College PhysicalCollege Physical Chemistry (For Second Year B.Sc.) (According to Revised Syllabus...

College PhysicalChemistry

(For Second Year B.Sc.)(According to Revised Syllabus 2017-2018)

K.B. BaligaEx-Head, Dept. of Chemistry,Mithibai College of Arts and

Chauhan Institute of Science, Mumbai.

S.A. ZaveriEx-Head, Dept. of Chemistry,

Maharashtra College,Mumbai.

Dr. Shankar S. MangaonkarDept. of Chemistry,

Mithibai College of Arts andChauhan Institute of Science,

Mumbai.

Dr. Arti SawantVice-Principal, Dept. of Chemistry,S.S. & L.S. Patkar-Varde College of

Arts, Science & Commerce,Goregaon, Mumbai.

Deepak TeckchandaniDept. of Chemistry,

C.H.M. College, Ulhasnagar.

Asha MathewDept. of Chemistry,

S.I.W.S. College, Wadala, Mumbai.

ISO 9001:2008 CERTIFIED

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First Edition : 1978Fifteenth Edition : 2009(as per Revised Syllabus)Sixteenth Edition : 2010Reprint : 2011, 2012Reprint : 2014Seventeenth Edition : 2015(as per Revised Syllabus)Eighteenth Edition : 2016Nineteenth Edition : 2017(as per Revised Syllabus)

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Preface

It gives a sense of satisfaction to the authors to bring out thiseighteenth edition of the book written as per the revised syllabus forS.Y. B.Sc. class laid down by the University of Mumbai which is inforce from June 2017. The primary interest of the authors has been topresent the subject-matter with simplicity and clarity. Facts andconcepts have been illustrated with a large number of solved problems.

According to the revised syllabus, use of S.I. Units has been madethroughout. The outstanding feature of this book is a brief note on S.I.Units with their notations and their values which have been given beforethe first chapter. Also the questions and problems given at the end ofeach chapter will be most useful to the students to prepare for theirexaminations. All changes have been made after the discussion inworkshop.

We also express our deep sense of gratitude to our publisher,M/s. Himalaya Publishing House Pvt. Ltd., for their untiring efforts andco-operation in bringing out the book in time in such an elegant form.

Every effort has been made to avoid printing errors, though somemight have crept in inadvertently. Valuable suggestions and criticismsfor the improvement of the book from our colleagues who are teachingthis course will be highly appreciated.

AUTHORS

SyllabusS.Y. B.Sc. Mumbai University

(With effect from the Academic Year 2017-2018)

Semester IIIPaper I

Topics

1.1 Chemical Thermodynamics-II (8L)1.1.1 Free Energy Functions: Helmholtz Free Energy, Gibb’s Free

Energy, Variation of Gibb’s Free Energy with Pressure andTemperature. Gibbs-Helmholtz Equation.

1.1.2 Thermodynamics of Open System: Partial Molal Properties,Chemical Potential and Its Variation with Pressure andTemperature, Gibb’s Duhem Equation.

1.1.3 Concept of Fugacity and Activity.1.1.4 van’t Hoff Reaction Isotherm and van’t Hoff Reaction Isochore.

(Numericals expected).

1.2 Electrochemistry (7L)1.2.1 Conductivity, Equivalent and Molar Conductivity and their

Variation with Dilution for Weak and Strong Electrolytes.1.2.2 Kohlrausch Law of Independent Migration of Ions.1.2.3 Applications of Conductance Measurements: Determination of

Degree of Ionization And Ionization Constant of WeakElectrolyte, Solubility and Solubility Product of SparinglySoluble Salts, Ionic Product of Water. (Numericals expected).

1.2.4 Transference Number and Its Experimental Determination usingMoving Boundary Method. (Numericals expected). FactorsAffecting Transference Number.

Paper II

1.1 Chemical Kinetics-II (7L)1.1.1 Types of Complex Chemical Reactions: Reversible or Opposing,

Consecutive and Parallel Reactions (No derivations, onlyexamples expected),Thermal Chain Reactions: H. and Br. Reaction. (Only stepsinvolved, no kinetic expression expected).

1.1.2 Effect of Temperature on the Rate of Reaction, ArrheniusEquation, Concept of Energy of Activation (Ea). (Numericalsexpected).

1.1.3 Theories of Reaction Rates: Collision Theory and ActivatedComplex Theory of Bimolecular Reactions. Comparisonbetween the Two Theories (Qualitative treatment only).

1.2 Solutions (8L)1.2.1 Thermodynamics of Ideal Solutions: Ideal Solutions and

Raoult’s Law, Deviations from Raoult’s Law–Non-idealSolutions. Vapour Pressure-Composition and Temperature -Composition Curves of Ideal and Non-ideal Solutions.Distillation of Solutions. Lever Rule. Azeotropes.

1.2.2 Partial Miscibility of Liquids: Critical Solution Temperature;Effect of Impurity on Partial Miscibility of Liquids with Respectto Phenol-Water , Triethanolamine – Water and Nicotine – WaterSystems.

1.2.3 Immiscibility of Liquids- Principle of Steam Distillation.1.2.4 Nernst Distribution Law and its Applications, Solvent

Extraction.

