Jee - Module 1 - Chem - Physical Chemistry [PDF]

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TABLE OF CONTENTS

SOLID STATE Theory ................................................................................................................................................

8

Solved examples ...............................................................................................................................

20

Exercise - 1 : Basic Objective Questions.............................................................................................

30

Exercise - 2 : Previous Year JEE MAIN Questions ............................................................................... 35 Exercise - 3 : Advanced Objective Questions ..................................................................................

37

Exercise - 4 : Previous Year JEE Advanced Questions ....................................................................... 43 Answer Key ........................................................................................................................................

256

SOLUTIONS Theory ................................................................................................................................................

46

Solved examples ...............................................................................................................................

59

Exercise - 1 : Basic Objective Questions.............................................................................................

74

Exercise - 2 : Previous Year JEE MAIN Questions ............................................................................... 80 Exercise - 3 : Advanced Objective Questions ..................................................................................

85

Exercise - 4 : Previous Year JEE Advanced Questions ......................................................................... 92 Answer Key ........................................................................................................................................ 258

7 CHEMICAL KINETICS Theory ................................................................................................................................................

96

Solved examples ............................................................................................................................... 109 Exercise - 1 : Basic Objective Questions.............................................................................................

121

Exercise - 2 : Previous Year JEE MAIN Questions ............................................................................... 128 Exercise - 3 : Advanced Objective Questions ..................................................................................

133

Exercise - 4 : Previous Year JEE Advanced Questions ....................................................................... 144 Answer Key ........................................................................................................................................

260

ELECTROCHEMISTRY Theory ................................................................................................................................................ 150 Solved examples ............................................................................................................................... 163 Exercise - 1 : Basic Objective Questions.............................................................................................

177

Exercise - 2 : Previous Year JEE MAIN Questions .............................................................................. 185 Exercise - 3 : Advanced Objective Questions ..................................................................................

190

Exercise - 4 : Previous Year JEE Advanced Questions ....................................................................... 200 Answer Key ........................................................................................................................................ 262

SURFACE CHEMISTRY Theory ................................................................................................................................................ 206 Solved examples ............................................................................................................................... 230 Exercise - 1 : Basic Objective Questions............................................................................................. 238 Exercise - 2 : Previous Year JEE MAIN Questions ............................................................................. 243 Exercise - 3 : Advanced Objective Questions .................................................................................. 247 Exercise - 4 : Previous Year JEE Advanced Questions ....................................................................... 253 Answer Key ........................................................................................................................................ 264

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01 SOLID STATE

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Chapter 01

SOLID STATE

1. INTRODUCTION

2. CLASSIFICATION OF SOLIDS

Solid state is a state of matter besides liquid and gaseous state. In case of solids the inter molecular forces are very strong and empty spaces between the atoms/ions/ molecules is very less. That is why they have a fixed shape and volume.

Solids are broadly classified on the basis of following parameters. 

based on various properties



based on bonding present in building blocks. 2.1 On the basis of various properties

1.1 CHARACTERISTIC PROPERTIES OF SOLIDS

Based on their various properties solids can be classified as

Solids are characterised by the following properties 

High density



Crystalline solids



Low compressibility



Amorphous solids.



Rigidity



Definite shape and volume.

Crystalline solids have a regular structure over the entire volume and sharp properties whereas amorphous solids have irregular structure over long distances and properties are not that sharp. Various differences are listed in table below

Property

Crystalline solids

Amorphous solids

Shape

They have long range order

They have short range order.

Melting point

They have definite melting point

They do not have definite melting point.

Heat of fusion

They have a definite heat of fusion

They do not have definite heat of fusion

Compressibility

They are rigid and incompressible

They may be compressed to some extent

Cutting with a sharp

They are given cleavage

They give irregular cleavage i.e. they

edged tool

break into two pieces with plane surfaces break into two pieces with irregular surfaces

clevage property Isotropy and

They are anisotropy : It is the property of being independent on the direction

They are isotropy : It is the property of being dependonr on the direction

Volume change

There is a sudden change in volume when they melt.

There is no sudden change in volume on melting.

Symmetry

They possess symmetry

They do not possess any symmetry.

Interfacial angles

They possess interfacial angles.

They do not possess interfacial angles.

Anisotropy

SCAN CODE SOLID STATE

SOLID STATE

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2.2 Based on bonding There are various type of solids based on type of bonding present in their building blocks. Various types of solids along with their properties are given in the table below. The different properties of the four types of solids are listed in Different Types of Solids Types of Solid

(1) Molecular Solids

Constituent

Nature of

particls

Bonding

Examples

Physics

Electrical

Melting

conductivity

point

Molecules

(i) Non polar

Dispersion or Ar, CCl4, H2, London forces I2, CO2,

Soft

Insulator

Very low

(ii) Polar

Dipole-dipole Interactions

HCl, SO2

Soft

Insulator

Low

H2O (ice)

Hard

Insulator

Low

(2) Ionic solids Ions

Coulombic or electrostatic

NaCl, MgO, ZnS, CaF2

Hard but brittle Insulators in High solid state but conductors in molten state and in aqueous solutions

(3) Metallic solids

Metallic bonding

Fe, Cu, Ag, Mg

Hard but malleable and ductile

Conductors in solid state as well as in molten state

Covalent bonding

SiO2 (quartz) SiC, C (diamond) AIN, C(graphite)

Hard

Insulators

Soft

Conductor (exception)

(iii) Hydrogen bonded

Hydrogen bonding

Positive ions in a sea of delocalised electrons

(4) Covalent or Atoms network solids

Very high

SCAN CODE SOLID STATE

SOLID STATE

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3. STRUCTURE OF CRYSTALLINE SOLIDS 3.1 Crystal lattice and Unit Cell The regular array of the building blocks (atoms/ions/ molecules) inside the crystalline solid is called “Crystal Lattice”. The smallest part or crystal lattice which can be repeated in all directions to generate entire crystal lattice is called “Unit Cell”. In unit cell the atoms of ions or molecules are represented by small spheres. Various lattices are formed by variation in following parameters:

3.2 Primitive Unit Cells and Bravais Lattices

The edge length along 3 axes - a, b, c  The interfacial angles - ,  , .  Location of atom/ions w.r.t each other in crystal lattice.

In all, there are seven types of unit cells and there can be some sub types of unit cells. These seven unit cells are called Primitive Unit Cells or Crystal Habits. Which are listed in table below.

Crystal System

Axial distances

Axial angles

Examples

Cubic

abc

      90

Copper, Zinc blende, KCl

Tetragonal

abc

      90

Sn (White tin), SnO2, TiO2

Orthorhombic

abc

    90

Rhombic sulphur, CaCO3

Monoclinic

abc

    90;   90

Monoclinic sulphur, PbCrO2

Hexagonal

abc

    90;   120

Graphite, ZnO

Rhombohedral

abc

      90

CaCO3 (Calcite),HgS (Cinnabar)

Triclinic

abc

     90

K2Cr2O7, CuSO4.5H2O

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SOLID STATE

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Bravais Lattice Crystal System

Space lattice

Examples

Cubic

Simple : Lattice

Body Centered :

Face Centered : Points

Pb, Hg, Ag,Au,

a  b  c Here, a, b

Points at theeight

Points at the eight

at the eight cornersand

Cu, ZnS, diamond,

and c are parameters cornersof the

corners and at

at the sixface centres.

KCl, NaCl, Cu2O,

(dimensions of aunit

unitcells.

the body centre.

CaF2 and alums. etc.

Tetragonal

Simple : Points at

Body Centered :

SnO2, TiO2, ZnO2,

a  b  c.

the eightcorners

Points at the eight corners and at the

NiSO4 ZrSiO4,PbWO4,

     90

of the unit cells.

body centre.

white Sn etc.

cell along three axes)] size of crystals depend on parameters.

      90,  and  are sizes of three angles between the axes.

Orthorhombic

Simple : Points

End Centered :

Body

Face Centered :

KNO3, K2SO4,

(Rhombic)

at the eight

Also called side

Centered :

Points at the

PbCO3, BaSO4

a  b  c.

corners of the

centered or base

Points at the

eight corners

Rhomic sulplur.

      90

unit cells.

centered. Points

eight corners

and at the six

MgSO4.7H2O etc.

at the eight corners

and at the

face centres.

and at two face

body centre.

centres opposite to each other.

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SOLID STATE

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Rhombohedral

Simple : Points at the eight

NaNO3, CaSO4

or Triangonal

corners of the unit cells.

calcite, quartz

a  b  c,

As, Sb, Bi etc.

      90

Hexagonal

Simple : Points at the twelve or Points at the twelve

ZnO, PbS, CdS, HgS,

a  b  c,

corners of the unit cell out corners of the hexagonal

graphite, ice,

    90  120

lined by thick line. Prism and at the centres of top

Mg, Zn, Cd. etc.

and bottom faces.

Monoclinic

Simple : Points at the eight

End Centered : Points at the

Na2SO4, 10H2O,

a  b  c,

corners of the unit cells.

eight corners and at two face

Na2B4O7.10H2O,

centres opposite to each other.

CaSO4.2H2O, mono

    90,  90

clinic sulphur etc.

Triclinic

Simple : Points at the eight corners of the unit cells.

CuSO4.5H2O, K2Cr2O7, H3BO3

a  b  c,

      90

We will focus majority on cubic unit cells and their arrangements.

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3.3 Cubit Unit Cells

3.3.1 Types of cubic Unit Cells

This is the most common unit cell. In a cubic unit cell there are following locations for the atoms or spheres

* Corners * Body Centre * Face Centres Following are the contributions of a sphere kept at various locations.

These unit cells differ from each other in following factors. * Location of spheres inside the unit cell.

* Rank of the unit cell (effective number of spheres inside

Location

Contribution

Corners

1/8

Body Centre

1

Face Centre

1/2

Type of Cublic Crystal

a unit cell) (Z) * Relation between edge length and radius of one sphere.

* Packing fraction (fraction of volume occupied by spheres in a unit cell) The following parameters for all the 3 unit cells are listed is the table below:

No. of atoms at different locations (lattice points)

Structure Rank (Z)

Packing

Relation between atomic and radius edge length (a)

Corners Body Centre Face Centre

Simple Cubic

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

Body Centred

8

1

Face Centred

8









1

52%

2

6

4

68%

74%

r  a/2

r

r

SCAN CODE SOLID STATE

3a 4

2a 4

SOLID STATE 3.4 Density of cubic crystals Density of cubic crystal is given by the following formula

M Z Density   a 3  N A where, Z  rank of unit cell M  Molar mass of solid a  Edge length of unit cell NA  Avogadro Number. 3.5 Close packing in Solids : Origin of unit cells Suppose we have spheres of equal size and we have to arrange them in a single layer with the condition that spheres should come in close contact with each other. Two types of layers are possible:

15 If we put 2nd layer in depressions of first hexagonal layer A two types of voids are created. X type of voids are those which are hollow and through voids of layer A and layer B. while Y type of voids are those voids of layer B which are exactly above spheres of layer A. If we place the spheres of 2nd layer on Y voids then we are repeating layer 1 and ABABAB…. Type packing is obtained. In this arrangement hexagonal unit cell is obtained and packing is called Hexagonal Close Packing (HCP). The efficiency of this pacing is 74%. If the 3rd layer is placed on X-type of voids ten a new layer C is obtained and then the arrangement will be repeated. We will obtain ABCABCABC….. type of packing. The unit cell for this arrangement is FCC and the packing efficiency is 74%.

1. Square Packing 2. Hexagonal Packing In square packing spheres are placed in such a way that the rows have a horizontal as well as vertical arrangements. In this case Co-ordination Number is 4.

Hexagonal packing is more efficient. Its Co-ordination Number is 6 and voids in the packing are smaller than square packing. If we place another layer on square packing then there are following possibilities: 1. A similar layer placed just above foundation layer that is the spheres of the second layer coming just above the spheres of the first layer and layers get repeated. In first layer is termed A the packing in this case is AA… type and the unit cell is simple cubic. 2. On other hand if spheres of second layer are placed in depressions of first layer we get BCC unit cell and ABAB …. Type of packing. Arrangements based on hexagonal foundation layer are as follows :

SCAN CODE SOLID STATE

SOLID STATE

16 4.3.2 Tetrahedral Void It forms by contact of 4 spheres and is positioned at the centre of tetrahedron formed by contact of 4 spheres. Figure

Key Points

* Radius Ratio

r  0.225 R

* No. of voids in FCC crystal  8 * Position : at a distance

a 3 4

from every corner.

* Co-ordination no.  4 4.3.3 Octahedral Void Figure

4. VOIDS

Key Points

* Radius Ratio

4.1 Definition

r  0.414 R

The empty spaces inside a spheres are called “voids”. The size and shape of voids depends upon the type of unit cell and packing.

* No. of voids in FCC

4.2 Radius Ratio The size of void is expressed in terms of radius ratio of a sphere that can be exactly fir in the void to the radius of surrounding spheres. This expressed as:

* Rank  4 * Co-ordination no.  4

Radius ration

crystal  4

* Position : Body centre Edge Centres.

r R

4.3 Types of voids 4.3.1 Trigonal void It is a void formed of equal radii and touching each other as shown in figure. Figure

Key Points

* Radius Ratio

r  0.155 R

* Smallest void * Co-ordination no.  3

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4.3.4 Cubic Void

5.1 NaCl Type Structure

This void forms by close contact of 8 spheres Figure Key Points

Key Points

* Cl occupy corners and face 

r * Radius Ratio  0.732 R * No. of voids in cubic crystal  1

centres and Na  occupy octahedral voids in FCC crystal.

* Effective formula  Na4Cl4

* Position : at body centre * Co-ordination number  8 * Rank  1

*Co-ordination No. of Na  6 *Co-ordination No. of Cl  6 * Distance b/w nearest

It is clear from above details that Trigonal < Tetrahedral < Octahedral < Cubic void

void

void

a  neighbours  rNa   rCl   2 

void

5. CLASSIFICATION OF IONIC STRUCTURES

5.2 ZnS type Structure

Ionic compounds are formed by the simultaneous arrangement of cations and anions in lattice/unit cell. The larger of two species occupies major positions in a unit cell and the smaller ones occupy voids according to their size. Which is decided on the bases of radius ration (r/r). The various ratios are listed below. Limiting Radius Ratios,

x  r/r

C.N.

Figure

Key Points

* S ions occupy main positions and 2

Zn 2 ions are present in alternate tetrahe dral voids in FCC crystal. * Effective formula  Zn4S4

* Co-ordination No. of Zn2  4 * Co-ordination No. of S2  4

Shape Example

* rZn  rS  2

x < 0.155

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Linear

BeF2

0.155  x  0.225 0.225  x  0.414 0.414  x  0.732 0.732  x  0.999

3 4 6 8

Planar Triangular Tetrahedron Octahedron Body Centred Cubic

AlCl3 ZnS NaCl CsCl

Based on these ration ranges, ionic crystal are classified into 5 categories which are as follows

2

a 3 4

5.3 Fluorite Type Structure Figure

Key Points

* Ca 2 ions occupy main positions and F ions occupy tetrahedral voids in FCC crystal. * Effective formula  Ca4F8

* Co-ordination No. of Ca2  8 * Co-ordination No. of F  4 a 3

* rCa  rF  4 2

–

SCAN CODE SOLID STATE

SOLID STATE

18 6.2 Vacancies

5.4 Anti Fluorite Structure Figure

Key Points

* Common in alkali oxides like Na2O, Li2O etc. * O2 ions occupy FCC Li ions occupy the tetrahedral voids * C.N. of Li  4

These are defects that occur when positions that should contain atoms or ions are vacant. 6.3 Interstitial sites These are sites located between regular positions sometimes atoms or ions may occupy these positions.

* C.N. of O2  8 * rLi  rO  

2

a 3 4

5.5 CsCl Type Structure Figure

Key Points

* Cl ions simple cubic Locations (corners) and Cs ions occupy body centre in BCC lattice. * Effective formula  CsCl

* Co-ordination No. of Cs  8 * Co-ordination No. of Cl  8 * rCs  rCl  



a 3 . 2

6. IMPERFECTIONS IN SOLIDS

6.4 Stoichiometric Defects These defects do not disturb stoichiometry of solid substance. 6.4.1 Schottky defects It is a vacancy defect in ionic solids. No of missing cations and anions is equal so electrical neutrality is maintained. This defect decreases the density of the substance. The defect is shown by ionic substances in which cation and anion are of almost similar sizes. Eq. KCl, NaCl, AgBr etc.

Sometimes some defects or imperfections occur in crystal structure. 6.1 Classification of defects

6.4.2 Frenkel defect In ionic solids the smaller ion is dislocated from its normal position to an interstitial site. It creates a vacancy defect at its original site and interstitial defect at new location. It is also called as dislocation defect. It does not change the density of solid.

SCAN CODE SOLID STATE

SOLID STATE

19

This type of defect is shown by ionic substances in which there is a large difference in size of ions. Eq ZnS, AgCl, AgBr etc.

Due to the presence of extra cations in the interstitial sites. Note :AgBr shows both Schottky and Frenkel defects. 6.5 Non Stoichiometric Defects

An extra cation may be present in interstitial site and an electron is present in another interstitial site so that electrical neutrality is maintained. This is similar to Frenkel defect and if found in crystal having Frenkel defect.

The compounds having these defects contain combining elements in a ratio different from required by their stoichiometric formulae. 6.5.1 Metal Excess Defect Due to anionic vacancies : The anion may be missing from

its lattice site leaving an e behind so that charge remains balanced. The site containing electron is called F centre. They import colour to the crystal, F stands for Farbenzenter meaning colour. This defect is similar to schottky defect and is found is

6.5.2 Metal Deficiency Defect

crystals having schottky defect eq. NaCl, KCl etc. This defect occurs when metal shows variable valency. eq. FeO is mostly found is varying compositions between Fe0.93O to Fe0.96O. in crystals of FeO some Fe2 cations are missing and the loss of posiive charge is made up by the presence of required number of Fe3 ions.

SCAN CODE SOLID STATE

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20

SOLVED EXAMPLES

Example-1

Sol.

(a) Metallic solids (e.g., zinc, chromium, etc.)

Why are solids rigid ?

(b) Ionic solids (e.g., sodium chloride, potassium bromide)

Solids are rigid because the intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids cannot move from their positions, that is, they have fixed positions.However, they can oscillate about their mean positions.

(c) Covalent solids (e.g., diamond, graphite) (d) Molecular solids (e.g., iodine, naphthalene) Example - 5 What difference in behavior between the glass and

Example-2

sodium chloride would you expect to observe if you break off a piece of either cube ?

Why does the window glass of the old buildings look milky? Sol. Sol.

Due to continuous heating in the day and cooling at night, that is, annealing over a long period, glass acquires some crystalline characteristics. Glass is generally amorphous by nature, but due to annealing it becomes crystalline in nature. Hence, window glass of the old buildings looks

If we break or cut a piece of cube of sodium chloride, then we will observe that the surfaces on the two portions of the cube after breaking will be smooth but in case of glass crystals the two surfaces of the cube after breaking will be irregular. This is because of the cleavage property of crystalline solids according to which if we break or cut a

milky.

crystalline solid then we get two smooth surfaces, but Example - 3

amorphous solids do not show cleavage property. Therefore in case of amorphous solids after cutting we get

Why does urea have a sharp metling point but glass does not ? Sol.

Urea is a crystalline solid, whereas glass is an amorphous solid. Crystalline solids have sharp melting points, whereas amorphous solids do not possess sharp melting points.

two irregular surfaces. Example - 6 Identify the forces that must be overcome to cause melting in the following solids. Rank the compounds in the order

Example - 4

of expected melting points (lowest one first) : What are crystalline solids ? State the categories of

(a) Diamond ; (b) CF4 ; (c) CrF2 ; (d) SCl2.

crystalline solids with examples. Sol.

Crystalline solids are solids with highly ordered

Sol.

(a) Diamond Network (solid) covalent bonds

arrangement of constituent particles that is repeated many

(b) CF4 London dispersion forces

times throughout the structure. The pattern is so ordered

(c) CrF2 Electrostatic forces

and regular that arrangement of constituent particles at any site can be predicted, if arrangement at any one site is

(d) SCl2 London forces and dipole-dipole forces

known. They are characterized by rigid structure, high and

The order is : CF4 < SCl2 < CrF2 < diamond.

sharp melting points.

SOLID STATE

21 (b) A face-centered unit cell has one constituent particle present at the center of each face in addition to the particles present at the corners as shown in fig.

Example - 7 Compare and contrast the bonding in molecular solids and network covalent solids. Which would you expect to have a higher melting point, molecular solids or network covalent solids ? Sol.

An end-centered unit cell has one constituent particle each at the center of any two opposite faces in addition to the particles present at the corners as shown in fig.

In molecular solids, the constituent particles are molecules. The forces operating between them are weak dispersion forces or dipole-dipole forces of attraction or hydrogen bonding depending on the type of molecular solid. In covalent or network solids, the constituent particles are non-metal atoms linked to adjacent atoms by covalent bonds throughout the solid. Covalent or network solids have a higher melting point than molecular solids.

Example - 8 MgO has the structure of NaCl, and TiCl has the structure of CsCl. What are the coordination numbers of the ions in MgO and TiCl ? Sol.

(a) Example - 12

What relationship is there between a crystal lattice and a unit cell ?

The coordination number of ions in MgO is 6 and coordination number of ions in TiCl is 8.

Example - 9

Sol.

What type of solids are electrical conductors, malleable and ductile. Sol.

Example - 10

Sol.

Example - 13

The significance of a lattice point is that each lattice point represents one constituent particle of the solid which may be an atom, a molecule (group of atoms), or an ion.

A solid is made up 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 ?

Example - 11 Distinguish between (a) Hexagonal and monoclinic unit cells. (b) Face-centered and end-centered unit cells. Sol.

(a) For a hexagonal unit cell, a = b  c and  =  = 90º, = 120º For a monoclinic unit cell, a  b  c and  =  = 90º,  90º

Crystal lattice : A regular arrangement of the constituent particles of a crystal in a three dimensional space is called crystal lattice or space lattice. Unit cell : The smallest three-dimensional portion of a complete space lattice, which when repeated over and again in different directions produces the complete space lattice. The relation between unti cell and crystal lattice is that the unit cell is part of the lattice as shown in fig.

Metallic solids are electrical conductors, malleable and ductile.

Give the significance of a “lattice point”.

(b)

Sol.

Suppose number of atoms of element Q = N. So, numberof tetrahedral sites = 2N. Therefore, the number of atoms of element P = 2N. The ratio P : Q = 2N:N = 2:1. Hence, the formula of the compound is P2Q.

SOLID STATE

22

Example - 14

Example - 18

Name the parameters that characterize a unit cell. Sol.

A compound with fcc crystal structure has a density of 2.163 × 103 kg m–3. Calculate the edge length of its unit cell. The molar mass of the compound is 58.2 g mol–1.

The parameters that characterize a unit cell are as follows: (a) Its dimensions along the three edges a, b and c. These edges may or may not be equal.

Sol.

(b) The angles between the edges. These are represented by  (between edges b and c),  (between edges a and c) and  (between edges a and b).



A cubic solid is made up of two elements X and Y. Atoms Y are present at the corners of the cube and atoms X at the body center. What is the formula of the compound? What are the coordination numbers of X and Y ? Number of atoms of and number atoms of Y = 1. The formula of the compound is XY. The coordination number of the compound will be 8 because one X atom is surrounded by eight Y atoms at the corner of the cube.



2.163  103  103

Gold (atomic radius = 0.144 nm) crystallizes in a facecentered unit cell. What is the length of a side of the cell?

a3 

For a face-centerd unit cell, a = 2 2 r It is given that the atomic radius, r = 0.144 nm. So,



2.163  106 106

= 2.163 g cm–3

Z M 4  58.2 g mol 1    N A 2.163  6.02  10 23 mol 1g cm 3 232.8 g mol 1

13.02  10 23 g cm 3mol 1

 17.880  10  23 cm3

So, the edge length of the unit cell is

Hence, length of a side of the cell = 0.407 nm. Example - 17

(1 10 2 )3



= 178.8 × 10–24 cm3

a  2 2  0.144 = 0.407 nm

a = (178.8)1/3 × 10–8 = 5.634 × 10–8 cm or 563.4 pm. Example - 19

The atomic radius of nickel is 124 pm. Nickel crystallizes in face-centered cubic lattice. What is the length of the edge of unit cell expressed in pm and angstroms ? Sol.

a3  NA

For fcc crystal structure, Z = 4 and we know that NA = 6.02 × 1023 mol–1 M = 58.2 g mol–1. Rearranging Eq. (1) in terms of a, we get

Example - 16

Sol.

Z M

where Z is the number of particles present per unit cell; M is the atomic mass of the element ; a is the edge length of the unit cell; NA is the Avogadro constant ;  is density of the crystal given as 2.163 × 103 kg m–3. Since density is usually expressed in g cm–3, so

Example - 15

Sol.

For an element, the density of a crystal is given by

The figures given below show the location of atoms in three crystallographic planes in a fcc lattice. Draw the unit cell for the corresponding structure and identify these planes in your diagrams.

In a fcc, edge lenth = a and the radius of each atom be r. In this structure, atoms touch each other along the facediagonal, there fore Face diagonal =

2 a  4r

Sol. or

a

4r 2

 2 2  124 pm = 350.7 pm

thus, the unit cell edge length is 350.7 pm.

Face centred cube, face diagonal plane, diagonal plane.

SOLID STATE Example - 20 How many copper atoms are there within the facecentered cubic unit cell of copper ? Sol.

23 = 40 × 40 = 1600 mm2 = 16 cm2

( 40 mm = 4 cm)

= 4 × 4 cm2

In the face-centered cubic unit cell, there are eight atoms at the corners and six atoms on the faces. Therefore, the contribution of atom at the corners = 1/8 × 8 = 1, and contribution of atoms on the faces = 1/2 × 6 = 3. Thus the number of copper atoms present is four per unit cell.

Example - 21 A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centers. Derive the formula of the compound. Sol.

As a atoms are present at the 8 corners of the cube, therefore number of atoms of A in the unit cell = 1/8 × 8 = 1. As B atoms are present at the face centers of the cube, therefore number of atoms of B in the unit cell = 1/2 × 6 = 3. Therefore, ratio of atoms A:B = 1:3. Hence, the formula of compound is AB3.

Maximum no. of spheres = 14 (full) + 8 (half) = 18 Per unit area 

18 =1.125 16

Example - 22 Explain how much portion of an atom is located at (a) corner and (b) body center of a cubic unit cell is part of its neighboring unit cell. Sol.

(a) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore, 1/8th portion of the atoms belongs to one unit cell. (b) An atom located at the body center of a cubic unit cell is not shared by any other unit cell. Hence, it belongs fully to the unit cell.

Example - 23 You are given marbles of diameter 10 mm. They are to be placed such that their centers are lying in a square bound byfour lines and each ot length 40 mm. What will be the arrangments of marbles in plane so that maximum number of marbles can be placed inside the area ? Sketch the diagram and derive expression for the number of molecules per unti area. Sol.

Maximum numberof marble balls or spheres can be accommodated if they are arranged in a hexagonal close packing (hcp). Area of square having 4 lines each of 40 mm is

Length AB of square = 40 mm = 4 cm Length AC = 4 cm.

PZ 3  sin 60º  XZ 2

In XYZ

PZ  10

3 5 3 2

AC = 5 + 4 × 5 3 = 5 + 20 3 = 40 mm = 4.0 cm

face-centred cubic

face-diagonal plane

diagonal plane

SOLID STATE

24

Example - 24 A solid substance AB has a rock salt geometry. What are the coordination numbers of A and B ? How many atoms of A and B are present in the unit cell ? Sol.

The coordination number of A and B is 6. Four atoms each of A and B are present in the unit cell.

Sol.

Sol.

Sol.

Number of cations (A ) per unit cell = 8 × 1/8 = 1

 . The shortest distance between barium atoms will be 2rBa Thus, Ba-Ba distance = 4.352 × 10–8 cm.

Example - 30

Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of primitive cubic and body-centered cubic lattices are 52.4% and 68%, respecitvely.

An element occurs in bcc structure with cell edge 300 pm. The density of the element is 5.2 g cm–3. How many atoms of the element does 200 g of the element contain ?

Example - 27 Sol.

The ccp lattice is formed by the atoms of the element N. Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N. According to the questions, the atoms of element M occupy 1/3rd of the ratio of the number of atoms of M to that of N is M : N = (2/3) : 1 = 2:3. Thus, the formula of the compound is M2N3.

Example - 28 The mineral haematite, Fe2O3 consists of a cubic closepacked array of oxide ions with Fe3+ ions occupying interstitial positions. Predict whether the iron ions are in the octahedral holes. Radius of Fe3+ = 65 pm and that of O2– = 145 pm.

 3a  4rBa

3  0.5025 10 7 cm = 2176 × 10–8 cm 4

 rBa 

Example - 26

Sol.

65 pm  0.45 145 pm

From the geometry of the body-centered cubic structure, we find the shortest distance between barium atoms will occur down the body diagonal where Diagonal =

Hence, the ratio of A+ : B– is 1 : 1 and the formula of the compound is AB.

A compund form is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?



Barium crystallizes in a body-centered cubic structure in which the cell-edge length is 0.5025 nm. Calculate the shortest distance between neighboring barium atoms in the crystal.

Number of anions (B–) per unit cell = 1

Sol.

r



Example - 29

+

Which of the following lattices has the highest packing efficiency : (a) Primitive cubic (b) body-centered cubic and (c) hexagonal close-packed lattice ?

r

This lies in the range 0.414 – 0.732. Hence, Fe3+ ions will be in the octahedral holes.

Example - 25 The unit cell of a substance has cations A+ at the corners of the unit cell and the anions B– in the center. What is the simplest formula of the substance ?

The radius ratio is

The edge length of unit cell a = 300 pm = 300 × 10–10 cm, density of element  = 5.2 cm–3, mass of the element m = 200 g. Let the molecular mass of the element be M and number of moles be n then we have

n

m m M  M n

For bcc structure, Z = 2. Now using the relation



ZM 3

a  NA

m Z  Zm n  3   n  3 a  NA a  NA   2  200

 (300 10

10 3

)  6.02  10 23  5.2

The required number of atoms will be N = NA × n = 6.02 × 1023 × = 2.85 × 1024

2  200 (300  10

10 3

)  6.02  10 23  5.2

SOLID STATE

25

Example - 31

Sol.

mass, M = 2.7 × 10–2 kg mol–1 ;

A metal with atomic mass 27 has a density of 2.7 g cm–3 and its unit cell has an edge of 4 Å, what is the nature of the crystal lattice for the metal ? Sol.

edge length, a = 405 pm = 405 × 10–12 m = 4.05 × 10–10 m. Applying the relation, for density, we get (using NA = 6.023 × 1023 mol–1)

The density is given by



Z M



a3  NA

Rearranging in terms of Z, and substituting values, we get

Z

It is given that density of the element,  = 2.7 × 103 ; molar

Z M a3  NA

Therefore,

  a 3  N A 2.7  (4  108 )3  6.02 10 23  M 27

Z

  a 3N A M

= 3.85  4 which means that the nature of crystal lattice is facecentered cubic.



Example - 32

This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centered cubic (fcc) or cubic close-packed (ccp). Example - 34

Edge length of unit cell a = 289 pm = 289 ×10 –10 cm, molecular mass of chromium M = 52 g mol–1, density of chromium metal  = 7.2 g cm–3. Let the number of atoms be Z. The density can be calculated using the relation



2.7  10 2 kg mol1

= 4.004  4

The density of chromium metal is 7.2 g cm–3. If the unit cell has edge length of 289 pm, determine the type of unit cell. (Atomic mass of Cr = 52 u; NA = 6.02 × 1023). Sol.

2.7 103 kgm 3  ( 4.05 10 10 m)3  6.022 10 23 mol1

Silver crystallizes in a cubic close-packed structure. The radius of a silver atom is 1.44 Å. Calculate the density of Ag. Sol.

Z M

The density is calculated as

3

a NA 

Therefore,

a 3 N A 7.2  (289  1010 )3  6.023 10 23 Z  2 M 52

Z M a3  NA

where for ccp structure, Z = 4 and r = 1.44 Å. The relationship between radius of an atom with the edge length a for ccp crystal structure is r = 0.3535a.

Since the unit cell has 2 atoms, hence it is body-centered cubic (bcc).

a

1.44  4.07Å  4.07  108 cm 0.3535

Example - 33 Substituting these values and using M for Ag = 107.87 g –2

–1

An element with molar mass 2.7 × 10 kg mol forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m–3, what is the nature of the cubic unit cell ?

mol–1, we get



4  107.87 (4.07  10  8)3  6.02  10 23

= 10.63 g cm–3

SOLID STATE

26

Example - 35

Example - 37

A metal (atomic mass = 50) has a body-centered cubic

KBr has fcc structure. The density of KBr is 2.75 g cm–3.

crystal structure. The density of the metal is

Find the distance between K + and Br – .

5.96 g cm–3 . Find the volume of the unit cell

(Atomic mass of Br = 80.0).

23

–1

(NA = 6.023 × 10 atoms mol ). Sol.

Sol.

For KBr which has fcc structure, Z = 4 ; Avogadro’s

Atomic mass of metal M = 50 g mol–1 ; density of the

constant NA = 6.02 × 1023 mol–1 ; density of KBr is

metals is  = 5.96 g cm–3. For body-centered cubic crystal,

 = 2.75 g cm–3 ; the moelcular mass of KBr = 119 g mol–1.

Z =2. Avogadro’s cosntant NA = 6.023 × 1023 atoms mol–1.

Let the distance between K+ and Br– be a. Now using the relation for density, we get

Let the volume of the crystal be V. Now, using relation for density





Z M 3

 a3 

a  NA

Z M Z M 2  50 V  V  NA   N A 5.96  6.023 103

Z M  4  119      N A  2.75  6.02 10 23  1/ 3

 4  119  a    23   2.75  6.02  10 

 2.786  10  23 cm 3

Hence, volume of the unit cell = 2.786 × 10–23 cm3.

 a = 6.6×10–8 cm

In an fcc, the ions/atoms touch each other along the face diagonal.

Example - 36 Iron (II) oxide has a cubic structure and each unit cell has

Face diagonal = 2 a = 2 × 6.6 × 10–8 cm = 9.334 × 10–8 cm

side 5 Å. If the density of the oxide is 4 g cm–3, calculate

The distance between K+ and Br– = 1/2 (face dagonal)

the number of Fe2+ and O2– ions present in each unit

= 1/2 × 9.33 × 10–8 cm = 4.667 × 10–8 cm

cell (molar mass of FeO = 72 g mol –1 ,

Example - 38

NA= 6.02 × 1023 mol–1.) Sol.

A unit cell of sodium chloride has four formula units. The edge length of the unit cell is 0.564 nm. What is the density of sodium chloride ?

Edge length of each unit cell of iron (II) oxide is a = 5 Å = 5 × 10–8 cm. The density of iron (II) oxide is  = 4 g cm–3; the molecular mass of iron (II) oxide is M = 72 g mol–1 ; Avogadro’s constant is NA = 6.02 × 1023 mol–1. Let the

Sol.

No. of atoms in fcc unit cell = 4

number of Fe2+ and O2– ions present in each unit cell be Z.

Mol. wt. of Nacl = 23 + 35.5 = 58.5

Now using the relation for density we get

Length of edge of unit cell = 0.564 × 10–9 m = 0.564 × 10–7 cm

  a3  NA  3 Z  M a  NA Z M

8 3



4  (5 10 )  6.02  10 72

Volume of unit cell = (0.564 × 10–7)3

Density of NaCl =

23

4 = 2.165 g/cm3

2+

Hence the number of Fe unit cell will be Z = 4.

2–

and O ions present in each

4  58.5 (0.564  10

7 3

)  6.023 10 23

SOLID STATE

27

Example - 39

Example - 42

The density of mercury is 13.6g/mL. Calculate approximately the diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of mercury atom.

What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ? Sol.

Sol.

d = 13.6 g/mL At. mass of Hg = 200 = M 1 g atomic wt. of Hg = 200 g = wt. of 6.023 × 1021 No. of atoms present in 1 g of Hg





Example - 43

6.023 10 23  3.0115  1021 200

Volume of 1 atom of Hg or unit cell

What are the two ways by which non-stoichiometric defects due to metal deficiency may occur ? Sol.

1 g of Hg No. of atoms  density 1



13.6  3.0115  10

21

 2.44  10  23 cm 3

Edge length = Diameter of Hg atom Diameter of 1 atom of Hg = (Volume of unit cell)1/3 = (2.44 × 10–23)1/3

When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant. Vacancy defect leads to a decrease in the density of the solid.

Non-stoichiometry implies that either metal or non-metal atoms are present in excess. When non-metal atoms are in excess, two types of defects may arise due to metal deficiency. In the first type of defect, a positive ion is absent from its lattice site and the charge is balanced by adjacent metal (positive) ion having higher oxidation state, for example, FeO, FeS, etc. In the second type of defect, an extra negative ion is present in an interstitial position and the charge balance is maintained by the adjacent metal atom having higher charge.

Example - 44

d = (24.40 × 10–24)1/3 = 2.91 × 10–8 cm = 2.91 Å Ionic solids, which have anionic vacancies due to metal excess defect, develop color. Explain with the help of a suitable example.

Example - 40 Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it. Sol.

When a cation of higher valence is added to an ionic solid as an impurity in it, the cation of higher valence repalces more than one cation of lower valence so as to keep the crystal electrically neutral. As a result, some sites become vacant. For example, when Sr2+ is added to NaCl, each Sr2+ ion replaces two Na+ ions. However, one Sr2+ ion occupies the site of one Na+ ion and the other site remains vacant. Hence, vacancies are introduced.

Sol.

The color develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited. For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions leave their extra electrons at the lattice site and diffuse to the surface to form NaCl with the deposited Na atoms. The electron occupying the anionic lattice site is known as an F-center.

Example - 45

Example - 41 What are the possible type of defects that can occur if a Ca2+ ion replaces a Na+ in a crystal lattice of NaCl ?

Why does Frenkel defect not change the density of AgCl crystals ? Sol. Sol.

Frenkel defect does not change the density of AgCl crystals because in the Frenkel defect, the ions are not removed from the crystal. So there will be no change in the crystal structure, that is, there is no decrease in the number of ions. All the ions are inside the crystal, they are only dislocated.

When a Ca2+ ion replaces a Na+ in a crystal lattice of NaCl, one extra positive charge in introduced. The Na+ vacancy or Cl– interstitial will be formed to balance the charge. Hence, the possible defects that can occur are Schottky defects.

SOLID STATE

28

Example - 46

field and a strong magnetic effect is produced. The ordering of the domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.

Which point defect in its crystal units alters the density of a solid ? Sol.

Schottky defect decreases the density of the solid.

Example - 51

Example - 47 What type of stoichiometric defect is shown by (a) ZnS and (b) AgBr ? Sol.

What is the origin of the magnetic properties of an atom ? Sol.

(a) ZnS shows Frenkel defect. (b) AgBr shows Frenkel as well as Schottky defects.

Example - 48 Mention one property which is caused due to the presence of F-center in a solid. Sol.

Example - 52

One property which is caused due to the presence of Fcenter in a solid is appearance of color.

A Group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?

Example - 49 What other element may be added to silicon to make electrons available for conduction of an electric current ? Sol.

Silicon is used to create most semiconductors commercially. To make electrons available for conduction of an electric current, we add Group 15 elements (phosphorus, arsenic, antimony) to silicon which is tetravalent by nature. With the addition of each Group 15 element, an excess electron is created which results in conduction of the current.

Sol.

Sol.

Ferromangetic substances would make better permanent magnets. In solid state, the metal ions of ferromagnetic substansces are grouped together into small regions. These regions are called domains and each domain acts as a tiny magnet. In an unmagnified piece of a ferromagnetic substance, the domains are randomly oriented. As a result, the magnetic moments of the domain get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic

An n-type semiconductor conducts electricity because of the presence of extra electrons. Therefore, a Group 14 element can be converted to n-type semiconductor by doping it with a Group 15 element.

Example - 53 Why does germanium act as an n-type semiconductor ? Sol.

Example - 50 What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

The origin of magnetic properties of an atom lies in the magnetic moment that arises from two types of motion of the electrons in the atom. An electron spins around its axis and moves in its orbital around the nucleus. These motions of the electron (charge particle) can be considered as a loop of current and impart permanent spin and an orbital magnetic moment to it. The magnitude of this magnetic moment is small and measured in the unit called Bohr magneton.

Germanium acts as n-type semiconductor when a pentavalent impurity atom is added to pure germanium crystal. With addition of each impurity atom, one extra electron is created in the crystal lattice of germanium and hence the majority charge carriers are electrons in n-type semiconductors.

Example - 54 Classify each of the following as being either a p-type semiconductor : (a) Ge doped with In ; (b) S doped with B. Sol.

(a) Ge (a Group 14 element) is doped In (a Group 13 element). We know that germanium is tetravalent, and that it is surounded by four other germanium atoms in closepacked structure. Now when an indium atom is added, which is trivalent, a hole is created due to the absence of an electron in lattice site as shown in fig. and the semiconductor generated will be a p-type semiconductor.

SOLID STATE

29 Radius of octahedral void rA = 0.414 × rB = 0.414 × 241.5 pm = 100.0 pm Radius of tetrahedral void

= 0.225 × rB = 0.225 × 241.5 pm = 54.3 pm +

(b) When Si (a Group 14 element) is doped with B (a Grouop 15 element). We know that silicon is tetravalent by nature which means each silicon atom is surrounded by four other silicon atoms. When one boron atom (pentavalent in nature) is added to silicon atoms, then one extra electron is created in crystal lattice and the semiconductor generated will be called as n-type semiconductor.

As the radius of the cation C (50 pm) is smaller than the size of the tetrahedral void, it can be placed into the tetrahedral void (but not exactly fitted into it). Example - 57 Predict the structure of MgO crystal and coordination number of its cation in which cation and anion radii are equal to 65 pm and 140 pm, respectively. Sol.

Radius of the cation is r+ = 65 pm and radius of anion is r– = 140 pm. Therefore, the radius ratio is

Example - 55

r

Where are in the periodic table the atoms that are most likely to be involved in the formation of semiconductors located ?

r





65 pm  0.464 140 pm

Since the radius ratio (0.464) is in between 0.414 and 0.732, Sol.

Semiconductors are typically formed by elements of Group 13–16, for example B, Si, Al, Ge.

Example - 56 A solid A+B– has NaCl type close-packed structure. If the anion has a radius of 241.5 pm, what should be the ideal radius of the cation ? Can a cation C+ having radius of 50 pm be fitted into the tetrahedral hole of the crystal A+B– ? Sol.

As A+B– has NaCl structure, A+ ions will be present in the octahedral voids. Ideal radius of the cation will be equal to the radius of the octahedral void because in that case, it will touch the anions and the arrangement will be closepacked. Hence,

Mg2+ ions occupy octahedral voids. Therefore, the coordination number of the cation is 6. Example - 58 Agl crystallizes in cubic close-packed ZnS structure. What fraction of tetrahedral sites is occupied by Ag+ ions ? Sol.

In the face-centered unit cell of Agl, there are four I– ions. As there are four I– ions in the packing, therefore there are eight tetrahedral voids. Of these, half are occupied by silver cations.

SOLID STATE

30

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Characterstic Properties of Solid 1.

Amorphous substances show

Crystal Lattice and Unit Cell 9.

Glass is

(1) Short and long range order (2) Short range order (3) Long range order 10.

(4) Have no sharp M.P.

2.

(a) 1 and 3 are correct

(b) 2 and 3 are correct

(c) 3 and 4 are correct

(d) 2 and 4 are correct

Amorphous solids are (a) Solid substance in real sense

11.

(b) Liquid in real sense (c) Super cooled liquid (d) Substance with definite melting point 3.

4.

12.

Crystalline solids are (a) Glass

(b) Rubber

(c) Plastic

(d) Sugar

A crystalline solid (a) Changes abruptly from solid to liquid when heated (b) Has no definite melting point

13.

(c) Undergoes deformation of its geometry easily (d) Has an irregular 3-dimensional arrangements 5.

6.

7.

8.

Which of the following is a molecular crystal

(a) Microcrystalline solid

(b) Super cooled liquid

(c) Gel

(d) Polymeric mixture

The number of spheres contained (i) in one body centred cubic unit cell and (ii) in one face centred cubic unit cell, is (a) In (i) 2 and in (ii) 4

(b) In (i) 3 and in (ii) 2

(c) In (i) 4 and in (ii) 2

(d) In (i) 2 and in (ii) 3

The number of atoms or molecules contained in one body centered cubic unit cell is (a) 1

(b) 2

(c) 4

(d) 6

An alloy of Cu, Ag and Au is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centre and gold is present at body centre, the alloy has a formula (a) Cu4 Ag2 Au

(b) Cu4 Ag2 Au

(c) Cu4 Ag3 Au

(d) CuAgAu

A compound is formed by elements A and B. This crystallizes in the cubic structure when atoms A are the corners of the cube and atoms B are at the centre of the body. The simplest formula of the compounds is

(a) SiC

(b) NaCl

(a) AB

(b) AB2

(c) Graphite

(d) Ice

(c) A2B

(d) AB2

Solid CO2 is an example of

14.

(a) Molecular crystal

(b) Ionic crystal

(c) Covalent crystal

(d) Metallic crystal

In graphite, carbon atoms are joined together due to (a) Ionic bonding

(b) Vander Waal's forces

(c) Metallic bonding

(d) Covalent bonding

Mostly crystals show good cleavage because their atoms, ions or molecules are (a) Weakly bonded together (b) Strongly bonded together (c) Spherically symmetrical (d) Arranged in planes

15.

In a face centred cubic arrangement of A and B atoms when A atoms are at the corner of the unit cell and B atoms of the face centres. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is (a) A7B3

(b) AB3

(c) A7B24

(d) A7/8B3

In a cubic structure of compound which is made from X and Y, where X atoms are at the corners of the cube and Y at the face centres of the cube. The molecular formula of the compound is (a) X2Y

(b) X3Y

(c) XY2

(d) XY3

SOLID STATE 16.

17.

18.

19.

20.

21.

22.

31

Number of atoms in the unit cell of Na (bcc type crystal) and Mg (fcc type crystal) are, respectively (a) 4, 4

(b) 4, 2

(c) 2, 4

(d) 1, 1

An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compounds would be (a) A3B

(b) AB3

(c) A2B

(d) AB

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is (a) A2B

(b) AB2

(c) A2B2

(d) A2B5

Packing and Voids 23.

24.

25.

The two dimensional coordination number of a molecule in square close packed layer is (a) 4 (b) 2 (c) 3 (d) 6 Which of the following statements is not true about the voids? (a) Octahedral void is formed at the centre of six spheres which lie at the apices of a regular octahedron. (b) There is one octahedral site for each sphere. (c) There are two tetrahedral sites for each sphere. (d) Octahedral voids are formed when the triangular voids in second layer exactly overlap with similar voids in the first layer. A solid A+B– has the B– ions arranged as below. If the A+ ions occupy half of the octahedral sites in the structure. The formula of solid is

If the number of atoms per unit in a crystal is 2, the structure of crystal is (a) Octahedral

(b) Body centred cubic

(c) Face centred cubic

(d) Simple cubic

26.

Percentage of free space in cubic close packed structure and in body centred packed structure are respectively (a) 30 % and 26%

(b) 26% and 32%

(c) 32% and 48%

(d) 48% and 26%

(a) AB

(b) AB2

(c) A2B

(d) A3B4

For the structure of solid given below if the lattice points represent A+ ions & the B– ions occupy all the tetrahedral voids then coordination number of A is

In fcc lattice, the neighbouring number of atoms for any lattice point is (a) 6

(b) 8

(c) 12

(d) 14

A crystal plane makes intercepts 2a, 2b and 3c on the crystallographic axes where a, b, c represent the intercepts of the unit plane. The Miller indices of the plane will be (a) 2,2,3

(c)

1 1 1 , , 3 3 2

(b) 3,3,2

(d)

1 1 1 , , 2 2 3

27.

(a) 2

(b) 4

(c) 6

(d) 8

The total volume of atoms present in face-centred cubic unit cell of a metal is (r is atomic radius) (a)

20r 3 3

(b)

24r 3 3

(c)

12r 3 3

(d)

16r 3 3

32

SOLID STATE 28.

29.

30.

31.

32.

Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ?

36.

(a) 109 pm

(b) 127 pm

(a) 0.732 r

(b) 0.414 r

(c) 157 pm

(d) 181 pm

(c) 0.225 r

(d) 0.155 r

The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is (a) 288 pm

(b) 398 pm

(c) 618 pm

(d) 144 pm

37.

38.

(a) n

(b) n/2

(c) n/4

(d) 2n

34.

35.

In NaCl lattice the radius ratio is

rNa  rCl 

(a) 0.225

(b) 0.115

(c) 0.5414

(d) 0.471

=

For an ionic crystal of the general formula AX and coordination number 6, the value of radius ratio will be (a) Greater than 0.73

In ABC ABC packing if the numberof atoms in the unit cell is n then the number of tetrahedral voids in the unit cell is equal to

(b) In between 0.73 and 0.41 (c) In between 0.41 and 0.22 (d) Less than 0.22 39.

In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be (a) X4Y3

(b) X2Y3

(c) X2Y

(d) X3Y4

40.

The unit cell of a NaCl lattice (a) Is body centred cube

(b) Has 3 Na+ ions

(c) Has 4 NaCl units

(d) Is electrically charged

Which of the following statements is not true about NaCl structure (a) Cl– ions are in fcc arrangement

In ccp arrangement the pattern of successive layers can be designated as

(b) Na+ ions has coordination number 4

(a) AB AB AB

(b) ABC ABC ABC

(d) Each unit cell contains 4NaCl molecules

(c) AB ABC AB

(d) ABA ABA ABA

(c) Cl– ions has coordination number 6 41.

In CsCl structure, the coordination number of Cs+ is (a) Equal to that of Cl–, that is 6

Ionic Structures 33.

The maximum radius of sphere that can be fitted in the tetrahedral hole of cubical closed packing of sphere of radius r is.

(b) Equal to that of Cl–, that is 8

In which compound 4 : 4 coordination is found (a) ZnS

(b) CuCl

(c) AgI

(d) All of these

In A+ B– ionic compound, radii of A+ and B– ions are 1pm and 187 pm respectively. The crystal structure of this compound will be

(c) Not equal to that of Cl–, that is 6 (d) Not equal to that of Cl–, that is 8

Defects 42.

What type of crystal defect is indicated in the diagram shown below ?

(a) NaCl type

(b) CsCl type

Na+ Cl– Na+ Cl– Na+ Cl–

(c) ZnS type

(d) Similar to diamond

Cl– Cl– Na+ Na+

 rc If the value of ionic radius ratio   ra

  is 0.52 in an ionic  

compound, the geometrical arrangement of ions in crystal is (a) Tetrahedral

(b) Planar

(c) Octahedral

(d) Pyramidal

Na+ Cl– Cl–, Na+ Cl– Cl– Na+ Cl– Na+ Na+ (a) Frenkel defect (b) Schottky defect (c) Interstitial defect (d) Frenkel and Schottky defects

SOLID STATE 43.

33

Schottky defect in crystals is observed when

50.

(a) Density of crystal is increased

(a) Electron are held in the voids of crystals

(b) Unequal number of cations and anions are missing from the lattice

(b) F-centre produces colour to the crystals (c) Conductivity of the crystal increases due to F- centre

(c) An ion leaves its normal site and occupies an interstitial site (d) Equal number of cations and anions are missing from the lattice 44.

The correct statement regarding F-centre is

(d) All of these 51.

Schottky defect in crystals is observed when (a) an ion leaves its normal site and occupies an interstitial site

If a electron is present in place of anion in a crystal lattice, then it is called (a) Frenkel defect

(b) Schottky defect

(c) Interstitial defect

(d) F-centre

(b) few cations are missing from the crystal (c) equal number of cations and anions are missing from the crystal

52.

Zinc oxide loses oxygen on heating according to the reaction,

(d) few anions are missing from the crystal. 45.

46.

1 heat ZnO   Zn2   O2  2e 2

Which defect causes decrease in the density of crystal (a) Frenkel

(b) Schottky

It becomes yellow on heating because

(c) Interstitial

(d) F-centre

In a Schottky defect,

(a) Zn ions and electrons move to interstitial sites and Fcentres are created

(a) an ion moves to interstitial position between the lattice points

(b) Oxygen and electrons move out of the crystal and ions become yellow

2+

(b) electrons are trapped in a lattice site

2+

(c) Zn again combine with oxygen to give yellow oxide

(c) some lattice sites are vacant 2+

(d) Zn are replaced by oxygen

(d) some extra cations are present in interstitial spaces. 47.

48.

In a solid lattice the cation has left a lattice site and is located at an interstitial position, the lattice defect is (a) Interstitial defect

(b) Valency defect

The substance which possesses zero resistance at zero kelvin is called

(c) Frenkel defect

(d) Schottky defect

(a) Conductor

(b) Superconductor

(c) Insulator

(d) Semiconductor.

53.

Alkali halides do not show Frenkel defect because (a) cations and anions have almost equal size

54.

Which substance acts as superconductor at 4 K?

(b) there is a large difference in size of cations and anions.

(a) Hg

(c) Cu

(c) cations and anions have low coordination number

(c) Na

(d) Mg

(d) anions cannot be accommodated in voids. 49.

Magnetic and Electrical Properties

55.

A solid with high electrical and thermal conductivity from the following is

What is the effect of Frenkel defect on the density of ionic solids?

(a) Si

(b) Li

(a) The density of the crystal increases.

(c) NaCl

(d) Ice

(b) The density of the crystal decreases.

56.

Which substance shows antiferromagnetism?

(c) The denisty of the crystal remains unchanged.

(a) ZrO2

(b) CdO

(d) There is no relationship between density of a crystal and defect present in it.

(c) CrO2

(d) Mn2O3

34

SOLID STATE 57.

58.

59.

60.

61.

Which one among the following is an example of ferroelectric substance? (a) Quartz

(b) lead chromate

(c) Barium titanate

(d) Rochelle salt.

The electricity produced on applying stress on the crystals is called (a) Pyroelectricity

(b) Piezoelectricity

(c) Ferroelectricity

(d) Anti-ferroelectricity

Crystals where dipoles may align themselves in an ordered manner so that there is a net dipole moment, exhibit (a) Pyro electricity

(b) Piezo electricity

(c) Ferro electricity

(d) Antiferro electricity

On heating some polar crystals, weak electric current is produced. It is termed as : (a) Piezo electricity

(b) Pyro electricity

(c) Photoelectric current

(d) Superconductivity

65.

Lithium forms body-centred cubic structure. The length of the side of its unit cell is 351 pm. What will be the atomic radius of lithium?

66.

A metal crystallises into a lattice containing a sequence of layer as AB AB AB ....... . What percentage of voids are left in the lattice?

67.

Sodium metal crystallizes as a body centred cubic lattice with the cell edge 4.29Å. What is the radius of sodium atom?

68.

Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. What will be its density?

69.

The unit cell of aluminum is a cube with an edge length of –3 405 pm. The density of aluminium is 2.70 g cm . What is the structure of unit cell of aluminium?

70.

The number of atoms in 100 g of an FCC crystal with density –3 d = 10 g cm and cell edge as 200 pm is equal to?

71.

Copper crystallizes in face centred cubic structure. The edge length of a unit cell is found to be 3.61108 cm.

Pure silicon and germanimum behave as (a) Conductors

(b) Semiconductors

(c) Insulators

(d) Piezoelectric crystals.

Calculate the density of copper it the molar mass of copper is 63.5 g m ol  1 . 72.

How many lithium atoms are present in a unit cell with –3 edge length 3.5 Å and density 0.53 gcm ? (Atomic mass of Li = 6.94)

73.

An element with atomic mass 100 has a bcc structure and

Numerics 62.

Experimentally it was found that a metal oxide has formula M0.98O. Mental M, present as M2+ and M3+ in its oxide. 3+

Fraction of the metal which exists as M would be 63.

How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g ? (Atomic mass : Na = 23, Cl = 35.5)

64.

Copper crystallises in fcc lattice with a unit cell edge of 361 pm. What will be the radius of copper atom?

edge length 400 pm. What will be its density? 74.

A unit cell of sodium chloride has four formula units. The edge length of unit cell is 0.564 nm. What is the density of sodium chloride?

SOLID STATE

35

EXERCISE - 2 : PREVIOUS YEARS JEE MAIN QUESTIONS 1.

Sodium metal crystallizes in a body centred cubic lattice

8.

with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately: (a) 5.72 Å 2.

3.

(b) 0.93 Å

(c) 1.86 Å (d) 3.22 Å Which of the following exists as covalent crystals in the solid state? Online 2015 SET (1) (a) Silicon (b) Sulphur (c) Phosphorous (d) Iodine An element (atomic mass 100g/mol) having bcc structure has until cells edge 400 pm. Then density of the element is : Online 2016 SET (1) 3 (a) 10.376 g/cm (b) 5.188 g/cm3 (c) 7.289 g/cm3

4.

(2015)

9.

Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is : (08-04-2019) (a) A2B2O

(b) AB2O4

(c) A4B2O

(d) A4B2O

Consider the bcc unit cells of the solid 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2? (08-04-2019)

(d) 2.144 g/cm3

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be : (a) 2 2a (c)

5.

6.

a 2

(b)

(2017)

2a

(d) 2a

Which type of 'defect' has the presence of cations in the interstitial sites ? (2018) (a) Frenkel defect (b) Metal deficiency defect (c) Schottky defect (d) Vacancy defect Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance ? Online 2018 SET (1)

10.

(a) 90%

(b) 75%

(c) 65%

(d) 45%

The statement that is INCORRECT about the interstitial compound is: (08-04-2019) (a) they have metallic conductivity (b) they have high melting points (c) they are chemically reactive (d) they are very hard

11.

An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is: (12-04-2019)

(a) (b) (c) (d) 7.

12.

All of the following share the same crystal structure except : Online 2018 SET (2) (a) LiCl

(b) NaCl

(c) RbCl

(d) CsCl

(a)

a 2

(b)

(c)

3 a 2

(d) a

2a

The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively : (12-04-2019) (a) 1 : 2 : 4

(b) 4 : 2 : 3

(c) 4 : 2 : 1

(d) 8 : 1 : 6

36

SOLID STATE 13.

At 100C , copper(Cu) has FCC unit cell structure with

19.

0

cell edge length x A . What is the approximate density of Cu(in g cm3 ) at this temperature? [Atomic mass of Cu = 63.55 u] (a)

205 x3

(b)

(09-01-2019)

105 x3

211 422 (d) 3 x3 x Which primitive unit cell has unequal edge length

20.

(c) 14.

(a  b  c) and all axial angles different from 90 ? (10-01-2019) (a) Triclinic (b) Hexagonal (c) Monoclinic (d) Tetragonal 15.

A compound of formula A 2 B3 has the hcp lattice. Which

(c) 2a 21.

atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms : (10-01-2019)

(d) hcp lattice  B, 16.

1 Tetrahedral voids – A 3

A solid having density of 9  103 kg m 3 forms face centred

a 2

A crystal is made up of metal iron ' M1 ' and ' M 2 ' and oxide ions. Oxide ions form a ccp lattice structure. The cation

' M 2 ' occupies 12.5% of tetrahedral voids of oxide lattice.

The oxidation number of ' M1 ' and ' M 2 ' are, respectively: (06-Sept-2020) (a) +2, +4 (b) +3, +1 (c) +4, +2 (d) +1, +3

1 (b) hcp lattice  A, Tetrahedral voids - B 3 2 Tetrahedral voids - A 3

(d)

' M1 ' occupies 50% of octahedral voids and the cation

2 (a) hcp lattice  A, Tetrahedral voids - B 3

(c) hcp lattice  B,

A diatomic molecule X2 has a body-centred cubic (bcc) structure with a cell edge of 300 pm. The density of the molecule is 6.17 g cm–3. The number of molecules present in 200 g of X 2 is : (Avogadro constant (N A) = 6 × 10 23 mol –1) (05-Sept-2020) (a) 8 NA (b) 2 NA (c) 40 NA (d) 4 NA An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the cyrstal lattice is : (05-Sept-2020) a (a) a (b) 2

22.

Two liquids of density 1 and  2  2  2 1  are filled up behind a square wall of side 10 as shown in figure. Each liquid has a height of 5 . The ratio of forces due to these liquids exerted on the upper part MN to that at the lower part NO is (Assume that the liquids are not mixing)

cubic crystals of edge length 200 2 pm. What is the molar mass of the solid? (11-01-2019) [Avogadro constant  6  10 23 and mol1 ,   3 ]

17.

18.

(a) 0.0432 kg mol–1 (b) 0.0216 kg mol–1 –1 (c) 0.0305 kg mol (d) 0.4320 kg mol–1 The radius of the largest sphere which fits properly at the centre of the edge of a body centered cubic unit cell is: (Edge length is represented by ‘a’) (11-01-2019) (a) 0.0027a (b) 0.047 a (c) 00137a (c) 0.07a An element with molar mass 2.7  102 kg mol1 forms a cubic unit cell with edge length 405 pm. If its density is

2.7  103 kg m 3 , the radius of the element is approximately ………….. × 10–12 m (to the nearest integer). (03-Sept-2020)

(08-Jan-2020) (a)

2 3

(b)

1 2

(c)

1 4

(d)

1 3

SOLID STATE

37

EXERCISE - 3 ADVANCE OBJECTIVE QUESTIONS Single Choice Questions 1.

2.

3.

4.

5.

8.

Which of the following is an example of covalent crystal solid (a) Si

(b) NaF

(c) Al

(d) Ar

9.

Which of the following is an example of metallic crystal solid (a) C

(b) Si

(c) W

(d) AgCl

Which of the following is an example of ionic crystal solid (a) Diamond

(b) LiF

(c) Li

(d) Silicon

10.

(b) 4

(c) 8

(d) 2

(a) 17.7%

(b) 7.84%

(c) 11.5%

(d) 9.6%

Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its density (in kg m–3) will be (a) 454

(b) 804

(c) 852

(d) 911 –3

The density of KBr is 2.75 g cm . The length of the unit cell is 654 pm. Atomic mass of K = 39, Br = 80. Then the solid is (a) Face centred cubic (b) Simple cubic system

What is the coordination number of sodium in Na2O (a) 6

The percentage of iron present as Fe(III) in Fe0.93O1.0 is

(c) Body centred cubic system (d) None of these

For a solid with the following structure, the co-ordination number of the point B is

11.

In octahedral holes (voids) (a) A simple triangular void surrounded by four spheres (b) A bi-triangular void surrounded by four spheres (c) A bi-triangular void surrounded by six spheres (d) A bi-triangular void surrounded by eight spheres

12.

6.

(a) 3

(b) 4

(c) 5

(d) 6

If an atom is present in the centre of the cube, the participation of that atom per unit cell is

1 (a) 4 (c) 7.

For the structure given below the site marked as S is a

1 2

(b) 1 13. (d)

1 8

A metallic element has a cubic lattice. Each edge of the unit cell is 2 Å. The density of the metal is 2.5 g cm–3 . The unit cells in 200 g of the metal are (a) 1 × 1025

(b) 1 × 1024

(c) 1 × 1022

(d) 1 × 1020

14.

(a) Tetrahedral void

(b) Cubic void

(c) Octahedral void

(d) None of these

The number of octahedral voids in a unit cell of a cubical closest packed structure is (a) 1

(b) 2

(c) 4

(d) 8

The coordination number of a metal crystallizing in a hexagonal close packed structure is (a) 4

(b) 12

(c) 8

(d) 6

38

SOLID STATE 15.

16.

17.

18.

The number of tetrahedral voids in a unit cell of cubical closest packed structure is

1. Density of NaCl crystal assuming all sites are occupied = 2.178 × 103 kg m–3

(a) 1

(b) 2

(c) 4

(d) 8

2. Density of NaCl crystal by not considering the unoccupied sites but only the occupied sites = 2.165 × 103 kg m–3.

A binary solid x+ y– has a zinc blende structure with y– ions forming the lattice and x+ ion occupying 25% tetrahedral sites. The formula of the solid is (a) xy

(b) x2y

(c) xy2

(d) xy4

The percentage of unoccupied sites in NaCl crystal is (a) 5.96 × 10–2 (c) 5.96

20.

(a) 0.205 nm

(b) 0.290 nm

(a) 3.01 × 10

(c) 0.145 nm

(d) 0.578 nm

(c) 9.03 × 10

The following structure drawn is of

25.

23.

23

(b) 6.02 × 10

23

23

(d) 12.04 × 10

23

The structure of MgO is similar to NaCl. What would be the coordination number of magnesium? (a) 2

(b) 4

(c) 6

(d) 8

(a) Fluorite

(b) Caesium chloride

In the crystals, which of the following ionic compounds would you expect maximum distance between centres of cations and anions?

(c) Wurtzite

(d) Zinc blende

(a) LiF

(b) CsF

(c) CsI

(d) LiI

The number of formula unit in unit cell of CsCl, type is (a) 1

(b) 2

(c) 3

(d) 4

27.

The interionic distance for caesium chloride crystal will be

(c)

22.

+

A metal crystallizes in a face-centered unti cell. Its edge length is 0.410 nm. The radius of the metal is

(a) a

21.

(d) 8.68

3+

If Al ions replace Na ions at the edge centres of NaCl lattice, then the number of vacancies in 1 mole of NaCl will be

24.

26.

19.

(b) 5.96 × 10–1

(b)

3a 2

a 2

A solid A+ B– has a body centred cubic structure. The distance of closest approach between the two ions is 0.767 Å. The edge length of the unit cell is

(a)

3

pm

(b) 142.2 pm

2 pm

(d) 88.56 pm.

2

2a (d)

If the coordination number of Ca2+ in CaF2 is 8, then the coordination number of F– ion would be (a) 3

(b) 4

(c) 6

(d) 8

(c)

3 28.

(a) Four Fe2+ and four O2– (b) Two Fe2+ and four O2–

Potassium fluoride has NaCl type structure. What is the distance between K+ and F– ions if cell edge is a cm? (a) 2 a cm

(b) a/2 cm

(c) 4a cm

(d) a/4 cm

A sample of electrically neutral NaCl crystal is analysed for its density which has some unoccupied sites. Two readings where taken.

Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Å. Assuming density of the oxide as 4.0 g cm–3, then the number of Fe2+ and O2– ions present in each unit cell will be

(c) Four Fe2+ and two O2– 29.

(d) Three Fe2+ and three O2–

Which one of the following defects does not affect the density of the crystal ? (a) Schottky defect

(b) Interstitial defect

(c) Frenkel defect

(d) Both b and c

SOLID STATE 30.

39

Frenkel defect is caused due to

39.

(a)\ An ion missing from the normal lattice site creating a vacancy (b) An extra positive ion occupying an interstitial position in the lattice (c) An extra negative ion occupying an interstitial position in the lattice

(d) It is as closely packed as body-centered cubic packing 40.

(d) The shift of a positive ion from its normal lattice site to an interstitial site 31.

(d) The edge length of the crystal A+B– is equal to the

(c) Cation vacancies only

distance between A+ and B– ions.

An oxide of transition metal that has conductivity as well as appearance like that of copper is (a) Ti2O3 (b) V2O3 (c) ReO3

34.

(c) The shape of the octahedral void is octahedral.

(b) Anion vacancies and interstitial anions (d) Cation vacancies and interstitial cations

33.

(d) Mn2O3

41.

42.

36.

37.

38.

Which of the following are true ?

(b) In ZnS (zinc blende), Zn2+ ions are present in alternate tetrahedral voids. (c) In CaF2, F– ions occupy all the octahedral voids. (d) In Na2O, O2– ions occupy half of the octahedral voids. 43.

Graphite is

The coordination number of 8 for cation is found in (a) CsCl (b) NaCl (c) CaF2

(a) a good conductor

(b) sp3 hybridized

(c) an amorphous solid

(d) soft and slippery

(d) calcium fluoride

(a) In NaCl crystals, Na+ ions are present in all the octahedral voids.

The oxide that is insulator is (a) VO (b) CoO (c) ReO3 (d) Ti2O3 Which of the following is ferromagnetic? (a) Ni (b) Co (c) Fe3O4 (d) All are correct.

The coordination number of anion is 4n (a) sodium chloride (b) zinc chloride (c) caesium chloride

Multiple Choice Questions 35.

Which of the following statements are true ? (a) An element with bcc structure has two atoms per unit cell. (b) An ionic compound A+B– with bcc structure has one AB formula unit per unit cell.

Ionic solids, with Schottky defects, contain in their structure (a) Equal number of cation and anion vacancies

32.

Which of the following are not true about hexagonal close packing ? (a) It has a coordination number of 6. (b) It has 26% empty space. (c) It has ABAB..... type of arrangement

(d) Na2O

(a) triclinic

(b) orthorhombic

Which of the following statement are correct ? (a) The coordination number each type of ion in CsCl crystal is 8. (b) A metal that crystallizes in bcc structure has coordination number of 12. (c) A unit cell of an ionic crystal shares some of its ions with other unit cells. (d) The length of the edge of unit cell of NaCl is 552 pm

(c) monoclinic

(d) tetragonal

( rNa  = 95 pm, rCl = 181 pm)

44.

Diamond is (a) a covalent solid

(b) non-conductor

(c) soft and slippery

(d) sp2 hybridized

Crystal systems in which no two axial lengths are equal are

Which of the following systems do not give correct description of axial lengths and axial angles ? (a) Hexagonal : a = b  c,     90º ,   120º (b) Trigonal : a = b  c,     90º ,   90º (c) Monoclinic : a  b  c,       90º (d) Cubic : a = b  c,       90º

45.

The correct statement(s) regarding defects in solids is (are) (a) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion. (b) Frenkel defect is a dislocation defect (c) Trapping of an electron in the lattice leads to the formation of F-center (d) Schottky defects have no effect on the physical properties of solids.

40

SOLID STATE 46.

Which of the following statements are not true ?

51.

(a) The conductivity of metals increases with increase in temperature.

(a) 407 pm.

(b) 189 pm.

(b) Electrical conductivity of semiconductors increases with increase in temperature.

(c) 814 pm.

(d) 204 pm.

(c) A diode is a combination of conductor and a semiconductor.

52.

(d) V2O5 behave as an insulator.

COMPREHENSION BASED QUESTION Comprehension



48.

49.

An element X crystallizes in a structure having an fcc unit cell of an edge 100 pm. If 24 g of the element contains 24 × 1023 atoms, the density is (c) 4 g cm–3

(d) 24 g cm–3

The number of atoms present in 100 g of a bcc crystal (density = 12.5 g cm–3) having cell edge 200 pm is (a) 1 × 1025

(b) 1 × 1024

(c) 2 × 1024

(d) 2 × 1026

A metal A (atomic mass = 60) has a body - centered cubic crystal structure. The density of the metal is 4.2 g cm–3. The volume of unit cell is (a) 8.2 × 10–23 cm3

(b) 4.72 × 10–23 cm3

(c) 3.86 × 10–23 cm3

(d) None of these

53.

Comprehension X-ray studies show that the packing of atoms in a crystal of a metal is found to be in layer such that starting from any layer, every fourth layer is exactly identical. The density of the metal is found to be 19.4 g cm–3 and its atomic mass is 197 u. 50.

The fraction occupied by metal atoms in the crystal is (a) 0.52

(b) 0.68

(c) 0.74

(d) 1.0

(b) 143.9 pm.

(c) 146.5 pm.

(d) 267.8 pm.

Each point in a crystal lattice represents a constituent particle which can be an atom, ion or a molecule. The smallest repeating subunit in the lattice is known as unit cell. The unit cells are described as simple (points at all the corners), body-centered (points at all the corners and in the center), face-centered (points at all the corners and center of all faces), and end-centered (points at all the corners and centers of two opposite end faces) unit cells. For the stable ionic crystalline structures, the radius ratio limit for a cation to fit perfectly in the lattice of anions is determined by the radius ratio rule. This also defines the number of nearest neighbors of opposite charges surrounding the ion, which is known as the coordination number. This depends upon the ratio of radii of two types of ions, r+/r–. The radius ratios for coordination numbers 3, 4, 6 and 8 are 0.155 – 0.225–0.414, 0.414–0.732 and 0.732–1, respectively. The coordination number of ionic solids increases with increase in pressure and decreases with increase in temperature.

a3  NA

(b) 40 g cm–3

(a) 103.5 pm.

Crystalline solids have an ordered internal arrangement of atoms due to which they exhibit symmetry. The atoms are arranged in a symmetrical fasion in the three-dimensional network known as space lattice. Each point in a crystal lattice is known as lattice point or lattice site.

Z M

(a) 2.40 g cm–3

Assuming the metal atom to be spherical, its radius will be

Comprehension

Density of a unit cell is the same as the density of the substances. So, if the density of the substance is known, we can calculate the number of atoms or dimesions of the unit cell. The density of the unit cell is related to its mass (M), number of atoms per unit cell (Z), edge length (a in cm), and Avogadro’s constant NA as :

47.

The length of the edge of the unit cell will be

54.

In a cubic lattice of ABC, A atoms are present at all corners except one corner which is occupied by B atoms. C atoms are present at face centers. The formula of the compound is (a) A8BC24

(b) ABC3

(c) A7B24C

(d) A7BC24

The ionic radii of Na+, K+ and Br– are 137, 148 and 195 pm, respectively. The coordinaiton number of cation in KBr and NaBr structures are, respectively, (a) 8, 6

(b) 6, 4

(c) 6, 8

(d) 4, 6

SOLID STATE

41 59.

Comprehension Crystals as such are never perfect and in general have some defect, which simply refers to a disruption in the periodic order of a crystalline material. Based on their nature, these are classified into three types : stoichiometric, impurity and non-stoichiometric defects. In non-stoichiometric defects, the ratio of the number of atoms of one kind to the number of atoms of the other kind does not correspond exactly to the ideal whole number ratio implied by the formula. For example, the ideal formula of ferrous oxide shoul dbe FeO, but in actually found to be Fe0.93O in a sample, because some ferric ions have replaced ferrous ions in the crystal. These defects affect the properties of the crystals. Sometimes such defects are introduced to improve the properties of the crystals for a specific use, such as in the field of electronics. Elements of Group 14 are commonly doped with impurities of elements of Group 13 or 15. In ionic crystals, impurities are introduced with cations having higher valence than the cations in the pure crystal structure. 55.

60.

61.

62.

Which one of the following doping will produce p-type semi-conductor ? (a) Silicon doped with arsenic (b) Germanium doped with phosphorus (c) Germanium doped with aluminium

63.

(d) Silicon doped with phosphorus 56.

NaCl was doped with 10–3 mol % SrCl2. The concentration of cation vacancies is (a) 6.02 × 1018 mol–1

(b) 6.02 × 1015 mol–1

(c) 6.02 × 1021 mol–1

(d) 6.02 × 1012 mol–1

58.

(a) A

(b) B

(c) C

(d) D

(d) D –

Assertion : In CaF2, F ions occupy all the tetrahedral sites.

65.

Reason : The number of Ca2+ is double the number of F– ions. (a) A (b) B (c) C (d) D Assertion : Due to Frenkel defect there is no effect on density of a solid. Reason : Ions shift from lattice sites to interstitial sites in Frenkel defect. (a) A (b) B

Assertion : Crystalline solids are anisotropic. Reason : The constituent particles are very closely packed.

(c) C (d) D Assertion : Size of tetrahedral void is much larger than an octahedral void. Reason : The cations may occupy more space than anions in crystal packing. (a) A (b) B (c) C (d) D Assertion : Zinc blende and wurtzite both have fcc arrangement of sulphide ions. Reason : There are four formula units of ZnS in both. (a) A (b) B

64.

(D) If ASSERTION is false but REASON is true. 57.

