
substance in ordinary
condition is non-conductor,
but if heated it becomes
conductor due to transition
of electrons into
conduction band. Such
conductors
are
semiconductors.
(c) Semiconductors and their types :
1. Semiconductors : The substances like silicon, germanium which have poor electrical
conductance at low temperature but the conductance increases with the increase in
temperature are called semiconductors.
2. Their conductivity lies between metallic conductors and insulators.
3. The energy difference between valence band and conduction band
is relatively small, hence, the electrons from valence band can be
excited to conduction band by heating.
4. Types of semiconductors :
There are two types of semiconductors as follows :
(i) Intrinsic semiconductors
(ii) Extrinsic semiconductors obtained by doping Si.
The extrinsic semiconductors are of three types.
They are (a) n-type semiconductors, (b) p-type semiconductors
and (c) n-p type semiconductors.
5.
Intrinsic semiconductor : Silicon, 14Si, is a semiconductor
having electronic configuration 1s2, 2s2, 2p6, 3s2, 3p2,
3d0, 4s0.
All the 4 electrons in Si are involved in the covalent bonding
in the tetrahedrally situated neighbouring Si atoms.
On heating some of valence band electrons go to conduction
band and silicon becomes electric conductor.
Silicon and germanium are the examples of intrinsic
semiconductors.
Chapter - 1 Solid State
50
6.
7.
n-type semiconductors :
(a) n-type semiconductor is an extrinsic
semiconductor obtained by doping
silicon with external electron rich
impurity like fifth group element. For
example, arsenic.
(b) Arsenic has five valence electrons, out
of which four are involved in covalent
bonding with neighbouring Si atoms
while one electrons remains free and
delocalised.
(c) These free electrons increase the electrical conductivity of the semiconductor.
(d) The semiconductors with extra non-bonding free
electrons are called n-type semiconductors.
p-type semiconductors.
(a) p-type semiconductor is an extrinsic
semiconductor obtained by doping Silicon
with external electron deficient impurity like
third group element. For example, boron.
(b) Boron (5B) has only 3 valence electrons
which form covalent bonds with the
neighbouring Si atoms, while one bond has
shortage of one electron.
(c) This creates a vacancy or a hole, hence the electron from neighbouring Si atom jumps
into this hole creating a vacancy in itself. This process continues, i.e., positive holes
move in one direction while electrons moves in opposite direction.
(d) Due to electron deficient positions, this semiconductor is called p-type semiconductor.
(e) When p-type semiconductor is connected to the external source of electricity, electrons
from neighbouring silicon atoms jump into the holes so that electrons move towards
positive electrode and holes migrate towards negative electrode.
(f) Hence, electrical conduction in p-type semiconductor is due to electrons and holes.
(d) Uses of semiconductors :
(a) They are used in transistors, digital computers and cameras.
(b) They are used in solar cells and television sets.
(c) By combining n-type and p-type semiconductors, n-p junctions are formed which are
effectively used in rectifiers or to convert light energy into electrical energy.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
51
.
(e) Origin of magnetic properties in solids :
(a) While electrons are revolving round the
nucleus in various orbits, they are also
(b) A spinning charge generates a magnetic
field. Hence, spinning electrons act like tiny
magnets.
(c) If an orbital contains only one electron then
it may spin either clockwise or anticlockwise.
This generates magnetic field.
(d) However, if the orbital contains two electrons then one electron spins in clockwise
direction and other spins in anticlockwise direction, then their spins are balanced and
magnetic moments and magnetic properties are cancelled and the substance is said to
be diamagnetic.
(e) When a substance contains one or more unpaired
electrons spinning in same direction, then their
magnetic moments and magnetic properties add
and the substance is said to be paramagnetic.
(f) Diamagnetism :
1. The magnetic properties of a substance arise due to
presence of the electrons.
2. An electron while revolving around the nucleus, also
spins around its own axis and generates a magnetic
moment and a magnetic property.
3. If an atom or a molecule of the substance contains all electrons paired, spinning clockwise
and anticlockwise, their magnetic moments and magnetic properties get cancelled. Hence
they oppose and repel the applied magnetic field. This phenomenon is called diamagnetism
and the substance is said to be diamagnetic.
For example : Zn, Cd, H2O, NaCl, etc.
(g) Paramagnetism :
1. The magnetic properties of a substance arise due to
the presence of electrons.
2. An electron while revolving around the nucleus, also spins around its own axis and generates
a magnetic moment and magnetic properties.
Chapter - 1 Solid State
52
3.
If an atom or a molecule contains one or more unpaired electrons spinning in same direction,
clockwise or anticlockwise, then the substance is associated with net magnetic moment
and magnetic properties. They experience a net force of attraction when placed in the
magnetic field. This phenomenon is called paramagnetism and the substance is said to be
paramagnetic.
For example, O2, Cu2+, Fe3+, Cr3+, NO, etc.
(h) Ferromagnetism :
1. The magnetic properties of a substance arise
due to the presence of electrons.
2. An electron while revolving around the nucleus
also spins around its own axis and generates
a magnetic moment and magnetic properties.
3. The substances which possess unpaired electrons and high paramagnetic character and
when placed in a magnetic field are strongly attracted and show permanent magnetic moment
even when the external magnetic field is removed are said to be ferromagnetic. They can
be permanently magnetised.
4. In the solid state, the metal ions of ferromagnetic substance are grouped together into small
regions called domains, where each domain acts as a tiny magnet.
For example : Fe, Co, Gd, CrO2, etc.
(i) Antiferromagnetism :
Substances like MnO showing anti-ferromagnetism have domain structure similar to
ferromagnetic substance, but their domains are oppositely oriented and cancel out each
other’s magnetic moment.
(j) Ferrimagnetism
Ferrimagnetism is observed when the magnetic moments of the domains in the substance
are aligned in parallel and antiparallel directions in unequal number. They are weakly attracted
by magnetic field as compared to ferromagnetic substances. Eg: Fe3O4 etc.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
53
(k) Gouy’s method for determining magnetic properties of the substances :
Principle of Gouy’s method :
1. An electron in an atom while revolving
around nucleus also spins around its own
axis generating a magnetic moment and
magnetic field.
2. The resultant magnetic properties depend
upon, the number of paired and unpaired
electrons.
(a) If all the electrons are paired, their
magnetic moments and magnetic
properties are cancelled and the
substance is said to be diamagnetic.
Such substance repels the applied
magnetic field.
(b) If a substance contains one or more unpaired electrons, then the substance is associated
with a net magnetic moment and magnetic properties. Such a substance is paramagnetic
in nature, which is attracted when placed in the applied external magnetic field.
Procedure :
1. The substance is weighed in the absence and presence of the magnetic field,
2. A diamagnetic substance is pushed upward when placed in the magnetic field between two
magnets. Hence, the weight of the substance appears less in the presence of magnetic field
than in the absence of magnetic field,
3. A paramagnetic substance is pulled down due/to attraction when placed in the magnetic
field, hence the weight of the substance appears more in the presence of magnetic field.
4. If the substance is ferromagnetic, it is pulled down more weighing still higher in the presence
of magnetic field.
Hence, from the changes in weights* in and out of magnetic field, magnetic properties of
the substances can be decided.
•
*1.
Ans.
What are semiconductors? Describe the two main types of semiconductors?
Refer 1.6 (c)
*2.
Ans.
Explain Band theory?
Refer 1.6 (b)
*3.
Ans.
Explain the origin of electrical properties in solids.
Refer 1.5(e)
Chapter - 1 Solid State
54
4.
Ans.
Explain n-type semiconductors.
Refer 1.6(c)6
5.
Ans.
Explain p-type semiconductors.
Refer 1.6(c)7
*6.
Classify the following semiconductors into n or p-type.
(a) B doped with Si (b) As doped with Si (c) P doped with Si (d) Ge doped with In.
Ans.
(a)
(b)
(c)
(d)
Semiconductor
Type
B doped with Si
As doped with Si
P doped with Si
Ge doped with In
p-type
n-type
n-type
p-type
7.
Ans.
What are the uses of semiconductors ?
Refer 1.6 (d)
*8.
Ans.
Distinguish between conductor, insulator and semiconductor.
Conductor
1. A substance which conducts
heat and electricity to a greater
extent is called conductor.
2. In this, conduction bands and
valence bands overlap or are
very closely spaced.
3. There is no energy difference
or very less energy difference
between valence bands and
conduction bands.
4. There are free electrons in the
conduction bands.
Insulator
Semiconductor
1. A substance which cannot 1. A substance which has poor
conduct heat and electricity
electrical conductance at low
under any conditions is called
temperature but higher
insulator.
conductance at higher
temperature is called
semiconductor.
2. In this, conduction bands and 2. In this, conduction bands
valence bands are far apart.
and valence bands are spaced
closely.
3. The energy difference 3. The energy difference
between conduction bands
between conduction bands
and valence bands is very
and valence bands is small.
large.
4. There are no free electrons 4. The electrons can be easily
in the conduction bands and
excited from valence bands to
electrons can't be excited
conduction bands by heating.
from valence bands to
conduction bands due to
large energy difference.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
55
5. The conductance decreases
with the increase in
temperature.
6. E.g., Metals, alloys.
7. The conducting properties
third substance.
8.
9.
Ans.
5. No effect of temperature on 5. Conductance increases
conducting properties.
with
the increase in
temperature.
6. Wood, rubber, plastics.
6. E.g. Si, Ge
7. No effect of addition of any 7. By doping, conducting
substance.
properties improve.
E.g. n-type, semiconductors.
8.
8.
Identify the following substances as paramagnetic, diamagnetic or ferromagnetic :
(b) Magnesium 12Mg
(c) 20Ca2+ cation
(a) Sodium 11Na
–
(d) 17Cl anion
(e) Iron 26Fe
(f) 27Co atom
(g) 28Ni atom
Paramagnetic
(a) Sodium 11Na
Diamagnetic
(b) Magnesium 12Mg
(c) 20Ca2+ cation
(d) 17C1– anion
Ferromagnetic
(e) Iron 26Fe
(f) 27Co atom
(h) 28Ni atom
*10.
Ans.
Explain the origin of magnetic properties in solids.
Refer 1.5 (e)
*11.
Ans.
Explain diamagnetism.
Refer 1.5 (f)
*12.
Ans.
Explain paramagnetism.
Refer 1.5 (g)
13.
Ans.
Explain ferromagnetism.
Refer 1.5 (h)
Chapter - 1 Solid State
56
14.
Ans.
Explain Antiferromagnetism.
Refer 1.5 (i)
15.
Ans.
Explain Ferrimagnetism
Refer 1.5 (j)
16.
Ans.
Explain Gouy's method for determining magnetic properties of the substances.
Refer 1.5 (k)
•
1.
2.
*3.
*4.
*5.
•
6.
7.
Multiple Choice Questions :
Theoretical MCQs :
To get n–type doped semi-conductor, the impurity to be added to silicon should have
following number of valance electron?
a) 1
b) 2
c) 3
d) 5
A p-type material is electrically .......
a) positive
b) negative
c) neutral
d) depends on the concentration of p-impurities
Semiconductors are manufactured by addition of impurities of .......
a) p-block elements b) s-block elements c) lanthanoids
d) actinoids
p-type semi conductor is formed when trace amount of impurity is added to silicon. The number
of valence electrons in the impurity atom must be .......
a) 3
b) 5
c) 1
d) 2
n-type semiconductor is formed when trace amount of impurity is added to silicon. The number
of electrons in the impurity atom must be .......
a) 3
b) 5
c) 1
d) 2
Which group element is added to germanium in order to get n-type semiconductor?
a) 12
b) 13
c) 14
d) 15
The addition of traces of Aluminium in the crystal lattice of Germanium gives rise to which type
of semiconductor?
a) none of the above
b) p-type
c) n-p junction
d) n-type
HOURS BEFORE EXAM
Isomorphous : Solids which have same crystal structure. Their constituent particles have
same ratios. e.g. NaF and MgO, NaNO3 and CaCO3 etc.
Anisotropy : The substances which have physical properties different in different direction.
e.g. refractive index, electric conductance, dielectric constant, thermal conductivity etc.
Isotropy : The solids in which physical properties do not change with the direction of
measurement. eg. amorphous solids.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
57
Classification of solids
Crystalline
e.g. NaCl, ZnS, Copper, Gold etc.
Ionic solids
e.g. NaCl
Molecular solids
Polar molecular
solids e.g. HCl,
NH3, solid SO2
Amorphous
e.g. Glass, rubber, plastic
Covalent or
Network solids
e.g. diamond, graphite
Metallic solids
e.g. metals
Hydrogen bonded
molecular solid
e.g. solid water (ice)
Non-polar
molecular solid
e.g. solid CO2,
solid Ar
Crystal Lattice : Regular three dimensional arrangement of points in space are called
space point or crystal lattice.
Unit cell : Smallest part of crystal lattice which when repeated in different direction generates
the entire lattice.
1
Cubic
2
Tetragonal
3
Orthorhombic
Body centered
Primitive
a = b ≠ c,
a = b ≠ c,
α = β = γ = 90º α = β = γ = 90º
4
Monoclinic
Primitive
End-centered
a ≠ b ≠ c, α
a ≠ b ≠ c,
α = β = 90º, = γ = 90º α = β = 90º, γ = 90º
6
Simple or
Body centred Face centered
a = b = c
a = b = c
primitive
a = b = c α = β = γ = 90º α = β = γ = 90º
α = β = γ = 90º
Body centred
a ≠ b ≠ c
α = β = γ = 90º
5
Triclinic
a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90º
7
Hexagonal
a = b ≠ c
α = β = 90º
γ = 120º
Primitive
a ≠ b ≠ c
α = β = γ = 90º
End-centred
a ≠ b ≠ c
α = β = γ = 90º
Chapter - 1 Solid State
Rhombohedral
a = b = c,
α = β = γ ≠ 90º
Face-centred
a ≠ b ≠ c
α = β = γ = 90º
58
Coordination number : Number of particles/atoms/spheres surrounding a single particle.
Number of atoms per unit cell of cubic lattice
Simple or primitive
1) 8 corner atoms ×
1
8
1) 8 corner atoms ×
= 1 atom/unit cell
2) r =
a
2
3) Packing efficiency = 52.4%
4) Density of unit cell (ρ) =
mass of unit cell
n×M
=
volume of unit cell a 3 ⋅ N A
∴ density (P) =
Body centred
1
=1
8
2) No. of atoms at body
center position = 1
3) Total No. of atoms/unit
cell = 2
4) r =
Face centred
1) No. of corner atoms
=8×
1
=1
8
2) No. of atoms at face
centre = 6 ×
1
= 3
2
3) Total No. of atoms/unit cell
=3+1=4
a
2 2
5) Packing efficiency = 68%
n×M
4) r =
3a
4
5) Packing efficiency = 74%
a 3 ⋅ NA
Defects in solids : Some irregularity arises during crystallisation of solids, known as
imperfections or defects. These are.
Point defects
Line defects
arises due to faculty arrangement
of constituent particles.
arises due to deviation from the ideal arrangement
of entire row of lattice point
Interstitial defect
Vacancy defect
Cations and anions
remain unoccupied in
stoichiometric proportion.
Frankel defect
Electrical neutrality is
Some constituent particle
maintained.
occupy interstitial position.
Result in decrease in
Shown by ionic solids.
density of solid.
Large difference in ionic size
i.e. between cation & anion.
e.g. AgCl, ZnS, AgBr, AgI.
Unique Solutions ®
Impurity defect
♦ Regular cation in crystal
is substituted by other
cation.
Schottky defect
Shown in by ionic solid only.
Cation and anion have almost
same size.
Cation and anion remain absent in
stoichiometric ratios.
Electrical neutrality is maintained.
Decreases the density of solid.
S.Y.J.C. Science - Chemistry - Part I
59
Electrical Conductivity :
Electrical conductivity in solid is due to free mobile electrons which can be explained on the
basis of band theory. Based on electrical conductivity solids are classified as
Conductors
The energy difference
between the valence band and
conduction band i.e. band gap
is negligibly small.
Conducts at room temperature
or even at slightly lower temp.
Insulators
Semiconductors
Band gap is slightly greater than Band gap is large.
conductors but lower than that of They do not conduct at
insulators.
any temperature.
Conductivity increases with increase
in temperature.
Basically. they are insulator at an
absolute zero. e.g. Si, Ge.
n-type
Prepared by doping silicon or Germanium with
group 15 element like As, P etc.
Conductivity increases due to presence of
excess of electrons.
p-type
Prepared by doping Si or Ge with group 13
elements like Ga or In.
Conductivity increases due to (creation of
holes (+ve) in the crystal structure.
Magnetic properties of solids :
Magnetic property in solids arises due to presence of unpaired electrons as each electron acts
as tiny bar magnet. Based on magnetic behaviour solids are classified as
Diamagnetic
All electrons are paired i.e. their spin is paired ( ).
Their magnetic moment is zero. They are weakly repelled by
external magnetic field.
e.g. water, benzene, NaCl etc.
Paramagnetic
All electrons are not spin paired.
They have net magnetic moment.
They are weakly attracted
towards external magnetic field.
Loose their magnetism in
absence of external magnetic
field. e.g. O2–,Cu2+, Fe+3 etc.
Ferromagnetic
All spins are aligned in one
direction only.
Very strongly attracted by
applied magnetic field.
They can be permanently
magnetised.
They produce strong
magnetic field.
Magnetic moment is high.
Anti-ferromagnetic
Ferrimagnetism
Alignment of spin is in opposite direction. Alignment of spin is parallel and anti parallel
direction i.e. two electrons spin in parallel while
Very strongly attracted towards applied
3rd electron has spins in opposite direction.
magnetic field.
Net magnetic moment is zero due to They have net magnetic moment.
Weakly attracted by applied magnetic field.
compensation of domains.
Becomes paramagnetic on heating. e.g. Fe2O4,
CuFe2O4, MgFe2O4 etc.
Chapter - 1 Solid State
60
FORMULAS
Type of atom
At the corner of unit cell
At the body centre
At the face centre
Volume
contribution
1/8
1
½
Property
Type of packing
Available space occupied
Coordination number
hcp
AB AB
74%
12
Unit Cell
Simple cubic
Body centered cubic
Face Centered cubic
Hexagonal
A...
ccp
ABC ABC A...
74%
12
No.
per
1
1+
1+
1+
of atoms
unit cell
1=2
3=4
2=3
bcc
AB AB A...
68%
8
Relationship between radius of constituent particle (r) and edge length (a) :
a. Simple cubic unit cell : a = 2r
b.
c.
1.
2.
3.
4.
Face centred unit cell : a = 2 2r
4r
Body centred unit cell : a =
3
Crystal structure
SCC
BCC
FCC
HCP
Packing fraction
0.524
0.68
0.74
0.74
Packing efficiency
52.4%
68%
74%
74%
Void space
47.6 %
32%
26%
26%
DENSITY OF THE UNIT CELL (ρ ) : It is defined as the ratio of mass per unit cell
z×M
to the total volume of unit cell. d = 3
a × NA
Where Z = Number of particles per unit cell
M = Atomic mass or molecular mass
NA = Avogadro number (6.022 × 1023 mol–1)
α = Edge length of the unit cell = α pm = α × 10–10 cm
a3 = volume of the unit cell
The density of the substance is same as the density of the unit cell.
0.155 − 0.225
0.225 − 0.414
0.414 − 0.732
0.732 − 1
Coordination number
3
4
6
8
Unique Solutions ®
Crystalline structure
Plane triangular
Tetrahedral
Octahedral
cubic
S.Y.J.C. Science - Chemistry - Part I
Example
B2O3
ZnS
NaCl
CsCl
61
NUMERICALS WITH SOLUTIONS :
*1.
Niobium is found to crystallize with bcc To find :
structure and found to have density of
density (d) = ?
–3
8.55g cm . Determine the atomic radius Solution :
of niobium if its atomic mass is 9.
n⋅M
Given :
Formula = d = V . N
A
Crystal structure of Niobium = bcc
molar mass = 63.5 g mol
density (d) = 8.55 g cm–3
No. of atoms per unit cell (n) in fcc = 4
Atomic mass of Niobium = 93
Total volume of unit cell = a3 = (3.61 ×
To find :
10–8 cm)3 = 47.045 × 10–24 cm3
the atomic radius of niobium = ?
Solution :
4 × 63.5
∴ d =
n⋅M
47.045 × 10 –24 × 6.022 × 10 23
Formula = d = V ⋅ N
A
d = 8.93 g cm-3.
No. of atoms per unit cell (n) in bcc = 2
Total volume of unit cell = a3
*3. Silver crystallises in fcc structure with edge
2 × 93
length of unit cell, 4.07 × 10–8 cm and if
∴ d =
3
23
a × 6.023 × 10
density of metallic silver is 10.5 g cm–3.
Calculate the molecular mass of silver.
a3 = 36.12 × 10–24
Given :
a = 3 36.12 ×10 –24
Type of crystal structure = fcc
edge length (a) = 4.07 × 10–8 cm
3 × 3.304 × 10 –8
density (2δ) = 10.5 g cm3–
=
4
To find :
= 1.431 × 10–8 cm
molar mass of silver = ?
or 14.31 × 10 cm
Solution :
a = 14.31 nm
n⋅M
Formula = d = V ⋅ N
A
*2. Copper crystallises into a fcc structure and
d ⋅ v ⋅ NA
the unit cell has length of edge
M=
n
3.61 × 10–8 cm. Calculate the density of
No. of atoms per unit cell (n) in fcc = 4
copper if the molar mass of Cu is 63.5 g
Total volume of unit cell
mol.
= (v) = a3
Given :
= (4.07 × 10–8 cm)3
Crystal structure = fcc
= 47.045 × 10–24 cm3
edge length (a) = 3.61 × 10–8 cm
Chapter - 1 Solid State
62
∴
Atomic mass of iron = 56
Type of crystal lattice = ?
Solution :
10.5 × 67.419 × 6.022 × 10 23
M =
4
M = 106.5
Determine the density of cesium chloride
which crystallises in a bcc structure with
the edge length of unit cell, 4.07 × 10–8 cm
and if density of metallic silver is 10.5 g
cm–3. Calculate the molecular mass of
silver.
Given :
edge length (a) = 412.1 pm
= 4.12 × 10–8 cm
Type of crystal structure = bcc
Atomic mass of cesium = 133
Atomic mass of chlorine = 35.5
∴ Molar mass of CsCl = 168.5
To find :
Density of cesium chloride (CsCl) = ?
Solution :
volume of one unit cell =
*4.
n⋅M
Formula = d = V ⋅ N
A
No. of atoms per unit cell of bcc = 2
∴
d
=
2 × 168.5
(4.12 × 10 –8 ) 3 (6.022 × 10 23 )
=
56
= 7.125g cm3–
7.86
=
=
=
Total volume of unit cell
a3
(2.88 5 10–8)3
23.888 × 10–24 cm3
No. of unit cell in 56g metal
=
Total volume
Vol. of one unit cell
=
23.888 × 10 –24
7.125
=
No. of unit cells
=
6.022 × 10 23
2.982 × 10 –24
=
= 8.001 g cm .
Unit cell of iron crystal has edge length of
288 pm and density of 7.86 g cm –3.
Determine the type of crystal lattice (Fe =
56) (Hint : Determine number of atoms of
unit cell. It comes to 2, hence, bcc type)
Given :
edge length (a) = 288 pm = 2.88 × 10–8 cm
density (d) = 7.869 cm–3
Unique Solutions ®
2.982 × 1023 unit cells.
No. of atoms per unit cell
=
3
*5.
M
d
2.019
Since the number of atoms per unit cell
is 2, it is bcc type.
*6.
An atom crystallises in fcc crystal lattice
and has a density of 10 g cm–3 with unit
cell edge length of 100 pm. Calculate
number of atoms present in 1 g of crystal.
Given :
Density (d) = 10 g cm3
edge length (a) = 100 pm = 1 × 10–8 cm
S.Y.J.C. Science - Chemistry - Part I
63
To find :
No. of atoms in 1 g crystal =
Solution :
?
mass
volume of 1 g of metal = density
=
1
10
= 0.1 g cm3
Total volume of unit cell = a 3
= (1 × 10–8)3
= 1 × 10–24 cm3
No. of atoms in limits volume
=
=
∴
0.1
1 × 10 –24
1 × 1023 unit cells
Face entered cube contains 4 atoms per
unit cell.
Number of atoms in 1g = 4 × 1 × 1023
Number of atoms in 1g = 4 × 1023 atoms.
*7.
An element A and B constitute bcc type
crystalline structure. Element A occupies
body centre position and B is at the corners
of cube. What is the formula of the
compound? What are the coordination
numbers of A and B? (Formula AB,
coordination numbers of A = 8, and B = 8).
Solution :
No. of atoms occupying body centered
position = 1
No. of atoms at the corners per unit cell
=
∴
8×
1
=1
8
Formula of compound is AB
∴
Since, in body centred cube, each atom is
surrounded by eight nearest neighbour, its
coordination number is 8.
Coordination No. of A = 8
coordination No. of B = 8
*8.
Atoms C and D form fcc crystalline
structure. Atom C is present at the corners
of the cube and D is the faces of the cube.
What is the formula of the compound.
Solution:
As, atom C, occupies corners of the cube,
no. of atoms present at the corners per unit
1
cells = 8 corners × = 1
8
No. of atoms present at the faces per unit
1
cell = 6 × = 3
2
∴ Formula of compound is CD3.
9.
A cubic unit cell contains atoms A at the
corners, atoms B at face centres and atom
C at the body centre. What is the formula
of the crystalline compound ?
Solution :
Atoms A are at 8 corners, atoms B at the 6
face centres and one atom C at body centre.
1
Total number of atoms of A = ×8 = 1
8
1
Total number of atoms of B = × 6 = 3 .
2
One atom of C at the body centre.
Therefore the unit cell contains one atom
of A, three atoms of B and one atom of C.
Hence, the formula of the compound is
AB3C.
∴ The formula of the crystalline
compound is AB3C.
Chapter - 1 Solid State
64
10.
In a cubic crystalline structure of zinc
blende (ZnS), sulphide ions are at corners
and face centres while zinc ions occupy
half of tetrahedral voids. Find in the unit
cell :
(a) Number of Zn2 + ions
(b) Number of S2− ions
(c) Number of ZnS molecules
(d) Molecular formula of zinc blende.
Solution :
S2− ions are at 8 corners and 6 face
centres. Zn 2+ ions occupy half of
tetrahedral voids.
Number of S2− ions in unit cell
1
1
× 8 +
× 6
8
2
(corners) (face centres)
∴
6 face centres where atoms A are present.
Number of A atoms
=
1
1
× 8 +
× 6
8
2
(corners) (face centres)
=1+3=4
In cubic unit cell, there are 8 tetrahedral
voids. Hence,
Number of B atoms =
∴
2
16
×8 =
3
3
Hence, the formula should be A4B16/3 or
A12B16 or A3B4.
The formula of the compound is A3B4.
Atoms of elements A and B form a molecule
AxBy and crystallises in hcp lattice. The
element A occupies 2/3 of tetrahedral
= 1+ 3=4
voids. What is the formula of AxBy ?
Cubic unit cell has 8 tetrahedral voids. Since
Solution :
half of them are occupied by Zn2+ ions,
Atoms A are present in 2/3 tetrahedral
there are 4Zn2+ ions in the unit cell.
voids while the atoms B are present at 12
Hence, number of zinc sulphide (ZnS)
corners and 2 face centres of hcp crystal
molecules is 4. Molecular formula of zinc
lattice.
blende is ZnS.
Hence, total number of B atoms is,
(a) Number of S2−− ions = 4
1
1
(b) Number of Zn2+ ions = 4
× 12 + × 2 = 2 + 1 = 3
6
2
(c) Number of ZnS molecules = 4
(d) Molecular formula of zinc blende
For each B atom, there will be 2 tetrahedral
= ZnS.
voids. Hence, total tetrahedral voids will
be 3 × 2 = 6.
11. In a crystalline compound, atoms A occupy
The number of A atoms in voids will be,
ccp lattices while atoms B occupy 2/3 rd
2
×6 = 4.
tetrahedral voids. What is the formula of
3
the compound ?
Thus, hcp unit cell contains 4 atoms of A
Solution :
and 3 atoms of B, hence the formula of
In ccp unit, lattice points are 8 corners and
AxBy is, A4B3.
=
Unique Solutions ®
12.
S.Y.J.C. Science - Chemistry - Part I
65
[Alternative Method :
In hcp crystal lattice, the number of
tetrahedral voids is twice the number of
2
anions of element B. Hence for
3
occupancy of tetrahedral voids by A
indicates half the occupancy of lattice points
by anion B. Hence, the ratio of cation of
A and anion B will be
14.
o
A+ is 1.3 A while that of anion B– is 1.76
0
A . Predict the structure of AB and
coordination number of A+.
Solution :
o
A+ = r+ = 1.285 A
2 1
: or 4 : 3.
3 2
o
Radius of anion, B− = r_ = 1.76 A
Structure of AB = ?
Coordination number of A+ = ?
Therefore, the molecular formula will
be A4B3.
A compound AmBn with A in the form of
cation and B in the form of anion form ccp
type crystal lattice. The cations of A occupy
all tetrahedral voids. Predict the formula
of compound.
Solution :
(a) In the ccp type crystalline lattice, cations of
A occupy all the tetrahedral voids while anions
of B occupy 8 corners and 6 face centres.
(b) Total number of anions of B
13.
=
(c)
∴
(d)
1
1
× 8 + × 6 = 1 + 3 =4
8
2
Number of octahedral voids = Number of
anions = 4
Number of cations of A = 4.
Hence, ccp unit cell will contain 4 cations
of A and 4 anions of B.
Hence, AmBn = A4B4.
Therefore, the formula of the
compound is AB.
In the ionic solid AB, the radius of cation
r+ 1.285
=
0.730
r–
1.76
∴
Since the radius ratio is 0.730 ≅ 0.732, the
crystalline structure is cubic and
coordination number of A+ is 8.
Crystalline structure is cubic.
Coordination number of A+ = 8.
15.
If the coordination number of a cation in
an ionic solid is 4. What is the type of void
occupied by cation ?
Solution :
Coordination number of cation is 4. Hence,
the void must be tetrahedral.
16.
An organic compound crystallises in an
orthorhombic system with two molecules
per unit cell. The unit cell dimensions are
12.05 ×10–8 and 15.05 × 10–8 cm and 2.69
× 10–8 cm. If density of crystal is 1.419 g
cm 3 . Calculate the molar mass of
compound.
Chapter - 1 Solid State
66
Given :
Cell dimension = 12.05 × 10–8
and 15.05 × 10–8 cm
and 2.69 × 10–8 cm
d = 1.149 g cm3
To find :
M=?
Solution :
Molar mass (M)
18.
= density × Total × No. of molecules per unit cell
d × V × NA
M =
n
=
=
(1.419)(12.05×15.05×2.69×10 –24 )(6.022×10 23 )
2
=
209 g mo1–1
Lithium borohydride, LiBH 4 in an
orthorhombic system with four molecules
per unit cell. The unit cell dimensions are,
a = 6.81ºA, b = 4.43ºA and c = 7.14ºA. If the
molar mass of LiBH4 is 21.76 g mol–1.
Calculate the density of the crystal.
Given :
Molar mass = 21.76 g mole–1.
unit cell dimensions are, a = 6.81ºA,
b = 4.43ºA and c = 7.14ºA
To find :
d=?
Solution :
1–A = 10–8 cm
N⋅M
= V ⋅ NA
=
mass
100
=
= 13.9 cm3.
density
7.20
No. of unit cells in this volume
17.
d
A metallic element exists as a cubic lattice.
Each edge of the unit cell is 2.88 × 10–8 cm.
The density of metal is 7.20 g cm3– How
many unit cell there will be in 100g of metal?
Given :
d = 7.20 g cm–3
Total volume of unit cell = 2.88 × 10–8 cm
To find :
no. of unit cell = ?
Solution :
Total volume of unit cell
= a3 = (2.88 × 10–8) = 23.9 × 10–24 cm3.
Volume of 100 g of metal
4 × 21.76
(6.81 × 4.43 × 7.17 × 10
3–
–24
) × (3.022 × 10
= 0.668 g cm .
Unique Solutions ®
23
)
=
Volume of 100g of metal
Volume of unit cell
=
13.9
23.9 × 10 –24
=
∴
5.82 × 1023
Number of unit cells in 100g of metal
is 5.82 × 1023.
19.
Sodium crystallises in a bcc unit cell.
Calculate the approximate number of unit
cells in 9.2g of sodium. (at mass of Na =
23g)
[CBSE sample paper 2011]
Given :
unit cell = 9.2 g of sodium
To find :
number of unit cells = ?
Solution :
No. of atoms in 9.2 g of sodium
S.Y.J.C. Science - Chemistry - Part I
67
=
wt. of sodium × Avogadro's number
Atomic mass
=
9.20× 6.022 × 10 23
= 2.4088 × 1023
23
To find :
density = ?
Solution :
=
A bcc unit cell contains 2 atoms
∴
No. of unit cells
=
No. of unit cells =
2.4088 × 10
2
23
1.2044 × 1023
20.
Sodium metal crystallises in bcc lattice with
cell edge, 4.29ºA. What is radius of sodium
atom?
Given :
a = 4.29ºA
22.
To find :
r=?
Solution :
For a bcc, the radius of atoms is
Given
r
=
=
r
21.
n × formula mass of substance
a3 × NA
AB has rock salt structure i.e. f.c.c. type.
Number of atoms per unit cell (n) is 4.
formula mass = 6.023 yN03– kg
a = 2yr3nm
= 2yr3 × 10–9m
4 × 6.023y × 10 –3
1
d =
6.022 × 10 23 × (2y 3 × 10 –9 ) 3
d = 5.0 kg m3–.
Silver has cubic unit cell with a cell edge
of 408 pm. It’s density is 10.6 g cm–3. How
many atoms of silver are there in unit cell?
What is the structure of silver?
:
edge length = 408 pm = 408 × 10–8 cm
density = 10.6 g cm3.
Atomic mass of silver = 108
To find :
density = ?
Solution :
3
a
4
1.732 × 4.29
4
= 1.86ºA
A compound AB has rock salt type
structure. The formula wt. of AB is 6.023
y amu and the closest A–B distance is
∴
d =
nM
a 3N A
n =
d × a3 × NA
M
y 1 3 nm. Where y is an arbitrary number.
=
Find the density of.
(I.I.T. 2004)
Given :
formula wt. of AB = 6.023 y amu
distance =
1
3
nm
∴
10.6×(408×10 –8 ) 3 ×108×6.022×10 23
108
= 4.02
No. of atoms per unit cell = 4
No. of atoms per unit cell of silver is
4, it has face centred cubic crystal.
Chapter - 1 Solid State
68
23.
Copper crystallises in fcc entered cubic
lattice and has density of 8.930 g cm–3 at
293K. Calculate the radius of copper atom.
(at mass of Ca = 63.55)
Given :
density = 8.930 g cm3.
Atomic mass of Ca = 63.55
To find :
Solution :
nM
d= 3
a NA
a3
=
4 × 63.55
8.93 × 6.022 × 10 23
= 4.73 × 10–23cm3
a
= (4.73 × 10–23)r3
= 3.616 × 10–8cm
for fcc
r =
=
Number of atoms (n) in fcc unit cell = 4
2a
2 × 3.616 × 10 –8
=
4
4
1.141 × 3.616 × 10 –8
4
= 127.8 × 10–8cm or 127.8 pm
radius = 127.8 × 10–8cm or 127.8 pm
4 × 63.55
a × 6.022 × 10 23
∴
8.93 =
NUMERICALS FOR PRACTICE
3
1.
Potassium iodide has unit cell with cell edge of 7.5 pm. The density of KI is
3.12 g cm3–. How many potassium and iodide ions are contained in it?
(Ans. : 4K+8 4I–)
2.
Rubidium atomic mass 85.51 crystallises in a body centred cubic lattice with density of 1.51 g
cm3–. If the radius of rubidium atom is 248 pm, calculate Avagadro’s number.
(Ans. : 6.023 × 1023)
3.
Lithium metal has body centred cubic close packing. It’s density is 0.53 g cm3– and its molar
mass is 6.94 g mol. Calculate the volume of the unit cell of lithium.
(Ans. : 4.35 × 1023 cm3)
4.
The density of copper metal is 8.95 of cm3–. If the radius of atoms is 127.8 pm. is the copper
unit cell simple cubic body centred or face centred cubic.
(At mass of Ca = 63.54)
5.
A metallic element ‘x’ exists as a cubic lattice. Each edge of the unit cell is 2.90ºA and density
of metals is 7.20 g cm3. How many unit cell will be present in 100 g of the metal?
(Ans. : 5.7 × 1023)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
69
6.
A cubic solid is made of two elements X and Y. Atoms Y are present at the corners of cube
and atoms X at the body centre. What is the formula of the compound? What are the coordinations
of X and Y? (CBSE sample paper 2008)
(Ans. : XY, 8 each)
7.
When atoms are placed at the corners of all 12 edges of a cube. How many atoms are present
per unit cell?
(Ans. : 1)
8.
A unit cell consists of a cube in which there are A atoms at the corners and B atoms at the
centres and A atoms are missing from two corners in each unit cell. What is the simplest formula
of the compound? (PB S.B.E. 2011)
(Ans. : AB4)
1.
Ans.
∴
∴
2.
Ans.
∴
∴
3.
Ans.
HIGHER ORDER THINKING SKILLS (HOTS)
Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids.
What is the formula of compound formed by the element A and B
(NCERT).
The number of tetrahedral voids formed is twice the number of element B and only 2/3rd of
it is occupied by atoms of element A.
2
The ratio of atoms of element and B = 2 × : 1 = 4 : 3
3
The formula of the compound is A4B3
A solid made of two elements P and Q. Atoms Q are in ccp arrangement, while atoms P occupy
all the tetrahedral sites. What is the formula of the compound?
(PB.S.B.E 2002).
The ccp lattice is formed by the element Q. The number of octahedral voids generated would
be equal to the number of atoms Q. The number of tetrahedral voids is double the number of
octahedral voids.
The ratio of P : Q = 2 : 1
The formula of compound = P2Q
How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00 g?
(a) 2.57 × 1021
(b) 5.14 × 1021
(c) 1.28 × 1021
(d) 1.71 × 1021
Amount of NaC =
No. of unit cells in 1.0 g of NaCl =
1.00g
58.5g/mol
6.02 × 10 23 mol –1
1.00g
×
4
58.5g/mol
= 2.57 × 1021
Chapter - 1 Solid State
70
4.
Ans.
How many Cs+ ions occupy second nearest neighbour locations of a Cs+ ion in the structure
of cesium chloride crystals?
The cesium chloride structure, each Cs+ is surrounded by eight Cl– as the nearest neighbour
3
r , where r = d –
Cl
2
at a distance of
– Cl –
The structure shows that each Cs+ finds six Cs+ as the next nearest neighbours at a distance
of r, where r = d
5.
Ans.
Cl – – Cl –
The compound Y Ba2Cu3O7, which shows superconductivity, has copper in oxidation state .......
Assume that the rare earth element yttrium is in its usual + 3 oxidation state.
The oxidation state of Y = +3
Barium exists as Ba2+ and oxygen as O2– (as in oxides) if x is the charge on copper atom, then
(+3) + 2 × (+2) + 3 × x + 7 × (–2) = 0
This gives, x =
7
3
Thus, the oxidation state of copper in the given compound is
7
3
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
Chapter
2
Solutions and
Colligative Properties
"Imagination is more important than knowledge."-Einstein
Introduction :
The Study of Chemistry is incomplete without the study of solutions. Solutions are mixtures
of two or more components. Depending on sizes of the components, the mixtures are classified
into three types :
1. A coarse mixture
2. A colloidal dispersion
3. A true solution
SYLLABUS
2.1
(a)
(b)
(c)
SOLUTIONS
Various terms involved in solutions
Types of solutions
Ways of expressing concentrations of
solutions
(d) Solubility of solute in solvent
(e) Henry’s law
(f) Solid solutions
Theoretical MCQs
Numerical MCQs
2.2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
COLLIGATIVE PROPERTIES AND
LOWERING OF VAPOUR PRESSURE
Colligative properties
Vapour pressure of liquids
Factors which affect vapour pressure
Lowering of vapour pressure
Relative lowering of vapour pressure
Raoult’s law
Raoult’s law for a solution of non-volatile
solute in a volatile solvent
Expression for relative lowering of vapour
pressure / Expression for lowering of vapour
pressure
(i) Expression for molar mass of a solute and
relative lowering of vapour pressure
Theoretical MCQs
Numerical MCQs
2.3
(a)
(b)
(c)
(d)
(e)
(f)
ELEVATION OF BOILING POINT AND
DEPRESSION OF FREEZING POINT
(1) Boiling point
(2) Elevation in boiling point
Relation between elevation in boiling point
and lowering of vapour pressure
Molal elevation constant or Ebullioscopic
constant
Relation between elevation of boiling point
and molar mass of the solute
(1) Freezing point
(2) Depression in freezing point
(3) Cause for depression of freezing point
Depression of freezing point and lowering
of vapour pressure graphically
72
(g) Molal depression constant or Cryoscopic
constant
(h) Relation between molecular mass of solute
and depression in freezing point
(i) Appplications of depression of freezing point
Theoretical MCQs
Numerical MCQs
2.4 OSMOTIC PRESSURE, ABNORMAL
MOLECULAR MASS AND VAN’T HOFF
FACTOR
(a) Semi permeable membrane
(b) Osmosis
(c) Abbe Nollet experiment
(d) Types of osmosis and osmotic pressure
(1) Osmotic pressure
(2) Isotonic solutions
(3) Hypertonic solutions
(4) Hypotonic solutions
(e) Osmosis in day to day life
1.
2.
3.
(f) Laws of osmotic pressure
(1) van’t Hoff - Boyle law
(2) van’t Hoff - Charles law
(3) van’t Hoff - Avogadro’s law
(4) van’t Hoff’s equation for dilute solutions
(g) Determination of molecular mass from
osmotic pressure
(h) Reverse Osmosis
(i) Abnormal molecular mass
(j) van’t Hoff factor
(k) van’t Hoff factor (i) and degree of
dissociation (α)
Theoretical MCQs
Numerical MCQs
HOURS BEFORE EXAM
NUMERICALS WITH SOLUTION
NUMERICALS FOR PRACTICE
HIGHER ORDER THINKING SKILLS
A coarse mixture : It is formed when the sizes of the constituent components are relatively
bigger, e.g. mixture of salt and sugar.
A colloidal dispersion : It is formed when the sizes of the particles dispersed in solvent
are in the range of 10-7 cm to 10-4 cm. Colloidal particles carry positive or negative charge
which stabilizes colloidal dispersion e.g. ferric hydroxide sol, arsenic sulphide sol, etc. Colloidal
solutions are heterogeneous and can be easily separated.
A true solution : A true solution is formed when soluble substances are dissolved in the solvent.
The sizes of the particles dissolved are very small of the order of 10-8 cm. True solutions are
homogeneous and cannot be separated into components by simple mechanical methods.
The extent to which solute dissolves in solvent to form homogeneous solution depends on nature
of solute and solvent, because the solute particles are very small. The components of true
solutions cannot be separated by simple physical methods like centrifugation, filtration etc.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
73
2.1
SOLUTIONS :
Concept Explanation :
(a) Various terms involved in solutions :
1. Solution : A true solution is defined as a homogeneous mixture of two or more substances,
the composition of which is not fixed and may be varied within certain limits.
2. Solute : The component that constitutes smaller part of solution is called solute.
3. Solvent : The component that constitutes larger part of the solution is called solvent.
4. Solvation : The process of solvation is the interaction of solvent molecules with solute
particles to form aggregates.
5. Hydration or Aquation : When water is a solvent then the process of solvation is called
hydration or Aquation.
6. Binary solution : Solution with two components in it are called binary solutions.
Similarly for ternary and quaternary solutions.
7. Aqueous solutions : Solutions in which water is used as a solvent are called aqueous solutions.
8. Non – Aqueous solutions : Solutions in which solvent used is other than water are called
non – aqueous solutions.
9. Concentration of Solution : It is defined as the amount of solute dissolved in a specific
amount of solvent.
10. Dilute solutions : Solutions containing relatively less amount of solute are called dilute solutions.
11. Concentrated solutions : Solutions containing relatively more amount of solute are called
concentrated solutions.
(b) Types of solutions :
Solute
Solvent
Gaseous Solutions
Gas
Gas
Liquid
Gas
Common Examples
Solid
Liquid Solutions
Gas
Liquid
Solid
Solid Solutions
Gas
Gas
Mixture of non reacting gases, air.
Chloroform vapours mixed with nitrogen gas, water
vapour in air (humidity).
I2 in air, dust or smoke particles in air.
Liquid
Liquid
Liquid
Oxygen dissolved in water, CO2 dissolved in water.
Ethanol dissolved in water.
Sucrose or salt in water, benzoic acid in benzene.
Solid
Liquid
Solid
Solid
Solid
Solution of hydrogen in palladium, pumice stone
(phenomenon of adsorption of gases over metals).
Mercury with sodium (amalgam) and crystalline
salts (CuSO4;5H2O).
Copper dissolved in gold, alloys like, brass , bronze.
Chapter - 2 Solutions and Colligative Properties
74
(c) Ways of expressing concentrations of solutions :
1. Percentage by mass (W/W) :
The mass of solute in gram dissolved in solvent to form 100 gram of solution is called
percentage by mass.
Mass percentage (% by weight) =
mass of solute
× 100
mass of solution
2. Percentage by volume (V/V) :
It is defined as the ratio of number of parts by volume of the solute to one hundred parts
by volume of the solution.
Percentage by Volume of solute =
volume of solute
× 100
volume of solution
Note : Volume is temperature dependent quantity, hence, percentage by volume
changes with change of temperature.
3. Mole fraction (x) :
The mole fraction of any component of a solution is defined as the ratio of number of
moles of that component present in the solution to the total number of moles of all components
of the solution.
Let us suppose that a solution contains n2 moles of solute and n1 moles of the solvent.
Then,
n2
Mole fraction of solute (x2) = n + n
1
2
n1
Mole fraction of solvent (x1) = n + n
1
2
x1 + x2 = 1
(For binary solution containing 1 solute and 1 solvent)
Note :
(1) It may be noted that the mole fraction is independent of temperature.
(2) It does not have any units.
(3) The sum of mole fractions of all the constituents of a solution is unity.
4. Molarity (M) :
It is defined as the number of moles of solute present in 1 dm3 volume of the solution.
Number of moles of solute
Molarity (M) =
Volume of solution in litre or dm 3
mass of the substance
No. of moles of a substance (n) = Molar mass of the substance
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
75
Note :
(1) Molarity has one disadvantage. It changes with temperature because of
expansion or contraction of the liquid with temperature.
(2) Units of Molarity : mol dm–3 OR Mol lit–1
(3) Molar = 1M, Decimolar = 0.1M , CentiMolar = 0.01M, Millimolar = 0.001M
5. Molality (m) :
(a) It is defined as the number of moles of solute dissolved in 1 kg of solvent.
Molality =
No. of moles of solute in kg
Mass of solvent in kg
(b) Units of Molality : Mol / Kg.
(c) Molality is considered better as compared to molarity because the molarity changes
with temperature because of expansion or contraction of the liquid with temperature.
However, molality does not change with temperature because mass of the solvent
does not change with change in temperature.
6. Parts per million :
It is the mass or volume of solute in gram or cm3 per 106 g or 106cm3 of the solution.
Mass or volume of Solute
Parts per Million (ppm) = Total Mass or volume of solution × 10
6
(d) Solubility of solute in solvent :
1. Solubility : The maximum amount of solute that dissolves completely in a given amount
of solvent at a constant temperature is called solubility.
It is expressed in terms of mol dm-3. Solubility changes with temperature.
2. Saturated solution : A solution in which no more solute can be dissolved at the given
temperature and pressure is called as saturated solution.
In a saturated solution an equilibrium exists between dissolution and crystallization.
3. Unsaturated solution : It contains less amount of solute than required for forming saturated
solution.
4. Super saturated solution : It contains excess of solute than required for formation of
saturated solution.
5. Factors affecting the solubility of a solid in a liquid :
(a) Nature of solute and the solvent : “Like dissolves like” is the general rule followed.
Polar solutes are soluble in polar solvents. E.g.: NaCl in water.
Non polar solutes are more soluble in non-polar solvents. Eg: Iodine in CCl4.
(b) Temperature : The solubility of a solid in liquid is affected by temperature.
Chapter - 2 Solutions and Colligative Properties
76
Generally solubility increases with increase in temperature if the dissolution process
is endothermic. If the dissolution process is exothermic then solubility increases with
decrease in temperature. However, this is not always true.
(c) Pressure : Solids are incompressible hence, change in pressure has no effects on
solubility of solids in liquids.
6. Factors affecting solubility of a gases in liquids :
(a) Nature of solute and solvent : The principle of “like dissolves like ” is also applicable
here. Polar gases like HCl and NH3 are highly soluble in polar solvent like water.
While O2, H2 and N2 being non polar is less soluble in water.
(b) Effect of temperature : According to Charles law, volume of a given mass of a
gas increases with increase in temperature.
Therefore, the volume of a given mass of dissolved gas in a solution also increases
with increase in temperature. So that it becomes impossible for the solvent to
accommodate gaseous solute in it and gas bubbles out.
Hence, solubility of gas in liquid decreases with increase in temperature.
(c) Effect of pressure : Gases are highly compressible in nature.
Hence, solubility of gases is greatly influenced by change of external pressure of the
gas. Increase of external pressure increases the solubility of gas.
In case ‘b’ when pressure of the gas is increased, its solubility also increases.
(d) Effect of addition of soluble salt : Solubility of dissolved gas is suppressed when
a soluble salt is added to the solution of gas. Eg : When table salt is dissolved in carbonated
drink, the solubility of CO2 in soft-drink decreases and CO2 escapes out producing
effervescence.
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(e) Henry’s law :
1. The law states that “the solubility of a gas in a liquid at constant temperature is proportional
to the pressure of the gas above the solution”.
If ‘S’ is the solubility of the gas in mol dm-3 and ‘P’ is the pressure of the gas in atmosphere,
then according to Henry’s law
S∝ P
S = KH × P
(KH = proportionality constant called as Henry’s constant and has unit mol dm–3 atm–1)
2. Henry’s Constant : It is defined as solubility of gas in mol dm–3 at 1 atmosphere pressure
at reference temperature. Henry’s constant depends upon the nature of the gas, the nature
of the solvent and the temperature.
3. Marine life like fish prefer to stay at lower level in the sea.
(a) In summer during a hot day, temperature at the surface of water is relatively high
and therefore, the solubility of oxygen in the upper layer is minimum.
(b) At the same time the temperature of water at lower level is much less.
(c) Water at lower level therefore, has more dissolved oxygen.
(d) Hence, marine life prefer to stay at lower level in the sea.
(f) Solid solutions :
1. A solid solution of two or more metals or of a metal or metals with one or more non
metals is called an alloy or solid solutions.
2. Duralumin is an alloy of mainly aluminium, copper, magnesium and manganese. It is light;
strong as steel and is used in aircraft construction.
3. Lead hardened by 10–20% antimony is used in bearings, bullets etc. It is resistant to acids
and is therefore, used for manufacturing of lead storage battery plates.
4. Alloy of antimony with tin and copper is called Babbitt metal and it is antifriction alloy
used in machine bearings.
5. Chromium is used in steel alloys which are very hard and strong. Stainless steel contains
chromium which makes it resistant to corrosion.
6. Alloy spiegeleisen containing 5 to 20% manganese in iron and ferromanganeous containing
70 to 80% manganese are used for making very hard steels for rails, safes and heavy machinery.
7. Alloy manganin containing 84% Cu, 12%Mn and 4% Ni has almost zero temperature coefficient
of electrical resistance and used in instruments used for making electrical measurements.
8. Alloys of mercury with other metals are called amalgams. This property of mercury to
form amalgams is used for extracting metals from ores.
Chapter - 2 Solutions and Colligative Properties
78
•
*1.
Ans.
Define the terms : (a) Solution, (b) Solute, (c) Solvent, (d) Concentration.
(a) Solution : A true solution is defined as a homogeneous mixture of two or more substances,
the composition of which is not fixed and may be varied within certain limits.
(b) Solute : The component that constitutes smaller part of solution is called solute.
(c) Solvent : The component that constitutes larger part of the solution is called solvent.
(d) Concentration : It is defined as the amount of solute dissolved in a specific amount of solvent.
*2.
Ans.
What are the different types of solutions?
Solute
Solvent
Common Examples
Gaseous Solutions
Gas
Gas
Mixture of non reacting gases, air.
Liquid
Gas
Chloroform vapours mixed with nitrogen gas, water
vapour in air (humidity).
Solid
Gas
I2 in air, dust or smoke particle in air.
Liquid Solutions
Gas
Liquid Oxygen dissolved in water, CO2 dissolved in water.
Liquid
Liquid Ethanol dissolved in water.
Solid
Liquid Sucrose or salt in water, benzoic acid in benzene.
Solid Solutions
Gas
Solid
Solution of hydrogen in palladium, pumic stone
(phenomenon of adsorption of gases over metals).
Liquid
Solid
Mercury with sodium (amalgams) and crystalline
salts (CuSO4. 5H2O).
Solid
Solid
Copper dissolved in gold (alloys), brass, bronze.
*3.
Ans.
Explain the different ways of expressing concentration of a solution.
Refer 2.1(c) 1, 2, 3, 4, 5, 6.
*4.
Ans.
Define : (a) Percentage by weight, (b) Mole fraction, (c) Molality, (d) Molarity.
(a) Percentage by mass (W/W) : The mass of solute in gram dissolved in solvent to form
100 gram of solution is called percentage by mass.
mass of solute
Mass percentage (% by weight) = mass of solution × 100
(b) Mole fraction : The mole fraction of any component of a solution is defined as the ratio
of number of moles of that component present in the solution to the total number of moles
of all components of the solution.
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S.Y.J.C. Science - Chemistry - Part I
79
Let us suppose that a solution contains n2 moles of solute and n1 moles of the solvent.
Then,
n2
Mole fraction of solute (x2) = n + n
1
2
n1
Mole fraction of solvent (x1) = n + n
1
2
x1 + x2 = 1
(For binary solution containing 1 solute and 1 solvent)
(c) Molality (m) :
(1) It is defined as the number of moles of solute dissolved in 1 kg of solvent.
Molality =
Number of moles of the solute in kg
Mass of solvent in kg
(2) Units of Molality : Mol / Kg .
(3) Molality is considered better as compared to molarity because the molarity changes
with temperature because of expansion or contraction of the liquid with temperature.
However, molality does not change with temperature because mass of the solvent
does not change with change in temperature.
(d) Molarity (M) : It is defined as the number of moles of solute present in 1 dm3 volume
of the solution.
Number of moles of solute
Molarity =
Volume of solution in litre or dm 3
mass of the substance
No. of moles of a substance (n) =
Molar mass of the substance
*5.
Ans.
Explain why concentration in terms of molality is preferred in comparison to molarity.
Molality is considered better as compared to molarity because the molarity changes with
temperature because of expansion or contraction of the liquid with temperature. However,
molality does not change with temperature because mass of the solvent does not change with
change in temperature.
6.
Ans.
Define solubility and explain the factors affecting solubility of a solid in a liquid.
(a) Solubility : The maximum amount of solute that dissolves completely in a given amount
of solvent at a constant temperature is called solubility.
Solubility is expressed in grams of solute per 100cm3 of solvent.
It is also expressed in terms of mol dm-3. Solubility changes with temperature.
(b) Factors affecting the solubility of a solid in a liquid :
(1) Nature of solute and the solvent : “Like dissolves like” is the general rule followed.
Polar solutes are soluble in polar solvents. E.g.: NaCl in water.
Chapter - 2 Solutions and Colligative Properties
80
Non polar solutes are more soluble in non-polar solvents. Eg: Iodine in CCl4.
(2) Temperature : The solubility of a solid in liquid is affected by temperature.
Generally solubility increases with increase in temperature if the dissolution process
is endothermic. If the dissolution process is exothermic then solubility increases with
decrease in temperature. However, this is not always true.
(3) Pressure : Solids are incompressible hence, change in pressure has no effects on
solubility of solids in liquids.
7.
Ans.
Define solubility and explain the factors affecting solubility of a gas in a liquid.
(a) Solubility : The maximum amount of solute that dissolves completely in a given amount
of solvent at a constant temperature is called solubility.
Solubility is expressed in grams of solute per 100 cm3 of solvent.
It is also expressed in terms of mol dm-3. Solubility changes with temperature.
(b) Factors affecting solubility of a gases in liquids :
(1) Nature of solute and solvent : The principle of “like dissolves like ” is also applicable
here. Polar gases like HCl and NH3 are highly soluble in polar solvent like water.
While O2, H2 and N2, being non polar is less soluble in water.
(2) Effect of temperature : According to Charles law, volume of a given mass of a
gas increases with increase in temperature.
Therefore, the volume of a given mass of dissolved gas also increases with increase
in temperature. So that it becomes impossible for the solvent to accommodate gaseous
solute in it and gas bubbles out.
Hence, solubility of gas in liquid decreases with increase in temperature.
(3) Effect of pressure : Gases are highly compressible in nature.
Hence, solubility of gases is greatly influenced by change of external pressure of the
gas. Increase of external pressure increases the solubility of gas.
In case ‘b’ when pressure of the gas is increased, its solubility also increases.
(4) Effect of addition of soluble salt : Solubility of dissolved gas is suppressed when
a soluble salt is added to the solution of gas. Eg : When table salt is dissolved in carbonated
drink, the solubility of CO2 in soft-drink decreases and CO2 escapes out producing
effervescence. [Refer fig. 2.1 d (6)]
*8.
Ans.
State and explain Henry’s law.
The law states that “the solubility of a gas in a liquid at constant temperature is proportional
to the pressure of the gas above the solution”.
If ‘S’ is the solubility of the gas in mol dm-3 and ‘P’ is the pressure of the gas in atmosphere,
then according to Henry’s law
S ∝P
S = KH × P
(KH = proportionality constant called as Henry’s constant and has unit mol dm–3 atm–1)
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S.Y.J.C. Science - Chemistry - Part I
81
9.
Ans.
10.
Ans.
*11.
Ans.
Define Henry’s constant.
It is defined as solubility of gas in mol dm–3 at 1 atmosphere pressure at reference temperature.
Henry’s constant depends upon the nature of the gas, the nature of the solvent and the temperature.
Give reason : Marine life like fish prefer to stay at lower level in the sea.
Marine life like fish prefer to stay at lower level in the sea, because :
(a) In summer during a hot day, temperature at the surface of water is relatively high and
therefore, the solubility of oxygen in the upper layer in minimum.
(b) At the same time the temperature of water at lower level is much less.
(c) Water at lower level therefore, has more dissolved oxygen.
Explain solid solution.
Refer 2.1 (f)
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
A molal solution is one that contains one mole of a solute in .......
a) 1 L of solvent
b) 1000g of solvent
c) 1 L of solution
d) 22.4 litres of solution
Which of the following is independent of temperature?
(MHT-CET 2008)
a) Normality
b) Molarity
c) Molality
d) Formality
Molarity of solution depends on .......
a) Temperature
b) Nature of solute dissolved
c) Mass of solvent
d) Pressure
*2.
*3.
•
*4.
Ans.
Numerical MCQs
20g of NaOH (Molar mass = 40gmol–1)
is dissolved in 500cm3 of water. Molality
of resulting solution is .......
a) 0.1m
b) 0.5m
c) 1.5m
d) 1.0m
Molality
=
moles of NaOH
Mass of solvent in kg
=
20
40 × 0.5kg
5.
Ans.
A solution is 0.25% by weight. The weight
of solvent containing 1.25g of solute would
be .......
a) 500g
b) 498.75g
c) 500.25g
d) 501.25g
% by weight =
mass of solute
× 100
mass of solution
1.25
0.25 = mass of solution × 100
= 1m
∴
mass of solution = 500g
mass of solvent = 500 – 1.25 = 498.75g
Chapter - 2 Solutions and Colligative Properties
82
6.
Ans.
=
=
=
7.
Ans.
8.
Ans.
∴
Calculate the mole fraction of glucose in
solution (molecular weight = 180g) when
3.6g of glucose is added to 90ml water at
25ºC to produce glucose solution .......
a) 4.0
b) 0.4
c) 0.04
d) 0.004
mole fraction of glucose
moles of glucose
moles of glucose + moles of water
3.6 /180
3.6 /180 + 90 /18
3.6
903.6
= 0.004
9.
Ans.
10.
A centi-molar solution is diluted 10 times.
Its molarity would become .......
a) 0.01
b) 0.1
c) 0.001
d) 0.005
original molarity
Resulting molarity =
dilution
0.01
=
= 0.001M
10
An aqueous solution of non-electrolyte ‘A’
with molecular mass 60 contains 6g in
500mL and has a density equal to 1.05g
mL–1. The molality of solution is .......
a) 1.25
b) 0.19
c) 0.25
d) 0.30
Mass of solution = density × volume
= 1.05 × 500
= 525g
Mass of solvent = 525 – 6 = 519g
= 0.519kg
Ans.
∴
11.
Ans.
12.
moles of solute
Molality (m) = mass of solvent in kg
=
6
= 0.19m
60 × 0.519
Unique Solutions ®
Ans.
100ml of 0.1M solution A is mixed with
20ml of 0.2M solution B. The final molarity
of the solution is .......
a) 0.12M
b) 0.15M
c) 0.18M
d) 0.21M
M =
M 1V1 × M 2 V2
V1 + V2
=
0.1 × 100 + 0.2 × 20
100 + 20
=
0.12M
The amount of solute required to prepare
10L of a decimolar solution is .......
a) 0.01mol
b) 0.2mol
c) 0.1mol
d) 1mol
n
n
M =
,
0.1 =
v
10
n = 1 mol
What is molality of the solution prepared
by dissolving 18g of glucose.
(mol. mass – 180) in 500g of water?
a) 1.2m
b) 0.4m
c) 0.1m
d) 0.2m
n
m = W (kg)
1
18
=
180 × 500 × 10 –3 (kg)
= 0.2m
Ten grams of potassium chloride are
dissolved in 103 kg of solution, its strength
may be expressed as .......
a) 1 ppm
b) 10 ppm
c) 100 ppm
d) 1000 ppm
ppm =
mass of solute
× 10 6
mass of solution
S.Y.J.C. Science - Chemistry - Part I
83
=
10 × 10 –3 (kg)
10 3
× 10
6
•
15.
= 10 ppm
13.
Ans.
14.
Ans.
A sugar syrup of weight 214.2 contains
34.2g of sugar (molar mass – 342). The
molality of the solution is .......
a) 0.0099
b) 0.56
c) 0.28
d) 0.34
n
m =
W1 (kg)
34.2
=
342 × (214.2 – 34.2) × 10 –3 kg
1
=
10 × 180 × 10 –3
1
=
= 0.56m
1.8
The mole fraction of the solute in one molal
aqueous solution is .......
a) 0.027
b) 0.06
c) 0.018
d) 0.009
Let the mass of H2O be 1kg
n2
m
= W (kg)
1
n2
1
=
, n2 = 1 mole
1
1000
= 55.55 mol
=
18
n2
= n +n
x2
1
2
x2
1
=
1 + 55.55
=
1
≈ 0.018
56.55
Ans.
The molecular weight of KOH is 56. What
is the molarity of solution prepared by
dissolving 84.0 gram of KOH in 500ml of
solution?
(Sept. 2009)
a) 3
b) 5
c) 2
d) 2.5
M =
n2
V
84
=
16.
Ans.
∴
17.
Ans.
∴
56 × 500 × 10 –3
= 3M
If 100ml of 1M NaOH solution is diluted
to 1 dm3, the molarity of the resulting
solution is ........
(March 2010)
a) 1M
b) 0.1M
c) 10M
d) 0.05M
= M2 V2
M 1V 1
1 × 100 = M2 × 1000
1
M2 =
= 0.1M
10
Number of moles present in 6dm3 of
solution to form centimolar solution is .......
(MHT-CET 2008)
–2
b) 6 × 10–3
a) 6 × 10
c) 6 × 10–1
d) 6 × 10–4
n2
M
=
V
n2
0.01 =
6
n2
= 0.06 mole
= 6 × 10–2
18.
If molality of NaOH (molar mass = 40)
solution is 1.25m, then calculate W/W%.
(MHT-CET 2009)
a) 2.35%
b) 1.18%
c) 4.76%
d) 9.52%
Chapter - 2 Solutions and Colligative Properties
84
Ans.
∴
n2
W1 (kg)
Let the mass of solvent be 1 kg
W2
1.25 =
40 × 1
W2 = 40 × 1.25 × 1 = 50g
m
20.
=
mass of solute×100
% W/W = mass of solute + mass of solution
50
× 100
50 + 1000
50
× 100 = 4.76%
=
1050
Ans.
=
19.
Ans.
∴
2.2
3
45g of glucose (Molar mass = 180) is
dissolved in 100g of water. What is molality
of the solution?
(MHT-CET 2010)
a) 0.25m
b) 0.025m
c) 2.5m
d) 0.0025m
n2
m = W (kg)
1
45
=
180 × 100 × 10 –3
=
0.1dm of 2M acidic solution is diluted
such that the concentration becomes 0.1M,
then the final volume of the solution.
(MHT-CET 2009)
3
b) 10 dm3
a) 1 dm
c) 2 dm3
d) 20 dm3
M 1 V 1 = M 2V 2
2 × 0.1 = 0.1 × V2
V 2 = 2 dm3
21.
Ans.
∴
2.5m
25cm3 of 0.2M solution of H2SO4 is diluted
to 0.5 dm3, the resulting solution is .......
(MHT-CET2010)
a) 0.01M
b) 0.001M
c) 0.2M
d) 0.002M
M 1 V 1 = M 2V 2
0.2 × 25 × 10–3 (dm3)
= M2 × 0.5 (dm3)
M2
0.2 × 25 × 10 –3
=
0.5
= 0.01 M
COLLIGATIVE PROPERTIES AND LOWERING OF VAPOUR
PRESSURE :
Concept Explanation :
(a) Colligative properties :
1. “Colligative properties are defined as the properties of solutions that depend only on the
number of solute particles in solution and not on the nature of solute particles.”
2. Examples :
(a) Lowering of vapour pressure
(b) Elevation of boiling point of solvent in solution
(c) Depression of freezing point of solvent in solution
(d) Osmotic pressure.
3. Explanation :
(a) All colligative properties are connected together because these properties are based
on the same principle.
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85
(b) The properties depend upon number of solute particles, the particles may be atoms,
molecules, ions or aggregates of molecules.
(c) The actual number of solute particles present in solution decides the value of colligative
properties.
(d) It is convenient to restrict the discussion to solutions of non electrolytes in liquid solvent.
(e) The non electrolytes that neither dissociate nor associate in solution.
(f) Colligative properties can be used to determine molar masses of non electrolyte solutes.
(g) Colligative properties are applicable only to dilute solutions with concentrations less
than or equal to 0.2M.
(b) Vapour pressure of liquids :
1. The vapour pressure of a
substance is defined as the
pressure exerted by the gaseous
state of that substance when it
is in equilibrium with the solid
of liquid phase.
2. At room temperature, the
molecules of liquid are in
constant random motion.
3. The molecules at the surface of the liquid escape from the surface by absorbing energy
from the surrounding.
This process is called evaporation. If the vessel containing the liquid is closed with a lid,
then the molecules in vapour phase are trapped in the empty space and are in continuous
random motion.
4. During this motion, molecules collide with each other and also with the walls of the container,
thereby losing their energy and returning back to liquid state. This process is called
condensation.
5. Eventually the rate of evaporation and condensation becomes equal i.e. equilibrium is attained.
At this point the number of molecules in gaseous phase remains constant. The pressure
exerted by these molecules on the surface of liquid is called as vapour pressure of the liquid.
6. The vapour pressure of the liquid depends on the nature of the liquid and the temperature.
With increase of intermolecular forces of attraction, vapour pressure of liquid decreases
and with rise of temperature vapour pressure of liquid increases.
(c) Factors which affect vapour pressure :
1. Nature of solvent :
(a) All liquids exhibit tendency for evaporation.
(b) If the intermolecular forces of attraction are weak the liquids evaporate readily and
Chapter - 2 Solutions and Colligative Properties
86
are called volatile liquids. Eg: Ethyl acetate is the most volatile liquid. Other Egs. Ethyl
alcohol, Acetone etc.
(c) If intermolecular forces are strong, the liquids are less volatile and hence, exert lower
vapour pressure. Eg: Oil, mercury etc.
2. Nature of solute :
(a) Non-volatile solute : The vapour pressure of a solvent is lowered when a non–volatile
solute is dissolved to form a solution. This is due to the fact that when a non–volatile
solute is added, the surface area available for evaporation of volatile solvent is decreased.
Eg: Sugar in water.
(b) Volatile solute : In case of volatile solute, the vapour pressure of the resulting solution
is generally increased. Eg: Ethyl alcohol mixed with water.
3. Temperature :
(a) As temperature increases, vapour pressure of a
liquid increases.
(b) The temperature at which the vapour pressure of
the liquid becomes equal to the atmospheric pressure,
the liquid starts to boil. It is boiling point of the liquid
( Tb0 ).
(d) Lowering of vapour pressure :
The difference between the vapour
pressure of pure solvent ( p10 ) and the
vapour pressure of solvent from
solution (p) containing a non-volatile
solute is called the lowering of vapour
pressure. Thus, p10 – p = Δp =
Lowering of vapour pressure.
It is found that, the vapour pressure
of the solution (p) containing a non
volatile solute is always less than that
of the pure solvent p10 , i.e. p < p10 .
(e) Relative lowering of vapour pressure :
The relative lowering of vapour pressure of a solution is the ratio of the lowering of vapour
pressure of solvent from solution to the vapour pressure of the pure solvent.
Thus, Relative lowering of vapour pressure =
Unique Solutions ®
p10 – p
p10
S.Y.J.C. Science - Chemistry - Part I
87
(f) Raoult’s law :
The law states that, the partial vapour pressure of any volatile component of a solution
is the product of vapour pressure of that pure component and the mole fraction of the
component in the solution. Consider a solution containing two volatile components A1 and
A2 with mole fractions x1 and x2 respectively Let p10 and p 02 be the vapour pressures
of the pure components A1 and A2 respectively.
According to Raoult’s law, the partial vapour pressures p1 and p2 of two components are
given by,
0
p 10 × x1 and
p1 =
p2 = p 2 × x 2
p T = p1 + p 2
pT =
p 10 × x1 + p 02 × x 2
Note : The solution which obeys Raoult’s law over the entire range of concentration
is called ideal solution. If the solution does not obey Raoult’s law then the
solution is non–ideal.
(g) Raoult’s law for a solution of non-volatile solute in a volatile solvent
Raoult’s law :
For a solution containing a non-volatile solute in the volatile solvent, the law states that,
the vapour pressure of a solution is equal to the product of vapour pressure of the pure
solvent ( p 10 ) and mole fraction of the solvent (x1).
Explanation : Consider a binary solution of a non-volatile solute in a volatile solvent.
Let, p 10 = Vapour pressure of pure solvent.
p = vapour pressure of solution
x1 = Mole fraction of the solvent,
∴ p ∝ x1
OR p = p10 × x1
The above graph illustrates that as mole fraction of
solvent (x1) in solution increases the vapour pressure
of the solvent in solution (p) also increases.
(h) Expression for relative lowering of vapour pressure
OR
Expression for lowering of vapour pressure :
According to Raoult’s law, vapour pressure of the solution is equal to the product of
vapour pressure of the pure solvent and its mole fraction.
1. Since p = p10 × x1
Where,
p 10 = vapour pressure of pure solvent.
p = vapour pressure of solution
x 1 = Mole fraction of the solvent,
Chapter - 2 Solutions and Colligative Properties
88
2.
Lowering of vapour pressure = Δp = p10 – p
3.
∴ Δp = p10 – p10 × x1
4.
∴ Δp = p10 (1 – x1 )
5.
∴ Δp = p10 × x 2 (since, x1 + x2 = 1, therefore, 1 – x1 = x2 )
Hence, lowering of vapour pressure is the product of vapour pressure of pure solvent
( p10 ) and mole fraction of solute (x2).
p
6.
7.
From (5) we get,
p10 – p
∴
p10
= x2
= x2
p10
Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute.
which proves that it is a colligative property.
(i) Expression for molar mass of a solute and relative lowering of vapour
pressure :
1. Relative lowering of vapour pressure is given by the formula,
p10 – p
p10
2.
= x2
Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1.
Hence, number of moles of solvent, n1 and number of moles of solute, n2 in solution are
given as, n1 =
3.
W2 / M 2
n2
=
W1 / M 1 + W2 / M 2
n1 + n 2
........(ii)
Combining (i) and (ii). We get;
p10 – p
W2 / M 2
.........(iii)
W1 / M 1 + W2 / M 2
The relation (iii) is applicable only for dilute solutions. Hence, n1 >> n2 . Hence, n2 may
be neglected in comparison with n1 in the above equation. Thus, eqn. (iii) can be written
as follows :
p10
5.
W2
W1
and n2 =
M2
M1
The mole fraction of solute x2 is given by
x2 =
4.
.........(i)
= x2 =
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
89
p10 – p
p10
6.
*1.
Ans.
Therefore,
W2 / M 2
= W /M
1
1
p10 – p
p10
W2 M1
= W M
1
2
Define colligative properties. Give examples.
Colligative properties are defined as the properties of solutions that depend only on the number
of solute particles in solution and not on the nature of solute particles.
Examples : a) Lowering of vapour pressure;
b) Elevation of boiling point of solvent in solution;
c) Depression of freezing point of solvent in solution;
d) Osmotic pressure.
*2.
Ans.
What is a vapour pressure of liquid?
(a) The vapour pressure of a substance is defined as the pressure exerted by the gaseous
state of that substance when it is in equilibrium with the solid of liquid phase.
(b) At room temperature, the molecules of liquid are in constant random motion.
(c) The molecules at the surface of the liquid escape from the surface by absorbing energy
from the surrounding.
This process is called evaporation. If the vessel containing the liquid is closed with a lid,
then the molecules in vapour phase are tapped in the empty space and are in continuous
random motion.
(d) During this motion, molecules collide with each other and also with the walls of the container,
thereby losing their energy and returning back to liquid state. This process is called
condensation.
(e) Eventually the rate of evaporation and condensation becomes equal i.e. equilibrium is attained.
At this point the number of molecules in gaseous phase remains constant. The pressure
exerted by these molecules on the surface of liquid is called as vapour pressure of the
liquid.
(f) The vapour pressure of the liquid depends on the nature of the liquid and the temperature.
With increase of intermolecular forces of attraction, vapour pressure of liquid decreases
and with rise of temperature vapour pressure of liquid increases. [Refer Fig. 2.2 (b)]
*3.
Ans.
Explain the effect of temperature on the vapour pressure of a liquid.
The effect of temperature on the vapour pressure of a liquid.
(a) As temperature increases, vapour pressure of a liquid increases.
Chapter - 2 Solutions and Colligative Properties
90
(b) The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric
pressure, the liquid starts to boil. It is boiling point of the liquid ( Tb0 ). [Refer Fig. 2.2 (c) (3)]
*4.
Ans.
What are the factors which affect vapour pressure?
(a) Nature of solvent :
(1) All liquids exhibit tendency for evaporation.
(2) If the intermolecular forces of attraction are weak the liquids evaporate readily and
are called volatile liquids. Eg: Ethyl acetate is the most volatile liquid. Other Examples.
Ethyl alcohol, Acetone etc.
(3) If intermolecular forces are strong, the liquids are less volatile and hence exert lower
vapour pressure. Eg: Oil, mercury etc.
(b) Nature of solute :
(1) Non-volatile solute : The vapour pressure of a solvent is lowered when a non–volatile
solute is dissolved to form a solution. This is due to the fact that when a non–volatile
solute is added, the surface area available for evaporation of volatile solvent is decreased.
Eg: Sugar in water.
(2) Volatile solute : In case of volatile solute, the vapour pressure of the resulting solution
is generally increased. Eg: Ethyl alcohol mixed with water.
(c) Temperature :
(1) As temperature increases, vapour pressure of a liquid increases.
(2) The temperature at which the vapour pressure of the liquid becomes equal to the
atmospheric pressure, the liquid starts to boil. It is boiling point of the liquid ( Tb0 ).
[Refer Fig. 2.2 (c) (3)]
*5.
Ans.
What is lowering of vapour pressure of a solution?
The difference between the vapour pressure of pure solvent ( p10 ) and the vapour pressure
of solvent from solution (p) containing a non-volatile solute is called the lowering of vapour
pressure. Thus, p10 – p = ∆p = Lowering of vapour pressure.
It is found that, the vapour pressure of the solution (p) containing a non-volatile solute is always
less than that of the pure solvent p10 , i.e. p < p10 . [Refer Fig.. 2.2 (d)]
*6.
Ans.
What is relative lowering of vapour pressure?
The relative lowering of vapour pressure of a solution is the ratio of the lowering of vapour
pressure of solvent from solution to the vapour pressure of the pure solvent. Thus,
Relative lowering of vapour pressure =
7.
Ans.
p10 – p
p10
State and explain Raoult’s law of vapour pressure.
The law states that, the partial vapour pressure of any volatile component of a solution is the
product of vapour pressure of that pure component and the mole fraction of the component
in the solution. Consider a solution containing two volatile components A1 and A2 with mole
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91
fractions x1 and x2 resp. Let p10 and p 02 be the vapour pressures of the pure components
A1 and A2 resp.
According to Raoult’s law, the partial vapour pressures p1 and p2 of two components are given
0
0
by, p1 = p1 × x1 and p2 = p 2 × x 2
p T = p1 + p2 = p10 × x1 + p 02 × x 2
*8.
∴
State Raoult’s law and obtain expression for lowering of vapour pressure when non -volatile
solute is dissolved.
According to Raoult’s law, vapour pressure of the solution is equal to the product of vapour
pressure of the pure solvent and its mole fraction
Where,
p = p10 × x1
p 10 = vapour pressure of pure solvent.
p = vapour pressure of solution
x 1 = Mole fraction of the solvent,
0
Lowering of vapour pressure = Δp = p1 – p
∴
Δp = p10 – p10 × x1
∴
Δp = p10 (1 – x1 )
∴
Δp = p10 × x 2 (since, x1 + x2 = 1, therefore, 1 – x1 = x2 )
...(i)
Hence, lowering of vapour pressure is the product of vapour pressure of pure solvent
Ans.
∴
9.
Ans.
∴
( p10 ) and mole fraction of solute (x2).
Δp
From (i) we get, 0 = x2
p1
0
p1 – p
= x2
p10
Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute; which
proves that it is a colligative property.
Derive an expression for relative lowering of vapour pressure. OR
Show that the relative lowering in vapour pressure of an ideal solution is equal to mole fraction
of the solute.
According to Raoult’s law, vapour pressure of the solution is equal to the product of vapour
pressure of the pure solvent and its mole fraction
p = p10 × x1
Where,
p 10 = vapour pressure of pure solvent.
p
x1
= vapour pressure of solution
= Mole fraction of the solvent,
Lowering of vapour pressure = Δp = p10 – p
∴
Δp = p10 – p10 × x1
Chapter - 2 Solutions and Colligative Properties
92
∴
Δp = p10 (1 – x1 )
∴
Δp = p10 × x 2 (since, x1 + x2 = 1, therefore, 1 – x1 = x2 )
Hence, lowering of vapour pressure is the product of vapour pressure of pure solvent
( p10 ) and mole fraction of solute (x2).
Δp
p10 – p
= x2
From (v) we get, 0 = x2 ∴
p1
p10
Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute this
proves that it is a colligative property.
*10.
Ans.
Derive the relationship between relative lowering of vapour pressure and molar mass of solute.
Relative lowering of vapour pressure is given by the formula,
p10 – p
= x2
...(i)
p10
Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1.
Hence, number of moles of solvent, n1 and number of moles of solute, n2 in solution are given
W1
W2
as, n1 = M and n2 = M
1
2
The mole fraction of solute, x2 is given by
n2
W2 / M 2
x2 = n + n = W / M + W / M
...(ii)
1
2
1
1
2
2
Combining (i) and (ii) We get;
p10 – p
W2 / M 2
=
x
=
...(iii)
2
W1 / M1 + W2 / M 2
p10
The relation (iii) is applicable only for dilute solutions. Hence, n1 >> n2 . Hence, n2 may be
neglected in comparison with n1 in the above equation. Thus, eqn. (iii) can be written as follows
W2 / M 2
p10 – p
:
= W /M
0
p1
1
1
0
W
p1 – p
2 M1
Therefore,
=
W1 M 2
p10
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
According to the Raoult’s law, the relative lowering of vapour pressure is equal to the .......
a) Mole fraction of solvent
b) Mole fraction of solute
c) Independent of mole fraction of solute
d) Molality of solution
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*2.
*3.
Ans.
*4.
*5.
*6.
•
7.
Ans.
Partial pressure of solvent in solution of non-volatile solute is given by equation, .......
a) p0 = xp
c) p = x1p0
d) p0 = x1p
a) p = x2p0
When partial pressure of solvent in solution of non-volatile solute is plotted against its mole
fraction, nature of graph is .......
a) A straight line passing through origin
b) A straight line parallel to mole fraction of solvent
c) A straight line parallel to vapour pressure of solvent
d) A straight line intersecting vapour pressure axis
∴ The graph of p v/s x1 will be a straight line
passing through origin.
Lowering of vapour pressure of solution .......
a) is a property of solute
b) is a property of solute as well as solvent
c) is a property of solvent
d) is a colligative property
Vapour pressure of solution of a non-volatile solute is always .......
a) Equal to the vapour pressure of pure solvent
b) Higher than vapour pressure of pure solvent
c) Lower than vapour pressure of pure solvent
d) Constant
Relative vapour pressure lowering depends only on .......
a) Mole fraction of solute
b) Nature of solvent
c) Nature of solute
d) Nature of solute and solvent
Numerical MCQs :
In a very dilute solution the no. of moles
of solvent are 10 times more than that of
solute. The vapour pressure of the solution
would be .......
(Given vapour pressure of pure solvent =
80 mm)
a) 800 mm
b) 88 mm
c) 72 mm
d) 720 mm
p
=
p 10 × n 1
=
80 ×
=
80 ×
n1
n1 + n 2
10n 2
10n 2 + n 2
8.
Ans.
10n 2
11n 2 ≈ 72 mm
The vapour pressure of a pure liquid ‘A’
is 70 torr at 27º C. It forms an ideal solution
with another liquid ‘B’. The mole fraction
of B is 0.2 at 27ºC. Find the vapour
pressure of pure liquid B if vapour pressure
of solution is 84 torr.
a) 14
b) 56
c) 140
d) 70
=
80 ×
p
=
p 0A × x A + p 0B × x B
84
=
70 × 0.8 + p 0B × 0.2
Chapter - 2 Solutions and Colligative Properties
94
=
p 0B × 0.2
28
0.2
=
p 0B
p 0B
= 140
84 – 56
∴
9.
11.
2.3
Ans.
p
=
p 10 × n 1
= 23.8 × 0.9
[n1 = 1 – n2 = 1 – 0.1 = 0.9]
= 21.42 mm Hg
The vapour pressure of water at room
temperature is 23.8 mm Hg. The vapour
pressure of an aqueous solution of sucrose
with mole fraction 0.1 is equal to .......
a) 2.39 mm Hg
b) 2.42 mm Hg
c) 21.42 mm Hg
d) 21.44 mm Hg
•
10.
Which is not a colligative property?
(MHT-CET 2009)
a) Freezing point
b) Lowering of Vapour pressure
c) Osmotic pressure
d) Elevation in boiling point
The addition of non volatile solute into the pure solvent
a) Increases the vapour pressure of solvent
b) Decreases the boiling point of solvent
c) Decreases the freezing point of solvent
d) Increases the freezing point of solvent
(Sept. 2008)
ELEVATION OF BOILING POINT AND DEPRESSION OF
FREEZING POINT :
Concept Explanation :
(a) Boiling Point Elevation :
1. Boiling Point : The boiling point of a liquid is the temperature at which its vapour pressure
becomes equal to the atmosphere pressure.
2. Elevation in Boiling Point (ΔTb) : The difference in the boiling points of the solution
(T) and pure solvent (Tº) is called the elevation in boiling point (ΔTb).
∴ ΔTb = T – Tº
Note : The vapour pressure of the solution containing non-volatile solute is less
than that of the solvent. Therefore, the solution has to be heated to a higher
temperature so that its vapour pressure becomes equal to the atmospheric
pressure. Thus, the boiling point of the solution is always higher than that
of the pure solvent. (T > Tº)
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(b) Relation between Elevation in boiling point and lowering of vapour
pressure :
1. The boiling point of a liquid is the temperature
at which its vapour pressure becomes equal
to the atmospheric pressure.
2. The vapour pressure of the solution containing
a non volatile solute is less than that of the
pure solvent.
3. Therefore, the solution has to be heated to a
higher temperature so that its vapour pressure
becomes equal to the atmospheric pressure.
4. Thus, the boiling point of the solution is always
higher than that of pure solvent.
(T > To)
5. The curve AB gives the vapour pressures for the pure solvent and the curve CD gives
the vapour for the solution at different temperatures.
6. At point P the vapour pressure of solvent becomes equal to atmospheric pressure. At point
P the temperature is To. Hence by definition, To is the boiling point of pure solvent.
7. At point Q the vapour pressure of solution becomes equal to atmospheric pressure. At
point Q the temperature is T. Hence by definition, T is the boiling point of Solution.
8. Mathematically, elevation in boiling point, ΔTb may be expressed as:
ΔTb = T – To
9. From the graph it is evident that ΔTb ∝ Δp = p10 – p (Lowering of vapour pressure.)
10. It has been experimentally found that the elevation in the boiling point ΔTb of a solution
is proportional to the molal concentration of the solution, i.e.
ΔTb ∝ m
or ΔTb = Kb × molality(m)
Note : ΔTb ∝ m, Thus, elevation in boiling point is directly proportional to the
molal concentration of the solute (i.e., number of molecules) and therefore,
it is a colligative property.
.
(c) Molal elevation constant or Ebullioscopic constant :
Molal boiling point elevation constant, Kb is defined as the elevation in boiling point produced
by dissolving one mole of solute in 1kg of solvent (i.e. 1 molal solution)
Units of Kb (S.I) = K kg mol–1 or K/m.
e.g. : Water (Boiling point = 373K), Kb = 0.52 K kg mol–1
The value of Kb depends only on the nature of solvent and not on the nature of solute.
Chapter - 2 Solutions and Colligative Properties
96
(d) Relation between elevation of boiling point and molar mass of the
solute :
1. The elevation in boiling point (ΔTb) is useful in determining the molecular mass of the solute.
2. Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar
mass of the solute in kg mol-1.
moles of solute
=
Weight of solvent in Kg
3.
molality (m) =
4.
m
5.
ΔTb
6.
ΔTb =
K b × W2
M 2 × W1
7.
M2
K b × W2
ΔTb × W1
W2 / M 2
W1
W2
W2 / M 2
=
M 2 × W1
W1
= Kb × molality(m)
=
=
[From (4) and (5)]
From the relation, molecular mass of the solute can be calculated when the values of other
quantities are known.
Note : The experimental method to determine molecular mass of a non volatile solute
by determining boiling point of pure solvent and solution of known concentration
is called ebullioscopy.
(e) Freezing Point Depression :
1. Freezing point : The freezing point of a liquid may be defined as the temperature at
which, the vapour pressure of solid is equal to the vapour pressure of liquid.
2. Depression in freezing point : The depression in freezing point may be defined as the
decrease in freezing point of the solution (difference in freezing points of the pure solvent
and the solution). It may be denoted by ΔTf. Mathematically, ΔTf = T – T°, where,
T° and T represent the freezing points of pure solvent and the solution respectively.
3. Cause for depression of freezing point :
(i) The freezing point of a liquid is the temperature at which the liquid and the solid have
the same vapour pressure.
(ii) Addition of a non-volatile solute to a liquid decreases the freezing point, i.e., the freezing
point of the solution is less than that of the pure solvent.
(iii) This happens due to lowering of vapour pressure of the solvent by the addition of
the non volatile solute.
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(f) Depression of freezing point and lowering of vapour pressure
graphically :
1. The freezing point of a liquid is the temperature at which the liquid and the solid have
the same vapour pressure.
2. Addition of a non-volatile solute to a liquid decreases the freezing point, i.e., the freezing
point of the solution is less than that of the pure solvent.
3. This happens due to lowering of vapour pressure of the solvent by the addition of the
non volatile solute.
4. The curves BC, DE and AB represent the vapour pressures of the pure solvent, solution
and the solid respectively at different temperatures.
5. From the curves, it is clear that at point B the solid and liquid states of pure solvent have
same vapour pressure.
6. At point B the temperature is Tº which will be freezing point of pure solvent, Tº.
7. When solute is dissolved in the solvent, the vapour pressure of solvent is lowered and
resulting solution can no longer freeze at temperature Tº
8. A new equilibrium is established at point D, where
vapour pressure of solvent of the solution and solid
solvent becomes equal.
9. At point D the temperature is T which will be
freezing point of solution.
10. The vapour pressure curve DE of the solution,
always lies below the vapour pressure curve of
pure solvent. Hence, intersection of vapour
pressure curve of solution and solid solvent can
occur only at a point lower than Tº.
11. Therefore, any solution must have freezing point
lower than that of pure solvent.
12. Thus, the depression in freezing point, ΔTf = Tº - T
13. It is evident from the graph that ΔTf α pº – ps.
14. But, ΔTf ∝ m (experimentally)
∴ ΔTf = Kf x m (molality)
15. Kf is molal freezing point depression constant or cyroscopic constant.
(g) Molal depression constant or cyroscopic constant :
Molal depression constant or cyroscopic constant is defined as depression in freezing point
produced by dissolving 1 mole of solute in 1 kg of solvent (i.e., 1 molal solution).
1. Units of Kb (S.I) = K kg mol-1 or K/m.
Chapter - 2 Solutions and Colligative Properties
98
2.
3.
e.g : Water (Freezing point 0oC.) Kf = 1.86 K kg mol-1
The value of Kf depends only on the nature of solvent.
(h) Relation between molecular mass of solute and depression in freezing
point :
1. The depression in freezing point (ΔTf) is useful in determining the molecular mass of
the solute.
2. Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar
mass of the solute in kg mol-1.
3.
molality (m) =
4.
m
5.
ΔTf
6.
ΔTf
7.
M2
moles of solute
Weight of solvent in kg
=
W2 / M 2
M1
W2
= M ×W
2
1
Where
W2 = Weight of solute in kg,
W1= Weight of solvent in kg,
= Kf × molality(m)
M2 = Molecular wt of solute in kg mol–1
K f × W2
= M ×W
[From (4) and (5)]
2
1
K f × W2
Tf × W1
From the relation, molecular mass of the solute can be calculated.
=
Note : The experimental method to determine molecular mass of a non volatile solute
by determining freezing point of pure solvent and solution of known
concentration is called cryoscopy.
(i) Applications of depression of freezing point :
1. In cold countries ice frozen on roads and footpaths, is melted by spraying salts like sodium
chloride or calcium chloride, ice melts into water as the salt depresses the freezing point
of water.
2. Also in cold countries de-icing of aeroplanes is done by depressing the freezing point of
water.
3. Ethylene glycol is a common automobile antifreezer. It is water soluble and non volatile.
It is also used as coolant in automobile radiators.
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*1.
Ans.
*2.
Ans.
∴
What is boiling point of liquid?
Boiling Point : The boiling point of a liquid is the temperature at which its vapour pressure
becomes equal to the atmosphere pressure.
What is elevation of boiling point?
Elevation in Boiling Point (ΔTb) : The difference in the boiling points of the solution (T) and
pure solvent (Tº) is called the elevation in boiling point (ΔTb).
ΔTb = T – Tº
*3.
Ans.
What is molal elevation constant? Does it depend on nature of the solute?
Molal boiling point elevation constant, Kb is defined as the elevation in boiling point produced
by dissolving one mole of solute in 1kg of solvent (i.e. 1 molal solution)
Units of Kb (S.I) = K kg mol-1 or K/m.
E.g. : Water (Boiling point = 373K), Kb = 0.52 K kg mol-1
The value of Kb depends only on the nature of solvent and not on the nature of solute.
4.
Ans.
Explain graphically Relation between Elevation in boiling point and lowering of vapour pressure.
[Refer Fig. 2.3 b]
(a) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal
to the atmospheric pressure.
(b) The vapour pressure of the solution containing a non volatile solute is less the that of the
pure solvent.
(c) Therefore, the solution has to be heated to a higher temperature so that its vapour pressure
becomes equal to the atmospheric pressure.
(d) Thus, the boiling point of the solution is always higher than that of pure solvent.
(T > To)
(e) The curve AB gives the vapour pressures for the pure solvent and the curve CD gives
the vapour for the solution at different temperatures.
(f) At point P the vapour pressure of solvent becomes equal to atmospheric pressure. At point
P the temperature is To. Hence, by definition, To is the boiling point of pure solvent.
(g) At point Q the vapour pressure of solution becomes equal to atmospheric pressure.
At point Q the temperature is T. Hence by definition, T is the boiling point of Solution.
(h) Mathematically, elevation in boiling point, ΔTb may be expressed as : ΔTb = T – To
(i) From the graph it is evident that ΔTb ∝ Δp = p10 – p (Lowering of vapour pressure.)
(j) It has been experimentally found that the elevation in the boiling point ΔTb of a solution
is proportional to the molal concentration of the solution, i.e.
ΔTb ∝ m or ΔTb = Kb × molality(m)
Chapter - 2 Solutions and Colligative Properties
100
*5.
Ans.
Derive the relation between elevation of boiling point and molar mass of the solute.
The elevation in boiling point (ΔTb) is useful in determining the molecular mass of the solute.
Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar mass
of the solute in kg mol-1.
molality (m) =
=
m
=
ΔTb =
ΔTb =
moles of solute
Weight of solvent in Kg
W2 / M 2
W1
W2
W2 / M 2
= M ×W
W1
2
1
Kb × molality(m)
K b × W2
M 2 × W1
Where
W2 = Weight of solute in kg,
W1= Weight of solvent in kg,
M2 = Molecular wt of solute in kg mol–1
...(i)
...(ii)
[From (i) and (ii)]
K b × W2
ΔTb × W1
From the relation, molecular mass of the solute can be calculated when the values of other
quantities are known.
M2
=
*6.
Ans.
Define freezing point of liquid.
The freezing point of a liquid may be defined as the temperature at which, the vapour
pressure of solid is equal to the vapour pressure of liquid.
7.
Ans.
Define freezing point depression.
Depression in freezing point : The depression in freezing point may be defined as the decrease
in freezing point of the solution (difference in freezing points of the pure solvent and the solution).
It may be denoted by ΔTf.
Mathematically, ΔTf = T0 – T, where,
T0 and T represent the freezing points of pure solvent and the solution respectively.
*8
Ans.
What causes depression in freezing point?
Cause for depression of freezing point :
(a) The freezing point of a liquid is the temperature at which the liquid and the solid have
the same vapour pressure.
(b) Addition of a non volatile solute to a liquid decreases the freezing point, i.e., the freezing
point of the solution is less than that of the pure solvent.
(c) This happens due to lowering of vapour pressure of the solvent by the addition of the
non volatile solute.
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101
9.
Ans.
Explain relation between Depression of freezing point and lowering of vapour pressure graphically.
[Refer Fig. 2.3 (f)]
(a) The freezing point of a liquid is the temperature at which the liquid and the solid have
the same vapour pressure.
(b) Addition of a non volatile solute to a liquid decreases the freezing point, i.e., the freezing
point of the solution is less than that of the pure solvent.
(c) This happens due to lowering of vapour pressure of the solvent by the addition of the
non volatile solute.
(d) The curves BC, DE and AB represent the vapour pressures of the pure solvent, solution
and the solid respectively at different temperatures.
(e) From the curves, it is clear that at point B the solid and liquid states of pure solvent have
same vapour pressure
(f) At point B the temperature is To which will be freezing point of pure solvent, To.
(g) When solute is dissolved in the solvent, the vapour pressure of solvent is lowered and
resulting solution can no longer freeze at temperature To
(h) A new equilibrium is established at point D, where vapour pressure of solvent of the solution
and solid solvent becomes equal
(i) At point D the temperature is T which will be freezing point of solution.
(j) The vapour pressure curve DE of the solution, always lies below the vapour pressure
curve of pure solvent. Hence, intersection of vapour pressure curve of solution and solid
solvent can occur only at a point lower than To.
(k) Therefore, any solution must have freezing point lower than that of pure solvent.
(l) Thus, the depression in freezing point, ΔTf = To - T
(m) It is evident from the graph that ΔTf α po – ps.
(n) But, ΔTf ∝ m (experimentally)
∴ ΔTf = Kf x m (molality)
(o) Kf is molal freezing point depression constant or cyroscopic constant.
*10.
Ans.
Define molal depression constant or cryoscopic constant.
Molal depression constant or cyroscopic constant is defined as depression in freezing point
produced by dissolving 1 mole of solute in 1 kg of solvent (i.e., 1 molal solution).
(a) Units of Kb (S.I) = K kg mol-1 or K/m.
(b) Eg : Water (Freezing point 0oC.) Kf = 1.86 K kg mol-1
(c) The value of Kf depends only on the nature of solvent.
*11.
How is molar mass of a non volatile solute related to the depression of freezing point? Derive
an equation.
