ESSENTIALS OF PHYSICAL CHEMISTRY - KrishiKosh

ESSENTIALS OF PHYSICAL CHEMISTRY

tl.

J.

Heyrovsky

~ ..li'lAo.~d Nobel Prize in 1959 for devising "Polarograph ic ~i~" , I g-mercury electrode. " Oscillographic polarograph y"

bas proved very use' ul in ore analysis a nd in the determination of t he purity of samples or pharmaceutical products : such as vitamins. hormones and antihiol ics.

ESSENTIALS OF

PHYSICAL

CHEMIS~rRY

(A Text Book .for B.Sc. Students of Indian Universities)

By

B. S. BAHL, M.Sc., Principal, D. A. V. College, Amritsar, Fellow and Syndic, Panjab University I..l'iD

G. D. TULI, M.Sc., Ph.D. Priflcipal and Head o/the Deptt. 0/ Chemistry Multanimal Modi Degree College, Modinagar

-.f a certain volume of that substance to the weight of the same vol~tme of hydrogen, both t'o!umes being measured under identical conditions oj temperature and pressure. V.D. = Wt. of certain volume of the_1l_ubstan(l€l (under same Wt. of the same volume of Hydrogen ture and pressure).

tempera-

I

Applying Avogadro's Hypothesis we have, vVt. of n mol~cules of the substance V.D.= d Wt. of n molecules of Hy rogen Wt. of 1 molecule of the substance Wt. of 1 molecule of Hydrogen . Now by definition: Wt. of 1 molecule of the substance M. Wt. = Wt. of 1 atom of Hydrogen--

(i) "

'"

(ii)

Dividing (ii) by (i) M. Wt. 'Vt. of I molecule of Hydrogen V.ri.--' = Wt. of 1 atom of Hydrogen Since, 1 molecule of Hydrogen contains 2 atoms

~. Wt.

= 2, or M. Wt. = 2 V.D. V.D. This relation is very useful in the determination of molecular weights. *Atomicity of an cule.

element is the number of atoms present in its mole·

FUNDAMEN't'AL PRINCIPL]]S

(5) Avogadro's Law has led to the deduction: 22'4 litres of any gas at N.T.P. will contain its molecular weight in grams.

It has bf>en found experimentally that 22'41itres of hydrogl'n at N.T.P. weigh equal to 2 grams, i.e., its molecular weight in grams. According to Avogadro's law, .. qual volumes of all gases under the ~ame conditions of temperature and pressure contain equa.l number of moleculeFl, and hence it follows that:

22'4 litres of any gas at N. T.P. will contain I gram molecule or its mole'(;ular weight in grams. This relation also helps us in the determination of mole·cular ,veights.

(6) 'Phis law helps in the calculation of the molecular formulae of gase" and is useful in Gas Analysis. If the volumes of the reactants and the products are known, the molecular formula can be found out without further data. For Exampie :~ I volume of Nitrogen +3 volumes of Hydrogen -0> 2 volumes of ammonia. If n be the number cf molecules in 1 volume, according to Avogadro's law:

n molecules of Nitrogen + 3n molecules of Hydrogen -0> 2n molecules of ,ammonia. Or, 1 molecule of Nitrogen + 3 molecules of Hydrogen -0> 2 molecules of .ammonia. Since Nitrogen and Hydrogen are diatomic, 2 atoms of Nitrogen + 6 atoms of Hydrogen -"'.>- 2 molecules of ammonia . . . . . 1 molecule of ammonia contains 1 atom of Nitrogen and 3 atoms 2 ---

+

+ (1 d

x I )M2

,: ,

'

where n is the observed refractive index and d is the density of the mixture, x J is molar fraction of the component ,1' and Ml and M2 the molecular weight,s of component 1 and 2 respectively. R}, 2 is then related to individual molar refractions Rl and R2 by the expression R I , 2 = x1R 1 + (1 - x1 JR 2 • This relation may be used to study solid-liquid mixtures as well. VISCOSITY

