EQUILIBRIUM 185C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6UNIT 7EQUILIBRIUMChemical equilibria are important in numerous biologicaland environmental processes. For example, equilibriainvolving O2 molecules and the protein hemoglobin play acrucial role in the transport and delivery of O2 from ourlungs to our muscles. Similar equilibria involving COmolecules and hemoglobin account for the toxicity of CO.When a liquid evaporates in a closed container,molecules with relatively higher kinetic energy escape theliquid surface into the vapour phase and number of liquidmolecules from the vapour phase strike the liquid surfaceand are retained in the liquid phase. It gives rise to a constantvapour pressure because of an equilibrium in which thenumber of molecules leaving the liquid equals the numberreturning to liquid from the vapour. We say that the systemhas reached equilibrium state at this stage. However, thisis not static equilibrium and there is a lot of activity at theboundary between the liquid and the vapour. Thus, atequilibrium, the rate of evaporation is equal to the rate ofcondensation. It may be represented byH2O (l) H2O (vap)The double half arrows indicate that the processes inboth the directions are going on simultaneously. The mixtureof reactants and products in the equilibrium state is calledan equilibrium mixture.Equilibrium can be established for both physicalprocesses and chemical reactions. The reaction may be fastor slow depending on the experimental conditions and thenature of the reactants. When the reactants in a closed vesselat a particular temperature react to give products, theconcentrations of the reactants keep on decreasing, whilethose of products keep on increasing for some time afterwhich there is no change in the concentrations of either ofthe reactants or products. This stage of the system is thedynamic equilibrium and the rates of the forward andAfter studying this unit you will beable toidentify dynamic nature ofequilibrium involved in physicaland chemical processes;state the law of equilibrium;explain characteristics ofequilibria involved in physicaland chemical processes;write expressions forequilibrium constants;establish a relationship betweenKp and Kc;explain various factors thataffect the equilibrium state of areaction;classify substances as acids orbases according to Arrhenius,Bronsted-Lowry and Lewisconcepts;classify acids and bases as weakor strong in terms of theirionization constants;explain the dependence ofdegree of ionization onconcentration of the electrolyteand that of the common ion;describe pH scale forrepresenting hydrogen ionconcentration;explain ionisation of water andits duel role as acid and base;describe ionic product (Kw ) andpKw for water;appreciate use of buffersolutions;calculate solubility productconstant. NCERTnot to be republished186 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6reverse reactions become equal. It is due tothis dynamic equilibrium stage that there isno change in the concentrations of variousspecies in the reaction mixture. Based on theextent to which the reactions proceed to reachthe state of chemical equilibrium, these maybe classified in three groups.(i) The reactions that proceed nearly tocompletion and only negligibleconcentrations of the reactants are left. Insome cases, it may not be even possible todetect these experimentally.(ii) The reactions in which only small amountsof products are formed and most of thereactants remain unchanged atequilibrium stage.(iii) The reactions in which the concentrationsof the reactants and products arecomparable, when the system is inequilibrium.The extent of a reaction in equilibriumvaries with the experimental conditions suchas concentrations of reactants, temperature,etc. Optimisation of the operational conditionsis very important in industry and laboratoryso that equilibrium is favorable in thedirection of the desired product. Someimportant aspects of equilibrium involvingphysical and chemical processes are dealt inthis unit along with the equilibrium involvingions in aqueous solutions which is called asionic equilibrium.7.1 EQUILIBRIUM IN PHYSICALPROCESSESThe characteristics of system at equilibriumare better understood if we examine somephysical processes. The most familiarexamples are phase transformationprocesses, e.g.,solid liquidliquid gassolid gas7.1.1 Solid-Liquid EquilibriumIce and water kept in a perfectly insulatedthermos flask (no exchange of heat betweenits contents and the surroundings) at 273Kand the atmospheric pressure are inequilibrium state and the system showsinteresting characteristic features. We observethat the mass of ice and water do not changewith time and the temperature remainsconstant. However, the equilibrium is notstatic. The intense activity can be noticed atthe boundary between ice and water.Molecules from the liquid water collide againstice and adhere to it and some molecules of iceescape into liquid phase. There is no changeof mass of ice and water, as the rates of transferof molecules from ice into water and of reversetransfer from water into ice are equal atatmospheric pressure and 273 K.It is obvious that ice and water are inequilibrium only at particular temperatureand pressure. For any pure substance atatmospheric pressure, the temperature atwhich the solid and liquid phases are atequilibrium is called the normal melting pointor normal freezing point of the substance.The system here is in dynamic equilibrium andwe can infer the following:(i) Both the opposing processes occursimultaneously.(ii) Both the processes occur at the same rateso that the amount of ice and waterremains constant.7.1.2 Liquid-Vapour EquilibriumThis equilibrium can be better understood ifwe consider the example of a transparent boxcarrying a U-tube with mercury (manometer).Drying agent like anhydrous calcium chloride(or phosphorus penta-oxide) is placed for afew hours in the box. After removing thedrying agent by tilting the box on one side, awatch glass (or petri dish) containing water isquickly placed inside the box. It will beobserved that the mercury level in the rightlimb of the manometer slowly increases andfinally attains a constant value, that is, thepressure inside the box increases and reachesa constant value. Also the volume of water inthe watch glass decreases (Fig. 7.1). Initiallythere was no water vapour (or very less) insidethe box. As water evaporated the pressure inthe box increased due to addition of water NCERTnot to be republishedEQUILIBRIUM 187C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6molecules into the gaseous phase inside thebox. The rate of evaporation is constant.However, the rate of increase in pressuredecreases with time due to condensation ofvapour into water. Finally it leads to anequilibrium condition when there is no netevaporation. This implies that the number ofwater molecules from the gaseous state intothe liquid state also increases till theequilibrium is attained i.e.,rate of evaporation= rate of condensationH2O(l) H2O (vap)At equilibrium the pressure exerted by thewater molecules at a given temperatureremains constant and is called the equilibriumvapour pressure of water (or just vapourpressure of water); vapour pressure of waterincreases with temperature. If the aboveexperiment is repeated with methyl alcohol,acetone and ether, it is observed that differentliquids have different equilibrium vapourpressures at the same temperature, and theliquid which has a higher vapour pressure ismore volatile and has a lower boiling point.If we expose three watch glassescontaining separately 1mL each of acetone,ethyl alcohol, and water to atmosphere andrepeat the experiment with different volumesof the liquids in a warmer room, it is observedthat in all such cases the liquid eventuallydisappears and the time taken for completeevaporation depends on (i) the nature of theliquid, (ii) the amount of the liquid and (iii) thetemperature. When the watch glass is open tothe atmosphere, the rate of evaporationremains constant but the molecules aredispersed into large volume of the room. As aconsequence the rate of condensation fromvapour to liquid state is much less than therate of evaporation. These are open systemsand it is not possible to reach equilibrium inan open system.Water and water vapour are in equilibriumposition at atmospheric pressure (1.013 bar)and at 100C in a closed vessel. The boilingpoint of water is 100C at 1.013 bar pressure.For any pure liquid at one atmosphericpressure (1.013 bar), the temperature atwhich the liquid and vapours are atequilibrium is called normal boiling point ofthe liquid. Boiling point of the liquid dependson the atmospheric pressure. It depends onthe altitude of the place; at high altitude theboiling point decreases.7.1.3 Solid Vapour EquilibriumLet us now consider the systems where solidssublime to vapour phase. If we place solid iodinein a closed vessel, after sometime the vessel getsfilled up with violet vapour and the intensity ofcolour increases with time. After certain time theintensity of colour becomes constant and at thisstage equilibrium is attained. Hence solid iodinesublimes to give iodine vapour and the iodinevapour condenses to give solid iodine. Theequilibrium can be represented as,I2(solid) I2 (vapour)Other examples showing this kind ofequilibrium are,Camphor (solid) Camphor (vapour)NH4Cl (solid) NH4Cl (vapour)Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperatureC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republished188 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.67.1.4 Equilibrium Involving Dissolutionof Solid or Gases in LiquidsSolids in liquidsWe know from our experience that we candissolve only a limited amount of salt or sugarin a given amount of water at roomtemperature. If we make a thick sugar syrupsolution by dissolving sugar at a highertemperature, sugar crystals separate out if wecool the syrup to the room temperature. Wecall it a saturated solution when no more ofsolute can be dissolved in it at a giventemperature. The concentration of the solutein a saturated solution depends upon thetemperature. In a saturated solution, adynamic equilibrium exits between the solutemolecules in the solid state and in the solution:Sugar (solution) Sugar (solid), andthe rate of dissolution of sugar = rate ofcrystallisation of sugar.Equality of the two rates and dynamicnature of equilibrium has been confirmed withthe help of radioactive sugar. If we drop someradioactive sugar into saturated solution ofnon-radioactive sugar, then after some timeradioactivity is observed both in the solutionand in the solid sugar. Initially there were noradioactive sugar molecules in the solutionbut due to dynamic nature of equilibrium,there is exchange between the radioactive andnon-radioactive sugar molecules between thetwo phases. The ratio of the radioactive to nonradioactivemolecules in the solution increasestill it attains a constant value.Gases in liquidsWhen a soda water bottle is opened, some ofthe carbon dioxide gas dissolved in it fizzesout rapidly. The phenomenon arises due todifference in solubility of carbon dioxide atdifferent pressures. There is equilibriumbetween the molecules in the gaseous stateand the molecules dissolved in the liquidunder pressure i.e.,CO2(gas) CO2(in solution)This equilibrium is governed by Henryslaw, which states that the mass of a gasdissolved in a given mass of a solvent atany temperature is proportional to thepressure of the gas above the solvent. Thisamount decreases with increase oftemperature. The soda water bottle is sealedunder pressure of gas when its solubility inwater is high. As soon as the bottle is opened,some of the dissolved carbon dioxide gasescapes to reach a new equilibrium conditionrequired for the lower pressure, namely itspartial pressure in the atmosphere. This is howthe soda water in bottle when left open to theair for some time, turns flat. It can begeneralised that:(i) For solid liquid equilibrium, there isonly one temperature (melting point) at1 atm (1.013 bar) at which the two phasescan coexist. If there is no exchange of heatwith the surroundings, the mass of the twophases remains constant.