(c) C (d) D Assertion : Triclinic system is the most unsymmetrical system. Reason : All axial lengths are different in a triclinic system. (a) A (b) B (c) C (d) D Assertion : The number of tetrahedral voids is double the number of octahedral voids. Reason : The size of the tetrahedral void is half of that of the octahedral void. (a) A (b) B

(c) C

Assertion–Reason Type Questions (A) If both ASSERTION and REASON are true and reason is the correct explanation of the assertion. (B) If both ASSERTION and REASON are true but reason is not the correct explanation of the assertion. (C) If ASSERTION is true but REASON is false.

Assertion : Graphite is an example of tetragonal crystal system. Reason : For a tetragonal system, a = b  c,  =  =  = 90º. (a) A (b) B

(c) C 66.

(d) D

Assertion : In ZnO, the excess Zn2+ ions are present in interstitial sites.

Assertion : Ionic crystals have the highest melting point.

Reason : Metal excess crystals have either missing cation or anion in interstitial site.

Reason : Covalent bonds are stronger than ionic bonds.

(a) A

(b) B

(a) A

(b) B

(c) C

(d) D

(c) C

(d) D

42

SOLID STATE 67.

Assertion : Antiferromagnetic substances become paramagnetic on heating to high temperature.

71.

atoms (r) is related to edge of unit cell (a) as r = 0.3535a. The

Reason : Heating results in spins of electrons becoming random. (a) A

(b) B

(c) C

(d) D

total number of atoms present per unit cell is _____. 72.

unit cell is _____ . 73.

number of A and B per formula unit is ______.

Column II

(A) Molecular solid

(p) Dry ice

(B) Covalent solid

(q) Coppe

(C) Metallic solid

(r) Generally behave as

74.

(s) Generally have low

75.

76.

Column II

Chromium metal crystallizes with a body centred cubic lattice. The length of the unit edge is found to be 287 pm. Calculate the atomic radius. What would be the density of

Match the type of packing with the metal possessing it/space occupied. Column I

Sodium crystallizes in a bcc cubic lattice with the cell edge, a = 4.29 Å. What is the radius of sodium atom ?

melting points 69.

In hexagonal close packing, the difference in the number of tetrahedral and octahedral voids in a unit cell is ____.

insulators (D) Ionic solid

Atoms of element A form hcp arrangement and those of element B occupy 2/3rd of tetrahedral voids. The total

Match the types of solid with their examples/properties. Column I

A cubic unit cell has one atom on each corner and one atom on each body diagonal. The number of atoms in the

Matrix–Match Type Qestions 68.

A metal X crystallizes in a unit cell in which the radius of

chromium in g/cm3 ? 77.

Al3+ ions replace Na+ ions at the edge centers of NaCl

(A) Body-centered cubic (bcc)

(p) Iron

lattice. The number of vacancies in one mole Nacl is

(B) Hexagonal cubic

(q) 52%

calculated as n × 1023. The value of n approximately is

(C) Cubic close packing

(r) 68%

(D) Simple cubic

(s) 74%

Integer Type Questions

__. 78.

The radius ratio of an ionic solid r+/r– is 0.524. The coordination number of this type of structure is ____.

70.

A metal crystllaizes into two cubic phases, face centred cubic (fcc) and body centred cubic (bcc), whose unit cell

79.

NH4+ and Br– ions have ionic radii of 139 pm and 186 pm,

lengths are 3.5 and 3.0 A, respectively, Calculate the ratio

respectively. The coordination number of NH4+ ion in NH4Br

of densities of fcc and bcc.

is _____ .

SOLID STATE

43

EXERCISE - 4 PREVIOUS YEARS JEE ADVANCED QUESTIONS 1.

In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is (2001)

7.

(a) AB2

(b) A2B

8.

(c) A4B3

(d) A3B4

(c) Na2O

Assertion and Reason (A) If both ASSERTION and REASON are true and reason is the correct explanation of the assertion.

9.

If both ASSERTION and REASON are true but reason is not the correct explanation of the assertion. (C) If ASSERTION is true but REASON is false. (D) If ASSERTION is false but REASON is true. 2. Assertion : In any ionic soid (MX) with Schottky defects, the number of positive and negative ions are same. Reason : Equal numbers of cation and anion vacancies are present. (2001) (a) A (b) B 3.

(d) D

A substance AxBy crystallizes in a face centred cubic (fcc) lattice in which atoms ‘A’ occupy each corner of the cube and atoms ‘B’ occupy the centres of each face of the cube. Identify the correct composition of the substance AxBy (2002)

(d) composition cannot be specified (a) Marbles of diameter 10 mm are to be put in a square area of side 40 mm so that their centres are within this area. Find the maximum number of marbles per unit area

6.

(2003)

The crsytal AB (rock salt structure) has molecular weight 6.023 y u. where, y is an arbitrary number in u. If the minimum distance between cation and anion is y1/3 nm and the observed density is 20 kg/m3. Find the (a) density in kg/m3 and (b) type of defect (2004) An element crystallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter of atom which can be placed in interstitial site without distorting the structure. (2005)

crystal system

In hexagonal systems of crystals, a frequently encountered arragenement of atoms is described as a hexagonal prism. Here the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A spacefilling model of this structure, called hexagonal closepacked (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ‘r’. (2008)

(c) A3B

5.

Match the crystal system/unit cells mentioned in ColumnI with their characteristic features mentioned in Column-II. (2007) Column I Column II (A) Simple cubic and (p) have these cell face-centred cubic parameters a = b = c and = =  (B) Cubic and rhombohedral (q) are two crystal systems (C)Cubic and tetragonal (r) have only two crystallography angles of 90º (D) Hexagonal and (s) belong to same

Comprehension (Ques. 10 to 12)

(b) A4B3

(b) Deduce an expression for calculating it.

The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 Å which crystallizes in cubic lattice. If the density is 2 g/cc then find the radius of metal atom. (NA = 6 × 1023). Give the answer in pm. (2006)

monoclinic

(a) AB3

4.

(d) CaF2

Match the Column

(B)

(c) C

Which of the following fcc structure contains cations in alternate tetrahedral voids ? (2005) (a) NaCl (b) ZnS

10.

The number of atoms on this HCP unit cell is : (a) 4

(b) 6

(c) 12

(d) 17

44

SOLID STATE 11.

The volume of this HCP unit cell is : (a) 24 2 r 3

(b) 16

(c) 12 2 r 3

(d)

16. 2r3

64r 3

Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions (X) except the central one

3 3

(ii) Replace all the face centered cations (M) by anions (X) 12.

The empty space in this HCP unit cell is : (a) 74%

(b) 47.6%

(c) 32%

(d) 26%

(iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M)

13.

If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are (2015)

1 1 (a) , 2 8 (c) 14.

1 1 , 2 2

(b) 1,

(d)

1 4

1 1 , 4 8

The value of

 number of anions    in Z is .......... .  number of cations  (2018)

17.

The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s). (2020/Paper 1)

The Correct statement(s) for cubic close packed (ccp) three dimensional structure is (are) (2016) (a) The number of the nearest neighbours of an atom present in the topmost layer is 12 (b) The efficiency of atom packing is 74% (c) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively (d) The unit cell edge length is 2 2 times the radius of the atom

15.

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm–3, then the number of atoms present in 256 g of the crystal is N × 1024. The value of N is (2017)

(a) The empirical formula of the compound in MX. (b) The cation M and anion X have different coordination geometries. (c) The ratio of M-X bond length to the cubic unit cell edge length is 0.866. (d) The ratio of the ionic radii of cation M to anion X is 0.414.

SOLID STATE

45

Note:

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46

02 SOLUTIONS

47

Chapter 02

SOLUTIONS INTRODUCTION

For example, common salt in water.

In normal life we rarely come across pure substances. Most of these are mixtures containing two or more pure substances. Their utility or importance in life depends on their composition. The air around us is a mixture of gases primarily oxygen and nitrogen; the water we drink contains very small amounts of various salts dissolved in it. Our blood is a mixture of different components. Alloys such as brass, bronze, stainless steel, etc. are also mixtures. In this Unit, we will consider mostly liquid solutions and their properties.

1.1 Type of Solutions Solutions which contain two components in it are called Binary Solutions. Substances which are used to prepare a solution are called as Components. The component that is present in the largest quantity is known as Solvent. Solvent determines the physical state in which solution exists.

1. SOLUTIONS

The other component present in lesser quantity in the solution is termed as Solute.

Definition

Each component may be solid, liquid or in gaseous state. A solution is a homogeneous mixture of two or more than two components.

1.2 Concentration term

Mass % of a component  The amount of solute dissolved per unit solution or solvent is called Strength of solution. There are various methods of measuring strength of a solution. : 1.

2.

Mass of component in the sol.  100 Total Mass of sol.

Volume percentage (%v/v): “It represents volume of a component in 100 mL of solution”

Mass Percentage (%w/w): “It represents mass of a component present in 100 g of solution”

Vol. % of a component =

Vol. of component ×100 Total vol. of solution

SCAN CODE SOLUTIONS

SOLUTIONS 3.

48

Mass by volume percentage (%w/v): “It represents mass of solute in grams present in 100 mL of solution”

Mass by vol. percent = 4.

Mass of solute in g 100 Vol. of sol. in mL

8.

Normality, (N) It represents no. of equivalents of solute present in 1 L of solution.

Normality 

No. of Equivalents of solute Vol. of sol. in L

Parts per Million (ppm)

No. of equivalents 

Parts per Million 

No. of parts of the component 106 Total no. of all the componens of sol.

Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume. 5.

“It represents the moles of a solute present in one mole of solution”

No. of moles of the component Total no. of moles all the components

For example, in a binary mixture, if the number of moles of A and B are nA and nB respectively, the mole fraction of A will be

xA  6.

M (z is the valency factor) z

SOME IMPORTANT RELATIONSHIPS

Mole Fraction (x)

Mole fraction 

E

Weight Equivalent weight (W / E)

nA nA  nB

Molarity, (M)

Dilution Law If a solution is diluted by adding solvent to it, then the amount of solute remains constant and we can write: M1V1 = M2V2 and N1V1 = N2V2 Molarity and Normality Normality = z × Molarity NOTE : Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity & normality are a function of temperature. This is because volume depends on temperature and the mass does not.

“It represents moles of solute present in 1 L of solution”

2. VAPOUR PRESSURE OF LIQUID SOLUTION Moles of solute Molarity, M  Vol. of sol. in L Units of Molarity are mol/L also represented by ‘M’ or ‘Molar’. “Density of a solution is mass of the solution per unit volume”

Density, d  7.

Mass of sol.  m/V Vol. of sol.

Molality, m

2.1 Definition Vapour pressure of a liquid/solution is the pressure exerted by the vapours in equilibrium with the liquid/solution at a particular temperature. Vapour pressure  escaping tendency 2.2 Vapour pressure of liquid solutions and Raoult’s Law : (Raoult’s law for volatile solutes)

“It represents moles of solute present per kg of solvent”

Moles of solute Molality  Mass of solvent in kg

Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Units of molality are mol/kg which is also represented by ‘m’ or ‘molal’.

Consider a solution containing two volatile components 1 and 2 with mole fractions x1 and x2 respectively. Suppose

SCAN CODE SOLUTIONS

49

SOLUTIONS at a particular temperature, their partial vapour pressures are 0 1

In general

0 2

p1 and p2 and the vapour pressure in pure state are p and p .

pi  yi p total

Thus, according to Raoult’s Law, for component 1 2.3 Vapour pressures of solutions of solids in liquids

p1  x1 and p1 = p10 x1 Similarly, for component 2 p 2  p 02 x 2

According to Dalton’s law of partial pressure, the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as : ptotal = p1 + p2 Substituting the values of p1 and p2, we get ptotal = x1 p10  x 2 p 02  (1  x 2 ) p10  x 2 p 02  p10  (p 02  p10 ) x 2

and Raoult’s Law



Raoult’s law for non volatile solutes If a non-volatile solute is added to a solvent to give a solution, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapour pressure is also reduced. The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature. Raoult’s law in its general form can be stated as, for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. In a binary solution, let us denote the solvent by 1 and solute by 2. When the solute is non-volatile, only the solvent molecules are present in vapour phase and contribute to vapour pressure. Let p1 be the vapour pressure of the solvent, x1 be its mole fraction, p10 be its vapour pressure in the pure state. Then according to Raoult’s law

p1  x1 and p1 = x1 p10 = ptotal

The plot of vapour pressure and mole fraction of an ideal solution at constant temperature. The dashed line I and II represent the partial pressure of the components. It can be seen from the plot that p1 and p2 are directly proportional to x1 and x2, respectively. The total vapour pressure is given by line marked III in the figure. Mole fraction in vapour phase If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures:

If a solution obeys Raoult’s law for all concentrations, its vapour pressure would vary linearly from zero to the vapour pressure of the pure solvent.

p1  y1p total p 2  y 2 p total

SCAN CODE SOLUTIONS

SOLUTIONS 2.4 Ideal and Non-ideal solutions

50 

Solvent-Solute(A-B) type of force is stronger than the other two.



The vapour pressure is lower than predicted by the law



HMIXING < 0



VMIXING< 0

Ideal solutions : An ideal solution is the solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations. Properties of Ideal solutions : 

HMIXING = 0



VMIXING = 0



Intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B.

For example,phenol and aniline, chloroform and acetone etc

Eg. solution of benzene and toluene, solution of n-hexane and n-heptane

Pressure composition curves for solution showing negative deviation

Non – ideal solutions : When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution. Solutions showing positive deviation from Raoult’s Law : 

Solvent-Solute(A-B) type of force is weaker than SoluteSolute(B-B) & Solvent-Solvent(A-A) forces.



The vapour pressure is higher than predicted by the law.



HMIXING > 0



VMIXING > 0

Eg. ethanol and acetone, carbon disulphide and acetone Pressure composition curve for solution showing positive deviation

3. Concept of Boiling a. Azeotropes Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. Minimum boiling azeotrope The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. For example, ethanol-water mixture containing approximately 95% of ethanol forms an azeotrope with boiling point 351.15 K.

Solutions showing negative deviations from Raoult’s law :

SCAN CODE SOLUTIONS

51

SOLUTIONS

anthracene do not. On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not. 2.

Temperature : In a nearly saturated solution, If ( solH > 0), the solubility increases with rise in temperature and

Boiling temperature - composition Diagram for solutions showing large positive deviations. (Minimum boiling azeotrope) Maximum boiling azeotrope :

If ( solH < 0) the solubility decreases with rise in temperature. Effect of pressure : Does not have any significant effect as solids and liquids

The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. Nitric acid and water mixture containing 68% nitric acid forms an azeotrope with a boiling point of 393.5 K.

are highly incompressible. 4.2 Henry’s law Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”. This is expressed as:

p  KH x Boiling temperature - composition Diagram for solutions showing large negative deviations. (Maximum boiling azeotrope)

Here K H is the Henry’s law constant. Characteristics of KH :

4. SOLUBILITY 

K H is a function of the nature of the gas.



Higher the value of K H at a given pressure, the lower is the

4.1 Solubility of a solid in liquid Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. Factors affecting the solubility of a solid in liquid : 1.

Nature of solute and solvent : Like dissolves like. For example, While sodium chloride

solubility of the gas in the liquid. 

KH values increase with increase of temperature indicating that the solubility of gases increases with decrease of temperature.

and sugar dissolve readily in water, naphthalene and

SCAN CODE SOLUTIONS

SOLUTIONS

52

Applications of Henry’s law

Knowing that x2 = 1 – x1, equation reduces to

1.

In the production of carbonated beverages.

p1  x 2 p10

2.

In the deep sea diving.

3.

For climbers or people at high altitudes.

Raoult’s Law as a special case of Henry’s Law According to Raoult’s law, pi = xi p i0 In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Its solubility according to Henry’s law,

Equation can be wirtten as p1 p10  p1   x2 p10 p10

The expression on the left hand side of the equation as mentioned earlier is called relative lowring of vapour pressure and is equal to the mole fraction of the solute. The above equation can be written as :

p10 -p1 n2  n2  =  since x 2 =  0 p1 n1 +n 2  n1 +n 2 

p  K H x. Thus, Raoult’s law becomes a special case of Henry’s law in which K H becomes equal to p i0 .

5. COLLIGATIVE PROPERTIES The properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are called colligative properties. There are four colligative properties: 1.

Relative Lowering of vapour Pressure

2.

Elevation in Boiling Point

3.

Depression in freezing point

4.

Osmotic pressure 5.1 Relative Lowering of vapour Pressure When a non-volatile solute is added to a solvent, the vapour pressure decreases. The lowering of vapour pressure w.r.t. the vapour pressure of the pure solvent is called “Relative lowering in vapour pressure”. According to Raoult’s Law : p1  x1 p10

Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 1.

If a pressure larger than the osmotic pressure is applied to the solution side, the solvent will flow from the solution into the pure solvent through the semi permeable membrane. This phenomenon is called reverse osmosis.

An  nA Initial particle = 1

Application :

Final particles = 1 –  + n i = 1 –  + n

Desalination of sea water : When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane.

Modified Expressions : Relative lowering of vapour pressure of solvent, p10  p1 n  i. 2 0 p1 n1

6. ABNORMAL MOLAR MASSES

Elevation of Boiling point Tb = i Kb m

When the molecular mass of a substance determined by studying any of the colligative properties comes out to be different than the theoretically expected value, the substance is said to show abnormal molar mass.

Depression of Freezing point, Tf = i Kf m Osmotic pressure of solution,  = i n2 RT/V

7. VAPOUR PRESSURE

Abnormal Molar Masses are observed: 1.

When the solute undergoes association in the solution.

2.

When the solute undergoes dissociation in the solution.



Effect of temperature on vapour pressure On increasing the temperature of the liquid the escaping tendency of the molecules increases and the vapour pressure increases.

van’t Hoff Factor : To calculate extent of association or dissociation, van’t Hoff introduced a factor i, known as the van’t Hoff Factor.



i

Normal molar mass Abnormal molar mass

i

Observed colligative property Calculated colligative property

Total no. of moles of particles after association (dissociation) No. of moles of particles before association (dissociation)

Association : Number of particles will always decrease due to association therefore i < 1. nA  An Let initial particles (ni) = 1 Final number( nf) = 1 –  + /n van’t Hoff factor, i = nf/ni = 1 –  + /n

The distribution of molecular kinetic energies in a liquid 

Claussius-Clapeyron Equation In p2/p1 = (HVAP/R) (1/T1 – 1/T2) where HVAP represents the enthalpy of vaporisation of the liquid.

8. THERMODYNAMICS OF DISSOLUTION If the interactions grow stronger the process is exothermic and if they go weaker during the formation of solution the process becomes endothermic. In general S is positive in dissolution process. If the mixing process is spontaneous/natural then G has to be negative.

SCAN CODE SOLUTIONS

SOLUTIONS

56

8.1 Boiling Point Elevation GVAP = 0 Tb = HVAP/SVAP The non-volatile solute increases the randomness of the solution phase and the entropy of the vapours remains the same. Due to this, SVAP decreases thus giving rise to the boiling point.

The lower freezing point of a solution relative to that of a pure solvent is due to a difference in their entropies of fusion, Sfusion.

9. OSTWALD WALKER METHOD

The higher boiling point of a solution relative to that of a pure solvent is due to a difference in their entropies of vapourization, Svap.

This is a typical method to measure the relative lowering in vapour pressure of a solution. Dry air is passed successively through three systems: solution, pure solvent and then a drying agent.

GFUS = 0

w1 and w2 represent the decrease in weight of the vessels and w3 represents the increase in weight of the third vessel due to absorption.

Tf = HFUS/SFUS

w1  PSOLUTION w2  PSOLVENT – PSOLUTION

The entropy difference will increase in this case due to the increase in the entropy of solution. This increase in entropy will result in decrease of the freezing point according to the above relation.

(as the air was already saturated)

8.2 Freezing Point Depression

w3  PSOLVENT Using the above relations the relative lowering in vapour pressure can be calculated.

SCAN CODE SOLUTIONS

57

SOLUTIONS

IMPORTANT FORMULAE In the formulae given below A represents solvent and B represents solute, also

6.

The normality (N) and molarity (M) of a solution are related as follows :

MA = Molar mass of solvent MB = Molar mass of solute

1.

WA = Mass of solvent

VB = volume of solute

V = Volume of solution

d = density of solution.

WB Mass of percentage (w/w) = W  W  100 A B

Normality × Equivalent. mass (solute) = Molarity × Molar mass (solute) 7.

Relationship between Molarity and Mass percentage (p) If p is the mass percentage and d is the density of the solution then

VB Volume percentage (v/v) = V  V  100 A B

Molarity 

 w  WB  100 Mass by volume percentage  v   V (mL)  

Normality 

WB 6 Parts per million (ppm) = W  W  10 A B

2.

Relationship between Molarity and Normality

8.

p  d  10 Eq. mass (solute)

Dilution formula : If the solution of some substance is diluted by adding solvent from volume V1 to volume V2, then

nA Mole fraction of A, xA = n  n A B

M1V1 = M2V2 Similarly, N1V1 = N2V2 9.

nB mole fraction of B, xB = n  n A B

p  d  10 Mol. mass (solute)

Molarity of a mixture : If V1 mL of a solution of molarity M1 is mixed with another solution of same substance with volume V2 and molarity M2, then molarity of the resulting mixture of solution (M) can be obtained as :

xA + xB = 1 3.

Molarity (M)

Moles of solute nB WB = Volume of solution in litre  V  M  V (L) B (L) 4.

M

M1V1  M 2 V2 V1  V2

10. Raoult’s law for volatile solute.

Molality (m)

pA = p 0A x A and pB = p 0B x B

Moles of solute nB WB  1000 = Mass of solvent in kg  W 1000  M  W A B A

where pA and pB are partial vapour pressures of component ‘A’ and component ‘B’ in the solution. p 0A and p0B are

5.

Normality (N)

Gram Equivalents of solute WB = Volume of solution in litre  GEM of solute  V (L) GEM = Gram Equivalent Mass

vapour pressures of pure components ‘A’ and ‘B’ respectively. Total vapour pressure = p = pA + pB = p 0A x A  p B0 x B .

SOLUTIONS 11.

58

Raoult’s law for non-volatile solute.

15. Van’t Hoff factor.

i

Normal molecular mass Observed molecular mass

or

i

Observed colligative property Calculated colligative property

or,

Total number of moles of particles after association / dissociation i Number of moles of particles before association / dissociation

0 A

p p nB n W M  xB   B  B A 0 pA n A  n B n A M B WA

(For a dilute solution nB 1, solute undergoes dissociation. i < 1, solute undergoes association. i = 1, neither association nor dissociation. associated molecule,

1 i  , solute is dimer.. 2

59

SOLUTIONS

SOLVED EXAMPLES Binary Solutions Example - 1 Why is an increase in temperature observed on mixing chloroform with acetone ?

Example - 4 A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11). Calculate : (i) molal concentration, and (ii) mole fraction of sugar in the syrup.

Sol. The bonds between chloroform molecules and molecules of acetone are dipole-dipole interactions but on mixing, the chloroform and acetone molecules start forming hydrogen bonds which are stronger bonds resulting in the release of energy. This gives rise to an increase in temperature.

Sol. (i)

Weight of sugar syrup = 214.2 g Weight of sugar in syrup = 34.2 g Weight of water in syrup = 214.2 – 34.2 = 180.0 g Moles of sugar =

34.2  0.1 (Molar mass = 342) 342

Example - 2 CCl4 and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. Sol. CCl4 is non-polar covalent compound, whereas water is a polar compound. CCl4 can neither form H-bonds with water molecules nor can it break H-bonds between water molecules, therefore, it is insoluble in water. Ethanol is a polar compound and can form H-bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions. Concentration terms Example - 3

Molality =

0.1 1000 = 0.56 m. 180 34.2  0.1 342

(ii) Moles of sugar =

Moles of water =

180  10 18

Example - 5 What is the mole fraction of the solute in 2.5 m aqueous solution ? Sol. 2.5 m aqueous solution means that 2.5 moles of solute are present in 1000 g of water. Thus, Moles of solute = 2.5

Differentiate between molality and molarity of solution. What is the effect of change in temperature of a solution on its molality and molarity ?

Moles of water =

1000  55.6 18

Mole fraction of solute = Sol. Molarity is defined as number of moles of solute dissolved in 1 L of the solution. Molality is defined as number of moles of solute dissolved per kg of the solvent. Molarity is a function of temperature as volume depends on temperature. With increase in temperature, volume of solution increases, hence molarity decreases. Molality is not dependent on temperature because mass does not change with temperature.

2.5 = 0.043. 2.5  55.6

Example - 6 A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g mol–1).

SOLUTIONS

60

Sol. 6.90 M solution of KOH contains 6.90 moles of KOH in Sol. (a)  Eq. of Ca(OH)2 

1000 mL of solution. Wt. of KOH in solution = 6.90 × 56 = 386.4 g

0.74  1000  2 74  5

N

Since the solution is 30% by weight, it means that 30 g of KOH are present in 100 g of solution.

100  386.4 30

= 1288 g of solution

w  Eq.   E 

Volume of solution = 5 / 1000 litre

Wt. of KOH in 1000 mL solution = 386.4 g

 386.4 g of KOH is present in =

0.74 74 / 2

N=4  M 

N 4  2 Valency 2

3.65 (b)  Eq. of HCl  36.5

and

Volume of solution = 200/1000 litre

Weight 1288 Density   = 1.288 g mL–1 Volume 1000

 N

3.65  1000  0 .5 36.5  200

Example - 7 and M 

Calculate the moles of methanol in 5 litres of its 2 m

N 0.5   0 .5 Valency 1

solution, if the density of the solution is 0.981 kg L–1 (Molar mass of methanol = 32.0 g mol–1). (c)

Eq. of H2SO4 

1  2 ( Eq. = mole × valency) 10

Sol. Mass of 5L solution = 5L × 0.981 kg L–1 = 4.905 kg Volume of solution = 500/1000 litre = 4905 g Mass of 2 m solution = 1000 g + 2 moles of methanol



N

= 1000 + 2 × 32 = 1000 + 64 = 1064 g Now 1064 g of solution contains methanol = 2 mol

2  1000  0.4 and 10  500

M

0.4  0.2 2

Mole Fraction Example - 9

4905 g of solution contains methanol

2  4905 = 9.22 mol. = 1064

Calculate the mole fraction of methanol in a solution containing 100 g of water and 50 g of methanol. Sol. Mole fraction of solute,

Example - 8

xB  Calculate normality and molarity of the following : (a) 0.74g of Ca (OH)2 in 5 mL of solution.

Moles of solute Moles of solute  Moles of solvent

Mole fraction of methanol,

(b) 3.65g of HCl in 200 mL of solution. (c) 1/10 mole of H2SO4 in 500 mL of solution.

x CH3OH 

Moles of CH 3OH Moles of CH 3OH  Moles of H 2 O

Mass of CH 3OH Moles of CH3OH = Molar mass of CH OH 3

SOLUTIONS

61

Molar mass of CH3OH = 12 + 1 × 3 + 16 + 1 = 32 g mol–1 Moles of CH3OH 

50  1.56 mol 32

Molarity Example - 11 Consider 18 g of glucose (molar mass 180 g mol–1) to be present in 500 cm3 of its aqueous solution. What is the molarity of the solution ?

Now, moles of H2O

Mass of H 2 O 100 = Molar mass of H O  18  5.56 mol 2 So, the mole fraction of methanol,

x CH3OH 

1.56 1.56  = 0.219 1.56  5.56 7.12

Sol.

Molarity 

Number of moles of solute  1000 Volume in mL

Here, solute is glucose whose mass = 18 g and molar mass = 180 g mol–1. Therefore,

Example - 10 Number of moles of glucose = A solution is 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component. Sol. Let the total mass of solution = 100 g

Given mass  18    moles Molar mass  180  Also, the volume of solution = 500 cm3. Therefore,

Mass of water = 25 g

Molarity 

Mass of ethanol = 25 g

18  1000 = 0.2 M 180  500

Mass of acetic acid = 50 g Molality Moles of water =

25  1.388 18

Example - 12

( Molar mass of H2O = 18) Moles of ethanol =

25  0.543 46

( Molar mass of C2H5OH = 46) Moles of acetic acid =

50  0.833 60

The solubility of Ba (OH)2.8H2O in water at 288 K is 5.6 g per 100 g of water. What is the molality of hydroxide ion in saturated solution of Ba (OH)2.8H2O at 288 K. Sol. The molar mass of Ba (OH)2.8H2O is = 137 + 2 × 17 + 8 × 18 = 137 + 34 + 144 = 315

( Molar mass of CH3COOH = 60)

Ba (OH)2  Ba2+ + 2OH–

Total number of moles = 1.388 + 0.543 + 0.833 = 2.764

Suppose x is molality of Ba2+ ions, molality of (OH)– is 2x.

1.388 Mole fraction of water = = 0.502 2.764

100 g water has 5.6 g of Ba (OH)2.8H2O  1000 g water will have

Mole fraction of ethanol =

5.6 1000  56 100

0.543 = 0.196 2.764 Moles of Ba (OH)2.8H2O =

Mole fraction of acetic acid =

0.833 = 0.302 2.764

56  0.178 mol 315

 Molality of hydroxyl ion = 0.178 × 2 = 0.356 m

SOLUTIONS

62

Normality

Sol.

Example - 13

p oC7 H16  105.2 kPa, p oC8H18  46.8 Kpa M C7H16  100g mol 1 , M C8H18  114 g mol1

Calculate the normality of the resulting solution made by adding 2 drops (0.1 mL) of 0.1N H2SO4 in 1 litre of distilled water.

n C7 H16 

25  0.25 100

n C8 H18 

28.5  0.25 114

Sol.  Meq. of solute does not change on dilution Meq. of H2SO4 (conc.) = Meq. of H2SO4 (dil.) 0.1 × 0.1 = N × 1000 ( Meq. = N × V in mL) N = 10–5

x C7H16 

n C7 H16 n C7H16  n C8H18



0.25  0.5 0.25  0.25

Raoult's Law

x C8H18  1  x C7 H16  1  0.5  0.5

Example - 14 State Raoult’s law for solutions of volatile liquids. Taking suitable examples explain the meaning of positive and negative deviations from Raoult’s law. Sol. Raoult’s law : It states that for a solution of volatile liquids the partial pressure of each component is directly proportional to its mole fraction. Mathematically,

pA  x A

pB  x B

p A  p Ao x A

p B  p oB x B

(i)

p C7H16  pCo 7H16 x C7H16 = 105.2 × 0.5 = 52.60 kPa

(ii)

p C8H18  pCo 8H18 x C8H18  46.8  0.5

p C8H18 = 23.4 kPa p  p C7 H16  p C8 H18 = 52.6 + 23.4 = 76.0 kPa (iii)

Mole fraction of octane (y C8 H18 ) in the vapour phase.

Positive deviation from Raoults law : In this type of deviation the partial pressure of each component of solution is greater than that calculated from Raoults’s law,

p C8H18  yC8 H18 p total

i.e., p A  p oA x A & p B  p Bo x B .

23.4  yC8H18  76.0

Example : A solution of water and ethanol. Negative deviation from Raoult’s : In this type of deviation the partial pressure of each component of solution is less

yC8H18 

than that expected from Raoult’s law, i.e., p A  p oA x A & p B  p oB x B .

Example : A solution of acetone and chloroform. Example - 15 Heptane and octane form an ideal solution at 373 K, The vapour pressures of the pure liquids at this temperature are 105.2 kPa and 46.8 kPa respectively. If the solution contains 25g of heptane and 28.5g of octane, calculate

Example - 16 At 300 K, the vapour pressure of an ideal solution containing one mole of A and 3 mole of B, is 550 mm of Hg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of solution increases by 10 mm of Hg. Calculate the V.P. of A and B in their pure state. Sol. Initially, PM  PA0 . X A  PB0 . X B 550  PA0 .

(i) vapour pressure exerted by heptane (ii) vapour pressure exerted by solution (iii) mole fraction of octane in the vapour phase.

23.4 = 0.308 76.0

or

1 3  PB0 . 1 3 1 3

PA0  3PB0  2200 ..........(1)

63

SOLUTIONS When, one mole of B is further added to it

PM  PA0 . X A  PB0 . X B 560  PA0 .

1 4  PB0 . 1 4 1 4

Sol. Wt. of benzene, C6H6 = 10.0g; g. mol. wt. of C6H6 = (6 × 12) (6 × 1) = 78 g; wt. of ethyl alcohol, C2H5OH = 10.0g; g. mol. wt. of C2H5OH = (2 × 12) + (5 × 1) + 16 + 1 = 46.0g. Thus : No. of moles of C6H6 =

 PA0  4PB0  2800 ......... (2) Solving Eqs. (1) and (2), we get

PA0  400 mm, PB0  600 mm.

No. of moles of C2H5OH = X C6 H 6 

Example - 17 Benzene and toluene form two ideal solutions A and B at 313K. Solution A contains 4 mole of the toluene and one mole of C6H6. Solution B contains equal masses of toluene and benzene. Calculate total pressure in each case. The vapour pressure of C6H6 and toluene are 160 and 60 mm respectively at 313 K. Sol. A :

PM  PB  PT  PB0 . X B  PT0 . X T  XB 

1 4 and X T  1 4 1 4

PM  160 

1 4  60  ; 1 4 1 4

= 32 + 48 = 80 mm

B:

PM  160 

w / 78 w / 92  60  w w w w   78 92 78 92

(Given, equal weights are mixed) 92 78  160   60  170 170

10 = 0.128; 78

10 = 0.217 46

0.128  0.371; 0.128  0.217

X C2 H5OH  1  0.371  0.629 PC 6 H 6  PC06 H 6 X C 6 H 6  (0.371) (100) = 37.1 torr.; PC 2 H 5OH  PC0 2 H 5OH X C 2 H 5OH  (0.629) (44) = 28 torr..

Total pressure = 37.1 + 28 = 65.1 torr. NON IDEALSOLUTION SHOWING POSITIVE DEVIATION FROM RAOULT'S LAW Example - 19 Why does a solution of ethanol and cyclohexane show positive deviation from Raoult’s law ? Sol. On adding cyclohexane, its molecules get in between the molecules of ethanol thus breaking the hydrogen bonds and reducing ethanol-ethanol interactions. This will increase the vapour pressure of the solution and result in positive deviation from Raoult’s law. EFFECT OF TEMPERATURE ON SOLUBILITY OF SOLIDS IN LIQUIDS Example - 20 What is the effect of temperature on the solubility of a solid solute in a liquid solvent ?

At 298 K, the vapour pressure of pure ethyl alcohol is

Sol. Solubility of solid in liquid is significantly affected by temperature changes. This is because of dynamic equilibrium which follow Le Chatelier’s principle. For saturated solution such as NaNO3, KNO3, etc. the solubility increases with increase of temperature because in these substances, the process of dissolution is endothermic.

44 torr and that of pure benzene is 100 torr. Assuming

Solute + Solvent + Heat  Solution

= 86.588 + 27.529 = 114.117 mm Example - 18

ideal behaviour, calculate the vapour pressure at 298 K of a solution which contains 10.0g of each substance (At. wt., C = 12, H = 1, O = 16)

There are few substances such as cerium sulphate, lithium carbonate, etc. whose solubility decreases with increase of temperature. This is because the process of dissolution is exothermic (sol. H < 0).

SOLUTIONS

64 Henry's Law

Solubility

Example - 24

Example - 21 Define the following terms :(i) Dissolution

(ii) Crystallisation

(iii) Saturated solution Sol. (i) When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution. (ii) Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation. (iii) A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. Example - 22

State Henry’s Law. What is the significance of KH ? Sol. Henry’s law : It state that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as p = KHx where, KH is the Henry’s law constant. Significance of KH. As pA = KHxA. Thus at constant temperature for the same partial pressure of different gases, x A  1/K H. In other words solubility is inversely proportional to Henry’s constant of the gas. Higher the value of KH, lower is the solubility of the gas. As H2 is more soluble than helium, H2 will have lower value KH than that of helium. APPLICATION OF HENRY'S LAW Example - 25

What is the effect of pressure on the solubility of (a) gases in liquids and (b) solids in liquids ? Sol. (a) Solubility of gases in liquid increases with the increase of pressure at a particular temperature. If the system is in dynamic equilibrium, then the rate of dissolution is equal to the rate of evaporation. If the pressure increases over the system, the number of gaseous particles per unit volume increases. As a result, more particles strike the surface of the solution and enter it. Thus, a new equilibrium is established. Hence, on increasing the pressure, the solubility of gases in liquid also increases. (b) Pressure has very little effect on the solubility of a solid in a liquid because solids have definite shape and volume; whereas liquids have definite volume but not shape. Hence, solids and liquids cannot be compressed much by increase in pressure.

Why are the aquatic species more comfortable in cold water rather than warm water ? Sol. The aquatic species are more comfortable in cold water rather than warm water because the solubility of gas in liquid decreases with increase in temperature, and increases with decrease in temperature. Therefore, the percentage of oxygen present in cold water will be more because in cold water, solubility of oxygen gas in water will be more than in hot water. Now, we know that oxygen is essential for respiration or breathing. Hence, the aquatic species are comfortable in cold water; whereas in hot water, the concentration of oxygen is less which makes the breathing process difficult for the aquatic species. Example - 26 Why oxygen mixed with helium is used by deep sea divers ?

Example - 23 Give an example of a solution containing a liquid solute in a solid solvent. Sol. An example of a solution containing a liquid solute in a solid solvent is amalgam of mercury (Hg) with sodium (Na). In this solution, sodium is solid solvent and mercury is liquid solute.

Sol. Deep sea divers must cope with high concentration of dissolved gases while breathing air at high pressure under water as according to Henry’s law, solubility of gases increases with pressure. However, when the divers come towards the surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries

65

SOLUTIONS and creates a medical condition known as bends. To avoid bends and also the toxic effect of high concentration of nitrogen in the blood, the tanks used by divers are filled with air diluted with helium. Value of KH for He is 144.97 kbar and N2 is 76.08 kbar, therefore He has a lower solubility. Colligative Properties Of Dilute Solutions

Expressing concentration in terms of moles per liter, we have



nB RT V

where nB is the number of moles of the solute. Relative Lowering of Vapour Pressure

Example - 27 What is meant by colligative properties ? Give expressions for any two. Sol. Colligative properties are the properties of solutions which depend only on the relative numbers of solute and solvent particles, but not on the nature of the solute. The four important colligative properties are as follows : (a)

Relative lowering of vapor pressure.

(b)

Elevation in boiling point.

(c)

Depression in freezing point.

(d)

Osmotic pressure. Relative lowering of vapor pressure : In the presence of a non-volatile solute, the vapor pressure of the solution becomes lower than the vapor pressure of pure solvent. The relative lowering of vapor pressure is

Example - 28 Why does vapour pressure of a liquid decrease when a non-volatile solute is added into it ? Sol. It is because some liquid molecules at the surface are replaced by the molecules of the solute which are nonvolatile. Example - 29 Derive an equation to express that relative lowering of vapour pressure for a solution is equal to the mole fraction of the solute in it when the solvent alone is volatile. Sol. For a solution of volatile liquids Raoult’s law, is given as p = pA + p B If solute (component B) is nonvolatile then

p  (p oA  p A ) / p Ao

p  p A  p oA x A

o A

where p is the vapor pressure of pure solvent. The relative p  p oA (1  x B )

lowering depends only on the concentration of the solute and not in nature of solute

p  p oA  p oA x B

p  p oA x B

p oA x B  p oA  p

where xB is the mole fraction of the solute. p oA  p  xB p oA

Osmotic pressure : When two solutions of different concentration are separated by a semiperimeable membrane, the solvent shows a tendency to move from dilute to concentrated solution. This property is known as osmosis.

Thus, relative lowering of vapour pressure is equal to the mole fraction of non-volatile solute.

This uneven passage of the solvent creates pressure called osmotic pressure. It can be measured by determining the amount of counter pressure needed to prevent osmosis. It is proportional to concentration and temperature of the solution and is given by

  CRT

Example - 30 A solution is made by dissolving 30 g of a non-volatile solute in 90 g of water. It has a vapour pressure of 2.8 kPa at 298 K. At 298 K, the vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of solute.

SOLUTIONS

66

Sol. Let the molar mass of non-volatile solute be MB Moles of water 

30 90  5, Moles of solute = MB 18

Applying Raoult’s law,

p0  ps w B M A   p MB wA

3.64  2.8 30 18   3.64 M B 90 0.84 6  3.64 M B

MB 

6  3.64 0.84

MB = 26 g mol–1

Example - 32 What weight of the non-volatile urea (NH2 – CO – NH2) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 25% ? What will be molality of the solution ?

po  ps 25  Sol. Here, po 100

For urea, MB = 60 g mol–1 For water, MA = 18 g mol–1 p o  ps n B w B / M B   ps nA w A / MA



25 w B / 60  75 100 /18

or

wB = 111 g

Molality of the solution =

temperature is 0.850 bar. A non-volatile, non-electrolyte

solution is 0.845 bar. What is the molar mass of the solid substance ? Sol. The various quantities known to us are as follows : PAo  0.850 bar; p = 0.845 bar; MA = 78 g mol–1;

wB = 0.5 g; wA = 39 g Substituting these values in equation poA  p w B  M A  , we have p1o MB  w A 0.850 bar  0.845 bar 0.5 g  78 g mol1  0.850 bar M B  39 g

Thus, MB = 170 g mol–1.

111g 1   1000g kg 1 1 60 g mol 100 g

= 18.5 mol kg–1 or 18.5 m

solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol–1), then vapour pressure of the

or

wA = 100 g, WB = ?

Example - 31 The vapour pressure of pure benzene at a certain

p o  p s 25  ps 75

Elevation in Boiling Point Example - 33 A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31ºC. Determine the molar mass of this compound. (Boiling point of pure benzene = 80.10 ºC and Kb for benzne = 2.53 ºC molal–1.) Sol. The molar mass MB of solute can be calculated from the relation,

MB 

K b  w B  1000 Tb  w A

Mass of solute, wB = 1.25 g, mass of solvent, wA = 99 g, elevation in boiling point, (Tb) = 80.31 – 80.10 = 0.21 ºC. Kb for benzene = 2.53 ºC molal–1.

MB 

2.53 1.25  1000 = 152.11 g mol–1 0.21 99

67

SOLUTIONS Example - 34

Sol. Molal boiling point elevation constant : It may be defined

The boiling point of chloroform was raised by 0.325ºC when 0.5141 g of anthracene was dissolved in 35g of chloroform. Calculate the molecular weight of anthracene. (K b = 39.0 per 100 g of chloroform) Sol. wt. of solute, W2 = 0.5141g ; wt. of solvent,

solution is unity, that is, 1 mol of solute is dissolved in 1 kg (1000 g) of the solvent. The units of molal elevation constant is degree molality–1, that is, K molal–1 or ºC molal–1 or K kg mol–1. It is denoted by Kb. Molal freezing point elevation constant : It may be defined

35 W1 = kg = 0.035 kg; 1000

as the depression in freezing point when the molality of the solutions is unity, that is, 1 mol of the solute is

Tb = 0.325ºC ; Kb = 39ºC (100 g mol–)

dissolved in 1000 g (1 kg) of the solvent. It is denoted by

100 kg mol  = 39ºC × 1000

Kf. The unit of Kf is degree molality–1, that is, K molal–1 or K kg mol–1.

= 3.9ºC kg mol– ; mol. wt. of anthracene, M2 = ? We know that Tb 

as the elevation in boiling point when the molality of

Cause of elevation in boiling point : Vapor pressure of the

K b  W2 in g ; M 2  W1 in kg

solution is lower than that of pure solvent. Hence, when the non-volatile solute is added to pure solvent, the

K b  W2 in g M2  Tb  W1 in kg

solution has to be heated more to make vapor pressure equal to the atmosphric pressure.

1

Or

M2 

3.9º C kg mol  0.5141g  176.26g mol 1 0.325º C  0.035 kg

Cause of depression in freezing point : Vapor pressure of the solution is less than that of pure solvent. Freezing point of a substance is the temperature at which the solid

MOLAL BOILING POINT ELEVATION CONSTANT

and the liquid forms of the substance are in equilibrium, Example - 35

that is, the solid and the liquid forms of substance have

Calculate the molal elevation constant of water, it being given that 0.1 molal aqueous solution of a substance boiling at 100.052 ºC. Sol. Molality of solution, m = 0.1 molal; the boiling point of

the same vapor pressure. Therefore, for the solution, this will occur at a lower temperature. Depression in Freezing Point Example - 37

o b

solution Tb = 100.052 ºC. Therefore, Tb  Tb  T is the elevation in the boiling point, Tb  100.052 – 100 = 0.052 ºC. By applying the relationship Tb  K b m we get

Kb 

Tb 0.052 º C –1  m 0.1 molal = 0.52 ºC molal

Molal elevation constant Kb = 0.52 ºC molal–1. Example - 36 Define molal boiling point elevation constant and molal freezing point depression constant. Explain the cause for elevation in boiling point and depression in freezing point when non-volatile solute is added to a solvent.

Explain why the freezing point of a solution is lower than the freezing point of the pure solvent. Sol. Freezing point of a substance may be defined as the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase. The freezing point of a solution is lower than that of the pure solvent because the vapor pressure of solution is less than that of pure solvent. So according to Raoult’s law, when a non-volatile solid is added to the solvent, its vapor pressure decreases and it becomes equal to that of solid solvent at a lower temperature. The decrease in the freezing point is known as depression of freezing point, and it is denoted by Tf.

SOLUTIONS

68

Example - 38 Ethylene glycol (molar mass = 62 g mol–1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. Would it be advisable to keep this substance in the car radiator during summer ? Given : Kf for water = 1.86K kg/mol Kb for water = 0.512K kg/mol Sol.

Tf  K f 

w B 1000  mB w A

12.4 1000   3.76 K 62 100 since water freezes at 0ºC, so freezing point of the solution containing ethylene glycol will be – 3.76º C.  1.86 

Tb  K b 

w B 1000  MB w A

 0.512 

12.4 1000   1.024 K 62 100

Tb  Tbo  Tb  100  1.024 = 101.024 ºC

since water boils at 100ºC, so a solution containing ethylene glycol will boil at 101.024ºC, so it is advisable to keep this substance in car radiator during summer. Example - 39 Ethylene glycol (C2H6O2) is used as an antifreeze. What mass of ethylene glycol should be added to 5.00 kg of water to lower the freezing point to –5.0 ºC ? Given that Kf = 1.86 ºC molal–1. Consider ethylene glycol to be a non-electrolyte. Sol. Given that Tf = 5.0 ºC and Kf = 1.86 ºC molal–1.

Molar mass of ethylene glycol is = 2 × 12 + 4 × 1 + 2 × 16 = 62 g mol–1. Hence, the mass of ethylene glycol to be added to water is

13.5 mol  Example - 40

A 0.1539 molal aqeuous solution of cane sugar (molar mass = 342 g mol–1) has a freezing point of 271 K, whereas the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (molar mass = 180 g mol–1) per 100 g of solution ? Sol. Molality of cane sugar solution = 0.1539 molal and the depression of freezing point, Tf = 273.15 – 271 = 2.15 K. from Tf = Kfm, we have Kf 

Tf 2.15K o –1  m 0.1539 mol kg -1 = 13.97 C molal

Amount of glucose in 100 g of solution = 5 g and amount of water in 100 g of solution = 100 g – 5 g = 0.095 kg. Number of moles of glucose = 5/180 mol = 0.0278 mol and the molality of solution m = 0.0278 mol/0.095 kg = 0.2926 mol kg–1. So, Tf = Kf × m = 13.97K kg mol–1 × (0.2926) = 4.08 K Therefore, freezing point of glucose solution = 273.15 – 4.08 = 269.07K. Example - 41 Calculate the temperature at which a solution containing 54 g glucose, (C6H12O6) in 250 g of water will freeze. (Kf for water = 1.86 K molal–1) Sol. Molar mass of glucose MB = 12 × 6 + 12 × 1 + 6 × 16 = 180 g mol–1. According to depression in freezing point,

Therefore, molality is

Molality 

T 5.0   2.7 molal K f 1.86

Therefore, the number of moles of ethylene glycol is found from the relation

Molality 

Number of moles of solute Mass of solvent

n  2.7  n  13.5 mol 5.00

62 g = 837 g mol

Tf 

K f  w B  1000 MB  w A

where Kf is the molal depression constant, wB is the given mass of solute, MB is the molar mass of solute, wA is the given mass of solvent (water). Substituting given values, we get

Tf 

1.86  54 1000 100440   2.23K 180  250 45000

Therefore, the freezing point of the solution = 0 – 2.23 = – 2.23 K.

69

SOLUTIONS Example - 42

Theoretical molar mass (MTh) of CdI2

A sample of naphthalene has freezing point, 80.6ºC.

= 112 + (2 × 127) = 366 g mol–1

When 0.512g of a solute is dissolved in 7.03g of naphthalene, the solution has a freezing point of 75.2ºC. Calculate the molecular mass of the solute. (K f for naphthalene = 6.8ºC kg/mol) Sol. Tf = (80.6 – 75.2)ºC = 5.4ºC ; wt. of solute, W2 = 0.512 g; wt. of solvent, W1 = 7.03 g ; mol. wt. of solute, M2 = ?, Kf = 6.8ºC kg mol–1. We know that:

K (in º C kg mol 1 )  W2 (in g ) Tf  f ;  W1 in g  1 M 2 (in g mol )    kg  1000 

CdI2    Cd2+ + 2I–

For Initial conc.

1

At equilibrium

0

0

1 – 2

i = 1 – + 2 = 1 + 2 ...... (1)

M Th 366g mol 1  But i = M 311.6g mol 1 obs

.......(2)

Substituting the value of i from (2) in (1), we get

366  1  2 or 311.6



1  366   1  2  311.6 

or = 0.087 or 0.087 × 100 = 8.7 % dissociated Osmotic Pressure

5.4º C 

 M2 

6.8º C kg mol 1  0.512g 7.03 M2  Kg 1000

6.8º C kg mol 1  0.512g  91.7 g mol 1 5.4º C  00703 kg

Example - 43 The freezing point of a solution of 4.0g CdI2 in 95g of water was – 0.25ºC. Calculate the degree of dissociation of salt. (K f for water = 1.85ºC kg/mol (At. wt. Cd = 112; I = 127). Sol. Wt. of solute, W2 = 4.0g ; wt. of solvent, W1 = 0.095 kg; Tf = 0.25 ºC; M 2 

K f  W2 in g Tf  W1 in g

Example -44 Define osmotic pressure and explain its origin. What is the effect of temperature on osmotic pressure ? Sol. Osmosis is the phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution. Osmotic pressure is the pressure applied on the solution side to just stop the flow of the solvent from its side to the solution side across a semipermeable membrane. The increase in temperature will increase the osmotic pressure as   T. Example - 45 Consider 100 mg of a protein to be dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25 ºC, what is the molar mass of the protein ? (Given that R = 0.0821 L atm mol –1 K –1 and 760 mm Hg = 1 atm.) Sol. We know that osmotic pressure (  ) is given by

1.85º C kg mol 1  4g M   311.6 mol 1  2 0.25º C  0.095 kg



nRT V

and M B 

w BRT V

(1)

SOLUTIONS

70

where MB is the molar mass of protein;  = 13.3 mm Hg = 13.3/760 = 0.0175 atm; V = 10 mL = 10/1000 = 0.01 L; R = 0.0821 L atm mol–1K–1; T = 25 ºC = (273 + 25) K = 298 K; wB = 100 mg = 100/1000 g = 0.1 g. Substituting all the values in Eq. (1), we get

MB 

0.1 0.0821 298 = 13980 g mol–1 0.0175  0.01

Sol. The osmotic pressure method has the advantage over rise in boiling point or fall in freezing point for determining molar masses of macromolecules because Pressure measurement around the temperature and the molarity of solution is used instead of molality. Compared to other colligative properties, its magnitude is large even for very dilute solutions.

The vapour pressure of water at 293 K is 2338 Pa and the vapour pressure of an aqueous solution is 2295.8 Pa. If density of solution is 1010 Kg/m3 at 313 K, calculate the osmotic pressure at 313 K. Molecular weight of solute = 60.



P 0  Ps n n  18   Ps N W

2338  2295.8 n  18  2295.8 W n

P = 2.53 × 106 Pa Example - 48

of 2g/dm3, what is mol. wt. of solute ? Sol. Given that,  = 20.66 Nm–2,

w 2 10 3  2g / dm 3  kg / m 3 3 V 10 S = 8.314 JK–1 mol–1, T = 273 + 25 = 298 K w w ST V  ST  m  m V  m

2 10 3  8.314  298 10 3  20.66

= 239.84 kg mol–1

REVERSE OSMOSIS Example - 49

Example - 47

Sol. At 293 K ;

1.0393n  n  8.314  313 1010

25ºC is 20.66 Nm–2. If the solution had a concentration

How is it that measurement of osmotic pressures is more widely used for determining molar masses of macromolecules than the rise in boiling point or fall in freezing point of their solutions ?

(ii)

P

The osmotic pressure of a non-volatile solute in C6H6 at

Example - 46

(i)

Now, PV = nRT

42.2

1

 W  2295.8  18 or W = 979.25 × ng

 Weight of solution = 979.25n + wt. of solute = 979.25n + 60n = 1039.25 ng = 1.0393n kg

Weight 1.0393n 3 of solution   m  Volume density 1010 ( at 313 K )

State the condition resulting in reverse osmosis. Sol. Reverse osmosis will occur if a pressure higher than the osmotic pressure is applied on the solution. ABNORMAL MOLAR MASS Example - 50 What is van’t Hoff factor and how is it related to the molar mass of the solute. Sol. van’t Hoff factor is defined as the ratio of the observed value of colligative property to the calculated value of colligative property. It is denoted by i.

i

Observed value of the colligative property Calculated value of the colligative property

van’t Hoff factor is inversely proportional to the molecular mass of the solute. Therefore,

i

Normal molecular mass M  Calculated Abnormal molecular mass M Observed

71

SOLUTIONS Example - 51 of water, the amount of NaCl to be added is Expalin why in case of sodium chloride, the observed values of colligative properties are greater than the calculated values. Sol. In case of NaCl, the observed values of colligative properties are greater than the calculated values because in a dilute solution NaCl is expected to dissociate completely. Therefore, NaCl has observed values twice the calculated values of colligative properties.

for 65g of water, it is

(125.775  65)  8.175g 1000

Hence, mass of NaCl to be added = 8.175 g in 65 g of water. Example - 54 (a) Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 40ºC. (Vapour pressure of water at 40ºC = 55.3 mm of Hg) (b) Why is freezing point depression of 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution ?

NaCl  Na+ + Cl– Van't Hoff Factor Example - 52 A 0.118 molal solution of LiCl has a freezing point of –0.415 ºC. What is the van’t Hoff factor for this solute at this concentration ?

125.775 and 1000

Sol. (a)

po  p  x urea po

Tf (not ionized) = Kf × m

Let the mass of solution be 100 g  Mass of Urea = 10 g 10 1 n urea   60 6 90 n H2 O  18 n H2O  5.

= (1.86 ºC molal–1) × 0.118 molal = 0.219 ºC

x urea 

Sol. Molality of solution m = 0.118 molal. For LiCl, Kf = 1.86ºC molal–1 and Tf (measured) = 0.415ºC. Now, the depression in freezing point is found as

The van’t Hoff factor can be calculated using relation

n urea 1/ 6 1/ 6   n urea  n H2O 1/ 6  5 31/ 6

1 31 55.3  p 1  55.3 31 55.3 55.3  p  31 55.3 55.3  p 31 1  55.3 1    p  31  x urea 

i

Tf (measured) 0.415 º C  = 1.89 Tf (not ionized) 0.219 º C

Example - 53 What mass of NaCl (molar mass = 58.5 mol–1) must be dissolved in 65 g of water to lower the freezing point by 7.5 ºC ? The freezing point depression constant Kf for water is 1.86 ºC molal–1. Assume van’t Hoff factor for NaCl to be 1.87. Sol. Given that molar mass, M = 58.5 g mol–1; van’t Hoff factor for NaCl, i = 1.87; freezing point depression constant, Kf for water = 1.86 ºC molal–1 and Tf = 0 – (–7.5) = 7.5 ºC. Now, applying the relation and substituting values, we get T = i × Kf × m 7.5 = 1.87 × 1.86 × m m = 2.15 molal Grams of NaCl = 2.15 × 58.5 = 125.775 g per kg. If 125.775 g of NaCl is added to 1000 g of water, then for 1g





p  55.3 

30 31

p = 53.52 mm Hg. (b)

NaCl being an electrolyte, dissociates almost completely to give Na+ and Cl– ions in solution whereas glucose being non-electrolyte, does not dissociate. Therefore, the number of particles in 0.1 M NaCl solution is nearly double than that in 0.1 M glucose solution. Freezing point depression, being a colligative property, is therefore, nearly twice for NaCl solution than for glucose solution of same molarity.

SOLUTIONS

72

Example - 55

yA = Mole fraction of component – 1 in vapour phase.

(i) Benzoic acid completely dimerises in benzene. What will be the vapour pressure of a solution containing 61g of benzoic acid per 500g benzene when the vapour pressure of pure benzene at the temperature of experiment is 66.6 torr ? (ii) What would have been the vapour pressure in the absence of dimerisation ? (iii) Derive a relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase.

Sol. (i)

p  iX B po

yB = Mole fraction of component – 2 in vapour phase.

yA 

pA pA  p total p A  p B

yB 

pB pB  p total p A  p B

yA 

x A p oA x A p oA  o o o x A p A  x B p B x A p A  (1  x A ) p oB

yB 

x B p oB x A p oA  x B p Bo

Example - 56

1 i 2 XB 



In an aqueous solution, KCl undergoes complete dissociation into K+ and Cl– ions. What is the value of osmotic pressure of a 0.525M solution of KCl at 300 K ? (Use R = 0.0821 L atm K–1)

nB 61/122  n A  n B 61/122  500 78

0.5 0.5  0.5  6.41 6.91

Sol. As KCl dissociates K+ and Cl– ions,  i = 2.

p 1 50   66.6 2 691 p 

Osmotic pressure ( ) can be determined using the formula

50  66.6  2.41 691 2

 = iCRT = 2 × 0.525 × 0.0821 × 300 = 25.86 atm Example - 57

po – p = 2.41 p = 66.6 – 2.40

When 0.5 g KCl was dissolved in 100 g water, the solution originally at 20 ºC, froze at – 0.24 ºC. Calculate the percentage ionization of salt. Kf per 1000 g of water = 1.86 ºC molal–1.

= 64.20 torr (ii) In the absence of dimerisation. i=1 p  XB po

Sol. Mass of solute KCl, w2 = 0.5 g; mass of solute H2O, w1 = 100g. Molar mass of H2O, M1 = 1 × 2 + 16 = 18 g mol–1 and Tf = 0 – (–0.24) = 0.24 ºC. According to the equation,

50  66.6  4.82 691 P = 66.6 – 4.82 = 61.78 torr p 

(iii) From Raoult’s law, p A  x A p oA and

p B  x B p oB

where, xA = mole fraction of liquid A

p A  x A p oA

xB = mole fraction of liquid B

p B  x B p Bo

M2 

1000  K f  w 2 1000  1.86 º C molal 1  0.5g  Tf  w1 0.24 º C  100g



930  38.75g mol1 24

Molar mass of KCl = 39 + 35.5 = 74.5 g mol–1.

73

SOLUTIONS Therefore, van’t Hoff factor is M2 (observed) =

Calculate molecular mass 74.5 i   1.92 Observed molecular mass 38.75



Now, KCl dissociates as KCl  K+ + Cl– Initial moles

1 mol

Moles after dissolution

0 1–

1000g  5.12K kg mol 1  2.0  102 kg 10.240  102  1.0 kg  0.69 K 0.69

= 14840.58 × 10–2 = 148.4 g mol–1

0

Also, M2 (calculated) for C6H5OH



= 12 × 6 + 1 × 5 + 16 + 1 = 94 g mol–1. Therefore,

Total number of moles after dissociation = 1 –  +  +  = 1 + . Therefore,

i

1  i    i  1  1.92  1  0.92 1

M 2 (calculated) 4   0.633 M 2 (observed) 148.4 2C6H5OH  (C6H5OH)2

Therefore, percentage ionization = 0.92 × 100 = 92%. Example - 58

1000 K f w 2 w1Tf

Initial moles

1 mol

Moles after dissociation Phenol associates in benzene to a certain extent to form dimers. A solution containing 2.0 × 10–2 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. Calculate the degree of association of phenol (Kf for benzene = 5.12 K molal–1) Sol. Mass of phenol, w2 = 2.0 × 10–2 kg; Kf = 5.12 K molal–1; mass of benzene, w1 = 1.0 kg; Tf = 0.69 K. Therefore, according to the equation

1–

Total moles after dissociation = 1   

i or

1  ( / 2) 1

 = 2 (1 – i) = 2 (1 – 0.633) = 0.734

/2

   1  . Therefore, 2 2

SOLUTIONS

74

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Concentration terms 1.

2.

3.

4.

5.

6.

7.

9.

What is the mass percentage of carbon tetrachloride if 22g of benzene is dissolved in 122g of carbon tetrachloride ? (a) 84.72%

(b) 15.28 %

(c) 50%

(d) 44%

10.

A solution is obtained by mixing 200 g of 30% and 300 g of 20% solution by weight. What is the percentage of solute in the final solution ? (a) 50%

(b) 28%

(c) 64%

(d) 24%

11.

What is the mole fraction of glucose in 10% w/w glucose solution ? (a) 0.01

(b) 0.02

(c) 0.03

(d) 0.04

12.

Rectified spirit contains 95% ethanol by weight. The mole fraction of ethanol will be (a) 0.881

(b) 0.99

(c) 0.118

(d) 0.81

13.

The mole fraction of C2H5OH (Mol. wt. = 46) in 5 molal aqueous ethyl alcohol solution is (a) 0.082

(b) 0.82

(c) 5

(d)

5 55.55

(b) 0.617 m

(c) 0.668 m

(d) 1.623 m

14.

15.

An aqueous solution of urea containing 18g urea in 3

1500 cm of the solution has a density equal to 3

1.052 g/cm . If the molecular weight of urea is 60, then the molality of the solution is

8.

(a) 0.200

(b) 0.192

(c) 0.100

(d) 1.200

(a) 0.3 M

(b) 0.03 M

(c) 3 M

(d) 0.103 M

To a 4 L of 0.2 M solution of NaOH, 2 L of 0.5 M NaOH are added. The molarity of resulting solution is : (a) 0.9 M

(b) 0.3 M

(c) 1.8 M

(d) 0.18 M –1

150 mL of C2H5OH (density = 0.78g mL ) is diluted to one litre by adding water ; molarity of the solution isof (a) 2.54

(b) 11.7

(c) 2.99

(d) 29.9

How many grams of a dibasic acid (Mol. wt. = 200) should be present in 100 mL of its aqueous solution to give decinormal strength ? (a) 1g

(b) 2g

(c) 10g

(d) 20g. 3

The normality of 10% H2SO4 solution (d = 1.1g/cm ) is (a) 2.04

(b) 1.02

(c) 1.85

(d) 2.25

Raoult's Law

What will be the molality of a solution of glucose in water which is 10 % w/w ? (a) 0.01 m

What will be the molarity of 30 mL of 0.5 M H2SO4 solution diluted to 500 mL ?

The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is (a) 0.50

(b) 0.6

(c) 0.27

(d) 0.73

The vapour pressure of a pure liquid A is 70 torr at 27ºC. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at 27ºC. The vapour pressure of pure liquid B at 27ºC. (a) 14

(b) 56

(c) 140

(d) 70.

What is the molarity of a solution containing 10g of NaOH in 500 mL of solution ?

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) 0.25 mol L–1

(b) 0.75 mol L–1

(a) 23.9 mm Hg

(b) 24.2 Hg

(c) 0.5 mol L–1

(d) 1.25 mol L–1

(c) 21.42 mm Hg

(d) 21.44 mm Hg

16.

75

SOLUTIONS 17.

Liquid A and B form an ideal solution. If vapour pressure –2

22.

–2

of pure A and B are 500 Nm and 200 Nm respectively, the vapour pressure of a solution of A in B containing 0.2 mole fraction would be –2

(a) 700 N m

–2

(c) 260 N m 18.

19.

(a) On mixing acetone and chloroform will form an ideal solution

–2

(b) 300 N m

–2

(d) 140 N m

(b) On mixing acetone and chloroform positive deviation is shown since the vapour pressure increases.

3 moles of P and 2 moles of Q are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are 80 and 60 torr respectively ? (a) 80 torr

(b) 140 torr

(c) 72 torr

(d) 70 torr

Intermolecular forces between n-hexane and n-heptane are nearly same as between hexane and heptane individually. When these two are mixed, which of the following is not true about the solution formed ?

When acetone and chloroform are mixed together, hydrogen bonds are formed between them. Which of the following statements is correct about the solution made by mixing acetone and chloroform ?

(c) On mixing acetone and chloroform negative deviation is shown since there is decrease in vapour pressure. (d) At a specific composition acetone and chloroform will form minimum boiling azeotrope. Azeotropes 23.

(a) It obeys Raoult’s law, i.e. pA = x A p oA and p B  x B p oB

Given below are few mixtures formed by mixing two components. Which of the following binary mixtures will have same composition in liquid and vapour phase ? (i) Ethanol + Chloroform (iii) Benzene + Toluene

(c) Vmixing is zero.

(iv) Ethyl chloride + Ethyl bromide

(d) It forms minimum boiling azeotrope. 20.

(ii) Nitric acid + Water

(b) H mixing is zero.

Study the figures given below and mark the correct statement.

(a) (i) and (iii)

(b) (i) and (ii)

(c) (i), (ii) and (iii)

(d) (iii) and (iv)

Colligative Properties Of Dilute Solutions 24.

2g of sugar is added to one litre of water to give sugar solution. What is the effect of addition of sugar on the boiling point and freezing point of water ? (a) Both boiling point and freezing point increase. (b) Both boiling point and freezing point decrease. (c) Boiling point increases and freezing point decreases.

(a) (i) Nitric acid + Water, (ii) Acetone + Ethyl alcohol

(d) Boiling point decreases and freezing point increases.

(b) (i) Water + Ethyl alcohol, (ii) Acetone + Benzene (c) (i) Acetone + Ethyl alcohol (ii) Acetone + Chloroform (d) (i) Benzene + Chloroform (ii) Acetone + Chloroform 21.

Two liquids HNO3 (A) and water (B) form a maximum boiling azeotrope when mixed in the ratio of 68% and 32% respectively. It means (a) A – B interactions are stronger than A – A and B – B interactions (b) A – B interactions are weaker than A – A and B – B interactions (c) vapour pressure of solution is more than the pure components (d) vapour pressure of solution is less since only one component vaporises.

Relative Lowering of Vapour Pressure 25.

In three beakers labelled as (A), (B) and (C), 100 mL of water, 100 mL of 1M solution of glucose in water and 100 mL of 0.5 M solution of glucose in water are taken respectively and kept at same temperature.

Which of the following statements is correct ?

SOLUTIONS

76

(a) Vapour pressure in all the three beakers is same.

32.

(b) Vapour pressure of beaker B is highest. (c) Vapour pressure of beaker C is highest. (d) Vapour pressure of beaker B is lower than that of C and vapour pressure of beaker C is lower than that of A. 26.

27.

28.

29.

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1g of Y in 20g of liquid X. If molar mass of X is 200, what is the molar mass of Y ? (a) 20

(b) 10

(c) 100

(d) 30

33.

34.

The vapour pressure of benzene at a certain temp. is 640 mm Hg . A non-volatile-non-electrolyte solid weighing 2.175g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance ? (a) 6.96

(b) 60

(c) 63.8

(d) none of the above

The vapour pressure of solution of 5g of non-electrolyte –2 in 100 g of water at a particular temperature is 2985 Nm . The vapour pressure of pure water at that temperature is –2 3000 Nm . The molecular weight of the solute is (a) 180

(b) 90

(c) 270

(d) 200.

35.

36.

37.

CCl4. Find the vapour pressure of the solution. (Density

(c) 50.14ºC

(d) none of these

The boiling point of 0.1 molal aqueous solution of urea is 100.18ºC at 1 atm. The molal elevation constant of water is (a) 1.8

(b) 0.18

(c) 18

(d) 18.6

At certain hill-station, pure water boils at 99.725ºC. If Kb

(a) 100.079ºC

(b) 103ºC

(c) 100.359ºC

(d) unpredictable –1

The molal b.p. constant for water is 0.513ºC kg mol . When 0.1 mole of sugar is dissolved in 200g of water, the solution boils under a pressure of 1 atm at (a) 100.513ºC

(b) 100.0513ºC

(c) 100.256ºC

(d) 101.025ºC

An aqueous solution containing 1g of urea boils at 100.25ºC. The aqueous solution containing 3g of glucose in the same volume will boil at (a) 100.75ºC

(b) 100.5ºC

(c) 100ºC

(d) 100.25ºC

If the elevation in boiling point of a solution of 10gm of solute (mol. wt. = 100) in 100 gm of water is Tb , the

3

of CCl4 = 1.58 g/cm ).

ebullioscopic constant of water is

(a) 141.93 mm

(b) 94.39 mm

(c) 199.34 mm

(d) 143.99 mm

Elevation in boiling point A solution containing 12.5g of non-electrolyte substance in 185g of water shows boiling point elevation of 0.80 K. Calculate the molar mass of the substance.

38.

(b) 25.3 g mol–1

(c) 16.08g mol–1

(d) 43.92g mol–1

If 1g of solute (molar mass = 50g mol–1) is dissolved in 50g of solvent and the elevation in boiling point is 1K. The molar boiling constant of the solvent is (a) 2

(b) 3

(c) 2.5

(d) 5

(b) 10Tb

(c) Tb

(d) Tb /10

What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10º C ? (Kf = 1.86º C m–1). (Molecular mass of glycerol is 92)

(Kb = 0.52 K kg mol ) (a) 53.06g mol–1

(a) 10

Depression in Freezing Point

–1

31.

(b) 100.28ºC

–1

3

30.

(a) 10.028ºC

for water is 0.513ºC kg mol , the boiling point of 0.69m solution of urea will be

Vapour pessure of CCl4 at 25ºC is 143 mm Hg. 0.5g of a non-volatile solute (mol. wt. 65) is dissolved in 100 cm of

The molal elevation constant for water is 0.56ºC per kg of water. The boiling point of solution made by dissolving 6.0g of water (NH2CONH2) in 200g of water is

39.

(a) 496 g

(b) 297 g

(c) 310 g

(d) 426 g

Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is (a) lowering in vapour pressure of snow (b) depression in freezing point of snow (c) increase in freezing point of snow (d) melting of ice due to increase in temperature by putting salt.

77

SOLUTIONS 40.

A solution of 1.25g of non-electrolyte in 20g of water freezes

Osmotic Pressure

–1

at 271.94 K. If Kf = 1.86 K Kg mol , then the molecular

47.

weight of the solute will be (Tfo  0º C) –1

(b) 207.8g mol

–1

–1

(c) 209.6g mol 41.

42.

If beaker X contains water, Y and Z contain

–1

(a) 179.79g mol

(d) 109.6g mol

The amount of urea to be dissolved in 500 cc of water (K = 1.86) to produce a depression of 0.186ºC in the freezing point is (a) 0.3 gm

(b) 3 gm

(a) Y-hypotonic solution, Z-hypertonic solution

(c) 6 gm

(d) 9 gm.

(b) Y-hypertonic solution, Z-hypotonic solution

The molar freezing point constant for water is 1.86º C/m. If

(c) Y and Z-isotonic solution

342g of cane sugar (C12H22O11) is dissolved in 1000 g of

(d) Y and Z-hypotonic solutions 48.

water, the solution will freeze at

43.

(a) –1.86º

(b) 1.86ºC

(c) –3.92ºC

(d) 2.42ºC

Pure benzene freezes at 5.45ºC at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at –1

(a) 5.08 K kg mol

–1

(c) 0.508K kg mol

49.

45.

(b) 508 K kg mol

(b) –2.0ºC

(c) 0ºC

(d) 2.0ºC

50.

(c) 1.86ºC

(d) None of the above.

46.

A solution is made by dissolving 20g of a substance in 500 mL of water. Its osmotic pressure was found to be 600mm of Hg at 15ºC. Find the molecular weight of the substance. (a) 1198

(b) 500

(c) 1200

(d) 1000

Sea water is desalinated to get fresh water by which of the following methods ?

(b) When excess pressure is applied on sea water pure water moves in by osmosis. (c) Water moves out from sea water due to osmosis. (d) Salt is precipitated from sea water when kept undisturbed for sometime. 51.

Osmotic pressure

The osmotic pressure of a solution can be increased by

(a) When pressure more than osmotic pressure is applied pure water is squeezed out of sea water by reverse osmosis.

–1

(b) –1.86ºC

(d) 7368.4 g mol–1

(d) removing semipermeable membrane.

1.86 K Kg mol , Kb = 0.52 K kg mol ) (a) 0ºC

(c) 3692.1 g mol–1

(c) decreasing the temperature

An aqueous solution of a non-electrolyte solute boils at 100.52º C. The freezing point of the solution will be (Kf = –1

(b) 12315.5 g mol–1

(b) increasing the number of solute molecules

–1

(d) 50.8ºC kg mol

If 15 gm of a solute in 100 gm of water makes a solution that freezes at –1.0ºC, then 30 gm of the same solute in 100 gm of water will make a solution that freezes at

(a) 6239.6g mol–1

(a) increasing the volume

–1

(a) –0.5ºC

Osmotic pressure of a solution containing 2g dissolved protein per 300 cm3 of solution in 20 mm of Hg at 27ºC. The molecular mass of protein is

3.55ºC. The Kf for benzene is

44.

Grapes placed in three beakers X, Y and Z containing different type of solutions are shown in figures.

Osmotic pressure is generally preferred for determining the molecular masses of protein because (a) it is difficult to find out mole fraction of protein for calculations by other methods (b) at elevated temperature the proteins are likely to decompose and osmotic pressure is measured around room temperature (c) the apparatus involved in finding out osmotic pressure is simpler than other methods (d) it is easy to boil or freeze a solution containing proteins.

SOLUTIONS 52.

78

What will be the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 150,000 in 500 mL of water at 37ºC ? (a) 30.96 Pa

(b) 34.36 Pa

(c) 68.72 Pa

(d) 48.25 Pa

59.

solution if van’t Hoff factor is 2.74 ?

60.

Van't Hoff Factor 53.

Which of the following will have same value of van’t Hoff factor as that of K4 [Fe (CN)6] ?

54.

55.

(a) Al2(SO4)3

(b) AlCl3

(c) Al(NO3)3

(d) Al(OH)3

A solute X when dissolved in a solvent associated to form a pentamer. The value of van’t Hoff factor (i) for the solute will be

61.

62.

(c) 100%

(d) 92%

The van’t Hoff factor of 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is (a) 0.95

(b) 0.97

(c) 0.94

(d) 0.96

For which of the following solutes the van’t Hoff factor is not greater than one ? (a) NaNO3

(b) BaCl2

(c) K4[Fe(CN)6]

(d) NH2CONH2

The elevation in boiling point of a solution of 9.43g of MgCl2 in 1 kg of water is (Kb = 0.52 K kg mol–1, Molar mass

(c) 0.2

(d) 0.1

of MgCl2=94.3g mol–1)

Why is the molecular mass determined by measuring colligative property in case of some solutes is abnormal ?

(c) Due to decomposition of solute molecules.

(b) 1m KCl solution

(c) 1m AlCl3 solution

(d) 1m C6H12O6 solution

(b) 9.24 g

(c) 2.834 g

(d) 1.825 g

(c) 0.17

(d) 0.94

A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution ?

64.

Density of a 2.05 M solution of acetic acid in water is 1.02g/mL. The molality of the solution is

65.

The density of a solution prepared by dissolving 120g of urea (mol. mass = 60 u) in 1000 g of water is 1.15g/ml. The molarity of this solution is

66.

The density in g/ml of a 3.60 M sulphuric acid solution

What amount of CaCl2 (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at 27ºC ? (a) 3.42 g

(b) 0.52

63.

Which of the following has the highest freezing point ? (a) 1m NaCl solution

(a) 0.156

Numeric-Type questions

(d) Due to large size of solute molecules.