Flowing is one of the characteristic properties of liquids. Some liquids flow more readily than others. Thus ether will move over a glass plate much mor~ quickly than glycerol. The rate of flow depends on the nature of the liquid and on the force which produces the flow. Let us examine the motion of a liquid on a glass plate . . The liquid may be considered to be consisting of a number of molecular layers arranged one over the other. The layer in contact •••.• _ ..• 4") La.yers of the with the glass plate remains stationary, • • . . . . . 3 Lmolecules of the second layer moves slowly, the . • . . • 2 (the liquid on third one moves a little quicker than • . . . 1) glass plate. the second and the velocity of the Fig. 19. Motion of a liquid fourth is maximum. There is thus a on a glass plate. movemcltt of different layers past one another in the direction of the flow. Th,e displacement of different layers relative to one another is opposed by the internal friction 01' viscosity of the liquid. When the flow of tH.-eJiquid becomes steady, I

PHYSICAL PROPERTIES AND CHEMICAL CONSTITUTION

35

there will be a constant difference i n thevelocity between two different layers. It has been found that force per unit area required to maintain this condition is directly proportional to the difference of velocity V, of two adjacent layers, and inversely proportional to their distance, x, apart, i.e., v v Force ex:: ~ or F = "fJ X x x where"fJ is the Co-efficient of Viscosity which may be defined as the force per unit area required to maintain unit difference of velocity between two parallel layers in the liquid, one centimetre apart. The co· efficient of viscosity is expressed in dynes per square centimetre. The reciprocal of the co-efficient of viscosity is called Fluidity. Fluidity

= _1_. 'IJ

Determination of Viscosity The direct experimental measurement of absolute viscosity offers considerable difficulty. The relative viscosity of a liquid with respect to, say water, is all that is required for most purposes. It can be determined conveniently with the help of an apparatus called Ostwald Viscometer. The apparatus is shown in Fig. 20. A definite quantity of the liquid under examination is put into the wide limb and sucked up the right limb C, slightly above the etched mark A. The quantity of the liquid should be enough to fill the viscometer between C and C' approximately. The liquid is now allowed to flow back and the time taken for the Capillary :meniscus to fall from A to B is noted. The liquid passes through the capillary tube BD and it is clear that the time of flow t, will be directly proportional to the co-efficient of viscosity. It is also inversely proportional to the density, d, of the liquid,

t ex:: ~,"fJ ex:: dt, or "fJ = k dt.

'"

(1)

The whole process is then repeated for water, .exactly under the same Fig. 20. The Ostwald conditions and taking about. the same volume. Then Viscometer. YJw = k d w fw ••• (2) where Y)w, d w and tw are the co-efficient of viscosity, density and time of flow of water respectively. From (1) anti (2) •.. (3)

36


Taking the viscosity of water under the experimental conditions as unity 'Y) dt

T = dw tw 'Y) is then the Relative Viscosity of the liquid. The absolute value can be calculated by substituting the value of absolute viscosity of water in the equation (3). Ostwald Viscometer is a very convenient apparatus for determination of viscosity at higher temperatures as it can easily be suspended in a thermostat. »l

Viscosity and Chemical Constitution Since viscosity depends on resistance offered when a molecular layer moves over another, some relationship between viscosity and constitution is to be anticipated. The following general rules have been discovered' (i) In a homologous series, the increase in viscosity per OHa group is approximately constant. (ii) An interesting relationship between viscosity and moleculal volume discovered by Dunstan (1909) may be stated as d M X 1) X 106 = 40 to 60, where .d is the density and M is the molecular weight. This expression is tIue only for non· associated liquids 1 whereas for associated liquids the number is t:'onsiderably greattr tban 60. Dunstan's relation bas been employed to know wbether a given liquid is associated or not. For example, tbe value of the expression for benzene and water is 73 and 559 respectively, wbich is taken to mean that benzene is a normal liquid whereas water is an associated on'il. Since

~

represents the molecular volume, (

~ )'~ represent~

the area over which one gram-molecule of the liquid is distributed, ie, molecular surface. The product of molecular Eurface and viscosity is termed Molecular Viscosity. Thus: Molecular viscosity = Molecular surface X Viscosity

I. The molecules of non-asso Ve, and 8Tc respectivelyin Van der Waal's equation.

( p we have

+ ~2) (V -

b)=RT.

64


Substituting the value of Pc, Vo and To in terms of a, b, and R asgiven in equations (8), (9) and (10), we get (

7T •

2;b 2

+ 9¢~b2- )

(3 ¢ b - b)

=

Ri1.

