Equilibrium - Unit 7 PDF

EQUILIBRIUM 185 UNIT 7 EQUILIBRIUM d he...

EQUILIBRIUM 185 UNIT 7 EQUILIBRIUM d he Chemical equilibria are important in numerous biological pu T and environmental processes. For example, equilibria is involving O2 molecules and the protein hemoglobin play a able to re ER After studying this unit you will be crucial role in the transport and delivery of O2 from our lungs to our muscles. Similar equilibria involving CO bl ide ntify dynamic nature of molecules and hemoglobin account for the toxicity of CO. equilibrium involved in physical and chemical processes; When a liquid evaporates in a closed container, state the law of equilibrium; molecules with relatively higher kinetic energy escape the explain characteristics of be C liquid surface into the vapour phase and number of liquid equilibria involved in physical molecules from the vapour phase strike the liquid surface and chemical processes; write expressions for and are retained in the liquid phase. It gives rise to a constant o N equilibrium constants; vapour pressure because of an equilibrium in which the establish a relationship between number of molecules leaving the liquid equals the number Kp and K c; returning to liquid from the vapour. We say that the system explain various factors that affect the equilibrium state of a has reached equilibrium state at this stage. However, this © reaction; is not static equilibrium and there is a lot of activity at the classify substances as acids or boundary between the liquid and the vapour. Thus, at bases according to Arrhenius, equilibrium, the rate of evaporation is equal to the rate of Bronsted-Lowry and Lewis concepts; condensation. It may be represented by classify acids and bases as weak H2O (l) ⇌ H2O (vap) or strong in terms of their ionization constants; The double half arrows indicate that the processes in explain the dependence of both the directions are going on simultaneously. The mixture degree of ionization on of reactants and products in the equilibrium state is called concentration of the electrolyte an equilibrium mixture. tt and that of the common ion; describe pH scale for Equilibrium can be established for both physical representing hydrogen ion processes and chemical reactions. The reaction may be fast concentration; or slow depending on the experimental conditions and the explain ionisation of water and no nature of the reactants. When the reactants in a closed vessel its duel role as acid and base; describe ionic product (Kw ) and at a particular temperature react to give products, the pK w for water; concentrations of the reactants keep on decreasing, while appreciate use of buffer those of products keep on increasing for some time after solutions; which there is no change in the concentrations of either of calculate solubility product constant. the reactants or products. This stage of the system is the dynamic equilibrium and the rates of the forward and 186 CHEMISTRY reverse reactions become equal. It is due to and the atmospheric pressure are in this dynamic equilibrium stage that there is equilibrium state and the system shows no change in the concentrations of various interesting characteristic features. We observe species in the reaction mixture. Based on the that the mass of ice and water do not change extent to which the reactions proceed to reach with time and the temperature remains the state of chemical equilibrium, these may constant. However, the equilibrium is not be classified in three groups. static. The intense activity can be noticed at (i) The reactions that proceed nearly to the boundary between ice and water. d completion and only negligible Molecules from the liquid water collide against concentrations of the reactants are left. In ice and adhere to it and some molecules of ice escape into liquid phase. There is no change he some cases, it may not be even possible to detect these experimentally. of mass of ice and water, as the rates of transfer of molecules from ice into water and of reverse (ii) The reactions in which only small amounts transfer from water into ice are equal at of products are formed and most of the pu T atmospheric pressure and 273 K. reactants remain unchanged at is equilibrium stage. It is obvious that ice and water are in re ER (iii) The reactions in which the concentrations equilibrium only at particular temperature and pressure. For any pure substance at bl of the reactants and products are comparable, when the system is in atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium. equilibrium is called the normal melting point The extent of a reaction in equilibrium or normal freezing point of the substance. be C varies with the experimental conditions such The system here is in dynamic equilibrium and as concentrations of reactants, temperature, we can infer the following: etc. Optimisation of the operational conditions (i) Both the opposing processes occur o N is very important in industry and laboratory so that equilibrium is favorable in the simultaneously. direction of the desired product. Some (ii) Both the processes occur at the same rate important aspects of equilibrium involving so that the amount of ice and water physical and chemical processes are dealt in remains constant. © this unit along with the equilibrium involving 7.1.2 Liquid-Vapour Equilibrium ions in aqueous solutions which is called as ionic equilibrium. This equilibrium can be better understood if we consider the example of a transparent box 7.1 EQUILIBRIUM IN PHYSICAL carrying a U-tube with mercury (manometer). PROCESSES Drying agent like anhydrous calcium chloride The characteristics of system at equilibrium (or phosphorus penta-oxide) is placed for a are better understood if we examine some few hours in the box. After removing the physical processes. The most familiar drying agent by tilting the box on one side, a tt examples are phase transformation watch glass (or petri dish) containing water is processes, e.g., quickly placed inside the box. It will be observed that the mercury level in the right solid ⇌ liquid limb of the manometer slowly increases and no liquid ⇌ gas finally attains a constant value, that is, the solid ⇌ gas pressure inside the box increases and reaches a constant value. Also the volume of water in 7.1.1 Solid-Liquid Equilibrium the watch glass decreases (Fig. 7.1). Initially Ice and water kept in a perfectly insulated there was no water vapour (or very less) inside thermos flask (no exchange of heat between the box. As water evaporated the pressure in its contents and the surroundings) at 273K the box increased due to addition of water EQUILIBRIUM 187 d he Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature molecules into the gaseous phase inside the dispersed into large volume of the room. As a box. The rate of evaporation is constant. pu T consequence the rate of condensation from is However, the rate of increase in pressure vapour to liquid state is much less than the decreases with time due to condensation of rate of evaporation. These are open systems re ER vapour into water. Finally it leads to an and it is not possible to reach equilibrium in bl equilibrium condition when there is no net an open system. evaporation. This implies that the number of Water and water vapour are in equilibrium water molecules from the gaseous state into position at atmospheric pressure (1.013 bar) the liquid state also increases till the and at 100°C in a closed vessel. The boiling equilibrium is attained i.e., be C point of water is 100°C at 1.013 bar pressure. rate of evaporation= rate of condensation For any pure liquid at one atmospheric H2O(l) ⇌ H 2O (vap) pressure (1.013 bar), the temperature at o N which the liquid and vapours are at At equilibrium the