The depression in freezing point (ΔTf) is useful in determining the molecular mass of the solute.
Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar mass
of the solute in kg mol-1.
Ans.
Chapter - 2 Solutions and Colligative Properties
102
molality (m) =
W2 / M 2
moles of solute
=
M1
Weight of solvent in kg
ΔTf =
W2
M 2 × W1
Kf × molality(m)
ΔTf =
K f × W2
M 2 × W1
m
=
...(i)
...(ii)
[From (i) and (ii)]
K f × W2
ΔTf × W1
From the relation, molecular mass of the solute can be calculated.
M2
12.
Ans.
=
Give the applications of freezing point depression.
(a) In cold countries ice frozen on roads and footpaths, is melted by spraying salts like sodium
chloride or calcium chloride, ice melts into water as the salt depresses the freezing point
of water.
(b) Also in cold countries de-icing of aeroplanes is done by depressing the freezing point of
water.
(c) Ethylene glycol is a common automobile antifreezer. It is water soluble and non volatile.
It is also used as coolant in automobile radiators.
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
When NaCl is added to water .......
a) Freezing point is raised
b) Boiling point is depressed
c) Freezing point does not change
d) Boiling point is raised
Molal elevation constant is elevation in boiling point produced by .......
a) 1g of solute in 100g of solvent
b) 100g of solute in 1000g of solvent
c) 1 mole of solute in one litre of solvent
d) 1 mole of solute in one kg of solvent
The determination of molar mass from elevation in boiling point is called as
a) Cryoscopy
b) Osmometry
c) Ebullioscopy d) Spectroscopy
If the mass is expressed in grams then Kb is given by .......
*2.
*3.
*4.
a)
M 2 × ΔTb × W2
1000 × w 1
b)
W2 × 1000
ΔTb × W1 × M 2
c)
M 2 × Δ Tb × W1
1000 × W2
d)
W1 × 1000
ΔTb × W2 × M 2
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
103
5.
Which of the following aqueous solutions will have the lowest value of freezing point in the
Ans.
∴
∴
6.
a) 1% Urea
b) 1% Glucose
c) 1% Sucrose
1
1
1
1
nurea =
nglucose =
nsucrose =
nstarch =
very large
60
180
342
nurea is highest.
It will have lowest value of freezing point.
Kf is equal to depression in freezing point produced by ................. .
a) 1 molar solute
b) 1 normal solute c) 1 molal solute d)
•
Numerical MCQs :
7.
Which one of the following aqueous
solutions will have highest boiling point?
a) 0.1 M urea
b) 30g of glucose per dm3
c) 3.42g of sucrose in 100ml
d) 0.2 M glucose
Ans.
Molarity in option B =
9.
30
1
=
180 × 1
6
= 0.16 M
Molarity in option C =
∴
8.
Ans.
d)
3.42
342 × 0.1
Ans.
= 0.1 M
Molarity in option in highest i.e. 0.2 M
glucose.
The boiling point of a decimolal aqueous
solution of glucose would be
....... (Given Kb = 0.52K kg mol–1)
a) 100.52ºC
b) 99.48ºC
c) 99.94ºC
d) 100.052º C
ΔTb = Kb × m
= 0.52 × 0.1 = 0.052
T
= Tº + ΔTb
=
100 + 0.052
=
100.052ºC
1% Starch
all are correct
The boiling point of a solution containing
2.62g of a substance A in 100g of water
is higher by 0.0512ºC than the boiling point
of pure water. The molar mass of the
substance(Kb = 5.12 Km–1) is .......
b) 262 g mol–1
a) 131 g mol–1
–1
c) 26.2 g mol
d) 2620g mol–1
M2 =
=
Kb
W
× 2
ΔTb W1
5.12
2.62 × 10 –3
×
0.0512 100 × 10 –3
= 2.62 kg mol–1 = 2620 g mol–1
10.
Ans.
The rise in the boiling point of a solutions
containing 1.8g of glucose in 100g of a
solvent is 0.1ºC. The molal elevation
constant of the liquid is .......
a) 0.01 k/m
b) 0.1 k/m
c) 1 k/m
d) 10 k/m
ΔTb = Kb × m
Chapter - 2 Solutions and Colligative Properties
104
11.
Ans.
0.1
= Kb ×
Kb
= 1 k/m
Ans.
180 × 100 × 10 –3
The molal freezing point constant for water
is 1.86. If 342 g of cane sugar (molar mass
= 342 g mol–1) is dissolved in 1000g of
water, the solution will freeze at ...........
a) 1.86ºC
b) –1.86ºC
c) –3.92ºC
d) 3.92ºC
ΔTf = Kf × m
T
12.
Ans.
15g of non-volatile solute is dissolved in
250g of solvent, shows elevation in boiling
point 0.01K. If gram molecular weight of
solute is 60, the molal elevation constant
for solvent is ..... (MHT-CET 2010)
a) 0.1 Km–1
b) 1 Km–1
c) 10 Km–1
d) 0.01 Km–1
ΔTb = Kb × m
342
= 1.86 × 342 × 1000 × 10 –3
15
0.01 = K b × 60 × 250 × 10 –3
= 1.86
= Tº – ΔTf
= 0 – 1.86
= –1.86ºC
Kb = 0.01 k/m
14.
The amount of urea (molar mass = 60g
mol–1) to be dissolved in 500g of water to
produce a depression of 0.186º in the
freezing point is ....... .
(Kf for water = 1.86 Km–1)
a) 0.6g
b) 60g
c) 3g
d) 6g
ΔTf = Kf × m
0.186 = 1.86 ×
∴
•
13.
1.8
Ans.
= 1.86 ×
mass of urea
=
60 × 500 × 10 –3
Mass of urea = 3g
∴
Unique Solutions ®
What will be freezing point of solution
when 0.08kg of ethylene glycol (molar
mass = 62) is added to 400g of water. (Kf
of water = 1.86K kg/mol)
(MHT-CET 2008)
a) 269 K
b) 267 K
c) 279 K
d) 297 K
ΔTf = Kf × m
=
ΔTf =
6
=
T
=
0.08 × 10 3
62 × 400 × 10 –3
186 × 8
62 × 4
6K
Tº – T
273– T
273 – 6 = 267 K
S.Y.J.C. Science - Chemistry - Part I
105
2.4
OSMOTIC PRESSURE, ABNORMAL MOLECULAR MASS AND
VANT HOFF FACTOR :
Concept Explanation :
(a) Semi permeable membrane :
A membrane which allows the solvent
molecules to pass through it but not the
solute particles is called a semi-permeable
membrane.
Eg : Natural : Animal bladder, Animal and
Plant cell wall, Potato membrane,
parchment paper, etc.
Artificial : Copper ferrocyanide membrane
Cu2 [Fe(CN) 6], Cellulose, Cellulose
nitrate, etc.
(b) Osmosis :
1. Osmosis was first studied by Abbe Nollet.
2. Osmosis : When a solution is separated from a pure solvent by a semi-permeable membrane
(such as an animal bladder or copper ferrocyanide) then it is observed
that the solvent flows from the region of pure solvent into solution.
3. Similarly, if two solutions of different concentrations are separated by a semi-permeable
membrane then the solvent flows from solution of lower concentration into solution of higher
concentration. This process is called osmosis.
4. Definition : The spontaneous and unidirectional flow of the solvent molecules from
pure solvent into the solution or from a solution of lower concentration to the solution
of higher concentration through a semipermeable membrane is called osmosis.
(c) Abbe Nollet experiment :
1. The apparatus consists of a long stem thistle funnel.
2. The mouth of the thistle funnel is closed by a semipermeable
3. Sucrose solution of some concentration is filled in the thistle
funnel.
4. The thistle funnel is then placed in beaker containing water
in inverted position.
5. The solvent and solution are separated from each other
by a semipermeable membrane.
Chapter - 2 Solutions and Colligative Properties
106
6.
7.
8.
9.
10.
There occurs a net flow of solvent molecules into the solution.
As a result the original volume of solution increases and liquid level of solution increases.
Hydrostatic pressure is developed due to liquid column in thistle funnel.
The liquid level in thistle funnel rises until the excess pressure so produced is sufficient
to stop the flow of solvent molecules.
This equilibrium hydrostatic pressure which stops the net flow of solvent into the solution
is equal to Osmotic pressure.
(d) Types of Osmosis and osmotic pressure :
1. Osmotic pressure : The osmotic pressure may be defined as the excess mechanical
pressure which is applied on the solution to stop the flow of solvent molecules into the
solution through a semi-permeable membrane.
It is generally, denoted by π. It is a colligative property.
2. Isotonic solutions : Solutions which have the same osmotic pressure at the same
temperature are called Isotonic solutions.
3. Hypertonic solutions : A solution having more osmotic pressure than some other solutions
is called hypertonic solutions.
4. Hypotonic solutions :
A solution having less osmotic pressure than the other solutions is called hypotonic solutions.
Note : It is interesting to note that a 0.91% (w/v) solution of sodium chloride (known
as saline water) is isotonic with human blood corpuscles. In this solution, the
corpuscles neither swell nor shrink. Therefore, the medicines are mixed with
saline water before being injected into the veins. However, a 5% NaCl solution
is hypertonic solution and when red blood cells are placed in this solution,
water comes out of the cells and they shrink. On the other hand, when red
blood cells are placed in distilled water (hypotonic solution) water flows into
the cells and they swell or burst.
(e) Osmosis in day to day life :
1. A raw mango kept in a concentrated salt solution loses water due to osmosis and shrivel
(shrinks and wrinkles) into pickle.
2. A limp carrot can be placed in water making it firm once again. Water moves in carrot
due to osmosis.
3. Blood cells placed in water containing less than 0.91% salt collapse due to loss of water
by osmosis.
4. People eating lot of salt experience edema i.e. swelling of tissue cells due to water retention.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
107
5.
6.
The fruits can be preserved by adding sugar. Since bacterium cell loses water due to osmosis,
shrivels and dies.
In plants the leaves loose water by transpiration. The solute concentration in leaf
therefore, increases. Due to osmosis water is pulled from soil and transported in upward
direction.
(f) Laws of osmotic pressure :
1. Boyle Van’t Hoff law : It states that “Temperature remaining constant, the osmotic pressure
of a dilute solution is directly proportional to the molar concentration of the solution or
inversely proportional to the volume of the solution.”
If π represents the osmotic pressure (in atmospheres) and C is the molar concentration
in moles/litre, we have
π ∝ C at constant temperature
If n moles of the substance are dissolved in V litres of the solution.
n
1
i.e. C ∝
V
V
The above results may, therefore, be written as
C=
1
(at constant temperature)
V
or πV = constant
(at constant temperature)
2. van’t Hoff Charle’s law : Similar to Charles law for gases, it state that ‘for a given
concentration of a solution, the osmotic pressure is directly proportional to the absolute
temperature’.
Mathematically, π ∝ T (for a given concentration)
π
= constant
(for a given concentration)
T
3. van’t Hoff Avogadro’s law :
Statement : This law states that, Equal volumes of isotonic solutions at the same temperature
contain equal number of moles of solute.
Van’t Hoff equation for two different dilute solutions can be written as follows:
π1V1 = n1RT1 and π2V2 = n2RT2
If π1 = π2 (Isotonic Solutions) and T1 = T2,
π∝
n1
n2
Then , V = V
1
2
Hence, When osmotic pressure and temperature are same, equal volumes of solutions
would contain equal number of moles of solute.
Chapter - 2 Solutions and Colligative Properties
108
4. Vant-Hoff’s equation for dilute solutions : According to Van’t Hoff-Boyle’s law
1
π∝
(at constant temperature)
V
According to van’t Hoff-Charle’s law,
π ∝ T
(at given concentration)
Combining the above results, we get
T
π∝
or πV = kT
… (i)
V
where k is called general solution constant.
vant Hoff further proved that the value of k is same as R,
Hence, πV = RT
Further if the solution contains n moles of the solute dissolved in V litres of solution, then
n
C=
V
Substituting the value of C in equation (1), we get;
n
π= R T
V
πV = n RT
... (ii)
(g) Determination of molecular mass from osmotic pressure :
According to van’t Hoff equation,
π = CRT
n
V
where n is the number of moles of solute dissolved in V litre of the solution.
n
RT
∴ π=
V
W2
The number of moles of solute n may be given as M . Here W2 is the weight of the
2
solute and M2 is its molar mass
Substituting the value of n in the above expression.
W2 RT
W2 RT
π = VM or M2 =
Vπ
2
Thus, the molar mass of the solute, M2, can be calculated.
But C =
(h) Reverse osmosis :
1. Definition : If, a solution is separated from the pure solvent by a semi-permeable
membrane and the pressure applied on the solution (more than the osmotic pressure)
the solvent starts flowing from the solution towards the pure solvent. This phenomenon
is known as reverse osmosis.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
109
2. Application of reverse osmosis :
Desalination :
(a) Sea water contains 3.5%w/w of dissolved salts.
(b) Drinking water is produced from sea water by
a process called as desalination by reverse
osmosis.
(c) In reverse osmosis high pressure greater than
osmotic pressure of sea water i.e., 30 atm. is
applied to force water from concentrated
aqueous solution like sea water to pure water
side through a semi permeable membrane.
(d) The method can be used provided a suitable semipermeable membrane is developed
which withstands the high pressure condition over a prolonged period.
(i) Abnormal molecular mass :
1. Colligative properties of solutions depend on actual number of solute particles present in
the solution.
2. In Dilute solution of non- electrolytes like urea, glucose etc. the solute remains in normal
molecular condition and does not undergo dissociation or association.
3. In case of solutions of electrolytes, it has been observed that, observed colligative properties
and thus, molar mass determined by these methods do not agree with the expected or
theoretical values.
4. This is because of association or dissociation of the solute molecules in the solution.
(a) Dissociation :
(i) Molecules of certain substances (acids, bases and salts) dissociate or ionize in
a solvent to give two or more particles. For example, AB dissociates to give
double number of particles as :
A + + B –
AB (ii) Consequently, the total number of particles increases in solution and, therefore,
the colligative property of such solutions will be large for solutes undergoing
dissociation.
(iii) Since the colligative property is inversely proportional to the molar mass of the
solute, the molar mass of solute in such cases will be less than normal. The value
of colligative property depends on the degree of dissociation of the solute in the
solution.
(iv) For example :
NaCl
⎯⎯
→
K 2SO 4
⎯⎯
→ 2K +
: 3 particles
+ SO 2–
4
AlCl 3
⎯⎯
→ Al 3+
+ 3Cl – : 4 particles
Na +
+ Cl – : 2 particles
Chapter - 2 Solutions and Colligative Properties
110
(b) Association :
(i) In some non-polar solvents, the solute molecules undergo association, i.e. two,
three or even more molecules exist in combination with each other to form bigger
molecules.
(ii) Therefore, the total number of molecules in solution become less than the
number of molecules of the substance added. This leads to a decrease in
the value of colligative properties because colligative properties is proportional
to no. of solute particles.
(iii) Since the colligative property is inversely proportional to the molar mass of the
solute, the molar mass of solute in such cases will be greater than normal.
(iv) For example, in benzene solvent solutes like acetic acid, benzoic acid etc. exist
as dimers. This is due to hydrogen bonding between these molecules.
(CH 3COOH) 2
2CH 3COOH (C 6 H 5COOH) 2
2C 6 H 5COOH (j) vant Hoff factor :
1. This factor is used to express the extent of dissociation or association of the solutes in
their solutions.
2. It defines as the ratio of observed colligative property produced by a given concentration
of electrolyte solution to the property observed for the same concentration of non electrolyte
solution.
3. Therefore,
observed value of colligative property
i = theoretical value of colligative property
Since colligative property α number of solute particles present in the solution. Hence,
i=
number of solute particles present in solution
theoretical number of solute particles
Similarly, colligative property α
1
molecular mass
theoretical molecular mass
4.
5.
6.
i = observed molecular mass
If i = 1, solute molecules are in normal state and no association or dissociation has taken
place.
If i > 1 , the solute molecules are dissociated in the solution.
If i < 1, the solute molecules are associated in the solution.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
111
(k) vant Hoff factor (i) and Degree of dissociation (α
α) :
Consider ‘m’ moles of an electrolyte AxBy dissolved in 1kg of solvent. Let α be the
degree of dissociation of the electrolyte.
∴
∴
xA y+ + yB x–
A x B y Initial moles
m
0
0
moles at equilibrium
m – mα
x(mα)
y(mα)
Total no. of moles at equilibrium =
mt = (m – mα) + x(mα) + y(mα)
= m(1 – α) + x(mα) + y(mα)
= m [1 – α + xα + yα]
= m[1 + xα + yα – α]
= m[1 + α (x + y – 1)]
Let the total no. of ions produced by dissociation by one mole of eletrolyte AxBy be x + y = n′
∴ mt = m (1 + α (n1 – 1))
observed number of moles
Now vant Hoff factor, i =
theoretical number of moles
mt
m (1 + (n ′ –1) α)
i =
=
m
m
i = 1 + (n′ – 1) α
Thus, degree of dissociation α =
i
=
M (Theoretical)
M (observed)
i –1
n′ – 1
= 1 + (n ′ – 1) α
M (Theoretical)
– 1 = (n ′ – 1) α
M (observed)
α =
•
1.
Ans.
M (Theoretical) – M (observed)
M (observed) (n ′ – 1)
Define : *a) Semi permeable membrane b) Osmosis.
(a) Semi permeable membrane : A membrane which allows the solvent molecules to pass
through it but not the solute particles is called a semi-permeable membrane.
e. g : Natural : Animal bladder, Animal and Plant cell wall, Potato membrane, parchment
paper, etc.
Artificial : copper ferrocyanide membrane Cu2 [Fe(CN)6], Cellulose, cellulose nitrate, etc.
Chapter - 2 Solutions and Colligative Properties
112
Definition : The spontaneous and unidirectional flow of the solvent molecules from pure solvent
into the solution or from a solution of lower concentration to the solution of higher concentration
through a semipermeable membrane is called osmosis.
*2.
Ans.
Explain the process of osmosis. or What is Osmosis?
(a) Osmosis was first studied by Abbe Nollet.
(b) Osmosis : When a solution is separated from a pure solvent by a semi-permeable membrane
(such as an animal bladder or copper ferrocyanide) then it is observed that the solvent
flows from the region of pure solvent into solution. [Ref. Fig. 2.4 (b)]
(c) Similarly, if two solutions of different concentrations are separated by a semi-permeable
membrane then the solvent flows from solution of lower concentration into solution of higher
concentration. This process is called osmosis.
(d) Definition : The spontaneous and unidirectional flow of the solvent molecules from pure
solvent into the solution or from a solution of lower concentration to the solution of higher
concentration through a semipermeable membrane is called osmosis.
3.
Ans.
Explain Abbe Nollet experiment.
(a) The apparatus consists of a long stem thistle funnel.
(b) The mouth of the thistle funnel is closed by a semipermeable membrane like pig’s bladder
(c) Sucrose solution of some concentration is filled in the thistle funnel.
(d) The thistle funnel is then placed in beaker containing water in inverted position.
(e) The solvent and solution are separated from each other by a semipermeable membrane.
(f) There occurs a net flow of solvent molecules into the solution.
(g) As a result the original volume of solution increases and liquid level of solution increases.
(h) Hydrostatic pressure is developed due to liquid column in thistle funnel.
(i) The liquid level in thistle funnel rises until the excess pressure so produced is sufficient
to stop the flow of solvent molecules.
(j) This equilibrium hydrostatic pressure which stops the net flow of solvent into the solution
is equal to Osmotic pressure.[Refer Fig. 2.4 (c)]
*4.
Ans.
Define : a) Osmotic pressure b) Isotonic solution c) Hypertonic solution d) hypotonic solution.
(a) Osmotic pressure : The osmotic pressure may be defined as the excess mechanical
pressure which is applied on the solution to stop the flow of solvent molecules into the
solution through a semi-permeable membrane. It is generally, denoted by π. It is a colligative
property.
(b) Isotonic solutions : Solutions which have the same osmotic pressure at the same
temperature are called Isotonic solutions.
(c) Hypertonic solutions : A solution having more osmotic pressure than some other solutions
is called hypertonic solutions.
(d) Hypotonic solutions : A solution having less osmotic pressure than the other solutions
is called hypotonic solutions.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
113
5.
Ans.
Write any four applications of osmosis in day to day life.
(a) A raw mango kept in a concentrated salt solution loses water due to osmosis and shrivel
(shrinks and wrinkles) into pickle.
(b) A limp carrot can be placed in water making it firm once again. Water moves in carrot
due to osmosis.
(c) Blood cells placed in water containing less than 0.91% salt collapse due to loss of water
by osmosis.
(d) People eating lot of salt experience edema i.e. swelling of tissue cells due to water retention.
(e) The fruits can be preserved by adding sugar. Since bacterium cell loses water due to osmosis,
shrivels and dies.
(f) In plants the leaves loose water by transpiration. The solute concentration in leaf therefore,
increases. Due to osmosis water is pulled from soil and transported in upward direction.
*6.
Ans.
State van’t Hoff’s-Boyle’s law.
Boyle Van’t Hoff law : It states that “Temperature remaining constant, the osmotic pressure
of a dilute solution is directly proportional to the molar concentration of the solution or inversely
proportional to the volume of the solution.”
If π represents the osmotic pressure (in atmospheres) and C is the molar concentration in
moles/litre, we have
π∝C
at constant temperature
If n moles of the substance are dissolved in V litres of the solution.
n
1
C∝
i.e.
C=
V
V
The above results may, therefore, be written as
1
π∝
(at constant temperature)
V
or πV = constant
(at constant temperature)
*7.
Ans.
State van’t Hoff-Charles law.
van’t Hoff Charle’s law : Similar to Charles law for gases, it state that ‘for a given concentration
of a solution, the osmotic pressure is directly proportional to the absolute temperature’.
Mathematically, π ∝ T
(for a given concentration)
π
= constant
T
*8.
Ans.
(for a given concentration)
Statement : This law states that, Equal volumes of isotonic solutions at the same temperature
contain equal number of moles of solute.
Van’t Hoff equation for two different dilute solutions can be written as follows:
Chapter - 2 Solutions and Colligative Properties
114
π 1V 1 = n1RT1
If π1 = π2
and
T1 = T2,
π2V2 = n2RT2
(Isotonic Solutions)
n1
n2
Then , V = V
1
2
Hence, when osmotic pressure and temperature are same, equal volumes of solutions would
contain equal number of moles of solute.
*9.
Ans.
Derive van’t Hoff equation for osmotic pressure of a solution.
Vant-Hoff’s equation for dilute solutions :
According to Van’t Hoff-Boyle’s law
1
π∝
(at constant temperature)
V
According to Van’t Hoff-Charle’s law,
π ∝ T
(at given concentration)
Combining the above results, we get
T
π∝
V
or πV = kT
…(i)
where k is called general solution constant.
Vant Hoff further proved that the value of k is same as R,
Hence, πV = RT
Further if the solution contains n moles of the solute dissolved in V litres of solution, then
n
C =
V
Substituting the value of C in equation (1), we get;
n
T
V
πV = n RT
π
*10.
Ans.
∴
= R
...(ii)
How is molar mass of a solute determined from osmotic pressure measurements.
According to van’t Hoff equation,
π = CRT
n
But C =
V
where n is the number of moles of solute dissolved in V litre of the solution.
n
RT
π=
V
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
115
W2
The number of moles of solute n may be given as M . Here W2 is the weight of the solute
2
and M2 is its molar mass
Substituting the value of n in the above expression.
W2 RT
W2 RT
π = VM or M2 =
Vπ
2
Thus, the molar mass of the solute, M2, can be calculated.
11.
Ans.
Explain reverse osmosis.
(a) Definition : If, a solution is separated from the pure solvent by a semi-permeable
membrane and the pressure applied on the solution (more than the osmotic pressure)
the solvent starts flowing from the solution towards the pure solvent. This phenomenon
is known as reverse osmosis.
(b) Application of reverse osmosis :
Desalination :
(1) Sea water contains 3.5% w/w of dissolved salts.
(2) Drinking water is produced from sea water by a process called as desalination by
reverse osmosis.
(3) In reverse osmosis high pressure greater than osmotic pressure of sea water i.e.,
30 atm. is applied to force water from concentrated aqueous solution like sea water
to pure water side through a semi permeable membrane.
(4) The method can be used provided a suitable semipermeable membrane is developed
which withstands the high pressure condition over a prolonged period.
[Refer Fig. 2.4 (h)]
*12.
Ans.
Explain Abnormal molecular masses.
(a) Colligative properties of solutions depend on actual number of solute particles present in
the solution.
(b) In Dilute solution of non-electrolytes like urea, glucose etc. the solute remains in normal
molecular condition and does not undergo dissociation or association.
(c) In case of solutions of electrolytes, it has been observed that, observed colligative properties
and thus, molar mass determined by these methods do not agree with the expected or
theoretical values.
(d) This is because of association or dissociation of the solute molecules in the solution.
(1) Dissociation :
(i) Molecules of certain substances (acids, bases and salts) dissociate or ionize in
a solvent to give two or more particles. For example, AB dissociates to give
double number of particles as :
A + + B –
AB Chapter - 2 Solutions and Colligative Properties
116
(ii) Consequently, the total number of particles increases in solution and, therefore,
the colligative property of such solutions will be large for solutes undergoing
dissociation.
(iii) Since the colligative property is inversely proportional to the molar mass of the
solute, the molar mass of solute in such cases will be less than normal. The value
of colligative property depends on the degree of dissociation of the solute in the
solution.
(iv) For example :
NaCl
⎯⎯
→ Na + + Cl – : 2 particles
K 2SO 4
⎯⎯
→ 2K +
: 3 particles
+ SO 2–
4
AlCl 3
⎯⎯
→ Al 3+ + 3Cl – : 4 particles
(2) Association :
(i) In some non-polar solvents, the solute molecules undergo association, i.e. two, three
or even more molecules exist in combination with each other to form bigger molecules.
(ii) Therefore, the total number of molecules in solution become less than the number
of molecules of the substance added. This leads to a decrease in the value of colligative
properties because colligative properties is proportional no. of solute particles.
(iii) Since the colligative property is inversely proportional to the molar mass of the
solute, the molar mass of solute in such cases will be greater than normal.
(iv) For example, in benzene solvent solutes like acetic acid, benzoic acid etc. exist
as dimers. This is due to hydrogen bonding between these molecules.
(CH 3COOH) 2
2CH 3COOH (C 6 H 5COOH) 2
2C 6 H 5COOH *13.
Ans
Write a note on van’t Hoff factor or Explain van’t Hoff factor.
(a) This factor is used to express the extent of dissociation or association of the solutes in
their solutions.
(b) It defines as the ratio of observed colligative property produced by a given concentration
of electrolyte solution to the property observed for the same concentration of non electrolyte
solution.
(c) Therefore, i =
observed value of colligative property
theoretical value of colligative property
Since colligative property α number of solute particles present in the solution. Hence,
number of solute particles present in solution
theoretical number of solute particles
1
Similarly, colligative property α molecular mass
i=
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
117
theoretical molecular mass
i = observed molecular mass
(d) If i = 1, solute molecules are in normal state and no association or dissociation has taken
place.
(e) If i > 1 , the solute molecules are dissociated in the solution.
(f) If i < 1, the solute molecules are associated in the solution.
*14.
Derive and expression for vant Hoff factor and degree of dissociation.
Derive the equations : (a) α =
Ans.
OR
i –1
M (theoretical) – M (observed)
(b) α =
n′ – 1
M (observed) (n ′ – 1)
Consider ‘m’ moles of an electrolyte AxBy dissolved in 1kg of solvent. Let α be the degree
of dissociation of the electrolyte.
⎯⎯
→ xA y+ + yB x –
A x B y ←⎯
⎯
∴
Initial moles
m
0
0
moles at equilibrium
m – mα
x(mα)
y(mα)
Total no. of moles at equilibrium = mt = (m – mα) + x(mα) + y(mα)
= m(1 – α) + x(mα) + y(mα)
= m[1 – α + xα + yα]
= m[1 + xα + yα – α]
= m[1 + α (x + y – 1)]
Let the total no. of ions produced by dissociation by one mole of eletrolyte AxBy be
x + y = n′
mt = m (1 + α (n′ – 1))
Now vant Hoff factor,
∴
∴
Thus, degree of dissociation α =
i
=
α =
observed number of moles
i = theoretical number of moles
mt
m (1 + (n ′ –1) α)
i =
=
m
m
i = 1 + (n′ – 1) α
i –1
n′ – 1
M (Theoretical)
= 1 + (n′ – 1) α
M (observed)
M (Theoretical)
– 1 = (n′ – 1) α
M (observed)
M (Theoretical) – M (observed)
M (observed) (n ′ – 1)
Chapter - 2 Solutions and Colligative Properties
118
*15.
Ans.
Explain Abnormal osmotic pressure.
Osmotic pressure is a colligative property, thus, depends upon the number of particles present
in the solute. Therefore, the calculated osmotic pressure is different from its experimental valie
in some cases.
This is because of association or dissociation of the solute particles in the solution.
a) Dissociation:
i) Molecules of certain substances (acids, bases and salts) dissociate or ionize in a solvent
to give two or more particles. For example, AB dissociates to give double number
A + + B –
of particles as : AB ii)
Consequently, the total number of particles increases in solution and, therefore, the
osmotic presure of such solutions will be large for solutes undergoing dissociation.
iii) For example: NaCl
NaCl ⎯⎯
→ N a+ + Cl –
:
2 particles
K 2SO 4 ⎯⎯
→ 2K + + SO 24 –
:
3 particles
: 4 particles
AlCl 3 ⎯⎯
→ Al 3+ + 3Cl –
b) Associaton :
i) In some non-polar solvents, the solute molecules undergo association, i.e. two, three
or even more molecules exist in combination with each other to form bigger molecules.
ii) Therefore, the total number of molecules in solution become less than the number
of molecules of the substance added. This leads to a decrease in the value of osmotic
pressure because osmotic pressure ∝ no. of solute particles.
iii) For example, in benzene solvent solutes like acetic acid, benzoic acid etc exist as
dimmers as. This is due to hydrogen bonding between these molecules.
(CH 3COOH) 2
2CH 3COOH (C 6 H 5COOH) 2
2C 6 H 5COOH Multiple Choice Questions :
•
*1.
Theoretical MCQs :
In osmosis .......
a) Solvent molecules pass from high concentration of solute to low concentration.
b) Solvent molecules pass from a solution of low concentration of solute to a solution
of high concentration of solute.
c) Solute molecules pass from low concentration to high concentration.
d) Solute molecules pass from high concentration to low concentration.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
119
*2.
*3.
*4.
The two solutions with same osmotic pressure are called .......
a) Isotonic
b) Isomeric
c) Hypotonic
d) Hypertonic
The values of gas constant and solution constant .......
a) Are different
b) Almost identical
c) Gas constant is greater than solution constant
d) Gas constant is smaller than solution constant
Which of the following 0.1M aqueous solutions will exert highest osmotic pressure?
c) MgSO4
d) Al2(SO4)3
a) NaCl
b) BaCl2
*5.
Abnormal molar mass is produced by .......
a) Association of solute
b) Dissociation of solute
c) Both association and dissociation of solute
d) Separation by semipermeable membrane
*6.
Which of the following aqueous solutions will have minimum elevation in boiling point?
a) 0.1M KCl
b) 0.05M NaCl c) 1M AlPO4
d) 0.1M MgSO4
*7.
Which of the following will have maximum depression in freezing point? (MHT-CET 2009)
b) 1M KCl
c) 0.5M Al2(SO4)3 d) 0.5M BaCl2
a) 0.5M Li2SO4
•
8.
Ans.
Numerical MCQs :
A solution has an osmotic pressure of 0.821 10.
atm at 300K. Its concentration would be
.......
a) 0.66M
b) 0.32M
c) 0.066M
d) 0.033M
π
= CRT
Ans.
0.821 = C × 0.0821 × 300
1
=
0.033M
30
1 molar solution of a non-electrolyte
compound will produce an osmotic pressure
.......
a) 1 atm
b) 44.8 atm
c) 10.0 atm
d) 22.4 atm
π = CRT
= 1 × 0.0821 × 273
= 22.4 atm
C
9.
Ans.
=
The weight of glucose to be dissolved in
250cm3 of water, to get a solution which is
isotonic with 0.2M solution of urea, is .......
a) 36g
b) 9g
c) 18g
d) 4.5g
πglucose = πurea Cglucose = Curea
n glucose
= 0.2
V
mass of glucose
∴
11.
= 0.2
180 × 250 × 10 –3
Mass of glucose = 0.2 × 180 × 250 × 10–3
= 9g
The Vant Hoff factor for non-electrolytes
like, urea, glucose etc. will be .......
a) 2
b) 0.5
c) 1.0
d) –1
Chapter - 2 Solutions and Colligative Properties
120
12.
Ans.
∴
•
13.
Ans.
The freezing point of 1 molal NaCl solution assuming NaCl to 100% dissociated in water is
.........
a) –1.86ºC
b) –3.72ºC
c) +1.86ºC
d) +3.72ºC
ΔTf = i × Kf × m
= 2 × 1.86 × 1 = 3.72ºC
ΔTf = Tº – T
3.72 = 0 – T
T
= 0 – 3.72
= –3.72ºC
Among the following equimolar aqueous solutions identify the one having highest boiling point.
(March 2008)
a) Sodium chloride
b) Sucrose
c) Sodium sulphate d) Urea
urea i = 1,
sucrose i = 1
NaCl i = 2,
Na2SO4 i = 3
HOURS BEFORE EXAM
Per cent by Weight (% w/w) =
Mass of solute
× 100
Mass of solution
Per cent weight by Volume (% w/v) =
Per cent by Volume (% v/v) =
Parts per million (ppm) =
No. of moles (n) = Molecular weight
Mass of solute
× 100
Volume of solution
Volume of solute
× 100
Volume of solution
Mass of solute
× 10 6
Total mass of solution
Weight in gms
No. of moles =
No. of moles =
Vol. of gas in dm 3 at S.T.P
22.4
No.of particles (Atoms, molecules or ions)
Unique Solutions ®
6.023×10 23
S.Y.J.C. Science - Chemistry - Part I
121
n2
Mole fraction of solute = x2 = n + n
2
1
n1 = No. of moles of solvent,
n2 = No. of moles of solute.
x1 + x2 = 1
Molarity =
Resulting molarity of mixed solution =
[For Binary solution]
No. of moles of solute
Volume of solution in litre or dm 3
Unit of molarity = mol/lit or mol/dm3
It depends upon temperature.
Molarity of pure water = 55.55
[M 1V1 + M 2 V2 ]
V1 + V2
For Dilution: M1V1 = M2V2
[If solution is diluted x times, then its molarity decrease by x times]
For Neutralization : Ma × Va = Mb × Vb (For acids or bases containing same no. of
H and OH)
Normality (N) =
No. of gm equi. of solute
gm / lit. of solute
eq. wt. of solute
Volume of solution in dm
weight in gms of solute
No. of gm eq.wt. of solute =
eq. wt. of solute
Molecular weight
3
OR =
Atomic weight
Molecular weight
Eq. wt. = No. of H or OH atoms =
= Total charge on cation or Anion
Valency
Units = gm Equivalent per litre.
Normality changes with temperature.
If solution is diluted x time, its normality decreases by x times.
N 1V1 + N 2 V2
Resulting Normality of mixed solution : N =
V1 + V2
For dilution : N1V1 = N2V2
For Neutralization: Na × Va = Nb × Vb (Irrespective of No. of H and OH present in acid
and base)
1N
= Normal solution
N/10
= Decinormal solution
N/100 = Centinormal solution
N/1000
= Millinormal solution
Relationship between Normality and Molarity :
Normality = Basicity × Molarity (for acid) Normality = Acidity × Molarity (for base)
Chapter - 2 Solutions and Colligative Properties
122
Basicity for acid = No. of H+ present
Acidity for base = No. of OH present
Molality =
[eg. HCl – 1, H2SO4 – 2, CH3COOH = 1]
[eg. NaOH = 1, Ca(OH)2 = 2]
No. of moles of solute
Weight of solvent in kg
Molality is independent of temperature.
Units = moles/kg.
Molality of pure water = 55.55
Lowering of vapour pressure = p10– p (where, p10 = vapour pressure of pure solvent)
(p = vapour pressure of solution containing a non volatile solute.)
p10 – p
Relative lowering of vapour pressure =
Raoults law (for 2 miscible volatile liquids)
p10
p1 =
p10 × x1
p1
p2 =
p 02 × x 2
p 10 = vapour pressure of pure solvent
p
=
p1 + p 2
x1
p
=
Total vapour pressure of solution
= Partial vapour pressure of Solvent
= mole fraction of solvent
Similarly for solute
Raoults law (for non-volatile solute in volatile solvent)
p0 – p
= x2 (x2 = mole fraction of solute)
p = po × x1 (x1 = mole fraction of solvent) and
p
Molecular weight of solute using Raoults law M2 =
(for dilute solutions)
M2 = Molecular weight of solute,
W2 = Weight of solute in gms,
M1 =
W1 =
W2 × M 1
p0
× 0 1
W1
p1 – p
Molecular weight of solvent.
Weight of solvent in gms
Boiling Point Elevation :
Elevation in B.P. = Increase in B.P. of solvent is non-volatile solute is added to it.
Where Tb = B.P. of solution,
ΔTb = T – T0
0
T = B.P. of pure solvent.
ΔTb ∝ molality (m) (Raoults law)
ΔTb = Kb × m
(Kb = Molal elevation B.P. constant = Ebullioscopic constant).
Units of Kb = Kelvin kg/mole.
Kb for water = 0.52 Kelvin kg/mol.
Molecular weight of solute.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
123
K b × W2
M 2 = ΔT × W
b
1
M2 =
Molecular weight of solute in kg mol-1
Kb
ΔTb
W2 =
W1 =
Weight of solute in kg
Weight of solvent in kg.
= Ebullioscopic constant
= T – Tº
Freezing Point Depression :
Depression in F.P. : When a non-volatile solute is added to volatile solvent, the F.P. of
solvent decreases.
ΔT = Tº – T
= F.P. of pure solvent.
T0
T
= F.P. of solution
(molality)
ΔTf ∝ m
(Raoults law)
ΔTf = Kf x m
= Molal F.P. depression constant
[Cryoscopic constant]
Kf
= Kelvin kg/mol
Kf
e.g. Kf of water = 1.86 Kelvin kg/mol.
M2
K f × W2
= ΔT × W
f
1
W2
W1
M2
= Weight of solute in kg
= Weight of solvent in kg
= Molecular weight of solute in kg mol-1
Important formulae related to Osmotic Pressure
π1
C1
=
π1
n1
=
π2
n2 ;
πV
=
w
RT
M
–2
Nm
π
or Pa
kPa
Atm
π2
C2 ;
π1
π2
=
T1
T2 ;
πV = nRT ;
V
m3
dm3
dm3
π
= C RT
R
8.314
8.314
0.082
Chapter - 2 Solutions and Colligative Properties
124
NUMERICALS WITH SOLUTION
*1.
Type-I-Concentration of solutions
34.2 g of glucose is dissolved in 400 g of
water. Calculate percentage by mass of
glucose solution.
Given :
mass of glucose = 34.2g
mass of water = 400 g
To find :
% by mass of glucose
Solution :
2.38(x + 500) =
2.38x + 500 × 2.38 =
500 × 2.38 =
500 × 2.38 =
100x
100x
100x – 2.38x
97.62x
500 × 2.38
97.62
mass of solute = 12.19 g.
mass of solute = x =
4.6 cm3 of methyl alcohol is dissolved in
25.2 g of water. Calculate (i) % by mass
of methyl alcohol (ii) mole fraction of
% by mass of glucose
methyl alcohol and water. (Given density
of methyl alcohol = 0.7952 gcm3 and
mass of glucose
×
100
= masss of glucose × mass of H O
C = 12, H = 1, O = 16)
2
Given :
34.2
× 100
=
Volume of methyl alcohol = 4.6 cm3
(34.2 + 400)
mass of water = 25.2g
34.2
× 100 = 7.87% w/w
=
density of methyl alcohol = 0.7952 g/cm3
434.2
% by mass of glucose is 7.87 w/w
To find :
i) % by mass of methyl alcohol
*2. A solution is prepared by dissolving certain
ii) mole fraction of methyl alcohol and water
amount of solute in 500g of water. The Solution :
percentage by mass of a solute in solution
Density of methyl alcohol
is 2.38. Calculate mass of solute. (12.19g)
mass of methyl alcohol
Given :
= volume of methyl alcohol
mass of water = 500g
mass of methyl alcohol
% by mass of solute = 2.38%
0.7952 =
To find :
4.6
mass of solute
∴ mass of methyl alcohol = 0.7952 × 4.6
Sol.
:
mass of methyl alcohol = 3.6579 g
Let the mass of solute be xg % by mass
Molar mass of methyl alcohol (CH3OH)
of solute
= 12 + 3 + 16 + 1 = 32
moles of methyl alcohol
mass of solute
= mass of solute × mass of water × 100
mass of methyl alcohol
=
x
× 100
2.38 =
x + 500
Unique Solutions ®
*3.
=
molar mass of methyl alcohol
3.658
= 0.1143 moles
32
S.Y.J.C. Science - Chemistry - Part I
125
mass of water
mole of water = molar mass of water
25.2
=
= 1.4 moles
18
mole fraction of methyl alcohol (x2)
=
=
=
=
=
=
=
=
12.8
12.8
× 100 =
× 100
12.8 + 16.8
29.6
43.24 % w/w
% by volume of benzene = 43.24% w/w
Calculate mole fraction of HCl in solution
of HCl containing 24.8% of HCl by mass.
(H = 1, Cl = 35.5)
0.1143
0.1143
=
= 0.0754
Given :
0.1143 + 1.4
1.5143
mole fraction of methyl alcohol = 0.0754
% by mass of HCl = 24.8%
mole fraction of water (x1)
To find :
1 – x2
mole fraction of HCl = ?
1 – 0.0754 = 0.9245
Solution:
mole fraction of water = 0.9245
HCl solution is 24.8% by mass.
% by mass of methyl alcohol
Hence, 100g HCl solution contain 24.8g .
HCl and (100 – 24.8) = 75.2g
mass of methyl alcohol
×100
mole of HCl =
mass of water + mass of methyl alcohol
moles of methyl alcohol
moles of methyl alcohol × moles of water
3.658
× 100
(25.2 + 3.658)
3.658
× 100 = 12.68% w/w
28.858
% by mass of methyl alcohol =
12.68% w/w
12.8 cm3 of benzene is dissolved in 16.8
cm3 of xylene. Calculate % by volume of
benzene.
Given :
volume of benzene = 12.8 cm3
volume of xylene = 16.8 cm3
To find :
% by volume of benzene
Solution:
% by volume of benzene
*4.
=
=
volume of benzene
×100
volume of benzene + volume of xylene
*5.
mass of HCl
n2 = molar mass of HCl
24.8
=
= 0.6794 mol
36.5
mole of water =
mass of water
n1 = molar mass of water
75.2
= 4.178 moles
18
mole fraction of HCl =
=
x2 =
moles of HCl
moles of HCl + moles of water
0.6794
0.6794 + 4.178
0.6794
=
4.8574
=
= 0.1398
mole fraction of HCl = 0.1398
Chapter - 2 Solutions and Colligative Properties
126
*6.
Calculate mole fraction of solute in its 2
molal aqueous solution.
Given :
molality = 2m
To find :
mole fraction of solute
Solution:
Let the mass of water be 1 kg = 1000g
∴
∴
molality
=
2
=
no. of moles of solute
mass of water in kg
moles of solute
1
=
=
mass of water
molar mass of water
1000
= 55.55 mole
18
=
moles of solute
moles of solute + moles of water
2
= 0.0347
2 + 55.55
=
=
mole fraction of solute = 0.0347
*7.
Calculate the mole fractions, molality and
molarity of HNO3 in a solution containing
12.2% HNO3. Given density of HNO3 =
1.038g cm–3, H = 1, N = 14, O = 16.
Given :
(i) % by mass of HNO3 = 12.2%
(ii) Density of HNO3 = 1.038 g cm–3
To find :
(i) mole fraction of HNO3
(ii) mole fraction of H2O
(iii) molality = ?
(iv) molarity = ?
Unique Solutions ®
mass of H 2O
= molar mass of H O
2
87.8
=
= 4.878 mol
18
mole fraction of HNO3 (x2)
moles of HNO 3
moles of HNO 3 + moles of H 2 O
n1
mole fraction of solute (x2)
=
moles of H2O
moles of solute (n2) = 2 mol
moles of water =
n1
=
Solution :
HNO3 solution is 12.2% by mass
Hence 100g of HNO3 solution containing
12.2g HNO3 and (100 – 12.2)
= 87.8g of H2O
Moles of HNO3
mass of HNO 3
n2 = molar mass of HNO
3
12.2
n2 =
= 0.1936 mol
63
=
=
0.1936
0.1936
=
0.1936 + 4.878
5.0713
0.0382
mole fraction of HNO3 (x2)= 0.0382
Mole fraction of H2O (x1)
1 – mole fraction of HNO3 (x2)
1 – 0.0382 = 0.9618
Mole fraction of H2O (x1) = 0.9618
moles of HNO 3
molality (m) = mass of H O in kg
2
0.1936
=
= 2.205m
87.8 × 10 –3 kg
molality (m) = 2.205m
Density of HNO3 solution
=
mass of HNO 3 solution
volume of HNO 3 solution
S.Y.J.C. Science - Chemistry - Part I
127
1.038 =
∴
100
volume of HNO 3 solution
100
vol. of HNO3 soln. =
1.038
= 96.34 cm3
= 96.34 × 10–3 dm3
Moles of HNO 3
Molarity = Volume of HNO solution
3
0.1936
=
96.34 × 10 –3
= 2.01 M
Molarity = 2.01 M
=
=
=
=
moles of water (n1)
mass of water
molar mass of H 2O
4.2
= 0.2333 moles
18
mole fraction of H2SO4 (x2)
moles of H 2SO 4
moles of H 2SO 4 + moles of H 2O
0.9775
0.9775
=
= 0.8071
0.9775 + 0.2333
1.211
mole fraction of H2SO4 (x2) = 0.8071
Density of H2SO4 solution
mass of H 2SO 4 solution
volume of H 2SO 4 solution
=
Sulphuric acid is 95.8% by mass. Calculate
100
mole fraction and molarity of H2SO4 of
1.91 = volume of H SO solution
3
density 1.91 cm .
2
4
(H = 1, S = 32, O = 16.)
100
∴
vol.
of
H
SO
soln.
=
2
4
Given :
1.91
(i) % by mass of sulphuric acid = 95.8%
= 52.36cm3
3
= 52.36 × 10–3 dm3
(ii) Density of H2SO4 solution = 1.91 g/cm
moles of H 2SO 4 (n 2 )
To find :
Molarity
=
volume of H 2SO 4 solution
(i) mole fraction of H2SO4
(ii) molarity = ?
0.9775
= 18.67 M
Solution :
52.36 × 10 –3
H2SO4 solution is 95.8% by mass
Molarity = 18.67 M
Hence, 100g of H2SO4 solution contains
95.8% of H2SO4 and 100 – 95.8 = 4.2g
*9. Aqueous solution of NaOH is marked 10%
of water.
(w/w). The density of the solution is 1.070g
moles of H2SO4 (n2)
cm–3. Calculate (i) molarity (ii) molality and
mass of H 2SO 4
(iii) mole fraction of NaOH and water. Na
= molar mass of H SO
= 23, H = 10, O = 16.
2
4
95.8
Given :
=
98
(i) % by mass of NaOH = 10% w/w
= 0.9775 moles
(ii) Density = 1.07 g/cm3
moles of H2SO4 (n2) = 0.9775 moles To find :
(i) Molarity = ?
*8.
Chapter - 2 Solutions and Colligative Properties
128
(ii) Molality = ?
(iii) Mole fraction of NaOH (x2) = ?
(iv) Mole fraction of water (x1) = ?
Solution :
NaOH solution is 10% by mass
Hence, 100g of NaOH solution contains
10 g of NaOH in 100 – 10 = 90 g of water
=
=
=
=
=
=
∴
=
=
=
volume of NaOH solution
100
1.07
93.46 cm3
93.46 × 10–3 dm3
molesof NaOH
molarity =
0.25
mass of NaOH
moles of NaOH(n2) =
molar mass of NaOH
=
10
= 0.25 mole
40
moles of H2O (n1)
mass of water
molar mass of water
90
= 5 moles
18
mole fraction of NaOH (x2)
moles of NaOH
moles of NaOH + moles of H 2 O
0.25
0.25
=
= 0.0476
0.25 + 5.0
5.25
mole fraction of NaOH (x2) = 0.0476
mole fraction of H2O (x1)
1 – mole fraction of NaOH (x2)
moles of NaOH
molality = mass of H O in kg
2
= 1 – 0.0476
= 0.9524
0.25
=
90 × 10 –3
= 2.777 m
molality = 2.777 m
Density of NaOH solution
mass of NaOH solution
volume of NaOH solution
100
1.07 =
volume of NaOH solution
Unique Solutions ®
valumeof NaOHsolutionindm 3
=
93.46 × 10 –3
= 2.675 M
Battery acid is 4.22 M aqueous H2SO4
solution, and has density of 1.21 g cm–3.
What is the molality of H2SO4?
(H = 1, S = 32, O = 16)
Given :
molarity of H2SO4 aqueous = 4.22 M,
density = 1.21 g cm–3
To find :
molality = ?
Solution:
let the volume of H2SO4 solution
1 dm3 = 1000 cm3
molarity
moles of H 2SO 4 (n 2 )
=
volume of H 2SO 4 solution in dm 3
moles of H 2SO 4
4.22 =
1
∴ moles of H2SO4 (n2) = 4.22 moles
moles of H2SO4 (n2)
mass of H 2SO 4
= molar mass of H SO
2
4
*10.
4.22 =
∴
mass of H 2SO 4
98
mass of H2SO4 = 98 × 4.22
= 413.5 g
mass of H 2SO 4 solution
Density = volume of H SO solution
2
4
S.Y.J.C. Science - Chemistry - Part I
129
1.21
∴
=
=
=
=
=
*11.
mass of H 2SO 4 solution
=
1000
mass of H2SO4 solution = 1.21 × 1000
= 1210 g
mass of solvent (H2O)
mass of H2SO4 solution – mass of H2SO4
1210 – 413.5 = 796.5g
796.5 × 10–3 kg
molality (m)
moles of H 2SO 4
mass of solvent (H 2 O)in kg
4.22
= 5.298 m
796.5 × 10 –3
molality = 5.298 m
=
=
=
=
=
=
moles of water (n1)
mass of water
molar mass of water
65
= 3.611 moles
18
mole fraction of ethylene glycol (x2)
moles of ethylene glycol
moles of ethylene glycol + moles of water
0.5645
0.5645
=
= 0.1352
0.5645 + 3.611
4.1755
mole fraction of ethylene glycol (x2)
= 0.1352
mole fraction of water (x1)
1 – mole fraction of ethylene glycol (x2)
1 – 0.1352 = 0.8648
mole fraction of water = 0.8648
35% (W/W) solution of ethylene glycol in
water, an anti-freezer used in automobiles
12. Find ppm (parts per million) of solute if 0.5g
in radiators as a coolant. It lowers freezing
of solute is present in 107 g of solution.
point of water to –17.6ºC. Calculate the
Given :
mole fraction of the components.
mass of solute = 0.5g,
Given :
mass of solution = 107 g
% w/w of ethylene glycol solution = 35%
To find :
To find :
ppm
mole fraction of ethylene glycol and water
Solution:
Solution :
parts per million (ppm)
Ethylene glycol solution is 35% by mass.
mass of solute
Hence, 100g of ethylene glycol
× 10 6
=
mass of solution
solution contain 35g of ethylene glycol
0.5
in 100 – 35 = 65g of water.
× 10 6 = 0.05 ppm
=
7
molar mass of ethylene glycol
10
–1
parts
per million of solute = 0.05 ppm
[HO – CH2 – CH2 – OH] = 62g mol
moles of Ethylene glycol(n2)
*13. The tolerable level of contamination of
mass of ethylene glycol
=
chloroform in a given sample of drinking
molar mass of ethylene glycol
water is found to be 15 ppm by mass.
35
=
= 0.5645 moles
(i) Express it in percentage by mass
62
(ii) Determine molality of chloroform in the
Chapter - 2 Solutions and Colligative Properties
130
given sample of water containing 15 ppm
of chloroform.
Given :
(i) ppm by mass of chloroform in water
= 15 ppm
To find :
(i) % by mass
(ii) molality
Solution :
ppm by mass of chloroform in water is 15
ppm. Hence, 106 g of solution contains 15g
of chloroform.
mass of water = 106 – 15 ≈ 106 g
% by mass of chloroform
=
=
=
=
=
=
=
/////
////
////
////
////
////
//
*14.
What is the concentration of dissolved
oxygen at 25ºC at 1 atmospheric pressure
if partial pressure of oxygen is 0.22 atm?
The Henry’s law constant for oxygen is
1.3 × 10–3 mol dm–3 atm–1.
Given :
(i) partial pressure of oxygen = Po2
= 0.22 atm
(ii) Henry Law constant for oxygen = KH
= 1.3 × 10–3, mol dm–3 atm–1
To find :
Concentration of dissolved oxygen (s)
Solution:
According to Henry’s law, S = KH × p O 2
mass of chloroform
× 100
mass of solution
15
× 100 = 1.5 × 10–3 %
10 6
% by mass of chloroform = 1.5 × 10–3 %
molar mass of chloroform (CHCl3)
12 + 1 + 3 (35.5)
13 + 106.5
119.5 g mol–1
moles of chloroform
mass of chloroform
molar mass of chloroform
15
= 0.1255 mole
119.5
moles of chloroform
molality = mass of H O in kg
2
0.1255
=
10 6 × 10 –3 (kg)
= 1.255 × 10–4 m
molality = 1.255 × 10–4 m
K = 1.3 × 10–3 mol dm–3 atm–1
p = 0.22 atm
Hence,
S = 1.3 × 10–3 mol dm–3 atm–1 × 0.22 atm
S = 2.86 × 10–4 mol dm–3
*15.
The solubility of nitrogen gas at 1 atm
pressure at 25ºC is 6.8 × 10–4 mol dm–3.
Calculate the solubility of N2 gas from
atmosphere at 25ºC if atmospheric pressure
is 1 atmosphere and partial pressure of N2
gas at this temperature and pressure is 0.78
atm.
Given :
(i) solubility of N2 gas at 1 atm
at 25ºC = 6.8 × 10–4 mol dm–3
To find :
solubility of N2 gas at 0.78 atm at 25ºC
=?
Solution :
According to Henry’s law
S = Kp N 2 , S = 6.8 × 10–4 mol dm–3,
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
131
p N 2 = 1 atm
p10 – p
Hence,
S = 6.8 × 10–4 mol dm–3 = K.1 atm
Hence,
K = 6.8 × 10–4 mol dm–3
= 6.8 × 10–4 mol dm–3 atm–1
p = 0.78 atm, S = ?
Therefore,
S = 6.8 × 10–4 mol dm–3 atm–1 × 0.78 atm
= 5.304 × 10–4 mol dm–3
Solubility of N2 is reduced to = 5.304
× 10–4 mol dm–3
W2 M 1
= M ×W
2
1
p10
760 – 755
=
760
M2 =
2.1
18
×
M 2 97.9
2.1 × 18 × 760
= 58.69 g mol–1
5 × 97.9
molar mass of the solute (M2) = 58.6g
mol–1
*17.
The vapour pressure of water at 20ºC is
17 mm Hg. Calculate the vapour pressure
of a solution containing 2.8g of urea
(NH2CONH2) in 50g of water N = 14, C
= 12, O = 16, H = 1. (16.71 mm Hg)
Given :
TYPE - II - PROBLEM BASED ON COLLIGATIVE
PROPERTIES, LOWERING OF VAPOUR PRESSURE,
RAOULTS LAW, MOLECULAR WEIGHT BY
LOWERING IN VAPOUR PRESSURE :
*16. The vapour pressure of 2.1% solution of
(i) vapour pressure of water = p10 = 17 mm
a non-electrolyte in water at 100ºC is 755
Hg
mm Hg. Calculate the molar mass of the
(ii) mass of urea = W2 = 2.8 g
solute.
(iii) mass of water = W1 = 50 g
Given :
To find :
(i) % by mass of non-electrolyte in water = 2.1%
vapour pressure of solution (p) = ?
(ii) Temp = 100ºC
Solution :
(iii) Vapour pressure of solution = p = 755 mm kg
molar mass of urea (NH2CONH2)
To find :
M2 = 60 g mol–1
molar mass of the solute = ?
W2 M 1
p10 – p
×
Solution :
=
0
M 2 W1
p1
100ºC pure water will boil.
∴
17 – p
17
v.p of pure water ( p10 ) = Atmospheric
pressure = 760 mm of kg
Solution is 2.1% by mass
Hence 100 g of solution contains 2.1 g of
solute (w2) in 100 – 2.1 = 97.9 g of water
(w1)
∴
2.8 18
×
60 50
2.8 × 18 × 17
17 – p =
60 × 50
17 – p = 0.2856
p = 17 – 0.2856 = 16.71 mm Hg
Vapour pressure of solution is 16.71
mm Hg
=
Chapter - 2 Solutions and Colligative Properties
132
*18.
In an experiment, 18.04g of mannitol were
dissolved in 100g of water. The vapour
pressure of water was lowered by 0.309
mm Hg. from 17.535 mm Hg. Calculate the
molar mass of mannitol. (184.3 g mol–1)
Given :
(i) mass of mannitol = W2 = 18.04 g
(ii) mass of water
= W1 = 100 g
(iii) lowering of vapour pressure
= p10 – p
= 0.309 mm Hg
(iv) vapour pressure of pure water
=
p 10
=
=
= 17.535 mm Hg
To find :
molar mass of mannitol (M2) = ?
Solution:
W2 M 1
p10 – p
= M ×W
0
p1
2
1
18.04 18
0.309
×
=
M 2 100
17.535
∴
Solution :
vapour pressure reduces to 80%
If vapour pressure of pure octane ( p10 ) = 40
then vapour pressure of solution (p) = 80
molar mass of octane (C8H18)
= M2 = 8 (12) + 18 (1)
= 96 + 18 = 114 g mol–1
moles of octane (n1)
M2 =
114 × 10 –3
114 × 10 –3
n1 = 1 mole
p10 – p
=
x2
100 – 80
100
=
n2
1 + n2
20
100
=
n2
1 + n2
p10
∴
18.04×18×17.535
= 184.3 g mol–1
0.309×100
molar mass of mannitol (M2) =184.3 g
mol–1
*19.
Calculate the mass of a nonvolatile solute
(molar mass 40 × 10–3 kg/mol), dissolved
in 114 × 10–3 kg of octane to reduce its
vapour pressures to 80%.
Given :
(i) vapour pressure reduces to 80%
(ii) mass of octane = 114 × 10–3 kg (W1)
(iii) molar mass of non-volatile solute M2
= 40 × 10–3 kg/mol
To find :
mass of non-volatile solute = W2 = ?
Unique Solutions ®
mass of octane
molar mass of octane
∴
1
5
1 + n2
1
4n2
=
=
=
n2
=
=
n2
1 + n2
5n2
5n2 – n2
1
moles of solute =
1
= 0.25 mol
4
mass of solute
molar mass of solute
mass of solute
0.25 =
∴
40 × 10 –3
mass of solute = 0.25 × 40 × 10–3
= 10 × 10–3 kg
mass of solute (W2) = 10 × 10–3 kg
S.Y.J.C. Science - Chemistry - Part I
133
(ii) Boiling point of solution
TYPE III - PROBLEM BASED ON ELEVATION
= T = 80.256ºC
OF BOILING POINT :
(iii) Mass of solute = W2
*20. Calculate the mass in grams of an impurity
= 0.419 g = 0.419 × 10–3 kg
of molar mass 100g mol–1 which would be
(iv) Mass of solvent
required to raise the boiling point of 50g of
= W1 = 75g = 75 × 10–3 kg
Chloroform by 0.30 K. (Kb for chloroform =
(v) molar mass of solute
3.63 K kg mol–1). (0.4132 g)
= M2 = 252.4 g mol–1
Given :
= 252.4 × 10–3 kg mol –1
(i) molar mass of Impurity
To find :
= M2 = 100 g mol–1
Molar elevation constant = Kb = ?
= 100 × 10–3 kg mol–1
Solution :
(ii) mass of Chloroform
ΔTb = T – Tº
= W1 = 50 g = 50 × 10–3 kg
= 80.256 – 80.2 = 0.056ºC
(iii) Elevation of boiling point
Kb
W2
= ΔTb = 0.3 K
M2 = ΔT × W
b
1
(iv) Kb of chloroform = 3.63 K kg mol–1
Kb
0.419 × 10 –3
–3
×
To find :
252.4 × 10 = 0.056
75 × 10 –3
mass of Impurity = W2 = ?
252.4 × 10 –3 × 0.056 × 75 × 10 –3
Solution :
∴ Kb =
–3
M2
100 × 10–3
∴
100 × 10 –3 × 50 × 10 –3 × 0.3
3.63
= 0.4132 × 10–3 kg = 0.4132g
mass of Impurity = 0.4132g
W2 =
0.419 × 10
= 2.53 K kg mol–1
Kb
W2
= ΔT × W
b
1
W2
3.63
= 0.3 ×
50 × 10 –3
Molar elevation constant = Kb = 2.53
K kg mol –1
*22.
A solution containing 0.5126g of
naphthalene (molar mass = 128.17 g mol–1)
in 50.0g of CCl4 gives a boiling point
elevation of 0.402K. While a solution of
0.6216g of unknown solute in the same
*21. Boiling point of a solvent is 80.2ºC. When
mass of the solvent gives a boiling point
0.419 g of the solute of molar mass 252.4
elevation of 0.647K. Find the molar mass
g mol–1 was dissolved in 75g of the solvent,
of the unknown solute.
the boiling point of the solution was found
(Kb for CCl4 = 5.03K kg mol–1 of solvent)
to be 80.256ºC. Find the molal elevation
Given :
constant.
For naphthalene solution :
Given :
mass of Naphthalene = W2
(i) Boiling point of solvent
= 0.5126 g = 0.5126 × 10–3 kg
= Tº = 80.2ºC
Chapter - 2 Solutions and Colligative Properties
134
(ii)
(iii)
(iv)
Solution :
ΔTb =
=
ΔTb =
0.36 =
molar mass M2
= 128.17 g mol–1
= 12.8.17 × 10–3 kg mol–1
mass of CCl4 = W1
= 50 g = 50 × 10–3 kg
ΔTb = 0.402 K
For unknown solute :
mass of unknown solute
= W2 = 0.6216 g
= 0.6216 × 10–3 kg
(ii) mass of CCl4 = W2 = 50g = 50 × 10–3 kg
(iii) ΔTb = 0.647 K
(iv) Kb for CCl4 = 5.03 K kg mol–1
To find :
molar mass of unknown solute = M2 = ?
Solution :
For the solution of Unknown solute
Kb
W2
M2 = ΔT × W
b
1
5.03 × 0.6216 × 10 –3
=
0.647 × 50 × 10 –3
= 0.09665 kg mol–1
M 2 = 96.65 g mol–1
T – Tº
353.71 – 353.35 = 0.36 K
Kb × m
2.53 × m
ΔTb
0.36
=
= 0.1423 mol kg–1
Kb
2.53
molality = 0.1423 mol kg–1
m
(i)
=
3.795g of sulphur is dissolved in 100g of
CS2. This solution boils at 319.81 K. What
is the molecular formula of sulphur in
solution? The boiling point of CS2 is 319.45
K. Given that Kb for CS2 = 2.42 K kg
mol–1 and atomic mass of S = 32.
Given :
(i) mass of sulphur
= W2 = 3.795 g = 3.795 × 10–3 kg
(ii) mass of CS2
= W1 = 100 g = 100 × 10–3 kg
(iii) boiling of CS2
= Tº = 319.45 K
(iv) boiling point solution
= T = 319.81 K
(v)
Kb for CS2 = 2.42 K kg mol–1
*23. Calculate molality of a solution of a
(vi) Atomic mass of S = 32
substance which on dissolving in benzene
boils at 353.71 K. Kb for benzene is 2.53 To find :
molecular formula of sulphur = ?
K kg mol–1 and boiling point of pure
Solution :
benzene is 353.35K.
ΔTb = T – T°
Given :
= 319.81 – 319.45 = 0.36 K
(i) Boiling point of pure benzene
= Tº = 353.35 K
K b × W2
M2 = ΔT × W
(ii) Boiling point of solution= 353.71 K
b
1
(iii) Kb for benzene = 2.53 K kg mol–1
2.42 K kg mol –1 ×3.795×10 –3 kg
To find :
=
0.36 K×100×10 –3 kg
molality = ?
Unique Solutions ®
*24.
S.Y.J.C. Science - Chemistry - Part I
135
∴
= 0.2551 kg mol–1
= 255.1 g mol–1
Atomic mass of S = 32
Number of atoms in a molecule of sulphur
=
molar mass of S
atomic mass of S
=
255.1
= 7.97 ≈ 8
32
Hence, molecular formula of S is S8 in
CS2
*25.
Boiling point of water at 750 mm of Hg
is 99.63ºC. How much sucrose must be
added to 500 g of water so that it boils at
100ºC?
Given :
(i) Boiling point of water = Tº = 99.63ºC
(ii) Boiling point of solution = T = 100ºC
(iii) Mass of water = W1 = 500 g = 0.5 kg
(iv) Kb of water = 0.52 K kg mol–1
To find :
mass of sucrose = W2 = ?
Solution :
molar mass of sucrose (C12H22O11)
M2 =12(12) + 22(1) + 11(16)
=342 g mol–1
=342 × 10–3 kg mol–1
ΔTb =T – Tº
=100 – 99.63
= 0.37ºC
Kb
W2
M2 = ΔT × W
b
1
W2
0.52
342 × 10–3 = 0.37 ×
500 × 10 –3
∴
W2 =
342×10 –3 ×0.37×500×10 –3
0.52
=121.7 × 10–3 kg = 121.7 g
mass of sucrose = 121.7 g
TYPE IV - PROBLEM BASED ON FREEZING
POINT DEPRESSION :
*26. An aqueous solution containing 12.5 × 10–3
kg of non-volatile compound in 0.1 kg of
water freezes at 272.49 K. Determine
molar mass of the compound.
(Kf for water = 1.86 K kg mol–1, B.P. of
water = 273.15 K)
Given :
(i) mass of solute = W2 = 12.5 × 10–3 kg
(ii) mass of water = W1 = 0.1 kg
(iii) freezing point of solution = T = 272.49K
(iv) freezing point of water = Tº = 273.15 K
(v) Kf of water = 1.86 K kg mol–1
To find :
molar mass of solute = M2 = ?
Solution:
ΔTf = Tº – T
= 273.15 – 272.49 = 0.66K
M2
Kf
W2
= ΔT × W
f
1
=
1.86 × 12.5 × 10 –3
0.66 × 0.1
= 352.2 × 10–3 kg mol–1
= 352.2 g mol–1
molar mass of solute = 352.2g mol–1
27.
Calculate the freezing point of solution
prepared by dissolving 4.5g of glucose
(Molar mass = 180g mol–1) in 250g of
bromoform. Given, freezing point of
Chapter - 2 Solutions and Colligative Properties
136
bromoform = 7.8ºC and Kf for bromoform
= 14.4 K kg mol–1.
Given :
(i) mass of glucose
= M2 = 4.5g = 4.5 × 10–3 kg
(ii) molar mass of glucose
= 180 g mol–1
= 180 × 10–3 kg mol–1
(iii) mass of bromoform
= W1 = 250g = 250 × 10–3 kg
(iv) freezing point of bromoform = Tº = 7.8ºC
(v) Kf of bromoform = 14.4 K. kg mol–1
To find :
freezing point of solution = T = ?
Solution :
K f × W2
ΔTf = M × W
2
1
ΔTf =
∴
=
=
*28.
14.4 K kg mol –1 × 4.5×10 –3 kg
180×10
–3
kg mol
–1
–3
× 250×10 kg
= 1.44 K = 1.44ºC
ΔTf = freezing point of solvent – freezing
point of solution
Hence, freezing point of solution
freezing point of solvent – ΔTf
7.80ºC – 1.44ºC = 6.36ºC
freezing point of solution = T = 6.36ºC
1.02 g of urea when dissolved in 98.5 g
of certain solvent decreases its freezing
point by 0.211 K. 1.60 g of unknown
compound when dissolved in 86.0 g of the
same solvent depresses the freezing point
by 0.34 K. Calculate the molar mass of
the unknown compound.
(Urea NH2CONH2, N = 14, C = 12, O
= 16, H = 1)
Unique Solutions ®
Given :
(i) For urea
mass of urea
= W2 = 1.02 g = 1.02 × 10–3 kg
mass of solvent
= W1 = 98.5 g = 98.5 × 10–3 kg
ΔTf = 0.211 K
(ii) For unknown compound
mass of unknown compound
= W2 = 1.6 g = 1.6 × 10–3 kg
mass of solvent
= W1 = 86g = 86 × 10–3 kg
ΔTF = 0.34 K
To find :
molar mass of unknown compound
= M2 = ?
Solution :
molality of urea solution,
mass of urea in kg
m = molar mass urea ×mass of solvent in kg
=
1.02 × 10 –3 kg
60 × 10 –3 kg mol –1 × 98.5 × 10 –3 kg
= 0.17 mol kg–1
Kf =
ΔTf
m
0.211K
= 0.17 mol kg –1 = 1.241 K kg mol–1
K f × W2
M2 = ΔT × W
f
1
=
1.241K kg mol –1 × 1.60 × 10 –3 kg
0.34 K × 86.0 × 10 –3 kg
= 67.90 × 10–3 kg mol–1
= 67.90 g mol –1
molar mass of unknown compound
= M2 = 67.90 g mol –1
S.Y.J.C. Science - Chemistry - Part I
137
TYPE V - PROBLEMS BASED ON OSMOTIC To find :
concentration in Case II = C2 = ?
PRESSURE :
29. Calculate the Osmotic pressure of 3.6 g Solution :
π 1 = C1 RT and π2 = C2 RT
of glucose (molar mass = 180 g mol–1)
dissolved in 100 ml of water at 298 K.
π1
C1
Hence, π = C
(R = 0.0821 L atm mol–1 K–1)
2
2
Given :
π 1 = 4.91 atm, π2 = 1.5 atm
(i) mass of glucose = 3.6 g
C 1 = mol L–1
(ii) molar mass of glucose = 180 g mol–1
(molar mass of glucose = 180 g mol–1)
–3
3
(iii) volume of water = 100 ml = 100 × 10 dm
π2
1.5atm
30
(iv) T = 298 K
Hence, C2 = π C1 = 4.91atm × 180 M
1
(v) K = 0.0821 L atm mol–1 K–1
To find :
moles of glucose
⎡
⎤
Osmotic pressure π = ?
⎢ C1 = volume of solution
⎥
⎢
⎥
Solution:
mass
of
glucose
⎢ =
⎥
⎢⎣
π V = nRT
molar massof glucose + volumeof solution ⎥⎦
3.6
× 0.0821 × 298
180
3.6 × 0.0821 × 298
π = 180 × 100 × 10 –3 = 4.893 atm
π × 100 × 10–3 =
∴
Osmotic pressure π = 4.893 atm
*30.
30 g of glucose dissolved in one litre of
water has an osmotic pressure 4.91 atm
at 303 K. If the osmotic pressure of the
glucose solution is 1.5 atm at the same
temperature, what would be its
concentration? (Molar mass of glucose is
180 g mol–1)
Given :
Case - I
(i) mass of glucose = 30 g
(ii) volume of water = 1 L
(iii) osmotic pressure = π1 = 4.91 atm
(iv) T = 303 K
Case - II
=
1.5 atm
π2
= 0.0509 M
The concentration in 2nd Case =
0.0509 M.
*31.
What is the mass of sucrose in its 1 L
solution (molar mass = 342 g mol–1) which
is isotonic with 6.6 × 10–3 kg L–1 of urea
(NH2CONH2)? (Given molar masses, H =
1, C = 12, N = 14, O = 16 in g mol–1).
Given :
Sucrose solution :
(i) Volume = 1 L
(ii) Molar mass of sucrose = 342 g mol–1
Urea solution
(i) Mass of urea = 6.6 × 10–3 kg
(ii) Volume of solution = 1 L
(iii) Molar mass of urea (NH2CON2) = 60 g
mol–1 solution are isotonic
To find :
mass of sucrose = ?
Chapter - 2 Solutions and Colligative Properties
138
Solution:
π = CRT
For isotonic solution π1 = π2
Hence, C 1 = C 2; C 1 and C 2 are
concentrations in mol L–1
Molarity of urea solution =
=
=
=
Solution :
πV =
πV =
3.02 × 103× 1 × 10–4 =
mass of urea
molar mass
Molar mass of urea
(14 + 2 +12 + 16 + 14 + 2) g mol–1
60 g mol–1 = 60 × 10–3 kg mol–1
Concentration of urea
M2 =
=
6.8×10 –3
×8.314×310.15
M2
6.8×10 –3 ×8.314×310.15
3.02×10 3 ×1 ×10 –4
58.06 kg mol–1
= 58060 g mol–1
molar mass of proteins = 58060 g mol–1
6.6 × 10 –3 kg
= 0.11M = C2
60 × 10 –3 kg/mol
Since urea solution is isotonic with sucrose
solution, C1 = C2
mass of sucrose
Hence, C1 =
= 0.11
342 × 10 –3
Mass of sucrose = 0.11 × 342 × 10–3 kg
Hence,
mass of sucrose = 37.62 × 10–3 kg
n RT
W2
M 2 RT
At 298 K, 1000 cm3 of a solution containing
4.34g of solute shows osmotic pressure of
2.55 atm. What is the molar mass of
solute? (R = 0.0821 L atm K–1 mol–1)
Given :
(i) T = 298 K
(ii) volume of solution = 1000 cm3 = 1 L
(iii) mass of solute = W2 = 4.34 g
(iv)
osmotic pressure = π = 2.55 atm
*32. Osmotic pressure of a solution containing
(v) R = 0.0821 L atm K–1 mol–1
6.8 × 10–3 kg of protein per 1 × 10–4 m3
of solution is 3.02 × 103 Pa at 37ºC. To find :
molar mass of solute = M2 = ?
Calculate the molar mass of protein. (R
–1
–1
Solution:
= 8.314 J K mol ).
πV = n RT
Given :
W2
(i) mass of proteins
πV = M RT
2
= W2 = 6.8 × 10–3 kg
4.34
(ii) volume of solution
2.55 × 1 = M × 0.021 × 298
2
= 1 × 10–4 m3
4.34
×
0.0821
× 298
(iii) osmotic pressure = π = 3.02 × 103 Pa
M2 =
2.55
(iv) R2 = 8.314J K–1 mol–1
= 41.64 g mol–1
(v) T = 37ºC + 273.15 = 310.15 K
To find :
molar mass of solute = 41.64 g mol–1
molar mass of protein
Unique Solutions ®
*33.
S.Y.J.C. Science - Chemistry - Part I
139
π (observed)
TYPE VI - PROBLEM BASED ON VANT
i
= π (Theoretical)
HOFF FACTOR :
*34. 0.2m aqueous solution of KCl freezes at –
π (observed)
1.83 =
0.680ºC. Calculate vant Hoff factor and
4.477
∴ π (observed) = 1.83 × 4.477 = 8.19 atm
observed osmotic pressure of solution at 0ºC.
Given :
(i) molality of Aqueous KCl solution = 0.2m *35. 0.01 molal aqueous solution of K3Fe(CN)6
freezes at –0.062ºC. Calculate the
(ii) freezing point of solution = T = –0.680ºC
–1
percentage dissociation of solute, Kf for
(iii) Kf = 1.86 K kg mol
water is 1.86 K kg–1 mol–1.
To find :
Given :
(i) vant Hoff factor
(i) molality of K3Fe(CN)6 = 0.01 M
(ii) observed osmotic pressure at 0ºC
(ii) freezing point of solution = T = –0.062ºC
Solution :
(iii) Kf = 1.86 K kg–1 mol–1
ΔTf (observed) = Tº – T
To find :
= 0 – (– 0.680)
% dissociation of solute
= 0.680ºC
Solution :
ΔTf (Theoretical) = Kf × m
= 1.86 × 0.2
K 3Fe(CN) 6 ⎯⎯
→ 3K + + Fe(CN) 3–
6
= 0.372º C
∴ n′ = 4
ΔTf (observed)
ΔTf (Theoretical) = Kf × m
i = ΔT (theoretical)
= 1.86 × 0.01
f
0.680
= 0.0186 K
=
= 1.83
0.372
ΔTf (observed) = Tº – T
i = 1.83
= 0 – (–0.062)
Let the volume of solution = 1 L = 1 kg
= 0.062ºC
(for dilute Aqueous solution)
ΔTf (observed)
i
=
moles of solute
ΔTf (theoretical)
molality (m) = mass of solvent in kg
0.062
=
= 3.333
0.0186
n2
0.2 =
i –1
1
α =
n′ – 1
∴
n solute = 0.2 moles
π (Theoretical) = CRT
3.333 – 1
=
= 0.78
4 –1
n
RT
=
% dissociation
= α × 100
V
0.2
× 0.082 × 273
=
1
= 4.477 atm
= 0.78 × 100 = 78%
percentage dissociation (% α) = 78%
Chapter - 2 Solutions and Colligative Properties
140
*36.
19.5g of monofluoro acetic acid was
dissolved in 0.5 kg of water. The freezing
point of solution was observed to be –1.0ºC.
Calculate the vant Hoff factor, degree of
dissociation and dissociation constant of acid.
(Atomic masses C = 12, H = 1, F = 19,
O = 16 and Kf =1.86 K kg–1 mol–1)
Given :
(i) mass of monofluoro acetic acid
(CH2FCOOH) = 19.5 g
(ii) mass of water = 0.5 kg
(iii) freezing point of solution = T = –100ºC
(iv) Kf = 1.86 K kg–1 mol–1
To find :
(i) vant Hoff factor
(ii) Degree of dissociation
(ii) Dissociation constant of acid
Solution:
ΔTf = Tº – T
= 0 – (–1) = 1ºC
=1K
molar mass of CH2FCOOH = 78 g mol–1
moles of CH2FCOOH
=
mass of CH 2 FCOOH
molar mass of CH 2 FCOOH
=
19.5
= 0.25 moles in 0.5 kg water.
78
molality =
m (observed)
i = m(theoretical)
=
–CH 2 FCOOH ⎯⎯
→ CH 2 FCOO – + H +
Q
0.25
= 0.5 mol kg–1
0.5
ΔTf (observed) = Kf × m
= 1.86 × m(observed)
= m(observed)
Unique Solutions ®
n′ = 2
Degree of dissociation (α)
=
i –1
1.076 – 1
=
= 0.076
n′ – 1
2 –1
Degree of dissociation (α
α ) = 0.076
CH 2FCOOH ⎯⎯
→ CH 2 FCOO – + H +
Initialmoles
m
o
o
Equlb moles m – mα
mα
mα
Ka =
=
a
(CH 2 FCOO – ) (H + )
(CH 2 FCOO – )
m α × mα
m – mα
mα 2
=
1–α
=
=
0.538
= 1.076
0.5
vant Hoff factor = 1.076
0.5 (0.076) 2
1 – 0.076
0.05 × (0.0760) 2
=
0.924
Ka = 3.126 × 10–3
Dissociation constant Ka = 3.126 × 10–3
moles of CH 2 FCOOH
mass of water in kg
1
1.86
= 0.538 mol kg–1
∴
*37.
10g of monochlorobutyric acid was
dissolved in 250g of water. If dissociation
constant of the acid is 1.45 × 10–3 and Kf
= 1.86 K kg mol–1, calculate depression
of the freezing point.
(C = 12, Cl = 35, H = 1, O = 16)
S.Y.J.C. Science - Chemistry - Part I
141
Given :
(i) mass of monochlorobutyric acid = 10g
(ii) mass of water = 250g
(iii) dissociation constant = Ka = 1.45 × 10–3
(iv) Kf = 1.86 K kg mol–1
To find :
Depression in freezing point = ?
Solution :
molar mass of monochlorobutyric acid
= 122 g mol–1
[CH3CH2CHCICOOH]
moles of monochlorobutyric acid
=
=
mass of acid
molor mass of acid
10
= 0.082 moles
122
mass of acid
molality = mass of H O in kg
2
0.082
= 250 × 10 –3
= 0.328 mol kg–1
CH 3CH 2CHClCOOH → CH 3CH 2CHClCOO – + H +
Initial moles
m
0
0
moles at eqlb m – mα
mα
mα
∴
∴
(CH 3CH 2CHClCOO – ) (H + )
Ka =
(CH 3CH 2CHClCOOH)
mα × mα
1.45 × 10–3 =
m (1 – α)
mα2
1.45 × 10–3 =
1– α
[Q α is very small 1 – α ≈ 1]
1.45 × 10–3 = mα2
1.45 × 10–3 = 0.328 × α2
1.45 × 10 –3
0.328
= 4.42 × 10–3
α2 =
α2
∴
α
=
α
=
–2
44.2 × 10 –4 = 6.648 × 10
i –1
n′ – 1
(n′ = 2)
6.648 × 10–2 =
i –1
2 –1
6.648 × 10–2 + 1 = i
i = 1.06648
ΔTf (theoretical) = Kf × m
= 1.86 × 0.328
= 0.61 K
∴
ΔTf (observed)
i = ΔT (theoretical)
f
ΔTf (observed) = i × ΔTf (theoretical)
= 1.06648 × 0.61
= 0.65 K
Depression of freezing point = 0.65 K
*38.
8g of benzoic acid when dissolved in 100g
of benzene lowers its freezing point by
1.62K. Calculate degree of association of
benzoic acid if it, forms dimers in benzene
Kf for benzene is 4.9K kg mol–1
(C = 12, H = 1, O = 16)
Given :
(i) mass of benzoic acid = W2 = 8g
(ii) mass of benzene = W1 = 100g
(iii) ΔTf = 1.62 K
(iv) Kf = 4.9 K kg mol–1
To find :
Degree of association of benzoic acid = ?
Solution :
Kf
W
× 2
. M2 =
Tf W1
4.9 × 8 × 10 –3
=
1.62 × 100 × 10 –3
= 241.9 × 10–3 kg mol–1
Chapter - 2 Solutions and Colligative Properties
142
∴
∴
M2 (observed) = 241.9 g mol–1
M 2 (theoretical) of benzoic acid
(C6H5COOH) is 122.
Observed Molecular mass of benzoic acid
is almost double the theoretical molecular
mass.
Benzoic acid must associate to form a
dimer.
(C 6H 5COOH)
2 C 6 H 5COOH Initialmoles
m
molesat equlb m – mα
=
=
o
mα
2
Total moles in solution
mα
m – mα +
2
α
m ⎛1 – ⎞
⎝
2⎠
*39.
0.6 ml of glacial acetic acid with density 1.06
gm L–1 is dissolved in 1 kg water and the
solution froze at –0.0205ºC. Calculate Vant
Hoff factor and Kf for water is 1.86 K kg
mol–1 (i = 1.041, Ka = 1.86 × 10–5)
Given :
(i) volume of acetic acid = 0.6 ml
(ii) density = 1.06 g mL–1
(iii) mass of water = 1 kg
(iv) freezing point of solution = T = –0.0205ºC
(v) Kf = 1.86 K kg mol–1
To find :
(i) vant Hoff factor (i) = ?
(ii) dissociation constant (Ka) = ?
Solution :
Density =
observed moles in solution
i = theoretical moles in solution
Theoretical molecular mass
Observed molecular mass
122
=
= 0.5041
241.98
α
1–
=i
2
α
0.5041 = 1 –
2
α
= 1 – 0.5041 = 0.4958
2
∴
=
i =
∴
∴
=
=
=
α
= 2 × 0.4958
α
= 0.9918
% association of benzoic acid = 99.18%
Unique Solutions ®
mass of acetic acid
0.6
mass of acetic acid = 1.06 × 0.6g
moles of acetic acid
mass of acetic acid
molar mass of acetic acid
1.06 × 0.6
= 0.0106 moles
60
molality (theoretical)
1.06
α
m ⎛1 – ⎞
α
⎝
2⎠
=
= 1–
2
m
mass of acetic acid
volume of acetic acid
=
mass of acetic acid
mass of H 2O in kg
0.0106
= 0.0106 mol kg–1
1
ΔTf (observed) = Tº – T
= 0 – (–0.0205)
= 0.0205ºC
ΔTf (theoretical) = Kf × M(Theoretical)
= 1.86 × 0.0106
= 0.0197ºC
S.Y.J.C. Science - Chemistry - Part I
143
i
ΔTf (observed)
= ΔT (theoretical)
f
0.0205
=
0.0197
= 1.041
mα × mα
=
CH 3COO + H
CH 3COOH –
∴
∴
dissociation constant (Ka) = 1.86 × 10–5
*40.
1.041 – 1
i –1
=
= 0.041
2 –1
n′ – 1
Degree of dissociation (α) = 0.041
CH 3COO – + H +
CH 3COOH Initial moles m
molesat equlb m – mα
0
mα
0
mα
Henry’s law constant for the molality of
methane in benzene at 298 K is 4.27 × 105
mm Hg. Calculate the solubility of methane
in benzene at 298 K under 760 mm Hg.
Given :
Henry’s law constant (K) = 4.27 × 105 mm
Hg Pressure (P) = 760 mm Hg
To find :
Solubility (s) = ?
Solution :
p = KHm
∴
(CH 3COO – ) (H + )
Ka =
(CH 3COOH)
NUMERICALS FOR PRACTICE
*1.
*2.
*3.
*4.
*5.
0.0106 × (0.041) 2
= 1.86 × 10–
1 – 0.041
+
n′ = 2
Degree of dissociation (α)
=
mα 2
= m – mα =
1–α
p
m = k
H
760 mm Hg
=
4.27 × 10 5 mm Hg
= 1.78 × 10–3
Solubility (s) = 1.78 × 10–3
TYPE - I - PROBLEM BASED ON CONCENTRATION OF SOLUTION :
6 g of urea was dissoved in 500g of water. Calculate percentage by mass of urea in solution.
[Ans : 1.186% by mass]
58 cm3 of ethyl alcohol was dissloved in 400 cm3 of water to form 454 cm3 of solution of
ethyl alcohol. Calculate percentage by volume of ethyl alcohol in water.
[Ans : 12.78 % by volume]
23g of ethyl alcohol (Molar mass 46g mol–1) is dissolved in 54 g of water (molar mass 18g
[Ans : 1]
mol–1). Calculate the mole fraction of ethyl alcohol and water in solution.
–1
A solution of NaOH (Molar mass 40g mol ) was prepared by dissolving 1.6g of NaOH in
[Ans : 0.08 mol dm–3]
500 cm3 of water. Calculate molarity of NaOH solution.
11.11 g of urea was dissolved in 100 g of water. Calculate the molality of solution. (N = 14,
H = 1, C = 12, O = 16)
[Ans : 1.852 mol kg–1]
Chapter - 2 Solutions and Colligative Properties
144
*6.
34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality
and mole fraction of sugar in the syrup (C = 12, H = 1, O = 16)
[Ans : 0.556 mol kg–1, 0.0099]
*7.
Calculate molarity and molality of sulphuric acid solution of density 1.198g cm–3 containing
27% by mass of sulphuric acid (molar mass of H2SO4 = 98g mol–1)
[Ans : 3.77 mol kg–1, 3.301 mol dm–3]
*8.
Commercially available concentrated hydrochloric acid is an aqueous solution containing 38%
HCl gas by mass. If its density is 1.1 g cm–3, calculate the molarity of HCl solution and also
calculate mole fractions of HCl and H2O.
[Ans : 11.45 mol dm–3, 0.232, 0.768]
9.
*10.
TYPE - II - PROBLEMS BASED ON HENRY’S LAW :
What is the solubility of oxygen if partial pressure of oxygen is 0.012 atm. The Henry’s law
constant for oxygen is 1.3 × 10–3 mol dm–3 atm–1. [Hint : S = KH × PO2]
[Ans : 1.56 × 10–5 mol dm–3]
TYPE - III - PROBLEMS BASED ON LOWERING OF VAPOUR PRESSURE :
The vapour pressure of a solution containing 13 × 10–3 kg of solute in 0.1 kg of water at
298 K is 27.371 mm Hg. Calculate the molar mass of the solute. Given that the vapour pressure
of water at 298 K is 28.065 mm Hg.
[Ans : 94.63 g mol–1]
*11.
The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non volatile
solute of mass 2.175 × 10–3 kg is added to 39.0 × 10–3 kg of benzene. The vapour pressure
of the solution is 600 mm Hg. What is the molar mass of the solute? (Given : atomic masses
C = 12, H = 1)
[Ans : 69.6 g mol–1]
*12.
A solution is prepared from 26.2 × 10–3 kg of an unknown substance and 112.0 × 10–3 kg
acetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. Calculate
the vapour pressure of solution if the molar mass of substance is 273.52 × 10–3 kg mol–1.
(Given the atomic masses C = 12, H = 1, O = 16)
[Ans : 0.500 atm]
*13.
TYPE IV - PROBLEMS BASED ON BOILING POINT ELEVATION :
0.15 molal solution of a substance boils at 373.23 K. Calculate molal elevation constant of
water. (Given boiling point of water = 373.15 K)
[Ans : 0.53 K kg mol–1]
*14.
A solution is prepared by dissolving 1.9 × 10–2 kg ammonia in 400 g of water. Calculate elevation
in boiling point if Kb for water is 0.52 K kg mol–1 (N = 14, H = 1, O = 16)[Ans : 1.45 K]
*15.
A solution containing 1.21g of camphor (molar mass 152g mol–1) in 26.68g of acetone boils
at 329.95 K. The boiling point of pure acetone is 329.45 K. Calculate molal elevation constant
for acetone.
[Ans : 1.676 K kg mol–1]
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
145
*16.
*17.
A solution was prepared by dissloving certain amount of compound in 31.8g of CCl4 has a
boiling point of 0.392 K higher than that of pure CCl4. If the molar mass of compound is 128g
mol–1, calculate the mass of the solute dissolved. (Given Kb for CCl4 = 5.02 K kg mol–1)
[Ans : 0.317 g]
TYPE V - PROBLEMS BASED ON FREEZING POINT DEPRESSION :
5.08g of a substance dissolved in 50g of water lowered the freezing point by 1.2 K. Calculate
the molar mass of the substance. The molar depression constant for water is
[Ans : 157.48g mol–1]
1.86 K kg mol–1.
*18.
0.440 × 10–3 kg of a substance (molar mass = 178.9 × 10–3 kg mol–1) dissolved in 22.2 ×
10–3 kg of benzene lowered the freezing point of benzene by 0.567 K. Calculate the molal
depression constant for benzene.
[Ans : 5.12 K kg mol–1]
*19.
The observed depression in the freezing point of water for a particular solution is 0.087 K.
Calculate the molality of the solution if molal depression constant for water is
1.86 K kg mol–1.
[Ans : 0.0467 mol kg–1]
*20.
A solution of glucose (C6H12O6) was prepared by dissolving certain amount in 100g of water.
The depression in freezing point was 0.0410 K. If molal depression constant for water is 1.86
K kg mol–1, calculate the mass of glucose dissolved. (C = 12, H = 1, O = 16)
[Ans : 0.396 g of glucose]
TYPE VI - PROBLEMS BASED ON OSMOTIC PRESSURE :
Calculate the osmotic pressure of 4.5g of glucose (molar mass = 180g mol–1) dissolved in
[Ans : 6.116 atmospheres]
100 ml of water at 298 K (R = 0.0821 L atm mol–1K–1)
*21.
*22.
A solution of cane sugar (molar mass 342g mol–1) containing 17.8g L–1 has an osmotic pressure
1.2 atm. Calculate the temperature of the solution. (R = 0.082 L atm mol–1K–1)
[Ans : 281.4 K]
*23.
A solution has an osmotic pressure of 3.90 × 105Nm–2 at 300 K. Calculate its volume containing
1 mole of solute if solution of same solute has osmotic pressure of 2.82 × 105Nm–2 and contains
1 mole of solute in 10.5 m3.
[Ans : 7.59 m3]
*24.
Calculate the volume of a solution containing 34.2g of cane sugar (Molar mass = 342g
mol–1) which has an osmotic pressure 2.42 atm at 20ºC.
[Ans : 0.9945 L]
*25.
A solution of particular amount of organic substance of molar mass 196g mol–1 dissolved in
2 litres of water gave an osmotic pressure of 0.54 atm at 12ºC. Calculate the mass of solute
[Ans : 9.04g]
dissolved. (R = 0.0821 L atm K–1 mol–1)
*26.
TYPE VII - PROBLEMS BASED ON VANT HOFF FACTOR :
Calculate the Vant Hoff factor of CdSO4 (molecular mass 208.4 g mol–1) if the dissolution
Chapter - 2 Solutions and Colligative Properties
146
of 5.21g of CdSO4 in 500 ml of water gives a depression in freezing point of 0.168ºC
(Kf of water = 1.86 K kg mol–1)
Hint :
K f × W2
M 2 = ΔT
f (observed) × W1
∴ ΔTf = 0.093ºC
(Theoretical)
i
=
ΔTf (observed)
ΔTf (Theoretical)
0.168
= 1.8
0.093
Vant Hoff Factor (i = 1.8)
i
=
HIGHER ORDER THINKING SKILLS (HOTS)
1. Calculate the molarity of water if its density is 1000kg/m3.
Sol. : Since density of water is 1000kg/m3
∴ 1000 kg of water = 1 m3
∴ 1 kg of water = 10–3m3
∴ 1000 g of water = 10–3 × 103dm3
= 1 dm3
No. of moles of water = Mass of water / molar mass of water
= 1000/18
= 55.55moles
moles of water 55.55
= 55.55M
Molarity = Molarity =
Volume
1
2. Among the 0.1M solutions of urea, NaCl, BaCl2, Na3PO4 and Al2(SO4)3, find solutions having
highest and lowest vapour pressure, boiling point, freezing point. Find the one having highest
and lowest value of colligative properties.
Sol.
Vapour pressure and freezing point will be lowest while boiling point highest fo Al2(SO4)3.
Vapour pressure and freezing point will be highest while boiling point lowest for NaCl.
NaCl will have lowest value of all colligative properties.
Al2(SO4)3 will have highest value of all colligative properties.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
Chapter Chemical
3
Thermodynamics And
Energetics
"A person starts to live when he can live outside himself." -Einstein
Introduction :
The energy of any body is defined as its capacity to do work. eg. a gas at higher temperature
has more energy and hence, has greater capacity to do work than the same gas at lower
temperature.
SYLLABUS
3.1
LIMITATIONS OF THERMODYNAMICS
3.2
BASIC CONCEPTS IN
THERMODYNAMICS
Types of system
Properties of system
State and state functions
Properties of state functions
Thermodynamic equilibrium
Types of processes
Reversible process
Irreversible process
Distinction between Isothermal and
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
3.3
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
NATURE OF HEAT AND WORK
Work
Work during expansion
Work during contraction
Pressure-volume work
Expression for pressure-volume work
Free expansion
Concept of maximum work
Conditions for maximum work
Expression for maximum work
Path dependence nature of work
(k) Heat (q)
( l ) Units of energy and work
(m) Sign conventions of W and q
Theoretical MCQs
Numerical MCQs
3.4
INTERNAL ENERGY
3.5
FIRST LAW OF THERMODYNAMICS
(a) Statements of first law of thermodynamics
(b) Mathematical equation of first law of
thermodynamics
(c) Modified first law of thermodynamics
(d) First law of thermodynamics for various
processes
Theoretical MCQs
Numerical MCQs
3.6
ENTHALPY
(a) Relationship between ΔH, ΔU & PΔ V
(b) Relationship between ΔH and ΔU for
chemical reactions
(c) Work done in chemical reaction
(d) Conditions under which ΔH = ΔU
148
3.7
(a)
(b)
(c)
(d)
ENTHALPIES OF PHYSICAL CHANGES
Enthalpy of phase transition
Enthalpy of atomic or molecular changes
Enthalpy of solution
Enthalpy of dilution
Theoretical MCQs
Numerical MCQs
3.8
(a)
(b)
(c)
(d)
(e)
(f)
(g)
THERMOCHEMISTRY
Enthalpy of chemical reactions
Standard enthalpy of formation
Standard enthalpy of combustion
Bond enthalpy
Reaction enthalpy from Bond enthalpy
Hess’s law of constant heat summation
Applications of Hess’s law
Theoretical MCQs
Numerical MCQs
3.9
SECOND LAW OF THERMODYNAMICS
3.10
SPONTANEOUS PROCESS
(IRREVERSIBLE PROCESS)
(a) Energy and spontaneity
(b) Entropy
→
→
→
→
(c) Quantitative definition of entropy
(d) Entropy and spontaneity (2nd law of
thermodynamics)
Theoretical MCQs
Numerical MCQs
3.11 GIBBS ENERGY
(a) Gibbs energy and spontaneity
(b) Predicting spontaneity of a process in
terms of ΔG
(c) Temperature of equilibrium between
spontaneous and nonspontaneous
Process
(d) ΔG and equilibrium constant
3.12 THIRD LAW OF THERMODYNAMICS
(a) Standard molar entropy (Sº)
(b) Usefulness standard molar entropy (Sº)
Theoretical MCQs
Numerical MCQs
HOURS BEFORE EXAM
NUMERICALS WITH SOLUTION
NUMERICALS FOR PRACTICE
HIGHER ORDER THINKING SKILLS
Energy has different forms such as kinetic energy, potential energy, heat energy, radiant energy,
electrical energy, chemical energy.