58.

(b) 87%

(b) 5

(b) Due to insolubility of solute molecules.

57.

(a) 75%

(a) 0.5

(a) Due to association or dissociation of solute molecules.

56.

What will be the degree of dissociation of 0.1 M Mg(NO3)2

which is 29% H2SO4 (molar mass = 98 g/mol) by mass will be

Which of the following statements is not correct ? (a) Osmotic pressure () of a solution is given by the relation  = MRT where M is the molarity of the solution.

67.

of phosphrous acid (H3PO3), the volume of 0.1 M aqueous

(b) The correct order of osmotic pressure for 0.2 M aqueous solution of each solute is CaCl2 > NaCl > CH3COOH > glucose. (c) Two solutions of sucrose of same molality prepared in different solvents will have same elevation boiling point. (d) Relative lowering in vapour pressure of a solution containing non-volatile solute is directly proportional to mole fraction of solute is Raoult’s law.

To neutralize completely 20 mL of 0.1 M aqueous solution solution of KOH required in mL is

68.

20

6.02 × 10 molecules of urea are present in 100 ml of its solution. The concentration of urea is

69.

Benzene and toluene form nearly ideal solutions. At 20ºC, the vapour pressure of benzene is 75 Torr and that of toluene is 22 Torr. The partial vapour pressure of benzene at 20ºC for a solution containing 78 g of benzene and 46 g of toluene in Torr is

79

SOLUTIONS 70.

71.

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be –1

73.

–1

CuCl2 (Molecular weight = 134.4, Kb = 0.52 K molarity ) in 1 kg water using the following information will be 74.

Kf for water is 1.86 K kg mol . If your automobile radiator

72.

A 5.25% solution of a substance is isotonic with a 1.5%

In 0.2m aqueous solution of weak acid HX, the degree of ionization is 0.3. Taking Kf for water as 1.85, the freezing

holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC ?

The elevation in boiling point of a solution of 13.44 g of

point of solution (in ºC) will be nearest to 75.

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,

solution of urea (molar mass = 60 g mol ) in the same

the change in freezing point of water in K (Tf), when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is

solvent. If the densities of both the solutions are assumed

(kf = 1.86 K Kg mol )

–1

–3

to be equal to 1.0 g cm , molar mass of the substance in g/ mol will be

–1

SOLUTIONS

80

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTION 1.

2.

The vapour pressure of acetone at 20°C is 185 torr. When

7.

1.2 g of a non-volatile substance was dissolved in 100 g

protic strong acid must be mixed with a 20% solution of

of acetone at 20°C, its vapour pressure was 183 torr. The

the same acid to produce 800 mL of a 29.875% acid

molar mass (g mol–1) of the substance is:

solution ?

(2015)

(b) 488

(a) 320

(b) 325

(c) 32

(d) 64

(c) 316

(d) 330

Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by

8.

freezing point was found to be 3.82°C. If Na2SO4 is 81.5% ionized, the value of x Online 2017 SET (1)

(b) dissociation

(Kf for water = 1.86°C kg mol–1) is approximately :

(c) complex formation

(d) association

(molar mass of S = 32 g mol–1 and that of Na = 23 g mol–1)

The solubility of N2 in water at 300 K and 500 torr partial 1

pressure is 0.01 gL .The solubility (in gL ) at 750 torr partial pressure is :

5.

6.

5g of Na2SO4 was dissolved in x g of H2O. The change in

(a) partial ionization

1

4.

(2017)

(a) 128

Online 2015 SET (1)

3.

What quantity (in mL) of a 45% acid solution of a mono-

Online 2016 SET (1)

(a) 0.0075

(b) 0.015

(c) 0.02

(d) 0.005

9.

(b) 752.4

(c) 759.0

(d) 7.6

(c) 45 g

(d) 65 g

A solution is prepared by mixing 8.5 g of CH2Cl2 and

(Molar mass of Cl = 35.5 g mol–1)

10.

An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is Online 2016 SET (2)

(b) 25 g

11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is :Online 2017 SET (2)

18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is: (2016) (a) 76.0

(a) 15 g

(a) 0.162

(b) 0.675

(c) 0.325

(d) 0.486

Two 5 molal solutions are prepared by dissolving a nonelectrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are MX

3 4

(a) 0.33

(b) 0.50

and MY, respectively where MX = MY . The relative

(c) 0.67

(d) 0.80

0.2 g of acetic acid is added to 20 g of benzene. If acetic

lowering of vapour pressure of the solution in X is “m” times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of “m” is :

acid associates to form a dimer in benzene, percentage

Online 2018 SET (2)

The freezing point of benzene decreases by 0.45°C when

association of acetic acid in benzene will be : (Kf for benzene = 5.12 K kg mol–1) (a) 80.4% (c) 94.6%

(a)

4 3

(b)

3 4

(c)

1 2

(d)

1 4

(2017)

(b) 74.6% (d) 64.6%

SOLUTIONS 11.

81

For 1 molal aqueous solution of the following compounds,

15.

which one will show the highest freezing point? (2018)

12.

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation

(a) [Co(H2O)4Cl2]Cl.2H2O

(b) [Co(H2O)3Cl3].3H2O

of the given ionic compounds in water, the concentration

(c) [Co(H2O)6]Cl3

(d) [Co(H2O)5Cl]Cl2.H2O

of XY (in mol L1 ) in solution is

The mass of a non-volatile, non-electrolyte solute (molar

(a) 4 104

(b) 6 102

mass =50 g mol-1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75 percent is :

(c) 16  104

(d) 4  102 (2019-04-09/Shift-1)

Online 2018 SET (3)

13.

(a) 37.5 g

(c) 75 g

(b) 16.6 g

(d) 50 g

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are:

16.

4.0 K kg mol 1 The depression in the freezing point of the

solvent for 0.03mol kg 1 solution K 2SO4 is: (Assume complete dissociation of the electrolyte) (2019-04-09/Shift-2)

17.

14.

(c) 5 00 mmHg, 0.5, 0.5

(b) 450 mmHg, 0.5, 0.5

(d) 450 mmHg, 0.4, 0.6

Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is:

(a) 0.12 K

(b) 0.36 K

(c) 0.18 K

(d) 0.24 K

What would be the molality of 20% (mass/mass) aqueous solution of KI? (molar mass of KI  166g mol1 )

(2019-04-08/Shift-1) (a) 500 mmHg, 0.4, 0.6

Molal depression constant for a solvent is

(2019-04-09/Shift-2)

18.

( xM  Mole fraction of ‘M’ in solution;

(a) 1.08

(b)1.48

(c) 1.51

(d) 1.35

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, Calculate the lowering of vapour pressure will be in mm of Hg. (molar mass of urea= 60 g mol-1) (2019-04-10/Shift-1)

xN  Mole fraction of ‘N’ in solution; yM Mole fraction of ‘M’ in vapour phase; yN  Mole fraction of ‘N’ in vapour phase) (2019-04-09/Shift-1)

XM yM (a) X  y N N

XM yM (b) X  y N N

19.

(a) 0.027 mmHg

(b) 0.031 mmHg

(c) 0.028 mmHg

(d) 0.017 mmHg

1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1: 5. The ratio

Tb  A  of the elevation in their boiling points,  B , is: b  (2019-04-10/Shift-2)

X M yM (c)  X M  y M    x N  YN  (d) X  y N N

(a) 5: 1

(b) 10: 1

(c) 1: 5

(d) 1: 0.2

SOLUTIONS 20.

82

The mole fraction of a solvent in aqueous solution of a

25.

1

solute is 0.8.The molality (in mol kg ) of the aqueous solution is (2019-04-12/Shift-1) (a) 13.88  10

2

(c) 13.88 21.

(b) 13.88  10

(a) 1 cup of water to 2 cups of pure milk

1

(b) 3 cups of water to 2 cups of pure milk (c) 1 cup of water to 3 cups of pure milk

(d) 13.88  103

A solution is prepared by dissolving 0.6g of urea (molar mass = 60 g mol-1) and 1.8 g of glucose (molar mass =180 g mol-1) in 100 mL, of water at 27°C. The osmotic pressure of the (R = 0.08206 L atm K-1 mol-1) solution is:

(d) 2 cups of water to 3 cups of pure milk 26.

22.

(b) 1.64 atm

(c) 4.92 atm

(d) 2.46 atm

K2HgI4 is 40% ionised in aqueous solution. The value of its van’t Hoff factor (i) is: (2019-01-11/Shift-2)

(2019-04-12/Shift-2) (a) 8.2 atm

The freezing point of diluted milk sample is found to be 0.2C , while it should have been 0.5C for pure milk. How much water been added to pure milk to make the diluted sample? (2019-01-11/Shift-1)

27..

Which one of the following statements regarding Henry’s law is not correct?

(a) 1.6

(b) 1.8

(c) 2.0

(d) 2.2

Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is: (2019-01-12/Shift-1)

(2019-01-09/Shift-1) (a) Higher the value of KH at a given pressure, higher is the solubility of the gas in liquids. (b) Different gases have different K H (Henry’s law constant) values at the same temperature.

28..

(c) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. (d) The value of KH increases with increase of temperature and KH is function of the nature of the gas. 23.

A solution containing 62 g ethylene glycol in 250 g water is cooled to –10°C. If Kf for water is 1.86 K kg mol-1, the amount of water (in g) separated as ice is:

(b) 2A

(c) A

(d) 4A

Molecules of benzoic acid (C6H5COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is:(Given that Kf = 5 K kg mol-1, Molar mass of benzoic acid = 122 g mol-1). (2019-01-12/Shift-2) (a) 2.4g

(b) 1.0g

(c) 1.5g

(d) 1.8g

(a) 48

(b) 32

8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH in solution and molality (in mol kg- 1) of the solution respectively are:

(c) 64

(d) 16

(2019-01-12/Shift-2)

(2019-01-09/Shift-2)

24.

(a) 3A

29.

Liquids A and B form and ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are 7  103 Pa and 12  103 Pa , respectively..

(a) 0.2, 22.20

(b) 0.2, 11.11

(c) 0.167, 11.11

(d) 0.167, 22.20

The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is

Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is:

(2019-01-10/Shift-1)

(2019-01-10/Shift-2)

30.

(a) x A  0.37; x B  0.63

(b) x A  0.28; x B  0.72

(a) K b  1.5 K f

(b) K b  K f

(c) x A  0.4; x B  0.6

(d) x A  0.76; x B  0.24

(c) K b  0.5 K f

(d) K b  2 K f

83

SOLUTIONS 31.

The amount of sugar (C12H22O11) required to

37.

prepare 2 L of its 0.1 M aqueous solution is: (2019-01-10/Shift-2)

32.

(a) 136.8 g

(b) 17.1 g

(c) 68.4 g

(d) 34.2 g

The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following process can explain this ?

(2020-09-04/Shift-1)

(a) Diffusion

(b) Osmosis

The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution with 2 L of the glucose solution is x  103 atm. x is _______. (nearest integer)

(c) Reverse osmosis

(d) Dialysis

(2020-09-04/Shift-2)

38.

(2020-09-02/Shift-2)

33.

At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state _____?

Henry’s constant (in kbar) for four gases , ,  and 

39.

in water at 298 K is given below :

(density of water = 103 kg m–3 at 298 K)

A solution of two components containing n1 moles of the 1st component and n2 moles of the 2 nd component is prepared. M 1 and M 2 are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g mL-1, C2 is the molarity and x2 is the mole fraction of the 2nd component, then C2 can be expressed as :

This table implies that :

(2020-09-06/Shift-1) (2020-09-03/Shift-1)

(a) solubility of

 at 308 K is lower than at 298 K.

(b) The pressure of a 55.5 molal solution of (c)



The mole fraction of glucose

dx2 (c) C2  M  x  M  M  2 2 2 1

 is 1 bar..

(C6H12O6 ) in an aqueous

binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is ………. . (2020-09-03/Shift-1) 35.

1000x2 (b) C2  M  x  M  M  1 2 2 1

 is 250 bar..

has the highest solubility in water at a given pressure.

(d) The pressure of a 55.5 molal solution of 34.

dx1 (a) C2  M  x  M  M  2 2 2 1

1000dx2 (d) C2  M  x  M  M  1 2 2 1

40..

The strengths of 5.6 volume hydrogen peroxide (of density 1 g/mL) in terms of mass percentage and molarity (M), respectively, are:(Take molar mass of hydrogen peroxide as 34 g/mol).

The elevation of boiling point of 0.10 m aqueous CrCl3xNH3 solution is two times that of 0.05 m aqueous CaCl2 solution. The value of x is …………….. . [Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere]

(2020-09-03/Shift-2)

36.

(a) 0.85 and 0.5

(b) 0.85 and 0.25

(c) 1.7 and 0.25

(d) 1.7 and 0.5

If 250 cm3 of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular masses of A and B is ……… × 10–2 (to the nearest integer). (2020-09-03/Shift-2)

(2020-09-06/Shift-1) 41.

A set of solution is prepared using 180g of water as a solvent and 10g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A = 100g mol-1 ; B =200g mol-1 ; C = 10,000g mol-1] (2020-09-06/Shift-2) (a) A > C > B

(b) B > C > A

(c) C > B > A

(d) A > B > C

SOLUTIONS 42.

84

At 35oC, the vapour pressure of CS2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is :

The following inferences are made: (A) X has higher intermolecular interactions compared toY (B) X has lower intermolecular interactions compared toY

(a) CS2 and acetone are less attracted to each other than to themselves

(C) Z has lower intermolecular interactions compared toY

(b) heat must be absorbed in order to produce the solution at 35oC

The correct inference(s)is/are: Shift-1)

(c) Raoult’s law is not obeyed by this system (d) a mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL (2020-01-07/Shift-1) 43.

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a nonvolatile solute are together sealed in a container. Over time:

(2020-01-08/

(a) C

(b) A

(c) B

(d) A and C

(2020-01-07/Shift-2) (a) the volume of the solution and the solvent does not change (b) the volume of the solution increases and the volume of the solvent decreases (c) the volume of the solution decreases and the volume of the solvent increases

45.

(Molecular weight of HNO3= 63).

(d) the volume of the solution does not change and the volume of the solvent decreases 44.

A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below:

The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is—

(2020-01-09/Shift-1) 46.

How much amount of NaCl should be added to 600 g of water (ñ=1.00 g/mL) to decrease the freezing point of water to -0.2°C? (The freezing point depression constant for water = 2 K Kg mol–1) (2020-01-09/Shift-1)

47.

3

One litre of sea water (d =1.03g/cm ) contains 10.3 mg of O2 gas. Determine the concentration of O2 in ppm:

(2020-01-09/Shift-2)

85

SOLUTIONS

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS 9.

Single Choice Questions 1.

2.

3.

4.

5.

6.

7.

If 0.50 mole of BaCl2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (a) 0.1

(b) 0.2

(c) 0.5

(d) 0.7

In a solution of 7.8 g of benzene (C6H6) and 46 g of toluene (C6H5CH3), the mole fraction of benzene is (a) 1/6

(b) 1/5

(c) 1/2

(d) 1/3

11.

To 5.85 g of NaCl, one kg of water is added to prepare a solution. What is the strength of NaCl in this solution ? (Molecular weight of NaCl = 58.5) (a) 0.1 Normal

(b) 0.1 Molal

(c) 0.1 Molar

(d) 0.1 Formal

12.

The formula weight of H2SO4 is 98. The weight of the acid in 400 ml of 0.1 M solution is (a) 2.45 g

(b) 3.92 g

(c) 4.90 g

(d) 9.8 g

The molarity of a solution of Na2CO3 having 10.6 g/500 ml of solution is (a) 0.2 M

(b) 2 M

(c) 20 M

(d) 0.02 M

13.

Molarity of a solution containing 1g NaOH in 250 ml of solution is (a) 0.1 M

(b) 1 M

(c) 0.01 M

(d) 0.001 M

Which one of the following solutions of sulphuric acid will exactly neutralise 25 ml of 0.2 M sodium hydroxide solution ? (a) 12.5 ml of 0.1 M solution (b) 25 ml of 0.1 M solution (c) 25 ml of 0.2 M solution (d) 50 ml of 0.2 M solution

8.

10.

The normality of 10% (weight/volume) acetic acid is (a) 1 N

(b) 10 N

(c) 1.7 N

(d) 0.83 N

30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength of the acid solution is (a) 0.1 N

(b) 0.15 N

(c) 0.3 N

(d) 0.4 N

10 ml of 3N-HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. The normality of the resulting solution is (a) N/2

(b) N/10

(c) N/20

(d) N/40

50 ml of 0.2 N HCl, 50 ml of 0.1 N H2SO4 and 100 ml 0.2 N HNO3 are mixed. The normality of resulting solution is (a) 0.10 N

(b) 0.15 N

(c) 0.175 N

(d) 0.20 N

A solution is prepared containing a 2 : 1 mol ratio of dibromo ethane (C2 H 4Br 2) and dibromo propane (C3H6Br2) what is the total vapour pressure over the solution assuming ideal behaviour ? Vapour pressure

(mm Hg)

C2H4Br2

173

C3H6Br2

127

(a) 300 mm Hg

(b) 158 mm Hg

(c) 150 mm Hg

(d) 142 mm Hg

Which of the following graphs represent the behaviour of ideal binary liquid mixture ? (a) Plot of 1/PTotal against yA is linear (b) Plot of 1/PTotal against yB is non-linear (c) Plot of PTotal against yA is linear (d) Plot of PTotal against yB is linear

14.

Which of the following statements is correct for a binary solution ? (a) A solution in which heat is evolved exhibits positive deviations from Raoult's law. (b) A solution in which heat is absorbed shows negative deviations from Raoult's law. (c) When one component in solution shows negative deviation from Raoult's law, the other exhibits positive deviation. (d) When one component in solution shows positive deviation from Raoult's law, so does the other.

SOLUTIONS 15.

16.

17.

18.

19.

Raoult's law states that for a dilute solution, (a) the lowering of vapour pressure is equal to the mole fraction of the solute (b) the relative lowering of vapour pessure is proprotional to the amount of solute in the solution (c) the relative lowering of vapour pressure is equal to the mole fraction of the solute (d) the vapour pressure of the solution is equal to the mole fraction of the solvent Which statement about the composition of vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume the temperature is constant at 25ºC. Vapour pressure data (25ºC) : Benzene 75 mm Hg Toluene 22 mm Hg (a) The vapour will contain higher percentage of benzene (b) The vapour will contain higher percentage of toluene (c) The vapour will contain equal amount of benzene and toluene (d) Not enough information is given to make a prediction Which of the following pairs shows a negative deviation from Raoult's law ? (a) Acetone–benzene (b) Acetone–ethanol (c) Acetone–chloroform (d) Benzene–methanol 100 ml of a liquid A was mixed with 25 ml of a liquid B to give a non-ideal solution of A-B mixture. The volume of this mixture would be (a) 75 ml (b) 125 ml (c) close to 125 ml (d) just more than 125 ml The diagram given below is a vapour pressurecomposition diagram for a binary solution of A and B. In the solution, A–B interactions are (a) similar to A–A and B–B interactions (b) greater than A–A and B–B interactions (c) smaller than A–A and B–B interactions

86 20.

An azeotropic solution of two liquids has boiling point lower than either of them when it (a) shows negative deviation from Raoult’s law (b) shows no deviation from Raoult’s law (c) show positive deviation from Raoult’s law (d) is saturated

21.

Mixture of volatile components A and B has total vapour pressure (in torr) : P = 254 – 119XA where, XA is mole fraction of A in mixture. Hence p0A and p0B are (in torr) :

22.

(a) 254, 119

(b) 119, 254

(c) 135, 254

(d) 154, 119

1 1 The plot of y against x is linear with slope and A A intercept respectively : PA0 PA0  PB0 and (a) 0 PB PB0

(c) 23.

24.

PB0 P0  P0 and A 0 B 0 PA PA

PA0 PB0  PA0 and (b) 0 PB PB0

(d)

PB0 P0  P0 and B 0 A 0 PA PB

If PA is the vapour pressure of a pure liquid A and the mole fraction of A in the mixture of two liquids A and B is x, the partial vapour pressure of A is : (a) (1  x) PA

(b) xPA

x (c) (1  x) PA

(d)

1 x PA x

The millimoles of N2 gas that will dissolve in 1L of water at 298 K. when it is bubbled through water and has a partial pressure of 0.96 bar will be (Given that at 298 K KH = 76.8 k bar)

(d) unpredictable.

25.

(a) 0.59

(b) 0.69

(c) 0.79

(d) 0.89

Air contains O2 and N2 in the ratio of 1 : 4. The ratio of their solubilities in terms of mole fractions at atmospheric pressure and room temperature will be (Given Henry’s 7 7 constant for O2 = 3.30 × 10 torr, for N2 = 6.60 × 10 torr) (a) 1 : 2

(b) 2 : 1

(c) 4 : 1

(d) 1 : 4

87

SOLUTIONS 26.

Consider the following statements

32.

1. Isotonic solutions have the same molar concentration at a given temperature. 2. The molal elevation constant Kb is characteristic of a solvent, and is independent of the solute added. 3. The freezing point of a 0.1 M aqueous KCl solution is more than that of a 0.1 M aqueous AlCl3 solution.

33.

Which of these statements is correct. (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3 27.

0

An aqueous solution freezes at –2.55 C. What is its

34.

boiling point ( K b of water  0.52 K / m; K f of water

 1.86 K / m) ? 0

(a) 107.0 C 0

(c) 100.1 C 28.

29.

30.

31.

0

(b) 100.6 C

Elevation in boiling point was 0.520C when 6 gm of a compound X was dissolved in 100 gm of water. Molecular weight of X is (Kb of water is 0.52 K-kg/mol) (a) 120

(b) 60

(c) 600

(d) 180

The molal b.p. constant for water is 0.5130C kg mol–1. When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at (a) 100.5130C kg/mol

(b) 100.05130C kg/mol

(c) 100.2560C kg/mol

(d) 101.0250C kg/mol

The latent heat of vaporisation of water is 9700 cal/mole and if the b.p. is 1000C, the ebullioscopic constant of water is (a) 0.5130C kg/mol

(b) 1.0260C kg/mol

(c) 10.260C kg/mol

(d) 1.8320C kg/mol

0

(d) 100.7 C

The relative lowering of the vapour pressure is equal to the ratio between the number of (a) solute molecules to the solvent molecules (b) solute molecules to the total molecules in the solution (c) solvent molecules to the total molecules in the solution (d) solvent molecules to the total number of ions of the solute Vapour pressure of a solution of non-volatile solute is (a) directly proportional to the mole fraction of the solvent (b) Independent of mole fraction of the solute (c) inversely proportional to the mole fraction of the solvent (d) directly proportional to the mole fraction of the solute Vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985 N/ m2. The vapour pressure of pure water is 3000 N/m2, the molecular weight of the solute is (a) 60 (b) 120 (c) 178.2 (d) 380 The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A nonvolatile electrolyte solute weighing 2.175 g is added to 30.08 g of benzene. If the vapour pressure of the solution is 600 mm of Hg, what is the molecular weight of the solid substance ? (a) 79.82

(b) 84.46

(c) 59.60

(d) 49.50

35.

36.

37.

The normal boiling point of toluene is 110.70C, and its boiling point elevation constant is 3.32 K kg mol–1. The enthalpy of vapourization of toluene is nearly (a) 17.0 kJ mol–1

(b) 34.0 kJ mol–1

(c) 51.0 kJ mol–1

(d) 68.0 kJ mol–1

The freezing point of a 0.05 molal solution of a nonelectrolyte in water is (a) –1.860C

(b) –0.930C

(c) –0.0930C

(d) 0.930C

During depression of freezing point in a solution the following are in equilibrium (a) liquid solution, solid solvent (b) liquid solvent, solid solute (c) liquid solute, solid solute (d) liquid solute, solid solvent

38.

39

Calculate the molecular weight of a substance if the freezing point of a solution containing 100 g of benzene and 0.2g of the substance is 0.17 K below that of benzene. The cryoscopic constant of benzene is 5.16 K kg mol–1. (a) 70.46

(b) 85.66

(c) 60.23

(d) 178.25

On freezing an aqueous solution of sugar, the solid that starts separating out is (a) sugar (b) ice (c) solution with the same composition (d) solution with a different composition.

SOLUTIONS 40.

88

Given that Tf is the depression in freezing point of the solvent in a solution of a non volatile solute of molality

47.

 T f    is equal to 1, the quantity lim m 1  m 

42.

43.

44.

(b) Kb (ebullioscopic constant)

(d) greater than 1 and less than 1 48.

Which one has the highest b.p. ?

(d) Hfus (enthalpy of fusion)

(a) 0.1 N Na2SO4

(b) 0.1 N MgSO4

10 g of glucose (1), 10 g of urea (2) and 10 g of sucrose (3) are dissolved in 250 mL of water at 300 K ( = osmotic pressure of solution). The relationship between the osmotic pressure of the solutions is

(c) 0.1 M Al2(SO4)3

(d) 0.1 M BaSO4

(a) 1 > 2 > 3

(b) 3 > 1 > 2

(c) 2 > 1 > 3

(d) 2 > 3 > 1

Which statement is incorrect about osmotic pressure (), volume (V) and temperature (T) ? (a) 1/v when T is constant. (b) T when V is constant. (c)  V when T is constant. (d) V is constant when T is constant. Which of the following pairs of solutions is isotonic ? (a) 6% urea and 6% glucose (w/v) (b) 18% urea and 18% glucose (w/v) (c) 6% urea and 18% fructose (w/v) (d) 34.2% sucrose and 60% glucose (w/v) Two aqueous solutions S1 and S2 are separated by a semi-permeable membrane. S2 has lower vapour pressure than S1. Then

49.

50.

51.

Which of the following aqueous solutions has the highest boiling point ? (a) 0.1 M KNO3

(b) 0.1 M Na3PO4

(c) 0.1 M BaCl2

(d) 0.1 M K2SO4

0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 K –1 Kg mol . The freezing point of the solution is : (a) – 0.45

(b) – 0.90

(c) – 0.31

(d) – 0.53

The freezing point of equimolal aqueous solutions will be highest for (a) C6H5NH3Cl (aniline hydrochloride) (b) Ca(NO3) (c) La (NO3)3 (d) C6H12O6 (glucose)

52.

0.01 M solution each of urea, common salt and Na2SO4 are taken, the ratio of depression of freezing point is

(a) More solvent will flow from S1 to S2

(a) 1 : 1 : 1

(b) 1 : 2 : 1

(b) More solvent will flow from S2 to S1

(c) 1 : 2 : 3

(d) 2 : 2 : 3

(d) No flow will take place.

The osmotic pressures of equimolar solutions of Al2(SO4)3, KCl and sugar will be in the order

A plant cell shrinks when it is kept in

(a) KCl < Al2(SO4)3< sugar

(a) Hypotonic solution

(b) sugar < KCl < Al2 (SO4)3

(b) A hypertonic solution

(c) sugar > KCl > Al2 (SO4)3

(c) A solution isotonic with cell sap

(d) KCl < sugar < Al2 (SO4)3

(d) Water. 46.

(b) less than 1 and greater than 1 (c) less than 1 and less than 1

(c) Solvent from S1 and S2 will flow at equal rates 45.

(a) greater than 1 and greater than 1

(a) Lf (latent heat of fusion)

(c) Kf (cryoscopic constant)

41.

The van't Hoff factors i for an electrolyte which undergoes dissociation and association in solvents are respectively

53.

If a solute undergoes dimerization and trimerization, the minimum values of the van't Hoff factors are

Solutions A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl2 and 0.1 M AlCl3. Which of the following pairs is isotonic ?

(a) 0.50 and 1.50

(b) 1.50 and 1.33

(a) A and B

(b) B and C

(c) 0.50 and 0.33

(d) 0.25 and 0.67

(c) A and D

(d) A and C

54.

89

SOLUTIONS 55.

56.

57.

58.

59.

0.004 M Na2SO4 is isotonic with 0.01 M glucose. The degree of dissociation of Na2SO4 is (a) 75% (b) 50% (c) 25% (d) 85% The average osmotic pressure of human blood is 7.8 bar at 370C. What is the concentration of an aqueous NaCl solution that could be introduced in the bloodstream ? (a) 0.16 mol L–1

(b) 0.32 mol L–1

(c) 0.45 mol L–1

(d) 0.65 mol L–1

62.

(a) have V (mixing) = + ve (b) have H (mixing) = – ve (c) form minimum boiling azeotropes. (d) have lower vapour pressure of each component in the solution than their pure vapour pressure. 63.

(b) –3.720C

(c) +1.860C

(d) +3.720C

(d) TSSol’n is always positive

The molal freezing point constant for water is 1.86 C/ mole. Therefore, the freezing point of 0.1M NaCl solution in water is expected to be (b) –0.1860C

(c) –0.3720C

(d) +0.3370C

(b) behaves as an ideal solution of A into B when XA0 (c) HSol’n is always positive

0

(a) –1.860C

Among 0.1 M solutions of urea, Na3PO4 and Al2 (SO4)3, which is incorrect (a) the vapour pressure and freezing point are the lowest for urea (b) the vapour pressure and freezing point are the highest for urea

64.

(b) chlorobenzene and bromobenzene (c) chloroform and benzene (d) n-butyl chloride and n-butyl bromide. 65.

(a) 0.05 (c) 0.15

(b) alters on changing the external pressure (c) remains unchanged during distillation at a constant external pressure (d) fluctuates even at constant pressure 66.

(b) inversely proportional to molecular mass of solute

(b) 0.30 (d) 0.015

In binary liquid mixture of components, A and B, the former has greater tendency to escape into the vapour state than demanded by Raoult’s law. Indicate the correct statement(s) (a) Component A shows positive deviation and the component B negative deviation (b) Both components show positive deviations (c) Component A shows negative deviation and the component B positive deviation (d) The component B has also greater tendency to escape into the vapour state than demanded by Raoult’s law

Colligative properties of a solution are (a) independent of the nature of solute (c) Proportional to concentration of solute (d) independent of the amount of solvent.

Multiple Choice Questions 61.

Composition of an azeotrope (a) is independent of external pressure because it is a compound

(d) the depression in freezing point is the highest for Al2(SO4)3 When 1.345 g of CuCl2 is dissolved in 1 kg of water, the elevation in boiling point will be (Kb = 0.52, molar mass of CuCl2 = 134.5)

Which pair (s) of liquids on mixing are expected to show no net volume change and no heat effect (a) acetone and ethanol

(c) the boiling point is the highest for Al2(SO4)3

60.

A binary liquids mixture of two liquid A and B showing the departure from the ideal behaviour : (a) behaves as an ideal solution of B into A when XB0

The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is (a) –1.860C

The solution showing positive deviation

67.

Which of the following is/are correct statements (s) about the Raoult’s law applied to a solution of non-volatile solute ? (a) Vapour pressure of solution is proportional to mole fraction of solute (b) V.P. of solution is proportional to the mole fraction of solvent (c) Relative lowering of V.P. = mole fraction of solute (d) Relative lowering of vapour pressure is proportional to the mole fraction of solute.

SOLUTIONS 68.

90

In the depression in freezing point experiment, it is observed that (a) the vapour pressure of the solution is less than that of pure solvent.

71.

72.

(b) the vapour pressure of the solution is more than that of pure solvent (c) only solute molecules solidify at the freezing point

Determine molecular weight of sodium salt of protein. (a) 2300 (b) 4600 (c) 9200 (d) 11500 Determine molecular weight of acidic form of protein : (a) 11390 (b) 2278 (c) 9112 (d) 4556

Assertion–Reason Type Questions

(d) only solvent molecules solidify at the freezing point. 69.

Study the figure given aside and pick out the correct option(s) of the following :

(A) If both ASSERTION and REASON are true and reason is the correct explanation of the assertion. (B) If both ASSERTION and REASON are true but reason is not the correct explanation of the assertion. (C) If ASSERTION is true but REASON is false. (D) If ASSERTION is false but REASON is true. 73. Assertion (A) : The sum of mole fractions of all components of a solution is unity. Reason : Mole fraction is independent of temperature. (a) A (b) B (c) C (d) D 74.

(a) A white precipitate of AgCl is formed on AgNO3 side (b) A white precipitate of AgCl is formed on BaCl2 side (c) No precipitate is formed on either side (d) Meniscus of BaCl2 solution rises and that of AgNO3 solution falls in due course of time

75.

Comprehension Based Questions Comprehension A Protein has been isolated as sodium salt with their molecular form Na 2 P (this notation means xNa  ions are associated with a negatively charged protein P  x ). A solution of this salt was prepared by dissolving x y

76.

of this sodium salt of protein in 10g of water and ebullioscopic analysis revealed that solution boils at temperature 6.78×10-3 o C higher than the normal boiling point of pure water. K b of water of

0.52k kg mol1 . Also elemental analysis revealed that the salt contain 1% sodium metal by weight. 70.

Deduce molecular formula of protein. (a) NaP

(b) Na2P

(c) Na4P

(d) Na5P

77.

Assertion (A) : Hmix and Vmix are zero for the ideal solution. Reason : The interactions between the particle of the components of a solution are almost identical as between particles in the liquids. (a) A (b) B (c) C (d) D Assertion (A) : The vapour pressure of a liquid decreases if some nonvolatile solute is dissolved in it. Reason (R) : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution. (a) A (b) B (c) C (d) D Assertion (A) : If more and more non-volatile solute is added to a solvent, the freezing point of the solution keeps on reducing. Reason (R) : Presence of large amount of the solid solute does not allow the solution to freeze. (a) A (b) B (c) C (d) D Assertion (A) : The boiling point of 0.1 M urea solution is less than that of 0.1 M KCl solution. Reason (R) : Elevation of boiling point is directly proportional to the number of species present in the solution. (a) A

(b) B

(c) C

(d) D

91

SOLUTIONS 78.

Assertion (A) : NaCl in water and organic acids in benzene show abnormal molecular mass

80.

Mo = Observed molecular mass of solute from colligative

Reason (R) : Abnormal molecular mass is obtained when the substance in the solution undergoes dissociation or association. (a) A

(b) B

(c) C

(d) D

property measurement Match the following : COLUMN - I

COLUMN - II

(A) Mo < Mn

(P) 0.1 M CH3COOH is benzene

(B) Mo  Mn/3

(Q) 0.1 M urea in water

fusion of ice (g )

(C) Mo > Mn

(R) 0.05 M barium chloride in water

Hv = Molar heat of vaporisation of water ; Lv = Latent –1 heat of vaporisation of water (g )

(D) Mo = Mn

(S) 0.1 M CH3COOH in water

Matrix–Match Type Questions 79.

Mn = Normal molecular mass of solute

Hf = Molar heat of fusion of ice ; Lf = Latent heat of –1

Match the following appropriately COLUMN - I (A) Molal depression

COLUMN - II

Integer Type Questions 81.

When 1.0 g of urea is dissolved in 200 g of an unknown

18  373  373  R (P) 1000H v

solvent X, the X freezing point is lowered by 0.25ºC.

constant of water

dissolved in 125 g of same solvent X, freezing point is

When 1.5g of an unknown, non-electrolytic solute Y is lowered by 0.2 ºC and vapour pressure is lowered by

(B) Molal elevation

373  373  R (Q) 1000 L v constant of water

(C) Tf of solution

(R)

18  273  273  R 1000 H f

containing 9.0 g of glucose in 50g of water (D) Tb of solution

(S)

273  273  R 1000 L f

containing 3.0g of urea in 50 g of water

1%. If freezing point of X, is 12ºC, determine molar enthalpy of fusion of X in kJ. 82.

A non-volatile organic compound X was used to makeup two solution. Solution A contains 5.0 g of X in 100 g of water and solution B contains 2.0 g of X in 100 g of benzene. Solution A has vapour pressure of 754.5 mm of Hg at normal boiling point of water and solution B has the same vapour pressure at the normal boiling point of

SOLUTIONS

92

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTION 6.

Integer Type Questions 1.

2.

3

–3

To 500 cm of water, 3.0 × 10 kg of acetic acid is added. if 23% of acetic acid is dissociated, what will be the depression –1 in freezing point ? Kf and density of water are 1.86 K kg –1 –3 mol and 0.997 g cm , respectively. (2000) The vapour pressure of two miscible liquids (A) and (B) are 300 and 500 mm of Hg respectively. In a flask 10 moles of (A) is mixed with 12 moles of (B). However, as soon as (B) is added, (A) starts polymerizing into a completely insoluble solid. The polymerization follows first-order kinetics. After 100 min, 0.525 mole of a solute is dissolved which arrests the polymerization completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate constant of the polymerization reaction. Assume negligible volume change on mixing and polymerization and ideal behaviour for the final solution. (2001)

Subjective Type Question 3.

4.

Consider the three solvents of identical molar masses. Match their boiling point with their kb values (2003)

8.

(a) 75%

(b) 50%

(c) 25%

(d) 85%

The elevation in boiling point, when 13.44 g of freshly prepared CuCl2 are added to one kilogram of water, is. –1 [Some useful data, kb = 0.52 kg K mol , molecular weight of CuCl2 = 134.4 g]. (2005) (a) 0.05

(b) 0.1

(c) 0.16

(d) 0.21

The elevation in boiling point of a solution of 13.44g of CuCl2 in 1 kg of water using the following information will b e (M o l e cu la r weight of CuCl 2 = 134.4 and –1 Kb = 0.52 K molal ) (2005) (a) 0.16

(b) 0.05

(c) 0.1

(d) 0.2

Integer Type Questions 9.

75.2 g of C6H5OH (phenol) is dissolved in 1kg of solvent of kf = 14. If the depression in freezing point is 7K, then find the percentage of phenol that dimerises. (2006)

10.

Solvents

Boiling point

kb values

X

100ºC

0.92

Y

27ºC

0.63

When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of –1 benzene (Kf = 1.72 K kg mol ), a freezing point depression of 2K is observed. The Van’t Hoff factor (i) is (2007)

Z

283ºC

0.53

(a) 0.5

(b) 1

(c) 2

(d) 3

During depression of freezing point in a solution the following are in equilibrium (2003) (a) liquid solvent, solid solvent (b) liquid solvent, solid solute (c) liquid solute, solid solute (d) liquid solute, solid solvent

Subjective Type Question 5.

7.

0.004 M Na2SO4 is isotonic with 0.01 M glucose. Degree of dissociation of Na2SO4 is (2004)

1.22 g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17ºC, while that in the benzene increases by 0.13ºC; Kb for acetone and benzene is 1.7 K –1 –1 Kg mol and 2.6 K kg mol respectively. Find molecular weight of benzoic acid in two cases and justify your answer. (2004)

Comprehension-I Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given, Freezing point depression constant of water (k fwater )=1.86K kg mol -1

Freezing point depression constant of ethanol (k ethanol )=2.0K kg mol -1 f

93

SOLUTIONS Boiling point elevation constant of water (k

water b

17.

)=0.52K kg mol

-1

Boiling point elevation constant of ethanol (k ethanol )=1.2K kg mol-1 b

Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K

For a dilute solution containing 2.5g of a non-volatile nonelectrolyte solute in 100g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb –1 = 0.76 K kg mol ). (2012) (a) 724

(b) 740

(c) 736

(d) 718

Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K

Multiple Choice Question

Vapour pressure of pure water = 32.8 mm Hg

18.

Vapour pressure of pure ethanol = 40 mm Hg

Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statements is (are) (2013)

Molecular weight of water = 18 g mol1 (a) ΔG is positive

(b) ΔSsystem is positive

(c) ΔSsurroundings =0

(d) ΔH=0

Molecular weight of ethanol = 46g mol1

11.

12.

13.

14.

In answering the following questions, 11 to 13 consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. (2008) The freezing point of the solution M is (a) 268.7 K (b) 268.5 K (c) 234.2 K (d) 150.9 K The vapour pressure of the solution M is (a) 39.3 mm Hg (b) 36.0 mm Hg (c) 29.5 mm Hg (d) 28.8 mm Hg Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is (a) 380.4 K (b) 376.2 K (c) 375.5 K (d) 354.7 K The Henry’s law constant for the solubility of N2 gas in 5 water at 298 K is 1.0 × 10 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure is (2009) –4

(a) 4.0 × 10

–4

15.

–5

(b) 4.0 × 10

–6

(c) 5.0 × 10 (d) 4.0 × 10 The freezing point (inº C) of solution containing 0.1g of K3[Fe(CN)6] (mol. wt. 329) in 100g of water (Kf = 1.86 K –1 kg mol ) is (2011) –2 –2 (a) – 2.3 × 10 (b) – 5.7 × 10 –3 –2 (c) – 5.7 × 10 (d) – 1.2 × 10

29.2% (w/W) HCl stock solution has density of 1.25 g mL–1. The molecular weight of HCl is 36.5g mol–1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is

19.

Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA/xB in the new solution is ............. . (given that the vapour pressure of pure liquid A is 20 Torr temperature T)

Match the column 20.

Match the thermodynamic processes given under Column-I with the expressions given under Column-II. (2015) Column–I

Column–II

(A) Freezing of water at 273 K and 1 atm

(P) q = 0

(B) Expansion of 1 mol of an ideal gas into a

(Q) w = 0

vacuum under isolated conditions (C) Mixing of equal volumes of two ideal

(R) Ssys < 

gases at constant temperature and pressure in an isolated container (D) Reversible heating of H2(g) at 1 atm from (S) U = 0

Integer Type Questions 16.

Integer Type Questions

(2012)

300K to 600 K, followed by reversible cooling to 300 K at 1 atm. (T) G = 0

SOLUTIONS Multiple Choice Questions 21.

Mixture(s) showing positive deviation from Raoult’s law at 35ºC is (are) (2016)

94 Integer Type Questions 23.

39 g of benzene, its vapor pressure decreases from 650

(a) carbon tetrachloride + methanol

mm Hg to 640 mm Hg. The depression of freezing point of

(b) carbon disulphide + acetone

benzene (in K) upon addition of the solute is _____

(c) benzene + toluene

(Given data: Molar mass and the molal freezing point

(d) phenol + aniline 22.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol-1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol-1]

On dissolving 0.5 g of a non-volatile non-ionic solute to

depression constant of benzene are 78 g mol–1 and 5.12 K kg mol–1, respectively) 24.

(2019/Shift-1)

The Mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g/cm3, the molarity of urea solution is_____. (Given data: Molar masses of urea and water are 60 g mol–

Among the following, the option representing change the freezing point is (2017)

1

and 18 g mol–1, respectively) (2019/Shift-2)

25.

5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated NaOH from a burette using phenolphthalein indicator. The volume of NaOH required

(a)

for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution ? (2020/Shift-1) Exp. No.

(b)

Vol. of NaOH (mL)

1

12.5

2

10.5

3

9.0

4

9.0

5

9.0

(c)

26.

Liquids A and B form ideal solution for all compositions of A and B at 25ºC. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of

(d)

pure liquid B in bar? (2020/Shift-2)

95

Note:

SOLUTIONS Please share your valuable feedback by scanning the QR code.

96

03 CHEMICAL KINETICS

Chapter 03

CHEMICAL KINETICS

97

THEORY 1.2 Instantaneous rate

INTRODUCTION Chemical Kinetics (Kinesis : Movement) The branch of chemistry which deals with the study of the rates of chemical reactions, the factors affecting the rates of the reactions and the mechanism by which the reactions proceed is called Chemical Kinetics.

Rate of change of concentration of any one of the reactants or products at that particular instant of time is called instantaneous rate. As t  0 or rinst 

d [R] d [P]  dt dt

IMPORTANT :

Classification of reactions On the basis of rates :

Rate of a reaction is always positive.



Since,  [R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make the rate of the reaction a positive quantity.

Very fast reactions e.g. precipitation of AgCl



Very slow reactions 1.3 Units of rate of a reaction

e.g. rusting of iron



Reactions taking place at moderate speeds

Units of rate are concentration time–1

e.g. hydrolysis of starch

or mol L–1s–1; atm s–1 (for gaseous reactions) 1.4 Overall rate of a reaction

1. RATE OF A CHEMICAL REACTION

When there are several reactants and products the

The rate of a reaction can be defined as the change in concentration of a reactant or a product in unit time.

individual rates of the various components may differ as

1.1 Average rate

For a reaction,

The rate of reaction measured over a definite time interval is called average rate of a reaction.

they would depend on the stoichiometric coefficients.

A + 2B

3C + 4D

Rate of disappearance of B = 2 × Rate of disappearance of

Consider a hypothetical reaction,

A (2:1)

R

Rate of formation of C = 3 × Rate of disappearance of A (3:1)

P

Average rate of reaction 



Decreasein concentration of R Time taken

  R  t

or 

Rate of formation of D = 4 × Rate of disappearance of A (4:1) To define a unique value for the overall rate of the reaction we divide the individual rates by the respective coefficients and equate their signs.

Increase in concentration of P Time taken

Overall Rate 

A / t   1/ 2  B / t

 1/ 4  D / t   1/ 3 C / t

 P  t

SCAN CODE CHEMICAL KINETICS

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98

Remember!!!

2.2 Order of a Reaction

aA + bB

cC + dD

aA + bB

 1/ a  A / t   1/ b  B / t Overall Rate =  1/ c C / t  1/ d D / t      1/ a  dA / dt   1/ b  dB / dt Instantaneous Rate  1/ c dC / dt  1/ d dD / dt     2. DEPENDENCE OF RATE ON CONCENTRATION

cC + dD

Rate= k [A]x [B]y Sum of these exponents, i.e., x + y gives the overall order of a reaction where x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction.

Factors Influencing Rate of a Reaction



concentration of reactants (pressure in case of gases),



temperature and



catalyst.

2.3 Units of rate constant Rate = k [A]x [B]y

Dependence on Concentration :

k

Rate concentration 1   [A]x [B]y time (concentration)n

2.1 Rate law Reaction

Order

Units of rate constant

Consider a general reaction aA + bB

cC + dD

The rate expression for this reaction is

Zero order reaction

0

mol L1 1  s (mol L1 )0

Rate [A]x [B]y = mol L–1 s–1

where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Rate = k [A]x [B]y

d  R  dt

x

 k  A  B 

First order reaction

1

mol L1 1   s 1 s (mol L1 )1

Second order reaction

2

mol L1 1  s (mol L1 )2

y

Above equation is known as differential rate equation, k is a proportionality constant called rate constant. Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. IMPORTANT Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally. e.g 2NO(g) + O2(g)  2NO2 (g) Rate = k [NO]2[O2] CHCl3 + Cl2  CCl4 + HCl

= mol–1 L s–1

3. INTEGRATED RATE EQUATIONS 3.1 Zero order reaction The rate of the reaction is proportional to zero power of the concentration of reactants. R P

Rate 

d  R  dt

 kR0

Rate = k [CHCl3] [Cl2]1/2 SCAN CODE CHEMICAL KINETICS

99

CHEMICAL KINETICS

dA dt

k K

 R    R  0

t

Rate

e.g.



The decomposition of gaseous ammonia on a hot platinum surface at high pressure. 1130K 2NH3 (g)   N 2 (g)  3H 2 (g) Pt catalyst

0

time

RP

Rate = 



Thermal decomposition of HI on gold surface 3.2 First order

Consider the reaction,

Rate = 

Rate = k [NH3]0 = k

The rate of the reaction is proportional to the first power of the concentration of the reactant R.

d [R]  k [R]0 dt

Rate 

d [R]  k 1 dt

d  R  dt

 k  R

d [R] = – k dt

K[A]0

Integrating both sides [R] = – k t + I

Rate of first order reaction exhibits exponential decay

where, I is the constant of integration.

Rate

At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation [R]0 = – k × 0 + I [R]0 = I Substituting the value of I in the equation

0

[R] = –kt + [R]0

Concentration of R

[R0]

time

R P

k = – slope

Rate  

or

d [R]  k [R] dt

d [R]   kdt [R]

Integrating this equation, we get In [R] = –kt + I ........ (1)

0

Time

SCAN CODE CHEMICAL KINETICS

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100

Again, I is the constant of integration and its value can be determined easily.

ln[R0] k = – slope ln[R]

When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation can be written as In [R]0 = – k × 0 + I In [R]0 = I Substituting the value of I in equation In [R] = –kt + ln [R]0

The first order rate equation (3) can also be written in the form

......... (2)

Rearranging this equation

ln

[R]  kt [R]0

or k 

[R]0 1 ln t [R]

t

0

k

[R]0 2.303 log t [R]

log

[R]0 kt  [R] 2.303

......... (3)

At time t1 from equation (2) In [R]1 = – kt1 + ln [R]0 At time t2 ln [R]2 = – kt2 + ln [R]0 where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively.

e.g.



Subtracting

C2 H 4 (g)  H 2 (g)  C2 H 6 (g)

ln [R]1 – ln [R]2 = – kt1 – (–kt2)

[R]1 ln [R]  k (t 2  t1 ) 2

k=

1 [R]1 ln (t 2 - t1 ) [R]2

Comparing equation (2) with y = mx + c, if we plot ln [R] against t, we get a straight line with slope = –k and intercept

Hydrogenation of ethane,

Rate = k [C2H4]



Decomposition of N2O5 and N2O 3.3 Half-Life of a Reaction The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. Represented as : t1/2.

equal to ln [R]0

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101

For a zero order reaction, rate constant is given by equation.

k

The concentration of water does not get altered much during the course of the reaction. So, in the rate equation the term [H2O] can be taken as constant.

[R]0  [R] t

Rate = k  [CH3COOC2H5]

1 At t = t1/2, [R] = [R]0 2

where k  = k [H2O]

The rate constant at t1/2 becomes

k

Cane Sugar

[R]0  1/ 2 [R]0 t1/ 2

t1/ 2 





H C12 H 22O11  H 2O    C6 H12O6  C6 H12O6

3.5 Practical Analysis of First Order Reactions

[R]0 2k

[R]0 2.303 log t [R]

Case - 1 : In gaseous phase reactions we prefer to measure the pressure of the gases or volume. For example the following reactions :



k

or t1/ 2 

t1/ 2

B(g) + C(g)

Let pi be the initial pressure of A and pt the total pressure at time ‘t’.

[R]0 2

So, the above equation becomes

t1/ 2 

For a first order gas phase reaction of the type : A(g)

at t1/2

[R] 

Fructose

Rate = k [C12H22O11]

For the first order reaction,

k

Glu cose

[R]0 2.303 log t1/ 2 [R]/ 2

2.303 log 2 k

Integrated rate equation for such a reaction can be derived as : Total pressure pt = pA + pB + pC (pressure units) pA, pB and pC are the partial pressures of A, B and C, respectively. If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each.

2.303  0.301 k

A(g)  B(g)  C(g) At t  0 pi atm 0 atm 0 atm At time t (pi  x) atm x atm x atm

0.693  k

3.4 Pseudo First Order Reactions where, pi is the initial pressure at time t = 0. Reactions which are not truly of the first order but under certain conditions become reactions of the first order.

pt = (pi – x) + x + x = pi + x x = (pt – pi)

e.g. H

CH 3COOC2 H 5  H 2 O  CH 3COOH  C 2 H 5OH

pA = pi – x = pi – (pt – pi) = 2pi – pt

Rate = k [CH3COOC2H5] [H2O]

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102 The above expression can be used to evaluate the value of k from the pressure data and also verify that the reaction is of first order by checking 2-3 data points. In the above analysis the expression would have been same for the following reactions as well as the constants will cancel out.

pi   2.303   k      log pA   t  





2.303 pi log t (2pi  p t )

A(s)  B(s) + 2C(g)

For a first order gas phase reaction of the type :

A(s)  2B(g) + C(g) (if total pressure is given)

A(s)  B(s)  C(g)

And the results will be same if the similar data is given in terms of volume.

The data given to us is: Time

Case 2 : If one of reactants is titrated with a red/ox reagent:

Pressure of gas C/Total Pressure

0

0

t

Pt



P

If we have to find the expression for k or verify that its a first order reaction then we will use the expression for k: ln

a  kt ax

Suppose we have a reaction of the type: AB+C And suppose we detect the amount of A left by titrating it with some reagent and volume of that reagent reacting with the left over A is given at different time intervals: Time

Volume of the reagent

0

V0

t

Vt

But we don’t know the values for a or a – x but we can find the above ratio by relating the given data with concentration values.

Now the volume of the reagent will be proportional to the moles of A present. Therefore:

For gases, P  number of moles

V0  a

A(s)  B(s) + C(g)

Vt  a – x

t=0

a

0

0

We can evaluate k:

t=t

a–x

x

x

t= 

0

a

a

Now we can write:

kt  In

V0 Vt

Pt  x

If the same reagent reacts with all the reactants and products:

P  a

V0  a

P – Pt  a – x a/(a –x) = P /( P – Pt) Now we can substitute this into the expression for k.

Vt  a + x 2V0 – Vt  a – x

kt  In

V0 2V0  Vt

P 1 k  In t P  Pt

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Reaction

Expression for rate constant

N 2 O 5  2NO 2 

2.303 V log t V  Vt Here V t = volume of O 2 after time t and V  volume of O2 after infinite time. Same as above, here Vt and V are volumes of N2 at time t and at infinite time respectively.

1 O2 2

k

NH4 NO2 (aq)  2H2O + N2

V .303 log 0 t Vt Here V o and V t are the volumes of KMnO 4 solution used for titration of same volume of reaction mixture at zero time (initially) and after time t.

1 H 2 O2  H 2 O  O2 2

k



H CH 3COOC 2 H 5  H 2 O   CH 3COOH

C2H5OH

k

V  V0 2.303 log  t V  Vt

Here V 0 ,V t and V  are the volumes of NaOH solution used for titration of same volume of reaction mixture after time, 0, t and infinite time respectively. 

H C12 H 22 O11  H 2 O    C 6 H12O 6  C6 H12 O6 d sucrose

d Glucose

  Fructose

(After the reaction is complete the

equimolar mixture of glucose and fructose obtained is laevorotatory)

r r 2.303 log 0  t rt  r Here, r 0 ,r t and r  are the polarimetric 0, t readings after time and infinity respectively. k

3.6 Practical Methods of determining order of a reaction i.

Initial Rate Data Method: We take different set of initial concentration and measure the initial rate. Then by keeping the concentration of one of the reactants constant and varying the other one we can study the effect on the rate and hence find out the order.

ii.

Logarithmic data method: For any order, be it fractional or integral, if we plot log (rate) vs log (concentration) graph it will always be a straight line for the reactions of the type: A  products Rate, r = k[A]

n

log r = log k + n log [A] We can take various data points and convert them to log values and plot them. We will obtain a straight line after curve-fitting with slope n and intercept log k. And hence we can find out the order and rate constant from the graph.

iii.

Half Life Method: If we take various concentrations of reactant and measure half life for all of them then we can find out the order of the reaction by mere observation or with the help of some calculations. t1/2  [A]01-n If simple observation is not possible then we can calculate the order of the reaction by taking two data points and using log for calculating n.

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By integrated rate Equations: If we have simple data of concentration and time we can use the integrated rate equations to find out the order. For this we will have to try and fit the data into the equation at various intervals and calculate the value of rate constants. If the values come out to be the same in all intervals then the data fits into the equation taken and we will know the order. For example, we have the following data: Time:

0

t1

t2

t3 ....... t

Conc:

A0

A1

A2

A3.......At

And if we assume that it can be of first order then we will calculate the values of k at minimum three data points by using the equation for first order:

1 A0 In K t At Let these values be k1, k2 and k3. If k1 = k2 = k3 then it means that this data fits into the above equation hence the order is 1. If it doesn’t we will have to try other equations as well. v.

Isolation method: In this method we try and eliminate one of the two reactants from the rate equation by taking it in excess. What happens is when the amount of a reactant is in excess its effect on the rate becomes marginal or negligible and then we can vary the concentration of the other reactant and observe its effect on rate and find out the order.

4. MOLECULARITY AND MECHANISM



Trimolecular or termolecular reactions : involve simultaneous collision between three reacting species for example, e.g.,

2NO + O2   2NO2

Reactions with molecularity greater than three are very rare. 4.2 Mechanism The reactions taking place in one step are called elementary reactions. When a sequence of elementary reactions (called mechanism) gives us the products, the reactions are called complex reactions. The different steps in which the complex reaction takes place is called the mechanism of the reaction. Rate determining step : The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step. A complex reaction can be represented as a series of elementary steps. For example 2NO2(g) + F2(g)  2NO2F(g)

The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.

Experimentally, Rate of reaction = k[NO2][F2]

Unimolecular reactions : when one reacting species is involved, for example, decomposition of ammonium nitrite. NH4NO2  N2 + 2H2O



2HI  H2 + I2

4.1 Molecularity

Classification of reactions on the basis of Molecularity :



for example, dissociation of hydrogen iodide.

Bimolecular reactions : involve simultaneous collision between two species

Probable mechanism : Step-1: NO2 + NO2  NO + NO3

(slow)

Step-2 : NO3 + CO  NO2 + CO2

(fast)

Slow step : bimolecular Hence, a bimolecular reaction.

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Reaction intermediates :

Threshold energy = Activation energy + Energy possessed

There are some species which are formed during the course of the reaction but do not appear in the overall reaction. They are called reaction intermediates.

by the reactants Less is the activation energy, faster is the reaction.

e.g. NO3 in the above example.

In order that the reactants may change into products, they

Distinction between Order and Molecularity of a reaction

have to cross an energy barrier (corresponding to threshold

Order

energy). Reactant molecules absorb energy and form an

1.

Order is the sum of the powers of the concentration of the reactants in the rate law expression.

intermediate called activated complex which immediately

2.

It can be zero and even a fraction.

3.

It is applicable to elementary as well as complex reactions.

4.

It can be determined experimentally only and cannot be calculated.

5.

For complex reaction, order is given by the slowest step.

Molecularity 1.

Molecularity is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

2.

It cannot be zero or a non integer.

3.

It is applicable only for elementary reactions. For complex reaction molecularity has no meaning.

4.

It can be calculated by simply adding the molecules of the slowest step.

5.

Generally, molecularity of the slowest step is same as the

dissociates to form the products. e.g.

Reaction profile of an exothermic reaction

order of the overall reaction.

5. TEMPERATURE DEPENDENCE

Reaction profile of an endothermic reaction

5.1 Activation Energy According to collision theory, a reaction takes place because the reactant molecules collide with each other. The minimum energy which the colliding molecules must have in order that the collision between them may be effective is called threshold energy. The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. SCAN CODE CHEMICAL KINETICS

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106

5.2 Temperature Dependence of the Rate of a Reaction For a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Temperature coefficient 

Rate constant at T  100 Rate constant at T 0

The area under the curve remains constant since total probability must be one at all times. At (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. Arrhenius equation Quantatively, the temperature dependence of the rate of a chemical reaction can be explained by Arrhenius equation

Explanation : At a particular temperature, if fractions of molecules are plotted versus corresponding kinetic energies, a graph of the type shown is obtained. The peak of the curve represents the kinetic energy possessed by the maximum fraction of molecules and is called most probable kinetic energy.

k  Ae Ea / RT where A is the Arrhenius factor or the frequency factor or pre-exponential factor. R is gas constant and Ea is activation energy measured in joules/mole. The factor e Ea / RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Thus, it has been found from Arrhenius equation that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. Taking natural logarithm of both sides of equation ln k  

Distribution curve showing energies among gaseous molecules

Ea  ln A RT

The plot of ln k vs 1/T gives a straight line with slope



Ea and intercept = ln A. R

At temperature T1, equation ln k1 

Ea  ln A RT1

At temperature T2, equation is ln k 2  Distribution curve showing temperature dependence of rate of a reaction

Ea  ln A RT2

(since A is constant for a given reaction)

With increase in temperature :

k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively.

(i)

maximum of the curve moves to the higher energy value i.e., most probable kinetic energy increases

Substracting equation form, we obtain

(ii)

the curve spreads to the right i.e., there is a greater proportion of molecules with much higher energies.

ln k 2  ln k1 

Ea E  a RT1 RT2

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CHEMICAL KINETICS k 2 Ea  1 1  ln k  R  T  T   1 1 2

log

Ea  1 1  k2     k1 2.303R  T1 T2 

log

E a  T2  T1  k2    k1 2.303R  T1T2 

6. EFFECT OF CATALYST

107 Important characteristics of catalyst : 

A small amount of the catalyst can catalyse a large amount of reactants.



A catalyst does not alter Gibbs energy, G of a reaction.



It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions.



A catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster.

7. COLLISION THEORY OF CHEMICAL REACTIONS According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other.

A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. e.g.

Rate of reaction depends on the number of effective collisions which in turn depends on :

MnO2 2KClO3  2 KCl  3O2

(i)

Energy factor : colliding molecules must have energy more than threshold energy.

(ii)

Steric or probability factor (P) : colliding molecules must have proper orientations at the time of collision.

Action of the catalyst According to intermediate complex theory, reactants first combine with catalyst to form intermediate complex which then decomposes to form the products and regenerating the catalyst.

Thus, the Arrhenius equation is modified to

k = PZAB eEa /RT

8. IMPORTANT FORMULAE 8.1 Rate of Reactions Effect of catalyst on activation energy

(aA + bB  cC + dD) Instantaneous Rate = –

1 dA 1 dB 1 dc 1 dD =– = = a dt b dt c dt d dt

Average Overall Rate =

1 A 1 B 1 C 1 D    a t b t c t d t

8.2 Arrhenius Equation

k  Ae



Ea RT

.

Catalyst provides an alternate pathway by reducing the activation energy between reactants and products and hence lowering the potential energy barrier.

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108

Zero-Order d[ A] k dt

Rate Law



Integrated

[A] = [A]0 – kt

First Order



d[ A]  k[A ] dt

[A] = [A]0e–kt

Second Order



d[ A]  k[A ]2 dt

1 1   kt [ A ] [ A ]0



d[ A]  k[ A ]n dt

1 [A ]

n 1



1 [A ]0n 1

 (n  1) kt

(Except first Order)

Rate Law

Units of Rate

nth-Order

1

M s

1 s

1 M.s

[A] vs. t

ln([A]) vs. t

1 vs. t [ A]

M

n 1

.s

Constant (k)

Linear Plot to

[A]n1

vs. t

(Except first Order)

determine k

Half-life

1

t1/ 2 

[ A ]0 2k

t1 / 2 

ln(2) k

t1/ 2 

1 [ A ]0 k

t1 / 2 

2 n 1  1 (n  1)k [A ]0n 1

(Except first Order)

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SOLVED EXAMPLES

Example-1

Rate of disappearance of Y2

What do you understand by ‘Rate of reaction’ ?

=

Sol. The change in the concentration of any one of the reactants or products per unit time is termed as the rate of reaction.

d [Y2 ] d [X 2 ] 3 d [XY3 ]  3  dt dt 2 dt

Example-5

Example-2

For an elementary reaction 2A + B  3C the rate of appearance of C at time ‘t’ is 1.3 × 10–4 mol L–1 s–1.

In a reaction R  P the concentration of R was observed. Find the rate of the reaction

Calculate at this time Conc., Mol/L

50

25

15

10 (i) rate of the reaction.

Time (s)

0

10

20

30

Sol: In 1st time interval rate = –2.5 mol/Ls. In 2nd time interval rate = –1 mol/Ls. In 3rd time interval rate = – 0.5mol/Ls. The rate comes out as negative as the concentration of A is reducing with time.

(ii) rate of disappearance of A.

Sol. (i)

Rate 

1 d [C] 1   1.3  104 3 dt 3

= 0.43 × 10–4 mol L–1 s–1

Example-3 In a reaction H2 + I2  2HI the rate of disappearance of I2 is found to be 10–6 mole per litre per second. What would be the corresponding rate of appearane of HI ?

(ii) Rate 

d [A] 2 d [C]   dt 3 dt

= 0.86×10-4 mol L-1 s-1

Sol. In the given reaction rate = rate of disappearance of H2 per mole

Example-6

= rate of disappearance of I2 per mole

Differentiate between rate of reaction and reaction rate constant.

= rate of appearance of HI per mole

  



d[ H 2 ] d[ I ] 1 d[ HI]  2    10 6 dt dt 2 dt

d[HI]  2 × 10–6 mole L–1s–1. dt

Example-4 For the assumed reaction X2 + 3Y2  2XY3, write the rate of equation in terms of rate of disappearance of Y2.

Sol. Rate of Reaction 1.

Rate of reaction is the change in concentration of a reactant or product in a unit interval of time.

2.

The rate of reaction at any instant of time depends upon the molar concentrations of the reactants at that time.

3.

Its units are always mol litre–1 time–1.

Reaction Rate Constant

d [X 2 ] 1 d [Y2 ] 1 d [XY3 ]   Sol. Rate = – dt 3 dt 2 dt

1.

It is the rate of reaction when the molar concentration of each of the reactants is unity.

CHEMICAL KINETICS 2.

The rate constant does not depend upon the concentrations of the reactants.

3.

Its units depend upon the order of reaction.

110 Sol. Rate = k [NO]2 [O2] Let initially, moles of NO = a, moles of O2 = b, volume of the vessel = V. Then

Example-7

[NO]  What is meant by an elementary reaction ?

a b M, [O2 ]  M V V 2

Sol. A reaction that takes place in one step is called an elementary reaction. Say, for example, reaction between H2 and I2 to form 2HI is an elementary reaction. Different steps of a complex reaction are each an elementary reaction.

a2b a b Rate (r1 )  k      k 3 V V V



Now, new volume =

Example-8 Define the following :



V . 3

New concentrations : [NO] 

(i) Elementary step in a reaction (ii) Rate of a reaction

[O2 ] 

Sol. (i) Elementary step : Each step of a complex reaction is called an elementary step.

Example-9



The specific reaction rate of a reaction is 6.2 × 10–3 mol lit–1 s–1. What is the order of reaction ?

 2NO 2NO(g) + O2(g)  2(g)

How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume ? Will there be any change in the order of the reaction with reduced volume ?

....... (ii)

r2  27 or r = 27 r = rate becomes 27 times. 2 1 r1

Distinguish between order and molecularity of reaction. Sol. Order 1.

i.e., zero order reaction.

Following reaction takes place in one step :

2  3a   3b  27ka b New rate (r2) = k      3 V V V

Example-11

mol1–n = mol1

Example-10

b 3b  V/3 V

Thus, there is no effect on the order of reaction.

Sol. Comparing with (mol L–1)1–n s–1, we get

n=0

a 3a  V/3 V

2



(ii) Rate of reaction : It is the change in the concentration of any of the reactants or products per unit time.

1–n=1

....... (i)

2. 3. 4. 5. 1.

2. 3.

Order is the sum of the powers of the concentration of the reactants in the rate law expression. It can be zero and even a fraction. It is applicable to elementary as well as complex reactions. It can be determined experimentally only and cannot be calculated. For complex reaction, order is given by the slowest step. Molecularity Molecularity is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. It cannot be zero or a non integer. It is applicable only for elementary reactions. For complex reaction molecularity has no meaning.

CHEMICAL KINETICS 4. 5.

111

It can be calculated by simply adding the molecules of the slowest step. Generally, molecularity of the slowest step is same as the order of the overall reaction. Example-12 Calculate the overall order of a reaction which has the rate expression (a) Rate = k [A]1/2 [B]3/2

Sol. (a)

Example-14 When could order and molecularity of a reaction (i) be the same and (ii) be different ? Sol. (i) If the rate determining step involves all the reactants taking part in the reaction, i.e., it is a one-step reaction, then order and molecularity of the reaction is same. (ii) If reaction involves more than one step, then order and molecularity are different.

(b) Rate = k [A]3/2 [B]–1

Rate = k [A]x [B]y

Example-15

order = x + y So,

order 

Define the following terms giving an example for each :

1 3   2, i.e., second order 2 2

(i) The order of a reaction (ii) The molecularity of a reaction.

(b)

Order 

3 1  (1)  i.e., half order.. 2 2

Example-13

Sol. (i) Order of reaction may be defined as the sum of powers of the concentration of the reactants in the rate law expression. For example consider the reaction

The rate of reaction, 2NO + Cl2  2NOCl is doubled when concentration of Cl2 is doubled and it becomes eight times when concentration of both NO and Cl2 are doubled. Deduce the order of the reaction. Sol. Let

NH4NO2   N2 + 2H2O Experimentally, it is observed that the rate law for this reaction is

r = k [NO]x [Cl2]y

Rate = k [NH4NO2]

2r = k [NO]x [2Cl2]y ...... (i)

Hence, the order of reaction is 1.

8r = k [2NO]x [2Cl2]y ........ (ii)

(ii) Molecularity of a reaction may be defined as the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction.

Dividing (ii) by (i), we get 8r k[2NO]x [2Cl2 ]y  2r k [NO]x [2Cl2 ]y

22 = [2]x

For example molecularity of the reaction

x=2

 H2 + I2 is 2 as it involves simultaneous 2HI  collision between two HI molecules.

Putting the value of x in (i), we get 2r = k [NO]2 [2Cl2]y Also

r = k [NO]2 [Cl2]y

Example-16

2r [2Cl2 ]y  r [Cl2 ]y

For the reaction N2(g) + 3H2(g)  2NH3(g), how are the

2 = [2]y

reaction expressions

y=1

interrelated ?

Rate = k [NO]2 [Cl2]1 Overall order of reaction = x + y = 2 + 1 = 3.

Sol.



1 d [H 2 ] 1 d [NH3 ]  3 dt 2 dt



d[H 2 ] dt

and

d [NH 3 ] dt

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112

Example-17

Example-19

The rate of decomposition of a substance A becomes eight times when its concentration is doubled. What is the order of this reaction ?

Define specific reaction rate or rate constant. Sol. Specific reaction rate is the rate of reaction when the molar concentration of each of the reactants is unity.

Sol. Rate law is, Example-20

Rate = k [A]n r = k [A]n

........ (i)

8r = k [2A]n

....... (ii)

Rate of a reaction is given by the equation : Rate = k [A]2 [B] What are the units for the rate and the rate constant for this reaction ?

Dividing (ii) by (i), we get 8r k [2A]n  r k [A]n

Sol. Units of rate = mol L–1 s–1

8 = [2]n

Units of k =

Rate mol L1s 1  2 [A] [B] (mol L1 ) (mol L1 )2

n=3 = L2 mol–2 s–1.

The order of reaction is 3. Example-21

Example-18 A + 2B  3C + 2D. The rate of disappearance of B is 1 × 10–2 mol L–1 s–1. What will be (i) Rate of the reaction (ii) Rate of change in concentration of A and C ? Sol. (i)

Rate = 

1 d [B] 1   1 102 2 dt 2

What is meant by the ‘rate constant, k’ of a reaction ? If the concentration is expressed in mol L–1 units and time in seconds, what would be the units of k (i) for a zero order reaction and (ii) for a first order reaction ? Sol. Rate constant is the rate of the reaction when the concentration of each reactant is taken as unity.



rate = k [A]n



 mol.  General unit of k =    lit. 

= 0.5 × 10–2 mol L–1 s–1 (ii) Rate = 

1 n

d [A] 1 d [B] 1 d [C]   dt 2 dt 3 dt

Rate of change in concentration of A



d [A] 1 d [B]  dt 2 dt –2

–1

= 0.5 × 10 mol L s

(i)

s 1

For a zero order reaction n = 0

 Unit of k = mol lit–1s–1 (ii) For a first order reaction n = 1

 Unit of k = s–1 –1

Example-22 Rate of change in concentration of C At 25ºC, the second order rate constant for the reaction I–

d [C] 3 d [B] 3    102 dt 2 dt 2 = 1.5 × 10–2 mol L–1 s–1

+ ClO–  IO– + Cl– is 0.0606 M–1  s–1. If solution is initially 3.50 × 10–3M with respect to each reactant, what will be the concentration of each species present after 300 s?

CHEMICAL KINETICS

113

Sol. Since the concentrations of the two reactants are equal at the start and remain so throughout the entire reaction, the reaction can be treated as a simple second order

Example-25 A reaction : Reactant  Product is represented by

reaction. k = 0.0606 M–1  s–1 [B0] = [A0] = 3.50 × 10–3 M [R0]

1 1   kt [ A ] [A 0 ]

Conc. of reactant (R)

1 1  = (0.0606) (300) = 18.18 [A] 3.5  10 3 Time (t)

(i) the order of the reaction in this case.

1  18.18 + 285.7 = 303.9 [A]

(ii) What does the slope of the graph represent ?

[A] = 3.29 × 10–3 M = [B]

Sol. (i) The reaction is of the zero order.

Example-23

(ii) Slope of the straight line graph =  k 

Why is it that instantaneous rate of reaction does not change when a part of the reacting solution is taken out ?

d [R] . dt

Example-26

Sol. Instantaneous rate is measured over a very small interval of time, hence, it does not change when a part of solution is taken out. Example-24 The reaction 2NO(g) + Cl2(g)  2NOCl was studied at – 10 ºC, and the following data were obtained : Initial concentration, mol/L R×n

NO

Cl2

1 2 3

0.10 0.10 0.20

0.10 0.20 0.20

Initial rate of formation of NOCl, mol/L. min 0.18 0.35 1.45

What is the order of reaction with respect to NO and with respect to Cl2 ? Sol. When the Cl2 concentration is doubled, holding the NO concentration constant (compare reactions 1 and 2), the initial rate doubles. Hence the reaction is first order with respect to Cl2. When the NO concentration is doubled (compare reactions 2 and 3), the initial rate quadruples. Hence the reaction is second order with respect to NO.

Answer the following questions on the basis of the above curve for a first order reaction A  P : (a) What is the relation between slope of this line and rate constant ? (b) Calculate the rate constant of the above reaction if the slope is 2 × 10–4 s–1. Sol. (a) slope = k/2.303 (b) k = 2.303 × slope = 2.303 × 2 × 10–4 = 4.606 × 10–4 s–1

CHEMICAL KINETICS Example-27 What are zero order reactions ? Derive integrated rate equation for zero order reaction.

114 Sol. A reaction which is of higher order but follows the kinetics of first order under special conditions is called a pseudo first order reaction. Example, Acid hydrolysis of ethyl acetate.

Sol. Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction,

H CH3 – COOC2H5 + H2O    CH3 – COOH + C2H5 – OH

Here, the rate law is given by expression

R P Rate = 

Rate = k [CH3 – COOC2H5]

As any quantity raised to power zero is unity

Rate  

The concentration of H 2 O is so large that it hardly undergoes any change during the reaction, therefore, it does not appear in the rate law.

d [R]  k [R]0 dt

Example-30

d [R]  k 1 dt

d [R] = – k dt

The optical rotation of cane sugar in 0.5 N lactic acid at 25ºC at various time intervals are given below : Time (min)

0

1435

11360



Rotation (º)

34.50º

31.10º

13.98º

–10.77º

Integrating both sides [R] = – k t + I Show that the reaction is of first order.

where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation

Sol.

C12 H 22 O11  H 2 O Lactic  acid  C 6 H12 O 6  C 6 H12 O 6 Sucrose (excess)

Glucose Fructose

Dextrorotatory

Dextro

Laevo

[R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation [R] = –kt + [R]0 Example-28 State any one condition under which a bimolecular reaction may be kinetically of first order. Sol. A bimolecular reaction may become kinetically of first order if one of the reactants is present in excess. Example-29 What is meant by a pseudo first order reaction ? Give an example of a pseudo first order reaction and write the rate equation for the same.

Laevorotatory Since in this reaction dextro form changes to laevo form, the optical rotation decreases with the progress of the reaction. Thus change in rotation is proportional to the amount of sugar remained after different time intervals. We now have, Time (min)

0

1435

11360



Change in

34.50 –

31.10–

13.98 –

–10.77 –

rotation (º)

(–10.77)

(–10.77)

(–10.77)

(–10.77)

= 45.27

= 41.87

= 24.75

=0

(a)

(a – x)

(a – x)

Substituting the data in Equation (5),

CHEMICAL KINETICS

for t = 1435 min k 1 

2.303 45.27 log  5.442  10 5 1435 41.87

and, for t = 11360 min 2.303 45.27 k1  log  5.311 10 5 11360 24.75

115 Example-32 What is Arrhenius equation to describe the effect of temperature on rate of a reaction ? Sol. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.