(ii) For liquid vapour equilibrium, thevapour pressure is constant at a giventemperature.(iii) For dissolution of solids in liquids, thesolubility is constant at a giventemperature.(iv) For dissolution of gases in liquids, theconcentration of a gas in liquid isproportional to the pressure(concentration) of the gas over the liquid.These observations are summarised inTable 7.1Liquid Vapour H2O p constant at givenH2O (l) H2O (g) temperatureSolid Liquid Melting point is fixed atH2O (s) H2O (l) constant pressureSolute(s) Solute Concentration of solute(solution) in solution is constantSugar(s) Sugar at a given temperature(solution)Gas(g) Gas (aq) [gas(aq)]/[gas(g)] isconstant at a giventemperatureCO2(g) CO2(aq) [CO2(aq)]/[CO2(g)] isconstant at a giventemperatureTable 7.1 Some Features of PhysicalEquilibriaProcess Conclusion NCERTnot to be republishedEQUILIBRIUM 189C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.67.1.5 General Characteristics of EquilibriaInvolving Physical ProcessesFor the physical processes discussed above,following characteristics are common to thesystem at equilibrium:(i) Equilibrium is possible only in a closedsystem at a given temperature.(ii) Both the opposing processes occur at thesame rate and there is a dynamic butstable condition.(iii) All measurable properties of the systemremain constant.(iv) When equilibrium is attained for a physicalprocess, it is characterised by constantvalue of one of its parameters at a giventemperature. Table 7.1 lists suchquantities.(v) The magnitude of such quantities at anystage indicates the extent to which thephysical process has proceeded beforereaching equilibrium.7.2 EQUILIBRIUM IN CHEMICALPROCESSES DYNAMICEQUILIBRIUMAnalogous to the physical systems chemicalreactions also attain a state of equilibrium.These reactions can occur both in forward andbackward directions. When the rates of theforward and reverse reactions become equal,the concentrations of the reactants and theproducts remain constant. This is the stage ofchemical equilibrium. This equilibrium isdynamic in nature as it consists of a forwardreaction in which the reactants give product(s)and reverse reaction in which product(s) givesthe original reactants.For a better comprehension, let usconsider a general case of a reversible reaction,A + B C + DWith passage of time, there isaccumulation of the products C and D anddepletion of the reactants A and B (Fig. 7.2).This leads to a decrease in the rate of forwardreaction and an increase in he rate of thereverse reaction,Eventually, the two reactions occur at theFig. 7.2 Attainment of chemical equilibrium.same rate and the system reaches a state ofequilibrium.Similarly, the reaction can reach the stateof equilibrium even if we start with only C andD; that is, no A and B being present initially,as the equilibrium can be reached from eitherdirection.The dynamic nature of chemicalequilibrium can be demonstrated in thesynthesis of ammonia by Habers process. Ina series of experiments, Haber started withknown amounts of dinitrogen and dihydrogenmaintained at high temperature and pressureand at regular intervals determined theamount of ammonia present. He wassuccessful in determining also theconcentration of unreacted dihydrogen anddinitrogen. Fig. 7.4 (page 191) shows that aftera certain time the composition of the mixtureremains the same even though some of thereactants are still present. This constancy incomposition indicates that the reaction hasreached equilibrium. In order to understandthe dynamic nature of the reaction, synthesisof ammonia is carried out with exactly thesame starting conditions (of partial pressureand temperature) but using D2 (deuterium)in place of H2. The reaction mixtures startingeither with H2 or D2 reach equilibrium withthe same composition, except that D2 and ND3are present instead of H2 and NH3. Afterequilibrium is attained, these two mixtures NCERTnot to be republished190 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Dynamic Equilibrium A Students ActivityEquilibrium whether in a physical or in a chemical system, is always of dynamicnature. This can be demonstrated by the use of radioactive isotopes. This is not feasiblein a school laboratory. However this concept can be easily comprehended by performingthe following activity. The activity can be performed in a group of 5 or 6 students.Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubeseach of 30 cm length. Diameter of the tubes may be same or different in the range of3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for thispurpose add a crystal of potassium permanganate to water) and keep second cylinder(number 2) empty.Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder1, close its upper tip with a finger and transfer the coloured water contained in itslower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the colouredwater in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferringcoloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till younotice that the level of coloured water in both the cylinders becomes constant.If you continue intertransferring coloured solution between the cylinders, there willnot be any further change in the levels of coloured water in two cylinders. If we takeanalogy of level of coloured water with concentration of reactants and products in thetwo cylinders, we can say the process of transfer, which continues even after the constancyof level, is indicative of dynamic nature of the process. If we repeat the experiment takingtwo tubes of different diameters we find that at equilibrium the level of coloured water intwo cylinders is different. How far diameters are responsible for change in levels in twocylinders? Empty cylinder (2) is an indicator of no product in it at the beginning.Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after theequilibrium is attained. NCERTnot to be republishedEQUILIBRIUM 191C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.62NH3(g) N2(g) + 3H2(g)Similarly let us consider the reaction,H2(g) + I2(g) 2HI(g). If we start with equalinitial concentration of H2 and I2, the reactionproceeds in the forward direction and theconcentration of H2 and I2 decreases while thatof HI increases, until all of these becomeconstant at equilibrium (Fig. 7.5). We can alsostart with HI alone and make the reaction toproceed in the reverse direction; theconcentration of HI will decrease andconcentration of H2 and I2 will increase untilthey all become constant when equilibrium isreached (Fig.7.5). If total number of H and Iatoms are same in a given volume, the sameequilibrium mixture is obtained whether westart it from pure reactants or pure product.(H2, N2, NH3 and D2, N2, ND3) are mixedtogether and left for a while. Later, when thismixture is analysed, it is found that theconcentration of ammonia is just the same asbefore. However, when this mixture isanalysed by a mass spectrometer, it is foundthat ammonia and all deuterium containingforms of ammonia (NH3, NH2D, NHD2 and ND3)and dihydrogen and its deutrated forms(H2, HD and D2) are present. Thus one canconclude that scrambling of H and D atomsin the molecules must result from acontinuation of the forward and reversereactions in the mixture. If the reaction hadsimply stopped when they reachedequilibrium, then there would have been nomixing of isotopes in this way.Use of isotope (deuterium) in the formationof ammonia clearly indicates that chemicalreactions reach a state of dynamicequilibrium in which the rates of forwardand reverse reactions are equal and thereis no net change in composition.Equilibrium can be attained from bothsides, whether we start reaction by taking,H2(g) and N2(g) and get NH3(g) or by takingNH3(g) and decomposing it into N2(g) andH2(g).N2(g) + 3H2(g) 2NH3(g)Fig 7.4 Depiction of equilibrium for the reaction( ) ( ) ( ) 2 2 3 N g + 3H g 2NH gFig.7.5 Chemical equilibrium in the reactionH2(g) + I2(g) 2HI(g) can be attainedfrom either direction7.3 LAW OF CHEMICAL EQUILIBRIUMAND EQUILIBRIUM CONSTANTA mixture of reactants and products in theequilibrium state is called an equilibriummixture. In this section we shall address anumber of important questions about thecomposition of equilibrium mixtures: What isthe relationship between the concentrations ofreactants and products in an equilibriummixture? How can we determine equilibriumconcentrations from initial concentrations? NCERTnot to be republished192 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6What factors can be exploited to alter thecomposition of an equilibrium mixture? Thelast question in particular is important whenchoosing conditions for synthesis of industrialchemicals such as H2, NH3, CaO etc.To answer these questions, let us considera general reversible reaction:A + B C + Dwhere A and B are the reactants, C and D arethe products in the balanced chemicalequation. On the basis of experimental studiesof many reversible reactions, the Norwegianchemists Cato Maximillian Guldberg and PeterWaage proposed in 1864 that theconcentrations in an equilibrium mixture arerelated by the following equilibriumequation,[ ][ ][ ][ ]C DA B= c K (7.1)where Kc is the equilibrium constant and theexpression on the right side is called