pressure exerted by the equilibrium is called normal boiling point of water molecules at a given temperature the liquid. Boiling point of the liquid depends remains constant and is called the equilibrium on the atmospheric pressure. It depends on vapour pressure of water (or just vapour © pressure of water); vapour pressure of water the altitude of the place; at high altitude the boiling point decreases. increases with temperature. If the above experiment is repeated with methyl alcohol, 7.1.3 Solid – Vapour Equilibrium acetone and ether, it is observed that different Let us now consider the systems where solids liquids have different equilibrium vapour sublime to vapour phase. If we place solid iodine pressures at the same temperature, and the in a closed vessel, after sometime the vessel gets liquid which has a higher vapour pressure is filled up with violet vapour and the intensity of more volatile and has a lower boiling point. colour increases with time. After certain time the If we expose three watch glasses intensity of colour becomes constant and at this tt containing separately 1mL each of acetone, stage equilibrium is attained. Hence solid iodine ethyl alcohol, and water to atmosphere and sublimes to give iodine vapour and the iodine repeat the experiment with different volumes vapour condenses to give solid iodine. The of the liquids in a warmer room, it is observed no equilibrium can be represented as, that in all such cases the liquid eventually I2(solid) ⇌ I2 (vapour) disappears and the time taken for complete evaporation depends on (i) the nature of the Other examples showing this kind of liquid, (ii) the amount of the liquid and (iii) the equilibrium are, temperature. When the watch glass is open to Camphor (solid) ⇌ Camphor (vapour) the atmosphere, the rate of evaporation remains constant but the molecules are NH4Cl (solid) ⇌ NH4Cl (vapour) 188 CHEMISTRY 7.1.4 Equilibrium Involving Dissolution pressure of the gas above the solvent. This of Solid or Gases in Liquids amount decreases with increase of Solids in liquids temperature. The soda water bottle is sealed under pressure of gas when its solubility in We know from our experience that we can water is high. As soon as the bottle is opened, dissolve only a limited amount of salt or sugar some of the dissolved carbon dioxide gas in a given amount of water at room escapes to reach a new equilibrium condition temperature. If we make a thick sugar syrup required for the lower pressure, namely its solution by dissolving sugar at a higher d partial pressure in the atmosphere. This is how temperature, sugar crystals separate out if we the soda water in bottle when left open to the cool the syrup to the room temperature. We air for some time, turns ‘flat’. It can be he call it a saturated solution when no more of generalised that: solute can be dissolved in it at a given temperature. The concentration of the solute (i) For solid ⇌ liquid equilibrium, there is in a saturated solution depends upon the only one temperature (melting point) at pu T temperature. In a saturated solution, a 1 atm (1.013 bar) at which the two phases is dynamic equilibrium exits between the solute can coexist. If there is no exchange of heat re ER molecules in the solid state and in the solution: with the surroundings, the mass of the two phases remains constant. Sugar (solution) ⇌ Sugar (solid), and bl the rate of dissolution of sugar = rate of (ii) For liquid ⇌ vapour equilibrium, the crystallisation of sugar. vapour pressure is constant at a given Equality of the two rates and dynamic temperature. (iii) For dissolution of solids in liquids, the be C nature of equilibrium has been confirmed with the help of radioactive sugar. If we drop some solubility is constant at a given radioactive sugar into saturated solution of temperature. (iv) For dissolution of gases in liquids, the o N non-radioactive sugar, then after some time radioactivity is observed both in the solution concentration of a gas in liquid is and in the solid sugar. Initially there were no proportional to the pressure radioactive sugar molecules in the solution (concentration) of the gas over the liquid. but due to dynamic nature of equilibrium, These observations are summarised in © there is exchange between the radioactive and Table 7.1 non-radioactive sugar molecules between the Table 7.1 Some Features of Physical two phases. The ratio of the radioactive to non- Equilibria radioactive molecules in the solution increases Process Conclusion till it attains a constant value. Gases in liquids Liquid ⇌ Vapour p H2 O constant at given When a soda water bottle is opened, some of H 2O (l) ⇌ H 2O (g) temperature the carbon dioxide gas dissolved in it fizzes Solid ⇌ Liquid Melting point is fixed at out rapidly. The phenomenon arises due to tt H 2O (s) ⇌ H 2O (l) constant pressure difference in solubility of carbon dioxide at Solute(s) ⇌ Solute Concentration of solute different pressures. There is equilibrium (solution) in solution is constant between the molecules in the gaseous state Sugar(s) ⇌ Sugar at a given temperature no and the molecules dissolved in the liquid (solution) under pressure i.e., Gas(g) ⇌ Gas (aq) [gas(aq)]/[gas(g)] is CO2(gas) ⇌ CO2(in solution) constant at a given This equilibrium is governed by Henry’s temperature law, which states that the mass of a gas CO2(g) ⇌ CO2 (aq) [CO 2 (aq)]/[CO 2 (g)] is dissolved in a given mass of a solvent at constant at a given temperature any temperature is proportional to the EQUILIBRIUM 189 7.1.5 General Characteristics of Equilibria Involving Physical Processes For the physical processes discussed above, following characteristics are common to the system at equilibrium: (i) Equilibrium is possible only in a closed system at a given temperature. d (ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition. he (iii) All measurable properties of the system remain constant. (iv) When equilibrium is attained for a physical pu T process, it is characterised by constant is Fig. 7.2 Attainment of chemical equilibrium. value of one of its parameters at a given re ER temperature. Table 7.1 lists such quantities. same rate and the system reaches a state of bl (v) The magnitude of such quantities at any equilibrium. stage indicates the extent to which the Similarly, the reaction can reach the state physical process has proceeded before of equilibrium even if we start with only C and reaching equilibrium. D; that is, no A and B being present initially, be C 7.2 EQUILIBRIUM IN CHEMICAL as the equilibrium can be reached from either PROCESSES – DYNAMIC direction. EQUILIBRIUM The dynamic nature of chemical o N Analogous to the physical systems chemical equilibrium can be demonstrated in the reactions also attain a state of equilibrium. synthesis of ammonia by Haber’s process. In These reactions can occur both in forward and a series of experiments, Haber started with known amounts of dinitrogen and dihydrogen backward directions. When the rates of the © maintained at high temperature and pressure forward and reverse reactions become equal, and at regular intervals determined the the concentrations of the reactants and the amount of ammonia present. He was products remain constant. This is the stage of successful in determining also the chemical equilibrium. This equilibrium is concentration of unreacted dihydrogen and dynamic in nature as it consists of a forward dinitrogen. Fig. 7.4 (page 191) shows that after reaction in which the reactants give product(s) a certain time the composition of the mixture and reverse reaction in which product(s) gives remains the same even though some of the the original reactants. reactants are still present. This constancy in tt For a better comprehension, let us composition