All types of energy can be converted from one form to other. Eg. when water stored in a
dam falls down, its potential energy is converted into kinetic energy. In galvanic cells, the chemical
energy is converted into electrical energy.
Energy can neither be created nor destroyed. When one form of energy disappears, the other
form of energy appear in equal magnitude. Thus, different forms of energy are related
quantitatively to each other.
Definition : Thermodynamics is the branch of science that deals with the different forms
of energy, the quantitative relationships between them and the energy changes that occur
in physical and chemical processes.
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3.1
LIMITATIONS OF THERMODYNAMICS :
Concept Explanation :
(a)
(b)
(c)
(d)
It gives no information regarding the rates at which physical and chemical processes occur.
It deals only with initial and final states of the system.
It does not tell anything about the mechanism of the process.
It deals with the properties of macroscopic systems (i.e. the systems containing large number
of atoms, ions or molecules) such as temperature, pressure, volume etc.
(e) Thermodynamics does not deals with the microscopic properties of the system such as
electronic structure of atom, constitution of molecule from atoms etc.
3.2
BASIC CONCEPTS IN THERMODYNAMICS :
Concept Explanation :
Universe is divided into two parts :
(1) System : A portion of the universe chosen to study the thermodynamic properties is called
a system.
(2) Surroundings : The portion of the universe other than system is called surroundings. A
real or imagenery line which separates system from surroundings is called boundary. An
exchange of matter and energy between the system and its surroundings can take place
through this boundary.
Example :
A reaction mixture in a stoppered flask in a system. The surroundings may be a constant
temperature both in which the flask is immersed. The walls of the flask act as boundary.
(a) Types of system :
The systems are classified into three types, on the basis of exchange of matter and energy
between the system and its surroundings.
1. Open system
2. Closed system
3. Isolated system
1. Open system : A system that can exchange both matter
and energy with its surrounding is called an open system.
Example :
(a) A sample of liquid water kept in an open beaker
(b) A chemical reaction carried out in a test tube
(c) Hot tea in an open cup.
2. Closed system : A system that can exchange only
energy but not matter with the surroundings is called a
closed system.
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Example :
(a) A sample of liquid water kept in a closed beaker
(b) A chemical reaction carried out in a closed vessel.
(c) Hot tea in a cup covered with saucer.
3. Isolated system : A system that neither exchange matter
nor energy with its surroundings is called an isolated
system.
Example :
(a) A sample of liquid water kept in a closed container
insulated from the surroundings.
(b) A chemical reaction carried out in a thermos flask
(c) Hot tea in a thermos flask.
(b) Properties of system :
A macroscopic system has certain thermodynamic properties like temperature, pressure,
volume, energy, density.
These macroscopic properties are classified into two types according to their dependence
on the amount of the substance present in the system.
1. Extensive property : The property of a system whose magnitude depends on the amount
of matter present in the system is called extensive property.
Examples : Mass, volume, internal energy, heat capacity, work done, enthalpy, entropy,
internal energy, heat of reaction etc.
Note : The volume of one mole of a gas is 22.4 dm3 at STP, but two moles of the
gas occupy 44.8 dm3 at STP.
2. Intensive property : The property of a system whose magnitude is independent of the
amount of matter present in the system is called its intensive property.
Examples :
Temperature, density, surface tension, viscosity, refractive index, melting point, boiling point,
specific heat.
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Note : The boiling point of water is 100ºC at 1 atm. pressure, no matter whether the
volume is 1L, 2L or 100 Litres.
The ratio of two extensive properties is an intensive property.
For example the ratio of two extensive properties, mass and volume (m/v) is density, which
is an intensive property.
(c) State and State Functions :
1. State of system :
(a) Every system has certain measurable thermodynamic properties such as temperature,
pressure and volume.
These properties are called state variables.
(b) A known set of state variables describes the state (condition) of the system at any
time.
2. Change of state : When there is a change in one or more state variables, then it is said
that system is changed from one state to another state.
3. State function :
(a) Definition : Any property of a system whose value depends on the current state
of the system and is independent of the path followed to reach that state is called
the state function.
(b) For a particular state of the system, the state function has fixed value in the present
condition of the system. It does not depend on previous history of the system.
(c) For example, 1g of water is placed in a container and maintained at 25ºC and 1 atm.
pressure. The volume of water will be approximately 1cm3.
All the quantities, 1g, 25ºC, 1 atm. and 1 cm3 describe the current state of water
and are independent of its previous history.
(d) Hence, the properties mass, temperature, pressure and volume are state functions.
Other examples–enthalpy (H), internal energy (U), number of moles, entropy (S) etc.
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(d) Properties of state functions :
The state functions have two very important properties.
1. To describe the state of a system, it is not necessary to specify the values of all the state
functions. When the values of a certain minimum number of state functions are assigned,
then the values of other state functions are automatically fixed.
Example : Consider a gaseous system, in which equation of state is, PV = n RT. This
equation shows relationship between different state functions.
Here only three out of four state functions. (n, P, V and T) are independent and one is dependent.
For a gaseous system of constant mass i.e. for fixed ‘n’, if we specify only two state
functions, the third is automatically fixed.
2. When the state of a system is changed, the change in any state function depends only
on the initial and final states of the system and not on the path.
Consider a system which in changed from initial state A (P1,V1) to final state B (P2V2)
by three different paths as shown in figure.
The change in pressure ΔP = P2 – P1 and the change in volume, ΔV = V2 – V1 is the
same irrespective of the path.
Note : Suppose the height of the terrace of
a building is 100 feet. We can reach the
terrace by different ways such as through
staircase or lift.
The distance travelled depends on the path
chosen. The distance travelled is therefore,
a path function. But the attitude (100 feet)
of the terrace does not depend on the path
chosen. Hence, altitude is a state function
(e) Thermodynamic equilibrium :
1. A system is said to be in thermodynamic equilibrium when the state functions
of the system do not change with time.
2. When a gas is enclosed in a cylinder fitted with a piston, it has a definite temperature,
pressure and volume. These properties remain constant as long as piston is motionless.
This in an equilibrium state.
3. During motion of the piston temperature, pressure and volume keeps on changing. This
state in which the piston in moving is a non-equilibrium state.
4. When the motion of piston is stopped; temperature, pressure and volume again becomes
constant. Now the system will be in a new-equilibrium state.
(f) Types of processes :
The transition of a system from one equilibrium state to another is called a process.
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Some common types of processes are :1. Isothermal process :
(a) Definition : A process in which the temperature of the system remains constant
throughout the transformation is called an isothermal process.
(b) In isothermal process heat can flow from the system to the surroundings or vice versa
in order to keep the temperature constant.
(c) In isothermal process since the temperature is constant therefore, the internal energy
(U) of the system remains constant i.e. ΔU = 0.
(d) Example - Boiling of water at 100ºC.
2. Isobaric process :
(a) Definition : A process in which the pressure of the system remains constant is
called isobaric process.
(b) In laboratory most of the chemical reactions are carried out in test tube at constant
atmospheric pressure i.e. ΔP = 0
3. Isochoric process :
(a) Definition : A process in which volume of the system remains constant is called
isochoric process.
(b) A chemical reaction carried out in closed vessel is an example of isochoric process
i.e. ΔV = 0
1. Definition : A process in which there is no exchange of heat between the system and
its surroundings is called an adiabatic process (q = 0).
2. In this process, the system is insulated from the surroundings.
3. When the process is exothermic, the temperature of the system rises.
4. When the process is endothermic, the temperature of the system falls.
5. In adiabatic process, since the temperature of the system rises or falls, and hence, its
internal energy also increases or decreases. i.e. ΔU ≠ 0.
6. Example any chemical reaction carried out in closed and insulated vessel.
Note : All the above mentioned processes can be conducted reversibly or irreversibly.
(h) Reversible process :
Definition : Any process conducted in such a manner that at every stage the driving
force is only infinitesimally greater than the opposing force and the process can be
reversed by a slight increase in the opposing force is called a reversible process.
1.
2.
Features of reversible process :
The driving and opposing forces are only infinitesimally different from each other.
The process can be reversed at any point during the process by making infinitesimal
change in conditions.
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3.
4.
5.
6.
7.
8.
(i)
1.
2.
3.
4.
5.
6.
7.
8.
9.
The process takes place in an infinite number of steps.
At the end of every step of the process the system attains mechanical equilibrium.
Maximum work is obtained during a reversible process.
The process takes place so slowly that the system is always in temperature-pressure
equilibrium with its surroundings.
Reversible process is not a real process but it is an imaginary process.
Example : Expansion of an ideal gas by very slightly lowering of external pressure.
Irreversible process :
In this process driving force is much more greater than opposing force.
It is real process & practical process.
It is spontaneous process.
The direction of the process cannot be reversed by applying small change in conditions.
It takes place in only one step.
It is fast process and takes place in finite time.
The system is in equilibrium only at the end of the process.
Work obtained is not maximum.
Examples : (a) All natural process are irreversible
(b) Diffusion of gases.
(c) Flow of water from higher level to lower level.
(j) Distinction between Isothermal and Adiabatic processes
Isothermal process
1. The temperature of the system remains
constant (ΔT = 0).
2. There is an exchange of heat between the
system and its surroundings. (q ≠ 0)
3. The internal energy of the system remains
constant (ΔU = 0).
4. The system is not thermally isolated from its
surroundings. Example : Boiling of water at
100ºC.
3.3
1. The temperature of the system increases or
decreases. (ΔT ≠ 0)
2. There is no exchange of heat between the
system and its surroundings. (q = 0)
3. The internal energy of the system increases
or decreases. (ΔU ≠ 0)
4. The system is thermally isolated from its
surroundings. Example : Any chemical reaction
carried out in closed and insulated vessel.
NATURE OF HEAT AND WORK :
Concept Explanation :
(a) There are only two ways of changing the energy of a closed system :
(i) by transfer of energy as work (ii) by transfer of energy as heat.
(b) Heat and work are considered equivalent and are interconvertible into each other. They
have same units.
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(a) Work :
1. Work is one of the means (method) by which a system can exchange energy with the
surroundings.
2. In mechanics, the mechanical work is defined as the transfer of energy by which a body
is moved through a distance by the application of force (i.e. W = f × d)
3. In chemical thermodynamics, the type of work involved is pressure-volume work. Hence,
work is also defined as the transfer of energy that can be used to change the height of
mass in the surroundings.
(b) Work during expansion :
1. Consider a chemical reaction, 2H 2 O 2 (l ) ⎯⎯
→ 2H 2 O(l ) + O 2 (g) ,
that takes place in a cylinder. Suppose some mass is kept on the piston.
2. The gas (O2) pushes the piston upwards and hence, mass raised
in the surroundings.
3. In the process of lifting the mass, the system loses energy to the
surroundings.
4. If no heat transfer occurs, the loss of energy by the system is
equal to the work done by the system is equal to the work
done by the system on the surroundings.
(c) Work during contraction :
1. Consider a reaction, NH 3 (g) + HCl(g) ⎯⎯
→ NH 4Cl(s) that takes place
in a cylinder.
2. In this case volume of gas decreases.
3. The piston moves downwards and hence, decreases the height of the mass
4. In this process, surroundings loses energy to the system.
5. If no heat transfer occurs, the gain in energy by the system is equal
to the work done by the surroundings on the system.
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(d) Pressure-volume work :
1. The work (W), which is done due to expansion or compression of a gas against an external
opposing pressure (P) is called pressure-volume work.
2. The product of PV = W, can be explained as PV =
f
3
f
× V = 2 × d = f.d = W
d
A
(Where, d = distance)
(e) Expression for pressure-volume work :
1. Suppose a gas at pressure P, volume V1, and temperature T, is
enclosed in a cylinder fitted with a frictionless piston of area A.
2. As the gas expands, it pushes the piston upward through a distance
d against an external opposing force f.
Derivation :
Step - 1 : The work done by the gas during expansion is given by
W
= opposing force × distance
= –f ⋅ d
(negative sign shows that internal energy of the system during
expansion decreases).
Step - 2 : But, f = pex.A
∴ W = –pex × A × d
Step - 3 : But A × d = Change in volume = ΔV = V2 – V1
∴ W = –pex.ΔV = –pex (V2 – V1)
During expansion : work is done by the system on surroundings,
∴ W = –ve
V2 > V 1
During compression : work is done on the system by surroundings,
∴ W = +ve
V2 < V 1
(f) Free expansion :
Expansion of a gas against zero opposing force (as in vacuum) is called free expansion.
When pex = 0, then W = 0
(g)
1.
2.
3.
4.
5.
Concept of maximum work :
When opposing force is zero, no work is done.
When opposing force increases from zero, more and more work will be done by the system.
However, if opposing force becomes greater than the driving force the process gets reversed.
Therefore, opposing force must be infinitesimally smaller than driving force to obtain maximum
work.
If P – pex = ΔP, Or pex = P – ΔP, then
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W = –(p – ΔP).ΔV
Thus, W tends to maximum as ΔP tends to zero.
(h)
1.
2.
3.
4.
Conditions for maximum work :
The process must be thermodynamically reversible.
The process takes place in an infinite number of steps.
The driving force is infinitesimally greater than the opposing force.
The system is in mechanical equilibrium with its surroundings. (Since two pressures are
almost the same).
(i) Expression for maximum work (Here pressure is variable) :
1. Suppose ‘n’ mole of an ideal gas is enclosed in a cylinder fitted with frictionless piston.
2. The gas is expanded isothermally and reversibly in a number of infinitesimally small steps
from V1 to V2 at temperature T.
3. During each step external pressure (pex) is made infinitesimally smaller than the pressure
of the gas (P), by removing masses gradually from the piston.
4.
5.
6.
7
The volume of the gas is increased by an infinitesimal quantity dV in a single step.
This process is repeated until the final volume V2 is reached.
Thus, in a single step the small quantity of work done is given by, dW = –pexdV
Since the expansion is reversible, therefore, the pressure of the gas is greater by a very
small amount dP than pex.
Hence,
p – pex = dP or px = P – dP
Derivation :
Step - 1
∴ dW = – (P – dP).dV = –pdv + dv.dv
Neglecting, dP.dV, as it is very small
∴ dW = –P.dV
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158
The total amount of work done during expansion from V1 to V2 will be the sum of all
such infinitesimal amount of work done of all the steps.
Step - 2 : Mathematically, the total amount of work is obtained by integrating the differential
equation between limits V1 to V2.
This work is maximum work because the expansion is reversible. Thus,
V2
2
∫
– dW = –
1
V2
∫ P.dV
Wmax = –
V1
∫ PdV
V1
Step - 3 : The ideal gas equation for n moles is, PV = nRT or P =
V2
∴ Wmax
= –
∫ nRT
V
dV
V
1
V2
Wmax
= –nRT
∫
V1
dV
V
v
= –nRT ln (V) v12
= –nRT (ln V2 – ln V1)
V2
= –nRT ln V
1
V2
= –2.303 nRT log10 V
1
At constant temperature, according to Boyle’s law,
V2
P1
P1V1 = P2V2 or V = P
1
2
P1
∴
Wmax = –2.303 nRT log10 P
2
(j) Path dependence nature of work :
Work is a path function and not a state function. Since the amount
of work done is different in different paths although the initial
and final states are the same.
Example : Suppose a system (gas) goes from initial state (V1)
to final state (V2) by three different paths.
Part-I : If the system expands freely in vacuum, then W = 0
Part-II : If the system expands against constant external
pressure (pex)
then W = –pex (V2 – V1)
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V
159
Part-III : If the system expands under Isothermal reversible conditions, then,
V2
Wmax = –2.303 nRT log10 V
1
(k) Heat (q) :
1. Heat is another way by which a system can exchange energy with its surroundings.
2. When the system and its surroundings are at different temperatures, heat either flows
in or out of the system.
3.
Example : Consider a flask containing HNO3 kept in a water bath as shown in fig.
if KOH solution is added to the flask, an exothermic reaction takes place. The temperature
of water bath rises.
HNO 3 (aq) + KOH (aq) ⎯⎯
→ KNO 3 (aq) + H 2 O(l )
4.
5.
This indicates that energy is transferred as
heat from the flask (system) to the water bath
(Surroundings).
Heat like work is not the property of the
system and hence, is not a state function.
Example : Suppose a system changes by two different paths
Path - I : The change from state A to State B occurs under adiabatic conditions. There
is no exchange of heat between the system and its surroundings. q = 0
l
II r m a
the 0
Iso q ≠
Thus, heat transfer is different and it depends on path.
Hence, heat is a path function and not a state function.
A
c
a ti
ia b
0
I
Path - II : If the change from state A to B takes place
under isothermal conditions. A quantity of heat q enters
the system q ≠ 0.
B
(l) Units of energy and work :
1. The unit of energy as heat or work is Joule (J).
1 J = 1 kg m2s–2
1 J = 1 Pa m3
2. W = –pex.ΔV, if pressure is expressed in atm and volume in litres. Then unit of work
done will be litre-atmosphere (L.atm) where
1 atm = 101.3 × 103 kg m–1 s–2
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160
1 L ⋅ atm = dm3 × 101.3 × 103 kg m–1 s–2
= m3 × 10–3 × 101.3 × 103 kg m–1 s–2
= 101.3 kg m2 s–2
= 101.3 J
(m)
Sign conventions of W and q :
+q → Heat is absorbed by the system from the surroundings.
–q → Heat is released from the system to the surroundings.
+W → Work is done on the system by the surroundings.
–W → Work is done by the system on the surroundings.
•
*1.
Define the terms :
(a) system (b) surroundings (c) open system (d) closed system (e) isolated system
(a) system Refer 3.2 (1), (b) surroundings Refer 3.2 (2), (c) open system Refer 3.2 (a) 1,
(d) closed system Refer 3.2 (a) 2, (e) isolated system Refer 3.2 (a) 3
Ans.
2.
Ans.
Define the terms : *(a) extensive properties *(b) intensive properties. Give two examples of each.
(a) Extensive property : The property of a system whose magnitude depends on the amount
of matter present in the system is called extensive property.
Examples : Mass, volume, internal energy, heat capacity, work done, enthalpy, internal energy
heat of reaction etc.
(b) Intensive property : Any property of a system whose magnitude is independent of the
amount of matter present in the system is called its intensive property.
Examples : Temperature, density, surface tension, viscosity, refractive index, melting point,
boiling point, specific heat.
*3.
Explain the term state function. Give two examples of state functions and two examples of
path functions.
Definition : Any property of a system whose value depends on the current state of the system
and is independent of the path followed to reach that state is called the state function.
Ans.
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(a) For a particular state of the system, the state function has fixed value in the present condition
of the system. It does not depend on previous history of the system.
(b) For example, 1g of water is placed in a container and maintained at 25ºC and 1 atm. pressure.
The volume of water will be approximately 1cm3.
All the quantities, 1g, 25ºC, 1 atm. and 1 cm3 describe the current state of water and
are independent of its previous history.
(c) Hence, the properties mass, temperature, pressure and volume are state functions.
Other examples : enthalpy (H), internal energy (U), number of moles, entropy (S) etc.
ΔH, ΔU, ΔS, ΔP, ΔV
Hence, the properties mass, temperature, pressure and volume are state function.
Other examples - enthalpy (H), internal energy (H), number of moles entropy (S) etc.
Examples of path functions : Work (w) and heat (q).
*4.
Ans.
Explain thermodynamic equilibrium.
Refer 3.2 (e).
*5.
Ans.
Distinguish between Isothermal and Adiabatic processes.
Refer 3.2 (j).
*6.
Ans.
What is a reversible process? What are its features?
Refer 3.2 (h).
*7.
Ans.
*8.
Ans.
Show that pressure times volume is equal to work. OR Show that the product of pressure
and volume is equal to work.
(a) The work (W), that is done due to expansion or compression of a gas against an external
opposing pressure (P) is called pressure-volume work.
(b) The product of PV = W, can be explained as f
f
3
× V = 2 × d = f.d = W
PV =
(Where, d = distance)
A
d
Derive the expression for work when a gas expands against constant external pressure.
Derivation :
Step - 1: The work done by the gas during expansion is given by
W = opposing × distance =
∴
∴
–f ⋅ d
(negative sign shows that internal energy of the system during expansion decreases).
Step - 2 : But, f = pex.A
W = –Pex × A × d
Step - 3 : But A × d = Change in volume = ΔV = V2 – V1
W = –pex.ΔV = –Pex (V2 – V 1)
During expansion : work is done by the system on surroundings,
Chapter - 3 Chemical Thermodynamics And Energetics
162
V2 > V 1
∴ W = –Ve
During compression : work is done on the system by surroundings,
∴ W = +Ve
V2 < V 1
*9.
Ans.
Classify the following reactions according to work done by the system, on the system and
no work done if pressure is constant.
(a) H 2 (g) + Cl 2 (g) ⎯⎯
→ 2HCl(g)
no work is done as volume is constant.
(b) 3O 2 (g) ⎯⎯
→ 2O 3(g)
Ans.
Work is done on the system as volume decreases.
Ans.
(c) 2SO 2 (g) + O 2 (g) ⎯⎯
→ 2SO 3 (g)
work is done on the system as volume decreases.
Ans.
(d) MgCO 3 (s) ⎯⎯
→ MgO (s) + CO 2 (g)
work is done by the system as volume increases.
Ans.
*10.
Ans.
(e) NH 4 NO 3 (s) ⎯⎯
→ N 2 O (g) + 2H 2O (g)
work is done by the system as volume increases.
Explain the concept of maximum work.
Refer 3.3 (g)
*11.
Ans.
A free expansion of a gas results into no work. Explain.
Expansion of a gas against zero opposing force (as in vacuum) is called force expansion.
When Pex = 0, then W = 0
*12.
Ans.
Derive the expression for maximum work.
Refer 3.3 (i)
*13.
Ans.
What are the sign conventions for q and W?
+q → Heat is absorbed by the system from the surroundings.
–q → Heat is released from the system to the surroundings.
+W → Work is done on the system by the surroundings.
–W → Work is done by the system on the surroundings. [Refer 3.3 (m)]
Multiple Choice Questions :
•
1.
Theoretical MCQs :
When an ideal gas expands in vacuum, the work done is .......
a) R
b)
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c)
Zero
d)
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R
2
163
*2.
*3.
*4.
In
a)
b)
c)
d)
a chemical reaction work is done by the system when .......
number of moles of gaseous reactants is equal to the number of moles of gaseous products
total number of moles increases.
number of moles of gaseous substances decreases.
number of moles of gaseous products is greater than the number of moles of gaseous
reactants.
When a sample of an ideal gas is allowed to expand at constant temperature against an atmospheric
pressure, .......
a) surroundings does work on the system
b) Δ U = 0
c) no heat exchange takes place between the system and surroundings
d) internal energy of the system increases.
In what reaction of the following, work is done by the system on the surroundings?
b) 3O 2 (g) ⎯⎯
a) Hg (l) ⎯⎯
→ 2O 3 (g)
→ Hg (g)
c)
•
5.
*6.
*7.
8.
•
9.
10.
H 2 (g) + Cl 2 (g) ⎯⎯
→ 2HCl (g)
d)
N 2 (g) + 3H 2 (g) ⎯⎯
→ 2NH 3 (g)
Numerical MCQs :
The work done during the expansion of a gas from a volume of 4dm3 to 6dm3 against a constant
external pressure of 3atm is ....... (1L atm = 101.32 J)
(CBSE 2004)
a) +304J
b) –304 J
c) –6J
d) –608 J
A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings.
Hence, ΔU is .......
a) 440 kJ
b) 200 J
c) 120.32 J
d) –200J
A gas expands in volume from 2L to 5L against a pressure of 1atm at constant temperature.
The work done by the gas will be .......
a) 3 J
b) –303.9 J
c) –303.9 L.atm d) 303.9 L.atm
–3 3
2 3
An ideal gas expands from 1 × 10 m to 1 × 10 m at 3K against a constant pressure of
1 × 105 N/m2. The work done is .......
a) –900 J
b) 270 kJ
c) –900 kJ
d) +900 kJ.
(Hints : W = –PexΔ V)
Which among the following statment is false .......
(IIT 2001)
a) Work is a state function
b) Temperature is a state function
c) Change of state is completely defined when initial and final states are specified.
d) Work appears at the boundary of the system.
1 mole of each CaC2, Al4C3 and Mg2C3 reacts with water in separate open flask. Numerical
value of work done by the system is in order.
a) CaC2 < Al4C3 = Mg2C3
b) CaC2 < Al4C3 < Mg2C3
c) CaC2 = Mg2C3 < Al4C3
d) CaC2 = Mg2C3 = Al4C3
Chapter - 3 Chemical Thermodynamics And Energetics
164
3.4. INTERNAL ENERGY (U) :
Concept Explanation :
(1) Every substance (system) contains a definite amount of energy which is called internal
energy (U).
(2) Internal energy is the sum of kinetic energies of all the particles of the system and potential
energies due to various bonds between the particles.
(3) In thermodynamic, the change in internal energy is determined
i.e. ΔU = U2 – U1
Where U 1 = system’s energy in initial state
U 2 = system’s energy in final state
(4) When energy is transferred into the system by heating or doing work on it, the internal
energy increases. i.e. ΔU = +ve
(5) When energy is transferred from the system by cooling it or by doing work on the surroundings.
The internal energy decreases.
i.e. ΔU = –ve
Note : ΔU always carries a sign even if it is positive.
3.5. FIRST LAW OF THERMODYNAMICS :
Concept Explanation :
First law of thermodynamics is simply the law of conservation of energy. It can be stated
in different ways –
(a) Statements of first law of thermodynamics :
1. The total internal energy of an isolated system is constant.
2. Energy can be converted from one form to another but it cannot be created or destroyed.
3. The total quantity of energy of the universe is constant.
4. Whenever a quantity of energy of one kind disappears, an exactly equivalent amount of
energy of other kind must appear.
5. The total amount of energy of a system and its surroundings must remain constant although
it may change from one form to another.
(b) Mathematical equation of first law of thermodynamics
The internal energy of the system can be increased in two ways.
1. by supplying heat to the system and
2. by doing work on the system.
(a) If a system with internal energy ‘U1’, absorbs heat ‘q’, its internal energy will change
to U1 + q.
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(b) Now, if work ‘W’ is done on the system, than the internal energy will further increase
and become equal to U1 + q + W.
(c) Let this be find energy of the system, U2
Then,
U 2 = U1 + q + W
i.e.
U2 – U1 = q + W
∴
ΔU = q + W
This is mathematical expression of first law of thermodynamics.
(d) For an infinitesimal changes, the above equation is written as, dU = dq + dW.
(c) Modified first law of thermodynamics :
1. Mass and energy are equivalent and are mutually interconvertible. (E = mc2)
2. Thus, the principle of conservation of energy may be replaced by the principle of conservation
of mass plus energy.
3. Hence, the first law of thermodynamics is modified as “The sum of mass and energy
of an isolated system (Universe) remains constant”.
(d) First law of thermodynamics for various processes :
1. Isothermal process
T = constant,
∴ ΔU = 0
∴ ΔU = q + W
0=q+W
or
q = –W
or
W = –q
i.e. heat absorbed is entirely used for doing work by the system on the surroundings or
the work done on the system by the surroundings results in the release of heat by the
system.
q=0
∴ ΔU = q + W
ΔU = 0 + W
or
ΔU = W
or
–ΔU = –W
i.e. work done on the system is equal to increase is internal energy of the system or work
done by the system is due to the expense of system’s internal energy which decreases
in the process.
3. Isochoric process
ΔU = q + W
= q – pex × ΔV
(∵ W = pex × ΔV)
∴ ΔU = q – pex × 0
∵ ΔV = 0,
ΔU = qv
i.e. the increase in internal energy of the system is equal to the heat absorbed by the
system at constant volume. As ‘U’ is a state function, qv is also a state function.
Chapter - 3 Chemical Thermodynamics And Energetics
166
4.
Isobaric process
pex = constant, ΔV ≠ 0
ΔU = q + W
ΔU = qP – pex.ΔV
(Q W = – p ex ⋅ ΔV )
or qP = ΔU + pex.ΔV
Note : Most of the chemical reactions are carried out at constant atmospheric pressure
(pex = constant). A special symbol ΔH called enthalpy change is used for the heat changes
produced in such reactions. ΔH is also a state function.
Limitations of the First Law of Thermodynamics :
(i) It does not tell about the direction of heat flow.
(ii) It does not tell about the extent of convertibility of one form of energy into another i.e.
how much heat energy would be transferred from one form to the other.
•
*1.
Ans.
State first law of thermodynamics. Justify its mathematical equation.
(a) Statements of first law of thermodynamics :
(1) The total internal energy of an isolated system is constant.
(2) Energy can be converted from one form to another but it cannot be created or destroyed.
(3) The total quantity of energy of the universe is constant.
(4) Whenever a quantity of energy of one kind disappears, an exactly equivalent amount
of energy of other kind must appear.
(5) The total amount of energy of a system and its surroundings must remain constant
although it may change from one form to another.
(b) Mathematical equation of first law of thermodynamics :
The internal energy of the system can be increased in two ways.
(i) by supplying heat to the system and (ii) by doing work on the system.
(1) If a system with internal energy ‘U1’, absorbs heat ‘q’, its internal energy will change
to U1 + q.
(2) Now, if work ‘W’ is done on the system, then the internal energy will further increase
and become equal to U1 + q + W.
(3) Let this be final energy of the system, U2
Then,
U 2 = U1 + q + W
i.e. U2 – U1 = q + W
∴
ΔU = q + W
This is mathematical expression of first law of thermodynamics.
(c) For an infinitesimal changes, the above equation is written as, dU = dq + dW
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Multiple Choice Questions :
•
1.
Theoretical MCQs :
First law of thermodynamics is merely the law of .......
a) Conservation of mass
b) Conservation of energy
c) Conservation of both mass and energy
d) none of these
The mathematical expression for the first law of thermodynamics is .......
a) H = E + PV
b) ΔH = ΔU + P ⋅ ΔV
c) q = ΔU + W
d) ΔU = q + W
2.
•
3.
4.
•
5.
Numerical MCQs :
If a gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of
2 × 105 N/m2, then change in internal energy is .......
a) –300 J
b) +100 J
c) –100 J
d) +300 J
Change in internal energy when 4 kJ of work is done on the system and 1 kJ of heat is given
out by the system is .......
(Kerala PMT - 2008)
a) 3 kJ
b) 4 kJ
c) 7 kJ
d) 6 kJ
a) Expanding system gain work energy and does work on the surroundings.
b) Expanding system loses work energy and does work on the surrounding.
c) Expanding system gain work energy and does work on the system.
d) Expanding system loses work energy and does work by the surroundings.
3.6. ENTHALPY (H) :
Concept Explanation :
(a) The enthalpy of a system is defined as the sum of the internal energy of the system
and its pressure-volume energy.
(b) Mathematically, the enthalpy is defined by the equation, H = U + PV
(c) Since U, P and V are state functions, H is also a state function.
(a) Relationship between ΔH, ΔU, and P.ΔV :
Step - 1 : Change in enthalpy (ΔH) depends on initial and final states of the system and
is given by, ΔH = H2 – H1
Step - 2 : But, H1 = U1 + P1V1 and H2 = U2 + P2V2
∴ ΔH = U2 + P2V2 – U1 – P1V1
∴ ΔH = (U2 – U1) + (P2V2 – P1V1)
Chapter - 3 Chemical Thermodynamics And Energetics
168
∴ ΔH = ΔU + Δ(PV)
At constant pressure, P1 = P2 = P,
∴ ΔH = ΔU + P.ΔV
Step - 3 : If pex = P, then from Ist law of thermodynamics,
qp = ΔU + P.ΔV
Comparing the above two equations, we get;
ΔH = qp
Thus, increase in enthalpy of a system is equal to the heat absorbed at constant pressure
(isobaric process).
Note :
(i) Since ‘H’ is a state function, qp is also a state function.
(ii) q is not a state function but qp and qv are state functions because the path of the
process is defined and therefore, qp and qv can have only specific values.
(b)
1.
2.
∴
3.
∴
4.
∴
∴
∴
Relationship between ΔH and ΔU for chemical reactions :
We know, ΔH = ΔU + P.ΔV
For reactions involving solids and liquids, ΔV is very small,
P.ΔV is neglected, Hence ΔH = ΔU
For reaction involving gases, ΔV cannot be neglected,
ΔH = ΔU + P.ΔV
= ΔU + P (V2 – V1)
= ΔU + PV2 – PV1
Assuming ideal behaviour of reactant and product gases,
PV1 = n1 RT and PV2 = n2 RT
ΔH = ΔU + n2 RT – n1 RT
ΔH = ΔU + RT (n2 – n1)
ΔH = ΔU + RT.Δn
Where, Δn = number of moles of gaseous products – number of moles of gaseous reactants.
(c) Work done in chemical reactions :
The work done by a system at constant pressure and temperature is given by equation,
∴ W = – p ex ⋅ ΔV
Assuming Pex = P
∴ W
= –P.ΔV
= –P (V2 – V1)
= –PV2 + PV1
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Assuming ideal behaviour, PV1 = nRT and PV2 = n2 RT
∴ W
= –n2RT + n1RT
= –RT (n2 – n1)
W
= –RT.Δn
The above equation gives the work done by a system, in chemical reaction.
1. If n2 > n1, W is negative and work is done by the system.
2. If n2 < n1, W is positive and work is done on the system.
3. If n1 = n2, W is zero, no work is done.
(d) Conditions under which ΔH = ΔU :
1. If reaction is carried out in closed vessel, volume is constant, ΔV = 0.
∴ ΔH = ΔU + P.ΔV
ΔH = ΔU
2. If reaction involves only solids and liquids, ΔV is neglected, hence, ΔH = ΔU.
3. If n1 = n2, i.e. Δn = 0, then ΔH = ΔU
3.7. ENTHALPIES OF PHYSICAL CHANGES :
Concept Explanation :
They are four main types :
(a) phase transition.
(b) the breaking of individual atoms, molecules into their ions and other fragments.
(c) dissolution of a substance into a solvent and
(d) dilution of a solution.
(a) Enthalpy of phase transition :
These are,
1. Enthalpy of fusion (Δfus H)
(a) Definition : The enthalpy change that accompanies the fusion of one mole of a
solid without change in temperature at constant pressure is called its enthalpy
of fusion.
(b) It is denoted by the symbol ΔfusH.
(c) Example :
ΔfusH = + 6.01 kJ mol–1 at 0ºC & 1 atm.
H 2 O(s) ⎯⎯
→ H 2 O(l )
(1 mole ice)
ΔfreezH = –6.01 kJ mol–1 at 0ºC & 1 atm.
H 2 O(l ) ⎯⎯
→ H 2 O(s) ,
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170
2. Enthalpy of vaporization (ΔvapH) :
(a) Definition : The enthalpy change that accompanies the vaporization of one mole of liquid
without changing its temperature at constant pressure is called enthalpy of vaporization.
(b) It is denoted by ΔvapH,
(c) Example :
H 2O (l) ⎯⎯
→ H 2O (g) ;
ΔvapH = + 40.7 kJ mol–1 at 100ºC & 1 atm.
(1 mole)
H 2 O (l) ⎯⎯
→ H 2 O (g) ;
ΔvapH = + 44.0 kJ mol–1 at 25ºC & l atm.
(d) The reverse of vaporization is condensation of vapour which is accompanied with
the evolution of heat.
(e) Example : H 2O (g) ⎯⎯
→ H 2 O (l) ;
ΔconH = –40.7 kJ mol–1 at 100ºC
Standard enthalpies of fusion and vaporization
Substance
F.P./K
ΔfusHº/kJ mol–1
B.P./K
ΔvapHº/kJ mol–1
Acetone (CH3COCH3)
177.8
5.72
329.4
29.1
Ethanol (C2H5OH)
158.7
4.60
351.5
43.5
Methanol (CH3OH)
175.5
3.16
337.2
35.3
Methane (CH4)
90.7
0.94
111.7
8.2
Benzene (C6H6)
278.7
9.87
353.3
30.8
Water (H2O)
273.2
6.01
373.2
40.7
3. Enthalpy of sublimation (ΔsubH) :
(a) Definition : The enthalpy change that accompanies the conversion of one mole
of solid directly into its vapour at constant temperature and pressure is called
its enthalpy of sublimation.
(b) It is denoted by ΔsubH.
→ H 2O (g) ; ΔsubH = 51.08 kJ mol–1 at 0ºC & 1 atm.
(c) Example : H 2 O (s) ⎯⎯
Note : It should be noted that whether the conversion of solid to vapour takes place
directly in one step or in two steps; the enthalpy change is the same since enthalpy
is a state function. For example :
H 2 O (s) ⎯⎯
→ H 2 O (l) ,
ΔfusH = +6.01 kJ mol–1 at 0ºC
H 2 O (l) ⎯⎯
→ H 2 O (g) ,
Δvap H = +45.07 kJ mol–1 at 0ºC
H 2 O (s) ⎯⎯
→ H 2 O (g) ,
ΔsubH = (+6.01 + 45.07) kJ mol–1at 0ºC
= 51.08 kJ mol–1 at 0ºC.
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4. Δsub H = ΔfusH + ΔvapH
(b) Enthalpy of atomic or molecular changes :
1. Enthalpy of ionization (ΔionH) :
(a) Definition : The enthalpy change that accompanies the removal of an electron
from each atom or ion in 1 mole of gaseous atoms or ions is called an enthalpy
of ionization.
(b) It is denoted by ΔionH.
(c) Example : Na (g) ⎯⎯
→ Na +
(g)
+ e– ;
ΔionH = 494 kJ mol–1
(1 mole)
(d) The enthalpy change for the removal of one electron from each atom in one mole
is called first ionization enthalpy :
(e) The second ionization energy is larger than first e.g.
Ca(g) ⎯⎯
→ Ca + (g) + e –
ΔionH = 590 kJ mol–1
Ca + (g) ⎯⎯
→ Ca 2+ (g) + e –
ΔionH = 1150 kJ mol–1
(f) The reverse of ionization enthalpy is the electron-gain enthalpy, which is defined
as the enthalpy change when one mole of gas-phase atoms of an element accept
electrons so as to form gaseous anions.
(g) Example :
Cl(g) + e – ⎯⎯
→ Cl – (g)
ΔegH = –349 kJ mol–1
Chapter - 3 Chemical Thermodynamics And Energetics
172
(a) Ionization enthalpies
Atom
Δ ionH/kJ mol
–1
(b) Electron gain enthalpies
Δ ionH/kJ mol
–1
Atom
Δ egH/kJ mol–1
1 st
2 nd
Na
494
4560
B
–28
K
418
3070
C
–122
Li
519
7300
O
–141, +844
Mg
736
1450
Cl
–349
Ca
590
1150
Br
–325
Ba
502
966
F
–328
Al
577
1820
I
–295
Cl
1260
–
S
–200, +532
Br
1140
–
I
1010
–
2. Enthalpy of atomization (ΔatomH) :
(a) Definition - The enthalpy change accompanying the dissociation of all the
molecules in one mole of a gas phase substance into gaseous atoms is called
enthalpy of atomization.
(b) Example :
ΔatomH = 242kJ mol–1
Cl 2 (g) ⎯⎯
→ Cl(g) + Cl(g)
(1 mole)
(c) Enthalpy of solution (ΔsolnH) :
1. Definition - The change in enthalpy when one mole of a substance is dissolved in
a specified quantity of solvent at a given temperature so as to form a solution of
particular concentration is called enthalpy of solution.
Note : If the solution so obtained is further diluted, there will again be a change in
enthalpy. If we go on diluting the solution, a stage will come when further dilution
produces no thermal effect. This state is called the state of infinite dilution.
2.
Definition - The change in enthalpy when one mole of a substance is dissolved in
a solvent so that further dilution does not give any change in enthalpy is called
enthalpy of solution.
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3. Example : When 1 mole of HCl(g) is dissolved in 50 moles of water, the enthalpy of
solution is given as HCl(g) + 50H 2 O(l ) ⎯⎯
→ HCl(50H 2 O) , ΔsoluH = –73.26 kJ
When 1 mole of HCl(g) is dissolved in large quantity of water that is when the solution
is infinitely dilute, the enthalpy of solution is given by.
ΔsolnH = –75.14 kJ
HCl(g) + aq ⎯⎯
→ HCl(aq) ;
Causes of enthalpy of solution :
The exact value of ΔsolnH for a given substance is the sum of the contributions of three kinds
of interactions.
1. Solute - solute interactions :
(a) Energy is required to overcome attractive forces between solute particles to dissolve
the solute.
(b) For an ionic solid, energy is called crystal lattice energy.
(c) Crystal lattice energy is defined as the enthalpy change accompanying the formation
of one mole of formula units of a crystalline solid from its constituent ions in the gaseous
state.
(d) Crystal lattice enthalpies are always negative.
M + (g) + X – (g) ⎯⎯
→ MX
(s) ,
ΔH is negative
(e) Reverse of crystal formation reaction takes place, during formation of solution.
(f) The smaller the magnitude of crystal lattice enthalpy, the more readily the dissolution
of solute occurs.
2. Determination of crystal lattice enthalpy :
It is determined by Born - Haber cycle after
Max Born and Fritz Haber, who developed it.
The lattice energy of NaCl, for example, is the
change in enthalpy, ΔHº, When Na+ and Cl– ions
in the gas phase come together to form 1 mole
of NaCl crystal.
3. Born-Haber Cycle for the formulation of NaCl crystal from its elements :
(a) The lattice energy of an ionic crystal can be determined by applying Hess’s law.
(b) Enthalpy change for direct formation of NaCl :
Na(s) +
1
Cl (g) ⎯⎯
→ NaCl(s)
2 2
Δf Hº = –411 kJ
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174
(c) Enthalpy change by indirect steps :
Step : 1 Sublimation Na(s) ⎯⎯
→ Na(g)
Step : 2 Dissociation
1
Cl
⎯⎯
→ Cl (g) ;
2 2 (g)
Step : 3 Ionisation
Na(g) ⎯⎯
→ Na + (g) + e – ;
Step : 4
Step : 5
∴
ΔsubH = +109 kJ
ΔdH = +122 kJ
ΔionH = 496 kJ
ΔegH = –348 kJ
Formation of crystal lattice (one mole of solid NaCl)
Na + (g) + Cl – (g) ⎯⎯
→ NaCl (s) ;
ΔLH = –(Lattice energy)
ΔfH = ΔsubH + ΔdH + ΔionH + ΔegH + ΔLH
–411 kJ = 109kJ + 122 kJ + 496 kJ – 348 kJ + ΔLH
By solving this equation we get,
Crystal Lattice energy of NaCl (ΔLH) = –790 kJ mol–1
Cl(g) + e– ⎯⎯
→ Cl – (g) ;
Lattice enthalpies
Substance
ΔLH/
Substance
kJ mol–1
ΔLH/
Substance
kJ mol –1
ΔLH/
kJ mol–1
LiF
–1037
NaF
–926
KF
–821
LiCl
–852
NaCl
–786
KCl
–717
LiBr
–815
NaBr
–752
KBr
–689
LiI
–761
NaI
–705
KI
–649
4. Solvent - Solvent interactions :
During the dissolution process, energy is required for the separation of solvent molecules
in order to make room for the solute particles.
5. Solute - Solvent interactions :
The solvent molecules surround the solute particles.
This process is called solvation. Energy is released in solvation process.
(d) Enthalpy of dilution :
1. When already prepared solution is further diluted, then either heat is released or absorbed,
which is called enthalpy of dilution.
2. Definition : The enthalpy change that occurs when a solution of one concentration
is diluted to form the solution of another concentration is called enthalpy of dilution.
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3. Example :
Further diluted Soln. : HCl(g) + aq ⎯⎯
→ HCl(aq) ;
Ist prepared Soln. :
HCl(g) + 50H 2 O ⎯⎯
→ HCl (50H 2O) ;
–
–
–
HCl + 50H 2O + aq ⎯⎯
→ HCl (aq) ;
Hence, enthalpy of dilution = –1.88 kJ.
•
*1.
Ans.
*2.
Ans.
∴
∴
∴
∴
∴
*3.
Ans.
ΔsolnH = –75.14 kJ
ΔsolnH = –73.26 kJ
+
ΔdilH = –1.88
Arrange in order of increasing enthalpy; H2O(s), H2O(g), H2O(l)
Enthalpy increases in the order : H2O(s) < H2O(l) < H2O(g)
What is the difference between ΔH and ΔU? What is the sign of ΔH for exothermic and
endothermic reactions? Under what circumstances ΔH = ΔU?
ΔH → Heat of reaction at constant pressure
ΔU → Heat of reaction at constant volume
ΔH = +ve for endothermic reaction
ΔH = –ve for exothermic reaction
Relationship between ΔH and ΔU for chemical reactions.
(a) We know, ΔH = ΔU + P.ΔV
(b) For reactions involving solids and liquids, ΔV is very small,
P.ΔV is neglected, Hence ΔH = ΔU
(c) For reaction involving gases, ΔV cannot be neglected,
ΔH = ΔU + P.ΔV
= ΔU + P (V2 – V1)
= ΔU + PV2 – PV1
(d) Assuming ideal behaviour of reactant and product gases,
PV1 = n1 RT and PV2 = n2 RT
ΔH = ΔU + n2 RT – n1 RT
ΔH = ΔU + RT (n2 – n1)
ΔH = ΔU + RT.Δn
Where, Δn = number of moles of gaseous products – number of moles of gaseous reactants.
Define enthalpy. At constant pressure show that ΔH = ΔU + P.ΔU
Definition : Enthalpy is a thermodynamic property of a system. It reflects the capacity to do
non-mechanical work and the capacity to release heat. Enthalpy is denoted as H; specific enthalpy
denoted as h.
Relationship between ΔH, ΔU, and P.ΔV.
Chapter - 3 Chemical Thermodynamics And Energetics
176
∴
∴
∴
∴
*4.
Ans.
Step - 1 : Change in enthalpy (ΔH) depends on initial and final states of the system and
is given by, ΔH = H2 – H1
Step - 2 : But, H1 = U1 + P1V1 and H2 = U2 + P2V2
ΔH = U2 + P2V2 – U1 – P1V1
ΔH = (U2 – U1) + (P2V2 – P 1V1)
ΔH = ΔU + Δ(PV)
At constant pressure, P1 = P2 = P,
ΔH = ΔU + P.ΔV
Step - 3 : If pex = P, then from Ist law of thermodynamics,
qp = ΔU + P.ΔV
Comparing the above two equations, we get;
ΔH = qp
Thus, increase in enthalpy of a system is equal to the heat absorbed at constant pressure (isobaric
process).
It is difficult to calculate ‘U’, however ΔU can be easily determined. Why? Explain giving
examples.
The absolute value of internal energy (U) cannot be determined because it is impossible to
determine accurately the most of the quantities that contribute to the internal energy of a system.
For example it is difficult to determine the kinetic energies of all the molecules, ions and atoms
of the system. It is difficult to determine the potential energies associated with the forces between
the particles.
*5.
Ans.
Obtain the relationship between ΔH and ΔU for a chemical reaction.
Refer 3.6(b)
*6.
Ans.
What is the expression for work done in a chemical reaction? Explain the meaning of each term.
Work done in chemical reactions : The work done by a system at constant pressure and temperature
is given by equation,
W = –pex.ΔV
Assuming pex = P
W = –P.ΔV
= –P (V2 – V1)
= –PV 2 + PV 1
Assuming ideal behaviour, PV1 = n1RT and PV2 = n2 RT
W = –n2RT + n1RT
= –RT (n2 – n1)
W = –RT.Δn
The above equation gives the work done by a system, in chemical reaction.
(a) If n2 > n1, W is negative and work is done by the system.
∴
∴
∴
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(b) If n2 < n1, W is positive and work is done on the system.
(c) If n1 = n2, W is zero, no work is done.
*7.
Ans.
Define and explain each of the following with one example each :
(a) enthalpy of fusion
(b) enthalpy of vaporization
(c) enthalpy of sublimation
(d) enthalpy of atomization
(e) enthalpy of ionization
(a) Enthalpy of fusion (Δfus H) :
(1) Definition : The enthalpy change that accompanies the fusion of one mole of a
solid without change in temperature at constant pressure is called its enthalpy
of fusion.
(2) It is denoted by the symbol ΔfusH.
(3) Example :
H 2 O(s) ⎯⎯
→ H 2 O(l ) ;
ΔfusH = + 6.01 kJ mol–1 at 0ºC and 1 atm.
(1 mole ice)
H 2 O(l ) ⎯⎯
→ H 2 O(s) ;
ΔfreezH = –6.01 kJ mol–1 at 0ºC and 1 atm.
(b) Enthalpy of vaporization (ΔvapH) :
(1) Definition : The enthalpy change that accompanies the vaporization of one mole
of liquid without changing its temperature at constant pressure is called enthalpy
of vaporization.
(2) It is denoted by ΔvapH,
(3) Example :
H 2 O(l ) ⎯⎯
→ H 2 O(g) ;
ΔvapH = + 40.7 kJ mol–1 at 100ºC and 1 atm.
(1 mole)
H 2 O(l ) ⎯⎯
→ H 2 O(g) ;
ΔvapH = + 44.0 kJ mol–1 at 25ºC and l atm.
(4) The reverse of vaporization is condensation of vapour which is accompanied with
the evolution of heat.
(5) Example :
H 2 O(g) ⎯⎯
→ H 2 O(l ) ;
ΔconH = –40.7 kJ mol–1 at 100ºC
[Refer table 3.7 (a)]
(c) Enthalpy of sublimation (ΔsubH) :
(1) Definition : The enthalpy change that accompanies the conversion of one mole
of solid directly into its vapour at constant temperature and pressure is called
its enthalpy of sublimation.
(2) It is denoted by ΔsubH.
Chapter - 3 Chemical Thermodynamics And Energetics
178
(3) Example :
ΔsubH = 51.08 kJ mol–1 at 0ºC and 1 atm.
H 2 O(s) ⎯⎯
→ H 2 O(g)
(4) ΔsubH = ΔfusH + ΔvapH
[Refer table 3.7 (a)3]
(d) Enthalpy of atomization (ΔatomH) :
(i) Definition - The enthalpy change accompanying the dissociation of all the
molecules in one mole of a gas phase substance into gaseous atoms is called
enthalpy of atomization.
(ii) Example :
ΔatomH = 242 kJ mol–1
Cl 2 (g) ⎯⎯
→ Cl(g) + Cl(g) ;
(1 mole)
(e) Enthalpy of ionization (ΔionH) :
(i) Definition : The enthalpy change that accompanies the removal of an electron
from each atom or ion in 1 mole of gaseous atoms or ions is called an enthalpy
of ionization.
(ii) It is denoted by ΔionH.
(iii) Example :
Na (g) ⎯⎯
→ Na + (g) + e – ;
ΔionH = 494 kJ mol–1
(1 mole)
(iv) The enthalpy change for the removal of one electron from each atom in one mole
is called first ionization enthalpy :
(v) The second ionization energy is larger than first e.g.
ΔionH = 590 kJ mol–1
Ca(g) ⎯⎯
→ Ca + (g) + e –
ΔionH = 1150 kJ mol–1
(vi) The reverse of ionization enthalpy is the electron-gain enthalpy, which is defined as
the enthalpy change when one mole of gas-phase atoms of an element accept electrons
so as to form gaseous anions?
(vii) Example : Cl(g) + e – ⎯⎯
ΔegH = –349 kJ mol–1
→ Cl – (g)
[Refer table 3.7 (b)1]
Ca + (g) ⎯⎯
→ Ca 2+ (g) + e –
Multiple Choice Questions :
•
1.
Theoretical MCQs :
Which one does not describe a correct definition of enthalpy?
a) H = E + PV
b) H = E – PV
c) E = H – PV
Unique Solutions ®
(AIIMS 1997)
d) H – E – PV = 0
S.Y.J.C. Science - Chemistry - Part I
179
2.
•
*3.
The enthalpy change of a reaction does not depends on .......
a) State of reactants and products
b) Nature of reactants and products
c) Different intermediate reactions
d) Initial and final enthalpy change of reaction
Numerical MCQs
The heat evolved in the following reaction .......
2H 2 (g) + O 2 (g) ⎯⎯
→ 2H 2 O (l) , ΔH = –484 kJ to produce 9 g of water is .......
a) 484 kJ
*4.
b)
*6.
8.
9.
242 kJ
b)
–368 kJ
c)
Zero
For which of the following substances ΔfHº is not zero .......
a) Ca(s)
b) He(g)
c) P(red)
d)
968 kJ
d)
+368 kJ
d)
CH3OH(l)
Given the reaction, N 2 (g) + 3H 2 (g) ⎯⎯
→ 2NH 3 (g) , ΔH = –92.6 kJ. The enthalpy of
formation of NH3 is .......
a) –185.2 kJ mol–1
•
7.
c)
In the reaction, H 2 (g) + Cl 2 (g) ⎯⎯
→ 2HCl (g) ΔH = –184 kJ, if 2 moles of H2 react with
2 moles of Cl2, then ΔU is equal to .......
a) –184 kJ
*5.
121 × 133 J
b)
–46.3 kJ mol–1 c)
92.6 kJ mol–1
d)
–92.6 kJ
The enthalpy of which of the following substances in standard state is zero?
c) NH3
d) HNO3
a) Carbon
b) CaCO3
(H.S.C. Board, March 2008)
The enthalpies of formation of N2O and NO are 82 and 90 kJ mol–1. The enthalpy of reaction
2N 2O(g) + O 2(g) ⎯⎯
→ 4 NO(g) is .......
(Hints ΔH = ∑ H (Products) – ∑ H (reactants) )
a) 8 kJ
b) 88 kJ
c) –16 kJ
d) 196 kJ
Two moles of an ideal gas is expanded isothermally and reversibly from 1L to 10L at 300K.
The enthalpy change (in kJ) for the process is .......
(Hint : For an isothermal reversible reaction, ΔH = n Cp ΔT = 0 Q ΔT = 0) (IIT 2004)
a) 11.4 kJ
b) –11.4 kJ
c) 0 kJ
d) 4.8 kJ
3.8. THERMOCHEMISTRY :
Concept Explanation :
Thermochemistry is the branch of physical chemistry which deals with the enthalpy changes
during a chemical reactions.
Chapter - 3 Chemical Thermodynamics And Energetics
180
(a) Enthalpy of chemical reactions (Heat of reaction)
(a) Definition : The amount of heat released or absorbed in a chemical reaction at
constant temperature and pressure is called enthalpy of chemical reaction or
heat of reaction.
(b) Example : aA + bB ⎯⎯
→ cC + dD
∴
Enthalpy change (ΔH) = (cHc + dHD) – (aHA + bHB)
or ΔH = ∑ H products – ∑ H reactants
(c) Thus, the enthalpy of chemical reaction is the difference between the sum of the
enthalpies of products and that of reactants.
1. Exothermic reaction :
(a) When ∑ H products is less than ∑ H reactants , ΔH is negative and heat is released.
(b) The reactions in which heat is released from the system to the surroundings are called
exothermic reactions.
(c) Example : 2KClO 3 (s) ⎯⎯
→ 2KCl(s) + 3O 2 (g) , ΔHº = –78 kJ
2. Endothermic reaction :
(a) When ∑ H products is more than ∑ H reactants , ΔH is positive and heat is absorbed.
(b) The reaction in which heat is absorbed by the system from the surroundings are called
endothermic reaction.
(c) Example : N 2 (g) + 2O 2 (g) ⎯⎯
→ 2NO 2 (g) ; ΔHº = +66.4 kJ
3. Standard enthalpy of chemical reactions :
(a) The standard state of a substance is that form of the substance in which it is most
stable at a pressure of 1 bar (105 Pa = 0.987 atm) or 1 atm and at 25ºC.
(b) If the reaction involves species in solution, their standard states are 1M concentrations.
(c) Example : Standard states of elements : H2(g), Hg(l), Na(s), C(graphite) at 1 atm. pressure
and 25ºC. Standard states of compounds : C2H5OH(l), H2O(l), CaCO3(s), CO2(g)
at 1 atm pressure and 25ºC.
(d) The standard enthalpy of a chemical reaction (Standard heat of reaction) is defined
as ‘the enthalpy change accompanying the reaction when all the substances
involved are in their standard states’.
(e) It is denoted by ΔHº
4. Thermochemical equation :
(a) Definition : A balanced chemical equation in which the exact value of enthalphy
change, physical states and number of moles of reactants and products are specified;
is known as thermochemical equation.
(b) Rules for writing the thermochemical equation :
(i) The equation must be balanced for the number of moles of reactants and products.
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S.Y.J.C. Science - Chemistry - Part I
181
(ii) The numerical value and sign of enthalpy change (ΔHº) must be shown on the
right hand side of the equation.
(iii) The physical states of reactants and products must be specified by the letters
s (solid), l (liquid), g (gas) or aq (aqueous).
(c) For the reverse reaction the sign of ΔHº must be changed.
(d) If the coefficients indicating the number of moles of all the substances are multiplied
or divided by a numerical factor, the value of ΔHº also be multiplied or divided by
the same factor.
Example : CH 4 (g) + 2O 2 ⎯⎯
→ CO 2 (g) + 2H 2O(l ) , ΔHº = –890 kJ
(b) Standard enthalpy of formation (standard heat of formation) ΔfHº :
1. Definition : The enthalpy change that accompanies a reaction in which one mole of
a pure compound in its standard state is formed from its elements in their standard
states, is called standard enthalpy of formation.
2. It is denoted by Δf Hº.
3. Example :
H 2 (g) +
1
O (g) ⎯⎯
→ H 2O(l ) ,
2 2
Δf Hº = –286 kJ mol–1
(1 mole)
C (graphite) + 2H 2 (g) ⎯⎯
→ CH 4 (g)
4.
Δf Hº = –74.8 kJ mol–1
The standard enthalpy of formation of an element from itself is zero.
i.e.
ΔfHº (C) = Δf Hº (H2) = Δf Hº (Cl2) = 0
Standard molar enthalpies of formation
Substance
Δ fHº
Substance
kJ/mol
Δ fHº
Substance
kJ/mol
Δ fHº
Substance
kJ/mol
Δ fHº
kJ/mol
AlCl3(s)
–704.2
Cl2O(g)
80.3
MgCO3(s)
–1096
H3PO4(s)
–1279
Al2O3 (s)
–1676
CuCl2(s)
–220.1
NH3(g)
–46.1
KClO3(s)
–397.7
BaCl2 (s)
–855
CuO(s)
–157
N2H4(l)
50.6
KNO3(s)
–494.6
BaCO3(s)
–1216
CuS(s)
–79.5
N2H4(g)
95.4
NaCl(s)
–411.2
Br2(g)
30.9
HF(g)
–271
NO(g)
90.2
Na2O(s)
–414.2
HBr(g)
–36.4
H2O(l)
–285.8
NO2(g)
33.2
NaHCO3(s) –950.8
Chapter - 3 Chemical Thermodynamics And Energetics
182
CaO(s)
–635.1
H2O(g)
–241.8
N2O(g)
82.0
NaNO3(s)
–468
CaCO3(s)
–1207
H2O2(l)
–187.8
N2O4(g)
9.16
H2S(g)
–20.6
CO(g)
–110.5
H2O2(g)
–136.3
N2O5(g)
11.0
SO2(g)
–296.8
CO2(g)
–393.5
HI(g)
26.5
NOCl(g)
51.7
SO3(g)
–395.7
CS2(g)
–116.7
FeO(s)
–272
O3(g)
143
ZnO(s)
–348.3
HCl(g)
–92.3
Fe2O3(s)
–824.2
PCl3(l)
–320
ZnS(s)
–206
ClO2(g)
102.5
MgO(s)
–601.7
PCl3(g)
–375
ZnCl2(s)
–415.1
Compound
Δ fHº
kJ/mol
Compound
Δ fHº
kJ/mol
192.3
CCl4(l)
–135.4
HCHO(g)
–108.6
CH3CHO(g)
166.2
C2H6(g)
–84.7
HCOOH(l)
–424.7
CH3COOH(l)
–484.5
C2H5OH(l)
–277.7
C6H12O6(s)
–1260
C6H6(g)
82.93
C2H5OH(g)
–235.1
CH4(g)
–74.8
C6H6(l)
49.0
C2H4(g)
52.3
CH3OH(l)
–238.7
C2H2(g)
226.7
C2H4O(g)
–52.6
CH3OH(g)
–201.2
Compound
Δ fHº
kJ/mol
CH3CHO(l)
(c)
Standard enthalpy of combustion (ΔcHº)
1. Definition : The enthalpy change accompanying a reaction in which 1mole of a substance
in the standard state reacts completely with oxygen or is completely burnt is called
standard enthalpy of combustion.
2. It is denoted by ΔcHº
3. Example :
C(g) +
1
O (g) ⎯⎯
→ CO 2 (g) ,
2 2
ΔcHº = –283 kJ mol–1
(1mole)
C 2 H 2 (g) +
5
O (g) ⎯⎯
→ 2CO 2 (g) + H 2 O(l ) ,
ΔcHº = –1300 kJ mol–1
2 2
Standard molar enthalpies of combustion
Element/
Compound
Δ cHº/
kJ/mol
Element/
Compound
ΔcHº/
kJ/mol
C (graphite)
–393.5
CH3OH (g)
–764
C (diamond)
–395.4
CH3OH (l)
–726
CO (g)
–283.0
C2H5OH (l)
–1368
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S.Y.J.C. Science - Chemistry - Part I
183
CH4 (g)
–890.0
HCOOH (l)
–255
C2H6 (g)
–1560
CH3COOH (l)
–875
C2H4 (g)
–1411
HCHO (g)
–571
C2H2 (g)
–1300
C2H5OH (g)
–1409
C6H6 (l)
–3268
CH3CHO (l)
–1166
C6H6 (g)
–3302
H2 (g)
–285.8
(d) Bond enthalpy (Bond energy) :
1. Definition : The enthalpy change necessary to break a particular covalent bond in
1 mole of gaseous molecules to produce gaseous atoms and / or radicals is called
bond enthalpy.
2. The bond enthalpy is always positive
3. Example : H 2 (g) ⎯⎯
→ H (g) + H (g)
ΔHº = 436.4 kJ.
1 mole
∴ ΔHº(H–H) = 436.4 kJ mol–1
(4) Bond enthalpy for diatomic molecules is the same as enthalpy of atomization.
(5) For polyatomic molecules, the average bond enthalpy of a particular bond is calculated.
Example : H2O has two identical
O – H bonds.
H 2 O (g) ⎯⎯
→ OH (g) + H (g) ,
ΔHº = 499 kJ
H 2 O (g) ⎯⎯
→ H (g) + O (g)
ΔHº = 428 kJ
ΔHº = 927 kJ
H 2 O (g) ⎯⎯
→ 2H (g) + O (g)
∴ Average bond enthalpy of O – H bond =
∴ ΔHº(0 – H) = 463.5 kJ mol–1
927
= 463.5 kJ
2
(e) Reaction enthalpy from Bond enthalpy :
Chapter - 3 Chemical Thermodynamics And Energetics
184
1.
3.
In chemical reactions, bonds are broken in reactant molecules and bonds are formed in
product molecules.
Enthalpy of a chemical reaction is the difference between the sum of the reactant bond
enthalpies and the sum of the product bond enthalpies.
ΔHº (reaction) = ∑ Hº (reactant bonds) – ∑ Hº (Product bonds)
4.
Example : H 2 (g) + I 2 (g) ⎯⎯
→ 2HI(g)
2.
∴ ΔHº = [ΔHº(H – H) + ΔHº (I – I)] – [2 × ΔHº (H – I)]
Single bond enthalpies (kJ mol–1)
Multiple bond enthalpies
Bond
C
C
C
C
C
=
≡
=
≡
≡
C
C
N
N
O
ΔHº kJ/mol
Bond
ΔHº kJ/mol
620
812
615
891
754
C=S
N=N
N≡N
O=O
N=O
477
418
941.4
498.7
607
(f) Hess’s law of constant heat summation :
1. Statement : It states that the change in enthalpy for a reaction is the same whether
the reaction takes place in one or a series of steps. OR It states that the enthalpy
change for a chemical reaction is the same regardless of the path by which the reaction
occurs.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
185
2.
Example:
Conversion of A to C can take place directly in a single step.
A ⎯⎯
→ C;
ΔHº1
Or in two steps
Step (1) : A ⎯⎯
→ B , ΔHº2
3.
Step (2) B ⎯⎯
ΔHº3
→ C,
According to Hess’s law, ΔHº1 = ΔHº2 + ΔHº3
Example (2) Preparation of methylene chloride can take place by two different paths.
→ CH 2 Cl 2 (g) + 2HCl (g) , ΔHº1= –203.3 kJ
Path - I : CH 4 (g) + 2Cl 2 (g) ⎯⎯
Path - II : Step (1) - CH 4 (g) + 2Cl 2 (g) ⎯⎯
→ CH 2Cl(g) + 2HCl(g) ; ΔHº2 = –98.3 kJ
Step (2) - CH 3Cl(g) + Cl 2 (g) ⎯⎯
→ CH 2Cl 2 (g) + HCl(g) ; ΔHº3 = –104.0 kJ
ΔHº = –202.3 kJ
Thus , whether the reaction takes place in one step or several steps the enthalpy of reaction
in the same.
CH + 2Cl CH + Cl
CH 4 (g) + 2Cl 2 (g) ⎯⎯
→ CH 2 Cl 2 (g) + 2HCl(g) ;
4
2
4
2
CH 3Cl + Cl2
CH3Cl + HCl
CH 2Cl2 + 2HCl
Hess’s law
(g) Applications of Hess’s law :
1. It is useful for the determination enthalpies of those reaction, whose values cannot be
experimentally determined correctly.
Example :
1
C (graphite) + O 2 (g) ⎯⎯
→ CO (g)
ΔHº = ?
2
Here some C is oxidised to CO2. Hence, its value is determined from the following two
equation.
(a)
C (graphite) + O 2 (g) ⎯⎯
→ CO 2 (g) ;
ΔHº = –393.5 kJ.
Chapter - 3 Chemical Thermodynamics And Energetics
186
(b)
2.
3.
4.
5.
6.
•
*1.
Ans.
1
CO (g) + O 2 (g) ⎯⎯
→ CO 2 (g) ;
ΔHº = –283.0 kJ.
2
[reverse eq. (b) and add eq. (a)]
It is used for the calculation of enthalpies of those reaction which do no occur directly.
Example - Formation of CH4 from the elements.
The enthalpy of formation of any compound from its elements can be calculated.
The enthalpy of combustion can be calculated.
The thermochemical equation can be added, subtracted or multiplied by a numerical factor
like ordinary algebraic equation.
It is useful for the determination of enthalpies of extremely slow reactions.
Define enthalpy of chemical reactions.
(a) Definition : The amount of heat released or absorbed in a chemical reaction at constant
temperature and pressure is called enthalpy of chemical reaction or heat of reaction.
(b) Example : aA + bB ⎯⎯
→ cC + dD
∴ Enthalpy change (ΔH) = (cHc + dHD) – (aHA + bHB)
or ΔH = ∑ H products – ∑ H reactants
(c) Thus, the enthalpy of chemical reaction is the difference between the sum of the enthalpies
of products and that of reactants.
*2.
Ans.
What is meant by standard state of a substance and standard enthalpy of reactions?
Refer 3.8(a) 3.
*3.
Ans.
Define : (a) Exothermic reactions, (b) Endothermic reactions
(a) Exothermic reaction :
(1) When ∑ H products is less than ∑ H reactants , ΔH is negative and heat is released.
(2) The reactions in which heat is released from the system to the suroundungs are called
exothermic reactions.
(3) Example : 2KClO 3 (s) ⎯⎯
→ 2KCl(s) + 3O 2 (g), Hº = –78kJ
(b) Endothermic reaction :
(1) When ∑ H products is more than ∑ H reactants , ΔH is positive and heat is absorbed.
(2) The reaction in which heat is absorbed by the system from the surroundings are called
endothermic reaction.
(3) Example : N 2 (g) + 2O 2 (g) ⎯⎯
→ 2NO 2 (g) ,
ΔHº = +66.4 kJ
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S.Y.J.C. Science - Chemistry - Part I
187
*4.
Ans.
What are the guidelines for writing a thermochemical equation?
Refer 3.8 (a) 4 – b
*5.
Ans.
Define standard enthalpy of formation. How is it useful to calculate standard reaction enthalpy?
Standard enthalpy of formation (standard heat of formation) ΔfHº :
(i) Definition : The enthalpy change that accompanies a reaction in which one mole of
a pure compound in its standard state is formed from its elements in their standard
states, is called standard enthalpy of formation.
(ii) It is denoted by ΔfHº.
(iii) Example :
1
H 2 (g) + O 2 (g) ⎯⎯
→ H 2O (l) ,
ΔfHº = –286 kJ mol–1
2
(1 mole)
ΔfHº = –74.8 kJ mol–1
(iv) The standard enthalpy of formation of an element from itself is zero. i.e.
ΔfHº (C) = ΔfHº (H2) = ΔfHº (Cl2) = 0
C (graphite) + 2H 2 (g) ⎯⎯
→ CH 4 (g) ,
[Refer Table 3.8 (1)b]
*6.
Ans.
Write balanced chemical equation that have ΔHº value equal to ΔfHº for each of the following
substances.
(b) KClO3
(c) CH3COOH
(d) C 12H22O11
(e) CH3CH2–OH
(a) C2H2
We know, ΔHº = ∑ m Δ f Hº (products) – ∑ n Δ f Hº (reactants)
(a) 2C(s) + 2H 2 (g) ⎯⎯
→ C 2 H 2 (g)
1
3
→ KClO 3 (s)
(b) K(s) + Cl 2 (g) + O 2 (g) ⎯⎯
2
2
(c) 2C(s) + O 2 (g) + 2H 2 (g) ⎯⎯
→ CH 3COOH(l )
11
→ C12 H 22 O11 (s)
(d) 12C(s) + 11H 2 (g) + O 2 (g) ⎯⎯
2
1
→ CH 3CH 2OH (l )
(e) 2C(s) + 3H 2 (g) + O 2 (g) ⎯⎯
2
*7.
Ans.
*8.
Ans.
Explain standard enthalpy of combustion with one example.
Refer 3.8 (c).
→ O 2(g) + 2HF(g) ΔHº = –323 kJ.
Consider the chemical reaction : OF2 (g) + H 2 O(g) ⎯⎯
What is ΔHº of the reaction if : (a) the equation is multiplied by 3 (b) direction of reaction is reversed?
(a) ΔHº = 3 × –323 kJ = – 969 kJ
(b) ΔHº = + 323 kJ.
Chapter - 3 Chemical Thermodynamics And Energetics
188
*9.
Ans.
*10.
Ans.
What is meant by bond enthalpy? How is it useful to calculate reaction enthalpy? Explain with
one example.
Refer 3.8 (d, e).
State and explain Hess’s law of constant heat summation.
Hess’s law of constant heat summation :
(a) Statement : It states that the change in enthalpy for a reaction is the same whether
the reaction takes place in one or a series of steps. OR It states that the enthalpy
change for a chemical reaction is the same regardless of the path by which the reaction
occurs.
(b) Example (1) : Conversion of A to C can take place directly in a single step.
A ⎯⎯
→ C ; ΔHº1 Or in two steps
Step (1)
ΔHº2
A ⎯⎯
→ B,
Step (2) B ⎯⎯
[Refer 3.8 (f)]
→ C , ΔHº3
According to Hess’s law, ΔHº1 = ΔHº2 + ΔHº3
(c) Example (2) : Preparation of methylene chloride can take place by two different paths.
Path - I
CH 4 (g) + 2Cl 2 (g) ⎯⎯
→ CH 2 Cl 2 (g) + 2HCl(g) , ΔHº1 = 203.3 kJ
Path - II Step (1) CH 4 (g) + Cl 2 (g) ⎯⎯
→ CH 2 Cl 2 (g) + 2HCl(g) ; ΔHº2 = –98.3 kJ
Step (2) CH 3Cl(g) + Cl 2 (g) ⎯⎯
→ CH 3Cl 2 (g) + HCl(g) ; ΔHº3 = –104.0 kJ
CH 4 (g) + 2Cl 2 (g) ⎯⎯
→ CH 2 Cl 2 (g) + 2HCl(g) ; ΔHº = –202.3 kJ
Thus, whether the reaction takes place in one step or several step the enthalpy of reaction
is the same. [Refer 3.8 (f) flow chart diagram]
*11.
Ans.
∵
∴
*12.
The heat evolved in a reaction of 7.5g of Fe2O3 with enough CO is 1.164 kJ. Calculate ΔHº
for the reaction, Fe 2O 3 (s) + 3CO(g) ⎯⎯
→ 2Fe(s) + 3CO(g)
gram Mol. Wt. of Fe2O3 = 2 × 56 + 3 × 16 = 160g
7.5g of Fe2O3, heat evolved = 1.164 kJ.
160 × 1.164
160g of Fe2O3, heat evolved =
= 24.832 kJ.
7.5
Calculate using data of table (3.7), the enthalpy of the reaction.[Refer 3.7 Table Text Book.N.101]
CH 3COOH(g) + CH 3CH 2OH(g) ⎯⎯
→ CH 3COOCH 2CH 3 (g) + H 2O(g) [Refer 3.8 (e)]
Ans. H
H
O
C
C
H
O
H + H
H
H
C
C
H
H
Unique Solutions ä
O
H ⎯⎯→ H
H
O
C
C
H
O
H
H
C
C
H
H
S.Y.J.C. Science - Chemistry - Part I
H + H –O–H
189
∴
*13.
ΔHº = ∑ ΔHº (reactants bonds) – ∑ ΔHº (Product bonds)
ΔHº = [8 × ΔHºC – H + 2 × ΔHºC – C + 1 × ΔHºC = O + 2 × ΔHºO – H + 2 × ΔHºC – O] –
[8 × ΔHºC – H + 2 × ΔHºC – C + 2 × ΔHºC – O + 1 × ΔHºC = O + 2 × ΔHºO – H]
ΔHº = [8 × 414 + 2 × 347 + 1 × 754 + 2 × 464 + 2 × 351] –
[8 × 414 + 2 × 347 + 2 × 351 + 1 × 754 + 2 × 464]
ΔHº = 0
∴
The standard enthalpy change for the reaction H 2 (g) ⎯⎯
→ H(g) + H(g) is 436.4 kJ mol–1.
Calculate standard enthalpy of formation of atomic hydrogen.
When 2 moles of H atoms are formed, 436.4 kJ energy is required
2 × 6.022 × 1023 atoms of H are formed = 436.4 kJ
436.4
1 atom of H is formed = 2 × 6.022 × 10 23 = 3.623 × 10–22 kJ.
Multiple Choice Questions :
•
1.
Theoretical MCQs :
The standard heat of formation of diamond is .......
a) Same as that of graphite
b) Greater than that of graphite
c) Less than that of graphite
d) Taken as zero
The heat of combustion of compound is always .......
a) Positive
b) Negative
c) Zero
d) Uncertain
Ans.
Q
2.
•
3.
4.
•
5.
Numerical MCQs :
The enthalpy changes at 298K in successive breaking of O – H bond of HOH are .......
H 2 O(g) ⎯⎯
→ H(g) + OH(g) ;
ΔH = 498 kJ mol–1
OH(g) ⎯⎯
→ H(g) + O(g) ;
ΔH = 428 kJ mol–1
The bond enthalpy of O – H bond is .......
b) 463 kJ mol–1
c) 428 kJ mol–1 d) 70 kJ mol–1
a) 498 kJ mol–1
The enthalpies of combustion of carbon and carbon monoxide are –390 kJ mol–1 and –278
kJ mol–1 respectively. The enthalpy of formation of carbon monoxide is .......
a) 668 kJ mol–1
b) 112 kJ mol–1
c) –112 kJ mol–1 d) –668 kJ mol–1
The formation of water from H2(g) and O2(g) is an exothermic process because .......
(PMT - 1986)
a) The chemical energy of H2(g), and O2 (g) is more than that of water
b) The chemical energy of H2(g), and O2 (g) is less than that of water
c) The temperature of H2(g), and O2 (g) is higher than that of water
d) The temperature of H2(g), and O2 (g) is lower than that of water
Chapter - 3 Chemical Thermodynamics And Energetics
190
3.9. SECOND LAW OF THERMODYNAMICS :
Concept Explanation :
It is based on human experience.
It can be stated as :
(i) ‘The spontaneous flow of heat is always unidirectional, from higher temperature to lower
temperature.’
(ii) Heat cannot be completely converted into an equivalent amount of work without producing
permanent changes either in the system or its surroundings.
(iii) No machine has yet been made that has an efficiency unity.
(The efficiency of machine is the ratio of work done to the total heat absorbed by heat
engine).
If heat absorbed is completely converted into work without producing changes in the system,
the efficiency must be unity.
(iv) During the course of every spontaneous (natural) process, the entropy of the universe
(system and surroundings) increases.
3.10. SPONTANEOUS PROCESS (IRREVERSIBLE PROCESS) :
Concept Explanation :
A reaction that occurs under the specified set of conditions is a spontaneous reaction.
A reaction that does not occur under the specified set of conditions is a non-spontaneous
reaction.
3. Definition : A process that takes place on its own without the external influence is
called a spontaneous process.
4. Definition : A process that does not take place without the continuous external force
is called a non-spontaneous process.
5. A non-spontaneous process can be made to occur. For example a gas can be compressed
into a smaller volume by applying external pressure on it.
6. All the natural (spontaneous) processes proceed till equilibrium is reached.
Examples of spontaneous processes :
1. Water always flows on its own form higher level to lower level.
2. Heat always flows from hotter body to a colder body.
3. Solid KCl spontaneously dissolves in water.
4. Acid-base neutralization is a spontaneous reaction.
5. Consider two bulbs connected through stop-cock. One bulb
is filled with a gas and other is evacuated. As soon as the
stop-cock is opened the gas expands spontaneously.
1.
2.
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S.Y.J.C. Science - Chemistry - Part I
191
(a) Energy and spontaneity :
1. The stability of a substance is determined by its energy. The substances having more energy
are less stable whereas those that have less energy are more stable.
2. In a spontaneous reaction, substances with more energy are converted into the substances
having less energy.
3. A chemical reaction must be exothermic in order to take place spontaneously e.g.
neutralization of acid-base.
KOH(aq) + HNO 3 (aq) ⎯⎯
→ KNO 3 (aq) + H 2O , ΔH = –57 kJ.
4. Although exothermicity favours spontaneity but does not assure it.
There are many processes that are endothermic yet spontaneous.
(a) Ice melts spontaneously above 0ºC with the absorption of heat from the surroundings.
above
Ice ⎯⎯⎯
→ H 2 O(l) , ΔH = +6.0 kJ mol–1
0ºC
(b) NaCl dissolves spontaneously in water with the absorption of heat.
NaCl (s) + aq ⎯⎯
→ Na + (aq) + Cl – (aq) , ΔH = +3.9 kJ mol–1
Note : Energy is not the only criterion for the spontaneity of processes and hence,
there must be another factor (thermodynamic property) called Entropy that determines
the spontaneity of processes.
(b) Entropy :
1. Entropy is a thermodynamic property that determines the spontaneity of a processes.
2. Definition : Entropy is a thermodynamic state quantity that is a measure of the
randomness or disorder of the molecules of the system. It is denoted by S.
3. Example :
(a) When ice melts, the highly ordered arrangement of molecules collapses and molecules
become free to move in liquid, hence, entropy increases and ΔS is positive.
(b) When liquid water vaporizes, the gaseous water molecules move freely and randomly.
A less disordered state becomes highly disordered and hence, entropy further increases.
4. Thus, greater the disorder of a system, higher is its entropy.
Chapter - 3 Chemical Thermodynamics And Energetics
192
(a) Sign of ΔS different process : When solid I2 undergoes sublimation, the disorder
of the system increases and hence, entropy increases. Thus, ΔS is positive.
I 2 (s) ⎯⎯
→
I 2 (g)
ΔS positive
ordered state disordered state
(b) One mole of H2 gas is converted to two moles of gas phase atoms. The number
of particles increases. Thus, disorder increases and hence, entropy of the system
increases.
H 2 (g)
2H(g)
⎯⎯
→
ΔS positive
disordered state
more disordered state
(c) 2H 2 (g) + O 2 (g) ⎯⎯
ΔS negative
→ 2H 2 O(l )
more disordered state less disordered state.
In the above process, three gas-phase moles are converted into two moles of liquid.
The more disordered state becomes less disordered state. The entropy of the system
decreases. ΔS is negative.
(c) Quantitative definition of entropy :
When one substance changes to another in a chemical reaction, sometimes it is difficult
to identify the disorder, hence, it is necessary to define entropy quantitatively.
1. Definition - The entropy change of a system in a process is equal to the amount of
heat transferred to it in a reversible manner divided by the temperature at which
the transfer takes place.
q rev
T
’
Where, ‘qrev is the heat transferred to the system at temperature T.
The unit of ΔS is JK–1.
Entropy (ΔS ) is a state function.
ΔS that is qrev /T measures the change in the disorder.
Thus, ΔS =
2.
3.
4.
(d) Entropy and spontaneity (2nd Law of thermodynamics) :
1. Consider the following two spontaneous reactions :
2H 2 O 2 (l ) ⎯⎯
→ 2H 2 O(l ) + O 2 (g) ,
ΔS = +126JK–1
ΔS = –327JK–1
The above two reactions show that during spontancous process, the entropy may increases
or decreases.
This discrepaney is explain by the second law of thermodynamics.
Second law of thermodynamics (statement) : It states that the total entropy of the system
and its surroundings (universe) increases in a spontaneous process.
2H 2 (g) + O 2 (g) ⎯⎯
→ 2H 2 O(l )
2.
3.
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S.Y.J.C. Science - Chemistry - Part I
193
4.
5.
•
*1.
Ans.
*2.
Ans.
*3.
Ans.
2nd Law of thermodynamics is expressed as – For spontaneous process,
ΔSuniverse = ΔStotal = ΔSsys + ΔSsurr > O
(a) When ΔStotal > 0; the process is spontaneous
(b) When ΔStotal < 0; the process is non-spontaneous
(c) When ΔStotal = 0; the process is at equilibrium
What is spontaneous process? Give examples.
Refer 3.10.
Which of the following are spontaneous?
(a) dissolution of sugar in hot coffee.
(b) separation of Ar and Kr from their mixture.
(c) spreading of fragrance when a bottle of perfume is opened.
(d) flow of heat from cold object to hot object
(e) heat transfer from ice to room at 25ºC.
(a), (b), (c), (e) → spontaneous processes.
(d) → non-spontaneous
Define entropy. Give its units. What does entropy measure?
Definition : Entropy is a thermodynamic state quantity that is a measure of the randomness
or disorder of the molecules of the system. It is denoted by S.
Unit of entropy = JK–1
Entropy (S) measures the randomness or disorder of the molecules of the system.
Multiple Choice Questions :
•
1.
Theoretical MCQs :
Entropy change of a system and its surroundings in equilibrium .......
a) Increases
b) Decreases
c) Remains constant
d) Either increases or decreases
In a reversible process, ΔSsys + ΔSsurr is .......
a) > 0
b) < 0
c) ≥ 0
d) = 0
2.
•
3.
Numerical MCQs :
1g. ice absorbs 335 J of heat to melt at 0ºC, the entropy change will be .......
a) 1.2 JK–1 mol–1
b) 335 JK–1 mol–1 c) 22.1 JK–1 mol –1d) 0.8 JK–1 mol–1
Chapter - 3 Chemical Thermodynamics And Energetics
194
4.
•
5.
If enthalpy of vaporisation of benzene is 308 kJ mol–1 at boiling point (80ºC). Calculate
entropy in changing it from liquid to vapour in J mol–1 K–1 .......(Haryana PMT - 2004)
a) 308
b) 0.873
c) 0.308
d) 873
1
2
→ CO 2 (g) , ΔH and ΔS are –283 kJ and – 87 JK–1.
For a reaction CO(g) + O 2 (g) ⎯⎯
It was intended to carry out this reaction at 1000K, 1500K, 3000K and 3500K. At which
of these temperatures would this reaction be thermodynamically spontaneous .......
(Kerala PET-2006)
a) 1500 K and 3500 K
b) 3000 K and 3500 K
c) 1000 K, 1500 K and 3500 K
d) 1500 K, 3000K and 3500 K
3.11. GIBBS ENERGY (G) :
Concept Explanation :
According to 2nd of thermodynamics, for spontaneous process ΔStotal (ΔSsys + ΔSsurr) must
be positive. Thus, we have determine two entropy change, ΔSsys and ΔSsurr.
But chemists are interested only in system (reaction mixture).
This problem was solved by American theoretician J. G. Gibbs. He introduced a new
thermodynamics function called Gibbs energy (G).
1. Definition - Gibbs energy of a system is defined as the maximum amount of energy
available to a system during a process that can be converted into useful work.
2. In other words, it is a thermodynamic quantity which is a measure of capacity of a
system to do useful work.
3. It is denoted by symbol G is given by G = H – TS.
Where H is the enthalpy of the system, S is its entropy and T is the absolute temperature.
4. H, T and S are state functions, hence, G is also a state function.
5. The change in Gibbs energy (ΔG) depends on initial and final states of the system and
not on the path connecting the two states.
6. The change in Gibbs energy at constant temperature and pressure is defined as –
ΔG = ΔH – TΔS.
(a) Gibbs energy and spontaneity :
1. ΔStotal = ΔSsys + ΔSsurr = ΔS + ΔSsurr (The subscript ‘sys’ is dropped, but it refers
to system only).
2. According to 2nd law of thermodynamics, for spontaneous process, at constant temperature
and pressure.
ΔStotal > 0.
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S.Y.J.C. Science - Chemistry - Part I
195
If ΔH is the enthalpy change of the system, then enthalpy change of the surroundings
is –ΔH.
ΔH
4. ∴ ΔSsurr = –
T
ΔH
5. Hence, total entropy is given by ΔStotal = ΔS –
.
T
This equation is expressed in terms of properties of system only.
6. On rearrangement of above equation,
–T ΔStotal = –T ΔS + ΔH
–T ΔStotal = ΔH – TΔS
But, ΔG = ΔH – TΔS
∴ ΔG = – TΔStotal
This equation indicates that ΔG and ΔStotal have opposite signs because T is always
positive.
7. Thus, for a spontaneous process, carried out at constant temperature and pressure.
ΔStotal > 0 and hence, ΔG < 0. i.e. Gibbs energy of the system decreases. Thus if,
(a) ΔG < 0, the process is spontaneous
(b) ΔG > 0, the process is non-spontaneous
(c) ΔG = 0, the process is at equilibrium
3.
(b) Predicting spontaneity of a process in terms of ΔG :
(i) Gibb’s equation is : ΔG = ΔH – T.ΔS
(ii) The spontaneous process is favoured by decrease in enthalpy (–ΔH) and increase in entropy
(+ΔS).
(iii) A non-spontaneous reactions is favoured by increase in enthalpy (+ΔH) and decrease
in entropy (–ΔS).
(iv) Reactions are classified into four categories with respect to dependence of sign of ΔG
and hence, spontaneity on ΔH, ΔS and T.
(a) If ΔH and ΔS are both negative the ΔG will be negative and the reaction will be
spontaneous only when the term ΔH is larger in magnitude than the magnitude of
T ΔS. (This is possible only at low temperatures.)
(b) If ΔH and ΔS are both positive then ΔG will be negative and the reaction will be
spontaneous only when the TΔS term is larger in magnitude than ΔH. (This is possible
only at high temperatures.)
(c) If ΔH is negative and ΔS is positive, ΔG will be always negative regardless of
temperature. The reaction is therefore, spontaneous at all temperatures.
(d) If ΔH is positive and ΔS is negative the ΔG will be always positive regardless of
temperature.
The reaction is therefore, non-spontaneous at all temperatures. The factors affecting
sign of ΔG and spontaneity are summarized in table. (Factor affecting spontaneity)
Chapter - 3 Chemical Thermodynamics And Energetics
196
Factor affecting spontaneity
ΔH
–
(exothermic)
ΔS
+
ΔG
–
Spontaneity of reactions
Reactions are spontaneous at all the temperatures.
_
(exothermic)
_
– or +
Reactions become spontaneous at low temperatures
when |T ⋅ S| < | H|
+
(endothermic)
+
– or +
Reactions become spontaneous at high temperatures
when T ⋅ S > H .
+
(endothermic)
–
+
Reactions are nonspontaneous at all temperatures.
(c) Temperature of equilibrium between spontaneous and
nonspontaneous process :
At equilibrium, the process is neither spontaneous nor non-spontaneous.
At equilibrium, ΔG = 0, hence, Gibb’s equation can be written as ΔG = ΔH – T ⋅ ΔS = 0
ΔH
ΔS
‘T’ is the temperature at which change over between spontaneous and non-spontaneous
behaviour occurs.
∴ ΔH = TΔS or T =
(d) ΔG and equilibrium constant :
In a chemical reaction all the substances may not be in their standard states. Hence, change
in Gibbs energy (ΔG) is related to Standard Gibbs energy change (ΔGº) as ΔG = ΔGº
+ RT ln Q Where, ΔGº = Standard Gibbs energy change when all the substances are
in their standard states. Q = Reaction quotient.
Reaction quotient (Q) is in terms of initial concentration or partial pressure of the reactants
and final concentrations or pressures of the products.
Example : aA + bB ⎯⎯
→ cC + dD
[C] c [D] d
Qc =
pcC × pd D
or
Qp = a
[A] a [B] b
p A × pbB
Where the values of concentrations or partial pressures are other than equilibrium
concentrations or partial pressure values.
When the reaction reaches equilibrium, the concentration and partial pressure reach their
equilibrium values and at this stage, Q = K
At equilibrium, ΔG = 0 and Q = K, hence Gibb’s equation becomes,
0 = ΔGº + RT In K.
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S.Y.J.C. Science - Chemistry - Part I
197
∴ ΔGº = –RT ln K = –2.303 RT log10K.
This equation gives the relationship between Standard Gibbs energy change (ΔGº) of the
reaction and its equilibrium constant.
3.12. THIRD LAW OF THERMODYNAMICS :
Concept Explanation :
1.
2.
3.
4.
5.
6.
7.
Define : It states that the entropy of a perfectly ordered crystalline substance is zero
at absolute zero of temperature.
Thus, S = 0 at T = 0 for any perfectly ordered crystalline substance.
If crystal contains some impurities, then S > 0 at T = 0.
This entropy of solid greater than zero is called residual entropy of the substance.
Third law is used to determine the absolute entropy of any substance either in solid, liquid
of gaseous state at any desired temperature.
Example : If a perfectly ordered crystalline substance with S = 0 at T = 0 is heated from
(zero) 0K to any desired temperature T, the entropy increases due to the increased vibrational
motions of molecules.
The increase in entropy is given by ΔS = Sr – S0.
Where Sr is the absolute entropy of the substance at temperature.
T and S0 its value at T = 0.
∴ ΔS = Sr.
Because, S0 = 0,
The value of Sr can be determined by measuring heat capacity of the solid at various
temperatures by using the expression,
T
Cp ⋅ dT
ΔS = Sr – S0 = Sr =
T
∫
O
(a) Standard molar entropy :
1.
2.
Definition : The absolute entropy of one mole a pure substance at 1 atm. pressure
and 25ºC is called Standard molar entropy of the substance.
It is denoted by Sº.
(b) Usefulness of Standard molar entropy (Sº) :
1. By knowing the values of Sº of all reactants and products, it is possible to calculate ΔSº.
ΔSº = ∑ m Sº (products) – ∑ n Sº (reactants)
2.
Sº is useful to compare entropies of different substances under the same conditions of
temperature and pressure.
→ Sº increases with increasing complexity of molecules.
→ Heavier substances have larger Sº values than higher ones.
Chapter - 3 Chemical Thermodynamics And Energetics
198
•
*1.
Predict the signs of ΔH, ΔS and ΔG of the system when a solid melts at 1 atm. and at
(a) –55ºC
(b) –95ºC (C) –77ºC, if the normal melting point of solid is –77ºC.
Ans.
*2.
Ans.
*3.
Ans.
Explain how entropy changes in the following processes.
(a) freezing of liquid
(b) sublimation of solid
(c) dissolving of sugar in water
(d) condensation of vapour
(a) Freezing of liquid – Entropy decreases as disordered state becomes ordered state.
ΔS is negative.
(b) Sublimation of solid – Entropy increases as ordered state of the system becomes disordered
because solid substances goes into gas phase. ΔS is positive.
(c) Dissolving of sugar in water – Entropy increases as ordered state of sugar molecules
goes into disordered state due to separation of molecules of sugar in water. ‘ΔS’ is positive.
(d) Condensation of vapours – Entropy decreases as gas-phase molecules are converted
into liquid state.
The more disordered state becomes less disordered state. ΔS is negative.
Which member of the following pairs have larger entropy? Why?
(a) CO2(s) or CO2(g)
(b) CH3OH (l) or C2H5OC2H5(l)
(c) NO(g) or N2O4(g)
(d) Xe(g) or Kr(g)
(e) Na(s) or NaCl(s)
(a) CO2(g) will have larger entropy because gas phase is highly disordered.
(b) CH3OH(l) will have larger entropy because molecules are smaller in size.
(c) N2O4(g) will have larger entropy because it has greater complexity.
(d) Xe has greater atomic mass and hence, greater entropy than Kr.
(e) NaCl(s) has greater entropy since it is more complex than Na(s).
*4.
Why is it more convenient to predict spontaneity of reaction in terms of ΔGsys rather than
ΔStotal?
Ans.
(a) 2nd law of thermodynamics says that ΔStotal (ΔSsys + ΔSsurr) must be positive for all
spontaneous processes.
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199
(b) Therefore, to know the spontaneity of a process, it is necessary to determine two entropy
changes, i.e. ΔSsys and ΔSsurr.
(c) However, the chemists are more interested in the system (reaction mixture) than the
surroundings.
(d) Hence, spontaneity of reaction is expressed in terms of ΔGsys.
*5.
Ans.
Define Gibbs energy and change in Gibbs energy. What are the units of Gibbs energy?
Refer 3.11 (1, 5, 6).
Units of ΔG : ΔG has the units of energy (i.e. kJmol–1or Jmol–1) because both ΔH and
TΔS are energy terms.
ΔG (J mol–1) = ΔH (J mol–1) – TΔS (k JK–1mol–1)
*6.
Ans.
State second law of thermodynamics in terms of entropy and express it mathematically.
Quantitative definition of entropy : When one substance changes to another in a chemical reaction,
sometimes it is difficult to identify the disorder, hence, it is necessary to define entropy
quantitatively.
(a) Definition - The entropy change of a system in a process is equal to the amount of
heat transferred to it in a reversible manner divided by the temperature at which
the transfer takes place.
q rev
Thus, ΔS =
T
Where, ‘qrev’ is the heat transferred to the system at temperature T.
(b) The unit of ΔS is JK–1.
(c) Entropy (ΔS ) is a state function.
(d) ΔS that is qrev /T measures the change in the disorder.
*7.
Ans.
State third law of thermodynamics. What is its usefulness?
Refer Concept explanation 3.12 (1) (b)
*8.
Ans.
*9.
Ans.
Why is it impossible for substance to have an absolute entropy zero at temperature greater
than 0 K?
If a perfectly ordered crystalline substance with S = 0, at T = 0 is heated from 0 K to any
desired temperature T. The entropy increases due to increased vibrational motions of molecules.
Hence, it is impossible to have an absolute entropy zero at temperature greater than 0 K.
It is possible for a reaction to be spontaneous yet endothermic. Comment with example
Energy and spontaneity :
(a) The stability of a substance is determined by its energy. The substances having more energy
are less stable whereas those that have less energy are more stable.
(b) Although exothermicity favours spontaneity but does not assure it.
There are many processes that are endothermic yet spontaneous.
Chapter - 3 Chemical Thermodynamics And Energetics
200
(1) Ice melts spontaneously above 0ºC with the absorption of heat from the surroundings.
above
ΔH = +6.0 kJ mol–1
Ice ⎯⎯
⎯→ H 2 0 (l) ,
0ºC
(2) NaCl dissolves spontaneously in water with the absorption of heat.
NaCl (s) + aq ⎯⎯
→ Na + (aq) + Cl – (aq) , ΔH = +3.9 kJ mol–1
*10.
Ans.
*11.
Ans.
Is it possible for a reaction to be nonspontaneous yet exothermic? Explain with example.
Normally exothermic process are spontaneous but there are some processes which are nonspontaneous inspite of being exothermic. For example : H2O(l) at 1 atm. and 0ºC forms ice.
Here both the phases are in equilibrium. This process is non-spontaneous.
1atm,0ºC
H 2 O(l) H 2 O(s) ;
H < 0.
Predict the sign of ΔS in the following processes. Give reasons for your answer.
(a) N 2 O 4 (g) ⎯⎯
→ 2NO 2 (g)
sign of ΔS is +ve as disorder in the product are larger than in the reactants.
(b) Fe 2O 3 (s) + 3H 2 (g) ⎯⎯
→ 2Fe(s) + 3H 2 O(g)
Ans.
reaction is in equilibrium, ΔS = 0.
(c) N 2 (g) + 3H 2 (g) ⎯⎯
→ 2NH 3 (g)
Ans.
sign of ΔS is –ve, as disorder in the product are less than in the reactants.
(d) MgCO 3 (s) ⎯⎯
→ MgO(s) + CO 2 (g)
Ans.
ΔS is +ve, as disorder increases in the products.