')~b' -,

\1

e get,

Dividing this equation throughout by

2~~h' (11)

This is known as VlJll der Waal's reduced equation of state. In. this equation the quantities a, b, Pc, '1\, Vc which are characteristic of a given gas, have cancelled out, thus making it applicable to all substances in the liquid or gaseous state irrespective of their specific nature. From equation (11) it is clear that when two substances. have the same reduced temperature and pressure, they will have the same reduced volume. Thus when two or more substances are at the same reduced temperature and pressure, they are said to be in corresponding states. In practice this means that the properties of liquids should be determined at the same reduced temperature because pressure has very slight effect on them. Since it has been found that boiling points of liquids are approximately ~ of the critical temperature, it follows that liquids at their boiling points (in degrees absolute) are approximately in corresponding st'ltes. There· .. fore, in studying the relation between the physical properties of liquid and the chemical constitution, the physical properties may be conveniently determined at the boiling points of liquids. ~IQUEFACTION OF GASES Liquefaction is the reverse of the process of vaporisation. The two conditions which tend to change a gas to the liquid state are low' temperature and high pressure. (1) Faraday's Method. Faraday (1823) succeeded in liquefy. ing a number of gases such as sulphur dioxide, carbon dioxide, nitric· oxide, chlorine, etc. He employed a V·shaped tube, in one arm of which the gas was prepared while in the LIQUEFIED other it was liquefied underGAS I its own pressure and with I the help of external cooling. A certain number of gases; including hydrogen, oxygen, and nitrogen, however, refused to liquefy even when the Fig. 38. Faraday's method for the liquefac- pressure waR raised to 2.790' tion of gaseR. atmospheres. The reason for the failure to liquefy these gases became clear in 1869 when Andrews showed that each substance has a characteristic temperature, its critical temperature, above which it cannot be liquefierl, however great the pressure may be. Faraday's success in the liquefaction of gases was due to the fact that, the critical temperature of the gases examined by him happened to be

65

TilE GASEOUS STATE

above or just below the ordinary atmospheric temperature. The discovery of the critical temperature emphasised the necessity of more effective cooling agents to liquefy other gases which resisted liquefaction. (2) Linde's Method. Linde (1895) liquefied air making use of the fact that a compressed gas on free expansion produces intense cooling. In a gas at high COOLER pressure the 'molecules Ie:: - - - ~ ~" - _-~ are very close to each ~-- - - - - - - - - - - < 2=123. Vupour pres;oure of. wuter P w = 733 mm. of C.H.NO!, Po = 760 - 733 = 27 mm. Putting the values in the relation Nitrobenzene 123 X 27 1 Water = 18 X 733 = 3'!}7 . :. the proportion of water to nitrobenzene in the distilla te is approximately as 4 : 1. _ .

SOLUTIONS OF SOLIDS IN LIQUIDS

All solids are soluble in all liquids but their solubility varies greatly with the nature of the solute and the solvent. The most striking characteristic of solution of solids in liquids is that there is a limit to the wlubility of every solid in any liquid. The point at which a liquid cannot take up more of the solute at a given temperature is called the Saturation point and such a solution is known as Saturated Solution. A solution may sometimes hold more solute than even if it were saturated and such a solution is called a Supersaturated Solution. Determination of Solubility. The solubility of a substance is defined as the weight of it dissolved in 100 grams of its sat'urated solution

80


at a given temperature. Thus, solubility can be determined by prepar-ing a saturated solution of the solid and then analysing it by evaporation or by a suitable chemical method. ~

Saturated solution of a solid substance may be prepared by shaking excess of it with the solvent in a vessel placed in a constanttemperature-bath and filtering the clear solution. A known volume of this saturated solution is evaporated in a china dish and from the weight of the residue the solubility can easily be calculated. Thismethod, though simple, does not yield accurate results. During filtration, cooling would take place and thus some solid may be deposited on the filter paper or in the stem of the funnel. However, this method is quite good for the determination of solubility at room temperature. The evaporation of a liquid is a highly undesirableoperation as it is not possible to avoid loss of the liquid caused by spurting. This difficulty can, however, be overcome whenever a chemical method of analysis is available. Another defect in thismethod is that it takes a long time to establish the equilibrium between the solid and the solution so that thetpreparation of saturated solution by simple agitation with the solvent is delayed. This difficulty may be overcome by first preparing the saturated solution at a.. higher temperature and then to cool it to the desired temperature at which solubility is to be determined.