indicates that the reaction has consider a general case of a reversible reaction, reached equilibrium. In order to understand A+B ⇌ C+D the dynamic nature of the reaction, synthesis of ammonia is carried out with exactly the no With passage of time, there is same starting conditions (of partial pressure accumulation of the products C and D and and temperature) but using D2 (deuterium) depletion of the reactants A and B (Fig. 7.2). in place of H2. The reaction mixtures starting This leads to a decrease in the rate of forward either with H2 or D2 reach equilibrium with reaction and an increase in he rate of the the same composition, except that D2 and ND3 reverse reaction, are present instead of H2 and NH3. After Eventually, the two reactions occur at the equilibrium is attained, these two mixtures 190 CHEMISTRY Dynamic Equilibrium – A Student’s Activity Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the 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 tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of d 3-5mm. Fill nearly half of the measuring cylinder-1 with colour ed water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder he (number 2) empty. Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its pu T lower portion to cylinder 2. Using second tube, kept in 2 nd cylinder , transfer the coloured is water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you re ER notice that the level of coloured water in both the cylinders becomes constant. bl If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking be C two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning. o N © tt no Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. EQUILIBRIUM 191 2NH3 (g) ⇌ N2(g) + 3H2(g) Similarly let us consider the reaction, H2(g) + I 2(g) ⇌ 2HI(g). If we start with equal initial concentration of H2 and I2, the reaction proceeds in the forward direction and the concentration of H2 and I2 decreases while that of HI increases, until all of these become d constant at equilibrium (Fig. 7.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the he concentration of HI will decrease and concentration of H2 and I2 will increase until they all become constant when equilibrium is pu T reached (Fig.7.5). If total number of H and I is atoms are same in a given volume, the same equilibrium mixture is obtained whether we re ER Fig 7.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product. N 2 ( g ) + 3H2 ( g ) ⇌ 2 NH3 ( g ) bl (H2 , N2, NH 3 and D 2, N2 , ND3 ) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the be C concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found o N that ammonia and all deuterium containing forms of ammonia (NH3, NH2D, NHD2 and ND3) and dihydrogen and its deutrated forms (H2, HD and D2) are present. Thus one can © conclude that scrambling of H and D atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped when they reached equilibrium, then there would have been no Fig.7.5 Chemical equilibrium in the reaction mixing of isotopes in this way. H2(g) + I2(g) ⇌ 2HI(g) can be attained Use of isotope (deuterium) in the formation from either direction of ammonia clearly indicates that chemical 7.3 LAW OF CHEMICAL EQUILIBRIUM tt reactions reach a state of dynamic AND EQUILIBRIUM CONSTANT equilibrium in which the rates of forward A mixture of reactants and products in the and reverse reactions are equal and there equilibrium state is called an equilibrium no is no net change in composition. mixture. In this section we shall address a Equilibrium can be attained from both number of important questions about the sides, whether we start reaction by taking, composition of equilibrium mixtures: What is H2 (g) and N2(g) and get NH3(g) or by taking the relationship between the concentrations of NH 3(g) and decomposing it into N2(g) and reactants and products in an equilibrium H2 (g). mixture? How can we determine equilibrium N2(g) + 3H2 (g) ⇌ 2NH3(g) concentrations from initial concentrations? 192 CHEMISTRY What factors can be exploited to alter the Six sets of experiments with varying initial composition of an equilibrium mixture? The conditions were performed, starting with only last question in particular is important when gaseous H 2 and I 2 in a sealed reaction vessel choosing conditions for synthesis of industrial in first four experiments (1, 2, 3 and 4) and chemicals such as H2, NH3, CaO etc. only HI in other two experiments (5 and 6). To answer these questions, let us consider Experiment 1, 2, 3 and 4 were performed a general reversible reaction: taking different concentrations of H2 and / or A+B ⇌ C+D I 2, and with time it was observed that intensity d where A and B are the reactants, C and D are of the purple colour remained constant and the products in the balanced chemical equilibrium was attained. Similarly, for he equation. On the basis of experimental studies experiments 5 and 6, the equilibrium was of many reversible reactions, the Norwegian attained from the opposite direction. chemists Cato Maximillian Guldberg and Peter Data obtained from all six sets of pu T Waage pr oposed in 1864 that the experiments are given in Table 7.2. is concentrations in an equilibrium mixture are It is evident from the experiments 1, 2, 3 related by the following equilibrium and 4 that number of moles of dihydrogen equation, re ER reacted = number of moles of iodine reacted = [C ][D] bl ½ (number of moles of HI formed). Also, Kc = [ A ][B] (7.1) experiments 5 and 6 indicate that, where K c is the equilibrium constant and the [H2(g)]eq = [I2(g)] eq expression on the right side is called the be C equilibrium constant expression. Knowing the above facts, in order to The equilibrium equation is also known as establish a relationship between the law of mass action because in the early concentrations of the reactants and products, o N days of chemistry, concentration was called several combinations can be tried. Let us “active mass”. In order to appreciate their work consider the simple expression, better, let us consider reaction between [HI(g)] eq / [H2(g)] eq [I 2(g)] eq gaseous H2 and I 2 carried out in a sealed vessel © at 731K. It can be seen from Table 7.3 that if we H2(g) + I2(g) ⇌ 2HI(g) put the equilibrium concentrations of the 1 mol 1 mol 2 mol reactants and products, the above expression Table 7.2 Initial and Equilibrium Concentrations of H2 , I2 and HI tt no EQUILIBRIUM 193 Table 7.3 Expression Involving the The equilibrium constant for a general Equilibrium Concentration of reaction, Reactants H2(g) + I2 (g) 2HI(g) aA + bB ⇌ cC + dD is expressed as, Kc = [C] c[D]d / [A]a[B]b (7.4) where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products. d Equilibrium constant for the reaction, 4NH3(g) + 5O 2(g) ⇌ 4NO(g) + 6H2O(g) is he written as 4 6 4 5 Kc = [NO] [H2O] / [NH3] [O2] pu T Molar concentration of different species is is indicated by enclosing these in square bracket is far from constant. However, if we consider and, as mentioned above, it is implied that re ER the expression, these are equilibrium concentrations. While bl [HI(g)]2eq / [H2(g)]eq [I2(g)]eq writing expression for equilibrium constant, we find that this expression gives constant symbol for phases (s, l, g) are generally value (as shown in Table 7.3) in all the six ignored. cases. It can be seen that in this expression Let us write equilibrium constant for the be C the power of the concentration for reactants reaction, H2(g) + I2(g) ⇌ 2HI(g) (7.5) and products are actually the stoichiometric as, Kc = [HI]2 / [H 2] [I2] = x (7.6) coefficients in the equation for the chemical The equilibrium constant for the reverse o N reaction. Thus, for the reaction H2(g) + I2(g) ⇌ 2HI(g), following equation 7.1, the equilibrium reaction, 2HI(g) ⇌ H 2(g) + I2(g), at