The values of k1 are fairly constant and so the reaction is of first order. Example-31 What is half life period ? Show that the time required for the completion of half of the reaction is independent of intial concentration. Sol. The half-life of a reaction is the time in which the concentration of a reactant is reudced to one half of its initial concentration. It is represented ast t1/2. For the first order reaction,

k

[R]0 2.303 log t [R]

k = A e Ea / RT where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol–1). Example-33 A first order reaction has a rate constant of 0.0051 min–1. If we begin with 0.10 M concentration of the reactant, what concentration of reactant will remain in solution after 3 hours ? Sol. For a first order reaction

t at t1/2

[R] =

[R]0 2

So, the above equation becomes

k

or

[R]0 2.303 log t1/ 2 [R]/ 2

[R]0 2.303 log k [R]

t = 3, h = 3 × 60 min = 180 min k = 0.0051 min–1, [R]0 = 0.10 M, [R] = ?

2.303 0.10 180 min = 0.0051 min 1 log [R]

t1/ 2 

2.303 log 2 k

log

0.1 180 min  0.0051 min 1 918   [R] 2.303 2303

t1/ 2 

2.303  0.301 k

log

0.1  0.3986 [R]

t1/ 2 

0.693 k

0.1 [R]

= Anti log (0.3986) = 2.503

It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

[R] =

0.1  0.03995M 2.503

[R] = 0.04 M

CHEMICAL KINETICS Example-34

116 Example-36 Show that time required for the completion of threefourths of a first order reaction is twice the time required for the completion of half reaction.

A first order reaction is 20% complete in 10 minutes. Calculate the time for 75% completion of reaction.

Sol.

t 20% 

2.303 a log k a  0.2a

t 20% 

2.303 10 2.303 log   .0969 k 8 k

t 75% 

2.303 a log k a  0.75a



Sol.

t

2.303 a log k ax

a 2.303 log t 3/ 4 a  3a / 4  log 4  k 2.303 a t1/ 2 log 2 log k a a/2 t 3/ 4 2log 2  t1/ 2 log 2

2.303 a 2.303 2.303 log  log 4   0.6021 k 0.25a k k

t 3/ 4 2 t1/ 2

t 75% 2.303 k 0.6021    t 20% k 2.303 0.0969

t 3/ 4  2t1/ 2 Example-37

t 75% 0.6021  10 0.0969

t 75% 

Show that in case of a first order reaction, the time required for 99.9% of the reaction to take place is about ten times than that required for half the reaction.

10  0.6021 = 62.13 min 0.0969

Example-35 Sol. The reaction SO2Cl2  SO2 + Cl2 is a first order reaction with k = 2.2 × 10–5 at 320ºC. Calculate the percentage of SO2Cl2 that is decomposed on heating this gas for 30 minutes. Sol.

k

log

t 99.9%  10t 50% Example-38

2.303 a log t ax

2.2 105 

The decomposition of N2O5(g) is a first order reaction with a rate constant of 5 × 10–4 s–1 at 45ºC. i.e., 2N2O5(g)  4NO2(g) + O2(g). If initial concentration of N2O5 is 0.25M, calculate its concentration after 2 min. Also calculate half life for decomposition of N2O5(g).

2.303 a log 30  60 ax

a 2.2  105  1800   0.1719 (a  x) 2.303

a = antilog 0.1719 = 1.486 (a  x)

a = 1.486a – 1.486x 0.486a = 1.486x

x 0.486   0.327 = 32.7% a 1.486

2.303 100 log t 99.9% 0.1 log 103 3 log 10 30.0  k    2.303 100 t 50% log 2 0.3010 3.01 log k 50

Sol.

[R]0 2.303 log t [R]t 2.303 0.25 5  104  log 2  60 [R]t k

[R]t = 0.23 M

t1/ 2 

0.693 s = 1386s 5 104

CHEMICAL KINETICS

117

Example-39 t 3/ 4 

The half-life for decay of radioactive 14C is is 5730 years. An archaeological artefact containing wood has only 80% of the 14C activity as found in living trees. Calculate the age of the artefact. Sol.

t 3/ 4 

t1/2 = 5730 year

0.693 0.693   1.209 104 year 1  k t 5730 1/ 2 t

0.693 0.693 1  s  1.83  102 s 1 (ii) k  t 37.9 1/ 2

[R]0 2.303 2.303 100 log  log 4 k [R] 1.2 10 80

t = 1 minute = 60s

t

2.303  0.097  104  1.209

[A]0 2.303 log 1.83  102 s 1 [A]

log

[A]0 60 1.83 102   0.4768 [A] 2.303

log

[A]0   0.4768 1.5232 [A]

t = 1847.7 years Example-40

[A] [A]0 = Anti log 1.5232  0.3336

Decomposition of phosphine (PH3) at 120ºC proceeds according to the equation : 4PH3(g)  P4(g) + 6H2(g)

[A] = 0.334 [A]0

It is found that this reaction follows the following rate equation :

The half life of PH3 in 37.9 s at 120ºC.

[A]0 2.303 log k [A]

60s 

2.303 104 (1  3  0.3010) 1.2

Rate = k [PH3]

2.303  100  2  0.30105 138.64  s 1.83 1.83

t3/4 = 75.76 s

2.303  104 2.303  10 4 t (log 10  log 8)  (1  3 log 2) 1.2 1.2

t

[A]0 2.303 2.303 log   100 log 22 s [A]0 1.83  102 s 1 1.83 4

Example-41

(i) How much time will be required for 3/4 of PH3 to decompose ?

Hydrogen peroxide, H2O2 (aq) decomposes to H2O(l) and O2(g) in a reaction that is of first order in H2O2 and has a rate constant, k = 1.06 × 10–3 min–1.

(ii) What fraction of the original amount of PH3 will remain undecomposed after 1 minute ?

(i) How long will it take 15% of a sample of H2O2 to decompose ?

Sol. (i)

Given t1/2 = 37.95

t1/ 2 

(ii) How long will it take 85% of a sample of H2O2 to decompose ?

0.693 0.693 ,k  1.83  102 s 1 k t1/ 2 Sol. (i)

k

[A]0 0.693 1 2.303 s ,t log 37.9 k [A]

t

[A]0 2.303 log k [A]

CHEMICAL KINETICS

Given : k  1.06 103 min 1 ,

t

118 Example-43

[A]0 100  [A] 85

Rate constant k of a reaction varies with temperature according to the equation

2.303 100 2303 log  85 1.06 1.06  103 min 1

Ea log k = Constant  2.303 RT

[2 log 10 – log 85] min

t

where Ea is the activation energy. When a graph is plotted for log k vs 1/T, a straight line with a slope – 6670 K is obtained. Calculate the energy of activation for the reaction in proper units. (R = 8.314 JK–1 mol–1)

2303 2303  0.0706 [2  1  1.9294] =  153.39 min 1.06 1.06 t = 153.4 min.

3 1 (ii) Given : k  1.06 10 min ,

t

Sol.

[A]0 100  [A] 15

Slope  

Ea   6670 k 2.303 R

Ea = 6670 × 2.303 × 8.314 = 127711.43 J mol–1 = 127.711 kJ mol–1

2.303 100 2303 ,log  15 1.06 1.06  103 min 1

Example-44

[2 log 10 – log 15] min The decomposition of phosphine



2303 2303  0.8239 [2  1  1.1761] = min 1.06 1.06

4PH3(g)  P4(g) + 6H2(g) has the rate law, Rate = k [PH3]

t = 1790 min.

The rate constant is 6.0 × 10–4 s–1 at 300 K and activation energy is 3.05 × 105 J mol–1. What is the value of rate constant at 310 K. [R = 8.314 JK–1 mol–1]

Example-42 For a reaction, the energy of activation is zero. What is the value of rate constant at 300 K, if k = 1.6 × 106 s–1 at 280 K ? [R = 8.31 JK–1 mol–1] Sol. Given T1 = 280 K, k1 = 1.6 × 106 s–1, k2 = ?,

By Arrhenius equation,

T1 = 300 K

k2 = ?

T2 = 310 K

log

 T2  T1  Ea k2    k1 2.303 R  T1T2 



As, Ea = 0

k2  log k  0 1 k2 1 k1

k1 = 6.0 × 10–4 s–1

Ea = 3.05 × 105 J mol–1

Ea = 0, T2 = 300 K

log

Sol.

k2 3.05  105  310  300   k1 2.303  8.314  310  300 

3.05  105  10 2.303  8.314  310  300

log

k2  1.718 k1

k2  antilog 1.718 = 51.62 k1

(log 1 = 1og 10º = 0)

k2 = k1 6

–1

Thus, the rate constant at 300 K is 1.6 × 10 s .

k2 = 51.62 × k1 = 51.62 × 6.0 × 10–4 s–1 = 3.1 × 10–2 s–1

CHEMICAL KINETICS

119

Example-45 or x =

(a) A reaction is of first order in A and of second order in B. Write the differential rate equation for this reaction.

0.54  0.42  0.06. 2

Applying first-order rate equation,

(b) The rate constant k of a reaction increases four fold when the temperature changes from 300 K to 320 K. Calculate the activation energy for the reaction. (R = 8.314 J mol–1 k–1)

k

2.303 0.42 log  0.0237 min 1 6 .5 0.42  0.06

Example-47

dx  k [A] [B]2 dt

Sol. (a) Rate =

Define threshold energy of a reaction.

k2 i.e. k = 4 1

(b) k2 = 4k1

Sol. Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecule.

T1 = 300k T2 = 320 k

log

E a  T2  T1  k2    k1 2.303 R  T1T2 

log 4 

For a decomposition reaction, the values of rate constant k at two different temperatures are given below :

Ea  320  300    2.303  8.314  300  320 

2log 2 

Ea 

Example-48

k1 = 2.15 × 10–8 L mol–1 s–1 at 650 K k2 = 2.39 × 10–7 L mol–1 s–1 at 700 K

Ea  20    19.147  300  320 

Calculate the value of activation energy for this reaction. (R = 8.314 JK–1 mol–1)

 0.3010  19.147  300  320  55327J 20

Sol.

log

Ea = 55.327 kJ Example-46

Ea 

In the course of the reaction CH3COOCH3(g)  C2H4(g) + H2(g) + CO(g) the initial pressure was found to be 0.42 atm while after 6.5 minutes, it was 0.54 atm. If the reaction follows first order kinetics, find the rate constant. Sol. Let the initial number of moles of CH3COOCH3 be a. Initial moles : a 0 0 0 CH3COOCH3  C2H4 + H2 + CO Moles after t :

E a  T2  T1  k2    k1 2303R  T1 T2 

(a – x)

x

x

x

According to the given data, a = 0.42 (initial moles corresponds to 0.42 atm) and {a – x) + x + x + x} = 0.54 (Mole after 6.5 min corresponds to 0.54 atm) or a + 2x = 0.54

Ea 

2.303  R  T1  T2 k log 2 T2  T1 k1

2.303  8.314J mol1K 1  650K  700K 2.39  107 log 700K  650K 2.15  108

Ea 

19.147  650  700 log (23.9  log 2.15)J mol1 50

Ea = 174237.7 (1.3783 – 0.3324)J mol–1 Ea = 174237.7 × 1.0459 J mol–1 = 182235.2 J mol–1 Ea = 182.24 kJ mol–1. Example-49 In general it is observed that the rate of a chemical reaction doubles with every 10º rise in temperature, If this generalization holds good for a reaction in the temperature range 295 K and 305 K, what would be the value of activation energy for this reaction ? (R = 8.314 JK–1 mol–1)

CHEMICAL KINETICS Sol. T1 = 295 K

Ea  

k1 = k (say), T2 = 305K

log

 T2  T1  Ea k2    k1 2.303 R  T1T2 

log

E a [305  295] 2k  k 2.303 R 305  295

120 k2 = 2k

log 2  2.303  8.314  305  295 10

0.3010  2.303  8.314  305  295 10

= 51855.2 J mol–1 = 51.855 kJ mol–1

The rate of a particular reaction triples when temperatures changes from 50ºC to 100ºC. Calculate the activation energy of the reaction. [log 3 = 0.4771; R = 8.314 JK–1 mol–1]

k 2  3k1 

k2 3 k1

T2 = 50 + 273 = 323 K, T2 = 100 + 273 = 373 K

log

 T2  T1  Ea k2    k1 2.303 R  T1T2 

log 3 

0.477119.127 120479 50

 50  0.477119.127 120479  E a   120479   50  

 E a  22011.59J  Ea = 22.0122 kJ mol -1 Example-51 The decompositon of N2O into N2 and O in the presence of argon follows second-order kinetics with k = (5.0 × 1011) e–29000K/T. Calculate the energy of activation. Sol. Comparing the given equation with Arrhenius’s equation, we have,

Example-50

Sol.

 Ea 

 373  323  Ea   2.303  8.31  373 323 

0.4771 

E a  50    19.147  120479 

E 29000K  E = 2900 × 8.314 = 241 kJ/mole. RT T Example-52 Define collision frequency Sol. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Example-53 Define effective collision Sol. The collisions in which molecules collide with sufficient kinetic energy and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collisions.

121

CHEMICAL KINETICS

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Rate law

Rate of Reaction 1.

The rate of a reaction is expressed in different ways as follows :



5.

1 d[C] 1 d[D] 1 d[A] d[B]    . 2 dt 3 dt 4 dt dt

For the gaseous reaction 2A + B  C + D, the rate is given by k [A] [B]. The volume of the container containing the reaction mixture is suddenly reduced to one fourth of its original volume. with respect to the original rate, now the rate would be

The reaction is

2.

(a) 4 A + B  2 C + 3 D

(b) B + 3 D  4 A + 2 C

(c) A + B  C + D

(d) B + D  A + C

(a)

1 16

(b)

(c) 16 times

The rate of formation of ammonia by the reaction : 6.

d[NH3 ] –4 N2 + 3 H2  2 NH3 expressed as = 2.5 × 10 mol dt

1 8

(d) 8 times

In the reaction 2 A + B  Products, the order w.r.t. A is found to be one and w.r.t. B equal to 2. Concentration of A is doubled and that of B is halved, the rate of reaction will be

–1 –1

L s . The rate of consumption expressed in terms of H2 as

d[H 2 ] will be dt

7.

(a) double (b) Three times (c) same

(a) doubled

(b) halved

(c) remain unaffected

(d) four times.

For the reaction A + B  C + D, doubling the concentration of both the reactants increases the reaction rate by 8 time and doubling the initial concentration of only B simply doubles the reaction rate. The rate law for the reaction is 2

(a) r = k [A] [B]

(d) one and a half time of that expressed in terms of NH3. 3.

1 d[HI] For the reaction. 2 HI  H2 + I2, the expression  2 dt

8.

[B]

(b) r = k [A] [B] 2

(d) r = k [A] [B].

The half-life period of a first order reaction is

represents

(a) directly proportional to the initial concentration ‘a’

(a) The rate of formation of HI

(b) inversely proportional to ‘a’

(b) The rate of disappearance of HI

(c) independent of ‘a’

(c) The instantaneous rate of the reaction

(d) independent of the rate constant of the reaction

(d) The average rate of reaction. 4.

(c) r = k [A]

1/2

9.

For the reaction, 2N 2 O5  4NO2  O2 , the rate

In the reaction of formation of sulphur trioxide by contact process 2SO 2+O 2 2SO 3 the rate of reaction was measured as

d[O2 ] = –2.5 × 10–4 mol L–1 s–1. The rate of dt –1 –1

reaction in terms of [SO2] in mol L s will be:

equation can be expressed in two ways –

[N2O5] and +

d[N 2 O5 ] =k dt

d[NO2 ] = k [N2 O5] k and k are related as: dt

(a) –3.75 ×10–4

(b) –5.00 × 10–4

(a) 2k = k

(b) k = k

(c) –1.25 × 10–4

(d) –2.50 ×10–4

(c) k = 2k

(d) k = 4k

CHEMICAL KINETICS 10.

122

The rate equation for the reaction, 2A + B  C is found to be : rate = k [A] [B]... The correct statement in relation to this reaction is that the

15.

NO(g) + Br2(g)

–1

(a) unit of k must be s

It the second step is the rate determining step, the order of the reaction with respect to NO(g) is

(c) rate of formation of C is twice the rate of disappearance of A (d) t1/2 is a constant. Consider the reaction, –

Cl2 (aq) + H2S (aq)  S (s) + 2H (aq) + 2Cl (aq)

16.

The rate equation for this reaction is, rate = k [Cl2] [H2S]

Cl+ + HS–  H+ + Cl– + S (fast)

13.

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

–1

Time/s 0

0.5

2

100

0.6

–1

–3

(b) Both (A) and (B)

(c) Neither (A) nor (B)

(d) (A) only

The rate constant for the first and zero reaction in terms of molarity, M units respectively (b) sec , M

–1

(d) M, sec

–1

(c) 7.8 × 10 atm s 17.

–1

(b) 2.235 × 10 atm s

–4

(a) (B) only

–1

Total pressure/atm

1

(a) 0.35 atm s

Cl2 + HS–  2Cl– + H+ + S (slow)

–1

(d) 3

What is the rate of reaction when total pressure is 0.65 atm ?

(B) H2S  H+ + HS– (fast equilibrium)

(c) M sec , sec

(c) 0

Experiment

(A) Cl2 + H2S  H+ + Cl– + Cl+ + HS– (slow)

–1

(b) 1

SO 2Cl 2(g)  SO 2(g )  Cl 2(g)

Which of these mechanisms is/are consistent with this rate equation ?

(a) sec , M sec

(a) 2

Practical Analaysis of First Order Reaction +

12.

NOBr2(g)

NOBr2(g) + NO(g)  2NOBr(g)

(b) value of k is independent of the initial concentrations of A and B

11.

The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr,

–4

–1

(d) 1.55 × 10 atm s

For a first order reaction, (A)   product the concentration of A changes from 0.1 M to 0.025 M in 40 min. The rat of reaction when the concentration of A is 0.01 M is

–1

–1

For a general reaction X   Y, the plot of conc. of X vs time is given in the figure. What is the order of the reaction and what are the units of rate constant ?

18.

(a) 1.73 × 10–5 M/min

(b) 3.47 × 10–4 M/min

(c) 3.47 × 10–5 M/min

(d) 1.73 × 10–4 M/min

A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will (a) Double

(b) Remain unchanged

(c) Triple

(d) Increase by a factor of 4

Reaction Mechanism 19. –1 –1

(a) Zero, mol L s –1

(c) First, s 14.

–1 –1

(b) First, mol L s

–1 –1

(d) Zero, L mol s

For the reaction 2A + B  D + E the following mechanism has been proposed : A + B  C + D (slow) A + C  E (fast)

For a reaction, A + 2B  C rate is given by 2 R = K [A] [B] . The order of reaction is

The rate law expression for the reaction is (a) Rate = k [A] [B]

(b) Rate = k [A] [B]

(a) 3

(b) 6

(c) Rate = k [A] [C]

(d) Rate = k [A] [B] [C]

(c) 5

(d) 7

2

2

CHEMICAL KINETICS 20.

The units of rate of reaction are –1 –1

(a) litre mol L s –1

(c) mol s 21.

22.

123 the chemical reaction will be (log 2 = 0.301) –1 –1

(a) 230.3 minutes

(b) 23.03 minutes

–1

(c) 46.06 minutes

(d) 460.6 minutes

(b) mol L s (d) mol s

The rate of reaction is doubled for every 10ºC rise in temperature. The increase in rate as result of increase in temperature from 10ºC to 100ºC is (a) 112

(b) 512

(c) 400

(d) 256

27.

28.

1 A  2B, rate of disappearance of ‘A’ is 2 related to the rate of appearance of ‘B’ by the expression For a reaction

(a)

d[ A] 1 d[B]  dt 2 dt

(b) 

23.

d[ A] d[ B]  dt dt

(d) 

d[ A ] d[B] 4 dt dt

30.

25.

26.

(b) 0.5 h

(c) 0.25 h

(d) 1 h

In a first order reaction, the conc. of reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is (a) 30 min

(b) 60 min

(c) 7.5 min

(d) 15 min

The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of

5 6

(b) 3 (d)

11 6

H For the reaction CH3COCH3 + I2    Products, the rate is governed by the expression

(a) 0

(b) 1

(c) 2

(d) 3

The rate of a reaction 2X + Y  Products is given by d[Y] 2 = k [X] [Y] dt

if X is present in large excess, then order of the reaction is :

1 of the original amount 32

(a) 4 h

The rate of reaction A + B + C  Products is given by 1/2 1/3 Rate = k [A] [B] [C]

The order w.r.t. iodine is

1 (c) of the original amount 16

The time for half-like period of a certain reaction, A  products is 1 h. When the initial concentration of the reactant ‘A’, is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1, if it is a zero order reaction ?

(d) Unimolecular reaction

dx  k [Acetone] [H+] dt

1 (a) of the original amount 4

24.

(c) first order reaction

(c)

The half life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be:

(d)

(b) second order reaction

(a) 1

d[ A] 1 d[ B]  dt 4 dt

1 (b) of the original amount 8

(a) bi molecular reaction

The order of the reaction is :

29. (c) 

A reaction involving the different reactants can never be

31.

(a) zero

(b) two

(c) one

(d) three

For a reaction pA + qB  products, the rate law expression m n is r = k [A] [B] then (a) (p + q)  (m + n)

(b) (p + q) = (m + n}

(c) (p + q) may or not be equal to (m + n) (d) (p + q) > (m + n) 32.

33.

–2

–1

The rate constant of a reaction is 2.5 × 10 minutes . The order of the reaction is (a) one

(b) zero

(c) two

(d) three

In the presence of acid, the initial concentration of cane-sugar was reduced from 0.2 M to 0.1 M in 5 hrs and

CHEMICAL KINETICS

124

to 0.05 M in 10 hrs. The reaction must be of

will be

(a) zero order

(b) first order

(c) second order

(d) fractional order

(a) a/k (c) 2a/k 39.

Integrated rate law 34.

A reaction is of first order when

Which one of the following formulae represents the first order reaction ? (a) k 

(a) The amount of product formed increases linearly with time

2.303 [A] log t [A]0

(c) [A] = [A0] e–kt

(b) The rate decreases linearly with time

35.

(b) a/2k (d) k/a

(b) k 

2.303 ax log t a

(d) k 

2.303 a log t ax

(c) The rate is linearly related to the concentration of the reactant

Practical Methods to Determine Order of Reaction

(d) The concentration of the reactant decreases linearly with time.

40.

For the first order reaction A  Products, which one of the following is the correct plot of log [A] versus t time ?

41. (a)

(b)

Half life of a chemical reaction at a particular concentration is 50 min. When the concentration of the reactant is doubled, the half life becomes 100 min, then the order of the reaction is : (a) zero

(b) first

(c) second

(d) third

Which of the following statements is false ? (a) For a first order reaction, the rate of reaction doubles as the concentration of reactant (s) doubles (b) Active mass of 64g of HI present in a two-litre flask is 0.25 (c) For zero order reaction, the rate changes with temperature

(c)

(d) Both order and molecularity of a reaction are always the same.

(d) 42.

Which one of the following statements is incorrect about the molecularity of a reaction ? (a) Molecularity of a reaction is not the number of molecules of the reactants present in the balanced equation

36.

37.

Which of the following represents the expression for 3/4th life of a first order reaction (a)

k log 4 / 3 2.303

(b)

2.303 log 3/ 4 k

(c)

2.303 log 4 k

(d)

2.303 log 3. k

(c) Molecularity is always a whole number (d) There is no difference between order and molecularity of a reaction. 43.

The correct expression for the rate constant for reactions of zero order is (a) k = [A0]/2t 1 t

(c) k  {[A]  [A]0 } 38.

(b) Molecularity of a reaction is the number of molecules in the slowest step

A + B  C + D (slow) A + C  E (fast)

1 t

The rate law expression for the reaction is (a) Rate = k [A]2 [B]

(b) Rate = k [A] [B]

2.303 log {[A 0 ]  [A]} t

(c) Rate = k [A] [C]

(d) Rate = k [A]2 [B] [C]

(b) k  {[A 0 ]  [A]} (d) k 

For the reaction 2 A + B  D + E the following mechanism has been proposed :

If ‘a’ is the initial concentration of the reactant, the time taken for completion of the reaction, if it is of zero order,

44.

For a single step reaction ; A + 2 B  Products, the molecularity is

CHEMICAL KINETICS

45.

46.

125

(a) zero

(b) 1

(c) 2

(d) 3

For a reaction X + Y  Z, rate  [X]. What is (i) molecularity and (ii) order of reaction ? (a) (i) 2, (ii) 1

(b) (i) 2, (ii) 2

(c) (i) 1, (ii) 1

(d) (i) 1, (ii) 2

(a)

d[ H 2 ] d[ I 2 ] d[ HI]   dt dt dt

(b)

dH 2 d[ I 2 ] d[ H ]   dt dt dt

(c)

1 d[ H 2 ] 1 d[ I 2 ] d[ HI]   2 dt 2 dt dt

For a reaction, 2NO  2H 2  N 2  2H 2O, the possible (d)  2

mechanism is

2NO  N 2O 2

50.

slow N 2O 2  H 2   N 2O  H 2O fast N 2O  H 2O  N 2O  H 2O

The rate law for a reaction between the substance A and B n m is given by rate = k [A] [B] . On doubling the concentration of A and halving the conc. of B, the ratio of the new rate of the earlier rate of reaction will be

What is the rate law and order of the reaction ? (a)

(a) Rate = k [N2O2], order = 1

d[ H 2 ] d[ I ] d[ HI]  2 2  dt dt dt

1 2

m n

(b) m + n

2

(b) Rate = k [NO] [H2], order = 3

(c) n – m

2

(c) Rate = k [NO] , order = 2

51.

2

(d) Rate = k [N2O2] [H2], order = 3 47.

(fast)

[O]  O3  2O 2

(slow)

respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by –1 100 kJ mol . The enthalpy change of the reaction (A2 + B2  2AB) in the presence of catalyst will be (in kJ mol–1).

The rate law expression will be (a) Rate = k [O] [O3] 2

(c) Rate = k [O3] 48.

2

(b) Rate = k [O3] [O2] (d) Rate = k [O2] [O]

–1

(a) 300

(b) 120

(c) 280

(d) 20

3A + 2B  C + D

The rate of a chemical reaction doubles for every 10ºC rise of temperature. If the temperature is raised by 50ºC, the rate of the reaction increases by about

The differential rate law can be written as :

(a) 10 times

(b) 24 times

(c) 32 times

(d) 64 times

52.

For the reaction,

(a) 

d C  1 d  A n m   k  A   B 3 dt dt

53.

(c) 

d A dt



d C  dt

n

2HI is

the temperature at which k1 becomes equal to k2 is

m

The differential rate law for the reaction H2 + I2

C   D; k 2  1012 e 24,606 / T

 k  A  B 

1 d  A  d C  n m   k  A  B (d) 3 dt dt

For the two gaseous reactions, following data are given A   B; k1  1010 e 20,000 / T

1 d  A  d C  n m   k  A   B (b)  3 dt dt

49.

The energies of activation for forward and reverse reac–1 – tion for A2 + B2 2AB are 180 kJ mol and 200 kJ mol 1

The chemical reaction, 2O3  3O 2 proceeds as

O3  O 2  [O]

(d) 2 ( n  m )

54.

(a) 400 K

(b) 1000 K

(c) 800 K

(d) 1500 K

In a 1st order reaction the fraction of molecules at 450ºC having sufficient energy (or fraction of effective collisions)

CHEMICAL KINETICS is 1.92 × 10 reaction ?

126

–16

(a)

. What is activation energy value of this

2

–1

(a) 21.757 × 10 J mole 4

–1

(c) 21.75 × 10 J mole

3

–1

(b) 21.757 × 10 J mole (d) None

Collision Theory 55.

In a reaction A  B threshold energy of the reactant is –1 120 kJ mole and the normal energy of the reactant is –1

20 kJ mole what is the activation energy barrier of the reaction A  B. –1

–1

(b) 100 kJ mole –1

(c) 140 kJ mole

–1

(d) 2400 kJ mole

The rates of reactions increase with increase of temperature because

62.

(a) activating energy of the reacting molecules increases (b) kinetic energy of the product molecules increases (c) the fraction of the reacting molecules possessing an energy equal to the activation energy or more increases (d) the collisions between molecules decrease. 57.

Mark the correct statement.

63.

(a) The catalyst catalyses the forward reaction (b) the catalyst catalyses the backward reaction (c) The catalyst influences the direct and the reverse reaction to the same extent (d) The catalyst increases the rate of forward reaction and decreases the rate of backward reaction 58.

The rate of reaction increases with increase of temperature because

4

InK (sec ) = 14.34 – (1.25 × 10 )/T

–1

(a) 6 kJ mole 56.

61.

A catalyst is used in the rate determining step of a mechanism, but then is formed again in some subsequent step so that there is no change in the concentration of the catalyst during the course of the reaction. (b) Catalyst does not appear in the overall stoichiometric equation for the reaction. (c) The role of the catalyst is to provide an alternate activated complex, one with a lower activation energy. (d) All are correct. The rate constant of a reaction is given by What will be the energy of activation ? (a) 24.83 Kcal mole (b) 12.42 Kcal/mole (c) 49.96 Kcal/mole (d) None The reactions of higher order are rare because (a) many body collisions involved very high activation energy. (b) many body collisions have a low probability energy (c) many body collisions are not energetically favoured (d) many body collisions can take place only in the gaseous phase. According to collision theory of reaction rates, the rate of reaction depends (a) only upon the total number of collisions per second (b) upon the colliding molecules with energy greater than threshold energy (c) upon the orientation of molecules at the time of collision (d) both on (b) and (c)

64.

The temperature dependence of the rate constant k is expressed as k = Ae  Ea / RT . When a plot between logk and 1/T is plotted we get the graph as shown. What is the value of slope in the graph ?

(a) fraction of reactant molecules having sufficient energy increases (b) the average energy of the products increases (c) threshold energy changes (d) activation energy is lowered. 59.

60.

The activation energy for a reaction is 9.0 Kcal/mol. The increase in the rate constant when temperature is increased from 298 to 308 K (a) 10%

(b) 100%

(c) 50%

(d) 63%

(a)

Ea RT

Ea (c)  2.303 RT log A

Which of the following statement is/are correct ? 65.

(b) 

Ea 2.303R

(d) 

Ea R 2.303 T

Rate of a reaction can be expressed by Arrhenius equa-

CHEMICAL KINETICS tion as k = A e

–E/RT

127

In this equation, E represents (2006)

(a) the energy above which all the colliding molecules will react (b) the energy below which colliding molecules will not react

Numerics 70.

–1

66.

71.

–1 –1

72.

For a reaction R  P, the concentration of a reactant changes from 0.05 M to 0.04 M in 30 minutes. What will be the average rate of reaction in minutes ?

73.

For the reaction 4NH 3  5O 2   4NO  6H 2O, if the –3

67.

(b) 99

(c) 16

(d) 60

is the rate of formation of H2O ? 74.

2NO 2  2NO  O 2

(a) Ea is energy of activation

If the rate of decrease of concentration of NO 2 is

(b) R is Rydberg constant

6.0 × 10

(c) K is equilibrium constant

concentration of O2 ?

k2 A   C, Activation energy E a 2

(c) k 2  k1e

Ea 2 / RT

Ea 2 / RT

(b) k1  k 2 e

–1 –1

mol L s . What will be the rate of increase of

1.386 hours are required for the disappearance of 75% of a reactant of first order reaction. What is the rate constant of the reaction ?

76.

Half-life period of a first order reaction is 10 min. What percentage of the reaction will be completed in 100 min ?

77.

The rate of formation of a dimer in a second order –1 –1

If E a2  2E a1 , then k1 and k2 are (a) k1  2k 2 e

–12

75.

A reactant (A) forms two products k1 A   B, Activation energy Ea1

69.

Nitrogen dioxide (NO2) dissociates into nitric oxide (NO) and oxygen (O2) as follows :

Which one is correct for K = Ae  E a / RT ?

(d) A is adsorption factor 68.

–1 –1

rate of disappearance of NH3 is 3.6 × 10 mol L s , what

(R = molar gas constant = 8.314 JK–1 mol–1) (a) 132

Consider the reaction, 2N 2O5   4NO 2  O 2 . In the reaction NO2 is being formed at the rate of 0.0125mol L s . What is the rate of reaction at this time ?

The rate coefficient (k) for a particular reactions is 1.3 × 10–4 M–1 s–1 at 1000C, and 1.3 ×10–3 M–1 s–1 at 1500C. What is the energy of activation (EA) (in kJ) for this reaction?

–1

decreases from 0.5 mol L to 0.4 mol L in 10 minutes. What is the rate of reaction during this interval ?

(c) the total energy of the reacting molecules at a temperature, T (d) the fraction of molecules with energy greater than the activation energy of the reation.

In a reaction 2HI  H2 + I2, the concentration of HI

Ea1 / RT

(d) k1  A k 2 e

Ea1 / RT

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R = 8.314 JK–1 mol–1 and log 2 = 0.301)

(a) 53.6 kJ mol–1

(b) 48.6 kJ mol–1

(c) 58.5 kJ mol–1

(d) 60.5 kJ mol–1

–1

dimerisation reaction is 9.1 × 10 mol L s at 0.01 mol L monomer concentration. What will be the rate constant for the reaction ? –3 –1

78.

The rate constant of a first order reaction is 15 × 10 s . How long will 5.0 g of this reactant take to reduce to 3.0 g ?

79.

What will be the half-life of the first order reaction for –1

which the value of rate constant is 200 s ? 80.

The half-life of a radioactive isotope is three hours. If the initial mass of the isotope was 300 gm, the mass which remained undecayed in 18 hours would be

81.

The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. The age of the sample is

CHEMICAL KINETICS

128

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS 1.

collisions

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A. Online 2017 SET (1)

(b) loss of active species on collision

(a) 9.84 K

(b) 4.92 K

(c) low probability of simultaneous collision of all the

(c) 2.45 K

(d) 19.67 K

Higher order (>3) reactions are rare due to :

(2015)

5.

(a) shifting of equilibrium towards reactants due to elastic

6.

reacting species (d) increase in entropy and activation energy as more molecules are involved 2.

3.

(Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693;

Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be: (2016) (a) 6.93 × 10–4 mol min–1

(b) 2.66 L min–1 at STP

(c) 1.34 × 10–2 mol min–1

(d) 6.93 × 10–2 mol min–1

R = 8.314 J mol–1 K–1)

7.

The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below Online 2016 SET (1)

k i  5.2  109 L mol1 s 1 



ClO  g   O  g   O2  g   Cl  g  —(ii) k ii  2.6  1010 L mol1s 1

The closet rate constant for the overall reaction

(a) 107.2 kJ mol–1

(b) 53.6 kJ mol–1

(c) 26.8 kJ mol–1

(d) 214.4 kJ mol–1

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kj mol–1. If k1 and k2 are rate constants for reactions R1

k  and R2, respectively at 300 K, then ln  2  is equal to  k1  –1 –1 (R = 8.314 J mol K ) (2017)

O3  g   Cl  g   O2  g   ClO  g  —(i)



The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : Online 2017 SET (2)

8.

(a) 8

(b) 12

(c) 6

(d) 4

Which of the following lines correctly show the temperature dependence of equilibrium constant, K, foran exothermic reaction ? (2018)

O3  g   O  g   2O2  g  is

4.

(a) 5.2 × 109 L mol–1 s–1

(b) 2.6 × 1010 L mol–1 s–1

(c) 3.1 × 1010 L mol–1 s–1

(d) 1.35 × 1020 L mol–1 s–1

The rate law for the reaction below is given by the expression K [A] [B]

A  B  Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be : Online 2016 SET (2)

(a) C and D

(b) A and D

(a) k

(b) k/3

(c) A and B

(d) B and C

(c) 3k

(d) 9k

CHEMICAL KINETICS 9.

10.

11.

129

At 518ºC, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torrs s-1 when 5% had reacted and 0.5 Torr s-1 when 33% had reacted. The order of the reaction is (2018) (a) 1

(b) 0

(c) 2

(d) 3

15.

(a) 175.0 mmHg

(b) 116.25 mmHg

(c) 136.25 mmHg

(d) 106.25 mmHg

(In 2 = 0.693, In 3 = 1.1) (a) 5 (c) 4.1

14.

2

(b) Rate  k  A  B

(c) Rate  k  A  B

(d) Rate  k  A   B

2

The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are:

(2019-04-09/Shift-1)

For a first order reaction, A  P,t1/2 (half- life) is 10 days. th

13.

2

N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be: Online 2018 SET (1)

1 The time required for conversion of A (in days) is : 4

12.

2

(a) Rate  k  A   B

16.

(a) 1, 1

(b) 0, 2

(c) 0, 1

(d) 1, 0

A bacterial infection in an internal would grows as

N '(t)  N 0 exp(t) , where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down

Online 2018 SET (2) (b) 3.2

as

(d) 2.5

hour?

If 50 percent of a reaction occurs in 100 second and 75 percent of the reaction occurs in 200 second, the order of this reaction is : Online 2018 SET (3) (a) Zero

(b) 1

(c) 2

(d) 3

N dN  5N 2 . What will be the plot of 0 vs, t after 1 dt N

(a)

(b)

(c)

(d)

k1 k2 For a reaction scheme. A   B   C if the rate of formation of B is set to be zero then the concentration of B is given by (2019-04-08/Shift-2)

 k1  (a)  k   A   2

(b)  k1  k 2   A 

(c) k1k 2  A

(d)  k1  k 2   A 

For the reaction 2A  B  C the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is:

(2019-04-10/Shift-1) 17.

For the re action of H 2 w i t h I 2 , the rate constant is

2.5 104 dm3 mol1 s1 at 327°C and 1.0 dm3 mol1 s1 at 527°C. The activation energy for the reaction, in

kJ mol1 is:

(2019-04-10/Shift-2)

(R = 8.314 J K 1 mol1 )

(2019-04-08/Shift-1)

(a) 72

(b) 166

(c) 150

(d) 59

CHEMICAL KINETICS 18.

130

In the following reaction; xA  yB

.

Which one of the following statements is correct? (a) Total order of the reaction is 4

 d A    d  B  log10     log10    0.3010  dt   dt 

(b) Order of the reaction with respect to B is 2

‘A’ and ‘B’ respectively can be :

(d) Order of the reaction with respect to A is 2

(a) C2 H 2 and C6 H 6

(b) n-Butane and Iso-butane

(c) N 2 O 4 and NO2

(c) C2 H 2 and C4 H8

(c) Order of the reaction with respect to B is 1 22.

equation (0  C < T < 300  C) : (k and E a are rate constant and activation energy respectively) Choose the correct option:

(2019-04-12/Shift-1) 19.

Consider the given plots for a reaction obeying Arrhenius

NO 2 required for a reaction is produced by the decomposition of N 2 O5 in CCl4 as per the equation

2N 2 O5 (g)  4NO 2 (g)  O 2 (g) The initial concentration of N 2 O5 is 3.00 mol L1 and it is 2.75 mol L1 after 30 (2019-01-10/Shift-1)

minutes. The rate of formation of NO 2 is :

(a) I is right but II is wrong (b) Both I and Ii are correct

(2019-04-12/Shift-2)

(c) I is wrong but II is right (d) Both I and II are wrong

(a) 1.667 102 mol L1 min 1 23.

(b) 4.167 103 mol L1 min 1 (c) 8.333 103 mol L1 min 1

expression for

(d) 2.083 103 mol L1 min 1 20.

k1  2A .the For an elementary chemical reaction, A 2  k 1

d A dt

(a) k1  A 2   k 1  A 

The following results were obtained during kinetic studies of the reaction

2A  B  products

(2019-01-10/Shift-2)

2

(b) 2k1  A 2   k 1  A 

2

(d) 2k1  A 2   2k 1  A 

(c) k1  A 2   k 1  A  24.

is

2

2

If a reaction follows the Arrhenius equation, the plot Ink v 1/(RT) gives straight line with a gradient (–y) unit. The energy required to activate the reactant is: (2019-01-11/Shift-1)

The time(in minutes) required to consume half of A is (2019-01-09/Shift-1)

21.

(a) 5

(b) 10

(c) 1

(d) 100

25.

For the reaction, 2A  B  products , when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L1s 1 to 2.4 mol L1s 1 . When the concentration of A alone is doubled, the rate 1 1 1 1 increased from 0.3 mol L s to 0.6 mol L s

(2019-01-09/Shift-2)

26.

(a) y/R unit

(b) y unit

(c) yR unit

(d) –y unit

The reaction 2X  B is a zeroth order reaction. If the initial concentration of X is 0.2M, the half life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be: (a) 9.0 h

(b) 12.0 h

(c) 18.0 h

(d) 7.2 h

Decomposition of X exhibits a rate constant for 0.05  g/ year. How many years are required for the decomposition of 5  g of X into 2.5  g? (2019-01-12/Shift-1) (a) 50

(b) 25

(c) 20

(d) 40

CHEMICAL KINETICS

131 31.

27.

For a certain reaction consider the plot of nk versus 1/T given in the figure. If the rate constant of this reaction at 5 1 400 K is 10 s , then the rate constant at 500 K is:

(2019-01-12/Shift-2)

32.

33. (a) 106 s 1

28.

(b) 2  104 s 1

(c) 104 s 1 (d) 4  104 s 1 The results given in the below table were obtained during kinetic studies of the following reaction : 2A  B  C  D 34.

29.

30.

X and Y in the given table are respectively : (2020-09-02/Shift-2) (a) 0.4, 0.4 (b) 0.3, 0.4 (c) 0.4, 0.3 (d) 0.3, 0.3 It is true that : (2020-09-03/Shift-1) (a) A second order reaction is always a multistep reaction (b) A first order reaction is always a single step reaction (c) A zero order reaction is a multistep reaction (d) A zero order reaction is a single step reaction

3 For the reaction 2A  3B  C  3P, which statement 2 is correct ? (2020-09-03/Shift-2) (a)

dn A dn B dn C   dt dt dt

(b)

dn A 3 dn B 3 dn C   dt 2 dt 4 dt

(c)

dn A 2 dn B 4 dn C   dt 3 dt 3 dt

(d)

dn A 2 dn B 3 dn C   dt 3 dt 4 dt

35.

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _____. (take : log 2 = 0.30; log 2.5 = 0.40) (2020-09-04/Shift-1) The number of molecules with energy greater than the threshold energy for a reactionincreases five fold by a rise of temperature from 27ºC to 42ºC. Its energy of activationin J/mol is ________. (Take ln 5 = 1.6094 ; R = 8.314 J mol–1K–1) (2020-09-04/Shift-2) A flask contains a mixture of compound A and B. Both compounds decompose by first order kinetics. The halflives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is : (Use ln 2 = 0.693) (2020-09-05/Shift-1) (a) 180 (b) 300 (c) 120 (d) 900 The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol–1 is : (R is gas constant) (2020-09-05/Shift-1)

(a) R (b) 2/R (c) 1/R (d) 2R Consider the following reactions A  P1; B  P 2; C  P3; D  P4, The order of the above reactions are a,b,c and d, respectively. The following graph is obtained when log [rate] vs. log [conc.] are plotted:

Among the following the correct sequence for the order of the reactions is: (2020-09-06/Shift-1) (a) c > a > b > d (b) d > a > b > c (c) d > b > a > c (d) a > b > c > d

CHEMICAL KINETICS 36.

132 (a) Ea>Ec> Ed> Eb

The rate of a reaction decreased by 3.555 times when the temperature was changed from 40ºC to 30ºC. The activation energy (in KJ mol-1) of the reaction is ...

(b) Ec>Ea> Ed> Eb (c) Eb> Ed>Ec>Ea

[Take; R = 8.314 J mol-1 K-1 In 3.555 = 1.268]

(d) E >Ea> Ed>Ec

(2020-09-06/Shift-1) 37.

During the nuclear explosion, one of the products is 90Sr with half of 6.93 years. If 1 µg of 90Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically

40.

For following reactions 700 K A   Product Catalyst,700 K A   Product;

It was found that the Eais decreased by 30 kJ/mol in the

(2020-01-07/Shift-1)

presence of catalyst. If the rate remains unchanged , the 38.

The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in activation energy upon adding enzyme is:

activation energy for catalysed reaction is (Assume pre exponential factor is same) (2020-01-09/Shift-1) (a) 75 kJ/mol

(2020-01-08/Shift-1) (a) - 6RT

(b) – 6  2.303 RT T

(b) 135 kJ/mol

(c) + 6RT

(d) +6  2.303 RT T

(c) 105 kJ/mol (d) 198 kJ/mol

39.

1 Consider the following plots of rate constant versus T for four different reactions. Which of the following orders is correct for the activation energies of these reactions? (2020-01-08/Shift-2)

41.

A sample of milk splits after 60 min. at 300K and after 40 min at 400K when the population of lactobacillus acidophilus in it doubles . The activation energy (in kJ/ mol) for this process is closest to —— .

2 (Given, R = 8.3 J mol”1K”1), ln    0.4 , e”3 = 4.0) 3 (2020-01-09/Shift-2)

CHEMICAL KINETICS

133

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS Single Choice Questions 1. The rate of reaction is defined as (a) decreases in the concentration of a reactant (b) increase in the concentration of a product (c) change in the concentration of any one of the reactants or products per unit time. (d) all the above three are correct. 2. For the reaction A + 2B  C, the rate of reaction at a given instant can be represented by (a) 

6.

(a) increases with increase in temperature (b) decreases with increase in temperature (c) does not depend upon temperature (d) does not depend upon concentration. 7.

d[A] 1 d[B] d[C]   dt 2 dt dt

d[A] 1 d[B] d[C]   (b) dt 2 dt dt (c) 

d[A] 1 d[B] d[C]   dt 2 dt dt For a gaseous reaction, the units of rate of reaction are 

–1

(a) L atm s

–1 –1

(c) atm mol s 4.

5.

–1

H = – 80 kJ mol

–1

H = – 30 kJ mol

–1

H = + 40 kJ mol

–1

H = + 20 kJ mol

R1

Ea = 40 kJ mol ,

R2

Ea = 20 kJ mol ,

R3

Ea = 60 kJ mol ,

R4

Ea = 30 kJ mol ,

–1 –1 –1 –1

(a) R2 < R3 < R4 < R1

–1

(b) R4 < R3 < R2 < R1

–1

(c) R1 < R4 < R3 < R2

(b) atm s

(d) mol s

 dx   is found The rate of a reaction at different times    dt 

as follows : Time (in minute)

Ea and H values of reactions R1, R2, R3 and R4 carried out at the same temperature are as given below :

At a given temperature and assuming that the backward reactions of all these reactions have the same frequency factor, the rates of R1, R2, R3 and R4 in their respective backward reactions are in the increasing order of

d[A] 1 d[B] d[C]   dt 2 dt dt

(d)

3.

Rate of a reaction

(d) R1 < R2 < R3 < R4 8.

–1 –1

Rate in (mol L s )

What is the order of a reaction which has a rate expression 3/2 –1 rate = k [A] [B] ? (a) 3/2

(b) 1/2

(c) zero

(d) none of these.

–2

0

2.80 × 10

10

2.78 × 10

20

2.81 × 10

30

2.79 × 10

–2 –2

9.

–2

The order of reaction is (a) zero (b) one (c) two (d) three. Which one of the following statements is incorrect ? (a) Rate law expression cannot be written from the stoichiometric equation. (b) Law of mass action expression can be written from the balanced equation. (c) Specific reaction rate of a reaction is constant at constant temperature. (d) Rate of reaction and rate constant have same units.

10.

11.

Order of a reaction can be (a) fractional

(b) zero

(c) integer

(d) all the above. –2

–2

2

–1

The rate constant of a reaction is 2.1 × 10 mol litre min . The order of reaction is (a) zero

(b) 1

(c) 2

(d) 3.

When concentration of reactant in reaction A  B is increased by 8 times, the rate increases only 2 times. The order of the reaction would be (a) 2

(b) 1/3

(c) 4

(d) 1/2

CHEMICAL KINETICS 12.

13.

14.

15.

16.

17.

134

For a given reaction half life period was found to be directly proportional to the initial concentration of the reactant. The order is (a) Zero

(b) 1

(c) 2

(d) 3

20.

The half-life period for a reaction at initial concentrations –1 of 0.5 and 1.0 mol lit are 200 sec and 100 sec respectively. The order of the reaction is (a) zero

(b) 1

(c) 2

(d) 3

(b) first order

(c) second order

(d) fractional order

In the decomposition of Ammonia it was found that at 50 torr pressure T1/2 was 3.64 hour while at 100 torr T1/2 was 1.82 hours. Then order of reaction would be (a) 0

(b) 1

(c) 2

(d) 3

(b) 1/3

(c) 4

(d) 1/2

The rate of reaction between A and B increases by a factor of 100, when the concentration of A is changed from –1 –1 0.1 mol L to 1 mol L . The order of reaction with respect to A is :

(c) zero

(d) none of these.

A hypothetical reaction A2 + B2  2 AB follows the mechanism as given below :

A + B2  AB + B (slow) A + B  AB (fast) The order of the over all reaction is : (a) 2 (c) 1 21.

(b) 1

1 2

(d) 0

If order of reaction is zero. It means that (a) rate of reaction is independent of temperature (b) rate of reaction is independent of the concentration of the reacting species (c) the rate of formation of activated complex is zero

When concentration of reactant in reaction A  B is increased by 8 times, the rate increases only 2 times. The order of the reaction would be (a) 2

(b) two

A2   A + A (fast)

In the presence of acid, the initial concentration of canesugar was reduced from 0.2 M to 0.1 M in 5 hrs and to 0.05 M in 10 hrs. The reaction must be of (a) zero order

(a) one

(d) rate of decomposition of activated complex is zero 22.

23.

The rate constant of a reaction has same units as the rate of reaction. The reaction is of (a) zero order

(b) first order

(c) second order

(d) none of these.

The rate constant for a zero order reaction is –2

18.

(a) 10

(b) 1

(c) 3

(d) 2

(b) zero

(c) 3

(d) 2

24.

(a) 0.5 M

(b) 1.25 M

(c) 12.5 M

(d) 1.0 M

SO2Cl2  SO2 + Cl2 is the first order as gas reaction with –5

19.

A reaction involves two reactants. The rate of reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The overall order of reaction will be

–1

after 25 sec is 0.5 M, the initial concentration must have been

For the reaction, A  B, the rate of reaction is quadrupled when the concentration of A is doubled, the rate expression n of the reaction is r = k [A] , when the value of n is (a) 1

–1

2 × 10 mol L sec . If the concentration of the reactant

–1

K = 2.2 × 10 sec at 320ºC. The percentage of SO2Cl2 decomposed on heating for 90 minutes is (a) 1.118

(b) 0.1118

(c) 18.11

(d) 11.30

CHEMICAL KINETICS 25.

26.

135

Four vessels 1, 2, 3 and 4 contain respectively, 10 mol atom (t1/2 = 10 hours), 1mol atom (t1/2 = 5 hours), 5 mol atom (t1/2 = 2 hour) and 2 mol atom (t1/2 = 1 hour) of different radioactive nuclides. In the beginning, the maximum radioactivity would be exhibited by the vessel (a) 4 (b) 3 (c) 2 (d) 1 Two gases A and B are filled in a container. The experimental rate law for the reaction between them, has been found to 2 be Rate = k [A] [B] Predict the effect on the rate of the reaction when pressure is doubled :

32.

(a) 0.5

(b) 2.0

(c) 10.0

(d) 200.0

The rate constant for the reaction –5

–1

2N2O5  4NO2 + O2 is 3.0 × 10 sec –5

–1

–1

(a) the rate is doubled (b) the rate becomes four times

(a) 1.4

(b) 1.2

(c) the rate becomes six times

(c) 0.04

(d) 0.8

33.

The rate law expression for the hypothetical reaction 2 A + 3 B  2 C is

dx 2 = k [A] [B] dt

(a) 1

(b) 2

(c) 3

(d) 5

For a first order reaction : A  B, Whose concentration vs. time curve is as shown in the figure. The rate constant is equal to –1

(a) 41.58 h –3 –1 (c) 1.155 × 10 s

The order of reaction is

28.

The equilibrium constant of reaction is 20.0 At equilibrium, the rate constant of forward reaction is 10.0. The rate constant for backward reaction is

If the rate is 2.4×10 mol Lt sec the concentration of –1 N2O5 (in mol litre ) is

(d) the rate becomes eight times 27.

31.

–1

(b) 4.158 s –1 (d) 6.93 min

The rate law of gaseous reaction : A(g) + B(g)  Products 2

is given by k[A] [B]. If the volume of the reaction vessel is suddenly doubled, which of the following will happen ? (a) The rate w.r.t. A will decrease two times (b) The rate w.r.t. A will decrease four times (c) The rate w.r.t. B will decrease two times

34.

(d) The overall rate will decrease 8 times of the original value 29.

The chemical reaction, 2O3   3O2 proceeds as follows

For a first order reaction P(g)   Q(g) + R(g). After 10 minutes the volume of R gas is 10 L and after complete reaction 50L. –1 Hence rate constant will be (in min ). (a)

1 ln 5 10

(b)

1 1 ln 10 5

(c)

1 ln 4 10

(d)

1 ln 1.25 10

O3  O2 + O (Fast) O + O3   2O2 (slow) The rate law expression should be 2

30.

35. 2

(a) Rate = K [O3]

(b) Rate = K[O3] [O2]

(c) Rate = K [O3] [O2]

(d) None

–1

(a) increases with increase of temperature

The unit of rate constant for a zero order reaction is –1

–1

(a) litre sec

–1

–1

(b) litre mol sec –1

(c) mol litre sec

The activation energy of a reaction is zero. The rate constant of the reaction

–1

(d) mol sec .

(b) decreases with increase of temperature (c) decreases with decrease of temperature (d) is nearly independent of temperature.

CHEMICAL KINETICS Milk turns sour at 40ºC three times faster at 0ºC. Hence Ea in calories of turning of milk sour is (a)

42.

(a) proportional to initial conc. of reactants

2.303  2  313  273 log 3 40

(b) independent of initial concentration of reactants (c) inversely proportional to initial concentration of reactants

2.303  2  313  273 log(1 / 3) (b) 40

(c)

37.

(d) inversely proportional to square of initial concentration of reactants.

2.303  2  40 log 3 273  313

43.

(a) directly proportional to initial concentration

The ratio of the rate constant of a reaction at any temperature T to the rate constant T   is equal to

(b) inversely proportional to initial concentration (c) independent of initial concentration (d) proportional to the square of initial concentration.

(b) Fraction of molecules in the activated state

44.

(c) Average life of the reaction (d) Pre-exponential factor in the Arrhenius equation The rate constant of a reaction A  B + C at 27ºC is –5 –1 –4 3.0 × 10 s and at this temperature 1.5 × 10 percent of the reactant molecules are able to cross-over the P.E. barrier. The maximum rate constant of the reaction is –9 –1

(a) 4.5 × 10 s –1 (c) 0.2 s 39.

40.

(b) 4.5 × 10 –1 (d) 20 s

45.

(a) 6.25 M

(b) 1.25 M

(c) 0.125 M

(d) 0.625 M

s

The half-life period of a radioactive element is 120 days. Starting with 1 gm, the amount of element decayed in 600 days will be

(c)

1 g 32

As the reaction progresses, the rate of reaction (a) increases (b) decreases (c) remains constant (d) first increases, then decreases. For the reaction 4 A + B  2 C + 2D, which of the following statement is not correct ? (a) the rate of disappearance of B is one fourth of the rate of disappearance of A (b) the rate of formation of C is one-half of the rate of consumption of A

–11 –1

The half-life a first order reaction is 24 hours. If we start with 10M initial concentration of the reactant then conc. after 96 hours will be

1 g (a) 16

41.

The half-life period of a zero order reaction is

(d) none

(a) Energy of activation of the reaction

38.

Half life period of 2nd order reaction is :

(c) the rate of appearance of D is half of the rate of disappearance of B (d) the rates of formation of C and D are equal. 46.

For the reaction A + B  C + D, the variation of the concentration of the product with time is given by the curve.

II IV

15 g (b) 16

(d)

31 g 32

Conc

36.

136

III

The half-life period of a radio active element is 30 min. One sixteenth the original quantity of the element will remain unchanged after

I Time

(a) one hour

(b) sixteen hour

(a) I

(b) II

(c) four hour

(d) two hour

(c) III

(d) IV

CHEMICAL KINETICS 47.

137 52.

Which one of the following statements is incorrect ? (a) The temperature coefficient of a reaction is the ratio of the rate constant at any two temperatures

O3  O2 + O (fast)

(b) The temperature coefficient of a reaction is the ratio of the rate constants at 298 K and 308 K

O + O3  O2 (slow) When the concentration of O2 is increased, then the rate

(c) The temperature coefficient of most of the reactions lies between 2 and 3 (d) In an endothermic reaction, activation energy of reactants is more than that of the products 48.

The decomposition of ozone is believed to occur by the mechanism :

53.

Under a given set of experimental conditions, with increase of concentration of the reactants, the rate of a chemical reaction

(a) increases

(b) decreases

(c) remains same

(d) cannot be predicted

In the accompanied diagram, ER, EP and EX represent the energy of the reactants, products and activated complex respectively. Which of the following is the activation energy for the backward reaction ?

(a) Decreases

EX

(b) increases A

(c) Remains unaffected 49.

The thermal decomposition of A  B + C has rate constant x mole 

–1/2

1/2

lit

–1

min at a given temperature. How would

temperature constant

50.

C

EP

ER

dA will change if concentration of A is doubled keeping dt

54.

(a) will increase by 2.828 times

D

B

ENERGY

(d) First decreases, then increases.

(a) A

(b) B

(c) C

(d) D

For the hypothetical reaction

(b) will increase by 4 times

A2 + B2

(c) will increase by 11.313 times

A2

(d) will not change

A + B2

AB + B (slow reaction)

A+ B

AB (fast reaction)

The thermal decomposition of acetaldehyde :

2AB, the mechanism is given as below A + A(fast reaction)

–3

CH3CHO  CH4 + CO, has rate constant of 1.8 × 10 –1/2 1/2 –1 mole L min at a given temperature. How would

then (a) the rate determining step is A + B2

d[CH 3CHO]  will change if concentration of acetaldehyde dt is doubled keeping the temperature constant ?

(b) the order of the reaction is 3/2 (c) the overall molecularity is 4

(a) will increase by 2.828 times (b) will increase by 11.313 times (c) will not change

(d) the rate expression is Rate = k [A] [B2] 55.

In the sequence of reaction k1 k2 k3 A   B     C  D

(d) will increase by 4 times 51.

AB + B

For a single step reaction; A + 2B  Products, the molecularity is

k3 > k2 > k1, then the rate determining step of the reaction is :

(a) Zero

(b) 1

(a) A  B

(b) B  C

(c) 2

(d) 3

(c) C  D

(d) A  D

CHEMICAL KINETICS 56.

138

If Ef and Er are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic, then

61.

(a) Ef < Er

(a) 30 kJ/mol

(b) 40 kJ/mol

(b) Ef > Er

(c) 70 kJ/mol

(d) 100 kJ/mol

62.

(c) Ef = Er (d) No relation can be given between Ef and Er as data are not sufficient. 57.

–3

–1

The rate constant of a reaction at 27ºC is 2.3 × 10 min and at this temperature 0.002% of the reactant molecules are able to cross over the energy barrier existing between the reactants and products. By increasing the temperature, the rate constant of reaction will increase to a maximum of –2

–1

–2

(a) 2.3 × 10 min

–1

(b) 4.6 × 10 min

–1

63.

64.

–1

(c) 2.30 min 58.

(d) 115 min

as x kJ mol of A. If energy change of the reaction is y kJ, the activation energy of the reverse reactions is :

(a) zero-order reaction

(b) first-order reaction

(c) second order reaction

(d) third order reaction

For an endothermic reaction, where H represents the enthalpy of reaction in kJ mol, the minimum value for the energy of activation will be (a) Less than H

(b) zero

(c) More than H

(d) Equal to H

A sample of rock from the moon contains equal number of 9 atoms of U and Pb (t½ for uranium = 4.5×10 years). The age of rock would be (a) 4.5 × 10 years 9 (c) 13.5 × 10 years

9

(b) 90 × 10 years 9 (d) 2.25 × 10 years

(a) – x(b) x – y

For the decomposition of N2O5 at a particular Temperature according to the equations

(c) x + y

2N2O5   4NO2 + O2

(d) y – x.

N2O5  2NO2 +

65.

For the reaction –

CH3Cl(aq) + OH

(aq)

–

–1

[CH3Cl]

[OH ]

+ d [CH3OH]/dt (M min )

0.2

0.1

2 × 10

0.4

0.1

4 × 10

0.4

0.2

8 × 10

–3 –3 –3 14

If Kc for the above reaction is 1 × 10 , then the specific –1 –1 reaction rate (M min ) for the replacement of –OH group of methanol by Cl atom is : –18

(a) 10 15 (c) 10

(b) 10 (d) 10

–15

Energy of activation of an exothermic reaction is

1 O 2 2

the activation energies are E1 and E2 respectively, then

– (aq)

 CH3OH(aq) + Cl

The kinetic data are as given below :

60.

Radioactive decay is a

9

An endothermic reaction A  B has an activation energy –1

59.

If a reaction A + B  C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy 70 kJ/mol, the activation energy for the reverse reaction is

66.

(a) E1 > E2

(b) E1 < E2

(c) E1 = 2E2

(d) E1 = E2\

According to the collision model of kinetics, certain activation energy must be overcome before a reaction can proceed. Based on the data given below, what is a reasonable estimate of the activation energy for the decomposition of NOCl ? 2 NOCl(g)  2 NO(g) + Cl2(g) temperature (K)

rate constant, k (L/mol s) –4

400

6.6 × 10

500

2.9 × 10

600

1.63 × 10

–1 1

2

(b) 1.23 × 10 J/K mol

5

(d) 1.34 × 10 J/K mol

(a) zero

(b) negative

(a) 1.00 × 10 J/K mol

(c) positive

(d) can’t be predicted

(c) 1.05 × 10 J/K mol

3 6

139

CHEMICAL KINETICS Multiple Choice Question 67.

71.

Which of the following statements about the Arrhenius equation is/are correct ?

(CH3CO)2 O + C2H5OH  CH3COOC2H5 + CH3COOH

(a) The pre-exponential factor becomes equal to the rate constant of the reaction at extremely high temperature

Select the correct statements of the following : (a) When reaction is carried out in dilute hexane solution, the rate law is given by k [Anhydride] [Alcohol]

(b) When the activation energy of the reaction is zero, the rate becomes independent of temperature

(b) When ethanol is the solvent, the rate law is given by k [Anhydride]

–Ea/RT

(c) The term e represents the fraction of the molecules having energy in excess of threshold value

(c) The values of k in the two cases are the same

(d) On raising temperature, rate constant of the reaction of greater activation energy increases less rapidly than that of the reaction of smaller activation energy. 68.

+

Hydrolysis of an ester is catalysed by H ion.

(d) Using ethanol as the solvent, its concentration changes significantly during the course of the reaction 72.

Using equimolar concentrations of two acids HX and HY, both being strong acids, the rate constants of the reaction –3

–1

–3

–1

(b) Time needed for a definite fraction of first order reaction does not vary with the initial concentration

(a) Rate constant may be taken as the measure of degree of ionization of the acid used as catalyst

(c) Time for 25% reaction is one-third of half-life in second order process

(b) HX is a stronger acid than HY, their relative strength being 1.7

(d) Rate of zero order reaction gets doubled if the concentration of the reactant is increased to a two fold value

(c) HX is a weaker acid than HY, their relative strength being 0.6 (d) none is correct

73.

When the temperature of a reaction is changed from T1 to T2 half-life is found to decrease. Thus :

Which of the following statements are correct ? (a) Time required for 75% completion is 1.5 times of halflife for zero order reaction

are found to be 3 × 10 min and 5 × 10 min respectively at a fixed temperature. It can be concluded that

69.

Estrification of acetic anhydride by ethanol takes place as

Which of the following graphs plotted are true ? (a) For a zero order reaction

(a) T2 > T1 (b) The reaction is exothermic (c) The reaction is endothermic (d) The reaction can be exothermic or endothermic 70.

+

Hydrolysis of a sugar is catalysed by H ion. Half-life of the reaction is independent of initial concentration of sugar at a particular pH. At a constant concentration of sugar rate increases 10 times when pH is decreased by one unit. Pick out the correct statements of the following :

(b) For a zero order reaction

(a) Rate  [sugar] +

(b) Rate  [H ] (c) Rate law : rate = k (sugar) +

(d) Rate law : rate = k [sugar] [H ]

(c) For a 3rd order reaction

CHEMICAL KINETICS

140

(d)

(d) For a 1st order reaction 75.

An increase in the rate of reaction with rise in temperature is due to (a) an increase in the number of collisions (b) an increase in the number of activated molecules (c) lowering of activation energy (d) shortening of mean free path.

Comprehension Based Questions Comprehension 74.

Which of the following graphs are properly represented

Concentrations measured as a function of time when gaseous N 2O 5 at initial concentration of 0.0200 M decomposes to gaseous NO2 and O2 at 55ºC. The change in concentration with time is given by the following graph.

(a)

(b)

76.

The instantaneous rate of reaction at the beginning of the reaction is –5

(a) 2.2 × 10 M/s –5

(c) 6.3 × 10 M/s

–5

(b) 1 × 10 M/s (d) zero

(c) 77.

The rate of formation of NO2 during the period 600 – 700 s is –5

(a) 3.7 × 10 M/s –5

(c) 4.8 × 10 M/s

–5

(b) 2.2 × 10 M/s –5

(d) 1.6 × 10 M/s

CHEMICAL KINETICS 78.

141

The rate of decomposition of N2O5 during the period 300 – 400 s is –4

(a) 3.7 × 10 M/s –5

(c) 1.9 × 10 M/s

–5

(b) 2.6 × 10 M/s –8

(d) 3.6 × 10 M/s

a

Hence, Rate = k  A 

b

 B

a

= k  A

b

 B0

a

= k'  A 

(k   k[B]0b ) or log  initial rate  = log  0  log k   a log [A ] ...... (iii) A plot of log (rate) against log  A values will be a straight

Comprehension Measurement of rate of reactions provides as insight into the series of elementary steps by which a reaction takes place. In most cases, the rate depends on the concentrations of the reactants which decreases with the progress of the reaction. In consequence, the rate invariably decreases with time. In chemical kinetics, the rate at any instant of time i.e., instantaneous rate is

line which enables to calculate both k and a. Similarly orders with respect to other reactants. taken in much smaller concentrations turn by turn, can be determined. Consider the reaction : 2I (g )  Ar(g )   I 2(g )  Ar(g) The following figures show the variation of log 0 against (a) log I0 for a given  Ar 0 and (b) log  Ar 0 for a given  I 0

d[R ] d[ P ] or , where R and P are the dt dt

expressed as

reactants and products respectively. For example, for the reaction

A + 2B  3C + D   i  The rate w.r.t. A, B, C and D would be d[A ] d[ B] d[C] d[D] , , and respectively. Obviously,, dt dt dt dt

these rates will be different . However, the unique or rationalized rate of reaction is defined as the rate of change of extent of reaction as : Rate (unique) of reaction =

d[ A] 1 d[ B] 1 d[C] d[D]    dt 2 dt 3 dt dt

The dependence of rate on concentrations of reactants is expressed in terms of rate law, which is established experimentally. a

Rate = k  A 

 Bb

(Rate law) ...............(ii)

The exponents a, b, etc. (determined experimentally) may or may not be equal to the respective stoichiometric coefficients. k is the velocity constant of the reaction. The determination of rate law is simplified by the isolation method in which the concentration of all the reactants except one are in large excess. The concentrations of the reactants in large excess remain practically constant throughout the reaction. If B is in large excess. we can approximate  B by  B0

The rate constants of most reactions increase as the temperature is increased. The rate constant increases by about 100-200% for a temperature rise of 10K . It is found experimentally for many reactions that a plot of n k against 1 T gives a straight line. This behaviour is expressed in the form of equation.

n k  n A 

Ea RT

.............. (iv)

A and E a are called pre-exponential factor and activation energy respectively. From the plot of n k versus 1 T , the values of A and E a can be calculated.

CHEMICAL KINETICS 79.

142

The rate of change of molar concentration of C in Reaction – 1, –3 –1 –1 is found to be 3.0 × 10 mol L s . The rate of reaction and rate of disappearance of the reactant B are respectively. –3

–1 –1

–3

–1 –1

–3

–1 –1

84.

(a) 3.0 × 10 mol L s each (b) 1.0 × 10 mol L s each

80.

(b) B

(c) C

(d) D

Assertion : If in a zero order reaction, the concentration of the reactant is doubled, the half-life period is also doubled.

(c) 1.0 × 10 mol L s and 2.0 × 10 mol L s

Reason : For a zero order reaction, the rate of reaction is independent of initial concentration.

(d) None of these.

(a) A

(b) B

According to the fig-1 (a) and 1 (b), the rate law for the reaction

(c) C

(d) D

–3

–1 –1

85. 2I(g) + Ar(g)  I2(g) + Ar(g) is given by 2

(a) Rate = k [I] [Ar]

(b) Rate = k[I]

Assertion : According to steady state hypothesis, in a multistep reaction, the change in concentration with time for reactive intermediates is zero.

The greater slope of the plot of n k against 1/T for a reaction shows the rate constant to be

Reason : The intermediates are so reactive that after a brief initial period their concentrations rise from zero to a small value and remains constant for most of the duration of the reaction.

(a) Independent of temperature

(a) A

(b) B

(c) C

(d) D

2

2

2

(c) Rate = k[I] [Ar] 81.

(a) A

(d) Rate = k[I] [Ar]

(b) Strongly dependent on temperature (c) Poorly dependent on temperature (d) Insufficient information to predict 82.

86.

Which of the following statements is not correct ? Reason : Lower the activation energy, faster is the reaction.

(a) Rate of effective collisions between reacting molecules is equal to A (b) Rate of the effective collisions between reacting molecules is Ae

(a) A

(b) B

(c) C

(d) D

 Ea /RT

87.

(c) Rate of total collisions between reacting molecules is equal to A

= e Ea /RT

Assertion–Reason Type Questions (a) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’ (b) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’ (c) Assertion is true but Reason is false

Assertion : For the 2N2O5  4NO2 + O2 ; Rate = K[N2O5]. Reason : Rate of decomposition of N2O5 is determined by slow step.

(d) Fraction of collisions between reacting molecules having energy in excess of activation energy

83.

Assertion : If the activation energy of a reaction is zero, temperature will have no effect on the rate constant.

a) A

(b) B

(c) C

(d) D

Matrix–Match Type Questions 88.

C0 = Initial concentration of reactant; C = Concentration of reactant at any time t; k = rate constant Match the following : Plots

Slopes

(d) Assertion is false but Reason is true

(A) C vs (abscissa) for zero order

(P)

Assertion : The rate of the reaction is the rate of change of concentration of a reactant or a product.

(B) log C vs t (abscissa) for first order

(Q) Zero

Reason : Rate of reaction remains constant during the complete reaction.

Unity

CHEMICAL KINETICS

143 then half-life of the radioactive drug is

 dc  (C)    vs C for zero order  dt 

(R) – k

(D) ln (–dc/dt) vs ln C for first order

(S) -k/2.303

92.

Temperature coefficient is defined as the factor by which the rate of reaction increases on increasing the temperature by 10ºC at a given temperature. If activation energy of a

Integer Type Questions

reaction is 85 kJ, determine the temperature coefficient at 300 K (Rounded off to the nearest whole number).

89.

For the reaction : A + B  P, if concentration of B is doubled, maintaining concentration of A constant, rate of reaction is also doubled. If concentration of A is tripled. maintaining concentration of B constant, rate of reaction increases by a factor of nine. If concentrations both A and B are doubled simultaneously, the rate of reaction will increase by a factor of

90.

For the reaction : A  P when initial concentration of reactant is halved, the half-life increases by a factor of eight. Order of reaction is

91.

4.0 kg of a radioactive drug is supplied from a reactor to a laboratory but laboratory receives only

2 kg of the

radioactive substance due to its rapid decay during transportation. If 6.0 hrs are elapsed in transportation,

CHEMICAL KINETICS

144

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS 1.

The rate constant for the reaction 2N2O5  4NO2 + O2, is –5

–1

–5

–1

8.

–1

3.0 × 10 sec . If the rate is 2.40 × 10 mol litre sec , –1

then the concentration of N2O5 (in mol litre ) is (2000)

2.

3.

(a) 1.4

(b) 1.2

(c) 0.04

(d) 0.8

9.

A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers –1 the activation barrier by 20 kJ mol . (2000) –1 –1

The rate of first order reaction is 0.04 mol L s at 10 min

10.

(b) I

(c) I 5.

(d) C.I

Consider the chemical reaction, N2(g) + 3H2(g)  2NH3(g). The rate of this reaction can be expressed in terms of time derivative of concentration of N2(g), H2(g) or NH3(g). Identify the correct relationship amongst the rate expressions. (2002)

11.

Cu (half-life = 12.8 h) decays by  emission (38%),  emission (19%) and electron capture (43%). Write the decay products and calculate partial half-lives for each of the decay processes. (2002)

7.

In a first order reaction the concentration of reactant 3 3 4 decreases from 800 mol/dm to 50 mol/dm in 2 × 10 sec.

(a) 2 × 10

(b) 3.45 × 10 –4

(c) 1.386 × 10

–4

(d) 2 × 10

(d) 1.73 × 10 M min

–5

–1

(m/L)

–1 –1

(mL s )

[A]0 [B]0 0.1 0.1 0.05 0.2 0.1 0.1 0.1 0.2 0.05 (a) Write the rate equation. (b) Calculate the rate constant. The rate constant of a reaction depends upon : (a) extent of reaction (b) time of reaction (c) temperature of the system (d) concentration of the system

(2004)

t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is k, the t1/4 can be written as (2005)

+

6.

–5

–1

13.

(d) Rate = –d[N2]/dt = –d[H2]/dt = d[NH3]/dt

4

–4

2X (g)  3Y (g) + 2Z (g) Time (in min) 0 100 200 Partial pressure of X (in mm of Hg) 800 400 200 Assuming ideal gas condition. Calculate (a) Order of reaction (b) Rate constant (c) Time taken for 75% completion of reaction (d) Total pressure when px = 700 mm (2005)

(c) Rate = –d[N2]/dt = 1/3d[H2]/dt = 1/2d[NH3]/dt

–1

–1

12.

(b) Rate = –d[N2]/dt = –3d[H2]/dt = 2d[NH3]/dt

The rate constant of reaction in sec is :

–4

(2004)

(a) Rate = –d[N2]/dt = –1/3d[H2]/dt = 1/2d[NH3]/dt

64

(b) 3.47 × 10 M min

For the given reaction, A + B  Products Following data are given Initial conc. Initial conc. Initial rate (m/L)

If ‘I’ is the intensity of absorbed light and ‘C’ is the concentration of AB for the photochemical process, * AB + hv AB , the rate of formation of AB* is directly proportional to (2001) 2

–1

(c) 3.47 × 10 M min

–1 –1

(a) C

–4

(a) 1.73 × 10 M min

and 0.03 mol L s at 20 min after initiation. Find the halflife of the reaction. (2001) 4.

The rate constant of a reaction depends on : (2003) (a) the temperature of a reaction (b) the time of a reaction (c) the extent of reaction (d) the initial conc. of the reactant The reaction, X  Product, follows first order kinetics. In 40 minutes the concentration of X changes from 0.1 to 0.025 M. The rate of reaction, when concentration of X is 0.01 M is (2004)

(2003)

(a) 0.75/k

(b) 0.69/k

(c) 0.29/k

(d) 0.10/k

145

CHEMICAL KINETICS 14.

Consider an endothermic reaction X  Y with the

Comprehension - I (Ques 19 to 21) Carbon-14 is used to determine the age of organic material. 14 The procedure is based on the formation of C by

activation energies Eb and Ef for the backward and forward reactions respectively. In general

(2005)

neutron capture in the upper atmosphere.

(a) there is no definite relation between Eb and Ef

14 7

(b) Eb = Ef

14

C is absorbed by living organisms during 14 photosynthesis. The C content is constant in living organism once the plant or animal dies, the uptake of 14 carbon dioxide by it ceases and the level of C in the

(c) Eb > Ef (d) Eb < Ef 15.