theequilibrium constant expression.The equilibrium equation is also known asthe law of mass action because in the earlydays of chemistry, concentration was calledactive mass. In order to appreciate their workbetter, let us consider reaction betweengaseous H2 and I2 carried out in a sealed vesselat 731K.H2(g) + I2(g) 2HI(g)1 mol 1 mol 2 molSix sets of experiments with varying initialconditions were performed, starting with onlygaseous H2 and I2 in a sealed reaction vesselin first four experiments (1, 2, 3 and 4) andonly HI in other two experiments (5 and 6).Experiment 1, 2, 3 and 4 were performedtaking different concentrations of H2 and / orI2, and with time it was observed that intensityof the purple colour remained constant andequilibrium was attained. Similarly, forexperiments 5 and 6, the equilibrium wasattained from the opposite direction.Data obtained from all six sets ofexperiments are given in Table 7.2.It is evident from the experiments 1, 2, 3and 4 that number of moles of dihydrogenreacted = number of moles of iodine reacted =_ (number of moles of HI formed). Also,experiments 5 and 6 indicate that,[H2(g)]eq = [I2(g)]eqKnowing the above facts, in order toestablish a relationship betweenconcentrations of the reactants and products,several combinations can be tried. Let usconsider the simple expression,[HI(g)]eq / [H2(g)]eq [I2(g)]eqIt can be seen from Table 7.3 that if weput the equilibrium concentrations of thereactants and products, the above expressionTable 7.2 Initial and Equilibrium Concentrations of H2, I2 and HI NCERTnot to be republishedEQUILIBRIUM 193C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6is far from constant. However, if we considerthe expression,[HI(g)]2eq / [H2(g)]eq [I2(g)]eqwe find that this expression gives constantvalue (as shown in Table 7.3) in all the sixcases. It can be seen that in this expressionthe power of the concentration for reactantsand products are actually the stoichiometriccoefficients in the equation for the chemicalreaction. Thus, for the reaction H2(g) + I2(g) 2HI(g), following equation 7.1, the equilibriumconstant Kc is written as,Kc = [HI(g)]eq2 / [H2(g)]eq [I2(g)]eq (7.2)Generally the subscript eq (used forequilibrium) is omitted from the concentrationterms. It is taken for granted that theconcentrations in the expression for Kc areequilibrium values. We, therefore, write,Kc = [HI(g)]2 / [H2(g)] [I2(g)] (7.3)The subscript c indicates that Kc isexpressed in concentrations of mol L1.At a given temperature, the product ofconcentrations of the reaction productsraised to the respective stoichiometriccoefficient in the balanced chemicalequation divided by the product ofconcentrations of the reactants raised totheir individual stoichiometriccoefficients has a constant value. This isknown as the Equilibrium Law or Law ofChemical Equilibrium.The equilibrium constant for a generalreaction,a A + b B c C + d Dis expressed as,Kc = [C]c[D]d / [A]a[B]b (7.4)where [A], [B], [C] and [D] are the equilibriumconcentrations of the reactants and products.Equilibrium constant for the reaction,4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) iswritten asKc = [NO]4[H2O]6 / [NH3]4 [O2]5Molar concentration of different species isindicated by enclosing these in square bracketand, as mentioned above, it is implied thatthese are equilibrium concentrations. Whilewriting expression for equilibrium constant,symbol for phases (s, l, g) are generallyignored.Let us write equilibrium constant for thereaction, H2(g) + I2(g) 2HI(g) (7.5)as, Kc = [HI]2 / [H2] [I2] = x (7.6)The equilibrium constant for the reversereaction, 2HI(g) H2(g) + I2(g), at the sametemperature is,Kc = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc (7.7)Thus, Kc = 1 / Kc (7.8)Equilibrium constant for the reversereaction is the inverse of the equilibriumconstant for the reaction in the forwarddirection.If we change the stoichiometric coefficientsin a chemical equation by multiplyingthroughout by a factor then we must makesure that the expression for equilibriumconstant also reflects that change. Forexample, if the reaction (7.5) is written as,_ H2(g) + _ I2(g) HI(g) (7.9)the equilibrium constant for the above reactionis given byKc = [HI] / [H2]1/2[I2]1/2 = {[HI]2 / [H2][I2]}1/2= x1/2 = Kc1/2 (7.10)On multiplying the equation (7.5) by n, we getTable 7.3 Expression Involving theEquilibrium Concentration ofReactantsH2(g) + I2(g) 2HI(g) NCERTnot to be republished194 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6nH2(g) + nI2(g) 2nHI(g) (7.11)Therefore, equilibrium constant for thereaction is equal to Kc n. These findings aresummarised in Table 7.4. It should be notedthat because the equilibrium constants Kc andKc have different numerical values, it isimportant to specify the form of the balancedchemical equation when quoting the value ofan equilibrium constant.800K. What will be Kc for the reactionN2(g) + O2(g) 2NO(g)SolutionFor the reaction equilibrium constant,Kc can be written as,[ ][ ][ ]22 2NO=N O c K( )( )( )-3 23 32.8 10 M=3.0 10 M 4.2 10 M = 0.6227.4 HOMOGENEOUS EQUILIBRIAIn a homogeneous system, all the reactantsand products are in the same phase. Forexample, in the gaseous reaction,N2(g) + 3H2(g) 2NH3(g), reactants andproducts are in the homogeneous phase.Similarly, for the reactions,CH3COOC2H5 (aq) + H2O (l) CH3COOH (aq)+ C2H5OH (aq)and, Fe3+ (aq) + SCN(aq) Fe(SCN)2+ (aq)all the reactants and products are inhomogeneous solution phase. We shall nowconsider equilibrium constant for somehomogeneous reactions.7.4.1 Equilibrium Constant in GaseousSystemsSo far we have expressed equilibrium constantof the reactions in terms of molarconcentration of the reactants and products,and used symbol, Kc for it. For reactionsinvolving gases, however, it is usually moreconvenient to express the equilibriumconstant in terms of partial pressure.The ideal gas equation is written as,pV = nRT = n Rp TVHere, p is the pressure in Pa, n is the numberof moles of the gas, V is the volume in m3 andProblem 7.1The following concentrations wereobtained for the formation of NH3 from N2and H2 at equilibrium at 500K.[N2] = 1.5 102M. [H2] = 3.0 102 M and[NH3] = 1.2 102M. Calculate equilibriumconstant.SolutionThe equilibrium constant for the reaction,N2(g) + 3H2(g) 2NH3(g) can be writtenas,( )( ) ( )2332 2NH g=N g H g c K( )( )( )2 22 2 31.2 10=1.5 10 3.0 10 = 0.106 104 = 1.06 103Problem 7.2At equilibrium, the concentrations ofN2=3.0 103M, O2 = 4.2 103M andNO= 2.8 103M in a sealed vessel atTable 7.4 Relations between EquilibriumConstants for a General Reactionand its Multiples.Chemical equation Equilibriumconstanta A + b B c C + dD Kc C + d D a A + b B Kc =(1/Kc )na A + nb B ncC + ndD Kc = (Kcn )C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republishedEQUILIBRIUM 195C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6T is the temperature in KelvinTherefore,n/V is concentration expressed in mol/m3If concentration c, is in mol/L or mol/dm3,and p is in bar thenp = cRT,We can also write p = [gas]RT.Here, R= 0.0831 bar litre/mol KAt constant temperature, the pressure ofthe gas is proportional to its concentration i.e.,p [gas]For reaction in equilibriumH2(g) + I2(g) 2HI(g)We can write either( )( ) ( )2c2 2HI g=H g I g K( )( )( ) 2 22or = HIcH IpKp p (7.12)( )( )( )22HII 2H 2Further, since HI g RI g RH g R= = = p Tp Tp TTherefore,( )( )( )( ) [ ]( ) ( ) 2 22 2 2HIH I 2 2HI g R=H g R . I g R = pp TKp p T T( )( ) ( )22 2HI g=H g I g = c K (7.13)In this example, Kp = Kc i.e., bothequilibrium constants are equal. However, thisis not always the case. For example in reactionN2(g) + 3H2(g) 2NH3(g)( )( )( )32 223 = NHpN HpKp p( ) [ ]( ) ( ) ( )2 233 32 2NH g RN g R . g R = TT H T( ) [ ]( ) ( )( )2 23 232 2NH g RRN g g = = cTK TH( ) 2 or R = p c K K T (7.14)Similarly, for a general reactiona A + b B c C + d D( )( )( )( )[ ] [ ] ( )( )[ ] [ ] ( )( )C D RA B R++ = =c d c d c dC Dp a b a b a bA Bp p TKp p T[ ] [ ][ ] [ ]C D ( )( ) ( )RA B+ + =c dc d a ba b T[ ] [ ][ ] [ ]( ) ( ) C DR RA B = =c da b cn n T K T (7.15)where n = (number of moles of gaseousproducts) (number of moles of gaseousreactants) in the balanced chemical equation.It is necessary that while calculating the valueof Kp, pressure should be expressed in barbecause standard state for pressure is 1 bar.We know from Unit 1 that :1pascal, Pa=1Nm2, and 1bar = 105 PaKp values for a few selected reactions atdifferent temperatures are given in Table 7.5Table 7.5 Equilibrium Constants, Kp for aFew Selected ReactionsProblem 7.3PCl5, PCl3 and Cl2 are at equilibrium at500 K and having concentration 1.59MPCl3, 1.59M Cl2 and 1.41 M PCl5.C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republished196 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Calculate Kc for the reaction,PCl5 PCl3 + Cl2SolutionThe equilibrium constant Kc for the abovereaction can be written as,[ ][ ][ ]( )( )23 2c5PCl Cl 1.591.79PCl 1.41K = = =Problem 7.4The value of Kc = 4.24 at 800K for the reaction,CO (g) + H2O (g) CO2 (g) + H2 (g)Calculate equilibrium concentrations ofCO2, H2, CO and H2O at 800 K, if only COand H2O are present initially atconcentrations of 0.10M each.SolutionFor the reaction,CO (g) + H2O (g) CO2 (g) + H2 (g)Initial concentration:0.1M 0.1M 0 0Let x mole per litre of each of the productbe formed.At equilibrium:(0.1-x) M (0.1-x) M x M x Mwhere x is the amount of CO2 and H2 atequilibrium.Hence, equilibrium constant can bewritten as,Kc = x2/(0.1-x)2 = 4.24x2 = 4.24(0.01 + x2-0.2x)x2 = 0.0424 + 4.24x2-0.848x3.24x2 0.848x + 0.0424 = 0a = 3.24, b = 0.848, c = 0.0424(for quadratic equation ax2 + bx + c = 0,( b b2 4ac )x2a =x = 0.848(0.848)2 4(3.24)(0.0424)/(3.24 2)x = (0.848 0.4118)/ 6.48x1 = (0.848 0.4118)/6.48 = 0.067x2 = (0.848 + 0.4118)/6.48 = 0.194the value 0.194 should be neglectedbecause it will give concentration of thereactant which is more than initialconcentration.Hence the equilibrium concentrations are,[CO2] = [H2-] = x = 0.067 M[CO] = [H2O] = 0.1 0.067 = 0.033 MProblem 7.5For the equilibrium,2NOCl(g) 2NO(g) + Cl2(g)the value of the equilibrium constant, Kcis 3.75 106 at 1069 K. Calculate the Kpfor the reaction at this temperature?SolutionWe know that,Kp = Kc(RT)nFor the above reaction,n = (2+1) 2 = 1Kp = 3.75 106 (0.0831 1069)Kp = 0.0337.5 HETEROGENEOUS EQUILIBRIAEquilibrium in a system having more than onephase is called heterogeneous equilibrium.The equilibrium between water vapour andliquid water in a closed container is anexample of heterogeneous equilibrium.H2O(l) H2O(g)In this example, there is a gas phase and aliquid phase. In the same way, equilibriumbetween a solid and its saturated solution,Ca(OH)2 (s) + (aq) Ca2+ (aq) + 2OH(aq)is a heterogeneous equilibrium.Heterogeneous equilibria