the same constant K c is written as, temperature is, 2 K c = [HI(g)]eq / [H2(g)]eq [I2(g)]eq (7.2) K′c = [H 2] [I2] / [HI] 2 = 1/ x = 1 / K c (7.7) © Generally the subscript ‘eq’ (used for Thus, K′c = 1 / Kc (7.8) equilibrium) is omitted from the concentration Equilibrium constant for the reverse terms. It is taken for granted that the reaction is the inverse of the equilibrium concentrations in the expression for Kc are constant for the reaction in the forward equilibrium values. We, therefore, write, direction. K c = [HI(g)]2 / [H2(g)] [I2(g)] (7.3) If we change the stoichiometric coefficients in a chemical equation by multiplying The subscript ‘c’ indicates that K c is throughout by a factor then we must make expressed in concentrations of mol L–1. sure that the expression for equilibrium tt At a given temperature, the product of constant also reflects that change. For concentrations of the reaction products example, if the reaction (7.5) is written as, raised to the respective stoichiometric no coefficient in the balanced chemical ½ H2(g) + ½ I2(g) ⇌ HI(g) (7.9) equation divided by the product of the equilibrium constant for the above reaction concentrations of the reactants raised to is given by their individual stoichiometric K″c = [HI] / [H2] 1/2 1/2 2 1/2 [I2] = {[HI] / [H2][I2]} coefficients has a constant value. This is known as the Equilibrium Law or Law of = x1/2 = K c1/2 (7.10) Chemical Equilibrium. On multiplying the equation (7.5) by n, we get 194 CHEMISTRY nH 2(g) + nI2(g) ⇌ 2nHI(g) (7.11) 800K. What will be Kc for the reaction Therefore, equilibrium constant for the N2(g) + O2(g) ⇌ 2NO(g) reaction is equal to K cn. These findings are Solution summarised in Table 7.4. It should be noted that because the equilibrium constants K c and For the reaction equilibrium constant, K ′c have different numerical values, it is K c can be written as, important to specify the form of the balanced chemical equation when quoting the value of [NO ]2 d Kc = an equilibrium constant. [N2 ][O 2 ] Table 7.4 Relations between Equilibrium he (2.8 × 10 M ) 2 Constants for a General Reaction -3 and its Multiples. = (3.0 × 10 M )(4.2 × 10 −3 −3 M) Chemical equation Equilibrium pu T constant = 0.622 is a A + b B ⇌ c C + dD K cC+dD ⇌ aA+bB re ER K′c =(1/Kc ) 7.4 HOMOGENEOUS EQUILIBRIA bl In a homogeneous system, all the reactants K′″c = ( Kc ) and products are in the same phase. For n na A + nb B ⇌ nc C + nd D example, in the gaseous reaction, N2(g) + 3H2(g) ⇌ 2NH3(g), reactants and Problem 7.1 be C products are in the homogeneous phase. The following concentrations were Similarly, for the reactions, obtained for the formation of NH3 from N2 CH3COOC2H5 (aq) + H2O (l) ⇌ CH3COOH (aq) o N and H2 at equilibrium at 500K. + C2H 5OH (aq) [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and [NH3] = 1.2 ×10–2M. Calculate equilibrium and, Fe3+ (aq) + SCN–(aq) ⇌ Fe(SCN)2+ (aq) constant. all the reactants and products are in © Solution homogeneous solution phase. We shall now The equilibrium constant for the reaction, consider equilibrium constant for some homogeneous reactions. N2(g) + 3H2(g) ⇌ 2NH3(g) can be written as, 7.4.1 Equilibrium Constant in Gaseous Systems  NH3 ( g )  2 So far we have expressed equilibrium constant Kc = of the reactions in terms of molar  N 2 ( g )   H2 ( g )  3 concentration of the reactants and products, and used symbol, Kc for it. For reactions tt = (1.2 ×10 ) −2 2 involving gases, however, it is usually more convenient to express the equilibrium (1.5 × 10 )(3.0 ×10 ) −2 −2 3 constant in terms of partial pressure. no The ideal gas equation is written as, = 0.106 × 104 = 1.06 × 103 pV = nRT Problem 7.2 n At equilibrium, the concentrations of ⇒ p= RT N2=3.0 × 10 –3M, O2 = 4.2 × 10–3M and V NO= 2.8 × 10–3M in a sealed vessel at Here, p is the pressure in Pa, n is the number of moles of the gas, V is the volume in m 3 and EQUILIBRIUM 195 T is the temperature in Kelvin NH3 ( g )  [ RT ] 2 −2 =  = K c ( RT ) Therefore, −2  N 2 ( g )   H 2 ( g )  3 n/V is concentration expressed in mol/m3 If concentration c, is in mol/L or mol/dm3, or K p = K c ( RT ) −2 and p is in bar then (7.14) p = cRT, Similarly, for a general reaction We can also write p = [gas]RT. aA + bB ⇌ cC + dD d Here, R= 0.0831 bar litre/mol K At constant temperature, the pressure of ( p )( p ) [ C] [ D] ( RT ) ( c d c d c +d ) he Kp = C = D the gas is proportional to its concentration i.e., ( p )( p ) [ A ] [B ] ( RT )( a A b B a b a +b ) p ∝ [gas] [C] [D] RT ( c +d ) − (a +b ) c d For reaction in equilibrium pu T = ( ) is H2(g) + I2(g) ⇌ 2HI(g) re ER [ A ]a [B]b We can write either [C] [D] RT ∆n c d bl ∆n HI ( g )  ( ) = Kc ( RT ) 2 = (7.15) Kc =  H2 ( g )   I2 ( g )  [ A ]a [B]b where ∆n = (number of moles of gaseous or K c = ( pH I )2 products) – (number of moles of gaseous be C ( p )( p ) H2 I2 (7.12) reactants) in the balanced chemical equation. It is necessary that while calculating the value Further, since p HI =  HI ( g )  RT of K p, pressure should be expressed in bar o N because standard state for pressure is 1 bar. p I2 =  I2 ( g)  RT We know from Unit 1 that : pH 2 =  H2 ( g )  RT 1pascal, Pa=1Nm–2, and 1bar = 105 Pa © Therefore, Kp values for a few selected reactions at different temperatures are given in Table 7.5 ( p HI ) HI ( g )  [ RT ] 2 2 2 Kp = = Table 7.5 Equilibrium Constants, Kp for a ( p )( p ) H2 I2 H 2 ( g )  RT.  I 2 ( g )  RT Few Selected Reactions  HI ( g )  2 = = Kc (7.13) H2 ( g )  I2 ( g )  In this example, K p = K c i.e., both tt equilibrium constants are equal. However, this is not always the case. For example in reaction N2(g) + 3H2(g) ⇌ 2NH3(g) no (p ) 2 NH3 Kp = ( p )( p ) 3 N2 H2 Problem 7.3 PCl5, PCl3 and Cl2 are at equilibrium at  NH3 ( g )  [ RT ] 2 2 500 K and having concentration 1.59M = PCl3, 1.59M Cl2 and 1.41 M PCl5.  N 2 ( g )  RT.  H2 ( g )  ( RT ) 3 3 196 CHEMISTRY Calculate Kc for the reaction, the value 0.194 should be neglected PCl5 ⇌ PCl3 + Cl2 because it will give concentration of the Solution reactant which is more than initial The equilibrium constant K c for the above concentration. reaction can be written as, Hence the equilibrium concentrations are, [CO 2] = [H2- ] = x = 0.067 M Kc = [PCl 3 ][Cl 2 ] = (1.59)2 = 1.79 [CO] = [H2O] = 0.1 – 0.067 = 0.033 M [PCl5 ] (1.41) d Problem 7.5 Problem 7.4 he For the equilibrium, The value of Kc = 4.24 at 800K for the reaction, CO (g) + H2O (g) ⇌ CO 2 (g) + H2 (g) 2NOCl(g) ⇌ 2NO(g) + Cl2(g) Calculate equilibrium concentrations of the value of the equilibrium constant, K c pu T is 3.75 × 10 –6 at 1069 K. Calculate the K p CO2, H2, CO and H2O at 800 K, if only CO is for the reaction at this temperature? and H 2O are present initially at re ER concentrations of 0.10M each. Solution Solution We know that, bl For the reaction, K p = Kc(RT)∆ n CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) For the above reaction, Initial concentration: ∆n = (2+1) – 2 = 1 be C 0.1M 0.1M 0 0 K p = 3.75 ×10–6 (0.0831 × 1069) Let x mole per litre of each of the product K p = 0.033 be formed. o N At equilibrium: 7.5 HETEROGENEOUS EQUILIBRIA (0.1-x) M (0.1-x) M xM xM Equilibrium in a system having more than one where x is the amount of CO2 and H2 at phase is called heterogeneous equilibrium. © equilibrium. The equilibrium between water vapour and Hence, equilibrium constant can be liquid water in a closed container is an written as, example of heterogeneous equilibrium. K c = x2/(0.1-x)2 = 4.24 H 2O(l) ⇌ H2O(g) x2 = 4.24(0.01 + x2-0.2x) In this example, there is a gas phase and a x2 = 0.0424 + 4.24x2-0.848x liquid phase. In the same way, equilibrium 3.24x2 – 0.848x + 0.0424 = 0 between a solid and its saturated solution, a = 3.24, b = – 0.848, c = 0.0424 Ca(OH) 2 (s) + (aq) ⇌ Ca2+ (aq) + 2OH–(aq) tt (for quadratic equation ax2 + bx + c = 0, is a heterogeneous equilibrium. (− b ± b2 − 4ac ) Heterogeneous equilibria often involve pure solids or liquids. We can simplify equilibrium no x= 2a expressions for the heterogeneous equilibria x = 0.848±√(0.848)2– 4(3.24)(0.0424)/ involving a pure liquid or a pure solid, as the (3.24×2) molar concentration of a pure solid or liquid x = (0.848 ± 0.4118)/ 6.48 is constant (i.e., independent of the amount x1 = (0.848 – 0.4118)/6.48 = 0.067 present). In other words if a substance ‘X’ is x2 = (0.848 + 0.4118)/6.48 = 0.194 involved, then [X(s)] and [X(l)] are constant, whatever the amount of ‘X’ is taken. Contrary EQUILIBRIUM 197 to this, [X(g)] and [X(aq)] will vary as the This shows that at a particular amount of X in a given volume varies. Let us temperature, there is a constant concentration take thermal dissociation of calcium carbonate or pressure of CO2 in equilibrium with CaO(s) which is an interesting and important example and CaCO3(s). Experimentally it has been of heterogeneous chemical equilibrium. found that at 1100 K, the pressure of CO2 in CaCO3 (s) ∆⇀  CaO (s) + CO2 (g) (7.16) equilibrium with CaCO3(s) and CaO(s), is ↽   2.0 ×105 Pa. Therefore, equilibrium constant On the basis of the stoichiometric equation, at 1100K for the above reaction is: d we can write, K p = p CO2 = 2 ×105 Pa /105 Pa = 2.00 CaO ( s )  CO 2 ( g)  Similarly, in the equilibrium between he Kc =  nickel, carbon monoxide and nickel carbonyl CaCO 3 ( s )  (used in the purification of nickel), Since [CaCO3(s)] and [CaO(s)] are both Ni (s) + 4 CO (g) ⇌ Ni(CO)4 (g), pu T constant, therefore modified equilibrium the equilibrium constant is written as is constant for the thermal decomposition of calcium carbonate will be  Ni ( CO) 4  re ER Kc =  K´c = [CO2(g)] (7.17) [CO]4 bl or K p = p CO2 (7.18) It must be remembered that for the existence of heterogeneous equilibrium pure Units of Equilibrium Constant solids or liquids must also be present be C The value of equilibrium constant Kc can (however small the amount may be) at be calculated by substituting the equilibrium, but their concentrations or concentration terms in mol/L and for Kp partial pressures do not appear in the o N partial pressure is substituted in Pa, kPa, expression of the equilibrium constant. In the bar or atm. This results in units of reaction, equilibrium constant based on molarity or Ag2O(s) + 2HNO3(aq) ⇌ 2AgNO3(aq) +H2O(l) pressure, unless the exponents of both the numerator and denominator are same. [AgNO3 ] 2 © For the reactions, Kc = H 2(g) + I2 (g) ⇌ 2HI, Kc and Kp have no unit. [HNO3 ]2 N2 O4 (g) ⇌ 2NO 2 (g), Kc has unit mol/L and Kp has unit bar Problem 7.6 Equilibrium constants can also be The value of K p for the reaction, expressed as dimensionless quantities if CO2 (g) + C (s) ⇌ 2CO (g) the standard state of reactants and is 3.0 at 1000 K. If initially products are specified. For a pure gas, the standard state is 1bar. Therefore a pressure p CO2 = 0.48 bar and p CO = 0 bar and pure tt of 4 bar in standard state can be expressed graphite is present, calculate the equilibrium as 4 bar/1 bar = 4, which is a partial pressures of CO and CO2. dimensionless number. Standard state (c0 ) for a solute is 1 molar solution and all Solution no concentrations can be measured with For the reaction, respect to it. The numerical value of equilibrium constant depends on the let ‘x’ be the decrease in pressure of CO2, standard state chosen. Thus, in this then system both Kp and Kc are dimensionless CO2(g) + C(s) ⇌ 2CO(g) quantities but have different numerical Initial values due to different standard states. pressure: 0.48 bar 0 198 CHEMISTRY At equilibrium: 5. The equilibrium constant K for a reaction (0.48 – x)bar 2x bar is related to the equilibrium constant of the corresponding reaction, whose equation is 2 p obtained by multiplying or dividing the Kp = CO p CO2 equation for the original reaction by a small integer. K p = (2x)2/(0.48 – x) = 3 Let us consider applications of equilibrium 4x 2 = 3(0.48 – x) constant to: d 4x 2 = 1.44 – x predict the extent of a reaction on the basis 4x 2 + 3x – 1.44 = 0 of its magnitude, he a = 4, b = 3, c = –1.44 predict the direction of the reaction, and (− b ± ) calculate equilibrium concentrations. b2 − 4ac x= 7.6.1 Predicting the Extent of a Reaction pu T 2a is The numerical value of the equilibrium = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 constant for a reaction indicates the extent of re ER = (–3 ± 5.66)/8 the reaction. But it is important to note that bl = (–3 + 5.66)/ 8 (as value of x cannot be an equilibrium constant does not give any negative hence we neglect that value) information about the rate at which the x = 2.66/8 = 0.33 equilibrium is reached. The magnitude of Kc or K p is directly proportional to the The equilibrium partial pressures are, concentrations of products (as these appear be C p CO = 2x = 2 × 0.33 = 0.66 bar in the numerator of equilibrium constant expression) and inversely proportional to the p CO = 0.48 – x = 0.48 – 0.33 = 0.15 bar concentrations of the reactants (these appear o N 2 in the denominator). This implies that a high value of K is suggestive of a high concentration 7.6 APPLICATIONS OF EQUILIBRIUM of products and vice-versa. CONSTANTS We can make the following generalisations © Before considering the applications of concerning the composition of equilibrium constants, let us summarise the equilibrium mixtures: important features of equilibrium constants as follows: If Kc > 103, products predominate over reactants, i.e., if Kc is very large, the reaction 1. Expression for equilibrium constant is proceeds nearly to completion. Consider applicable only when concentrations of the the following examples: reactants and products have attained constant value at equilibrium state. (a) The reaction of H2 with O2 at 500 K has a 2. The value of equilibrium constant is very large equilibrium c o n s t a n t , tt independent of initial concentrations of the K c = 2.4 × 1047. reactants and products. (b) H 2(g) + Cl2(g) ⇌ 2HCl(g) at 300K has 3. Equilibrium constant is temperature K c = 4.0 × 1031. no dependent having one unique value for a particular reaction represented by a (c) H 2(g) + Br2(g) ⇌ 2HBr (g) at 300 K, balanced equation at a given temperature. K c = 5.4 × 10 18 4. The equilibrium constant for the reverse If K c < 10–3, reactants predominate over reaction is equal to the inverse of the products, i.e., if K c is very small, the reaction equilibrium constant for the forward proceeds rarely. Consider the following reaction. examples: EQUILIBRIUM 199 (a) The decomposition of H2O into H2 and O2 If Qc = Kc , the reaction mixture is already at 500 K has a very small equilibrium at equilibrium. constant, K c = 4.1 × 10– 48 Consider the gaseous reaction of H2 (b) N2(g) + O2(g) ⇌ 2NO(g), with I2, at 298 K has K c = 4.8 ×10 – 31. H2(g) + I2(g) ⇌ 2HI(g); Kc = 57.0 at 700 K. Suppose we have molar concentrations If K c is in the range of 10 – 3 to 10 3, appreciable concentrations of both [H2]t=0.10M, [I 2] t = 0.20 M and [HI]t = 0.40 M. reactants and products are present. (the subscript t on the concentration symbols d Consider the following examples: means that the concentrations were measured (a) For reaction of H2 with I 2 to give HI, at some arbitrary time t, not necessarily at he equilibrium). K c = 57.0 at 700K. Thus, the reaction quotient, Qc at this stage (b) Also, gas phase decomposition of N2O 4 to of the reaction is given by, NO2 is another reaction with a value pu T Qc = [HI]t2 / [H2] t [I2] t = (0.40)2/ (0.10)×(0.20) of Kc = 4.64 × 10 –3 at 25°C which is neither is too small nor too large. Hence, = 8.0 Now, in this case, Q c (8.0) does not equal re ER equilibrium mixtures contain