Which one of the following statement for order of reaction is not correct ? (2005)

dead being, falls due to the decay which C-14 underoges

(a) Order can be determind experimentally

14 6

(b) Order of reaction is equal to sum of the powers of concentration terms in differential rate law.

16.

17.

The half-life period of

C 14 7 N+β

14

C is 5770 yr .

(c) It is not affected with the stoichiometric coefficient of the reactants

The decay constant (λ) can be calculated by using the

(d) Order cannot be fractional

following formula λ=

For the reaction A + B ........ C, it found that doubling the concentration of A increases the rate by 4 times, and doubling the concentration of B doubles reaction rate. What is the overall order of the reaction ? (2006) (a) 4

(b) 3/2

(c) 3

(d) 1

+

+

Ag NH3  [Ag(NH3) ] ; k1 = 6.8 × 10 +

+

+

–6 –6

(c) 1.08 × 10

The comparison of the β - activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over 14 periods longer than 30,000 yr . The proportion of C to 12

(2006) 19.

(2006)

14

–5

(b) 1.08 × 10

(b) Carbon dating can be used to find out the age of earth crust and rocks

–5

(d) 6.8 × 10

(c) Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbons content remains constant in living organisms

reaction of NO with Br2 to form NOBr NO(g) + Br2(g)  NOBr2(g)

(d) Carbon dating cannot be used to determine 14 concentration of C in dead beings.

2NOBr(g)

If the second step is the rate determining step, the order of the reaction with respect to NO (g) is (2006) (a) 1

(b) 0

(c) 3

(d) 2

Which of the following option is correct ? (a) In living organisms, circulation of C from atmosphere is high so the carbon content is constant in organism

The following mechanism has been proposed for the

NOBr2(g) + NO(g)

C in living matter is 1 : 1012 .

–3

then the formation constant of [Ag(NH3)2] is (a) 6.8 × 10

0.693 t1 2

–3

[Ag(NH3)] + NH3  [Ag(NH3)2] ; k2 = 1.6 × 10

18.

1 N+ 0 n1 14 6 C+ 1 p

20.

What should be the age of fossil for meaningful determination of its age ? (a) 6 yr (b) 6000 yr (c) 60,000 yr (d) It can be used to calculate any age

146

CHEMICAL KINETICS 21.

A nuclear explosion has taken place leading to increase in 14 14 concentration of C in nearby areas. C concentration is Cl in nearby areas and C2 in areas far away. If the age of

25.

tion k  Ae  E a / RT . Activation energy (Ea) of the reaction can be calculated by ploting (2007)

the fossil is determined to be T1 and T2 at the places respectively then

C1 1 ln C  2 26.

C1 1 ln C  2

(c) the age of fossil will be determined to be the same 27.

T1 C1 (d) T  C 2 2

(d) 100 days

T 50 of first-order reaction is 10 min. Starting with (2008)

–1

–1

–1

(b) 0.0693 × 2.5 mol L min –2

–1

(c) 0.0693 × 5 mol L min –1

–1

(d) 0.0693 × 10 mol L min 28.

Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substance becomes half in 40 s and 20 s through first order and zero order kinetics respec-

 k0  tively. Ratio  k  of the rate constants for first order (k1)  1

Consider the reaction, 2A + B  products When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is (2007)

24.

(c) 10 days

(a) 0.0693 mol L min

(log10 2 = 0.3)

–1 –1

(b) no unit

–1 –1

(d) s

(c) mol L s

(b) 300 days

–1

An organic compound undergoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10

(a) L mol s

(a) 1000 days

–1

[t1/8 ] respectively. What is the value of [t ] × 10 ? 1/10

23.

A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room ? (2007)

10 mol L , rate after 20 min is

Subjective Questions 22.

(b) log k vs

(c) k vs T

(b) the age of fossil will decrease at the place where explosion has taken place and T1  T2 =

1 T 1 (d) k vs log T

(a) log k vs T

(a) the age of fossil will increase at the place where explosion has taken place and T1–T2 =

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equa-

–1

Consider a reaction aG + bH  Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is (2007)

and zero order (k0) of the reaction is –1

29.

3

(2008) –3

(a) 0.5 mol dm

(b) 1.0 mol dm

(c) 1.5 mol dm–3

(d) 2.0 mol–1 dm

3

For a first order reaction A  P, the temperature (T) dependent rate constant (k) was found to follow the equation :

log k 

2000  6.0 T

The pre-exponential factor A and the activation energy Ea, respectively, are (2009) (a) 1.0 × 106 s–1 and 9.2 kJ mol–1

(a) 0

(b) 1

(b) 6.0 s–1 and 16.6 kJ mol–1

(c) 2

(d) 3

(c) 1.0 × 106 s–1 and 16.6 kJ mol–1 6

(d) 1.0 × 10 s–1 and 38.3 kJ mol–1

CHEMICAL KINETICS 30.

147

Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is (2010)

34.

In a bimolecular reaction, the steric factor P was experimentaly determined to be 4.5. the correct option(s) among the following is (are) (2017) (a) The activation energy of the reaction is unaffected by the value of the steric factor

(a)

(b) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation

(b)

(c) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally (d) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used 35. (c)

For a first order reaction A(g)  2B(g)  C(g) at constant volume and 300 K, the total pressure at the begin-

(d)

ning (t - 0) and at time t are P0 and Pt , respectively. Ini31.

tially, only A is present with concentration [A]0 , and t1/3

In the reaction,

is the time required for the partial pressure of A to reach 1/ 3rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases)

P + Q   R+S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is (2013)

32.

33.

(a) 2

(b) 3

(c) 0

(d) 1

(2018)

For the elementary reaction M  N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is (2015) (a) 4

(b) 3

(c) 2

(d) 1

According to the Arrhenius equation,

36. (2016)

(a) a high activation energy usually implies a fast reaction. (b) rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy. (c) higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant. (d) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.

(a)

(b)

(c)

(d)

Consider the following reversible reaction,

A(g)  B(g)  AB(g).

(2018)

The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J mol–1). If the preexponential factor of the forward reaction is 4 times that of –1

the reverse reaction, the absolute value of G  (in J mol ) for the reaction at 300 K is ........... . (Given; ln(2) = 0.7, RT = 2500 J mol–1 at 300 K and G is the Gibbs energy)

CHEMICAL KINETICS 37.

148

In the decay sequence,  x1 238  234 92 U  90Th

40.  x4

 x2  x3   234  234 Z   230 91 Pa  90Th

238 92 U

206 82 Pb

x1, x2, x3 and x4 are particles/radiation emitted by the respective isotopes. The correct option(s) is(are):

is known to undergo radioactive decay to form

by emitting alpha and beta particles. A rock initially

contained 68 × 10–6 g of 238 92 U . If the number of alpha particles that it would emit during its radioactive decay of

(2019/Shift-1) (a) x1 will deflect towards negatively charged plate.

238 92 U

(b) x2 is  

the value of Z ?

18 to 206 82 Pb in three half-lives is Z × 10 , then what is

(c) x3 is  -ray

(2020/Shift-1) 41.

(d) z is an isotope of uranium 38.

Consider the kinetic data given in the following table for the reaction A + B + C  product.

Which of the following plots is(are) correct for the given reaction? ([P]0 is the initial concentration of P)

(2020/Shift-2) The rate of the reaction for [A] = 0.15 mol dm–3, [B] = 0.25 mol dm–3 and [C] = 0.15 mol dm–3 is found to be Y × 10–5 mol dm–3s–1. The value of Y is _________ (2019/Shift-1) 39.

(a)

(b)

(c)

(d)

The decomposition reaction Δ 2N2O5(g)   2N2O4(g) + O2(g) is started in aclosed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y × 103 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 × 10–4 s–1, assuming ideal gas behavior, the value of Y is _____. Truncate after 2 decimals

(2019/Shift-2)

149

CHEMICAL KINETICS

Note:

CHEMICAL KINETICS Please share your valuable feedback by scanning the QR code.

150

04 ELECTROCHEMISTRY

151

Chapter 04

ELECTROCHEMISTRY 1.

4.

ELECTROCHEMISTRY

In this, we make non spontaneous process to occur by suppliying Electrical energy.

Electrochemistry is the study of production of electricity from the energy released during a spontaneous chemical reaction and the use of electrical energy to bring about non-spontaneous chemical transformations.

2.

In this, electrodes an dipped in an electrolytic solution containing cations and anions. On supplying current, the ions move towards electrodes of opposite polarity and simultaneous reduction and oxidation takes place.

ELECTROCHEMICAL CELLS A spontaneous chemical process is the one which can take place on its own and in such a process the Gibb’s energy of the system decreases. It is this energy that gets converted to electrical energy.The reverse process is also possible in which we can make non-spontaneous processes occur by supplying external energy in the form of electrical energy. These inter conversions are carried out in equipments called Electrochemical Cells.

4.1

Preferential Discharge of ions Where there are more than one cation or anion the process of discharge becomes competitive in nature. Discharge of any ion requires energy and in case of several ions being present, the discharge of that ion will take place first which requires less energy.

5.

2.1 TYPES

ELECTRODE POTENTIAL It may be defined as the tendency of an element, when it is placed in contact with its own ions to either lose or gain electrons and in turn becomes positively or negatively charged.

Electrochemical Cells are of two types: 2.1.1 Galvanic Cells/Voltaic Cells

The electrode potential will be named as oxidation or reduction potential depending upon whether oxidation or reduction has taken place.

Converts chemical energy into electrical energy 2.1.2 Electrolytic Cells Converts electrical energy into chemical energy.

3.

ELECTROLYTIC CELL

n   M  s    M  aq   ne reduction oxidation

GALVANIC CELL/VOLTAIC CELLS  M n   aq   ne    M s oxidation reduction

Cell energy is extracted from a spontaneous chemical process or reaction and it is converted to electric current. For example, Daniell Cell is a Galvanic Cell in which Zinc and Copper are used for the redox reaction to take place. Zn (s) + Cu2+ (aq)

5.1

(a) Both oxidation and reduction potentials are equal in magnitude but opposite in sign.

Zn2+ (aq) + Cu(s)

Oxidation Half : Zn (s)

Zn2+ (aq) + 2e–

Reduction Half : Cu2+(aq) + 2e–

Characteristics

(b) It is not a thermodynamic property, so values of E are

Cu(s)

not additive.

2+

Zn is the reducing agent and Cu is the oxidising agent.The half cells are also known as Electrodes. The oxidation half is known as Anode and the reduction half is known as Cathode. Electrons flow from anode to cathode in the external circuit. Anode is assigned negative polarity and cathode is assigned positive polarity. In Daniell Cell, Zn acts as the anode and Cu acts as the cathode.

6.

STANDARD ELECTRODE POTENTIAL (EO) It may be defined as the electrode potential of an electrode determined relative to standard hydrogen electrode under standard conditions. The standard conditions taken are :

SCAN CODE ELECTROCHEMISTRY

152

ELECTROCHEMISTRY (i) 1M concentration of each ion in the solution. (ii) A temperature of 298 K. (iii) 1 bar pressure for each gas.

7. ELECTROCHEMICAL SERIES The half cell potential values are standard values and are represented as the standard reduction potential values as shown in the table at the end which is also called Electrochemical Series.

8.

CELL POTENTIAL OR EMF OF A CELL The difference between the electrode potentials of two half cells is called cell potential. It is known as electromotive force (EMF) of the cell if no current is drawn from the cell.

(f) If an inert electrode like platinum is involved in the construction of the cell, it may be written along with the working electrode in bracket, say for example, when a zinc anode is connected to a hydrogen electrode.

Zn  s  | Zn 2   C1  || H   C2  | H 2 |  Pt  s 

10. SALT BRIDGE Salt bridge is used to maintain the charge balance and to complete the circuit by facilitating the flow of ions through it. It contains a gel in which an inert electrolyte like Na2SO4 or KNO3 etc are mixed. Negative ions flow to the anode and positive ions flow to the cathode through the salt bridge and charge balance is maintained and cell keeps on functioning.

Ecell = Ecathode + Eanode For this equation we take oxidation potential of anode and reduction potential of cathode. Since anode is put on left and cathode on right, it follows therefore, = ER + EL ECell  ECathode/ERed  EAnode/Red For a Daniel cell, therefore o E ocell  E Cu  E oZn / Zn 2  0.34   0.76   1.10 V 2 / Cu

9. CELL DIAGRAM OR REPRESENTATION OF A CELL The following conventions or notations are applied for writing the cell diagram in accordance with IUPAC recommendations. The Daniel cell is represented as follows : Zn(s) | Zn2+ (C1) || Cu2+ (C2) | Cu (s) (a) Anode half cell is written on the left hand side while cathode half cell is written on right hand side. (b) A single vertical line separates the metal from aqueous solution of its own ions. Zn  s  | Zn 2   aq  ; Anodic chamber

Cu 2   aq  | Cu  s  Cathodic chamber

(c) A double vertical line represents salt bridge (d) The molar concentration (C) is placed in brackets after the formula of the corresponding ion. (e) The value of e.m.f. of the cell is written on the extreme right of the cell. For example, Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu EMF = +1.1V

11. SPONTANEITY OF A REACTION G = – nFECELL For a spontaneous cell reaction, ΔG should be negative and cell potential should be positive. If we take standard value of cell potential in the above equation we will obtain standard value of ΔG as well. ΔGo = – nFE0CELL

12. TYPES OF ELECTRODES 12.1 Metal-Metal Ion electrodes A metal rod/plate is dipped in an electrolyte solution containing metal ions. There is a potential difference between these two phases and this electrode can act as a cathode or anode both. Anode: M

Mn+ + ne– SCAN CODE ELECTROCHEMISTRY

153

ELECTROCHEMISTRY Cathode: Mn+ + ne–

M

12.2 Gas Electrodes Electrode gases like H2, Cl2 etc are used with their respective ions. For example, H2 gas is used with a dilute solution of HCl (H+ ions). The metal should be inert so that it does not react with the acid.

Cathode : Hg2Cl2(s) + 2e–

2Hg(l) + 2Cl–(aq)

Anode : 2Hg(l) + 2Cl–(aq) Anode: H2

+

 2H + 2e

Cathode: 2H+ + 2e–

This electrode is also used as another standard to measure other potentials. Its standard form is also called Standard Calomel Electrode (SCE). Reg. For Saturated Calomel

–

H2

Electrode Potential  0.244 V at 25C

The hydrogen electrode is also used as the standard to measure other electrode potentials. Its own potential is set to 0 V as a reference. When it is used as a reference the concentration of dil HCl is taken as 1 M and the electrode is called “Standard Hydrogen Electrode (SHE)”. 12.3 Metal-Insoluble salt electrode We use salts of some metals which are sparingly soluble with the metal itself as electrodes. For example, if we use AgCl with Ag there is a potential gap between these two phases which can be identified in the following reaction: AgCl(s) + e–

Hg2Cl2(s) + 2e–

Ag(s) + Cl–

This electrode is made by dipping a silver rod in a solution containing AgCl(s) and Cl– ions. 12.4 Calomel Electrode Mercury is used with two other phases, one is a calomel paste (Hg2Cl2) and electrolyte containing Cl– ions.

12.5 Redox Electrode In these electrodes two different oxidation states of the same metal are used in the same half cell. For example, Fe2+ and Fe3+ are dissolved in the same container and an inert electrode of platinum is used for the electron transfer. Following reactions can take place: Anode: Fe2+ Fe3+ + e– Cathode: Fe3+ + e–

Fe2+

13. NERNST EQUATION It relates electrode potential with the concentration of ions. Thus, the reduction potential increases with the increase in the concentration of ions. For a general electrochemical cell reaction of the type. 

ne aA  bB  cC  dD

Nernst equation can be given as c

E cell  E

o cell

d

RT  C  D  ln nF  A a  Bb

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154

ELECTROCHEMISTRY c

E cell  E

o cell

d

 C  D  2.303  RT log nF  Aa  Bb

Substituting the values of R and F we get

concentrations and the solutions are connected through salt bridge, such cells are known as concentration cells. For example H2 | H+(c1) || H+ (c2) | H2 Cu | Cu+2 (c1) || Cu2+(c2) | Cu

c

o E cell  E cell 

d

C  D , at 298 K 0.0591 log n  Aa  Bb

In concentraction cell, Ecell  0 These are of two types :

14. APPLICATIONS OF NERNST EQUATION 14.1 Equilibrium Constant from Nernst Equation For a Daniel cell, at equilibrium, Ecell  0

E cell  0  E

or

E

o cell

o cell

 Zn 2   2.303RT  log 2F Cu 2  

 Zn 2   2.303RT  log  2   2F Cu 

15.1 Electrode concentration cells Where Electrolyte Conc are same, pressure at electrode is different H2 (P1) | H+ (C) || H+ (C) | H2(P2) Ecell = 0 –

P 0.059 log 2 n P1

where p 2  p1 for spontaneous reaction 15.2 Electrolyte concentration cell Where conc. of Electrolyte is different

2

But at equilibrium,

 Zn   Kc  Cu 2  

E ocell 

2.303RT log K c 2F

E ocell 

2.303  8.314  298 log K c 2  96500

0.0591  log K c 2

In general,

or,

E ocell 

0.0591 log K c n

n E ocell log KC = 0.0591

15. CONCENTRATION CELLS If two electrodes of the same metal are dipped separately into two solutions of the same electrolyte having different

The EMF of concentration cell at 298 K is given by Zn | Zn2+ (c1) || Zn2+ (c2) | Zn

Ecell  0 

c 0.0591 log 1 n c2

E cell 

c 0.0591 log 2 , n1 c1

where c2 > c1 for spontaneous reaction

16. CASES OF ELECTROLYSIS 16.1 Electrolysis of molten sodium chloride 2NaCl (l )  2Na+ (l ) + 2Cl– (l ) The reactions occurring at the two electrodes may be shown as follows : At cathode : 2Na+ + 2e–  2Na

E0 = – 2.71 V

At anode : SCAN CODE ELECTROCHEMISTRY

ELECTROCHEMISTRY

155

2Cl–  Cl2 + 2e–

E0 = – 1.36V

Anode : Zn

Overall reaction : 2Na+ (l ) + 2 Cl– (l ) 

2Na (l ) + Cl2 (g)

or 2NaCl (l )

2Na (l ) + Cl2 (g) At cathode At anode

16.2 Electrolysis of an aqueous solution of sodium chloride +

NaCl (aq)

Reaction :

–

Na (aq) + Cl (aq)

Cathode : MnO2 + NH 4 e–

The standard potential of this cell is 1.5 V and it falls as the cell gets discharged continuously and once used it cannot be recharged. 17.3 Mercury cells

Anode : Zn – Hg Amalgam

–

H2O (l )  H (aq) + OH (aq)

Cathode : Paste of HgO and carbon

(only slightly ionized)

Electrolyte : Paste of KOH and ZnO

At cathode :

Anode : Zn (Hg) + 2OH–

2Na+ + 2e–

2Na

2H2O + 2e–

H2 + 2OH– E0 = – 0.83V

MnO (OH) + NH3

These are used in small equipments like watches, hearing aids.

(almost completely ionized) +

Zn2+ + 2e–

E0 = – 2.71V

ZnO (s) + H2O + 2e–

Cathode : HgO (s) + H2O + 2e–

Hg (l) + 2OH–

Overall Reaction : Zn (Hg) + HgO (s)

ZnO (s) + Hg (l)

+

Thus H2 gas is evolved at cathode value Na ions remain in solution. At anode : 2H2O –

2Cl

The cell potential is approximately 1.35V and remains constant during its life. 17.4 Secondary Batteries

+

O2 + 4H + 4e Cl2 + 2e

–

–

0

E = – 1.23V 0

E = – 1.36V

Thus, Cl2 gas is evolved at the anode by over voltage concept while OH– ions remain in the solution.

17. BATTERIES When Galvanic cells are connected in series to obtain a higher voltage the arrangement is called Battery. 17.1 Primary Batteries Primary cells are those which can be used as long as the active materials are present. Once they get consumed, the cell will stop functioning and cannot be re-used. Example Dry Cell or Leclanche cell and Mercury cell. 17.2 Dry cell Anode : Zn container Cathode : Carbon (graphite) rod surrounded by powdered MnO2 and carbon. Electrolyte : NH4Cl and ZnCl2

Secondary cells are those which can be recharged again and again for multiple uses. e.g. lead storage battery and Ni – Cd battery. 17.5 Lead Storage Battery Anode : Lead (Pb) Cathode : Grid of lead packed with lead oxide (PbO2) Electrolyte : 38% solution of H2SO4 Discharging Reactions Anode: Pb(s) + SO42–(aq)

PbSO4(s) + 2e–

Cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O(l) Overall Reaction : Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l) To recharge the cell, it is connected with a cell of higher potential and this cell behaves as an electrolytic cell and the reactions are reversed. Pb(s) and PbO2(s) are regenerated at the respective electrodes.These cells deliver an almost consistent voltage. SCAN CODE ELECTROCHEMISTRY

156

ELECTROCHEMISTRY Recharging Reaction : 2PbSO4(s) + 2H2O(l)

Pb(s) +

PbO2(s) + 2H2SO4(aq) Electrolysis using Attachable Electrodes e.g CnSO4 using Cn Electordes Elecrtrolysis using Inest Electrodes e.g. PbBr2 by using Pt Electrodes

18. FUEL CELLS A fuel cell differs from an ordinary battery in the sense that the reactants are not contained inside the cell but are externally supplied from an external reservoir. Fuel cell is used in space vehicles and in this cell the two gases are supplied from external storages. In this cell carbon rods are used as electrodes with KOH as the electrolyte. Cathode : O2 (g) + 2H2O (l) + 4e– Anode : 2H2 (g) + 4OH– (aq)

Rusting of iron can be avoided by painting it or by coating it with some other metals like Zinc. The latter process is known as Galvanisation. As the tendency of Zn to get oxidised is more than iron it gets oxidised in preference and iron is protected. This method of protecting one metal by the other is also called Cathodic Protection.

4OH– (aq) 4H2O (l) + 4e–

overall Reaction : 2H2(g) + O2 (g)

2H2O (l)

20. CONDUCTANCE (G) It is the reciprocal of resistance and may be defined as the ease with which the electric current flows through a conductor.

G

1 R

SI unit is Siemen (S).

19. CORROSION It involves a redox reaction and formation of an electrochemical cell on the surface of iron or any other metal. At one location oxidation of iron takes place (anode) and at another location reduction of oxygen to form water takes place (cathode). First Fe gets oxidised to Fe2+ and then in the presence of oxygen it forms Fe3+ which then reacts with water to form rust which is represented by Fe2O3.xH2O. Anode : 2Fe (s)  2 Fe2+ + 4e– Eº = + 0.44 V Cathode : O2 (g) + 4H+ + 4e–  2H2O (l) Eº = 1.23 V

1 S = 1 ohm–1 (mho)

21. CONDUCTIVITY () It is the reciprocal of resistivity (). 1 1        G  R A A Now if  = 1 cm and A = 1 cm2, then  = G.. Hence, conductivity of an electrolytic solution may be defined as the conductance of a solution of 1 cm length with area of cross-section equal to 1 cm2.

Overall R × N : 2Fe (s) + O2 (q) + 4H+  2Fe2+ + + 2H2O Eºcell = 1.67 M

SCAN CODE ELECTROCHEMISTRY

ELECTROCHEMISTRY 22. FACTORS AFFECTING ELECTROLYTIC CONDUCTANCE 22.1 Electrolyte

157 It consists of two fixed resistance R3 and R4, a variable resistance R1 and the conductivity cell having the unknown resistance R2. The bridge is balanced when no current passes through the detector. Under these conditions,

An electrolyte is a substance that dissociates in solution to produce ions and hence conducts electricity in dissolved or molten state. Examples : HCl, NaOH, KCl (Strong electrolytes).

R1 R 3  R2 R4

or

R2 

R 1R 4 R3

24. MOLAR CONDUCTIVITY (m)

CH3–COOH, NH4OH (Weak electrolytes). The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The following factors govern the flow of electricity through a solution of electrolyte. (i)

(ii)

(iii)

(iv)

Nature of electrolyte or interionic attractions : Lesser the solute-solute interactions, greater will be the freedom of movement of ions and higher will be the conductance. Solvation of Ions : Larger the magnitude of solute-solvent interactions, greater is the extent of solvation and lower will be the electrical conductance. The nature of solvent and its viscosity : Larger the solventsolvent interactions, larger will be viscosity and more will be the resistance offered by the solvent to flow of ions and hence lesser will be the electrical conductance. Temperature : As the temperature of electrolytic solution rises solute-solute, solute-solvent and solvent-solvent interactions decreases, this results in the increase of electrolytic conductance.

It may be defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte placed between two large electrodes at one centimeter apart. Mathematically,  m    V,  m 

  1000 C

where, V is the volume of solution in cm3 containing 1 mole of electrolyte and C is the molar concentration.

Units :

m 

  1000 S cm 1  C mol cm 3

= ohm–1 cm2 mol–1 or S cm2 mol–1

25. EQUIVALENT CONDUCTIVITY (eq) It is conducting power of one equivalent of electrolyte placed between two large electrodes at one centimeter apart. Mathematically :

23. MEASUREMENT OF CONDUCTANCE 1   The value of could be known, R A if we measure l, A and R. The value of the resistance of the solution R between two parallel electrodes is determined by using ‘Wheatstones’ bridge method (Fig.)

 eq    v 

As we know,  

 eq 

  1000 N

Where, v is the volume of solution in cm3 containing 1 equivalent of electrolyte and N is normality. Units :   1000 N S cm1  equivalent cm 3

 eq 

or Ohm 1 cm 2 equivalent 1 or S cm 2 equivalent 1

26. VARIATION OF CONDUCTIVITY AND MOLAR CONDUCTIVITY WITH DILUTION Conductivity decreases with decrease in concentration, this is because the number of ions per unit volume that carry the current in the solution decreases on dilution. SCAN CODE ELECTROCHEMISTRY

158

ELECTROCHEMISTRY Molar conductivity   m    V  increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that the decrease in on dilution of a solution is more than compensated by increases in its volume. Graphic representation of the variation of  m vs c

electrolyte on dissociation gives v+ cations and v– anions then its limiting molar conductivity is given by  m  v   o  v   o

Here,  o and  o are the limiting molar conductivities of cations and anions respectively.

29. APPLICATIONS OF KOHLRAUSCH’S LAW 29.1 (i) Calculation of molar conductivities of weak electrolyte at infinite dilution For example, molar conductivity of acetic acid at infinite dilution can be obtained from the knowledge of molar conductivities at infinite dilution of strong electrolyte like HCl, CH3COONa and NaCl as illustrated below.

  oCH  COO  oNa    oH   Cl   oNa  oCl   3  o i.e.  m  CH3  COOH    om  CH3  COONa    om  HCl    om  NaCl

27. LIMITING MOLAR CONDUCTIVITY (m) The value of molar conductivity when the concentration approaches zero is known as limiting molar conductivity or molar conductivity at infinite dilution. It is possible to o determine the molar conductivity at infinite dilution   m  in case of strong electrolyte by extrapolation of curve of  m vs c. On contrary, the value of molar conductivity of weak electrolyte at infinite dilution cannot be determined by extapolation of the curve as the curve becomes almost parallel to y-axis when concentration approaches to zero. The mathematical relationship between  m and  om for strong electrolyte was developed by Debye, Huckel and Onsagar. In simplified form the equation can be given as  m   m  b c1/ 2

where  m is the molar conductivity at infinite dilution and b is a constant which depends on the nature of the solvent and temperature.

28. KOHLRAUSCH’S LAW It states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. In general, if an

29.2 (ii) Determination of Degree of Dissociation of Weak Electrolytes Degree of dissociation    

 cm  om

29.3 (iii) Determination of Dissociation Constant (K) of Weak Electrolytes:

also



K

c 2 1 



 cm  m

K

c   cm /  m  1   cm /  m

2



C   cm 

2

 m   m   cm 

30. USE OF G IN RELATING EMF VALUES OF HALF CELL REACTIONS When we have two half cell reactions such that on adding them we obtain another half cell reaction then their emfs cannot be added directly. But in any case thermodynamic functions like G can be added and emf values can be related through them. Consider the following three half cell reactions: SCAN CODE ELECTROCHEMISTRY

ELECTROCHEMISTRY Fe2+ + 2e–  Fe Fe3+ + 3e–  Fe

159 E1 E2

5.

Fe3+ + e–  Fe2+ E3 We can easily observe that the third reaction can be obtained by subtracting the first reaction from the second. But the same relation does not apply on the emf values. That is, E3  E2 – E1. But the G values can be related according to the reactions. That is,



 cm  m

where,  = Degree of dissociation  cm =

Molar

6.

– n3FE3 = – n2FE2 + n1FE1

where, K = Dissociation constant

– E3 = – 3E2 + 2E1

o o E ocell  E cathode  E anode

E3 = 3E2 – 2E1

= Eo Right + Eo left Nernst equation for a general electrochemical reaction 

NOTE:-

ne aA + bB   cC + dD

We should always remember that emf values are additive only when two half cell reactions are added to give a complete balanced cell reaction. In any other case we will be using G values to obtain relations between emf values.

c

E cell  E

RT  C  D  In nF  A a  Bb

E cell  E

o cell

 C  D 2.303RT  log nF  A a  Bb

c

a

o E cell  E cell 

 R       Cell constant A 8.

A = Area of cross-section of the electrodes.



3.

m 

M = Molarity of the solution. 4.

9.

 m (Ax By) = x  m (Ay+) + y  m (Bx–)

where,  m = Molar conductivity at infinite dilution x and y are the number of cations and anions produced by one formula unit of the electrolyte on complete dissociation.

b

 A  B 0.059 log n  Cc  Dd

at 298 K

n E ocell 0.0591

o (Creterion of spontaneity)  r G o   nFE cell

 r G o  2.303RT log K c

where, rGo = Standard Gibbs energy of the reaction. 10.

  1000 M

where,  m = Molar conductivity

log Kc =

d

where, Kc = Equilibrium constant.

1  cell constant R

where,  = Conductivity or specific conductance

d

o cell

31. FORMULAE

2.

given

2

G3 = G2 – G1

= Resistivity

a

For a weak binary electrolyte AB

c   cm  c 2 K    1    m   m   mc 

where, R = Resistance

at

concentration

7.

1.

conductivity

Q=I×t where Q = Quantity of charge in coulombs I = Current in amperes t = Time in seconds

11.

m=Z×I×t where m = mass of the substance liberated at the electrodes Z = Electrochemical equivalent. where E = Equivalent weight Z 

E 96500

SCAN CODE ELECTROCHEMISTRY

160

ELECTROCHEMISTRY

STANDARD REDUCTION POTENTIALS AT 298 K. IN ELECTROCHEMICAL ORDER Hg 2SO 4  2e   2Hg  SO 24

+ 0.62

MnO24  2H 2 O  2e   MnO2  4OH 

+ 0.60

+ 2.05

MnO4  e   MnO24

+ 0.56

+ 1.98

I2 + 2e–  2I–

+ 0.54

H4XeO6 + 2H+ + 2e–  XeO3 + 3H2O F2 + 2e–  2F– O3 + 2H+ + 2e–  O2 + H2O

+ 3.0 + 2.87 + 2.07

S2 O82  2e   2SO 24 Ag2+ + e–  Ag+

+

–

Co3+ + e–  Co2+

+ 1.81

Cu + e  Cu

+ 0.52

H2O2 + 2H+ + 2e–  2H2O

+ 1.78

I 3  2e   3I 

+ 0.53

+ 1.69

NiOOH + H2O + e–  Ni(OH)2 + OH–

+ 0.49

Ag 2 CrO 4  2e   2Ag  CrO 24

+ 0.45

+

–

Au + e  Au 4+

–

Pb + 2e  Pb

2+

+ 1.67

2HClO + 2H+ + 2e–  Cl2 + 2H2O

+ 1.63

Ce4+ + e–  Ce3+

+ 1.61

O2 + 2H2O + 4e–  4OH–

+ 0.40

2HBrO + 2H+ + 2e–  Br2 + 2H2O

+ 1.60

ClO4  H 2 O  2e   ClO3  2OH 

+ 0.36

MnO4  8H   5e   Mn 2  4H 2 O

+ 1.51

[Fe(CN)6]3– + e–  [Fe(CN)6]4–

+ 0.36

Mn3+ + e–  Mn2+

+ 1.51

Au3+ + 3e–  Au

+ 1.40

Cl2 + 2e–  2Cl–

+ 1.36

Cr2 O 72  14H   6e   2Cr 3  7H 2 O

+ 1.33

O3 + H2O + 2e–  O2 + 2OH–

+ 1.24

O2 + 4H+ 4e–  2H2O

+ 1.23

ClO4





 ClO3

2+

–

Cu + 2e  Cu –

Hg2Cl2 + 2e  2Hg + 2Cl –

AgCl + e  Ag + Cl

+ 0.22

Bi + 3e–  Bi

+ 0.20

2+

–

+

Cu + e  Cu 4+

–

Sn + 2e  Sn

AgBr + e  Ag + Br 4+

–

+ 1.23

2H+ + 2e–  H2

0.00 0, by definition

–

–

+ 1.09

Fe + 3e  Fe

– 0.04

–

3+

+ 0.97

O2 H2O + 2e–   HO2  OH 

– 0.08



+ 0.96

Pb2+ + 2e–  Pb

– 0.13

Pu + e  Pu 

 4H  3e  NO  2H 2 O

+

2Hg 2  2e   Hg 22

+ 0.92

ClO– + H2O + 2e–  Cl– + 2OH–

+ 0.89

Hg2+ + 2e–  Hg

+ 0.86

NO3  2H   e   NO 2  H 2 O

+ 0.80

Ag + e  Ag

+ 0.80



 2e  2Hg

Fe3+ + e–  Fe2+ –

–

In + e  In 2+

–

–

+ 0.07

–

Br2 + 2e  2Br

Hg 22

3+

–

3+

MNO2 + 4H+ + 2e–  Mn2+ + 2H2O

+

+ 0.15

–

Ti + e  Ti

NO3

+ 0.16

2+

+ 1.23

4+

+ 0.27

–

 H 2O

 2H  2e

+ 0.34

–

–

–

BrO + H2O + 2e  Br + 2OH

– 0.14

–

Sn + 2e  Sn –

AgI + e  Ag + I 2+

– 0.14 –

– 0.15

–

Ni + 2e  Ni

– 0.23

Co2+ + 2e–  Co

– 0.28

3+

–

In + 3e  In

– 0.34

+ –

+ 0.79

Tl e  Tl

– 0.34

+ 0.77

PbSO4 + 2e–  Pb + SO 24

– 0.36

+ 0.76

SCAN CODE ELECTROCHEMISTRY

ELECTROCHEMISTRY

161

Ti3+ + e–  Ti2+

– 0.37

V2+ + 2e–  V

– 1.19

Cd2+ + 2e–  Cd

– 0.40

Ti2+ + 2e–  Ti

– 1.63

In2+ + e–  In+

– 0.40

Cr3+ + e–  Cr2+

– 0.41

Fe2+ + 2e–  Fe

– 0.44

3+

–

Al + 3e  Al 3+

– 1.66

–

U + 3e  U 3+

– 1.79

–

Sc + 3e  Sc 2+

– 2.09

–

In3+ + 2e–  In+

– 0.44

Mg + 2e  Mg

– 2.36

S + 2e–  S2–

– 0.48

Ce3+ + 3e–  Ce

– 2.48

In3+ + e–  In2+

– 0.49

U4+ + e–  U3+

– 0.61

Cr3+ + 3e–  Cr

– 0.74

3+

–

La + 3e  La +

– 2.52

–

Na + e  Na 2+

– 2.71

–

Ca + 2e  Ca 2+

– 2.87

–

Zn2+ + 2e–  Zn

– 0.76

Sr + 2e  Sr

– 2.89

Cd(OH)2 + 2e–  Cd + 2OH–

– 0.81

Ba2+ + 2e–  Ba

– 2.91

2H2O + 2e–  H2 + 2OH–

– 0.83

Cr2+ + 2e–  Cr

– 0.91

Mn2+ + 2e–  Mn

– 1.18

2+

–

Ra + 2e  Ra

– 2.92

+

–

– 2.92

+

–

– 2.93

Cs + e  Cs Rb + e  Rb +

–

K +e K

– 2.93

Li+ + e–  Li

– 3.05

REDUCTION POTENTIALS IN ALPHABETICAL ORDER Ag+ + e–  Ag 2+

–

+ 0.80 +

Ag + e  Ag

+ 1.98

AgBr + e–  Ag + Br– –

–

+ 0.0713

Ca2+ + 2e–  Ca –

– 2.87 –

Cd(OH)2 + 2e  Cd + 2OH

– 0.81

Cd2+ + 2e–  Cd

– 0.40

3+

–

AgCl + e  Ag + Cl

+ 0.22

Ce + 3e  Ce

– 2.48

Ag2CrO4 + 2e–  2Ag + CrO 24

+ 0.45

Ce4+ + e–  Ce3+

+ 1.61

–

–

AgF + e–  Ag + F–

+ 0.78

Cl2 + 2e  2Cl

+ 1.36

AgI + e–  Ag + I–

– 0.15

ClO– + H2O + 2e–  Cl– + 2OH–

+ 0.89

Al3+ + 3e– Al

– 1.66

ClO4  2H   2e   ClO3  H 2 O

+ 1.23

Au+ + e–  Au

+ 1.69

ClO4  H 2 O  2e   ClO3  2OH 

+ 0.36

Co2+ + 2e–  Co

– 0.28

3+

–

Au + 3e Au 2+

–

+ 1.40

Ba + 2e  Ba

+ 2.91

Be2+ + 2e–  Be

– 1.85

Bi3+ + 3e–  Bi

+ 0.20

Br2 + 2e–  2Br– BrO– + H2O + 2e–  Br– + 2OH–

3+

–

Co + e  Co

2+

+ 1.81

Cr2+ + 2e–  Cr

– 0.91

+ 1.09

Cr2 O 72  14H   6e   2Cr 3  7 H 2 O

+ 1.33

+ 0.76

Cr3+ + 3e–  Cr

– 0.74

Cr3+ + e–  Cr2+

– 0.41

SCAN CODE ELECTROCHEMISTRY

162

ELECTROCHEMISTRY

Cs+ e–  Cs +

–

Cu + e  Cu 2+

–

– 2.92

MnO4  2H 2 O  2e   MnO2  4OH 

+ 0.60

+ 0.52

Na+ + e–  Na

– 2.71

Cu + 2e  Cu

+ 0.34

Cu2+ + e–  Cu+

+ 0.16

NiOOH + H2O + e  Ni(OH)2 + OH

+ 0.49

F2 + 2e–  2F–

+ 2.87

Fe2+ + 2e–  Fe

NO3  2H   e   NO 2  H 2 O

– 0.80

– 0.44

Fe3+ + 3e–  Fe

– 0.04

NO3  4H   3e   NO  2H 2 O

+ 0.96

+ 0.77

NO3  H 2 O  2e   NO 2  2OH 

+ 0.10

Fe3+ + e–  Fe2+ 3–

–

4–

[Fe(CN)6] + e  [Fe(CN)6] 2H+ + 2e–  H2

0, by definition

–

–

2H2O + 2e  H2 + 2OH +

+ 0.36

–

– 0.83

2HBrO + 2H + 2e  Br2 + 2H2O

+ 1.60

2HClO + 2H+ + 2e–  Cl2 + 2H2O

+ 1.63

H2O2 + 2H+ + 2e–  2H2O H4XeO6 + 2H+ + 2e–  XeO3 + 3H2O

+ 1.78 + 3.0

Hg 22  2e   2Hg

+ 0.79

Hg2Cl2 + 2e–  2Hg + 2Cl–

+ 0.27

Hg2+ + 2e–  Hg

+ 0.86

2Hg2+ + 2e–  Hg 22

+ 0.92

Hg2SO4 + 2e–  2Hg + SO 24

+ 0.62

I2 + 2e–  2I–

+ 0.54

I 3



 2e  3I



In+ + e–  In 2+

–

+

In + e  In 3+

–

+

2+

–

Ni + 2e  Ni

– 0.23 –

–

–

–

O2 + 2H2O + 4e  4OH +

+ 0.40

–

O2 + 4H + 4e  2H2O

+ 1.23

O 2  e   O 2

– 0.56

O2 + H2O + 2e–  HO 2 + OH–

– 0.08

+

–

O3 + 2H + 2e  O2 + H2O –

+ 2.07 –

O3 + H2O + 2e  O2 + 2OH 2+

–

4+

–

Pb + 2e  Pb Pb + 2e  Pb

+ 1.24 – 0.13

2+

+ 1.67

PbSO4 + 2e–  Pb + SO 24

– 0.36

Pt2+ + 2e–  Pt

+ 1.20

4+

–

3+

Pu + e  Pu 2+

+ 0.97

–

Ra + 2e  Ra +

– 2.92

–

Rb + e  Rb –

– 2.93

2–

+ 0.53

S + 2e  S

– 0.48

– 0.14

S2 O82  2e   2SO 24

+ 2.05

– 0.40

SC3+ + 3e–  Sc

– 2.09

In + 2e  In

– 0.44

In3+ + 3e–  In

– 0.34

In3+ + e–  In2+

– 0.49

K+ + e–  K

– 2.93

La3+ + 3e–  La

– 2.52

Li+ e–  Li

– 3.05

Mg2+ + 2e–  Mg

– 2.36

Mn2+ + 2e–  Mn

– 1.18

Mn3+ + e–  Mn2+

+ 1.51

MnO2 + 4H+ + 2e–  Mn2+ + 2H2O

+ 1.23

2+

–

4+

–

Sn + 2e  Sn Sn + 2e  Sn 2+

–

2+

–

+ 1.51

MnO4  e   MnO24

+ 0.56

+ 0.15

Sr + 2e  Sr

– 2.89

Ti + 2e  Ti

– 1.63

3+

–

2+

– 0.37

4+

–

3+

0.00

Ti + e  Ti Ti + e  Ti +

–

Tl + e  Tl 3+

– 0.34

–

U + 3e  U 4+

–

– 1.79

3+

U +e U 2+

– 0.61

–

V + 2e  V 3+

MnO4  8H   5e   Mn 2  4H 2 O

– 0.14

2+

–

– 1.19

2+

V +e V

– 0.26

Zn2+ + 2e–  Zn

– 0.76

SCAN CODE ELECTROCHEMISTRY

157

ELECTROCHEMISTRY

SOLVED EXAMPLES SOLVED EXAMPLES

Example – 1

Sol.

A cell is prepared by dipping copper rod in 1 M copper sulphate solution and zinc rod in 1 M zinc sulphate solution. The standard reduction potentials of copper and zinc are 0.34 V and – 0.76 V respectively. (i) What will be the cell reaction? (ii) What will be the standard electromotive force (EMF) of the cell ? (iii) Which electrode will be positive ?

(a)

(i) The alkalinity of the solution prevents the availability of H+ ions. (ii) Zinc is more electropositive than iron. Therefore, zinc coating acts anode and the exposed iron portions act as cathode. If zinc coating is broken, zinc undergoes corrosion, protecting iron from rusting. No attack occurs on iron till all the zinc is corroded.

(b) At anode At cathode

Cu  Cu2+ + 2e– [Ag+ + e–  Ag] × 2

(iv) How will the cell be represented ? Sol.

Cu + 2 Ag+ Cu2+ + 2Ag

(i) The cell reaction can be

Zn  Cu 2    Zn 2  Cu Cu  Zn 2    Cu 2  Zn The EMF comes out to be positive for the 1st reaction. Hence, the cell reaction is or

Cell representation Cu | Cu2+ (conc.) || Ag+ (conc.) | Ag

Zn  Cu 2    Zn 2  Cu

E cell  E ocell 

 Cu 2  0.059 log  2 n  Ag  

o o o o o (ii) E cell  E cathode  E anode  E Cu 2 / Cu  E Zn / Zn 2

= 0.34 + 0.76 = 1.10 V (iii) reduction takes place on copper electrode. Hence it is positive (iv) Zn | Zn2+ (1 M) || Cu2+ (1 M) | Cu

O = (0.80 – 0.34) 

 0.01  15.59  log  2   x 

Example – 2 (a) Account for the following (i) Alkaline medium inhibits the rusting of iron (ii) Iron does not rust even if the zinc coating is broken in a galvanized iron pipe. (b) Cu2+ + 2e–  Cu ; Eo = + 0.34 V Ag+ + e–  Ag; Eo = + 0.80 V (i) Construct a galvanic cell using the above data. (ii) For what concentration of Ag+ ions will the emf of the cell be zero at 25oC, if the concentration of Cu2+ is 0.01 M? [log 3.919 = 0.593]

0.059  0.01 log  2  2  x 

x = 1.597 × 10 –9 M

[Ag+] = 1.597 × 10–9M

Example – 3 Calculate the standard free energy change and maximum work obtainable for the reaction occurring in the cell : (Daniell cell). Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) [Given E oZn 2 / Zn  0.76 V, E oCu 2 / Cu   0.34 V, F  96, 500 C mol 1 ]

Also calculate the equilibrium constant for the reaction.

158

ELECTROCHEMISTRY E ocell  E oCu 2 / Cu  E oZn / Zn 2  0.34  0.76

(i)

Sol.

(b) At anode :

= 1.10 V

 Mg2+ + 2e– Mg   Ag Ag+ + e– 

A cathode :

]×2

The reaction taking place in the Daniell cell is

 Mg2+ + 2Ag Mg + 2Ag+ 

 Cu(s) + Zn2+(aq) Zn(s) + Cu2+(aq) 

For this reaction, n = 2 Ecell = E

Go = – nFE ocell

o cell

 Mg 2   0.059  log 2 n  Ag  

= – 2 × 96500 C mol–1 × 1.10 V

 3.17 

= – 212300 CV mol–1 = – 212300 J mol–1 (1 CV = 1 J)

= 3.17 – 0.0295 log 107

= – 212.300 kJ mol–1 Thus, the maximum work that can be obtained from the Daniel cell = 212.3 kJ.

= 3.17 – 0.0295 × 7 = 3.17 – 0.21 Ecell = 2.96 V

(ii) Go = – RT In Kc = – 2.303 RT log Kc 

0.059 0.1 log 2 2 104 

(c) (i) Mg | Mg2+ (0.1 M) || Ag+ (0.0001M) | Ag

–212300 = – 2.303 × 8.14 × 298 × log Kc

(ii) Yes, as the cell potential is positive. or

log K c 

212300  37.2704 2.303  8.314  298

Example – 5

Kc = Antilog 37.2074 = 1.6 ×1037



A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the

Example – 4

o electrode potential  E Zn 2 / Zn  0.76 V  .

The following chemical reaction is occurring in an electrochemical cell.

Sol.

The electrode reaction written as reduction reaction is Zn2+ + 2e–  Zn (n = 2)

 Mg2+ (0.10M) + 2Ag(s) Mg(s) + 2 Ag+ (0.0001 M) 

Applying Nernst equation, we get

The Eo electrode values are Mg2+ / Mg = – 2.36 V

E Zn 2 / Zn  E oZn 2 / Zn 

Ag+ / Ag = 0.81 V For this cell calculate/write (a)

(i) Eo value for the electrode 2Ag+/2Ag. (ii) Standard cell potential E ocell .

(b)

Cell potential (E)cell

(c)

(i) Symbolic representation of the above cell. (ii) Will the above cell reaction be spontaneous ?

Sol.

(a)

(i) E oAg / Ag  0.81V (ii) E

o cell

E

o cathode

E

As 0.1 M ZnSO4 solution is 95% dissociated, this means that in the solution, 95  Zn 2     0.1M  0.095 M 100



E Zn 2 / Zn  0.76 

0.0591 1 log 2 0.095

= – 0.76 – 0.02955 (log 1000 – log 95) o anode

 E oAg / Ag  E oMg / Mg2  0.81  2.36 E ocell  3.17 V

0.0591 1 log 2  Zn 2  

= – 0.76 – 0.02955 (3 – 1.9777) = – 0.76 – 0.03021 = – 0.79021 V

159

ELECTROCHEMISTRY Example – 6

(b) Calculate the EMF of the cell of Zn / Zn2+ (0.1 M) || Cd2+ (0.01 M) / Cd at 298 K,

Calculate the potential (emf) of the cell [ Given E oZn 2 / Zn  0.76 V and E oCd2 / Cd  – 0.40 V]

Cd | Cd2+ (0.10 M) || H+ (0.20 M) | Pt, H2 (0.5 atm) (Given Eo for Cd2+ / Cd = – 0.403 V, R = 8.14 JK–1 mol–1, F = 96,500 C mol–1). Sol.

The cell reaction is

Sol.

(a) Because of higher reduction potential of water, water is reduced in preference to sodium at therefore instea of deposition of sodium metal, hydrogen is discharged at cathode. H 2 O   2e    H 2 g   2OH 

Cd + 2H+ (0.20 M)  Cd2+ (0.10 M) + H2 (0.5 atm)

2Cl    Cl2 g 

E ocell  E oH  /1/ 2H  E oCd 2 / Cd  0 –(– 0.403) = 0.403 V 2

H 2 O    2Cl   H 2 g   Cl 2 g   2OH 

Applying Nernst equation to the cell reaction,

E cell

At anode Cl2 gas is liberated because of overpotential of oxygen.

Cd 2    PH2 2.303 RT o  E cell  log  2 nF  H  

(b)

2.303  8.314  298 0.1 0.5  0.403  log 2 2  96,500  0.2 

Zn   Zn 2   2e    Half cell reactions Cd 2   2e   Cd 

Zn + Cd2+  Zn2+ + Cd} cell reaction

= 0.403 – 0.003 = 0.400 V E ocell = E ocathode + E oanode = 0.76 – 0.40 = 0.36 V

Example – 7

0.0591 log Q n  Zn 2  0.0591  0.36  log  2  2  Cd  E = E ocell 

Represent the cell in which the following reaction takes place Mg(s) + 2Ag+ (0.0001 M)  Mg2+ (0.130 M) + 2 Ag 9(s) o Calculate its E(cell) if E  cell  = 3.17 V..

Sol.

 0.36 

The cell can be written as Mg | Mg2+ (0.130 M) || Ag+ (0.0001 M) | Ag

E  cell   E ocell 

 Mg 2    Mg 2   RT 2.303RT o In   E  log 2  cell nF 2F  Ag    Ag  

0.0591  0.1  log   = 0.33V 2  0.01 

Example – 9 (a) Depict the galvanic cell in which the following reaction takes place : Zn(s) + 2Ag+ (aq)   Zn2+(aq) + 2Ag(s)

0.059 V 0.130  3.17 V  log 2 2  0.0001

Also indicate that in this cell

= 3.17 V – 0.21 V = 2.96 V

(i) which electrode is negatively charged.

Example – 8

(ii) what are the carrier of the current in the cell.

(a) Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode and Cl2 at anode. Write overall reaction. o

( E Na



/ Na

o  2.71 V; E Ho 2 O / H 2   0.83 V, E Cl

1.36 V; E oH   O

2

/ H2O

2

/ 2Cl

(iii) what is the individual reaction at each electrode. (b) Write the Nernst equation and determine the e.m.f. of the following cell at 298 K:



 1.23 V)

Mg (s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu (s) (Given : E o Mg

2

/ Mg

 2.375 V, E o Cu 2 / Cu  0.34 V )

ELECTROCHEMISTRY Sol.

160

(a) Zn | Zn2+ (conc.) || Ag+ (conc) | Ag

Example – 11

(i) Zn electrode is negatively charged. Calculate the equilibrium constant, Kc for the reaction.

(ii) Current carriers of cell are

3Sn4+ + 2Cr 3Sn2+ + 2Cr3+

- electrons in external wire

Given Eo = 0.885 V.

- Zn2+ ions in anodic half cell. - Ag+ ions in cathodic half cell. +

–

- Ions of salt bridge, i.e., K and Cl .

Sol.

(iii) At anode Zn   Zn 2   2e

E ocell 

0.059 log K c , n  6 n

0.885 

At cathode 2Ag   e    2Ag (b) Mg   Mg 2   2e 

0.059 log K c 6

log K c 

Cu 2   2e   Cu

6  0.885 0.059

Kc = Antilog 90 = 1 × 1090

Mg  Cu 2    Cu  Mg 2 

Example – 12 Nernst equation (a) Calculate the equilibrium constant for the reaction E cell

 Mg 2   0.059  E o cell  log  2   n  Cu 





E cell  E o Cu 2 / Cu  E o Mg / Mg2 

 0.34   2.375  

Cd2+(aq) + Zn(s)   Zn2+(aq) + Cd(s) If E oCd 2 / Cd  0.403V

 Mg   0.059 log  2   2 Cu 

E oZn 2 / Zn  0.763V (b) When a current of 0.75A is passed through a CuSO4 solution for 25 min, 0.369 g of copper is deposited at the cathode. Calculate the atomic mass of copper.

0.059 103 log 4 2 10

(c) Tarnished silver contains Ag2S. Can this tarnish be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolytic solution such as NaCl. The standard electrode potential for half reaction:

= 0.34 + 2.375 – 0.0295 log 10

Ecell = 2.6855 V Ecell = 2.685 V

Ag2S(s) + 2e–   2Ag(s) + S2– is –0.71V

Example – 10

and for

Calculate the equilibrium constant of the reaction Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) E ocell  0.46 V

E

Sol. or

o cell

0.059 V  log K c  0.46 V 2

log K c 

0.46 V  2  15.6  Kc = Antilog 15.6 0.059 V

Kc = 3.92 × 1015

Sol.

Al3+ + 3e–   2Al(s) is – 1.66 V

(a) E ocell  E oc  E oa  0.403  0.763  0.360 V As

 nE o cell   2  0.360  log Kc =     0.059   0.059 

 0.720     12.20  0.059  Kc = antilog (12.20) = 1.585 × 1012

161

ELECTROCHEMISTRY (b) M = Z I t

0.369 

Example – 14

x  0.75  25  60 2  96500

Can nickel spatula be used to stir a copper sulphate solution ? Support your answer with a reason

(x = molar mass of copper) x = 63.3 g/mol

E oNi2 / Ni  0.25 V, E oCu 2 / Cu  0.34 V. Sol.

(c) E ocell for reaction of tarnished silver ware with

o o E ocell  E cathode  E anode

E ocell  E oCu 2 / Cu  E oNi2 / Ni  0.34 V   0.25   0.59 V

aluminium pan is (– 0.71 V) + 1.66 V i.e., + 0.95 V As E ocell is +ve, G = – ve, because G = – nEo F, i.e,

Tarnished silver ware, therefore, can be cleaned by

reaction will take place. Therefore, we cannot stir a copper sulphate solution with nickel spatula.

placing it in an aluminium pan as E ocell is positive.

Example – 13

Example – 15 Given,

Two half cell reactions of an electrochemical cell are given below :

The value of standard electrode potentail for the charge,  Fe2+(aq) will be Fe3+ (aq) + e–  (a) –0.072 V (b) 0.385 V (c) 0.770 V (d) –0.270 V Ans. (c)

MnO 4  aq   8H   aq   5e   Mn 2   aq   4 H 2 O    , E

o

 1.51 V

Sn2+ (aq)  Sn4+ (aq) + 2e–, Eo = + 0.15 V Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the quation Sol.

 Mn 2  At cathode : MnO 4  aq   8H   5e 

 aq   4H2 O   ]  2 2+

At anode: Sn   Sn

4

 aq   2e

EoFe3 / Fe  0.336 V, EoFe2 / Fe  0.439 V



E o  1.15V o

]  5 E = + 0.15 V

Sol.

Fe3+ +3e-  Fe ; E10 = -0.036 V 0 Fe2+ +2e-  Fe ; E 2 = -0.439 V Required equation is

Fe3+ + e-  Fe2+ ; E30 = ? Applying G = nFE0

G 30 =G10 - G 02

 -n FE  =  -n FE  -  -n FE  3

Overall reaction :

2MnO4  aq   5Sn 2  aq   16H   aq    2Mn 2 (aq) + 5Sn (aq) + 8H2O    4

o o E ocell  E ocathode  E anode  E oMnO / Mn 2  E Sn 4 / Sn 2 4

= 1.51 – (– 0.15) E

1

0 1

2

0 2

E 30 = 3E10 - 2E 20 = 3 ×  -0.036  -2×  -0.439 

E30 = -0.108+ 0.878 = 0.77 V

Example – 16

o o ESn  ESn  0.15V 4 2 / Sn 2 / Sn 4

o cell

0 3

 1.66 V

As E ocell is +ve therefore the reaction will take place in forward direction, i.e., favours the formation of products.

The Gibbs energy for the decomposition of Al2O3 at 500ºC is as follows 2 4 –1 Al2O3   Al + O2,  r G = + 966 kJ mol 3 3

The potential difference needed for electrolytic reduction of Al2O3 at 500ºC is at least (a) 4.5 V

(b) 3.0 V

(c) 2.5 V

(d) 5.0 V

162

ELECTROCHEMISTRY  0  v   0  v   0

Ans. (c) Sol.

The ionic reaction are :

where,  0 and  0 are the limiting molar conductivities

2 3+ 4 Al2 + 4e-  Al 3 3

of the cation and anion respectively and v+ and v– are the number of cations and anions formed from a formula unit of the electrolyte. For example, one formula unit of Al2(SO4)3 gives two Al3+ ions and three sulphate ions. Therefore,

2 2O3  O 2 + 4e3

Thus, no.of electrons transferred = 4 = n

o  om  Al2 (SO4  3  2 oAl3  3 SO 2 4

G = -nFE = - 4×96500 ×E

Application : It can be used to determine molar conductivity of weak electrolytes at infinite dilution :

or 966×103 = - 4×96500× E

Consider acetic acid as the example of a weak electrolyte.

3

966×10  E== -2.5 V 4×96500

o  om  CH3 COONa    CH   oNa  COO 3



Example – 17 and Fe2+/Fe are – 0.76, –0.23 and – 0.44 V respectively.

(c) X = Fe, Y = Zn

(d) X = Zn, Y = Ni



o Cl

From (i) + (ii) – (iii) we get

+2 The reaction X + Y+2   X + Y will be spontaneous when

(b) X = Ni, Y = Zn



o H

o  om  NaCl   oNa    Cl 

The standard reduction potentials for Zn2+/Zn, Ni2+/Ni

(a) X = Ni, Y = Fe

o m  HCl 

o o  oCH COO   oNa    oH   Cl   oCl    Na  3

  oCH COO   oH   oCH3 COOH 3

Example – 19

Ans. (d) Sol.

The element with high negative value of standard reduction potential are good reducing agents and can be easily oxidised .

The conductivity of 0.0011028 mol L–1 acetic acid is

Thus X should have high negative value of standard potentinal than Y so that it will be oxidised to X 2+ by reducing Y 2+ to Y

for acetic acid is 390.5 S cm2 mol–1.

4.95 × 10–5 S cm–1. Calculate its dissociation constant if  om

Sol.

m 

X = Zn , Y =Ni Alternatively, for a spontaneous reaction E 0 must be positive

E 0 =E 0 reduced specise -E 0 oxidised species

= -0.23-  -0.76   E 0 = + 0.53V

Example – 18

Sol.

Explain Kohlrausch’s law of independent migration of ions. Mention one application of Kohlrausch’s law. Kohlrausch’s law of independent migration of ions: The molar conductivity of an electrolyte at infinite dilution is the sum of the individual contributions of the anion and cation of the electrolyte.

 4.95  105 Scm 1 1000 cm3   c 0.001028 mol L1 L

= 44.88 S cm2 mol–1



 m 44.88 Scm 2 mol 1   0.115  om 390.5 Scm 2 mol1

0.001028 mol L1   0.115  c 2 K  0.115 1   

= 1.65 × 10–5 mol L–1

2

ELECTROCHEMISTRY

163

Example-20 1 = Cell constant A

Give the relationship between equivalent and molar conductance ?

R = Resistance Sol.

1000 1000 m    and  eq    Molarity Normality m   m Normality   eq Molarity



where,  m = Molar conductivity

Example – 21

 = Conductivity

The electrical resistance of a column of 0.05 mol L–1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. Sol.

M = Molarity of Solution (b) At anode : At cathode :

A = r2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2, l = 50 cm = 0.5 m

R

  1000 M

RA 5.55  103   0.785    or    50 cm A

Al  s    Al3  aq   3e]  2  Ni  s ]  3 Ni2+ (aq) + 2e 

 2Al3+ (aq) + Ni(s) 2Al (s) + 3Ni2+ (aq) 

= 87.135 cm 1 

1  1 1  S cm  0.01148 S cm  87.135 

 Conductivity =    

Molar conductivity,  m 

  1000 c

0.01148 S cm 1  1000 cm3 L1  = 229.6 S cm2 mol–1 0.05 mol L1

Example – 22 (a) State the relationship amongst cell constant of cell, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solute related to conductivity of its solution ?

2

E cell  E

 Al3  0.0591  log  3 n  Ni 2  

Here, n = 6, [Al3+] = 0.001 M = 1 × 10–3 M, [Ni2+] = 0.5M E ocell  1.41V 3 2

E

o cell

10   1.41  0.0591 log 106 0.0591  1.41V  log 3 6 6 0.125  0.5

 1.41 

(b) A voltaic cell is set up at 25oC with the following halfcells: Al | Ag3+ (0.001 M) and Ni | Ni2+ (0.50 M)

o cell

 1.41 

0.0591 0.0591 log 106  8   1.41   log106  log 23  6 6

0.0591 0.0591  6 log10  3log 2   1.41   6  3  0.3010  6 6

Calculate the cell voltage

 E o 2  0.25 V, E o 3  1.66 V  Al |Al  Ni |Ni  Sol.

(a)  

1 l    R A

where,   Conductivity

 1.41 

0.0591 0.3012  5.097   1.41  6 6

= 1.41 + 0.0502 = 1.4602V Ecell = 1.46 V

164

ELECTROCHEMISTRY Example – 23

Sol.

(a) Express the relationship amongst cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solute related to conductivity of its solution.

(a) Molar conductivity   m  : It may be defined as the conductivity of one molar electrolytic solution placed between two electrodes one centimetre apart and have enough area of cross section to hold entire volume.

(b) Calculate the equilibrium constant for the reaction.  Fe 2   aq   Cd  s  Fe s   Cd 2   aq   o (Given :E o Cd2 |Cd  0.40 V, E Fe  0.44 V). 2 |Fe

Sol.

(a) Conductivity    

1  Cell constant (G) Resistance  R 

  1000 m  , where,  m = Molar conductivity M  Fe2+ + Cd (b) Fe(s) + Cd2+(aq)  (aq) (s)

log k c  n

E ocell 0.059

Here, n = 2 o o E ocell  E cathode  E anode

 c

m 

 = Conductivity

where,

c = Concentration of solution in mol L–1  Zn 2   aq   2e  (b) At anode : Zn  s  

At cathode : Ag+(aq) + e–   Ag(s)] × 2 2+ Zn(s) + 2Ag+ (aq)   Zn + 2Ag(s)

E cell  E o cell 

Here, n = 2, [Zn2+] = 1 M o o Eocell = E Ag / Ag  E Zn / Zn 2  0.80V  0.76V

Eocell = 1.56 V

 E o Cd2 / Cd  E o Fe / Fe2 = – 0.4 + 0.44 E

o cell

1.48  1.56   0.04 V

log k c 

2  0.04 0.08  0.059 0.059

0.08  

log k c  1.3536 kc = Antilog 1.3536 kc = 22.57

Example – 24

0.0591 [Zn 2  ] log 2 n  Ag  

log

0.0591 1 log 2 2  Ag  

1 

0.0591 1 log 2 2  Ag  

 Ag 

2



0.16  2.7072  2.7072 0.0591 2

(a) Define the term molar conductivity. How is it related to conductivity of the related solution? (b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used. (E oZn 2 / Zn  0.76 V, E oAg2 / Ag  0.80 V)

log1  log  Ag    2.7072 0  2 log  Ag    2.7072 log  Ag    1.3536  2.6464  Ag    Anti log  2.6464   4.43  10 2 M  Ag    0.044 M

165

ELECTROCHEMISTRY Example – 25

Sol.

The measured resistance of a conductance cell containing 7.5 × 10–3 M solution of KCl at 25oC was 1005 ohms. Calculate (a) specific conductance (b) molar conductance of the solution. Cell constant = 1.25 cm–1. Sol.

Specific conductance    



1  cell constant R

1  1.25 cm 1  0.001244  1 cm 1 1005 

Molar conductance   m  



  1000 Molarity

0.001244  1 cm 1  1000 cm 3 L1 7.5  103 mol L1

(a) Eocell = (– 2.87 V) – (1.50 V) = – 4.37 V Gocell = – 6 × 96500 × – 4.37 V = + 2530.230 kJ/mol Since r Go is positive, reaction is non-spontaneous. Au3+/Au half cell will be a reducing agent, Ca2+/Ca half cell will be an oxidising agent.

(b)  cm  K 

1000 molarity

K = Specific conductance



4  105 S / cm  1000  40S cm2 mol1 0.001

 165.87  1 cm 2 mol1 .



Example – 26  m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate  0m for HAc. Sol.

 0m  HAc    oH   oAc   oH    oCl   oAc   oNa    oCl   oNa 

  om HCl    om NaAc    om  NaCl  = (425.9 + 91.0 – 126.4) S cm2 mol–1 = 390.5 S cm2 mol–1.

Example – 27 (

a) Calculate the standard free energy change for the following reaction at 25oC. Au(s) + Ca2+ (aq, 1M)   Au3+ (aq, 1M) + Ca(s) o E oAu3 / Au  1.50 V, E Ca  2.87 V 2 / Ca

Predict whether the reaction will be spontaneous or not at 25oC. Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent? (b)

The conductivity of 0.001 M acetic acid is 4 × 10–5 S/cm. Calculate the dissociation constant of acetic acid, if  om for acetic acid is 390. 5 S cm2/mol.

m 40   0.103 o  m 390.5 2

Kc 

C 2 0.001  0.103   1.19  105 1  1  0.103

Example – 28 (a) Define molar conductivity of a substance and describe how weak and strong electrolytes’ molar conductivity changes with concentration of solute. How is such change explained ? (b) A voltaic cell is set up at 25 oC with the following half cells: Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu What would be the voltage of this cell ? (Eocell = 0.46 V) Sol. Molar Conductivity   m  : It may be defined as the conductance of a solution containing 1 mole of electrolyte such that the entire solution is placed in between two electrodes one centimetre apart.

166

ELECTROCHEMISTRY Sol.

m  k  v or

m 

k  1000 M

Molar conductivity increases with decrease in concentration or increase in dilution as number of ions as well as mobility of ions increased with dilution.

Example – 30 (a) Current of 1.50 A was passed through an electrolytic cell

For strong electrolytes, the number of ions do not increase appreciably on dilution and only mobility or ions increases due to decrease in interionic attractions.

containing AgNO3 solution with inert electrodes. The weight of Ag deposited was 1.50g. How long did the

Therefore,  m increases a little as shown in graph by a straight line.

current flow ? (b) Write the reactions taking place at the anode and

For weak electrolytes, the number of ions as well as mobility of ions increases on dilution which results in a very large increase in molar conducvity especially near infinite dilutuion as shown by curve in the figure. At anode :

Cu(s)   Cu

2+ (aq)

+ 2e

cathode in the above cell if inert electrodes are used. (c) Give reactions taking place at the two electrodes if these are made up of Ag. Sol.

At cathode : 2Ag+(aq) + 2e   2Ag(s)

Here,

(a) According to Faraday’s first law, charge required to deposit 1.50 g.

Cu s  2Ag   aq    Cu 2   aq   2Ag s 

E cell  E ocell 

Higher the acidity,higher will be the tendency to release protons and hence lighter will be the electrical conductivity.Difluoroacetic acid will be strongest acid due to electron withdrawing effect of two fluorine atoms so as it will show maximum electrical conductivity.

Ag 

Cu 2   0.0591 log  2 n  Ag  

96500  1.50  1331.70 coulombs 108

Time taken 

Here, E ocell  0.46 V, n  2

(b) Inert electrodes

[Ag+] = 0.001M = 1 × 10–3 M, [Cu2+] = 0.1 M Anode:

E cell  0.46 

E cell  0.46 

0.0591 0.1 log 2 2 103 

(c) Ag electrodes Anode :

Ecell = 0.46 – 0.0591 × 2.5 × 1 = 0.46 – 0.14775 = 0.31225V

Cathode :

0.1 M difluoroacetic acid, 0.1 M fluoroacetic acid

+ – Ag(s)   Ag (aq) + e

Ag+ (aq) + e–   Ag(s)

Example – 31

Example – 29 The highest electrical conductivity of the following aqueous solutions is of

2H 2 O     O 2 g   4H   aq   4e 

 Ag  s  Cathode : Ag+ (aq) + e– 

0.0591 0.0591 log 105  0.46  × 5 log 10 2 2

Ecell = 0.312 V

1331.70  893.5s 1.50

Sol.

State two advantages of H2—O2 fuel cell over ordinary cell. The two advantages of H2—O2 fuel cell over ordinary cell are : (i) They do not cause any pollution. (ii) They have high efficiency of 60-70%.

0.1 M chloroacetic acid, 0.1 M acetic acid

167

ELECTROCHEMISTRY Example – 32

Ans. (b)

(a) State advantages of H2-O2 fuel cell over ordinary cell. (b) Silver is electrodeposited on a metallic vessel of total surface area 500 cm2 by passing a current of 0.5 amp for two hours. Calculate the thickness of silver deposited.

Sol.

cathode =

Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is–

(a) Advantages Fuels Cells:

Al3+ + 3e–

1. It is a pollution-free device since no harmful products are formed.

(a) 5.49 × 101 C of electricity (b) 5.49 × 104 C of electricity

3. These cells are light in weight as compared to electrical generators to produce corresponding quantity of power.

m = z I t. 108  0.5  2  3600 96500 m = 4.029 g 

d

m m v v d

4.029  0.3837 cm 3 10.5 Let the thickness of silver deposited be x cm. V

 V =A× x  x

(c) 1.83 × 107 C of electricity (d) 5.49 × 107 C of electricity Ans. (d) Sol.

(b) Mass of silver deposited

V A

0.3837 x 500  x = 7.67 × 10–4 cm.

Example – 33

Alº

To prepare 5.12 kg of aluminium metal by this method we require

2. This is very efficient cell. Its efficiency is about 75% which is considerably higher than conventional cells.

4. It is a continuous source of energy if the supply of gases is maintained.

108×9650 =10.8 g 96500

Example – 34

[Given: Density of silver = 10.5 g cm–3, Atomic mass of silver = 108 amu, F = 9,500 C mol–1] Sol.

The mass of silver deposited on the

From Faraday’s 1st law,,

W = Z× Q (W=weight,Z=electrochemical equivalent, Q = quantity of electricity) Now E = Z × F (E = Equivalent weight, F = faraday) or W =

E ×Q F

W×F A n (A= Atomic weight, n = valency of ion)

or Q =

Q= W×F or E

or Q =

n × w×F 3× 5.12×103 ×96500 = = 5.49×107 C A 27

Example – 35 A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X . This result in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt

When during electrolysis of a solution of AgNO3, 9650 C of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be

is : (F = 96, 500 C)

(a) 1.08 g

(b) 10.8 g

(a) + 1

(b) + 2

(c) 21.6 g

(d) 108g

(c) + 3

(d) + 4

168

ELECTROCHEMISTRY Ans. (c) Sol.

Ans. (c) Sol.

t = 2×60×60 =7200

Larger the standard reduction potential that deposit first sequence is as C > B > D > A

E w= ×I×t 96500 n×0.25=

Example – 38

10×7200 96500

(a) Give reasons for the following: (i) Rusting of iron is quicker in saline water than inordinary water.

n=3

Example – 36 How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper = 63.5 u, NA = Avogadro’s constant):

(ii) Resistance of a conductivity cell filled with 0.1 M KCl solution is 100 ohm.If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. (Conductivity of 0.1 M KCl solution is 1.29 Sm–1). Sol. (a) (i) It is because in saline water, there is more H+ ions. Greater the number of H+ ions, quicker the rusting.

(a)

NA 2

(b)

NA 5

(c)

NA 10

(d)

NA 20

(ii) Due to higher reduction potential of hydrogen we get hydrogen at cathode. 1 (b)    cell constant R  cell constant =  R = 1.29 Sm–1 × 100 ohm

Ans. (b) Sol.

= 129 m–1 = 1.29 cm–1 2



Cu  2e   Cu  s  63.5 g

For second solution 1    cell constant R 1   1.29  2.48  103 S cm 1 520

2mol

63.5 g copper require = 2 mol electron = 2 NA

electron

2N A 1g copper ion require  63.5

m 

2N A N  6.35  A 6.35 g copper ion require  63.5 5



Example – 37

1000 M

2.48 103 1000 248  0.02 2

 m  124 S cm 2 mol 1

Example – 39



o The standard electrode potentials E M  /M

 four metals

A, B, C and D are – 1.2 V, 0.6 V, 0.85 V and –0.76 V, respectively. The sequence of deposition of metals on applying potential is: (a) A, C, B, D

(b) B, D, C, A

(c) C, B, D, A

(d) D, A, B, C

Three iron sheets have been coated separately with three metals A, B and C whose standards reduction potentials are given below. metal

A

B

C

iron

E oValue – 0.46 V –0.66V –0.20V –0.44V

Identify in which case rusting will take place faster when coating is damaged.

169

ELECTROCHEMISTRY Sol.

As iron (–0.44V) has lower standard reduction potential than C (– 0.20V) only therefore when coating is broken, rusting will take place faster.

E o cell  E o cathode — E o anode

 E o Cd2 / Cd  E o Fe2 / Fe  40   0.44 

Example – 40

Eocell = 0.04V

(a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere.

2  0.04 0.08  0.059 0.059

(b) Calculate the equilibrium constant for the equilibrium reaction.

log kc = 1.3536

 Fe 2   aq   Cd  s Fes   Cd 2   aq  

kc = Antilog 1.3536

(Given : E oCd2 / Cd  0.40 V, E oFe2 / Fe  0.44V) Sol.

log k c 

kc = 22.57

Example – 41

(a) At anode : Oxidation of Fe atoms takes place

 Fe2+ + 2e– Fe 

Which type of a metal can be used in cathodic protection of iron against rusting ?

E o Fe2 / Fe  0.44 V

At cathode : Reduction of oxygenin the presence of H+ ions. The H+ ions are produced by either H2O or H2CO3 (formed by dissolution of CO2 in moisture)

Sol.

2H   aq   2e     2H

2H 

1 O 2   H 2O 2 g 

Example – 42 Chromium metal can be plated out from an acidic solution containing CrO3 according to the following equation:

Net reaction at cathodic area

2H   aq   E o H / O

2

1 O 2  2e    H 2O 2

/ H2 O

CrO3(aq) + 6H+(aq) + 6e– Cr(s) + 3H2O Calculate (i) how many grams of chromium will be plated out by 24,000 coulombs and (ii) how long will it take to plate out 1.5 g of chromium by using 12.5 amp current ? (At. mass of Cr = 52).

 1.23 V

The overall reaction 1 Fe s   2H   aq   O 2 g    Fe 2   aq   H 2 O   2

A metal which is more electropositive than iron such as Al, Zn, Mg can be used in cathode protection of iron against rusting.

Sol.

(i) 6 × 96, 500 coulomb deposit Cr = 1 mole = 52 g

E o cell  1.67 V

24,000 coulomb deposit Cr 

52  24000 g  2.1554 g 6  965000

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3 . xH2O).

(ii) 52 g of Cr is deposited by electricity = 6 × 96500 C

 Fe2+ (aq) + Cd(s) (b) Fe(s) + Cd2+(aq) 

1.5 g require electricity 

log kc = n

E o cell 0.059

Here, n = 2

6  96500  1.5 C  16071C 52

Time for which the current is required to be passed



16071.9  1336 s. 12.5 A

170

ELECTROCHEMISTRY Example – 43

Example – 46

What is galvanisation ? Sol.

The process of coating zinc over iron is called galvanisation.