often involve puresolids or liquids. We can simplify equilibriumexpressions for the heterogeneous equilibriainvolving a pure liquid or a pure solid, as themolar concentration of a pure solid or liquidis constant (i.e., independent of the amountpresent). In other words if a substance X isinvolved, then [X(s)] and [X(l)] are constant,whatever the amount of X is taken. ContraryC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republishedEQUILIBRIUM 197C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6to this, [X(g)] and [X(aq)] will vary as theamount of X in a given volume varies. Let ustake thermal dissociation of calcium carbonatewhich is an interesting and important exampleof heterogeneous chemical equilibrium.CaCO3 (s) CaO (s) + CO2 (g) (7.16)On the basis of the stoichiometric equation,we can write,( ) ( )( )23CaO s CO gCaCO s = c KSince [CaCO3(s)] and [CaO(s)] are bothconstant, therefore modified equilibriumconstant for the thermal decomposition ofcalcium carbonate will beKc = [CO2(g)] (7.17)2or = p CO K p (7.18)This shows that at a particulartemperature, there is a constant concentrationor pressure of CO2 in equilibrium with CaO(s)and CaCO3(s). Experimentally it has beenfound that at 1100 K, the pressure of CO2 inequilibrium with CaCO3(s) and CaO(s), is2.0 105 Pa. Therefore, equilibrium constantat 1100K for the above reaction is:25 5CO = = 2 10 Pa /10 Pa = 2.00 p K pSimilarly, in the equilibrium betweennickel, carbon monoxide and nickel carbonyl(used in the purification of nickel),Ni (s) + 4 CO (g) Ni(CO)4 (g),the equilibrium constant is written as( )[ ]44Ni COCO = c KIt must be remembered that for theexistence of heterogeneous equilibrium puresolids or liquids must also be present(however small the amount may be) atequilibrium, but their concentrations orpartial pressures do not appear in theexpression of the equilibrium constant. In thereaction,Ag2O(s) + 2HNO3(aq) 2AgNO3(aq) +H2O(l)[ ][ ]2323AgNO=HNO c KProblem 7.6The value of Kp for the reaction,CO2 (g) + C (s) 2CO (g)is 3.0 at 1000 K. If initially2= 0.48 CO p bar and CO p = 0 bar and puregraphite is present, calculate the equilibriumpartial pressures of CO and CO2.SolutionFor the reaction,let x be the decrease in pressure of CO2,thenCO2(g) + C(s) 2CO(g)Initialpressure: 0.48 bar 0Units of Equilibrium ConstantThe value of equilibrium constant Kc canbe calculated by substituting theconcentration terms in mol/L and for Kppartial pressure is substituted in Pa, kPa,bar or atm. This results in units ofequilibrium constant based on molarity orpressure, unless the exponents of both thenumerator and denominator are same.For the reactions,H2(g) + I2(g) 2HI, Kc and Kp have no unit.N2O4(g) 2NO2 (g), Kc has unit mol/Land Kp has unit barEquilibrium constants can also beexpressed as dimensionless quantities ifthe standard state of reactants andproducts are specified. For a pure gas, thestandard state is 1bar. Therefore a pressureof 4 bar in standard state can be expressedas 4 bar/1 bar = 4, which is adimensionless number. Standard state (c0)for a solute is 1 molar solution and allconcentrations can be measured withrespect to it. The numerical value ofequilibrium constant depends on thestandard state chosen. Thus, in thissystem both Kp and Kc are dimensionlessquantities but have different numericalvalues due to different standard states. NCERTnot to be republished198 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6At equilibrium:(0.48 x)bar 2x bar22= COpCOpKpKp = (2x)2/(0.48 x) = 34x2 = 3(0.48 x)4x2 = 1.44 x4x2 + 3x 1.44 = 0a = 4, b = 3, c = 1.44( b b2 4ac )x2a == [3 (3)2 4(4)(1.44)]/2 4= (3 5.66)/8= (3 + 5.66)/ 8 (as value of x cannot benegative hence we neglect that value)x = 2.66/8 = 0.33The equilibrium partial pressures are,CO p = 2x = 2 0.33 = 0.66 barCO2 p = 0.48 x = 0.48 0.33 = 0.15 bar7.6 APPLICATIONS OF EQUILIBRIUMCONSTANTSBefore considering the applications ofequilibrium constants, let us summarise theimportant features of equilibrium constants asfollows:1. Expression for equilibrium constant isapplicable only when concentrations of thereactants and products have attainedconstant value at equilibrium state.2. The value of equilibrium constant isindependent of initial concentrations of thereactants and products.3. Equilibrium constant is temperaturedependent having one unique value for aparticular reaction represented by abalanced equation at a given temperature.4. The equilibrium constant for the reversereaction is equal to the inverse of theequilibrium constant for the forwardreaction.5. The equilibrium constant K for a reactionis related to the equilibrium constant of thecorresponding reaction, whose equation isobtained by multiplying or dividing theequation for the original reaction by a smallinteger.Let us consider applications of equilibriumconstant to:predict the extent of a reaction on the basisof its magnitude,predict the direction of the reaction, andcalculate equilibrium concentrations.7.6.1 Predicting the Extent of a ReactionThe numerical value of the equilibriumconstant for a reaction indicates the extent ofthe reaction. But it is important to note thatan equilibrium constant does not give anyinformation about the rate at which theequilibrium is reached. The magnitude of Kcor Kp is directly proportional to theconcentrations of products (as these appearin the numerator of equilibrium constantexpression) and inversely proportional to theconcentrations of the reactants (these appearin the denominator). This implies that a highvalue of K is suggestive of a high concentrationof products and vice-versa.We can make the following generalisationsconcerning the composition ofequilibrium mixtures:If Kc > 103, products predominate overreactants, i.e., if Kc is very large, the reactionproceeds nearly to completion. Considerthe following examples:(a) The reaction of H2 with O2 at 500 K has avery large equilibrium c o n s t a n t ,Kc = 2.4 1047.(b) H2(g) + Cl2(g) 2HCl(g) at 300K hasKc = 4.0 1031.(c) H2(g) + Br2(g) 2HBr (g) at 300 K,Kc = 5.4 1018If Kc < 103, reactants predominate overproducts, i.e., if Kc is very small, the reactionproceeds rarely. Consider the followingexamples:C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republishedEQUILIBRIUM 199C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6(a) The decomposition of H2O into H2 and O2at 500 K has a very small equilibriumconstant, Kc = 4.1 1048(b) N2(g) + O2(g) 2NO(g),at 298 K has Kc = 4.8 1031.If Kc is in the range of 103 to 103,appreciable concentrations of bothreactants and products are present.Consider the following examples:(a) For reaction of H2 with I2 to give HI,Kc = 57.0 at 700K.(b) Also, gas phase decomposition of N2O4 toNO2 is another reaction with a valueof Kc = 4.64 103 at 25C which is neithertoo small nor too large. Hence,equilibrium mixtures contain appreciableconcentrations of both N2O4 and NO2.These generarlisations are illustrated inFig. 7.6If Qc = Kc, the reaction mixture is alreadyat equilibrium.Consider the gaseous reaction of H2with I2,H2(g) + I2(g) 2HI(g); Kc = 57.0 at 700 K.Suppose we have molar concentrations[H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M.(the subscript t on the concentration symbolsmeans that the concentrations were measuredat some arbitrary time t, not necessarily atequilibrium).Thus, the reaction quotient, Qc at this stageof the reaction is given by,Qc = [HI]t2 / [H2]t [I2]t = (0.40)2/ (0.10)(0.20)= 8.0Now, in this case, Qc (8.0) does not equalKc (57.0), so the mixture of H2(g), I2(g) and HI(g)is not at equilibrium; that is, more H2(g) andI2(g) will react to form more HI(g) and theirconcentrations will decrease till Qc = Kc.The reaction quotient, Qc is useful inpredicting the direction of reaction bycomparing the values of Qc and Kc.Thus, we can make the followinggeneralisations concerning the direction of theFig.7.6 Dependence of extent of reaction on K reaction (Fig. 7.7) : c7.6.2 Predicting the Direction of theReactionThe equilibrium constant helps in predictingthe direction in which a given reaction willproceed at any stage. For this purpose, wecalculate the reaction quotient Q. Thereaction quotient, Q (Qc with molarconcentrations and QP with partial pressures)is defined in the same way as the equilibriumconstant Kc except that the concentrations inQc are not necessarily equilibrium values.For a general reaction:a A + b B c C + d D (7.19)Qc = [C]c[D]d / [A]a[B]b (7.20)Then,If Qc > Kc, the reaction will proceed in thedirection of reactants (reverse reaction).If Qc < Kc, the reaction will proceed in thedirection of the products (forward reaction).Fig. 7.7 Predicting the direction of the reactionIf Qc < Kc, net reaction goes from left to rightIf Qc > Kc, net reaction goes from right toleft.If Qc = Kc, no net reaction occurs.Problem 7.7The value of Kc for the reaction2A B + C is 2 103. At a given time,the composition of reaction mixture is[A] = [B] = [C] = 3 104 M. In whichdirection the reaction will proceed? NCERTnot to be republished200 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6The total pressure at equilbrium wasfound to be 9.15 bar. Calculate Kc, Kp andpartial pressure at equilibrium.SolutionWe know pV = nRTTotal volume (V ) = 1 LMolecular mass of N2O4 = 92 gNumber of moles = 13.8g/92 g = 0.15of the gas (n)Gas constant (R) = 0.083 bar L mol1K1Temperature (T ) = 400 KpV = nRTp 1L = 0.15 mol 0.083 bar L mol1K1400 Kp = 4.98 barN2O4 2NO2Initial pressure: 4.98 bar 0At equilibrium: (4.98 x) bar 2x barHence,ptotal at equilibrium = N2O4 NO2 p + p9.15 = (4.98 x) + 2x9.15 = 4.98 + xx = 9.15 4.98 = 4.17 barPartial pressures at equilibrium are,N2O4 p = 4.98 4.17 = 0.81barNO2 p = 2x = 2 4.17 = 8.34 bar( ) 2 2 42= / p NO N O K p p= (8.34)2/0.81 = 85.87Kp = Kc(RT)n85.87 = Kc(0.083 400)1Kc = 2.586 = 2.6Problem 7.93.00 mol of PCl5 kept in 1L closed reactionvessel was allowed to attain equilibriumat 380K. Calculate composition of themixture at equilibrium. Kc= 1.80SolutionPCl5 PCl3 + Cl2Initialconcentration: 3.0 0 0SolutionFor the reaction the reaction quotient Qcis given by,Qc = [B][C]/ [A]2as [A] = [B] = [C] = 3 104MQc = (3 104)(3 104) / (3 104)2 = 1as Qc > Kc so the reaction will proceed inthe reverse direction.7.6.3 Calculating EquilibriumConcentrationsIn case of a problem in which we know theinitial concentrations but do not know any ofthe equilibrium concentrations, the followingthree steps shall be followed:Step 1. Write the balanced equation for thereaction.Step 2. Under the balanced equation, make atable that lists for each substance involved inthe reaction:(a) the initial concentration,(b) the change in concentration on going toequilibrium, and(c) the equilibrium concentration.In constructing the table, define x as theconcentration (mol/L) of one of the substancesthat reacts on going to equilibrium, then usethe stoichiometry of the reaction to determinethe concentrations of the other substances interms of x.Step 3. Substitute the equilibriumconcentrations into the equilibrium equationfor the reaction and solve for x. If you are tosolve a quadratic equation choose themathematical solution that makes chemicalsense.Step 4. Calculate the equilibriumconcentrations from the calculated value of x.Step 5. Check your results by substitutingthem into the equilibrium equation.Problem 7.813.8g of N2O4 was placed in a 1L reactionvessel at 400K and allowed to attainequilibriumN2O4 (g) 2NO2 (g) NCERTnot to be republishedEQUILIBRIUM 201C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Let x mol per litre of PCl5 be dissociated,At equilibrium:(3-x) x xKc = [PCl3][Cl2]/[PCl5]1.8 = x2/ (3 x)x2 + 1.8x 5.4 = 0x = [1.8 (1.8)2 4(5.4)]/2x = [1.8 3.24 + 21.6]/2x = [1.8 4.98]/2x = [1.8 + 4.98]/2 = 1.59[PCl5] = 3.0 x = 3 1.59 = 1.41 M[PCl3] = [Cl2] = x = 1.59 M7.7 RELATIONSHIP BETWEENEQUILIBRIUM CONSTANT K,REACTION QUOTIENT Q ANDGIBBS ENERGY GThe value of Kc for a reaction does not dependon the rate of the reaction. However, as youhave studied in Unit 6, it is directly relatedto the thermodynamics of the reaction andin particular, to the change in Gibbs energy,G. If,G is negative, then the reaction isspontaneous and proceeds in the forwarddirection.G is positive, then reaction is considerednon-spontaneous. Instead, as reversereaction would have a negative G, theproducts of the forward reaction shall beconverted to the reactants.G is 0, reaction has achieved equilibrium;at this point, there is no longer any freeenergy left to drive the reaction.A mathematical expression of thisthermodynamic view of equilibrium can bedescribed by the following equation:G = G0 + RT lnQ (7.21)where, G0 is standard Gibbs energy.At equilibrium, when G = 0 and Q = Kc,the equation (7.21) becomes,G = G0 + RT ln K = 0G0 = RT lnK (7.22)lnK = G0 / RTTaking antilog of both sides, we get,K =eG /RTV (7.23)Hence, using the equation (7.23), thereaction spontaneity can be interpreted interms of the value of G0.If G0 < 0, then G0/RT is positive, andeGV /RT >1, making K >1, which impliesa spontaneous reaction or the reactionwhich proceeds in the forward direction tosuch an extent that the products arepresent predominantly.If G0 > 0, then G0/RT is negative, andeGV /RT < 1, that is , K < 1, which impliesa non-spontaneous reaction or a reactionwhich proceeds in the forward direction tosuch a small degree that only a very minutequantity of product is formed.Problem 7.10The value of G0 for the phosphorylationof glucose in glycolysis is 13.8 kJ/mol.Find the value of Kc at 298 K.SolutionG0 = 13.8 kJ/mol = 13.8 103J/molAlso, G0 = RT lnKcHence, ln Kc = 13.8 103J/mol(8.314 J mol1K1 298 K)ln Kc = 5.569Kc = e5.569Kc = 3.81 103Problem 7.11Hydrolysis of sucrose gives,Sucrose + H2O Glucose + FructoseEquilibrium constant Kc for the reactionis 2 1013 at 300K. Calculate G0 at300K.SolutionG0 = RT lnKcG0 = 8.314J mol1K1300K ln(21013)G0 = 7.64 104 J mol17.8 FACTORS AFFECTING EQUILIBRIAOne of the principal goals of chemical synthesisis to maximise the conversion of the reactants NCERTnot to be republished202 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Fig. 7.8 Effect of addition of H2 on change ofconcentration for the reactants andproducts in the reaction,H2(g) + I2 (g) 2HI(g)to products while minimizing the expenditureof energy. This implies maximum yield ofproducts at mild temperature and pressureconditions. If it does not happen, then theexperimental conditions need to be adjusted.For example, in the Haber process for thesynthesis of ammonia from N2 and H2, thechoice of experimental conditions is of realeconomic importance. Annual worldproduction of ammonia is about hundredmillion tones, primarily for use as fertilizers.Equilibrium constant, Kc is independent ofinitial concentrations. But if a system atequilibrium is subjected to a change in theconcentration of one or more of the reactingsubstances, then the system is no longer atequilibrium; and net reaction takes place insome direction until the system returns toequilibrium once again. Similarly, a change intemperature or pressure of the system mayalso alter the equilibrium. In order to decidewhat course the reaction adopts and make aqualitative prediction about the effect of achange in conditions on equilibrium we useLe Chateliers principle. It states that achange in any of the factors thatdetermine the equilibrium conditions of asystem will cause the system to changein such a manner so as to reduce or tocounteract the effect of the change. Thisis applicable to all physical and chemicalequilibria.We shall now be discussing factors whichcan influence the equilibrium.7.8.1 Effect of Concentration ChangeIn general, when equilibrium is disturbed bythe addition/removal of any reactant/products, Le Chateliers principle predicts that:The concentration stress of an addedreactant/product is relieved by net reactionin the direction that consumes the addedsubstance.The concentration stress of a removedreactant/product is relieved by net reactionin the direction that replenishes theremoved substance.or in other words,When the concentration of any of thereactants or products in a reaction atequilibrium is changed, the compositionof the equilibrium mixture changes so asto minimize the effect of concentrationchanges.Let us take the reaction,H2(g) + I2(g) 2HI(g)If H2 is added to the reaction mixture atequilibrium, then the equilibrium of thereaction is disturbed. In order to restore it, thereaction proceeds in a direction wherein H2 isconsumed, i.e., more of H2 and I2 react to formHI and finally the equilibrium shifts in right(forward) direction (Fig.7.8). This is inaccordance with the Le Chateliers principlewhich implies that in case of addition of areactant/product, a new equilibrium will beset up in which the concentration of thereactant/product should be less than what itwas after the addition but more than what itwas in the original mixture.The same point can be explained in termsof the reaction quotient, Qc,Qc = [HI]2/ [H2][I2]C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd 27.7.6, 16.10.6 (reprint) NCERTnot to be republishedEQUILIBRIUM 203C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Addition of hydrogen at equilibrium resultsin value of Qc being less than Kc . Thus, in orderto attain equilibrium again reaction moves inthe forward direction. Similarly, we can saythat removal of a product also boosts theforward reaction and increases theconcentration of the products and this hasgreat commercial application in cases ofreactions, where the product is a gas or avolatile substance. In case of manufacture ofammonia, ammonia is liquified and removedfrom the reaction mixture so that reactionkeeps moving in forward direction. Similarly,in the large scale production of CaO (used asimportant building material) from CaCO3,constant removal of CO2 from the kiln drivesthe reaction to completion. It should beremembered that continuous removal of aproduct maintains Qc at a value less than Kcand reaction continues to move in the forwarddirection.Effect of Concentration An experimentThis can be demonstrated by the followingreaction:Fe3+(aq)+ SCN(aq) [Fe(SCN)]2+(aq) (7.24)yellow colourless deep red( ) ( )( ) ( )2+3+ Fe SCN aq=Fe aq SCN aq c K (7.25)A reddish colour appears on adding twodrops of 0.002 M potassium thiocynatesolution to 1 mL of 0.2 M iron(III) nitratesolution due to the formation of [Fe(SCN)]2+.The intensity of the red colour becomesconstant on attaining equilibrium. Thisequilibrium can be shifted in either forwardor reverse directions depending on our choiceof adding a reactant or a product. Theequilibrium can be shifted in the oppositedirection by adding reagents that remove Fe3+or SCN ions. For example, oxalic acid(H2C2O4), reacts with Fe3+ ions to form thestable complex ion [Fe(C2O4)3]3, thusdecreasing the concentration of free Fe3+(aq).In accordance with the Le Chateliers principle,the concentration stress of removed Fe3+ isrelieved by dissociation of [Fe(SCN)]2+ toreplenish the Fe3+ ions. Because theconcentration of [Fe(SCN)]2+ decreases, theintensity of red colour decreases.Addition of aq. HgCl2 also decreases redcolour because Hg2+ reacts with SCN ions toform stable complex ion [Hg(SCN)4]2. Removalof free SCN (aq) shifts the equilibrium inequation (7.24) from right to left to replenishSCN ions. Addition of potassium thiocyanateon the other hand increases the colourintensity of the solution as it shift theequilibrium to right.7.8.2 Effect of Pressure ChangeA pressure change obtained by changing thevolume can affect the yield of products in caseof a gaseous reaction where the total numberof moles of gaseous reactants and totalnumber of moles of gaseous products aredifferent. In applying Le Chateliers principleto a heterogeneous equilibrium the effect ofpressure changes on solids and liquids canbe ignored because the volume (andconcentration) of a solution/liquid is nearlyindependent of pressure.Consider the reaction,CO(g) + 3H2(g) CH4(g) + H2O(g)Here, 4 mol of gaseous reactants (CO + 3H2)become 2 mol of gaseous products (CH4 +H2O). Suppose equilibrium mixture (for abovereaction) kept in a cylinder fitted with a pistonat constant temperature is compressed to onehalf of its original volume. Then, total pressurewill be doubled (according topV = constant). The partial pressure andtherefore, concentration of reactants andproducts have changed and the mixture is nolonger at equilibrium. The direction in whichthe reaction goes to re-establish equilibriumcan be predicted by applying the Le Chateliersprinciple. Since pressure has doubled, theequilibrium now shifts in the forwarddirection, a direction in which the number ofmoles of the gas or pressure decreases (weknow pressure is proportional to moles of thegas). This can also be understood by usingreaction quotient, Qc. Let [CO], [H2], [CH4] and[H2O] be the molar concentrations atequilibrium for methanation reaction. When NCERTnot to be republished204 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Fig. 7.9 Effect of temperature on equilibrium forthe reaction, 2NO2 (g) N2O4 (g)volume of the reaction mixture is halved, thepartial pressure and the concentration aredoubled. We obtain the reaction quotient byreplacing each equilibrium concentration bydouble its value.( ) ( )( ) ( )4 232CH g H O gCO g H g = c QAs Qc < Kc , the reaction proceeds in theforward direction.In reaction C(s) + CO2(g) 2CO(g), whenpressure is increased, the reaction goes in thereverse direction because the number of molesof gas increases in the forward direction.7.8.3 Effect of Inert Gas AdditionIf the volume is kept constant and an inert gassuch as argon is added which does not takepart in the reaction, the equilibrium remainsundisturbed. It is because the addition of aninert gas at constant volume does not changethe partial pressures or the molarconcentrations of the substance involved in thereaction. The reaction quotient changes onlyif the added gas is a reactant or productinvolved in the reaction.7.8.4 Effect of Temperature ChangeWhenever an equilibrium is disturbed by achange in the concentration, pressure orvolume, the composition of the equilibriummixture changes because the reactionquotient, Qc no longer equals the equilibriumconstant, Kc. However, when a change intemperature occurs, the value of equilibriumconstant, Kc is changed.In general, the temperature dependence ofthe equilibrium constant depends on the signof H for the reaction.The equilibrium constant for an exothermicreaction (negative H) decreases as thetemperature increases.The equilibrium constant for anendothermic reaction (positive H)increases as the temperature increases.Temperature changes affect theequilibrium constant and rates of reactions.Production of ammonia according to thereaction,N2(g) + 3H2(g) 2NH3(g) ;H= 92.38 kJ mol1is an exothermic process. According toLe Chateliers principle, raising thetemperature shifts the equilibrium to left anddecreases the equilibrium concentration ofammonia. In other words, low temperature isfavourable for high yield of ammonia, butpractically very low temperatures slow downthe reaction and thus a catalyst is used.Effect of Temperature An experimentEffect of temperature on equilibrium can bedemonstrated by taking NO2 gas (brown incolour) which dimerises into N2O4 gas(colourless).2NO2(g) N2O4(g); H = 57.2 kJ mol1NO2 gas prepared by addition of Cuturnings to conc. HNO3 is collected in two5 mL test tubes (ensuring same intensity ofcolour of gas in each tube) and stopper sealedwith araldite. Three 250 mL beakers 1, 2 and3 containing freezing mixture, water at roomtemperature and hot water (363K),respectively, are taken (Fig. 7.9). Both the testtubes are placed in beaker 2 for 8-10 minutes.After this one is placed in beaker 1 and theother in beaker 3. The effect of temperatureon direction of reaction is depicted very wellin this experiment. At low temperatures inbeaker 1, the forward reaction of formation ofN2O4 is preferred, as reaction is exothermic, andthus, intensity of brown colour due to NO2decreases. While in beaker 3, hightemperature favours the reverse reaction of NCERTnot to be republishedEQUILIBRIUM 205C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6formation of NO2 and thus, the brown colourintensifies.Effect of temperature can also be seen inan endothermic reaction,[Co(H2O)6]3+(aq) + 4Cl(aq) [CoCl4 ]2(aq) +6H2O(l)pink colourless blueAt room temperature, the equilibriummixture is blue due to [CoCl4]2. When cooledin a freezing mixture, the colour of the mixtureturns pink due to [Co(H2O)6]3+.7.8.5 Effect of a CatalystA catalyst increases the rate of the chemicalreaction by making available a new low energypathway for the conversion of reactants toproducts. It increases the rate of forward andreverse reactions that pass through the sametransition state and does not affectequilibrium. Catalyst lowers the activationenergy for the forward and reverse reactionsby exactly the same amount. Catalyst does notaffect the equilibrium composition of areaction mixture. It does not appear in thebalanced chemical equation or in theequilibrium constant expression.Let us consider the formation of NH3 fromdinitrogen and dihydrogen which is highlyexothermic reaction and proceeds withdecrease in total number of moles formed ascompared to the reactants. Equilibriumconstant decreases with increase intemperature. At low temperature ratedecreases and it takes long time to reach atequilibrium, whereas high temperatures givesatisfactory rates but poor yields.German chemist, Fritz Haber discoveredthat a catalyst consisting of iron catalyse thereaction to occur at a satisfactory rate attemperatures, where the equilibriumconcentration of NH3 is reasonably favourable.Since the number of moles formed in thereaction is less than those of reactants, theyield of NH3 can be improved by increasingthe pressure.Optimum conditions of temperature andpressure for the synthesis of NH3 usingcatalyst are around 500C and 200 atm.Similarly, in manufacture of sulphuricacid by contact process,2SO2(g) + O2(g) 2SO3(g); Kc = 1.7 1026though the value of K is suggestive of reactiongoing to completion, but practically the oxidationof SO2 to SO3 is very slow. Thus, platinum ordivanadium penta-oxide (V2O5) is used ascatalyst to increase the rate of the reaction.Note: If a reaction has an exceedingly smallK, a catalyst would be of little help.7.9 IONIC EQUILIBRIUM IN SOLUTIONUnder the effect of change of concentration onthe direction of equilibrium, you haveincidently come across with the followingequilibrium which involves ions:Fe3+(aq) + SCN(aq) [Fe(SCN)]2+(aq)There are numerous equilibria that involveions only. In the following sections we willstudy the equilibria involving ions. It is wellknown that the aqueous solution of sugardoes not conduct electricity. However, whencommon salt (sodium chloride) is added towater it conducts electricity. Also, theconductance of electricity increases with anincrease in concentration of common salt.Michael Faraday classified the substances intotwo categories based on their ability to conductelectricity. One category of substancesconduct electricity in their aqueous solutionsand are called electrolytes while the other donot and are thus, referred to as nonelectrolytes.Faraday further classifiedelectrolytes into strong and weak electrolytes.Strong electrolytes on dissolution in water areionized almost completely, while the weakelectrolytes are only partially dissociated.For example, an aqueous solution ofsodium chloride is comprised entirely ofsodium ions and chloride ions, while thatof acetic acid mainly contains unionizedacetic acid molecules and only some acetateions and hydronium ions. This is becausethere is almost 100% ionization in case ofsodium chloride as compared to lessthan 5% ionization of acetic acid which isa weak electrolyte. It should be notedthat in weak electrolytes, equilibrium is NCERTnot to be republished206 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Fig.7.10 Dissolution of sodium chloride in water.Na+ and Cl ions are stablised by theirhydration with polar water molecules.established between ions and the unionizedmolecules. This type of equilibrium involvingions in aqueous solution is called ionicequilibrium. Acids, bases and salts comeunder the category of electrolytes and may actas either strong or weak electrolytes.7.10 ACIDS, BASES AND SALTSAcids, bases and salts find widespreadoccurrence in nature. Hydrochloric acidpresent in the gastric juice is secreted by thelining of our stomach in a significant amountof 1.2-1.5 L/day and is essential for digestiveprocesses. Acetic acid is known to be the mainconstituent of vinegar. Lemon and orangejuices contain citric and ascorbic acids, andtartaric acid is found in tamarind paste. Asmost of the acids taste sour, the word acidhas been derived from a latin word acidusmeaning sour. Acids are known to turn bluelitmus paper into red and liberate dihydrogenon reacting with some metals. Similarly, basesare known to turn red litmus paper blue, tastebitter and feel soapy. A common example of abase is washing soda used for washingpurposes. When acids and bases are mixed inthe right proportion they react with each otherto give salts. Some commonly knownexamples of salts are sodium chloride, bariumsulphate, sodium nitrate. Sodium chloride(common salt ) is an important component ofour diet and is formed by reaction betweenhydrochloric acid and sodium hydroxide. Itexists in solid state as a cluster of positivelycharged sodium ions and negatively chargedchloride ions which are held together due toelectrostatic interactions between oppositelycharged species (Fig.7.10). The electrostaticforces between two charges are inverselyproportional to dielectric constant of themedium. Water, a universal solvent, possessesa very high dielectric constant of 80. Thus,when sodium chloride is dissolved in water,the electrostatic interactions are reduced by afactor of 80 and this facilitates the ions to movefreely in the solution. Also, they are wellseparateddue to hydration with watermolecules.Faraday was born near London into a family of very limited means. At the age of 14 hewas an apprentice to a kind bookbinder who allowed Faraday to read the books hewas binding. Through a fortunate chance he became laboratory assistant to Davy, andduring 1813-4, Faraday accompanied him to the Continent. During this trip he gainedmuch from the experience of coming into contact with many of the leading scientists ofthe time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories,and in 1833 he also became the first Fullerian Professor of Chemistry. Faradays firstimportant work was on analytical chemistry. After 1821 much of his work was onelectricity and magnetism and different electromagnetic phenomena. His ideas have led to the establishmentof modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modestand kind hearted person. He declined all honours and avoided scientific controversies. He preferred towork alone and never had any assistant. He disseminated science in a variety of ways including hisFriday evening discourses, which he founded at the Royal Institution. He has been very famous for hisChristmas lecture on the Chemical History of a Candle. He published nearly 450 scientific papers.Michael Faraday(17911867)Comparing, the ionization of hydrochloricacid with that of acetic acid in water we findthat though both of them are polar covalent NCERTnot to be republishedEQUILIBRIUM 207C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6molecules, former is completely ionized intoits constituent ions, while the latter is onlypartially ionized (< 5%). The extent to whichionization occurs depends upon the strengthof the bond and the extent of solvation of ionsproduced. The terms dissociation andionization have earlier been used with differentmeaning. Dissociation refers to the process ofseparation of ions in water already existing assuch in the solid state of the solute, as insodium chloride. On the other hand, ionizationcorresponds to a process in which a neutralmolecule splits into charged ions in thesolution. Here, we shall not distinguishbetween the two and use the two termsinterchangeably.7.10.1 Arrhenius Concept of