appreciable concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g) bl is not at equilibrium; that is, more H2(g) and These generarlisations are illustrated in I2(g) will react to form more HI(g) and their Fig. 7.6 concentrations will decrease till Qc = K c. The reaction quotient, Qc is useful in be C predicting the direction of reaction by comparing the values of Qc and K c. Thus, we can make the following o N generalisations concerning the direction of the Fig.7.6 Dependence of extent of reaction on Kc reaction (Fig. 7.7) : 7.6.2 Predicting the Direction of the © Reaction The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Q c with molar concentrations and QP with partial pressures) Fig. 7.7 Predicting the direction of the reaction is defined in the same way as the equilibrium constant K c except that the concentrations in If Qc < Kc, net reaction goes from left to right tt Qc are not necessarily equilibrium values. If Q c > K c, net reaction goes from right to For a general reaction: left. aA+bB ⇌ cC+dD (7.19) If Qc = K c, no net reaction occurs. no Qc = [C] c[D]d / [A]a[B]b (7.20) Problem 7.7 Then, The value of Kc for the reaction If Qc > Kc , the reaction will proceed in the 2A ⇌ B + C is 2 × 10 –3. At a given time, direction of reactants (reverse reaction). the composition of reaction mixture is If Qc < Kc , the reaction will proceed in the [A] = [B] = [C] = 3 × 10 –4 M. In which direction of the products (forward reaction). direction the reaction will proceed? 200 CHEMISTRY Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc found to be 9.15 bar. Calculate K c, K p and is given by, partial pressure at equilibrium. Q c = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10 – 4M We know pV = nRT Q c = (3 ×10 –4)(3 × 10 –4) / (3 ×10 –4)2 = 1 Total volume (V ) = 1 L as Qc > K c so the reaction will proceed in Molecular mass of N2O 4 = 92 g d the reverse direction. Number of moles = 13.8g/92 g = 0.15 he 7.6.3 Calculating Equilibrium of the gas (n) Concentrations Gas constant (R) = 0.083 bar L mol–1K –1 In case of a problem in which we know the Temperature (T ) = 400 K initial concentrations but do not know any of pu T pV = nRT is the equilibrium concentrations, the following p × 1L = 0.15 mol × 0.083 bar L mol–1K –1 three steps shall be followed: × 400 K re ER Step 1. Write the balanced equation for the p = 4.98 bar bl reaction. N2O 4 ⇌ 2NO2 Step 2. Under the balanced equation, make a Initial pressure: 4.98 bar 0 table that lists for each substance involved in the reaction: At equilibrium: (4.98 – x) bar 2x bar Hence, be C (a) the initial concentration, p total at equilibrium = p N2 O4 + p NO2 (b) the change in concentration on going to equilibrium, and 9.15 = (4.98 – x) + 2x o N (c) the equilibrium concentration. 9.15 = 4.98 + x In constructing the table, define x as the x = 9.15 – 4.98 = 4.17 bar concentration (mol/L) of one of the substances Partial pressures at equilibrium are, that reacts on going to equilibrium, then use © the stoichiometry of the reaction to determine p N 2O4 = 4.98 – 4.17 = 0.81bar the concentrations of the other substances in p NO2 = 2x = 2 × 4.17 = 8.34 bar terms of x. K p = ( p NO2 ) / p N 2O4 2 Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to = (8.34)2/0.81 = 85.87 ∆n solve a quadratic equation choose the K p = K c(RT) mathematical solution that makes chemical 85.87 = K c(0.083 × 400) 1 sense. K c = 2.586 = 2.6 tt Step 4. Calculate the equilibrium concentrations from the calculated value of x. Problem 7.9 Step 5. Check your results by substituting 3.00 mol of PCl 5 kept in 1L closed reaction no them into the equilibrium equation. vessel was allowed to attain equilibrium at 380K. Calculate composition of the Problem 7.8 mixture at equilibrium. K c= 1.80 13.8g of N 2O 4 was placed in a 1L reaction Solution vessel at 400K and allowed to attain PCl5 ⇌ PCl 3 + Cl 2 equilibrium Initial N 2O4 (g) ⇌ 2NO2 (g) concentration: 3.0 0 0 EQUILIBRIUM 201 Let x mol per litre of PCl5 be dissociated, K = e – ∆G V / RT (7.23) At equilibrium: (3-x) x x Hence, using the equation (7.23), the reaction spontaneity can be interpreted in K c = [PCl3][Cl2]/[PCl5] terms of the value of ∆G0. 1.8 = x2/ (3 – x) If ∆G0 < 0, then –∆G0/RT is positive, and x2 + 1.8x – 5.4 = 0 V e– ∆G /RT >1, making K >1, which implies x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 a spontaneous reaction or the reaction d x = [–1.8 ± √ 3.24 + 21.6]/2 which proceeds in the forward direction to such an extent that the products are x = [–1.8 ± 4.98]/2 he present predominantly. x = [–1.8 + 4.98]/2 = 1.59 If ∆G0 > 0, then –∆G0/RT is negative, and V [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M e– ∆G /RT < 1, that is , K < 1, which implies [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction pu T which proceeds in the forward direction to is 7.7 RELATIONSHIP BETWEEN such a small degree that only a very minute re ER EQUILIBRIUM CONSTANT K, quantity of product is formed. bl REACTION QUOTIENT Q AND Problem 7.10 GIBBS ENERGY G The value of ∆G0 for the phosphorylation The value of K c for a reaction does not depend of glucose in glycolysis is 13.8 kJ/mol. on the rate of the reaction. However, as you Find the value of K c at 298 K. be C have studied in Unit 6, it is directly related to the thermodynamics of the reaction and Solution in particular, to the change in Gibbs energy, ∆G0 = 13.8 kJ/mol = 13.8 × 103J/mol ∆G. If, o N Also, ∆G0 = – RT lnK c ∆G is negative, then the reaction is Hence, ln K c = –13.8 × 103J/mol spontaneous and proceeds in the forward (8.314 J mol –1K –1 × 298 K) direction. ln Kc = – 5.569 © ∆G is positive, then reaction is considered Kc = e–5.569 non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the Kc = 3.81 × 10 –3 products of the forward reaction shall be Problem 7.11 converted to the reactants. Hydrolysis of sucrose gives, ∆G is 0, reaction has achieved equilibrium; at this point, there is no longer any free Sucrose + H2O ⇌ Glucose + Fructose energy left to drive the reaction. Equilibrium constant K c for the reaction A mathematical expression of this is 2 ×1013 at 300K. Calculate ∆G0 at tt thermodynamic view of equilibrium can be 300K. described by the following equation: Solution 0 ∆G = ∆G + RT lnQ (7.21) 0 ∆G = – RT lnK c where, G0 is standard Gibbs energy. no ∆G0 = – 8.314J mol–1K –1× At equilibrium, when ∆G = 0 and Q = K c, 300K × ln(2×1013) the equation (7.21) becomes, ∆G = – 7.64 ×10 J mol–1 0 4 ∆G = ∆G0 + RT ln K = 0 0 ∆G = – RT lnK (7.22) 7.8 FACTORS AFFECTING EQUILIBRIA lnK = – ∆G / RT 0 One of the principal goals of chemical synthesis Taking antilog of both sides, we get, is to maximise the conversion of the reactants 202 CHEMISTRY to products while minimizing the expenditure “When the concentration of any of the of energy. This implies maximum yield of reactants or products in a reaction at products at mild temperature and pressure equilibrium is changed, the composition conditions. If it does not happen, then the of the equilibrium mixture changes so as experimental conditions need to be adjusted. to minimize the effect of concentration For example, in the Haber process for the changes”. synthesis of ammonia from N2 and H2, the Let us take the reaction, choice of experimental conditions is of real H 2(g) + I2(g) ⇌ 2HI(g) d economic importance. Annual world production of ammonia is about hundred If H 2 is added to the reaction mixture at equilibrium, then the equilibrium of the he million tones, primarily for use as fertilizers. Equilibrium constant, Kc is independent of reaction is disturbed. In order to restore it, the initial concentrations. But if a system at reaction proceeds in a direction wherein H2 is equilibrium is subjected to a change in the consumed, i.e., more of H2 and I2 react to form pu T concentration of one or more of the reacting HI and finally the equilibrium shifts in right is (forward) direction (Fig.7.8). This is in substances, then the system is no longer at accordance with the Le Chatelier’s principle re ER equilibrium; and net reaction takes place in some direction until the system returns to which implies that in case of addition of a bl reactant/product, a new equilibrium will be equilibrium once again. Similarly, a change in temperature or pressure of the system may set up in which the concentration of the reactant/product should be less than what it also alter the equilibrium. In order to decide was after the addition but more than what it what course the reaction adopts and make a was in the original mixture. be C qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a o N change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to © counteract the effect of the change. This is applicable to all physical and chemical equilibria. We shall now be discussing factors which can influence the equilibrium. 