(e) CO 2 (g) ⎯⎯
→ CO 2 (s)
Ans.
ΔS is –ve as disorder decreases in the product.
(f) Cl 2 (g) ⎯⎯
→ 2Cl(g)
Ans.
ΔS is +ve, as one mole of Cl2(g) gives two moles of Cl(g) atoms and hence, disorder increases.
*12.
What can be said about the spontaneity of reactions when
(a) ΔH and ΔS are both positive.
Reactions become spontaneous at high temperature When T ⋅ ΔS > ΔH
(b) ΔH and ΔS are both negative.
Reactions become spontaneous at low temperatures When T ⋅ ΔS < ΔH
(c) ΔH is positive and ΔS is negative.
Reactions are nonspontaneous at all temperatures.
(d) ΔH is negative and ΔS is positive.
Reactions are spontaneous at all the temperatures.
Ans.
Ans.
Ans.
Ans.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
201
*13.
Ans.
Ans.
Ans.
Ans.
*14.
Ans.
Identify which member of the following pairs has larger entropy. Why?
(a) He (g) in a volume of 1L or He (g) in a volume of 5L both at 25ºC.
He (g) in a volume of 5L will have larger entropy become disorder will be more due to large
volume.
(b) O2 (g) at 1 atm or O2 (g) at 10 atm both at the same temperature.
O2(g) at 10 atm. will have lesser disorder due to high pressure and hence, entropy will be less.
(c) C2H5OH (l) or C2H5OH (g)
C2H5OH(g) will have more disorder and hence, entropy will be greater.
(d) 5mol of Ne or 2 mol of Ne.
5 mole of Ne will be more, Ne particles and hence, disorder will be more. Therefore, entropy
will be more.
What are the signs of ΔS and ΔH for the following reaction? Explain with reasons.
2H(g) ⎯⎯
→ H 2 (g)
2 moles of gas-phase atoms are converted into 1mole of H2(g). Number of particles decrease
and disorder is less and hence, entropy of the system decreases. ΔS is negative.
The above reaction shows formation of bond, which is exothermic process, hence, ΔH is –ve.
*15.
Ans.
Derive the relationship between ΔG and ΔStotal.
Refer 3.11 (a)
*16.
Ans.
Justify the inclusion of qrev and T in the definition of entropy, ΔS = qrev/T.
(a) When heat is added to a system the molecular motions are increased due to an increase
in the kinetic energies of the molecules.
(b) This results in increasing molecular disorder, hence, entropy increases. i.e. ΔS is proportional
to qrev.
(c) If a given quantity of heat is added to a system at higher temperature, then the additional
disorder created is less than if the same heat is added at lower temperature.
(d) This shows that ΔS is inversely proportional to the temperature.
*17.
The increase in entropy of a system alone does not guarantee the spontaneity of a process.
Explain.
(a) There are many spontaneous processes in which system’s entropy decreases.
(b) For example, 2H 2 (g) + O 2 (g) ⎯⎯
→ 2H 2 O(l ) , ΔS = – 327 JK–1
(c) In the above reactions, 3 moles of gaseous substances are converted to 2 moles of liquid
substance.
(d) The disorder in the reactants is larger than that in the product, hence, entropy decreases.
(e) The above reaction is spontaneous once initiated it proceeds with explosive violence.
Ans.
*18.
Ans.
The criterion of spontaneity in terms of Gibbs energy is the same as that laid down by second
law of thermodynamics. How?
(a) Gibbs energy (ΔG ) increases in a nonpontanceous process (ΔG > 0).
Chapter - 3 Chemical Thermodynamics And Energetics
202
(b) Thus, the criterion of spontaneity according to second law of thermodynamics
ΔStotal > 0 remains the same.
The way to express the spontaneity of a process is different. In other words, the second
law of thermodynamics is not violated.
(c) For a spontaneous process ΔG < 0 that is Gibbs energy of the system decreases.
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
If for a reaction ΔH is negative and ΔS is positive
a) spontaneous at all temperatures
b)
c) spontaneous only at high temperatures
d)
Which of the following has highest entropy?
c)
a) Al(s)
b) CaCO3(s)
*2.
*3.
*4.
*5.
*6.
*7.
H2O(l)
d)
CO2(g)
For the process, H 2 O(l ) ⎯⎯
→ H 2 O(g) at 100ºC, ΔS is ........
a) positive
b) negative
c) zero
d) unpredictable
For the conversion of liquid to solid below the melting point of solid ........
a) ΔSsys is negative and the process is spontaneous
b) ΔSsys is positive and the process is spontaneous
c) ΔSsurr is positive and the process is non-spontaneous
d) ΔSsys is zero and the process is at equilibrium.
Which of the following conditions guarantee that a reaction is spontaneous at constant T and P?
a) entropy of system increases
b) enthalpy of system decreases
c) entropy of system decreases and that of surroundings increases.
d) Gibbs energy of the system decreases.
Which of the following processes is non-spontaneous?
a) dissolving KCl in water
b) mixing of iodine vapour and nitrogen gas
c) decomposition of NaCl to Na (s) and Cl2 (g)
d) freezing of water at 270K.
For which of the following reactions ΔS is negative?
a)
8.
then the reaction is ........
non-spontaneous at all temperatures
spontaneous only at low temperatures
Mg(s) + Cl 2 (g) ⎯⎯
→ MgCl 2 (s)
b)
H 2 O(l ) ⎯⎯
→ H 2 O(g)
c) CaCO 3 (s) ⎯⎯
d) I 2 (g) ⎯⎯
→ CaO(s) + CO 2 (g)
→ 2I(g)
Heat energy cannot completely transformed into work without producing some change
somewhere, is the statement of ........
a) Hess’s law
b) First law of thermodynamics
c) Kirchoff’s law
d) Second law of thermodynamics
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
203
9.
Considering entropy (s) as a thermodynamic parameter, the criterion for the spontaneity of
any process is ........
(CBSE 2004)
b) ΔSsurr > 0 only
a) ΔSsys > 0 only
d) ΔSsys – ΔSsurr > 0
c) Δ Ssys + Δ Ssurr > 0
•
10.
Numerical MCQs :
If k < 1. What will be the value of ΔGº of the following .......
a) 1
b) Zero
c) Negative
11.
→ CO 2 (g)
What is ΔGº of the following reaction CO(g) + O 2 (g) ⎯⎯
(Gujarat CET 2007)
d) Positive
1
2
(CBSE 1990)
ΔHº = –282.84 kJ mol–1, SºCO (g) = 213.8 JK–1 mol–1, SºCO(g) = 197.9 JK–1 mol–1, SºO
2
2
(g) = 205.0 JK–1 mol–1 .......
b) –250 kJ mol–1
c) +25 kJ mol–1 d) +300 kJ mol–1
a) –257 kJ mol–1
•
12.
For a process H 2O (l ) (1bar,373K) ⎯⎯
→ H 2 O (g) (1 bar, 373 K). The correct set of
thermodynamic parameters is .......
(IIT 2007)
a) ΔG = 0, ΔS = +ve
b) ΔG = 0, ΔS = –ve
c) ΔG = +ve, ΔS = 0
d) ΔG = –ve, ΔS = +ve
HOURS BEFORE EXAM
Energy is capacity to do work. It has many forms such as kinetic energy potential energy,
heat, work, chemical energy and so on.
System - is the part of the universe chosen for thermodynamic consideration.
Surroundings - The part of universe other than system is called surroundings.
Extensive property - depends on the amount of matter present in the system.
Intensive property - does not depend on the amount of the substance.
A state function is a property whose value depends on the present state of the system
and not on the path.
A reversible process is any process in which the system is always in temperaturepressure equilibrium with its surroundings.
Expansion (–W) → work is done by the system.
Compression (+W) → work is done on the system.
First law of thermodynamics - its mathematical expression is ΔU = q + W
Work = force × displacement, but in chemistry we deal with the PV work done, which
is due to volume change.
Chapter - 3 Chemical Thermodynamics And Energetics
204
Work done during expansion, W = – Pex ⋅ ΔV
V2
Maximum work done during expansion, Wmax = – 2.303 n RT log 10
V1
P1
Wmax = – 2.303 n RT log10
P2
Internal energy (U) : of a system is the sum of all kinetic and potential energies of
all the particles in the system.
Enthalpy (H) : of a system is defined as H = U + PV.
Change in enthalpy at constant pressure is given by ΔH = ΔU + P ⋅ ΔV
For reactions involving solids and liquids only, ΔH = ΔE
Work done in a chemical reaction is given by,
W = – Δn ⋅ RT
Enthalpy change (ΔH) in a chemical reaction is given by, ΔH = ΔU + RT ⋅ Δn
ΔH = ∑ H (products) – ∑ H (Reactants)
Hess’s law - the change in enthalpy for a reaction is the same whether the reaction
takes place in one or a series of steps.
Spontaneous process - A process which occurs on its own without external help.
Entropy - is a measure of molecular disorder or randomness in a system it determines
q rev
spontaneity of a process. ΔS =
T
Second law of thermodynamics the total entropy of the universe (system + surroundings)
increases in a spontaneous process. ∴ ΔStotal > 0
Change in Gibbs energy is given by, ΔG = ΔH – T.ΔS
ΔG is negative for a spontaneous change, positive for a non-spontaneous change and
Zero for equilibrium.
Gibbs energy change for a reaction is given by, ΔG = ΔGº + RT ln Q.
Third law of thermodynamics : the entropy of a pure perfectly ordered crystalline
substance is zero at 0 K.
º
º
The standard entropy of a reaction is given by, ΔSº = ∑ S (products) – ∑ S (Reactants)
Conversions :
1 L.atm = 101.32J = 1.013 × 109erg = 24.206 cal
1 cal
= 4.184 J and 1 J = 0.2390 Cal
1 atm = 1.01325 × 105 Pa (Nm–2)
1J
= 1 Nm = 1kg m2 s–2
1 Pa
= 1 Nm–2 = 1 kg m–1 s–2
1L
= 10–3 m3 = 1dm3 = 103 cm3
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
205
NUMERICALS WITH SOLUTION
Three moles of an ideal gas are expanded Solution :
isothermally from a volume of 300 cm3 to
W = –pex (V2 – V1)
2.5 L at 300 K against a pressure of 1.9
36.50 J = –1.216 × 105 Nm–2
atm. Calculate the work done in L atm and
× (V2 – 500 × 10–6 m3)
joules.
= –1.216 × 105 Nm–2 × V2 +
Given :
1.216 × 105 Nm–2 × 500 ×
n = 3 moles,
10–6 m3
3
V1 = 300 cm = 0.300 L,
= –1.216 × 105 × V2 + 60.8
V2 = 2.5 L
∴ 1.216 × 105 V2 = 60.8 – 36.50
T = 300 K,
24.3
V2 = 1.216 × 10 5
pex = 1.9 atm.
To find :
= 200 × 10–6m3 = 200cm3
Work done (W) = ?
∴ V2 = 200 cm3
(in L and J)
Solution :
*3. Calculate the maximum work when
W = –pex (V2 – V 1)
24 g of oxygen are expanded isothermally
= –1.9 atm (2.5 L – 0.3 L)
and reversibly from a pressure of
= –1.9 × 2.2 L atm.
1.6 × 105 Pa to 100 kPa at 298 K.
W = –4.18 L atm
Given :
Q 1 L atm = 101.3 J
W O 2 = 24g,
∴ W = –4.18 × 101.3
P 1 = 1.6 × 105 Pa
W = –423.4J
= 1.6 × 102 kPa
= 160 kPa
*2. One mole of an ideal gas is compressed
3
=
100 kPa
P
2
from 500 cm against a constant pressure
T = 298 K.
of 1.216 × 105 Pa. The work involved in
the process is 36.50 J. Calculate the final To find :
Wmax = ?
volume.
Solution
:
Given :
WO 2 24
n = 1 mole,
no. of moles (n) =
=
= 0.75 mol
32
M
V1 = 500 cm3 = 500 × 10–6m3
P1
pex = 1.216 × 105 Pa (Nm–2),
Wmax = –2.303 × n × R × T × log10 P
2
W = 36.50 J.
= –2.303 × 0.75mol × 8.314 JK–1
To find :
160 kPa
V2 = ?
mol–1 × 298 K ×log
*1.
10
Chapter - 3 Chemical Thermodynamics And Energetics
100 kPa
206
= –2.303 × 0.75 × 8.314 × 298 ×
0.2041
Wmax = –873.4 J
*4.
Three moles of an ideal gas are
compressed isothermally and reversibly
to a volume of 2L. The work done is
2.983kJ at 22ºC. Calculate the initial
volume of the gas.
Given :
n = 3 mol,
V2 = 2L,
Wmax = 2.983 kJ = 2983 J
T = 22ºC + 273 = 295K
To find :
V1 = ?
Solution :
Wmax = –2.303 × n × R × T × log10
V2
V1
Given :
W N 2 = 2.8 × 10–2 kg
T = 300 K
P1 = 15.15 × 105 Nm–2
Wmax = – 17.33 kJ = –17330 J
To find :
P2 = ?
Solution :
W
2.8 × 10 –2
n=
=
=1
M
28 × 10 –3
Wmax = –2.303 × n × R × T × log10
–17330 J = –2.303 × 1mol × 8.314 JK –1mol –1
×300K× log10
∴
∴
*5.
17330J
= 5744.1 × log10
= 1.500
V1 = 1.5 × 2L = 3.0 L
15.15 × 10 5
P2
17330
5744.1
= log10
15.15 × 10 5
P2
3.0170
= log10
15.15 × 10 5
P2
Al (3.0170) =
= AL (0.1760)
15.15 × 10 5
P2
15.15 × 10 5
1.040 × 10 =
P2
3
2.8 × 10–2 kg of nitrogen is expanded
isothermally and reversibly at 300K from
15.15 × 105 Nm–2 when the work done
is found to be –17.33 kJ. Find the final
pressure.
Unique Solutions ®
15.15×10 5 Nm –2
P2
⎡
15.15×10 5 N ⎤
– ⎢ 2303× 1×8.314×300×log10
⎥
= ⎢
P2
⎣
⎦⎥
2983 J = –2.303 × 3 mol × 8.314 JK–1
2L
mol–1 × 295K × log10
V1
V
2983 = + 16945.2 × log10 1
2L
2983
V1
log10
=
16945.2
2L
= 0.1760
V1
2L
P1
P2
15.15 × 10 5
∴
P 2=
∴
= 14.5673 × 102 Nm–2≈ 1456.7 Nm–2
P2= 1456.7 Nm –2
1.040 × 10 3
S.Y.J.C. Science - Chemistry - Part I
207
2SO 2 (g) + O 2 (g) ⎯⎯
→ 2SO 3 (g)
Decomposition of 2 moles of NH4NO3 at
100ºC.
∴
∴
= –8.314 (JK–1 mol–1)
× 373 (K) × (6 – 0) (mol)
= –18610 J
= –18.610 kJ
W = –18.61 kJ
Work is done by the system
NH 4 NO 3 (s) ⎯⎯
→ N 2 O(g) + 2H 2 O(g)
Given :
1
(a) one mole of SO2 reacts with mole of
2
O2 to form 1mole of SO3
Hence, n1 = 1.5 mol, n2 = 1mole,
R = 8.314 JK–1 mol–1
T = 50ºC + 273 = 323 K
To find :
W=?
Solution :
W = –Δn RT
= –RT (n2 – n1)
= –8.314 (JK–1mol–1) × 323 (K) × (1 – 1.5) (mol)
= –8.314 × 323 × (–0.5) J
∴ W = +1343 J
∴ Work is done on the system.
*7.
The enthalpy change for the reaction
*6.
(a)
(b)
Calculate the work done in each of the
following reactions. State whether work is
done on or by the system.
The oxidation of one mole of SO2 at 50ºC.
Given :
(b) 2 moles of NH4NO3(s) gives 2 moles of
N2O(g) and 4 moles of H2O(g)
Hence, n1 = 0,
n = 6,
R = 8.314 JK–1 mol–1
T = 100ºC + 273K = 373K.
To find :
W=?
Solution :
W = –RT (n2 – n1)
C 2 H 4 (g)+H 2 (g) ⎯⎯
→ C 2 H 6 (g) is –620J
when 100 mL of ethylene and 100 mL of
H2 react at 1 atm pressure. Calculate the
pressure-volume work and ΔU.
Given :
Initial volume (V1) = 100 mL + 100 mL
= 200 mL.
Final volume (V2) = 100 mL, Δ H = –620J
∴ ΔV = V2 – V1
= 100 – 200
= –100 mL
= –0.1 L
To find :
Work = ? and ΔU = ?
Solution :
W = –p ex ⋅ ΔV
= –1 atm × (–0.1L)
= +0.1 L atm.
But, 1 L atm = 101.3 J
∴
W = +0.1 (L.atm) × 101.3
J
L ⋅ atm
W = +10.13 J
We know, ΔH = ΔU + P ⋅ ΔV
or ΔU = ΔH – P ⋅ ΔV
= –620J – (–10.13 J)
ΔU = –609.87 J.
Chapter - 3 Chemical Thermodynamics And Energetics
208
*8.
∴
*10.
*9.
a)
b)
FIRST LAW OF THERMODYNAMICS
A sample of gas absorbs 4000 kJ of heat.
If volume remains constant, what is ΔU?
Suppose that in addition to absorption
of heat by the sample, the surroundings
does 2000 kJ of work on the sample,
what is ΔU?
c) Suppose that as the original sample absorbs
heat, it expands against atmospheric
pressure and does 600 kJ of work on its
surroundings. What is ΔU?
Given :
q = 4000KJ, W = 2000 KJ
W = 600KJ
W = –p ex ⋅ ΔV
To find :
ΔU = ?
W = –1atm × 22.4 L = –22.4 L –atm.
Solution
:
W = –22.4 × 101.3 J
i) ΔU = q – p ex ⋅ ΔV
(Q 1 atm = 101.3J)
∴ ΔV = 0,
= –2269.12 J
∴ ΔU = q = 4000 kJ.
= –2.269 kJ
ii) ΔU = q + W
W = –2.27 kJ
= 4000 kJ + 2000 kJ
ΔU = qp + W
ΔU=
6000 kJ
at constant pressure,
iii) ΔU = q + W
ΔHº = qp = –2043 kJ
= 4000 kJ + (–W)
ΔU = –2043 kJ + (–2.27 kJ)
(Q work is done on surroundings)
ΔU = –2045.27 kJ
= 4000 kJ – 600 kJ
ΔU= 3400 kJ.
Oxidation of propane is represented
as C 3H 8 (g)+ 5O 2 (g) ⎯⎯
→ 3CO 2 (g) + 4H 2O(g) ,
o
ΔH = –2043 kJ. How much pressure volume work is done and what is the value
of ΔU at constant pressure of 1 atm when
the volume change is +22.4 L.
Given :
ΔHº = –2043 kJ
P = 1atm
ΔV = +22.4L
To find :
ΔU = ?
Solution :
ENTHALPY CHANGE
→ 2Fe(s) + 3CO2(g) from
Calculate standard enthalpy of the reaction Fe2O3(s) + 3CO(g) ⎯⎯
the following data :
Δf Hº (Fe2O3) = –824.2 kJ mol–1, ΔfHº (CO) = –110.5 kJ mol–1, ΔfHº (CO2) = –393.5 kJ mol–1.
Given :
Δf Hº (Fe2O3) = –824.2 kJ mol–1,
Δf Hº (CO) = –110.5 kJ mol–1,
Δf Hº (CO2) = –393.5 kJ
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
209
To find :
ΔHº = ?
Solution :
ΔHº = ⎡⎣ ∑ m ⋅ Δ f Hº (Products) – ∑ n ⋅ Δ f Hº (reactants) ⎤⎦
º
ΔHº = ⎡⎣ 2 × Δ f H (Fe) + 3 × Δ f Hº (CO 2 ) ⎤⎦ – ⎡⎣1 × Δ f Hº (Fe 2 O 3 ) + 3 × Δ f Hº (CO) ⎤⎦
ΔHº = [2mol × 0 + 3 mol × (–393.5 kJ mol–1)] – [1 mol × (–824.2 kJ mol–1)
+ 3mol × (–110.5 kJ mol–1)]
= [–1180.5 kJ] – [–824.2 kJ – 331.5 kJ]
= [–1180.5 kJ] – [–1155.7 kJ]
ΔHº = –24.8 kJ.
*11.
Calculate the standard enthalpy of formation of C2H6 from the following data :
ΔHº = –3119 kJ
2C 2 H 6 (g) + 7O 2 (g) ⎯⎯
→ 4CO 2 (g) + 6H 2 O(l ) ,
–1
Δf Hº (CO2) = –393.5 kJ mol
Δf Hº (H2O) = –285.8 kJ mol–1
Given :
ΔHº = –3119 kJ, ΔfHº (CO2) = –393.5 kJ mol–1 , ΔfHº (H2O) = –285.8 kJ mol–1
To find :
f H (C 2 H 6 )
=?
Solution :
ΔHº = ⎡⎣ ∑ m ⋅ Δ f Hº (Products) – ∑ n ⋅ Δ f Hº (reactants) ⎤⎦
ΔHº = ⎡⎣ 4mol×Δ f H (CO 2 ) + 6mol×Δ f H (H 2O) ⎤⎦ – ⎡⎣ 2mol×Δ f H (C 2 H 6 ) + 7mol×Δ f H O 2 ⎤⎦
–1
–3119 kJ = ⎡⎣ 4(mol) × (–393.5) (kJ mol ) + 6(mol)
– ⎡⎣ 2(mol) × Δ f H (C 2 H 6 ) + 7(mol) × 0 ⎤⎦
–3119 kJ = [ –1574kJ + (–1714.8kJ) ] – ⎡⎣ 2(mol) × Δ f H (C 2 H 6 ) ⎤⎦
∴
2(mol) × Δ f H (C 2H 6 ) = –3288.8 kJ + 3119 kJ
= –169.8 kJ
∴
169.8 (kJ)
Δ f H (C 2 H 6 ) = –
2 (mol)
Δ f H (C 2 H 6 ) = –84.9 kJmol–1.
Chapter - 3 Chemical Thermodynamics And Energetics
210
*12.
How much heat is evolved when 12g of
CO react with NO2 according to the
following
reaction,
2NO 2 (g) ⎯⎯
→ 4CO 2 (g) + N 2 (g)
hence, calculate the bond enthalpy of
O–H bond in H2O from the following data :
ΔvapH (H2O) = 44.0 kJ mol–1
=
38.55 kJ of heat is absorbed when 6.0g
of O2 react with CIF according to the
Calculate the enthalpy change for the
reaction H 2 O(g) ⎯⎯
→ 2H(g) + O(g) and
4CO (g) +
ΔHº = –1198 kJ.
Given :
Mol. Wt of CO = 12 + 16 = 28
4CO = 4 × 28 = 112g.
To find :
Heat evolved = ?
Solution :
When 112 g of CO reacts with NO2,
Heat Consumed = –1198 kJ
∴ 12g of CO reacts with NO2 heat
–12 × 1198
consumed =
112
ΔH = –128.4 kJ.
∴ Heat evolved = 128.4 kJ.
*13.
*14.
ΔfH (H2O) = –285.8 kJ mol–1
ΔaH (H2) = 436.0 kJ mol–1, ΔaH (O2)
= 498.0 kJ mol –1 where Δ aH is the
enthalpy of atomization.
Given :
i)
H 2 O(l ) ⎯⎯
→ H 2 O(g) ;
ΔvapH = +44 kJ mol–1
ii)
iii)
1
H 2 (g) + O 2 (g) ⎯⎯
→ H 2O(l )
2
ΔfH = –285.8 kJ mol–1
H 2 (g) ⎯⎯
→ 2H(g)
ΔaHH2 = 436 kJ mol–1
iv)
1
O (g) ⎯⎯
→ O(g)
2 2
Δ a H O 2 = 249 kJmol–1
reaction, 2CIF(g) + O2(g) ⎯⎯
→
Cl 2 O(g) + OF2 (g) . What is the standard
enthalpy of the reaction?
Given :
ΔHº = 38.55KJ
To find :
Heat abosrbed = ?
Solution :
When 6g of O2 reacts with CIF,
heat absorbed = 38.55 kJ.
∴ 32g of O2 reacts with ClF,
32 × 38.55
heat absorbed =
= +205.6 kJ
6
∴ Δ Hº = + 205.6 kJ.
To find :
ΔH = ?
and Bond enthalpy of O – H bond = ?
Solution :
Treatment : Reverse eq. (i) + Reverse eq.
(ii) + eq. (iii) + eq. (iv)
H 2O(g) ⎯⎯
→ H 2O(l ) ;
ΔvapH = –44KJ mol–1
H 2O(l ) ⎯⎯
→ H 2 (g) +
1
O (g) ;
2 2
ΔfH = +285.8 KJ mol–1
H 2 (g) ⎯⎯
→ 2H(g) ;
aH H 2 = 436.0KJ mol–1
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
211
1
O (g) ⎯⎯
→ O(g) ;
2 2
aH O 2 = 249.0kJ mol–1
Solution :
180g
heat
heat
ice ⎯⎯⎯⎯
→ liquid ⎯⎯⎯
→ vapours
(m.c.Δt)
H 2 O(g) ⎯⎯
→ 2H(g) + O(g) ;
∴
ΔH = (–44.0 + 285.8 + 436.0 +
249.0)KJ mol–1
ΔH = 926.8k]
H–O–H molecule contains two (O–H)
bonds
926.8
Bond enthalpy of O–H bonds =
=
2
463.4 kJ mol–1 ≈ Bond enthalpy of
O – H bond = 4634 kJmol–1 in H2O.
0ºC
100ºC
100ºC
180 × 4.18 × 100
1000
= 75.24 kJ ...(iii)
Total heat required
60.1 kJ + 407 kJ + 75.24 kJ
542.34 kJ.
Total heat required = 542.34 kJ.
q = m.c. Δt =
∴
=
=
*16.
6.24 g of ethanol are vaporized by supplying
*15. Calculate the total heat required to melt
5.89 kJ of heat energy. What is enthalpy
180 g of ice at 0ºC, heat it to 100ºC and
of vaporization of ethanol?
then vaporize it at that temperature. Given :
Δ fusH(ice) = 6.01 kJ mol –1 at 0ºC,
ΔHº = 5.89kJ (Heat of vaporisation),
–1
ΔvapH(H2O) = 40.7 kJ mol at 100ºC.
mass of ethanol = 6.24g
–1 –1
Specific heat of water = 4.18 J g K . To find :
Given :
ΔvapH = ?
(i) number of moles of (ice) H2O(s)
Solution :
gram molecular weight of C2H5OH
Wt
180
=
=
= 10 moles.
= 2 × 12 + 1 × 5 + 16 + 1 = 46g
Mol.Wt.
18
∴ 6.24 g ethanol are vaporised
(ii) ΔfusH (ice) = 6.01 kJ–1 at 0ºC
= 5.89 kJ of heat energy
∴ For 10 moles, ΔfusH ice = 60.1 J at 0ºC
∴ 46g of ethanol are vaporised
...(i)
46 × 5.89
(iii) ΔvapH (H2O) = 40.7 kJ mol–1 at 100º C
=
6.24
∴ For 10 moles, ΔvapH(H2O) = 407 kJ
...(ii)
= 43.5 kJ mol–1
(iv) Heat absorbed to raise the temperature
46g of ethanol are vaporised
from 0ºC to 100ºC
= 43.5 kJ mol –1.
–1
–1
sp. heat of water = 4.18 Jg K
To find :
Total heat required = ?
Chapter - 3 Chemical Thermodynamics And Energetics
212
Enthalpy of fusion of ice is 6.01 kJ mol–1. The enthalpy of vaporization of water is
45.07 kJ mol–1. What is enthalpy of sublimation of ice?
Given :
ΔvapH = 45.07 kJ mol–1
ΔfusionH = +6.01kJmol–1
To find :
ΔsubH = ?
Solution :
*17.
∴
H 2 O(s) ⎯⎯
→ H 2 O(l ) ;
ΔfusH = +6.01 kJ mol–1
H 2 O(l ) ⎯⎯
→ H 2 O(g) ;
ΔvapH = +45.07 kJ mol–1
H 2 O(s) ⎯⎯
→ H 2 O(g) ;
ΔsubH = (+6.01 + 45.07) kJmol–1
Δ subH = 51.08 kJ mol –1.
*18.
BOND ENERGY
Calculate ΔHº of the reaction CH 4 (g) + O 2 (g) ⎯⎯
→ CH 2 O (g) + H 2 O(g) from the following
data :
Bond
C–H
O=O
C=O
O–H
–1
414
499
745
464
ΔHº/kJ mol
Given :
ΔHº(O – H) = 464 kJmol–1
ΔHº(C – H) = 414kJmol–1,
ΔH (O – O) = 499 kJmol–1,
ΔHº(C = O) = 745 kJmol–1
To find :
ΔHº of the reaction = ?
Solution :
H –
H
H
|
C
|
C = O + H
– H + O = O
⎯⎯
→ H
–
– O –
H
|
H
ΔHº = ∑ ΔHº (reactant bonds) – ∑ ΔHº (product bonds)
∴
ΔHº = [4(mol) × ΔHº(C – H) + 1(mol) × ΔHº(O = O)] – [2(mol) × ΔHº(C – H)
+ 1(mol) × ΔHº(C = O) + 2 (mol) × ΔHº(O – H)]
ΔHº = [4 (mol) × 414 (kJ mol–1) + 1 (mol) × 499 (kJ mol–1)] –
[2 (mol) × 414 (kJmol–1) + 1 (mol) × 745 (kJmol–1) + 2 (mol) × 464 (kJmol–1)]
ΔHº = [1656 kJ + 499 kJ] – [828 kJ + 745 kJ + 928 kJ]
ΔHº = 2155 kJ – 2501 kJ
= –346 kJ
Δ Hº = –346 kJ.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
213
*19.
Calculate C–Cl bond enthalpy from the following data :
CH 3Cl (g) + Cl 2 (g) ⎯⎯
→ CH 2 Cl 2 (g) + HCl(g) ΔHº = –104 kJ
Bond
C–H
–1
ΔHº/kJ mol
414
Given :
ΔHº(C – H) = 414kJmol–1,
Cl–Cl
243
H–Cl
431
ΔHº(Cl – Cl) = 243 kJmol–1
ΔHº (H – Cl) = 431 kJmol–1
To find :
ΔHº(C – Cl) = ?
H
H
|
|
H – C – Cl + Cl – Cl ⎯⎯
→ H – C – Cl + H – Cl
Solution :
|
|
H
Cl
ΔHº = ∑ ΔHº (reactant bonds) – ∑ ΔHº (Product bonds)
º
º
º
ΔHº = ⎡⎣ 3(mol) × ΔH (C–H) + 1(mol) × ΔH (C–Cl) + 1(mol) × ΔH (Cl–Cl) ⎤⎦ –
º
º
º
⎡ 2(mol) × ΔH (C
⎤
– H) + 2(mol) × ΔH (C – Cl) + 1(mol) × ΔH (H – Cl) ⎦
⎣
–1
º
–1
ΔHº = ⎡⎣ 3(mol) × 414 (kJmol ) + 1(mol) × ΔH (C–Cl) + 1(mol) × 243(kJmol )⎤⎦
–1
º
–1
– ⎡⎣ 2(mol) × 414(kJmol ) + 2(mol) × ΔH (C–Cl) + 1(mol) × 431(kJmol )⎤⎦
–104 kJ
º
º
= ⎡⎣1242kJ + (1mol) × ΔH (C–Cl) + 243kJ ⎤⎦ – ⎡⎣828kJ + 2(mol) × ΔH C–Cl + 431kJ ⎤⎦
–104 kJ
º
º
= ⎡⎣1485kJ + 1(mol) × ΔH (C–Cl) ⎤⎦ – ⎡⎣1259kJ + 2(mol) × ΔH (C–Cl) ⎤⎦
º
= 1485kJ + 1(mol) × ΔH º(C–Cl) – 1259kJ – 2(mol) × ΔH (C–Cl)
–104 kJ
º
= 226kJ – 1(mol) × ΔH (C–Cl)
or 1(mol) × ΔH ºC–Cl = 226 kJ + 104 kJ
–104 kJ
∴
º
= 330 kJmol–1.
Δ H C–Cl
PROBLEMS (HESS’S LAW)
*20.
Calculate the standard enthalpy of the reaction 2C (graphite) + 3H 2 (g) ⎯⎯
→ C 2 H 6 (g) from
the following ΔHº values :
a)
C 2 H 6 (g) + 7/2O 2 (g) ⎯⎯
→ 2CO 2 (g) +3H 2 O (l ) ,
ΔHº = –1560 kJ
b)
H 2 (g) +1/2O 2 (g) ⎯⎯
→ H 2 O (l ) ;
ΔHº = –285.8 kJ
Chapter - 3 Chemical Thermodynamics And Energetics
214
c)
ΔHº = –393.5 kJ
C(graphite) + O 2 (g) ⎯⎯
→ CO 2 (g) ,
Given :
i)
C 2 H 6 (g) +7/2O 2 (g) ⎯⎯
→ 2CO 2 (g)+3H 2O (l ) ,
ΔHº = –1560 kJ
ii)
H 2 (g) + 1/2O 2 (g) ⎯⎯
→ H 2 O(l ) ,
ΔHº = –285.8 kJ
ΔHº = –393.5 kJ
iii) C(graphite) + O 2 (g) ⎯⎯
→ CO 2 (g) ,
To find :
ΔHº(reaction) = ?
Solution :
→ C 2 H 6 (g) ; ΔHº = ?
Required equation : 2C (graphite) + 3H 2 (g) ⎯⎯
Treatment : Reverse eq. (i) + 3 × eq. (ii) + 2 × eq. (iii)
7
2CO 2 (g) + 3H 2O(l ) ⎯⎯
→ C 2 H 6 + O 2 (g) , ΔHº = + 1560 kJ
2
3
3H 2 (g) + O 2 (g) ⎯⎯
→ 3H 2 O(l ) ,
ΔHº = –857.4 kJ
2
2C(graphite) + 2O 2 (g) ⎯⎯
→ 2CO 2 (g) ,
ΔHº = –787.0 kJ
2C(graphite) + 3H 2 (g) ⎯⎯
→ C 2 H 6 (g) , ΔHº = +1560 kJ – 857.4 kJ – 787 kJ
Δ Hº = –84.4 kJ.
*21.
Given the following equations calculate the standard enthalpy of the reaction,
2Fe(s) + 3/ 2 O 2 (g) ⎯⎯
→ Fe 2O 3 (s) ,
ΔHº = ?
a)
2Al(s) + Fe 2O 3 (s) ⎯⎯
→ 2Fe(s) + Al 2 O 3 (s) , ΔHº = –847.6 kJ
b)
2Al(s) + 3/2 O 2 (g) ⎯⎯
→ Al 2 O 3 (s) ,
ΔHº = –1670 kJ
Given :
i)
2Al(s) + Fe 2O 3 (s) ⎯⎯
→ 2Fe(s) + Al 2 O 3 (s) , ΔHº = –847.6 kJ
ii) 2Al(s) + 3/2 O 2 (g) ⎯⎯
→ Al 2 O 3 (s) ,
To find :
ΔHº(reaction) = ?
Solution :
Required equation : 2Fe(s) +
ΔHº = –1670 kJ
3
O (g) ⎯⎯
→ Fe 2O 3 (s)
2 2
Treatment : Reverse eq. (i) + eq. (ii)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
215
2Fe(s) + Al 2O 3 (s) ⎯⎯
→ 2Al(s) + Fe 2 O 3 (s)
3
2Al(s) + O 2 (g) ⎯⎯
→ Al 2O 3 (s)
2
3
2Fe(s) + O 2 (g) ⎯⎯
→ Fe 2 O 3 (s)
2
*22.
a)
;
ΔHº = +847.6 kJ
;
ΔHº = –1670 kJ
;
ΔHº = +847.6 kJ – 1670 kJ
Δ Hº = –822.4 kJ
Given the following equations and ΔHº values at 25ºC,
ΔHº = –911 kJ
Si(s) + O 2 (g) ⎯⎯
→ SiO 2 (s) ;
b)
2C(graphite) + O 2 (g) ⎯⎯
→ 2CO(g) ;
ΔHº = –221 kJ
c)
Si(s) + C(graphite) ⎯⎯
→ SiC(s) ;
ΔHº = –65.3 kJ
Calculate ΔHº for the following reaction, SiO 2 (s) + 3C(graphite) ⎯⎯
→ SiC(s) + 2CO(g) .
Given :
i)
Si(s) + O 2 (g) ⎯⎯
→ SiO 2 (s) ;
ΔHº = –911 kJ
ii)
2C(graphite) + O 2 (g) ⎯⎯
→ 2CO(g) ;
ΔHº = –221 kJ
iii) Si(s) + C(graphite) ⎯⎯
→ SiC(s) ;
To find :
ΔHº(reaction) = ?
Solution :
ΔHº = –65.3 kJ
→ SiC(s) + 2CO (g)
Required equation : SiO 2 (s) + 3C(graphite) ⎯⎯
Treatment : Reverse eq. (i) + eq. (ii) + eq. (iii)
∴
*23.
SiO 2 (s) ⎯⎯
→ Si(s) + O 2 (g) ;
ΔHº = +911 kJ
2C(graphite) + O 2 (g) ⎯⎯
→ 2CO(g) ;
ΔHº = –221 kJ
Si(s) + C(graphite) ⎯⎯
→ SiC(s) ;
ΔHº = –65.3 kJ
SiO 2 (s) + 3C(graphite) ⎯⎯
→ SiC(s) + 2CO(g) ;
ΔHº = +911 kJ –221 kJ –65.3 kJ
Δ Hº = +624.7 kJ
Given the following equations and ΔHº values at 25ºC,
a)
2H 3BO 3 (aq) ⎯⎯
→ B 2 O 3 (s) + 3H 2O(l ) ;
ΔHº = +14.4 kJ
b)
H 3BO 3 (aq) ⎯⎯
→ HBO 2 (aq) + H 2O(l ) ;
ΔHº = –0.02 kJ
c)
H 2 B 4 O 7 (s) ⎯⎯
→ 2B 2 O 3 (s) + H 2 O (l ) ;
ΔHº = 17.3 kJ calculate ΔHº for the following
reaction H 2 B 4 O 7 (s) + H 2O (l ) ⎯⎯
→ 4HBO 2 (aq)
Chapter - 3 Chemical Thermodynamics And Energetics
216
Given :
a)
2H 3BO 3 (aq) ⎯⎯
→ B 2 O 3 (s) + 3H 2O(l )
;
ΔHº = +14.4 kJ
b)
H 3BO 3 (aq) ⎯⎯
→ HBO 2 (aq) + H 2 O (l )
;
ΔHº = –0.02 kJ
;
ΔHº = 17.3 kJ
c) H 2 B 4 O 7 (s) ⎯⎯
→ 2B 2O 3 (s) + H 2O(l )
To find :
ΔHº(reaction) = ?
Solution :
Required equation : H 2 B 4 O 7 (s) + H 2O(l ) ⎯⎯
→ 4HBO 2 (aq)
Treatment : eq (iii) + 2 × reverse eq (i) + 4 ×eq (ii)
H 2 B 4O 7 (S) ⎯⎯
→ 2B 2O 3 (s) + H 2 O(l )
; ΔHº = 17.3 kJ
2B 2O 3 (s) + 6H 2 O(l ) ⎯⎯
→ 4H 3BO 3 (aq)
; ΔHº = –28.8 kJ
4 H 3BO 3 (aq) ⎯⎯
→ 4HBO 2 (aq) + 4H 2O(l ) ; ΔHº = –0.08 kJ
H 2 B 4 O 7 (s) + H 2O(l ) ⎯⎯
→ 4HBO 2 (aq)
Δ Hº = –11.58 kJ
*24.
; ΔHº = +17.3 kJ – 28.8 kJ – 0.08 kJ
ENTROPY
Calculate ΔStotal and hence, show whether the following reaction is spontaneous at
→ Hg(l ) + SO 2 (g) ;ΔHº = –238.6 kJ and ΔSº = +36.7 JK–1
25ºC. HgS (s) + O 2 (g) ⎯⎯
Given :
ΔHº = –238.6 kJ ΔSº = +36.7 JK–1
To find :
ΔS–total = ?
Solution :
The heat evolved in the reaction is 238.6 kJ. The same quantity of heat is absorbed by the
surroundings. Hence, entropy change of the surroundings is given by.
–(238.6) (kJ)
ΔHº
ΔSsurr = –
=
Log calculations
298 (K)
T
2.3777
ΔSsurr = +0.8007 kJ K–1
–1
–2.4742
= + 800.7 JK
ΔStotal =
ΔSsys =
ΔStotal =
ΔStotal =
Δ Stotal is
1.9035
ΔSsys + ΔSsurr
AL
–1
ΔSº = +36.7 JK
0.8007
36.7 (JK–1) + 800.7 (JK–1)
837.4 JK–1
positive, the reaction is spontaneous at 298 K.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
217
Determine whether the reactions with the
following ΔH and ΔS values are
spontaneous or nonspontaneous. State
whether they are exothermic or
endothermic.
i) ΔH = –110 kJ and ΔS = +40 JK–1 at
400 K.
ii) ΔH = +50 kJ and ΔS = –130 JK–1 at
250 K.
Given :
i) ΔH = –110 kJ and ΔS = +40 JK–1 at
400 K
ii) ΔH = +50 kJ and ΔS = –130 JK–1 at
250 K.
To find :
ΔG = ?
Solution :
For a certain reaction, ΔHº = –224 kJ and
ΔSº = –153 JK–1. At What temperature
will it change from spontaneous to
nonspontaneous? (Spontaneous below
1464 K.)
Given :
ΔHº = 224 kJ,
ΔSº = 153 JK–1
To find :
T=?
Solution :
The temperature at which the reaction
change from spontaneous to
nonspontaneous is given by,
ΔG = ΔH – T ⋅ ΔS
= –110 kJ – (400 K × 0.04 kJk–1)
= –110 – 16.0
= –126 kJ
ΔG = –126 kJ., spontaneous and
exothermic
= 1.4640 × 103 K
= 1464 K
As both Δ Hº and Δ
ΔSº are negative,
therefore, the reaction will be
spontaneous below 1464 K.
*25.
i)
Given :
ΔH = +50 kJ, ΔS = –130 JK–1 = –0.130
kJk–1, T = 250 K
To find :
ΔG = ?
Solution :
ii)
ΔG = ΔH – T ⋅ ΔS
= 50 kJ – (250k × (–0.130 kJ K–1)
= 50 kJ + 32.50 kJ.
= 82.50 kJ
Δ G = +82.5 kJ, nonspontaneous
and endothermic.
*26.
T=
*27.
ΔHº
=
ΔSº
– 224 (kJ)
–153 × 10 –3 (kJ K –1 )
Determine whether the following reaction
will be spontaneous or nonspontaneous
under standard conditions.
Zn(s) + Cu 2 + (aq) ⎯⎯
→ Zn 2+ (aq) + Cu(s)
ΔHº = –219kJ, ΔSº = –21JK –1 .
(spontaneous)
Given :
ΔHº = –219kJ, ΔSº = –21JK–1
To find :
T=?
Solution :
T =
ΔHº
ΔSº
–219 (kJ)
= –21 × 10 –3 (kJ K –1 )
Chapter - 3 Chemical Thermodynamics And Energetics
218
= 10.428 × 103 K
= 10428 K
As both Δ Hº and Δ Sº are negative
therefore, the reaction will be
spontaneous below 10428K.
Determine whether the following reaction
is spontaneous under standard conditions.
2H 2 O(l ) + O 2 (g) ⎯⎯
→ 2H 2 O 2 (l ) ΔHº
= +196 kJ, ΔSº = –126 JK–1
Does it have a cross-over temperature?
Given :
ΔHº = +196 kJ, ΔSº = –126o JK–1
To find :
State the reaction condition = ?
Solution :
As ΔHº is positive and ΔSº is negative
therefore, the reaction will be
nonspontaneous at all temperatures.
No, it has no cross over temperature.
=
∴
*28.
*29.
What is the value of ΔSsurr for the following
reaction at 298 K?
→ C6H12O6(s) +
6CO2(g) + 6H2O(l) ⎯⎯
6O2(g), ΔGº = 2879 kJ mol–1, ΔSº = –210
JK–1 mol–1.
Given :
ΔGº = 2879 kJmol–1,
ΔSº = –210 JK–1 mol–1
= –0.210 kJ K–1mol–1
T
= 298 K
To find :
ΔSsurr = ?
Solution :
ΔGº = ΔHº – T ⋅ ΔSº
2879 kJmol–1
= ΔHº – 298 K × (–0.210kJ K–1 mol–1)
Unique Solutions ®
ΔHº + 62.58 kJ mol–1
ΔHº = (2879 – 62.58) kJ = 2816.42 kJ.
ΔSsurr = –
ΔHº
T
ΔSsurr = –
2816.42
= –9.45 kJ K–1
298
Δ Ssurr = –9.45 kJ K –1
Calculate ΔSsurr when one mole of methanol
(CH3OH) is formed from its elements
under standard conditions if Δ f Hº
(CH3OH) = –238.9 kJmol–1
Given :
The heat evolved in the reaction is 23.89
kJ. The same quantity of heat is absorbed
by the surroundings. Hence, entropy
change for the surroundings is given by
T = 25ºC = 289K
To find :
ΔSsurr = ?
Solution :
*30.
ΔSsurr = –
–238.9 (kJ)
ΔHº
=–
298 (K)
T
= 0.8017 kJ K–1
= 801.7 JK–1
Δ Ssurr = 801.7 JK–1
*31.
GIBBS FREE ENERGY
Calculate K p for the reaction,
C 2 H 4 (g) + H 2 (g) ⎯⎯
→ C 2 H 6 (g) , Δ Gº
= –100 kJ mol–1 at 25ºC.
Given :
ΔGº = –100kJmol–1 = –100000 Jmol–1
T = 25ºC + 273K = 298K
S.Y.J.C. Science - Chemistry - Part I
219
R = 8.314 JK–1 mol–1
To find :
Kp = ?
Solution :
ΔGº = –2.303 RT log10 Kp
*33.
CO(g) + 2H 2 (g) ⎯⎯
→ CH 3OH(g) ,
ΔGº = –24.8 kJ mol–1 if Pco = 4 atm, p H 2
= 2 atm, p CH 3OH = 2 atm
–ΔGº
log10 Kp =
2.303RT
=–
∴
*32.
Kp
Kp
Given :
ΔGº = –24.8 kJmol–1, Pco = 4 atm,
–100000 Jmol –1
–1
2.303×8.314(JK mol
log10 kp =
Calculate ΔG for the reaction at 25ºC,
–1
)×298K
100000
5705.8
= 17.5260
= AL.[17.52]
= 3.356 × 1017
Kp for the reaction, MgCO3(s) ⎯⎯
→
–10
MgO(s) + CO2(g) is 9 × 10 . Calculate
ΔGº for the reaction at 25ºC.
Given :
Kp = 9 × 10–10,
T = 25ºC + 273K = 298 K.
To find :
ΔG = ?
Solution :
ΔGº = –2.303 RT + log10 Kp
= –2.303 × 8.314 JK–1 mol–1
× 298 K × log10(9 × 10–10)
= –2.303 × 8.314 JK–1 mol–1
× 298 K × [–10 + 0.9542]
= –2.303 × 8.314 JK–1 mol–1
× 298 K × [–9.0458]
= +2.303 × 8.314 × 298 × 9.0458 J.
= 51613 J
ΔGº= 51.613 kJ.
p H 2 = 2 atm, p CH 3OH = 2 atm, T = 25ºC
+ 273K = 298K, R = 8.314 JK–1mol–1
To find :
ΔG = ?
Solution :
QP =
=
PCH 3OH
PCO ×
PH2 2
=
2
4 × 22
2
1
2
=
=
16
8
4×4
ΔG = ΔGº + 2.303 RT log10 Qp
ΔG = –24.8 (kJmol–1) + 2.303
× 8.314 (JK–1mol–1)
1
× 298 (K) × log10
8
–1
ΔG = –24.8 (kJmol ) – 2.303
× 8.314 (JK–1mol–1)
× 298 (K) × log108
ΔG = –24.8 × 103Jmol–1
– [2.303 × 8.314 (JK –1mol–1)
× 298(K) × 0.9031
ΔG = –24.8 kJmol–1 – 5.15 × 103 kJmol–1
ΔG = –29.95 × 103kJmol–1
= –29.95 kJmol–1
ΔG = –29.95 kJmol–1.
Chapter - 3 Chemical Thermodynamics And Energetics
220
NUMERICALS FOR PRACTICE
*1.
Two moles of an ideal gas are expanded isothermally from a volume of 15.5 L to the volume
of 20 L against a constant external pressure of 1 atm. Estimate the amount of work in
L atm and J.
(Ans. : W = –4.5 L. atm and –455.8 J)
*2.
Calculate the constant external pressure required to compress one mole of an ideal gas from
a volume of 20 L to 8 L when the work obtained is 44.9 L atm.
(Ans. : Pex = 3.74 atm)
*3.
100 mL of ethylene (g) and 100 mL of HCl(g) are allowed to react at 2 atm pressure according
to the reaction, C 2 H 4 (g) + HCl(g) ⎯⎯
→ C 2 H 5Cl(g). Calculate the pressure-volume work
(Ans. : W = 20.26J)
in joules.
*4.
3 moles of an ideal gas are expanded isothermally and reversibly from 10 m3 to 20 m3 at 300
K. Calculate the work done. (R = 8.314Jk–1mol–1)
(Ans. : –5.187kJ)
*5.
4.4 × 10–2 kg of CO2 are compressed isothermally and reversibly at 293 K from the initial
pressure of 150 kPa when the work obtained is 1.245 kJ. Find the final pressure.
(Ans. : final pressure 250kPa)
(R = 8.314Jk–1mol–1)
*6.
280 mmol of a perfect gas occupies 12.7 L at 310 K. Calculate the work done when the gas
expands (a) isothermally against a constant external pressure of 0.25 atm (b) isothermally and
reversibly (c) into vacuum, until its volume has increased by 3.3 L.
(Ans. : (a) W = –83.6J, (b) Wmax –166.9J, (c) W = 0)
*7.
For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on its
surroundings. What are the changes in internal energy and enthalpy of the system?
(Ans. : Internal energy 4.5kJ and enthalpy 6 kJ)
*8.
In a particular reaction, 1 kJ of heat is released from the system and 4 kJ of work is done
on the system. What are the changes in internal energy and enthalpy of the system?
(Ans. : Internal energy = > 3kJ and enthalpy = < 1 kJ)
*9.
An ideal gas expands from a volume of 6 dm3 to 16 dm3 against a constant external pressure
(Ans. : ΔH = qp2444J)
of 2.026 × 105 N m–2. Find ΔH if ΔU is 418 J.
*10.
Calculate the work done in the following reaction when 2 moles of HCl are used at constant
pressure and 423 K. 4HCl(g) + O 2 (g) ⎯⎯
→ 2Cl 2 (g) + 2H 2 O (g) . State whether work is done
on the system or by the system. (Ans. : W is positive, work done on the system is 1758J)
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*11.
CO reacts with O2 according to the reaction 2CO (g) + O 2 (g) ⎯⎯
→ 2CO 2 (g) ΔH = –566kJ
How much pressure volume work is done and what is the value of ΔU for the reaction of
7.0g of CO at 1 atm pressure, if the volume change is –2.8L?
(Ans. : –70.47kJ)
*12.
Calculate standard enthalpy of formation of benzene from the following data :
15
C 6 H 6 (l) + O 2 (g) ⎯⎯
→ 6CO 2 (g) + 3H 2O(l),
ΔHº = ––3267 kJ
2
Δf Hº(CO2) = –393.5 kJ mol–1,
Δf Hº(H2O) = –285.8 kJ mol–1
(Ans. : ΔformHº 48.6 kJ mol–1)
*13.
Estimate the standard enthalpy of combustion of acetylene from the following data :
Δf Hº(CO2) = –393.5 kJ mol–1,
Δf Hº(H2O) = –285.8 kJ mol–1
Δf Hº(C2H2) = 227.3 kJ mol–1
(Ans. : ΔHº = –1300 kJ; ΔcHº = –1300 kJ/ mol)
*14.
Calculate the standard enthalpy of the reaction,
4CO (g) + 2NO 2 (g) ⎯⎯
→ 4CO 2 (g) + N 2 (g) from the following data.
ΔfHº(CO) = –110.5 kJ mol–1,
ΔfHº(CO2) = –393.5 kJ mol–1
*15.
ΔfHº(NO2) = 33.2 kJ mol–1
(Ans. : ΔHº –1198.4kJ)
→ CO 2 (g) + H 2 O (g) , ΔH = –527 kJ
Given the reaction, CH 2 O (g) + O 2 (g) ⎯⎯
How much heat will be evolved in the formation of 30 g of CO2?
(Ans. : Heat evolved = 359.4 kJ)
*16.
Calculate the enthalpy change for the reaction CH 4 (g) + Cl 2 (g) ⎯⎯
→ CH 3Cl(g) + HCl(g)
The bond enthalpies are :
Bond
C–H
Cl – Cl
C – Cl
H – Cl
ΔHº/kJ mol–1
414
243
330
431
(Ans. : ΔHº –104kJ)
*17.
Calculate the N – N bond enthalpy in the reaction :
N 2 H 4 (g) + H 2 (g) ⎯⎯
→ 2NH 3 (g) , ΔHº = –184 kJ if ΔHº(N–H) = 389 kJ mol–1,
(Ans. : N – N bond enthalpy is 159 kJ/mol–1)
ΔHº (H – H) = 435 kJ mol–1
*18.
Calculate the standard N – H bond enthalpy from the following data :
N 2 (g) + 3H 2 (g) ⎯⎯
→ 2NH 3 (g) , ΔHº = –83 kJ
ΔHº (N ≡ N) = 946 kJ mol–1, ΔHº (H – H) = 435 kJ mol–1
(Ans. : ΔHº(N – H) 389 kJ/mol–1)
Chapter - 3 Chemical Thermodynamics And Energetics
222
*19.
Calculate ΔHº for the reaction 2ClF(g) + O 2 (g) ⎯⎯
→ Cl 2 O (g) + OF2 (g) from the following
data :
(a) F2 (g) + ClF(g) ⎯⎯
→ ClF3 (l )
ΔHº = 139.2 kJ
(b) 2ClF3 (l) + 2O 2 (g) ⎯⎯
→ Cl 2O(g) + 3OF2 (g)
ΔHº = + 533.4 kJ
1
→ OF2 (g)
(c) F2 (g) + O 2 (g) ⎯⎯
2
ΔHº = 24.7 kJ
(Ans. : ΔHº = +205.6 kJ)
*20.
Calculate ΔHº for the reaction between ethene with water to form ethanol from the following
data :
(a) C 2 H 5OH (l) + 3O 2 (g) ⎯⎯
→ 2CO 2 (g) + 3H 2 O(l)
ΔHº = –1368 kJ
(b) C 2 H 4 (g) + 3O 2 (g) ⎯⎯
→ 2CO 2 (g) + 2H 2 O (l )
ΔHº = –1410 kJ
Is ΔHº calculated the enthalpy of formation of liquid ethanol?
(Ans. : ΔHº = – 42 kJ, No it is not)
*21.
Calculate the standard enthalpy of formation of CH3COOH (l) from the following data :
ΔfHº (CO2) = –393.3 kJ mol–1,
ΔfHº (H2O) = –285.8 kJ mol–1
ΔfHº (CH3COOH) = –875 kJ mol–1
(Ans. : ΔHº – 483.2 kJ mol–1)
*22.
For a certain reaction ΔH = –25 kJ and ΔS = –40 JK–1. At what temperature will it change
from spontaneous to nonspontaneous?
(Ans. : T = 625 K)
*23.
Determine ΔStotal and decide whether the following reaction is spontaneous at 298 K.
Fe 2O 3 (s) + 3CO(g) ⎯⎯
→ 2Fe(s) + 3CO 2 (g) ,
ΔHº = –24.8 kJ, ΔSº = 15 JK–1
(Ans. : ΔStotal = 98.2 JK –1)
*24.
Determine whether the reaction, N 2 O 4 (g) ⎯⎯
→ 2NO 2 (g) is spontaneous at 298 K from
the following data, ΔfHº = (N2O4) = 9.16 kJ mol–1, ΔfHº (NO2) = 32.2 kJ mol–1
ΔSº = 175.8 J K–1. At what temperature will the reaction become spontaneous?
(Ans. : T = 325.6 K)
*25.
Kp for the reaction, 2SO 2 (g) + O 2 (g) ⎯⎯
→ 2SO 3 (g) is 7.1 × 1024 at 298 K. Calculate ΔGº
(Ans. : ΔGº –141.8 KJmol–1)
for the reaction. (R = 8.314 J K–1 mol–1)
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223
*26.
Calculate kp for the reaction at 513 K, 2NOCl(g) ⎯⎯
→ 2NO(g) + Cl 2 (g) , ΔGº = 17.38
(Ans. : kp = 0.017)
kJ mol–1.
*27.
Calculate ΔG for the reaction at 298 K, 2SO 2 (g) + O 2 (g) ⎯⎯
→ 2SO 3 (g) , ΔGº = –142 kJ
mol–1 if partial pressures of SO2, O2 and SO3 are 50 atm, 75 atm and 2 atm respectively
at 298 K.
(Ans. : qp = 2.13 × 10–5, ΔG – 168.7 kJ mol–1)
1.
Ans.
2.
HIGHER ORDER THINKING SKILLS
Which of the following processes are accompanied by increase of entropy.
(a) Dissolution of iodine in a solvent
(b) HCl is added to AgNO3 and a precipitate of AgCl is obtained
(c) A partition is removed to allow two gases to mix
Increase of entropy : (a) and (c)
(NCERT)
Predict the sign of entropy change for each of the following changes of state.
(NCERT)
(a) Hg (l ) ⎯⎯
→ Hg(g)
Ans.
ΔS = +ve because liquid changes to more disordered gaseous state.
(b) AgNO 3 (s) ⎯⎯
→ AgNO 3 (aq)
Ans.
ΔS = +ve ; because aqueous solution has more disorder than solid.
(c) I 2 (g) ⎯⎯
→ I 2 (s)
Ans.
ΔS = –ve ; because gas is changing to less ordered solid.
(d) C (graphite) ⎯⎯
→ C (diamond)
Ans.
3.
Ans.
ΔS = +ve ; because graphite has more disorder more disorder than diamond.
How is the concept of coupling of reactions useful in explaining the occurrence of a nonspontaneous reaction?
(NCERT)
When a non-spontaneous reaction is coupled with a highly spontaneous reaction, having high
negative ΔG value, the overall free energy of the two combined reactions becomes negative.
Thus, both the coupled reaction take place spontaneonsly.
Chapter - 3 Chemical Thermodynamics And Energetics
224
4.
Ans.
Ans.
Comment on the following statements.
(NCERT)
(a) An exothermic reaction is always thermodynamically spontaneous.
(i) They are accompanied by decrease of energy.
(ii) The heat released is absorbed by the surroundings so that the entropy of the surroundings
increases and therefore, ΔStotal is positive. But, if ΔS system is –ve and T is very high,
then ΔG (ΔG = ΔH – T.ΔS) may be positive and the reaction may be non-spontaneous.
(b) Reaction with ΔGº < 0 always have an equilibrium constant greater than 1.
ΔGº < 0. Shows that process is spontaneous, equilibrium constant may not have value > 1
(ΔGº = –RT in K).
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S.Y.J.C. Science - Chemistry - Part I
Chapter
4
Electrochemistry
"The Sun, with all the planets revolving around it, and depending on it,
can still ripen a bunch of grapes as though it had nothing else in the Universe
to do." - Galileo
Introduction :
→
→
Electrochemistry is a branch of physical chemistry, which deals with the relationship
between chemical energy and electrical energy.
The energy accompanying spontaneous chemical reactions is utilised to produce electric
current. The cells used are called electrochemical cells. e.g. Daniel cell, Dry cell etc.
Here chemical energy is converted into electrical energy.
SYLLABUS
4.1
REDOX REACTIONS
4.3 ELECTROCHEMICAL CELLS
4.2
CONDUCTANCE IN ELECTROLYTIC
SOLUTIONS
Conductivity
Molar conductivity of an electrolyte
Variation of conductivity with concentration
Variation of molar conductivity with
concentration
Kohlrausch law of independent migration of
ions
(a) Electrochemical cells
(b) Types of cells
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Application of Kohlrausch law
Relation between molar conductivity and
degree of dissociation of week electrolytes
(α)
Measurement of conductivity
Theoretical MCQs
Numerical MCQs
(c) Distinguish between electrolytic cell and
electrochemical cell
4.4 ELECTROLYTIC CELLS
(a) Electrolysis of molten NaCl
(c) Quantitative aspects of Faraday’s laws of
electrolysis
Theoretical MCQs
Numerical MCQs
4.5 GALVANIC OR VOLTAIC CELLS
(a) Salt Bridge
(b) Representation or formulations of galvanic
cells
226
(c) Writing of cell reactions
(d) Daniell cell
(e) Types of electrodes
Theoretical MCQs
Numerical MCQs
4.6 REFERENCE ELECTRODES
4.9 COMMON TYPES OF CELLS
(a) Dry cell
(a) Standard hydrogen electrode
(b) Calomel electrode
4.7 ELECTROCHEMICAL SERIES
4.10 FUEL CELLS
4.8 ELECTRODE POTENTIALS AND CELL
POTENTIAL
4.11 CORROSION
(a) Standard potentials
(b) Nernst equation
Theoretical MCQs
(c) Cell potentials and Gibbs energy changes
for cell reactions
(d) Standard cell potentials and equilibrium
constants
(e) Measurement of cell potentials
→
→
→
→
Hours Before Exam
Numericals with Solution
Numericals for Practice
Higher Order Thinking Skills (HOTS)
When a non-spontaneous redox reaction is made to take place by supplying electrical
energy. The process is called electrolysis. The cells used are called electrolytic cells.
Here electrical energy is converted into chemical energy.
Electrochemistry is also related with the physical properties such as resistance and
conductance of electrolytic solutions.
The study of electrochemical cells is very important in science and technology, because
it makes possible the manufacture of many important chemicals such as NaOH. It is
used in the manufacturing of soaps, detergents, papers etc.
The metals like Al and Cu are useful for electrical wiring. They are prepared by electrolysis
from their ores.
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S.Y.J.C. Science - Chemistry - Part I
227
4.1
REDOX REACTIONS :
Concept Explanation :
1.
A reaction in which oxidation and reduction reactions occur simultaneously, is called redox
reaction. e.g..
Zn(s) + Cu 2+ (aq) ⎯⎯
→ Zn 2+ (aq) + Cu(s)
2.
In the above reaction, one species (Zn) loses electrons and undergoes oxidation. e.g..
Zn(s) ⎯⎯
→ Zn 2+ (aq) + 2e – (oxidation)
3.
The other species (Cu2+) accepts the electrons and undergoes reduction. e.g..
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s) (reduction)
4.
5.
6.
7.
8.
9.
10.
11.
12.
4.2
Thus, redox reactions are electron transfer reactions.
Oxidation is defined as the loss of one or more electrons from a substance.
Reduction is defined as the gain of one or more electrons by a substance.
Oxidizing agent (oxidant) is a species that accepts electron(s) and causes other substance
to lose electron(s).
Reducing agent (Reductant) is a species that donates electron(s) and causes other
substance to accept electron(s).
Oxidation is also defined as a process in which oxidation number of an element increases.
In reduction; oxidation number of an element decreases.
The total increase and decrease in oxidation number are equal in all balanced redox reactions.
Thus, redox reaction form the basis of all electrochemical processes.
CONDUCTANCE IN ELECTROLYTIC SOLUTIONS :
Concept Explanation :
1.
2.
The substances which allow the flow of electricity through them are called conductors.
The flow of electricity through a conductor involves the transfer of electron from one
point to the other.
Chapter - 4 Electrochemistry
228
3.
The mechanism of the transfer of electron is not the same for all the conductors, hence,
conductors are classified into two types :
(a) Electronic conductors :
Definitions : The conductors through which the conduction of electricity occurs
by a direct flow of electrons under the influence of applied potential are known
as electronic conductors. e.g.. Al, Cu etc.
(b) Electrolytic conductors :
(1) Definition : The conductors in which the conduction takes place by the
migration of positive and negative ions are known as electrolytic conductors.
e.g.. Solution of electrolytes.
(2) The current flow in these conductors is accompanied with the chemical changes
at the electrodes.
(3) A measurement of conductivity of solution provides an information regarding the
nature of solutions.
(4) If the conductivity of the solution is same as that of water, they are called nonelectrolytes. e.g.. sucrose, urea etc.
(5) If the conductivity of the solution is much higher than that of water, they are
called electrolytes. e.g. KCl, CH3COOH etc.
(6) The solutions of all the electrolytes of the same concentration (molarity) do not
have the same conductivity.
(7) Thus, electrolytes are classified as strong electrolytes and weak electrolytes.
Distinguish between electronic and electrolytic conductors.
Electronic conductors
1.
2.
3.
4.
Electrolytic conductors
The flow of electricity occurs by the
migration of electrons through the
conductor.
The conduction do not involve the transfer
of matter.
The conduction process do not involve the
chemical changes.
The resistance of the conductor increases
and conductivity decreases with increasing
temperature.
Unique Solutions ®
1.
2.
3.
4.
The electron transfer takes place by
migration of positive and negative ions
towards the electrodes.
The conduction involves the transfer of
matter.
The current flow is always accompanied
with chemical changes.
The resistance decreases and conductivity
increases with increasing temperature.
S.Y.J.C. Science - Chemistry - Part I
229
(a) Conductivity :
1. (a) According to Ohms law the electrical resistance of a conductor i.e. the passage of
current is equal to the potential difference, (V) divided by the electric current (I).
V
...(i)
I
(b) The electrical conductance (G) of a solution is the reciprocal of its resistance.
1
G=
...(ii)
R
(c) The electrical resistance (R) of a conductor is linearly proportional to its length (l)
and inversely proportional to its area of cross section (a),
l
a
l
∴
or R = ρ or ρ = R
...(iii)
R∝
a
l
a
The proportionality constant (ρ) is called resistivity of the conductor.
(d) If l = 1m and a = 1m2 then R = ρ
(e) Thus, the resistivity of a conductor (ρ) is defined as the resistance of the conductor
that is 1m long and 1m2 in cross-sectional area.
(f) From eq. (ii) and (iii);
a
1 a
G = × = k×
ρ l
l
The proportionality constant k is called conductivity of the conductor.
1 l
l
×
or k = G =
...(iv)
R a
a
(g) If l = 1m and a = 1m2 then k = G. Thus, conductivity (k) is defined as the
conductance of unit cube of a solution of an electrolyte.
1
(h) Combinations of eq. (iii) and (iv) gives, k =
ρ
Thus, conductivity (k) is the inverse of resistivity (ρ).
2. Unit of various terms :
(a) Unit of (Resistance) R : (SI unit)
V
Volt
R =
=
= ohm (Ω)
Ampere
I
(b) Unit of electrical conductance (G) :
1
1
G =
=
= ohm–1 (Ω–1) or mho
R
ohm
= (reciprocal ohm) Siemens (s)
...(SI unit)
C⎤
⎡
1S = 1Ω–1 = A V–1 = C V–1s–1 ⎢Coulomb (C) = A.s or A = ⎥
s⎦
⎣
∴
R=
Chapter - 4 Electrochemistry
230
(c) Unit of resistivity the conductor (ρ) :
m2
a
= ohm
= ohm.m (SI unit) = Ωm
m
l
(d) Unit of conductivity of the conductor (k) :
1
m
1 ρ
× 2 = ohm–1. m–1
× =
k =
ohm m
R a
–1
= S. m (SI unit)
ρ = R
Note : Common unit of conductivity (k) are Ω–1 cm–1 or S cm–1.
(b) Molar conductivity of an electrolyte (L) :
In 1880’s the German physicist George Kohlrausch introduced the concept of molar
conductivity (L).
Suppose, a cell consists of two parallel plates (electrodes) of 1 sq.cm. area of cross section,
and 1 cm apart from each other. If 1 cm3 of solution containing, 1 mole of an electrolyte
is placed between the plates, then the conductance shown by the solution is equal to its
molar conductivity.
1. Definition : It is defined as the conducting power of all the ions produced by dissolving
one mole of an electrolyte in solution.
2. Relation between conductivity (k) and molar conductivity (L) :
L=k×V
V = volume of solution containing one mole of the electrolyte or L = k ×
C = molar concentration, i.e. mol–1 (or mol dm–3)
3. Unit molar conductivity (L) :
(a) In SI units :
k
Sm –1
L =
=
= S m2 mol–1
C
mol m –3
(b) If concentration is expressed in mol L–1 and volume (V) in cm3
π =
1000 (cm 3 L–1 ) × k (Ω –1 cm –1 )
1000 × k
=
C
C(mol L–1 )
= Ω–1 cm2 mol–1
= S cm2 mol–1
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1000
C
231
(c) Variation of conductivity (k) with concentration :
1. The conductivity of an electrolytic solution always decreases with decrease in
concentration (i.e. on dilution). Reason :
(a) The conductivity is the conductance of unit volume of solution (say 1 cm3) and depends
on the number of ions present in unit volume.
(b) On dilution i.e. on decreasing the concentration of solution, the number of ions per
unit volume decreases. Hence, conductivity of solution decreases.
2. The conductivity of an electrolyte with concentration differs for strong and
weak electrolyte :
(a) For strong electrolytes, the conductivity increases rapidly with increasing concentration.
(b) For weak electrolytes, the conductivity is very low in dilute solution and increases
(c) In case of strong electrolytes, the increase in number of ions per unit volume increases
rapidly with concentration, because they are completely dissociated.
(d) In case of weak electrolytes, the increase in number of ions per unit volume is not
so large because they are partially dissociated.
(d) Variation of molar conductivity with concentration :
1. Unlike conductivity (k), the molar conductivity (L) of both strong and weak electrolytes
increases with dilution that is with decrease in concentration. Reason :
(a) The molar conductivity is the conductance of all the ions produced from 1 mole of
the electrolyte.
(b) As solution is diluted, the total number of ions increases due to increase in degree
of dissociation. Hence, molar conductivity increases.
k
= k.V
C
Although k decreases on dilution, but the increase in V is more than compensated
by the decrease in the value of k.
2. The molar conductivities is of strong and weak electrolytes show different behaviour
on decrease in concentration that is dilution :
(a) The molar conductivity of strong electrolytes increases rapidly and soon approaches
a maximum limiting value.
(b) The limiting value of molar conductivity is the value at zero concentration (L0) or
at infinite dilution.
(c) The molar conductivity of strong electrolytes approaches close to L0 value in just
0.001 M or 0.0001 M solutions.
(c) We know, L =
Chapter - 4 Electrochemistry
232
(d) The molar conductivity of weak electrolytes also increases on dilution. But at
concentration of 0.001 M or 0.0001 M, it is very much than the maximum limiting
value L0.
(e) Kohlrausch, on the basis of experiments, showed that the molar conductivity (L), of
strong electrolytes varies linearly with the square root of concentration ( C ) and
established the relation L = L 0 – a C
(where a = constant for a particular solvent at a particular temperature.)
(f) The molar conductivity of weak electrolytes does
not show such linear variation with C .
The figure shows how L values of strong and
weak electrolytes show different behaviour with
decrease in concentration.
Note :
(a) L0 is the molar conductivity at zero concentration or at infinite dilution. The infinite
dilution means the solution is so dilute that further dilution does not increase the
molar conductivity. In such a solution, the ions are far apart and do not interact
with one another.
(b) The L0 value of strong electrolytes can be obtained by extrapolating linear part
of L versus C graph to zero concentration.
(c) But this method cannot be used for weak electrolyte because L versus C curve
does not approach linearity.
(d) The L0 value of weak electrolytes can be calculated by Kohlrausch law.
(e) Kohlrausch law of independent migration of ions :
1. Statement : The law states that at infinite dilution, each ion migrates independently
of its co-ion and makes its own contribution to the total molar conductivity of an
electrolyte irrespective of the nature of other ion with which it is associated.
2. At infinite dilution (i.e. at zero concentration), L0 is the sum of molar conductivity of cation
( λ 0+ ) and anion ( λ 0– ) that is
L0 = λ 0+ and λ 0–
for NaCl, L0 = λ 0
Na +
+ λ0
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Cl –
S.Y.J.C. Science - Chemistry - Part I
233
(f) Application of Kohlrausch law :
1. Molar conductivity of infinite dilution (L0) for weak electrolyte : The law can be
used to calculate the molar conductivity of any electrolyte at zero concentration. [However,
this law is particularly useful to calculate L0 of weak electrolytes for which the extrapolation
method (by graph) is not useful.]
Example :
L0 value of weak electrolyte, CH3COOH can be calculated from the L0 values of strong
electrolytes HCl, CH3COONa and NaCl.
L0 (HCl) + L0 (CH3COONa) – L0 (NaCl)
=
λ0 + + λ0
=
λ0 + + λ0
H
H
Cl –
+ λ0
CH 3COO –
CH 3COO –
+ λ0
–λ 0
Na +
Na +
– λ0
Cl –
= L0 (CH3COOH)
Thus, L0 (CH3COOH) = L0 (HCl) + L0 (CH3COONa) – L0 (NaCl).
The L0 values of strong electrolytes can be calculated by extrapolation method and L0
of weak electrolytes can be evaluated.
(g) Relation between molar conductivity (L) and degree of dissociation
α) :
of weak electrolytes (α
1.
2.
L
L0
Where L is the molar conductivity of the weak electrolyte at the given concentration
C, and L0 is its molar conductivity at zero concentration.
The dissociation constant of a weak electrolyte (K) is given as,
α=
2
K
∴ K
α 2C
=
1– α
=
⎛ L ⎞
⎜⎝ L ⎟⎠ C
0
=
L
1–
L0
2
⎛ L ⎞
⎜⎝ L ⎟⎠ C
0
L0 – L
L0
=
L2
L20
×
L0
×C
( L 0 – L)
L2 C
=
L 0 ( L 0 – L)
(h) Measurement of conductivity :
The determination of conductivity (k) and molar conductivity (L) of a solution consists
of measurement of resistance of the solution by Wheatstone Bridge principle using
conductivity cell.
Chapter - 4 Electrochemistry
234
1. Conductivity cell : It consists of a glass tube
with two platinum plates coated with platinum
black. The cell is to be dipped in a solution
whose resistance is to be measured.
2. Cell constant :
(a) The conductivity (k) of an electrolytic solution is given by,
1 l
×
k=
...(i)
R a
l
(b) For any given cell, the ration is fixed quantity and is called the cell constant denoted
a
by b.
(c) Definition – Cell constant is defined as ‘the ratio of the distance between the
electrodes divided by the area of cross section of the electrode.’
l
Thus, cell constant = b =
...(ii)
a
from equation (i) and (ii)
b
k=
...(iii)
R
Unit of cell constant : m–1 or cm–1.
3. Determination of cell constant :
Wheatstone bridge principle
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(a)
(b)
(c)
(d)
(e)
(f)
The cell constant is determined by using 1M, 0.1M or 0.01M KCl solution.
The conductivity (k) of KCl is known at various temperature.
The resistance of KCl solution is determined by Wheatstone bridge principle.
AB is the uniform slide wire.
Rx is the variable known resistance placed in one arm of Wheatstone bridge.
The conductivity cell dipped in a solution of KCl of unknown resistance placed
in the other arm.
(g) D is a current detector, F is the sliding contact. A.C. represents the source of attempting
current.
(h) The sliding contact is moved along AB until no current flows. Suppose a balance point
is obtained at C.
(i) According to Wheatstone bridge principle,
R(Solution)
R(AC)
=
∴
R(Solution)
l (AC)
=
∴
R(solution)
=
∴
Resistance (R) ∝ length (l)
The cell constant (b)
Rx
R(BC)
Rx
l (BC)
l (AC)
Rx
l (BC)
=
=
k.R (solution)
Conductivity of solution × resistance offered
by the solution.
The conductivity (k) of KCl is known, here b can be calculated.
4. Determination of conductivity of the given solution :
KCl solution is replaced by the given solution and its resistance is measured by Wheatstone
bridge principle as described earlier.
b
∴ Conductivity (k) = R(given solution)
k
C
Since concentration c of solution is known, hence, L can be calculated.
Molar conductivity of the given solution = (L) =
•
*1.
Ans.
Define the term (a) resistivity, (b) conductivity, (c) molar conductivity
(a) Resistivity : The resistivity of a conductor is defined as the resistance of the conductor
that is 1m long and 1m2 in cross-sectional area.
(b) Conductivity : Conductivity (k) is defined as the conductance of unit cube of a solution
of an electrolyte.
Chapter - 4 Electrochemistry
236
(c) Molar conductivity : It is defined as the conducting power of all the ions produced by
dissolving one mole of an electrolyte in solution.
*2.
Ans.
Give SI units of (a) resistivity, (b) conductivity, (c) molar conductivity.
l
a
l
or R = ρ or ρ = R
a
l
a
The proportionality constant ρ is called resistivity of the conductor.
(a) Resistivity : R ∝
(b) Conductivity : G =
a
1 a
× = k×
ρ l
l
The proportionality constant k is called conductivity of the conductor.
or k = G
1 l
l
×
=
R a
a
(c) Molar Conductivity : In SI units :
L
*3.
Ans.
k
Sm –1
=
=
=
C
mol m –3
S m2 mol–1
Explain the terms conductivity and molar conductivity. How are they interrelated?
G=
a
1 a
× = k×
l
ρ l
1 l
l
= ×
R a
a
(a) The conductivity (k) is defined as the conductance of unit cube of a solution of an electrolyte.
(b) Molar conductivity is defined as the conducting power of all the ions produced by dissolving
one made of an electrolyte in solution.
Relation between conductivity (k) and molar conductivity (L) :
L = k × V
The proportionality constant k is called conductivity of the conductor or k = G
V = volume of solution containing one mole of the electrolyte or L = k ×
1000
C
c = molar concentration, i.e. mol–1 (or mol dm–3)
*4.
Ans.
Why does conductivity of a solution decreases on dilution of the solution?
(a) The conductivity is the conductance of unit volume of solution (say 1 cm3) and depends
on the number of ions present in unit volume.
(b) On dilution i.e. on decreasing the concentration of solution, the number of ions per unit
volume decreases. Hence, conductivity of solution decreases.
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*5.
Ans.
*6.
Ans.
*7.
How does molar conductivity of an electrolyte vary with concentration?
Refer 4.2 (d) 1.
How does the variation in molar conductivity of an electrolyte with concentration differs for strong
and weak electrolytes? OR How is the molar conductivity of strong electrolytes at zero concentration
determined by graphical method? Why is this method not useful for weak electrolytes?
Refer 4.2 (d) 2.
Ans.
Explain the different behaviour of strong and weak electrolytes towards the variation of
conductivity with concentration.
Refer 4.2 (c) 2.
State and explain Kohlrausch law of independent migration of ions. How is it useful to determine
the molar conductivity of weak electrolytes at zero concentration?
Refer 4.2 (e) and (f).
*9.
Ans.
What is Cell constant? What is its units? How is it determined?
Refer 4.2 (h) 2 and 3
Ans.
*8.
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
The SI unit of molar conductivity is ........
a) S cm2 mol–1
b) S dm2 mol–1
c) S m2
d) S m2 mol–1
The molar conductance of solution of an electrolyte is measured in ........
a) ohm cm mol–1
b) ohm–1 cm–1 mol–1
c) ohm cm–1 mol–1
d) ohm–1 cm2 mol–1
2.
•
3.
Numerical MCQs
The distance between two electrodes of a cell is 3.0 cm and area of each electrode is 6.0
cm. The cell constant is ........
a) 2.0
b) 1.0
c) 0.5
d) 18
•
4.
Conductivity (units Siemen’s S) is directly proportional to the area of the vessel and the
concentration of the solution in it and is inversely proportional to the length of the vessel, then
the unit of constant of proportionality is ......
(A.I.E.E.E. 2002)
–1
2
–1
–2
2
a) S m mol
b) S m mol
c) S m mol
d) S2 m2 mol–2
⎡
⎤
Area×Conc.
conductivity×length
S×m
or k =
∴k = 2
=Sm 2 mol –1 ⎥
⎢ Hint:Conductivity = k×
–3
length
area×conc.
m ×mol
⎣
⎦
Chapter - 4 Electrochemistry
238
5.
Electrolyte
KCl
KNO3
HCl
NaOAc
NaCl
L∞ 2
(Scm mol –1 )
∞
149.9
145.0
426.2
91.0
126.5
Calculate L HOAc using appropriate molar conductance of the electrolytes listed at infinite
(A.I.E.E. 2005)
dilution of H2O at 25ºC ......
a) 517.2
b) 552.7
c) 390.7
d) 217.5
Hint : L∞HOA c
= L∞NaOAc + L∞HCl – L∞NaCl
= 91.0 + 426.2 – 126.5
= 390.7
6.
0
The molar conductivites L NaOAc and L0HCl at infinite dilution in water at 25ºC are 91.0 and
0
426.2 S cm2/mol respectively. To calculate L HOAc the additional value required in ......
(A.I.E.E.E.2006)
a)
L0NaOH
b)
L0NaCl
c)
L0H 2O
d)
L0KCl
[Hint : L0HOAc = L0NaOAc + L0HCl – L0NaCl ]
4.3
ELECTROCHEMICAL CELLS :
Concept Explanation :
(a) Electrochemical cells :
1. The reactions which occur in electrochemical cells are called electrochemical reactions.
These reactions are redox reactions.
2. An oxidation half reaction occurs at one electrode and a reduction half reaction occurs
at the other electrode.
3. Cathode : It is defined as an electrode at which reduction occurs as electrons are
gained by some species.
4. Anode : It is defined as an electrode at which oxidation occurs as electrons are lost
by some species.
(b) Types of cells :
There are two types of cells :
1. Electrolytic cells – A cell in which nonspontaneous reaction is forced to occur by
passing a direct current into the solution, is called an electrolytic cell.
2. Electrochemical cells or voltaic or galvanic cells - A cell in which a spontaneous
chemical reaction produces electricity is called electrochemical cell.
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(c) Distinguish between electrolytic cell and electrochemical cell.
Electrolytic cell
1.
2.
3.
4.
5.
6.
4.4
Electrochemical cell (Voltaic cell)
It is a device in which electrical energy
is converted into chemical energy.
The redox reaction is non-spontaneous and
takes place only when electrical energy is
supplied.
It is used to bring about electrolysis.
In this cell cathode has negative polarity
and anode has positive polarity.
No salt bridge is used in this cell.
1.
Example : Voltmeter used for electrolysis
or water, electroplating, cells used for
purification of metals.
6.
2.
3.
4.
5.
It is a device in which chemical energy
is converted into electrical energy.
It is based upon the redox reaction which
is spontaneous.
It is used to produce electricity.
In this cell cathode has positive polarity
and anode has negative polarity.
To set up this cell a salt bridge or porous
pot is used.
Example : Daniel cell, Dry cell, Nickel
ELECTROLYTIC CELLS :
Concept Explanation :
1.
2.
The process which occurs in the electrolytic cell is called electrolysis.
Electrolysis : It is defined as a ‘process in which an electric current is used to bring
(a) Electrolysis of molten NaCl :
Construction :
1. It consists of a vessel with two graphite (carbon)
electrodes dipped in fused NaCl (electrolyte).
2. The electrode connected to the negative
terminal of a battery is called cathode.
3. The electrode connected to the positive
terminal of a battery is called anode.
Working :
1. When an electric current is passed through
fused NaCl then its electrolysis takes place.
2. Sodium is deposited at cathode and chlorine
is liberated at anode.
Chapter - 4 Electrochemistry
240
3.
4.
2NaCl ⎯⎯
→ 2Na + + 2Cl –
(Dissociation)
Reduction half reaction at cathode : Na+ ions migrate to the cathode. At cathode, each
Na+ ion accepts an electron which is supplied by the battery.
2Na + (l ) + 2e – ⎯⎯
→ 2Na(l )
5.
(reduction)
...(i)
Oxidation half reaction of anode : Cl– ions migrate to anode. At anode, each Cl– ion
gives one electron to the anode and is converted into neutral Cl atom in the primary process.
Two Cl atoms then combine to form Cl2 gas in the secondary process.
(Oxidation)
2Cl – (l ) ⎯⎯
→ Cl(g) + Cl(g) + 2e –
Cl(g) + Cl(g) ⎯⎯
→ Cl 2 (g)
(secondary process)
2Cl – (l ) ⎯⎯
→ Cl 2 (g) + 2e –
6.
..(ii)
Net cell reaction : on adding eq. (i) and (ii)
2Na + (l ) + 2Cl – (l ) ⎯⎯
→ 2Na(l ) + Cl 2 (g)
7.
Results of electrolysis :
(a) A pale green Cl2 gas is released at anode and (b) A molten silvery-white Na deposits
at cathode.
(b) Faraday’s Laws of Electrolysis :
First Law : It selects that the amount of substance that undergoes oxidation or reduction
at each electrode during electrolysis is directly proportional to the amount of electricity
that passes through the cell.
Second law : It states that when the same amount of electricity is passed through different
cells containing different electrolytes and arranged in series, the amounts of substances
oxidized or reduced at the respective electrodes are directly proportional to their chemical
equivalent masses.
Electrical units of charge :
1. Coulomb (C) : Coulomb is the smaller unit of charge or electricity. It is defined as
the amount of charge (electricity) that passes a given point when 1A current flows
for 1s.
2. Faraday (F) : Faraday is the bigger unit of charge. It is defined as the amount of
electric charge on one mole of electrons.
The charge of one electron is 1.602 × 10–19C. Hence, the charge of Avogadro’s number
of electrons that is of 1 mole of electrons will be 1.602 × 10–19(C) × 6.022 × 1023 = 96473
C/mol e– ≈ 96500 C/mole– = 1F.
Thus, Faraday is the charge of one mole of electrons and we write 1F = 96500 C/mol e–.
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(c) Quantitative aspects of Faraday’s laws of electrolysis :
1. Faraday’s first law of electrolysis : The mass of the reactant consumed or of the product
formed at an electrode during electrolysis is calculated by using the stoichiometry of the
half reaction and the molar mass of the substance. The following procedure is used in
the calculations.
(a) Quantity of electricity passed : The quantity of electricity (charge) passed through a
cell is calculated by measuring the quantity of electric current (I) passed through the cell
and the time (t) of the passage of current. The quantity of electricity Q is given by Q(C)
= I(A) × t(s)
(b) Number of electrons passed in moles : The charge of one mole electrons is
Q (C)
96500 C. Hence, number of moles of electrons actually passed =
96500(C/mol e – )
(c) Number of moles of product formed : The balanced equation of the half reaction
(stoichiometry) indicates the number of moles of electrons used and number of moles
of product formed. From these quantities the number of moles of product actually
formed can be calculated as Number of moles of the product formed = number moles
of electrons actually passed × mole ration
moles of product formed in half reaction
Where mole ratio = moles of electrons required in half reaction
(d) Mass of product formed : Mass of product formed = number of moles of product
formed x its molar mass.
From the above steps, the following general formulae can be used to calculate the
mass of the substance produced.
(i) Moles of the substance produced
I(A) × t(s)
× mole ratio from the stoichiometry
=
96500 (C/mol e – )
(ii) Mass of the substance produced
I(A) × t(s)
× mole ratio × molar mass of the substance
=
96500(C/mol e – )
2. Faraday’s second law of electrolysis : Two cells containing different electrolytes are
connected in series. The same quantity of electricity that is the same number of moles of
electrons are actually passed through them.
Moles of A produced in one cell = moles of electrons actually passed × mole ratio of
A half reaction
Moles of B produced in other cell = moles of electrons actually passed × mole ratio of
B half reaction.
moles of A produced
mole ratio of A half reaction
Hence, moles of B produced = mole ratio of B half reaction
Chapter - 4 Electrochemistry
242
•
*1.
Sketch the cell for electrolysis of molten MgCl2. Indicate anode, cathode with their signs. Show
the direction of electron and ion flow. Write electrode reactions and net cell reaction.
Ans.
MgCl 2 ⎯⎯
→ Mg
2+
Electrolysis of molten MgCl2
(Dissociation)
+ 2Cl –
2Cl – (l ) ⎯⎯
→ Cl(g) + Cl – (g) + 2e –
At anode :
(Oxidation)
Cl(g) + Cl(g) ⎯⎯
→ Cl 2 (g)
2Cl – (l ) ⎯⎯
→ Cl 2 (g) + 2e –
At cathode : Mg 2+ (l ) + 2e – ⎯⎯
→ Mg(l )
Net cell reaction : on adding eq. (i) and (ii);
...(i)
(reduction)
...(ii)
Mg 2+ (l ) + 2Cl – (l ) ⎯⎯
→ Mg(l ) + Cl 2 (g)
*2.
Ans.
Define anode and cathode.
Cathode : It is defined as an electrode at which reduction occurs as electrons are gained by
some species.
Anode : It is defined as an electrode at which oxidation occurs as electrons are lost by some
species.
*3.
Predict the half cell reactions that occur when fused KCl is electrolyzed in a cell with inert
electrodes. What is the overall cell reaction?
Ans.
2KCl ⎯⎯
→ 2K + + 2Cl –
(Dissociation)
At cathode
2K + (l ) + 2e – ⎯⎯
→ 2K(l )
(reduction)
...(i)
At anode
2Cl – (l ) ⎯⎯
→ Cl 2 (g) + 2e –
(oxidation)
...(ii)
over all reaction 2K + (l ) + 2Cl – (l ) ⎯⎯
→ 2K(l ) + Cl 2 (g)
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*4.
Ans.
(a) First Law : It states that the amount of substance that undergoes oxidation or reduction
at each electrode during electrolysis is directly proportional to the amount of electricity
that passes through the cell.
(b) Second law : It states that when the same amount of electricity is passed through different
cells containing different electrolytes and arranged in series, the amounts of substances
oxidized or reduced at the respective electrodes are directly proportional to their chemical
equivalent masses.
*5.
How will you calculate the number of moles of electrons actually passed and mass of the
substance produced during electrolysis of a salt solution using reaction stoichiometry?
Refer 4.4 (c).
Ans.
*6.
Ans.
∴
*7.
Ans.
How many electron will have a total charge of 1 coulomb?
1.602 × 10–19C charge is present on = 1 electron
1
= 6 × 1020 electron
1 C charge is present on =
1.602 × 10 –19
How many Faradays would be required to plate out 1.00 mole of free metal from the following cations?
(a) Mg2+
(b) Cr3+
(c) Pb2+
(d) Cu+
(a) Mg 2+ + 2e – ⎯⎯
→ Mg(s)
The reaction indicates that 1 mole of Mg is produced by the passage of 2 moles of electrons.
The charge of 2 mole electrons is 2 Faradays.
(b) Cr 3+ + 3e – ⎯⎯
→ Cr(s) ;
2+
–
(c) Pb + 2e ⎯⎯
→ Pb(s) ;
+
–
(d) Cu + e ⎯⎯
→ Cu(s) ;
*8.
Ans.
*9.
Ans.
Refer 4.4 (b) 1 and 2.
Describe the electrolysis of molten NaCl using inert electrodes.
Refer 4.4 (a)
*10.
Ans.
*11.
What is the difference between electrolytic cell and voltaic cell?
Refer 4.3 (c).
Sketch the cell for the electrolysis of molten NaCl. Indicate cathode, anode and their signs.
Show the flow of electrons and ions.
Refer 4.4 (a)
Why is cathode in an electrolytic cell considered to be negative and anode positive?
(a) In electrolytic cell, cathode is connected to the negative terminal of the battery and anode
in connected to the positive terminal of the battery,
(b) During electrolysis, at negative electrode, reduction reaction takes place hence, it acts as
cathode, while at positive electrode oxidation reaction takes place, hence, it acts as anode.
Ans.
12.
Ans.
Chapter - 4 Electrochemistry
244
•
*1.
*2.
*3.
4.
5.
6.
7.
8.
•
*9.
*10.
Multiple Choice Questions (MCQs)
Theoretical MCQs :
In the electrolysis of molten Al2O3 with inert electrodes .........
b) O2 gas is produced at anode
a) Al is oxidised at anode to Al3+
–
c) O is oxidised at anode
c) O is reduced at cathode
The number of electrons that have a total charge of 965 coulombs is .........
b) 6.022 × 1022
c) 6.022 × 1021 c) 3.011 × 1023
a) 6.022 × 1023
The time required to produce 2F of electricity with a current of 2.5 amperes is .........
a) 13.4h
b) 1200 min
c) 50000s
d) 1.5h
When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate
electrodes are .........
cathode
anode
a) Pure Zn
Pure Cu
b) Impure sample
Pure Cu
c) Impure Zn
Impure sample
d) Pure Cu
Impure sample
The electroplating with chromium is undertaken because .........
a) electrolysis of chromium is easier
b) chromium can form alloys with other metals
c) chromium gives a protective and decorative coating to the base metal
d) of high reactivity of chromium metal.
The storage battery acts as .........
a) Electrolytic cell
b) Voltaic cell
c) Fuel cell
d) both a and b
An example of a simple fuel cell is .........
c) Lead storage cell d) Dry cell
a) Daniel cell
b) H2 – O2 cell
The cell reaction in Daniell cell is .........
a)
Zn + Cu ⎯⎯
→ Zn 2+ + Cu 2+
b)
Zn 2+ + Cu ⎯⎯
→ Zn + Cu 2+
c)
Zn + Cu 2 + ⎯⎯
→ Zn 2+ + Cu
d)
Zn 2+ + Cu 2+ ⎯⎯
→ Zn + Cu
Numerical MCQs :
The number of Faradays required to produce 0.5 mol of free metal from Al3+ is .........
a) 3
b) 2
c) 6
d) 1.5
The number of coulombs necessary to deposit 1 g of potassium metal (molar mass 39 g mol–1)
from K+ ions is .........
c) 1237 C
d) 2474 C
a) 96500 C
b) 1.93 × 105 C
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*11.
*12.
•
13.
During electrolysis, 2A current is passed through an electrolytic solution for 965s. The number
of moles of electrons passed will be .........
a) 0.02
b) 0.01
c) 200
d) 0.037
The same quantity of electricity is passed through two cells one containing AlCl3 solution
and the two other containing ZnCl2 solution. If 0.04 mole of Al is produced in the first cell,
the number of moles of Zn produced in the second cell will be .........
a) 0.08
b) 0.0267
c) 0.06
d) 0.02
What is the ratio of the weights liberated at the cathode when the same current is fused through
two solutions of ferric and ferrous salts arranged in series for a given time interval?
(MHT - CET 2001)
a) 3 : 2
b) 1 : 3
c) 2 : 3
d) 3 : 1
⎛
⎞
WFe from ferric
56/3 2
⎜⎝ Hint : W from Ferrous = 56/2 = 3 = 2 : 3 ⎟⎠
Fe
14.
For the electrochemical cell, M / M+ || X–/X, Eº(M+/M) = 0.44V and Eº (X/X–) = 0.33V
From this data, one can deduce that, .........
(IIT 2000)
a)
15.
16.
M + X ⎯⎯
→ M + + X – is the spontaneous reaction
b) M + + X – ⎯⎯
→ M + X is the spontaneous reaction
c) Ecell = 0.77V
d) Ecell = –0.77V
[Hint : (b), Eºcell = +0.11V, here it is spontaneous]
The amount of electricity required to deposit 1 mol of aluminium from a solution of AlCl3 will
be ..........
(AIIMS 1992)
a) 0.33F
b) 1F
c) 3F
d) 1 Ampere
The change required for the reduction of 1 mol of MnO 4– to MnO2 is ..........
a) 1F
–
b)
3F
c)
5F
d)
(AIIMS 2006)
6F
4+
⎛ 7+
⎞
→ MnO 2
Hint : ⎜ Mn O 4 ⎟ ⎯⎯
⎝
⎠
–
i.e. MnO 4 + 3e – ⎯⎯
→ MnO 2
–
⎛
⎞
–
→ MnO 2 + 4OH – ⎟ Thus, for reduction of 1 mol of MnO – to MnO
⎜⎝ MnO 4 + 2H 2 O + 3e ⎯⎯
2
4
⎠
Charge required = 3F
Chapter - 4 Electrochemistry
246
4.5
GALVANIC OR VOLTAIC CELLS :
Concept Explanation :
1.
2.
3.
4.
5.
6.
In this cell, a spontaneous chemical reaction occurs, to produce electricity.
Each galvanic cell is made up of two half cells.
Each half-cell consists of a metal rod immersed in a solution of its own ions of known
concentration.
e.g. a rod of Zn metal is dipped in 1M solution of Zn2+ ions. It forms a half-cell.
The metal rods (electrodes) are connected by a wire.
The solutions of both the half-cells are connected by a salt bridge.
The anode is a negative electrode and cathode is a positive electrode.
(a) Salt Bridge :
1. It is a U-shaped glass tube containing a saturated solution of an
electrolyte such as KCl or NH4NO3 and 5% agar gel.
2. The ions of the electrolyte do not react with the ions of electrode
solutions.
3. Both the ends of the U tube are plugged with glass wool.
Functions of the salt bridge :
1. It provides an electrical contact between the two solutions and
thereby completes the electrical circuit.
2. It prevents the mixing of electrode solutions.
3. It maintains electrical neutrality in both the solutions by the flow
of ions.