Acids andBasesAccording to Arrhenius theory, acids aresubstances that dissociates in water to givehydrogen ions H+(aq) and bases aresubstances that produce hydroxyl ionsOH(aq). The ionization of an acid HX (aq) canbe represented by the following equations:HX (aq) H+(aq) + X (aq)orHX(aq) + H2O(l) H3O+(aq) + X (aq)A bare proton, H+ is very reactive andcannot exist freely in aqueous solutions. Thus,it bonds to the oxygen atom of a solvent watermolecule to give trigonal pyramidalhydronium ion, H3O+ {[H (H2O)]+} (see box).In this chapter we shall use H+(aq) and H3O+(aq)interchangeably to mean the same i.e., ahydrated proton.Similarly, a base molecule like MOHionizes in aqueous solution according to theequation:MOH(aq) M+(aq) + OH(aq)The hydroxyl ion also exists in the hydratedform in the aqueous solution. Arrheniusconcept of acid and base, however, suffersfrom the limitation of being applicable only toaqueous solutions and also, does not accountfor the basicity of substances like, ammoniawhich do not possess a hydroxyl group.7.10.2 The Brnsted-Lowry Acids andBasesThe Danish chemist, Johannes Brsted andthe English chemist, Thomas M. Lowry gave amore general definition of acids and bases.According to Brsted-Lowry theory, acid isa substance that is capable of donating ahydrogen ion H+ and bases are substancescapable of accepting a hydrogen ion, H+. Inshort, acids are proton donors and bases areproton acceptors.Consider the example of dissolution of NH3in H2O represented by the following equation:Hydronium and Hydroxyl IonsHydrogen ion by itself is a bare proton withvery small size (~1015 m radius) andintense electric field, binds itself with thewater molecule at one of the two availablelone pairs on it giving H3O+. This specieshas been detected in many compounds(e.g., H3O+Cl) in the solid state. In aqueoussolution the hydronium ion is furtherhydrated to give species like H5O2+, H7O3+ andH9O4+. Similarly the hydroxyl ion is hydratedto give several ionic species like H3O2, H5O3and H7O4 etc.The basic solution is formed due to thepresence of hydroxyl ions. In this reaction,water molecule acts as proton donor andammonia molecule acts as proton acceptorand are thus, called Lowry-Brsted acid andH9O4+ NCERTnot to be republished208 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Svante Arrhenius(1859-1927)Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivitiesof electrolyte solutions, to the University of Uppsala in 1884. For the next five years hetravelled extensively and visited a number of research centers in Europe. In 1895 he wasappointed professor of physics at the newly formed University of Stockholm, serving itsrector from 1897 to 1902. From 1905 until his death he was Director of physical chemistryat the Nobel Institute in Stockholm. He continued to work for many years on electrolyticsolutions. In 1899 he discussed the temperature dependence of reaction rates on thebasis of an equation, now usually known as Arrhenius equation.He worked in a variety of fields, and made important contributions toimmunochemistry, cosmology, the origin of life, and the causes of ice age. He was thefirst to discuss the green house effect calling by that name. He received Nobel Prize inChemistry in 1903 for his theory of electrolytic dissociation and its use in the developmentof chemistry.base, respectively. In the reverse reaction, H+is transferred from NH4+ to OH. In this case,NH4+ acts as a Bronsted acid while OH actedas a Brsted base. The acid-base pair thatdiffers only by one proton is called a conjugateacid-base pair. Therefore, OH is called theconjugate base of an acid H2O and NH4+ iscalled conjugate acid of the base NH3. IfBrsted acid is a strong acid then itsconjugate base is a weak base and viceversa.It may be noted that conjugate acid hasone extra proton and each conjugate base hasone less proton.Consider the example of ionization ofhydrochloric acid in water. HCl(aq) acts as anacid by donating a proton to H2O moleculewhich acts as a base.ammonia it acts as an acid by donating aproton.Problem 7.12What will be the conjugate bases for thefollowing Brsted acids: HF, H2SO4 andHCO3 ?SolutionThe conjugate bases should have oneproton less in each case and therefore thecorresponding conjugate bases are: F,HSO4 and CO32 respectively.Problem 7.13Write the conjugate acids for the followingBrsted bases: NH2, NH3 and HCOO.SolutionThe conjugate acid should have one extraproton in each case and therefore thecorresponding conjugate acids are: NH3,NH4+ and HCOOH respectively.Problem 7.14The species: H2O, HCO3, HSO4 and NH3can act both as Bronsted acids and bases.For each case give the correspondingconjugate acid and conjugate base.SolutionThe answer is given in the following Table:Species Conjugate Conjugateacid baseH2O H3O+ OHHCO3 H2CO3 CO32HSO4 H2SO4 SO42NH3 NH4+ NH2It can be seen in the above equation, thatwater acts as a base because it accepts theproton. The species H3O+ is produced whenwater accepts a proton from HCl. Therefore,Cl is a conjugate base of HCl and HCl is theconjugate acid of base Cl. Similarly, H2O is aconjugate base of an acid H3O+ and H3O+ is aconjugate acid of base H2O.It is interesting to observe the dual role ofwater as an acid and a base. In case of reactionwith HCl water acts as a base while in case of NCERTnot to be republishedEQUILIBRIUM 209C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.67.10.3 Lewis Acids and BasesG.N. Lewis in 1923 defined an acid as aspecies which accepts electron pair and basewhich donates an electron pair. As far asbases are concerned, there is not muchdifference between Brsted-Lowry and Lewisconcepts, as the base provides a lone pair inboth the cases. However, in Lewis conceptmany acids do not have proton. A typicalexample is reaction of electron deficient speciesBF3 with NH3.BF3 does not have a proton but still actsas an acid and reacts with NH3 by acceptingits lone pair of electrons. The reaction can berepresented by,BF3 + :NH3 BF3:NH3Electron deficient species like AlCl3, Co3+,Mg2+, etc. can act as Lewis acids while specieslike H2O, NH3, OH etc. which can donate a pairof electrons, can act as Lewis bases.Problem 7.15Classify the following species into Lewisacids and Lewis bases and show howthese act as such:(a) HO (b)F (c) H+ (d) BCl3Solution(a) Hydroxyl ion is a Lewis base as it candonate an electron lone pair (:OH ).(b) Flouride ion acts as a Lewis base asit can donate any one of its fourelectron lone pairs.(c) A proton is a Lewis acid as it canaccept a lone pair of electrons frombases like hydroxyl ion and fluorideion.(d) BCl3 acts as a Lewis acid as it canaccept a lone pair of electrons fromspecies like ammonia or aminemolecules.7.11 IONIZATION OF ACIDS AND BASESArrhenius concept of acids and bases becomesuseful in case of ionization of acids and basesas mostly ionizations in chemical andbiological systems occur in aqueous medium.Strong acids like perchloric acid (HClO4),hydrochloric acid (HCl), hydrobromic acid(HBr), hyrdoiodic acid (HI), nitric acid (HNO3)and sulphuric acid (H2SO4) are termed strongbecause they are almost completelydissociated into their constituent ions in anaqueous medium, thereby acting as proton(H+) donors. Similarly, strong bases likelithium hydroxide (LiOH), sodium hydroxide(NaOH), potassium hydroxide (KOH), caesiumhydroxide (CsOH) and barium hydroxideBa(OH)2 are almost completely dissociated intoions in an aqueous medium giving hydroxylions, OH. According to Arrhenius conceptthey are strong acids and bases as they areable to completely dissociate and produce H3O+and OH ions respectively in the medium.Alternatively, the strength of an acid or basemay also be gauged in terms of Brsted-Lowry concept of acids and bases, wherein astrong acid means a good proton donor and astrong base implies a good proton acceptor.Consider, the acid-base dissociationequilibrium of a weak acid HA,HA(aq) + H2O(l) H3O+(aq) + A(aq)conjugate conjugateacid base acid baseIn section 7.10.2 we saw that acid (or base)dissociation equilibrium is dynamic involvinga transfer of proton in forward and reversedirections. Now, the question arises that if theequilibrium is dynamic then with passage oftime which direction is favoured? What is thedriving force behind it? In order to answerthese questions we shall deal into the issue ofcomparing the strengths of the two acids (orbases) involved in the dissociation equilibrium.Consider the two acids HA and H3O+ presentin the above mentioned acid-dissociationequilibrium. We have to see which amongstthem is a stronger proton donor. Whicheverexceeds in its tendency of donating a protonover the other shall be termed as the strongeracid and the equilibrium will shift in thedirection of weaker acid. Say, if HA is astronger acid than H3O+, then HA will donateprotons and not H3O+, and the solution willmainly contain A and H3O+ ions. Theequilibrium moves in the direction offormation of weaker acid and weaker base NCERTnot to be republished210 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6because the stronger acid donates a protonto the stronger base.It follows that as a strong acid dissociatescompletely in water, the resulting base formedwould be very weak i.e., strong acids havevery weak conjugate bases. Strong acids likeperchloric acid (HClO4), hydrochloric acid(HCl), hydrobromic acid (HBr), hydroiodic acid(HI), nitric acid (HNO3) and sulphuric acid(H2SO4) will give conjugate base ions ClO4, Cl,Br, I, NO3 and HSO4 , which are much weakerbases than H2O. Similarly a very strong basewould give a very weak conjugate acid. On theother hand, a weak acid say HA is only partiallydissociated in aqueous medium and thus, thesolution mainly contains undissociated HAmolecules. Typical weak acids are nitrous acid(HNO2), hydrofluoric acid (HF) and acetic acid(CH3COOH). It should be noted that the weakacids have very strong conjugate bases. Forexample, NH2, O2 and H are very good protonacceptors and thus, much stronger bases thanH2O.Certain water soluble organic compoundslike phenolphthalein and bromothymol bluebehave as weak acids and exhibit differentcolours in their acid (HIn) and conjugate base(In ) forms.HIn(aq) + H2O(l) H3O+(aq) + In(aq)acid conjugate conjugateindicator acid basecolour A colourBSuch compounds are useful as indicatorsin acid-base titrations, and finding out H+ ionconcentration.7.11.1 The Ionization Constant of Waterand its Ionic ProductSome substances like water are unique in theirability of acting both as an acid and a base.We have seen this in case of water in section7.10.2. In presence of an acid, HA it accepts aproton and acts as the base while in thepresence of a base, B it acts as an acid bydonating a proton. In pure water, one H2Omolecule