Solubility Curve. A curve drawn between solubility and temperature is termed a Solubility Curve. It shows the effect of temperature on the solubility of a substance. The solubility curves of substances like calcium aBetate and calcium butyrate show decrease in:solubility with increase of temperature while there are others like those of potassium nitrate and lead nitrate which show a considerableincrease of solubilitv with temperature. The sdlubility curve of sodium chlorideshows very little rise with increase of temperature. In general the solubility curves are of two types. (1) Continuous solubility' TEMPERATURE curves, and Fig. 51. Different types of solubility (2) Discontinuous solucurves. bility curves. Solubility curves of calcium salts of fatty acids, potassium chlorate, lead nitrate, and sodium chloride are Continuous Solubility Curves as they show no sharp breaks anywhere. In case of CaS04 .2H2 0, no doubt, the curve first shows a rise and then a faU but it remains continuous at the maximum point. Sometimes thesolubility curves exhibit sudden changes of direction and these-

SOLUTIONS

8l

curves are, therefore, called Discontinuous Solubility Curves. The popular examples of substances which show discontinuous solubility curves are sodium sulphate, ferric chloride, ammonium nitrate, etc. In fact at the break a new solid phase appears and another solubility curve of that new phase begins. The breitk in a solubility curve thus shows a point where the two different curves meet each other. The solubility curves of sodium sulphate and ferric chloride will be discussed in detail in Chapter XVIII. SOLUTIONS OF SOLIDS IN SOLIDS

Solution of a solid in another solid can be prepared by melting them together and subsequent cooling of the mixture. For example, gold and silver when mixed together, melted, and then cooled, yield solid solutions which are perfectly homogeneous. Sometimes solid solutions may be obtained by simply pressing together the two metals and thus establishing better cuntact when one metal would diffuse into the other. Solutions of gold and lead have been obtained by this method. The study of solutions of solids in solids is of great practical importance in metallurgy . . The formation of solid solutic)ll% is not limited to metals only> Organic substances like naphthalene and ~-naphthol when melted together form the so-called 'mixed crystals' on cooling which are a solid solution of one of them in the other. QUESTIONS AND PROBLEMS 1. Describe the various kinds of solutions giving two examples in each case. 2. State and illustrate : (a) Dalton's Law of Partial Pressure~. (b) Henry'S Law. 3. State and explain Henry's Law regarding the solubility of a gas in a liquid. (Jammu and KaRhmir B.Sc., 1953) 4. Enunciate Henry's Law relating to solubility of gc1ses in liquids and apply it to the case of soda-water bottle. A mixture of nitrobenzene and water boil, at 99°C. If the vapour pressure, of water at thii temperaturo is 733 mm. of mercllry, calculate the proportion of nitrobenzene in the distillate. (Osmania B.Se., 195.'3) 5. Write notes on : (i) Critical solution temperature. (ii) Az€otropic mixture. (Travaneore B.Se., 1954) 6. Describe the methods you wo,uld use to separate two liquids from one another in a solution. State their limitations. llltlstrate with t'xamples and sketehes of the apparatus employed. 7. How do mixtures of liquids' behave when they are subjected to distillation? Give examples. 8. Discuss the vapour-pressure-composition curves of systems containing mixtures of liquids which are (i) immiscible, and (ii) miscible in all proportions. (Madras B.Sc., 1947) 9_ Discuss the distillation under constant pressure of completely miscible binary liquid mixtures having a boiling point ma)(imum. (Osmania B.Sc .• 1954) 10. Di3cuss the importance of 'Distillation' as a method of separating binary liquid mixtures with special reference to phase diagrams. (Karnatak B.Se , 1954)

82

ESSENTlALS OF PHYf'ICAL CHEMlSTRY

11.

What is fractional distillation?

Discuss the theory on which it is

based. 12.