7.8.1 Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ tt products, Le Chatelier’s principle predicts that: The concentration stress of an added reactant/product is relieved by net reaction Fig. 7.8 Effect of addition of H2 on change of no in the direction that consumes the added concentration for the reactants and substance. products in the reaction, The concentration stress of a removed H 2(g) + I2 (g) ⇌ 2HI(g) reactant/product is relieved by net reaction in the direction that replenishes the The same point can be explained in terms removed substance. of the reaction quotient, Qc, 2 or in other words, Q c = [HI] / [H2][I2] EQUILIBRIUM 203 Addition of hydrogen at equilibrium results replenish the Fe 3+ ions. Because the in value of Qc being less than K c. Thus, in order concentration of [Fe(SCN)]2+ decreases, the to attain equilibrium again reaction moves in intensity of red colour decreases. the forward direction. Similarly, we can say Addition of aq. HgCl2 also decreases red that removal of a product also boosts the colour because Hg2+ reacts with SCN– ions to forward reaction and increases the form stable complex ion [Hg(SCN)4]2–. Removal concentration of the products and this has – of free SCN (aq) shifts the equilibrium in great commercial application in cases of equation (7.24) from right to left to replenish d reactions, where the product is a gas or a SCN – ions. Addition of potassium thiocyanate volatile substance. In case of manufacture of on the other hand increases the colour ammonia, ammonia is liquified and removed he intensity of the solution as it shift the from the reaction mixture so that reaction equilibrium to right. keeps moving in forward direction. Similarly, in the large scale production of CaO (used as 7.8.2 Effect of Pressure Change pu T important building material) from CaCO3, A pressure change obtained by changing the is constant removal of CO2 from the kiln drives volume can affect the yield of products in case the reaction to completion. It should be of a gaseous reaction where the total number re ER remembered that continuous removal of a of moles of gaseous reactants and total bl product maintains Qc at a value less than Kc number of moles of gaseous products are and reaction continues to move in the forward different. In applying Le Chatelier’s principle direction. to a heterogeneous equilibrium the effect of Effect of Concentration – An experiment pressure changes on solids and liquids can be C be ignored because the volume (and This can be demonstrated by the following concentration) of a solution/liquid is nearly reaction: independent of pressure. Fe3+(aq)+ SCN –(aq) ⇌ [Fe(SCN)]2+(aq) (7.24) o N Consider the reaction, yellow colourless deep red CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)  Fe ( SCN ) 2+ ( aq )  Here, 4 mol of gaseous reactants (CO + 3H2) Kc =   become 2 mol of gaseous products (CH4 + © (7.25)  Fe3+ ( aq )   SCN – ( aq )  H2O). Suppose equilibrium mixture (for above reaction) kept in a cylinder fitted with a piston A reddish colour appears on adding two at constant temperature is compressed to one drops of 0.002 M potassium thiocynate half of its original volume. Then, total pressure solution to 1 mL of 0.2 M iron(III) nitrate will be doubled (according to solution due to the formation of [Fe(SCN)]2+. pV = constant). The partial pressure and The intensity of the red colour becomes therefore, concentration of reactants and constant on attaining equilibrium. This products have changed and the mixture is no equilibrium can be shifted in either forward longer at equilibrium. The direction in which tt or reverse directions depending on our choice the reaction goes to re-establish equilibrium of adding a reactant or a product. The can be predicted by applying the Le Chatelier’s equilibrium can be shifted in the opposite principle. Since pressure has doubled, the direction by adding reagents that remove Fe3+ equilibrium now shifts in the forward no or SCN – ions. For example, oxalic acid direction, a direction in which the number of (H2C2O 4), reacts with Fe3+ ions to form the moles of the gas or pressure decreases (we stable complex ion [Fe(C 2O 4) 3] 3 – , thus know pressure is proportional to moles of the decreasing the concentration of free Fe3+ (aq). gas). This can also be understood by using In accordance with the Le Chatelier’s principle, reaction quotient, Qc. Let [CO], [H2], [CH4] and the concentration stress of removed Fe3+ is [H 2O] be the molar concentrations at relieved by dissociation of [Fe(SCN)] 2+ to equilibrium for methanation reaction. When 204 CHEMISTRY volume of the reaction mixture is halved, the Production of ammonia according to the partial pressure and the concentration are reaction, doubled. We obtain the reaction quotient by N2(g) + 3H2(g) ⇌ 2NH3(g) ; replacing each equilibrium concentration by double its value. ∆H= – 92.38 kJ mol –1 is an exothermic process. According to  CH4 ( g )  H2O ( g )  Le Chatelier’s principle, raising the Qc =  temperature shifts the equilibrium to left and CO ( g )   H2 ( g )  3 d decreases the equilibrium concentration of As Q c < Kc , the reaction proceeds in the ammonia. In other words, low temperature is favourable for high yield of ammonia, but he forward direction. In reaction C(s) + CO2(g) ⇌ 2CO(g), when practically very low temperatures slow down pressure is increased, the reaction goes in the the reaction and thus a catalyst is used. reverse direction because the number of moles Effect of Temperature – An experiment pu T of gas increases in the forward direction. Effect of temperature on equilibrium can be is demonstrated by taking NO 2 gas (brown in 7.8.3 Effect of Inert Gas Addition re ER If the volume is kept constant and an inert gas colour) which dimerises into N 2O 4 gas (colourless). bl such as argon is added which does not take 2NO2(g) ⇌ N2O4(g); ∆H = –57.2 kJ mol–1 part in the reaction, the equilibrium remains undisturbed. It is because the addition of an NO 2 gas prepared by addition of Cu inert gas at constant volume does not change turnings to conc. HNO 3 is collected in two be C the partial pressures or the molar 5 mL test tubes (ensuring same intensity of concentrations of the substance involved in the colour of gas in each tube) and stopper sealed reaction. The reaction quotient changes only with araldite. Three 250 mL beakers 1, 2 and o N if the added gas is a reactant or product 3 containing freezing mixture, water at room involved in the reaction. temperature and hot water (36 3K ), respectively, are taken (Fig. 7.9). Both the test 7.8.4 Effect of Temperature Change tubes are placed in beaker 2 for 8-10 minutes. © Whenever an equilibrium is disturbed by a After