(b) Representation or formulation of galvanic cells :
A galvanic cell is represented by a short notation or diagram which includes electrodes,
aqueous solutions of ions and the other species that may or may not be involved in the
cell reaction. The following conventions are used to write the cell notation :
1. The metal electrodes or the inert electrodes (used as a conducting device in case of gas
electrodes) are placed at the ends of the cell formula. The anode (–) is written at the
extreme left and cathode (+) is placed at the extreme right of the cell notation.
2. The insoluble substances if any or gases are placed in the interior positions adjacent to
the metal electrodes.
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3.
4.
5.
6.
7.
8.
1.
2.
The aqueous solutions of ions are placed at the middle of the cell formula.
A single vertical line is used between two phases to indicate the phase boundary, such
as that between the solid electrode and the aqueous solution of ions. It represents the
direct contact between them.
A double vertical line is used between two solutions to indicate that they are connected
by a salt bridge.
The additional information such as concentration of solutions and the gas pressure is also given.
The ions in the same phase are separated by a comma. For example Fe3+, Fe2+ | Pt
A single half cell is written in the order, aqueous solution of ions first and after that the
solid electrode. For example, Zn2+(1M) | Zn. This order is reversed when the electrode
acts as an anode in the galvanic cell.
The following examples will illustrate these conventions :
The cell composed of Mg (anode) and Cu (cathode) consists of two half cells, Mg2+(1M)|Mg
and Cu2+(1M)|Cu. It is formulated as
Mg(s) | Mg2+(1M) || Cu2+(1M) | Cu
The notation for the cell consisting of H+(1M) | H2(g 1atm) | Pt and Ag+(1M) | Ag(s)
half cells with hydrogen gas electrode as anode and Ag electrode as cathode will be,
Pt | H2(g, 1atm) | H+(1M) || Ag+(1M) | Ag(s)
(c) Writing of Cell reaction :
1. The cell reaction corresponding to the cell diagram is written on the assumption that the
right hand electrode is cathode (+) and left hand electrode is anode (–).
2. In all the voltaic cells the electrons flow spontaneously from left to right that is from negative
electrode to positive electrode through external circuit. This implies that electrons are released
at left hand side electrode (anode) and consumed at right hand side electrode (cathode).
In other words oxidation half reaction takes place at left hand side electrode and reduction
half reaction at right hand side electrode.
3. The overall cell reaction is the redox reaction which is the sum of oxidation half reaction
at anode and reduction half reaction at cathode.
4. While adding the half reactions the electrons must cancel. For this purpose it may be
necessary to multiply one or both the half reactions by a suitable numerical factor(s).
5. No electrons appear in the overall cell reaction because all the electrons lost by the species
undergoing oxidation are gained by the species that undergoes reduction.
6. It is important to note that the individual half reactions may be written with one or more
electrons. For example, half reaction for H2 gas whether written as
Chapter - 4 Electrochemistry
248
2H+(aq) + 2e– → H2(g) or as H+(aq)+e– ½H2(g)
makes no difference. However, in writing the overall cell reaction, obviously, the electrons
must be balanced. The following examples will illustrate the above arguments :
(i) Consider the cell,
The oxidation half reaction at anode is,
The reduction half reaction at cathode is,
The overall cell reaction is obtained by multiplying the reduction half reaction by 2
to balance the electrons and then adding to oxidation half reaction.
Thus,
Pb(s) ⎯⎯
→ Pb 2+ (1M) + 2e –
(oxidation half reaction)
2Ag + (1M) + 2e – ⎯⎯
→ 2Ag(s)
(reduction half reaction)
Pb(s) + 2Ag + (1M) ⎯⎯
→ Pb 2+ (1M) + 2Ag
(overall cell reaction)
(ii) Consider the cell
Pt(s) | H2 (g, 1atm) | H+(1M) || Cr3+ (1M) | Cr(s)
The oxidation half reaction at anode is
H 2 (g, 1atm) ⎯⎯
→ 2H + (1M) + 2e –
The reduction half reaction at cathode is Cr 3+ (1M) + 3e – ⎯⎯
→ Cr(s)
To write the overall cell reaction, oxidation half reaction is multiplied by 3 and reduction
half reaction is multiplied by 2 and then the modified half reactions are added together.
Thus,
3H 2 (g, 1atm) ⎯⎯
→ 6H + (1M) + 6e –
(Oxidation half reaction)
2Cr 3+ (1M) + 6e – ⎯⎯
→ 2Cr(s)
(reduction half reaction)
3H 2 (g, 1atm) + 2Cr 3+ (1M) ⎯⎯
→ 6H + (1M) + 2Cr(s) (Overall cell reaction)
(d) Daniell Cell :
Principle : It is an electrochemical cell in which a spontaneous chemical reaction occurs
to produce electricity.
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1.
2.
3.
4.
Construction :
It consists of two half cells.
One half-cell is a beaker containing a strip of
metallic Zn that is immersed in 1M ZnSO4
solution.
The second half-cell is a beaker containing a
strip of metallic Cu that is immersed in 1M
CuSO4 solution.
These two half-cells are connected by a salt
bridge.
Formulation (Representation) :
1.
2.
3.
4.
5.
Cell Reaction :
Zn electrode is anode (–) : Here Zn atoms are oxidised to Zn2+ ions which go into
the solution and the electron are left on the metal strip.
(Oxidation half reaction)
...(i)
Zn(s) ⎯⎯
→ Zn 2+ (1M) + 2e –
The electrons produced at anode are transferred through the wire to Cu electrode.
Cu electrode is a cathode (+), where Cu2+ ions are reduced to metallic Cu; which is
deposited on the Cu strip.
(Reduction half reaction)
...(ii)
Cu 2+ (1M) + 2e – ⎯⎯
→ Cu(s)
–
Salt bridge maintains electrical neutrality of both the solutions. Two Cl ions from the salt
bridge migrate into the anode solutions for every Zn2+ ions formed. Two K+ ions from
the salt bridge migrate into the cathode solution to replace every Cu2+ ion reduced.
Overall cell reaction : It is obtained by adding eq. (i) and eq. (ii)
Zn(s) + Cu 2+ (1M) ⎯⎯
→ Zn 2+ (1M) + Cu(s)
1.
Observations :
The following are the experimental observations when Daniell cell operates :
The potential of the cell is 1.1V, initially.
Chapter - 4 Electrochemistry
250
2.
3.
The mass of Cu electrode increases due to the deposition of metallic Cu on it. The
concentration of Cu2+ ions decreases in the solution around Cu electrode as they are consumed
in the reduction reaction.
The zinc atoms are oxidized to Zn2+ ions which pass into the solution. This results in increasing
the concentration of Zn2+ ions in the anode solution and decreasing the mass of Zn electrode.
Salt bridge
Note : Always remember the following pairs in alphabetical order :
Electrode
Electrode
Left
Right
Anode
Cathode
Oxidation
Reduction
Negative
Positive
(e) Types of Electrodes :
The electrodes are classified according to their composition in the following types :
1. Metal - metal ion electrodes : It consists of a metal strip immersed into the solution
of its own ions. It is formulated as M+n(aq) | M(s)
The half reaction for the electrode is M +n (aq ) + ne – ⎯⎯
→ M(s)
For example, Zn2+(aq) | Zn for which the half reaction is Zn 2+ (aq) + 2e – ⎯⎯
→ Zn(s)
2.
Non metal-non metal ion electrode (Gas electrode) : It consists of a gas about an
inert metal electrode immersed in a solution containing ions of the gas. For example,
(i) Hydrogen gas electrode
The electrode reaction is
(ii) Chlorine gas electrode
The electrode reaction is
(iii) Oxygen gas electrode
The electrode reaction is
Unique Solutions ®
H + (aq) | H 2 (g, PH 2 ) | Pt
H + (aq) + e – ⎯⎯
→
1
H (g, PH 2 )
2 2
Cl – (aq) | Cl 2 (g, PCl 2 ) | Pt
1
Cl 2 (g,PCl 2 ) + e – ⎯⎯
→ Cl – (aq)
2
OH – (aq) | O 2 (g, PO 2 ) | Pt
1
O (g, PO 2 ) + H 2O(l ) + 2e – ⎯⎯
→ 2OH – (aq)
2 2
S.Y.J.C. Science - Chemistry - Part I
251
3.
Metal - Sparingly soluble salt electrode :
It consists of a metal coated with one of its
sparingly soluble salts immersed in a solution
of soluble salt that contains the same anion as
that of the sparingly soluble salt.
Example :
Silver-silver chloride electrode that consists of
a silver wire coated with AgCl and immersed
in a solution of a soluble salt such as KCl or
HCl or any other soluble salt that contains Cl–
ions. It is represented as
Cl–(aq) | AgCl(s) | Ag
The electrode reaction is written as AgCl(s) + e – ⎯⎯
→ Ag(s) + Cl – (aq)
4.
Redox electrode : It consists of metal wire (Pt) serving as an inert electrode immersed
in a solution containing the ions of the same substance in two valency states. For example,
(a) Fe2+ (aq), Fe3+ (aq) | Pt
The order of writing the two ions is immaterial. The reduction reaction for the electrode
is Fe 2+ (aq) + e – ⎯⎯
→ Fe 3+ (aq)
(b) Cu+(aq), Cu2+(aq) | Pt
Reduction reaction is Cu 2+ (aq) + e – ⎯⎯
→ Cu + (aq)
(c) Sn 3+ (aq), Sn 4+ (aq) | Pt
Reduction reaction is Sn 4+ (aq) + e – ⎯⎯
→ Sn 3+ (aq)
4.6
REFERENCE ELECTRODE :
Concept Explanation :
Definition : An electrode whose potential is arbitrarily taken as zero, is exactly known
as known as reference electrode.
1.
2.
3.
Primary reference electrode :
Standard hydrogen electrode (SHE) is used as a primary reference electrode. In this electrode
H2 gas at 1 atm. pressure, is in contact with 1M solution of H+ ions.
The half-cell reaction of SHE can be written either as oxidation or reduction.
The potential of SHE is considered as zero volt.
Chapter - 4 Electrochemistry
252
Note : SHE is not very convenient electrode, therefore, several other electrodes are
used as a secondary reference electrode.
1.
2.
Secondary reference electrode :
The potentials of these electrodes are exactly determined with respect to SHE. These electrode
are most convenient e.g. standard calomel electrode (SCE), silver–silver chloride electrode,
glass electrode.
These electrodes are used to determine the potentials of all other electrodes.
(a) Standard Hydrogen Electrode (SHE) :
1. Definition : An electrode in
which pure and dry hydrogen
gas is bubbled at 1 atm
pressure around a platinised
platinum plate immersed in 1M
H + ion solution, is called
standard hydrogen electrode.
2. It is a primary reference
electrode.
3. By convention, SHE is arbitrarily
assigned, a potential of exactly
zero volt at all temperatures.
Construction :
1. It consists of a platinum plate coated with platinum black.
2. The platinum plate is seated into a thin glass tube.
3. The glass tube contains little Hg and a Cu-wire.
4. The glass tube is surrounded by outer glass jacket.
5. The platinum plate is immersed in a 1M H+ ion solution.
6. Pure and dry H2 gas at 1 atm. pressure is bubbled through the side arm, over the surface
of platinum plate.
7. Platinum does not take part in the reaction. It serves only as the site of electron transfer.
Representation :
H + (1M) | H 2 (g,1atm) | Pt
1.
Electrode reaction (working) :
The platinum black is capable of absorbing large quantities of H2 gas. It allows the change
from gaseous to ionic form and the reverse process to occur.
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2.
At anode,
H 2 (g, 1atm) ⎯⎯
→ 2H + (1M) + 2e – ; Eº = 0.0000V (oxidation)
3.
At cathode
2H + (1M) + 2e – ⎯⎯
→ H 2 (g, 1atm) ; Eº = 0.000V (reduction)
1.
2.
Application of SHE :
When an electrode of unknown potential is combined with SHE to form a cell. Then the
measured cell potential is the potential of unknown electrode.
Example :
Zn | Zn 2+ (1M) || H + (1M) | H 2 (g,1atm) | Pt
By experiment, cell potential is found to be 0.763V
EºCell = EºSHE – EºZn
0.763V = 0 – EºZn
∴
EºZn = –0.763V
1.
2.
3.
Difficulties in setting SHE :
It is difficult to obtain pure and dry H2 gas.
The pressure of H2 gas cannot be maintained at exactly 1 atm. throughout the measurement.
The concentration of H+ ion solution cannot be exactly maintained at 1M. Due to the bubbling
of H2 gas into the solution, evaporation of water may take place. This results in changing
the concentration of solution.
(b) Calomel electrode :
B
1. It is a metal-sparingly soluble salt
Glass tube
electrode.
KCl
2. It is a secondary reference electrode.
solution
Bent side
Construction :
tube
1. It consists of a glass tube provided with
a bent side tube and another side tube ‘B’.
2. A little Hg is placed at the bottom of the Paste of
Hg + Hg2Cl2
dry glass tube.
Hg
3. A rubber bung (stopper) carrying a thin
Pt
Glass
glass tube with a platinum wire is then
wool plug
inserted, taking care that the platinum wire
Calomel Electrode
dips into Hg.
Chapter - 4 Electrochemistry
254
4.
5.
The Hg is then covered with a layer of Hg and Hg2Cl2 (Calomel) paste.
The glass tube is then filled with KCl solution of definite concentration.
Potential of Calomel Electrode :
The potential of the Calomel electrode depends on as the concentration of KCl solution.
Concentration of KCl
Reduction potential at 298 K.
1.
0.1M (Decimolar)
0.337 V
2.
1 M (Molar)
0.280 V
3.
Saturated Calomel electrode
0.242 V
Formulation :
The electrode is represented 2as
KCl(sat) | Hg2Cl2 | Hg(l)
Electrode reactions :
If the electrode is cathode (+) in the
galvanic cell, the half reaction that occurs
on it will be reduction.
Hg 2 Cl 2 (s) + 2e – ⎯⎯
→ 2Hg(l ) + 2Cl – (sat)
If it serves as anode (–) in the galvanic
cell the half reaction will be oxidation.
2Hg(l ) + 2Cl – (sat) ⎯⎯
→ Hg 2 Cl 2 (s) + 2e –
Application of calomel electrode :
It is used as a secondary reference electrode to determine the standard potentials of the
electrodes. For example, to determine the standard potential of Zn2+ (1M) | Zn electrode,
the following cell is constructed :
Zn | Zn2+ (1M) || KCl (sat) | Hg2Cl2(s) | Hg
As shown in the figure. The emf of this cell is measured experimentally from the measured
cell potential, the potential of the given electrode can be calculated as Ecell = Ec –EºZn.
Hence, EºZn = Ecell – Ec where Ec is the potential of calomel electrode.
1.
2.
3.
It is easy to construct and transport and convenient to handle.
The potential of the electrode is reproducible and remains constant.
No separate salt bridge is required for its combination with other electrodes.
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•
*1.
A voltaic cell consisting of Fe2+(aq) | Fe(s) and Bi3+ (aq) | Bi(s) electrodes is constructed.
When the circuit is closed mass of Fe electrode decreases and that of Bi electrode increases.
(a) Write cell formula (b) Which electrode is cathode and which is anode?
(d) Write electrode reactions and overall cell reaction.
When the circuit is closed, the mass of Fe electrode decreases that of Bi electrode increases,
it means that oxidation takes place at Fe electrode and reduction takes place at Bi electrode.
(a) Cell formula :
Ans.
(b) Fe electrode, is an anode since oxidation takes place.
Bi electrode, is a cathode since reduction takes place.
(c) Electrode reactions
⎡ Fe(s) ⎯⎯
→ Fe 2+ (1M) + 2e – ⎤ × 3
(Oxidation half-reaction)
⎣
⎦
3+
–
⎡ Bi (1M) + 3e ⎯⎯
→ Bi(s) ⎤ × 2
(Reduction half-reaction)
⎣
⎦
3Fe(s) + 2Bi 3+ (1M) ⎯⎯
→ 3Fe 2+ (1M) + 2Bi(s)
*2.
Ans.
*3.
Ans.
*4.
Ans.
(Overall cell reaction)
What is Salt Bridge? What are its functions in a galvanic cell?
1. It is a U-shaped glass tube containing a saturated solution of an electrolyte such as KCl
or NH4NO3 and 5% agar gel.
2. The ions of the electrolyte do not react with the ions of electrode solutions.
3. Both the ends of the U tube are plugged with glass wool.
Functions of the salt bridge :
1. It provides an electrical contact between the two solutions and thereby completes the electrical
circuit.
2. It prevents the mixing of electrode solutions.
3. It maintains electrical neutrality in both the solutions by the flow of ions. [Refer 4.5 (a)]
What are the conventions used to write cell diagram (cell formula)?
Refer 4.5 (b).
Describe the construction of Daniell cell. Write electrode half reactions and net cell reaction
in Daniell cell.
Refer 4.5 (d).
Chapter - 4 Electrochemistry
256
*5.
Ans.
Describe the following types of electrodes giving one example, with reference to formulation,
electrode reaction and Nernst equation for electrode potential.
(a) metal-sparingly salt electrode, (b) gas electrode.
Refer 4.5 (e).
6.
Ans.
What is reference electrode? Give one example.
Refer 4.6
7.
Ans.
Describe the construction and working of Standard Hydrogen Electrode.
Refer 4.6 (a)
8.
Ans.
Describe the construction and working of Calomel Electrode.
Refer 4.6 (b)
9.
Ans.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Negative polarity :
(a) According to latest IUPAC convention, the electrode where de-electronation or oxidation
takes place is called anode, for example zinc electrode, Zn(s) ⎯⎯
→ Zn 2+ (aq) + 2e –
(b) The electrons, thus produced are accumulated on the electrode hence, it acquires negative
charge. Due to above reasons the anode in galvanic cell in considered to be negative.
Positive polarity :
(c) According to latest IUPAC convention, the electrode where electronation or reduction
takes place is called cathode, for example copper electrode,
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s)
(d) Here, electrons are removed from the electrode and Cu2+(aq) ions are deposited on the
electrode, hence, it acquires positive charge. Due to above reason the cathode in galvanic
cell is considered to be positive.
4.7
ELECTROCHEMICAL SERIES (ELECTROMOTIVE SERIES)
Concept Explanation :
1.
2.
3.
4.
5.
Definition : A series (table) of the arrangement of electrodes (metal or non-metals
in contact with their ions) with the electrode half reactions in decreasing order of
standard potentials, is called electrochemical series. OR A series in which electrodes
or half-cells of elements are arranged in the decreasing order of their standard
reduction potential, is called electrochemical series.
The E0 value is a measure of the tendency of the species to be reduced.
The greater E0 value means greater tendency of the species to accept electrons and undergo
reduction.
The positive E0 values show the tendency of half reactions to occur in the forward direction.
The negative E0 values indicate the tendency of the half reactions to occur in the reverse
direction.
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The standard aqueous electrode potentials at 298K.
Electrochemical series
INCREASING
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
2F–
Au
Ce3+
Au
2Cl–
Pt
2Br –
Hg
Ag
Hg2
Fe2+
2I–
Cu
Ag + Cl–
Cu+
Sn2+
H2
Pb
Sn
Ni
Co
Cd
Fe
Cr
Zn
Al
Mg
Na
Ca
K
Li
REDUCING AGENT
F2 + 2e–
Au+ + e–
Ce4+ + e–
Au3+ + 3e–
Cl2 + 2e–
Pt2+ + 2e–
Br2 + 2e–
Hg2+ + 2e–
Ag+ + e–
Hg22+ + 2e–
Fe3+ + e–
I2 + 2e–
Cu2+ + 2e–
AgCl(s) + e–
Cu2+ + e–
Sn4+ | 2e–
2H+ + 2e–
Pb2+ + 2e–
Sn2+ + 2e–
Ni2+ + 2e–
Co2+ + 2e–
Cd2+ + 2e–
Fe2+ + 2e–
Cr3+ + 3e–
Zn2+ + 2e–
Al3+ + 3e–
Mg2+ + 2e–
Na+ + e–
Ca2+ + 2e–
K+ + e–
Li+ + e–
Right side species
(reducing agents)
STRENGTH AS
STRENGTH AS
O X I D I S I NG A G E N T
Left side species
(oxidizing agents)
F– | F2 | Pt
Au+ | Au
Ce4+, Ce+3 | Pt
Au3+ | Au
Cl– | Cl2 | Pt
Pt2+ | Pt
Br– | Br2 | Pt
Hg2+ | Hg
Ag+ | Ag
Hg22+ | Hg
Fe3+, Fe2+ | Pt
I– | I2(s) | Pt
Cu2+, Cu
Ag|AgCl(s)|Cl–
Cu2+, Cu+ | Pt
Sn4+, Sn2+ | Pt
H+ | H2 | Pt
Pb2+ | Pb
Sn2+ | Sn
Ni2+ | Ni
Co2+ | Co
Cd2+ | Cd
Fe2+ | Fe
Cr3+ | Cr
Zn2+ | Zn
Al3+ | Al
Mg2+ | Mg
Na+ | Na
Ca2+ | Ca
K+ | K
Li+ | Li
E0/V
Half reaction
INCREASING
Electrode
+2.87
+1.68
+1.61
+1.50
+1.36
+1.20
+1.08
+0.854
+0.799
+0.79
+0.771
+0.535
+0.337
+0.222
+0.153
+0.15
0.00
–0.126
–0.136
–0.257
–0.280
–0.403
–0.440
–0.740
–0.763
–1.66
–2.37
–2.714
–2.866
–2.925
–3.045
Note the following points :
*All ions are at 1M concentration in water.
*All gases are at 1 atm pressure.
*Fe3+, Fe2+ | Pt, Cu2, Cu+ | Pt, Sn4+,Sn2+ | Pt, the order of writing ions in solution is immaterial
Chapter - 4 Electrochemistry
258
Applications of electrochemical series :
1. Relative strength of oxidising agents in terms of Eº values :
(a) The species on left side of half reactions are oxidising agent.
(b) The substances in the upper left side of half reactions are strong electron acceptors,
as they have large positive Eº values and have greater tendency to be reduced. Hence,
they are stronger oxidising agents e.g. F2, Au+, Ce4+ etc.
(c) Thus, the species located in the upper left side of half reaction can be chosen as
oxidising agents.
(d) The strength as oxidisng agent decreases from top to bottom.
2. Relative strength of reducing agents in terms of Eº values :
(a) The species on the right side of half reactions are reducing agent.
(b) The substances in the bottom right side of half reaction are strong electron donors,
as they have large negative Eº values and have greater tendency to be oxidised. Hence,
they are stronger reducing agents e.g. Li, K, Ca etc.
(c) Thus, the species located at the bottom right side of half-reaction can be chosen as
reducing agents.
(d) The strength as reducing agent decreases from bottom to top.
3. Identifying the spontaneous direction of reaction :
(i) E0cell is calculated from the E0 value given in the series for half-reactions.
(a) If E0cell is positive, then overall cell reaction is spontaneous.
(b) If E0cell is negative, then it is nonspontaneous.
(ii) Example : Mg + 2Ag + ⎯⎯
→ Mg 2+ + 2Ag
By convention, the cell is represented as Mg | Mg 2+ || Ag + | Ag
Given, E0 Mg2+ | Mg = –2.37V ; E0Ag+| Ag = + 0.799V
E0cell = E0red (Ag) – E0red(Mg)
= R.H.E.
L.H.E.
= +0.799V – (–2.37V)
= 3.169V
Since, emf of the cell comes out to be positive. Hence, the reaction is spontaneous,
that is Ag+ can oxidize Mg.
(v) In general, any oxidising agent (the species on the left of half reaction) can oxidize
any reducing agent (the species on the right of half reaction) that appears below it
but cannot oxidize the species to located above it in the electrochemical series. This
is called diagonal rule.
4. Calculate of E0 cell : The electrochemical series can be used to calculate the standard
cell potential from the Eº values for half reactions given in it. For example, consider the
cell, Zn(s) | Zn 2+ (1M) || Pb 2+ (1M) | Pb
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The E0cell is given by, E0cell
=
E0cathode – E0anode
= E0pb – E0Zn
E 0Pb
E0cell
4.8
= –0.126V and E0Zn = –0.763V
= –0.126V – (–0.763V)
= 0.637V
ELECTRODE POTENTIALS AND CELL POTENTIALS :
Concept Explanation :
Electrode Potentials :
1. Definition : Electrode
potential is defined as
‘the difference of
electrical potential
between metal electrode
and solution around it
under
equilibrium
condition.
2. When a metal plate is
dipped in its salt solution,
the metal atoms tend to
go into the solution as
positive ions leaving
electrons on the metal
plate.
3. This tendency is called De-electronation or oxidation. It tries to make the plate negatively
charged.
4. As the same time, the positive ions present in the solution tend to deposit on the metal
plate by removing electrons from the metal plate.
5. This tendency is called electronation or reduction. It tries to make the plate positively
charged.
6. At equilibrium the two tendencies are matched.
7. If de-electronation is faster than electronation, the mutual electrode acquires negative charge.
8. If electronation is faster metal layer electrode acquires positive charge.
9. Thus, an electrical double layer is formed at the interface between metal plate and solution.
10. There is a difference of electrical potential metal solution at equilibrium, which is called
electrode potential.
Chapter - 4 Electrochemistry
260
(a)
(b)
(c)
(d)
(e)
(f)
Cell Potentials :
When a galvanic cell operates, a spontaneous redox chemical reaction occurs inside it.
The half-cell reaction at anode is oxidation. The potential associated with this, is called
oxidation potential.
The half cell reaction at cathode is reduction. The potential associated with this, is called
reduction potential.
The cell potential is the algebraic sum of the oxidation potential and reduction potential.
Formula : Ecell = Eoxi (anode) E red(cathode)
Definition : The cell potential or electromotive force (emf) of the cell is defined as
‘the difference of potential between the electrodes corresponding to an external flow
of electrons from anode to cathode. (OR) ‘The difference of potential between the
two electrodes of the cell in open circuit (i.e. when no current is allowed to flow
in the circuit) is called cell potential or emf of a cell.’
Note : The flow of current through the circuit is determined by the push of electrons
at the anode and attraction of electrons at the cathode. These two forces constitute
the driving force or electrical pressure that sends electrons through the circuit.
This driving force is called emf or cell potential or cell voltage. It is measured in Volts.
(a) Standard Potentials :
1. Definition : It is defined as the potential developed when a particular electrode is
dipped in the solution of its ions at unit activity. (i.e. 1 molar concentration or 1 atm.
pressure in case of gas electrode) and at 25ºC (298K).
2. It is designated as E0.
(a) Standard oxidation potential (E0oxi) : It is defined as the potential developed due
to oxidation when the particular electrode is dipped in the solution of its ions at unit
activity (1 molar concentration) and at 298 K.
(b) Standard reduction potential (E0red) : It is defined as the potential developed due
to reduction when the particular electrode is dipped in the solution of its ions at unit
activity (1 molar concentration) and at 298 K. Eºred = –Eºoxi
Note : According to IUPAC, standard reduction potential is now called standard electrode
potential (Eº). It is determined relative to the SHE molar standard conditions.
Standard cell potential or standard emf of a cell is the algebraic sum of the standard oxidation
potential of anode (left hand electrode) and standard reduction potential of cathode. (Right
hand electrode)
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E 0 cell
=
E 0 oxi
+
E 0 red
(old convention)
(L.H.E.)
(R.H.E.)
(Anode)
(Cathode)
But according to latest IUPAC conventions, the emf of the cell is the difference between
the reduction electrode potentials of the two electrodes (or half-cells) and it is given as
E 0 cell =
E 0 red
–
E 0 red
(R.H.E.)
(Cathode)
(L.H.E.)
(Anode)
(b) Nernst Equation :
The cell potential and electrode potential depends on temperature, concentration of solutes
and partial pressures of gases. This dependence of potential is given by Nernst equation.
E = E0 –
RT
lnQ
nF
[products]
Where Q is reaction quotient given by Q = [reactants]
0
E = E –
0
= E –
RT
[products]
ln
nF
[reactants]
2.303RT
[products]
log 10
nF
[reactants]
Where
Eº = standard potential of electrode or cell
n = moles of electrons used in the reaction
F = Faraday = 96500 C/mol e–
(products) and (reactants) represent the concentrations of products and reactants
respectively.
T = temperature in K
R = gas constant = 8.314 J K–1 mol–1
At 25ºC,
2.303 RT
= 0.0592 V mol e–
F
Hence, at 25ºC, the Nernst equation becomes
0
E= E –
0.0592
[products]
log 10
n
[reactants]
...(i)
The first term in Nernst equation represents standard state electrochemical conditions.
The second term is the correction for non-standard state electrochemical conditions. The
cell potential or electrode potential is equal to the standard potential if the concentrations
Chapter - 4 Electrochemistry
262
of reactants and products are unity. Thus, if we substitue (products) = (reactants) = 1
in equation (i), we see that,
0.0592
log 101 = Eº
E = Eº –
n
Calculation of cell potential using Nernst equation :
Consider the cell Zn | Zn 2+ (aq) || H + (aq) (g, PH 2 ) | Pt
The Oxidation at anode, reduction at cathode and overall cell reaction are as follows :
Zn(s) ⎯⎯
→ Zn 2+ (aq) + 2e –
(Oxidation at anode)
2H + (aq) + 2e – ⎯⎯
→ H 2 (g, PH 2 ) (reduction at cathode)
Zn(s) + 2H + (aq) ⎯⎯
→ Zn 2+ (aq) + H 2 (g, PH 2 )
The emf of the cell is given by Ecell
= Eº cell –
(Overall cell reaction)
[Zn 2+ ] × PH 2
0.0592
log10
At 25ºC
n
[H + ] 2
The concentration of solute are molar concentrations and gas pressure is in atm.
Calculation of electrode potential : Consider the electrode Zn2+(aq) | Zn. The reduction
reaction of the electrode is Zn 2+ (aq) + 2e – ⎯⎯
→ Zn(s) . The potential of the electrode
at 25ºC is given by Nernst equations.
1.
2.
3.
4.
Eel =
Eº el –
0.0592
1
log 10
2
[Zn 2+ ]
=
Eº el +
0.0592
log10 [Zn 2+ ]
2
The following conventions are followed while calculating the cell potentials (emf) :
In the formula of a galvanic cell, the right hand side electrode is cathode (+) and left hand
side electrode is anode (–).
All standard potentials are reduction potentials that is they refer to the reduction reactions.
The cathode has a higher standard potential than the anode. Hence, to formulate a cell
from the given electrodes, the electrode with higher standard potential must be written
on the right hand side and that with lower potential on the left hand side.
For a spontaneous reaction as written (from left to right) the cell potential should be positive,
for this purpose, the cathode is written on the right hand side and anode on the left hand
side.
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(c) Cell potentials and Gibbs energy changes for cell reactions :
1. When a cell produces a current, the current can be used to do work - to run a motor
for example [it means that electrical work is done by the system (cell)]. Hence,
thermodynamic principles can be used.
2. Electrical work done = amount of charge flowing × cell potential
3. For every one mole of electron transferred in the cell reaction, the quantity of electricity
that flows through the cell is one Faraday. (1 F = 96500C)
4. Hence, if n moles electron are transferred in any cell reaction, the quantity of electricity
5. ∴ Electrical work done = nF × Ecell
= nFEcell ...(i)
6. Maximum electrical work is done by the cell when the process occurs reversibly.
7. Electrical work done by the system (cell) results in the corresponding decrease in free
energy (ΔG) of the system.
∴ Electrical work done = –ΔG
...(ii)
8. From equation (i) and (ii);
–ΔG = nFEcell or ΔG = –nFEcell
Thus, the standard Gibbs energy change (ΔGº) for cell reaction and standard emf of the
cell (Eºcell) are related as, ΔGº = –nFEºcell.
E0cell values are intensive properties :
We have, ΔGº = –nFEºcell.
Gibbs energy (ΔGº) is an extensive property since it depends on the amount of substances.
If stoichiometric equation of redox reaction is multiplied by 2 that is the amount of substances
oxidised and reduced are doubled, ΔGº doubles. The number of electron transferred ‘n’
also doubles.
Hence, the ratio,
E0cell
= –
Becomes, –
ΔGº
nF
2Δ Gº
ΔGº
= –
2nF
nF
Thus, E0cell remains constant. This indicates that the electrical potential is an intensive
property which does not depend on the amount of substances. Therefore, whenever, the
half reaction is multiplied by a numerical factor to balance the no. of electrons, the
corresponding Eº value is not to be multiplied by that factor.
Chapter - 4 Electrochemistry
264
(d) Standard cell potentials and equilibrium constants :
The relation between standard free energy change of cell reaction and emf of the cell
is given by the equation.
ΔGº = –nFEºcell
The relation between standard Gibbs energy change for a reaction and its equilibrium constant
is given by the equation.
ΔGº = –RT ln K
Combining these two equations we obtain,
–nFEºcell= –RT ln K
Eº
RT
ln K
nF
2.303 RT
log 10 K
=
nF
0.0592
log 10 K at 25ºC
=
n
=
(e) Measurement cell potential (EMF) :
(Using potentiometer) (Poggendorff’s compensation method)
1. Circuit and apparatus :
(a) The potentiometer consists of
a thin conducting wire AB of
uniform cross section. It is
stretched over a meter scale.
(b) A two-volt accumulator
(battery) C is connected in
series with a rheostat and with
the terminals of wire AB.
(c) S is the standard cell of known
emf.
(d) X is a cell, the emf of which
is to be determined.
(e) The standard cell ‘S’ is connected to one end A of wire AB and through a galvanometer
G, to a sliding contact (jockey) J.
(f) The jockey (J) can be moved along wire AB.
(g) A special double switch K is provided to permit the cell S or X to be connected in
the circuit.
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(h) In connecting the accumulator and the cell ‘S’ or cell X to wire AB, it is essential
that the positive poles should be connected to the same end (A) of the wire.
(i) Thus, the cell ‘S’ or X will send a current through the circuit in the direction opposite
to that supplied by accumulator C.
2. Measurement of emf :
(a) The standard cell S is placed in the circuit by means of the switch K.
(b) The sliding contact J is moved along AB and its position is adjusted to a point say
D. So that no current passes through the galvanometer.
(c) The length AD is measured. The emf of the cell S, Es ∝ l(AD)
(d) Now, the cell X is placed in the circuit by means of the switch K.
(e) The sliding contact is again moved along AB and the point of balance on the wire
is determined as before. Let this be at D′.
(f) The length AD′ is measured. The emf of the cell X, Ex ∝ l(AD′)
3. Calculations : Dividing Ex, by Es, we can write,
Hence Ex = E s 1(AD)
Knowing all the quantities on the right hand side of the equation the emf of the cell X
can be calculated.
•
*1.
The galvanic cells give voltages not more than 2.5V. Suggest a galvanic cell that will have
maximum potential which is greater than 2.5V.
Ans.
Mg | Mg 2+ (1M) || (1M)Cu 2+ | Cu
E0red(Mg) = –2.37V, E0red(Cu) = +0.337V
E0cell = E0red – E0red
R.H.E.
L.H.E.
= +0.337 – (–2.37)
= +0.337 + 2.370
0
E cell = 2.707 volts
*2.
Ans.
Define the terms : (a) Oxidation potential, (b) Reduction potential, (c) Cell potential.
(a) Oxidition potential : It is defined as the potential developed due to oxidation when the
particular electrode is dipped in the solution of its ions at unit activity (1 molar concentration)
and at 298 K.
Chapter - 4 Electrochemistry
266
(b) Reduction potential : It is defined as the potential developed due to reduction when
the particular electrode is dipped in the solution of its ions at unit activity (1 molar
concentration) and at 298 K. E0red = – E0oxi
(c) Cell potential : The cell potential or electromotive force (emf) of the cell is defined as
‘the difference of potential between the electrodes corresponding to an external flow of
electrons from anode to cathode. (OR) ‘The difference of potential between the two
electrodes of the cell in open circuit (i.e. when no current is allowed to flow in the circuit)
is called cell potential or emf of a cell.’
*3.
Ans.
What conditions are required for a cell potential to be called standard cell potential?
Following conditions are required for the cell potential to be called as standard cell potential :
(a) The concentration of the solutions of cathode and anode must be 1 molar.
(b) In case of gas electrode, the pressure of the gas must be 1 atmosphere.
(c) The working temperature of the cell must be 25ºC (298K). The standard cell potential
is calculated as :
=
E0red(cathode) –
E0red(anode)
E0cell
RHE
LHE
*4.
Ans.
Formulate a cell for each of the following reactions :
(a) Sn 2+ (aq) + 2AgCl(s) ⎯⎯
→ Sn 4+ (aq) + 2Ag(s) + 2Cl – (aq)
Sn 2+ (aq) | Sn 4+ (aq) || Cl – (aq) | AgCl (s) | Ag
(b) Mg(s) + Br2 (l ) ⎯⎯
→ Mg 2+ (aq) + 2Br – (aq)
Ans.
*5.
Mg | Mg 2+ (aq) || Br2 (l ) Br – (aq) | Pt
Formulate a cell from the following electrode reactions :
(a) Cl 2 (g) + 2e – ⎯⎯
→ 2Cl – (aq)
(b) 2I – (aq) ⎯⎯
→ I 2 (s) + 2e –
Ans.
*6.
Ans.
*7.
Pt | I 2 (s) | I – (aq) || Cl – (aq) | Cl 2 (g) | Pt
Write Nernst equation and explain the terms involved in it. What part of the equation represents
the correction factor for non-standard state conditions?
Refer 4.8 (b)
Write Nernst Equation for the following reactions :
(a) Cr(s) + 3Fe 3+ (aq) ⎯⎯
→ Cr 3+ (aq) + 3Fe 2+ (aq)
Ans.
Ecell
0
= E cell –
0.0592
[Cr 3+ ]
. log 10
3
[Fe 3+ ]3
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(b) Al 3+ (aq) + 3e – ⎯⎯
→ Al(s)
Ans.
*8.
(a)
Ans.
(b)
Ans.
*9.
(a)
Ans.
0
Eel = E el –
0.0592
1
log10
3
[Al 3+ ]
Consider the following E0 values and half reactions :
I 2 (s) + 2e – ⎯⎯
→ 2I – (aq)
E0 = 0.535V
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s)
E0 = 0.337V
Cd 2+ (aq) + 2e – ⎯⎯
→ Cd(s)
E0 = –0.403V
Which of the metals or non-metals or ions is the strongest oxidising agent and which is the
strongest reducing agent?
I2(s) will be the strongest oxidising agent.
Cd(s) will be the strongest reducing agent.
The half reactions can be used to construct three galvanic cells. Which will have the highest
cell potential?
Cd(s) | Cd 2+ (aq) || I – (aq) | I 2 (s) | Pt
Predict whether,
Ag+ can oxidize Pb to Pb2+ under standard state conditions.
E0Ag = 0.799 V and E0Pb = –0.126 V
The half-reactions are;
(i) Ag + (aq) + e – ⎯⎯
E0Ag = 0.799V
→ Ag(s) ;
(ii) Pb 2+ (aq) + 2e – ⎯⎯
E0Pb = –0.126V
→ Pb(s)
for the oxidation of Pb, the half reaction is reverse of equation (ii)
(iii) Pb(s) ⎯⎯
→ Pb 2+ (aq) + 2e –
E0Pb = 0.126V
now add equation (i) and (iii),
Ag + (aq) + e – ⎯⎯
→ Ag(s) (reduction)
E0Ag = 0.799V
Pb(s) ⎯⎯
→ Pb 2+ (aq) + 2e – (oxidation)
E0Pb = 0.126V
0
Pb(s) + 2Ag + (aq) ⎯⎯
→ 2Ag(s) + Pb 2+ (aq) E = 0.925V
E0 for the overall reaction is positive. Hence, the overall reaction is spontaneous. This
indicates that Ag+ can oxidize Pb.
(b)
Ans.
Cu2+ can oxidize Cd to Cd2+ under standard state conditions.
E0Cu = 0.337 V, E0Cd = –0.403V
E0Cu = 0.337V
(i) Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s) ;
(ii) Cd 2+ (aq) + 2e – ⎯⎯
→ Cd(s) ;
E0Cd = –0.403V
Chapter - 4 Electrochemistry
268
∴
for oxidation Cd(s), the equation is;
E0Cd = +0.403V
(iii) Cd(s) ⎯⎯
→ Cd 2+ (aq) +2e – ;
Now add eq. (i) and (iii);
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s) (reduction) ;
E0Cu = 0.337V
Cd(s) ⎯⎯
→ Cd 2+ (aq) + 2e – (oxidiation) ;
E0Cd = 0.403V
(overall) Cd(s) + Cu 2+ (aq) ⎯⎯
→ Cu(s) + Cd 2+ (aq) ;
E0Cd = 0.740V
ΘΘ
ΘΘ∵ E0 for the overall reaction is positive. Hence, the overall reaction is spontaneous. This indicates
that Cu2+ can oxidize Cd.
*10.
Ans.
How can Nernst equation be used to show that electrode potential is equal to the standard
electrode potential by putting appropriate values in Nernst equation? Use Cd2+|Cd half cell
for illustration.
The cell potential or electrode potential is equal to the standard potential if the concentrations
of reactants and products are unity.
Cd 2+ (aq) + 2e – ⎯⎯
→ Cd(s)
E = E0 –
0.0592
[products]
log 10
n
[reactants]
0.0592
[Cd(s)]
log 10
2
[Cd 2+ ]
The concentration of solid phase [Cd(s)] is taken to be unity.
If [Cd2+] = 1 as in 1 M solution
0
E= E –
0
E= E –
*11.
Ans.
0.0592
. log 101 = E0
2
How are ΔG0, E0cell and equilibrium constant related for a particular reaction?
The relation between standard free energy change of cell reaction and emf of the cell is given
by the equation.
ΔGº = –nFE0cell
The relation between standard Gibbs energy change for a reaction and its equilibrium constant
is given by the equation.
ΔGº = –RT ln K
Combining these two equations we obtain,
–nFE0cell = –RT ln K
RT
ln K
E0 =
nF
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*12.
Ans.
=
2.303 RT
log 10 K
nF
=
0.0592
log 10 K at 25ºC
n
How does the equation ΔG0 = –nFEº explain that an electrical potential is an intensive property?
E0cell values are intensive properties
We have, ΔG0 = –nFE0cell
Gibbs energy (ΔG0) is an extensive property since it depends on the amount of substances.
If stoichiometric equation of redox reaction is multiplied by 2 that is the amount of substances
oxidised and reduced are doubled, ΔGº doubles. The number of electron transferred ‘n’ also
doubles.
ΔG 0
2ΔG 0
ΔG 0
becomes = –
= –
nF
2nF
nF
Thus, E0cell remains constant. This indicates that the electrical potential is an intensive property
which does not depend on the amount of substances. Therefore, whenever, the half reaction
is multiplied by a numerical factor to balance the no. of electrons, the corresponding E0 value
is not to be multiplied by that factor.
Hence the ratio, E0cell = –
*13.
Ans.
*14.
Ans.
*15.
Ans.
Arrange the following oxidizing agents in order of increasing strength under standard state
conditions. The standard potentials for the reduction half reactions are given.
Ag+(aq) (0.8V), Al3+(aq) (–1.66V), F2(g) (2.87V), Cl2(g) (1.36V), I2(s) (0.54V), Cd2+(aq) (–0.4V).
Oxidising strength decreases as :
F2(g) (2.87V) > Cl2(g) (1.36V) > Ag+(aq)(0.8V) > I2(s)(0.54V) > Cd2+(aq) (0.4V) > Al3+(aq)(–1.66V)
Arrange the following reducing agents in order of increasing strength under standard state
conditions, the standard potentials for the reduction half reactions being given.
Al(s) (–1.66V), Cl–(aq) (1.36V), Cu(s) (0.34V), Fe(s) (–0.44V), Br–(aq) (1.09V), Ni(s) (–0.26V)
Increasing strength as reducing agent :
Cl–(aq) (1.36V) < Br–(aq) (1.09V) < Cu(s) (0.34V) < Ni(s) (–0.26V) < Fe(s) (–0.44V) <
Al(s) (–1.66V)
Which species in each of the following pairs is better oxidizing agent under standard state
(a) Br2(l) (1.09V) or Au3+(1.4 V)
The greater E0 value means greater tendency of the species to accept electrons and undergo
reduction. Hence, they are better oxidising agents.
Au3+ (1.4 V) has greater E0 value than Br2(l) (1.09 V),
Hence, it is a better oxidising agent.
Chapter - 4 Electrochemistry
270
(b)
Ans.
H+(aq) or Ag+(aq) (0.8 V)
Ag+(aq) (0.8 V) has greater Eº value than H+(aq)
Hence, Ag+ is a better oxidising agent.
(c)
Ans.
Pb2+(aq) (–0.13 V) or Co2+ (–0.28 V)
Pb2+(aq) (0.13 V) has greater Eº value than Co2+(–0.28 V)
Hence, Pb2+ is a better oxidising agent.
(d)
Ans.
Cl2(g) (1.36 V) or Cr3+ (–0.74 V)
Cl2(g) (1.36 V) has greater Eº value than Cr3+ (–0.74 V)
Hence, Cl2 is a better oxidising agent.
*16.
Which species in each of the following pairs is better reducing agent under standard state
K(s) (–2.93 V) or Mg(s) (–2.36V)
E0 values for K(s) is more negative, it has less tendency to accept electron, on the contrary
it has greater tendency to donate electron and hence, K is a better reducing agent.
Co2+(aq) (1.81 V) or In(s) (–0.14 V)
E0 values for In is negative, it has less tendency to accept electron, on the contrary it has
greater tendency to donate electron and hence, it is a better reducing agent.
(a)
Ans.
(b)
Ans.
(c)
Ans.
Ce3+(aq) (1.61 V) or Ti2+ (–0.37 V)
E0 value for Ti2+ is less hence, it is a better reducing agent.
(d)
Ans.
Hg(l) (0.86 V) or Ni(s), (–0.23 V)
E0 value of Ni(s) is less hence, it is a better reducing agent.
*17.
Ans.
What is redox electrode? Give an example with formulation, electrode reaction and Nernst
equation for electrode potential.
Redox electrode :
(a) The electrode at which oxidation reduction takes place is called redox electrode.
(b) This electrode consists of platinum wire dipped in a solution containing the ions of the
same metal in two different oxidation states. e.g. Fe2+, Fe3+ ion or Sn2+, Sn4+ ions
Formulation : Fe2+(aq), Fe3+(aq) | Pt
Electrode reaction :
Reduction reaction : Fe 3+ (aq) + e – ⎯⎯
→ Fe 2+ (aq)
Nernst equation for electrode potential : In the above reduction reaction n = 1
E
Fe 3+ , Fe 2+
= E0
Fe 3+ , Fe 2+
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0.0592
[Fe 2+ ]
log 10
n
[Fe 3+ ]
S.Y.J.C. Science - Chemistry - Part I
271
*18.
Ans.
What are standard electrode potentials and standard emf of the galvanic cell?
Standard electrode potential :
(a) Definition : It is defined as the difference of electrical potential between metal electrode
and the solution around it, when all the substance involved in electrode reaction are in
their standard states, having 1 M concentrations, the gases at 1 atm pressure at 25ºC.
The standard electrode potential is represented as E0.
(b) Standard electrode potential can be standard oxidation potential or standard reduction potential.
(c) If electrode half reaction is oxidation, than the potential developed is called standard oxidation
potential. (E0oxi)
(d) If electrode half-reaction is reduction, then the potential developed is called standard reduction
potential. (E0red)
Standard emf of galvanic cell :
(a) Definition : The standard emf of the cell is the difference between the standard potential
of cathode (RHE) and the standard potential of anode (LHE), when all the substances
involved in the cell reaction are in their standard state that is solutions are at 1M
concentration, gases at 1 atm pressures and solids and liquids are in pure form at 25ºC.
=
E0(cathode)
–
E0(anode)
(b) E0cell
RHE
LHE
*19.
With electrode reaction and over all cell reaction when lead accumulator behaves as an electrolytic
cell.
(a) During recharging lead accumulator behaves as an electrolytic cell.
(b) The cell must be recharged when its potential falls to 1.8V.
(c) To recharge the cell a potential greater than 2V is applied.
(d) All the cell reaction that occur during discharging are reversed and H2SO4 is regenerated
and its concentration is increased.
(e) During this process the roles of anode and cathode are reversed. Now, PbO2 electrode
is anode (+) and Pb is cathode (–).
(f) Oxidation reaction at anode (+) : It is the reverse of reduction at cathode during discharging.
Ans.
–
PbSO 4 (s) + 2H 2 O(l ) ⎯⎯
→ PbO 2 (s) + 4H + (aq) + SO 2–
4 (aq) + 2e
(g) Reduction reaction at cathode (–) : it is the reverse of oxidation at anode during discharging.
PbSO 4 (s) + 2e – ⎯⎯
→ Pb(s) + SO 2–
4 (aq)
(h) Overall cell reaction during recharging :
2PbSO 4 (s) + 2H 2 O(l ) ⎯⎯
→ Pb(s) + PbO 2 (s) + 2H 2SO 4 (aq)
*20.
Ans.
ΔG0 for a redox reaction depends on the number of electrons transferred. Explain.
(a) In redox reaction of the cell, the amount of electricity passed depends on the number of
moles of electron transferred (n).
Chapter - 4 Electrochemistry
272
(b) Standard free energy change = ΔG0 = –nF E0cell.
(c) If the stoichiometry of the redox reaction changes, (n) also changes, Since free energy
change is an extensive property, therefore ΔG0 also changes with the number of electrons
(n) transferred.
*21.
Write shorthand notation for the cell for each of the following reactions :
(a) Cu 2+ (aq) + 2Ag(s) + 2Br – (aq) ⎯⎯
→ Cu(s) + 2AgBr(s)
Ans.
Cu 2+ (aq) + 2Ag(s) + 2Br – (aq) ⎯⎯
→ Cu(s) + 2AgBr(s)
(i)
2Ag(s) + 2Br – (aq) ⎯⎯
→ 2AgBr(s) + 2e –
(oxidation half-reaction at anode)
(ii) Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s) (reduction half-reaction at cathode)
Shorthand notation of the cell :
Ag(s) | AgBr(s) | Br–(aq)(1M) || Cu2+(aq) (1M) | Cu
(b) Sn2+(aq) is oxidized by Br2(l)
Ans.
Sn 2+ (aq) + Br2 (l ) ⎯⎯
→ Sn 4+ (aq) + 2Br – (aq)
(i) Sn 2+ (aq) ⎯⎯
→ Sn 4+ (aq) + 2e – (oxidation half-reaction at anode)
(ii) Br2 (l) + 2e – ⎯⎯
→ 2Br – (aq) (reduction half-reaction at cathode)
Shorthand notation of the cell
Pt | Sn2+(aq), Sn4+(aq) || Br–(aq) | Br2(l) | Pt
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
Daniell cell operates under non-standard state conditions. If the equation of the cell reaction
is multiplied by 2 then .......
a) E and Eº remain unchanged
b) E is doubled
c) n remains unchanged in Nernst equation
d) Q is halved in Nernst equation
Consider the cell Pt | Cl 2 (g) | HCl(aq) || HBr(aq) | Br2 (l ) Pt . If concentration of HCl is
increased, the cell potential will .......
a) increase
b) decrease
c) remain the same
d) become maximum
Standard cell potential is .......
a) measured at a temperature of 25ºC
b) measured when ion concentrations of aqueous reactants are 1.00 M
c) measured the conditions of 1.00 atm. for gaseous reactants.
d) all of the above
*2.
3.
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273
4.
Zn ⎯⎯
→ Zn 2+ + 2e – , Eº = +0.76V, Cr 3+ + 3e – ⎯⎯
→ Cr , Eº = –0.74V the anode in this
cell is .......
a) Zn
5.
*7.
8.
Cr
c)
Zn2+
d)
Cr3+
Write the cell diagram for the reaction below : Cl 2 (g) + 2Ag(s) ⎯⎯
→ 2Ag + (aq) + 2Cl – (aq)
a)
6.
b)
Ag | Ag + (aq) | Cl 2 (g), Cl – (aq) | Pt
b)
Ag | Ag + (aq), Cl – (aq) | Cl 2 (g) | Pt
c) Pt Cl 2 (g) | Cl – (aq) || Ag + (aq) | Ag
c) Ag | Ag + (aq) || Cl 2 (g), Cl – (aq) | Pt
Chlorine cannot displace ;
a) Fluorine from NaF
b) Iodine from NaI
c) Bromine from NaBr
d) Iodine from KI
0
E of an electrode half reaction is related to ΔGº by the equation, Eº = –ΔGº/nF. If the amount
of Ag+ in the half reaction Ag + + e – ⎯⎯
→ Ag is tripled then .......
a) n is tripled
b) ΔGº reduces to one third
c) Eº reduces to one third
d) all the above
Calculate the potential (in volts) for the following voltaic cell at 25ºC.
Cr | Cr 3+ (0.10M) || Cu 2+ (0.0010M) | Cu
9.
a) 1.25V
b) 1.33V
Nernst equation is given by .......
a)
0
E= E –
c) E = E 0 +
•
*10.
*11.
12.
0.0592
[Reduced state]
log 10
n
[oxidised state]
0.0592
[oxidised state]
log 10
n
[Reduced state]
c)
1.41V
b)
0
E= E +
d)
d)
1.57V
2.303
[oxidised state]
log 10
n
[Reduced state]
2.303
[Reduced state]
0
E = E – n log10 [oxidised state]
Numerical MCQs
0
The reaction, 2Br – (aq) + Sn 2+ (aq) ⎯⎯
→ Br2 (l ) + Sn(s) with the standard potentials, E Sn=
–0.114V, E 0 Br2 = +1.09V, is .......
a) spontaneous in reverse direction
b) spontaneous in forward direction
c) at equilibrium
d) non-spontaneous in reverse direction
The strongest oxidizing agent among the species In3+ (E0 = –1.34V), Au3+ (E0 = 1.4V), Hg2+
(E0 = 0.86V), Cr3+ (E0 = –0.74V) is .......
b) Au3+
c) Hg2+
d) In3+
a) Cr3+
The standard reduction potential of Li+ | Li, Ba2+ | Ba; Na+ | Na and Mg2+ | Mg are –3.05,
–2.73, –2.71 and –2.37V respectively. Which of the following is the strongest oxidising agent
b) Li+
c) Mg 2+
d) Ba2+
a) Na+
(Hint : The species having the maximum value of standard reduction potential will be the strongest
oxidising agent. Greater the reduction potential (less negative) stronger the oxidising agent.)
Chapter - 4 Electrochemistry
274
13.
Which among the following is the strongest reducing agent.
Fe 2+ + 2e – ⎯⎯
→ Fe(–0.44V)
Ni 2+ + 2e – ⎯⎯
→ Ni (–0.25V)
Sn 2+ + 2e – ⎯⎯
→ Sn (–0.14V)
Fe 3+ + e ⎯⎯
→ Fe 2+ (0.77V)
a) Fe
b) Fe2+
c) Ni
d)
(Hint : smaller the reduction potential stronger is the reducing agent)
*14.
Sn
The standard potential for the cell reaction 2Tl (s) + Hg 2+ (1M) ⎯⎯
→ 2T l + (1M) + Hg(l )
where EºTl = –0.34V, EºHg = 0.86V is .......
a) 0.52V
b) –0.52V
c)
–1.2V
d)
+1.2V
*15.
0
The following reaction occurs in a galvanic cell 2Cu + ⎯⎯
→ Cu 2+ + Cu . If E
*16.
and E 0 +
= 0.52V the standard cell potential will be .......
Cu | Cu
a) 0.68V
b) 0.36V
c) –0.36V
d) –0.68V
Consider the following half reaction and choose the correct alternative .......
i)
ii)
*17.
= 0.16V,
Cl 2 (g) + 2e – ⎯⎯
→ 2Cl – (aq) ; Eº = 1.36V
Br2 (l ) + 2e – ⎯⎯
→ 2Br – (aq) ; Eº = 1.07V
iii) I 2 (s) + 2e – ⎯⎯
→ 2I – (aq) ; Eº = 0.53V
b) Cl2 can oxidize Br– but not I–
a) Br2 cannot oxidize I–
–
c) I2 can oxidize Cl
d) Br2 can oxidize I– but not Cl–
Maximum standard emf will be delivered by the cell consisting of the half cells .......
a) F– | F2 and Br– | Br2
b) F– | F2 and Li+ | Li
d) Br– | Br2 and Au+ | Au
c) Li+ | Li and Ca2+ | Ca
1
H (g) ⎯⎯
→ H + (aq) + Ag(s) , where standard potential for
2 2
silver half cell reaction is 0.8V, will be .......
a) –77.2 kJ
b) +77.2 kJ
c) 154.4 kJ
d) –38.6kJ
*18.
+
ΔGº for the reaction Ag (aq) +
*19.
0
The value of constant in Nernst equation E = E –
20.
Cu + | Cu
constant
ln Q at 25ºC is .......
n
a) 0.0592mV
b) 0.0592V
c) 25.7mV
d) 0.0296V
The standard reduction potential of Cu2+ and Ag+ in V are + 0.34 and + 0.80, respectively.
Determine the value of E in volts for the following cell at 25ºC.
Cu | Cu 2+ (1.00M) ||| Ag + (0.0010M) | Ag
a) 0.37V
b)
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0.55V
c)
–0.28V
d)
S.Y.J.C. Science - Chemistry - Part I
0.28V
275
*21.
Consider the cell, Pt | H 2 (g) | H + (aq) || I – (aq) | I 2 (s) . If the standard cell potential is 0.54V
then the standard potential for cathode half reaction will be .......
a) 0V
b) –0.54V
c) +0.54V
•
22.
d)
For the electrochemical cell,
M / M+ || X–/X, E0(M+/M) = 0.44V and E0 (X/X–) = 0.33V
From this data, one can deduce that,
a)
M + X ⎯⎯
→ M + + X – is the spontaneous reaction
b)
M + + X – ⎯⎯
→ M + X is the spontaneous reaction
1.08V
(IIT 2000)
c) Ecell = 0.77V
d) Ecell = –0.77V
(Hint : (b), E0cell = +0.11V, here it is spontaneous)
23.
0
The E M 3+ /M 2+ values for Cr, Mn, Fe and Co are –0.41, +1.57, 0.77 and +1.97V respectively.
For which one of these metals the change in oxidation state from +2 to +3 is easiest?
(AIEEE 2004)
a) Cr
b) Mn
c) Fe
d) Co
[Hint : Give values are reduction potentials M 2+ ⎯⎯
→ M 3+ + e – means oxidation. Oxidation
24.
potentials of Cr will be highest and hence, most easily oxidized]
The standard reduction potential for Fe2+ | Fe and Sn2+ | Sn electrodes are –0.44 and –0.14
volt respectively. For the cell reaction. Fe 2+ + Sn ⎯⎯
→ Fe + Sn 2+ , the standard emf is ........
a) +0.30V
25.
b)
–0.58V
c)
+0.58V
d)
(IIT 1990)
–0.30V
The standard reduction potential values of the three metallic cations, X, Y, Z are 0.52, –3.03
and –1.18V respectively. The order of reducing power of the corresponding metals is ..........
(IIT 1998)
a) Y > Z > X
26.
b)
X>Y>Z
c)
Z>Y>X
For a spontaneous reaction ΔG, equilibrium constant K and E0
d)
Z>X>Y
cell will be respectively ..........
(AIEEEE 2005)
a) –ve, > 1, +ve
b)
+ve, > 1, –ve
c)
–ve, < 1, –ve
d)
–ve, > 1, –ve
[Hint : –ΔGº = RT log K, –veΔG implies that log K will be the i.e. K > 1]
Chapter - 4 Electrochemistry
276
4.9
COMMON TYPES OF CELLS :
Concept Explanation :
The voltaic cells (electrochemical cells) are classified as :
1. Primary voltaic cells - The voltaic cells that cannot be recharged are called primary
voltaic cells e.g. Dry cell.
2. Secondary voltaic cells - The voltaic cells that can be recharged by reversing the direction
of current flow and thereby regenerating the original reactants (chemicals) are called
secondary voltaic cells. e.g. Lead accumulators.
(a) Dry cell (Leclanche’ cell) :
It is a primary voltaic cell which cannot be recharged. It is called dry cell because the
electrolyte is a viscous aqueous paste and not a liquid solution.
1. Construction :
(a) The container of cell is made up of zinc which also acts as anode (negative electrode)
(b) The Zn container is lined from inside with a porous paper to separate it from the
(+)
other materials of the cell.
Brass cap
(c) An inert graphite rod in the centre of
the cell, immersed in an electrolyte paste
Zn container
acts as cathode (positive electrode).
Paper spacer
(d) Graphite rod is surrounded by a paste
of MnO2 and carbon black.
Paste of MnO2+carbon
(e) The rest of the cell is filled with
Graphite rod
electrolyte which is a moist paste of
Paste of
NH4Cl and ZnCl2 (some starch is
NH4CI + ZnCl2
added to the paste to make it thick so
(-)
that it cannot be leaked out).
Dry cell
(f) The cell is sealed at the top to prevent the drying of the paste by evaporation of moisture.
2. Cell reaction :
(a) Oxidation of anode - When the cell operates that is current is drawn from the cell,
metallic zinc is oxidized to Zn2+ ions.
Zn(s) ⎯⎯
→ Zn 2+ (aq) + 2e –
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(b) Reduction at cathode - The electron liberated in oxidation reaction at anode flow
along the container and migrate to the cathode through the external circuit. At cathode
NH4+ ions are reduced.
2NH 4 + (aq) + 2e – ⎯⎯
→ 2NH 3 (aq) + H 2 (g)
The H2 gas formed is oxidised by MnO2 present at the cathode.
H 2 (g) + 2MnO 2 (s) ⎯⎯
→ Mn 2O 3 (s) + H 2O(l )
Thus, MnO2 prevents the collection of H2 gas on the cathode. Hence, net reduction
process is the combination of these two reactions.
2NH 4 + (aq) + 2MnO 2 (s) + 2e – ⎯⎯
→ Mn 2 O 3 (s) + 2NH 3 (aq) + H 2O(l )
(c) Net Cell reaction - The net cell reaction is the sum of oxidation at anode and reduction
at cathode.
Zn(s) ⎯⎯
→ Zn 2+ (aq) + 2e –
(oxidation)
2NH 4 + (aq) + 2MnO 2 (s) + 2e – ⎯⎯
→ Mn 2 O 3 (s) + 2NH 3 (aq) + H 2O(l ) (reduction)
Zn(s) + 2NH 4 + (aq) + 2MnO 2 (s) ⎯⎯
→ Zn 2+ (aq) + Mn 2O3 (s) + 2NH 3 (aq) + H 2O(l )
(overall)
The ammonia produced at cathode combines with Zn to form a soluble compound
containing complex ion
2+
Zn 2+ (aq) + 4NH 3 (aq) ⎯⎯
→ [Zn(NH 3 ) 4 ] 2+ (aq)
(d) Representation :
(e) Cell potential = 1.5V
(f) Cause of irreversibility of dry cell : Zn2+ ions produced at anode combine with ammonia,
NH3 produced at cathode and form very stable complex ions .
Zn 2+ (aq) + 4NH 3 (aq) ⎯⎯
→ [Zn (NH 3 ) 4 ] 2+ (aq)
Since Zn2+ ions are internally removed, the reaction at anode can't be reversed, hence
overall discharging reaction can't be reversed and results in poor life of the cell.
(i) Dry cell can't be recharged.
(ii) If the current from the dry cell is removed rapidly, gaseous H2 can be consumed
rapidly as a result the voltage drops rapidly.
Chapter - 4 Electrochemistry
278
(iii) Zn anode corrodes due to its action with H+ from NH +4 .
2+
Zn (S) + 2H +(aq) ⎯⎯
→ Zn (aq)
+ H 2(g)
3.
(a)
(b)
(c)
(h) Applications :
(i) Dry cell is used as a source of electric power in radios, flashlights, torches, clocks, etc.
(ii) Since they are available in small size, they can be used conveniently.
Alkaline dry cell :
The alkaline dry cell is a modified form of Leclanche's dry cell.
If the current from the dry cell is removed rapidly, gaseous H2 can be consumed rapidly
as a result the voltage drops rapidly.
Zn anode corrodes due to its action with H+ from NH +4 .
+
2+
Zn (S) + 2H (aq)
⎯⎯
→ Zn (aq)
+ H 2(g)
(d) To avoid these difficulties, an alkaline dry cell is developed. In this, alkali like NaOH or
KOH is used instead of NH4Cl.
(e) Reaction in alkaline dry cell :
(i) Oxidation at zinc anode :
–
Zn (S) + 2OH (aq)
⎯⎯
→ ZnO (S) + H 2 O (1) + 2e –
Reduction at graphite cathode :
2MnO 2(S) + H 2 O (1) + 2e – ⎯⎯
→ Mn 2 O 3(S) + 2OH –
(aq)
The overall cell reaction :
Zn (S) + 2MnO 2(S) ⎯⎯
→ ZnO (S) + Mn 2 O 3(S)
(f) The voltage of alkaline dry cell is 1.54 V which is more than Leclanche's dry cell which
has voltage of 1.5 volt.
(g) The alkaline dry cell can be represented as, Zn|ZnCl2(aq)NaOH(aq)MnO2(s)|C +
(h) Advantages of alkaline dry cell over Leclanche's cell :
(i) It has higher emf of 1.54 V.
(ii) Cell potential is independent of concentrations of alkali electrolytes.
(iii) No gases are formed as in Leclanche's dry cell where H2 gas is formed.
(iv) There is no decline in voltage even if high electric current is withdrawn.
(v) The alkaline dry cell has longer life than acidic dry cell, since Zn corrodes very slowly
in alkali medium than in acidic medium.
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1. Principle :
(+)
(a) It is a reversible
(-)
electrochemical cell.
(b) In this cell electrical energy
38% H2SO4
is not generated but it is
Acid proof container
previously stored from
external source. Hence, it is
called secondary cell or Pb plates
Pb plates
storage
battery
or packed with
packed with
PbO2
accumulator.
(c) It can be recharged and reused.
2. Construction :
(a) A group of lead plates packed with spongy lead acts as anode (negative electrode)
(b) Another group of lead plates packed with PbO2 acts as cathode (positive electrode)
(c) The electrodes are immersed in an aqueous solution of 38% (by mass) of H2SO4
of density equal to about 1.28 g/mL or 4.963 M which serves as an electrolyte.
3. Notation of the cell (Representation)
Working of the cell :
(i) Cell reactions during discharge (cell functions as galvanic cell) :
(a) Oxidation reaction at anode (negative electrode) : When the cell provides
current, spongy lead is oxidized to Pb2+ ions.
(oxidation)
Pb(s) ⎯⎯
→ Pb 2+ (aq) + 2e –
Pb 2+ (aq) + SO 2–
→ PbSO 4 (s)
4 (aq) ⎯⎯
(precipitation)
Pb(s) + SO 2–
→ PbSO 4 (s) + 2e – (overall oxidation at anode)...(i)
4 (aq) ⎯⎯
(b) Reduction reaction at cathode (positive electrode) : The electrons travel through
the external circuit from anode to cathode. Here PbO2 is reduced to Pb2+ ions.
(reduction)
PbO 2 (s) + 4H + (aq) + 2e – ⎯⎯
→ Pb 2+ (aq) + 2H 2 O(l )
Pb 2+ (aq) + SO 2–
→ PbSO 4 (s)
4 (aq) ⎯⎯
(precipitation)
–
PbO 2 (s) + 4H + (aq) + SO 2–
→ PbSO 4 (s) + 2H 2 O(l )
4 (aq) + 2e ⎯⎯
(overall reduction at cathode)
Chapter - 4 Electrochemistry
...(ii)
280
(c) Net cell reaction during discharge : On adding equation (i) and (ii);
Pb(s) + PbO 2 (s) + 2H 2SO 4 (aq) ⎯⎯
→ 2PbSO 4 (s) + 2H 2 O(l )
(d) The potential of the cell depends on concentration (density) of H2SO4. Density
discharging H2SO4 is consumed and its concentration (density) decreases.
(e) The emf of lead storage cell is 2 Volts.
(ii) Recharging of the cell (cell now functions as electrolytic cells)
(a) The cell must be recharged when its potential falls to 1.8 V.
(b) To recharge the cell a potential greater than 2V is applied.
(c) Thus, all the cell reactions are reversed.
discharging
Pb(s) + PbO 2 (s) + 2H 2SO 4 (aq) 2PbSO 4 (s) + 2H 2O(l )
charging
4. Application :
(a) It is used in laboratory as a source of direct current.
(b) It is used in automobiles.
(c) It is used in invertors.
5.
(a)
(b)
(c)
It can be recharged.
It consists of a cadmium electrode in contact with an alkali and acts as anode while nickel
(1V) oxide, NiO2 in contact with an alkali acts as cathode. The alkali used is moist paste
of KOH.
(d) Reactions in the cell :
(i) Oxidation at cadmium anode :
–
Cd (S) + 2OH (aq)
⎯⎯
→ Cd(OH) 2(S) + 2e –
(ii) Reduction at NiO2(s) cathode :
–
NiO 2(S) + 2H 2 O (1) + 2e – ⎯⎯
→ Ni(OH) 2(S) + 2OH (aq)
The overall cell reaction is the combination of above two reactions.
Cd (S) + NiO 2(s) + 2H 2O (1) ⎯⎯
→ Cd(OH) 2(S) + Ni(OH) 2(S)
(e) Since the net cell reaction doesn't involve any electrolytes but solids, the voltage is independent
of the concentration of alkali electrolyte.
(f) Since the reaction products at both the electrodes are solids, they adhere to the electrode
surface. Hence, these electrode reactions can be reversed during the charging process.
Therefore, this is a chargeable dry cell.
(g) Since no gases are produced, this cell can be sealed.
(h) The cell potential is about 1.4 V.
(i) This cell has longer life than other dry cells.
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4.10 FUEL CELLS :
H2gas
e
H 2O
O2gas
Concept Explanation :
1.
2.
3.
4.
Definition - The galvanic cells in which
the energy of combustion of fuels is
Anode
directly converted into electrical
(-)
energy are called fuel cells.
In these cells, one of the reactants is a
fuel such as H2 gas or CH3OH
In fuel cells, reactants are not placed
within the cell, but they are continuously
supplied to the electrodes.
Example : H2–O2 fuel cell.
cathode
(+)
Hot
KOH
solution
Porous carbon
(a) Hydrogen - Oxygen fuel cell :
electrodes with Pt
1. In this cell fuel used is H2 gas and O2 gas is an oxidizing
agent.
H2 – O2 fuel cell
2. Here, the energy of combustion of H2 is converted into electrical energy.
(b) Principle : It is an electro-chemical cell which converts the energy of combustion of H2
and O2 directly into electrical energy.
(c) Construction :
1. The anode consists of a porous carbon rod containing a small amount of finely divided
platinum that acts as a catalyst.
2. The H2 gas is continuously bubbled through the anode.
3. The cathode is also a porous carbon rod impregnated with finely divided platinum catalyst.
4. O2 gas is bubbled through the cathode.
5. Both the electrodes are immersed in hot KOH solution.
(d) Cell reactions :
1. Oxidation at anode (–) - At anode H2 gas is oxidised to H2O
2H 2 (g) + 4OH – (aq) ⎯⎯
→ 4H 2 O(l ) + 4e –
(oxidation)
...(i)
2. Reduction at cathode (+) - At cathode, O2 is reduced to OH–.
O 2 (g) + 2H 2 O(l ) + 4e – ⎯⎯
→ 4OH – (aq)
(Reduction reaction) ...(ii)
3. Net cell reaction - an adding eq. (i) and (ii)
2H 2 (g) + O 2 (g) ⎯⎯
→ 2H 2 O(l )
(e) Representation :
– Pt | H 2 (g) | OH– | O 2 (g) | Pt +
Chapter - 4 Electrochemistry
(Redox reaction)
282
EMF of the cell (cell potential) :
E0cell = E0red (cathode) – E0red (anode)
= 0.4V – (–0.83V)
= 1.23V
(f) Advantages of fuel cells :
1. The reacting substances are continuously supplied to the electrodes. Hence, unlike conventional
cells the fuel cells do not have to be discharged when the chemicals are consumed.
2. They are non-polluting because the only reaction product is water.
3. The fuel cells provide electricity with an efficiency of about 70% which is almost double
of the efficiency of thermal plants which is only about 40%.
(g) Drawbacks of H2–O2 fuel cell :
1. In practice voltage is less than 1.23V because of deviations from reversible behaviour.
2. H2 gas is hazardous to use and cost of preparing H2 is high.
1.
2.
3.
4.
5.
Application of fuel cells :
The fuel cells have been used in automobiles on experimental basis.
The cell was used for providing electrical power in the space programme.
In spacecrafts it is operated at such a high temperature that water evaporates at the same
rate as it is produced. The vapour is then condensed and pure water is used for drinking
for astronauts.
In future the fuel cells are also likely to be used as power generators for hospitals, hotels
and homes.
The applications of fuel cells using methanol as fuel might also become available in small
electronic products such as cell phones and laptop computers.
4.11 CORROSION :
Concept Explanation :
1.
2.
3.
Definition - ‘The process of
destructive attack on a metal
surface by its environment
through
chemical
or
electrochemical reaction which
leads to loss of metal in the form
of its compounds (oxides,
sulphides, carbonates, sulphates etc.) is called corrosion.
In case of iron the corrosion is called rusting (which is hydrated ferric oxide Fe2O3. H2O.
Corrosion is an electrochemical phenomenon, since on the surface of iron tiny galvanic
cells are formed.
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4.
5.
Rusting occurs in presence of water and air. Water vapours on the surface of the metal
dissolve CO2 from air to from H2CO3. This acts as an electrolyte. H2CO3 ionizes to small
extent producing H+ ions.
At a particular spot on the surface of iron oxidation takes place and it behaves as an
anode.
At anode : 2Fe(s) ⎯⎯
→ 2Fe 2+ (aq) + 2e – (oxidation) Eº = 0.44V
6.
Electron released at anodic spot move through the metal, reach another spot and reduce
O2 in presence of H+ ion and this spot behaves as a cathode.
At cathode : O 2 (g) + 4H + (aq) + 4e – ⎯⎯
→ 2H 2 O(l ) (reduction) Eº = 1.23V
7.
Net cell reaction : The overall reaction is
2Fe(s) + O 2 (g) + 4H + (aq) ⎯⎯
→ 2Fe 2+ (aq) + 2H 2 O(l ) ; Eºcell = 0.44 + 1.23 = 1.67V
8.
When Fe2+ ion migrate from anode region, they come in contact with O2 dissolved in
surface portion by water droplet. They are further oxidised to Fe3+ ions.
4Fe 2+ (aq) + O 2 (g) + 4H + (aq) ⎯⎯
→ 4Fe 3+ (aq) + 2H 2 O(l )
9.
Fe3+ ions form an insoluble hydrated oxide which is deposited as red-brown material called
rust.
2Fe 3+ (aq) + 4H 2O(l ) ⎯⎯
→ Fe 2O 3 .H 2O(s) + 6H + (aq)
(Rust)
Protection of metals from corrosion :
Following methods are used :
1. Coating metal surface by paint : It is the most common method to shield the Fe from
O2 and moisture to prevent rusting.
2. Galvanizing :
(a) Iron can be protected from rust formation by coating it with another metal, which
is more electropositive, such as Zn. The process of deposition of thin layer of Zn
on Fe is called galvanizing.
(b) Standard potential for the half reaction are :
Fe 2+ (aq) + 2e – ⎯⎯
→ Fe(s) ; Eº = –0.44 V
Zn 2+ (aq) + 2e – ⎯⎯
→ Zn(s) ; Eº = –0.763V
Thus, Zn is stronger reducing agent than ‘Fe’ and Zn can be more easily oxidised
than Fe. This shows that during corrosion, Zn is oxidized instead of Fe.
(c) Even if Fe is oxidised to Fe2+, Zn immediately reduces Fe2+ to Fe. Thus, as long
as Fe and Zn are in contact, Zn protects Fe from oxidation even if Zn layer is scratched.
Chapter - 4 Electrochemistry
284
3. Cathodic Protection : The process is which metal in protects from corrosion by
connecting it to a more easily oxidizable much is called Cathodic Protection.
(a) The iron object (pipe or tank) buried in the moist soil to be protected from corrosion
is connected to a more active metal either directly or through a wire.
(b) The iron object acts as cathode and the protecting metal (sacrificial metal) acts as anode.
(c) The anode is gradually used up due to the oxidation of metal to its ions due to loss
of electrons.
(d) These electrons are transferred through the wire to H+ ions present around the iron
object and thus, protect it from rusting.
(e) As iron object acts as cathode, it is called cathodic protection.
(f) For protecting iron objects, generally Mg or Zn are used, which are called sacrificial
anode. These metals oxidize more easily than Fe.
(g) A galvanic cell is formed in which Mg or Zn acts as anode and Fe object acts as
cathode. The moist soil serves as an electrolyte. The half reaction are :
Mg(s) ⎯⎯
→ Mg 2+ (aq) + 2e – ;Eº = 2.36V
O 2 (g) + 4H + (aq) + 4e – ⎯⎯
→ 2H 2 O(l ) ;
Eº = 1.23V
Because Mg oxidizes instead of Fe, the Fe object is protected from rusting.
4. Passivation :
(a) The process in which metal surface is made inactive is called passivation.
(b) The metal to be protected from corrosion is treated with a strong oxidizing agent such
as conc. HNO3. As a result a thin oxide layer is formed on the surface of metal.
5. Alloy formation :
(a) The tendency of Fe to oxidize can be reduced by forming its alloy with other metals.
(b) Example : stainless steel is an alloy of Fe and Cr.
(c) The formation of a layer of chromium oxide protects iron from rusting.
•
*1.
Ans.
*2.
Ans.
*3.
Ans.
Sketch and describe the operation of Dry cell.
Refer 4.9 (a)
Describe the construction and working of H2 – O2 fuel cell.
Refer 4.10
(a) The reacting substances are continuously supplied to the electrodes. Hence, unlike conventional
cells the fuel cells do not have to be discharged when the chemicals are consumed.
(b) They are non-polluting because the only reaction product is water.
(c) The fuel cells provide electricity with an efficiency of about 70% which is almost double
of the efficiency of thermal plants which is only about 40%.
Drawbacks of H2–O2 fuel cell :
(1) In practice voltage is less than 1.23V because of deviations from reversible behaviour.
(2) H2 gas is hazardous to use and cost of preparing H2 is high.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
285
*4.
Ans.
*5.
Ans.
*6.
Ans.
Refer 4.9 (b)
Write electrode reactions and overall cell reaction during the operation of lead storage cell.
Refer 4.9 (b) 3
Predict whether the following reactions occur under standard state conditions.
(a) Oxidation of Ag(s) by Cl2(g). E0Ag = 0.8V, E0Cl2 = 1.36V
(b) Reduction of Fe3+ to Fe2+ by Au(s).
E 0 3+ 2+ = 0.77V, E0Au = 1.4V..
Fe , Fe
0
(a) Since the standard reduction potential of Cl (g) is higher (E
– = 1.36V) than that
(
)
2
Cl 2 /Cl
0
of Ag(s) E Ag + /Ag = 0.8V hence chlorine will oxidise Ag(s) to Ag+, the redox reaction
taking place is 2Ag(s) + Cl 2 (g) ⎯⎯
→ 2Ag + + 2Cl – (aq)
(
)
0
(b) Since the standard reduction potential of Au(s) is higher E Au 3+ /Au = 1.4V than that
3+
2+
(
)
0
couple E Fe 3+ , Fe 2+ = 0.77V , hence Au(s) will not reduce Fe3+ to Fe2+
of Fe , Fe
but Au(s) will oxidise Fe2+ to Fe3+ has greater tendency to accept electrons.
Multiple Choice Questions :
•
1.
Theoretical MCQs :
In the lead storage battery during discharging .......
(a) pH decreases
(b) pH remains same
(c) pH increases
(d) pH increases or decreases depending upon the extent of discharging.
The efficiency of H2 – O2 fuel cell is about .......
(a) 70%
(b)
90%
(c)
40%
(d)
100%
When a lead storage battery is discharged ..........
(IIT 1987)
a) SO2 is evolved
d) sulphuric acid is consumed
The half-cell reactions for rusting of iron are ..........
(IIT 2005)
2.
•
3.
4.
1
2H + + O 2 + 2e – ⎯⎯
→ H 2O , E0 = +1.23V
2
Fe 2+ + 2e – ⎯⎯
→ Fe(s) , E0 = –0.44V
ΔGº (in kJ) for the reaction
a) –76
b) –322
c) –122
[Hint : For emf to be +ve, oxidation should occur at iron electrode
Ecell : 1.23 + 0.44V = 1.67V
ΔGº = –nF Eºcell = –2 × 96500 × 1.67J = –322kJ]
Chapter - 4 Electrochemistry
d)
–176
286
HOURS BEFORE EXAM
Electrochemistry is a branch of physical chemistry, which is concerned with the inter
conversion of chemical and electrical energy.
The conductance (G) of a solution of electrolytes is the inverse of resistance (R) i.e.
G=
1
R
The conductivity (k) of a solution is the conductance of unit cube of solution.
SI unit of conductivity is Sm–1.
The molar conductivity (L) is defined as the conductivity (k) divided by molar concentration
k
C
SI unit of molar conductivity is Sm2mol–1.
The molar conductivity of solution is maximum at zero concentration that is at infinite dilution.
Kohlrausch law states that the molar conductivity of a solution at zero concentration
is the sum of molar conductivities of cation (λº+) and anion (λº–) i.e.
L0 = λº+ + λº–
The cell constant (b) of the conductivity cell is the ratio of the distance between the
(C) i.e. L =
electrodes (l) divided by the area of cross section (a) of the electrode, i.e. b =
l
a
An electrochemical cell is a device to study chemical reactions electrically.
A galvanic cell is an electrochemical cell in which a spontaneous redox reaction is used
to produce an electric current.
Cell reaction consists of two half reaction, oxidation half reaction that occurs at anode
and reduction half reaction that occurs at cathode.
Daniell cell : .........
e–
2+
Anode – Zn | Zn (1M) || (1M) Cu | Cu + cathode
current
2+
E0cell = E0red
– E 0oxi
cathode
anode
R.H.E.
L.H.E.
Reaction in strage battery
discharging
⎯⎯⎯⎯⎯
→ 2PbSO 4 (s) + 2H 2 O (l )
Pb(s) + PbO 2 (s) + 2H 2SO 4 (l ) ←⎯⎯⎯⎯
⎯
charging
Gibbs energy changes are related with cell potential as.
∴ ΔGº = –nFEº
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
287
Faraday (F) is a charge on 1 mole of electrons.
1F = 96500 C/mole e–.
The electrochemical series is the arrangement of electrodes in order of their decreasing
standard reduction potentials.
The cell potentials are giving by Nernst equation,
0
Ecell = E cell –
0.0592
[products]
log10
at 298k
n
[reactants]
The equilibrium constant (K) of the cell is related to E0cell by the equation,
E0cell =
0.0592
log 10 K at 298k
n
Corrosion is an electrochemical process in which Fe is oxidised to form rust.
Galvanisation is a process in which Fe surface is covered with Zn, in order to prevent
Fe from rusting.
NUMERICALS WITH SOLUTION
*1.
TYPE - I - MOLAR CONDUCTIVITY
A conductivity cell filled with 0.01M KCl
gives at 25ºC the resistance of 604 ohms.
The conductivity of KCl at 25ºC is
0.00141Ω–1 cm–1. The same cell filled
with 0.001M AgNO3 gives a resistance
of 6529 ohms. Calculate the molar
conductivity of 0.001M AgNO3 solution
at 25ºC.
Given :
i) For KCl,
C = 0.01M,
R = 604Ω,
–1
k = 0.00141Ω cm–1
ii) For AgNO3,
C = 0.001M,
R = 6529Ω
To find :
i) Λ = ?
ii) b = ?
Solution :
i) The cell constant of the cell is given by
ii)
∴
iii)
b = k. R
For KCl, k = 0.00141Ω–1cm–1
and R = 604Ω
Here, b = 0.00141 (Ω–1cm–1) × 604(Ω)
= 0.8516 cm–1.
The conductivity of 0.001M AgNO3 is
given by,
b
k=
where, R = 6529Ω
R
k =
0.8516 (cm –1 )
6529(Ω)
= 1.304 × 10–4Ω–1cm–1
The molar conductivity of 0.001M AgNO3
is given by,
Λ =
Λ =
1000k
C
1000(cm 3L–1)×1.304×10 –4 (Ω –1cm –1)
0.001(mol L–1)
Λ = 130.4Ω
Ω –1cm2mol –1
Chapter - 4 Electrochemistry
288
*2.
A conductivity cell filled with 0.1M KCl
gives at 25ºC a resistance of 85.5 ohms.
The conductivity of 0.1M KCl at 25ºC is
0.01286 ohm–1cm–1. The same cell filled
with 0.005 M HCl gives a resistance of
529 ohms. What is the molar conductivity
of HCl solution at 25ºC?
Given :
i) For KCl,
C = 0.1M
R = 85.5Ω
–1
ii) k = 0.01286Ω cm–1
For HCl,
C = 0.005M
R = 529Ω
To find :
i) b = ?
ii) Λ = ?
Solution :
i) Cell constant (b) = k.R
For KCl,
k = 0.01286(Ω–1 cm–1)
R = 85.5Ω
∴
b = 0.01286(Ω–1 cm–1) × 85.5(Ω)
= 1.0995 cm–1.
(ii) The conductivity of 0.005M HCl is given
by,
k
∴
=
k
=
b
where, R = 529Ω
R
(iii)
= 2.078 × 10 Ω cm
The molar conductivity of 0.005 M–HCl
is given by,
1000k
Λ=
C
=
–1
–1
1000(cm 3L–1)×2.078×10 –3 (Ω –1cm –1)
Ω
Λ = 415.6Ω
0.005(mol L–1)
–1
The molar conductivity of 0.05M BaCl2
solution at 25ºC is 223 Ω –1 cm 2
mol–1. What is its conductivity?
Given :
C = 0.05M
L = 223Ω–1 cm2 mol–1
To find :
k=?
Solution :
1000k
L×C
Λ=
or k =
C
1000
∴
k =
∴
k =
cm2mol –1
Unique Solutions ®
223(Ω –1cm2 mol–1 ) × 0.005 (mol L–1 )
1000(cm3 L–1 )
0.01115 Ω –1cm–1
*4.
The conductivity of 0.02M AgNO3 at 25ºC
is 2.428 × 10–3 Ω–1 cm–1. What is its molar
conductivity?
Given :
C = 0.02M
k = 2.428 × 10–3 Ω–1 cm–1
To find :
L =?
Solution :
1000k
L =
C
L =
1.0995(cm –1 )
529Ω
–3
*3.
∴
1000(cm3L–1 ) × 2.428 × 10–3 (Ω –1cm –1 )
0.02(mol L–1 )
L = 121.4 Ω–1cm2mol–1
*5.
A conductivity cell filled with 0.02M
H2SO4 gives at 25ºC a resistance of 122
ohms. If the molar conductivity of 0.02M
H2SO4 is 618 Ω–1 cm2 mol–1. What is the
cell constant?
Given : C = 0.02M
R = 122Ω
S.Y.J.C. Science - Chemistry - Part I
289
L = 618 Ω–1 cm2 mol–1
b =?
To find :
Sol.
:
–1
2
–1
248.1 Ω–1 cm2 mol–1 and 126.5 Ω–1 cm2
618(Ω cm mol ) × 0.02(molL )
1000
∴
k =1.236 × 10–2 Ω2cm–1
Now, k =
∴
are respectively 149.7 Ω–1 cm2 mol–1,
–1
k =
The molar conductivities at zero
concentrations of NH4Cl, NaOH and NaCl
1000k
L×C
or k =
L=
C
1000
∴
∴
*7.
b
or b = k.R
R
R = 122Ω
Cell constant (b)
= 1.236 × 10–2 (Ω–1cm–1) × 122(Ω)
Cell constant (b) = 1.507cm –1
mol–1. What is the molar conductivity of
NH4OH at zero concentration?
Given :
L0(NH4Cl) = 149.7Ω–1 cm2 mol–1
L0 (NaOH) = 248.1 Ω–1cm2 mol–1
L0 (NaCl) = 126.5Ω–1cm2 mol–1
To find :
L0(NH4OH) = ?
*6.
A conductivity cell filled with 0.02M Solution :
AgNO3 gives at 25ºC a resistance of 947
According to Kohlrausch law of
ohms. If the cell constant is
independent migration of ions.
2.3 cm–1, what is the molar conductivity
0
0
L0(NH4Cl) = λ NH +4 + λ Cl –
of 0.02M AgNO3 at 25ºC?
0
0
Given :
L0(NaOH) = λNa + + λ OH –
0
0
C = 0.02M,
R = 947Ω,
L0(NaCl) = λNa + + λ Cl –
–1
b = 2.3 cm
Hence,
To find :
L0(NH4Cl) + L0(NaOH) – L0 (NaCl)
L =?
0
0
0
0
0
0
= λ NH +4 λ Cl – + λ Na + – λ OH – – λ Cl – – λ Na +
Solution :
0
0
= λ NH +4 + λ OH –
–1
b
2.3cm
b = k.R or k =
=
= Lº(NH4OH)
R
947Ω
Thus, L0(NH4OH)
= 2.428 × 10–3Ω–1cm–1.
=
∴
1000k
L=
C
=
∴
1000(cm3 L–1 )× 2.428×10–3 (Ω –1cm–1 )
0.02(mol L–1 )
L = 121.4 Ω–1cm2mol–1
∴
L0(NH4Cl) + L0(NaOH) – L0(NaCl)
Lº(NH4OH) = 149.7(Ω–1cm2 mol–1)
+ 248.1(Ω–1cm2 mol–1) – 126.5 (Ω–1cm2 mol–1)
= 271.3 Ω–1cm2 mol–1
The molar conductivity of CH3OOH
at zero
concentration = 271.3 Ω–1cm2 mol–1
Chapter - 4 Electrochemistry
290
What is the molar conductivity of AgI at
TYPE - II - FARADAY’S LAW
zero concentration if the L0 values of NaI,
*9. Estimate the mass of copper metal
AgNO3 and NaNO3 are respectively 126.9
produced during the passage of 5A current
Ω–1 cm2 mol–1, 133.4 Ω–1 cm2 mol–1 and
through CuSO4 solution for 100 minutes.
121.5 Ω–1 cm2mol–1?
The molar mass of Cu is 63.5 g mol–1.
Given :
Given :
–1
2
–1
L0(NaI) = 126.9 Ω cm mol
I = 5A,
t = 100min
L0(AgNO3) = 133.4 Ω–1 cm2 mol–1
i.e. 100 min × 60sec = 6000s,
L0(NaNO3) = 121.5 Ω–1 cm2 mol–1
molar mass of Cu = 63.5 g mol–1.
To find :
To find :
L0(AgI) = ?
mass of Cu metal produced = ?
Solution :
Solution :
According to Kohlrausch Law of
The quantity of electricity passed,
independent migration of ions
Q = I(A) × t(s)
0
0–
∴ Q = 5(A) × 6000(s)
= λ Na + + λ I
L0(NaI)
Q = 30000C
0
0
L0(AgNO3) = λ Ag + + λ NO –
i)
Moles of electrons actually passed
3
*8.
0
0
L0(NaNO3) = λ Na + + λ NO 3–
Q (C)
Hence,
L0(NaI) + L0(AgNO3) – L0(NaNO3)
=
0
λ Na
+
=
0
0
λAg
+ – λ I–
=
∴
∴
+
λ 0I –
+
λ 0Ag +
+
λ 0NO –
3
=
ii)
30000(C)
–
9500(C/mole )
=
96500(C/mole– )
0.3108 mole–
The half reaction at cathode is,
Cu2+ (l ) + 2e– ⎯⎯
→ Cu(s)
λº (AgI)
Thus,
λº(AgI) = λº(NaI) + λº(AgNO 3 ) –
λ0(NaNO3)
Lº(AgI) = 126.9 (Ω–1 cm2 mol–1)
+ 133.4 (Ω–1 cm2 mol–1)
– 121.5 (Ω–1 cm2 mol–1)
= 138.8 Ω–1 cm2 mol–1
The molar conductivity of AgI at zero
Ω–1 cm2 mol–1
concentration = 138.8Ω
Unique Solutions ®
=
∴
moles of Cu produced
mole ratio = moles of electrons required
1 (mol Cu)
=
iii)
=
=
=
iv)
=
2 (mol e – )
= 0.5 mol Cu/mole–
Moles of Cu produced
moles of electrons actually passed × mole
ratio
0.3108 (mol e–) × 0.5 (mol Cu/ mol e–)
0.1554 mol Cu
mass of Cu formed
moles of Cu formed × molar mass of Cu
S.Y.J.C. Science - Chemistry - Part I
291
=
=
0.1554 (mol Cu) × 63.5 (g mol–1 Cu)
9.87 g Cu
Mass of Cu formed = 9.87 g Cu
∴
Q
2160
=
= 720s = 12 min
I
3
Times of electrolysis = 12 min
t=
*10.
How long will it take to produce 2.415 g *11. What current strength in ampere will be
of Ag metal from its salt solution by passing
required to produce 2.369 × 10–3 kg of Cu
a current of 3A? How many moles of
from CuSO4 solution in one hour? How
many moles of electrons are required?
electrons are required? Molar mass of Ag
Molar mass of Cu is 63.5 g mol–1.
is 107.9 g mol–1.
Given :
Given :
I = 3A, mass of Ag produced = 2.415g,
mass of Cu produced = 2.369 ×10–3 kg
–1
= 2.369g
molar mass of Ag = 107.9g mol
To find :
t = 1 hr = 1 × 60 × 60 = 3600s
t = ? no. of moles of electrons = ?
molar mass of Cu = 63.5 g mol–1.
Solution :
To find :
Reduction half reaction at cathode,
I = ? no. of moles of electrons = ?
Solution :
Ag + (aq) + e – ⎯⎯
→ Ag(s)
1 faraday = 96500C = 1 mol electrons
1 mole 1 mole
1 mole
no. of moles of Ag formed
1 mol Cu = molar mass of Cu = 63.5 g
Reduction half reaction;
mass of Ag formed
=
molar mass of Ag
Cu 2+ + 2e – ⎯⎯
→ Cu
1mole 1mole
1mole
2.415
=
mol = 0.02238mol
moles of Cu deposited
107.9
no. of moles of Ag formed =
2.369
=
= 0.0373 mol Cu
0.02238mol
63.5
From the reaction
From the reaction,
∵ 1 mol of Cu requires 2 mole electrons.
∵ 1 mole of Ag requires 1 mole of electrons
∴ 0.02238 mole of Ag will require = 0.02238
∴ 0.0373 mol Cu will require = 2 × 0.0373
= 0.0746 mol electrons
mol electrons
moles of electrons required =
∵ 1 mole of electrons carries a charge of
0.0746mol
96500C,
∴ 0.02238 mole of electron will carry a
Now,
∵ 1 mole of electron = 96500C
charge
∴ 0.0746 mol electron = 96500 × 0.0746
= 0.02238 × 96500
= 7199C
= 2160C
∴ Quantity of electricity passed = Q = 2160C
∴ Quantity of electricity required = Q = 7199C
∵ Q=I×t
Q=I×t
Chapter - 4 Electrochemistry
292
Given :
moles of Zn = 3 moles,
moles of Cr = 1 mole
Current strength (I) = 2 Ampere
To find :
*12. A current of 6 amperes is passed through
(i) Faraday of electricity required for Zn = ?
AlCl3 solution for 15 minutes using Pt
(ii) Faraday of electricity required for Cr = ?
electrodes, when 0.504g of Al is produced. Solution :
What is the molar mass of Al?
(i)
Given :
a) Zn2+ + 2e – ⎯⎯
→ Zn
I = 6A, t = 15 min = 15 × 60 = 900s
mole ratio
mass of Al produced = 0.504g
moles of Zn produced
To find :
=
moles of electrons
molar mass of Al = ?
1 (mol Zn)
Solution :
= 2 (mol e– )
Reduction half reaction,
= 0.5 mol Zn/ mol e–
Al 3+ (aq) + 3e – ⎯⎯
→ Al(aq)
b) moles of electrons required
Quantity of electricity passed = Q
moles of Zn
= I × t = 6 × 900 = 5400C
=
mole ratio
Q
5400
3(mol)
no. of moles of electrons =
=
F
96500
=
= 6 mol e–
–
0.5
(mol
Zn/mol
e
)
= 0.05596mol
c) 1 mole of electron requires = 1F electricity
From half-reaction,
∴ 6 moles of electron requires = 6 F
∵ 3 moles of electrons deposit 1 mole Al
electricity
∴ 0.05596 moles of electron will deposit
(ii)
0.05596
=
= 0.01865 mol Al
a) Cr3+ + 3e– ⎯⎯
→ Cr
3
∴
∵
∴
I=
7199
Q
=
= 2A
t
3600
Now,
0.01865 mole Al weighs = 0.504g
0.504
1 mole Al will weigh =
= 27g
0.01865
Thus, molar mass of Al = 27 g mol–1.
mole ratio =
=
(b)
*13.
(i)
(ii)
How many moles of electrons are required
for the reduction of
3 moles of Zn2+ to Zn
1 mole of Cr3+ to Cr?
How many faradays of electricity will be
required in each case?