Semester IVPaper I

Topics

1.1 Electrochemistry-II (8L)1.1.1 Electrochemical Conventions, Reversible and Irreversible Cells.1.1.2 Nernst Equation and Its Importance, Types of Electrodes,

Standard Electrode Potential, Electrochemical Series(Numericals expected).

1.1.3 Thermodynamics of a Reversible Cell, Calculation ofThermodynamic Properties: G, H and S from EMF Data.(Numericals expected).

1.1.4 Calculation of Equilibrium Constant from EMF Data.(Numericals expected).

1.1.5 Concentration Cells with Transference and without Transference.Liquid Junction Potential and Salt Bridge.

1.1.6 pH Determination using Hydrogen Electrode and QuinhydroneElectrode.(Numericals expected).

1.2 Phase Equilibria (7L)1.2.1 Phases, Components and Degrees of Freedom of a System,

Criteria of Phase Equilibrium. Gibbs Phase Rule and itsThermodynamic Derivation.

1.2.2 Derivation of Clapeyron Equation, Derivation of Clausius –Clapeyron Equation and its Importance in Phase Equilibria.(Numericals expected).

1.2.3 Phase Diagrams of one-Component Systems (Water andSulphur).

1.2.4 Two Component Systems Involving Eutectics (Lead-silverSystem)., Congruent Melting Point Zinc Magnesium System.Incongruent Melting Point Sodium-potassium System.

Paper II1.1 Solid State (7L)

1.1.1 Recapitulation of Laws of Crystallography and Types ofCrystals.

1.1.2 Characteristics of Simple Cubic, Face Centered Cubic and BodyCentered Cubic Systems, Interplanar Distance in Cubic Lattice(Only expression for ratio of interplanar distances are expected).

1.1.3 Use of X-rays in the Study of Crystal Structure, Bragg’sEquation (Derivation expected), X-rays Diffraction Method ofStudying Crystal Lattice Structure, Structure of NaCl and KCl.Determination of Avogadro’s Number (Numericals expected).

1.2 Catalysis (8L)1.2.1 Types of Catalysis, Catalytic Activity, Specificity and

Selectivity, Inhibitors, Catalyst Poisoning and Deactivation1.2.2 Mechanisms and Kinetics of Acid-base Catalyzed Reactions,

Effect of pH.1.2.3 Mechanisms and Kinetics of Enzyme Catalyzed Reactions

(Michaelis-Menten equation).1.2.4 Effect of Particle Size and Efficiency of Nanoparticles as

Catalyst.

Contents

Semester IIIPaper Chapter Topics Page No.

I 1.1 Chemical Thermodynamics-II 1 - 25

1.2 Electrochemistry 26 - 63

II 1.1 Chemical Kinetics-II 64 - 88

1.2 Solutions 89 - 116

Semester IVI 1.1 Electrochemistry-II 117 - 158

1.2 Phase Equilibria 159 - 181

II 1.1 Solid State 182 - 210

1.2 Catalysis 211 - 229

Semester III Paper I

Chapter 1.1

CHEMICAL THERMODYNAMICS-II

Chemical thermodynamics deals with the applications of laws ofthermodynamics to chemical systems. The first law of thermodynamicsalso known as the law of conservation of energy does not give any ideaabout the direction of a chemical process. According to second law ofthermodynamics, entropy can be said to be measure of unavailableenergy. The concept of free energy is very important as it gives theamount of available energy to perform useful work, that is, to carry outphysical or chemical process. The free energy change can be used topredict the spontaneous nature of a chemical process and also can beused to study both physical and chemical equilibria.1.1.1 FREE ENERGY FUNCTIONS(1) Helmholtz Free Energy

The concept of Helmholtz free energy was given by Germanphysicist Hermann von Helmholtz (1821-1894). The symbol A forHelmholtz free energy is taken from German word ‘Arbeit’ whichmeans work. The function A is also known as work function and itrepresents the maximum amount of energy which is available for beingconverted into work.

Helmholtz free energy ‘A’ is extensive property and a statefunction.

Hemholtz free energy is given by the relationA = E – TS … (1)If A1 and A2 are the Helmholtz free energies of system in its initial

and final states respectively, thenA1 = E1 – T1S1 … (2)A2 = E2 – T2S2 … (3)

2 College Physical Chemistry

Where, E1 and E2 are the internal energies of system in its initialand final states and S1 and S2 are the entropies of system in its initial andfinal states respectively.

The change in Helmholtz free energy of system is given by therelation

A = A2 – A1 … (4) A = (E2 – T2S2) – (E1 – T1S1) … (5)At constant temperature T1 = T2 = T A = (E2 – E1) – T (S2 – S1) … (6) A = E – TS … (7)For a reversible process at constant temperature

S =T

qrev … (8)

TS = qrev … (9)Substituting in equation (7) givesA = E – qrev … (10)But by first law of thermodynamicsE = q + W … (11)It follows that,Wrev = E – qrev … (12)Since process is carried out reversibly, W represents maximum

work Wmax

Wmax = E – qrev … (13)Comparing equations (10) and (13) A = Wmax … (14)or – A = – Wmax … (15)This indicates that decrease in Helmholtz free energy (i.e., –A)

gives the maximum work that can be done by the system during thegiven change.(2) Gibbs Free Energy (Net Work)

American physicist J.W. Gibbs (1839-1903) introduced anotherthermodynamic function known as Gibbs free energy. It relates to network done by the system.