Example – 44 Write the chemical equations for all the steps involved in the rusting of iron, Give any one method to prevent rusting of iron. Sol.

 Fe 2   aq   2e  , E oFe Anode: Fe  s   

2

 0.44 V

/ Fe

   2H 2 O, Cathode : O 2 g   4H  aq   4e 

E oH  / O

2

/ H2 O

 1.23V

During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’. These are (a) Fe and Ni (b) Ag and Au (c) Pb and Zn (d) Sn and Ag Ans. (b) Sol. In electrolytic refining, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. Copper is refned by using electrolytic method. Impurities blister copper deposit as anode mud which contains antimany selenium, tellurium, silver, gold and platinum.

Example – 47

Overall reaction Fe(s)   Fe2+(aq) + 2e– ] × 2 O2(g) + 4H+ + 4e–   2H2O

Sol.

2Fe + O2 + 4H+(aq)   2Fe2+ + 2 H2O E ocell  1.67V

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is : Ca, Cu, Cr, Ag Metal cannot obtained by electrolysis of aq. solution of its salt is Ca . This is because standard reduction potential of water is higher than reduction of Ca 2  . So water will reduce to form hydrogen gas & OH - .

Example – 48

Further 4Fe2+(aq) + O2(g) + 4H2O(l)   2Fe2O3 + 8H+(aq) Fe2O3 + xH2O   Fe2O3 . xH2O Hydrated ferric oxide (Rust)

Galvanisation is used to prevent rusting of iron.

Example – 45

(a) What type of a cell is the lead storage battery ? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery while operating. (b) A voltaic cell is set up at 25oC with the half-cells Al | Al3+ (0.001 M) and Ni | Ni2+ (0.50 M). Write the equation for the reaction that occurs when the cell genrates an electric current and determine the cell potential. (Given :E oNi | Ni  0.25 V, E oAl | Al  1.66V). 2

Consider the following reaction at 1100ºC o –1 (I) 2C + O2   2CO, G = – 460 kJ mol

Sol.

o –1 (II) 2Zn + O2   2ZnO, G = – 360 kJ mol Based on these, determine whether zinc oxide can be reduced by carbon or not Yes

1 ZnO  s   Zn  s  + O 2 ,ΔG=360 kJ mol 2

Sol. (a) The lead storage battery is a secondary cell. The cell reactions when the battery is in use are given below  PbSO 4 s  2e  At anode: Pb s  SO 24  aq   At cathode: PbO 2s   SO 24  aq   4H   aq   2e 

Because net ΔG is-ve.

PbSO4s  2H 2 O  

Overall cell reaction: Pb(s) + 2H2SO4(aq)  PbSO4(s) + 2H2O(  ) (b) 2Al(s) + 3Ni+2 (aq)  3Ni(s) + 2Al+3 (aq) E ocell = 1.41 V

1 C  s  + O2  g   CO  g  ,ΔG=-460kJ mol 2

ZnO  s  +C  s   Zn  s  +CO  g  ,ΔG=-100 kJ mol

3

E cell  1.41 

 (103 )2  0.0591 log  3  = 1.46 V 6  (0.5) 

177

ELECTROCHEMISTRY

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS 6.

Basics of electrochemical Cell 1.

The reduction potential of hydrogen half-cell will be negative if

The reaction

(a) p(H2) = 1 atm and [H+] = 2.0 M

1 H 2  g   AgCl  s   H   aq   Cl   aq   Ag  s  2 occurs in the galvanic cell :

(b) p(H2) = 1 atm and [H+] = 1.0 M

(a) Ag | AgCl(s) | KCl(sol.) || AgNO3(sol.) | Ag

(d) p(H2) = 2 atm and [H+] = 2.0 M

(c) p(H2) = 2 atm and [H+] = 1.0 M

(b) Pt | H2(g) | HCl(sol). || AgNO3 (sol) | Ag

7.

Which of the following reaction is possible at anode ?

(c) Pt | H2(g) | HCl (sol.) || AgCl(s) | Ag 2.

3.

(d) Pt | H2 (g) | KCl (sol.) || AgCl(s) | Ag

(a) F2 + 2e–  2F–

The standard reduction electrode potential values of the element A, B and C are + 0.68, –2.50, and –0.50 V respectively. The order of their reducing power is :

1   (b) 2H  O 2  2e  H 2 O 2

(a) A > B > C

(b) A > C > B

(c) 2Cr 3  7H 2 O   Cr2 O72   14H   6e 

(c) C > B > A

(d) B > C > A

(d) None of these

A metal having negative reduction potential when dipped in the solution of its own ions, has a tendency :

8.

Zn(s) + Cu2+ (0.1 M)  Zn2+ (1 M) + Cu (s)

(a) to pass into the solution

taking place in a cell, E ocell is 1.10V. E ocell for the cell will

(b) to be deposited from the solution

RT    0.0591 be  2.303 F  

(c) to become electrically positive (d) to remain neutral 4.

The EM3 / M2 values for Cr, Mn, Fe and Co are – 0.41, + 1.57, + 0.77 and + 1.97 V respectively. For which one of these metals the change in oxidation state from +2 to +3 is easiest?

5.

(a) Co

(b) Mn

(c) Fe

(d) Cr 3+

9.



/ Ag

2+

= 0.80 V, ECu 2 /Cu = 0.34 V and EAu3 / Au = 1.50 V,

ELi / Li = – 3.03 V

10.

with increasing voltage, the sequence of deposition of metals on the cathode will be : (a) Li, Cu, Ag, Au

(b) Cu, Ag, Au

(c) Au, Ag, Cu

(d) Au, Ag, Cu, Li

(a) 2.14 V

(b) 1.80 V

(c) 1.07 V

(d) 0.82 V

The cell, Zn  Zn2+ (1M) Cu2+ (1M)  Cu (Eºcell = 1.10v) was allowed to be completely discharged at 298 K. The

 [ Zn 2 ]   relative concentration of Zn2+ to Cu2+, i.e.,  2  is  [Cu ] 

An aqueous solution containing 1 M each of Au , Cu , + + Ag , Li is being electrolysed by using inert electrodes. The value of standard potentials are :

EAg

For the redox reaction

(a) 9.65 × 104

(b) antilog (24.08)

(c) 37.3

(d) 1037.3

0 o Given E Cr 3 / Cr  0.72 V, E Fe 2  / Fe  0.42 V.

The potential for the cell Cr Cr3+ (0.1M)  Fe2+ (0.01M)  Fe is (a) – 0.26 V

(b) 0.26 V

(c) – 0.339 V

(d) 0.336 V

ELECTROCHEMISTRY 11.

178

Which of the following statement is wrong about galvanic cell ?

17. The emf of the cell +

2+

Ti | Ti (0.0001 M) | | Cu (0.01 M)/Cu is 0.83 V

(a) cathode is positively charged

The emf of this cell will be increased by :

(b) anode is negatively charged

++

12.

(c) reduction takes place at the anode

(a) Increase the concentration of Cu ions

(d) reduction takes place at the cathode

(b) Decreasing the concentration of Ti

For a cell given below :

(c) Increasing the concentration of both

Ag | Ag+ || Cu2+ | Cu

(d) (a) and (b) both





13.

+

18. The EMF of the cell

Ag+ + e–   Ag;

Eº = x

Mg | Mg (0.01 M)| |Sn (0.1M)| Sn at 298 K is (Given

Cu2+ + 2e–   Cu;

Eº = y

EMg

2+

2

,Mg

2+

 2.34 V, ESn

2

, Sn

 0.14 V )

Eº cell is

(a) 2.17 V

(b) 2.23 V

(a) x + 2y

(b) 2x + y

(c) 2.51 V

(d) 2.45 V

(c) y – x

(d) y – 2x

19.

In a cell that utilises the reaction

Consider the following Eo values :

Zn (s) + 2H+ (aq)  Zn2+ (aq) + H2(g)

E oFe3 / Fe2   0.77V

addition of H2SO4 to cathode compartment will (a) lower the E and shift equilibrium to the left

o ESn  0.14 V 2 / Sn

Under standard conditions the potential for the reaction

(b) lower the E and shift the equilibrium to the right

Sn(s) + 2Fe3+ (aq)  2Fe2+ (aq) + Sn2+ (aq) is

(c) increase the E and shift the equilibrium to the right

(a) 1.68 V

(b) 1.40 V

(d) increase the E and shift the equilibrium to the left

(c) 0.91 V

(d) 0.63 V

14. Which of the following represents the potential of silver wire dipped in to 0.1 M AgNO3 solution at 25°C ? (a) E°red

(b) (E°red + 0.059)

(c) (E°ox – 0.059)

(d) (E°red – 0.059)

15. The reduction electrode potential E, of 0.1 M solution of M ions (E°RP = – 2.36 V) is : (a) – 2.41

(b) + 2.41

(c) – 4.82

(d) None

16. Consider the cell

+

H 2 (Pt) H 3 O  (aq) Ag  Ag. The measured 1atm pH  5.5 x M

20. The standard electrode potentials (reduction) of 2+ 3+ 4+ 2+ Pt/Fe , Fe and Pt/Sn , Sn are + 0.77 V and 0.15 V respectively at 25°C. The standard EMF of the reaction 4+ 2+  Sn2+ + 2Fe3+ is Sn + 2Fe  (a) – 0.62 V

(b) – 0.92 V

(c) + 0.31 V

(d) + 0.85 V

21. Strongest reducing agent is : (a) K

(b) Mg

(c) Al

(d) I

22. The oxidation potential of Zn, Cu, Ag, H2 and Ni are 0.76, –0.34, – 0.80, 0, 0.55 volt respectively. Which of the following reaction will provide maximum voltage ? 2+

EMF of the cell is 1.023 V. What is the value of x ?

  Cu + Zn2+

(a) Zn + Cu

+

E 0 Ag



2+

–2

(a) 2 × 10 M

–3

(b) 2 × 10 M

(c) H2 + Cu

2+

–3

(c) 1.5 × 10 M

2+

 2Ag + Zn (b) Zn + 2Ag 

, Ag = + 0.799 V [T = 25°C]

–2

(d) 1.5 × 10 M

(d) H2 + Ni

  2H+ + Cu   2H+ + Ni

ELECTROCHEMISTRY

179

23. E° for the half cell reactions are as, 2+ – (a) Zn  Zn + 2e ; E° = + 0.76 V 2+ – (b) Fe  Fe + 2e ; E° = + 0.41 V The E° for half cell reaction, 2+

2+

Fe + Zn   Zn + Fe is : (a) – 0.35 V (b) + 0.35 V (c) + 1.17 V (d) – 0.17 V 24. The standard electrode potential for the reaction +

–

Ag (aq) + e   Ag(s) 2+

–

 Sn(s) Sn (aq) + 2e  at 25°C are 0.80 volt and – 0.14 volt, respectively. The emf of the cell. 2+ + Sn | Sn (1 M) | | Ag (1M) | Ag is (a) 0.66 volt (b) 0.80 volt (c) 1.08 volt (d) 0.94 volt 25. The equation representing the process by which standard reduction potential of zinc can be defined is : 2+ –  Zn (a) Zn (s) + 2e   Zn2+(g) + 2e– (b) Zn (g)  2+ –  Zn (c) Zn (g) + 2e  2+ –  Zn (s) (d) Zn (aq.) + 2e 

26. What will be the emf for the given cell ? + + Pt | H2 (g, P1) | H (aq) || H2 (g, P2) | Pt|H (aq)

RT P1 (a) F ln P 2

RT P1 (b) 2F ln P 2

29. The potential of the cell containing two hydrogen electrodes as represented below +

–6

+

(a) –0.118 V

(b) –0.0591 V

(c) 0.118 V

(d) 0.0591 V

30. Calculate the standard free energy change for the reaction, + + 2 Ag + 2H  H2 + 2 Ag +

–

E° for Ag + e  Ag is 0.80 V (a) + 154.4 kJ

(b) + 308.8 kJ

(c) –154.4 kJ

(d) –308.8 kJ +

(a) 1

(b) 4

(c) 7

(d) 10

32. The standard EMF of Daniell cell is 1.10 volt. The maximum electrical work obtained from the Daniell cell is (a) 212.3 kJ

(b) 175.4 kJ

(c) 106.15 kJ

(d) 53.07 kJ

33. The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The potential of the cell would be (the value of 2.303 RT/F is 0.059 V) (a) 0.177 V

(b) 0.087 V

(c) –0.177 V

(d) 0.059 V

34. What is the free energy change for the half reaction + – Li + e  Li? –1

Given ELi /Li  3.0V, F = 96500 C mol and T = 298 K. –1

(d) None of these

27. If the pressure of hydrogen gas is increased from 1 atm. to 100 atm., keeping the hydrogen ion concentration constant at 1 M, the voltage of the hydrogen half-cell is at 25°C will be (a) 0.059 V (b) –0.059 V (c) 0.295 V (d) 0.118 V.

Relationship between emf and gibb’s free energy change 28. The standard free energy change for the following reaction is – 210 kJ. What is the standard cell potential ?

–1

–1

(b) –298.5 kJ mol –1

(c) 32.166 CV mol

–1

(d) –289500 CV mol

35. The emf of Daniell cell is 1.1 volt. If the value of Faraday is 96500 coulombs per mole, the change in free energy in kJ is (a) 212.30 (b) –212.30 (c) 106.15 (d) –106.15

Relating half cell potential using dG 36. When two half-cells of electrode potential of E1 and E2 are combined to form a cell of electrode potential E3, then (when n1, n2 and n3 are no. of electrons exchanged in first, second and combined half-cells) : (a) E3 = E2 – E1

(b) E3 =

E1 n 1  E 2 n 2 n3

2H2O2 (aq)   2H2O(l) + O2(g) (a) + 0.752

(b) + 1.09

(c) + 0.420

(d) + 0.640

+

31. The emf of the cell H2(1 atm) Pt | H (a = x) | | H (a = 1) | H2(1 atm) Pt at 25°C is 0.59 V. The pH of the solution is

(a) 289.5 kJ mol

RT P2 (c) 2F ln P 1

–4

Pt, H2(g) | H (10 M) | |H (10 M)| H2(g), Pt at 298 K is

(c) E3 =

E1 n 1  E 2 n 2 n 32

(d) E3 = E1 + E2

180

ELECTROCHEMISTRY 37. If EAu  / Au is 1.69 V and EAu3 / Au is 1.40 V, then EAu  / Au3

38.

(a) 0.19 V

(b) 2.945 V

(c) 1.255 V

(d) None of these

Given the data at 25ºC, Ag + I–  AgI + e–; Eo = 0.152 V +

–

o

Ag  Ag + e ; E = – 0.800 V What is the value of log Ksp for AgI ?

RT    0.059V   2.303 F   (a) – 8.12

(b) + 8.612

(c) – 37.83

(d) –16.13

2+

2+

39. Co | Co (C2) || Co (C1) | Co for this cell, G is negative if : (a) C2 > C1

(b) C1 > C2

(c) C1 = C2

(d) unpredictable

40. Which are used as secondary reference electrodes ?

43. A fuel cell is : (a) The voltaic cells in which continuous supply of fuels are send at anode to give oxidation (b) The votalic cell in which fuels such as : CH4, H2, CO are used up at anode (c) It involves the reactions of H2 – O2 fuel cell such as : –

 4H2O(l) + 4e Anode : 2H2 + 4OH  Cathode : O2 + 2H2O(l) + 4e   4OH

(d) All of the above 44. In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to (a) generate heat (b) create potential difference between the two electrodes (c) produce high purity water (d) remove adsorbed oxygen from electrode surfaces. 45. Adding powdered Pb and Fe to a solution containing 1.0 M 2+ 2+ is each of Pb and Fe ions would result into the formation of 2+

(a) More of Pb and Fe ions 2+

(b) More of Fe and Pb ions (c) More of Fe and Pb

(a) Calomel electrode

2+

2+

(d) More of Fe and Pb ions

(b) Ag/AgCl electrode

46. Zn can not displace following ions from their aqueous solution :

(c) Hg/Hg2Cl2 – KCl electrode

+

41. Which of the following has been universally accepted as a reference electrode at all temperatures and has been assigned a value of zero volt ? (a) platinum electrode

(b) copper electrode

(c) graphite electrode

(d) standard hydrogen electrode

For the following cell with hydrogen electrodes at two different pressure p1 and p2

Pt (H 2 ) | H  (aq) | Pt (H 2 ) emf is given by p1

1M

(a)

p RT log e 1 F p2

(c)

p RT log e 2 F p1

2+

(a) Ag

(d) All of the above

42.

–

p2

(b)

p RT log e 1 2F p2

(d)

p RT log e 2 2F p1

(c) Fe

(b) Cu

2+

(d) Na

+

47. Which of the following displacement does not occur : +

2+

(a) Zn + 2H  Zn + H2  +

2+

(b) Fe + 2Ag  Fe + 2Ag  2+

2+

2+

2+

(c) Cu + Fe  Cu + 2Fe  (d) Zn + Pb  Zn + Pb  48. The position of some metals in the electrochemical series in decreasing electropositive character is given as Mg > Al > Zn > Cu > Ag. What will happen if a copper spoon is used to stir a solution of aluminium nitrate ? (a) The spoon will get coated with aluminium (b) An alloy of copper and aluminium is formed (c) The solution becomes blue (d) There is no reaction

181

ELECTROCHEMISTRY 49.

50.

Standard reduction electrode potentials of three metals A, B and C are + 0.5V, –3.0V and –1.2V respectively. The reducing power of these metals are (a) B > C > A

(b) A > B > C

(c) C > B > A

(d) A > C > B

2

(a) Fe

(b) Mn

(c) Cr

(d) Co

(a)

56.

o of them is expected to have the highest E M3 / M2 value ?

57.

52.

(b) Mn (Z = 25)

(c) Fe (Z = 26)

(d) Co (Z = 27)

For a cell given below : Ag | Ag+ || Cu2+ | Cu 



1000x y

1000 (c) xy

Four successive members of the first row transition elements listed below with atomic numbers. Which one

(a) Cr (Z = 24)

–1

and y is the molarity of the solution, then  m (in S cm mol ) is given by :

o 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 ?

51.

–1

55. If x is specific resistance (in S cm) of the electrolyte solution

x (b) 1000 y

(d)

xy 1000

Conductivity (Seimen’s S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then constant of proportionality is expressed in (a) S m mol–1

(b) S2 m2 mol–2

(c) S m2 mol–1

(d) S2 m2 mol

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100. The conductivity of this solution is 1.29S m–1. Resistance of the same cell when filled with 0.02 M of the same solution is 520. The molar conductivity of 0.02 M solution of electrolyte will be (a) 1.24 × 10–4S m2 mol–1

(b) 12.4 × 10–4S m2 mol–1

(c) 124 × 10–4S m2 mol–1

(d) 1240 × 10–4S m2 mol–1

Ag+ + e–   Ag;

Eº = x

Cu2+ + 2e–   Cu;

Eº = y

Resistance of 0.2 M solution of an electrolyte is 50. The specific conductance of the solution is 1.3 S m–1. If resistance of the 0.4M solution of the same electrolyte is 260, its molar conductivity is

(a) x + 2y

(b) 2x + y

(a) 6250 S m2 mol–1

(b) 6.25 × 10–4S m2 mol–1

(c) y – x

(d) y – 2x

(c) 625 × 10–4S m2 mol–1

(d) 62.5 S m2 mol–1

58.

Eº cell is

53. Which of the following solutions of NaCl will have the highest specific conductance ?

59.

Resistance of 0.2M solution of an electrolyte is 50 .

(a) 0.001 N

(b) 0.1 N

The specific conductance of the solution is 1.4S m–1. The resistance of 0.5M solution of the same electrolyte is

(c) 0.01 N

(d) 1.0 N

280 . The molar conductivity of 0.5 M solution of the

54. The specific conductance of a salt of 0.01 M concentration –4 1 is 1.061 × 10 S cm molar conductance of the same solution will be : –4

(a) 1.061 × 10

(b) 1.061

(c) 10.61

(d) 106.1

electrolyte in S m2 mol–1 is : (a) 5  10 3

(b) 5  103

(c) 5  102

(d) 5  10 4

182

ELECTROCHEMISTRY –1

60. The conductivity of 0.1 N NaOH solution is 0.022 S cm . When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreases to 0.0055 S–1 2 –1 cm . The equivalent conductivity in S cm equivalent of NaCl solution is (a) 0.0055

(b) 0.11

(c) 110

(d) none

62.

(b) 105.2

(c) 130.6

(d) 150.2

Electrolyte : 2

-1

^ (S cm mol ) :

KCl

KNO3

HCl

NaOAc

NaCl

149.9

145

426.2

91

126.5

of the electrolytes listed above at infinite dilution in H2O at 25ºC (a) 217.5

(b) 390.7

(c) 552.7

(d) 517.2

The molar conductivities oNaOAc , and oHCl at infinite dilution in water at 25ºC are 91.0 and 426.2S cm2/mol respectively. To calculate

oHOAc , the

additional value

required is

64.

(a) oNaOH

(b) oNaCl

o (c)  H 2O

(d) oKCl

(c) one amp/hour

(d) 96,500 C

23

(a) 6.022 × 10

23

(c) 12.044 × 10

23

(b) 3.011 × 10

22

(d) 6.022 × 10

(a) 0.50

(b) 1.00

(c) 1.50

(d) 2.00

68. An ion is reduced to the element when it absorbs 6 × 10 electrons. The number of equivalents of the ion is : (a) 0.10

(b) 0.01

(c) 0.001

(d) 0.0001

20

69. Electrolysis can be used to determine atomic masses. A current of 0.550 A deposits 0.55 g of a certain metal in 100 minutes. Calculate the atomic mass of the metal if n = 3 : (a) 100

(b) 45.0

(c) 48.25

(d) 144.75

70. An electrolysis of a oxytungsten complex ion using 1.10 A for 40 min produces 0.838 g of tungsten. What is the charge of tungsten in the material ? (Atomic weight : W = 184)

The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25ºC are given below : oCH 3COONa

(b) 96,500 C/sec

67. 13.5 g of Al get deposited when electricity is passed through the solution of AlCl3. The number of faradays used are :

Calculate HOAc using appropriate molar conductances

63.

(a) one amp/sec

66. Number of electrons involved in the electrodeposition of 63.5 g of Cu from a solution of CuSO4 is :

61. The molar conductance at infinite dilution of AgNO3, AgCl and NaCl are 116.5, 121.6 and 110.3 respectively. The molar conductances of NaNO3 is : (a) 111.4

65. The electric charge for electrode deposition of one gram equivalent of a substance is :

2

= 91.0 S cm /equiv..

oHCl = 426.2 S cm2/equiv.. What additional information/quantity one needs to calculate º of an aqueous solution of acetic acid ?

(a) 6

(b) 2

(c) 4

(d) 1

71. When molten lithium chloride (LiCl) is electrolyzed, lithium metal is formed at the cathode. If current efficiency is 75% then how many grams of lithium are liberated when 1930 C of charge pass through the cell : (Atomic weight : Li = 7) (a) 0.105

(b) 0.120

(c) 0.28

(d) 0.240

72. A current of 9.65 ampere is passed through the aqueous solution NaCl using suitable electrodes for 1000 s. The amount of NaOH formed during electrolysis is

(a) º of chloroacetic acid (ClCH2COOH) (b) º of NaCl (c) º of CH3COOK 

o

(d) The limiting equivalent conductance of H ( H  ).

(a) 2.0 g

(b) 4.0 g

(c) 6.0 g

(d) 8.0 g

183

ELECTROCHEMISTRY 73. How many electrons are delivered at the cathode during electrolysis by a current of 1A in 60 seconds ? 20

(b) 6.0 × 10

21

(d) 6.0 × 10

(a) 3.74 × 10 (c) 7.48 × 10

23

20

74. The moles of electrons required to deposit 1 gm equivalent aluminium (at. wt. = 27) from a solution of aluminium chloride will be (a) 3

(b) 1

(c) 4

(d) 2

75. Time required to deposit one millimole of aluminium metal by the passage of 9.65 amperes through aqueous solution of aluminium ion is : (a) 30 s

(b) 10 s

(c) 30,000 s

(d) 10,000 s

76. How many coulomb of electricity are consumed when 100 mA current is passed through a solution of AgNO3 for 30 minute during an electrolysis experiment.

81. A current of 2 ampere was passed through solutions of CuSO4 and AgNO3 in series. 0.635 g of copper was deposited. Then the weight of silver deposited will be : (a) 0.59 g

(b) 3.24 g

(c) 1.08 g

(d) 2.16 g

82. The weight ratio of Al and Ag deposited using the same quantity of current is : (a) 9 : 108

(b) 2 : 12

(c) 108 : 9

(d) 3 : 8

83. The weight of silver (eq. wt. = 108) displaced by the quantity of current which displaced 5600 mL of hydrogen at STP is : (a) 54 g

(b) 108 g

(c) 5.4 g

(d) None of these

84. When one coulomb of charge is passed through an electrolytic solution for one sec, the mass deposited on the electrode is equal to : (a) equivalent weight (b) molecular weight

(a) 108

(b) 18000

(c) electrochemical equivalent

(c) 180

(d) 3000

(d) one gram

77. A current of 9.65 amp. flowing for 10 minute deposits 3.0 g of a metal. The equivalent wt. of the metal is :

85. By the electrolysis of aqueous solution of CuSO4, the products obtained at both the electrodes are

(a) 10

(b) 30

(a) O2 at anode and H2 at cathode

(c) 50

(d) 96.5

(b) H2 at anode and Cu at cathode

78. 108 g fairly concentrate solution of AgNO3 is electrolyzed using 0.1 F of electricity. The weight or resulting solution is: (a) 94 g

(b) 11.6 g

(c) 96.4 g

(d) None

79. When the same electric current is passed through the solution of different elecrolytes in series the amounts of elements deposited on the electrodes are in the ratio of their: (a) atomic number

(b) atomic masses

(c) specific gravities

(d) equivalent masses

(c) O2 at anode and Cu at cathode (d) H2S2O8 at anode and O2 at cathode 86. During the electrolysis of fused NaCl, the reaction that occurs at the anode is : (a) Chloride ions are oxidized (b) Chloride ions are reduced (c) Sodium ions are oxidized (d) Sodium ions are reduced

80. The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous H2SO4 and fused MgSO4 are : (a) 1 : 8

(b) 1 : 12

(c) 1 : 16

(d) None of these

Electroytic cell 87. Which reaction occur at cathode during electrolysis of fused lead bromide ?

 Pb2+ + 2e– (a) Pb 

–

–

 Br (b) Br + e 

ELECTROCHEMISTRY –  Br + e– (c) Br 

184 If the molar cond. of acetic acid at infinite dilution is 380.8 –1 2 –1 ohm cm mol , the value of dissociation constant will be

2+ –  Pb (d) Pb + 2e 

88. On electrolysing a solution of dilute H2SO4 between platinum electrodes, the gas evolved at the anode is (a) SO2

(b) SO3

(c) O2

(d) H2

98. At infinite dilution, the eq. conductances of CH3COONa, 2 HCl and CH3COOH are 91, 426 and 391 mho cm respectively at 25°C, The eq. conductance of NaCl at infinite dilution will be :

89. Reaction that takes place at graphite anode in dry cell is 2+

–

(a) Zn + 2e  Zn(s) 2+

(b) Zn(s)  Zn + 2e 2+

97. The equivalent conductivity of 0.1 N CH3COOH at 25°C is 80 and at infinite dilution 400. The degree of dissociation of CH3COOH is

–

–

(c) Mn + 2e  Mn(s) +

99.

–

(d) Mn(s)  Mn + e + 1.5 V.

KCl are 126, 152 and 150 S cm2 mol–1 respectively, The

Batteries, Fuel Cells and Corrosion

 o for NaBr is

90. When a lead storage battery is discharged (a) PbSO4 is formed

(b) Pb is formed

(c) SO2 is consumed

(d) H2SO4 is formed

The limiting molar conductivities  o for NaCl, KBr and

100. The specific conductance of a saturated solution of silver –1 + bromide is k S cm . The limiting ionic conductivity of Ag –1 and Br ions are x and y, respectively. The solubility of –1 silver bromide in gL is : (molar mass of AgBr = 188)

91. As lead storage battery is charged (a) lead dioxide dissolves

101. The specific conductivity of a saturated solution of AgCl is  –6 –1 –1 –1 2 –1 3.40 × 10 ohm cm at 25 °C. If  Ag = 62.3 ohm cm mol 

(b) sulphuric acid is regenerated

–1

2

–1

(c) lead electrode becomes coated with lead sulphate

& Cl = 67.7 ohm cm mol , the solubility of AgCl at 25°C

(d) the concentration of sulphuric acid decreases.

is :

92. In electroplating the article to be electroplated is made :

102. How many minutes will it take to plate out 0.50 g of Cr from a Cr2(SO4)3 solution using a current of 1.50 A ? (Atomic weight : Cr = 52.0)

(a) cathode (b) anode (c) either cathode or anode

103. Three faradays of electricity was passed through an aqueous solution of iron (II) bromide. The mass of iron metal (at. mass 56) deposited at the cathode is -

(d) simply suspended in the electrolytic bath. 93. A spoon to be electroplated with gold should be : (a) cathode

(b) anode

(c) electrolyte

(d) none of these

94. The specific conductance of a N/10 KCl at 25°C is 0.0112 –1 –1 ohm cm . The resistance of cell containing solution at the same temperature was found to be 55 ohms. The cell constant will be 95. Resistance of 0.1 M KCl solution in a conductance cell is –1 300 ohm and conductivity is 0.013 Scm . The value of cell constant is : –1

2

–1

96. Molar conductance of 0.1 M acetic acid is 7 ohm cm mol .

104. W g of copper deposited in a copper voltameter when an electric current of 2 ampere is passed for 2 hours. If one ampere of electric current is passed for 4 hours in the same voltameter, copper deposited will be : 105. Several blocks of magnesium are fixed to the bottom of a ship to make the ship lighter

ELECTROCHEMISTRY

185

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTION 1.

Two Faraday of electricity is passed through a solution of

6.

CuSO4. The mass of copper deposited at the cathode is :

What will happen if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO4 ? Online 2016 SET (1)

(at. mass of Cu =63.5 amu)

2.

(2015)

(a) 2g

(b) l27g

(c) 0 g

(d) 63.5 g

Galvanization is applying a coating of : (a) Cr

(b) Cu

(c) Zn

(d) Pb

(a) The copper metal will dissolve and zinc metal will be deposited. (b) The copper metal will dissolve with evolution of hydrogen gas. (c) The copper metal will dissolve with evolution of oxygen gas.

(2016)

(d) No reaction will occur. 7. 3.

Given, E oCl

2 /Cl



 1.36 V, E oCr3 /Cr  0.74 V

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is : Online 2016 SET (2)

E oCr O2 /Cr3  1.33 V, E oMnO / Mn 2  1.51 V

(a) 2.24 L

(b) 4.48 L

Among the following, the strongest reducing agent is

(c) 6.72 L

(d) 8.96 L

2

7

4

(2017) (a) Cr

(b) Mn 2

(c) Cr 3

(d) Cl–

8.

Identify the correct statement : Online 2016 SET (2) (a) Iron corrodes in oxygen-free water.

4.

5.

(b) Iron corrodes more rapidly in salt water because its electrochemical potential is higher.

How long (apporoximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane ? (Atomic weight of B = 10.8u) (2018) (a) 3.2 hours

(b) 1.6 hours

(c) 6.4 hours

(d) 0.8 hours

(c) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential. (d) Corrosion of iron can be minimized by forming an impermeable barrier at its surface. 9.

What is the standard reduction potential (E°) for

At 298 K, the standard reduction potentials are 1.51 V for

Fe3+  Fe ?

MnO4 |Mn2+, 1.36 V for Cl2| Cl, 1.07 V for Br2|Br–, and

Given that

Online 2017 SET (1)



0.54 V for I2|I . At pH = 3, permanganate is expected to

 RT   0.059 V  oxidize :   F  Online 2015 SET (1)

Fe2  2e  Fe;

Eo

Fe3  e  Fe2 ;

Eo

Fe2 /Fe

 0.47 V

Fe3  /Fe2

(a) Cl, Brand I

(b) Brand I

(a) –0.057 V

(b) + 0.057 V

(c) Cland Br

(d) Ionly

(c) +0.030 V

(d) –0.30 V

 0.77 V

186

ELECTROCHEMISTRY 10.

To find the standard potential of M3+/M electrode, the following cell is constituted : 3+

–1

+

15.

Which one of the following graphs between molar conductivity (  m ) versus

–1

Pt/M/M (0.001 mol L )/Ag (0.01 mol L )/Ag The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e–  M at 298 K will be : Online 2017 SET (2)

(2019-04-10/Shift-1) (a)

(Given E  Ag /Ag at 298 K = 0.80 Volt)

11.

12.

13.

(a) 0.38 Volt

(b) 0.32 Volt

(c) 1.28 Volt

(d) 0.66 Volt

When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is: Online 2018 SET (1) (a) 1.0

(b) 0.5

(c) 0.1

(d) 2.0

(b)

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is : Online 2018 SET (3) (a) 9.81 g

(b) 10.9 g

(c) 98.1 g

(d) 109.0 g

(c)

The standard Gibbs energy for the given cell reaction in KJ mol 1 at 298 K is: Zn(s)  Cu 2  (aq)  Zn 2  (aq)  Cu(s), E 0  2 V at 298 K

[Faraday’s constant F = 96500 C mol1 ] (2019-04-09/Shift-1)

14.

(a) -192

(b) 384

(c) -384

(d) 192

A solution of Ni  NO3 2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? (2019-04-09/Shift-2) (a) 0.20

(b) 0.05

(c). 0.10

(d) 0.15

C is correct?

(d)

187

ELECTROCHEMISTRY 16.

Given :

20.

Co3 e   Co 2  ; E 0  1.81V

Pt(s) | H 2 (g,1bar) | HCl(aq) | AgCl | Ag(s) | Pt(s) the cell

Pb 4  2e  Pb 2  ; E 0  1.67

potential is 0.92 V when a 106 molal HCl solution is used.The standard electrode potential of

Ce 4   e   Ce3 ; E 0  1.61V

(AgCl / Ag, Cl1 )

Bi3  3e   Bi; E 0  0.20V

electrode is :

Oxidizing power of the species will increase in the order (2019-04-12/Shift-1)

(a) 0.94 V (c) 0.40 V

(b) Co3  Pb4  Ce4  Bi3 (c) Bi3  Ce4  Pb4  Co3

(2019-01-10/Shift-2)

2.303RT    0.06V at 298K  Given, F  

(a) Ce4  Pb3  Bi3  Co3

21.

(d) Co3  Ce4  Bi3  Pb4 17.

In the cell

(b) 0.76 V (d) 0.20 V

For the cell Zn(s) | Zn 2  (aq) || M x  (aq) | M(s) , different half cells and their standard electrode potentials are given below:

The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is: (Molar mass of PbSO4 = 303 g mol1 ) (2019-01-09/Shift-1) (a) 22.8 (c) 7.6

18.

0 If E zn2 / zn  0.76 V , which cathode will given a maxi-

(b) 15.2 (d) 11.4

mum value of E 0cell per electron transferred? (2019-01-11/Shift-1)

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction

(a) Ag  / Ag

(b) Fe3 / Fe2

 Zn 2  (aq)  Cu(s) Zn(s)  Cu 2  (aq) 

(c) Au3  /Au

(d) Fe2 / Fe

22.

At 300 K is approximately

Given the equilibrium constant:

(R = 8 JK 1 mol1 , F = 96000 C mol1 )

K C of the reaction:

(a) e80

(b) e 160

Cu(s)  2Ag  (aq)  Cu 2  (aq)  2Ag(s) is

(c) e320

(d) e160

0 10  1015 calculate the Ecell of the reaction of 298 K

(2019-01-09/Shift-2) 19.

RT    2.303 F at 298K  0.059V   

Consider the following Zn 2   2e   Zn(s); E 0  0.76 V

Ca 2   2e   Ca(s); E 0  2.87V

(a) 0.04736 mV (c) 0.4736 V

Mg 2   2e   Mg(s); E 0  2.36V Ni 2   2e   Ni(s); E 0  0.25V

The reducing power of the metals increases in the order (2019-01-10/Shift-1) (a) Ca  Zn  Mg  Ni

(b) Ni  Zn  Mg  Ca

(c) Zn  Mg  Ni  Ca

(d) Ca  Mg  Zn  Ni

23.

(2019-01-11/Shift-2) (b) 0.4736 mV (d) 0.04736 V

 0m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S

cm 2 mol1 ,respectively. If the conductivity of 0.001 M HA is 5  10 5 Scm 1 , degree of dissociation of HA is : (2019-01-12/Shift-2) (a) 0.50 (c) 0.125

(b) 0.25 (d) 0.75

188

ELECTROCHEMISTRY 24.

The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is :

29.

Cu(s)  Sn 2  (aq.)  Cu 2 (aq.)  Sn (s)  Eo   0.16V, E o 2  sin 2  |Sn  Cu |Cu     0.34V, Take F  96500 C mol1    25.

For

the

(2020-09-02/Shift-1) disproportionation reaction

2Cu  (aq)  Cu(s)  Cu 2  (aq) at K, In K (where K is the equilibrium constant) is ………… × 10–1. (2020-09-02/Shift-2)

Given : E

0 Cu 2  /Cu 

E0

Cu  /Cu

26.

 0.52 V

Eo

= -0.76 V

Zn 2+ Zn

RT  0.025) F

Identify the incorrect statement form the options below for the above cell: (2020-09-04/Shift-1)

Let C NaCl and CBaSO4 be the conductances (in S) measured for saturated aqueous solution of NaCl and BaSO4, respectively, at a temperature T. Which of the following is false ? (2020-09-30/Shift-1)

(a) If Eext=1.1 V, no flow of e- or current occurs

(a) C NaCl (T2 )  C NaCl (T1 ) for T2  T1

(d) If Eext< 1.1 V, Zn dissolves at anode and Cu doposits at cathode

(b) If Eext> 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode (c) If Eext> 1.1 V, e- flows from Cu to Zn

30.

(d) CNaCl  CBaSO4 at a given T The photoelectric current from Na (Work function, w0 = 2.3 eV) is stopped by the output voltage of the cell Pt (s) H2 (g, 1 Bar) HCl (aq. pH = 1) | AgCl (s) | Ag(s). The pH of aq. HCl required to stop the photoelectric current form K (w0 = 2.25 eV), all other conditions remaining the same, is ……….. × 10–2 (to the nearest integer). (2020-09-30/Shift-1) Given,

2.303 28.

= + 0.34 V

Cu 2+ Cu

 0.16 V

(b) CBaSO4 (T2 )  CBaSO4 (T1 ) for T2  T1 (c) Ionic mobilities of ions from both salts increase with T. 27.

Eo

 0  0  E Ag  /Ag  0.80 V, E Au  /Au  1.69 V   

(2020-09-04/Shift-2) (a) Silver and gold in proportion to their atomic weights

RT  0.06 V; E o  0.22 V F AgCl|Ag|Cl

(b) Silver and gold in equal mass proportion (c) only silver

An acidic solution of dichromate is electrolyzed for 8 minutes using 2A current. As per the following equation Cr2 O 72   14H   6e   2Cr 3  7H 2 O

The amount of Cr 3 obtained was 0.104 g. The efficiency of the process (in %) is (Take : F = 96000 C, At. Mass of chromium = 52) ………….. . (2020-09-30/Shift-2)

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 MAgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be:

(d) only gold 31.

An oxidation-reduction reaction in which 3 electrons are transferred has a G 0 of 17.37 kJ mol–1 at 25ºC. The value of E 0cell

(in V) is …………… × 10–2.(1 F = 96,500 C mol–1) (2020-09-05/Shift-1)

189

ELECTROCHEMISTRY 32.

The variation of molar conductivity with concentration of an electrolyte (x) in aqueous solution is shown in the given figure.

35.

Given that the standard potential; (Eo) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the Eo of Cu2+/ Cu+ is : (2020-01-07/Shift-1) (a) +0.158 V (c) 0.182 V

36.

(b) -0.158 V (d) -0.182 V

The equation that is incorrect is: (2020-01-07/Shift-2) (a)  0m NaBr   0m NaI  0m KBr  0m NaBr (b)  0m NaBr   0m NaCl   0m KBr   0m KCl (c)  0m KCl   0m NaCl   0m KBr   0m NaBr

The electrolyte X is :

33.

(2020-09-05/Shift-2)

(a) HCl

(b) CH3COOH

(c) NaCl

(d) KNO3

(d)  0m H 2 O   0m HCl   0m NaOH   0m NaBr 37.

Potassium chlorate is prepared by the electrolysis of KCl

2 H 2 O  O2  4 H   4e  ; E   1.23V

in basic solution 6OH   Cl   ClO3  3H 2O  6e If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10g of KClO3using a current of 2A is

(R=8.314 Jmol–1K–1; temp.=298 K; oxygen under std. atm. Pressure of 1bar.) (2020-01-08/Shift-1) 38.

(2020-09-06/Shift-1)

 Sn 2    Pb 2   when this cell attains equilibrium is ______.

For the Given Cell ;

(Given)

Cu  s  Cu 2  C1 M  Cu 2   C2 M  Cu  s 

0 0 ESn  0.14V ; EPb  0.13V , 2 2 / Sn / Pb

Change in Gibbs energy (DG) is negative, if : (2020-09-06/Shift-2) (a) C2  2C1

(b) C2 

C1 2

For an electrochemical cell

Sn  s  Sn 2   aq.,1M  Pb 2   aq.,1M  Pb  s  the ratio

(Given : F = 96,500 C mol-1; molar mass of KClO3 = 122g mol-1)

34.

What will be the electrode potential for the given half cell reaction at pH= 5?

39.

2.303RT  0.06V F

Amongst the following, the form of water with lowest ionic conductance at 298 K is : (a) distilled water (b) sea water

(c) C1  2C2

(d) C1  C2

(c) saline water used for intra venous injection (d) water from a well

(2020-01-09/Shift-2)

190

ELECTROCHEMISTRY

EXERCISE -3 : ADVANCED OBJECTIVE QUESTIONS Single Choice Questions 1.

2.

3.

Which one of the following reaction occurs at the cathode? (a) 2OH– H2O + O + 2e– (b) Ag Ag+ + e– (c) Fe2+ Fe3+ + e– 2+ – (d) Cu + 2e Cu Reaction taking place at anode is (a) ionisation (b) reduction (c) oxidation (d) hydrolysis Which of the following reaction is possible at anode ? (a) 2Cr3+ + 7H2O

4.

5.

8.

Cr2 O 72 + 14H+

(b) F2 2F– (c) ½O2 + 2H+ H2O (d) None of these The electrochemical cell stops working after some time because (a) Electrode potentials of both the electrodes become zero (b) Electrode potentials of both the electrodes become equal (c) One of the electrode is eaten way (d) The reaction starts proceeding in opposite direction Cell reaction for the cell Zn | Zn2+ (1.0 M) || Cd2+ (1.0 M) | Cd is given by

ZnSO4 Zn (0.01M)

9.

(a) E1 < E2

(b) E1 = E2

(c) E2 = 0 E1 Eº of a cell aA + bB

(d) E1 > E2 cC + dD is

(a) E + RT ln

(c) E  10.

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

(b) E 

For the cell reaction, ( aq ) ] 

Zn (s )

 Zn 2  C 2 (aq)  Cu (s)

of an electrochemical cell, the change in free energy, G at a given temperature is a function of (a) ln (C1)

 C2  (b) ln  C   1

(c) ln (C1+ C2)

(d) ln (C2)

(d) Zn + Cd2+  Zn2+ + Cd.

Which graph correctly correlates Ecell as a function of concentrations for the cell

If the salt bridge is removed suddenly from a working cell, the voltage

Zn(s) + 2Ag (aq)   Zn (aq) + 2Ag(s), E°cell = 1.56 V

(a) increases

[Zn 2 ] Y-axis : Ecell, X-axis : log10 [Ag  ]

(b) decreases

11.

+

2+

(c) drops to zero

7.

RT [C]c [ D]d ln nF [A]a [B]b

RT [C]C [D]D RT [a ]A [B]B ln E  ln (d) nF [A]A [B]B nF [C]C [d]D

Cu 2  [C1

(b) Zn2+  Zn – 2e–

6.

CuSO4 (1.0M) Cu

When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2 ?

(a) Cd  Cd2+ + 2e– (c) Cd + Zn2+  Zn + Cd2+

The e.m.f. of a Daniel cell at 298K is E1

(d) may increases or decrease depending upon cell reaction. Which one of the following is not a function of a salt bridge ? (a) To allow the flow of cations from one solution to the other (b) To allow the flow of anions from one solution to the other (c) To allow the electrons to flow from one solution to the other (d) To maintain electrical neutrality of the two solutions.

(a)

(b)

(c)

(d)

191

ELECTROCHEMISTRY 12.

17.

In the reaction : Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) the reduction half cell reaction is (a) Cu + 2e– Cu2+ (b) Cu – 2e– Cu2+ + (c) Ag + e Ag

Zn  Zn2+(aq)  Cu2+(aq)  Cu is 1.10 V at 25ºC. The equilibrium constant for the cell reaction Zn + Cu2+(aq)  Cu + Zn2+(aq) is of the order of

+

13.

(d) Ag – e Ag In the cell, Zn  Zn2+  Cu2+  Cu, the negative terminal is (a) Cu (b) Cu2+ (c) Zn (d) Zn2+

18.

Applications of electrochemical series 14.

The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are PbSO4 + 2e–  Pb + SO 24

(Eº = – 0.31 V)

Ag+ (aq) + e–  Ag (s)

19.

(Eº = 0.80 V)

(b) 1037

(c) 10–17

(d) 1017

For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295V at 25ºC. The equilibrium constant of the reaction at 25ºC will be (a) 1 × 10–10

(b) 29.5 × 10–2

(c) 10

(d) 1 × 1010

If a strip of copper metal is placed in a solution of ferrous sulphate (a) Copper will precipitate out (b) Iron will precipitate out

(a) Pb + SO 24 + 2 Ag+ (aq)  2 Ag (s) + PbSO4

(c) Copper and iron both will be dissolved

(c) Pb +

SO 24

(d) No reaction will take place. 20.

+

+ Ag (s)  Ag (aq) + PbSO4

(d) PbSO4 + Ag (s)  Ag+ (aq) + Pb + SO 24 .

E oAg  / Ag  + 0.80 V..

Zn, Eº = – 0.762 V

Mg2+ + 2e–

Mg, Eº = – 2.37 V

When zinc dust is added to a solution of magnesium chloride

To a mixture containing pieces of zinc, copper and silver, 1 M H2SO4 was added. H2 gas was found to be evolved. Which of the metal/metals do you think has/have reacted ? o o Given E Zn / Zn 2   0.76 V, E Cu 2  / Cu  0.34 V,

The standard potentials at 25ºC for the following half reactions are given against them Zn2+ + 2e–

(a) All the metals

(b) Only Zn

(c) Both Zn and Cu

(d) Only Ag.

(b) Zinc chloride is formed

In a simple electrochemical cell, which is in standard state, half cell reactions with their appropriate oxidation potentials are

(c) Zinc dissolves in the solution

Pb (s) – 2e–

(d) Magnesium is precipitated.

Ag (s) – e–

The relationship between standard reduction potential of a cell and equilibrium constant is shown by

Which of the following reaction takes place ?

(a) No reaction will take place

16.

(a) 10–37

The feasible reaction will be

(b) PbSO4 + 2 Ag+ (aq)  Pb + SO 24 + 2 Ag (s)

15.

Eº for the cell,

21.

Pb2+ (aq) E0 = + 0.13 volt Ag+ (aq) E0 = – 0.80 volt

(a) Pb2+ (aq.) + 2 Ag (s)  2 Ag+ (aq.) + Pb (s) o (a) E cell 

n 0.059 log k c (b) E ocell  log k c 0.059 n

(b) Pb2+ (aq.) + Ag (s)  Ag+ (aq.) + Pb (s) (c) Ag+ (aq.) + Pb (s)  Ag (s) + Pb2+ (aq.)

(c) E ocell  0.059 n log k c

o (d) E cell

log k c  n

(d) 2 Ag+ (aq.) + Pb (s)  2Ag (s) + Pb2+ (aq.)

192

ELECTROCHEMISTRY 22.

23.

The standard reduction potential values of the three metallic cations X, Y and Z are 0.52, – 3.03 and – 1.18V respectively. The order of reducing power of the corresponding metals is (a) Y > Z > X

(b) X > Y > Z

(c) Z > Y > X

(d) Z > X > Y

27.

Electrode potential data are given below : 3  Fe (aq ) e

2  Fe ( aq ) ; Eº = + 0.77V

  Al 3( aq )  3e

Al(s); Eº = – 1.66 V

(a) A, D, B and C

(b) C, B, D and A

(c) B, D, C and A

(d) C, A, D and B

An unknown metal M displaces nickel from nickel (II) sulphate solution but does not displace manganese from manganese sulphate solution. Which order represents the correct order of reducing power ? (a) Mn > Ni > M

(b) Ni > Mn > M

(c) Mn > M > Ni

(d) M > Ni > Mn

Based on the data, the reducing power of Fe2+, Al and Br– will increase in the order

I = – 3.04 V

II = – 1.90V

(a) Br– < Fe2+ < Al

(b) Fe2+ < Al < Br–

III = 0 V

IV = 1.90 V

(c) Al < Br– < Fe2+

(d) Al < Fe2+ < Br–

(a) III

(b) II

(c) I

(d) IV

28.

2Br–(aq); Eº = + 1.08 V

The standard reduction potentials at 298K for the following half reactions are given against each

29.

Zn2+(aq) + 2e  Zn(s); – 0.762 V

(b) Y will oxidise X and not X

2H+(aq) + 2e  H2 (g); 0.00V 3+

Fe

(aq)

2+

+ e  Fe 

(aq)

(c) Y will oxidise both X and Z

; 0.770 V

(d) Y will reduce both X and Z 30.

Which is the strongest reducing agent ? (a) Zn(s)

(b) Cs(s)

(c) H2 (g)

(d) Fe3+(aq)

Standard potentials (Eº) for some half-reactions are given below : (I) Sn4+ + 2e

31.

Sn2+ ; Eº = + 0.15 V

(II) 2Hg2+ + 2e (III) PbO2 + 4H+ + 2e

Hg 22 ;

Eº = 0.92 V

(a) Sn4+ is a stronger oxidising agent than Pb4+ (b) Sn

is a stronger reducing agent than Hg2

Standard reduction electrode potentials of three metals A, B and C are respectively – 0.5, – 3.0 V and – 1.2 V. The reducing powers these metals are (a) B > C > A

(b) A > B > C

(c) C > B > A

(d) A > C > B

I2 and Br2 are added to a solution containing 1 M each of I– and Br– ions. Which of the following reaction will take place ? (Given : standard reduction potentials of I2 and Br2 are 0.53 and 1.09 volts respectively) (a) Iodine will reduce bromide ions

Pb2+ + 2H2O; Eº = + 1.45 V

(b) Bromine will reduce iodide ions

based on the above, which one of the following statements is correct ?

2+

A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y– and 1 M Z – at 25ºC. If the reduction potential is Z > Y > X, then (a) Y will oxidise X and not Z

Cr3+(aq) + 3e  Cr(s); – 0.740 V

25.

Electrode potentials (Eºred) of four elements, A, B, C, D are –1.36, –0.32, 0, –1.26V respectively. The decreasing reactivity order of these elements is

The standard reduction potentials of four elements are given below. Which of the following will be the most suitable reducing agent ?

Br2 (aq) + 2e–

24.

26.

2+

(c) Pb2+ is a stronger oxidising agent than Pb4+ (d) Pb2+ is a stronger reducing agent than Sn2+

(c) Iodide ions will reduce bromine (d) Bromide ions will reduce iodine. 32.

When the cell reaction attains a state of equilibrium, the EMF of the cell is (a) zero

(b) positive

(c) negative

(d) not definite.

ELECTROCHEMISTRY 33.

193

The standard reduction potentials Eº for the half reactions are as

39.

Zn2+ + 2e– ; Eº = 0.76V

Zn

2+

Fe

Fe

–

+ 2e ; Eº = 0.41V

40.

The EMF for the cell reaction will be

34.

(a) – 0.3V

(b) 0.35V

(c) 1.17V

(d) –1.17V 2+

The standard reduction potential for Fe /Fe and Sn2+/Sn electrodes are – 0.44 and – 0.14 volt respectively. For the cell reaction Fe2+ + Sn

41.

Fe + Sn2+

the standard emf will be

35.

(a) + 0.30 V

(b) – 0.58 V

(c) + 0.58 V

(d) – 0.30 V

42.

The emf of the cell Ni/Ni 2+ (1.0M)  Au 3+ (1.0M)/Au is [Eº for Ni2+/Ni = – 0.25V ; Eº for Au3+/Au = 1.5 V]

36.

(a) + 1.25 V

(b) +1.75V

(c) – 1.25V

(d) – 1.75 V

43.

Cu + Zn2+ is 1.1 volt at 25ºC.

The EMF for the cell reaction, when 0.1M Cu2+ and 0.1 2+

M Zn

37.

solutions are used, at 25ºC is

(a) –1

(b) –1 cm–1

(c) cm–1

(d)  cm.

Conductance (unit Siemen’s S) is directly proportional to 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) S m mol–1

(b) S m2mol–1

(c) S–2m2mol

(d) S2m2 mol–2

The unit of specific conductivity is (a) ohms cm–1

(b) ohms cm–2

(c) ohms–1 cm

(d) ohms–1 cm–1

The cell constant of a given cell is 0.47 cm–1. The resistance of a solution placed in this cell is measured to be 31.6 ohm. The conductivity of the solution (in Scm–1 where S has usual meaning) is (a) 0.15

(b) 1.5

(c) 0.015

(d) 150

Molar conductivity of a solution is 1.26 × 102 –1 cm2 mol–1. Its molarity is 0.01. Its specific conductivity will be

The standard EMF for the cell reaction, Zn + Cu2+

The units of cell constant are

44.

(a) 1.26 × 10–5

(b) 1.26 × 10–3

(c) 1.26 × 10–4

(d) 0.0063



–1

–1

m

is the unit of

(a) 1.10 V

(b) 0.10 V

(a) Molar conductivity

(c) –1.10 V

(d) – 0.110 V

(b) Specific conductivity (c) Equivalent conductivity

For the redox reaction : Zn (s) + Cu2+ (0.1M)

(d) Molar conductivity at inifinite dilution.

Zn2+ (1M) + Cu (s) 45.

taking place in a cell,

Eocell

is 1.10 volt. Ecell for the cell

The value of specific conductance is equal to the conductance of the solution when (a) The cell constant is zero

38.

RT    0.0591 will be  2.303 F  

(b) The cell constant is one

(a) 2.14 volt

(b) 1.80 volt

(d) The size of the vessel is very large

(c) 1.07 volt

(d) 0.82 volt

(c) The electrodes are made of copper

46.

The unit of equivalent conductivity is

The specific conductivity of N/10 KCl solution at 20ºC is

(a) ohm cm

0.212 ohm–1 cm–1 and the resistance of the cell containing this solution at 20ºC is 55 ohm. The cell constant is

(b) ohm–1 cm+2 (g equivalent)–1

(a) 4.616 cm–1

(b) 11.66 cm–1

(c) 2.173 cm–1

(d) 3.324 cm–1

(c) ohm cm2 (g equivalent) (d) S cm–2

194

ELECTROCHEMISTRY 47.

The resistance of 0.1 N solution of a salt is found to be 2.5 × 103 ohm. The equivalent conductance of the solution

54.

in Scm2/eq is (cell constant = 1.15 cm–1)

48.

(a) 4.6

(b) 5.6

(c) 6.6

(d) 7.6

The conductance of 0.1 M HCl solution is greater than that of 0.1 M NaCl. This is because

55.

(a) HCl is more ionized than NaCl (b) HCl is an acid whereas NaCl solution is neutral +

If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm–1 then its

50.

(c) 102

(d) 10–6

51.

(b) 1.06 × 103

(c) 1.06 × 104

(d) 5.3 × 102

The equivalent conductance at infinite dilution of a weak acid such as HF

57.

Molar ionic conductivities of a two-bivalent electrolytes x2+ and y2– are 57 and 73 respectively. The molar conductivity of the solution formed by them will be (a) 130 S cm2 mol–1

(b) 65 S cm2 mol–1

(c) 260 S cm2 mol–1

(d) 187 S cm2 mol–1

Equivalent conductances of NaCl, HCl and CH3COONa at

Which of the following is wrong about molar conductivity ?

infinite dilution are 126.45, 426.16 and 91 ohm–1 cm2

(a) The solution contains Avogadro’s number of molecules of the electrolyte

respectively. The equivalent conductance of CH3COOH at infinite dilution would be

(b) It is the product of specific conductivity and volume of solution in cc containing 1 mole of the electrolyte

(a) 101.38 ohm–1cm2

(b) 253.62 ohm–1cm2

(c) Its units are ohm–1 cm2 mol–1

(c) 390.71 ohm–1cm2

(d) 678.90 ohm–1cm2

(d) Its value for 1 M NaCl solution is same as that of 1M glucose solution. 52.

56.

cm2 mol–1 is

(a) 1.06 × 102

(d) 384

(d) is an undefined quantity

Specific conductance of 0.1 M sodium chloride solution is 1.06 × 10–2 ohm–1 cm–1. Its molar conductance in ohm– 1

(c) 392

(c) can be determined from measurements on dilute solutions of NaF, NaCl and HCl

molar conductance in ohm–1 cm2 mol–1 will be (b) 103

(b) 196

(b) can be determined by measurement on very dilute HF solutions

(d) Interionic forces in HCl are weaker than those in NaCl.

(a) 104

(a) 250

(a) can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HI

+

(c) H ions have greater mobility than Na ions

49.

–6

The ionization constant of a weak electrolyte is 25 × 10 while the equivalent conductance of its 0.01 M solution is 19.6 S 2 –1 cm eq . The equivalent conductance of the electrolyte at 2 –1 infinite dilution (in S cm eq ) will be

58.

The increase in the molar conductivity of HCl with dilution is due to

According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte, A2B is   (a)  ( A  )   ( B )

  (b)  ( A  )   ( B )

(a) increase in the self ionisation of water (b) hydrolysis of HCl

 (c) 2 ( A  ) 

(c) decrease in the self ionisation of water (d) decrease in the interionic forces. 53.

The increase in the value of molar conductivity of acetic acid with dilution is due to (a) decrease in interionic forces (b) increase in degree of ionisation (c) increase in self ionisation of water (d) none of these

59.

1    2 (B )

  (d) 2 ( A  )   ( B ) .

The limiting conductivity of NaCl, KCl and KBr are 126.5, 2 –1 150.0 and 151.5 S cm eq , respectively. The limiting – 2 –1 equivalent ionic conductance for Br is 78 Scm eq . The + limiting equivalent ionic conductance for Na ions would be : (a) 128

(b) 125

(c) 49

(d) 50

195

ELECTROCHEMISTRY 60.

61.

The reference electrode is made by using (a) ZnCl2

(b) CuSO4

(c) HgCl2

(d) Hg2Cl2

68.

Which one of the following represents a standard hydrogen electrode correctly ?

solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5, Faraday constant = 96500 C mol–1).

(a) Pt, H2 (1 atm) | H+ (1 M), 298 K

(a) 0.07 M

(b) 0.2 N

(b) Pt, H2 (1 atm) | H+ (0.1 M), 298 K

(c) 0.005M

(d) 0.02N

(c) Pt, H2 (0.1 atm) | H+ (1 M), 273 K

69.

+

(d) Pt, H2 (0.1 atm) | H (0.1 M), 273 K. 62.

The standard hydrogen electrode potential is zero, because (a) there is no potential difference between the electrode and the solution

63.

64.

65.

66.

67.

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the

In a solution of CuSO4 how much time will be required to precipitate 2 g copper by 0.5 ampere current ? (a) 12157.48 sec

(b) 102 sec

(c) 510 sec

(d) 642 sec

(b) hydrogen ions acquire electrons from a platinum electrode

What is the amount of chlorine evolved when 2 amperes of current is passed for 30 minutes in an aqueous solution of NaCl ?

(c) it has been measured accurately

(a) 66 g

(b) 1.32 g

(d) it has been defined that way

(c) 33 g

(d) 99 g

70.

(a) 96500

(b) 48250

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic mass of Ag = 108 g mol–1), the amount of silver deposited is

(c) 193000

(d) 10000

(a) 16.2 mg

(b) 21.2 mg

(c) 10.8 mg

(d) 6.4 mg

The number of coulombs required for the deposition of 107.870g of silver is

1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 Faraday of electricity was passed using 0.15 volt above the oxidation potential of silver. The silver content of the beaker after the above shall be (a) 0 g

(b) 0.108g

(c) 0.108 g

(d) 1.08 g

71.

72.

73.

96500 C of electricity liberates from CuSO4 solution (a) 63.5g of Cu

(b) 31.75 g of Cu

(c) 96500 g of Cu

(d) 100 g of Cu

74.

A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is (a) + 1

(b) + 2

(c) + 3

(d) + 4

The charge required to deposit 9 g of Al from Al3+ solution is (At. wt. of Al = 27.0) (a) 3216.3 C

(b) 96500 C

(c) 9650 C

(d) 32163 C

The quantity of electricity needed to deposit 127.08 g of copper is (a) 1 Faraday

(b) 4 Coulombs

(c) 4 Faraday

(d) 1 Ampere

By passing 0.1 Faraday of electricity through fused sodium chloride, the amount of chlorine liberated is (a) 35.45 g

(b) 70.9 g

(c) 3.545 g

(d) 17.77 g

(a) 40.65 g

(b) 4.065 g

Silver is monovalent and has atomic mass of 108. Copper is divalent and has an atomic mass of 63.6. The same electric current is passed for the same length of time through a silver coulometer and a copper coulometer. If 27.0 g of silver is deposited, then the corresponding amount of copper deposited is

(c) 0.4065 g

(d) 65.04 g

(a) 63.60 g

(b) 31.80 g

(c) 15.90 g

(d) 7.95 g

A 5 ampere current is passed through a solution of zinc sulphate for 40 minutes. Find the amount of zinc deposited at the cathode

75.

196

ELECTROCHEMISTRY 76.

The electrolysis of silver nitrate solution is carried out using silver electrodes. Which of the following reaction occurs at the anode? +

(a) Ag  Ag + e +

82.

Which of the following reaction occurs at the anode during the charging of lead storage battery ? (a) Pb2+ + 2e– (b) Pb2+ + SO 24

–

(b) Ag + e  Ag +

(c) Pb

–

–

(d) PbSO4 + 2H2O

–

(d) 4 OH  2H2 + O2 + 4e

78.

During electrolysis of a concentrated aqueous solution of NaCl, what is the product at cathode? (a) Na

(b) Cl2

(c) O2

(d) H2

79.

(a) lead dioxide dissolves (c) lead electrode becomes coated with lead sulphate (d) the concentration of sulphuric acid decreases 84.

+

Which of the following material is not present in mercury cell ?

(b) Reduction of Na ion

(a) HgO

(b) KOH

(c) Reduction of H2O

(d) Oxidation of H2O

(c) Zinc

(d) HgCl2.

Which one of the following reactions occurs at the anode when CuSO 4 solution is electrolysed using platinum electrodes? 2+

85.

86.

–

(b) 2SO 24   2H 2 O  2H 2SO 4  O 2  4e  +

–

(c) 2H2O  O2 + 4H + 4e (d) SO

2 4

 SO 2  O 2  2e

87. 

Which one of the following reactions takes place at the anode when an aqueous solution of CuSO4 is electrolysed using copper electrodes? 2+

(a) Cu  Cu + 2e (b) SO

2 4

88.

 SO 2  O 2  2e





(d) 2H 2 O  O 2  4H  4e

89.

(a) Leclamche cell

(b) Electrolytic cell

(c) Mercury cell

(d) Daniell cell

Which of the following material is not present in a dry cell ? (a) MnO2

(b) NH4Cl

(c) ZnCl2

(d) KCl.

Which of the following cell is a secondary cell ? (a) Mercury cell

(b) Ni cell

(c) Dry cell

(d) Fuel cell.

When a lead storage battery is charged, it acts as (a) a primary cell

(b) an electrolytic cell

(c) a galvanic cell

(d) a concentration cell.

(a) Copper chloride in water (b) NaCl in water (c) Ferric chloride in water

The electroplating with chromium is undertaken because (a) Electrolysis of chromium is easier (b) Chromium can form alloys with other metals (c) Chromium gives protective and decorative coating to the base metal



The passage of current liberates H2 at cathode and Cl2 at anode. The solution is

(d) AuCl3 in water.

Which cell has a constant voltage throughout its life ?

–

(c) 2SO 24   2H 2 O  2H 2SO 4  O 2  4e 

81.

As lead storage battery is charged

(a) Oxidation of Cl ion

(a) Cu  Cu + 2e

80.

PbO2 + 4H+ + SO 24 + 2e–

(b) sulphuric acid is regenerated

During the electrolysis of aqueous sodium chloride, the cathodic reaction is –

83.

PbSO4

Pb2+ + 2e–

(c) 2H2O  4H + O2 + 4e

77.

Pb

–

(d) Of the high reactivity of metallic chromium 90.

Prevention of corrosion of iron by Zn coating is called (a) Galvanization

(b) Cathodic protection

(c) Electrolysis

(d) Photoelectrolysis

ELECTROCHEMISTRY 91.

197

For the cell :2+

2Ag + Pt

2Ag + F2

Multiple Choice Questions +

2Ag + Pt Eº = 0.4 volt +

2Ag + 2F

–

Eº = 2.07 volt

If the potential for the reaction Pt assigned zero. Then

92.

93.

Co(s) | CoCl2 (M1) HCl (M2) | H2(g) (p1)| Pt(s)

Pt2+ + 2e– is

(a) EAg /Ag   0.4V

(b) EAg  /Ag  0.4V

(c) EF2 /F   1.67 V

(d) EF2 /F   2.74 V

Which of the following changes will increase the emf of cell ?

(a) Increase volume of CoCl2 solution from 500 ml to 1000 ml (b) increase M2 from 0.1 to 0.5 M (c) Decrease pressure of H2(g) from 0.2 to 0.1 atm

The variation of ^m of acetic acid with concentration is correctly represented by

(d) increase mass of cobalt electrode 94.

During the working of the cell, with the passage of time (a) spontaniety of the cell reaction decreases, E cell decreases

(a)

(b) Q decreases, Ecell increases (c) Wuseful increases (d) At equilibrium Q = Kc, Ecell = 0 95.

Which one of the following statements is incorrect regarding an electrochemical cell ? (a) The electrode on which oxidation takes place is called anode.

(b)

(b) Anode is a negative pole (c) The direction of current is same as that of flow of electrons (d) The flow of current is partly due to flow of electrons and partly due to flow of ions. (c)

96.

A calomel electrode is represented as Hg, Hg2Cl2,KCl. If in such a half cell (a) reduction takes place then Cl – ion concentration increases (b) oxidation takes place then Cl – ion concentration decreases

(d)

(c) the electrode reaction may be represented as Hg2Cl2(s) + 2e–

2Hg (l ) + 2Cl– (aq)

(d) the electrode reaction taking place is Hg2Cl2(s)

Hg 22 (aq) + 2Cl– (aq)

198

ELECTROCHEMISTRY 97.

Given that E oNi 2  / Ni  0.25 V,

E oCu 2  / Cu  0.34 V,

E oAg  / Ag  0.80 V,

E oZn 2  / Zn  0.76 V

102. Iron tanks are protected from rusting by connecting them with magnesium wire. Which of the following statements (s) is/are correct ? (a) Mg acts as anode and iron acts as cathode (b) Moist soil acts as electrolyte

Which of the following reactions under standard conditions will not take place in the specified directions ? (a) Ni2+ (aq) + Cu (s)  Ni (s) + Cu2+ (aq) (b) Cu (s) + 2Ag+ (aq)  Cu2+ (aq) + 2Ag (s) +

2+

(c) Cu (s) + 2H (aq)  Cu

Faraday’s law of electrolysis are not related to the (a) atomic number of cation (b) atomic number of anion (c) equivalent weight of the cation as well as anion (d) speed of cation

99.

(d) Corrosion prevention is spontaneous phenomenon Assertion and Reason (A) If both ASSERTION and REASON are true and reason

(aq) + H2 (g)

(d) Zn (s) + 2H+ (aq)  Zn2+ (aq) + H2(g) 98.

(c) Corrosion prevention is electrochemical phenomenon

If 9 g of H2O is electrolysed completely with 50% current efficiency (a) 1F of electricity will be needed (b) 2F of electricity is needed (c) 5.6 L of O2 at STP will be formed (d) 11.2 L of O2 will be formed at STP.

100. At an anode in an electrolytic cell where electrolysis is taking place, which of the following processes must occur ? (a) Oxidation (b) Loss of electrons by anions (c) Formation of cations by anode (d) Electron density is higher 101. On passing electricity through an aqueous solution of copper sulphate using copper electrodes then (a) copper is deposited at cathode (b) copper is dissolved at anode (c) O2 is liberated at anode (d) the concentration of the solution does not change

is the correct explanation of the assertion. (B)

If both ASSERTION and REASON are true but reason is not the correct explanation of the assertion. (C) If ASSERTION is true but REASON is false. (D) If ASSERTION is false but REASON is true. 103. Assertion (A) : If standard reduction potential for the reaction

Ag+ + e–  Ag is 0.80 volts then for the reaction 3 Ag+ + 3 e–  3 Ag Eº = 2.4V Reason (R) : If concentration is increased, reduction electrode potential is increased. (a) A (b) B (c) C (d) D 104. Assertion (A) : We cannot add the electrode potentials in order to get the cell potential if number of moles of electrons exchanged are not same. Reason (R) : Because the potentials are non thermodynamic properties. (a) A

(b) B

(c) C

(d) D

105. Assertion (A) : Increasing the concentration increases the value of conductance. Reason (R) : Increasing the concentration increases interionic forces of attraction. (a) A (b) B (c) C (d) D 106. Assertion (A) : The correct order of equivalent conductance at infinite dilution is KCl > NaCl > LiCl Reason (R) : KCl is stronger electrolyte than NaCl which is stronger than LiCl. (a) A (b) B (c) C (d) D

199

ELECTROCHEMISTRY 107. Assertion (A) : The voltage of mercury cell remains constant for longer period of time.

Comprehension Electrochemical series is a series of elements arranged in increasing order of their reduction potential.

Reason (R) : It is because net cell reaction does not involve any ion. (a) A

(b) B

(c) C

(d) D

E oH  / H  0. The metals above H have –ve reduction 2 potential, they are more reactive than hydrogen whereas metals below hydrogen are less reactive than H. Reduction potential of metal depends upon (i) sublimation energy (ii) ionisation energy and (iii) hydration energy of ions.

Comprehension For the reaction Zn(s) + Cu 2+ (aq) + Zn 2+ (aq)

Cu(s)

111. Which of the following is best oxidising agent ?

2

Reaction Quotient =

[ Zn ] [Cu 2 ]

, variation of E with cell

(a) Cu2+

(b) Na+

(c) Ag+

(d) Al3+

112. Which of the following is weakest reducing agent among alkali metals in aqueous ?

Q is given by (where Q = concentration quotient) A

(a) Na

(b) K

(c) Rb

(d) Cs

113. Which of the following cannot evolve H2 from dil acid ? Ecell

(a) Pt

(b) Zn

(c) Mg

(d) Pb

Match the Column O

log Q

114.

OA = 1.10 volts, hence 108. When Ecell is 1.1591 volts. It implies, 2

(a)

[Cu ] 2

[ Zn ]

2

 0.01

(b)

2

(c)

[ Zn ] 2

[Cu ]

[ Zn ] [Cu 2 ]

 0.01

Column I (A) Electrolytic cell

(p) G = – ve

(B) Galvanic cell

(q) G = + ve

(C) Faraday’s First Law

(r) Salt bridge

(D) Faraday’s Second Law

(s) m = Z × I × t

2

 0.1

(d)

[ Zn ] [Cu 2  ]

1

W1 W2 (t) E  E 1 2

109. The G for the process will be –ve if, (a)

(c)

[Cu 2 ] [ Zn 2 ] [ Zn 2 ] [Cu 2 ]

 10

 10

2

(b)

(d)

[ Zn 2 ] [Cu 2 ] [ Zn 2  ] [Cu 2 ]

115.

 10

3

 10

5

[Cu 2  ]

Column I (A) Charge on one mole

Column II (p) 1 F

of electrons

110. When E cell is 1.1591 and concentration ratio is,

[ Zn 2 ]

Column II

 10  2 it implies,

(B) 108g of silver deposited

(q) 96500 C

at electrode from (C) 22.4L of hydrogen

(r) 2 F

at STP collected from

(a) T = 273 ºC

(b) T = 298 ºC

(c) T = 298 K

(d) T = 300 K

(D) 8g of oxygen collected from

(s) 5.6 L at STP

200

ELECTROCHEMISTRY

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTION Objective Questions (Only one correct option) 1.

For the following cell,

5.

The half cell reactions for rusting of iron are :

1   0 Catnode 2 H  aq   2e  O2  g   H 2 O  l  ; E  1.23V 2

(2017)

Anode  Fe s   Feaq2   2e E 0  0.44V

Zn(s) | ZnSO 4 (aq) || CuSO 4 (aq) | Cu(s)

The Gº (in kJ) for the reaction is

When the concentration of Zn 2 is 10 times the concentration 2 of Cu , the expression for G (in J mol–1) is

[F is Faraday constant; R is gas contant; T is temperature; Eº (cell) = 1.1 V]

2.

(b) 1.1 F

(c) 2.303 RT – 2.2F

(d) – 2.2 F

For the following electrochemical cells at 298K, 4

2

Pt(s)|H2(g,1bar)|H (aq,1M)||M (aq), M (aq)|Pt(s)

Ecell

7.

(c) 1

(d) 2

Zn | Zn2+ (a = 0.1 M) || Fe2+ (a = 0.01 M) | Fe.

(a) 100.32/0.059

(b) 100.32/0.0295

(c) 100.26/0.0295

(d) 100.32/0.295

In the electrolytic cell, flow of electrons is from (2003)

(c) cathode to anode through internal supply

RT  0.059V F (2016)

(b) –1

(d) – 176

(b) cathode to anode through external supply

The value of x is (a) –2

(c) – 122

(a) cathode to anode in solution

[M2 (aq)]  0.092V when 4  10X. [M (aq)]

0 Given : EM4 /M2  0.151V; 2.303

(b) – 322

The emf of the above cell is 0.2905V. Equilibrium constant for the cell reaction is (2004)

(a) 2.303 RT + 1.1F



3.

6.

(a) – 76

Consider the following cell reaction,

(d) anode to cathode through internal supply. 8.

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reactions and their standard potentials are given below :

MnO 4 (aq)  8H  (aq)  5e    Mn 2  (aq)  4H 2 O(l ),

2Fe(s)  O 2 (g)  4H  (aq)   2Fe 2 (aq)  2H 2O(l ), E o  1.67 V

E o  1.51 V Cr2 O 72  (aq)  14H  (aq)  6e    2Cr 3 (aq)  7H 2 O(l ),

At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH =3, the cell potential at 25ºC is (2011)

4.

(2005)

(a) 1.47 V

(b) 1.77 V

(c) 1.87 V

(d) 1.57 V

E o  1.38 V 2+ o Fe3+ (aq) + e–   Fe (aq) E = 1.77V – o Cl2 (g) + 2e–   2Cl (aq) E = 1.40 V

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required

Identify the incorrect statement regarding the quantitative estimation of aqueous Fe (NO3)2 (2002)

to liberate 0.01 mol of H2 gas at the cathode is

(a) MnO 4 can be used in aqueous HCl

(1 Faraday = 96500 C mol–1)

(2008)

(a) 9.65 × 104sec

(b) 19.3 × 104sec

(c) 28.95 × 104sec

(d) 38.6 × 104sec

(b) Cr2 O72  can be used in aqueous HCl (c) MnO 4 can be used in aqueous H2SO4 (d) Cr2 O 72  can be used in aqueous H2SO4

201

ELECTROCHEMISTRY 9.