Unique Solutions ä
moles of Cr produced
moles of electrons
1 (mol Cr)
=
=
=
3 (mol e – )
= 0.333 mol Cr/mol e–
Moles of electron required
moles of Cr
mole ratio
1 (mol)
0.333 mol Cr/mol e –
3 mole–
S.Y.J.C. Science - Chemistry - Part I
= 3 mole–
293
(c)
∴
∴
Calculate the amounts of Na and chlorine
gas produced during the electrolysis of
fused NaCl by the passage of 1 ampere
current for 25 minutes. Molar masses of
*14. In the electrolysis of AgNO3 solution 0.7g
Na and chlorine gas are 23 g mol–1 and
of Ag is deposited after a certain period
71g mol–1 respectively.
of time. Calculate the quantity of electricity
Given :
required in coulomb. Molar mass of Ag is
I = 1A,
107.9 g mol–1.
t = 25 mins i.e. 25 × 60 = 1500s,
Given :
molar masses of Na = 23g mol–1,
mass of Ag = 0.7g,
molar mass of Cl2 gas = 71g mol–1
–1
molar mass of Ag = 107.9 g mol
To find :
To find :
(i) mass of Na produced = ?
(i) Q = ? (in coulmb)
(ii) mass of Cl2 produced = ?
Solution :
Solution :
(i) moles of Ag produced
Half reaction at the electrodes are
=
1 mole of electron requires = 1 F electricity
3 moles of electron requires = 3 F electricity
mass of Ag
molar mass of Ag
(i)
107.9 (g mol–1 )
=
(ii)
0.0064 mol Ag
=
=
(iii)
=
=
=
(iv)
Na + (aq) + e – ⎯⎯
→ Na(s) (red at cathode)
Cl– (aq) ⎯⎯
→
0.7(g)
=
*15.
(ii)
=
Ag + (aq) + e– ⎯⎯
→ Ag(s)
mole ratio
moles of Ag
1 (mol Ag)
– =
moles of e
1(mole – )
1 mole Ag/mol e–
moles of electron actually passed
moles of Ag
mole ratio
0.0064 (mol Ag)
(iii)
=
∴
1 (mole Ag)/mol e
0.0064 mol e–
Q = moles of electrons actually passed
× 96500 (C/mole–)
C
= 0.0064 (mole–) × 96500
(mol e– )
Q = 626 C
Na
(7iv)
=
=
+ e– (oxi at anode)
mole ratio of Na
1 mol Na
= 1 mol Na/mol e–
1 mol e–
mass of Na produced
I(A) × t(s)
96500(C/mol e– )
× mole ratio of Na ×
molar mass of Na
I = 1A, t = 25 × 60 = 1500s
mass of Na = 23g mol–1
mass of Na produced
=
–
1 Cl (g)
2
2
1(A)×1500(s)
96500(C/mol e– )
×1mol Na/mole– ×23g mol–1
= 0.3575 g
mole ratio of Cl2
0.5 mol Cl 2
1 mol e –
0.5 mol Cl2/mol e–
Chapter - 4 Electrochemistry
294
(v)
=
=
=
mass of Cl2 produced
I(A) × t(s)
× mole ratio of Cl2
96500(C/mole – )
× molar mass of Cl2
1(A) × 1500(s)
× 0.5 mol Cl2/mol e–
96500(C/mole – )
× 71 g mol–1
0.5517 g
*16.
Calculate the mass of Mg and volume of
chlorine gas at STP produced during
electrolysis of molten MgCl2 by the passage
of 2 amperes of current for 1 hour. Molar
masses of Mg and Cl2 are respectively 24
g mol–1 and 71 g mol–1.
Given :
Hence, mass of Mg produced
=
=
Mg
(aq) + 2e
–
–
×
1 mol Mg
96500 (C/mole ) 2 mol e
=
∴
=
=
⎯⎯
→ Mg(s)
–
×24 g mole–1
0.8953g
Moles of Cl2 produced
I(A) × t(s)
=
I = 2A, t = 1hr i.e. 1 × 3600 sec = 3600sec
molar mass of Mg = 24g mol–1
molar mass of Cl2 = 71g mol–1
To find :
(i) mass of Mg = ?
(ii) volume of Cl2 gas = ?
Solution :
∴V =
Half reaction at the electrodes are
2+
2(A) × 3600(s)
96500(C/mol e – )
mole ratio of Cl2
× mole ratio of Cl2
1 mol Cl2
2mol e–
moles of Cl2 produced
1 mol Cl2
2(A) × 3600(s)
×
–
96500 (C/mole ) 2 mol e –
0.0373 mol Cl2
Volume of Cl2 is given by ideal gas equation
nRT
P
n = 0.0373 mol,
R = 0.08205L atm K–1mol–1
T = 298K
P = 1 atm
V=
–1
–1
0.0373(mol)×0.08205(L atm.K mol )×298(K)
1atm
V = 0.9/20 L = 912.0 cm3
...(reduction at cathode)
2Cl – (aq) ⎯⎯
→ Cl 2 (g) + 2e –
=
*17.
...(oxidation at anode)
Mass of Mg produced at cathode
I(A) × t(s)
× mole ratio of Mg
96500(C/mole – )
× molar mass of Mg
I = 2 A, t = 1hr × 60 × 60
= 3600s
molar mass of Mg = 24 g mol–1
1 mol Mg
Mole ratio of Mg =
2 mol e –
Unique Solutions ä
In a certain electrolysis experiment 0.561
g of Zn is deposited in one cell containing
ZnSO4 solution. Calculate the mass of Cu
deposited in another cell containing CuSO4
solution in series with ZnSO4 cell. Molar
masses of Zn and Cu are 65.4 g mol–1 and
63.5 g mol–1 respectively.
Given :
molar mass of Zn = 65.4 mol–1
molar mass of Cu = 63.5 g/mol
mass of Zn deposited = 0.561g
S.Y.J.C. Science - Chemistry - Part I
295
To find :
mass of Cu deposited = ?
Solution :
The half reaction for the formation of Zn
at cathode of ZnSO4
Cell is Zn2+ + 2e – ⎯⎯
→ Zn(s)
mole ratio =
=
=
1 mol– Zn
2 mol e–
moles of Zn deposited
mass of Zn deposited
0.561(g)
=
molar mass of Zn
65.4(g mol–1 )
0.008578 mol Zn
Now,
moles of
mole ratio of
Cu deposited Cu half reaction
=
moles of
mole ratio Zn
Zn deposited
half reaction
Hence,
1(mol Cu)
moles of
Cu deposited
2(mol e– )
=
0.00857 mol Zn 1(mol Zn)
2(mol e– )
∴
∴
=
=
=
*18.
moles of Cu deposited = 0.008578 mol Cu
mass of Cu deposited
moles of Cu deposited × molar mass of Cu
0.00857(mol) × 63.5 (g mol–1)
0.5447g
Two electrolytic cells one containing AlCl3
solution and other containing ZnSO 4
solution are connected in series. The same
quantity of electricity is passed between
the cells. Calculate the amount Zn
deposited in ZnSO4 cell if 1.2g of Al are
deposited in AlCl3 cell. The molar masses
of Al and Zn are 27g mol–1 and 65.4g mol–1
respectively.
Given :
molar masses of Al = 27 g mol–1
molar masses of Zn = 65.4 g mol–1
mass of Al deposited = 1.2g
To find :
mass of Zn deposited = ?
Solution :
Al 3+ (aq) + 3e – ⎯⎯
→ Al(s) ;
1 mol Al
mole ratio = 3 mol e –
Zn 2+ (aq) + 2e – ⎯⎯
→ Zn(s) ;
1 mol Zn
mole ratio =
2 mol e –
moles of Al deposited
=
=
=
mass of Al deposited
molar mass of Al
1.2 g
27g mol –1
0.0444 mol Al
Now,
moles of
mole ratio
Zn deposited
of Zn
= mole ratio
moles of
Aldeposited
of Al
∴
∴
=
=
∴
=
=
=
molesof Zn deposited
0.0444 mol Al
–
=
1(mol Zn) / 2(mole )
–
1(molAl) / 3(mole )
moles of Zn deposited
0.0444(mol Al) ×
3 mol Zn
2 mol Al
0.0666 mol Zn
mass of Zn deposited
moles of Zn deposited ×molar mass of Zn
0.0666 (mol) × 65.4 (g mol–1)
4.36 g
Chapter - 4 Electrochemistry
296
*19.
=
0.463 – 0.0296 × 2.6990
=
0.463 – 0.0799
=
0.3831 V
Ecell = 0.3831 V
(c) ΔG0 = –nFE0cell
= –2 × 96500 × 0.463
= –89360J
= –89.36kJ
0
ΔG = –89.36kJ
ΔG = –nFEcell
= –2 × 96500 × 0.3831
= 73940J
= –73.94kJ.
ΔG = –73.94kJ.
Consider the following cell
2+
Pb(s)| Pb (0.5M)||Cu
2+
(0.001M)|Cu
(a) Write the cell reaction.
(b) Calculate Ecell and E0cell at 25ºC
(c) Calculate ΔG and ΔGº for the cell
reaction.
Given :
Pb(s) | Pb 2+ (aq)(0.5M) || Cu 2+ (aq)
(0.001M) | Cu(s)
0
E 2+
= –0.126V
Pb
E0
/Pb
= 0.337V
[Pb ] = 0.5M; [Cu2+] = 0.001M
To find :
E0cell = ?, Ecell = ?, ΔG0 = ?, ΔG = ?
Solution :
(a) Pb(s) ⎯⎯
→ Pb(aq) + 2e –
(Oxidation at LHE)
∴
(b)
Cu 2+ /Cu
2+
In the electrolysis of water, one of the half
+
–
→ H 2 (g)
reactions is 2H (aq)+ 2e ⎯⎯
Calculate the volume of H2 gas collected at
2+
–
25ºC
and 1 atm pressure by passing 2A for
Cu (aq) + 2e ⎯⎯
→ Cu(s)
1h through the solution. R = 0.08205 L. atm
(Reduction at RHE)
K–1 mol–1. (Hint : Use equation 4.19 and
Pb(s) + Cu 2+ (aq) ⎯⎯
→ Pb 2+ (aq) + Cu(s)
then PV = nRT.)
(Overall cell reaction)
Given
:
n=2
2H + (aq) + 2e – ⎯⎯
→ H 2 (g) ,
0
0
0
(s) E cell = E Cu 2+ /Cu – E Pb 2+ /Pb
I = 2A, t = 1 hr i.e. 1 × 3600 sec. = 3600s,
R = 0.08205 L.atm K–1mol–1, T = 25ºC,
= 0.337 – (–0.126)
i.e. 25 + 273 = 298K
= 0.463V
0
To find :
E cell = 0.463V
volume of H2 gas collected V = ?
Solution :
0.0592
[Pb 2+ ]
2H + (aq) + 2e – ⎯⎯
→ H 2 (g)
. log 10
E 0cell –
Ecell =
2+
n
[Cu ]
1 mol H 2
mole ratio of H2 =
2 mol e –
0.0592
0.5
0.463
–
.
log
=
10
= 0.5 mol H2/mol e–
2
0.001
moles of H2 produced
=
0.463 – 0.0296 log10 500
Unique Solutions ®
*20.
S.Y.J.C. Science - Chemistry - Part I
297
I(A) × t(s)
=
(
96500 C/mol e
–
)
2(A) × 3600(s)
=
=
–
96500(C/mol e )
0.0373 mol H2
V=
...(overall cell reaction)
× 0.5 mol H 2 /mol e –
0.0373(mol)×0.08205(L atm K –1 mol –1)×298(K)
1atm
V = 0.917L
*21.
0
E0cell = E Cl
2
TYPE - 2 - CELL POTENTIALS AND
GIBBS ENERGY CHANGE
Write the cell reaction and calculate the
standard potential of the cell.
Ni(s)| Ni 2+ (1M)||Cl– (1M)|Cl 2 (g,1atm)|Pt
– E0Ni
Cathode – Anode
= 1.36V – (–0.25V)
E0cell = 1.36 + 0.25 = 1.61V
E 0 cell = 1.61V
nRT
P
n = 0.0373 mol,
R= 0.08205 L atm K–1 mol–1,
T = 25ºC + 273 = 298K,
P = 1 atm
V =
Ni(s) + Cl 2 (g) ⎯⎯
→ Ni 2+ (1M) + 2Cl – (1M)
× mole ratio of H 2
*22.
Write the cell reaction and calculate the
emf of the cell,
Pb(s)|Pb2+ (1M)|| KCl(sat)|Hg2Cl2 (s)|Hg
E0anode = – 0.126V, E0cathode = 0.242V
Identify anode and cathode. Name the right
hand side electrode.
Given :
E0Pb = –0.126V, E0Cl = 0.242 V
To find :
Cell reaction = ?
Right hand side electrode = ?
Solution :
Pb(s) ⎯⎯
→ Pb 2+ (1M) + 2e –
E0Cl2 – 1.36V and E 0Ni = –0.25V
...(Oxidation at anode)
Hg 2 Cl 2 (s) + 2e – ⎯⎯
→ 2Hg(l ) + 2Cl – (sat)
Given :
E 0 Cl 2 = 1.36V,
E0Ni = –0.25V
To find :
Cell reaction = ?
E0cell = ?
Solution :
...(reduction at cathode)
Pb(s) + Hg 2 Cl 2 (s) ⎯⎯
→
Pb 2+ (1M) + 2Hg(l ) + 2Cl – (sat)
Ni(s) ⎯⎯
→ Ni 2+ (1M) + 2e –
...(Oxidation at anode)
Cl 2 (g) + 2e – ⎯⎯
→ 2Cl – (1M)
...(reduction at cathode)
...(overall reaction)
Right hand electrode is cathode
– E0Pb
E0cell = E0Cl
Cathode – Anode
= 0.242V – (–0.126V)
E0cell = 0.242 + 0.126 = 0.368V
E0cell = 0.368V
Chapter - 4 Electrochemistry
298
*23.
The following redox reaction occurs in a
galvanic cell.
2Al(s) +3Fe2+ (1M) ⎯⎯
→ 2Al3+ (1M) +3Fe
(a) Write the cell notation.
(b) Identify anode and cathode.
(c) Calculate E0cell if E0anode
= –1.66V and E0cathode = –0.44V
(d) Calculate ΔGº for the reaction.
Given :
E0Fe = –0.44V
E0Al = –1.66V,
To find :
i) E0cell = ?
ii) ΔG = ?
Solution :
i) Al(s) | Al 3+ (1M) || (1M) Fe 2+ | Fe(s)
ii) Al electrode is anode and Fe electrode is
cathode
– E0Al
iii) E0cell = E0Fe
= cathode – anode
= –0.44V – (–1.66)
0
E cell = 1.22(V)
(iv) From equation, n = 6 mol e–
ΔGº = – nFE
= –6(mol e–) × 96500 (C/mol e–) × 1.22V
= –706380 (VC) = –706380J
= –706.38 kJ
(The emf of the cell is positive, the
reaction is spontaneous)
*24.
Calculate the potential of the following cell
at 25ºC.
Sn(s)|Sn2+ (0.025M)|| Ag+ (0.015M) Ag(s)
E0Sn = –0.136V,
E0Ag = 0.799V
Given :
E0Ag = 0.799V,
E0Sn = –0.136V
Unique Solutions ®
To find :
E0cell = ?
Nernst equation = Ecell = ?
Solution :
– E0Sn
E0cell = E0Ag
= (cathode)
(anode)
= 0.799V – (–0.136V)
E0cell = 0.799V + 0.136 = 0.935V
E0cell =
E 0cell –
0.0592(V mole – )
log 10
(Sn 2+ )
2(mole – )
(Ag + ) 2
...because n = 2mol e–
0.025
= 0.935V – 0.0296(V) × log 10 (0.015)
2
=
=
=
∴
0.025
0.000225
0.935V – 0.0296(V) × log10 111.111
0.935V – 0.0296(V) × 2.0457
0.935V – 0.0605V
Ecell = 0.8745V
*25.
Consider the following redox reaction
= 0.935(V) – 0.0296(V) ×log 10
Mg(s) + Sn 2+ (aq) ⎯⎯
→ Mg2+ (aq) + Sn(s)
(a) Write the cell formula
(b) Calculate Ecell for the reaction at
25ºC if
[Mg2+] = 0.035M, [Sn2+] = 0.025M
E0Mg = –2.37V
and E0Sn = –0.136V
(c) Calculate ΔG for the cell reaction.
Given :
Mg(s) + Sn 2+ (aq) ⎯⎯
→ Mg 2+ (aq) +Sn(s)
To find :
i) Cell formula = ?
iii) Ecell = ?
ii) E0cell = ?
iv) ΔG = ?
S.Y.J.C. Science - Chemistry - Part I
299
Solution :
Solution :
i)
Mg(s)|Mg 2+ (0.035M)||(0.025M)Sn 2+ |Sn(s)
Cr(s) ⎯⎯
→ Cr 3+ (0.0065M) + 3e–] × 2
ii)
E0cell =
Co 2+ (0.012)M + 2e – ⎯⎯
→ Co(s) ] × 3
iii)
E 0 cell
E0Sn
– E0Mg
Cathode
Anode
= –0.136V – (–2.37V)
= –0.136V + 2.37V
= 2.234V
Ecell =
E 0 cell –
0.0592(V mole – )
–
2(mol e )
.log 10
2Cr(s) + 3Co 2+ ⎯⎯
→ 2Cr 3+ + 3Co(s)
(0.012M)
(0.0065M)
E0cell = E0Co
– E0Cr
= cathode
anode
= –0.280V – (–0.74V)
= –0.280V + 0.74V
E0cell = 0.460V
(Mg 2+ )
(Sn 2+ )
n=2
=
2.234V – 0.0296(V) × log 10 0.035
0.025
=
=
=
2.234V – 0.0296(V) × log10 1.4
2.234V – 0.0296(V) × 0.1461
2.234V – 0.00432V
E 0 cell = 2.229V
ΔG = –nFE
= –2(mol e–) × 96500 (C/mol e–)
× 2.229V
= –430197 (VC) = –430197J
= –430.197kJ
ΔG= –430.197kJ
iv)
*26.
Ecell = E 0 cell –
6(mole – )
[Cr ]
× log 10
[Co 2+ ]
= 0.460V – 0.00986(V)
(0.0065) 2
× log 10
(0.012) 3
= 0.460V – 0.00986(V)
3+
Ecell
× log 10
4.225 × 10 –5
0.1728 × 10 –5
= 0.460V – 0.00986(V)
× log10 24.450
= 0.460V – 0.00986(V) × 1.3883
= 0.460V – 0.01368(V)
E cell = 0.4463V
ΔG = –nFE
= 6(mol e–) × 96500(C/mol e–)
× 0.4463V
= –258407.7 (VC)
ΔG = –258407.7J = –258.407kJ
Write the cell reaction and calculate the
emf of the cell at 25ºC.
Cr(s) | Cr3+ (0.0065M) || Co2+ (0.012M) | Co(s)
E0Co = –0.280V, E0Cr = –0.74V
What is ΔG for the cell reaction?
Given :
Cr(s)|Cr 3+ (0.0065M)||Co 2+ (0.012)M|Co(s)
E0Co = –0.280V, EºCr = –0.74V
To find :
i) Cell reaction = ? ii) Ecell = ?
iii) ΔG = ?
0.0592(V mol e – )
*27.
(i)
(ii)
Construct a cell consisting of Ni2+ | Ni half
cell and H+ | H2 (g, 1atm) | Pt half cell.
Write the cell reaction.
Calculate emf of the cell if [Ni2+]
= 0.1M, [H+] = 0.05M
and EºNi = –0.257
Chapter - 4 Electrochemistry
300
Given :
(i) Half cell : Ni2+ | Ni, H+ | H2 (g, 1atm)|Pt
(ii) [Ni2+] = 0.1M, [H+] = 0.05M,
E0Ni = –0.257V
To find :
(i) Cell reaction = ?
(ii) E0cell = ?
(iii) Ecell = ?
Solution :
Ni(s) | Ni2+(0.1M) || (0.05M)H+
| H2(g, 1atm) | Pt
E0Ni = –0.257V
E0cell = E 0H 2
– E0Ni
= cathode
Anode
= 0.00
– (–0.257 V)
E 0 cell = 0.257V
Ni(s) ⎯⎯
→ Ni
2+
(0.1M) + 2e
*28.
Sr(s) + Mg2+ (aq) ⎯⎯
→ Sr2+ (aq) + Mg(s)
that occurs in a galvanic cell. Write the cell
formula.
E0Mg = –2.37V and E0Sr = –2.89V
Given :
Sr(s) + Mg 2+ (aq) ⎯⎯
→ Sr2+ (aq) + Mg(s)
E0Mg = –2.37V and E0Sr = –2.89V
To find :
(i) E0cell = ?
(iii) K = ? (equilibrium constant)
Solution :
Sr(s) + |Sr 2+ (aq) ||Mg 2+ (aq)| Mg(s)
E0Mg = –2.37V and E0Sr
–
E0cell = E0Mg
= cathode
= –2.37V –
E0cell = 0.52V
–
...(Oxidation at anode)
2H + (0.05M) + 2e – ⎯⎯
→ H 2 (g, 1atm)
–
0
Ecell =
E cell –
Ecell =
E 0cell –
0.0592(Vmole )
n
0.0592 (Vmole – )
2 (mol e – )
[Ni
× log10
× log 10
2+
]×PH
Calculate the equilibrium constant for the
redox reaction at 25ºC,
2
+ 2
[H ]
Ecell
=
0.1 ×1
(0.05) 2
log10K =
Ni(s) + 2H + (0.05M) ⎯⎯
→
Ni 2+ (0.1M) + H 2 (g, 1atm)
...(reduction at cathode)
0.1
0.0025
= 0.257V – 0.0296 (V) × log10 40
= 0.257V – 0.0296 (V) × 1.6021
= 0.257V – 0.04742V
E cell = 0.2095V
*29.
Ecell = 0.257V – 0.0296(V) × log 10
Unique Solutions ®
(a)
= –2.89V
E0Sr
Anode
(–2.89 V)
0.0592 log K
10
n
0.52 × 2
0.0592
= 17.56
K
= Antilog (17.56)
K = 3.690 × 1017
Using Nernst equation, calculate the
potentials for the following half
reactions :
I2 (s) + 2e– ⎯⎯
→ 2I – , [I – ] = 0.03M
E 0I2 = 0.535V
(b)
Fe 3+ (aq) + e – ⎯⎯
→ Fe 2+ (aq),
S.Y.J.C. Science - Chemistry - Part I
301
[Fe 2+ ]= 0.1M , [Fe3+] = 0.01M,
0.0592(V mol e – )
–
1(mol e – )
E 0Fe 2+ , Fe 3+ = 0.771V
Given :
(i)
(ii)
(Fe2+) = 0.1M
E 0I2 = 0.535V
E
Fe 3+ (aq) + e – ⎯⎯
→ Fe 2+ (aq),
[Fe 2+ ]= 0.1M , [Fe3+] = 0.01M,
0
E Fe
2+ ,Fe 3+
= 0.771V
To find :
(i) E I – /I 2 = ?
(Fe 3+ )
Eel =
E 0el –
E
=
I – /I 2
E0 –
–
I /I 2
0.0592(V mole – )
.log 10[I – ] 2
n
= 0.535V –
0.0592(V mol e – )
2(mole – )
.log10 (0.03) 2
= 0.535V – 0.0296(V) . log10 9 × 10–4
= 0.535V + 4 × 0.0296(V) × 0.9542
= 0.535V + 0.1129V
E I/I 2 = 0.6479V
Fe 3+ (aq) + e – ⎯⎯
→ Fe 2+ (aq)
E 2+ , 3+ = E 0
Fe
Fe
Fe 2+ , Fe 3+
,
Fe 3+
0.0592(V)
. log10 0.1
1
0.01
0.771V –
=
=
0.771V – 0.0592(V) × log10 10
0.771(V) – 0.0592(V) × 1
,
Fe 2+ Fe 3+
= 0.7118V
Consider a galvanic cell that uses the half
reactions.
2H + (aq) + 2e – ⎯⎯
→ H 2 (g)
Mg 2+ (aq) + 2e – ⎯⎯
→ Mg(s)
0.0592(V mole – )
[products]
log10
n
[reactants]
E 0 I 2 = 0.535V, n = 2 mol e–, [I–] = 0.03M
I – /I 2
*30.
Fe 2+
=
E
(ii) E Fe 2+ , Fe 3+ = ?
Solution :
(i) The Nernst equation for electrode potential
is :
(ii)
(Fe 2+ )
0
3+
E Fe
2+ , Fe 3+ = 0.771V, (Fe ) = 0.01M,
I2 (s) + 2e– ⎯⎯
→ 2I – , [I – ] = 0.03M
E
log
Write balanced equation for the cell
reaction. Calculate E0cell, Ecell and ΔGº
if concentrations are 1M each and PH 2 =
10 atm.
E0Mg = –2.37V
Given :
(i) 2H + (aq) + 2e – ⎯⎯
→ H 2 (g) ,
(ii)
Mg 2+ (aq) + 2e – ⎯⎯
→ Mg(s) ,
E 0Mg = –2.37V, PH 2 =10atm
To find :
(i) E0cell= ?
(ii) Ecell= ?
(iii) ΔGº = ?
Solution :
(i) 2H + (aq) + 2e – ⎯⎯
→ H 2 (g)
Chapter - 4 Electrochemistry
302
(ii)
Mg 2+ (aq) + 2e – ⎯⎯
→ Mg(s)
Reverse equation (ii) and then add in eq. (i)
Mg(s) ⎯⎯
→ Mg 2+ (aq) + 2e –
2H + (aq) + 2e – ⎯⎯
→ H 2 (g)
Mg(s) + 2H + (aq) ⎯⎯
→ Mg 2+ (aq) + H 2 (g)
...(Balanced equation)
E0cell
=
E 0H 2
+ E0Mg
cathode
Anode
= 0.00
– (–2.37V)
= 2.37V
E 0 cell
Ecell
2+
[Mg ]×PH 2
0.0592(V mole – )
.log 10
n
[H + ] 2
=
Eº cell –
=
2.37V –
=
2.37V – 0.0296(V) × log10 10
E cell = 2.34V
ΔGº = –nFEº
= –2(mol e–) × 96500 (C/mole–)
× 2.37(V)
= –457410 (VC)
ΔGº = –457410J = –457.410 kJ
0.0592(V mole – )
(1)
log10
×10
2
(1) 2
Solution :
The standard potential of Cu+ | Cu is
greater. Hence, this electrode is cathode.
E0cell = E 0cathode – E0anode
= 0.52V
– 0.16V
0
E cell = 0.36V
ΔGº = –nFEº
= –1(mol e–) × 96500 (C/mol e–)
× 0.36V
= –34740(VC)
ΔGº = –34740J = –34.740kJ
0.0592 . log K
E0cell =
10
n
0.0592 . log K
0.36(V)=
10
1
0.36 × 1
log10K =
= 6.081
0.0592
∴ K = antilog (6.081)
K = 1.205 × 106
Calculate the potential of the following cell
at 25ºC.
Zn | Zn 2+ (0.6M) || H + (1.2M) | H 2
(g, 1 atm) | Pt
E0Zn = –0.763V
Given :
*31. Calculate E0cell, ΔGº and equilibrium
Zn | Zn2+(0.6M) || H+ (1.2M) | H2(g, 1 atm) | Pt,
constant for the reaction.
E0Zn = –0.763V
+
2+
2Cu (aq) ⎯⎯
→ Cu (aq) + Cu(s)
To find :
0
0
E Cu
+ /Cu = 0.52V, and E +
=
2+
Ecell = ?
Cu , Cu
Solution :
0.16V
Zn2+ | Zn half cell has lower potential than
Given :
0
0
hydrogen gas electrode, whose standard
+ /Cu = 0.52V, E Cu + , Cu 2+ = 0.16V
E Cu
potential is zero. Hence, hydrogen gas electrode
2Cu + (aq) ⎯⎯
→ Cu 2+ (aq) + Cu(s)
cathode and Zn2+ | Zn electrode will be anode.
To find :
Zn(s) ⎯⎯
→ Zn 2+ (0.6M) + 2e –
ii) ΔGº = ?
i) E0cell = ?
...Oxidation at anode
iii) K = ? (Equilibrium constant)
Unique Solutions ®
*32.
S.Y.J.C. Science - Chemistry - Part I
303
2H + (1.2M) + 2e – ⎯⎯
→ H 2 (g, 1atm)
...reduction at cathode
Zn(s) + 2H + (1.2M) ⎯⎯
→
Zn 2+ (0.6M) + H 2 (g, 1atm)
E 0cell
=
0
EH
2
...(overall cell reaction)
0
– E zn
= 0.00 – (–0.763V) =
0.763V
The emf of the cell is given by nernst
equation,
Ecell
2+
=
E 0cell –
[Zn ] × PH
0.0592(V mole)
2
log 10
n
[H + ] 2
n = 2 mol e–, [Zn2+] = 0.6M, [H+] = 1.2M,
PH 2 = 1 atm
Ecell
=
0.763(V) –
0.0592 (V mol e – )
–
2(mol e )
log10
0.6×1
t = 4 hrs = 4 × 60 × 60 = 14400 s
To find :
mass of Cu deposited (WCu) = ?
I=?
Solution :
(i) No. of moles of Ag deposited
mass of Ag
4
=
=
molar mass of Ag
107.9
= 0.03707 mol of Ag
Reactions
(a) Ag + + e – ⎯⎯
→ Ag
(half reaction in AgNO3 cell)
(b) Cu 2+ + 2e – ⎯⎯
→ Cu
(half reaction is CuCl2 cell)
mole ratio of Ag
moles of Ag produced
1
= No. of moles of electrons = = 1
1
(1.2) 2
Ecell = 0.763(V) – 0.0296(V) × (–0.3803)
= 0.763(V) + 0.01125
Ecell = 0.7742(V)
A constant electric current flows for 4
hours through two electrolytic cells
connected in series. One contains two
electrolytic cells connected in series. One
contains AgNO3 solution and second
contains CuCl2 solution. During this time
4 grams of Ag are deposited in the first
cell.
(i) How many grams of Cu are deposited in
the second cell?
(ii) What is the current flowing in amperes?
Given :
mass of Ag deposited = 4g
molar mass of Cu = 63.5 g mol–1.
molar mass of Ag = 107.9 g mol–1
mole ratio of Cu =
1
= 0.5
2
moles of Cu produced
mole ratio of Cu
=
moles of Ag produced
mole ratio of Ag
moles of Cu produced
*33.
=
=
b)
∵
∴
∵
∴
∵
∴
mole ratio of Cu
× moles of Ag produced
mole ratio of Ag
0.5
× 0.03707
1
mass of Cu produced = 0.01854 × 63.5
= 1.177g
mass of Cu produced = 1.177g
From the reaction,
1 mol Ag+ requires = 1 mole electrons
0.03707 molAg will require = 0.03707 molAg
1 mol electrons = 1 Faraday
0.03707 mol electron = 0.03707 Faraday
0.03707 Faraday = 0.3707 × 96500
Chapter - 4 Electrochemistry
304
=
∴
Quantity of electricity (Q) 3577C
Q=I×t
3577
I=
= 0.25A
14400
Current passed = 0.25A
*34.
The passage of 0.95A for 40 minutes
deposited 0.7493g of Cu from CuSO4
solution. Calculate the molar mass of Cu.
Given :
I = 0.95A, t = 40 min = 40 × 60 = 2400s
mass of Cu deposited = 0.7493g
To find :
molar mass of Cu = ?
Solution :
Reduction half reaction;
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s)
Q
∴
*35.
∵
2280
96500
= 0.02362 mol
2 mol electron deposit = 1 mol Cu
0.02362 mol electron will deposit
0.02362
=
= 0.01181 mol Cu
2
Now,
0.01181 mol Cu weighs = 0.7493g
0.7493 × 1
1 mol Cu will weigh =
= 63.44g
0.01181
Thus, molar mass of Cu = 63.44 g mol–1.
Molar mass of Cu = 63.44 g mol–1.
A quantity of 0.3 g of Cu was deposited
from CuSO4 solution by passing 4A through
the solution for 3.8 min. Calculate the value
Unique Solutions ®
Cu 2+ (aq) + 2e – ⎯⎯
→ Cu(s)
∴
= I×t
= 0.95 × 2400
= 2280C
no. of moles of electron =
∵
∴
Given :
mass of Cu deposited = 0.3g
I = 4A
t = 3.8 min = 3.8 × 60 = 228s
To find :
Solution :
Q =
I×t
=
4 × 228 = 912C
Reduction half-reaction;
∴
∵
∴
No. of moles of Cu deposited
0.3
=
= 0.004724 mol
63.5
From reduction half reaction;
1 mol Cu = 2 mole electrons
0.004724 and Cu = 2 × 0.004724
= 0.009448 mol electrons
Now,
0.009448 mol electron = 912C
912
1 mol electrons =
= 96528C
0.009448
1 Faraday charge is equal to charge on 1
mol electrons
Set up the cell consisting of H+(aq)| H2(g)
and Pb2+(aq) | Pb(s) electrodes. Calculate
the emf at 25ºC of the cell if [Pb2+] = 0.1M,
[H+] = 0.5M and hydrogen gas is at 2atm
pressure.
E0Pb = –0.126V.
Given :
*36.
H + (aq) | H 2 (g) and Pb 2+ (aq) | Pb(s) ,
E0Pb = –0.126V
S.Y.J.C. Science - Chemistry - Part I
305
[Pb2+] = 0.1M, [H+] = 0.5M,
To find :
PH 2 = 2 atm pressure
i) E Cl – /Cl 2 = ?
Solution :
To find :
(i) Ecell = ?
Solution :
Pb(s)|Pb 2+(aq)(0.1M)||(0.5M)H +(aq) |
H2(g, 2atm)Pt
0
E Pb
= –0.126V
E0cell =
EH0 2
=
E cell =
(cathode)
0.00
–
+0.126V
0
Ecell=
–
0
E Pb
(Anode)
(–0.126V)
(i)
Fe 2+ /Fe
=?
2Cl – (1.2M) ⎯⎯
→ Cl 2 (g,3.6atm) + 2e –
... Eº = 1.36V
The Nernst equation for electrode potential
is,
Eel = E 0 el –
E
Cl – /Cl 2
=
0.0592(Vmole – )
products ⎞
⋅ log 10 ⎛⎜
⎝ reactants ⎟⎠
n
E0
Cl – /Cl
0
E Cl
– /Cl
[Pb 2+ ]× P H 2
0.0592(V mole – )
E 0 cell –
i log 10
n
[H + ] 2
–
ii) E
2
–
2
p Cl 2
0.0592(V mol e – )
.log10
2
n
⎡Cl – ⎤
⎣
⎦
= 1.36V,
n = 2 mol e–,
[Cl–] = 1.2M
2+
n = 2 mol e . [Pb ] = 0.1M,
[H+] = 0.5M,
0
E Cl
– /Cl
PH 2 = 2 atm
=
1.36(V) –
=
=
=
=
1.36(V) – 0.0296(V) × log10 2.4305
1.36(V) – 0.02966 (V) × 0.3856
1.36 (V) – 0.01143
1.348 V
ii)
Fe 2+ (2M) + 2e – ⎯⎯
→ Fe(s)
Ecell = 0.126(V) –
0.0592(V mole – )
0.1×2
.log10
–
2(mole )
(0.5) 2
Ecell = 0.126(V) – 0.0296(V) . log10 0.8
= 0.126 (V) – 0.0296(V) × (–0.0969)
= 0.126 (V) + 0.0029(V)
E cell =0.1289V
Write balanced equations for the half
reactions and calculate the reduction
potentials at 25ºC for the following half
cells :
i) C1– (1.2M) | Cl2 (g, 3.6atm) E0 = 1.36V
ii) Fe2+ (2M) | Fe(s) E0 = –0.44V
Given :
Cl– (1.2M) | Cl2 (g, 3.6 atm) E0 = 1.36V
Fe2+ (2M) | Fe(s) E0= –0.44V
2
0.0592(V mol e – )
2 (mol e – )
×log 10 3.50
(1.2) 2
... Eº = –0.44V
*37.
*38.
E
=
E
=
=
=
=
Eº – 0.0592 log 10 1
2
2
–0.44 – 0.02966 × log.10 0.5
0.44 – 0.02966 × (–0.3010)
0.44 + 0.00892
–0.431 V
Device galvanic cell for each of the
following reactions and calculate ΔGº for
the reaction.
Chapter - 4 Electrochemistry
306
Zn dissolves in HCl to produce Zn2+ and
ΔGº = –nFE0
H2 gas E0Zn = –0.763V
= –6 × 96500 × 0.74
3+
ii) Cr dissolves in dil. HCl to produce Cr
= –428460 J
and H2 gas E0Cr = –0.74V
ΔGº= –428.460 kJ
Given :
Zn dissolves in HCl to produce Zn2+ and *39. The equilibrium constant for the following
H2 gas ,
reaction at 25ºC is 2.9 × 109. Calculate
EºZn = –0.763V
standard voltage of the cell.
3+
Cr dissolves in dil. HCl to produce Cr
Br2 (l) + 2Cl – (aq)
Cl 2 (g) + 2Br – (aq) and H2 gas,
Given :
EºCr = –0.74V
(i) K = 2.9 × 109,
Solution :
Br(l ) + 2Cl – (aq)
Cl 2 (g) + 2Br – (aq) i) Zn(s) | Zn2+(aq) || H+(aq) | H2(g) || Pt
To find :
E0Zn = –0.763V
(i) E0cell = ?
0
E0cell = E 0H 2
– E Zn
Solution :
cathode
Anode
Br2 (l ) + 2Cl – (aq)
Cl 2 (g) + 2Br – (aq) = 0.00
– (–0.763V)
n = 2, k = 2.9 × 109
E0cell = +0.763V
0.0592 . log k
ΔGº = –nFEº
E0cell =
10
n
= –2 × 96500 × 0.763
0.0592 . log (2.9 × 10 9 )
= 147259 J
E0cell =
10
2
ΔGº= 147.259 kJ
0.0592 × 9.4624
(ii) Cr(s) | Cr3+(aq) || H+(aq) | H2(g) | Pt
E0cell =
2
E0Cr = –0.74V
0
E
=
0.280V
cell
0
E0cell = E 0H 2
– E Cr
= cathode
Anode
= 0.00
– (–0.74V)
0
E cell = +0.74V
i)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
307
*1.
NUMERICALS FOR PRACTICE
TYPE - I - MOLAR CONDUCTIVITY AND CELL CONSTANT
The distance between the electrodes of a conductivity cell is 0.98 cm and area of cross section
of the electrode is 1.3 cm2. Calculate the cell constant for the cell. [Ans. 0.7538 cm–1]
*2.
The conductivity of 0.005M NaI solution at 25ºC is 6.065 × 10–4 Ω–1. Calculate its molar
conductivity.
[Ans. 121.3 Ω–1cm2 mol–1]
*3.
The molar conductivity of 0.002 M HCl solution is 407.2 Ω–1cm2 mol–1 at 25ºC. Calculate
its conductivity.
[Ans. 8.144 × 10–3 Ω–1cm1]
*4.
A conductivity cell dipped in 0.0005 M NaCl gives at 25ºC a resistance of 13710 ohms. If
the electrodes of the conductivity cell are 0.7 cm apart and the area of cross section of the
electrode is 0.82 cm2, what is the molar conductivity of the solution at 25ºC?
[Ans. 124.6 Ω–1cm2 mol–1]
*5.
A conductivity cell dipped in 0.01 M AgNO3 solution gives at 25ºC a resistance of 1442 ohms.
If the molar conductivity of 0.01m AgNO3 solution is 124.8 Ω–1cm2 mol–1, what is the cell
constant?
[Ans. 1.8 cm–1]
*6.
A conductivity cell filled with 0.01 M KCl gives at 25ºC a resistance of 484 ohms. The conductivity
of 0.01 M KCl at 25ºC is 0.00141Ω–1cm–1. 0.001M of NaCl solution when filled in the same
cell gives a resistance of 5490 ohms at 25ºC. Calculate the molar conductivity of NaCl solution.
[Ans. 124.3 Ω–1cm2 mol–1]
*7.
*8.
*9.
*10.
TYPE - II - KOHLRAUSCH LAW
The molar conductivities of potassium acetate, HBr and KBr are respectively 114.4Ω–1cm2
mol–1, 428.2 Ω–1cm2 mol–1 and 151.9 Ω–1cm2 mol–1 at zero concentration. Calculate the molar
conductivity of acetic acid at zero concentration.
[Ans. 390.7 Ω–1cm2 mol–1]
The molar conductivity of 0.01 M solution of acetic is 16.6 Ω–1cm2 mol–1. If its molar conductivity
at infinite dilution is 390.7 Ω–1cm2 mol–1, what is its degree of dissociation is 0.01 M solution?
Calculate dissociation constant of acetic acid.
[Ans. 1.89 × 10–5]
TYPE - III - FARADAY’S LAWS
During electrolysis of molten CaCl2, 0.8 A current is passed through the cell for 1 h. Calculate
the mass of product formed at cathode. (Molar mass of Ca = 40 g mol–1)
[Ans. 0.5968 g]
How many Faradays of electricity are required to produce 5 g of Mg from MgCl2? (molar
mass of Mg = 24 g mol–1)
[Ans. 0.4166 F]
Chapter - 4 Electrochemistry
308
*11.
In the electrolysis of AgNO3 solution 0.42 g of Ag is produced at cathode after a certain period
of time. Calculate the quantity of electricity required. (Molar mass of Ag = 108 g mol–1)
[Ans. 375.4 C]
*12.
Calculate the mass of copper metal produced during the passage of 2.5A of current through
a solution of CuSO4 for 40 minutes. Molar mass of Cu is 63.5 g mol–1. [Ans. 1.975 g.]
*13.
What current strength will be required to deposit 2.692 × 10–3 kg of Ag in 7 minutes from
AgNO3 solution? Molar mass of Ag is 108 × 10–3 kg mol–1.
[Ans. 5.72 A]
*14.
How much time will be required to liberate 3 × 10–2 kg of iodine from KI solution by the
passage of 4A current through it? Molar mass of iodine is 127 g mol–1. [Ans. 1.583 h]
*15.
Calculate the amount of Cu and Br2 produced in 1.5 hours at inert electrodes in a solution
of CuBr2 by a current of 3.5A. Molar masses of Cu and Br2 are 63.5 g mol–1 and 159.8
[Ans. 15.65 g]
g mol–1 respectively.
*16.
Write half reactions for the electrolysis of molten BaCl2. How many grams of Ba are produced
by current of 1A passed for 15 min? What volume of Cl2 gas will be liberated at 298K and
1 atm pressure? R = 0.08205 L. atm K–1 mol–1. Molar mass of Ba is 137 g mol–1.
[Ans. 0.114 L.]
*17.
In a certain electrolysis experiment 1.55 g of Ag were deposited in one cell containing aqueous
AgNO3 solution. Calculate the mass of Zn deposited in another cell containing aqueous ZnSO4
solution in series with AgNO3 cell. Molar masses of Zn and Ag are 65.4 g mol–1 and 107.9
[Ans. 0.47g.]
g mol–1 respectively.
*18.
In a certain electrolysis experiment 0.79g of Ag was deposited in one cell containing aqueous
AgNO3 solution, while 0.231 g of an unknown metal X was deposited in another cell containing
aqueous XCl2 solution is series with AgNO3 cell. Calculate the molar mass of X. Molar mass
[Ans. 63.11 g mol–1]
of Ag is 107.9 g mol–1.
*19.
In an electrolysis experiment 0.18 g of Al was deposited in a cell containing aqueous AlCl3
solution. Calculate the mass of Cu deposited in another cell containing aqueous CuSO4 solution,
in series with AlCl3 cell. Molar masses of Al and Cu are 27 g mol–1 and 65.4 g mol–1 respectively.
[Ans. 0.654 g]
*20.
TYPE - IV - CELL POTENTIAL
Write the cell reaction and calculate the standard emf of the cell.
Cd | Cd2+ (1M) || Ag+(1M) | Ag E0Cd = –0.403V and E0Ag = 0.799V
*21.
[Ans. 1.202V]
Construct a cell from Ni2+ | Ni and Cu2+ | Cu half cells. Write the cell reaction and calculate
the standard potential of the cell.
[Ans. 0.573V]
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
309
*22.
The standard emf of the following cell is 0.463V Cu | Cu2+(1M) || Ag+(1M) | Ag
If the standard potential of Cu electrode is 0.337V, what is the standard potential of Ag electrode?
[Ans. 0.800V]
*23.
Calculate equilibrium constant for the reaction.
Ni(s) + 2Ag + (aq) ⎯⎯
→ Ni 2+ (aq) + 2Ag(s) at 25ºC EºNi = –0.25V and EºAg = 0.799V
[Ans. 2.754 × 1035]
*24.
Predict whether Cd2+(aq) ion can oxidize Zn(s) or Cu(s) under standard state conditions. Eº
values for half reactions of Cd2+, Zn(s) and Cu(s) are –0.403V,–0.763V and 0.337V respectively.
[Ans. It means Cd2+ ion cannot oxidize metallic Cu]
TYPE - V - NERNST EQUATION
*25.
Construct a cell containing Zn2+ | Zn half cell and hydrogen gas electrode. What will be the
emf of the cell at 25ºC if [Zn2+] =0.24M, [H+] = 1.6M and PH 2 = 1.8 atm? EºZn = –0.763V.
[Ans. 0.786V]
*26.
Calculate Eºcell, Ecell and ΔG for the following reaction at 25ºC.
Mg(s) + Sn 2+ (0.003M) ⎯⎯
→ Mg 2+ (0.04M) + Sn(s)
EºMg = –2.37V and EºSn = – 0.14V
*27.
[Ans. –429.6 kJ.]
Write the cell reaction and calculate the emf of the cell.
Pt | H2(g, 1atm) | H+(0.5M) || KCl (1M) | Hg2Cl2(s) | Hg(l)
at 25ºC. Ecalomel = 0.28V.
*28.
[Ans. 0.2978V]
Calculate the emf of the cell Zn(s) | Zn2+(0.008M) || Cr3+ (0.01M) | Cr
at 25ºC, EºZn = –763V, EºCr = –7.4V
*29.
(a)
(b)
[Ans. 0.0456]
Using Nernst equation, calculate the potentials of the following half reactions at 25ºC.
0
–
Br (l) + 2e – ⎯⎯
→ 2Br – (aq) [Br ] = 0.01M and E Br = 1.08V
2
Cu
2+
2
(aq) + e
–
0
⎯⎯
→ Cu (aq) [Cu ] = 0.1M, [Cu ] = 0.05M, E Cu + , Cu 2+ = 0.153V
+
+
2+
[Ans. 0.1352V]
Chapter - 4 Electrochemistry
310
1.
Ans.
2.
Ans.
Ans.
3.
Ans.
4.
Ans.
HIGHER ORDER THINKING SKILLS (HOTS)
What would happen if no salt bridge is used in an electrochemical cell such as Zn-Cu cell?
The metal ions (Zn2+ ions) which are produced by the loss of electron are accumulated in
one electrode and the negative ions (SO2–4 ions) are accumulated in the other electrode. Thus,
both the solution will develop charges and the current will stop. Further, inner circuit is also
not completed.
What is the free energy change for
(a) Galvanic cell
For galvanic cell, ΔG < 0, since redox reaction is spontaneous.
(b) Electrolytic cell
(b) For electrolytic cell, ΔG > 0, since redox reaction is non-spontaneous.
The standard reduction potential values of three metallic cations X, Y, Z are 0.52, –3.03,
–1.18V respectively. What will be the order of reducing power of the corresponding metals.
(a) The standard oxidation potentials of the metals X, Y, Z would be –0.52, + 3.03 and +
1.18V respectively.
(b) Higher is the oxidation potential, more easily the metal is oxidised and therefore, greater
is the reducing power.
(c) Hence, the reducing powers will be in the order Y > Z > X.
State the products of electrolysis obtained on the cathode and the avode when a dilute solution
of H2SO4 with platinum electrodes is electrolysed.
H 2SO 4 (aq) ⎯⎯
→ 2H + (aq) + SO 2–
4 (aq)
+
–
H + OH
H 2 O At cathode : H + + e – ⎯⎯
→ H;
H + H ⎯⎯
→ H 2 (g)
At anode : OH ⎯⎯
→ OH + e ;
4OH ⎯⎯
→ 2H 2 O(l ) + O 2 (g)
Hence, H2 (g) is liberated at the cathode and O2 (g) is berated at the anode.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
Chapter
5
Chemical Kinetics
"If I have seen further than others, it is by standing upon the shoulders
of giants"-Isaac Newton
Introduction :
There are three most important characteristics of chemical reactions : position of equilibrium,
feasibility of chemical reactions and rates of reactions.
SYLLABUS
5.1
Rate of Reaction (Average rate and
(i) Molecularity
instantaneous rate)
(j) Distinquish between order and molecularity
(a) Chemical kinetics
(b) Rate of reaction
Theoretical MCQs
(c) Average rate of reaction
Numerical MCQs
(d) Instantaneous rate of reaction
Theoretical MCQs
Numerical MCQs
5.2
Rate law, Rate constant and order of
reaction and molecularity
5.3
Integrated rate laws
(a) Integrated rate laws
(b) Zero order reaction
(c) First order reaction
(d) Pseudo first order reaction
(e) Experimental determination of rate laws
and order
(a) Rate law
(b) Rate constant or specific reaction rate
Theoretical MCQs
constant (k)
Numerical MCQs
(c) Order of a reaction
(d) Reaction with no order
(e) Elementary reaction
(f) Complex reaction
5.4
Collision theory, activation energy and
catalyst
(g) Rate determining step
(a) Collision theory requirements for a
(h) Reaction intermediate
bimolecular chemical reaction to take place
312
(b) Arrhenius equation for collision theory of
(i) Difference between catalyst and reaction
intermediate
bimolecular reactions
(c) Temperature dependence of reaction rates
(d) Arrhenius equation and temperature variation
Theoretical MCQs
(e) Arrhenius equation and energy of activation
Numerical MCQs
(f) Determination of activation energy (Ea)
Hours before exam
(g) Effect of catalyst on rates of reactions
Numericals with Solution
(h) Effect of catalyst
Numericals for Practice
Higher Order Thinking Skills
In unit 3 of the present text, we have seen how thermodynamic properties, ΔS and ΔG tell
us whether a chemical reaction represented by an equation can occur under the given set of
conditions. The reactions between strong acids and strong bases are thermodynamically favoured
but occur with rapid rates, the study of reaction rates helps to predict how quickly a reaction
mixture approaches equilibrium. Another view to the study of rates gives an information regarding
the mechanisms of chemical reactions. Many reactions occur into a sequence of steps. These
steps can be discovered only after the study of rates at which the reactions take place.
Therefore, chemical kinetics is a branch of science that deals with the study of rates of chemical
reactions, the factors affecting these rates and the mechanisms by which the reactions occur.
In chemical industries the chemical processes are profitable if reactions rates are fast.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
313
5.1. RATE OF REACTION (Average rate and instantaneous rate) :
Concept Explanation :
(a) Chemical kinetics :
1. Definition : “Chemical kinetics is a branch of science that deals with the study of
rates of reactions, the factors affecting these rates and the mechanisms by which
the reactions occur.”
2. Importance of chemical kinetics :
(a) The study of reaction rates helps to predict how quickly a reaction mixture approaches
equilibrium.
(b) The study of reaction rates gives information regarding the mechanisms of chemical
reactions.
(c) Chemical kinetics helps us to determine conditions by which reaction rates could
be altered.
(d) It helps us to study the effect of various factors like temperature, pressure, catalyst
and concentration of reactants on the rate of reaction.
(b) Rate of reaction :
1. Definition : “The change in molar concentration of any one of the reactants or products
per unit time is called rate of reaction”.
2. Explanation : Consider a simple hypothetical reaction,
A→B
In above reaction, the concentration of reactant A decreases and that of product B increases
with respect to time.
Decreasein molar conc.of rectant A
Time taken for decrease
∆[A]
= –
∆t
Rate of disappearance (consumption) of A =
OR
Rate of formation of B =
Increasein molar conc.of product B
time taken for increase
= +
∆[B]
Δt
Because the concentration of A decreases, ∆[A] is a negative quantity. Therefore, the
negative sign in the expression for rate of disappearance of the reactant makes the rate
a positive quantity.
Chapter - 5 Chemical Kinetics
314
In this reaction, the rate of decrease in concentration of reactant, A is same as the rate
of increase in concentration of product B. Therefore,
Rate of reaction = –
∆[A]
∆[B]
=
∆t
∆t
In some chemical reactions, the coefficient of reactants and products are not same for
example,
A + B → 2C
In this reaction the rate of disappearance of A and B is same but the rate of appearance
of C is twice as compared to A and B.
∴
2 × rate of
=
2 × rate of
=
rate of
disappearance of A
disappearance of B
appearance of C
OR
– ∆[A]
– ∆[B]
1 ∆[C]
=
=
Rate of Reaction =
∆t
∆t
2 ∆t
3. Units of rate of reaction :
Unit of Rate of Reaction for liquid state reaction, can be given as,
Change in concentration
Rate of reaction =
time
Since the concentration of reactants or products are normally expressed in moles per litre
(Mole/dm3) and time is expressed in minutes or seconds.
Thus, the unit of rate of reaction is
(Where M = molarity)
Mol L−1. time −1. Or M time−1.
For gaseous reactions, the concentration of reactants and products are expressed in terms
of partial pressures. Therefore, in such cases the unit of rate of reaction will
be atm. time-1.
4. Factors affecting the rate of reaction.
The rate of reaction depends upon the following factors;
(a) Concentration of reactants : Greater the concentration of the reactants greater
is the rate of reaction.
(b) Temperature : Increase in temperature increases the rate of reaction.
(c) Catalyst : Catalyst generally increases the rate of reaction.
(d) Nature of reactants : Reactants in gaseous state react faster as compared to
liquid or solid reactants. In case of solids, powdered solids react faster than solids
in lump form.
(e) Pressure : In case of gaseous reactants an increase in pressure increases the rate
of reaction.
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(c) Average rate of reaction :
1. Definition : The average rate of a reaction is defined as the change in concentration
of a reactant or product divided by the time interval over which the change occurs.
2. Explanation :
→ 2C
Eg: A + B ⎯⎯
Average Rate of Reaction =
3. Example:
– Δ[A]
– Δ[B]
1 Δ[C]
=
=
Δt
Δt
2 Δt
N 2 + 3H 2 ⎯⎯
→ 2NH 3
Average Rate of Reaction =
–
Δ[N 2 ]
–1 Δ[H 2 ]
1 Δ[NH 3 ]
=
=
Δt
3 Δt
2 Δt
4. Determination of average rate of reaction :
Consider the reaction, A → B
(a) The average rate of this reaction
can be determined graphically.
(b) A graph is plotted between different
values of the molar concentration
of the reactant A, on Y-axis and
corresponding values of time on
X-axis.
(c) Let x1 be the molar concentration
of reactant A, at time t1 and Let
x2 be it’s molar concentration at
time t2.
Change in molar conc. = x 2 – x 1
Change in time = t2 – t1
Average Rate of Reaction =
x 2 – x1
Δx
= t –t
Δt
2
1
(i) The average rate of reaction does not give the true rate of reaction.
(ii) As reaction proceeds the molar concentration of reactant decreases hence, the
rate of reaction also decreases.
(iii) Therefore, the rate of reaction does not remain constant over a long interval of
time and hence, it is not the true rate of reaction.
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316
(d) Instantaneous rate of reaction :
1. Definition : “The rate of chemical reaction, measured at a specific instant is called
instantaneous rate of reaction”.
2. Explanation :
A + B ⎯⎯
→ 2C
Instantaneous Rate of Reaction =
– d[A]
– d[B]
1 d[C]
=
=
dt
dt
2 dt
Note : The change in sign from Δ to d in the rate equation.
3.
Example :
N 2 + 3H 2 ⎯⎯
→ 2NH 3
Average Rate of Reaction =
4.
– d[N 2 ]
–1 d[H 2 ]
1 d[NH 3 ]
=
=
dt
3 dt
2
dt
Determination of instantaneous rate of reaction :
Consider the reaction : A → B
In order to determine the
instantaneous rate of reaction,
the progress of a reaction is
followed by measuring the
concentrations of a reactant or
product at different time
intervals. The concentrations of
a reactant or of a product are
plotted against time .
In order to find out, the rate of reaction at very small interval of time say dt, a tangent
is drawn to the curve at time t1, as shown in graph.
Instantaneous rate of consumption of reactant at time t1 = slope of the tangent
– d[A]
=
dt
Instantaneous rate of formation of product at time t1
= slope of the tangent
d[B]
=
dt
– d[A]
d[B]
Instantaneous rate of reaction at time t1 =
=
dt
dt
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•
1.
Ans.
Define Chemical kinetics.
Chemical kinetics is a branch of science that deals with the study of rates of reactions, the
factors affecting these rates and the mechanisms by which the reactions occur.
2.
Ans.
What is the importance of chemical kinetics?
(a) The study of reaction rates helps to predict how quickly a reaction mixture approaches
equilibrium.
(b) The study of reaction rates gives information regarding the mechanisms of chemical reactions.
(c) Chemical kinetics helps us to determine conditions by which reaction rates could be altered.
(d) It helps us to study the effect of various factors like temperature, pressure, catalyst and
concentration of reactants on the rate of reaction.
3.
Ans.
4.
Ans.
5.
Ans.
*6.
Ans.
*7.
Ans.
Define and explain rate of a reaction.
Refer 5.1 (b) 1 and 2.
Give the units of rate of reactions.
Refer 5.1 (b) - (3)
What are the factors affecting rate of a reaction?
Refer 5.1 (b) - (4)
Define average rate of reaction. How is it determined graphically?
Refer 5.1 (c).
Define Instantaneous rate of reaction. How is instantaneous rate evaluated (determined)?
Refer 5.1 (d) 1 and 4
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
The rate of chemical reaction can be expressed in terms of .......
a) rate of consumption of catalyst
b) rate of consumption of reactants only
c) rate of consumption of reactants and formation of products both
d) rate of formation of products only
The rate of a reaction is expressed in the units .......
a) L mol–1t–1
b) mol dm–3t–1
c) Ms
d) M–1s–1
Chemical kinetics, a branch of physical chemistry, deal with .......
a) heat changes in a reaction
b) physical changes in a reaction
c) rates of reactions
d) structure of molecules
For a gaseous reaction the unit of rate of reaction is .......
a) mol dm–3 time–1
b) mol dm–3
c) atm time–1
d) mols–1
*2.
3.
4.
Chapter - 5 Chemical Kinetics
318
5.
6.
7.
•
*8.
*9.
For the reaction, A + B ⎯⎯
→ 2C + D which one is the incorrect statements?
a) Rate of disappearance of A = Rate of disappearance of B
b) Rate of disappearance of A = Rate of appearance of D
c) Rate of disappearance of B = 2 × Rate of appearance of C
1
d) Rate of disappearance of B = × Rate of appearance of C
2
dx
The term
in the rate expression refers to the .......
dt
a) instantaneous rate of reaction
b) average rate of reaction
c) increase in the concentration of reactants
d) concentration of reactants
For the reaction, A + 2B ⎯⎯
→ C , the rate of the reaction at given instant of time is represented
by .......
d[A]
1 d[B]
d[C]
d[A]
1 d[B]
d[C]
a) +
= ⋅
= +
b) –
= – ⋅
= +
dt
2 dt
dt
dt
2 dt
dt
d[A]
1 d[B]
d[C]
d[A]
1 d[B]
d[C]
c) +
= + ⋅
= +
d) –
= + ⋅
= +
dt
2 dt
dt
dt
2 dt
dt
Numerical MCQs :
In the reaction A + 3B ⎯⎯
→ 2C , the rate of formation of C is .......
a) the same as rate of consumption of A
b) the same as the rate of consumption of B
c) twice the rate of consumption of A
d) 3/2 times the rate of consumption of B
The instantaneous rate of reaction 2A + B ⎯⎯
→ C + 3D is given by .......
a)
*10.
dA
dt
b)
c)
d[B]
dt
→ 2NO 2 (g). If
Consider the reaction 2NO(g) + O 2 (g) ⎯⎯
–
d[O 2 ]
will be .......
dt
a) 0.052 M/s
•
11.
1 d[A]
2 dt
b)
0.114 M/s
c)
d)
d[NO 2 ]
= 0.052 M/s then
dt
0.026 M/s
d)
Some statements are given below .......
a) Instantaneous rate of a reaction is the rate at a given point
b) Average rate is over a given time interval
c) As the reaction proceeds it becomes faster and faster
Unique Solutions ®
1 d[D]
3 dt
S.Y.J.C. Science - Chemistry - Part I
–0.026 M/s
319
d) The average rate =
12.
Increase in conc. of reactant
Among the above, the false statement (s)
Time taken
is/are .......
a) only C
b) A and C
c) C and D
d) B, C and D
The average rate and instantaneous rate of a reaction may be same if the time interval is
a) infinitesimally small b) moderate
c) small
d) large
5.2. RATE LAW, RATE CONSTANT AND ORDER OF REACTION AND
MOLECULARITY :
Concept Explanation :
(a) Rate Law :
1. Definition : “The rate law is defined as an experimentally determined equation which expresses
the rate of a chemical reaction in terms of the molar concentration of the reactants.”
2. Explanation : Consider the general reaction aA + bB ⎯⎯
→ cC + dD
a
b
According to the law of mass action, Rate = k[A] [B] ... (i)
But according to the rate law, expression can be written as Rate = k [A]p [B]q ... (ii)
In the above rate law equation, the value of p and q are determined experimentally and
they may or may not be equal to the coefficients of a and b in the reaction.
3. Examples :
→ 2H 2 O(l ) + O 2 (g) : For the above reaction, rate of the reaction is
(a) 2H 2 O 2 (g) ⎯⎯
experimentally found to be proportional to the concentration of H2O2. Hence, the rate
law for the reaction is Rate = k[H2O2].
→ NO(g) + CO 2 (g) : For the above reaction, rate of the reaction
(b) NO 2 (g) + CO(g) ⎯⎯
4.
is experimentally found to be proportional to [NO2]2 and independent of [CO].
Hence, the rate law for the reaction is Rate = k[NO2]2.
Applications of rate law :
(a) By knowing the rate law and the rate constant, the rate of reaction for any given
composition of the reaction mixture can be estimated.
(b) It can be used to estimate the concentration of the reactants and products at any
instant after the start of the reaction.
(c) By knowing the rate law, the mechanism of the reaction can be predicted.
(b) Rate constant or specific reaction rate constant (k) :
1. Definition : It is a constant of proportionality in the rate law expression and is equal
to the rate of reaction when the molar concentration of each of the reactants is unity.
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320
2.
3.
Characteristics of rate constant :
(a) Greater the value of rate constant, faster is the reaction.
(b) Each reaction has a definite value of rate constant at particular temperature.
(c) Value of rate constant increases with increase in temperature.
(d) Value of rate constant does not depend upon the concentration of the reactants.
(e) For a given reaction, rate constant has a higher value in the presence of a catalyst
than in the absence of the catalyst.
(f) Units of rate constant depend upon order of reaction.
Units of rate constant :
The units of rate constant : (Mole Lit–1)1 – n sec–1, where n = order of reaction
Zero order
First order
Second order
Mole Lit–1sec–1
sec–1
Mole–1 Lit sec–1
(c) Order of a reaction :
1. Definition : “The order of a chemical reaction with respect to each reactant is defined
as the exponent to which the concentration term of that reactant, in the rate law
is raised.”
“The overall order of a reaction is defined as the sum of the exponents (powers)
to which the concentration terms in the rate law are raised.
2. Explanation :
Consider a chemical reaction,
aA + bB ⎯⎯
→ products
3.
According to rate law expression, Rates of Reaction = k [A]p[B]q.
Therefore, the order of reaction (n) is written as, n = p + q.
The order of reaction with respect to A is ‘p’ and that of with respect to B is ‘q’.
The overall order of reaction is p + q.
Examples :
(a) 2H 2 O 2 (g) ⎯⎯
→ 2H 2 O(l ) + O 2 (g) : For the above reaction, rate of the reaction is
experimentally found to be proportional to the concentration of H2O2. Hence, the rate
law for the reaction is Rate = k[H2O2].
The reaction is first order in terms of H2O2 and its overall order is also first.
(b) NO 2 (g) + CO (g) ⎯⎯
→ NO (g) + CO 2 (g) : For the above reaction, rate of the
reaction is experimentally found to be proportional to [NO2]2 and independent of
[CO]. Hence, the rate law for the reaction is Rate = k[NO2]2[CO]0.
The reaction is second order in terms of [NO2] and zero order in terms of [CO].
Overall order of reaction = 2 + 0 = 2.
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4.
Characterstics of order of reaction :
(a) It is the sum of powers of the concentration terms in the rate law expression.
(b) It is an experimentally determined quantity.
(c) Order of reaction may be integer, fraction or even zero.
(d) It is not related to the stoichiometric coefficients of the reaction, hence, cannot be
predicted from the stoichiometric balanced equation .
(e) It is defined only in terms of the concentrations of the reactants and not of products.
(f) It may change with change in the experimental conditions.
(g) Order is applicable to elementary as well as complex reactions. The molecularity of
the slowest step will be the order of a complex reaction.
(d) Reaction with no order :
1. There are certain reactions which do not have specific order. Hence, they are
reactions with no order.
2. Examples :
(a) H 2 (g) + Br2 (g) ⎯⎯
→ 2HBr (g)
The rate law for this reaction is found to be rate =
k[H 2 ] [Br2
1
]2
1 + [HBr] / [Br2 ] m
Where m is a constant. The above equation does not correspond to a simple order.
Therefore, the term order cannot be applied for the reaction.
(b) Many enzyme catalysed reactions have no proper order,
Kb
ES ⎯⎯⎯
E + S →P+E
K b [S] [E]
The rate law for this reaction is found to be rate = K + [S]
M
Where E is the enzyme S is the substrate and Kb and KM are constants. This equation
does not have any simple order.
(e) Elementary Reaction :
1. Definition : An elementary reaction is defined as a reaction that takes place in a single
step and cannot be broken down further into simpler chemical reactions.
2. Explanation :
(a) Many reactions that follow a simple rate law actually takes place in a series of steps.
Such reactions are called complex reactions.
(b) Each step in a complex reaction is called as elementary reaction.
(c) This implies that a complex reaction is broken down in a series of elementary reactions.
(d) The overall reaction is obtained by adding all the elementary steps in a reaction.
(e) The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(f) The concept of molecularity is applicable only for a elementary reaction.
Chapter - 5 Chemical Kinetics
322
3.
Example :
O 3 (g) ⎯⎯
→ O 2 (g) + O (g)
The above reaction is a single step elementary reaction.
(f) Complex reaction :
1. Definition : A reaction that follows a simple rate law but occurs in a series of steps
is termed as a complex reaction.
2. Explanation :
(a) A complex reaction occurs in multiple steps.
(b) Each step in a complex reaction is an elementary reaction.
(c) The overall reaction is obtained by adding all the elementary steps in a reaction.
(d) The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(e) Molecularity of a complex reaction has no meaning.
3. Example :
For example, the reaction
2NO 2 Cl(g)
⎯⎯
→ 2NO 2 (g) + Cl 2 (g)
The reaction takes place in two steps :
k1
NO 2Cl(g) ⎯⎯
→ NO 2 (g) + Cl(g)
k2
NO 2 (g) + Cl 2 (g)
(ii) NO 2Cl(g) + Cl(g) ⎯⎯→
(i)
2NO 2Cl(g) ⎯⎯
→ 2NO 2 (g) + Cl 2 (g)
(slow, unimolecular)
(fast, bimolecular)
(overall reaction)
The first step being slower than the second is the rate determining step. The rate law
predicted by slow step is rate = k1[NO2Cl]
It can be seen that Cl is formed in the first step and consumed in the second. Hence,
it is the reaction intermediate.
(g) Rate determining step :
1. The slowest step in the reaction mechanism is called a rate determining step.
2. Explanation :
(a) A complex reaction occurs in multiple steps.
(b) Each step in a complex reaction is an elementary reaction.
(c) The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(d) One of the steps is the slowest step compared to other steps. Such a step is called
rate determining step.
(e) The rate of the overall reaction depends on the rate determining step.
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3.
(f) The overall reaction can never occur faster than its rate determining step.
(g) The rate determining step can occur anywhere in the reaction mechanism.
(h) The rate law of a rate determining step can be obtained from its stoichiometric equation.
Example :
For example, the reaction
→ 2NO 2 (g) + Cl 2 (g)
2NO 2 Cl(g) ⎯⎯
The reaction takes place in two steps :
k1
NO 2Cl(g) ⎯⎯
→ NO 2 (g) + Cl(g)
k2
NO 2 (g) + Cl 2 (g)
(ii) NO 2Cl(g) + Cl(g) ⎯⎯→
(i)
2NO 2Cl(g) ⎯⎯
→ 2NO 2 (g) + Cl 2 (g)
(slow, unimolecular)
(fast, bimolecular)
(overall reaction)
The first step being slower than the second is the rate determining step. The rate law
predicted by slow step is rate = k1[NO2Cl]
It can be seen that Cl is formed in the first step and consumed in the second. Hence,
it is the reaction intermediate.
(h) Reaction intermediate :
1. Definition : “The additional species other than the reactants or products formed in
the reaction mechanism is called reaction intermediate.”
2. Explanation :
(a) The reaction intermediate appears in the mechanism but does not appear in the
overall reaction.
(b) It is always formed in one step of the mechanism and is consumed in the
subsequent step.
(c) The concentration of reaction intermediate does not appear in the rate law, since its
concentration is very small and undetermined.
(d) It is not present at the start of the reaction neither at the end of the reaction.
3. For example, the reaction
2NO 2 Cl(g)
⎯⎯
→ 2NO 2 (g) + Cl 2 (g)
The reaction takes place in two steps :
(i)
k1
NO 2Cl(g) ⎯⎯
→ NO 2 (g) + Cl(g)
k2
NO 2 (g) + Cl 2 (g)
(ii) NO 2Cl(g) + Cl(g) ⎯⎯→
2NO 2Cl(g) ⎯⎯
→ 2NO 2 (g) + Cl 2 (g)
(slow, unimolecular)
(fast, bimolecular)
(overall reaction)
The first step being slower than the second is the rate determining step. The rate law
predicted by slow step is rate = k1[NO2Cl].
Chapter - 5 Chemical Kinetics
324
It can be seen that Cl is formed in the first step and consumed in the second. Hence,
it is the reaction intermediate.
Note : The example given above can be used as an example for Complex reaction,
Rate determining step or for Reaction intermediate.
(i) Molecularity :
1. Definition : “The molecularity of an elementary reaction is defined as the number
of reactant molecules taking part in the reaction.”
2. Explanation :
(a) It is a theoretical concept.
(b) It is always an integer and never be a fraction or zero.
(c) It does not change with experimental conditions.
(d) It is the property of only elementary reactions and has no meaning for complex reactions.
3. On the basis of molecularity, chemical reactions are classified as follows:
(a) Unimolecular reaction: It is the reaction, in which only one molecule of reactant is
involved. for e.g. Br2 (g) ⎯⎯
→ 2Br (g)
(b) Bimolecular reaction: It is the reaction, in which two molecules of reactants are involved.
for e.g. H 2 + I 2 ⎯⎯
→ 2HI
(c) Trimolecular reaction : It is the reaction, in which three molecules of reactants are
involved.
(j) Distinguish between Order and Molecularity
Molecularity
Order
1. It is the sum of the exponents to which the
concentration terms in the rate law are raised.
2. It is purely an experimental property
indicating the dependence of observed
reaction rate on the concentration of reactants.
3. It may be an integer, fraction or zero.
4. It may change with experimental conditions.
5. It is the property of both elementary and
complex reactions.
Unique Solutions ®
1. It is the number of reactant molecules taking
part in an elementary reaction.
2. It is theoretical property indicating the
number of reactant molecules involved in each
3. It is always an integer and never be a fraction
or zero.
4. It does not change with experimental
conditions.
5. It is the property of only elementary reactions
and has no meaning for complex reactions.
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•
*1.
Ans.
Define rate law.
“The rate law is defined as an experimentally determined equation which expresses the rate
of a chemical reaction in terms of the molar concentration of the reactants.”
*2.
Ans.
Explain rate law with example.
Refer 5.2 (a) 1, 2 and 3.
*3.
Ans.
Define rate constant.
It is a constant of proportionality in the rate law expression and is equal to the rate of reaction
when the molar concentration of each of the reactants is unity.
4.
Ans.
Give characteristics of rate constant.
Refer 5.2 (b)
*5.
Ans.
Give the units of rate constants of first and zero order reactions.
The units of rate constant : (Mole Lit –1)1 – n sec–1, where n = order of reaction
Zero order
Mole Lit–1sec–1
*6.
Ans.
First order
sec–1
Explain the term order of a reaction with example.
Refer 5.2 (c) 1, 2 and 3.
7.
Ans.
Give the characteristics of order of a reaction.
(a) It is the sum of powers of the concentration terms in the rate law expression.
(b) It is an experimentally determined quantity.
(c) Order of reaction may be integer, fraction or even zero.
(d) It is not related to the stoichiometric coefficients of the reaction, hence, cannot be predicted
from the stoichiometric balanced equation .
(e) It is defined only in terms of the concentrations of the reactants and not of products.
(f) It may change with change experimental conditions.
(g) Order is applicable to elementary as well as complex reactions. The molecularity of the
slowest step will be the order of a complex reaction.
8.
Ans.
Explain reactions with no order.
Refer 5.2 (d)
*9.
Ans.
Explain the term elementary reaction.
(a) Definition : An elementary reaction is defined as a reaction that takes place in a
single step and cannot be broken down further into simpler chemical reactions.
(b) Explanation :
(1) Many reactions that follow a simple rate law actually takes place in a series of steps.
Such reactions are called complex reactions.
Chapter - 5 Chemical Kinetics
326
(2)
(3)
(4)
(5)
Each step in a complex reaction is called as elementary reaction.
This implies that a complex reaction is broken down in a series of elementary reactions.
The overall reaction is obtained by adding all the elementary steps in a reaction.
The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(6) The concept of molecularity is applicable only for an elementary reaction.
(c) Example :
O 3 (g) ⎯⎯
→ O 2 (g) + O(g)
The above reaction is a single step elementary reaction.
*10.
Ans.
What do you mean by a complex reaction?
(a) Definition : A reaction that follows a simple rate law but occurs in a series of steps
is termed as a complex reaction.
(b) Explanation :
(1) A complex reaction occurs in multiple steps.
(2) Each step in a complex reaction is an elementary reaction.
(3) The overall reaction is obtained by adding all the elementary steps in a reaction.
(4) The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(5) Molecularity of a complex reaction has no meaning.
(c) Example : Refer 5.2 (f).
*11.
Ans.
Explain the terms rate determining step.
(a) The slowest step in the reaction mechanism is called a rate determining step.
(b) Explanation :
(1) A complex reaction occurs in multiple steps.
(2) Each step in a complex reaction is an elementary reaction.
(3) The sequence of elementary reactions in a complex reaction help to determine the
mechanism of a chemical reaction.
(4) One of the steps is the slowest step compared to other steps. Such a step is called
rate determining step.
(5) The rate of the overall reaction depends on the rate determining step.
(6) The overall reaction can never occur faster than its rate determining step.
(7) The rate determining step can occur anywhere in the reaction mechanism.
(8) The rate law of a rate determining step can be obtained from its stoichiometric equation.
Example : Refer 5.2 (g).
*12.
Ans.
Explain the term reaction intermediate.
Refer 5.2 (h).
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327
*13.
Ans.
What is molecularity of an elementary reaction?
(a) Definition : The molecularity of an elementary reaction is defined as the number of
reactant molecules taking part in the reaction.
(b) Explanation :
(1) It is a theoretical concept.
(2) It is always an integer and never be a fraction or zero.
(3) It does not change with experimental conditions.
(4) It is the property of only elementary reactions and has no meaning for complex reactions.
(c) On the basis of molecularity, chemical reactions are classified as follows.
(1) Unimolecular reaction: It is the reaction, in which only one molecule of reactant is
involved. for e.g. Br2 (g) ⎯⎯
→ 2Br (g)
(2) Bimolecular reaction: It is the reaction, in which two molecules of reactants are involved.
for e.g. H 2 + I 2 ⎯⎯
→ 2HI
(3) Trimolecular reaction: It is the reaction, in which three molecules of reactants are
involved.
*14.
Ans.
*15.
Distinguish between order and molecularity.
Refer 5.2 (j)
The gas-phase reaction between NO and Br2 is represented by the equation.
2NO (g) + Br2 (g) ⎯⎯
→ 2NOBr (g)
(a) Write the expression for the rate of consumption of reactants and formation of products.
(b) Write the expression for the rate of overall reaction in terms of rates of consumption
of reactants and formation of products.
Ans.
2 NO (g) + Br2 (g) ⎯⎯
→ 2NOBr (g)
– d[NO]
dt
– d[Br2 ]
Rate of consumption or Br2 at time t =
dt
d[NOBr]
Rate of formation of NOBr at time t =
dt
– d[Br2 ]
– d[NO]
d[NOBr]
(b) Rate of reaction =
=
=
dt
dt
dt
(a) Rate of consumption of No at time t =
*16.
The decomposition of N2O5 is represented by the equation
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g)
(a) How is the rate of formation of NO2 related to the rate of formation of O2?
Chapter - 5 Chemical Kinetics
328
(b) How is the rate of formation of O2 related to the rate of consumption of N2O5?
Ans.
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g)
(a) Rate of formation of NO2 at time t =
d[NO 2 ]
dt
Rate of formation of O2
at time t =
d[O 2 ]
dt
They are related by rate of reaction as follows
Rate of reaction =
d[O 2 ]
1 d[NO 2 ]
=
4 dt
dt
Hence, rate of formation of NO2 = 4 × Rate of formation of O2
(b) Rate of formation of O2 =
d[O 2 ]
dt
– d[N 2 O 5 ]
dt
d[O 2 ]
–1 d[N 2O 5 ]
Rate of reaction =
=
dt
2
dt
1
Hence, rate of formation of O2 = × rate of consumption of N2O5.
2
Rate of consumption of N2O5 =
*17.
Nitric oxide reacts with H2 according to the reaction
2NO(g) + 2H 2 (g) ⎯⎯
→ N 2 (g) + 2H 2 O (g)
Ans.
What is the relationship among d[NO]/dt, d[H2]/dt, d[N2]/dt and d [H2O]/dt?
Rate of consumption of reactants and rate of formation of products are related as follows
Rate of Reaction =
*18.
+d[N 2 ]
–1 d[H 2 ]
+1 d[H 2O]
–1 d[NO]
=
=
=
2 dt
dt
2
dt
2 dt
2
The rate law for the gas-phase reaction 2NO(g) + O 2 (g) ⎯⎯
→ 2NO 2 (g) is rate = k[NO2] [O2].
What is the order of the reaction with respect to each of the reactants and what is the overall
order of the reaction?
Ans.
∴
∴
2NO(g) + O 2 (g) ⎯⎯
→ 2NO 2 (g)
Rate = k [NO]2[O2]
Order of the reaction with respect to NO = 2
Order of the reaction with respect to O2 = 1
Overall order of one reaction = 2 + 1 = 3
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*19.
Ans.
*20.
The reaction H 2 O 2 (aq) + 3I – (aq) + 2H + (aq) ⎯⎯
→ 2H 2 O(l ) + I 3– (aq) is first order in H2O2
and I–, zero order in H+ Write the rate law.
H 2 O 2 (aq) + 3I – (aq) + 2H + (aq) ⎯⎯
→ 2H 2 O(l ) + I 3– (aq)
Since the reaction is first order in H2O2 and I– and zero order in H+,
then Rate = k [H2O2]1 [I–]1 [H+]0
= k [H2O2] [I–]
Identify the molecularity and write the rate law for each of the following elementary reactions :
(a) NO (g) + O 3 (g) ⎯⎯
→ NO 3 (g) + O (g)
(b) H 2 I (g) + I (g) ⎯⎯
→ 2HI (g)
(c) Cl(g) + Cl(g) + N 2 (g) ⎯⎯
→ Cl 2 (g) + N 2 (g)
Ans.
(a) NO(g) + O 3 (g) ⎯⎯
→ NO 3 (g) + O(g)
Molecularity = 2
Rate k = [NO] [O3]
(b) H 2 I (g) + I (g) ⎯⎯
→ 2HI (g)
∴ Rate = k [H2I] [I]
Molecularity = 2
(c) Cl(g) + Cl(g) + N 2 (g) ⎯⎯
→ Cl 2 (g) + N 2 (g)
∴ Rate = k [O]2 [N2]
Molecularity = 3
*21.
Ans.
A complex reaction takes place in two steps as follows :
(i)
NO (g) + O 3 (g) ⎯⎯
→ NO 3 (g) + O (g)
(ii)
(a)
(b)
(c)
NO 3 (g) + O (g) ⎯⎯
→ NO 2 (g) + O 2 (g)
Write the overall reaction.
Identify reaction intermediate.
Identify the rate determining step if the predicted rate law is rate = K[NO] [O3]
(a) Step I
Step II
:
NO (g) + O 3 (g) ⎯⎯
→ NO 3 (g) + O (g)
:
NO 3 (g) + O (g) ⎯⎯
→ NO 2 (g) + O 2 (g)
Overall Reaction : NO(g) + O 3 (g) ⎯⎯
→ NO 2 (g) + O 2 (g)
(b) The reaction intermediates are [NO2] and [O] since they are formed in the first step and
consumed in the next step.
(c) Rate law for above reaction is Rate = k [NO] [O3]
∴ The rate determining step is 1st step i.e. NO(g) + O 3 ⎯⎯
→ NO 3 (g) + O(g) .
Chapter - 5 Chemical Kinetics
330
*22.
What is the order for the following reactions?
(a) 2NO 2 (g) + F2 (g) ⎯⎯
→ 2NO 2 F (g) , rate = k[NO2][F2]
(b) CHCl 3 (g) + Cl 2 (g) ⎯⎯
→ CCl 4 (g) + HCl(g) , rate = k[CHCl 3 ] [Cl 2
Ans.
(a) 2NO 2 (g) + F2 (g) ⎯⎯
→ 2NO 2 F (g) , rate = k[NO2][F2]
∴ Order = 1 + 1 = 2
(b) CHCl 3 (g) + Cl 2 (g) ⎯⎯
→ CCl 4 (g) + HCl(g) , rate = k[CHCl 3 ] [Cl 2
∴ Order = 1 +
*23.
Ans.
*24.
1
]2
1
]2
1
3
=
2
2
Write the rate law for the following reactions :
(a) A reaction that is zero order in A and second order in B.
(b) A reaction that is second order in NO and first order in Br2.
(a) If the reaction is zero order in A and second order is B
Then, Rate = k [A]0 [B]2
Rate = k [B]2.
(b) If the reaction is 2nd order in NO and first order in Br2, then Rate = k[NO]2[Br]1.
Consider the reaction
Tl + (aq) + 2Ce 4+ (aq) ⎯⎯
→ Tl 3+ (aq) + 2Ce 3+ (aq) . The rate law for the reaction is rate = k
[Ce4+] [Mn2+]. In the presence of Mn2+ the reaction occurs in the following elementary steps :
(a) Ce 4+ (aq) + Mn 2+ (aq) ⎯⎯
→ Ce 3+ (aq) + Mn 3+ (aq)
(b) Ce 4+ (aq) + Mn 3+ (aq) ⎯⎯
→ Ce 3+ (aq) + Mn 4+ (aq)
(c) Tl + (aq) + Mn 4+ (aq) ⎯⎯
→ Tl 3+ (aq) + Mn 2+ (aq)
Ans.
Identify (a) catalyst (b) intermediate (c) rate determining step.
(a) The catalyst in the above reaction will be Mn+2. Since Mn+2 is present at the start as
well as end of the reactions.
(b) The intermediate in the above reaction will be Mn3+ and Mn4+ since both Mn+3 and Mn+4
are formed in one step and consumed is the next step.
(c) The rate law given is Rate = k[Ce4+] [Mn2+].
∴ The rate determining step will be Ce 4+ (aq) + Mn 2+ (aq) ⎯⎯
→ Ce +3 (aq) + Mn 3+ (aq) .
*25.
Ans.
Comment on the relationship between coefficients of the balanced overall equation for a reaction
and the exponents to which the concentration terms in the rate law are raised.
What do these exponents represent?
(a) The rate of a reaction at a given temperature depends on the concentration of reactants.
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→ Products
(b) Consider the general reaction : aA + bB ⎯⎯
(c) The dependence of the reaction rate on the concentration of each reactant is given by
an equation called as differential rate law, which is determined experimentally.
(d) Rate ∝ [A]x[B]y, therefore, Rate = K[A]x[B]y, where K = rate constant
(e) The exponents x and y in the rate law are not necessarily equal to the stoichiometric
coefficients (a and b) of A and B in the balanced reaction.
(f) The values of x and y must be experimentally determined.
(g) The values of x and y can be negative, zero or fractions.
(h) For Example : 2H 2 O 2 (g) ⎯⎯
→ 2H 2 O (l ) + O 2 (g)
For the above reaction, rate of the reaction is experimentally found to be proportional to
the concentration of H2O2. Hence, the rate law for the reaction is Rate = k[H2O2].
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
A reaction is first order with respect to reactant A and second order with respect to reactant
B. The rate law for the reaction is given by .......
b) rate = [A][B]2
a) rate = k [A][B]2
c) rate = k [A]2[B]
d) rate = k [A]0[B]2
Molecularity of an elementary reaction .......
a) may be zero
b) is always integral
c) may be semi-integral
d) may be integral, fractional or zero.
The unit of rate constant for first order reaction is .......
b) s
c) s –1
d) min
a) min–2
The order of a reaction .......
a) is never equal to zero or fraction
b) can be predicted from chemical equation of reaction
c) is equal to sum of total number of moles of reactants taking part in reaction
d) always determined experimentally
Catalytic decomposition of PH3 is a reaction of .......
a) Zero order
b) Second order
c) First order
d) Third order
For a reaction, A + 2B → D + E following mechanism is suggested .......
ii) B + C → E (Fast step)
i) A + B → C + D (Slow step)
The rate expression for the reaction.
b) r = k [A]2
a) r = k [A] [B]2
c) r = k [A] [B]1/2
d) r = k [A] [B]
*2.
*3.
4.
5.
6.
Chapter - 5 Chemical Kinetics
332
7.
8.
•
*9.
10.
11.
•
*12.
*13.
5.3
Which of the following statements regarding molecularity of the reaction is wrong?
a) it may be either whole number or fractional
b) it is calculated from the reaction mechanism
c) it depends on the rate determining step
d) it is number of molecules of reactants taking part in a single step chemical reaction
Many reactions proceed in a sequence of steps, so the overall rate of reaction is determined
by .......
a) the order of different steps
b) the slowest step
c) molecularity of the steps
d) the fastest step
Numericals MCQ :
The rate of the first order reaction A ⎯⎯
→ products is 0.01 M/s, when reactant concentration
is 0.2 M. The rate constant for the reaction will be .......
b) 0.05 min–1
c) 0.1 s–1
d) 0.01 s–1
a) 0.05 s –1
2
1/2
For reaction A + B + C → Products, Rate k [A] [B] [C] . The order of the reaction is .......
a) 2.8
b) 2
c) 3
d) 3.5
For a reaction, A + B → Product it is observed that (i) [A] = constant, [B] is doubled, rate
becomes double (ii) [B] = constant, [A] is doubled, rate becomes four times. The order of
the reaction is .......
a) 1
b) 4
c) 2
d) 3
The reaction between H2(g) and ICl (g) occurs in the following steps :
(ii) H 2 + ICl ⎯⎯
(i) H 2 + ICl ⎯⎯
→ HI + HCl
→ I 2 + HCl
The reaction intermediate in the reaction is .......
a) HCl
b) HI
c) I2
d) ICl
The formation of SO3 from SO2 and O2 takes place in the following steps :
(ii) 2NO + O 2 ⎯⎯
(i) 2SO 2 + 2NO 2 ⎯⎯
→ 2SO 3 + 2NO
→ 2NO 2
a) NO2 is intermediate
b) NO is catalyst
c) NO2 is catalyst and NO is intermediate d) NO is catalyst and NO2 is intermediate
INTEGRATED RATE LAWS :
Concept Explanation :
(a) Integrated rate laws :
Definition : The equations which are obtained by integrating the differential rate laws
and which give a direct relationship between the concentrations of the reactants and
time are called integrated rate laws.
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(b) Zero Order Reaction :
1. Definition : “The reaction whose rate is independent of the reactant concentration and
remains constant throughout the course of the reaction is called zero order reaction”.
2. Differential rate law for zero order reaction :
Consider a general reaction,
A → products
The rate of disappearances of reactant, A is given by the relation,
Rate = –
d[A]
= k [A]0 = k
dt
∴ Rate = k
Thus, the rate of this reaction is proportional to the zero power of the concentration of
the reactant.
3. Examples of zero order reaction :
(a) The decomposition of gaseous ammonia on a hot platinum surface is a zero order
reaction at high pressure.
1130K
2NH 3 (g) ⎯⎯⎯⎯→
N 2 (g) + 3H 2 (g)
Pt catalyst
Rate = k [NH3]0 = k
(b) Decomposition of N2O to N2 and O2 on Platinum catalyst :
2N 2 O (g) ⎯⎯
→ 2N 2 (g) + O 2 (g)
4.
(c) Decomposition of PH3 on hot tungsten at high pressure.
Integrated Rate law for zero order reaction :
Consider the hypothetical zero order reaction; A → Products
– d[A]
The rate law for this zero order reaction will be, Rate =
= k[A]0 = k
dt
On rearranging, d[A] = – kdt
On integration of the equation between limits [A] = [A]0 at t = 0 and [A] = [A]t at
t = t, we get
[A] t
∫
[A]
t
∫
d[A] = –k dt
0
0
[A]
[A][A]t = – k(t) 0t
0
∫ d[A] = – ∫ kdt
[A]t – [A0] = –kt
[A]t = –kt + [A0]
Chapter - 5 Chemical Kinetics
334
5.
Graphical representation of zero order reaction :
(a) The differential rate law for zero order
reaction is given by
Rate =
–d[A]
= k[A]0 = k
dt
The plot of rate versus [A]t is a horizontal
line with zero slope.
(b) The integrated rate law for zero order
reaction is given by
[A]t = –kt + [A0]
The above equation is in the form of linear
equation y = mx + C. Hence, [A]t versus
t is a straight line with slope –k.
6. Half life of zero order reactions :
The rate law expression for zero order reaction is, [A]t = –kt + [A]0
Where [A]0 and [A]t are the concentrations of the reactant at time, t = 0 and after time
t respectively,
Half-life period, t1/2 is the time when the concentration reduces from [A]0 to [A]0/2. i.e.,
at t = t1/2, [A]t = [A]0 / 2.
[A] 0
∴
= –kt1/2 + [A]0
2
[A] 0
[A] 0
∴ kt1/2 = [A] 0 –
=
2
2
[A] 0
∴ t1/2 = 2k
Hence, for a zero order reaction, half-life period is directly proportional to the initial
concentration of the reactant.
7. Explanation of examples of zero order reaction :
(i) The decomposition of gaseous ammonia on a hot platinum surface is a zero order
reaction at high pressure.
1130K
2NH 3 (g) ⎯⎯⎯⎯→
N 2 (g) + 3H 2 (g)
Pt catalyst
The platinum surface gets completely covered by a layer of NH3 molecules.
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The number of NH3 molecules attached on the surface of platinum is very small
compared to the total number of ammonia molecules.
Most of the ammonia molecules remain in the gas phase and these do not react.
Consequently the rate of the reaction is independent of the total concentration of NH3
and hence, it remains constant. Hence, it is a zero order reaction.
(ii) The decomposition of nitrous oxide to N2 and O2 in the presence of Pt catalyst.
2N 2 O (g) ⎯⎯
→ 2N 2 (g) + O 2 (g)
The platinum surface gets completely covered by N2O molecules. Most of the molecules
remain in the gas phase.
Because only the N2O molecules on the surface react, the reaction rate is independent
of the total concentration of N2O. Hence, it a zero order reaction.
(c) First Order Reaction :
1. Definition : “The reaction whose rate depends on the single reactant concentration
raised to a first power is called the first order reaction”.
2. Differential rate law for first order reaction :
Consider a general reaction, A → products
d[A]
= k[A].
dt
Thus, the rate of this reaction is proportional to first power of the concentration of the
reactant.
Examples of first order reaction :
The examples of first order reaction are :
(a) Decomposition of H2O2 :
The rate of disappearances of reactant, A is given by the relation, Rate = –
3.
2H 2 O 2 (l ) ⎯⎯
→ 2H 2 O (l ) + O 2 (g)
Rate = k[H2O2]
(b) Decomposition of N2O5 :
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g)
Rate = k[N2O5]
(c) Isomerisation of cyclopropane to propene :
Rate = k[C3H6]
4.
Integrated Rate law for first order reaction :
Consider a general reaction, A → products
The rate of disappearances of reactant, A is given by the relation,
Chapter - 5 Chemical Kinetics
336
Rate = –
d[A]
= k[A]
dt
...(i)
(a) Where [A] is the concentration of the reactant A that remains unreacted at time
t and –d[A] / dt is the rate of the reaction measured at time t at which A is converted
to the products. By rearranging and integrating the equation (i) between the limits
[A] = [A]0 at t = 0 and [A] = [A]t at t = t, we write,
[A] t
∫
[A]0
t
d[A]
–k dt
A =
∫
0
where [A]0 is the initial concentration of A at t = 0 and [A]t is the concentration of A
that remains unreacted at t.
[A]
On performing the integration, we get {In[A]}[A]0t = –k(t) 0t
Or ln [A]t – ln[A]0 = –kt
[A] t
and ln [A] = –kt
0
[A] 0
Hence, kt = ln [A]
t
1
...(ii)
[A] 0
that is kt = t ln [A]
t
Converting ln to log10 the integrated rate law becomes
k=
[A] 0
2.303
log10
t
[A] t
...(iii)
Note :
The rate law equation can also be written in the following alternative forms :
[A] t
ln
(a) The equation (ii),
[A] 0 = –kt
can be written by taking antilog of both sides as
[A] t
–kt
[A] 0 = e
This is called as exponential form
of first order reaction.
Or
[A] t = [A]0e–kt
(b) If we let ‘a’ mol dm–3 (that is M) as the initial concentration of A at t = 0 and
x as the concentration of A that decreases during time t from the beginning of the
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reaction then the concentration of A that remains unreacted at t is given by
[A]t = [A]0 – x = a – x
where [A]0 = a mol dm–3. With these values the equation (iii) can be written as
k=
5.
2.303
a
log 10
t
a–x
Graphical representation of first order reaction :
Equation :
The differential rate law for first order reaction,
Rate = –
d[A]
= k[A]
dt
Rate = k [A]t (y = mx)
Nature of graph : When rate of a first order reaction
is plotted against concentration[A]t, a straight line is
obtained.
Slope : The slope of the line gives the value of K.
Equation :
Exponential form of first order reaction,
[A]t = [A] 0e–kt
Nature of graph : When the concentration of reactant
is plotted against time (t), a curve is obtained.
Equation :
The integrated rate law expression
[A] 0
2.303
k = t log 10 [A]
t
On rearrangement we get,
kt
= log10[A]0 – log10[A]t Hence,
2.303
log10[A]t =
–k
t + log10[A]0.
2.303
The above equation is of the form, y = mx + C
Chapter - 5 Chemical Kinetics
338
Nature of graph : When log10[A]t is plotted against time(t), we get a straight line
–k
Slope : The slope of the line is equal to
2.303
Equation :
k=
[A] 0
2.303
log10
. The above integrated rate
t
[A] t
law expression can be rearranged to give,
log10
[A] 0
k
=
t + 0. The above equation
[A] t
2.303
is of the form, y = mx + C
[A] 0
Nature of graph : When log 10 [A] is plotted
t
against time(t), we get a straight line
Slope : The slope of the line is equal to
6.
k
2.303
Half life of first order reaction (t 1/ 2 ) :
The half life of a reaction is defined as the time needed for the reactant concentration
to fall to one half of its initial value.
For the first order reaction the integrated rate law is given by the equation
k=
[A] 0
2.303
log10
t
[A] t
where [A]0 is the initial concentration of the reactant at t = 0. Its concentration falls to
[A]t at time t from the start of the reaction. The concentration of the reactant falls to
[A]0/2 at time t 1/2 , the half life period. Hence, at t = t 1/2 , [A]t = [A]0/2. With this condition,
the equation can be written as
k
=
=
=
[A] 0
2.303
log 10
[A] 0
t 1/2
2
2.303
log10 2
t 1/2
2.303
× 0.301 = 0.693
t 1/2
t 1/2
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=
t 1/2 =
0.693
k
...(iv)
Hence,
The equation (iv) implies that the half life of a first order reaction is constant and is independent
of the reactant concentration.
(d) Pseudo First Order reaction :
1. Definition : A reaction which has higher order true rate law but are found to behave
as first order are called pseudo first order reactions.
2. Examples :
(a) The hydrolysis of ester in presence of an acid catalyst, can be explained by taking
example of hydrolysis of methyl acetate in presence of dil. HCl.
dil HCl
CH 3COOCH 3 + H 2O ⎯⎯⎯⎯
→ CH 3COOH + CH 3 – OH
The rate of this reaction is given by Rate of reaction = k′[CH3COOCH3] [H2O].
According to the rate law, the above reaction is expected to have second order.
But practically, it is observed that the water present in this reaction is in large excess,
therefore, the concentration of water remains practically constant during the course
of reaction.
[H2O] = constant = k′′
∴ Rate of reaction = k′k′′ [CH3COOCH3]
= k [CH3COOCH3]
This indicates that, the rate of reaction is determined only by the concentration of
methyl acetate. Hence, this reaction is first order reaction. Such reactions are known
as pseudo-first order reaction.