donates proton and acts as an acidand another water molecules accepts a protonand acts as a base at the same time. Thefollowing equilibrium exists:H2O(l) + H2O(l) H3O+(aq) + OH(aq)acid base conjugate conjugateacid baseThe dissociation constant is represented by,K = [H3O+] [OH] / [H2O] (7.26)The concentration of water is omitted fromthe denominator as water is a pure liquid andits concentration remains constant. [H2O] isincorporated within the equilibrium constantto give a new constant, Kw, which is called theionic product of water.Kw = [H+][OH] (7.27)The concentration of H+ has been foundout experimentally as 1.0 107 M at 298 K.And, as dissociation of water produces equalnumber of H+ and OH ions, the concentrationof hydroxyl ions, [OH] = [H+] = 1.0 107 M.Thus, the value of Kw at 298K,Kw = [H3O+][OH] = (1 107)2 = 1 1014 M2(7.28)The value of Kw is temperature dependentas it is an equilibrium constant.The density of pure water is 1000 g / Land its molar mass is 18.0 g /mol. From thisthe molarity of pure water can be given as,[H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.Therefore, the ratio of dissociated water to thatof undissociated water can be given as:107 / (55.55) = 1.8 109 or ~ 2 in 109 (thus,equilibrium lies mainly towards undissociatedwater)We can distinguish acidic, neutral andbasic aqueous solutions by the relative valuesof the H3O+ and OH concentrations:Acidic: [H3O+] > [OH ]Neutral: [H3O+] = [OH ]Basic : [H3O+] < [OH]7.11.2 The pH ScaleHydronium ion concentration in molarity ismore conveniently expressed on a logarithmicscale known as the pH scale. The pH of asolution is defined as the negative logarithmto base 10 of the activity ( ) H a + of hydrogen NCERTnot to be republishedEQUILIBRIUM 211C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6ion. In dilute solutions (< 0.01 M), activity ofhydrogen ion (H+) is equal in magnitude tomolarity represented by [H+]. It shouldbe noted that activity has no units and isdefined as:H a + = [H+] / mol L1From the definition of pH, the followingcan be written,pH = log aH+ = log {[H+] / mol L1}Thus, an acidic solution of HCl (102 M)will have a pH = 2. Similarly, a basic solutionof NaOH having [OH] =104 M and [H3O+] =1010 M will have a pH = 10. At 25 C, purewater has a concentration of hydrogen ions,[H+] = 107 M. Hence, the pH of pure water isgiven as:pH = log(107) = 7Acidic solutions possess a concentrationof hydrogen ions, [H+] > 107 M, while basicsolutions possess a concentration of hydrogenions, [H+] < 107 M. thus, we can summarisethatAcidic solution has pH < 7Basic solution has pH > 7Neutral solution has pH = 7Now again, consider the equation (7.28) at298 KKw = [H3O+] [OH] = 1014Taking negative logarithm on both sidesof equation, we obtainlog Kw = log {[H3O+] [OH]}= log [H3O+] log [OH]= log 10 14pKw = pH + pOH = 14 (7.29)Note that although Kw may change withtemperature the variations in pH withtemperature are so small that we oftenignore it.pKw is a very important quantity foraqueous solutions and controls the relativeconcentrations of hydrogen and hydroxyl ionsas their product is a constant. It should benoted that as the pH scale is logarithmic, achange in pH by just one unit also meanschange in [H+] by a factor of 10. Similarly, whenthe hydrogen ion concentration, [H+] changesby a factor of 100, the value of pH changes by2 units. Now you can realise why the changein pH with temperature is often ignored.Measurement of pH of a solution is veryessential as its value should be known whendealing with biological and cosmeticapplications. The pH of a solution can be foundroughly with the help of pH paper that hasdifferent colour in solutions of different pH.Now-a-days pH paper is available with fourstrips on it. The different strips have differentcolours (Fig. 7.11) at the same pH. The pH inthe range of 1-14 can be determined with anaccuracy of ~0.5 using pH paper.Fig.7.11 pH-paper with four strips that mayhave different colours at the same pHFor greater accuracy pH meters are used.pH meter is a device that measures thepH-dependent electrical potential of the testsolution within 0.001 precision. pH meters ofthe size of a writing pen are now available inthe market. The pH of some very commonsubstances are given in Table 7.5 (page 212).Problem 7.16The concentration of hydrogen ion in asample of soft drink is 3.8 103M. whatis its pH ?SolutionpH = log[3.8 103]= {log[3.8] + log[103]}= {(0.58) + ( 3.0)} = { 2.42} = 2.42Therefore, the pH of the soft drink is 2.42and it can be inferred that it is acidic.Problem 7.17Calculate pH of a 1.0 108 M solutionof HCl. NCERTnot to be republished212 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Table 7.6 The Ionization Constants of SomeSelected Weak Acids (at 298K)Acid Ionization Constant,KaHydrofluoric Acid (HF) 3.5 104Nitrous Acid (HNO2) 4.5 104Formic Acid (HCOOH) 1.8 104Niacin (C5H4NCOOH) 1.5 105Acetic Acid (CH3COOH) 1.74 105Benzoic Acid (C6H5COOH) 6.5 105Hypochlorous Acid (HCIO) 3.0 108Hydrocyanic Acid (HCN) 4.9 1010Phenol (C6H5OH) 1.3 1010Solution2H2O (l) H3O+ (aq) + OH(aq)Kw = [OH][H3O+]= 1014Let, x = [OH] = [H3O+] from H2O. The H3O+concentration is generated (i) fromthe ionization of HCl dissolved i.e.,HCl(aq) + H2O(l) H3O+ (aq) + Cl (aq),and (ii) from ionization of H2O. In thesevery dilute solutions, both sources ofH3O+ must be considered:[H3O+] = 108 + xKw = (108 + x)(x) = 1014or x2 + 108 x 1014 = 0[OH ] = x = 9.5 108So, pOH = 7.02 and pH = 6.987.11.3 Ionization Constants of Weak AcidsConsider a weak acid HX that is partiallyionized in the aqueous solution. Theequilibrium can be expressed by:HX(aq) + H2O(l) H3O+(aq) + X(aq)Initialconcentration (M)c 0 0Let be the extent of ionizationChange (M)-c+c+cEquilibrium concentration (M)c-cccHere, c = initial concentration of theundissociated acid, HX at time, t = 0. = extentup to which HX is ionized into ions. Usingthese notations, we can derive the equilibriumconstant for the above discussed aciddissociationequilibrium:Ka = c22 / c(1-) = c2 / 1-Ka is called the dissociation or ionizationconstant of acid HX. It can be representedalternatively in terms of molar concentrationas follows,Ka = [H+][X ] / [HX] (7.30)At a given temperature T, Ka is ameasure of the strength of the acid HXi.e., larger the value of Ka, the stronger isthe acid. Ka is a dimensionless quantitywith the understanding that the standardstate concentration of all species is 1M.The values of the ionization constants ofsome selected weak acids are given inTable 7.6.Table 7.5 The pH of Some Common SubstancesName of the Fluid pH Name of the Fluid pHSaturated solution of NaOH ~15 Black Coffee 5.00.1 M NaOH solution 13 Tomato juice ~4.2Lime water 10.5 Soft drinks and vinegar ~3.0Milk of magnesia 10 Lemon juice ~2.2Egg white, sea water 7.8 Gastric juice ~1.2Human blood 7.4 1M HCl solution ~0Milk 6.8 Concentrated HCl ~1.0Human Saliva 6.4The pH scale for the hydrogen ionconcentration has been so useful that besidespKw, it has been extended to other species and NCERTnot to be republishedEQUILIBRIUM 213C:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6quantities. Thus, we have:pKa = log (Ka) (7.31)Knowing the ionization constant, Ka of anacid and its initial concentration, c, it ispossible to calculate the equilibriumconcentration of all species and also the degreeof ionization of the acid and the pH of thesolution.A general step-wise approach can beadopted to evaluate the pH of the weakelectrolyte as follows:Step 1. The species present before dissociationare identified as Brsted-Lowryacids / bases.Step 2. Balanced equations for all possiblereactions i.e., with a species acting both asacid as well as base are written.Step 3. The reaction with the higher Ka isidentified as the primary reaction whilst theother is a subsidiary reaction.Step 4. Enlist in a tabular form the followingvalues for each of the species in the primaryreaction(a) Initial concentration, c.(b) Change in concentration on proceeding toequilibrium in terms of , degree ofionization.(c) Equilibrium concentration.Step 5. Substitute equilibrium concentrationsinto equilibrium constant equation forprincipal reaction and solve for .Step 6. Calculate the concentration of speciesin principal reaction.Step 7. Calculate pH = log[H3O+]The above mentioned methodology hasbeen elucidated in the following examples.Problem 7.18The ionization constant of HF is3.2 104. Calculate the degree ofdissociation of HF in its 0.02 M solution.Calculate the concentration of all speciespresent (H3O+, F and HF) in the solutionand its pH.SolutionThe following proton transfer reactions arepossible:1) HF + H2O H3O+ + FKa = 3.2 1042) H2O + H2O H3O+ + OHKw = 1.0 1014As Ka >> Kw, [1] is the principle reaction.HF + H2O H3O+ + FInitialconcentration (M)0.02 0 0Change (M)0.02+0.02+0.02Equilibriumconcentration (M)0.02 0.02 0.02 0.02Substituting equilibrium concentrationsin the equilibrium reaction for principalreaction gives:Ka = (0.02)2 / (0.02 0.02)= 0.02 2 / (1 ) = 3.2 104We obtain the following quadraticequation:2 + 1.6 102 1.6 102 = 0The quadratic equation in can be solvedand the two values of the roots are:= + 0.12 and 0.12The negative root is not acceptable andhence,= 0.12This means that the degree of ionization,= 0.12, then equilibrium concentrationsof other species viz., HF, F and H3O+ aregiven by:[H3O+] = [F ] = c= 0.02 0.12= 2.4 103 M[HF] = c(1 ) = 0.02 (1 0.12)= 17.6 10-3 MpH = log[H+] = log(2.4 103) = 2.62Problem 7.19The pH of 0.1M monobasic acid is 4.50.Calculate the concentration of species H+, NCERTnot to be republished214 CHEMISTRYC:\Chemistry XI\Unit-7\Unit-7-Lay-5(reprint).pmd Reprint 27.7.6Table 7.7 The Values of the IonizationConstant of Some Weak Bases at298 KBase KbDimethylamine, (CH3)2NH 5.4 104Triethylamine, (C2H5)3N 6.45 105Ammonia, NH3 or NH4OH 1.77 105Quinine, (A plant product) 1.10 106Pyridine, C5H5N 1.77 109Aniline, C6H5NH2 4.27 1010Urea, CO (NH2)2 1.3 1014A and HA at equilibrium. Also, determinethe value of Ka and pKa of the monobasicacid.SolutionpH = log [H+]Therefore, [H+] = 10 pH = 104.50= 3.16 105[H+] = [A] = 3.16 105Thus, Ka = [H+][A-] / [HA][HA]eqlbm = 0.1 (3.16 10-5) 0.1Ka = (3.16 105)2 / 0.1 = 1.0 108pKa = log(108) = 8Alternatively, Percent dissociation isanother useful method for measure ofstrength of a weak acid and is given as:Percent dissociation= [HA]dissociated/[HA]initial 100% (7.32)Problem 7.20Calculate the pH of 0.08M solution ofhypochlorous acid, HOCl. The ionizationconstant of the acid is 2.5 105.Determine the percent dissociation ofHOCl.SolutionHOCl(aq) + H2O (l) H3O+(aq) + ClO(aq)Initial concentration (M)0.08 0 0Change to reachequilibrium concentration(M) x + x+xequilibrium concentartion (M)0.08 x xx Ka = {[H3O+][ClO] / [HOCl]}= x2 / (0.08 x)As x