Gibbs free energy ‘G’ is extensive property and a state function.

Chemical Thermodynamics-II 3

Gibbs free energy is given by the relationG = H – TS … (16)If G1 and G2 are the Gibbs free energies of system in its initial and

final states respectively, thenG1 = H1 – T1S1 … (17)G2 = H2 – T2S2 … (18)Where H1 and H2 are the enthalpies of system in its initial and final

states and S1 and S2 are the entropies of system in its initial and finalstates respectively.

The change in Gibbs free energy of system is given by the relationG = G2 – G1 … (19)

G = (H2 – T2S2) – (H1 – T1S1) … (20)At constant temperature T1 = T2 = T G = (H2 – H1) – T (S2 – S1) … (21) G = H – TS … (22)But at constant pressure, H = E + PV … (23)Substituting in equation (22) gives G = E + PV – TS … (24)But by equation (7) E – TS = A G = A + PV … (25)The above equation gives relation between the Gibbs free energy

change and Helmholtz free energy change.Since A = Wmax

G = Wmax + PV … (26) –G = – Wmax – PV … (27)But work done due to expansion of gas against constant pressure is

given by Wexp = – PV PV = –Wexp

G = – Wmax + Wexp = – (Wmax – Wexp) = – Wnet … (28)This indicates that decrease in Gibbs free energy (i.e., –G) gives

the net work that can be done by the system during the given change.Significance of Gibbs Free Energy Change

The change in Gibbs free energy enables to predict if the reactionis spontaneous or not.


(i) If G is negative, the reaction is spontaneous.(ii) If G is positive, the reaction is non-spontaneous.

(iii) If G = 0, the reaction is at equilibrium.The results of G = H – TS can be summarised as follows:

Sign of H Sign of S Sign of G Conclusion(i) – ve – ve – ve Always

spontaneous(ii) + ve – ve + ve Non-Spontaneous(iii) – ve – ve – ve (At low T)

+ ve (At high T)SpontaneousNon-Spontaneous

(iv) + ve + ve – ve (At high T)+ ve (At low T)

SpontaneousNon-Spontaneous

Variation of Free Energy with Temperature and PressureGibbs free energy (G) undergoes change with temperature and

pressure. The Gibbs free energy is give n by the relationG = H – TS … (29)But H = E + PV … (30) G = E + PV – TS … (31)Complete differentiation givesdG = dE + PdV + VdP – TdS – SdT … (32)By first law of thermodynamics idf work done in expansion is PV,

thendq = dE + PdV

Also S =T

dqrev

TS = dqrev = dE + PdV … (33)Substituting in the equation (32) givesdG = VdP – SdT … (34)

(i) At constant temperature dT = 0 dG = VdP …(35)

orTP

G

= V … (36)


(ii) At constant pressure dP = 0 dG = – SdT … (37)

orPT

G

= – S … (38)

Gibbs-Helmholtz EquationJ. Willard Gibbs and H. von. Helmholtz derived a relation between

free energy change (G), enthalpy change (H) and rate of change offree energy with temperature at constant pressure.

PTG

)(

This relationship is known as “Gibbs-Helmholtz equation”. It isused to evaluate important thermodynamic parameters such as H, E,S etc. It is widely used to study the physical and chemical changes.

Gibbs Free energy G of a system is given byG = H – TS … (39)But H = E + PV G = E + PV – TS … (40)Complete differentiation of equation (40) givesdG = dE + PdV + VdP – TdS – SdT … (41)According to first law of thermodynamicsdq = dE + PdV

Also dS =T

dqrev

TdS = dE + PdVor dE = TdS – PdV … (42)Substituting this in equation (41) givesdG = VdP – SdT … (43)At constant pressure dP = 0 dG = – SdT … (44)If G1 and G2 are the free energies of system in its initial and final

states respectively.


Let dG1 and dG2 be the changes in the free energies of systemwhen the temperature is changes by a very small amount dT, then byequation (44).

dG1 = – S1dT … (45)dG2 = – S2dT … (46)

where, S1 and S2 are the entropies of the system in its initial and finalstates respectively.

Subtraction of equation (45) from (46) givesdG2 – dG1 = – S2dT + S1dT

dG2 – dG1 = – (S2 – S1) dT d(G) = – (S) dT

At constant pressure,PT

G

)( = – S … (47)

But G = H – TS

–S =T

HG … (48)

From (47) and (48)

THG =

PTG

)( … (49)

G = H +PT

GT

)( … (50)

This equation is called as Gibbs-Helmholtz equation. The term

PTG

)( is called temperature coefficient of free energy change.

Example 1: Calculate the free energy change of a process whoseenthalpy change at 373 K is – 264.93 kJ and the temperature coefficientof the process is 19.58 J.