(b) The hydrolysis of cane sugar (sugar) in presence of an acid catalyst is an example
of pseudo–first order reaction.
C12 H 22 O11 (aq) + H 2 O(l ) ⎯⎯
→ C 6 H 12 O 6 (aq) + C 6 H 12 O 6 (aq)
sucrose
glucose
fructose
Rate = k′ [C12H22O11] [H2O]
The above rate law indicates that the reaction is second order. However, [H2O] is
in excess.
∴ [H2O] remain constant
∴ k′ × [H2O] = k
∴ rate = k [C12H22O11]
Thus, the rate of reaction is dependent on only concentration of sucrose. Hence, the
reactions is pseudo–first order reaction.
Chapter - 5 Chemical Kinetics
340
(e) Experimental Determination of rate laws and order :
1. Method I : Isolation Method :
(a) In this method all the reactants except one (isolated) are present in large excess.
(b) The concentrations of these reactants remain constant throughout the course of the reaction.
(c) The dependence of rate on the concentration of the isolated species is experimentally
determined.
(d) This will give the rate law and order of the reaction with respect to the isolated reactant
are measured.
(e) This procedure is repeated for other reactants in the reaction.
(f) Example : Consider an equation
aA + bB ⎯⎯
→ cC + dD
2.
∴ Rate law is rate = k [A]x [B]y
In one experiment, A is isolated by taking B in large excess.
∴ Initial concentration of B = [B]0 = remain constant
∴ Rate = k [A]x [B]0y
= k [A]x [∵ [B]0y × k = k′]
Hence, order x can be measured.
Similarly ‘y’ can also be determined
∴ Overall order of reaction = x + y.
Method II : Method of Initial rates :
(a) This method involes the measurement of rate at the begining of the reaction for different
initial concentrations of reactants by using isolation method.
(b) Consider a reaction aA + bB ⎯⎯
→ cC + dD
∴ Rate law is rate = k′ [A]x [B]y
(c) If B is taken in large excess, its concentration remains constant.
∴ The initial rate of the reaction with A isolated is given by r0 = k [A]x0
Hence,
log10 r0 = log10k + log10[A0]x
∴
log10 r0 = x log10[A]0 + log10k
[y = mx + c]
(d) For series of concentrations of isolated A, the rates r0 are measured. When the graph
of log10 r0 against log10 [A]0 is plotted, a straight line is obtained. The slope of the
line gives order x with respect to A.
(e) The procedure is repeated with B to give order ‘y’ with respect to ‘B’
(f) The overall order of the reaction is x + y.
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341
2.
Method III : Use of Integrated law.
(a) This is a trial and error methods. In this the concentration of the reactant is measured
at different intervals of time.
(b) These data are substituted in the integrated rate law of reactions of first order, second
order and so on.
(c) The equation which gives the constant value of the rate constant (k) gives the correct
order of reaction.
[A ]
2.303
log10 0
For example, for the first order reaction, k =
t
[A] t
Where
[A0] = Initial concentration of A.
and
[A]t = concentration that remains unreacted at time t.
(d) By substitution of time-concentration data in the integrated rate law equation the values
of k at different time intervals are calculated.
(e) In these k values are constant then reaction is a first order reaction.
•
*1
What is zero order reaction? Write the differential rate law for zero order reaction. Derive
its integrated rate law. How is it represented graphically?
Refer 5.3 (b) 1 to 5
Ans.
2.
Ans.
Derive an expression for half life of a zero order reaction.
The rate law expression for zero order reaction is, [A]t = –kt + [A]0
Where [A]0 and [A]t are the concentrations of the reactant at time, t = 0 and after time t
respectively,
Half-life period, t1/2 is the time when the concentration reduces from [A]0 to [A]0 / 2. i.e.,
at t = t1/2, [A]t = [A]0 / 2.
[A] 0
∴
= –kt1/2 + [A]0
2
[A] 0
[A] 0
∴ kt1/2 = [A] 0 –
=
2
2
[A] 0
∴ t1/2
=
2k
Hence, for a zero order reaction, half-life period is directly proportional to the initial concentration
of the reactant.
3.
Ans.
Decomposition of NH3 on platinum surface at high temperature is a zero order reaction. Explain.
The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction
at high pressure.
Chapter - 5 Chemical Kinetics
342
1130K
2NH 3 (g) ⎯⎯⎯⎯→
N 2 (g) + 3H 2 (g)
Pt catalyst
The platinum surface gets completely covered by a layer of NH3 molecules.
The number of NH3 molecules attached on the surface of platinum is very small compared
to the total number of ammonia molecules.
Most of the ammonia molecules remain in the gas phase and these do not react.
Consequently the rate of the reaction is independent of the total concentration of NH3 and
hence, it remains constant.
Hence, it is a zero order reaction.
4
Ans.
*5
Ans.
*6.
Ans:
*7.
What is a first order reaction. Explain with an example.
Refer 5.3 (c) 1 to 3.
Define integrated rate law. Derive the expression for integrated rate law for first order reaction,
A → products.
Integrated rate laws :
Definition : The equations which are obtained by integrating the differential rate laws and which
give a direct relationship between the concentrations of the reactants and time are called integrated
rate laws.
Expression for Integrated Rate law for first order reaction : Refer 5.3 (c) 4.
Describe the graphical representation of first order reaction.
Refer 5.3 (c) 5.
Ans.
Define half life of a reaction. Derive the relationship between half life and rate constant for
a first order reaction. OR Show that half life of a first order reaction is independent of initial
concentration of the reactant.
Refer 5.3 (c) 6.
*8.
Ans.
What is pseudo-first order reaction? Explain with one example.
Refer 5.3 (d).
*9.
Ans.
The reaction, CH 3 COOC 2 H 5 (aq) + H 2O (l ) ⎯⎯
→ CH 3COOH (aq) + C 2 H 5OH (aq) follows
the first order kinetics, the rate law being rate = k [CH3COOC2H5]. However, the reaction
shows that it is second order. Explain.
The hydrolysis of ester in presence of an acid catalyst, can be explained by taking example
of hydrolysis of methyl acetate in presence of dil. HCl.
dil HCl
CH 3COOCH 3 + H 2O ⎯⎯⎯⎯
→ CH 3COOH + CH 3 – OH
The rate of this reaction is given by Rate of reaction = k′[CH3COOCH3] [H2O].
According to the rate law, the above reaction is expected to have second order.
But practically, it is observed that the water present in this reaction is in large excess, therefore,
the concentration of water remains practically constant during the course of reaction.
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343
[H2O] = constant = k′′
∴ Rate of reaction = k′k′′ [CH3COOCH3]
= k [CH3COOCH3]
This indicates that, the rate of reaction is determined only by the concentration of methyl acetate.
Hence, this reaction is first order reaction. Such reactions are known as pseudo-first order reaction.
*10.
Ans.
Describe one method to determine rate law and order of a reaction.
Method I : Isolation Method :
(a) In this method all the reactants except one (isolated) are present in large excess.
(b) The concentrations of these reactants remain constant throughout the course of the reaction.
(c) The dependence of rate on the concentration of the isolated species is experimentally
determined.
(d) This will give the rate law and order of the reaction with respect to the isolated reactant
are measured.
(e) This procedure is repeated for other reactants in the reaction.
(f) Example : Consider an equation aA + bB ⎯⎯
→ cC + dD
∴ Rate law is rate = k [A]x [B]y
In one experiment, A is isolated by taking B in large excess.
∴ Initial concentration of B = [B]0 = remain constant
∴ Rate = k [A]x [B]0y
= k [A]x [∵ [B]0y × k = k′]
Hence, order x can be measured.
Similarly ‘y’ can also be determined
∴ Overall order of reaction = x + y.
*11.
Ans.
A certain reaction occurs in the following steps :
(i)
Cl(g) + O 3 (g) ⎯⎯
→ ClO(g) + O 2 (g)
(ii)
a)
b)
c)
ClO (g) + O (g) ⎯⎯
→ Cl(g) + O 2 (g)
Write the chemical equation for overall reaction.
Identify the reaction intermediate.
Identify the catalyst.
(a) Step I
Step II
Net reaction
:
Cl(g) + O 3 (g) ⎯⎯
→ ClO (g) + O 2 (g)
:
ClO (g) + O (g) ⎯⎯
→ Cl(g) + O 2 (g)
O 3 (g) + O (g) ⎯⎯
→ 2O 2 (g)
(b) Reaction intermediate is ClO(g) which is formed in the first step and removed in the
second step.
Chapter - 5 Chemical Kinetics
344
(c) Cl(g) acts as a catalyst. It is present at the start and end of the reaction but is not seen
in the overall reaction.
*12.
What is the rate law for the reaction NO 2 (g) + CO(g) ⎯⎯
→ NO(g) + CO 2 (g) . If the reaction
occurs in the following steps?
[(i) NO 2 + NO 2 ⎯⎯
→ NO 3 + NO (slow) and (ii) NO 3 + CO ⎯⎯
→ NO 2 + CO 2 (fast)]
What is the role of NO3?
Ans.
Step I
:
NO 2 + NO 2 ⎯⎯
→ NO 3 + NO
(slow)
Step II
:
NO 3 + CO ⎯⎯
→ NO 2 + CO 2
(fast)
NO 2 (g) + CO(g) ⎯⎯
→ NO(g) + CO 2 (g)
Overall reaction
NO3 is formed in the first step and consumed in the second step.
Hence, NO3 is an intermediate.
*13.
Ans.
The rate law for the reaction 2NO (g) + Cl 2 (g) ⎯⎯
→ 2NOCl(g) is given by rate = k[NO]
[Cl2]. The reaction occurs in the following steps :
(i) NO (g) + Cl 2 (g) ⎯⎯
→ NOCl 2 (g)
(ii) NOCl 2 (g) + NO(g) ⎯⎯
→ 2NOCl(g)
(a) Is NOCl2 a catalyst or reaction intermediate? Why?
(b) Identify the rate determining step
Step I
: NO (g) + Cl 2 (g) ⎯⎯
→ NOCl 2 (g)
Step II
:
NOCl 2 (g) + NO (g) ⎯⎯
→ 2NOCl(g)
Overall reaction
2NO(g) + Cl 2 (g) ⎯⎯
→ 2NOCl(g)
(a) Since NOCl2 is formed in the first step and consumed in the second step. It is an intermediate.
(b) ∴ The rate law is rate = k [NO] [O3]
NO and O3 are involed in the first step, it will be the rate determining step and must be
the slower step.
*14.
Ans.
The rate law for the reaction 2H 2 (g) + 2NO (g) ⎯⎯
→ N 2 (g) + 2H 2 O (g) is given by
rate = k [H2][NO]2. The reaction occurs in the following steps :
(i) H 2 + 2NO ⎯⎯
→ N 2 O + H 2O
(ii) N 2 O + H 2 ⎯⎯
→ N 2 + H 2O
What is the role of N2O in the mechanism? Identify the slow step.
Step I
:
H 2 + 2NO ⎯⎯
→ N 2 O + H 2O
Step II
:
N 2 O + H 2 ⎯⎯
→ N 2 + H 2O
Overall reaction 2H 2 (g) + 2NO (g) ⎯⎯
→ N 2 (g) + 2H 2 O (g)
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S.Y.J.C. Science - Chemistry - Part I
345
(a) N2O is formed in the first step and consumed in the second step it is an intermediate.
(b) The rate law is rate = k [H2] [NO]2
∴ H2 and NO are involed in the first step, it will be the rate determining step and must be
the slower step.
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
The integrated rate equation for first order reaction A ⎯⎯
→ products is given by .......
*2.
3.
4.
5.
6.
7.
a) k =
2.303 [A] 0
ln
t
[A] t
t
b) k = – t ln [A]
0
c) k =
[A] t
2.303
log 10
t
[A] 0
t
d) k = t ln [A]
0
1
1
[A]
[A]
The slope of the straight line obtained by plotting rate versus concentration of reactant for
a first order reaction is .......
a) –k
b) –k/2.303
c) k/2.303
d) k
If [A]0 is the initial concentration and [A] is the concentration at time t, then for a first order
reaction, .......
a) [A]t = [A] 0 ekt
b) [A]t = [A] 0 e – kt
c) [A]0 = [A]t ek/t
d) [A]t = [A] 0 e – k/t
For a first order reaction, the plot of log (a – x) against time(t) would give a straight line
with a slope equal to .......
2.303
k
2.303
a)
b) –
c) –
d) 2.303k.
k
2.303
k
Half life period of zero order reaction is given by equation t1/2 = .......
3
1
A
0.693
a) 0
b)
c) [A] k
d)
2k[A 0 ] 2
k
2k
0
Some statements, about a zero order reactions, are given below.
A) Its rate is independent of the conc. of reactant.
B) The plot of conc. of reactant with time is a straight line with slope = k.
C) It is generally influenced by some external agency.
D) The equation, [A]t = [A]o – kt, holds good for such a reaction.
Among the above, the incorrect statement(s) is/are
a) B
b) B and C
c) B and D
d) C and D
The half life period of a reaction is constant for .......
a) zero order
b) first order
c) second order
d) none
Chapter - 5 Chemical Kinetics
346
8.
9.
10.
11.
•
12.
*13.
*14.
15.
•
16.
17.
For a zero order reaction, the plot of concentration of reactants Versus time is linear with .......
a) +ve slope and passing through the origin
b) – ve slope and passing through the origin
c) +ve slope and not passing through the origin
d) – ve slope and not passing through the origin
The units of rate constant for a zero order reaction are ........
b) litre mole – 1sec – 1
a) litre sec – 1
c) mol litre – 1 sec – 1
d) mol sec – 1
th
The time required for the completion of 3/4 of the first order reaction is .......
2.303
2.303
1
2.303
4
2.303
3
log 4
log
log
log
a)
b)
c)
d)
k
k
4
k
3
k
4
For a zero order reaction, A → Products, the rate equations is .......
b) [A]t – [A]0 = kt c) [A]t – [A]0 = k d) [A]0 – [A]t = kt
a) [A]0 – [A]t = k
Numerical MCQs :
If half life period for a first order reaction is 2 min. The rate constant is .......
b) 1.386 min–1
c) 0.3465 min–1
d) 2 min – 1
a) 0.5 min–1
The half life of a first order reaction is 30 min and the initial concentration of the reactant
is 0.1 M. If the initial concentration of reactant is doubled, then the half life of the reaction
will be .......
a) 1800 s
b) 60 min
c) 15 min
d) 900 s
The slope of a graph ln[A]t versus t for a first order reaction is –2.5 × 10–3s–1. The rate
constant for the reaction will be .......
b) 1.086 × 10–3 s–1
a) 5.76 × 10–3 s–1
–3 –1
c) –2.5 × 10 s
d) 2.5 × 10–3 s –1
A first order reaction is 90% complete in 100sec. The rate constant is .......
b) 0.693 x 10 – 2 s – 1
a) 0 sec – 1
c) 2.303 x 10 – 2 s – 1
d) 0.01s – 1
For a first order reaction, the plot of rate versus conc. of reactant is .......
a) non-linear and passing through the origin
b) linear with the negative slope
c) linear with positive slope and passing through the origin
d) linear with positive slope but not passing through the origin
Some statements are given below .......
a) For a first order reaction, t1/2 = constant
b) For a zero order reaction, t1/2 ∝ Initial conc.
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S.Y.J.C. Science - Chemistry - Part I
347
c) For a zero order reaction, Rate ∝ conc. of reactant
d) For a first order reaction, the plot of rate versus conc. is linear with slope = - k.
Among the above the false statements are .......
a) B, C and D
b) B and C
c) A, B and D
d) C and D
5.4
COLLISION THEORY, ACTIVATION ENERGY AND CATALYST :
Concept Explanation :
(a) Collision theory requirements for a bimolecular chemical reaction to
take place :
1. Collision between reactant molecules :
(a) The basic requirement of collision theory is that the reacting species (atoms, ions or
molecules) must come together and collide in order for the reaction to occur.
(b) Therefore, the rate of reaction must be equal to the rate of collision.
(c) However, it has been found that the rate of reaction is much lower as compared to
the rate of collisions.
(d) It follows, therefore, that all the collisions between reactant molecules are not fruitful,
i.e. do not result in formation of product.
(e) This implies that the collision is not the sufficient cause for the chemical reaction
to occur.
2. Energy requirement (activation energy) :
(a) The activation energy [Ea] is defined as the
minimum kinetic energy required for a
molecular collision to lead to a reaction.
(b) This kinetic energy is required to arrange
outer electrons in breaking and making of
bonds.
(c) It can be seen from the graph that the fraction of molecules with kinetic energy more
or equal to activation energy is less, the rate of reaction is less than the rate or frequency
of collisions.
3. Orientation of reactant molecules :
(a) In reactions involving complex molecules, the above conditions are not sufficient for
the reaction to occur.
(b) Hence, apart from sufficient energy the colliding molecules must have proper
orientations.
(c) If the molecules have improper orientation they do not react even if they possess
the required kinetic energy.
Chapter - 5 Chemical Kinetics
348
(d) Consider a reaction.
AB + A 2 B ⎯⎯
→ AB 2 + A 2
4. Potential energy barrier :
(a) When molecules react, an intermediate or activated complex or a transition state is
formed.
(b) In the formation of activated complex, atoms, molecules require energy to overcome
the repulsion between reactant molecules.
(c) This energy is provided from the kinetic energy which
is converted into potential energy in the activated
complex.
(d) Due to high energy, activated complex is unstable,
short-lived and decomposed into the products.
(e) To form activated complex, the reactant molecules
must climb the energy barrier.
(f) Hence, only those molecules having sufficient kinetic
energy collide with proper orientation result in formation
of products.
(b) Arrhenius equation from collision theory of bimolecular reactions :
1. In order for a bimolecular reaction to occur the reactant molecules must collide and the
colliding molecules must have total kinetic energy equal to or greater than the activation
energy of the reaction.
2. The fraction of molecules having kinetic energy equal to or greater than the activation
energy [Ea] is given by
– E /RT
f= e a
Unique Solutions ®
...(i)
S.Y.J.C. Science - Chemistry - Part I
349
3.
Consider a second order reaction
AB + C ⎯⎯
→ AC + B
4.
5.
6.
Then collision rate for the above reaction is given by
Collision rate = Z × [AB] × [C]
Where Z = collision frequency and [AB] and [C] are concentration of reactants.
Reaction rate
= P.f × collision rate
= P.f.Z [AB] × [C]
...(ii)
Where f = fraction of collisions with suffcient kinetic energy [Ea]
and P = fraction of collisions with proper orientations.
Reaction rate for a second order reaction is also given by
Reaction rate = k [AB] [C]
...(iii)
Comparing (ii) and (iii) we get;
k = P.f.Z
∴ k = P.Z × e – E a /RT
[Form (i)]
– E /RT
7.
k = A.e a
Where A = P.Z is called a frequency factor or pre-exponential factor.
The Arrhenius equation can be written in following forms.
– E /RT
k = A.e a
∴ ln k = ln A –
Ea
RT
Ea
2.303RT
Where k = Rate constant at absolute temperature T.
Ea = Energy of activation
R = gas constant
A = Frequency factor or pre-exponential factor.
log10k = log10A –
(c) Temperature dependence of reaction rates :
1. The kinetic energy of the molecule increases with the increase in temperature.
2. The fraction of molecules possessing minimum energy (activation energy [Ea]) increases
with increase in temperature.
3. Hence, the fraction of colliding molecules that possess kinetic energy Ea also increases,
hence, the rate of reaction increases with increase in temperature.
4. The figure shows that the area that respresents the fraction of molecules with kinetic energy
exceeding Ea is greater at higher temperature T2 than at lower temp T1.
Chapter - 5 Chemical Kinetics
350
5.
Arrhenius equation indicates that larger the value of Ea, smaller is the rate constant and
slower is the rate of reaction.
E
As T increases a decreases.
RT
This causes an increase in –
Ea
– E /RT
, e a , k and hence, the rate of reaction.
RT
T
⇒ E a /RT ⇒ – E a /RT ⇒ e –E a /RT ⇒
k
⇒
rate
increases
decreases
increases
increases
increases
increases
(d) Arrhenius equation and temperature variation :
Relation showing variation in rate constant with temperature.
1. As per Arrhenius equation
k = A × e – E a /RT
Where k = rate constant at temp T
A = frequency factor
Ea = Energy of activation
2. Taking log we get
ln k = ln A –
Ea
RT
log10k = log10A –
Ea
2.303 RT
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
351
3.
lf k1 and k2 are the rate constants at temp T1 and T2 respectively, then
Ea
log10k1 = log10A – 2.303 RT
1
Ea
log10k2 = log10A – 2.303 RT
2
⎡
⎤
⎡
⎤
Ea
Ea
– ⎢ log 10 A –
⎥
2.303RT2 ⎦
2.303RT1 ⎥⎦
⎣
⎣
Ea
Ea
–
2.303 RT1
2.303 RT2
∴ log10k2 – log10K1 = ⎢ log 10 A –
k2
∴ log10 k =
1
=
Ea
⎡1
1⎤
–
⎢
2.303 R ⎣ T1 T2 ⎥⎦
k2
⎡ T2 – T1 ⎤
Ea
∴ log10 k =
⎢
⎥
2.303 R ⎣ T1 T2 ⎦
1
(e) Arrhenius equation and energy of activation :
1.
2.
3.
4.
Arrhenius equation is k = A.e – E a /RT
Where k = Rate constant at temp T.
A = frequency factor
Ea = energy of activation
As the activation energy Ea decreases, the
energy barrier decreases and more reactant
molecules can cross the activation energy
barrier.
The graph shows when Ea decreases, the
fraction of molecules with kinetic energy greater
than or equal to Ea increases.
As Ea decreases,
Ea
⇒ E a /RT
⇒ – E a /RT ⇒ e –E a /RT ⇒ k
⇒ rate
increases
increases
decreases
decreases
increases
increases
(f) Determination of activation energy (Ea) :
1.
Consider a first order reaction B ⎯⎯
→C
2.
3.
The rate constants for the above reaction is determined at various temperatures.
By Arrhenius equation
k = A × e – E a /RT
Chapter - 5 Chemical Kinetics
352
∴ ln k = ln A –
Ea
RT
∴ log10k = log10 A –
Ea
2.303RT
– Ea
+ log10 A
2.303RT
(y = mx + c)
4. log10 k is plotted against reciprocal of temperature T.
– Ea
5. The slope of the straight line graph is
from which energy of activation can be
2.303R
Ea
calculated as follows slope =
2.303R
∴ Ea = slope × 2.303 R
logk =
(g) Effect of catalyst on rates of reactions :
1. A catalyst is a substance, when added
to the reactants, increases the rate of
reaction without itself being consumed.
2. The catalyst enters the reaction but
does not appear in the balanced
equation. This is because it is
consumed in one step and regenerated
in the another.
3. The catalyst provides an alternative
pathway (mechanism) of lower
activation energy for the reaction to
occur.
4. The height of the barrier for
uncatalysed reaction is greater than
that in a catalysed reaction.
5. Therefore, the number of molecules
that possesses minimum kinetic energy
[Ea] increases.
6.
According to Arrhenius equation.
k = A.e – E a /RT
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S.Y.J.C. Science - Chemistry - Part I
353
At constant temperature (T) and
for the same concentration,
Ea
⇒ E a /RT
⇒ – E a /RT ⇒ e –E a /RT ⇒ k
⇒ rate
increases
increases
decreases
decreases
increases
increases
7.
eq : 2H 2 O 2 (l ) ⎯⎯
→ 2H 2 O (l ) + O 2 (g)
The activation energy for above reaction is 76kJ mol–1 Hence, it is very slow at room temperature.
In presence of I– as catalyst the activation energy decreases to 57 kJ mol–1 and the reaction
becomes considerably faster.
(h)
1.
2.
3.
Effect of catalyst :
Activation energy : It lowers the activation energy of the reaction.
Rate of forward reaction : A catalyst increases the rate of forward reaction.
Rate of backward reaction : A catalyst increases the rate of backward reaction to the
same extent as the rate of forward reaction.
Hence, a catalyst does not effect the reaction equilibrium it just helps to achieve the equilibrium
faster.
(i) Difference between catalyst and reaction intermediate :
Catalyst
1.
2.
3.
4.
Reaction Intermediate
A catalyst is present at the start of the
reaction.
A catalyst is consumed in one step of the
reaction but regenerated in next step.
It accelerates the rate of reaction.
The concentration of catalyst and reaction
intermediate may appear in the rate law.
1.
2.
3.
4.
It is produced during the mechanism of
the reaction.
Reaction intermediate is formed in one
step and consumed in the next step.
It has no effect on the rate of reaction.
The reaction intermediate does not
appear in the rate law.
•
*1.
Ans.
What are the requirements for a bimolecular reaction to take place?
The main requirements for a bimolecular reaction to occur are :
(a) The reactant molecules must collide.
(b) The colliding molecules must have total kinetic energy equal to or greater than the activation
energy of the reaction.
(c) The colliding molecules must have proper orientations relative to each other.
*2.
Ans.
Write Arrhenius equation and explain the terms involved.
The Arrhenius equation can be written in following forms.
Chapter - 5 Chemical Kinetics
354
∴
k = A.e – A.e –E a /RT
Ea
ln k = ln A –
RT
Ea
2.303RT
Where k = Rate constant at absolute temperature T.
Ea = Energy of activation
R = Gas constant
A = Frequency factor or pre-exponential factor.
log10k = log10A –
*3.
Ans.
Explain with the help of potential energy barrier, how does increase in temperature increases
the rate of reaction.
Refer 5.4 (c)
*4.
Ans.
Define Activation Energy.
Activation energy is defined as the minimum kinetic energy required for a molecular collision
*5.
Ans.
Using Arrhenius equation explain why the rate of reaction increases with increase in temperature.
Refer 5.4 (c).
6.
Derive an expression for Arrhenius equation and temperature variation. OR Derive an expression
showing variation in rate constant with temperature.
Refer 5.4 (d).
Ans.
*7.
Ans.
*8.
Ans.
*9.
Ans.
How will you determine activation energy with the help of a graph.
(Intext question Textbook page no : 213)
Refer 5.4 (f).
Explain, with the help of potential energy barrier, how does a catalyst increase the speed of
a reaction.
Refer 5.4 (g).
Comment on the effect of catalyst on each of the following :
a) Activation energy
b) Rate of forward reaction
c) Rate of backward reaction.
a) Activation energy : It lowers the activation energy of the reaction.
b) Rate of forward reaction : A catalyst increases the rate of forward reaction.
c) Rate of backward reaction : A catalyst increases the rate of backward reaction to the
same extent as the rate of forward reaction.
Hence, a catalyst does not effect the reaction equilibrium it just helps to achieve the equilibrium
faster.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
355
*10.
Ans.
How does a catalyst differs from a reaction intermediate?
Refer 5.4 (i).
*11.
Ans.
Explain briefly the collision theory of bimolecular reactions.
Refer 5.4 (a).
12.
Ans.
Derive Arrhenius equation from Collision theory of bimolecular reactions.
Refer 5.4 (b).
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
The rate constant of a reaction .......
a) decreases with increasing Ea
b) decreases with decreasing Ea
c) is independent of Ea
d) decreases with increasing temperature
A catalyst increases the rate of the reaction by .......
b) increasing T
c) decreasing Ea d) decreasing T
a) increasing Ea
The Arrhenius equation is
*2.
*3.
– E /RT
a) A = ke e a
4.
b)
A E a / RT
e
k
c)
k = Ae E a /RT
d) k = Ae – RT/E a
The units of pre-exponential factor or frequency factor are same as .......
a) Rate of reaction
b) Rate constant
c) Order of reaction
d) Molecularity of reaction.
HOURS BEFORE EXAM
Chemical kinetics is concerned with the rates of reactions factors affecting the rates and the
mechanisms by which they proceed.
The rate of reaction is defined as the decrease is molar concentration of a reactant or increase
in molar concentration of a product per unit time.
It can be expressed as average rate during a given time interval or the instantaneous rate
at a specific time.
The reaction rate depends on five factors :
The concentration of reactants, temperature, presence of a catalyst nature of reactant and
pressure.
The concentration dependence is given by the rate law. The rate law for a general reaction,
aA + bB ⎯⎯
→ cC + dD , is given by the expression, rate = k[A]x[B]y, where k is the rate
constant of the reaction and x and y represent the order of the reaction with respect to reactants
A and B respectively. The overall order of the reaction is x + y.
Chapter - 5 Chemical Kinetics
356
The order of the reaction is defind as the sum of the exponents to which the concentration
terms in the rate law are raised.
The values of x and y and hence, the order of the reaction must be determined by experiment
and cannot be predicted from the stoichiometric equation of the reaction.
The values of x and y are not related to the stoichiometric coefficients of the reactants in
the balanced equation of the reaction.
The integrated rate law is a direct relationship between time and concentration of the reactants.
It can be used to calculate concentrations at any time t and the time required for an initial
concentration of the reactants to reach any particular value. For the first order reaction the
[A] 0
2.303
log 10
t
[A] t
A graph of log10[A]t versus time is a straight line with slope equal to –k/2.303.
The zero order reaction are those for which the rate of the reaction is independent of the
concentration of reactants. The integrated rate law for zero order reaction is
[A]t = –kt + [A]0.
The graph of [A]t versus t is a straight line with a slope equal to –k. The half life of a reaction
is the time needed for the reactant concentration to fall to one half of its initial value. The
half lives of first and zero order reactions are given by t1/2 = 0.693/k and t1/2 = [A]02k respectively.
According to the collision theory a bimolecular reaction occurs when two properly oriented
molecules with sufficent kinectic energy collide.
The minimum kinetic energy necessary for a molecular collision to form products is called
activation energy.
According to collision theory the colliding molecules from an intermediate configuration called
activated complex or transition state.
The potential energy barrier exists between the reactants and products. The reactant molecules
must surmount this barrier before they are converted into products. The configuration at the
top of the barrier is an activated complex. The height of the barrier is the activation energy.
The temperature dependence of rate constant is describeal by Arrhenius equation,
k = Ae–Ea/RT, where Ea is the energy activation and A is the frequency factor. The graph
of log10k against reciprocal of temperature is a straight line with slope –Ea/2.303 R from which
activation energy can be calculated. The rate constant k1 and k2 at two different temperatures
T1 and T2 are related by the equation.
integrated rate law is K =
log 10
E a ⎡ T2 – T1 ⎤
k2
=
k 1 2.303R ⎢⎣ T1 T2 ⎥⎦
A catalyst is a substance that increases the rate of the reaction without being itself consumed
in the reaction. It functions by providing an alternative mechanism of lower activation energy
for the reaction to occur.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
357
An elementary reaction is one that occurs in a single step. The molecularity of an elementary
reaction is the number of reactant molecules that participate in the reactions. A complex reactions
occurs in a series of elementary reaction. The order of an elementary reaction is equal to
its molecularity.
The elementary reactions are classified as unimolecular bimolecular or trimolecular depending
on the number of reactant molecules involved. The observed rate law for a complex reaction
depends on the sequence of the elementary steps and their relative rates.
The slowest step in the mechanism is called rate determining step.
A chemical species that is formed in one elementary step in the mechanism of complex reaction
and consumed in the subsequent step is called a reaction intermediate.
NUMERICALS WITH SOLUTION
Type - I - Rate of reaction (Average rate and instantaneous rate)
Consider the reaction 2A + B ⎯⎯
→ 2C ,
Suppose that at a particular moment during
the reaction, rate of disappearance of A
is 0.076 M/s,
i) What is the rate of formation of C?
ii) What is the rate of consumption of B?
iii) What is the rate of the reaction?
Given :
Rate of disappearance of A is
d[A]
= 0.076 M/s
dt
To find :
d[C]
i) rate of formation of C is
dt
d[B]
ii) rate of consumption of B is
dt
Solution :
*1.
ii)
d[C]
dt
– d[A]
=
dt
= 0.076 M/s
=
Rate of consumption of B is
1
of rate
2
of consumption of A
Hence, Rate of consumption of B
=
– d[B]
dt
=
=
=
c)
Rate of reaction
=
2A + B ⎯⎯
→ 2C
i)
Rate of formation of A
The reaction shows that rate of formation
of C is same as rate of disappearance of
Hence,
Chapter - 5 Chemical Kinetics
– d[B]
dt
=
=
=
–1 d[A]
2
dt
–1
× 0.076
2
0.038 M/s
–1 d[A]
2
dt
1 d[C]
2 dt
0.038 M/s
358
*2.
Consider the reaction
c)
3I (aq) + S 2O 82– (aq) ⎯⎯
→ I 3– (aq) + 2SO 2–
4 (aq)
At a particular time t, –
d[SO 2–
4 ]
= 2.2 × 10–2 M/s.
What are the values of
i)
iii)
d[I – ]
–
dt
–
d[I –3 ]
Solution :
dt
d[S 2O 82– ]
ii) –
dt
*3.
at the same time?
∴
The given reaction shows that rate of
3
consumption of I– is times of rate of
2
formation of SO 2–
4
Rate of consumption of I–
=
=
=
=
– d[I – ]
dt
3 d[SO 2–
4 ]
2
dt
3
× 2.2 × 10 –2
2
3.3 × 10–2 M/s
b)
Rate of consumption of S 2 O 82– is
∴
rate of formation of SO 2–
4
Rate of consumption of S 2 O 82–
=
=
=
=
–
d[I –3 ]
dt
1 d[SO 2–
4 ]
=
2
dt
1
× 2.2 × 10 –2
=
2
= 1.1 × 10–2 M/s
Concentrations of various species at
different times for reaction
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g)
3I(aq) + S 2O 82– (aq) ⎯⎯
→ I 3– (aq) + 2SO 42– (aq)
a)
1
of rate of
2
formation of SO 2–
4
∴
dt
Rate of formation of I3 is
d[S 2O 82– ]
dt
1 d[SO 2–
4 ]
2
dt
1
× 2.2 × 10 –2
2
1.1 × 10–2 M/s
Unique Solutions ®
1
of
2
Time/s
[N2O5]/M
[NO2]/M
[O2]/M
0
200
400
600
0.0300
0.0213
0.0152
0.0108
0
0.0174
0.0296
0.0 384
0
0.00435
0.00740
0.00960
Calculate the average rate of
decomposition of N2O5 and the average
rates of formation of NO and O2 during
the time interval 200s to 400s. What is the
average rate of the reaction during the
same interval.
Given :
Time/s [N2O5]/M [NO2]/M
[O2]/M
0
0.0300
0
0
200
0.0213
0.0174
0.00435
400
0.0152
0.0296
0.00740
600
0.0108
0.0 384
0.00960
To find :
i) Average rate of decomposition of N2O5 = ?
ii) Average rate of formation of NO and O2
during the time interval 200s to 400s = ?
iii) Average rate of the reaction = ?
Solution :
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g)
S.Y.J.C. Science - Chemistry - Part I
359
a)
Average rate of decomposition of N2O5
=
– Δ[N 2O 5 ]
Δt
=
– (0.0152 – 0.0213)
400 – 200
*4.
– (–0.0061)
200
= 3.05 × 10–5 M/s
Average rate of formation of NO2
a) NO 2 (g) + F2 (g) ⎯⎯
→ NO 2 F (g) + F(g)
(slow, bimolecular)
=
b)
=
=
Δ [NO 2 ]
Δt
0.0296 – 0.0174
400 – 200
0.0122
200
= 6.1 × 10–5 Ms –1
Average of formation of O2
Δ [O 2 ]
=
Δt
0.00740 – 0.00435
=
400 – 200
= 1.525 × 10 –5 Ms –1
Average rate of reaction
=
c)
d)
=
–1 Δ[N 2 O 5 ]
2
Δt
=
1 Δ[NO 2 ]
4
Δt
=
=
=
=
=
Δ[O 2 ]
Δt
1
× 3.05 × 10 –5
2
1
× 6.1 × 10 –5
4
1.525 × 10–5
1.525 × 10 –5 Ms –1
Type - II - Rate law, Rate constant
and Order of reaction and
Molecularity
A reaction occurs by the following
mechanism :
→ NO 2 F (g)
b) F(g) + NO 2 (g) ⎯⎯
(Fast, bimolecular)
i) What is the overall reaction?
ii) What is the molecularity of each of the
elementary steps?
iii) What rate law is predicted by the
mechanism?
iv) Identify the intermediate.
Given :
i)
NO 2 (g) + F2 (g) ⎯⎯
→ NO 2 F (g) + F(g)
(slow, bimolecular)
ii)
F(g) + NO 2 (g) ⎯⎯
→ NO 2 F (g)
(Fast, bimolecular)
To find :
i) Overall reaction,
ii) Molecularity of each of the elementary steps
iii) Mechanism predicted by the rate law
iv) Intermediate identification.
Solution :
i) The overall reaction is obtained by adding
the two elementary steps of the
mechanism. Thus,
NO 2 (g) + F2 (g) ⎯⎯
→ NO 2 F(g) + F(g)
F (g) + NO 2 (g) ⎯⎯
→ NO 2 F (g)
2NO 2 (g) + F2 (g) ⎯⎯
→ 2NO 2 F (g)
(overall reaction)
Chapter - 5 Chemical Kinetics
360
ii)
iii)
The step (i) and (ii) both involve two
reactant molecules each. Hence, both the
steps are bimoecular.
The rate law is predicted by the
stoichiometric equation of the slow, rate
determining step (i)
NO 2 (g) + F2 (g) ⎯⎯
→ NO 2 F(g) + F(g)
iv)
(slow, rate determining)
The predicted rate law is rate = k [NO2] [F2]
The species F is a reaction intermediate
because it is produced in step (i) and
consumed in step (ii).
*5.
For the reaction,
A 2 + B + C ⎯⎯
→ AC + AB , it is found
that tripling the concentration of A2 triples
the rate, doubling the concentration of C
doubles the rate and doubling the
concentration of B has no effect.
i) What is the rate law?
ii) Why the change in concentration of B has
no effect.
Given :
i) Copy the reaction,
ii) Rate of reaction is tripled = 3A2,
iii) Rate of reaction is doubled = 2 C,
iv) Rate of reaction = 2B
To find :
i) Rate law,
ii) Change in concentration of B has no effect
on the rate of reaction.
Solution :
∴
∴
∴
ii)
∴
∴
iii)
∴
∴
∴
(d)
The new rate law
rate 2 = k {3[A2]}x [B]y[C]z
New rate is tripled
Rate2 = 3 × rate1
{3[A2]}x [B]y [C]z = 3 × [A2]x[B]y [C]z
3x [A2]x = 3 [A2]x
3x = 31
x=1
Now If [C] is doubled
The new rate law will be rate3 = k [A2]x
[B]y {2[C]}z
New Rate is doubled
Rate3 = 2 × Rate1
k [A2]x [B]y {2[C]}z = 2 × k [A2]x [B]y
{[C]}z
2z [C]z = 2 [C]z
2z = 2 1
Z=1
Now (B) is doubled
Rate4 = [A2]x {[2B]}y [C]z
The rate reaming unchanged
Rate4 = Rate1
k [A2]x [2y] [B]y [C]z = k [A2]x [B]y [C]z
2y = 1
y=0
Hence, the rate law is
Rate = k [A2]x [B]y [C]z
= k [A2]1 [B]0 [C]1
= k [A2] [C]
A 2 + B + C ⎯⎯
→ AC + AB
i)
The rate law may be represented as
Rate = k [A2]x [B]y [C]z
The concentration of [A2] is tripled.
Unique Solutions ®
*6.
(a)
The reaction A + B ⎯⎯
→ products is first
order in each of the reactants.
Write the rate law.
S.Y.J.C. Science - Chemistry - Part I
361
∴ Hence, new rate becomes 9 time by the
How does the reaction rate change if the
initial rate.
concentration of B is decreased by a factor 3?
(d)
If
concentration of [A] is doubled and that
(c) What is the change in the rate if the
of [B] is halved then new rate4
concentration of each reactant is tripled?
(d) What is the change in the rate if
1
= k 2 [A] [B]
concentration of A is doubled and that of
2
=
k
[A]
[B]
B is halved?
= rate1 [from (i)]
Given :
Hence, the rate remains same.
i) A + B ⎯⎯
→ products
ii) The reaction is a first order.
*7. Consider the reaction
To find :
i) Rate law,
2A + 2B ⎯⎯
→ 2C + D .
ii) Rate of reaction = ? when [B] is decreased
From the following data, calculate the order
by a factor 3,
and rate constant of the reaction.
iii) Change in a rate if 3[B],
[A]0/M
[B]0/M
r0/Ms–1
1
0.488
0.160
0.24
iv) Change in the rate if 2[A] and [B] .
2
0.244
0.160
0.06
Solution :
0.244
0.320
0.12
A + B ⎯⎯
→ products
Write the rate law of the reaction.
(a) The above reaction is first order in each
Given :
of the reactants
...(i)
Rate1 = k [A] [B]
i) 2A + 2B ⎯⎯
→ 2C + D
(b) If concentration of [B] is decrease by a
ii) [A]0/M
[B]0/M
r 0/Ms–1
factor of 3
0.488
0.160
0.24
1
0.244
0.160
0.06
Then new Rate2 = k [A]
[B]
3
0.244
0.320
0.12
1k
[A] [B]
=
To find :
3
i) Rate law of the reaction,
1
×
Rate
=
ii) Order and rate constant of the reaction.
1 ...[From i]
3
Solution
:
1 rd
of the initial
Hence, new rate becomes
i) Let the order of the reaction be x in A and
3
y in B, then
rate.
Rate = k [A]x [B]y
(c) If concentration of each reactant is tripled
Substituting the data given we get
Then new Rate3 = k 3[A] 3[B]
r 1 = 0.24 = k [0.488]x [0.160]y ...(i)
= 9 k [A] [B]
r 2 = 0.06 = k [0.244]x [0.160]y ....(ii)
∴ Rate3
= 9 × rate1
r 3 = 0.12 = k [0.244]x [0.320]y ....(iii)
...[From (i)]
(b)
Chapter - 5 Chemical Kinetics
362
Dividing (i) by (ii) we get;
k[0.488] x [0.160] y
0.24
=
k[0.244] x [0.160] y
0.06
What is the rate constant and order of the
reaction?
Given :
x
∴
∴
⎛ 0.488 ⎞
4 = ⎜
⎝ 0.244 ⎟⎠
4 = 2x
22 = 2x
x = 2
Now dividing (ii) by (iii) We get;
0.06
0.12
=
i)
ii)
k[0.244] x [0.160] y
k[0.244] x [0.320] y
1
⎡ 0.160 ⎤
= ⎢
2
⎣ 0.320 ⎥⎦
y
y
∴
∴
ii)
1
⎡ 1⎤
= ⎢ ⎥
2
⎣ 2⎦
y=1
Hence, the rate law will be rate = k[A]x [B]y
Rate = k [A]2 [B]1
∴ Order of the reaction = x + y
= 2+1
= 3
In order to find rate constant [k]
0.24 = k [0.488]x [0.160]y ...[From (i)]
0.24 = k [0.488]2 [0.160]1
k =
[A]/M
[B]/M
r/Ms–1
0.3
0.6
0.6
0.05
0.05
0.20
0.15
0.30
1.20
To find :
i) Rate law,
ii) Rate constant,
iii) Order of the reaction.
Solution :
Let the order of the reaction be x in A and
y in B then rate law is given by
Rate = k [A]x[B]y
Substituting the given data in above
equation we get ;
r1 = 0.15 = k [0.3]x [0.05]y
...(i)
x
y
...(ii)
r2 = 0.30 = k [0.6] [0.05]
x
y
r3 = 1.2 = k [0.6] [0.2]
...(iii)
Dividing (i) by (ii) we get ;
0.15
0.30
0.24
0.488 × 0.488 × 0.160
=
1
2
Order and rate constant of the reaction
= 6.3M –2 s –1
*8.
2A + B ⎯⎯
→ products
For the reaction 2A + B ⎯⎯
→ products,
find the rate law from the following data
[A]/M
[B]/M
r/Ms–1
0.3
0.6
0.6
0.05
0.05
0.20
0.15
0.30
1.20
Unique Solutions ®
=
k [0.3] x [0.05] y
k [0.6] x [0.05] y
⎡ 0.3 ⎤
⎢⎣ 0.6 ⎥⎦
x
x
∴
1
⎡1⎤
= ⎢ ⎥
2
⎣ 2⎦
x=1
Now dividing (ii) by (iii) we get;
0.3
=
1.2
1
4
=
k[0.6] x [0.05] y
k[0.6] x [0.2] y
⎡ 0.05 ⎤
⎢⎣ 0.2 ⎥⎦
y
S.Y.J.C. Science - Chemistry - Part I
363
1
4
=
⎡5⎤
⎢⎣ 20 ⎥⎦
y
y
∴
*9.
1
⎡1⎤
= ⎢ ⎥
4
⎣ 4⎦
y=1
Hence, the rate law will be Rate = k [A]x
[B]y
Rate = k [A] [B]
Order of the reaction = x + y
= 1+1
= 2
For calculation rate constant we can
substitute the values of x and y in Equ. ...(i)
r 1 = 0.15 = k [0.3]x [0.05]y
= 0.15 = k [0.3]1 [0.05]1
= 0.15 = k × 0.3 × 0.05
= 0.15 = k × 0.015
k = 10 M –1 s –1
The rate law for the reaction
C 2 H 4 Br2 +3I – ⎯⎯
→ C 2 H 4 + 2Br – + I 3–
∴
∴
is Rate = k [C2H4Br2] [I–]
If [C2H4Br2] = 0.12 M, [I–] = 0.18 M
Then rate = 1.1 × 10–4 M/s
1.1 × 10–4 = k [0.12] [0.18]
1.1 × 10 –4
0.12 × 0.18
k
=
k
= 5.1 × 10 –3 M –1 s –1
*10.
Consider the reaction,
A 2 + B ⎯⎯
→ products.
If the concentration of A2 and B are halved,
the rate of the reaction decreases by a
factor of 8. If the concentration of A is
increased by a factor of 2.5, the rate
increases by the factor of 2.5 What is the
order of the reaction? Write the rate law.
Given :
i)
A 2 + B ⎯⎯
→ products
ii)
[A2] and [B] =
iii)
Rate of reaction is decreased by
1
,
2
1
,
8
is rate = k [C2H4Br2] [I–]. The rate of the
iv) [A2] is increased by 2.5,
reaction is found to be 1.1 × 10–4 M/s when
the concentrations of C2H4Br2 and I– are
v) Rate increases by the factor of 2.5
0.12 M and 0.18 M respectively. Calculate
the rate constant of the reaction.
To find :
Given :
i) Order of the reaction,
ii)
Rate law
–
–
–
i) C 2 H 4 Br2 +3I ⎯⎯
→ C 2 H 4 + 2Br + I 3
Solution :
ii) Rate of reaction = 1.1 × 10–4 M/s,
A 2 + B ⎯⎯
→ products
iii) [C2H4Br2] = 0.12M, [I–] = 0.18M.
i) The rate law may be represented as
To find :
Rate = k [A2]x [B]y
Rate constant of the reaction
ii) If concentration of A2 is increased of a
Solution :
factor by 2.5, Then
C 2 H 4 Br2 +3I – ⎯⎯
→ C 2 H 4 + 2Br – + I 3–
New Rate r2 = k 2.5 [A2]x [B]y
Chapter - 5 Chemical Kinetics
364
∴
∴
∴
iii)
∴
∴
∴
The rate increases by a factor of 2.5
Hence, r2 = 2.5 r1
k [2.5]x [A2]x [B]y = 2.5 k [A2]x [B]y
(2.5)x = 2.5
x = 1
Now if the concentration of A2 and B are
halved then
x
y
⎧1
⎫ ⎧1
⎫
New rate, r3 = k ⎨ [A 2 ]⎬ ⎨ [B]⎬
⎩2
⎭ ⎩2
⎭
The rate decreases by a factor of 8,
1
r3 =
r
8 1
x
y
⎛1⎞
x ⎛1⎞
y
[A
]
k ⎜ ⎟
⎜⎝ ⎟⎠ [B]
2
⎝2⎠
2
1
= k [A2]x [B]y
8
x
∴
⎛1⎞ ⎛1⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠
2
2
∴
⎛1⎞
⎜⎝ ⎟⎠
2
∴
∴
∴
(d)
*11.
x+ y
y
=
1
8
⎛1⎞
= ⎜ ⎟
⎝2⎠
C + D ⎯⎯
→ products
i)
∴
∴
∴
∴
3
x+y=3
1+y=3
[∴ x =
y=3–1
y=2
Rate law is
Rate = k [A2]x [B]y
= k [A2]1 [B]2
Order of the reaction =
=
=
concentration of D is tripled. What is the
order of the reaction? Write the rate law.
Given :
i) C + D ⎯⎯
→ products
ii) Rate of reaction increases by 4, when [C]
is doubled.
iv) Rate of reaction is tripled, when[D] is
tripled.
To find :
i) Order of the reaction,
ii) Rate law
Solution :
1]
∴
∴
∴
∴
∴
x+y
1+2
3
ii)
Consider the reaction C + D products. The
rate of the reaction increases by a factor
of 4 when concentration of C is doubled.
The rate of the reaction is tripled when
Unique Solutions ®
*12.
The rate law may be representated as
Rate = k [C]x [D]y
If concentration of C is doubles than New
rate r2 = k [2C]x [D]y
The rate increases by a factor of 4
r2 = 4r1
k2x [C]x [D]y = 4 k [C]x [D]y
2x = 4
2x = 22
x = 2
If concentration of D is tripled, then New
rate r3 = k [C]x [3D]y
The rate is also tripled
r3 = 3r1
k [C]x 3y [D]y = 3 k [C]x [D]y
3y = 3
y = 1
Rate law will be rate = k [C]x [D]y
Rate = k [C] 2 [D]1
∴ Order of the reaction = x + y
= 2+1
= 3
The reaction
F2 (g) + 2ClO 2 (g) ⎯ ⎯
→ 2FClO 2 (g) is
S.Y.J.C. Science - Chemistry - Part I
365
first order in each of the reactants. The
rate of the reaction is 4.88 × 10–4 M/s when
[F2] = 0.015 M and [ClO2] = 0.025 M.
Calculate the rate constant of the reaction.
Given :
i) F2 (g) + 2ClO 2 (g) ⎯⎯
→ 2FClO 2 (g)
ii) The reaction is a first order,
iii) Rate of reaction = 4.88 × 10–4 M/s,
iv) [F2] = 0.015M, [ClO2] = 0.025M.
To find :
Rate constant of the reaction = ?
Solution :
F2 (g) + 2ClO 2 (g) ⎯⎯
→ 2FClO 2 (g)
iii) k = 0.42 M–2s–1,
iv) [H2] = 0.015M, [NO] = 0.025M
To find :
Rate of the reaction = ?
Solution :
2H 2 (g) + 2NO(g) ⎯⎯
→ 2H 2 O(g) + N 2 (g)
i)
∴
ii)
The above reaction is first order in H2 and
second order in NO
Rate law will be
Rate = k [H2] [NO]2
..(i)
If [H2] = 0.015 M, [NO] = 0.025M
k = 0.42 M–2 s–1
Then rate = k [H2] [NO]2
= 0.42 × 0.015 × [0.025]2
Rate = 3.94 × 10 –6 Ms–1
The above reaction is first order in each
of the reactants
∴ Rate = k [F2] [ClO2]
If [F2] = 0.015 M, [ClO2] = 0.025 M
then rate = 4.88 × 10–4 M/s
*14. The rate of reaction A ⎯⎯
→ products
∴ 4.88 × 10–4 = k [0.15] [0.025]
1.25 × 10–2 M/s when concentration of A
4.88 × 10 –4
k =
is 0.45M. Determine the rate constant if
0.015 × 0.025
∴ k = 1.3 M–1 s–1
the reaction is (a) first order in A (b) second
Rate constant of the reaction
order in A.
–1
–1
= 1.3 M s
Given :
i) Rate of reaction = 1.25 × 10–2 M/s,
*13. The reaction
ii) [A] = 0.45 M.
To
find :
2H 2 (g) + 2NO (g) ⎯⎯
→ 2H 2 O (g) + N 2 (g)
i) Rate constant (k) if reaction is first order
is first order in H2 and second order in NO.
in A.
The rate constant of the reaction at a certain
ii) Rate constant (k) if reaction is second
temperature is 0.42 M–2s–1. Calculate the
order in A.
rate when [H 2] = 0.015 M and
Solution :
[NO] = 0.025 M.
A ⎯⎯
→ products
Given :
i) If the reaction is first order in A then rate
i) 2H 2 (g) + 2NO (g) ⎯⎯
→ 2H 2 O (g) + N 2 (g)
law will be
ii) The reaction is a first order in H2 and
Rate = k [A]
second order in NO,
Chapter - 5 Chemical Kinetics
366
∴
ii)
∴
1.25 × 10–2 = k × 0.45
1.25 × 10 –2
k =
0.45
k = 2.778 × 10 –2 s –1
If the reaction is second order in A then
rate law will be
Rate = k [A]2
1.25 × 10–2 = k × (0.45)2
k
1.25 × 10 –2
= 0.45 × 0.45
[k = 6.173 × 10 –2 M–1 s–1]
k
k
ii)
[A] 0
2.303
. log 10
t
[A] t
2.303
20
. log10
=
38
8
2.303
=
× 0.3979
38
= 0.02411 min –1
=
t 1/2 =
=
0.693
k
0.693
0.02411
t 1/2 = 28.74 min
Type - III - Integrated rate laws
In acidic solution sucrose is converted to
*16. The half life of a first order reaction is
a mixture of glucose and fructose in pseudo
1.7 hours. How long will it take for 20%
first order reaction. It has been found that
of the reactant to disappear?
the concentration of sucrose decreased
Given :
from 20 mmol L–1 to 8 mmol L–1 in 38
i) (t1/2) = 1.7 hours,
minutes. What is the half life of the
ii) 20% of the reactant to disappear.
reaction?
To find :
Given :
t = ? (hours or min)
i) [sucrose] = [A]0 = 20 m mol L–1
Solution :
ii) [sucrose] = [A]t = 8 m mol L–1
i) For a first order reaction
iii) t = 38 min
0.693
iv) The reaction is a pseudo first order
t 1/2 =
k
reaction.
0.693
To find :
∴ k =
t 1/2
i) (k) Rate constant = ?
0.693
ii) half life of the reaction (t1/2) = ?
=
1.7
Solution :
k = 0.4076 h –1
i) Initial concentration of sucrose
ii) For 20% of the reactant to disappear initial
= [A]0 = 20 mmol L–1
concentration [A]0 of reactant = 100
Final concentration of sucrose
Final concentration remaining at time t =
= [A]t = 8 mmol L–1, t = 38 mim
[A]t = 100 – 20 = 80
Substituting the above values in the first
order equation
∴ k = 2.303 . log [A] 0
10
*15.
t
Unique Solutions ®
[A] t
S.Y.J.C. Science - Chemistry - Part I
367
= 2.303 ⋅ log 100
10
t
80
2.303
0.4076 =
⋅ log10 1.25
t
t
t
2.303
× 0.0969
=
0.4076
= 0.5483 hours
ii)
Now if [A]0 = 100, [A]t = 100 – 90 = 10
k=
[A] 0
2.303
⋅ log10
t
[A] t
0.035 =
2.303
100
⋅ log10
t
10
2.303 × 1.0
0.035
t = 65.8 min
For a first order reaction
t=
The gaseous reaction A 2 ⎯⎯
→ 2A is first
iii)
order in A2. After 12.3 minutes 65,% of
0.693
A2 remains undecomposed. How long will
t 1/2
=
k
it take to decompose 90% of A2? What
is the half life of the reaction?
0.693
=
Given :
0.035
i) [A]0 = 100,
t 1/2 = 19.8 min
ii) [A]t = 65,
iii) t = 12.3 mins
iv) The reaction is of first order.
*18. The rate constant of a first order reaction
To find :
is 6.8 × 10–4 s–1. If the initial concentration
i) t = ?,
of the reactant is 0.04 M, what is its molarity
ii) t1/2 = ?
after 20 minutes? How long will it take for
Solution :
25% of the reactant to react?
i) [A]0 = 100, [At] = 65, t = 12.3 mins
Given :
For a first order reaction
i) (k) Rate constant = 6.8 × 10–4 s–1.
ii) [A]0 = 0.04M,
k = 2.303 ⋅ log [A] 0
iii) t = 20 mins
10
t
[A] t
= 20 × 60 = 1200 sec.
iv) 20% of reactant react
Then k = 2.303 ⋅ log [A] 0
10
To find :
t
[A] t
i) [A]t = ?,
2.303
100
ii) Reactant remaining = ?
⋅ log10
k =
iii) t = ?
12.3
65
Solution :
2.303 × 0.1871
i) In a first order reaction
k =
12.3
If k = 6.8 × 10–4 s–1
–1
k =
0.0350 min
Initial conc [A]0 = 0.04 M.
*17.
Chapter - 5 Chemical Kinetics
368
t = 20 mins = 20 × 60 = 1200 ses.
Then k =
2.303
log10 [A] 0
t
[A] t
6.8 × 10–4
∴
∴
∴
∴
ii)
=
[A] 0
2.303
log10
[A] t
1200
68 × 10 –4 × 1200
=
log10
2.303
= 0.3543
Taking antilog an both sides we get
[A] 0
[A] t
[A] 0
[A] t
0.04
[A] t
= AL [0.3543] = 2.26
= 2.26
0.04
2.28
[A]t = 0.0177 M
If 25% of reactant reacts
Then Reactant reacted
= 25% of [A]0
[A]t =
*19.
The rate constant of a certain first order
reaction is 3.12 × 10–3 min–1.
i) How many minutes does it take for the
reactant concentration to drop to 0.02 M
if the initial concentration of the reactant
is 0.045 M?
ii) What is the molarity of the reactant after
1.5h?
Given :
i) (k) Rate constant = 3.12 × 10–3 min–1,
ii) [A]0 = 0.045 M,
iii) [A]t = 0.02 M,
iv) t = 1.5 hrs.
To find :
i) [A]t = ?
ii) t = (in mins)
Solution :
i) [A]0 = 0.045 M and [A]t = 0.02M
k = 3.12 × 10–3 min–1
[A] 0
2.303
⋅ log10
Then k =
t
[A] t
t
= [A]t
= 0.04 – 0.01
Reactant remaining = 0.03 M.
Now, k =
[A] 0
2.303
log10
[A] t
t
6.8 × 10–4 =
iii)
t
t
=
=
=
ii)
0.04
2.303
log10
0.03
t
3.12 × 10 –3
2.303 × 0.3522
= 259.9 min
3.12 × 10 –3
t ≈ 260 mins
If [A]0 = 0.045 M, t = 1.5 hrs. k = 3.12
× 10–3 min–1
k=
2.303 × 0.125
[A] 0
2.303
⋅ log10
t
[A] t
3.12 × 10–3 =
–4
6.8 × 10
= 423 secs.
= 7.05 min.
2.303
⎛ 0.045 ⎞
⋅ log 10 ⎜
⎝ 0.02 ⎟⎠
t
2.303 × log10 2.25
3.12 × 10–3 =
25
× 0.04
=
100
= 0.01 M
Reactant remaining
∴
log10
[A] 0
[A] t
=
=
Unique Solutions ®
[A] 0
2.303
⋅ log 10
1.5 × 60
[A] t
3.12 × 10 –3 × 1.5 × 60
2.303
0.122
S.Y.J.C. Science - Chemistry - Part I
369
∴
Taking Antilog on both sides
[A] 0
= Al [0.122]
[A] t
∴
[A] 0
[A] t
= 1.324
∴
[A]t
=
[A]t
(b)
[A] 0
1.324
0.045
=
= 0.034 M
1.324
t/min
[A]/M
Given
t/min
[A]/M
k
= 2.303 ⋅ log [A] 0
10
t
[A] t
k
= 2.303 ⋅ log ⎛ 0.08 ⎞
10 ⎜
⎝ 0.0359 ⎟⎠
20
k
= 2.303 . [log10 0.08 – log10 0.0359]
20
= 2.303 × ( 2.9031 – 2.5551)
20
[A]t = 0.034 M.
*20.
If [A]0 = 0.08M, [A]t = 0.0359, t = 20 min.
The concentration of a reactant in a first
→ products , varies
order reaction A ⎯⎯
with time as follows :
0
10
20
30
40
0.0800 0.0536 0.0359 0.0241 0.0161
Show that the reaction is first order.
:
0
10
20
30
40
0.0800 0.0536 0.0359 0.0241 0.0161
= 2.303 × 0.3480
20
k = 0.04 min–1
Since the value of rate constant using
first order rate law equation comes
constant, the reaction is of first order.
*21.
From the following data for the liquid phase
reaction A ⎯⎯
→ B , determine the order
To find :
of the reaction and calculate its rate
Rate of reaction is a first order.
constant.
Solution :
t/s
0
600
1200
1800
i) If [A]0 = 0.08 M, [A]t = 0.0536, t = 10
[A]/mol L–1 0.624 0.446 0.318 0.226
mins.
Given :
[A] 0
2.303
⋅ log10
k =
t/s
0
600
1200
1800
t
[A] t
–1
0.624 0.446 0.318 0.226
[A]/mol L
2.303
0.08
k = 10 ⋅ log10 0.0536
To find :
2.303
i) Order of the reaction
= 10 . (log10 0.08 – log10 0.0536)
ii) rate constant
Solution :
2.303
× (2.9031 – 2. 7292)
=
i) If [A]0 = 0.624 mol L–1
10
[A]t = 0.446 mol L–1
2.303
=
× 0.1739
t = 600s
10
–1
k = 0.04 min .
Chapter - 5 Chemical Kinetics
370
k
ii)
k
If
k
[A] 0
2.303
⋅ log10
t
[A] t
2.303
0.624
⋅ log 10
=
t
0.446
2.303
=
. (log100.624 – log100.446)
600
2.303
× (1.7952 – 1.6493)
=
600
2.303
× 0.1459
=
600
= 5.6 × 10–4 s –1
[A]0 = 0.624 mol L–1
[A]t = 0.318 mol L–1
t = 1200s
=
=
[A] 0
2.303
⋅ log10
t
[A] t
=
2.303
⋅ log 10
t
ii) [A]t = 100 – 99.9 = 0.1
To find :
t99.9% = 3t90%
Solution :
For 99.9% completion of first order reaction
[A]0 = 100, [A]t = 100 – 99.9 = 0.1
[A] 0
2.303
⋅ log10
t
[A] t
k
=
k
2.303
100
⋅ log10
= t
0.1
99.9%
k
=
2.303
⋅ log10 1000
t 99.9%
2.303 × 3
...(i)
t 99.9%
For 90% completion of first order reaction
[A]0 = 100, [A]t = 100 – 90 = 10
k
=
2.303
=
. (log10 0.624 – log10 0.318)
1200
k
=
[A] 0
2.303
⋅ log10
t
[A] t
2.303
× (1.7952 – 1.5024)
=
1200
k
=
2.303
⋅ log 10 10
t 90%
2.303 × 0.2928
=
1200
k
⎛ 0.624 ⎞
⎜⎝
⎟
0.318 ⎠
k = 5.61 × 10–4 s –1
Since the values of rate constant using first
order rate law equation comes constant.
The reaction is of first order
The rate constant k = 5.6 × 10–4 s–1.
Show that the time required for 99.9%
completion of a first order reaction is three
times the time required of 90% completion.
Given :
i) [A]0 = 100,
2.303
t 90%
Equating (i) and(ii) we get;
∴
*22.
Unique Solutions ®
*23.
=
...(ii)
2.303 × 3
2.303
=
t 99.9%
t 90%
t99.9% = 3 × t90%
Hence, time required for 99.9%
completion of a first order reaction is
3 times the time required for 90%
completion of the reaction.
A flask contains a mixture of A and B. Both
the compounds decompose by first order
S.Y.J.C. Science - Chemistry - Part I
371
kinetics. The half lives are 60 min for A
and 15 min for B. If initial concentrations
of A and B are equal, how long will it take
for the concentration of A to be three time
that of B?
Given :
i) [A]0 = [B]0 = M,
ii) [A]t = 3x,
iii) [B]t = x,
iv) t1/2 for A = 60 mins,
v) t1/2 for B = 15 mins,
To find :
t = ? when [A] = 3[B]
Solution :
For A : t 1/2 = 60 min; For B : t 1/2 = 15min
Let the initial concentration of
[A]0 = [B]0 = Mmol dm–3
After time t, let the concentration of
[B]t = x and then [A]t = 3x
0.693
0.693
kA = t
=
60
1/2
–1
= 0.01155 min
0.693
0.693
=
t 1/2
15
–1
= 0.0462 min
kB =
∴
[B] 0
2.303
. log10
t
[B] t
kB =
2.303
M
. log10
x
t
=
0.0462 × t
2.303
M
= 0.02 t.
x
...(i)
Now
kA =
[A] 0
2.303
. log10
[A] t
t
=
2.303 .
M
log10
t
3x
=
2.303
t
=
2.303
0.02t + (log 1 – log 3 )
10
10
t
M
1⎞
⎛
+ log10 ⎟
⎜⎝ log 10
x
3⎠
(
)
2.303
( 0.02 t – 0.4771)
t
= 2.303 × 0.02
kA =
kA
2.303 × 0.4771
t
1.1
0.01155 = 0.04606 –
t
1.1
= 0.04606 – 0.01155
t
1.1
= 0.03451
t
t = 31.8 min.
=
∴
∴
∴
kB =
log10 M
x
kB × t
=
2.303
log10
Type - IV - Collision theory,
activation energy and catalyst
*24. The rate constants of a first order reaction
are 0.58s–1 at 313 K and 0.045 s–1 at 293
K. What is energy of activation for the
reaction?
Given :
i) (k) Rate constant = 0.58 s–1
ii) T1 = 293k, T2 = 313k
iii) K1 = 0.045 s–1, K2 = 0.58 s–1
Chapter - 5 Chemical Kinetics
372
To find :
Ea = ?
Solution :
At Temp T1 = 293K k1 = 0.045s–1
Temp T2 = 313K k2 = 0.58s–1
By Arrhenius Equation,
log 10
k2
E a ⎛ T2 – T1 ⎞
=
k1
2.303R ⎜⎝ T2 T1 ⎟⎠
log 10
Ea
⎛ 313 – 293 ⎞
0.58
= 2.303×8.314 ⎜⎝ 313×293 ⎟⎠
0.045
To find :
k2 = ? at 30ºC.
Solution :
Energy of activation
Ea = 104 kJ mol–1
= 104 × 103 J mol–1
At Temp T1 = 25ºC
= 25 + 273 = 298K,
k1 = 3.7 × 10–5 s–1
At Temp T2 = 30ºC
= 30 + 273 = 303K
log100.58 – log100.045
E a × 20
=
2.303 × 8.314 × 313 × 293
E a (T2 – T1 )
log10 k 2 =
2.303R (T2 ⋅ T1 )
k1
104 × 10 3 (303 – 298)
= 2.303 × 8.314 × 303 × 298
(1.7634 – 2.6532)
=
E a × 20
2.303 × 8.314 × 313 × 293
1.1102 =
∴
Ea =
log10
∴
E a × 20
2.303 × 8.314 × 313 × 293
1.1102×2.303×8.314×313×293
20
4
Ea = 9.746 × 10 J mol–1
Ea = 97.47 kJ mol–1
Energy of activation = Ea = 97.47 kJ
mol–1.
k2
= 0.3
k1
k2
= Antilog (0.3)
k1
k2
= 1.995
k1
k2 = 1.995 × k1
= 1.995 × 3.7 × 10–5
k 2 = 7.4 × 10 –5 s –1
Rate Constant at 30ºC = 7.4 × 10–5 s–1.
*26.
*25.
The energy of activation for a first order
reaction is 104 kJ mol–1. The rate constant
at 25ºC is 3.7 × 10–5 s–1. What is the rate
constant at 30ºC?
Given :
i) Ea = 104 kJ mol–1,
ii) (k1) Rate consant at 25ºC = 3.7 × 10–5s–1,
iii) T1 = 25ºC,
iv) T2 = 30ºC.
Unique Solutions ®
What is the activation energy for a reaction
whose rate constant doubles when
temperature changes from 30ºC to 40ºC?
Given :
i) k2 = 2k1,
ii) T1 = 30ºC & T2 = 40ºC.
To find :
Ea = ?
Solution :
If Temp T1 = 30ºC = 30 + 273 = 303K
S.Y.J.C. Science - Chemistry - Part I
373
Then Rate constant = k1
If Temp T2 = 40ºC = 40 + 273
= 313K
Then Rate constant k2 = ?
Now k2 = 2k1
E a (T2 – T1 )
k
log10 2 =
2.303R (T2 .T1 )
k1
E a (313 – 303)
2k 1
log10
=
2.303 × 8.314 × 313 × 303
k1
E a × 10
log10 2
=
2.303 × 8.314 × 313 × 303
log10
k2
k1
∴
∴
∴
∴
k2
k1
= 0.4771
= Antilog (0.4771)
k2
= 3
k1
k2 = 3k 1
Rate constant increases by 3 times.
The rate of a reaction at 600K is 7.5 ×
105 times the rate of the same reaction at
0.3010 × 2.303 × 8.314 × 313 × 303
400K. Calculate the energy of activation
Ea =
10
for the reaction. (Hint : The ratio of rates
= 5.466 × 104 J mol–1
is equal to the ratio of rate constants.)
= 54.66 kJ mol–1
Given :
Theenergy of activation = 54.66 kJ
(i) Rate of a reaction = 7.5 × 105 (at 600k)
mol–1.
(ii) R2 = R1,
(iii) T1 = 400k, T2 = 600k
*27. The activation energy for a certain reaction To find :
is 334.4 kJ mol–1. How many times larger
(i) Ea = ?
is the rate constant at 610 K than the rate Solution :
constant at 600 K?
Let T2 = 600K,
Given :
T1 = 400K,
i) Ea = 334.4 kJ mol–1
Then Rate 2 = k 2 = 7.5 × 105
ii) T1 = 600k, T2 = 610 k
Rate1
k1
To find :
k2
=?
k1
Solution :
T2 = 610K
T1 = 600K,
–1
Ea = 334.4 kJ mol
= 334.4 × 103 J mol–1
Now
=
log10 k 2 = E a ⎛ T2 – T1 ⎞
k1
2.303R ⎜⎝ T2 T1 ⎟⎠
334.4 × 10 3 (610 – 600)
2.303 × 8.314 × 610 × 600
*28.
Now
log10
k 2 = E a ⎛ T2 – T1 ⎞ log 7.5 × 105
10
2.303R ⎜⎝ T2 . T1 ⎟⎠
k1
=
E a (600 – 400)
2.303 × 8.314 × 600 × 400
5.8751 =
Chapter - 5 Chemical Kinetics
E a × 200
2.303 × 8.314 × 600 × 400
374
∴
Ea =
Ea
5.8751 × 2.303 × 8.314 × 600 × 400
200
1
1
0.7781 × 2.303 × 8.314
–
=
T1 T2
88 × 10 3
1
1
–
= 1.7 × 10–4
T1 T2
1
1
–
= 1.7 × 10–4
T1 298
= 1.35 × 105 kJ mol–1
= 135 × 103Jmol–1
= 135 kJ mol–1.
*29.
The rate constant of a first order reaction
at 25ºC is 0.24 s –1. If the energy of
activation of the reaction is 88 kJ mol–1,
at what temperature would this reaction
have rate constant of 4 × 10–2 s–1?
Given :
i) (k2) Rate constant = 0.24 s–1,
ii) Ea = 88 kJ mol–1,
iii) (k1) Rate constant = 4 × 10–2 s–1,
iv) T2 = 25ºC.
To find :
T1 = ?
Solution :
Let T2 = 25ºC = 25 + 273 = 298K
Then k2 = 0.24 s–1
Then If k1 = 4 × 10–2 s–1
Then T1 = ?
E a (T2 – T1 )
k
Now log10 2 =
2.303R (T2 . T1 )
k1
88 × 10 3 ⎛ T2 – T1 ⎞
⎛ 0.24 ⎞
log10 ⎜
=
⎟
⎝ 0.04 ⎠ 2.303 × 8.314 ⎜⎝ T2 . T1 ⎟⎠
log10 (0.24) – log10 0.04
=
88 × 10 3 ⎛ 1
1 ⎞
–
⎜
2.303 × 8.314 ⎝ T1 T2 ⎟⎠
3
⎛1
88 × 10
1 ⎞
–
(1.3802 – 2.6021) = 2.303
⎜
× 8.314 ⎝ T1 T2 ⎟⎠
0.7781 =
88 × 10 3 ⎛ 1
1 ⎞
–
⎜
2.303 × 8.314 ⎝ T1 T2 ⎟⎠
Unique Solutions ®
1
298
–4
= 1.7 × 10 + 0.003356
= 0.00017 + 0.003356
1
T1
∴
= 1.7 × 10–4 +
1
T1
= 0.003526
T1
=
T1
1
0.003526
= 283.6 K.
The temp at which rate constant
4 × 10 –2 s–1 is 283.6 K.
*30.
The rate constant for a reaction at 500ºC
is 1.6 × 103 M–1 s–1. What is the
frequency factor of the reaction if its
energy of activation is 56 kJ mol–1.
Given :
i) (k) Rate constant = 1.6 × 103 M–1 s–1,
ii) T = 500ºC = 500 + 273 = 773k,
iii) Ea = 56 kJ mol–1
To find :
Arrhenius frequency factor (A) = ?
Solution :
At T = 500 + 273 = 773 K
k = 1.6 × 103 M–1 s–1
Ea = 56 kJ mol–1
= 56 × 103 J mol–1
Now by Arrhenius equation;
k = A × e –E a /RT
∴
log10k = log10A
– Ea
2.303 RT
S.Y.J.C. Science - Chemistry - Part I
375
log101.6 × 103
56 × 10 3
= log10A –
2.303 × 8.314 × 773
3.2041 +
∴
56 × 10 3
= log10A
2.303 × 8.314 × 773
log10A = 3.2041 + 3.784
log10A = 6.9881
A = Antilog (6.9881)
A = 9.729 × 106 M–1 s –1 .
Note : The unit of frequency factor is
same as rate constant.
Frequency factor =
A = 9.729 × 106 M–1 s –1
*31.
A first order gas-phase reaction has an
energy of activation of 240 kJ mol–1. If the
frequency factor of the reaction is 1.6 × 1013
s–1, calculate its rate constant at 600 K.
Given :
i) Ea = 240 kJ mol–1,
ii) (A) frequency factor = 1.6 × 1013 s–1,
iii) T = 600k.
To find :
(k) Rate constant = ?
Solution :
Energy of activation
Ea = 240 kJ mol–1
= 240 × 103 J mol–1
Frequency factor = A = 1.6 × 1013 s–1
T = 600K
By Arrhenius equation
Ea
log10k = log10A –
2.303RT
240 × 10 3
= log10 1.6×1013–
2.303 × 8.314 × 600
= 13.2041 – 20.9
∴
log10k = –7.6959
k = Antilog [–7.6959]
= AL[8.3041]
k = 2.01 × 10–8 s–1
Rate constant = k = 2.01 × 10–8 s–1.
*32.
The half life of a first order reaction is 900
min at 820 K. Estimate its half life at 720
K if the energy of activation of the reaction
is 250 kJ mol–1.
Given :
i) (t 1/2 ) 2 = 900 mins,
ii) T2 = 820 k,
iii) T1 = 720 k,
iv) Ea = 250 kJ mol–1,
To find :
(t 1/2 )1 = ?
Solution :
At Temp T2 = 820K t 1/2 = 900 min
If Temp T1 = 720K ( t 1/2 ) = ?
Ea = 250 kJ mol–1
= 250 × 103 J mol–1
For a first order reaction
1
t 1/2 ∝
k
( t 1/2 )1 k 2
∴
...(i)
( t 1/2 ) = k 1
2
⎛k2 ⎞
Ea
⎛ T2 – T1 ⎞
Now log10 ⎜⎝ k ⎟⎠ = 2.303R ⎜⎝ T × T ⎟⎠
1
2
1
log10
log10
log10
(820–720)
250×103
(t 1/2 )1
= 2.303×8.314 × (820×720)
(t 1/2 ) 2
( t 1/2 )1
900
( t 1/2 )1
Chapter - 5 Chemical Kinetics
900
=
250 × 10 3 × 100
2.303 × 8.314 × 820 × 720
= 2.2115
376
∴
( t 1/2 )1
900
( t 1/2 )1
900
=
Antilog (2.211)
=
1.626 × 102 × 900
(t 1/ 2 )1
= 1.463 × 10 5 min
900
The half life at 720K will be 1.463 × 105 min.
NUMERICALS FOR PRACTICE
*1.
TYPE - I - RATE OF REACTION AVERAGE RATE AND INSTANTANEOUS RATE :
Write the expressions for the following reaction in terms of rate of consumption of reactants
and formation of products. F2 (g) + 2ClO 2 (g) ⎯⎯
→ 2FClO 2 (g)
Solution :
d[F2 ]
dt
d[ClO 2 ]
Rate of consumption of ClO2 at time t = –
dt
d[FClO 2 ]
Rate of formation of FClO2 at time t = –
dt
d[F2 ]
1 d[ClO 2 ]
1 d[FClO 2 ]
Rate of reaction at time t = –
=
=
dt
2
dt
2
dt
Consider the reaction,
Rate of consumption of F2 at time t = –
*2.
2N 2O 5 (g) ⎯⎯
→ 4NO 2 (g) + O 2 (g) in liquid bromine. At a particular moment during the
reaction N2O5 disappears at a rate of 0.02 M/s. At what rates NO2 and O2 are formed?
What is the rate of the reaction?
Solution :
d[N 2 O 2 ]
= 0.02 M/s
dt
The reaction shows that the rate of formation of NO2 is twice the rate of consumption of
N2O5. Hence,
–
d[NO 2 ]
d[N 2 O 5 ]
= –2
dt
dt
=
–2 × (–0.02 M/s) = 0.04 M/s
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
377
The rate of consumption of N2O5 is twice the rate of formation of O2.
–
d[O 2 ]
d[N 2 O 5 ]
= –2
dt
dt
Hence,
d[O 2 ]
dt
or
=
–
1 d[N 2O 5 ]
2
dt
d[O 2 ]
1
= – × (– 0.02 M/s)
dt
2
= 0.01 M/s
Rate of reaction = –
1 d[N 2O 5 ]
2
dt
=
1 d[NO 2 ]
4 dt
=
d[O 2 ]
dt
1
= – × (– 0.02 M/s) = 0.01 M/s
2
*3.
Nitrogen dioxide decomposes to nitric oxide and molecular oxygen as
2NO 2 (g) ⎯⎯
→ 2NO (g) + O 2 (g)
The concentration time data for the consumption of NO2 at 300ºC are as follows :
Time/s
[NO2]/M
0
100
200
300
8.4 × 10–3
5.6 × 10–3
4.3 × 10–3
3.0 × 10–3
Calculate the average rate of decomposition of NO2 and the average rates of formation of
NO and O2 during the time interval 200 s to 300 s. What is the average rate of the reaction
during the same interval?
Solution :
Average rate of decomposition of NO2
=
3.0 × 10 –3 (M) – 4.3 × 10 –3 (M)
Δ[ΝΟ 2 ]
–
–
=
= 1.3 × 10–5 M/s
300(s) – 200(s)
Δt
Average rate of formation od NO
=
–
Δ[ΝΟ 2 ]
Δ[ΝΟ]
= –
= 1.3 × 10–5 M/s
Δt
Δt
Average rate of formation of O2
Chapter - 5 Chemical Kinetics
378
=
–
Δ[Ο 2 ]
1 Δ[ΝΟ 2 ]
1.3 × 10 –5 M/s
= –
=
= 6.5 × 10–6 M/s
Δt
2 Δt
2
Average rate of reaction = –
Δ[Ο 2 ]
1 Δ[ΝΟ 2 ]
1 Δ[ΝΟ]
=
=
= 6.5 × 10–6 M/s
2 Δt
Δt
2 Δt
TYPE - II - RATE LAW, RATE CONSTANT AND ORDER OF REACTION AND MOLECULARITY :
*4.
Write the rate law for the reaction A + B ⎯⎯
→ C , from the following data :
Initial rate/Ms–1
Initial [A]/M
Initial [B]/M
0.4
0.2
4.0 × 10–5
0.6
0.2
6.0 × 10–5
0.8
0.4
3.2 × 10–4
What is the value of k?
[Ans : k = 2.5 × 10–3 M–2 s–1]
*5.
The rate law for the reaction A + B ⎯⎯
→ products is rate = k [A][B]2. The rate of the reaction
at 25ºC is found to be 0.25 M/s when [A] = 1.0 M and [B] = 0.2 M. Calculate the rate constant
k at this temperature.
[Ans : 6.25 M–2 s–1]
*6.
*7.
The rate law for the reaction, 2NOBr (g) ⎯⎯
→ 2NO (g) is rate = k[NOBr]2. The rate constant
for the reaction at a certain temperature is 1.62 M–2 s–1. If the concentration of NOBr at
a certain time is 2.00 × 10–3 M, what will be the rate of reaction?
[Ans : 6.48 × 10–6 M–2 s–1]
Consider the reaction . If the concentration of A is doubled at constant [B], the rate of the
reaction increases by a factor 4. If the concentration of B is doubled at constant [A], the
rate is doubled. What is the rate law?
[Ans : rate = k [A]2[B]]
*8.
Consider the reaction 2H 2 (g) + 2NO (g) ⎯⎯
→ 2H 2 O (g) + N 2 (g) . The rate of the reaction
*9.
doubles if the concentration of H2 is doubled. If the concentration of NO is doubled the rate
increases by a factor 4. Write the rate law.
[Ans : rate = k [H2][NO]2
What is the order with respect to each of the reactants and overall order of the following
reaction?
(a) 2NO (g) + 2H 2 (g) ⎯⎯
rate = k [NO]2 [H2]
→ N 2 (g) + 2H 2 O (g)
(b) CHCl 3 (g) + Cl 2 (g) ⎯⎯
→ CCl 4 (g) + HCl(g)
Unique Solutions ®
rate = k [CHCl3] [Cl2]1/2
[Ans : 1 + 1/2 = 3/2]
S.Y.J.C. Science - Chemistry - Part I
379
*10.
The reaction, 2NO (g) + Cl 2 (g) ⎯⎯
→ 2NOCl(g) is first order in Cl2 and second order in NO.
Write the rate law for the reaction.
[Ans : rate = k [NO]2 [Cl2]
*11.
The rate of the reaction, A + B ⎯⎯
→ P is 3.6 × 10–2 M/s [A] = 0.2M and [B] = 0.1M. Calculate
k if the reaction is 2nd order in A and first order in B.
*12.
[Ans : 9.0 M–2 s–1]
The rate of reaction A + 2B ⎯⎯
→ C + 2D is 6 × 10–4 M/s when [A] = [B] = 0.3M. What
is the overall order of the reaction if the constant at a given temperature is 2 × 10–3 s–1.
[Ans : 1]
*13.
The rate of a first order reaction, A ⎯⎯
→ B is 5.4 × 10–6 Ms–1 when [A] is 0.3M. Calculate
the rate constant of the reaction.
*14.
[Ans : 1.8 × 10–5 s–1]
For the reaction, A + B ⎯⎯
→ P . If [B] is doubled at constant [A], the rate of the reaction
doubles. If [A] is tripled and [B] is doubled, the rate of the reaction increases by a factor
of 6. What is the order of the reaction with respect to each reactant and the overall order
of the reaction?
[Ans : It follows that the reaction is first order with respect to A and B each
and seconder overall.]
*15.
*16.
*17.
*18.
TYPE - III - INTEGRATED RATE LAWS :
The half life of a first order reaction is 0.5 min at a certain temperature. Calculate (a) the
rate constant of the reaction (b) the time required for 80% reactant to decompose.
[Ans : seconds are required for 80% decomposition of the reactant]
The rate constant for a first order reaction is 7.0 × 10–4 s–1. If the initial concentration of
the reactant is 0.080 M, what concentration will remain after 35 minutes?
[Ans : The concentration that remains after 35min is 0.0184 M]
The half life of a first order reaction is 6.0 h. How long will it take for the concentration of
reactant to decrease from 0.8 M to 0.25 M?
[Ans : Half life of reaction is 34 min]
The concentration of N2O5 in liquid bromine varied with time as follows :
t/s
0
200
400
600
[N2O5]/M
0.15
0.098
0.064
0.042
Show that the reaction is first order.
[Ans : This indicates the reaction obeys the integrated rate equation of first
order reaction. The reaction is a first order reaction.]
Chapter - 5 Chemical Kinetics
380
*19.
For the decomposition of ethylene oxide into CH4 and CO, the variation of total pressure (P)
of the reaction mixture with time is as given below :
t/s
0
300
600
900
P/mm
120
127.2
134.0
Show that the reaction is a first order reaction.
*20.
*21.
140.3
[Ans : k = 2.0066 × 10–4 s–1]
The kinetics of hydrolysis of methyl acetate in excess of dilute HCl at 298 K was followed
by withdrawing 5 mL of the reaction mixture at different time intervals and titrating against
standard alkali. The following results were obtained.
t/min
0
10
20
30
∞
Volume of alkali/mL
20.1
20.5
20.9
21.3
35.2
Show that the reaction follows first order kinetics and calculate the rate constant.
[Ans : k = 2.71 × 10–3 min–1]
The following results were obtained in the decomposition of H2O2 in pressure of I– ions at
298 K.
t/min
3.0
6.0
9.0
∞
3
6.2
11.8
16.8
60.6
Volume of O2 evolved/cm
–
1
I
Show that the reaction, H 2O 2 (g) ⎯⎯
→ H 2O(l) + O 2 (g) is a first reaction. What is the
2
rate constant?
[Ans : k = 0.0361 min–1]
TYPE - IV - COLLISION THEORY, ACTIVATION ENERGY AND CATALYST :
*22.
*23.
*24.
*25.
The rate constant for the reaction, CH 3CHO (g) ⎯⎯
→ CH 4 (g) + CO (g) is 0.035 M–1/2 s–1
at 730 K and 0.343 M–1/2 s–1 at 790 K. Calculate the activation energy of the reaction.
(R = 8.314 JK–1 mol–1).
[Ans : The energy of activation of the reaction is 182.4 kJ mol–1]
For a certain second order reaction energy of activation is 240 kJ mol–1. Calculate its rate constant
at 1023 K if the rate constant at 923 K is 0.0113 M–1 s–1. (R = 8.314 JK–1 mol–1]
[Ans : The rate constant at 1023 K is 0.24 M–1 s–1 ]
If the rate of a reaction increases by a factor of 2.36, when the temperature is raised from
303 K to 313 K. What is the energy activation? R = 8.314 JK–1 mol–1.
[Ans : 67.6 kJ mol–1]
For a first order reaction rate constant at 500 K is 7.66 × 10–4 s–1. Calculate frequency factor
if the energy of activation for the reaction is 160 kJ mol–1.
[Ans : 3.96 × 1013 s–1]
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
381
*26.
The energy of activation for a certain second order reaction is 85.2 kJ mol–1 and its frequency
factor is 3.1 × 1011 L mol–1s–1 at 310 K. Calculate the rate constant of the reaction.
[Ans : 1.372 × 10–3 L mol–1 s–1 ]
HIGHER ORDER THINKING SKILLS (HOTS)
*1.
The following results were obtained in the decomposition of H2O2 in KI solution at 30ºC.
t/s
Volume of O2 collected/cm
3
100
200
300
∞
7.3
13.9
19.6
65.0
Show that the reaction is first order. Calculate the rate constant of the reaction.
Given :
t/s
100
200
300
∞
Volume of O2 collected/cm3
7.3
13.9
19.6
65.0
To find :
i) Reaction is first order,
ii) Rate constant
Solution :
1
→ H 2O + O 2
i) H 2 O 2 ⎯⎯
2
Volume of H2O2 decomposed ∝ volume of O2 formed
when H2O2 is completely decomposed, at t = ∝
1
∝ Original amount of H2O2 = [A]0
∞
1
(volume of O2 liberated at time t) ∝ Amount of H2O2 decomposed at time t.
t
1 1
– = [A]t
∴ Amount of H2O2 remaining undecomposed at time t =
∞ t
ii) Hence, the first order Rate law empression.
[A] 0
2.303
⋅ log10
t
[A] t
k
=
k
⎛ 1 ⎞
2.303
⎜
⎟
⋅ log 10 ⎜ ∞ ⎟
=
1
1
t
⎜ – ⎟
⎝∞ t ⎠
=
2.303
⎛ 65 ⎞
⋅ log 10 ⎜
⎝ 65 – 7.3 ⎟⎠
t
Chapter - 5 Chemical Kinetics
382
k
iii)
2.303
⋅ log 10
t
=
2.303
(log10 65 – log10 57.7)
100
=
2.303
× (1.8129 – 1.7612) =
100
At t = 200s, [A0] =
=
=
∴ k =
*2.
2.303
× 0.0517
100
= 1.2 × 10 –3 s –1.
[A]t =
iv)
⎛ 65 ⎞
⎜⎝
⎟
57.7 ⎠
=
1
= 65cm3
∞
1 1
–
∞ t
65 – 13.9
51.1
[A] 0
2.303
⋅ log10
t
[A] t
=
2.303
⎛ 65 ⎞
⋅ log 10 ⎜
⎝ 51.1 ⎟⎠
t
=
2.303
. (log 10 65 – log 10 51.1)
200
= 2.303 (1.8129 – 1.7084) = 2.303 × 0.1045
200
200
–3 –1
k = 1.2 × 10 s .
∴ The values of rate constant using first order rate law equation comes constant it is a first
order reaction constant value of rate constant [k = 1.2 × 10–3 s–1]
From the following data for the decomposition of azoisopropane,
(CH 3 ) 2CHN= CH(CH 3 ) 2 ⎯⎯
→ N 2 +C 6H14 estimate the rate of the reaction when total pressure
is 0.75 atm.
Time/s
Total pressure/atm
0
200
0.65
1.0
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
383
Given :
i)
Time/s
0
200
Total pressure/atm
0.65
1.0
ii) (CH 3 ) 2CHN= CH(CH 3 ) 2 ⎯⎯
→ N 2 +C 6H14
iii) Total Pressure = 0.75 atm.
To find :
rate of reaction = ?
Solution :
(CH 3 ) 2CHN= CH(CH 3 ) 2 ⎯⎯
→ N 2 +C 6H14
∴
At time t = 0
0.65 atm
0 0
time t = 200s
0.65 – x
x x
Total pressure = 0.65 – x + x + x
1.00 atm = 0.65 + x
x = 1 – 0.65
x = 0.35 atm
Initial pressure = [A]0 = 0.65 atm
pressure at t = 200s =
= 0.65 – x
[A]t
= 0.65 – 0.35
= 0.30 atm
Now, for a first order reaction
k
=
[A] 0
2.303
2.303
0.65
⋅ log10
⋅ log 10
=
t
[A] t
200
0.30
k
=
2.303
2.303
(1.8129 – 1.4771)
(log 0.65 – log 0.30) =
200
200
=
2.303
× 0.3358
200
k
∴
∴
(CH 3 ) 2CHN= CH(CH 3 ) 2 ⎯⎯
→ N 2 +C 6H14
= 3.87 × 10–3 s–1
At time t = 0 0.65 atm
At time = t
0.65 – x
Ptotal = 0.65 – x + x + x
0.75 = 0.65 + x
x = 0.10 atm
0
x
0
x
Ptotal = 0.65 + x
Chapter - 5 Chemical Kinetics
384
∴
Pressure at time = t = [A]t
= 0.65 – x
= 0.65 – 0.10
= 0.55 atm
Now, rate of first order reaction at time ‘t’ is
Rate = k [A]t
= k × 0.55
= 3.87 × 10–3 × 0.55
Rate = 2.13 × 10–3 atm s–1
Rate of the reaction = 2.13 × 10–3 atms–1
*3.
The kinetics of hydrolysis of methyl acetate in excess of dilute HCl at 298 K was followed
by withdrawing 5 mL of the reaction mixture at different time intervals and titrating against
standard alkali. The following results were obtained.
t/min
Volume of alkali/mL
0
20.1
10
20.5
20
20.9
30
21.3
∞
35.2
Show that the reaction follows first order kinetics and calculate the rate constant.
Given :
i)
t/min
Volume of alkali/mL
0
20.1
10
20.5
20
20.9
30
21.3
∞
35.2
HCl
CH 3COOCH 3 (aq) + H 2O ⎯⎯⎯
→
CH 3COOH(aq) + CH 3OH (aq)
To find :
i) Rate of reaction is first order
ii) Rate constant (k) = ?
Solution :
HCl
The hydrolysis reaction is CH 3COOCH 3 (aq) + H 2O ⎯⎯⎯
→
CH 3COOH(aq) + CH 3OH (aq)
V 0 = volume of alkali required for titration at t = 0
= volume of alkali required for neutralization of HCl initially present
V∞ = volume of alkali required after completion of reaction
= volume of alkali requires for HCl initially present and for acetic acid produced after
completion of reaction
Hence,
V∞ – V0 = volume of alkali required for acetic acid produced from entire ester = [A]0 = ‘a’
V t = volume of alkali required at different time intervals
ii)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
385
= volume of alkali requried for acetic acid produced at different time intervals and
HCl present initially
V t – V 0 = volume of alkali required for acetic acid produced at different time intervals
=x
Hence, a – x = V∞ – V0 – Vt + V0 = V∞ – Vt
The rate constant of first order reaction is
k
=
2.303
a
. log
10
t
a–x
V∞ – V0
2.303
. log
10
t
V∞ – Vt
V∞ = 35.2 mL, V0 = 20.1 mL, Vt = 20.5
mL, t = 10 min
Hence, k
2.303
35.2 mL – 20.1 mL
⋅ log10
=
10(min)
35.2 mL – 20.5 mL
2.303
log10 1.027
=
10(min)
= 2.66 × 10–3 min–1
Vt = 20.9 mL,
t = 20 min
=
(i)
(ii)
k
(iii)
=
2.303
15.1 mL
⋅ log10
20(min)
35.2 mL – 20.9 mL
=
2.303
⋅ log10 1.056
20(min)
=
2.303
× 0.02366 = 2.72 × 10–3 min–1
20(min)
Vt = 21.3 mL, t = 30 min
k
k
=
2.303
15.1 mL
⋅ log10
30(min)
(35.2 – 21.3) (mL)
=
2.303
⋅ log 10 1.086
30(min)
=
2.303
× 0.03583
30(min)
= 2.75 × 10–3 min–1
All the k values are almost same, the reaction is of first order and mean
k = 2.71 × 10–3 min–1.
Chapter - 5 Chemical Kinetics
386
*4.
The rate constants for a first order reaction at various temperatures are as follows :
k/s–1 × 10–3
1.0
1.422
2.0
2.782
Temperature/ºC
25
30
35
40
3.829
45
Plot an appropriate graph of the data and calculate the energy of activation.
Given :
k/s–1 × 10–3
1.0
1.422
Temperature/ºC
25
30
To find :
i) Ea = ?
ii) A graph of log10k vs 1/T.
Solution :
1
Rate constant log10k Temp AbsoT
k
lute T
1.0 × 10–3
–30
1.422 × 10–3 –2.8471
2.0 × 10–3 –26990
2.782 × 10–3 –2.55336
3.829 × 10–3 –2.4169
25
30
35
40
45
298
303
308
313
318
2.0
2.782
35
40
3.36 × 10–3
3.30 × 10–3
3.25 × 10–3
3.20 × 10–3
3.14 × 10–3
By Arrhenius equation
log10k = log10A –
log10k =
Slope
∴
=
Ea
2.303RT
– Ea
+ log10 A
2.303RT
– Ea
2.303R
Ea = –slope × 2.303 R
Now slope of the above graph
–2.8471 – (–3.0)
y 2 – y1
=
=
x 2 – x1
(3.3 × 10 –3 – 3.36 × 10 –3 )
0.1529
–152.9
=
=
= –2548
– 0.06 × 10 –3
0.06
Ea = –(–2548) × 2.303 × 8.314
E a = 48.79 kJ mol–1
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3.829
45
Chapter General Principles and
6
Processes of Isolation
of Elements
"An investment in knowledge pays the best interest." -Benjamin Franklin
Introduction:
The metals which are unaffected by moisture, air, carbon dioxide from air, and non-metals occur
in free state. However, such metals are very few like gold, silver, platinum, copper etc. Most
of metals occur in the combined form as carbonates, silicates, phosphates, sulphides, oxides etc.
SYLLABUS
6.1
(a)
(b)
(c)
(d)
(e)
Concentration of ores
Hydraulic washing or gravity separation
Hydraulic classifier
Magnetic separation method
Froth Floatation method
Leaching
Theoretical MCQs
6.2
Extraction of crude metal from the
concentrated ore
Calcination
Roasting
Difference between roasting and calcination
Smeltings
Flux
Slag
Reduction by precipitation or displacement
method i.e. Hydrometallurgy
Electrolytic reduction or Electrometallurgy
Theoretical MCQs
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
6.3 Thermodynamic principles of
metallurgy (Ellingham diagram)
(a) Variation of ΔG with respect to temperature
(b) Ellingham diagram
(c) Importance of carbon and carbon monoxide
as reducing agents using Ellingham
Diagram
(d) Limitations of Ellingham diagram
Theoretical MCQs
Purification of refining of crude metal
Liquation
Polling
Zone-refining
Electrolytic refining
Vapour phase refining
Van-Arkel method for refining Zirconium or
titanium
(g) Chromatographic method
Theoretical MCQs
6.4
(a)
(b)
(c)
(d)
(e)
(f)
6.5
(a)
(b)
(c)
(d)
(e)
(f)
Extraction of Zinc from Zinc blende
Concentration of ore
Roasting
Reduction
Electrolytic refining of zinc
Physical properties
Uses of zinc
6.6
(a)
(b)
(c)
Extraction of Iron from Haemetite
Concentration of ore
Roasting
Smelting
388
Theoretical MCQs
6.7
(a)
(b)
(c)
1)
2)
Extraction of Aluminium from Bauxite
Purification of bauxite
Electrolysis of pure alumina
Refining of Aluminium (Hoope’s process)
Theoretical MCQs
6.8
(a)
(b)
(c)
(d)
Extraction of copper from copper pyrites
Concentration of ore
Smelting
Bessemerisation
Refining
Theoretical MCQs
Hours before exam
Higher Order Thinking Skills
Common terms used in metallurgy
Mineral : A naturally occurring substance obtained by mining which contains the metal in
free or combined state is called mineral.