Explain the principle underlying the process of steam-distillation. (Calcutta B.Sc., 1940) 13. \Vrite a note on steam distillation. ( Mysore B.Sc., 1954) 14. Naphthalene (C,oH s ) distills in steam at 98'3° C under a pressure of 753 mm. The vapour pressure of water at this temperature is 71[) mm. Calculate the proportions of naphthalene and water in the distillate. 15. A mixture of bromobenzene and wat,r boils at 9fj'2°C under normal atmospheric pressure. At this temperature, vapour pressure of bromo-benzene is 119 mm. of mercury. Calculate the proportion of the two liquids in the distillate. (Br=79'92; C= 12; H= I) (Baroda B.Sc., 1953) 11\. Explain the principle underlying the process of steam distillation. Nitrobenzene can be distilled with steam under one atmospheric pressure at a temperature of 99·2°C., CalCUlate the amount of steam necessary to distil 100 gms. of nitrobenzene. The vapour pressure of water at 99'2°C is 739 mm. of mercury. (Poona B.Sc., 19f4) 17. Define the solubility of a sub3tance. Illu3trate the different types of solubility curves and explain their shapes. 18. Discuss the principle of: (i) Steam distillation. (ii) FraC!tional distillation of miscible liquid pair. (Delhi B.Sc., J954} 19. DesMibe the vapour pressure properties of a system of two immiscible liquids. How are the facts utilised to estimate the molecular weight of a liquid by steam distillation? A mixture of cholorobenzene and water, which ale virtually immi8cible, boils at 90'3°C at an external pressure of 740'2 mm. The vapour pressure of pure water at 9(}'3'C is 53(}'1 mm. Calculate the weight composition of the distillate. (Molecular weights: CsH sCl=1l2'5 and H.O=18·02) (Ans. 9: 1 approx.) (Baroda B.Sc., 1955) 20. Describe the vapour pressure curves of mixture~ of completely miscible liquids and explain their behaviour on distillation. (Osmania B.Sc., 1955) . 21. Describe the behaviour of two liquids when a mixture of the two is distilled at atmospheric pressure. Indicate how you can determine the critical solution temperature of a mixture of phenol and water. (Travancore B.Sc., 1956) 22. Discuss the principles involved in the separation of the components of a mixture of two completely miscible liquids by distillation. Examine how far these principles are applicable to the separation of (a) nitrogen from liquid air, and (b) ethyl alcohol from its aqueous solution. (Mysore B.Se., 1956) 23. Discuss fully the distillation of a mixture of two liquids. Illustrate your answer with examples and the vapour-pressure composition curves. (Poona B.Sc., 1959) 24. Discmos the distillation of those binary liquids which exhibit a maximum in their boiling point curve. (Lucknow B.Sc., 1959) 25. What is fractional distillation? Discuss the principle underlying this process. (Marthwada B.Sc., J9/i9) 26. Explain Dalton's Law of Partial Pressures. Discuss the ~tram distillation of a non·miscible liquid like aniline. (Delhi B.Sc., 19/i9)

CHAPTER V

THEORY OF DILUTE SOLUTIONS When the solute is present in a very small quantity as compared

to the solvent, the system is c-Llled a Dilute Solution. Although. the general laws discussed in this ch'1pter are applica.ble to all types of dilute solutions, for a detailed study we select the ordinary popular type viz., the solution ofa solid in a liquid. THE DIFFUSION OF THE SOLUTE

One of the most important properties of gases is the property ,of diffusion by virtue of which a gas can distribute itself uniformly throughout the whole space offered. The molecules of a liquid or a .substance in solution also possess this property but the process of diffusion in a liquid takes place much more slowly than in a gas, for the molecules of a liqui:l are more closely crowded together. The diffusion of solute particles through a solvent can nicely be illustrated by slipping a layer of cone. solution of potassium -permanganate at the bottom of a cylinder full of water, with the 'help of a thistle funnel. The coloured p3tassium permanganate can actually be seen diffusing up through water. The process of diffusion of the solute in a solvent is due to (i) motion of solute molecules from a concentra,ted solution into the solvent, and (ii) motion of solvent molecules into the concentrated solution. The proa38S of diffusion continues till the solute is distributed uniformly throughout the solvent (Fig. 52). SEMI.PERMEABLE MEMBRANES

When a conc. solution and solvent are separated from each other by a speaial type of m3mryrane, the movement of the solute SEMIP[PM[ ABLE : ) M[MBRANE

"~',

50WT/ON

WATER

Fig. 52. Diffusion of s,lute in a solvent. (Diagrammatic)

WATER

fig, 53.

Solvent molecules pass through a semi-p"rmeable membrane whereas solute particles are held up by it. (Diagrammatic)

particles from the cone. solution into solvent i~ prevented but the flow of the solvent into it is allowed to continue. Such a membrane

84

ESSENTIALS 0]