Saturated solution of KNO3 is used to make ‘salt-bridge’ because (2001)

Ag( NH3 ) 2  e 

(a) velocity of K+ is greater than that of NO3

[Use 2.303 ×

(b) velocity of NO3 is greater than that of K+ +

13.

12.

+

2Ag + C6H12O6 + H2O  2Ag (s) + C6H12O7 + 2H+

(d) KNO3 is highly soluble in water

Find ln K of this reaction

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is (2001)

(a) 66.13

(b) 58.38

(c) 28.30

(d) 46.29

(a) LiCl > NaCl > KCl

(b) KCl > NaCl > LiCl

(c) NaCl > KCl > LiCl

(d) LiCl > KCl > NaCl

14.

Objective Questions II (One or more than one correct option) 11.

RT F = 0.0592 and = 38.92 at 298 K] F RT (2006)

 3

(c) velocities of both K and NO are nearly the same 10.

Ag (s) + 2NH3; E ored = 0.337V

When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much ? (a) Eoxi will increases by a factor of 0.65 from E ooxi (b) Eoxi will decrease by a factor of 0.65 from E ooxi

In a galvanic cell, the salt bridge (2014) (a) does not participate chemically in the cell reaction. (b) Stops the diffusion of ions from one electrode to another (c) in necessary for the occurrence of the cell reaction. (d) ensures mixing of the two electrolytic solutions. For the reduction of NO3 ion in an aqueous solution Eº is + 0.96V. Values of Eº for some metal ions are given below

(c) Ered will increase by a factor of 0.65 from E ored (d) Ered will decrease by a factor of 0.65 from E ored 15.

Ammonia is always added in this reaction. Which of the following must be incorrect ? (a) NH3 combines with Ag+ to form a complex

V2+ (aq) + 2e–   V

Eº = – 1.19V

(b) Ag(NH3)+2 is a stronger oxidising reagent than Ag+

Fe3+ (aq) + 3e–   Fe

Eº = – 0.04 V

(c) In absence of NH3 silver salt of gluconic acid is formed (d) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode

Au3+ (aq) + 3e–   Au Eº = + 1.40 V Hg2+ (aq) + 2e–   Hg Eº = + 0.86V

Comprehension

The pair (s) of metals that is (are) oxidized by NO3 in

(2007)

(2009)

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules

Tollen’s reagent is used for the detection of aldehydes. When a solution of AgNO3 is added to glucose with NH4OH, then gluconic acid is formed. (2006)

(approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.

aqueous solution is (are) (a) V and Hg (c) Fe and Au

(b) Hg and Fe (d) Fe and V

Comprehension Based Questions Comprehension

Ag+ + e–  Ag; E ored Alignment = 0.80 V C6H12O6 + H2O  C6 H12 O7 + 2H+ + 2e–; gluconic acid

E ooxidation = – 0.05 V

A 4.0 M aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200; 1 Faraday = 96500 coulombs)

ELECTROCHEMISTRY 16.

17.

18.

202

The total number of moles of chlorine gas evolved is (a) 0.5 (b) 1.0 (c) 2.0 (d) 3.0 If the cathode is a Hg electrode, the maximum weight (in g) of amalgam formed from this solution is (a) 200

(b) 225

(c) 400

(d) 446

(d) Fe3[Fe(CN)6]3 (2010)

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

(b) 48250

M(s) | M  (aq; 0.05 molar) || M  (aq;1 molar) M (s)

(c) 96500

(d) 193000

For the above electrolytic cell the magnitude of the cell potential |Ecell | = 70 mV. (2010) (2007)

I2 + 2e– – –

22.

23.

Eº = 0.54 –

Cl2 + 2e  2Cl

Eº = 1.36

Mn3+ + e–  Mn2+

Eº = 1.50

Fe3+ + e–  Fe2+

Eº = 0.77

O2 + 4H+ + 4e–  2H2O

Eº = 1.23

(a) E cell  0; G  0

(b) E cell  0; G  0

(c) E cell  0; Gº  0

(d) E cell  0; Gº  0

If the 0.05 molar solution of M+ is replaced by a 0.0025

(a) 35 mV

(b) 70 mV

(c) 140 mV

(d) 700 mV

Comprehension

(2012)

The electrochemical cell shown below is a concentration cell. M|M2+ (saturated solution of a sparingly soluble salt, MX2) | |M2+ (0.001 mol dm–3)|M.

Among the following, identify the correct statement

The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.

(b) Fe2+ is oxidised by iodine (c) iodide ion is oxidised by chlorine (d) Mn2+ is oxidised by chlorine

For the above cell

molar M+ solution, then the magnitude of the cell potential would be

(a) chloride ion is oxidised by O2

While Fe3+ is stable, Mn3+ is not stable in acid solution because

The solubility product (Ksp ; mol3 dm–9) of MX2 at 298 based on the information available the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(a) O2 oxidises Mn2+ to Mn3+

(a) 1 × 10–15

(b) 4 × 10–15

(b) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+

(c) 1 × 10–12

(d) 4 × 10–12

(c) Fe3+ oxidises H2O to O2 3+

(d) Mn 21.

(c) Fe4 [Fe(CN)6]2

(a) 24125

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (Eº) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their Eº (V with respect to normal hydrogen electrode) values.

20.

(b) Fe3[Fe(CN)6]2

Comprehension

The total charge (coulombs) required for complete electrolysis is

Comprehension

19.

(a) Fe4[Fe(CN)6]3

oxidises H2O to O2

Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and H2SO4 in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of

24.

25.

The value of G (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1) (a) – 5.7

(b) 5.7

(c) 11.4

(d) – 11.4

ELECTROCHEMISTRY

203

Subjective Types Questions 26.

27.

31.

Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 min. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with. (2000)

Ag  (aq)  Cl  (aq)   AgCl (s)

Given

The following electrochemical cell has been set-up : Pt (1) | Fe3+, Fe2+ (a = 1) | Ce4+, Ce3+ (a = 1) | Pt (2)

G of (Cl) 

– 129 kJ/mol

G of (Ag  )

77 kJ/mol

(b) 6.539 × 10–2 g of metallic Zn (u = 65.39) was added to 100 mL of saturated solution of AgCl. Calculate

The standard potential of the following cell is 0.23 V at 15ºC and 0.21 V at 35ºC.

log10

[Zn 2  ] . Given that [Ag  ]2

Pt | H2(g) | HCl (aq) | AgCl (s) | Ag (s)

Ag+ + e–   Ag

Eº = 0.80V

(i) Write the cell reaction.

Zn2+ + 2e–   Zn

Eº = – 76V

(ii) Calculate Hº and Sº for the cell reaction by assuming that these quantitites remain unchanged in the range 15ºC to 35ºC.

Also find how many moles of Ag will be formed ?

(iii) Calculate the solubility of AgCl in water at 25ºC.

Integer Types Questions 32.

Given : The standard reduction potential of the (Ag+ (aq)/Ag (s) is 0.80 V at 25ºC. (2001) (a) Will pH value of water be same at temperature 25ºC and 4ºC. Justify in not more than 2 or 3 sentences. (b) Two students use same stock solution of ZnSO4 and a solution of CuSO4. The emf of one cell is 0.03V higher than the other. The conc of CuSO4 in the cell with higher emf value is 0.5 M. Find out the conc of CuSO4 in the other cell. Given : 2.303 RT/F = 0.06V. (2003) 30.

– 109 kJ/mol

Calculate Eº of the cell. Find log10 Ksp of AgCl. (2005)

If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current, will the current increases or decreases with time ? (2000)

29.

G of (AgCl)

Represent the above reaction in form of a cell.

Eº (Fe3+, Fe2+) = 0.77 V and Eº (Ce4+, Ce3+) = 1.61 V

28.

(a) Calculate G or of the following reaction :

Find the equilibrium constant for the reaction 2+

Cu

2+

+ In

3+

Given

E oCu 2 / Cu   0.15V, E oI n 2 / In   0.4V, E oIn3 / In   0.42 V

o 3 2 1 Given  (Ag  )  6  10 Sm mol ,

 o(Br  )  8  10 3 Sm 2 mol1 ,  o( NO )  7  103 Sm 2 mol 1. 3

33.

For the electrochemical cell,

(2018)

Mg(s) | Mg 2 (aq, 1M) || Cu 2 (aq, 1M) | Cu(s)

the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is ............. .

 Cu + In  +

We have taken a saturated solution of AgBr, Ksp is 12 × 10–14. If 10–7 M of AgNO3 are added to 1L of this solution, find conductivity (specific conductance) of this solution in terms of 10–7 Sm–1 units. (2006)

(given, (2004)

F = 11500 K V–1, where F is the Faraday constant R

and R is the gas constant, ln (10) = 2.30)

ELECTROCHEMISTRY

204 Objective Questions (Only one correct option)

Subjective Types Questions 34.

Consider an electrochemical cell :

(2018)

A(s) | A n  (aq, 2M) || B2n  (aq, 1M) | B(s).

The value of H  for the cell reaction is twice of G  at 300 K. If the emf of the cell is zero, the S (in J K–1

35.

Molar conductivity ( m ) of aqueous solution of sodium stearate, which behaves as a strong electrolyte is recorded at varying concentrations (c) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution? (critical micelle concentration (CMC) is marked with an arrow in the figures) (2019/Shift-1)

mol–1) of the cell reaction per mole of B formed at 300 K is ............ . (Given : ln(2) = 0.7, R (universal gas constant)

(a)

(b)

(c)

(d)

= 8.3 J K–1 mol–1. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

205

ELECTROCHEMISTRY

Note:

ELECTROCHEMISTRY Please share your valuable feedback by scanning the QR code.

206

SURFACE CHEMISTRY

05 SURFACE CHEMISTRY

207

Chapter 05

SURFACE CHEMISTRY

2.1 Feature of Adsortion

1. GENERAL INTRODUCTION Surface Chemistry is that branch of chemistry which deals with the study of the phenomena occurring at the surface or interface, i.e., at the boundary separating two bulk phases. The two bulk phases can be pure compounds or solutions. The interface is represented by putting a hyphen or a slash between the two bulk phases involved, e.g., solid-liquid or solid/liquid. No interface exists between gases as they are completely miscible. Important phenomena occur at the interface, e.g., dissolution, crystallisation, corrosion, heterogeneous catalysis, electrode processes, etc.

Adsorbate and Adsorption The substance which gets adsorbed on any surface is called adsorbate for example, if a gas gets adsorbed on to the surface of a solid, then the gas is termed as the adsorbate. The substance on the surface of which adsorption takes place is called adsorbent. Adsorbent may be a solid or a liquid. Metal powders, powdered charcoal, animal charcoal silica powder etc. are commonly used as adsorbents.

2. ADSORPTION AND ABSORPTION

Adsorbate (gas or solute)

Adsorbent (solid)

Definition of Adsorption The phenomenon of attracting and retaining the molecules

Adsorbent and adsorbate

of a substance on the surface of a liquid or solid resulting in to higher concentration of the molecules on the surface is

Desorption

called adsorption. Adsorption of gases at metal surface is

The removal of the adsorbed substance from a surface is called desorption. This can be done by heating or reducing the pressure of the system.

called occlusion

× × × × × ×× × × × × × × × × × Adsorbent × × (solid) ×× × × × × × × × × ×× × × × ×

Removal of adsorbate

Desorption

Sorption

Definition of Absorption When the molecules of a substance are uniformly distributed throughout the body of a solid or liquid, this phenomenon is called absorption. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

Substance molecules Body of solid (liquid)

The phenomenon in which adsorption and absorption occur simultaneously is called sorption. Dyes are absorbed as well absorbed in cotton fibre. Adsorption is instantaneous i.e. a fast process while absorption is a slow process. 2.2 Difference between Adsorption and Absorption Main points of difference between adsorption and absorption are given below.

Absorption

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SURFACE CHEMISTRY DIFFERENCE BETWEEN ADSORPTION AND ABSORPTION [TABLE -1] Adsorption 1. 2.

3. 4.

Absorption

It is a surface phenomenon. In it, the substance is only retained on the surface and does not go into the bulk or interior of the solid or liquid. In it the concentration of the adsorbed molecules is always greater at the free phase. It is rapid in the begining and slows down to attain equilibrium

1. It concerns with the whole mass of the absorbent. 2. It implies that a substance is uniformly distributed, through the body of the solid or liquid. 3. In it the concentration is low. 4. It occurs at the uniform rate that is slow rate that is slow rate

Examples (i) (ii) (v) (vi)

Examples

Water vapours are adsorbed by CaCl2. (i) Water vapours are absorbed by anhydrous silica gel. NH3 is adsorbed by charcoal. (ii) NH3 is absorbed in water forming NH4OH. Decolourisation of sugar solution by activated or animal charcoal. Ink is adsorbed by blotting paper.

2.3 Mechanism of adsorption

Thermodynamics of Adsorption

Only the surface atoms of an adsorbent play an activrole in

Adsorption is spontaneous process, so for the process G is negative. According to Gibbs equation. G = H – TS For adsorption S (changes in entropy) is always negative becuase adsorption of molecules on the surface lowers the disorber. So for value of G to be negative, H must necessarily be negative and. H|S|. The process is exothermic because it involves forces of attraction between adsorbate and adsorbent. As the process of adsorption proceeds further H because less and less after certain time period. When H becomes equal TS,G = 0 and system attains equilibrium. At equilibrium, Rate of adsorption = Rate of desorption

adsorption. These atoms possess unbalanced forces of various types such as, Vander Waal’s forces and chemical bond forces (free valencies). Thus, due to residual or unbalanced inward forces of attraction or free valancies at the surface, liquids and solids have the property to attract and retain the molecules of a gas or a dissolved substance on to their surface.

(a)

(b)

Ni

Ni

Ni

Ni

2.5 Types of adsorption

(c)

Molecules at the surface experiencing a net inward force of attraction in case of



(a) liquid (b) solid (c) metal with free valencies 2.4 Thermodynamic of adsorption

 Adsorption refers to the existence of a higher concentration of any particular component at the surface of a liquid or a solid phase.

 Adsorption is invariably accompanied by evolution of heat,



i.e. it is an exothermic process. In other words, ΔH of adsorption is always negative. When a gas is adsorbed, the freedom of movement of its molecules becomes restricted. On account of it, decrease in the entropy of the gas after adsorption, i.e. ΔS is negative.



Adsorption can be classified into two categories as described below. Physical adsorption : If the forces of attraction existing between adsorbate and adsorbent are Vander Waal’s forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or Vander Waal’s adsorption. It can be easily reversed by heating or decreasing the pressure. Chemical adsorption : If the forces of attraction existing between adsorbate particles and adsorbent are almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also called as chemisorption or Langmuir adsorption. This type of adsorption cannot be easily reversed.

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SURFACE CHEMISTRY COMPARISON BETWEEN PHYSISORPTION AND CHEMISORPTION [Table -2]

1.

Physisorption

Chemisorption

(Vander Waal’s adsorption)

(Langmuir adsorption)

Low heat of adsorption usually in range

1.

High heat of adsorption in the range of

of 20-40 kJ/mol

50-400 kJ/mol.

2.

Force of attraction are Vander Waal’s forces.

2.

Forces of attraction are Chemical bond forces.

3.

It is reversible.

3.

It is irreversible.

4.

It usually takes place at low

4.

It takes place at high temperature.

5.

It is not related.

temperature and decreases with increasing temperature. 5.

It is related to the case of liquefication of the gas.

6.

It forms multimolecular layers.

6.

It forms monomolecular layers.

7.

It does not require any activation

7.

It requires high activation energy.

8.

High pressure is favourable. Decrease

energy. 8.

High pressure is favourable. Decrease of pressure causes desorption.

9.

of pressure does not cause desorption

It is not very specific.

9.

It is highly specific.

Note : Due to formation of multilayers physical adsorption decreases after some times. Chemisorption and physisorption both are exothermic. 2.5.1 Factors which affect the extent of adsorption on solid surface The following are the factors which affect the adsorption of gases on solid surface.  Nature of the adsorbate (gas) and adsorbent (solid)





Porous and finely powdered solid e.g. charcoal, fullers earth, adsorb more as compared to the hard non-porous materials. Due to this property powdered charcoal is used in gas masks used in coal mines. Gases with high critical temperature are adsorbed at higher extent as compared to gases with lower critical temperatures.

Volumes of gases at N.T.P., adsorbed by 1g of charcoal at 288 k

Gas

H2

N2

CO

CH4

CO2

HCl

NH3

SO2

Volume adsorbed (mL)

4.7

8.0

9.3

16.2

48

72

181

380

Critical temp (K)

33

126

134

190

304

324

406

430

Critical temperature increases  Ease of liquefaction increases  Adsorption increases 

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210

SURFACE CHEMISTRY  Surface area of the solid adsorbent : The extent of adsorption depends directly upon the surface area of the adsorbent, i.e. larger the surface area of the adsorbent, greater is the extent of adsorption.

 Freundlich adsorption isotherm : Freundlich adsorption isotherm is obeyed by the adsorptions where the adsorbate forms a monomolecular layer on the surface of the adsorbent.

 Effect of pressure on the adsorbate gas :

 An increase in the pressure of the adsorbate gas increases the extent of adsorption.

 At low temperature, the extent of adsorption increases rapidly with pressure.

 At low pressure, the extent of adsorption is found to be directly proportional to the pressure.

 At high pressure (closer to the saturation vapour pressure of the gas), the adsorption tends to achieve a limiting value.

1 x  kP n m

log

or

x 1  log k  log P m n

where, x is the weight of the gas adsorbed by m gm of the adsorbent at a pressure P, thus x/m represents the amount of gas adsorbed by the adsorbent per gm (unit mass), k and n are constant at a particular temperature and for a particular adsorbent and adsorbate (gas), n is always greater than one, indicating that the amount of the gas adsorbed does not increase as rapidly as the pressure. At low pressure, the extent of adsorption varies linearly with pressure

1 x  P1 ;  1 n m

 At high pressure, it becomes independent of pressure 1 x  P0 ;  0 n m

x depends upon pressure raised to m

 At moderate pressure powers

 Effect of temperature : As adsorption is accompanied by evolution of heat, so according to the Le-Chatelier’s principle, the magnitude of adsorption should decrease with rise in temperature.

Note :The amount of heat when one mole of the gas is adsorbed on the adsorbent is called the heat of adsorption. 2.6 Adsorption Isotherms A mathematical equation which describes the relationship between pressure (p) of the gaseous adsorbate and the extend of adsorption at any fixed temperature is called adsorption isotherms. Thus, if x g of an adsorbate is adsorbed on m g of the adsorbent, then Extent of adsorption 

x m

x P m

1 n

2.7 Adsorption Isobar A graph drawn between the amount of the gas adsorbed per gram of the adsorbent (x/m) and temperature ‘t’ at a constant equilibrium pressure of adsorbate gas is known as adsorption isobar.

P - Constant

P - Constant x/m

x/m

Temperature

Temperature

Physical adsorption

Chemical adsorption

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211

SURFACE CHEMISTRY The physical adsorption isobar shows a decrease in x/m throughout with rise in temperature, the chemisorption isobar

 Concentration of the solute in the solution. 2.10 Positive and Negative Adsorption

shows an initial increase with temperature and then the expected decrease. The initial increase is because of the fact

In case of adsorption by solids from the solutions, mostly

that the heat supplied acts as activation energy required in

the solute is adsorbed on the surface of the solid adsorbent

chemisorption (like chemical reactions).

so that concentration of solute on the surface of the adsrobent is greater than in the bulk. However in some cases.

2.8 Adsorption from Solutions

the solvent from the solution may be adsorbed by the

Solid surfaces can also adsorb solutes from solutions. A few

adsorbent so that the concentration of the solution increases

examples are :

as compared to initial concentration.

 When litmus solution is shaken with charcoal, it becomes

When the concentration of the adsorbate is more on the

colour less because the dye of the litmus solution is adsorbed

surface of the adsorbent than in the bulk, it is called positive adsorption.

by charcol.

If the concentration of the adsorbate increases in the bulk

 When the colourless Mg(OH)2 is precipitated in the presence

after adsorption, it is called negative adsorption.

of magneson reagent (a blue coloured dye). it acquires blue Blood solution + Conc. KCl solution  Positive adsorption

colour because the dye is adsorbed on the solid precipitate.

Blood solution + dilute KCl solution  Negative adsorption

 The extent of adsorption from solution depends upon the

concentration of the solute in the solution, and can be

2.11 Applications of Adsorption The phenomenon of adsorption finds a number of

expressed by the Freundlich isotherm.  The Freundlich adsorption isotherm for the adsorption from

x = kc1/ n where, x is the mass of the solute m adsorbed, m is the mass of the solid adsorbent, c is the solution is,

applications. Important applications are given as follows.



Production of high vacuum : A bulk of charcoal cooled in liquid air is connected to a vessel which has already been exhausted as for as possible by a vacuum pump. The

equilibrium concentration of the solute in the solution, n is a

remaining traces of air are adsorbed by the charcoal to

constant having value greater than one. k is the

produce a very high vacuum.

proportionality constant, (The value of k depends upon the



In Gas masks : It is a device which consists of activated

nature of solid, its particle size, temperature, and the nature

charcoal or a mixture of adsorbents. This apparatus is used to

of solute and solvent etc.)

adsorb poisonous gases (e.g. Cl2 , CO, oxide of sulphur etc.)

 The plot of x/m against c is similar to that Freundlich

adsorption isotherm. The above equations may be written x 1 = logk + log c where c, is in the following form, log m n the equilibrium concentration of the solute in the solution. 2.9 Factors affecting adsorption from solution The adsorption from solutions by solid adsorbents is found to depend upon the following factors :

 Nature of the adsorbate and the adsorbent.  The adsorption decreases with temperature.  Greater the surface area of the adsorbent greater is the adsorption.

and thus purify the air for breathing is coal mines.

 For desiccation or dehumidification : Certain substances have a strong tendency to absorb water such as silica and alumina (Al2O3). These substances can be used to reduce/ remove water vapours or moisture present in the air. Silica gel is also used for dehumidification in electronic equipment.

 Removal of colouring matter from solution : Animal charcoal removes colours of solutions by adsorbing coloured impurities. It is also used as decolouriser in the manufacture of cane sugar.

 Heterogeneous catalysis : Mostly heterogeneous catalytic reactions proceed through the adsorption of gaseous reactants on solid catalyst. For example, SCAN CODE SURFACE CHEMISTRY

212

SURFACE CHEMISTRY (a) Finely powdered nickel is used for the hydrogenation of

 Chromatographic analysis : The phenomenon of adsorption

oils.

has given an excellent technique of analysis known as chromatographic analysis. The technique finds a number of applications in analytical and industrial fields. Chromatographic technique is based on differential adsorption of different constituents of a mixture.

(b) Finely divided vanadium pentaoxide (V2O5) is used in the contact process for the manufacture of sulphuric acid. (c) Pt, Pd are used in many industrial processes as catalyst. (d) Manufacture of ammonia using iron as a catalyst.

 Separation of inert gases : Due to the difference in degree

 In dyeing : Many dyes get adsorbed on the cloth either directly or by the use of mordants. 3. CATALYSIS

of adsorption of gases by charcoal, a mixture of inert gases can be separated by adsorption on coconut charcoal at different low temperatures.

“Catalyst is a substance which speeds up and speeds down a chemical reaction without itself being used up at the end of the reaction and the phenomenon is known as catalysis. 3.1 Types of catalysis

 Softening of hard water : The hard water is made to pass through a column packed with zeolite (sodium aluminium silicate)

 De-ionisation of water : For softening water can be de-ionised by removing all dissolved salts with the help of cation and anion-exchanger resin.

Catalytic reactions can be broadly divided into the following types, (a) Homogeneous catalysis : When the reactants and the catalyst are in the same phase (i.e. solid, liquid or gas). The catalysis is said to be homogeneous. The following are some of the examples of homogeneous catalysis.

 Oxidation of sulphur dioxide into sulphur trioxide with oxygen in the presence of oxides of nitrogen as the catalyst in the lead chamber process.

 In curing diseases : A number of drugs are adsorbed on the germs and kill them or these are adsorbed on the tissues and

NO(g) 2SO2 (g)  O2 (g)   2SO3 (g)

heat them.

 Cleaning agents : Soap and detergents get adsorbed on the interface and thus reduce the surface tension between dirt and cloth, subsequently the dirt is removed from the cloth.

 Froth flotation process :  A low grade sulphide ore is concentrated by separating it

from silica and other earthy matter by this method.  The finely divided ore is added to water containing pine oil

and foaming agent.  The air is bubbled through the mixture.  The foam formed rises to the surface on which mineral

particles wetted with oil are adsorbed while earthy matter settle down at the bottom.

 In adsorption indicators : Surface of certain precipitates such as silver halide, have the property of adsorbing some dyes like eosin, fluorescein etc. In this case of precipitation titrations (for example AgNO 3 Versus NaCl) the indicator is adsorbed at the end point producing a characteristic colour on the precipitate.



The reactants, products and catalyst all are in gaseous state i.e. same phase. Hydrolysis of methyl acetate is catalysed by H+ ions furnished by hydrochloric acid . HCl 

  CH 3COOCH 3     H 2 O     

 Hydrolysis of sugar is catalysed by H  ions furnished by sulphuric acid. 

2 4  C12 H 22 O11     H 2 O       H SO

Sucrose solution 

C6 H12 O6  Glucose

    C2 H12 O6   

solution 

(b) Heterogeneous catalysis : The catalytic process in which the reactants and the catalyst are in different phases is known as heterogeneous catalysis. Some of the examples of heterogeneous catalysis are given below.

 Oxidation of sulphur dioxide into sulphur trioxide in the presence of platinum metal or vanadium pentaoxide as catalyst in the contact process for the manufacture of sulphuric acid. The reactants are in gaseous state while the Pt(s) catalyst is in solid state. SO2 (g)  O2 (g)   2SO3 (g)

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SURFACE CHEMISTRY  Combination between nitrogen and hydrogen to form

 When nitric acid is poured on copper, the reaction is very

ammonia in the presence of finely divided iron in Haber’s process.

slow in the beginning, gradually the reaction becomes faster due to the formation of nitrous acid during the reaction which acts as an auto-catalyst.

Fe(s) N 2 (g)  3H 2 (g)   2NH3 (g)

 Oxidation of ammonia into nitric oxide in the presence of platinum gauze as a catalyst in Ostwald’s process. Pt(s) 4NH3 (g)  5O2 (g)   4NO(g)  6H 2 (g)

 Hydrogenation of vegetable oils in the presence of finely divided nickel as catalyst. Ni(s) Vegetable oils (l) + H2 (g)   Vegetable Ghee (g)

 In hydrolysis of ethyl acetate, acetic acid and ethyl alcohol are formed. The reaction is initially very slow but gradually its rate increases. This is due to formation of acetic acid which acts as an auto-catalyst in this reaction. (f) Induced catalysis : When one reaction influences the rate of other reaction, which does not occur under ordinary conditions, the phenomenon is known as induced catalysis. Some examples are as follows,

 The reduction of mercuric chloride (HgCl2) with oxalic acid

(c) Positive catalysis : When the rate of the reaction is accelerated by the foreign substance, it is said to be a positive catalyst and phenomenon as positive catalysis. Some examples of positive catalysis are given below.

is very slow, but potassium permanganate is reduced readily with oxalic acid. If, however, oxalic acid is added to a mixture

 Decomposition of H 2 O 2 in presence of colloidal platinum.

thus, induces the reduction of mercuric chloride.

Pt

2 H 2 O 2 (l)  2 H 2 O(l)  O 2 (g) (d) Negative catalysis : There are certain, substance which, when added to the reaction mixture, retard the reaction rate instead of increasing it. These are called negative catalyst or inhibitors and the phenomenon is known as negative catalysis. Some examples are as follows.

 The oxidation of chloroform by air is retarded if some alcohol is added to it.

of potassium permanganate and HgCl2 both are reduced simultaneously. The reduction of potassium permanganate,

(g) Acid-base catalysis : According to the Arrhenius and Ostwald H+ or OH– ions act as catalysts.

 For example, Hydrolysis of an ester, CH3COOC2H5 (l) + H2O (l) 

H or   CH3COOH (l) + C2H5OH (l) OH 

 Inversion of cane sugar, +

H C12 H 22 O11 (l) +H 2 O  C6 H12 O6  l  + C6 H12 O6  l 

Alcohol(l) 2CHCl3 (l)  O2 (g)   2COCl2 (g)  2HCl(g)

(e) Auto-catalysis : In certain reactions, one of the product acts as a catalyst. In the initial stages the reaction is slow but as soon as the products come into existence the reaction rate increases. This type of phenomenon is known as autocatalysis. Some examples are as follows, I

 The rate of oxidation of oxalic acid by acidified potassium permanganate increases as the reaction progresses. This acceleration is due to the presence of Mn 2  ions which are formed during reaction. Thus Mn 2  ions act as auto-catalyst.

5 H 2 C2 O4  2KMnO4  3H 2SO4   2 MnSO 4  K 2SO 4  10CO 2  8H 2 O

Sugar

Fructose

Glucose

 Conversion of acetone into diacetone alcohol, CH3COCH3 (l) + CH3COCH3 (l)



OH  

CH3COCH2. C(CH3)2OH (l)

 Decomposition of nitramide, NH2NO2 (l)

OH    N2O (g) + H2O (l)

3.2 Characteristics of catalysis The following are the characteristics which are common to most of catalytic reactions.

 A catalyst remains unchanged in mass and chemical composition at the end of the reaction. SCAN CODE SURFACE CHEMISTRY

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SURFACE CHEMISTRY  The catalyst can not initiate the reaction: The function of the catalyst is to alter the speed of the reaction rather than to start it.

 The catalyst is generally specific in nature: A substance, which acts as a catalyst for a particular reaction, fails to catalyse the other reaction , different catalysts for the same reactant may for different products.

 The catalyst can not change the position of equilibrium: The catalyst catalyse both forward and backward reactions to the same extent in a reversible reaction and thus have no effect on the equilibrium constant.

 Catalytic promoters : Substances which themselves are not

(b) Adsorption theory of Heterogeneous Catalysis :

catalysts, but when mixed in small quantities with the

Heterogeneous catalytic reactions generally proceed via

catalysts increase their efficiency are called as promoters or

adsorption of reactants on the surface of the catalyst.

activators.

Mechanism of such surface reactions may be explained in

 Catalytic poisons : Substances which destroy the activity

terms of diffusion theory of catalysis. This theory postulates

of the catalyst by their presence are known as catalytic

the following sequence for gaseous reactions on a solid

poisons.

surface.

 Change of temperature alters the rate of catalytic reaction as it does for the same reaction in absence of catalyst : By increasing the temperature, there is an increase in the catalytic power of a catalyst but after a certain temperature its power begins to decrease. A catalyst has thus, a particular temperature at which its catalytic activity is maximum. This temperature is termed as optimum temperature.

Step: (i) Diffusion of the reactants to the surface. Step: (ii) Adsorption of the reactant molecules onto the surface. Step: (iii) Actual chemical reaction on the surface. Step: (iv) Desorption of the products from the surface. Step: (v) Diffusion of the products away from the surface. In generally, Step (iii) determines the rate of reaction. However step (ii) and (iv) may be rate determining.

A positive catalyst lowers the activation energy 3.3 Theories of Catalysis (a) Collision Theory of Homogeneous Catalysis : According to the collision theory, a reaction occurs on account of effective collisions between the reacting molecules. For effective collision, it is necessary that the molecules must possess a minimum amount of energy known as activation

Mechanism for the surface reactions

energy (Ea). After the collision molecules form an activated complex which dissociate to yield the product molecules. The catalyst provides a new pathway involving lower amount of activation energy. Thus,

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SURFACE CHEMISTRY (c) Advantages of modern adsorption theory : This theory can explain the following :

 Small quantity of the catalyst is sufficient because the catalyst is regenerated again and again.

 The catalyst takes part in the reaction but is produced back unchanged in mass and chemical composition at the end of the reaction.

 Catalytic poisons, if present, are preferentially adsorbed on the catalyst surface, thereby hindering the adsorption of the reactant molecules. Hence, they slow down the activity of the catalyst. This theory, however, cannot explain satisfactorily the action of catalytic promoters. 3.4 Important Features of Solid Catalysis (a)

Activity : Activity is the ability of catalysts to accelerate chemical reactions. The degree of acceleration can be as high as 10 10 times in certain reactions. For example reaction between H 2 and O 2 to form H 2 O in presence of platinum as catalyst takes place with explosive violence. In absence of catalyst, H2 and O2 can be stored indefinitely without any reaction.

(a) The sodalite (or b-) cage, (b) zeolite- A

(b) Selectivity : Is the ability of catalysts to direct reaction to yield particular products (excluding other). CH 3

Pt Example : (i) n-heptane   toluene

(c)

O BiMoO4  CH 2 = CHCH (ii) CH3 CH = CH 2  Acrolein

3.5 Zeolites (shape selective catalysis) Zeolites are alumino–silicates of the general formula, Mn [AlO2]x.(SiO2)y .mH2O, where, M may be simple cation like Na  ,

K  or Ca 2  , n is the charge on the simple cation, m is the number of molecules of water of crystallization. They have honey comb like structure called  cage. The reactions catalyzed by zeolites depend upon the shape and size of the reactant and product molecules. That is why these types of reactions are called shape selective-catalysis.

, (c) Faujasite. Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerization. An important zeolite catalyst used in the petroleum industry is ZSM–5 (Zeolite Sieve of Molecular Porosity 5). It converts alcohols directly into gasoline (petrol) by first dehydrating them so that a mixture of hydrocarbons is formed. ZSM 5 Alcohols  Dehydration  Hydrocarbons

hydrated zeolites are used as ‘ion-exchangers’ in softening of hard water.

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SURFACE CHEMISTRY 3.6 Catalysis used in Industries SOME INDUSTRIALCATALYTIC PROCESSES Process

Haber's process for the manufacture of ammonia Ostwald's process for the manufacture of nitric acid

Reactions with catalyst and other conditions Finely divided iron (catalyst ) N2 (g)  3 H2 (g)   2 NH3 (g) Molydenum (Pr omoter ) 200 bar pressure 723  773 K temp.

Platinised 4 NH3 (g)  5 O2 (g)   4 NO (g)  6 H2O (g) asbestos (catalyst )

573K 2 NO (g)  O2 (g)   2 NO2 (g)

4 NO2 (g)  2 H2O (l )  O2 (g)  4 HNO3 (l )

Contact process for the manufacture of sulphuric acid

Platinised asbestos   2 SO2 (g)  O2 (g)   2 SO3 (g) or V2O5 (catalyst )

 H O(l )

673 723K 2 SO3 (g)  H2SO4 (l )   H2S2O7 (l )   2H2SO4 (aq) Oleum

Bosch's process for the manufacture of hydrogen.

Fe2O3 (Catalyst) CO  H2  H2O(g)   CO2 (g)  2 H2 (g)  Cr2O3 (promoter )   673 873K

Deacon's process for the manufacture of chlorine

2 4 HCl (g)  O2 (g)  773K  2H2O (l )  2 Cl 2 (g)

Size

CO (g)  2 H2 (g)

Hydrogenation of vegetable oils

Finely divided Ni Oil (l )  H2 (g)    Vanaspati ghee (s) 423  473 K

Water gas

CuCl

ZnO  Cr2 O3

4. ENZYMES

 

Enzymes are complex nitrogenous substances secreted by low forms of vegetables & organisms. Enzymes are actually protein molecules of higher molecular mass. (ranging from 15,000 – 1,000,000 g/mol) Enzymes form colloidal solutions in water and are very effective catalysts. They catalyse numerous reactions, especially those connected with natural processes. Numerous reactions occur in the bodies of animals and plants to maintain the life process. These reactions are catalysed by enzymes. The enzymes are thus, termed as bio-chemical catalysts and the phenomenon is known as bio-chemical catalysis. eg. Nitrogenase is an enzyme present in bacteria on the root nodules of leguminous plants such as peas and beans, catalyses the conversion of atmospheric N2 to NH3. In the human body, the enzyme carbonic anhydrase catalyses the reaction of CO2 with H2O, CO2(aq) + H2O (l)  H+ (aq.) + HCO3 (aq.) The forward reaction occurs when the blood takes up CO2 in the tissues, and the reverse reaction occurs when the blood releases CO2 in lungs.

200 bar 432 K

CH3OH (l )

High pressure

4.1 Mechanism of enzyme catalysis

Thus, the enzyme catalysed reactions take place in two steps as follows : Step 1. Formation of enzyme-substrate complex.  ES E  S (Fast and Reversible)  Enzyme Substrate Enzyme  Substrate complex

Step 2. Dissociation of enzyme-substrate complex to form the products.     EP 

ES Enzyme

Substrate complex

 

Enzyme  Pr oduct

E Enzyme

 P Product

association

 Slow and Rate     determining 

 Re generated  The rate of formation of product depends upon the concentration of ES. SCAN CODE SURFACE CHEMISTRY

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SURFACE CHEMISTRY 5. COLLOIDAL STATE

Suspension

The colloidal state depends on the particle size. It is regarded as intermediate state between true solution and suspension. True solution In true solutions the size of the particles of solute is very small and thus, these can not be detected by any optical means and freely diffuse through membranes. It is a homogenous system.

The size of particles is large and, thus it can be seen by naked eye and do not pass through filter paper. It is a heterogeneous system. The size of different solutions are sometimes expressed in other units also as given below : True solutions

Colloids

Suspensions

10–7 m

< 1nm

1 nm –1000 nm > 1000 nm 1 nm = 10–9m

< 10 Å

10 Å – 10000 Å > 10000 Å 1 Å = 10–10m

100 nm

Size < 1 nm

Relation

1000 pm – 105 pm

> 105 pm

1 pm = 10–12m

Three types solutions

CHARACTERISTICS OF TRUE SOLUTIONS, COLLOIDAL SOLUTIONS AND SUSPENSIONS S.No Property

True Solutions

1. 2.

Homogeneous Less than 10–9 m or 1 nm (i.e., < 10 Å) Pass through ordinary filter paper as well as animal membrane.

3.

Nature Particle size (diameters) Filtrability

4. 5.

Settling Visibility

6. 7.

Diffusion Appearance

Colloidal Solutions

Heterogeneous Between 10–9 to 10–6m or 1 nm to 1000 nm Pass through ordinary filter paper but not through animal membrane. Do not settle. Do not settle. Particles are Scattering of light by invisible. the particles is observed under ultraDiffuse quickly. Diffuse slowly. Clear and Transluscent. transparent

Dispersed phase and Dispersion Medium

Suspensions Heterogeneous More than 10–6m or 1000 nm (i.e., > 10000Å Do not pass through filter paper and animal membrane. Settle on standing. Particles are visible to naked eye or under a microscope microscope. Do not diffuse. Opaque.

5.1 Classification of Colloids

Disper sed phase : (Discontinuous phase) : It is the

(a) Classification based on the physical state of the dispersed

component present in small proportion and is just like a

phase and dispersion medium Depending upon the physical

solute in a solution. For example in the colloidal solution of

state of dispersed phase and dispersion medium whether

silver in water (silver acts as a dispersed phase)

these are solids, liquids or gases, eight types of colloidal

Dispersion medium : (Continuous phase) : It is generally

systems are possible.

the component present in excess and is just like a solvent in a solution. For example, in the colloidal solution of silver in water. Water acts as a dispersion medium. SCAN CODE SURFACE CHEMISTRY

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SURFACE CHEMISTRY DIFFERENT TYPES OF COLLOIDAL SYSTEMS Sr. No

Dispersed phase

Dispersion

Colloidal System

Examples

1.

Liquid

Gas

Aerosol of liquids

Fogs, clouds, mists, fine insecticide sprays

2.

Solid

Gas

Aerosol of solids

Smoke, volcanic dust, haze Foam or froth Soap lather.

3.

Gas

Liquid

Foam or froth

4.

Liquid

Liquid

Emulsions

5.

Solid

Liquid

Sols

Lemonade froth, foam, whipped cream, soda water Milk, emulsified oils, medicines Most paints, starch in water, proteins, gold sol, arsenic sulphide sol, ink

6.

Gas

Solid

Solid foam

7.

Liquid

Solid

Gels

8.

Solid

Solid

Solid sols (coloured glass)

(b) Classification based on Nature of interaction between dispersed

Pumice stone, styrene rubber, foam rubber Cheese, butter, boot polish, jelly, curd Ruby glass, some gem stones and alloys



phase and dispersion medium : 

Lyophilic colloids (water loving) : “The colloidal solutions in which the particles of the dispersed phase have a great affinity for the dispersion medium, are called lyophilic colloids.”



Lyophobic colloids (water hating) : “The colloidal solutions in which there is no affinity between particles of the dispersed phase and the dispersion medium are called lyophobic colloids.” Lyophobic colloids (water hating) : “The colloidal solutions in which there is no affinity between particles of the dispersed phase and the dispersion medium are called lyophobic colloids.”

DISTINCTION BETWEEN LYOPHILIC AND LYOPHOBIC SOLS Sr. Property No 1. Surface tension 2. Viscosity 3. Reversibility 4. Stability 5. Visibility 6. Migration

7. Action of electrolyte 8. Hydration Examples

Lyophilic (suspensiod)

Lyophobic Sols (Emulsoid)

Lower than that of the medium Much higher than that of the medium Reversible More stable Particles can’t be detected even under ultramicroscope Particles may migrate in either direction or do not migrate in an electric field because do not carry any charge. Addition of smaller quantity of electrolyte has little effect Extensive hydration takes place Gum, gelatin, starch, proteins, rubber etc.

Same as that of the medium Same as that of the medium Irreversible Less stable Particles can be detected under ultramicroscope. Particles migrate either towards cathode or anode in an electric field because they carry charge. Coagulation takes place No hydration Metals like Ag and Au, hydroxides like Al(OH3), Fe(OH)3, metal sulphides

like As2S3 etc.

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SURFACE CHEMISTRY

These macromolecules forming the dispersed phase are generally polymers having very high molecular masses.

(c) Classification based on types of particles of dispersed phase : Depending upon the type of the particles of the dispersed phase, the colloids are classified as follows.

Naturally occurring macromolecules are starch, cellulose, proteins, enzymes, gelatin etc. Artificial macromolecules are synthetic polymers such as nylon, polythene, plastics, polystyrene etc. Their solutions are quite stable and resemble with true solution in many respects. They have usually lyophobic character. The molecules are flexible and can take any shape.

 Multimolecular colloids : When on dissolution, atoms or smaller molecules of substances (having diameter less than 1nm) aggregate together to form particles of colloidal dimensions,the particles thus formed are called multimolecular colloids. In these sols the dispersed phase consists of aggregates of atoms or molecules with molecular size less than 1 nm. For

 Associated colloids : These are the substances which

big size molecules (called macromolecules) which on

dissolved in a medium behave as normal electrolytes at low concentration but behave, as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregates particles thus formed are called micelles. Their molecules contain both lyophilic and lyophobic groups. The colloidal behaviour of such substances is due

dissolution form size in the colloidal range. Such substances

to the formation of aggregates or clusters in solutions. Such

are called macromolecular colloids.

aggregated particles are called micelles.

example, sols of gold atoms and sulphur (S 8 ) molecules. In these colloids, the particles are held together by Vander Waal’s forces. They have usually lyophilic character.

 Macromolecular colloids : These are the substances having

DIFFERENCE BETWEEN DIFFERENT TYPES OF COLLOIDS Multimolecular colloids

Macromolecular colloids

Associated colloids

They are molecules of large

They are formed by of

aggregation of a large number

size, e.g., polymers like rubber,

aggergation of a large

of atoms or molecules which

nylon, starch, proteins, etc.

no. of ions in concen-

1. They are formed by the

generally have diameters less

trated solution e.g., soap

than 1 nm, e.g., sols of gold,

sol.

sulphur, etc. 2. Their molecular masses are

They have high molecular

Their molecular masses

masses.

are generally high.

Due to long chain, the van der

Higher is the concen-

held together by weak van der

Waals forces holding them are

tration, greater are the

Waals forces.

comparatively stronger.

van der Waals forces.

not very high 3. Their atoms or molecules are

5.1.1 Micelles Micelles are the clusters or aggregated particles formed by association of colloid in solution. The common examples of micelles are soaps and detergents. The formation of micelles takes place above a particular temperature called Kraft temperature (T k) and above a particular concentration called critical micelle concentration (CMC). They are capable of forming ions. Micelles may contain as many as 100 molecules or more.

 

For example sodium stearate (C17H35 COONa) is a typical example of such type of molecules. When sodium stearate is dissolved in water, it gives Na+ and C17H35COO– ions. C 17 H 35 COO Na  C 17 H 35 COO   Na  Sodium stearate



Stearate ion

The stearate ions associate to form ionic micelles of colloidal size. It has long hydrocarbon part of C17H35 radical. Which is lyophobic and COO– part which is lyophilic.

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SURFACE CHEMISTRY List of surfactants and their critical micelle concentration (CMC) Surfactant Anionic CH 3(CH 2) 6COONa CH 3(CH 2) 10COONa CH 3(CH 2) 7OSO 3Na CH 3(CH 2) 11 OSO 3Na CH 3(CH 2) 5C 6H 4SO 3Na CH 3(CH 2) 11C 6H 4SO 3Na Cationic CH3(CH2)9NH2,HCl CH3(CH2)11NH2HCl CH 3(CH 2) 7N(CH 3) 3Br CH 3(CH 2) 11N(CH 3) 3Br Non-ionic CH 3(CH 2) 7 C 6H 11 O 6 C 12 H 2 O 9 (C 16 H 31 O 2) 2 CH 3(CH 2) 10COOC12H21O 10

CMC(g/1)

Temp.(°C)

6.5 × 10 1 5.6 3.0 × 10 1 2.6 9.8 4.0 × 10 –1

20 20 25 25 75 50

8.5 2.7 7.8 × 10 1 5.4

25 30 – 50 25 25

7.3 1.1 × 10 –2 7.1 × 10 –3

25 20 50

– 70 – 50 – 60 – 75

When oil droplet comes in contact with soap solution, the stearate ions arrange themselves around it in such a way that hydrophobic parts of the stearate ions are in the oil (or grease) and the hydrophilc. As hydrophilic part is polar, these polar groups can interact with the water molecules present around the oil droplet. As a result, the oil droplet is pulled away from the surface of the cloth into water to form ionic micelle which is then washed away with the excess of water. SCAN CODE SURFACE CHEMISTRY

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SURFACE CHEMISTRY

(a) Grease or oil on surface of cloth (b) Stearate ions arranged around the oil droplet (c) Ionic micelle formed surrounded by sheath of negative charge. In fact, the stearate ions of soap molecules help in making a



By reduction : A number of metals such as silver, gold and

stable emulsion of oil with water which is washed away with

platinum, have been obtained in colloidal state by treating

the excess of water. It may be noted that a sheath of negative

the aqueous solution of their salts, with a suitable reducing

charge is formed around the oil globule.

agent such as formaldehyde, phenyl hydrazine, hydrogen peroxide, stannous chloride etc.

Similarly, in case of detergents, e.g., sodium lauryl sulphate,  3

+

2AuCl3  3SnCl 2  3SnCl4  2Au

 3

viz, CH3(CH2)11 OSO Na , the polar group is OSO alongwith the long hydrocarbon chain. It is an example of an

GoldSol

2AuCl3  3HCHO  3H 2 O  2Au  3HCOOH  6HCl GoldSol

anionic detergent as anions associate together to form an ionic micelle, similar to that of soap. A well known example of

The gold sol, thus prepared, has a purple colour and is called

a cationic detergent forming an associated colloid is that of

purple of cassius.

cetyl trimethyl , CH3(CH2)15(CH3)3N+Br–. 5.2 General Methods of preparation of Colloids



By hydrolysis : Many salt solutions are rapidly hydrolysed by boiling dilute solutions of their salts. For example, ferric

Lyophilic and lyophobic colloidal solutions (or sols) are

hydroxide and aluminium hydroxide sols are obtained by

generally prepared by different types of methods. Some of

boiling solutions of the corresponding chlorides.

the common methods are as follows.

FeCl3  3H 2 O  Fe(OH)3  3HCl

(a) Preparation of Lyophilic colloids :

 The lyophilic colloids have strong affinity between particles of dispersed phase and dispersion medium.These colloidal solutions are readily formed by simply mixing the dispersed phase and dispersion medium under ordinary conditions.

 For example, the substance like gelatin, gum, starch, egg,

Colloidial Sol

Similarly silicic acid sol is obtained by the hydrolysis of sodium silicate.

 By double decomposition : A sol of arsenic sulphide is obtained by passing hydrogen sulphide through a cold solution of arsenious oxide in water.

As 2 O3 +3H 2S  As2S3 +3H 2 O

albumin etc. pass readily into water to give colloidal solution.

 Mechanical dispersion : In this method, the substance is

They are reversible in nature because these can be

first ground to coarse particles. It is then mixed with the

precipitated and directly converted into colloidal state.

dispersion medium to get a suspension. The suspension is

(b) Preparation of Lyophobic colloids :

then grinded in colloidal mill. It consists of two metallic discs

 By oxidation : A colloidal solution of sulphur can be obtained

nearly touching each other and rotating in opposite

by bubbling oxygen (or any other oxidising agent like

directions at a very high speed about 7000 revolution per

HNO3 ,Br etc.) through a solution of hydrogen sulphide in water.

minute. The space between the discs of the mill is so adjusted that coarse suspension is subjected to great shearing force giving rise to particles of colloidal size. Colloidal solutions

2H 2S  O 2 ( or any other oxidising agent)  2H 2 O  2S SCAN CODE SURFACE CHEMISTRY

222

SURFACE CHEMISTRY of black ink, paints, varnishes, dyes etc. are obtained by this

Important peptizing agents are sugar, gum, gelatin and electrolytes. Freshly prepared ferric hydroxide can be

method.

converted into colloidal state by shaking it with water containing Fe3+ or OH– ions, viz. FeCl3 or NH 4 OH respectively.. 3

Fe(OH)3  FeCl3   Fe  OH 3 Fe   3Cl – Pr ecipitat electrolyte Colloidal sol



By electrical dispersion or Bredig’s arc method : This method is used to prepare sols of platinum, silver, copper

5.3 Purification of colloidal solution

or gold.

The colloidal solutions prepared by the above methods

The metal whose sol is to be prepared is made as two

usually contain impurities especially electrolytes which can

electrodes which are immersed in dispersion medium such

destabilize the sols. These impurities must be eliminated to

as water etc. The dispersion medium is kept cooled by ice.

make the colloidal solutions stable. The following methods

An electric arc is struck between the electrodes. The

are commonly used for the purification of colloidal solutions.

tremendous heat generates by this method and give colloidal

 Dialysis :The process of separating the particles of colloid

solution. The colloidal solution prepared is stabilised by

from those of crystalloid, by means of diffusion through a

adding a small amount of KOH to it.

suitable membrane is called dialysis. It’s principle is based upon the fact that colloidal particles

Metal electrodes

can not pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it.

Arc Water

ice

Bredig's arc method

Note: (1) This method is not suitable when the dispersion medium

The colloidal solution is taken in a bag (parchment

is an organic liquid as consider occurs.

paper).The bag is suspended in fresh water. The impurities

(2) This method comprises both dispersion and condensation.

slowly diffused out of the bag leaving behind pure colloidal

 By Peptization : The process of converting a freshly

solution .

prepared precipitate into colloidal form by the addition of

The distilled water is changed frequently to avoid

suitable electrolyte is called peptization. The electrolyte is

accumulation of the crystalloids otherwise they may start

used for this purpose is called peptizing agent or stabilizing

diffusing back into the bag. Dialysis can be used for

agent. Cause of peptisation is the adsorption of the ions of

removing HCl from the ferric hydroxide sol.

the electrolyte by the particles of the precipitate.

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223

SURFACE CHEMISTRY 

Electrodialysis :The ordinary process of dialysis is slow.

(a) Physical properties :

To increase the pace of purification, the dialysis is carried

 Heterogeneous nature : Colloidal sols are heterogeneous in

out by applying electric field. This process is called

nature. They consists of two phases; the Dispersed Phase and the Dispersion Medium.

electrodialysis. Kidneys in the human body act as dialysers to purify blood which is colloidal in nature.

 Stable nature : The colloidal solutions are quite stable. Their particles are in a state of motion and do not settle down at the bottom of the container.

The important application of dialysis process in the artificial kidney machine used for the purification of blood of the



patients whose kidneys have failed to work. The artificial kidney machine works on the principle of dialysis.

Filterability : Colloidal particles are readily passed through the ordinary filter papers. However they can be retained by special filters known as ultrafilters (parchment paper).

(b) Colligative properties :

 Due to formation of associated molecules, observed values of colligative properties like relative decrease in vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure are smaller than expected.



For a given colloidal sol the number of particles will be very small as compared to the true solution.

(c) Mechanical properties :



Brownian movement : The colloidal particles are moving at random in a zig – zag motion. This type of motion is called Brownian movement.

 Ultra – filtration : Sol particles directly pass through ordinary filter paper because their pores are larger (more than 1 or 1000 m  ) than the size of sol particles (less than 200 m  ). If the pores of the ordinary filter paper are made smaller by soaking the filter paper in a solution of gelatin or colloidion and subsequently hardened by soaking in formaldehyde, the treated filter paper may retain colloidal particles and allow the true solution particles to escape. Such filter paper

The reason of brownian motion is that the molecules of the

is known as ultra - filter and the process of separating colloids

dispersion medium are constantly colloiding with the

by using ultra – filters is known as ultra – filtration.

particles of the dispersed phase.

 Ultra – centrifugation : The sol particles are prevented from setting out under the action of gravity by kinetic impacts of the molecules of the medium. The setting force can be



Diffusion : The sol particles diffuse from higher concentration to lower concentration region. However, due to bigger size, they diffuse at a lesser speed.

enhanced by using high speed centrifugal machines having

 Sedimentation : The colloidal particles settle down under

15,000 or more revolutions per minute. Such machines are

the influence of gravity at a very slow rate. This phenomenon

known as ultra–centrifuges.

is used for determining the molecular mass of the macromolecules.

5.4 Properties of colloidal solutions

(d) Optical properties

The main characteristic properties of colloidal solutions are

Tyndall effect :

as follows.

When light passes through a sol, its path becomes visible because of scattering of light by particles. It is called Tyndall SCAN CODE SURFACE CHEMISTRY

224

SURFACE CHEMISTRY effect. This phenomenon was studied for the first time by

when a small quantity of silver nitrate (AgNO3) solution is

Tyndall. The illuminated path of the beam is called Tyndall

added to a large quantity of potassium iodide (KI) solution,

cone. In a true solution, there are no particles of sufficiently large diameter to scatter light and hence no Tyndall effect is observed. The Tyndall effect has also been observed by an instrument called ultra – microscope.

 the colloidal particles of silver iodide adsorb I from the

solution to become negatively charged, (at this stage KI is in excess, and I  being common to AgI)

AgI  (Colloidal particle)

I

(in excess in the medium)

(AgI)I



(Colloidal particle becomes positively charged)

But, when a small quantity of potassium iodide (KI) solution is added to a large quantity of silver nitrate solution (AgNO3) ; the colloidal silver iodide particles adsorb Ag  from the solution to become positively charged, (at this stage AgNO3 is in excess and Ag+ is common to AgI),

AgI  (Colloidal particle)

Ag 



(in excess in the medium)

(AgI)Ag  (Colloidal particle becomes positively charged)

(e) Origin of the charge on colloidal particles The origin of the charge on the sol particles in most cases is due to the preferential adsorption of either positive or negative ions on their surface. The sol particles acquire electrical charge in any one or more of the following ways.

 Due to the dissociation of the surface molecules: Some colloidal particles develop electrical charge due to the dissociation / ionisation of the surface molecules. The charge on the colloidal particles is balanced by the oppositely charged ions in the sol. For example, an aqueous solution of soap (sodium palmitate) which dissociates into ions as, C15 H 31COON  C15 H 31COO  Na . 



Similarly, the ferric hydroxide colloidal particles develop positive charge due to the adsorption of Fe 3  ions from the solution.

Fe(OH)3 + Fe3+  Fe(OH)3 .Fe3+ (Colloidal particle)

Sodium palmitate

The cations (Na+) pass into the solution while the anions

(C15 H31COO ) have a tendency to form aggregates due to



weak attractive forces present in the hydrocarbon chains. Due to frictional electrification : It is believed that the frictional electrification due to the rubbing of the dispersed phase particles with that of dispersion medium results in some charge on the colloidal particles. The dispersion medium must also get some charge, because of the friction. Since it does not carry any charge, the theory does not seem to be correct.



Due to selective adsorption of ions : The particles constituting the dispersed phase adsorb only those ions preferentially which are common with their own lattice ions. For example,

(in exess in the (Colloidal particle of ferric hydroxide get positively charged due to the adsorption medium ) of Fe3+ ions on it)

Ferric hydroxide colloidal particles develop negative charge due to adsorption of either OH  or

Fe(OH)3 +OH–  Fe(OH)3  OH–

(Colloidal particle)

(in excess)

(Negatively charged ferric hydroxide collodial particle)

Fe(OH)3 +Cl –  Fe(OH)3  Cl –

(Collodial particle)

(in excess)

(Negatively charged ferric hydroxide collodial particle)

(f) Electrical properties : Colloidal particles carry an electric charge and the dispersion medium has an opposite and equal charge, the system as a whole being electrically neutral. The presence of equal and similar charges on colloidal particles is largely responsible in giving stability to the system because the mutual forces of repulsion between similarly charged particles prevent them from coalescing and coagulating when they come closer to one another.

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225

SURFACE CHEMISTRY Electrophoresis The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative. 5.5 Stability of sols

Sols are thermodynamically unstable and the dispersed phase (colloidal particles) tend to separate out on long standing due to Vander Waal’s attractive forces. However sals tend to exhibit some stability due to

 Electrical double layer theory :The electrical properties of colloids can also be explained by electrical double layer theory. According to this theory a double layer of ions appear at the surface of solid. The ion preferentially adsorbed is held in fixed part and imparts charge to colloidal particles. The second part consists of a diffused mobile layer of ions. This second layer consists of both the type of charges. The net charge on the second layer is exactly equal to that on the fixed part. The existence of opposite sign on fixed and diffuse parts of double layer leads to appearance of a difference of potential, known as zeta potential or electrokinetic potential. Now when electric field is employed the particles move (electrophoresis)



Stronger repulsive forces between the similarly charged particles : All colloidal particles in any sol possess similar charge. Therefore, due to the electrostatic repulsion these are not able to come closer and form aggregates. Thus stronger repulsive forces between the similarly charged particles in a sol promote its stability.



Particle-solvent interactions :Due to strong particle-solvent (dispersion medium) interactions, the colloidal particles get strongly solvated. Due to solvation, the effective distance between the colloidal particles increases, and therefore, the Vander Waal’s force of attraction decreases. As a result, the particles are not able to form aggregates. Lyophilic sols are mainly stabilized by solvation effects due to strong interactions between the sol particles and the dispersion medium.

5.6 Coagulation or Flocculation or Precipitation “The phenomenon of the precipitation of a colloidal solution by the addition of the excess of an electrolyte is called coagulation or flocculation.”

 By electrophoresis : In electrophoresis the colloidal particles move towards oppositely charged electrode. When these come in contact with the electrode for long these are discharged and precipitated.



Electro-osmosis : In it the movement of the dispersed particles is prevented from moving by semipermeable membrane. Electro-osmosis is a phenomenon in which dispersion medium is allowed to move under the influence of an electrical field, whereas colloidal particles are not allowed to move. The existence of electro-osmosis has suggested that when liquid is forced through a porous material or a capillary tube, a potential difference is setup between the two sides called as streaming potential.

 By mixing two oppositely charged sols : When oppositely charged sals are mixed in almost equal proportions, their charges are neutralised. Both sals may be partially or completely precipitated as the mixing of ferric hydroxide (+ve sal) and arsenious sulphide (–ve sal) brings them in precipitated form. This type of coagulation is called mutual coagulation or meterial coagulation.

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226

SURFACE CHEMISTRY 

By boiling : When a sal is boiled, the adsorbed layer is disturbed due to increased collisions with the molecules of dispersion medium. This reduces the charge on the particles and ultimately they settle down to form a precipitate.



By persistent dialysis : On prolonged dialysis, the traces of the electrolyte present in the sal are removed almost completely and the colloids become unstable.

the valency of the active ion or flocculating ion, greater will be its coagulating power”. Coagulating power of an electrolyte is directly proportional to the valency of the active ions (ions causing coagulation). For example to coagulate negative sol of As 2S3 , the coagulation power of different cations has been found to decrease in the order as, Al3  Mg 2  Na 

 By addition of electrolytes : The particles of the dispersed phase i.e., colloids bear some charge. When an electrolyte is added to sal, the colloidal particles take up ions carrying opposite charge from the electrolyte. As a result, their charge gets neutralised and this causes the uncharged, particles to come closer and to get coagulated or precipitated. For example, if BaCl2 solution is added to As2 S3 sol the Ba2+ ions are attracted by the negatively charged sol particles and their charge gets neutralised. This leads to coagulation.

Similarly, to coagulate a positive sol such as Fe(OH)3, the coagulating power of different anions has been found to decrease in the order :

[Fe(CN)6 ]4  PO34  SO24  Cl 

 Coagulation or flocculation value : “The minimum concentration of an electrolyte which is required to cause the coagulation or flocculation of a sol is known as flocculation value.” or

5.6.1 Hardy schulze rule : The coagulation capacity of different electrolytes is different. It depends upon the valency of the active ion are called flocculating ion, which is the ion carrying charge opposite to the charge on the colloidal particles. Greater

“The number of millimoles of an electrolyte required to bring about the coagulation of one litre of a colloidal solution is called its flocculation value.” Thus , a more efficient flocculating agent shall have lower flocculating value.

FLOCCULATION VALUES OF SOME ELECTROLYTES Sol

Electrolyte

Flocculation Sol Electrolyte

value (mM)

As2S3 (–vely charged)

Flocculation value (mM)

NaCl

51.0

KCl

9.5

KCl

49.5

BaCl2

9.3

CaCl2

0.65

Fe(OH)3

K2SO4

0.20

MgCl2

0.72

(+vely charged)

MgSO4

0.22

MgSO4

0.81

AlCl3

0.093

Al2(SO4)3

0.096

Al(NO3)3

0.095

Note : Coagulating value or flocculating value 

1 . co agu lating p ow er

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227

SURFACE CHEMISTRY  Coagulation of lyophilic sols : There are two factors which

 Gold number : Zsigmondy introduced a term called gold

are responsible for the stability of lyophilic sols. These

number to describe the protective power of different colloids. This is defined as, “weight of the dried protective agent in

factors are the charge and solvation of the colloidal particles.

milligrams, which when added to 10 ml of a standard gold

When these two factors are removed, a lyophilic sol can be

sol (0.0053 to 0.0058%) is just sufficient to prevent a

coagulated. This is done by adding electrolyte and by adding

colour change from red to blue on the addition of 1 ml of 10

suitable solvent. When solvent such as alcohol and acetone

% sodium chloride solution, is equal to the gold number of

are added to hydrophilic sols the dehydration of dispersed

that protective colloid.”

phase occurs. Under this condition a small quantity of electrolyte can bring about coagulation. 5.7 Protection of colloids and Gold number Lyophilic sols are more stable than lyophobic sols. Lyophobic sols can be easily coagulated by the addition of small quantity of an electrolyte. When a lyophilic sol is added to any lyophobic sol, it becomes less sensitive towards electrolytes. Thus, lyophilic colloids can prevent the coagulation of any lyophobic sol. “The phenomenon of preventing the coagulation of a

Thus, smaller is the gold number, higher is the protective

lyophobic sol due to the addition of some lyophilic colloid is called sol protection or protection of colloids.”

action of the protective agent.

The

Protective power 

protecting power of different protective (lyophilic) colloids

1 Gold number

is different. The efficiency of any protective colloid is expressed in terms of gold number.

GOLD NUMBERS OF SOME HYDROPHILLC SUBSTANCES Sr.

Hydrophilic

Gold

Sr. Hydrophilic

Gold

No.

substance

number

No. substance

number

1.

Gelatin

0.005-0.01

5

Sodium oleate

0.4 – 1.0

2.

Sodium caseinate

0.01

6

Gum tragacanth

2

3.

Haemoglobin

0.03-0.07

7

Patato starch

25

4.

Gum arabic

0.15-0.25

The protective colloids play very significant role in stabilisation of the non–aqueous dispersions, such as paints, printing inks etc.



Congo rubin number : Ostwald introduced congo rubin number to account for protective nature of colloids. It is defined as “the amount of protective colloid in milligrams which prevents colour change in 100 ml of 0.01 % congo rubin dye to which 0.16 g equivalent of KCl is added.”

5.8 Application of colloids  Purification of water by alum (coagulation) : Alum which



yield Al 3  ions, is added to water to coagulate the negatively charged clay particles. In rubber and tanning industry (coagulation and mutual coagulation) : Several industrial processes such as rubber plating, chrome tanning, dyeing, lubrication etc are of colloidal nature.

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228

SURFACE CHEMISTRY In rubber plating, the negatively charged particles of rubber (latex) are made to deposit on the wires or handle of various tools by means of electrophoresis. The article on which rubber is to be deposited is made anode. In tanning the positively charged colloidal particles of hides and leather are coagulated by impregnating them in negatively charged tanning materials (present in the barks of trees). Among the tanning agent chromium salts are most commonly used for the coagulation of the hide material and the process is called chrome tanning.



Artificial rains : It is possible to cause artificial rain by throwing the electrified sand or silver iodide from an aeroplane and thus coagulating the mist hanging in air.



Smoke precipitation (Coagulation) : Smoke is a negative sol consisting of carbon particles dispersed in air. Thus, these particles are removed by passing through a chamber provided with highly positively charged metallic knob. Formation of deltas (Coagulation) : River water consists of negatively charged clay particles of colloidal dimension. When the river falls into the sea, the clay particles are









coagulated by the positive Na  , K  , Mg 2  ions etc. present in sea water and new lands called deltas are formed. Clotting of blood : Blood consists of negatively charged colloidal particles (albuminoid substance). The colloidal nature of blood explains why bleeding stops by applying a ferric chloride solution to the wound. Actually ferric chloride solution causes coagulation of blood to form a clot which stops further bleeding. Colloidal medicine : Argyrol and protargyrol are colloidal solution of silver and are used as eye lotions. Colloidal sulphur is used as disinfectant. Colloidal gold, calcium and iron are used as tonics. Photographic plates : These are thin glass plates coated with gelatin containing a fine suspension of silver bromide. The particles of silver bromide are colloidal in nature.

6.1 Types of Emulsion Depending upon the nature of the dispersed phase, the emulsions are classified as; (a) Oil-in-water emulsions (O/W) : The emulsion in which oil is present as the dispersed phase and water as the dispersion medium (continuous phase) is called an oil-in-water emulsion. Milk is an example of the oil-in-water type of emulsion. In milk liquid fat globules are dispersed in water. Other examples are, vanishing cream etc.

(b) Water-in-oil emulsion (W/O) : The emulsion in which water forms the dispersed phase, and the oil acts as the dispersion medium is called a water-in-oil emulsion. These emulsion are also termed oil emulsions. Butter and cold cream are typical examples of this types of emulsions. Other examples are cod liver oil etc. 6.2 Preparation of Emulsions Emulsions are generally prepared by vigorously agitating a mixture of the relevant oil and water by using either a high speed mixer or by using ultrasonic vibrators. The emulsions obtained by simple mechanical stirring are unstable. The two components (oil and water) tend to separate out. To obtain a stable emulsion, a suitable stabilizing substance is generally added. The stabilizing substance is called emulsifier or emulsifying agent. The emulsifier is added along with the oil and water in the beginning. For Examples : substances which can act as emulsifiers are soaps, detergents, long chain sulphonic acid, lyophilic colloids like gelatin, albumin, casein etc.

6. EMULSIONS “Emulsions are the colloidal solutions in which both the dispersed phase and the dispersion medium are liquids.” A good example of an emulsion is milk in which fat globules are dispersed in water. The size of the emulsified globules is generally of the order of 10–6m. Emulsion resemble lyophobic sols in some properties.

Oil droplets

Oil

Oil Water

Oil Oil

Oil

Oil Oil

Oil in water

6.3 Nature of emulsifier

 

Different emulsifiers may act differently in the case of a particular emulsion. For example, Sodium oleate is used to prepare oil-in-water (O/W) emulsions. Magnesium and calcium oleates are used to prepare waterin-oil (W/O) emulsions. When calcium oleate is added to an emulsion stabilized by sodium oleate, the stability of the system decreases. At a certain ratio of Na + ;Ca 2+ , the oil-in-water emulsion becomes unstable. If the Ca 2+ ions concentration is increased further very quickly, then the reversal of the emulsion type occurs, that is the oil-in-water emulsion gets converted into a water-in-oil type. SCAN CODE SURFACE CHEMISTRY

229

SURFACE CHEMISTRY 6.4 Identification of emulsions

6.5 Properties of emulsion



Emulsions show all the characteristic properties of colloidal solution such as Brownian movement, Tyndall effect, electrophoresis etc.



These are coagulated by the addition of electrolytes containing polyvalent metal ions indicating the negative charge on the globules.



The size of the dispersed particles in emulsions in larger than those in the sols. It ranges from 1000 Å to 10,000 Å. However, the size is smaller than the particles in suspensions.



Emulsions can be converted into two separate liquids by heating, centrifuging, freezing etc. This process is also known as demulsification.

Several methods are available to find out whether an emulsion is of the oil-in-water type or of the water-in-oil type emulsion. An emulsion can be identified as follows. (a) Dilution test : Add water to the emulsion. If the emulsion can be diluted with water this means that water acts as the dispersion medium and it is an example of oil-in-water emulsion. In case it is not diluted, then oil acts as dispersion medium and it is an example of water-in-oil emulsion. (b) Dye test : An oil soluble suitable dye is shaken with the emulsion. If colour is noticed on looking at a drop of the

6.6 Applications of emulsions

emulsion, it is oil-in-water type emulsion. In case the entire

 Concentration of ores in metallurgy

background is coloured, it is an example of water-in-oil type.

 In medicine (Emulsion water-in-oil type)

(c) Conductivity test : Add small amount of an electrolyte (e.g. KCl) to the emulsion. If this makes the emulsion electrically conducting , then water is the dispersion medium. If water is not the dispersed phase.

 Cleansing action of soaps.  Milk, which is an important constituent of our diet an emulsion of fat in water.  Digestion of fats in intestine is through emulsification.

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230

SURFACE CHEMISTRY

SOLVED EXAMPLES Adsorption

Sol.

Example – 1 (a) Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy. Still it is a spontaneous process. Explain.

x (a) (i) Extent of adsorption   decreases with increase m in temperature at constant pressure as adsorption is an exothermic process. (ii) Extent of adsorption increases with increase in pressure at constant temperature. At low value of pressure.

(b) How does an increase in temperature affect both physical as well as chemical adsorption ? Sol.

x  P or m

(a) According to the equation

At high pressure

ΔG = ΔH - TΔS

x  P o or m

for a process to be spontaneous G should be negative. Even though G is negative here, G is negative because reaction is highly exothermic i.e., H is negative.

x  KPo m

i.e.,

x K m

In the intermediate range of pressure

x  P1/ n m

(b) On increasing temperature desorption occurs in physical adsorption. Chemical adsorpation increases first and then decreases with increase in temperature.

Example – 2 Consider the adsorption isotherms given below and interpret the variation in the extent of adsorption (x/m) when

x  KP m

or

x  KP1/ n m

(b)

Catalyst : Finely divided iron.

(n > 1)

Promoter: Molybdenum.

Example – 3

(a) (i) temperature increases at constant pressure Explain factors affecting adsorption. (ii) pressure increases at constant temperature. (b) Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia.

Sol.

Adsorption is generally affected by following factors: (i) Temperature : Adsorption is an exothermic process. Hence, according to Le chaterlier’s principle, the rate of adsorption decreases with the increase in temperature and vice versa. Thus, adsorption is inversely proportional to temperature at a constant pressure. In the graphs plotted, showing the relationship between the rate of adsorption and the temperature at a constant pressure, the physical adsorption is favoured at low and chemical adsorption at relatively high temperature.

231

SURFACE CHEMISTRY

(v) Surface area of adsorbent : Since, adsorption is the surface phenomenon, the rate of adsorption is directly proportional to the surface area of the adosrbent. Thus, greater the total surface area of the adsorbent, greater is the adsorption. Therefore, finely divided particles of the adsorbent show greater adsorption due to their large surface area.

(ii) Pressure : The curve shows the effect of temperature on the rate of the adsorption at constant pressure, is called Adsorption Isobar. In case of gases, adsorption increases with increase in the pressure of the gas. However, increase is not uniform. At low pressure the adsorption increases linearly while at moderate pressure it increases exponentially. At high pressure complete surface area of adsorbent is covered and hence adsorption becomes independent of pressure.

(vi) Concentration of adsorbate : The extent of adsorption depends upon the concentration of the absorbate, used in a solution. The rate of adsorption increases with the increase in concentration of solution of the adsorbate, but upto certain limit. After the equilibrium is achieved, further increase in the concentration has no effect on the rate of adsorption.

Example – 4 How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid ? Sol.

The influence is in the following ways : (i) Smaller the size of the particles of the adsorbent, greater is the surface area and greater is the adsorption. (ii) At constant temperature, adsorption first increases with increase of pressure and then attains equilibrium at a high pressure.

(iii) Nature of adsorbent : The gases which are easily liquified and highly soluble in water are easily adsorbed. NH3, HCl, Cl2, SO2, etc are adsorbed readily. These gases have greater Van der Waal’s forces of attraction and hence get easily adsorbed. As molecular weight increases, adsorption increases. e.g. SO2 is adsorbed more than CO2 at the same temperature. (iv) Nature of Adsorbent : Adsorption also depends on the nature of adsorbent. The adsorbent with more free valencies and unbalanced forces adsorb to greater extent. e.g. the transition metals. The same gas can be adsorbed to different extents on different adsorbents. e.g. H2 is adsorbed by different metals in the following order. Pd > Pt > Ni > Fe

(iii) In physical adsorption, it decreases with increase of temperature but in chemisorption, first it increases and then decreases.

Example – 5 1 g of charcoal adsorbs 100 mL 0.5 M CH3COOH to form a monolayer, and therby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.0 × 102 m2/g. Sol.

Number of moles of acetic acid which we present in 100 mL before adding charcoal = 0.05 Number of moles of acetic acid in 100 mL after adding charcoal = 0.049

232

SURFACE CHEMISTRY Number of moles of acetic acid adsorbed on the surface of charcoal = (005 – 0.049) = 0.001

Example – 8 Indicate a chemical reaction involving a homogeneous catalyst.

So, number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 × 6.02 × 1023 = 6.02 × 1020 2

2

 Surface area of charcoal = 3.01 × 10 m

Hence, area occupied by one molecule of acetic acid on the surface of charcoal

3.01  10 2   5  10  19 m 2 , 6.02  10 20

Sol.

NO g

Example – 9 Give an example of a shape selective catalyst. Sol.

So, area = 5 × 10–19 m2.

Zeolites are shape selective catalysts. A zeolite called ZSM-5 converts alcohol to gasoline.

Example – 10

Example – 6 Which has a higher enthalpy of adsorption, physisorption or chemisorption ?

What is meant by ‘shape selective’ catalysis? Sol.

Chemisorption

Sol.

  2SO 2  g   O 2  g    2SO3  g 

Catalyst

Example – 7 Distinguish between Homogenous and Heterogenous catalysis.

Shape selective catalysis is a chemical reaction in which the rate depends on the pore size of the catalyst and also on the shape and size of the reactant and product molecules. Zeolite acts as a shape selective catalyst. For example, ZSM-5 (Zeolite sieve of molecular porosity 5) is used as a shape selective catalyst to produce gasoline of high octane number from methanol.

Example – 11

Sol. No. Homogenous catalysis

No.