Ore : A mineral containing a high percentage of the metal, from which the metal can be
profitably extracted is called an ore.
Note : Every ore is a mineral but every mineral is not an ore.
Metal
1)
2)
3)
4)
Minerals
Ores
Aluminium (Al)
Bauxite, Al2O3.2H2O Cryolite, Na3AlF6
China clay, Al2O3.2SiO2.2H2O
Bauxite Al2O3.
Iron (Fe)
Haematite, Fe2O3 Magnetite, Fe3O4
Limonite, 2Fe2O3,3H2O Iron pyrite, FeS2
Siderite, FeCO3
Haematite Fe2O3
Zinc (Zn)
Zinc blende, ZnS Zincite, ZnO, Calamine, ZnCO3
Zinc blende, Zns
Magnesium
Magnesite, MgCO3 Dolomite, MgCO3, CaCO3
Epsum salt MgSO4,7H2O
Dolomite MgCO3,
CaCO3
Metallurgy : The process of extraction of a metal in a pure state from its ore is called metallurgy.
Different methods used in extraction are:
a) Pyrometallurgy : A process in which the ore is reduced to the metal at high temperature
using suitable reducing agent like carbon, hydrogen, aluminium etc. is called pyrometallurgy.
b) Hydrometallurgy : A process of extracting metals from aqueous solutions of their salts
using suitable reducing agents is called hydrometallurgy. c) Electrometallurgy : A process
of extraction of metals by electrolytic reduction of molten (fused) metallic compounds is
called electrometallurgy.
Gangue : A sandy, earthy and other unwanted impurities present in the ore are called gangue.
General principles used in metallurgy : The various steps involved in the extraction of pure
metals from their ores are : a) Concentration of ores b) Conversion of ores into oxides or
other desired compounds c) Reduction of ores to form crude metals d) Refining of metals.
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6.1. CONCENTRATION OF ORES :
Concept Explanation :
The process of removal of unwanted materials (gangue) from the ore is known as
concentration of ore or dressing or benefaction of the ore. The selection of method of
concentration depends on the physical properties of gangue, metal to be extracted, available
facilities and the environmental factors. Some of the important procedures are as follows.
(a) Hydraulic washing or gravity separation :
This method is based on differences in
the density of mineral and gangue. The
pulverised (powdered) ore is fed on to
the top of vibrating table (Wilfley’s table)
with a stream of water flowing across
the table. Due to jerks, the less dense
particles are carried away by water and
the heavier particles settle between the
wooden cleats or riffles fixed on the table.
e.g. Cassiterite, SnO2 is concentrated by
this method.
(b) Hydraulic classifier :
This is based on difference in gravities of
the ore and the gangue particles. It consist
of a large conical reservoir fitted with a
hooper at the top for the addition of
powdered ore and an inlet at the bottom.
There is provision to remove light gangue
particles from the side near the top and
concentrated ore from bottom. Water jet is
introduced from the inlet at the bottom in
upward direction to wash the powdered ore.
The lighter gangue particles are washed and
the heavier ore is left behind. This method
is also called levigation.
(c) Magnetic separation method :
This method is used when the impurities are of magnetic in nature. e.g. Cassiterite (an
ore of tin) which contains Wolframite or Ferrous tungstate (FeWO4) which is magnetic
Chapter - 6 General Principles and Processes of Isolation of Elements
390
in nature. The magnetic separator consist of
leather belt moving over two rollers, one of
which encloses magnets in it. The powdered
ore is dropped over this moving belt at one
end.
The impurities which are magnetic in nature
are attracted by the magnet and hence, falls
just below the magnet while the nonmagnetic ore falls away from it.
(d) Froth floatation method :
This method of concentration
of ore is used for the sulphide
ores e.g. Galena (PbS), Zinc
blende (ZnS) etc. It is based
on the principle that metallic
particles are wetted by oil
while gangue particles are
wetted by water. It consist
of big tank containing water
fitted with stirrer. The
tank along with certain oil
like pine oil, eucalyptus oil, fatty acids, xanthates etc. which produces froth when stirred by
passing a current of compressed air. Some amount of aniline or cresol is added which stabilizes
the froth. The ore particles are wetted by froth, being lighter, floats on the surface and skimmed
off while the impurities are wetted by water, becomes heavy and settles at the bottom. It is
possible to separate two sulphide ores by adjusting proportions of oil to water or by
using depressants. e.g. An ore containing ZnS and PbS, NaCN is added which selectively
prevents ZnS coming to froth but allows PbS to come with the froth.
(e) Leaching :
1. Define : It is the process of extracting a soluble material from an insoluble solid by
dissolving out in suitable solvent. Leaching is often used if the ore is soluble in some
suitable solvent.
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2.
For example, in the metallurgy of silver and gold (hydrometallurgy), the powdered ore
is leached with dilute solution of NaCN or KCN in presence of air when the respective
metal form soluble cyanides from which metal is obtained by the displacement reaction
using more electropositive or active metal live Zn.
4M(s) + 8CN – (aq) + 2H 2O(aq) + O 2 (g) ⎯⎯
→ 4[M(CN) 2 ] – (aq) + 4OH – (aq)
[M = Ag or Au]
metalcyanide of Ag or Au
–
2[M(CN) 2] – (aq) + Zn (s) ⎯⎯
→ [Zn (CN) 4 ] 2 (aq) + 2M (s)
3.
Alumina is obtained from bauxite by leaching it with NaOH (Bayer’s process) or with
Na2CO3 (Hall’s process).
(a) Bayer’s Process : Bayer’s process is used when ore contains ferric oxide as main
impurity. The ore is digested at about 423K with the concentrated solution of NaOH
in an autoclave. The Al2O3 in bauxite dissolves in NaOH to form sodium meta aluminate
NaAlO2. The impurities being insoluble remains behind.
Al 2 O 3 ⋅ 2H 2 O (s) + 2NaOH (aq) ⎯⎯
→ 2NaAlO 2(aq) + 3H 2 O (l)
The solution is filtered to remove insoluble impurities and agitated with freshly prepared
Al(OH)3, so that aluminium in NaAlO2 gets precipitated as Al(OH)3.
NaAlO 2 + 2H 2O ⎯⎯
→ NaOH + Al(OH) 3
(b) Hall’s process : In this process, the bauxite ore is fused Na2CO3 is to convert Al2O3
into soluble Sodium Aluminate NaAlO2.
Al 2O 3 ⋅ 2H 2O(s) + 2Na 2CO 3 ⎯⎯
→ 2NaAlO 2 (aq) + 3CO 2 (g) + 2H 2O(l )
The solution is filtered to remove insoluble impurities. The filtered solutions is neutralized
by passing carbon dioxide gas so that Al(OH)3 is precipitated.
2NaAlO 2 + CO 2 + 3H 2 O ⎯⎯
→ 2Al(OH) 3 + Na 2CO 3
The precipitated of Al(OH)3 is washed, dried and heated to obtain Al2O3.
Δ
2Al(OH) 3 ⎯⎯→ Al 2 O 3 + 3H 2O
•
*1.
Ans.
Identify the ores mentioned below which can be concentrated by magnetic separation method?
Fe2O3, FeCO3, ZnO, ZnS, CuFeS2
Fe2O3, FeCO3, CuFeS2
*2.
Ans.
Which process is generally used for the benefication of sulphide ores?
Froth floatation process.
*3.
Ans.
What is the role of depressants in froth floatation process?
Depressants are used to prevent certain types of particles from forming the froth with bubbles
Chapter - 6 General Principles and Processes of Isolation of Elements
392
in froth floatation process which helps in the separation of two sulphide ores. For example,
an ore containing ZnS and PbS, sodium cyanide is used as depressant. It prevents ZnS from
forming froth. Hence, lead sulphide can be separated.
*4.
Ans.
What is the difference between minerals and ores?
(a) Mineral: A naturally occurring substance obtained by mining which contains the metal in
free or combined state is called mineral.
(b) Ore : A mineral containing a high percentage of the metal, from which the metal can
be profitably extracted is called an ore.
Hence, every ore is a mineral but every minerals is not ore.
*5.
Ans.
Which are the different methods used in metallurgy?
Metallurgy is of three types (a) pyrometallurgy (b) Hydrometallurgy and (c) Electrometallurgy.
*6.
Ans.
Describe the magnetic separation process.
Refer 6.1 (c)
*7.
Ans.
Explain the working of froth floatation process.
Refer 6.1 (d)
*8.
Ans.
Explain the terms : Leaching
Refer 6.1 (e) 1 and 2
*9.
Ans.
Which are the various steps involved in the extraction of pure metals from their ores?
The various steps involved in the extraction of pure metals from their ores are broadly classified
into four groups.
(a) Concentration of ores.
(b) Conversion of ores into oxides or other desired compounds.
(c) Reduction of ores to form crude metals.
(d) Refining of metals.
*10.
Ans.
What are the different furnaces used for extraction of metals?
(a) Reverberatory furnace, (b) Blast furnace, (c) Bessemer converter (d) Hearth.
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
The ores that are concentrated by floatation method are .......
a) Carbonates
b) Sulphides
c) Oxides
d) Phosphates
Which of the following metal is obtained by leaching its ore with dilute cyanide solution .......
a) Silver
b) Titanium
d) Zinc
Rocky impurities present in the mineral are called .......
a) Flux
b) Gangue
c) Matte
d) Slag
In order to separate PbS and ZnS in froth floatation process ....... is used as depressants.
a) NaCN
b) KCN
c) AgCN
d) AuCN
*2.
3.
4.
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393
•
5.
6.
Which one contains both iron and copper?
a) Cuprite
b) Chalcolite
c) Copper pyrites d) Malachite
–
In the equation 4M + 8CN + 2H 2 O + O 2 ⎯⎯
→ 4 [M(CN) 2 ] – + 4OH – the metal M is .......
a) Copper
b) Iron
c) Gold
d) Zinc
6.2. EXTRACTION OF CRUDE METAL FROM THE CONCENTRATED
ORE :
Concept Explanation :
The extraction of metal from it’s concentrated ore depends on the nature of ore as well as
impurities present in it. Generally the concentrated ore is converted into a form which is suitable
for reduction i.e., it is converted into it’s oxide which can be easily reduced using suitable reducing
agent. Thus, it is oxidation-reduction which involves following process.
(a) Calcination :
1. Define : It is a process of heating of concentrated ore strongly below it’s melting
point in absence of air or in a limited supply of air.
2. Calcination is generally carried out for the ores containing carbonates and hydrated oxide.
3. During calcination
(a) Organic matter, moisture, volatile impurities like CO2 are expelled from the ore.
(b) Ore becomes porous.
(c) Carbonate ores are decomposed to give metal oxides and CO2.
(d) The hydrated ores lose water of hydration. e.g.
Δ
ZnCO 3 (s) ⎯⎯
→ ZnO (s) + CO 2 (g)
(Calamine ore)
CaCO 3
Δ
⎯⎯
→ CaO + CO 2 (g)
2Fe 2O 3 ⋅ 3H 2O (s)
Limonite ore
Δ
⎯⎯
→ 2Fe 2 O 3 (s) + 3H 2O(g)
(b) Roasting :
1. Roasting is carried out mainly for sulphide ores :
It is a process in which the concentrated ores are heated to high temperature,
below the melting point of the metal, in a reverberatory furnace in excess of air
or oxygen. The impurities like moisture, sulphur, arsenic etc. are expelled out as
their volatile oxide.
Chapter - 6 General Principles and Processes of Isolation of Elements
394
2.
→ SO 2 (g) ↑
S + O 2 ⎯⎯
4 As + 3O 2 ⎯⎯
→ 2As 2O 3 ↑
P4 + 5O 2 ⎯⎯
→ 2P2 O 5 ↑
3. Metal sulphides are converted to metal oxides e.g.
2 ZnS + 3O 2 ⎯⎯
→ 2 ZnO + 2SO 2 (g)
2 PbS + 3O 2 ⎯⎯
→ 2 PbO + 2SO 2 (g)
2Cu 2S + 3O 2 ⎯⎯
→ 2Cu 2 O + 2SO 2 (g)
4. Sometimes, the oxidation of sulphide takes place only to the sulphate stage. For example,
PbS + 2O 2 ⎯⎯
→ PbSO 4
ZnS + 2O 2 ⎯⎯
→ ZnSO 4
(c) Difference between roasting and calcination :
Roasting
1.
2.
3.
4.
Calcination
Ore is heated to a high temperature in the
presence of air.
The process is applied for sulphide ores.
Metal sulphide ores are converted to oxide
or sulphate.
1.
3.
Ore is heated in absence of air or in a
limited supply of air.
This process is applied for Carbonate and
Hydrated oxides.
Metal carbonate are converted to Oxides.
2ZnS + 3O 2 ⎯⎯
→ 2ZnO + SO 2 (g)
4.
ZnCO 3 ⎯⎯→ ZnO(s) + CO 2 (g)
2.
Δ
(d) Smelting :
Smelting is the process of extracting the molten crude metal from it’s concentrated ore
at high temperature along with reducing agent like carbon or coke. Using carbon as reducing
agent has two fold advantages. Carbon itself is reducing agent and on oxidation forms carbon
monoxide which is also a reducing agent. Smelting is generally carried out in a blast
furnace.
For example,
SnO 2 + 2C
Cassiterite
⎯⎯
→ Sn + 2CO ↑
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Δ
⎯⎯
→ Zn + CO↑
823 K
Fe 2 O 3 + 3CO ⎯⎯⎯
→ 2Fe + 3CO 2 ↑
During smelting some amount of flux is added to the concentrated ore which combine
with impurities present in the ore and forms fusible slag.
ZnO + C
Δ
FeO + SiO 2 ⎯⎯→ FeSiO 3
(flux)
(slag)
(e) Flux :
1. Flux is a chemical substance which is added to the concentrated ores during smelting in
order to remove the gangue of impurities by chemical reaction forming a fusible mass
called Slag.
2. Basically the selection of flux material depends on the nature of impurities (gangue) to
be removed. If the gangue is acidic in nature (e.g. SiO2) then basic flux is used.
SiO 2 + FeO
(gangue)
(flux)
3.
Δ
⎯⎯
→ FeSiO 3
(slag)
If the gangue is basic in nature (e.g. CaO), then acidic flux like SiO2 is used.
CaO + SiO 2
Δ
⎯⎯
→ CaSiO 3
(slag)
(f) Slag : It is a waste product formed by the combination of gangue and
flux during extraction of metal by smelting process :
It is lighter than molten metal and insoluble in it hence, can easily be skimmed of.
It prevents the molten metal from oxidation by air as it forms separate layer over the
molten metal. for e.g.
CaO + SiO 2 ⎯⎯
→ CaSiO 3
(slag)
(g) Reduction by precipitation or displacement method i.e.
Hydrometallurgy :
The process of extraction of metals by dissolving it in suitable chemical reagent and
the precipitations of metal by more electropositive metal is called Hydrometallurgy.
Some metals are reduced by displacement using more reactive metal from their complexes.
For example silver and gold are extracted from their complexes by more reactive Zinc
metal. In this process, the concentrated ore is dissolved in suitable solution to form their
soluble complexes. The metal ion is than precipitated by adding Zinc dust.
For example, in the metallurgy of silver and gold (hydrometallurgy), the powdered ore
is leached with dilute solution of NaCN or KCN in presence of air when the respective
metal form soluble cyanides from which metal is obtained by the displacement reaction
using more electropositive or active metal live Zn.
Chapter - 6 General Principles and Processes of Isolation of Elements
396
4M(s) + 8CN – (aq) + 2H 2O(aq) + O 2 (g) ⎯⎯
→ 4[M(CN) 2 ] – (aq) + 4OH – (aq)
[M = Ag or Au]
metalcyanide of Ag or Au
–
2[M(CN) 2] – (aq) + Zn (s) ⎯⎯
→ [Zn (CN) 4 ] 2 (aq) + 2M (s)
(h) Electrolytic reduction OR Electrometallurgy :
The process of extraction of metals by electrolytic process is known as electrometallurgy.
This process is employed for the extraction of highly electropositive metals like sodium,
potasisum, magnesium, calcium and aluminium. The ores of these metals cannot be reduced
by conventionals reducing agents like carbon, carbon monoxide etc. In this method, the
metallic components such as oxide, hydroxide, halides are electrolyzed in their fused state.
The metal ions are discharged at the cathode. For example, Sodium is extracted by the
electrolysis of it’s fused state NaCl.
+ +
NaCl(l ) ⎯⎯
Cl –
→ Na
Molten
–
At Anode : 2Cl – ⎯⎯
→ Cl 2 (g) + 2e
At Cathode : 2Na + + 2e – ⎯⎯
→ 2Na (s)
•
*1.
Ans.
Explain the terms (a) Roasting, (b) Smelting, (c) Calcination
(a) Roasting : Refer 6.2 (b), (b) Smelting : Refer 6.2 (e) (c) Calcination : Refer 6.2 (a)
*2.
Ans.
What is pyrometallurgy?
The process of extraction of metal by heating the metal oxide with suitable reducing agent at
high temperature is known as pyrometallurgy or thermal reduction. Depending on the nature of
oxide and metal, the extraction of metal can be carried out by using following reducing agent.
Reduction with coke and carbon monoxide : In the metallurgy Al, Fe, Pb, Zn, Mg, Co etc.
Reduction with Na, Al, Mg or hydrogen : In the metallurgy of Mn, Cr, As, Mo, W etc.
Reduction with water gas (CO + H2) : In the metallurgy of Ni.
Self reduction or Auto reduction : In the metallurgy of Pb, Hg, Cu etc.
*3.
Ans.
*4.
Ans.
What is the meaning of hydrometallurgy?
Refer 6.2 (h)
Explain the following terms. (a) Gangue, (b) Slag
(a) Gangue : A sandy, earthy and other unwanted impurities present in the ore are called
gangue. e.g. in the extraction of iron, silica is the gangue present in haematite ore.
(b) Slag : Slag is a waste product formed by the combination of gangue and flux during
smelting process.
Explain flux.
Refer 6.2 (e)
5.
Ans.
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*6.
Ans.
What is calcination? Write examples with reactions.
Refer 6.2 (a)
*7.
Ans.
What is smelting? Explain with an example.
Refer 6.2 (d)
*8.
Ans.
Explain Electrometallurgy.
Refer 6.2 (h)
*9.
Ans.
At 673K, which is the better reducing agent carbon or carbon monoxide?
At 673, carbon monoxide is the better reducing agent.
*10.
Using standard potentials write equation for the net reaction that you would predict in
the following experiments.
(a) Zinc metal is added to aqueous sodium triiodide.
= –0.76 Volt is lower than that of I3–
The standard reduction potential of Zn E 0 +2
Ans.
Ans.
(E 0 I 3 – /I – )
Zn
/ Zn
= 0.536 volts. Hence, Zn reduces I3– to I– and the reduction reaction is
Zn + NaI 3 ⎯⎯
→ ZnI 2 + NaI .
(b) Iodine is added to excess aqueous HClO3.
The reduction potential of I2 (E0 I–/I2 = 0.434V) is lower than that of HClO3 (E0ClO3–/ClO–
= 0.50V). Hence, I2 reduces HClO3 to HOCl and itself is oxidized.
2I 2 + HClO 3 + 2H 2O ⎯⎯
→ 4HOI + HOCl
*11.
Ans.
Why is permanganate not a suitable oxidizing agent for quantitative estimation of Fe2+
in presence of HCl?
(a) Fe2+ can be estimated quantitatively by titrating against a standard solution KMnO4 in
acidic medium.
5Fe 2+ + MnO 4+ ⎯⎯
→ 5Fe 3+ + Mn 2+ + 4H 2 O .
(b) For acidic medium H2SO4(aq) is used and not HCl solution.
(c) In case HCl is used, we will have to consider reduction potential of following reduction
reactions :
0
(1) MnO 4– + 8H + + 5e – ⎯⎯
→ Mn 2+ + 4H 2 O (E red = 1.51V)
(2) ½Cl 2 (g) + e – ⎯⎯
→ Cl(aq)E 0 red = 1.40V
0
(3) The reduction potential Fe3+ is Fe 3+ +e – ⎯⎯
→ Fe 2+ E red = 0.77V.
(4) Since the reduction potential of Cl2/Cl– and Fe2+/Fe3+ are less than for MnO4–. Both
HCl and Fe2+ will be oxidised by permanganate solution.
(5) Therefore, Fe2+ cannot be estimated quantitatively by permanganate in the presence
of HCl.
*12.
Ans.
In electrometallurgy of aluminium, why is the graphite rod used?
The process of electrometallurgy is specifically employed for the extraction of highly
Chapter - 6 General Principles and Processes of Isolation of Elements
398
electropositive metals like sodium, potassium, magnesium, calcium and aluminium. The ores
of these metals cannot be reduced by conventional reducing agents such as carbon, carbon
monoxide, hydrogen etc. Since, graphite is good conductor also the liberated oxygen combines
with graphite and forms carbon dioxide.
*13.
Ans.
Suggest a flux for the removal of silica gangue.
Limestone (CaCO3). Limestone on decomposition gives quick lime which combine with silica
to form calcium silicate (CaSiO3)
Δ
CaCO 3 ⎯⎯
→ CaO + CO 2
CaO + SiO 2 ⎯⎯
→ CaSiO 3
(Slag)
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
Electrochemical process (electrolysis of fused state) is used to extract .......
a) Iron
c) Sodium
d) Silver
In metallurgy, flux is a substance used to convert .......
a) Mineral into silicate
b) Fusible impurities to infusible impurities
c) Infusible impurities to soluble impurities
d) Soluble impurities into infusible impurities
Heating of iron pyrites in air to remove sulphur is called .......
a) Fusion
b) Calcination
c) Roasting
d) Smelting
In blast furnace iron oxide is reduced to iron by .......
a) Carbon monoxide b) Lime stone
c) Carbon
d) Zinc
Which one of the following metal is leached by cyanide process?
a) Ag
b) Na
c) Al
d) Cu
*2.
3.
4.
5.
•
6.
7.
The method chiefly used for the extraction of lead and tin from their ores are
(IIT 2004)
respectively .......
a) self reduction and carbon reduction
b) self reduction and electrolytic reduction
c) carbon reduction and self reduction
d) cyanide process and self reduction
In the process of extraction of Gold, .......
Roasted gold ore + CN –
O
2
+ H 2O ⎯⎯→
[X] + OH –
[X] + Zn ⎯⎯
→ [Y] + Au
Identify the complexes [X] and [Y]
a) X = [Au (CN)2]–, Y = [Zn (CN)4]2–
b)
X = [Au (CN)4]–3, Y = [Zn (CN)4]2–
c) X = [Au (CN)2]– , Y = [Zn (CN)4]2–
d)
X = [Au (CN)4]–, Y = [Zn (CN)4]–
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6.3. THERMODYNAMIC PRINCIPLES OF METALLURGY
(ELLINGHAM DIAGRAM)
(a)
1.
2.
3.
Variation of ΔG with respect to temperature for the following :
Formation of metal oxide from metal.
Formation of CO(g) from carbon
Formation of CO2(g) from carbon
The change in Gibbs free energy is given by ΔG = ΔH – TΔS
Where, ΔG = Change in Gibbs free energy,
ΔH = Heat of reaction,
ΔS = Change in entropy.
For a reaction to proceed in the required direction, the ΔG must be negative.
1. Formation of metal oxide from metal :
(a) Consider the formation of metal oxide, MO(s) from metal M(s).
2.
3.
1
M (s) + O 2(g) ⎯⎯
→ MO (s)
2
(b) For this reaction H is negative (since it is combustion) and ΔS is also negative (as
O2 is converted to MO(s)).
(c) Since, ΔG = ΔH – TΔS, at low temperature ΔG may be negative. As temperature
increases, ΔG increases or becomes less negative.
Formation of CO(g) from carbon :
(a) The oxidation of carbon, C(s) to CO, can be explained with the help of thermodynamic
concepts.
1
C (s) + O 2(g) ⎯⎯
→ CO (s)
2
(b) For this reaction H is negative (since it is combustion) and ΔS is positive (moles
of gaseous products> moles of gaseous reactants) .Therefore the term TΔS becomes
positive.
(c) At all temperatures ΔG < 0 and ΔG decreases with the increase in temperature.
Formation of CO2(g) from carbon :
(a) The oxidation of carbon, C(s) to CO2, can be explained with the help of thermodynamic
concepts.
C (s) + O 2(g) ⎯⎯
→ CO (2g)
(b) For this reaction H is negative (since it is combustion) and ΔS is zero (moles
of gaseous products= moles of gaseous reactants) .Therefore the term T ΔS
becomes zero.
(c) Hence, ΔG does not vary with temperature.
Chapter - 6 General Principles and Processes of Isolation of Elements
400
The corresponding ΔG° is
called standard free energy
change of combustion
reaction.
Ellingham diagram can also be
plotted for the formation of
sulphides, halides, etc.
+200
+
4Ag
0.0
g
2H
–1
–200
ΔGº / kJ mol
(b) Ellingham diagram :
1. Ellingham diagram : It is the
plot of Gibbs free energy
change ( Δ G°) with the
temperature for the reaction
of a metal and other elements
with one mole of gaseous
oxygen at one atmosphere
to form corresponding metal
oxides.
O
2Ag 2
gO
2H
O2
+O
2
O2
2Fe +
C + O2
CO2
–400
2 Zn
2 Zn
–600
–800
–1000
2FeO
O
+ O2
4 Al +
3
O2
g+
2M
O2
2 Al 2O 3
3
gO
2M
–1200
0
500
1000
1500
2000
2500
Temperature / K
Ellingham diagram for oxide formation
2.
3.
Features of Ellingham diagram :
(a) The graph of G° against temperature for the formation of metal oxide is a straight
line with positive slope.
(b) In case of some metal oxides like MgO, ZnO and HgO, there is a sudden change
in the slopes indicating change in the physical state.
(c) In case of metal oxides of Hg and Ag, the graphs are at the upper part of Ellingham
diagram indicating less tendency for oxide formation.
(d) For the formation of gaseous CO, the graph is a straight line with a negative slope.
This line intersects the lines of many metal oxides.
(e) For the formation of gaseous CO2, the graph is a straight line almost parallel to
temperature axis, indicating very less effect of temperature on ΔG° of combustion
reaction.
Significance of Ellingham diagram :
(a) Ellingham diagram is obtained by plotting standard free energy change ΔG° for the
formation of oxides of metals versus temperature.
(b) The positive slope of formation of metal oxide lines indicate that the stability of metal
oxides decreases due to increase in ΔG° with the increase in temperature.
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(c) The sudden change in the graphs and slopes indicates a phase change from solid to
liquid or from liquid to vapour.
(d) The negative slope for CO shows that it becomes more stable with the increase in
temperature while almost constant value of ΔG° for CO2 indicates, it is less stable
than CO at higher temperatures. Therefore, at high temperature carbon preferably
forms CO and not CO2.
(e) The variation in the negative values of free energy change ( ΔG°) in decreasing
order is,
MgO > Al2O3 > Cr2O3 > FeO > Ag2O
hence, the tendency to undergo oxidation of the corresponding metals will be
Mg > Al > Cr > Fe > Ag.
(f) The Ellingham diagram enables to select a suitable reducing agent for the reduction
of given metal oxide. An element can reduce the oxide of another element above
it in Ellingham diagram.
Eg : Al can be used as a reducing agent for the reduction of Cr2O3.
Cr2O 3 + 2Al ⎯⎯
→ Al 2O 3 + 2Cr
c) Importance of carbon and carbon monoxide as reducing agents using
Ellingham diagram.
1. Carbon is used extensively as a reducing agent for several metal oxides. For example,
blast furnance
2Fe 2 O 3 + 3C ⎯⎯⎯ ⎯⎯ ⎯⎯
→ 4Fe + 3CO 2
1500 K
ZnO + C ⎯⎯⎯⎯→ Zn + CO
Thus, carbon forms CO2 as well CO in the reduction reactions.
2.
3.
For the formation of CO 2(g) (C (s) + O 2(g) ⎯⎯
→ CO 2(g) ) Δ S is almost zero and the graph
in Ellingham diagram is horizontal showing that G° does not change with temperature.
For the formation of
CO (g) (2C (s) +O 2(g) ⎯⎯
→ 2CO) ΔS
is positive and increases with
temperature while ΔG° values
become more negative with the
increase in temperature and the
graph has negative slope at higher
temperature. These two graphs
(lines) intersect at 700 K.
4.
Therefore, below 700 K, the formation of CO2 is favoured while above 700 K, formation
of CO is favoured.
Chapter - 6 General Principles and Processes of Isolation of Elements
402
5.
CO can also be used as a reducing agent. For example,
MO (S) + CO (g) ⎯⎯
→ M (s) + CO 2(g)
CO gets oxidised to CO2 by oxygen, hence it is a strong reducing agent.
2CO (g) + O 2(g) ⎯⎯
→ 2CO 2(g)
6.
ΔH and ΔS are negative for this reaction. Hence, even at low temperature, CO can
reduce metal oxides and therefore, it is a better reducing agent. For example,
haematite, Fe2O3 is reduced by CO at low temperature.
At low temperature, CO is a better reducing agent than carbon.
(d) Limitations of Ellingham diagram :
1. The diagram simply indicates whether a reaction is possible or not. It does not predict
the kinetics of reduction i.e. the rate or time taken for the reduction to occur.
2. The interpretation of ΔGº is based on the assumption that the reactants and products are
in a state of equilibrium and ΔG is based on the equation ΔGº = –RT log K
However, this is not always true because the reactants or products may be solid.
*1.
Ans.
*2.
Ans.
*3.
What are the features of Ellingham diagram? What is its significance?
Refer 6.3 (b) 2, 3
Using an Ellingham diagram, indicate the lowest temperature at which ZnO can be reduced
to zinc metal by carbon. Write the overall reaction at this temperature.
The graph for the formation of CO in the Ellingham diagram is a straight line with downward
slope (ΔS increases and ΔGº decreases with temperature). Below the temperature of 1000K,
the formation of CO2 is favoured and above 1000 K formation of CO is formed. For example,
1000 K
ZnO + C ⎯⎯⎯⎯→ Zn + CO
The reduction of Zinc oxide by carbon occurs above 1000 K temperature.
Explain how ΔGº varies with temperature in the reaction.
2C (s) + O 2 (g) ⎯⎯
→ 2CO (g)
Ans.
Refer 6.3 (a) 2.
*4.
Ans.
What is the minimum temperature for reduction of MgO by carbon?
1873 K.
5.
Ans.
Which is better reducing agent C or CO?
Refer 6.3 (c).
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403
Multiple Choice Questions :
•
1.
Theoretical MCQs :
According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may
be used to reduce which one of the following oxides at the lowest temperature?
b) Cu2O
c) MgO
d) ZnO
a) Al2O3
•
2.
In view of the sign of Δr ΔGº for the following reaction .......
PbO 2
⎯⎯
→ 2PbO ΔGº < 0
⎯⎯
→ 2SnO ΔGº > O
Which oxidation state is more characteristic for lead and tin?
(AIEEE 2011)
a) For Pb +2, for tin +2
b) For Pb +4, for tin +4
c) For Pb +2, for tin +4
d) For Pb +4, for tin +2
SnO 2
3.
+ Pb
+ Sn
6.4. PURIFICATION OR REFINING OF CRUDE METAL :
Concept Explanation :
The crude metals obtained after smelting or by any other process contains impurities of
unreduced metal oxide, nonmetals, some other metals and gases. Removal of these impurities
from crude metal is known as purification or refining which can be carried by several
methods, depending upon the impurity to be removed. These are :
(a) Liquation :
This method is employed for the
metals having very low melting point
such as tin, Bismuth, lead etc. In this
method the impure metal is placed
on the sloping hearth of
reverberatory furnace and heated in
inert atmosphere of carbon
monoxide. The metal melts and
flows down which is collected at the
bottom of sloping hearth in receiver
leaving behind non-fusible impurities
on the hearth.
Charge
Sloping hearth
Screen
Impurities
Pure
molten
metal
Furnace
Liquation
Chapter - 6 General Principles and Processes of Isolation of Elements
404
(b) Polling :
This method is used for the metals which contain oxide as an impurity e.g. copper and
tin. In this method, molten metal is agitated vigorously with green pole (bamboo) or logs
of wood. The heat of molten metal makes the green logs to liberate hydrocarbon gases
which reduces metal oxide into metal e.g. molten impure copper becomes 99.5% pure
after polling process.
(c) Zone-refining :
This method is based on the principle
that impurities are more soluble in
the melt than in the solid state of
metal. The method is used for
metals which are required in very
high purity e.g. extremely pure
silicon, germanium, boron, gallium
and indium.
The metal to be purified is casted into thin bar. A circular mobile heater is fixed at one
end of impure metal. One zone of bar is melted by a circular mobile heater in the atmosphere
of an inert gas like argon. At the heated zone, metal melts as the heater moves slowly,
the impurities also move into the adjacent molten part. Thus, impurities are made to move
in to one end which is finally cut off and discarded. Mean time molten metal solidifies
as it is away from the heater. Thus, we get extremely pure metal.
(d) Electrolytic refining :
This is most common method of refining and is based on the principles of electrolysis.
The impure metal is made an anode and cathode is strip of pure metal which are placed
in electrolytic bath containing soluble salt of same metal on passing current, anode
undergoes oxidation and passes in solution.
An equivalent of metal cation from
solution get deposited on cathode
and grows in size.
The impurities fall down below the
anode as anode mud. The reactions
taking place at anode and cathode
are as follows.
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405
At anode :
M ⎯⎯
→ M n+
(impure)
–
+ ne
At cathode : M n+
+ ne – ⎯⎯
→M
Metals like Ag, Cu, Ni, Al and Zn are
refined by this method.
(e) Vapour phase refining :
This method is based on the fact that certain metals are converted to their volatile compound
which on heating decomposes to give metal while impurities are unaffected during compound
formation. The metal to be refined by this method should :
1. Form a volatile compound with available reagent and
2. The volatile compound formed should be easily decomposable so that metal can easily
be recovered.
e.g. Nickel is refined by this technique and is known as Mond process. Nickel is heated
in stream of carbon monoxide to form nickel tetra carbonyl complex which on the thermal
decomposition gives Nickel.
330-350 K
Ni
+ 4CO
impure
⎯⎯⎯⎯→
Ni (CO) 4
Nickel tetra
carbonyl
450-470K
+ 4CO
⎯⎯⎯⎯→ Ni
pure
(f) Van-Arkel method for refining Zirconium or titanium :
This method is similar to Mond process and used to obtain ultra pure Zirconium and Titanium
metal. In this method, the metal is converted to unstable volatile compound (e.g. Iodide)
taking care that impurities are not affected during compound formation. The compound
is then decomposed to get pure metal.
Ti(s)
+
Impure
2I 2 (g)
Zr
+ 2I 2
impure
523 K
⎯⎯⎯→ TiI 4 (g)
870 K
⎯⎯⎯→ ZrI 4
vapours
1700 K
⎯⎯⎯⎯
→ Ti(s) + 2I 2 (g)
pure
2075 K
+ 2I 2
⎯⎯⎯⎯
→ Zr
pure
(g) Chromatographic method :
The technique of chromatography is based on the principle that different component of
Chapter - 6 General Principles and Processes of Isolation of Elements
406
The mixture is put in a liquid or gaseous medium (called moving phase). Different component
of mixture are adsorbed at different levels of porous column (adsorbent called stationary
phase). The adsorbed components are removed (eluted) by using suitable solvents (eluant).
The mobile phase and stationary phase are chosen such that component of the sample
have different solubilities in two phases. A component which is quite soluble in stationary
phase takes longer time to travel through it than a component which is not very soluble
in stationary phase. Thus, component of sample are separated from each other as they move
through stationary phase. Depending upon the physical states of two phases and also on
the process of passage of moving medium, chromatographic technique is given different
names such as column chromatography, paper chromatography, thin layer chromatography,
gas chromatography etc.
•
*1.
Ans.
What is polling?
Refer 6.4 (c)
*2.
Ans.
How is zone refining process used to obtain ultra pure metals?
Refer 6.4 (c)
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407
•
*1.
*2.
3.
4.
•
5.
Multiple Choice Questions :
Theoretical MCQs :
Van Arkel method of purification of metals involves converting the metal to a .......
a) Volatile compound
`
b) Volatile unstable compound
c) Non-volatile stable compound
d) Non-volatile unstable compound
Zone refining is a method to obtain .......
a) Very high temperature
b) Ultra pure Al
c) Ultra pure Germanium
d) Ultra pure oxides
Zone refining has been employed for preparing ultrapure sample of .......
a) Cu
b) Na
c) Ge
d) Zn
Titanium can be obtained in a state of high purity by .......
a) Van Arkel method b) Poling
c) Cupellation
d) Electrorefining
Which of the following pairs of metal is purified by Van Arkel method?
(IIT 2011)
a) Ga and In
b) Zr and Ti
c) Ag and Au
d) Ni and Fe
6.5. EXTRACTION OF ZINC FROM ZINC BLENDE :
Concept Explanation :
Zinc is reactive metal hence, does not occur in free (native) form. The ores of Zinc ore :
(1) Zinc blende ZnS
(2) Calamine ZnCO3
(3) Zincite ZnO
(4) Franklinite
ZnO.Fe2O3
(5) Willemite Zn2SiO4
The chief ore of zinc is zinc blende which occurs in Zawar mines located near Udaypur
(Rajasthan).
(a) Concentration of ore :
The crushed ore is concentrated by :
1. Gravity separation : The powdered ore is washed with powerful stream of water.
The lighter gangue impurity particles are washed away and the heavier ore particles
remain behind.
2. Electromagnetic separation : The impurities of iron oxide from zinc blende are separated
by magnetic separation. This is possbile because iron oxide is magnetic in nature.
3. Froth Floatation process : Froth Floatation process is the important step in concentrating
zinc blende. In this process, finely powered ore of zinc blende is treated with water
containing pine oil in a tank. This oily mixture is stirred by passing compressed air into
it by stirrer. This process leads to froth formation. Sulphide particles wetted by pine
oil rise to the surface of tank in the form of foam. Earthy impurities wetted by water
settle down at the bottom which are removed. In this way sulphide particles are separated
from other impurities.
Chapter - 6 General Principles and Processes of Isolation of Elements
408
(b) Roasting :
The concentrated ore is roasted in excess of air at about 1200 K to convert zinc sulphide
into zinc oxide.
During roasting sulphur present in zinc blende is completely expelled out hence, the roasting
is known as dead roasting or sweat roasting.
2ZnS + 3O 2 ⎯⎯
→ 2ZnO + 2SO 2(g) ↑
Sometimes ZnS gets converted to ZnSO4.
ZnS + 2O 2 ⎯⎯
→ ZnSO 4
At 1200 K ZnSO4 decomposes to give ZnO.
2ZnSO 4 ⎯⎯
→ 2ZnO + 2SO 2 + O 2
(c) Reduction :
Zinc oxide is reduced to zinc by heating with crushed coke at 1673 K in vertical fire
clay retort.
Δ
ZnO + C ⎯⎯⎯
→ Zn + CO (g)
1673K
The vapours of zinc formed are
collected in condenser to form
molten zinc which on cooling
solidifies into zinc spelter. The
waste is continuously removed
from bottom by automatic
mechanism and fresh charge is
added from top of the retort.
This process gives 97% to 98%
Zinc contains impurities of iron,
refined by electrolytic method.
(d) Electrolytic refining of zinc :
To remove impurities the spelter is dissolved in dilute sulphuric acid. It forms solution of
ZnSO4. The solution of ZnSO4 should contain about 3% free H2SO4. The solution is filtered
and electrolyzed using aluminum cathode and lead anode at a voltage of 3.5V. Pure zinc
is deposited on the aluminium cathode.
(e) Physical properties :
1. It is silvery white metal in pure form. However, becomes grey coloured when exposed
to moist atmosphere.
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409
2.
3.
4.
5.
It’s atomic radius is 125 pm and ionic radius (Zn2+) is 74 pm.
The specific gravity of zinc metal is 7.14 g cm–3.
It melts at 692 K and it’s boiling point is 1193 K.
It is good conductor of heat and electricity, malleable and ductile in the temperature range
of 373 K to 423 K.
(f)
1.
2.
3.
Uses of Zinc :
It is used for galvanizing iron.
Zinc is used in the extraction of Gold and silver in cyanide process.
Zinc dust is used as a reducing agent in the laboratory. It is also used in the manufacture
of drugs, dyestuff, paints and other chemicals.
4. It is used as an electrode in dry cells.
5. It is also used in the manufacturing of alloys such as brass, german silver etc.
•
*1.
Ans.
How is zinc extracted from zinc blende?
Refer 6.5 (a, b, c and d)
Multiple Choice Questions :
•
1.
In electro-refining, the impure metal is made .......
a) anode
b) cathode
c) anode or cathode
Extraction of zinc from zinc blende is achieved by .......
a) Electrolytic reduction
b) Roasting followed by reduction with carbon
c) Roasting followed by reduction with other metal
d) Roasting followed by self reduction
2.
d) electrolyte
(IIT 2007)
6.6. EXTRACTION OF IRON FROM HAEMATITE :
Concept Explanation :
The common ores of iron are :
1. Haematite
Fe2O3 (red oxide of iron)
2. Limonite
2Fe2O3.3H2O (hydrated oxide of iron)
3. Magnetite
Fe3O4 (magnetic oxide of iron)
4. Siderite
FeCO3
5. Iron pyrite FeS2
Chapter - 6 General Principles and Processes of Isolation of Elements
410
Extraction of iron :
(a) Concentration of ore :
The ore is crushed in jaw crusher and is concentrated by gravity separation process
(discussed earlier) in which it is washed with water to remove sand, clay etc.
(b) Roasting :
The concentrated ore is then calcined. (heated in limited supply of air in reverberatory)
Impurities like sulphur, phosphorus and arsenic are converted to their oxide and are removed
as they are volatile.
S + O 2 ⎯⎯
→ SO 2 ↑
4As + 3O 2 ⎯⎯
→ 2As 2 O 3 ↑
4FeO + O 2 ⎯⎯
→ 2Fe 2O 3
The roasted ore is converted into small lumps.
(c) Smelting :
The roasted ore is reduced with coke in the blast furnace. A blast furnace is tall cylindrical
furnace having diameter of 5 to 10 meter and height of 25 meter. It is made of steel lined
with fire bricks. It is narrow at the top and has cup and cone arrangement for the introduction
of charge. At the base of furnace it is provided with :
1. tuyers arrangement for the introduction of hot air
2. taping hole for withdrawal of molten iron and
3. an outlet from which slag can be flown out.
The ore (charge) is mixed with coke and lime stone in the ratio 15 : 5 : 3 and introduced
through cup and cone arrangement and at the same time blast of hot air is blown upward
with the help of tuyers.
The burning of coke to carbon monoxide supplies most of the heat required for the working
temperature of the furnace and give temperature up to 2200 K at the bottom of furnace.
As the gases move up, they meet the descending charge and the temperature falls gradually.
At the bottom of furnace, the reducing agent is carbon itself, but at the top of the furnace,
reducing agent is carbon monoxide. Following reaction takes place in the furnace.
(i) Zone of combustion OR combustion zone : (5-10 m height from the bottom). The
hot air blown through the tuyers reacts with coke to form carbon monoxide.
1
C + O 2 ⎯⎯
→ CO ΔH = –220 kJ
2
The reaction is highly exothermic. As a result temperature in the zone of combustion
is around 2000 K. A part of carbon monoxide dissociates to give finely divided carbon.
2CO ⎯⎯
→ O 2 + 2C
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In the blast furance the hot gas rich in CO rises up the furnace and heats the charge
coming down and also reacts with charge. This carbon monoxide acts both as a fuel
and a reducing agent.
(ii) Zone of reduction : (22-25 m height near the top). Carbon monoxide reduces ferric
oxide to spongy iron (porous solid) at about 900 K.
Fe 2O 3 + 3CO ⎯⎯
→ 2Fe + 3CO 2
A small amount of ferric oxide is also reduced to iron by carbon.
Fe 2O 3 + 3C ⎯⎯
→ 2Fe + 3CO
(iii) Zone of slag formation :
(20 m in height). The main
constituents of gangue are silica,
alumina and phosphates. The
removal of gangue is effected
as flux.
Limestone decomposes to give
calcium oxide (quick lime) at about
1200 K.
CaCO 3 ⎯⎯
→ CaO + CO 2
Calcium oxide combines with silica
and alumina at about 1500 K to
form molten slag of calcium silicate
and calcium aluminate.
CaO + SiO 2 ⎯⎯
→ CaSiO 3
12CaO + 2Al 2 O 3 ⎯⎯
→ 4Ca 3AlO 3 + 3O 2
(iv) Zone of fusion : (15 m height) MnO2 and Ca3(PO4)2 present in the iron ore are
reduced to Mn and P. A part of SiO2 is also reduced to Si. The spongy iron coming
down from the top of the furnace melts and absorbs the impurities like carbon, silicon,
manganese, phosphorus and sulphur. The molten iron collects at the bottom of the
furnace. The lighter slag floats on the surface of molten iron. Molten slag and iron
are removed through separate outlets. Molten iron is cooled in moulds. The solids
blocks of iron are referred as pigs, hence, the name pig iron. It is also called cast
iron. It contains upto 4% carbon. The hot waste gases consisting of mostly N2,
CO, CO2 escapes through an outlet at the top of the furnace. They are used for
Chapter - 6 General Principles and Processes of Isolation of Elements
412
Hot gases
preheating the blast of air. The various changes taking place in the blast
as follows :
Temp K
Change taking place
Equation
500 K
Ore loses moisture
900 K
Reduction of ore by CO
Fe 2 O 3 + 3CO ⎯⎯
→ 2Fe + 3CO 2
1200 K
Limestone Decomposes
CaCO 3 ⎯⎯
→ CaO + CO 2
1500 K
Reduction of ore by C
Fe 2O 3 + 3C ⎯⎯
→ 2Fe + 3CO
1500 K
Fusion of iron and slag
Formation
CaO + SiO 2 ⎯⎯
→ CaSiO 3
2000 K
Combustion of coke
Fe 2 O 3 + 3CO ⎯⎯
→ 2Fe + 3CO 2
Cast iron is hard and brittle. It cannot be welded or tempered. It is used in the manufacture
of casted material and certain automobile parts.
Commercial varieties of iron :
(i) Cast or pig iron : It contains 2 to 4.5% carbon along with impurities such as sulphur, silicon,
phosphorous, manganese etc. It is brittle and cannot be welded.
(ii) Wrought iron : It is the purest form of iron and contains carbon and other impurities not
more than 0.5%. It is malleable and can easily be welded.
(iii) Steel : It contains 0.2 to 1.5% carbon. The mechanical properties of steel can be altered
by the addition of small amount of element like Mn, Cr and Ni. It’s properties are
intermediate between cast iron and wrought iron.
•
*1.
Ans.
*2.
Ans.
3.
Ans.
*4.
Ans.
How is iron extracted from haematite?
Refer 6.6 (a, b, c)
Write reactions involved at different temperature in the blast furnace.
Refer 6.6 C (iv) Table
Name the principal gangue associated with haematite?
SiO2 (silica) and Alumina (Al2O3)
Which is the effective reducing agent in the extraction of iron from haematite?
Carbon monoxide and carbon (coke).
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
413
*5.
Ans.
Why is carbon monoxide a better oxidising agent than carbon for the reduction of haematite
at lower temperature?
At lower temperature, carbon dioxide is thermodynamically more stable than CO. The Δ G
for the conversion of CO to CO2 is negative at lower temperture. Hence, CO can be used
to reduce Fe2O3 to Fe. It reduces Fe2O3 to Fe and itself gets oxidised to CO2.
Fe 2O 3 + 3CO ⎯⎯
→ 2Fe + CO 2
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
In blast furnace, iron oxide is reduced by .......
a) Silica
b) CO
c) C
d) Lime stone
In extraction of iron limestone is used for .......
a) Formation of slag
b) Reduction of Fe Ore
c) Purification of Fe formed
d) Oxidation of Fe Ore
The reduction of iron in a blast furnace involves all the steps except .......
a) Fusion
b) Reduction
c) Sublimation
d) roasting
*2.
3.
•
*4.
*5.
Fe2O3 is reduced to spongy iron near the top of blast furnace by .......
a) CO
b) CO2
c) C
d)
Highest carbon content iron is .......
a) Stainless steel
b) Wrought iron
c) Cast iron
d)
H2
Mild iron
6.7. Extraction of Aluminium from Bauxite :
Concept Explanation :
The minerals of aluminium are :
(ii) Cryolite : Na3AlF6
(i) Bauxite
: Al2O3.2H2O
(iii) Feldspar : KAlSi3O8
(iv) Mica
: KAlSi2O10(OH)2
Aluminium is normally extracted from bauxite which involves following steps :
1. Purification of bauxite and
2. Electrolysis of alumina.
(a) Purification of bauxite :
The bauxite ore can be purified by any one the following method, depending on the nature
of impurity present in it.
(i) Bayer’s process : This method is widely used for the purification of red bauxite which
contains iron as an impurity.
(ii) Hall’s process : This process is also used for red bauxite.
Chapter - 6 General Principles and Processes of Isolation of Elements
414
(iii) Serpeck’s process : This method of purification of bauxite is adopted for white bauxite
which contains silica as chief impurity.
1. Purification of bauxite by Hall’s process : In this method, the bauxite ore is fused
with sodium carbonate and lime. The fused mass is dissolved in water and filtered to remove
impurities. The clear solution of meta-Aluminate is then heated to 323 – 373 K and stream
of carbon dioxide is passed through solution when aluminium hydroxide precipitates out
which is filtered, washed, dried and ignited at 1473 K to get pure Al2O3.
Δ
Al 2O 3 .2H 2O + Na 2 CO 3 ⎯⎯
→ 2NaAlO 2
+ CO 2
+ 2H 2 O
2NaAlO 2 + CO 2 + 3H 2O ⎯⎯
→ 2Al(OH) 3 + Na 2 CO 3
1473 K
2Al(OH) 3 ⎯⎯⎯⎯
→ Al 2 O 3 + 3H 2O
(b) Electrolysis of pure alumina :
The electrolysis is carried out
in an iron tank having lining
of carbon which acts as
cathode. The anode consist of
number of carbon rods which
dip in the fused electrolyte.
The purified alumina is
dissolved in molten cryolite
(Na3AlF6 80-85%). Small
amount of flurospar
(5 – 6%) is added. The addition of cryolite and flurospar lowers the melting point of pure
alumina to 1143 K and is good conductor. The electrolyte is covered with a layer of carbon
powder (coke)
During electrolysis, following reactions occur :
3NaF + AlF3
Na 3 AlF6 Al +3 + 3F –
AlF3 At cathode : Al +3 + 3e – ⎯⎯
→ Al
At anode : F – ⎯⎯
→ F + le –
2Al 2O 3 + 6F2 ⎯⎯
→ 4AlF3 + 3O 2
2C + O 2 ⎯⎯
→ 2CO
2C + O 2 ⎯⎯
→ 2CO 2
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
415
The O2 liberated at anode reacts with carbon anode and forms CO & CO2. Thus, anode
is being eaten away (used up) and needs replacement from time to time.
Aluminium liberated at cathode is removed periodically from tapping hole provided at the
bottom of cell. This electrolytic method of reduction of aluminium from bauxite is known
as Hall-Herault’s process.
(c) Refining of Aluminium (Hoope’s electrolytic process) :
The aluminium obtained by Hall’s process is 99% pure. It is further refined by Hoope’s
process. The process is carried out in an iron tank lined inside with carbon. It has three
layers of molten liquids having different densities.
The top layer consist of pure aluminium with carbon electrodes dipping in it and acts as
cathode. The middle layer is mixture of cryolite and barium fluoride in molten state which
acts as an electrolyte.
The bottom layer is impure aluminium which along with carbon lining acts as anode.
On passing electric current aluminium ions from middle layer are discharged at the cathode
as pure aluminium. An equivalent amount of aluminium from bottom layer passes into middle
layer leaving behind the impurities. Thus, this method gives pure aluminium.
•
*1.
Ans.
*2.
Ans.
*3.
Ans.
4.
Ans.
What are the steps involved in the extraction of aluminium from bauxite?
Refer 6.7 (a. b and c)
Explain the refining of aluminium.
Refer 6.7 (c)
What is the role of CaF2 in metallurgy of aluminium?
The addition of CaF2 is to lower the melting point of pure alumina and is good conductor.
Mention names and formulas of two ores of aluminium.
The minerals of aluminium are :
(a) Bauxite
: Al2O3.2H2O
(b) Cryolite : Na3AlF6
(c) Feldspar : KAlSi3O8
(d) Mica
: KAlSi2O10(OH)2
(any two)
Multiple Choice Questions :
•
1.
Theoretical MCQs :
In alumino-thermic process, aluminium is used as .......
a) Reducing agent
b) Oxidising agent c) Catalyst
d) Electrolyte
A common method used for the extraction of metals from their oxides by reduction is .......
a) Al
b) Fe
c) Cr
d) Co
2.
Chapter - 6 General Principles and Processes of Isolation of Elements
416
6.8. EXTRACTION OF COPPER FROM COPPER PYRITES :
Concept Explanation :
1.
(i)
(iii)
(v)
The chief ores of copper are :
Copper glance Cu2S
Malachite
Cu(OH)2.CuCO3
Azurite
2CuCO3.Cu(OH)2
(ii) Copper pyrite
CuFeS2
(iv) Cuprite (Ruby copper) Cu2O
2. Extraction : Copper is mainly extracted from copper pyrite which contains iron sulphide,
gangue and smaller quantities of arsenic, selenium, tellurium, silver, gold and platinum :
(a) Concentration of ore :
The finely powdered ore is concentrated by froth floatation process. The concentrated
ore is then subjected to roasting in reverberatory furnace in limited supply of air. In
reverberatory furnace, copper pyrite, CuFeS2 get converted to Cu2S and FeS, while impurities
like arsenic and antimony are removed as their volatile oxide.
2 CuFeS 2
+ O 2 ⎯⎯
→ Cu 2S + 2FeS + SO 2
(b) Smelting :
The roasted mass is mixed with some powdered coke and sand and is heated strongly
in blast furnace. FeS is preferentially oxidised as iron is more reactive than copper. FeO
combines with silica (flux) to form fusible slag.
2FeS + 3O 2 ⎯⎯
→ 2FeO + 2SO 2
FeO + SiO 2 ⎯⎯
→ FeSiO 3
(slag)
The lower layer of molten mass contains cuprous sulphide and some traces of iron sulphide,
known as copper matte which is removed from bottom tapping hole.
(c) Bessemerisation :
The molten copper matte is then transferred to Bessemer converter and a blast of hot
air mixed with sand is blown. During this process Cu2S is partially oxidised to Cu2O which
further reacts with remaining Cu2S to form copper and sulphur dioxide (auto reduction).
2Cu 2S + 3O 2 ⎯⎯
→ 2Cu 2 O + 2SO 2 ↑
Cu 2S + 2Cu 2O ⎯⎯
→ 6Cu + SO 2 ↑
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
417
Traces of FeS present in the matte is oxidised to FeO which combines with silica to form slag.
2 FeS + 3O 2 ⎯⎯
→ 2 FeO + 2SO 2 ↑
FeO + SiO 2 ⎯⎯
→ FeSiO 3 (slag)
After the reaction is completed,
the converter is tilted and the
molten mass is poured in to
sand moulds. As the metal
solidifies, the dissolved SO2
escapes producing blisters on
the metal surface hence, known
as blister copper.
pure and gets deposited at the cathode.
CO2 gas
SO2 gas
Acidic/basic lining
Hot air
Slag
Bessemer converter
(d) Refining :
The blister copper thus, obtained is 99% pure and contains impurities of silver and
gold. It is further refined electrolytically to get 99.99% pure copper. In electrolytic
refining, the electrolyte is acidified CuSO4. The anode is impure copper while the
cathode is pure copper rod. On passing current Cu2+ ions from solution get deposited
at cathode while an equivalent amount of anode dissolves. The noble metals do not
dissolve in dilute acid present in the electrolyte and settles down as anode mud while
other impurities pass into solution.
•
*1.
Ans.
Describe the process of extraction of copper.
Refer 6.8 (b)
Multiple Choice Questions :
•
*1.
Theoretical MCQs :
During smelting silica is added to roasted copper ore to remove .......
a) Cuprous Sulphide
b) Ferrous sulphidex c) Ferrous oxide d)
The ore which contains copper and iron both .......
a) Malachite
b) Chalcopyrite
c) Chalcocite
d)
*2.
Ferrous sulphide
Azurite
Chapter - 6 General Principles and Processes of Isolation of Elements
418
HOURS BEFORE EXAM
Metallurgy : Is an art and science of extraction of metal of their ore commercially and
economically.
Ores : Naturally occurring substances (minerals) from which one or more metals can be extracted
economically.
Minerals : Naturally occurring substances which are obtained from mine and contains one or
more metals in it.
Metallurgy
Pyrometallurgy
Hydrometallurgy
The metal is obtained by the
reduction of their concentrated
ore by subjecting them to high
temperature using reducing
agent. like coke eg. Iron, Zinc,
Copper etc.
The metal is obtained by dissolving
concentrated ore in suitable
solvent and then by displacement
method eg. Silver, Gold etc.
Electrometallury
The concentrated ore is
subjected to electrolysis
using suitable electrolyte.
e.g. Aluminium.
Principle underlying metallurgy are :
1) Concentration of ore : Removal of unwanted impurities (gangue) from the ore is called
concentration. An ore can be concentrated by different methods depending upon the nature of
impurity to be removed.
Hand-picking
When the impurities
are in the form of
lumps.
Hydraulic washing or
gravity separation
When ore particles are highter
(having low density than
gangue) than impurity.
e.g. Cassiterite, SnO2
Leaching
Magnetic Separation method
When ore is non-magnetic and
impurities are magnetic.
e.g. wolframite (FeWO4) is
removed from casseterite ore.
When ore dissolves in some
suitable solvent but impurities
remain insoluble. e.g. Silver,
Gold ore is concentrated by
this method. Alumina from
Bauxite.
Unique Solutions ®
Hydraulic classifier
When impurities are
lighter than gangue
particle.
Froth floatation
Best suited for sulphide ore.
Even two sulphide ore can be
selectively concentrated using
depressant. e.g. PbS can be
selectively removed from ZnS
using NaCN as depressent.
S.Y.J.C. Science - Chemistry - Part I
419
2) Extraction of crude metal from concentrated ore :
Calcination
♦
♦
♦
♦
♦
♦
Carried out at about 200 – 300ºC.
Carried out in insufficient or absence of air.
To remove organic volatile impurities.
Moisture, hydrated water is removed.
Carbonates decomposes to form oxide.
To make concentrated ore porous.
Roasting
♦ Carried out at high temperature (below the m.p.
♦
♦
♦
♦
of metal)
Carried out in excess of air or O2.
Carried out in a reverberatory furnace.
Sulphide is expelled out as SO2.
Metal sulphide is converted into metal oxide.
3) Reduction of metal oxide into crude metal :
(a) Using reducing agent i.e. pyrometallurgy and process is called smelting. The concentrated ore is
mixed with coke and heated in blast furnace. The metal oxide is reduced to metal.
M x O y + YC ⎯⎯
→ xM + yCO 2
ZnO + C ⎯⎯
→ Zn + CO
eg. ZnO + CO ⎯⎯
→ Zn + CO 2
Δ
Fe 2 O 3 + 3C ⎯⎯
→ 2Fe + 3CO
823K
Fe 2 O 3 + 3CO ⎯⎯⎯→ 2Fe + 3CO 2
1123K
FeO + CO ⎯⎯⎯→ Fe + CO 2
To remove impurities (gangue) flux is added which form slag with it.
SiO 2 + FeO ⎯⎯
→ FeSiO 3
gangue flux
slag
CaO + SiO 2 ⎯⎯
→ CaSiO 3
(b) Reduction by precipitation OR displacement method i.e. Hydrometallurgy. The metal is obtained
from their complexes by adding to it more reactive metal. eg. Silver sulphide is dissolved in
NaCN solution to form diayanoargenta, a complex ion. From this complex, Silver is obtained by
adding Zn dust in sol. when Silver get precipitated.
Aq 2S + 4NaCN ⎯⎯
→ 2Na[Ag(CN) 2 ] + Na 2S
2Na[Ag (CN) 2 ] + Zn ⎯⎯
→ 2Ag + Na 2 [Zn(CN) 4 ]
(c) Electrolytic reduction or electrometallurgy : The metals like alkali and alkaline earth metals
and aluminium which cannot be reduced by using suitable reducing agent this method is
employed e.g. Electrolysis of fused NaCl.
NaCl
Na + + Cl –
At anode 2Cl – ⎯⎯
→ Cl 2 + 2 – e
At cathode 2Na + + 2e ⎯⎯
→ 2Na
Chapter - 6 General Principles and Processes of Isolation of Elements
420
4) Purification or refining of crude metal :
Removal of impurity from crude metal to get 99.99% pure metal is known as refining or
purification. The different methods of purification are :
(b) Liquation
(a) Distillation
It is used for low boiling
metals. The impure
metal is evaporated to
obtain pure metal as
distillate e.g. Zn and
Hg.
(d) Zone-refining
(c) Polling
This method is used to
remove impurity of
oxide. The molten
metal is agitated
vigorously with green
wooden logs or pole
when the hydrocarbon
gases liberated from
green log reduces
metal oxide to metal
e.g. copper.
Van-Arkelmethod
♦ Similar to mond
process.
♦ Zirconium and by
this method.
♦ They are converted
to their volatile
unstable iodide.
This method gives ultra
pure metal. e.g. Silicon.
The principle is,
impurities are more
soluble in the molten
liquid than in the solid.
Chromatographic
method
♦ Based
on the
principle that different component of
mixture are differently adsorbed on an
♦ Different component
of mixture are
levels of porous
column
♦ The adsorbed comp♦ On heating to high
onents are removed
♦ Nickel is refined by
temperature de(eluted) by using
this process (Mond
compose to give
suitable solvents
pure metal.
process).
The impure metal is heated
on the sloping hearth of
rever beratory furnace in
an inert atmosphere of CO.
The metal melts and gets
Electrolytic refining pure form e.g. Tin, Bismuth,
♦ Based on the prin- lead etc.
ciple of electrotro- Vapour phase refining
♦ Certain metals are
lysis.
converted into their
♦ Most common
volatile compound.
method.
♦ On heating they
♦ Metals like Ag, Cu,
decompose to give
Ni Zn are refined
pure metal while
by this method.
impurities
are
unaffected.
Extraction of metals
(a) Zinc
Ores :
1. Zinc blende, ZnS (chief ore)
4. Franklinite, ZnO.Fe2O3
2. Calamine, ZnCO3
5. Willemite, Zn2SiO4
3. Zincite, ZnO
Concentration of ORE (ZnS)
Froth floatation method
Roasting : 2ZnS + 3O 2 ⎯⎯
→ 2ZnO + 2SO 2
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
421
1673
Reduction : ZnO + C ⎯⎯⎯
→ Zn + CO
Purification : Electrolytic refining
1.
2.
3.
4.
5.
6.
Physical properties :
Silvery white metal, turns grey when exposed to most atmosphere.
Specific gravity is 7.14 g cm3.
Melting point : 692K, Boiling point : 1193 K.
Good conductor of heat and electricity.
Malleable and ductile in the temp. range 373 – 423K.
1.
2.
3.
4.
5.
6.
Uses :
In galvanising of iron.
In the extraction of gold and silver (cyanide process).
As reducing agent in laboratory (Zn-dust).
In the manufacture of drugs, dyestaft point and other chemicals.
As an electrode in dry cell.
As an alloy e.g. brass, german silver.
(b) Iron :
1. Ores
(i) Haemetite (chief ore), Fe2O3
(ii) Limonite, 2Fe2O3.3H2O
(iii) Magnetite, Fe3O4
(iv) Siderite, FeCO3
(v) Ironpyrite, FeS2
2. Concentration of ore : By gravity separation.
3. Calcination : Carried out in reverberatory furnace in limited supply of air. During the process
(i) Moisture is removed.
(ii) Sulphur, phosphorous, arsenic are converted to their volatile oxide.
(iii) Some FeCO3 present get converted to oxide.
S + O 2 ⎯⎯
→ SO 2 ↑
4As + 3O 2 ⎯⎯
→ 2As 2 O 3 ↑
P4 + 5O 2 ⎯⎯
→ P4O10 ↑
FeCO 3 ⎯⎯
→ FeO + CO 2 ↑
4FeO + O 2 ⎯⎯
→ 2Fe 2O 3
4. Smelting :
(i) Carried out in blast furnace.
(ii) Calcined ore is mixed with coke.
(iii) Reduction of Fe2O3 takes place in steps.
(a) Zone of combustion (2200K) – bottom of furnace.
C + O 2 ⎯⎯
→ CO 2
ΔH = –393.5 kJ
Chapter - 6 General Principles and Processes of Isolation of Elements
422
(b) Zone of fusion (1500K)
ΔH = +163.2 kJ
(c) Slag formation Zone (1270K)
CO 2 + C ⎯⎯
→ CO
CaCO 3 ⎯⎯
→ CaO + CO 2
CaO + SiO 2 ⎯⎯
→ CaSiO 3 (slag)
(d) Reduction Zone (875 k)
FeO + CO ⎯⎯
→ Fe + CO 2
3Fe 2O 3 + CO ⎯⎯
→ 2Fe 3O4 + CO 2
Fe 3O 4 + CO ⎯⎯
→ 3FeO + CO 2
Fe 2O 3 + 3CO ⎯⎯
→ 2Fe + 3CO 2
at the lower, hotter part the main reaction is
FeO + C ⎯⎯
→ Fe + CO
The metal obtained is called cast iron or pig iron.
Commercial varieties of iron :
(1) Cast iron : contains 2 – 4% Carbon, cannot be Welded
(2) Wrought iron : Purest form of iron and contains impurities of C, not more than 0.5% can
be welded.
(3) Steel : Contains 0.2 – 1.5% Carbon. The properties of steel is altered by adding small amount
of metal like Mn, Ni and Cr. It’s property is intermediate between cast iron and
wrought iron.
Metallurgy of Aluminium :
1.
2.
Ores : (i) Bauxite, Al2O3.2H2O (chief ore)
(ii) Cryolite, Na3AlF6
(iii) Feldspar, KAlSi3O8
(iv) Mica, K; AlS2O10 (OH)2.
Concentration : Bauxite ore is fesed with sodium carbonate and lime.
Fused mass of
bauxite + water ⎯⎯
→ dissolve ⎯⎯
→ Filtered ⎯⎯
→
+CO 2 (g)
filtered
→ Al(OH) 3 ↓
filtrate ⎯⎯⎯⎯⎯
Al O
and heated to 1473k 2 3
3.
Electrolysis of pure alumina : It is carried out in big iron tank. The Al2O3 is dissolved
in molten cryolite (Na3AF6) – 80 – 85% and 5 – 6% CaF2 because it is good conductor
and also lower the m.p. of Al2O3. The anode is graphite rod which remain dipping in the
electrolyte. The iron tank is lined with carbon which acts as cathode. During electrolysis following
reaction occur.
3NaF + AlF3
Metallurgy of : Na 3 AlF6 Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
423
Al 3+ + 3F –
AlF3 At anode : 2F – ⎯⎯
→ F2 + 2e –
At cathode : Al 3+ + 3e – ⎯⎯
→ Al
The fluorine liberated at anode, reacts with Al2O3 to form AlF3.
2Al 2O 3 + 6F2 ⎯⎯
→ 4AlF3 + 3O 2
2C + O 2 ⎯⎯
→ 2CO
2CO + O 2 ⎯⎯
→ 2CO 2
Thus, instead of fluorine, O2 is liberated at anode which react with it and forms CO2. Thus
anode is eaten away and need to be replaced from time to time.
4. Purification or refining of aluminium : Aluminium obtained by Hall’s Heraults process is
99% pure. It is further purified by Hoope’s process to get 99.99% Aluminium.
Copper :
1. Ores of Copper :
(i) Copper glane, Cu2S
(ii) Copper pyrite, CuFeS2 (Chief ore)
(iv) Cuprite (Ruby copper), Cu2O
(iii) Malachite Cu(OH)2.CuCO3
(v) Azurite, 2 CuCO3.Cu(OH)2
2. Concentration of ore : The powdered ore is by froth floatation process.
3. Roasting : Roasting is carried out in reverberatory furnace in limited supply of air.
→ Cu 2S + 2FeS + SO 2
2CuFeS 2 + O 2 ⎯⎯
→ 2FeO + 2SO 2
2FeS + 3O 2 ⎯⎯
→ 2Cu 2 O + 2SO 2
2Cu 2S + 3O 2 ⎯⎯
4. Smelting : Roasted mass is mixed with some coke and silica and heated strongly in blast
furnace. During roasting if at all any Cu2O is formed combines with FeS and forms Cu2S
while Fe is removed as slag.
Cu 2 O + FeS ⎯⎯
→ Cu 2S + FeO
FeO + SiO 2 ⎯⎯
→ FeSiO 3
5. Bessemerisation : The copper matte is transferred to Bessemer converted white metallic
copper is obtained due to auto oxidation. The copper obtained is known as blister copper because
SO2 escape through molten copper producing blister on the surface of metal.
Cu 2S + 2Cu 2 O ⎯⎯
→ 6Cu + SO 2
2Cu 2S + 3O 2 ⎯⎯
→ 2Cu 2 O + 2SO 2
6. Purification or refining : Blister copper is 99% pure and contains impurities of silver and
gold. It is further refined electrolytically using CuSO4 as an electrolyte to get 99.99% pure
copper.
Chapter - 6 General Principles and Processes of Isolation of Elements
424
Higher Order Thinking Skills
1.
Ans.
2.
Why can’t aluminium be reduced by carbon?
Aluminium cannot be reduced by carbon, because it is a stronger reducing agent than carbon.
Why the graphite rods in the extraction of aluminium from Al2O3 have to be replaced from
time to time?
During the reaction, oxygen is liberated at anode which reacts with carbon electrode to form
carbon monoxide and carbon dioxide, which results in slow corrosion of carbon electrode. Hence,
have to be replaced from time to time.
Thermite process is quite useful for repairing broken parts of machines. Explain.
In thermite process, oxides of metal like iron is reduced by aluminium which is highly exothermic
and large amount of heat is released during the reaction. As a result of this, metal will be
Ans.
3.
Ans.
heat
→ Al 2 O 3(s) + 2Fe + heat
in molten state. e.g. Fe 2 O 3 + 2Al ⎯⎯⎯
molten
The metal is allowed to fall between the broken parts which fills the gaps and machine is repaired.
Giving examples, differentiate between ‘Roasting’ and ‘Calcination.’
4.
Ans.
Calcination
Roasting
1. Moisture and organic impurities are removed. 1. Volatile impurities are removed as their oxides
e.g. SO2, As2O3 etc.
2. It is used for carbonate and oxide ores.
2. It is used for sulphide ores.
3. It is carried out in absence of air.
3. It is carried out in presence of air i.e. regular
4. It is carried out in the temperature range of
supply of air.
373K – 473 K. e.g.
4. It is carried out below the melting point of
heat
metal.
Fe O ⋅ xH O(s) ⎯⎯⎯
→ Fe O (s) + x.H O (g)
2
3
2
2
3
2
heat
ZnCO 3 (s) ⎯⎯⎯
→ ZnO (s) + CO 2 (g)
heat
CaCO 3 ⋅ MgCO 3 (s) ⎯⎯⎯
→ CaO (s) + MgO (s)
+ 2CO 2 (g)
5.
Ans.
2ZnS + 3O 2 ⎯⎯
→ 2ZnO + 2SO 2 ↑
2Cu 2S + 3O 2 ⎯⎯
→ 2Cu 2 O + 2SO 2 ↑
2PbS + 3O 2 ⎯⎯
→ 2PbO +2SO 2 ↑
Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO? What
are those conditions?
Below the m.p. of silicon (1693 K) ΔfGº curve for the formation of SiO2, lies above the ΔfGº
curve for formation of MgO. Therefore, below 1693 K, Magnesium can reduce SiO2 to Silicon.
However, above 1693 K, Δf Gº curve for MgO lies above Δf Gº curve for SiO2 and
therefore, at temperature above 1693K, Si can reduce MgO to Mg.
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Chapter
7
p – Block
Elements
"You cannot teach a man anything; you can only help him to discover
it in himself." - Galileo
Introduction :
The elements in which the last electron enters p orbitals of the atoms are called p-Block Element.
p – block elements are placed in groups 13 to 18 of the periodic table. Their valence shell
electronic configuration is n2np1–6 (except helium which has 1s2 configuration). The properties
of p-block elements like that of others are greatly influenced by ionization enthalpy, atomic
sizes, electron gain enthalpy and electro negativity. p-block elements include metals, non-metals
and metalloids
SYLLABUS
7.1
GENERAL INTRODUCTION
7.2
GROUP 16 ELEMENTS
(a) Group 15 elements
(a) General introduction
(b) Occurrence
(b) Trends in physical and chemical properties
(c) Electronic configuration
(c) Chemical properties
(d) Trends in physical and chemical properties
(d) Dioxygen
(e) Chemical properties
(e) Classification of oxides
(f) Dinitrogen
(f) Ozone O3
(g) Compounds of Nitrogen - Ammonia
(g) Sulphur
(h) Compounds of Nitrogen - Nitric acid
(h) Compounds of Sulphur
(i) Oxides of Nitrogen
(i) Sulphuric acid
(j) Allotropic forms of phosphorus
(j) Oxoacids of sulphur
(k) Compounds of phosphorus
(l) Oxyacids of phosphorus
Theoretical MCQs
Theoretical MCQs
426
7.3 GROUP 17 ELEMENTS (HALOGEN
7.4 GROUP 18 ELEMENTS (NOBLE GASES)
(a) Introduction and electronic configuration
FAMILY)
(a) General introduction
(b) Trends in physical properties
(b) Trends in physical properties
(c) Uses of noble gases
(c) Chemical reactivity
(d) Anomalous behaviour of fluorine
Theoretical MCQs
(e) Compounds of halogen - chlorine
(f) Hydrogen chloride
(g) Interhalogen compounds
(h) Oxyacids and oxoacids of halogens
Hours Before Exam
Higher Order Thinking Skills (HOTS)
Theoretical MCQs
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7.1
GENERAL INTRODUCTION :
Concept Explanation :
(a) Group 15 elements :
Group 15 elements are representative elements. Metallic character increases down the
group. N, P are non metals. As, Sb, are metalloids. Bi is metal.
(b) Occurrence :
N – (a) Nitrogen occurs in gaseous state as free N2 in the atmoshere to the extent of 78%
by volume and in the combined state as nitrate.
(b) It is present as constituent of protein and albumins.
(c) In the earth’s crust, it occurs as sodium nitrate NaNO3 (chile salt petre) and potassium
nitrate (Indian salt petre).
P – (a) Phosphorus occurs in minerals of the apatite family, Ca9[PO4]6.CaX2 (X=F, Cl or
OH) e.q. Fluorapatite, Ca9[PO4]6.CaF2, which is the main constituent of phosphate
rocks.
(b) Phosphorus is an essential constituent of animal and plant matter.
(c) It is present in bones as well as in living cells.
As, Sb, Bi – Present as oxides and with sulphide minerals.
(c) Electronic configuration :
1. Valence shell configuration is ns2np3, which is configuration of exact half filled state of
np subshell and is externally stable.
Hence, these elements are very stable and less reactive.
Electronic configuration of group 15 elements
Atomic
Element Symbol
No.
Nitrogen
Phosphorus
Arsenic
Antimony
Bismuth
N
P
As
Sb
Bi
7
15
33
51
83
Electronic Configuration
1s22s22p3
1s22s22p63s23p3
1s22s22p63s23p63d104s24p3
1s22s22p63s23p63d104s24p64d105s25p3
1s22s22p63s23p63d104s24p6
4d104f145s25p65d106s26p3
Chapter - 7 p – Block Elements
Brief
representation
of electronic
configuration
[He] 2s22p3
[Ne] 3s23p3
[Ar] 3d104s24p3
[Kr] 4d105s25p3
[Xe] 4f145d106s24p3
428
2.
Five valence electrons are preceded by 2 and 8 electrons in the case of N and P respectively.
In the case of As, Sb, Bi, they are preceded by 18 electrons.
Hence, As, Sb, Bi behave differently.
The np3 configuration being the exact half filled state is stable on the basis of Hund’s
rule of maximum multiplicity. This explains the greater stability and less reactivity of these
elements.
3.
(d) Trends in physical and chemical properties :
Among the members of group 15, there is regular gradation in properties with increase
in atomic number form the characteristics of true non-metal (nitrogen) to a distinct metal
(bismuth). This transition from non-metal to metallic character is evident from the steady
changes of physical properties of these elements. Some of the physical properties are shown
in Table.
No.
Property
N
P
8.
Atomic Number
7
Atomic Mass g/mol–1
14.01
70
171 (N3–)
Electronegativity
3.00
Ionisation enthalpy
I
1402
ΔH/kJmol–1
II
2856
III
4577
Density (g/cm3)
0.879 at
63 K
Melting point/K
63
9.
Boiling Point/K
1.
2.
3.
4.
5.
6.
7.
As
15
30.97
110
212 (P3–)
2.10
Sb
Bi
33
83
51
209.00
74.92
121.75
120
150
140
222 (As3–) 76 (Sb3+) 108 (Bi3+)
2.20
1.82
1.67
1012
1903
2910
1.823
947
1798
2736
5.778
(Grey ∝ form)
317.1
1087.5
(White phosphorus) (Grey ∝ form)
77.2
554 (white P)
883
834
1595
2443
6.580
703
1610
2466
9.808
904
544
1653
1813
Atomic and physical properties of group 15 elements
1. Atomic and ionic radii : Values increase from N to P considerably due to addition of
new shells but increase is small from As to Bi due to completely filled inner d or f orbitals.
2. Electronegativity : Value decreases down the group with increase in atomic size.
3. Ionisation enthalpy : Value decreases down the group due to increase in atomic size.
Due to stable ns2np3 configuration, ionisation enthalpy of group 15 elements is very high.
As seen from the table ΔH1 < ΔH2 < ΔH3.
4. Non metallic and metallic character : Electropositive character and metallic nature
increases down the group. N, P are non metals. As, Sb are metalloids. Bi is metal.
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429
5. Physical appearance and density : Nitrogen exists as diatomic molecule in N2 state.
N ≡ N.
P, As, Sb exist as tetraatomic molecules. Bi is monoatomic solid. Density increases down
the group.
6. Atomic volume and melting points : Atomic volume increases down the group. But
atomic volume of P is more than that of Sb. This is due to weak intermolecular attraction
between P4 units. In As4 – close packing results in greater intermolecular attraction and
smaller volume. Therefore, melting point of Phosphorus is less than that of Arsenic.
7. Allotropy : Except Bi all the other elements in this group exhibit allotropy. Following are
the common allotropes :
(i) Solid Nitrogen :
β – Nitrogen
α – Nitrogen
Cubical crystal structure Hexagonal crystal structure
(ii) Phosphorus :
White phosphorus
tetrahedron structure
Red phosphorus
complex chain structure
Black phosphorus
stable layer structure
Arsenic :
Grey As
Yellow As
Black As
Antimony :
Metallic Sb
Yellow as α Sb
Explosive Sb
Chapter - 7 p – Block Elements
430
8. Conductivity : Conductivity increases down the group N and P are non conductors of
heat and electricity. As is poor conductor. Sb and Bi are good conductors.
9. Oxidation state : Electron configuration is ns2np3. So common oxidation states are +3,
+5 and –3. Stability of +3 oxidation state increases and +5 oxidation state decreases down
the group due to ‘Inert pair effect’ of ns2 electrons. However, N does not exhibit +5 oxidation
state through covalence due to non availability of d orbitals in valence shell. N exhibits
all the oxidation states from –3 to +5.
Formation of PCl5, PF5
3s
3p
3d
P (Ground state)
P* (Excited state)
– five unpaired electrons.
NH3 N2H4 NH2OH
Compound
Oxidation state
–3
–2
–1
N2
N 2O
0
+1
NO HNO2 NO2 HNO3
+2
+3
+4
+5
(e) Chemical Properties :
1. Action of air : All elements react with oxygen of air to form oxides. Reactivity of the
elements with oxygen increases down the group. Nitrogen reacts with oxygen only at very
temperature, which is possible by striking an electric arc.
electric (2000K)
N 2 + O 2 ⎯⎯⎯⎯⎯⎯
→ 2NO
arc
Nitric oxide
2. Action of oxidising agents : Hot, conc. HNO3 and H2SO4. Nitrogen does not react.
Phosphorus and Arsenic form Oxyacids.
(a) Action of hot and conc. Nitric acid
P4 + 20HNO 3 ⎯⎯
→ 4 H 3 PO 4 + 20 NO 2 + 4H 2 O
conc.
Phosphoric acid
hot
As 4 + 20HNO 3 ⎯⎯→ 4 H 3 AsO 4 + 20NO 2 + 4H 2 O
Arsenic acid
hot
4Sb + 20 HNO 3 ⎯⎯→ Sb 4 O10 + 20 NO 2 + 10 H 2 O
Antimony oxide
hot
Bi + 6 HNO 3 ⎯⎯→ Bi(NO 3 ) 2 + 3H 2 O + 3NO 2
Bismuth nitrate
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431
(b) Action of hot and conc. H2SO4
hot
P4 + 10H 2SO 4 ⎯⎯→ 4 H 3 PO 4 + 10SO 2 + 4 H 2 O
Phosphoric acid
hot
As 4 + 10H 2SO 4 ⎯⎯→ 4H 3 AsO 4 + 10SO 2 + 4H 2 O
Arsenic acid
hot
2Sb + 6 H 2SO 4 ⎯⎯→ Sb 2 (SO 4 ) 3 + 3SO 2 + 6 H 2 O
Antimony sulphate
hot
2Bi + 6 H 2SO 4 ⎯⎯→ Bi 2 (SO 4 ) 3 + 3SO 2 + 6 H 2 O
3. Action of alkalies : Nitrogen has no action.
P4 + 3 NaOH + 3H 2 O ⎯⎯
→ PH 3
+ 3 NaH 2 PO 2
Phosphine Sodium hypophosphite
As 4 + 12 NaOH ⎯⎯
→ 4 Na 3 AsO 3 + 6 H 2
soluble (Sodium arsenite)
2Sb + 6 NaOH ⎯⎯
→ 2 Na 3SbO 3 + 3H 2
soluble (Sodium antimonite)
4. Action of metals : Group 15 elements react with all metallic elements. Nitrogen forms nitrides.
6 Li + N 2 ⎯⎯
→ 2 Li 3 N (Lithium Nitride)
3Mg + N 2 ⎯⎯
→ Mg 3 N 2 (Magnesium Nitride)
P, As, Sb, Bi form phosphides, arsenide and antimonate and bismuthide as respectively.
6Ca + P4 ⎯⎯
→ 2Ca 3P2 (Calcium phosphide)
5. Reactivity towards hydrogen :
All the elements of group 15 react with hydrogen to form gaseous hydrides EH3
(E = N, P, As, Sb, Bi)
Properties of hydrides of group 15 elements
Property
E-H distance (pm)
Boiling point/K
Melting Point/K
HEH angle (º)
Δdiss (E-H)/kJmol–1
ΔH–f /kJmol–1
NH 3
PH3
AsH 3
SbH3
BiH 3
101.7
239.6
195.3
107.8
389
– 46.29
141.9
185.5
139.5
93.6
322
13.4
151.9
210.6
156.7
91.8
297
66.4
170.7
256.0
185
91.3
255
145.1
–
290
–
–
–
278
Chapter - 7 p – Block Elements
432
(a) Down the group, bond dissociation energy decreases along with stability of hydrides.
(b) Down the group, reducing character increases.
(c) Down the group, basicity decreases. NH3 > PH3 > AsH3 > SbH3 > BiH3.
(d) Down the group, strong smell and poisonous nature increases of group 15 elements.
6. Tendancy to form hydrogen bonding : Among hydrides of group 15 elements, only
NH3 can form intermolecular hydrogen bond with itself and with water molecules. .
This is due to polarity of N – H bond caused due to large electronegativity difference between
N and H. Hydrogen bonding is responsible for solubility and high boiling point of NH3.
7. Anomalous nature of Nitrogen :
Nitrogen has
(a) small atomic size
(b) high ionisation enthalpy
(c) high electronegativity
(d) non availability of d-orbitals in valence shell
So it shows following properties different than other elements of its group.
(a) Nitrogen is a gas while all other elements of the group are solids at room
temperature.
(b) Nitrogen exits as diatomic molecule (N2), while phosphorus and the other elements
exit as tetraatomic molecules (As4, Sb4, P4, etc).
(c) Due to small size and high electronegativity, nitrogen is the only element of the
group which can form hydrogen bonds in its hydride compounds. Other elements
do not form hydrogen bonding in their hydride compounds.
(d) Nitrogen has tendency to form pπ-pπ multiple bonds. Other elements of the group
do not form pπ-pπ multiple bonding. Instead, they form multiple bonding through
dπ-pπ overlapping.
(e) It shows wide range of oxidation states.
(f) Except NF3, other trihalides are unstable.
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(f) Dinitrogen :
1. Preparation :
fractional
(a) Commercial method : Liquid air ⎯⎯⎯⎯→
Liquid Nitrogen distills over,
distillation
B.p. of liq. N2 = 77.2 K.
B.p. of liq. O2 = 90.0 K.
(b) Laboratory method :
NH 4 Cl (aq) + NaNO 2 (aq ) ⎯⎯
→ N 2 (g) + 2H 2 O (l) + NaCl (aq)
Impurities of NO, HNO3 are removed by passing over K2Cr2O7/H2SO4 mixture.
Δ
(c) From Ammonium dichromate : (NH 4 ) 2Cr2 O 7 ⎯⎯
→ N 2 + 4H 2O + Cr2O 3
(d) From bleaching powder : 3CaOCl 2 + 2NH 3 ⎯⎯
→ 3CaCl 2 + 3H 2 O + N 2
(e) From Sodium or Barium azide : Ba(N 3 ) 2 ⎯⎯
→ Ba + 3N 2
2NaN 3 ⎯⎯
→ 2Na + 3N 2 Nitrogen obtained by this method is poor.
2. Physical properties :
(a) It is a colourless, odourless and inert gas.
(b) Its isotopes are 147 N , 157 N .
(c) It is slightly soluble in water.
(d) It has low freezing point (195.3 K) and low boiling point (239.6 K).
(e) N ≡ N bond has large bond enthalpy.
3. Chemical properties :
At high temperature, Dinitrogen undergoes following reactions.
(a) It combines with metals and non metals to form ionic and covalent nitrides.
3Mg + N 2 ⎯⎯
→ Mg 3 N 2
6Li + N 2 ⎯⎯
→ 2Li 3 N
Ionic nitrides
2Al + N 2 ⎯⎯
→ 2AlN
Fe/Al O + K O Haber's process
2 3
2
N 2 + 3H 2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
→ 2NH 3
N 2 + O 2 ⎯⎯
→ 2NO
Covalent nitrides
(b) It reacts with calcium carbide to form Calcium cynamide.
1000 K
CaC 2 + N 2 ⎯⎯⎯→
CaCN 2 + C
1Atm
4. Uses :
N – (a) Dinitrogen dilutes the action of dioxygen in air and thus, makes combustion much
less rapid.
(b) It is used for filling electric lamps and provides an inert atmosphere in certain
metallurgical operations.
Chapter - 7 p – Block Elements
434
(c) It is used in the manufacture of NH3, HNO3, CaCN2 and other nitrogen
compounds.
(d) Liquid dinitrogen is used as a refrigerant to preserve biological materials, food
items and in cryosurgery.
(g) Compounds of Nitrogen - NH3 :
1. Preparation of Ammonia :
(a) Laboratory method :
NH 4 Cl + Ca(OH) 2 ⎯⎯
→ CaCl 2 + 2NH 3 + 2H 2 O
Slaked lime
CaO
Moist NH 3 ⎯⎯⎯⎯⎯
→ Dry NH 3
Quick lime
It cannot be dried over H2SO4, P2O5 or anhydrous CaCl2, as it reacts with them.
(b) Manufacturing method :
Haber’s process :
N 2 + 3H 2 2NH 3 ΔHf = – 46.1 kJ mol–1
Optimum conditions : (a) High pressure of 200 atm
(b) Temperature 700 K
(c) Catalyst finely divided Iron and Molybdenum orAl2O3, K2O
(d) Ratio N2 : H2 :: 1:3
(c) By hydrolysis of calcium cyanamide CaCN2.
1000 K, 1atm
CaC 2 + N 2 ⎯⎯⎯⎯ ⎯→ CaCN 2 + C
CaCN 2 + 3H 2 O ⎯⎯
→ CaCO 3 + 2NH 3
Steam
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435
2. Physical properties :
(a) It is colourless gas with pungent smell.
(b) Freezing point is 198.4 K; Boiling point is 239.7 K.
(c) Water soluble and basic in nature.
(d) Ammonia molecules are associated due to hydrogen bonding in solid and liquid state.
(e) Due to intermolecular hydrogen bonding, melting point and boiling point is high.
[Refer Fig. 7.1 (e) 6]
(f) It liquifies on cooling under pressure to form liquid ammonia, which is used as cooling
agent.
3. Chemical properties :
(a) It reacts with Air.
Δ
4 NH 3 + 3O 2 ⎯⎯→ 2 N 2 + 6H 2 O
Pt
4 NH 3 + 5O 2 ⎯⎯⎯⎯⎯
→ 4 NO + 6 H 2 O
500 K,9 bar
(b) It has reducing action on CuO and Cl2.
Δ
2 NH 3 + 3CuO ⎯⎯→ 3Cu + 3H 2O + N 2
8 NH 3 + 3Cl 2 ⎯⎯
→ 6 NH 4 Cl + N 2
excess
(c) It reacts with Halogens in excess.
NH 3 + Cl 2 ⎯⎯
→ NCl 3 + 3HCl
excess
Nitrogen trichloride (explosive)
8 NH 3 + 3Br2 ⎯⎯
→ 6 NH 4 Br + N 2
2 NH 3 + 3I 2 ⎯⎯
→ NH 3 NI 3 + 3HI
Explosive (Nitrogen triodide ammonia)
8 NH 3 NI 3 ⎯⎯
→ 5 N 2 + 9 I 2 + 6 NH 4 I
(d) It reacts with active metals like Na, K.
575k
2 Na + 2 NH 3 ⎯⎯⎯
→ 2 NaNH 2 + H 2
(e) It reacts with Sodium hypochloride.
glue/
2NH 3 + NaOCl ⎯⎯⎯→
NH 2 NH 2 + NaCl + H 2 O
gelatin
hydrazine
Chapter - 7 p – Block Elements
436
(f) It undergoes self ionisation in liquid state.
–
NH +4 + N H 2
2 NH 3 (g) It reacts with water to form alkaline solution.
NH 3 + H 2 O ⎯⎯
→ NH +4 + OH –
(h) Formation of coordination compounds.
Due to presence of lone pair of electrons of N; NH3 acts as a Lewis base. It links
with metal ions by donating lone pair.
Cu 2+ (aq) + 4 NH 3(aq) ⎯⎯
→ [Cu(NH 3 ) 4 ] 2+ (aq)
blue
deep blue
+
–
Ag (aq) + Cl (aq) ⎯⎯
→ AgCl (s) ppt
White
AgCl (s) + 2NH 3(aq) ⎯⎯
→ [Ag(NH 3 ) 2Cl] (aq)
ppt
colourless
Brown ppt
Nessler's reagent
→
(i) Ammonium salts ⎯⎯⎯⎯⎯⎯
K 2 HgI 4
Million's base
2K 2 HgI 4 + NH 3 + 3KOH ⎯⎯
→ NH 2 HgOHgI + 7 KI + 2H 2 O
Iodide of Million's reagent
This is a test for ammonia and ammonium salts.
4. Uses of Ammonia :
(a) For preparation of fertilisers.
(b) For preparation of nitric acid in Ostwald’s process.
(c) Liquid NH3 is used as refrigerant.
5. Structure of NH3 :
State of hybridisation of N is sp3. N – H bonds are formed
by sp3 – S overlap. Lone pair of electrons occupy sp3 hybrid
orbital. Bond angle is 107.50 to avoid repulsion between
bonded pair - lone pair. Ammonia molecule is trigonal pyramidal
(h) Compounds of Nitrogen – HNO3 :
1. Preparation :
(a) Laboratory method :
distill
NaNO 3 + H 2SO 4 ⎯⎯⎯
→ NaHSO 4 + HNO 3
Nitre
(b) Manufacture :
Ostwald’s process
Pt/Rh gauge catalyst
4 NH 3 (g) + 5O 2 (g) ⎯⎯⎯⎯⎯⎯⎯
→ 4 NO (g) + 6 H 2 O (g)
500 K 9 bar
air
Nitric oxide
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2 NO 2 Nitrogen dioxide
2 NO (g) + O 2 (g) (g)
3 NO 2 (g) + H 2 O ⎯⎯
→ 2 HNO 3 l + NO (g)
Nitric acid
NO is by product. Which is reused.
On concentration by distillation 68% by mass Nitric acid is obtained.
By dehydration with conc. H2SO4, 98% concentration is obtained.
It is aqua fortis i.e. strong water, because it attacks all metals.
2. Physical properties :
(a) Pure nitric acid is colourless liquid.
(f.p. 231.4 K and b.p. 355.6 K)
(b) In gaseous state molecule of HNO3
has a planar structure.
(c) The pure acid is unstable.
In aqueous solution nitric acid behaves as strong acid and undergo ionisation forming
nitrate ions. HNO 3 (aq) + H 2O (l) ⎯⎯
→ H 3O + (aq) + NO 3 – (aq)
3. Chemical properties :
(a) It is a strong acid in aqueous solution.
HNO 3(aq) + H 2O (l ) ⎯⎯
→ H 3O + (aq) + NO 3– (aq)
(b) Action on metals : Except Gold and Platinum all metals are attacked by nitric acid.
It is strong oxidising agent.
Nature of products depend on concentration of acid and temperature.
3Cu + 8HNO 3 ⎯⎯
→ 3Cu(NO 3 ) 2 + 2 NO + 4 H 2 O
dil.
Cu + 4 HNO 3 ⎯⎯
→ Cu(NO 3 ) 2 + 2 NO 2 + 2H 2 O
conc.
4 Zn + 10 HNO 3 ⎯⎯
→ 4 Zn(NO 3 ) 2 + N 2 O + 5H 2 O
dil.
Zn + 4 HNO 3 ⎯⎯
→ Zn(NO 3 ) 2 + 2NO 2 + 2H 2 O
conc.
Metals like cromium and aluminium are covered with a layer of oxide.
Hence, they do not react with Nitric acid.
(c) Action on non metals :
C + 4 HNO 3 ⎯⎯
→ CO 2 + 4 NO 2 + 2H 2 O
conc.
Chapter - 7 p – Block Elements
438
I 2 + 10HNO 3 ⎯⎯
→ 2 HIO 3 + 10 NO 2 + 4 H 2 O
iodic acid
S8 + 48HNO 3 ⎯⎯
→ 8H 2SO 4 + 48 NO 2 + 16 H 2O
P4 + 20 HNO 3 ⎯⎯
→ 4H 3 PO 4 + 20 NO 2 + 4 H 2 O
phosphoric acid
(d) Action of Aqua regia : (HCl + HNO3 : : 3:1) on gold and platinum.
conc. conc.
3HCl + HNO 3 ⎯⎯
→ NOCl + 2H 2O + 2[(Cl)]
HCl
Au + 3[Cl] ⎯⎯
→ AuCl 3 ⎯⎯⎯
→ HAuCl 4 Aurochloric acid
2HCl
Pt + 4[Cl] ⎯⎯
→ PtCl 4 ⎯⎯⎯
→ H 2 PtCl 6 Chloroplatinic acid
(e) Action on organic compounds :
(1) It introduces ‘nitro’ group in organic compounds.
conc.H SO
4 → C H NO + H O
C 6 H 6 + HONO 2 ⎯⎯ ⎯2⎯⎯
6 5
2
2
Benzene conc.
Nitrobenzene
conc.H SO
2 4 → C H (NO ) CH + 3H O
C 6 H 5 CH 3 + 3HONO 2 ⎯⎯ ⎯⎯⎯
6 2
2 3
3
2
Toluene
conc.
2, 4,6- trinitrotoluene
conc.H SO
2 ⎯
4 → C H (NO ) OH + 3H O
C 6 H 5 OH + 3HONO 2 ⎯⎯⎯⎯
6 2
2 3
2
conc.
2, 4,```