Solution: By Gibbs-Helmholtz equation

G = H +PT

GT

)(

= –264.93 + 373 ×1000

58.19

= – 264.93 + 7.3 = – 257.63 kJ


Example 2: The free energy change accompanying a given processis – 91.21 kJ at 293 K and – 89.12 kJ at 303 kJ. Calculate the enthalpychange for the process at 298 K.

Solution: By Gibbs-Helmholtz equation

G = H +PT

GT

)(

dG1 at 293 K = – 91.21 kJdG2 at 303 K = – 89.12 kJ

kJTTdGdG

TG

P209.0

10209

293303)21.91(12.89)(

12

12

G value at 298 K may be taken as the average of dG values at293 K and 303 K.

kJdGdG

G 165.902

)12.89(21.912

21

Substituting in Gibbs-Helmholtz equation gives– 90.165 = H + 298 × 0.209 H = – 90.165 – 62.282 = – 152.447 kJ.

1.1.2 THERMODYNAMICS OF OPEN SYSTEMPartial Molal Properties

The properties of system can be divided into two types extensiveand intensive properties. The extensive property is the property whichdepends on the total amount of the substance or the total amounts of thedifferent constituents of the system. Mass, volume, energy, entropy, freeenergy are extensive properties. On the other hand, an intensiveproperty does not depend on the total mass or the total amounts ofdifferent constituents of the system. Pressure, temperature, density,refractive index are the intensive properties.

The total mass of the system is extensive property but mass permole is intensive property. Volume per mole, energy per mole are alsointensive properties. The total mass of a closed and isolated systemsalways remains constant. The change in any thermodynamic property isdue to change in the state of the system and not due to addition orremoval of matter. But for open systems as the mass and thecomposition of the constituents present in the system can also vary, it isnecessary to take into consideration the contribution of each constituent


to the total property of the system. G.N. Lewis introduced the conceptof ‘partial molal properties’ for studying open systems.

Consider an open system composed of ‘i’ constituents. Let n1, n2,n3, …., ni be the number of moles of constituents 1, 2, 3,… i present in thesystem. Let X be the thermodynamic extensive property of the systemselected for study. Its value is determined by the state of the system (i.e.,temperature ‘T’ and pressure ‘P’) and the amounts of variousconstituents present in the system. It means the property X is a functionof temperature, pressure and amounts of various constituents present inthe system.

i.e., X = f (P,T, n1, n2, n3, …., ni) … (51)The change in property dX due to small change in pressure,

temperature and amounts of various constituents is given by

innTPNPNT nX

TX

PXdX

,...,,1,,2

iii nnTPinnnTPnX

nX

)1(131 ,....,,,,...,,,2

...

... (52)

The quantityinnTP

nX

,...,,,12

is called partial molal property of 1st

constituent and is denoted by 1X

The quantityinnnTP

nX

...,,,,231

is called partial molal property of

2nd constituent and is denoted by 2X

The quantity)1(21 ...,,,,

innnTPinX is called partial molal property of

ith constituent and is denoted by iX

Thus,innTP

nXX

,...,,,11

2

… (53)

dP dT dn1

dn2 dni


innnTPnXX

,...,,,,22

31

… (54)

)1(21 ,...,,,,

innnTPii n

XX … (55)

Substituting in equation (52) gives

iiNTNP

dnXdnXdnXTX

PXdX

...2211,,

... (56)

At constant pressure and temperature, the first two terms on R.H.S.of equation (56) are zero

dX = iidnXdnXdnX ...2211 … (57)

From equation (53) it can be concluded that the partial molalproperty signifies the change in the extensive property X of the systemwith the change in the number of moles of first constituent of the systemat constant pressure and temperature when the number of moles of otherconstituents remains unchanged. In other words, the partial molalproperty is the change in the property X of system at constantpressure and temperature when one mole of a particular constituentis added to such a large quantity of system so that the added moledoes not affect the composition of the system.Partial Molal Volume

The partial molal volume of ith component of system is given by therelation.

jnnnTPii n

VV,...,,,, 21

… (58)

Where nj indicates number of each components except ith are heldconstant.

The partial molal volume can be defined as the change in volumeof system when 1 mole of ith component is added to a large quantity ofsystem at constant pressure and temperature keeping number of molesof other components constant.Partial Molal Free Energy (Chemical Potential)

The partial molal free energy of ith component of system is givenby the relation.

dP dT


jnnnTPii n

GG,...,,,, 21

… (59)

The partial molal free energy is also known as chemical potentialand it is denoted by symbol .

i.e.,jnnnTPi

i nG

,...,,,, 21

… (60)

The chemical potential can be defined as the change in free energyof the system when one of ith component is added to a large quantity ofsystem at constant pressure and temperature keeping number of molesof other components constant.

Chemical potential ‘’ is an intensive property. It is a propertywhich determines the reactivity of the system. A change in the system atconstant pressure and temperature will proceed from state of a higherchemical potential to the lower chemical potential, until its value isuniform throughout the system.

For a binary mixture, the change in free energy is given by therelation

12 ,,2,,1,, nTPnTPNPNT nG

nG

TG

PGdG

… (61)

At constant pressure and temperature the above equation can bewritten as:

12 ,,2,,1,)(

nTPnTPTP n

GnGdG

… (62)

But2,,1

1nTP

nG

and1,,2

2nTP

nG

Substituting in equation (62) gives(dG)P,T = 1dn1 + 2dn2 … (63)On integrating (63) gives(G)P,T = 1n1 + 2n2 … (64)or in general (G)P,T = ini … (65)For a single (pure) component

dP dT dn1 dn2

dn1 dn2


(G)P,T = n = n G … (66)

nGG … (67)

Hence, for a pure substance, the chemical potential is given by itsfree energy per mole.Variation of Chemical Potential with Pressure and Temperature(i) Variation of chemical potential with pressure

jnnnTPii n

G

,...,,,, 21

Differentiating w.r.t. pressure at constant temperature andcomposition

jj nTinT

i

nG

PP,,

… (68)

The variation of free energy with pressure at constant temperatureis given by

dG = VdP or VPG

T

Differentiating w.r.t. ni at constant temperature and composition

inTinTi

VnV

PG

njj

,,… (69)

From (68) and (69)

inT

i VP

j

,

... (70)

This signifies that the rate of change of chemical potential withrespect to pressure of any component of the system is equal to partialmolal volume of that component at constant temperature andcomposition.


(ii) Variation of chemical potential with temperature

jnnnTPii n

G

,...,,,, 21

Differentiating w.r.t. temperature at constant pressure andcomposition.

jj nPinP

i

nG

TT,,

… (71)

The variation of free energy with temperature at constant pressureis given by

dG = – SdT orPT

G

= – S

Differentiating w.r.t. ni at constant pressure and composition

inPinPi

SnS

TG

njj

,,… (72)

From equations (71) and (72)

inP

i ST

j

,

… (73)

This signifies that the rate of change of chemical potential withrespect to temperature of any component of the system is equal to partialmolal entropy of that component at constant pressure and composition.Gibbs-Duhem Equation

Consider an open system composed of i constituents. Let n1, n2,n3, ..., ni be the number of moles of constituents 1, 2, 3,..., i present in thesystem. The free energy G of a system is an extensive property and it isa function of pressure, temperature and the amounts of variousconstituents present in the system.

i.e., G = f (P, T, n1, n2, n3, …, ni) … (74)The free energy change due to small change in pressure,

temperature and amounts of various constituents is given by,

innTPNPNT nG

TG

PGdG

,...,,,1,,2

dP dT dn1


ji nnTPinnnTPnG

nG

,...,,,,...,,,,2131

...

... (75)

ButjnnnTPi

i nG

,...,,,, 21

At constant pressure and temperature, equation (75) can be writtenas:

(dG)P,T = 1dn1 + 2dn2 + …. + idni … (76)Integrating equation (76) gives(G)P,T = 1n1 + 2n2 + … + ini … (77)General differentiation of (77) gives(dG)P,T = 1dn1 + n1d1 + 2dn2 + n2d2 + … + idni + nidi … (78)On rearranging(dG)P,T = (1dn1 + 2dn2 + … + idni) + (n1d1 + n2d2 + …

+ nidi) … (79)Comparing equations (76) ad (79) givesn1d1 + n2d2 + … + nidi = 0or nidi = 0 … (80)This equation is called Gibbs-Duhem equation. It is very

important in study of chemical potential of binary systems, solutions,liquid-vapour equilibria, phase equilibria etc.

For two component system equation (80) becomes,n1d1 + n2d2 = 0 … (81) n1d1 = – n2d2 … (82)

d1 = 21

2 dnn

… (83)

This equation shows that the change of chemical potential of onecomponent brings about change in chemical potential of othercomponent of system. Negative sign indicates that if chemical potentialof one component increases, then that of other decreases. But the totalsum of change of chemical potentials is always zero.

dnidn2


Example 3: 22.6 g of an aqueous solution of ethanol contains 4.6 gethanol. If change of chemical potential of ethanol is –10 kJ. Findchange of chemical potential for water.

Solution: n1 = 4.6/46 = 0.1 (mol. of ethanol)n2 = 18/18 = 1 (mol. of water)

d1 = –10 kJ (ethanol)d2 = ? (water)

n1d1 + n2d2 = 00.1 × (–10) + 1 × d2 = 0 d2 = + 1 kJChemical potential of water is increased by 1 kJ.

1.1.3 CONCEPTS OF FUGACITY AND ACTIVITYFugacity, Activity and Activity Coefficient

The chemical potential of ith component is given by,

inT

i VP

j

,

For a single (pure) component system

VP T

… (84)

For an ideal gas PV = nRT

PRTVRT

nVP … (85)

Substituting in (84) gives,

PdPRTd … (86)

Integrating equation (86) gives. = RT ln P + C … (87)

where, C is integration constant. At standard state when P = 1, then =0, where 0 is the chemical potential under standard conditions of unitatmosphere.

Hence equation (87) becomes, = 0 + RT ln P … (88)


The above equation (88) is applicable only for ideal gases. It is notapplicable for real gases.

To apply equation (88) for real gases, Lewis introduced the conceptof fugacity which is denoted by symbol f. Fugacity represents the‘escaping tendency’ and has the same dimensions as pressure.

Substituting in equation (88) gives, = 0 + RT ln f … (89)Since an real gas behaves ideally only when p 0, it follows that

f p, when p 0

I.e., 1lim0

p

fp

… (90)

The ratio f / p is called activity coefficient and is denoted by symbol .

Thuspf

… (91)

Thus, the activity coefficient can be defined as the ratio of fugacityto pressure of gas, which gives a direct measure if the extent to whichany real gas deviates from ideality at any given pressure andtemperature.

For an ideal gas = 1 … (92) For an ideal gas f = p … (93)In a similar way, the chemical potential of any ith component in an

ideal solution, is given byi = i0 + RT ln Ci … (94)

where, Ci is the concentration of ith component in ideal solution. But innon-ideal solution, there exists electrostatic attraction, ionisation,reaction of solute with solvent etc. Hence more exact results areobtained when concentration is replaced by activity in equation (94).The equation (94) thus changes as:

i = i0 + RT ln ai … (95)where, ai is the activity of ith component in the solution. The activity isrelated with the concentration of the solution by the equation,

a = C ×

orCa

... (96)


where, is called activity coefficient and can e defined as the ratio ofactivity to the concentration of a particular component in a solution andit is the measure of deviation of a solution from its ideal behaviour.1.1.4 VAN’T HOFF’S REACTION ISOTHERM

Every chemical reaction has free energy change (G) which isgiven by van’t Hoff’s reaction isotherm.

Consider a reversible reaction

aA + bB⇌ lL + mM

The free energies of reactants and products are given in terms ofchemical potentials at constant pressure, temperature and composition.

G(reactants) = aA + bB

G(products) = lL + mM

The change in free energy is given by,G = G(products) – G(reactants)

G = (lL + mM) – (aA + bB) … (97)The chemical potential of a substance is given by therelation, = 0 + RT ln a

where, 0 is the standard chemical potential and ‘a’ is the activity of thesubstance.

G = [l (0L + RT ln aL) + m (0M + RT ln aM)]– [a (0A + RT ln aA) + b (0B + RT ln aB)] … (98)

On rearranging equation (98) gives ,

G = [(l 0L + m0M) – (a0A + b0B)] + RT ln bB

aA

mM

lL

aaaa

… (99)

Replacing 0 by G0, i.e., free energies of reactants and products inequation (99)

G = [l G0L + mG0M) – (aG0A + bG0B)]

+ RT ln bB

aA

mM

lL

aaaa

… (100)

bB

aA

mM

lL

aaaa

RTGG

ln0 ... (101)


The term bB

aA

mM

lL

aaaa

= K, where, K is known as equilibrium constant

G = G0 + RT ln K … (102)At equilibrium G = 0 G0 = – RT ln K … (103)This equation is known as van’t Hoff reaction isotherm.With activities in any arbitrary state, equation (101) becomesG = – RT ln K + RT ln Qa … (104)Where Qa = activity ratio of products and reactants at arbitrary

state.For gaseous reaction, van’t Hoff’s isotherm reaction (104) is,G = – RT ln Kp + RT ln QP … (105)Example 4: The standard free energy change for a gaseous

reaction is 71.128 kJ. Calculate the equilibrium constant of reaction at2000 K (9 Given: R = 8.314 JK–1 mol–1)

Solution: G0 = – 2.303 RT log Kp

857.1–2000314.8303.2

71128303.2

log0

RTGK p

Kp = antilog = – 1.857 = 0.0139.Example 5: The standard free energy change for a gaseous

reaction is 1 × 105 J. Calculate the equilibrium constant for the reactionat 673 K. (Given R = 8.314 JK–1 mol–1)

Solution: Go = –2.303 RT log Kp

log10 Kp = 7603.7673314.8303.2

101303.2

50

RTG

KP = Antilog –7.7603 = 1.737 × 10–8.van’t Hoff’s Reaction Isochore (Variation of Equilibrium Constantwith Temperature)

The equilibrium constant of a reaction remains constant at a giventemperature but changes with change in the temperature. The variationof equilibrium constant with temperature is given by van’t Hoff’sreaction isochore.


The standard free energy change is related with equilibriumconstant by the relation,

G0 = – RT ln Kp … (106)Differentiating w.r.t. temperature at constant pressure

p

pp

PT

KRTKR

TG

ln

ln)( 0... (107)

Multiplying by T

p

pp

PT

KRTKRT

TGT

ln

ln)( 20

... (108)

Substituting G0 = – RT ln Kp

p

p

PT

KRTG

TGT

ln)( 20

0... (109)

The Gibb’s-Helmholtz equation for a particular case in which allthe substances taking part are in their standard states given by therelation.

G0 = H0 +P

TGT

)( 0

… (110)

orP

TGT

)( 0

= G0 – H0 … (111)

Comparing (109) and (111)

RT2 HHT

K

P

p

0ln… (112)

Since H0 does not change appreciably with pressure hence H0

in equation (112) can be replaced by H.

2

ln

RTH

dT

Kd p … (113)

Equation (113) is known as van’t Hoff’s reaction isochore.


On rearranging and integrating (113) between limits oftemperatures T1 and T2 at which the equilibrium constants are

1pK and

2pK gives

2

1

2

1

2)(lnT

T

K

Kp T

dTRHKd

p

p

… (114)

21

11ln1

2

TTRH

K

K

p

p ... (115)

Changing log10

21

11log303.21

2

TTRH

K

K

p

p ... (116)

Thus by knowing H, Kp at different temperatures can becalculated. Equation (116) can be used for reactions involving reactantsand products in gaseous state.

For reactions involving reactants and products in aqueous solutions,van’t Hoff’s reaction isochore can be derived as follows:

Kp = Kc (RT) n

Taking logarithm givesln Kp = ln Kc + n ln RTDifferentiation w.r.t. temperature T gives

Tn

dTKd

dT

Kdcp

lnln(Since R is constant)

Substitution fordT

Kd plnin equation (113) gives,

2

lnRT

HTn

dTKd c

222

lnRT

ERT

nRTHTn

RTH

dTKd c

... (114)


On rearranging and integrating (114) between limits oftemperatures T1 and T2 at which the equilibrium constants are

1CK and

2CK gives

2

1

2

1

2)(lnT

T

K

Kc T

dTREKd

C

C

... (115)

21

11ln1

2

TTRE

K

K

C

C ... (116)

Changing to log10

21

11log303.21

2

TTRE

K

K

C

C

Thus by knowing E, KC at different temperatures can becalculated.

Example 6: The equilibrium constant for the reactionN2(g) + 3H2(g) ⇌ 2NH3(g) is 1.64 × 104 at 673 K and 0.144 × 104 at 773 K.Calculate the heat of formation of one mole of ammonia from itselements in this temperature range. (Given: R = 8.314 JK1 mol1).

Solution:

2110

11303.2

log1

2

TTRH

K

K

p

p

21

121010 303.2

loglog12 TT

TTR

HKK pp

773673673773

314.8303.21064.1log10144.0log 4

104

10H

773673

100314.8303.2

)7852.3(8416.4 H

773673

100314.8303.2

0564.1 H

100)0564.1(773673314.8303.2

H

= –105226.93 J for two moles


466.526132

93.105226 H J for one mole.

Example 7: The equilibrium constant for a gaseous reaction is 169at 500 K and its heat of reaction is – 42.676 kJ. Find equilibriumconstant at 690 K. (Given: R = 8.314 JK1 mol1)

Solution:

2110

11303.2

log1

2

TTRH

K

K

p

p

21

121010 303.2

loglog12 TT

TTR

HKK pp

690500500690

314.8303.210676.42169loglog

3

1010 2pK

log102pK = 2.2279 – 1.2275 = 1.0004

2pK = Antilog 1.0004 = 10.01

QUESTIONS AND PROBLEMS1. Define Helmholtz free energy and Gibbs free energy.2. Show that decrease in Helmholtz free energy at constant temperature gives

maximum work.3. Show that decrease in Gibbs free energy at constant pressure and

temperature gives net work.4. Derive relation between Gibbs free energy and Helmholtz free energy.5. Explain variation of free energy with pressure and temperature.6. Derive Gibbs-Helmholtz equation.7. What is partial molal property?8. Define Chemical potential.9. Derive Gibbs-Duhem equation.

10. Explain variation of chemical potential with pressure and temperature.11. Derive Gibbs-Duhem equation.12. Define fugacity.13. Explain the terms fugacity, activity coefficient and activity with relevant

equations.14. Derive van’t Hoff’s reaction isotherm.15. Derive van’t Hoff’s reaction isochore.16. For an isobaric gaseous system, free energy change at 27C was –30.2 kJ

and at 28C was –30.28 kJ. Find enthalpy change for a gaseous system at27C. (Ans.: 54.2 kJ)


17. For a certain process G = – 5.412 kJ and H = – 7.562 kJ at 300 K. Find(i) entropy change and (ii) temperature coefficient of free energy changefor process. (Ans.: (i) –7.17 JK–1 (ii) 7.17 JK–1)

18. The equilibrium constant for a gaseous reaction is 500 at 102C. Calculateits standard free energy change at 102C (R = 8.314 JK–1 mol–1)

(Ans: –19.38 kJ)19. The standard free energy change for reaction:

)(2)(2 Cl21H

21

gg ⇌ HCl(g)

is 96.1 kJ at 298 K. Find equilibrium constant for the reaction (R = 8.314JK–1 mol–1]. (Ans.: 6.96 × 1016)

20. The equilibrium constant for a reaction N2(g) + O2(g) 2NO(g) is 1.21 ×10–4 at 1800 K and 4.08 × 10–4 at 2000 K. Find the heat of reaction. (Given:R = 8.314 JK–1 mol–1) (Ans.: 1.82 × 105 J)

21. The equilibrium constant for a gaseous reaction gets doubled when thetemperature is raised from 27C to 37C. Calculate the heat of reaction.(Given: R = 8.314 JK–1 mol–1) (Ans.: 53598.59 J)

22. For a gaseous reaction N2 + O2 2NO, heat of reaction is 9.097 × 104 J,If Kp for the reaction is 1.21 × 10–3 at at 1800 K, Find Kp at 2000 K(R = 8.314 JK–1 mol–1] (Ans.: 2.739 × 10–3)

(A) Multiple Choice Questions1. Gibbs free energy is

(a) Extensive property (b) intensive property(c) constitutive property

2. If G has negative value, then the reaction is(a) Non-sponteneous (b) Sponteneous(c) at equilibrium

3. Which of the following expression is true.(a) G = G + TS (b) G = E – TS(c) G = H – TS

4. Change of free energy with pressure at constant temperature is related(a) volume (b) entropy(c) volume and entropy

5. Change of free energy with temperature at constant pressure is related to(a) volume (b) entropy(c) equilibrium constant

6. Partial molal properties are studied for(a) Extensive properties, closed system(b) Intensive porperties, open system(c) Extensive properties, open system


7. The term fugacity was introduced by(a) J.W. Gibbs (b) Hermann von Helmholtz(c) G.N. Lewis

8. When a reaction is carried out in liquid phase Keq is designated as(a) Kp (b) KC

(c) Kl

9. The maximum work done by the system can be obtained at the expense of;(a) Gibb’s Free energy (b) Helmholtz free energy(c) Enthalpy

10. The net work done by the system can be obtained at the expense of;(a) Gibb’s Free energy (b) Helmholtz free energy(c) Enthalpy

11. For a spontaneous process:(a) G = 0 (b) G < 0(c) G > 0

12. For a non-spontaneous process:(a) G = 0 (b) G < 0(c) G > 0

13. For a system at equilibrium(a) G = 0 (b) G < 0(c) G > 0

14. As the pressure on the system increases, its free energy(a) decreases (b) increases(c) remains same

15. As the pressure on the system decreases, its free energy(a) decreases (b) increases(c) remains same

16. As the temperature of the system increases, its free energy(a) decreases (b) increases(c) remains same

17. As the temperature of the system decreases, its free energy(a) decreases (b) increases(c) remains same

18. Each substance in a given state has a tendency to escape from that stateand this escaping tendency is called:

(a) Activity (b) Fugacity(c) Chemical potential

19. For ideal gases, activity coefficient:(a) = 1 (b) < 1(c) > 1


20. For non-ideal gases, activity coefficient(a) = 1 (b) < 1(c) > 1

21. For an ideal gases, fugacity (f) is related to pressure (p) by the relation(a) f = p (b) f < p(c) f > p

22. For non-ideal gases, fugacity (f) is related to pressure (p) by the relation(a) f = p (b) f < p(c) f > p

Answers:1. (a), 2. (b), 3. (c), 4. (a), 5. (b), 6. (c), 7. (c), 8. (b), 9. (b), 10. (a), 11. (b),12. (c), 13. (a), 14. (b), 15. (a), 16. (a), 17. (b), 18. (b), 19. (a), 20. (b),21. (a), 22. (b)

(B) State True or False1. If G is negative, reaction is spontaneous.2. If G is positive, reaction is non-spontaneous.3. For reaction at equilibrium, G > 0.4. For a reaction at equilibrium, G = 0.5. As pressure on system increases, its free energy decreases.6. As temperature increases, free energy increases.7. Partial molal properties are studied for intensive properties in a closed

system.8. Chemical potential is an extensive property.9. The Gibbs-Duhem equation is given by the relation idni = 0

10. Fugacity is a measure of escaping tendency of a substance.11. For an ideal gas, = 112. For an ideal gas, f = p

Answers:True: 1., 2., 4., 10., 11., 12False: 3., 5., 6., 7., 8., 9

(C) Fill in the Blanks1. The maximum work done by the system can be obtained at the expense of

_______.2. Decrease in Helmholtz free energy at constant temperature gives _______

work.3. The net work done by the system can be obtained at the expense of

_________.4. Decrease in Gibbs free energy at constant pressure and temperature gives

_____ work.5. For spontaneous process, free energy change is __________.


6. For non-spontaneous process, free energy change is ________.7. As pressure on system increases, its free energy _________.8. As temperature of system increases, its free energy _______.9. The concept of partial molal property was introduced for studying ______

systems.10. Chemical potential is __________ property.11. For a pure substance, the chemical potential is equal to ________ per mole.12. _______ is the measure of escaping tendency for real gases.

Answers:1. Helmholtz free energy, 2. maximum, 3. Gibbs free energy, 4. net,5. negative, 6. positive, 7. increases, 8. decreases, 9. open, 10. intensive,11. free energy, 12. fugacity.

(D) Match the ColumnsColumn A Column B

1. For spontaneous process (p) G = 02. For non-spontaneous process (q) fugacity3. For system at equilibrium (r) G < 04. Escaping tendency (s) G > 0

Answers:1. (r), 2. (s), 3. (p), 4. (q)

College PhysicalCollege Physical Chemistry (For Second Year B.Sc.) (According to Revised Syllabus...

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