Heterogenesis catalysis

1.

The catalyst and the reactants form a single phase.

1.

The catalyst and the reactants are in different phases.

2.

The catalyst dissolves into the gas phase or solution (a reacting mixture).

2.

The catalyst does not dissolve in the reacting mixture.

3.

The reaction occurs in the liquid phase or gas phase.

3.

The reaction does not occur in gas or liquid phase but occurs on the surface of solid catalyst

4.

The catalyst is often involved in the chemical reaction.

4.

The catalyst adsorbs the reactant molecules to form a chemical bond between catalyst and the reactant molecules.

5.

The catalyst cannot be easily separated from the products of reaction.

5.

The catalyst can be easily seperated from reaction products.

6.

The rate of the reaction does 6. not depend on the surface area of catalyst

The rate of reaction is proportional to the surface area of catalyst

7.

The homogeneously catalysed 7. reactions are often little faster than those which are heterogeneously catalysed

The heterogeneously catalysed reactions are often slower than those which are homogeneously catalysed

Sol.

Give the mechanism of enzyme catalysis. An enzyme molecule has one or more active sites at which a specific substrate (reactant) molecule fits. In other words the active site acts like a lock into which fits only a specific key (substrate or reactant). This concept of lock and key was developed by German chemist Emil Fischer in 1894. Now it is known that the structure of active site is flexible and hence it can accomodate more than one type of substrate. The active site returns to its original state after the products are released.

233

SURFACE CHEMISTRY The binding of substrate (reactant) at the active site of the enzyme catalyst produces enzyme – substrate complex which then decomposes to give the product of reaction and the enzyme is recovered.  ES E + S 

Example – 12 What are the physical states of dispersed phase and dispersion medium of froth ? Dispersed phase

: Gas

Dispersion medium: Liquid

Example – 13 Write two differences between sols and emulsions. (i) Sols are dispersions of solids in liquids while emulsions are dispersions of liquids in liquids. (ii) Sols are quite stable whereas emulsions are less stable.

Example – 14 How are the following colloids different from each other in respect of dispersion medium and dispersed phase ? Give one example of each type. (i) An aerosol

(ii) A hydrosol

(iii) An emulsion

Sol.

Example – 15 Sol.

(i) Multimolecular colloids are formed by aggregation of small molecules (diameter < 1 nm) while macromolecular colloids are formed by macromolecules (polymers) and consist of single molecules. (ii) Multimolecular colloids are generally lyophobic whereas macromolecular colloids are generally lyophilic.

Colloids

Sol.

Write two differences between multimolecular colloids and macromolecular colloids. Sol.

 E+P

where, E and S are enzyme and substrate respectively. ES is an enzyme-substrate complex and P is the product.

Sol.

Example – 16

In what way is a sol different from a gel ? Colloidal system in which solid is dispersed in liquid is called sol and that in which liquid is dispersed in solid is called gel.

Example – 17 Classify colloids on the basis of number of molecules or atoms. Sol. On the basis of number of molecules or atoms colloidal systems are classified as Multimolecular and macromolecular colloids. (i) Multimolecular Colloids : (a) Colloidal particles are bigger than molecule of true solutions and smaller than the particles of suspensions. (b) When a substance is dispersed in a suitable dispersion medium, a larger number of atoms or molecules aggregate to form a particle of colloidal size. Example : Au sol, Ag sol and Sulphur sol. Gold sol consists of particles which are formed by aggregation of many gold atoms. (ii) Macromolecular Colloids Certain substances form larger molecules having the dimension of colloidal particles. Such molecules are called macromolecules. Macromolecules have very high molecular masses. Most lyophilic sols are macromolecular colloids. Example : gelatin, starch, nucleic acids. Characteristics of macromolecular colloids : (a) Some of the macromolecules dissolve in water and give homogenous solution. (b) Solution of macromolecules behave like lyophilic colloids. (c) There is considerable interaction between the solute and the solvent in macromolecular sols. (d) Due to the large size and shape of the macromolecules, the macromolecular colloids cause serious deviations from the solutions obeying ordinary laws.

234

SURFACE CHEMISTRY Example – 18

Example – 21

Distinguish between true solution, colloidal solution and suspension.

Give an example of an associated colloid. Sol.

Sol. No. True solution

Colloidal dispersion

Suspension

1.

It is homogenous.

It is heterogeneous.

It is heterogeneous.

2.

The particles are ions or molecules.

The particles are either single macromolecules or aggregates of molecules or atoms.

The particles are agregates of molecules or atoms.

3.

The particle size ranges between 0.1 nm to 1nm.

The particle size lies between 1nm to 103 nm.

The particle size is larger than 103 nm.

4.

The particles do not settle.

The particles do not settle.

The particles do not settle out.

5.

The particles readily pass through parchment.

The particles pass slowly through parachment.

The particles do not pass through parchment.

6.

The particles are invisible even under powerful microscope

The particles are invisible under microscope.

The particle are visible under microscope or sometimes with naked eye.

7.

The particles do not scatter light

The particles scatter light.

It does not show the scatter light.

8.

It does not show electrophoresis, Brownian movement etc.

It shows the properties like Brownian movement, electrophoresis etc.

It does not show the properties electrophoresis, Brownian movement, etc.

Example – 22 What is Kraft temperature ? Sol.

Sol.

Why is a colloidal sol stable ? Sol.

What are lyophilic and lyophobic sols ? Give one example of each type. Which one of these two types of sols is easily coagulated and why? Sol.

Lyophilic sols : Lyophilic sols are those sols in which the particles of dispersed phase have great affinity for the dispersion medium, e.g., sols of gum, gelatine, starch, etc. Lyophobic sols : In this type of sols the particles of dispersed phase have little or no affinity for the dispersion medium, e.g., gold sol, Fe (OH)3 sol, As2S3 sol., etc.

All the particles in a colloidal sol carry the same charge and hence keep on repelling each other and cannot aggregate together to form bigger particles.

Example – 24

The given statement is true. This is because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditons. For example, NaCl in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves like a crystalloid while concentrated solution behaves as a colloid (called associated colloid). It is the size of the particles which matters, i.e., the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1000 nm, it is in the colloidal state.

Example – 20

Kraft temperature is the minimum temperature above which the formulation of micelles takes place.

Example – 23

Example – 19 Comment on the statement that “Colloid is not a substance but state of a substance.”

Soaps and detergents are associated colloids.

What happens in the following activities and why? (i) An electrolyte is added to a hydrated ferric oxide sol in water. (ii) A beam of light is passed through a colloidal solution. (iii) An electric current is passed through a colloidal solution. Sol.

(i) The positively charged colloidal particles of Fe(OH)3 get coagulated by the negatively charged ions provided by electrolyte. (ii) The path of light becomes visible due to scattering of light by colloidal particles (Tyndall effect). (iii) Electrophoresis takes place in which colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

235

SURFACE CHEMISTRY Example – 25

Sol.

Write a note on optical properties of colloids (Tyndall effect). Sol.

(b) Fe (OH) 3 sol is positively charged which is coagulated by negatively charged Cl – present in Sodium Chloride solution.

It was investigated by British physicist John Tyndall. Colloidal system consist of discrete particles of dimensions comparable to the wavelength of the Ultraviolet and visible radiations So, whenever a beam of light is passed through a colloidal system, the colloidal particles scatter the light in all directions. Now if the particle is viewed through a microscope, against a dark background, it becomes visible as a speck of light. This Scattering of light by the colloidal particles is known as Tyndall effect. Tyndall effect is more prominent when the wavlength of the incident light and the dimensions of the scattering particles are of the same order. That is why light is not scattered by true solutions. If the colloidal particles are assumed to be small, and do not absorb light, it may be observed that (i) The intensity of the scattered light is greater in the direction parallel to the incident beam and smallest in the direction perpendicular to the incident beam. (ii) The intensity of the scattered light decreases and hence the tyndall beam assumes a blue colour imparting the complementary yellow colour to the transmitted light.

(a) It is due to greater force of attraction between dispersed phase and dispersion medium in lyophilic colloid than lyophobic colloid.

(c) Sky appears blue in colour due to scattering of light by colloidal particles. This is known as Tyndall effect.

Example – 27 What causes Brownian movement in a colloidal solution ? Sol.

This is due to the unequal bombardment of colloidal particles by the molecules of dispersion medium.

Example – 28 Define the term ‘Tyndall effect’. Sol.

The scattering of light by colloidal particles is known as Tyndall effect.

Example – 29 Sol.

Write a note on electrical properties of colloids. (i) Electrical properties of colloidal dispersions are due to existence of electrical charge an colloidal particles. (ii) All particles in a given dispersion carry same charge either + ve or – ve. (iii) The repulsion between the particles carying the same charge keeps them separated. Due to the same charge the particles do not come close together to form large particles that will precipitate out. thus stability of colloids is associated with an electrical charge present an the particles.

Example – 30 Example – 26 Explain the following observations : (a) Lyophilic colloid is more stable than lyophobic colloid. (b) Coagulation takes place when sodium chloride solution is added to a colloidal solution of ferric hydroxide. (c) Sky appears blue in colour.

What are the various reasons for the origin of electrical charge on colloidal particles ? Sol. The various reasons for the origin of electrical charge an the colloidal particles are as follows : (i) Frictional electrification caused by the mutual rubbing of the colloidal particles with molecules of the dispersion medium. (ii) Electron capture by particles from air and during electro-dispersion in Bredig’s method. (iii) Preferential adsorption of ions from solutions.

236

SURFACE CHEMISTRY Example – 31

Example – 35

What is the ‘coagulation’ process. Sol.

The process of settling of colloidal particles is called coagulation.

Example – 32 Why is ferric chloride preferred over potassium chloride in case of a cut leading to bleeding ? Sol.

Fe3+ ion has greater coagulating power than Na+ ion as it has higher charge.

What happens in electrophoresis ? Sol.

(i) The movement of colloidal particles under the influence of an electric field is called electrophoresis or cataphoresis. (ii) The term cataphoresis is used because most of the colloidal sol studied at that time contained positively charged colloidal particles and thus migrated towards cathode. (iii) The phenomenon of electrophoresis can be demonstrated by an apparatus shown below

Example – 33 Mention two ways by which lyophilic colloids can be coagulated. Sol.

This can be done (i) by adding an electrolyte (ii) by adding a suitable solvent.

Example – 34 Explain the following observations: (a) Ferric hydroxide sol gets coagulated on addition of sodium chloride solution. (b) Cottrell’s smoke precipitator is fitted at the mouth of the chminey used in factories. (c) Physical adsorption is multillayered, while chemisorption is monolayered. Sol.

(a) As ferric hydroxide, Fe (OH)3 is a positively charged sol, so it gets coagulated by chloride ions Cl–, released by NaCl solution.

Construction : The apparatus consist of an U tube filled with colloidal sol. Water is added slowly over the sol to form sharp boundaries between sol and water in the two arms of the tube. Two platinum electrodes are inserted in water in two arms of the U tube and are connected to a high voltage battery. Working : When an electric field is applied the

(b) Cottrell’s smoke precipitator, neutralises the charge on unburnt carbon particles, coming out of chimney and they get precipitated and settle down at the floor of the chamber.

boundary in one arm is seen to move down and in

(c) As physical adsorption, involves only weak van der waal’s force of interaction, so many layers of adsorbate get attached, while chemisorption involves chemical bond formation between adsorbate and adsorbents monolayer is formed.

migrate towards positive electrode and they are

other arm to move up. If it is observed that the boundary falls gradually on the negative electrode side and moves up the positive electrode side, then the particles negatively charged. If the reverse occurs, then the particles carry positive charge.

237

SURFACE CHEMISTRY Example – 36

Example – 37

Discuss Hardy – Schulze rule. Sol.

Hardy – Schulze rules : The precipitation action of ions of an electrolyte depends on the sign of charge and the valence of the ion. The effects of these factors on the coagulation of colloidal particles were stated by H. Schulze and Sir William Bate Hardly which are known as Schulze – Hardy rules. (i) The ions of opposite sign to those present on the surface of the particles cause the precipitation of the particles. For example, positively charged ions of an electrolyte will cause the precipitation of negatively charged ions will effect the precipitation of positive particles. (ii) The precipitation power of an electrolyte increases very rapidly with an increase in the valence of anion or cation. Thus, the precipitating power of ions decreases in the order Al 3+ > Mg 2+ > Na + and PO 34  SO 24   Cl  . Lyophilic colloids differ from lyophobic colloids in respect of coagulation by the addition of electrolytes. The lyophilic colloids require much larger amount for their precipitation than that required for the precipitation of lyophobic colloids. This is because the lyophilic particles are surrounded by layer of medium through which penetration of ions is difficult. The higher concentration of electrolytes remove this layer of medium surrounding the particles and then neutralises their charge resulting in their coagulation. (v) An addition of solvents is a method useful for the precipitation of hydrophilic colloids. If solvents like alcohol or acetone, which have affinity for water are added to hydrophilic colloids, the colloids, the colloidal particles are coagulated due to dehydration.

Define ultrafiltration. Sol.

In this process, colloidal solutions are purified by carrying out filtration through special types of graded filters called ultra-filters. Filter paper allows the passage of electrolyte but does not allow the passage of colloidal particles.

Example – 38 What is an emulsion? Sol.

Emulsion is a colloidal solution in which both the dispersed phase and dispersion medium are liquids e.g. milk, cod liver oil, etc.

Example – 39 Distinguish between oil in water and water in oil emulsions. Sol. No. Oil in water

Water in oil

1.

Oil is the dispersed phase and water is the dispersion medium

1.

Water is the dispersed phase and oil is the dispersion medium

2.

If water is added, it will be miscible with the emulsion but if oil is added, it will not be miscible.

2.

If water is added, it will not be miscible with the emulsion but if oil is added, it will be miscible.

3.

An addition of small amount of an electrolyte makes the emulsion conducting.

3.

An addition of small amount of an electrolyte has not effect on conducting power.

4.

Water is continuous phase.

4.

Oil is continuous phase.

5.

Basic metal sulphates, water soluble alkali metal soaps are used as emulsifiers.

5.

Water insoluble soaps such as those of Zn, Al, Fe, alkaline earth metals are used as emulsifiers.

SURFACE CHEMISTRY

238

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Adsorption 1.

8.

Adsorption is phenomenon in which a substance

(a) Physical adsorption

(a) Goes into the body of the other substance

2.

3.

4.

(b) Remains close the other substance

(b) Chemical adsorption

(c) Accumulates on the surface of the other substance

(c) Heat of adsorption

(d) None of these

(d) Enthalpy of adsorption

In adsorption of oxalic acid on activated charcoal, the activated charcoal is known as

9.

(a) Temperature is increased

(b) Adsorbate

(b) Temperature is decreased

(c) Adsorber

(d) Absorber

(c) Pressure is increased

Adsorption is always

(d) Concentration is increased

(a) Endothermic

(b) Exothermic

(c) Either (a) or (b)

(d) None of these

10.

Adsorption is

(a) Physical adsorption

(b) Chemisorption

(c) Both

(d) None of both

(c) An exothermic process hence increase in temperature increases adsorption (d) None of these

(d) Enthalpy of adsorption (Hadsorption) is slow and positive

Which one of the following statement is not correct

12.

Physical adsorption

(a) The extent of adsorption depends on the nature of the adsorbent and adsorbate

(a) Involves the weak attractive interaction between the adsorbent and adsorbate

(b) The extent of adsorption depends on the pressure of the gas

(b) Involves the chemical interactions between the adsorbent and adsorbate

(c) The extent of adsorption depends on the temperature

(c) Is irreversible in nature

(d) The extent of adsorption has no upper limit Sorption is the term used when

13.

(a) Adsorption takes place (b) Absorption takes place (c) Both (a) and (b) 7.

Adsorption is multilayered in the case of

Which of the following statements is incorrect regarding physissorptions ? (a) It occurs because of van der Waals’ forces (b) More easily liquefiable gases are adsorbed readily (c) Under high pressure it results into multi molecular layer on adsorbent surface

11.

(b) An endothermic process hence increase in temperature increases adsorption

6.

There is desorption of physical adsorption when

(a) Adsorbent

(a) Exothermic process hence increase in temperature decreases adsorption in cases where Vander Waal’s forces exist between adsorbate and adsorbent.

5.

If the adsorbate is held on a surface by weak Vander Waal’s forces, the adsorption process is called

(d) Desorption takes place

For adsorption, the thermodynamic requirement is that (a) H must be negative (b) S must be negative (c) G must be negative (d) H , S and G must be negative

14.

(d) Increases with increase of temperature Which one of the following characteristics is not correct for physical adsorption ? (a) Adsorption on solids is reversible (b) Adsorption increases with increase in temperature (c) Adsorption is spontaneous (d) Both enthalpy and entropy of adsorption are negative Physical adsorption is essentially quite appreciable (a) At room temperature (b) At higher temperature (c) At lower temperature (d) None of these

SURFACE CHEMISTRY 15.

239

Which of the following is not a characteristic of chemisorption (a) H is of the order of 400 kJ

20.

Which one is Freundlich’s equation (a)

x 1  log k  log P m n

(b)

(c)

x  kP 2 m

(d) log

x  exp (kP ) m

(b) Adsorption is irreversible (c) Adsorption may be multimolecular layer (d) Adsorption is specific 16.

21.

Chemisorption (a) Involves the weak attractive interactions between adsorbent and adsorbate (b) Is irreversible in nature (c) Decreases with increase of temperature (d) Involves multilayer formation of adsorbent on adsorbate

17.

22.

Chemisorption (a) Increases with temperature

In Langmuir’s model of adsorption of a gas on a solid surface (a) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered (b) the adsorption at a single site on the surface may involve multiple molecules at the same time (c) the mass of gas striking a given area of surface is proportional to the pressure of the gas (d) the mass of gas striking a given area of surface is independent of the pressure of the gas According to Freundlich adsorption isotherm which of the following is correct ? (a)

x  p0 m

(b)

x  p1 m

(b) Decreases with temperature (c) Remains unaffected by change of temperature (d) first increases then decreases with temperature 18.

x  p1/ n m (d) All of the above are correct for different range of pressure (c)

When the temperature is lowered and pressure is raised, the adsorption of a gas on a solid (a) Decreases

Catalysis

(b) Increases

23.

(c) Remains unaffected

Which one is false in the following statements (a) A catalyst is specific in its action

(d) Decreases first then increases

(b) A very small amount of the catalyst alters the rate of a reaction

Adsorption Isotherms 19.

x 1  log k  log C m n

(c) The number of free valencies on the surface of the catalyst increases on subdivision

Which plot is the adsorption isobar for chemisorption where x is the amount of gas adsorbed on mass m (at constant pressure) at temperature T

(d) Ni is used as catalyst in the manufacture of ammonia 24.

A catalyst is a substance which (a) Alters the equilibrium in a reaction

(a)

(b) Is always in the same phase as the reactants

x

x

(c) Participates in the reaction and provides easier pathway for the same

(b)

m

T

T

(d) Does not participate in the reaction but speeds it up 25.

(a) The catalyst and the reactants should be gases

x

x

(c)

(d)

m

In a homogeneous catalysis (b) The catalyst and the reactants should form a single phase (c) Catalyst and the reactants are all solids

T

T

(d) The catalyst and the reactants are all liquids

SURFACE CHEMISTRY 26.

240

Which of the following statements about a catalyst is true

32.

(a) It lowers the energy of activation (b) The catalyst altered during the reaction is regenerated

27.

(d) Carbonic anhydrate

33.

In shaving cream, the dispersion medium is

A catalyst

(a) Liquid

(b) Gas

(a) Increases the free energy change in the reaction

(c) Solid

(d) None of the above

34.

(c) Solid dispersed in gas (d) Solid dispersed in solid 35.

Addition of catalyst in a system

(b) No effect on equilibrium concentrations

36.

When dispersion medium is water, the colloidal system is called (a) Sol

(b) Aerosol

(c) Organosol

(d) Aquasol

Fog is an example of colloidal system (a) Liquid dispersed in gas(b) Gas dispersed in gas

(c) Decreases equilibrium concentrations (d) Increases rate of forward reaction and decreases rate of backward reaction

Smoke is an example of (a) Gas dispersed in liquid (b) Gas dispersed in solid

(a) Increases equilibrium concentrations

(c) Solid dispersed in gas (d) Gas dispersed in liquid 37.

Following are various types of colloids. In electric field. X (Colloids)

1. Following are the terms about activity and selectivity

II : Activity is the ability of catalysts to direct reaction to yield particular products and selectivity is the ability of catalysts to accelerate chemical reactions.

Y (Classification) -

A

sol

II Milk of magnesia -

B

aerosol

III Soap suds

-

C

gel

IV Butter

-

D

foam

I

I : Activity is the ability of catalysts to accelerate chemical reactions and selectivity is the ability of catalysts to direct reaction to yield particular products

Rain cloud

Correct matching is

Select correct term :

I

II

III

IV

(a) I

(b) II

(a) A

B

C

D

(c) I and II both

(d) None of these

(b) A

C

B

D

(c) B

A

D

C

(d) B

A

C

D

Enzymes are (a) Micro-organisms

(b) Proteins

(c) Inorganic compounds (d) Moulds 31.

(c) Nitrogenase

(d) All of these

(d) Can either increase or decrease the free energy change depending on what catalyst we use

30.

(b) Urease

Colloids

(c) Does not increase or decrease the free energy change in the reaction

29.

(a) Invertase

(c) It does not alter the equilibrium

(b) Decreases the free energy change in the reaction

28.

In the human body, enzyme that catalyses the reaction of CO 2 with H 2 O is

Enzymes are

38.

Which of the following statement is wrong for lyophobic sol (a) Dispersed phase is generally in organic material

(a) Substances made by chemists to activate washing powder

(b) Can be easily coagulated by small addition of electrolyte

(b) Very active vegetable catalysts

(c) Dispersed phase particles are poorly hydrated and colloid is stabilised due to charge on the colloidal particles

(c) Catalysts found in organisms (d) Synthetic catalysts

(d) Reversible in nature that is after coagulation can be easily set into colloidal form

SURFACE CHEMISTRY 39.

241

The difference between a lyophilic and lyophobic colloid is in their (a) Particle size

47.

(b) Behaviour towards dispersion medium (c) Filtrability

40.

41.

(d) None of these

48.

(b) Bredig’s arc method

(c) Exchange of solvent

(d) Oxidation method

(b) Peptization

(c) Electrodispersion

(d) Dialysis

(c) Colloidal particles from crystalloids (d) Colloidal particles from gel Colloidal solution cannot be obtained from two such substances which are

51.

(c) Ba2+ < Na2+ < Al3+

(d) Al3+ < Na+ < Ba2+

(a) D < A < C < B

(b) C < B < D < A

(c) A < C < B < D

(d) B < D < A < C

Milk is a colloid in which

(c) A gas is dispersed in liquid

(c) In different physical state (d) None of the above

46.

(b) Na+ < Ba2+ < Al3+

(b) A solid is dispersed in liquid

(b) In same physical state

45.

(a) Al3+ < Ba2+ < Na+

(a) A liquid is dispersed in liquid

(a) Insoluble in each other

44.

The coagulating power of electrolytes having ions Na+,

Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is

(b) Suspended particles from crystalloids

43.

(d) Fe(OH)3

50.

Dialysis is the process of separation of (a) Suspended particles from colloids

(c) Au

The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statement is not correct ? (a) Coagulation in both sols can be brought about by electrophoresis (b) Mixing the sols has no effect (c) Sodium sulphate solution causes coagulation in both sols (d) Magnesium chloride solution coagulates, the gold sol more readily than the iron (III) hydroxide sol

Dialysis : In this method, colloidal particles are retained by animal membrane or a parchment paper while electrolytes pass through them. 42.

(b) Gelatin

49.

When a freshly precipitated substance is converted into a colloidal solution with the help of a third substance, the process is known as (a) Coagulation

(a) As 2 S 3

Al3+ and Ba2+ for arsenic sulphide sol increases in the order

Colloidal solutions of metals like gold, silver and platinum are generally prepared by using (a) Peptization

Which one of the sols acts as protective colloid

Difference between colloids and crystalloids is of

(d) Some suger is dispersed in water 52.

When dispersed phase is liquid and dispersion medium is gas, then the colloidal system is called

(a) Particle composition

(b) Particle size

(a) Smoke

(b) Clouds

(c) Concentration

(d) Ionic character

(c) Emulsion

(d) Jellies

The shape of colloidal particles is (a) Sphere like

(b) Rod like

(c) Disc like

(d) All the above

53.

Butter is a colloid. It is formed when (a) Fat is dispersed in solid casein (b) Fat globules are dispersed in water

Which characteristic is true in respect of colloidal solutions

(c) Water is dispersed in fat

(a) They always have two phases

(d) Casein is suspended in H2O

(b) They are only in liquid state

54.

When sugar is added to a colloidal solution it brings about

(c) They can’t be electrolysed

(a) Ionization

(b) Coagulation

(d) They are only hydrophilic

(c) Peptization

(d) None of the above

SURFACE CHEMISTRY 55.

Milk is (a) Dispersed fats in oil

56.

61. (b) Dispersed fats in water

(c) Dispersed water in fats (d) Dispersed water in oil A cleared solution is again converted into colloidal solution. The process is called (a) Peptisation (b) Electrolytic addition (c) Electrophoresis

57.

242

(d) None of these

58.

The volume of a colloidal particle, VC as compared to the volume of a solute particle in a true solution VS, could be

59.

The mass adsorbed per gram of adsorbed O2 having pressure 10 atm at 400 K, if placed in contact with solid surface is 2 g in one litre vessel. The pressure of O2 after adsorption becomes 2 atm. Assume no change in temperature and R = 0.08 L-atm. K–1 mol–1.

60.



particle (micelle): N A  6  10

23



62.

When 9.0 ml of arsenius sulphide sol and 1.0 ml of 1.0 × 10–4 M BaCl2 are mixed, turbidity due to precipitation just appears after 2 hours the coagulating value is 10 –4 Ba 2   M = 10–2 M = 10 mmol/L 10  10 –3

63.

When a graph is plotted between log

64.

How many of the following are correctly matched?

A colloidal system in which gas bubbles are dispersed in a liquid is known as (a) Foam (b) Sol (c) Aerosol (d) Emulsion

Numeric Type Questions

A detergent (C14H28SO4– Na+) solution becomes a colloidal sol at a concentration of 10–3 M on an avareage 1015 colloidal particles are present in 100 mm3. What is the average number of ions are contains in one colloidal

x and log p, it is a m straight line with an angle 45° and intercept 0.3010 on y axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gm of adsorbent?

(I) Lyophilic colloids -reversible sols (II) Associated colloides – micelles (III) Tyndall effect – scattering of light by colloidal particle

x Graph between log   and log p is a straight line at m

(IV) Electrophoresis – movement of dispersion medium under the influence of electric field

angle of 45o with the intercept of 0.6020 as shown below: 65.

How many of the following are aerosols of liquids? Fog, Clouds, Smoke, Dust

66.

How many of the given are multimolecular colloids? Nylon, Gold sol, Sulphur sol, Dacron, Starch.

67.

How many of the given are non-elastic gels? Silica, Aluminium, Ferric hydroxide gel, Agar-Agar gel, Starch, Gelatin.

x  at a pressure of m

Calculate the extent of adsorption  1 atm.

68.

How many of the given are protective colloids? Gelatin, Haemoglobin, Gum, Starch, Ferric hydroxide sol, Gold sol.

SURFACE CHEMISTRY

243

EXERCISE - 2 : PREVIOUS YEARS JEE MAIN QUESTIONS 1.

3 g of activated charcoal was added to 50 mL of acetic acid

6.

solution (0.06N) in a flask. After an hour it was filtered and

x versus log p gives a m straight line with slope equal to 0.5, then : x ( is the mass of the gas adsorbed per gram of m adsorbent) (2017) adsorption isotherm. Plot of log

the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : (2015) (a) 42 mg

(b) 54 mg

(c) 18 mg

(d) 36 mg

Adsorption of a gas on a surface follows Freundlich

(a) Adsorption is independent of pressure. (b) Adsorption is proportional to the pressure. (c) Adsorption is proportional to the square root of

2.

Which one of the following is an incorrect statement for physisorption Online 2015 SET (1)

pressure.

(a) It is a reversible process (b) It requires less heat of adsorption

7.

(c) It requires activation energy (d) It takes place at low temperature 3.

For a linear plot of log (x/m) versus log p in a Freundlich adorption isotherm, which of the following statement is correct? (k and n are constants) (2016) (a) 1/n appears as the intercept

8.

(b) Only 1/n appears as the slope.

x versus log p gives a m straight line with slope equal to 0.5, then : adsorption isotherm. Plot of log

(c) log (1/n) appears as the intercept (d) both k and 1/n appear in the slope term. 4.

A particular adsorption process has the following characteristics : (i) It arises due to van der Waals forces and (ii) it is reversible. Identify the correct statement that describes the above adsorption process :

(a) Enthalpy of adsorption is greater than 100 kJ mol–1. (b) Energy of activation is low. 9.

(d) Adsorption increases with increase in temperature. 5.

Gold numbers of some colloids are : Gelatin : 0.005 – 0.01, Gum Arabic : 0.15 – 0.25; Oleate : 0.04 – 1.0; Starch: 15 – 25. Which among these is a better protective colloid ? Online 2016 SET (2) (a) Gelatin

(b) Gum Arabic

(c) Oleate

(d) Starch

x is the mass of the gas adsorbed per gram of m adsorbent) Online 2017 SET (2) (a) Adsorption is independent of pressure. (b) Adsorption is proportional to the pressure. (c) Adsorption is proportional to the square root of pressure. (d) Adsorption is proportional to the square of pressure. Which of the following statements about colloids is False ? Online 2018 SET (1) (a) Freezing point of colloidal solution is lower than true solution at same concentration of a solute. (b) Colloidal particles can pass through ordinary filter paper. (c) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed. (d) When excess of electrolyte is added to colloidal solution, colloidal particle will be precipitated. (

Online 2016 SET (1)

(c) Adsorption is monolayer.

(d) Adsorption is proportional to the square of pressure. Among the following, correct statement is : Online 2017 SET (1) (a) Brownian movement is less pronounced for smaller particles than for bigger–particles. (b) Sols of metal sulphides are lyophilic. (c) Hardy Schulze law states that bigger the size of the ions, the greater is its coagulating power. (d) One would expect charcoal to adsorb SO2 more than hydrogen CH4. Adsorption of a gas on a surface follows Freundlich

SURFACE CHEMISTRY 10.

244

If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log

14.

x versus log P is linear. The m

slope of the plot is :

A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation

x  kp 0.5 Adsorption of the gas increases with: m

Online 2018 SET (2)

(n and k are constants and n > 1)

11.

(a) 2k

(b) log k

(c) n

(d)

(2019-04-10/Shift-1) (a) Increase in p and decrease in T

1 n

(b) Increase in p and increase in T (c) Decrease in p and increase in T

Which one of the following is not a property of physical adsorption ? Online 2018 SET (3)

(d) Decrease in p and decrease in T 15.

The correct option among the following is:

(a) Higher the pressure, more the adsorption

(2019-04-10/Shift-2)

(b) Lower the temperature, more the adsorption

(a) Colloidal particles in lyophobic sols can be precipitated by electrophoresis.

(c) Greater the surface area, more the adsorption (d) Unilayer adsorption occurs 12.

(b) Brownian motion in colloidal solution is faster the viscosity of the solution is very high.

Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of log

(c) Colloidal medicines are more effective because they have small surface area.

x versus log p is shown m

x in the given graph, is proportional to: m

(d) Addition of alum to water makes it unfit for drinking. 16.

Among the following, the INCORRECT statement about colloids is : (2019-04-12/Shift-2) (a) The range of diameters of colloidal particles is between 1 and 1000 nm

(2019-04-08/Shift-1)

(b) The osmotic pressure of a colloidal solution is of higher order than the true solution at the same concentration (c) They can scatter light (d) They are larger than small molecules and have high molar mass 17.

Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of gas adsorbed on mass m of the adsorbent at pressure p.

x is m

proportional to

13.

2/3

(a)

P

(c)

P3

2

(b)

P

(d)

P3/ 2

The aerosol is a kind of colloid in which: (2019-04-09/Shift-1) (a) solid is dispersed in gas (2019-01-09/Shift-1)

(b) gas is dispersed in solid (c) liquid is dispersed in water

(a)

P2

(b)

P1/ 4

(d) gas is dispersed in liquid

(c)

P1/ 2

(d) P

SURFACE CHEMISTRY 18.

245

For coagulation of arsenious sulphide sol, which one of the following salt solution will be most effective?

22.

(2019-01-12/Shift-2)

(2019-01-09/Shift-2)

19.

(a) BaCl2

(b) AlCl3

(c) NaCl

(d) Na3PO4

(a) It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane. (b) Tyndall effect can be used to distinguish between a colloidal solution and a true solution.

Haemoglobin and gold sold are examples of: (2019-01-10/Shift-2)

(c) Lyophilic sol can be coagulated by adding an electrolyte.

(a) positively and negatively charged sols, respectively (b) positively charged sols

(d) Latex is a colloidal solution of rubber particles which are positively charged.

(c) negatively charged sols

20.

(d) negatively and positively charged sols, respectively

Integer-Type Question

The correct match between Item I and Item II is

23.

Item I

Item II

A. Allosteric effect

P. Molecule binding to the active site of enzyme

B. Competitive

Q. Molecule crucial for

inhibitor

communication in the body

C. Receptor

R. Molecule binding to a site

D. Poison

a pressure of 4 atm is : (Given log3 = 0.47771)

(a) A  R, B

 P, C  Q, D  S

(b) A  P, B

 R, C  Q, D  S

(c) A  R, B

 P, C  S, D  Q

(d) A  P, B

 R, C  S, D  Q

H 2 CH 4 CO 2 SO 2 190 304

630

Temp/K On the basis of data given, predict which of the following gases shows least adsorption on a definite amount of charcoal (2019-01-12/Shift-1)

(c)

Amongst the following statements regarding adsorption, those that are valid are : (2020-09-02/Shift-2)

(B) On a given adsorbent, ammonia is adsorbed more than nitrogen gas. (C) On adsorption, the residual force acting along the surface of the adsorbent increases. (D) With increase in temperature, the equilibrium concentration of adsorbate increases.

Given

(a)

(2020-09-02/Shift-1)

(A) H becomes less negative as adsorption proceeds.

(2019-01-11/Shift-2)

Critical: 33

x at m

S. Molecule binding to the enzyme covalently

Gas:

x and log p gives a straight line with slope equal m

to 2 and the intercept equal to 0.4771. The value of

24.

enzyme

The mass of gas adsorbed, x, per unit mass of adsorbate, m, was measured at various pressures, p. A graph between

log

other than the active site of

21.

Among the following, the false statement is :

SO 2

(b)

CO 2

(d)

CH 4 H2

25.

(a) (B) and (C)

(b) (C) and (D)

(c) (A) and (B)

(d) (D) and (A)

Tyndall effect is observed when : (2020-09-03/Shift-1) (a) The diameter of dispersed particles is much smaller than the wavelength of light used. (b) The diameter of dispersed particles is much larger than the wavelength of light used. (c) The refractive index of dispersed phase is greater than that of the dispersion medium. (d) The diameter of dispersed particles is similar to the wavelength of light used.

SURFACE CHEMISTRY 26.

246 (b)

An ionic micelle is formed on the addition of : (2020-09-03/Shift-2) (a) liquid diethyl ether to aqueous NaCl solution (b) sodium stearate to pure toluene (c) excess water to liquid

(c)

(d) excess water to liquid 27.

28.

29.

Match the following: (i) Foam (ii) Gel (iii) Aerosol (iv) Emulsion

(a) smoke (b) cell fluid (c) jellies (d) rubber (e) froth (f) milk (2020-09-04/Shift-1) (a) (i)-e; (ii)-c; (iii)-a; (iv)-f (b) (i)-b; (ii)-c; (iii)-e; (iv)-d (c) (i)-d; (ii)-b; (iii)-a; (iv)-e (d) (i)-d; (ii)-b; (iii)-e; (iv)-f A sample of red ink (a colloidal suspension) is prepared by mixing eosin dye, egg white,HCHO and water. The component which ensures stability of the ink sample is (2020-09-04/Shift-2) (a) HCHO (b) Water (c) Eosin dye (d) Egg white Identify the correct molecular picture showing what happens at the critical micellar concentration (CMC) of an aqueous solution of a surfactant ( polar head; non-polar tail; water).

(d)

31.

Kraft temperature is the temperature: (2020-09-06/Shift-1) (a) Above which the aqueous solution of detergents starts boiling (b) Below which the formation of micelles takes place (c) Above which the formation of micelles takes place (d) Below which the aqueous solution of detergents starts freezing

Integer-Type Question 32.

For Freundlich adsorption isotherm, a plot of log (x/m)(y axis) and log p (x - axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is ............

104 g .  log 3  0.4771 . (2020-09-05/Shift-1) (a) (B) 30.

(b) (A)

(c) (C) (d) (D) Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the gas adsorbed on mass m of the adsorbent, the correct plot of

x versus p is: m

(2020-09-05/Shift-2) (a)

(2020-09-06/Shift-2)

33.

The flocculation value of HCl for As2S3 sol is 30 mmolL– 1 . If H2SO4 is used for the flocculation of arsenic sulphide, the amount, in grams, of H2SO4 in 250 mL required for the above purpose is ____. (2020-01-07/Shift-2)

34.

As per Hardy-Schulze formulation, the flocculation values of the following for ferrichydroxide sol are in theorder: (2020-01-08/Shift-1) (a) AlCl8>K3[Fe(CN)6]>K2CrO4>KBr=KNO3 (b) K3[Fe(CN)6] KNO3

SURFACE CHEMISTRY

247

EXERCISE - 3 : JEE ADVANCE QUESTIONS 6.

Single Choice Questions 1.

Following is the variation of physical adsorption with temperature

Following are the events taking place to explain adsorption theory I : Desorption (a)

II : Diffusion of the reactants along the surface

(b)

III : Adsorption of the reactants IV : Formation of the activated surface complex These events are taking place in the following order

2.

(a) I, II, III, IV

(b) III, II, IV, I

(c) III, IV, I, II

(d) IV, III, II, I

Adsorption of gases on solid surface is generally exothermic because (a) enthalpy is positive (b) entropy decreases (c) entropy increases

3.

4.

5.

7.

(d)

Rate of physisorption increases with (a) decrease in temperature (b) increases in temperature

(d) free energy increases

Energy of activation of forward and backward reaction are equal in cases (numerical values) where (a) H  0

(c)

(c) decrease in pressure (d) decrease in surface area 8.

(b) No catalyst present

The adsorption of a gas on a solid surface varies with pressure of the gas in which of the following manner

(c) S  0

(a) Fast , slow, independent of the pressure (b) Slow , fast, independent of the pressure

(d) Stoichiometry is the mechanism

(c) Independent of the pressure , fast, slow

In neutralisation of KI by AgNO3 positive charge is due to absorption of

(d) Independent of the pressure, slow, fast

(a) Ag+ ions

(b) Ag

(c) I ions

(d) Both (b) and (c)

There are certain properties related to physical adsorption

9.

In the following isotherm

(a)

(b) Desorption may start along AB

I : Reversible II : Formation of unimolecular layer

x  p 0 when point A is reached m

(c)

x  p 1 / n along OA m

(d) All of these

III : Low heat of adsorption IV : Occurs at low temperature and decreases with increasing temperature Which of the above properties are for physical adsorption (a) I, II, III

(b) I, III, IV

(c) II, III, IV

(d) I, III

10. When saturation is attained in terms of adsorption, variation x

of  m  and C (concentration) is given by the portion of   the isotherm (a) OA

(b) OB

(c) AB

(d) BC

SURFACE CHEMISTRY 11.

Graph between log

248

x and log P is a straight line inclined m

at an angle   45 0 . When pressure is 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be (a) 1 g/g adsorbent

(b) 1.5 g/g adsorbent

(c) 2.5 g/g adsorbent

(d) 0.25 g/g adsorbent

12. Mark the correct statement about given graph

15. For the adsorption of solution on a solid surface, x  kC 1 / n . m

Adsorption isotherm of log  x  and log C was found of m  the type

log

x m log k log C

This is when (a) C = 0

(b) C = 1 M

(c) C = constant

(d) C = 2 M

16. In Freundlich adsorption isotherm, adsorption is proportional to pressure P as (a) P0 (b) P (c) Pn (a) X is threshold energy level (b) Y and Z are energy of activation for forward and backward reaction respectively (c) Q is heat of reaction and reaction is exothermic

(d) P1/n

17. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is (a) CH3(CH2)15N+ (CH3)3 Br– (b) CH3(CH2)11 OSO3Na 

(d) All of these 13. Oxidation of oxalic acid by acidified KMnO4 is an example of autocatalysis. Which of the following graph represents the variation of rate of this reaction with time?

(c) CH3(CH2)6COO– Na+ (d) CH3(CH2)11N+(CH3)3Br– Option (a) and (b) both are capable of forming micelle (a)  CH3(CH2)15N+ (CH3)3 Br– Cetyl trimethyl ammonium bromide

(a)

(c)

(b)

(d)

14. For the adsorption of a gas on a solid, the plot of log (x/ m) versus log P is linear with slope equal to (a) k

(b) log k

(c) n

(d) 1/n

(b)  CH3 (CH2)11 OSO3 Na  Sodium lauryl sulphate But critical concentration for micelle formation decreases as the molecular weight of hydrocarbon chain of surfactant grows because in this true solubility diminishes and the tendency of surfactant molecule to undergo association increases. 18. Cleansing action of soap occurs because (a) Oil and grease can be absorbed into the hydrophobic centres of soap micelles and washed away (b) Oil and grease can be absorbed into the hydrophilic centres of soap micelles and washed away (c) Oil and grease can be absorbed into both hydrophilic and hydrophobic centres but not washed away (d) Cleansing action is not related to micelles

SURFACE CHEMISTRY

249 31. Coagulation or demulsification can be done by some of the methods given below

19. Enzymes are (a) Micro-organisms (b) Protein

I : By addition of substance which would destory the emulsifier

(c) Inorganic compounds (d) Moulds 20. Which of the following statements is incorrect (a) Enzymes are in colloidal state

II : By addition of an electrolyte which would destroy the charge III : By heating, freezing and centrifuging Select correct methods

(b) Enzymes are catalysts (c) Enzymes can catalyse any reaction

(a) I, II

(b) I, II, III

(d) Urease is an enzyme

(c) II only

(d) III only

21. Enzymes are known to increase the rate of reaction by 2

(a) 10 times

(b) 10 times

5

12

(c) 10 times

–2

(d) 10 times

22. The ability of the catalyst to direct the reaction to yield particular product is called (a) Reactivity

(b) Selectivity

(c) Activity

(d) Fugacity

23. Blood may be purified by (a) Dialysis (c) Coagulation

32. Ferric hydroxide sol is positively charged colloid. The coagulating power of NO3 ,SO24 and PO34 ions would be in the order (a) NO3  SO24  PO34 (b) SO24  NO3  PO34 (c) PO34  SO24  NO3 (d) NO3  SO24  PO34 33. Peptising agent is

(b) Electro - osmosis

(a) Always an electrolyte

(d) Filtration

(b) Always a non - electrolyte

24. Colloidal solution is not purified by (a) Dialysis (b) Electrodialysis (c) Electrophoresis (d) Ultrafiltration 25. The simplest way to check whether a system is colloidal, is (a) Tyndall effect (b) Electro dialysis (c) Brownian movement (d) Finding out particle size 26. The colloidal particles can pass through (a) Filter paper as well as animal membrane (b) Animal membrane but not through filter paper (c) Filter paper but not through animal membrane (d) Semipermeable membrane 27. Sky looks blue due to (a) Dispersion effect (b) Reflection (c) Transmission (d) Scattering 28. Ultramicroscope is used for observing (a) Light reflection (b) Light absorption (c) Light scattering (d) Light polarisation 29. Colloidal paints are used to paint metallic portions of cars. This application is based on (a) Coagulation (b) Electro-osmosis (c) Peptization (d) Electrophoresis 30. Colloidal particles carrying charge is best shown by (a) Brownian movement (b) Tyndall effect (c) Electrophoresis (d) Dialysis

(c) Electrolyte or non - electrolyte (d) A lyophilic colloid 34. As2S3 sol has a negative charge. Capacity to precipitate it is highest in (a) AlCl3

(b) Na3PO4

(c) CaCl2

(d) K2SO4

35. Maximum coagulation power is in (a) Na+

(b) BA++

(c) Al+++

(d) Sn++++

36. Which of the following is most effective in coagulating a ferric hydroxide sol (a) KCl

(b) KNO3

(c) K2SO4

(d) K3[Fe(CN)6]

Multiple Choice Questions 37. Select the correct statements among following: (a) At 83 K, N2 is physisorbed on the surface of iron (b) At 773 K and above N2 is chemisorbed on the iron surface (c) Activation energy is +ve in case of physiosorption and zero in case of chemisorption (d) Activation energy is zero in case of physiosorption and +ve in case of chemisorption

SURFACE CHEMISTRY 38. Which of the following is/are correct about Freunlich’s adsorption isotherm for gases and solutions? (a)

x  KC1/ n m

(b)

x 1  bP  m a

250 45. Select the correct statements for Brownian movement: (a) It becomes more intense with increase in temperature (b) It increases with increase in particle size (c) It is due to collision of molecules of dispersion medium with the colloidal particles (d) It increases the stability of colloidal solution Assertion–Reason Type Questions

x (c) log    log K  n log P m 1 x (d) log    log K  log C n m 39. Which of the following are correct about the catalyst? (a) They participate in the reaction but recovered at last (b) It does not affect G (c) It does not affect H (d) It alters the mechanism of reaction 40. Which of the following is/are not affected by the catalyst? (a) Heat of reaction (b) Equilibrium constant (c) Amount of product (d) Rate constant of the reaction 41. Which of the following act as negative catalyst? (a) Ethanol in oxidation of chloroform (b) Tetra ethyl lead used as antiknocking agent (c) Glycerol in the decomposition (d) Fe in the formation of ammonia by Haber process 42. Select the correct statements about enzymes: (a) Enzymes are biological catalysts found in organisms (b) All enzymes are proteins (c) Enzymes can catalyse any reaction (d) Enzymes activity is optimum at 27oC 43. Select the correct statements among following: (a) Foam is a colloidal solution of gas in liquid (b) Aerosol is a colloidal solution of liquid in gas (c) Gel is a colloidal solution of solid in liquid (d) Emulsion is a colloidal solution of liquid in liquid 44. Which of the following statments are true? (a)Flocculation value is inversely proportional to coagulating power (b) Colloidal silica is protective colloid (c) Alum is used for cleaning muddy water (d) Gelatin is added in ice cream, it acts as emulsifier

Read the following questions and answer as per the direction given below: (A) If both ASSERTION and REASON are true and reason is the correct explanation of the assertion. (B)

If both ASSERTION and REASON are true but reason is not the correct explanation of the assertion.

(C) If ASSERTION is true but REASON is false. (D) If ASSERTION is false but REASON is true. 46. Assertion : For adsorption G, H, S all have -ve values. Reason : Adsorption is a exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate. (a) A

(b) B

(c) C

(d) D

47. Assertion : The extent of adsorption of CO2 is much more higher than of H2. Reason : CO2(g) has higher critical temperature and more van der Waals’ force of attraction as compare to H2(g). (a) A

(b) B

(c) C (d) D 48. Assertion : A reaction cannot become fast by itself unless a catalyst is added in the reaction mixture. Reason : Catalyst may increase or decrease the rate of a reaction. (a) A (b) B (c) C (d) D 49. Assertion : Positive catalysts increase the rate of chemical reaction. Reason : Catalysts increase the rate of reaction by lowering the activation energy of the reaction. (a) A (b) B (c) C (d) D 50. Assertion : Micelles are formed by surfactant molecules above the critical micelle concentration (CMC). Reason : The conductivity of a solution having surfactant molecules decrease sharply at the CMC. (a) A

(b) B

(c) C

(d) D

SURFACE CHEMISTRY

251

51. Assertion : For coagulation of positively charged sols, [Fe(CN)6]4– ion has higher coagulating power than that of PO 34 , SO 24  , Cl–.

Reason : Because according to Hardy Schulze rule, higher is the valency of ions for the oppositely charged sol particles, better will be the precipitation. (a) A

(b) B

(c) C

(d) D

Comprehension Based Questions Comprehension A graph between x/m and the pressure P of the gas at a constant temperature is called adsorption isotherm. Where x is the no. of moles of the adsorbate and m is the mass of the adsorbent. Adsorption isotherms of different shapes have been experimentally observed. According to Fnmdlich adsorption isotherm.

This is true when: (a) P = 0 (c)

1 =1 n

(b) P = 1 (d)

1  n

x 54. Graph between log   and log P is a straight line at m o angle 45 with intercept OA as shown.

x/m = kP1/n where k and n are constant parameters depending upon the nature of the solid and gas. 52. In the given isotherm select the incorrect statement:

x Hence,   at a pressure of 2 atm is: m (a) 2 (c) 8

(b) 4 (d) 1

Comprehension

(a)

x  P1/n along OA m

(b)

x  P 0 when point B is reached m

Coagulation is the process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase.The minimum concentration of an electrolyte in milli-moles per litre of the electrolyte solution which is required to cause the coagulation of colloidal sol is called coagulation value. Therefore higher is the coagulating power of effective ion, smaller will be the coagulation value of the electrolyte. Coagulation value 

1 coagulating power

(c)

x does not increase as rapidly with pressure along m BC due to less suraface area available for adsorption

The coagulation values of different electrolytes are different. This behaviour can be easily understood by Hardy-Schulze rule which states.

(d) nature of isotherm is different for two gases for same adsorbent

"The greater is the valency of the effective ion greater is its precipitating power."

x 53. Adsorption isotherm of log   and log P was found of m the type:

55. Which one has the highest coagulation power? (a) K+

(b) Ca2+

(c) Al3+

(d) Sn4+

SURFACE CHEMISTRY

252

56. As2S3 sol is negatively charged, capacity to precipitate it is highest in: (a) K2SO4

(b) Na3PO4

(c) AlCl3

(d) CaCl3

60. Select correct statement: (a) Water in oil emulsions are more viscous than the aqueous emulsions (b) Electrical conductance of aqueous emulsions is less than that of oil emulsions

57. The ability of an ion to bring coagulation of a given colloid depends upon: (a) the sign of its charge

(c) Deemulsification can be done by soap or detergent (d) An emulsion can be diluted with H2O then it is oil in water (O/W) type

(b) magnitude of its charge

(c) both magnitude and sign (d) none of these 58. The coagulation of colloidal particles of the sol can be caused by: (a) heating (b) adding electrolyte (c) adding oppositely charged sol (d) all of these Comprehension Emulsions are normally prepared by shaking the two components together vigorously although some kind to emulsifying agent usually has to added to stabilize the product. This emulsifying agent may be a soap or other surfactant (surface active) species or a lyophilic sol that forms a protective film around the dispersed phase. Emulsions broadly classified into two types: (i) Oil in water emulstions (O/W): Oil acts as dispersed phas and water acts as dispersion medium. (ii) Water in oil emulsions (W/O): Water acts as dispersed phase and oil acts as dispersion medium. Dye test, dilution test may be employed for identification of emulsions. 59. Read two statements: (1) Milk is an example of oil in water (O/W) type emulsion (2) Cold cream is an example of water in oil (W/O) type emulsions (a) Only statement 1 is correct

Matrix–Match Type Questions 61. Column–I

Column–II

(A) Chemisorption

(p) Exothermic

(B)

Physical adsorption

(q) Endothermic

(C)

Desorption

(r) Removal of adsorbed material

(D) Activation of adsorbent 62. Column-I

(s) Highly specific in nature

Column-II

(A) BaSO4

(p) Inhibitor for decomposition of H2 O2

(B) Acetamide

(q) Catalyst

(C) Zeolite

(r) Remove hardness of water

(D) Nickel

(s) Poison for Pd in Lindlar’s catalyst

63. Column–I

Column–II

(A) Liquid dispersed in gas

(p) Foam

(B) Gas dispersed in liquid

(q) Emulsion

(C) Liquid dispersed in solid

(r) Aerosol

(D) Liquid dispersed in liquid

(s) Gel

64. Column–I

Column–II

(A) Milk

(p) Aerosol

(B) Dust

(q) Emulsion

(C) Cheese

(r) Gel

(D) Froth

(s) Foam

65. Column–I

Column–II

(b) Only statement 2 is correct

(A) Peptization

(p) Preparation of sols

(c) Both are correct

(B) Ultra centrifugation

(q) Purification of sols

(d) None of these

(C) Electrodialysis

(r) Preparation of metal sols

(D) Bredig’s are method

(s) Movement of ion across the membrane in presence of electric field

SURFACE CHEMISTRY

253

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTION 1.

Rate of physisorption increases with

(2003, 1M)

6.

(a) decrease in temperature

Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 is

(b) increase in temperature

(2009, 1M)

(c) decrease in pressure (d) decrease in surface area 2.

Lyophilic sols are

(b) CaCl2

(c) Al2(SO4)3

(d) NH4Cl

(2005, 1M)

(a) irreversible sols (b) prepared from inorganic compounds (c) coagulated by adding electrolytes (d) self-stabilishing 3.

(a) Na2SO4

Multiple Choice Questions 7.

The correct statement(s) pertaining to the adsorption of a gas on a solid surface is(are)

Spontaneous adsorption of a gas on solid surface is an exothermic process because (2007, 3M)

(2011)

(a) Adsorption is always exothermic.

(a) H increases for system (b) Physisorption may transform into chemisorption at

(b) S increases for gas

high temperature

(c) S decreases for gas (d) G increases for gas

(c) Physisorption increases with increasing temperature

Read the following question and answer as per the direction given below:

but chemisorption decreases with increasing temperature

(a) Assertion is true; Reason is true; Reason is a correct explanation of Assertion.

(d) Chemisorption is more exothermic than physisorption

(b) Assertion is true; Reason is true; Reason is not the correct explanation of Assertion.

however it is very slow due to higher energy of activation.

(c) Assertion is true; Reason is false. (d) Assertion is false; Reason is correct. 4.

8.

Assertion : Micelles are formed by surfactant molecules above the critical micelle concentration(CMC).

(a) Preferential adsorption of ions on their surface from the solution

Reason : The conductivity of a solution having surfactant molecules decreases sharply at the CMC. (2007)

5.

(a) A

(b) B

(c) C

(d) D

(b) Preferential adsorption of solvent on their surface from the solution. (c) Attraction between different particles having opposite charges on their surface.

Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions, is (2008, 3M)

(d) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles.

(a) CH 3 (CH 2 )15 N  (CH 3 )3 Br  (b) CH 3 (CH 2 )11 OSO 3 Na  (c) CH 3 (CH 2 )6 COO  Na  (d) CH 3 (CH 2 )11 N  (CH 3 )3 Br 

Choose the correct reason(s) for the stability of the Lyophobic colloidal particles. (2012)

9.

The given graph/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct ? (2012)

SURFACE CHEMISTRY

254 Multiple Choice Questions 11.

When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The true statement(s) regarding this adsorption is (are) (2015) (a) O2 is physisorbed (b) Heat is released * (c) Occupancy of  2 of O2 is increased

(d) Bond length of O2 is increased 12.

The correct statement(s) about surface properties is(are) (2017) (a) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system

(a) I is physisorption and II is chemisorption (b) I is physisorption and III is chemisorption (c) IV is chemisorption and II is chemisorption

(b) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature

(d) IV is chemisorption and III is chemisorption 10.

Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25ºC. For this process, the correct statement is (2013) (a) the adsorption requires activation at 25ºC (b) the adsorption is accompanied by a decreases in ethalpy (c) the adsorption increases with increase of temperature (d) the adsorption is irreversible

(c) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium (d) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution

255

SURFACE CHEMISTRY

Note:

SURFACE CHEMISTRY Please share your valuable feedback by scanning the QR code.

256

CHAPTER -1 SOLID STATE

Answer Key CHAPTER -1 SOLID STATE EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

71. Density of copper is = 8 . 9 6 g / c m 72. 136.864 =

73. =

3

6.942 13.6865 Z= =2 6.94 10-1

200 = 5.18 g / cm 3 38.5472

74. 2.16 g cm-3

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (d) 2. (c) 3. 5. (a) 6. (a) 7. 9. (b) 10. (a) 11. 13. (a) 14. (c) 15. 17. (b) 18. (d) 19. 21. (c) 22. (b) 23. 25. (b) 26. (d) 27. 29. (d) 30. (d) 31. 33. (d) 34. (b) 35. 37. (c) 38. (b) 39. 41. (b) 42. (b) 43. 45. (b) 46. (c) 47. 49. (c) 50. (d) 51. 53. (b) 54. (a) 55. 57. (c) 58. (b) 59. 61. (b) 62. x = 0.04 63. 2.571021 unit cells 64. Given, a = 361 pm 65. 152pm 66. Space left = 100 - 74 = 26 % 67. Atomic radius = 1.85 × 10-8

(d) (d) (b) (d) (b) (a) (d) (a) (c) (c) (d) (c) (d) (b) (b)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60.

68. 0.91 g / cm3 69. 

6.023  66.43 = Z  Z = 4 (fcc) 100

70. 100 g =

6.023  1023  100 = 0.5  1025 atoms 12

(a) (d) (c) (c) (b) (d) (b) (b) (b) (b) (c) (a) (a) (d) (b)

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (c)

2.

(a)

3. (b)

4. (c)

5. (a)

6.

(c)

7. (d)

8. (b)

9. (a)

10. (a)

11. (a)

12. (a)

13. (d)

14. (a)

15. (a)

16. (c)

17. (d)

18. (143.00)

19. (d)

20. (d)

21. (a)

22. (c)

257

CHAPTER -1 SOLID STATE EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

74. 0006

75. 1.86 Å

76. 124.3 pm, 7.3

77. 0003

78. (0006)

79. (0008)

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (a)

2.

(c)

3. (b)

4.

(b)

5. (d)

6.

(b)

7. (a)

8.

(c)

9. (d)

10. (a)

11. (c)

12. (c)

13. (c)

14. (b)

15. (d)

16. (c)

17. (c)

18. (d)

19. (a)

20. (c)

21. (b)

22. (b)

23. (b)

24. (a)

25. (c)

26. (c)

27. (d)

28. (a)

29. (c)

30. (d)

31. (a)

32. (c)

33. (b)

34. (d)

35. (a, d)

36. (a, b)

37. (a, b, c) 38. (b,c,d)

39. (a, d)

40. (a, b)

41. (b, d)

42. (a, b)

43. (a, c)

44. (a, c, d)

45. (b, c)

46. (a, c)

47. (b)

48. (c)

49. (b)

50. (c)

51. (a)

52. (b)

10. (b)

11. (a)

12. (a)

13. (a)

53. (d)

54. (a)

55. (c)

56. (a)

14. b,c,d

15. 2

16. 3

17. (a,c)

57. (b)

58. (d)

59. (d)

60. (b)

61. (c)

62. (d)

63. (d)

64. (c)

65. (a)

66. (c)

67. (a)

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (d)

2. (a)

3. (a)

4. (c) 25,

68. A - (p, r, s); B - ( r); C - (q); D - (r) 69. A - (r); B - (s); C - (p, s); D - (q) 70. 1.26

71. 0004

72. 0005

73.0007

5. (a) 5Kg/m3,

(b) metal excess defect

6. 117 pm 7. (b) 8. 217 pm 9. A - p, s ; B - p, q ; C - q ; D - q, r

258

CHAPTER -2 SOLUTIONS

Answer Key CHAPTER -2 SOLUTIONS EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (a) 5. (a) 9. (b) 13. (d) 17. (c) 21. (a) 25. (d) 29. (a) 33. (a) 37. (c) 41. (b) 45. (b) 49. (b) 53. (a) 57. (d) 61. (d) 65. (2.05 M) 69. (50) 73. (0.16)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74.

(d) 3. (a) (b) 7. (b) (b) 11. (a) (c) 15. (c) (c) 19. (d) (c) 23. (b) (a) 27. (d) (d) 31. (c) (a) 35. (c) (b) 39. (b) (a) 43. (a) (a) 47. (a) (a) 51. (b) (c) 55. (a) (a, b, d) 59. (b) (a) 63. (0.086) (1.22) 67. (40 mL) (350) 71. (93 g) (– 0.480 ºC) 75. (0.0558 K)

4. (a) 8. (c) 12. (a) 16. (c) 20. (c) 24.(c) 28. (a) 32. (b) 36. (d) 40. (d) 44. (b) 48. (a) 52. (b) 56. (d) 60.(a) –1 64. (2.28molkg ) 68. (0.01 M) –1 72. (210.0gmol )

DIRECTION TO USE Scan the QR code and check detailed solutions.

1.

(d)

2. (d)

3.

(b)

4.

(b)

5.

(b)

6. (c)

7 . (c)

8.

(c)

9.

(c)

10. (b)

11. (b)

12. (c)

14. (A) 18. (d)

15. (B) 19. (c)

16. (b) 20. (c)

23. 27. 31. 35. 39.

24. 28. 32. 36. 40.

13. (A) 17. (c) 21. 25. 29. 33. 37.

(c) (b) (c) (a) (600.00)

41. (d) 45. 14.00

22. 26. 30. 34. 38.

(a) (b) (d) (047.00) (167.00)

42. (d) 46. 1.76

(c) (a) (c) (d) (d)

43. (b) 47. 10.00

(b) (a) (b) (177.00) 5

44. (c)

259

CHAPTER -2 SOLUTIONS EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (a) 2. (a) 3. (b) 5. (a) 6. (a) 7. (b) 9. (a) 10. (c) 11. (c) 13. (a) 14. (d) 15. (c) 17. (c) 18. (c) 19. (c) 21. (c) 22. (c) 23. (b) 25. (a) 26. (d) 27. (d) 29. (d) 30. (c) 31. (b) 33. (c) 34. (a) 35. (b) 37. (a) 38. (c) 39. (b) 41. (c) 42. (c) 43. (c) 45. (b) 46. (c) 47. (d) 49. (b) 50.(a) 51. (d) 53. (b) 54. (a) 55. (a) 57. (b) 58. (c) 59. (a) 61. (b,d) 62. (a,c) 63. (a,b,d) 65. (b,c) 66. (a,b,c,d) 67. (b, d) 69. (c,d) 70. (d) 71. (d) 73. (b) 74. (a) 75. (b) 77. (a) 78. (a) 79. (A–R,S; B–P,Q; C–R,S; D–P,Q) 80. (A–R,S; B–R; C–P; D–Q) 81. (0034) 82. (0085) 83. (0020)

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68. 72. 76.

(b) (c) (b) (a) (c) (b) (b) (b) (c) (c) (a) (c) (c) (a) (d) (b,d) (a,d) (a) (b)

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (0.23°C) 5. (122;244) 8. (a) 11. (d) 14. (a)

2. 6. 9. 12. 15.

17. (a)

18. (b,c,d)

-4

-1

(1.005×10 min ) (a) (0075) (a) (a)

4. 7. 10. 13. 16.

(a) (c) (a) (b) (0008)

19. (19)

20. (A–R,T; B–P,Q,S; C–P,Q,S; D–P,Q,S,T) 21. (a,b) 22. (c) 23. 1.02 24. 2.98 25. (000.11) 26. (000.20)

260

CHAPTER -3 CHEMICAL KINETICS

Answer Key CHAPTER -3 CHEMICAL KINETICS EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (b) 5. (c) 9. (a) 13. (a) 17. (b) 21. (b) 25. (a) 29. (a) 33. (b) 37. (b) 41. (d) 45. (a) 49. (d) 53. (b) 57. (c) 61. (a) 65. (b)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66.

(d) (b) (b) (a) (d) (b) (c) (c) (c) (a) (d) (b) (d) (c) (a) (b) (d)

69. (a)

70. (5 × 10 M min )

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. –3

–1 –1

71. (0.0031 mol L s ) –3

–1 –1

(c) (d) (d) (a) (b) (c) (d) (c) (b) (c) (b) (b) (d) (b) (d) (d) (a)

–4 –1

–4

–2

–1 –1

–3

79. (3.46 × 10 s) 81. (1853 years)

–1

72. ( 3.3 × 10 M min ) –12

76. (99.9%)

77. (9.1 × 10 L mol s )

(b) (c) (a) (c) (b) (c) (d) (a) (c) (a) (d) (b) (c) (c) (d) (b) (b)

–1

73. (5.4 × 10 mole L s ) 74. (3 × 10 75. (2.7 × 10 s )

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68.

78. (33.775) 80. (4.68 gm)

–1 –1

mol L s )

DIRECTION TO USE Scan the QR code and check detailed solutions.

1.

(c)

2. (a)

3.

(d)

4.

(a)

5.

(b)

6. (a)

7.

(d)

8.

(c)

9.

(c)

10. (d)

11. (3)

12. (b)

13. (A) 17. (B)

14. (B) 18. (D)

15. (D) 19. (A)

16. (D) 20. (B)

21. (B)

22. (A)

23. (D)

24. (B)

25. (C) 29. (c)

26. (A) 30. (c)

27. (C) 31. (060.00)

28. (b) 32. (84297.47)

33. (d)

34.

(d)

35. (c)

36. E a  100kJ / mol

37. (D)

39. (b)

41. 3.98

40. (c)

38. (b)

261

CHAPTER -3 CHEMICAL KINETICS EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (c)

2. (c)

3.

(b)

4.

(a)

5. (d)

6. (a)

7.

(d)

8.

(b)

9. (d)

10. (d)

11. (b)

12. (a)

13. (c)

14. (b)

15. (c)

17. (d)

18. (d)

21. (b)

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (d)

2. 100 kJ

5. (a)

6. T1  33.68h,T2  67.36h,T3  29.76h,

16. (b)

7. (c)

8. (a)

19. (c)

20. (c)

10. (a)Rate  k  A  ,(b)  0.5s-1

22. (a)

23. (d)

24. (d)

25. (b)

26. (d)

27. (c)

28. (d)

29. (b)

30. (c)

31. (a)

32. (d)

33. (c)

34. (d)

35. (d)

37. (b)

38. (d)

41. (d)

42. (c)

45. 49. 53. 57.

(c) (a) (a) (d)

46. 50. 54. 58.

(b) (a) (a, b) (b)

(b)

(c)

13. (c)

14. (d)

15. (d)

16. (c)

17. (b)

36. (a)

18. (d)

19. (c)

20. (b)

21. (b)

39. (d)

40. (d)

22. (0009)

23. (a)

24. (d)

25. (b)

43. (a)

44. (b)

26. (d)

27. (b)

28. (d)

29. (d)

47. 51. 55. 59.

48. 52. 56. 60.

(b) (b) (a) (c)

30. (a)

31. (d)

32. (b)

33. (b, c, d)

34. (a, b)

35. (a, d)

37. (a), (b), (d) 38. 6.75

39. 2.30

40. (001.20)

41. (a)

(a) (d) (a) (b)

63. (c)

64. (a)

65. (d)

66. (c)

67. (a, b, c)

68. (a, c)

69. (a, d)

70. (a, b, c)

71. (a, b)

72. (a, b, c)

73. (b, c)

74. (a, b)

75. (a, b, c)

76. (d)

77. (b)

78. (c)

79. (c)

80. (d)

81. (b)

82. (a)

83. (c)

84. (b)

85. (a)

86. (b)

87. (a)

 A-R; B-S;C-Q; D-P  90. (0004)

4.

12. (a)(b),(c)(d)

62. (b)

89. (0008)

9.

25 min.

11. (c)

61. (d)

88.

3.

91. 4 hours

92. (0003)

262

CHAPTER -4 ELECTROCHEMISTRY

Answer Key CHAPTER -4 ELECTROCHEMISTRY EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73. 77. 81. 85. 89. 93. 95. 97.

(c) (c) (d) (c) (d) (a) (d) (c) (c) (d) (d) (a) (a) (d) (c) (b) (d) (c) (a) (c) (d) (c) (b) (a) –1 (3.9 cm ) (0.2)

2. (d) 3. (a) 4. 6. (c) 7. (c) 8. 10. (b) 11. (c) 12. 14. (d) 15. (a) 16. 18. (b) 19. (c) 20. 22. (b) 23. (b) 24. 26. (b) 27. (b) 28. 30. (a) 31. (d) 32. 34. (a) 35. (b) 36. 38. (d) 39. (b) 40. 42. (b) 43. (d) 44. 46. (d) 47. (c) 48. 50. (c) 51. (d) 52. 54. (c) 55. (c) 56. 58. (b) 59. (d) 60. 62. (b) 63. (b) 64. 66. (c) 67. (c) 68. 70. (a) 71. (a) 72. 74. (b) 75. (a) 76. 78. (c) 79. (d) 80. 82. (a) 83. (a) 84. 86. (a) 87. (d) 88. 90. (a) 91. (b) 92. –1 94. (0.616 cm ) –5 –3 96. (3.442 × 10 mol dm ) 98. (126) 99. (128 S cm2 mol–1)

 k 1000188   100.  xy  

101. ( 2.6 × 10 M)

102. (30)

104. (w)

103. (84 g)

(d) (c) (c) (a) (a) (d) (b) (a) (b) (d) (b) (d) (c) (c) (c) (b) (c) (b) (c) (b) (c) (c) (a)

–5

105. (Text)

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (d)

2. (c)

3.

(a)

4.

(a)

5. (b)

6. (d)

7.

(d)

8.

(d)

9. (a)

10. (b)

11. (a)

12. (a)

13. (C)

14. (B)

15. (B)

16. (C)

17. (C)

18. (D)

19. (B)

20. (A)

21. (C)

22. (C)

23. (C)

24. (96500.00)

27. (058.00)

28. (060.00)

31. (006.00) 35. (a) 39. (a)

32. (b) 36. (a)

25.

(144.00) 26. (d)

29. (b) 33. 11 37. 1.52

30. (a) 34. (a) 38. 2.15

263

CHAPTER -4 ELECTROCHEMISTRY EXERCISE - 3 : ADVANCED OBJECTIVE QUESTION

109. (a,b,c,d) 110. (c)

111. (c)

112.(a)

113. (a) 114. (A  q; B  p,r; C  s; D  t) 115. (A  p, q; B  p,q; C  r; D  p,q,s) 116. (0009) 117. (0001) 118. (12.53 ltr Cl2, 44.76gm of NaOH) 119. (0035)

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. 5.

(d) (d)

2. (c) 6. (c)

3. 7.

(a) (c)

9.

(b)

10. (b)

11. (b)

12. (c)

13. (c)

14. (a)

15. (a)

16. (b)

17. (b)

18. (d)

19. (d)

20. (b)

21. (d)

22. (a)

23. (a)

24. (a)

25. (b)

26. (a)

27. (c)

28. (c)

29. (a)

30. (a)

31. (c)

32. (a)

33. (b)

34. (d)

35. (b)

36. (a)

37. (c)

38. (b)

39. (c)

40. (b)

41. (d)

42. (c)

43. (b)

44. (b)

45. (b)

46. (b)

47. (a)

48. (c)

49. (b)

50. (a)

51. (d)

52. (d)

53. (b)

54. (c)

55. (c)

56. (a)

57. (c)

58. (d)

59. (d)

60. (d)

61. (a)

62. (d)

63. (a)

64. (a)

65. (b)

66. (c)

67. (b)

68. (c)

69. (a)

70. (b)

71. (c)

73. (c)

74. (c)

77. (d)

4. 8.

(b) (d)

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (c) 5. (b) 9. (c) 13. (b) 17. (d) 21. (a)

2. 6. 10. 14. 18. 22.

72. (b)

25. (d)

26. C  8  105 M

75. (d)

76. (a)

28. (ii) H 0 =-49.987 kJ, S0 =-96.5 J (iii)1.6×10-10

78. (c)

79. (c)

80. (a)

29. 1.6 1010

81. (b)

82. (d)

83. (b)

84. (d)

30. K=1010

85. (c)

86. (d)

87. (b)

88. (b)

31. 2  103

89. (c)

90. (a)

91. (c)

92. (c)

32. 55 (in terms of 107 Sm-1 )

93. (a,b,c)

94. (a,d)

95. (c)

96. (a,b,c)

97. (a,c)

98. (a,b,d)

99. (b,d)

100. (a,b)

101. (a,b,d)

102. (a,b,c,d) 103.(d)

104. (a)

33. (10) 34. –11.62 J mol–4 k–1 35. (a)

105. (d)

106. (b)

108. (b)

107. (a)

(d) (b) (b) (a) (d) (b)

3. 7. 11. 15. 19. 23.

(d) (c) (a, c) (d) (c) (c)

4. 8. 12. 16. 20. 24.

(b) (a) (a, b, d) (b) (d) (b)

264

CHAPTER -5 SURFACE CHEMISTRY

Answer Key CHAPTER -5 SURFACE CHEMISTRY EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

EXERCISE - 2 : PREVIOUS YEAR JEE MAIN QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (c)

2.

(a)

3. (b)

4. (a)

5. (d)

6.

(c)

7. (c)

8. (a)

9. (a)

10. (a)

11. (d)

12.(a)

13. (b)

14. (c)

15. (c)

16.(b)

17. (d)

18. (b)

19 (c)

20.(d)

21. (a)

22. (d)

23. (d)

24.(c)

25. (b)

26. (d)

27. (c)

28.(b)

29. (a)

30. (b)

31. (c)

32.(d)

33. (a)

34. (c)

35. (d)

36.(a)

37. (c)

38. (d)

39. (b)

40.(b)

41. (b)

42. (c)

43. (d)

44.(b)

45. (d)

46. (a)

47. (b)

48.(b)

49. (b)

50. (c)

51. (a)

52.(b)

53. (c)

54. (d)

55. (b)

56.(d)

57. (a)

58. (1000)

59. 4.00

60.4.00

61. 60.00 65. 2.00

62. 10.00 66. 2.00

63. 0.60 67. 3.00

64.3.00 68.4.00

1. (c)

2.

(c)

3. (b)

4. (b)

5. (a)

6 . (c)

7. (b)

8. (c)

9. (a)

10. (d)

11. (d)

12.(A)

13. (A)

14. (B)

15. (A)

16.(B)

17. (C)

18. (B)

19. (A)

20.(A)

21. (A)

22. (C)

23. (048.00)

24.(c)

25. (d)

26. (c)

27. (a)

28.(d)

29. (d) 33. 0.3675

30. (b) 34. (c)

31. (c)

32.48.00

CHAPTER -5 SURFACE CHEMISTRY

265

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

DIRECTION TO USE Scan the QR code and check detailed solutions.

DIRECTION TO USE Scan the QR code and check detailed solutions.

1. (b)

2.

(b)

3. (a)

4. (a)

5. (b)

6.

(b)

7. (a)

8. (a)

9. (d)

10. (d)

11 (c)

12.(d)

13. (c)

14. (d)

15. (b)

16.(d)

17. (a)

18. (a)

19. (b)

20.(c)

21. (d)

22. (b)

23 (a)

24.(c)

25. (a)

26. (c)

27. (d)

28.(c)

29. (d)

30. (c)

31. (b)

32.(c)

33. (a)

34. (a)

35 (d)

36.(d)

37. (a,b,d)

38. (a,b,d)

39. (a,b,c,d)

40.(a,b,c)

41. (a,b,c)

42. (a,b,d)

43. (a,b,d)

44.(a,c,d)

45. (a,c,d)

46. (a)

47. (a)

48.(d)

49. (a)

50. (a)

51. (a)

52.(b)

53. (c)

54. (b)

55. (d)

56.(c)

57. (c)

58. (d)

59. (c)

60.(d)

61. (A – p, s; B – p; C – q, r; D – q) 62. (A – s; B – p; C – q, r; D – q) 63. (A – r; B – p; C – s; D – q) 64. (A – q; B – p; C – r; D – s) 65. (A – p; B – q; C – q, s; D – p, r)

1. (a)

2.

(d)

3. (c)

4. (b)

5. (a)

6.

(c)

7. (a,b,d)

8. (a,d)

9. (a,c)

10. (b)

11 (b,c,d)

12.(a,b)

266

Note