this one is placed in beaker 1 and the change in the concentration, pressure or other in beaker 3. The effect of temperature volume, the composition of the equilibrium on direction of reaction is depicted very well mixture changes because the reaction in this experiment. At low temperatures in quotient, Qc no longer equals the equilibrium beaker 1, the forward reaction of formation of constant, K c. However, when a change in N2O4 is preferred, as reaction is exothermic, and temperature occurs, the value of equilibrium thus, intensity of brown colour due to NO2 constant, Kc is changed. decreases. While in beaker 3, high temperature favours the reverse reaction of In general, the temperature dependence of tt the equilibrium constant depends on the sign of ∆H for the reaction. The equilibrium constant for an exothermic no reaction (negative ∆H) decreases as the temperature increases. The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases. Temperature changes affect the Fig. 7.9 Effect of temperature on equilibrium for equilibrium constant and rates of reactions. the reaction, 2NO2 (g) ⇌ N2 O4 (g) EQUILIBRIUM 205 formation of NO2 and thus, the brown colour Similarly, in manufacture of sulphuric intensifies. acid by contact process, Effect of temperature can also be seen in 2SO2(g) + O2(g) ⇌ 2SO3(g); Kc = 1.7 × 1026 an endothermic reaction, though the value of K is suggestive of reaction [Co(H2O) 6]3+ (aq) + 4Cl– (aq) ⇌ [CoCl4]2–(aq) + going to completion, but practically the oxidation 6H2O(l) of SO2 to SO3 is very slow. Thus, platinum or pink colourless blue divanadium penta-oxide (V2O5) is used as At room temperature, the equilibrium catalyst to increase the rate of the reaction. d mixture is blue due to [CoCl 4] 2–. When cooled Note: If a reaction has an exceedingly small in a freezing mixture, the colour of the mixture K, a catalyst would be of little help. he turns pink due to [Co(H2O)6]3+. 7.9 IONIC EQUILIBRIUM IN SOLUTION 7.8.5 Effect of a Catalyst Under the effect of change of concentration on A catalyst increases the rate of the chemical pu T the direction of equilibrium, you have reaction by making available a new low energy is incidently come across with the following pathway for the conversion of reactants to equilibrium which involves ions: re ER products. It increases the rate of forward and reverse reactions that pass through the same Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq) bl transition state and does not affect There are numerous equilibria that involve equilibrium. Catalyst lowers the activation ions only. In the following sections we will energy for the forward and reverse reactions study the equilibria involving ions. It is well by exactly the same amount. Catalyst does not known that the aqueous solution of sugar be C affect the equilibrium composition of a does not conduct electricity. However, when reaction mixture. It does not appear in the common salt (sodium chloride) is added to balanced chemical equation or in the water it conducts electricity. Also, the o N equilibrium constant expression. conductance of electricity increases with an Let us consider the formation of NH3 from increase in concentration of common salt. dinitrogen and dihydrogen which is highly Michael Faraday classified the substances into exothermic reaction and proceeds with two categories based on their ability to conduct © decrease in total number of moles formed as electricity. One category of substances compared to the reactants. Equilibrium conduct electricity in their aqueous solutions and are called electrolytes while the other do constant decreases with increase in temperature. At low temperature rate not and are thus, referred to as non- electrolytes. Faraday further classified decreases and it takes long time to reach at electrolytes into strong and weak electrolytes. equilibrium, whereas high temperatures give Strong electrolytes on dissolution in water are satisfactory rates but poor yields. ionized almost completely, while the weak German chemist, Fritz Haber discovered electrolytes are only partially dissociated. that a catalyst consisting of iron catalyse the tt For example, an aqueous solution of reaction to occur at a satisfactory rate at sodium chloride is comprised entirely of temperatures, where the equilibrium sodium ions and chloride ions, while that concentration of NH3 is reasonably favourable. of acetic acid mainly contains unionized no Since the number of moles formed in the acetic acid molecules and only some acetate reaction is less than those of reactants, the ions and hydronium ions. This is because yield of NH3 can be improved by increasing there is almost 100% ionization in case of the pressure. sodium chloride as compared to less Optimum conditions of temperature and than 5% ionization of acetic acid which is pressure for the synthesis of NH 3 using a weak electrolyte. It should be noted catalyst are around 500 °C and 200 atm. that in weak electrolytes, equilibrium is 206 CHEMISTRY established between ions and the unionized exists in solid state as a cluster of positively molecules. This type of equilibrium involving charged sodium ions and negatively charged ions in aqueous solution is called ionic chloride ions which are held together due to equilibrium. Acids, bases and salts come electrostatic interactions between oppositely under the category of electrolytes and may act charged species (Fig.7.10). The electrostatic as either strong or weak electrolytes. forces between two charges are inversely proportional to dielectric constant of the 7.10 ACIDS, BASES AND SALTS medium. Water, a universal solvent, possesses d Acids, bases and salts find widespread a very high dielectric constant of 80. Thus, occurrence in nature. Hydrochloric acid when sodium chloride is dissolved in water, the electrostatic interactions are reduced by a he present in the gastric juice is secreted by the lining of our stomach in a significant amount factor of 80 and this facilitates the ions to move of 1.2-1.5 L/day and is essential for digestive freely in the solution. Also, they are well- processes. Acetic acid is known to be the main separated due to hydration with water pu T constituent of vinegar. Lemon and orange molecules. is juices contain citric and ascorbic acids, and tartaric acid is found in tamarind paste. As re ER most of the acids taste sour, the word “acid” bl has been derived from a latin word “acidus” meaning sour. Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with some metals. Similarly, bases be C are known to turn red litmus paper blue, taste bitter and feel soapy. A common example of a base is washing soda used for washing o N purposes. When acids and bases are mixed in the right proportion they react with each other to give salts. Some commonly known Fig.7.10 Dissolution of sodium chloride in water. + – examples of salts are sodium chloride, barium Na and Cl ions are stablised by their © sulphate, sodium nitrate. Sodium chloride hydration with polar water molecules. (common salt ) is an important component of Comparing, the ionization of hydrochloric our diet and is formed by reaction between acid with that of acetic acid in water we find hydrochloric acid and sodium hydroxide. It that though both of them are polar covalent Faraday w as born near London into a family of very limited means. At the age of 14 he was an apprentice to a kind bookbinder who allowed Faraday to read the books he was binding. Through a fortunate chance he became laboratory assistant to Davy, and during 1813-4, Faraday accompanied him to the Continent. During this trip he gained tt much from the experience of coming into contact with many of the leading scientists of the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratorie