Equilibrium - Unit 7 - PDF

192 CHEMISTRY UNIT 7 EQUILIBRIUM Chemical equilibria are important in numero...

192 CHEMISTRY UNIT 7 EQUILIBRIUM Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O2 molecules and the protein hemoglobin play a After studying this unit you will be crucial role in the transport and delivery of O2 from our able to lungs to our muscles. Similar equilibria involving CO identify 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 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 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. 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 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 pKw 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 2022-23 EQUILIBRIUM 193 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 the boundary between ice and water. (i) The reactions that proceed nearly to Molecules from the liquid water collide against completion and only negligible ice and adhere to it and some molecules of ice concentrations of the reactants are left. In escape into liquid phase. There is no change some cases, it may not be even possible to of mass of ice and water, as the rates of transfer detect these experimentally. 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 atmospheric pressure and 273 K. reactants remain unchanged at It is obvious that ice and water are in equilibrium stage. equilibrium only at particular temperature (iii) The reactions in which the concentrations and pressure. For any pure substance at of the reactants and products are atmospheric pressure, the temperature at comparable, when the system is in 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. 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 is very important in industry and laboratory simultaneously. so that equilibrium is favorable in the (ii) Both the processes occur at the same rate direction of the desired product. Some so that the amount of ice and water important aspects of equilibrium involving remains constant. physical and chemical processes are dealt in this unit along with the equilibrium involving 7.1.2 Liquid-Vapour Equilibrium ions in aqueous solutions which is called as This equilibrium can be better understood if ionic equilibrium. we consider the example of a transparent box carrying a U-tube with mercury (manometer). 7.1 EQUILIBRIUM IN PHYSICAL Drying agent like anhydrous calcium chloride PROCESSES (or phosphorus penta-oxide) is placed for a The characteristics of system at equilibrium few hours in the box. After removing the are better understood if we examine some drying agent by tilting the box on one side, a physical processes. The most familiar watch glass (or petri dish) containing water is examples are phase transformation quickly placed inside the box. It will be processes, e.g., observed that the mercury level in the right solid liquid limb of the manometer slowly increases and liquid gas finally attains a constant value, that is, the pressure inside the box increases and reaches solid gas 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 2022-23 194 CHEMISTRY 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. consequence the rate of condensation from 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 vapour into water. Finally it leads to an and it is not possible to reach equilibrium in 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., 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) H2O (vap) pressure (1.013 bar), the temperature at At equilibrium the pressure exerted by the which the liquid and vapours are at water molecules at a given temperature equilibrium is called normal boiling point of remains constant and is called the equilibrium the liquid. Boiling point of the liquid depends vapour pressure of water (or just vapour on the atmospheric pressure. It depends on pressure of water); vapour pressure of water the altitude of the place; at high altitude the increases with temperature. If the above boiling point decreases. 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 more volatile and has a lower boiling point. filled up with violet vapour and the intensity of colour increases with time. After certain time the If we expose three watch glasses intensity of colour becomes constant and at this 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 equilibrium can be represented as, that in all such cases the liquid eventually disappears and the time taken for complete I2(solid) I2 (vapour) 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 NH4Cl (solid)  NH4Cl (vapour) remains constant but the molecules are 2022-23 EQUILIBRIUM 195 7.1.4 Equilibrium Involving Dissolution of pressure of the gas above the solvent. This 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 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 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 temperature. In a saturated solution, a 1 atm (1.013 bar) at which the two phases dynamic equilibrium exits between the solute can coexist. If there is no exchange of heat 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 (ii) For liquid vapour equilibrium, the the rate of dissolution of sugar = rate of vapour pressure is constant at a given crystallisation of sugar. temperature. Equality of the two rates and dynamic (iii) For dissolution of solids in liquids, the nature of equilibrium has been confirmed with solubility is constant at a given the help of radioactive sugar. If we drop some temperature. radioactive sugar into saturated solution of (iv) For dissolution of gases in liquids, the non-radioactive sugar, then after some time concentration of a gas in liquid is radioactivity is observed both in the solution proportional to the pressure and in the solid sugar. Initially there were no (concentration) of the gas over the liquid. radioactive sugar molecules in the solution These observations are summarised in but due to dynamic nature of equilibrium, Table 7.1 there is exchange between the radioactive and 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 pH2Oconstant at given H2O (l) H2O (g) temperature When a soda water bottle is opened, some of the carbon dioxide gas dissolved in it fizzes Solid Liquid Melting point is fixed at out rapidly. The phenomenon arises due to H2O (s) H2O (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 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 temperature This equilibrium is governed by Henry’s CO2(g) CO2(aq) [CO 2 (aq)]/[CO 2 (g)] is law, which states that the mass of a gas constant at a given dissolved in a given mass of a solvent at temperature any temperature is proportional to the 2022-23 196 CHEMISTRY 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. (ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition. (iii) All measurable properties of the system remain constant. (iv) When equilibrium is attained for a physical process, it is characterised by constant Fig. 7.2 Attainment of chemical equilibrium. value of one of its parameters at a given temperature. Table 7.1 lists such Eventually, the two reactions occur at the quantities. same rate and the system reaches a state of (v) The magnitude of such quantities at any equilibrium. stage indicates the extent to which the physical process has proceeded before Similarly, the reaction can reach the state of reaching equilibrium. equilibrium even if we start with only C and D; that is, no A and B being present initially, as the 7.2 EQUILIBRIUM IN CHEMICAL equilibrium can be reached from either direction. PROCESSES – DYNAMIC The dynamic nature of chemical EQUILIBRIUM equilibrium can be demonstrated in the Analogous to the physical systems chemical synthesis of ammonia by Haber’s process. In reactions also attain a state of equilibrium. a series of experiments, Haber started with These reactions can occur both in forward known amounts of dinitrogen and dihydrogen and backward directions. When the rates of maintained at high temperature and pressure the forward and reverse reactions become and at regular intervals determined the equal, the concentrations of the reactants amount of ammonia present. He was and the products remain constant. This is successful in determining also the the stage of chemical equilibrium. This concentration of unreacted dihydrogen and equilibrium is dynamic in nature as it dinitrogen. Fig. 7.4 (page 191) shows that after consists of a forward reaction in which the a certain time the composition of the mixture reactants give product(s) and reverse remains the same even though some of the reaction in which product(s) gives the reactants are still present. This constancy in original reactants. composition indicates that the reaction has For a better comprehension, let us reached equilibrium. In order to understand consider a general case of a reversible reaction, the dynamic nature of the reaction, synthesis of ammonia is carried out with exactly the A+B C+D same starting conditions (of partial pressure With passage of time, there is and temperature) but using D2 (deuterium) accumulation of the products C and D and in place of H2. The reaction mixtures starting depletion of the reactants A and B (Fig. 7.2). either with H2 or D2 reach equilibrium with This leads to a decrease in the rate of forward the same composition, except that D2 and ND3 reaction and an increase in the rate of the are present instead of H2 and NH3. After reverse reaction, equilibrium is attained, these two mixtures 2022-23 EQUILIBRIUM 197 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 3-5mm. Fill nearly half of the measuring cylinder -1 with coloured water (for this purpose 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 cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2 nd cylinder, transfer the coloured 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 notice that the level of coloured water in both the cylinders becomes constant. 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 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. Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. 2022-23 198 CHEMISTRY 2NH3(g) N2(g) + 3H2(g) Similarly let us consider the reaction, H2(g) + I2(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 constant at equilibrium (Fig. 7.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the concentration of HI will decrease and concentration of H2 and I2 will increase until they all become constant when equilibrium is reached (Fig.7.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we Fig 7.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product. N 2 ( g ) + 3H2 ( g )  2NH3 ( g ) (H 2, N2, NH3 and D2 , N2, ND3) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found 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 from either direction Use of isotope (deuterium) in the formation of ammonia clearly indicates that chemical 7.3 LAW OF CHEMICAL EQUILIBRIUM 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 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 NH3(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? 2022-23 EQUILIBRIUM 199 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 H2 and I2 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 I2, and with time it was observed that intensity 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 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 Waage proposed in 1864 that the experiments are given in Table 7.2. 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, reacted = number of moles of iodine reacted = Kc = [C ][D] ½ (number of moles of HI formed). Also, [ A ][B ] experiments 5 and 6 indicate that, (7.1) where Kc is the equilibrium constant [H2(g)]eq = [I2(g)]eq and the expression on the right side is called Knowing the above facts, in order to the equilibrium constant expression. 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, 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 [I2(g)]eq gaseous H2 and I2 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 2022-23 200 CHEMISTRY 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, c d a b Kc = [C] [D] / [A] [B] (7.4) where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products. Equilibrium constant for the reaction, 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) is written as 4 6 4 5 Kc = [NO] [H2O] / [NH3] [O2] Molar concentration of different species is indicated by enclosing these in square bracket is far from constant. However, if we consider and, as mentioned above, it is implied that these the expression, are equilibrium concentrations. While writing [HI(g)]2eq / [H2(g)]eq [I2(g)]eq expression for equilibrium constant, symbol for we find that this expression gives constant phases (s, l, g) are generally ignored. value (as shown in Table 7.3) in all the six Let us write equilibrium constant for the cases. It can be seen that in this expression reaction, H2(g) + I2(g) 2HI(g) (7.5) the power of the concentration for reactants 2 as, Kc = [HI] / [H2] [I2] = x and products are actually the stoichiometric (7.6) coefficients in the equation for the chemical The equilibrium constant for the reverse reaction. Thus, for the reaction H2(g) + I2(g) reaction, 2HI(g) H2(g) + I2(g), at the same 2HI(g), following equation 7.1, the equilibrium temperature is, constant Kc is written as, 2 Kc = [HI(g)]eq / [H2(g)]eq [I2(g)]eq (7.2) K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc (7.7) Thus, K′c = 1 / Kc (7.8) Generally the subscript ‘eq’ (used for Equilibrium constant for the reverse equilibrium) is omitted from the concentration reaction is the inverse of the equilibrium terms. It is taken for granted that the constant for the reaction in the forward concentrations in the expression for Kc are direction. equilibrium values. We, therefore, write, If we change the stoichiometric coefficients Kc = [HI(g)]2 / [H2(g)] [I2(g)] (7.3) 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 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 ½ H2(g) + ½ I2(g) HI(g) (7.9) coefficient in the balanced chemical equation divided by the product of the equilibrium constant for the above reaction concentrations of the reactants raised to is given by K″c = [HI] / [H2] 1/2 1/2 2 1/2 their individual stoichiometric coefficients [I2] = {[HI] / [H2][I2]} has a constant value. This is known as = x = Kc 1/2 (7.10) 1/2 the Equilibrium Law or Law of Chemical On multiplying the equation (7.5) by n, we get Equilibrium. 2022-23 EQUILIBRIUM 201 nH2(g) + nI2(g) D 2nHI(g) (7.11) 800K. What will be Kc for the reaction Therefore, equilibrium constant for the N2(g) + O2(g) 2NO(g) n reaction is equal to Kc. These findings are Solution summarised in Table 7.4. It should be noted that because the equilibrium constants Kc and For the reaction equilibrium constant, K ′c have different numerical values, it is Kc can be written as, important to specify the form of the balanced chemical equation when quoting the value of Kc = [NO]2 an equilibrium constant. [ N 2 ][O2 ] (2.8 × 10 M ) 2 Table 7.4 Relations between Equilibrium -3 Constants for a General Reaction = and its Multiples. (3.0 × 10 M) (4.2 × 10 −3 −3 M ) Chemical equation Equilibrium constant = 0.622 aA+bB c C + dD Kc 7.4 HOMOGENEOUS EQUILIBRIA cC+dD aA+bB K′c =(1/Kc ) In a homogeneous system, all the reactants and products are in the same phase. For K′″c = (Kc ) n na A + nb B ncC + ndD example, in the gaseous reaction, N 2(g) + 3H 2(g) 2NH3(g), reactants and products are in the homogeneous phase. Problem 7.1 Similarly, for the reactions, The following concentrations were CH3COOC2H5 (aq) + H2O (l) CH3COOH (aq) obtained for the formation of NH3 from N2 + C2H5OH (aq) and H 2 at equilibrium at 500K. – [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and and, Fe3+ (aq) + SCN (aq) Fe(SCN)2+ (aq) [NH3] = 1.2 ×10–2M. Calculate equilibrium all the reactants and products are in constant. homogeneous solution phase. We shall now Solution consider equilibrium constant for some homogeneous reactions. The equilibrium constant for the reaction, N2(g) + 3H2(g) 2NH3(g) can be written 7.4.1 Equilibrium Constant in Gaseous as, Systems So far we have expressed equilibrium constant  NH3 ( g ) 2 of the reactions in terms of molar Kc = concentration of the reactants and products, N 2 (g )  H2 ( g ) 3 and used symbol, Kc for it. For reactions involving gases, however, it is usually more = (1.2 × 10 ) −2 2 convenient to express the equilibrium (1.5 × 10 ) (3.0 × 10 ) −2 −2 3 constant in terms of partial pressure. The ideal gas equation is written as, = 0.106 × 104 = 1.06 × 103 pV = n RT Problem 7.2 n ⇒ p= RT At equilibrium, the concentrations of V N2=3.0 × 10 –3M, O2 = 4.2 × 10–3M and Here, p is the pressure in Pa, n is the number NO= 2.8 × 10–3M in a sealed vessel at of moles of the gas, V is the volume in m3 and T is the temperature in Kelvin 2022-23 202 CHEMISTRY Therefore,  NH3 ( g ) [RT ] 2 −2 =  n/V is concentration expressed in mol/m3 = K c ( RT ) −2  N 2 ( g )  H 2 ( g ) 3 If concentration c, is in mol/L or mol/dm3, and p is in bar then p = cRT, or K p = K c ( RT ) −2 (7.14) We can also write p = [gas]RT. Here, R= 0.0831 bar litre/mol K Similarly, for a general reaction At constant temperature, the pressure of aA + bB cC + dD the gas is proportional to its concentration i.e., p ∝ [gas] Kp = ( p )( p ) [C] [D] ( RT )( c C = d D c d c +d ) For reaction in equilibrium H2(g) + I2(g) 2HI(g) ( p )( p ) [ A ] [B] (RT )( a A b B a b a +b ) We can write either = [C ]c [D]d (RT )(c +d )−(a +b )  HI ( g ) 2 Kc = [ A ]a [B ]b H2 ( g ) I2 ( g ) [C ] [ D] c d ∆n ∆n b ( = RT ) = K c (RT ) (7.15) ( p HI ) [ A ] [B ] 2 a or K c = ( p )( p ) H2 I2 (7.12) where ∆n = (number of moles of gaseous products) – (number of moles of gaseous Further, since p HI =  HI (g ) RT reactants) in the balanced chemical equation. It is necessary that while calculating the value pI2 =  I2 ( g ) RT of Kp, pressure should be expressed in bar pH2 =  H2 ( g ) RT because standard state for pressure is 1 bar. We know from Unit 1 that : Therefore, –2 1pascal, Pa=1Nm , and 1bar = 105 Pa ( pHI )2  HI ( g ) [RT ] 2 2 Kp values for a few selected reactions at Kp = = ( p )( p ) different temperatures are given in Table 7.5 H2 I2  H2 ( g ) RT. I2 ( g ) RT Table 7.5 Equilibrium Constants, Kp for a Few Selected Reactions HI ( g ) 2 = = Kc (7.13) H2 (g ) I2 ( g ) In this example, K p = K c i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction N2(g) + 3H2(g) 2NH3(g) (p ) 2 NH 3 Kp = ( p )( p ) 3 N2 H2 Problem 7.3  NH3 ( g ) [RT ] 2 2 = PCl5, PCl3 and Cl2 are at equilibrium at  N 2 ( g ) RT.  H 2 ( g ) ( RT ) 3 3 500 K and having concentration 1.59M PCl3, 1.59M Cl2 and 1.41 M PCl5. 2022-23 EQUILIBRIUM 203 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 Kc for the above concentration. reaction can be written as, Hence the equilibrium concentrations are, Kc = [PCl ] [Cl ] = (1.59) 3 2 2 = 1.79 [CO2] = [H2-] = x = 0.067 M [CO] = [H2O] = 0.1 – 0.067 = 0.033 M [PCl ] 5 (1.41) Problem 7.5 Problem 7.4 For the equilibrium, The value of Kc = 4.24 at 800K for the reaction, CO (g) + H2O (g) CO2 (g) + H2 (g) 2NOCl(g) 2NO(g) + Cl2(g) Calculate equilibrium concentrations of the value of the equilibrium constant, Kc CO2, H2, CO and H2O at 800 K, if only CO is 3.75 × 10–6 at 1069 K. Calculate the Kp and H 2 O are present initially at for the reaction at this temperature? concentrations of 0.10M each. Solution Solution We know that, For the reaction, ∆n Kp = Kc(RT) CO (g) + H2O (g) CO2 (g) + H2 (g) For the above reaction, Initial concentration: ∆n = (2+1) – 2 = 1 0.1M 0.1M 0 0 Kp = 3.75 ×10–6 (0.0831 × 1069) Let x mole per litre of each of the product Kp = 0.033 be formed. 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. Kc = x2/(0.1-x)2 = 4.24 H2O(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) (for quadratic equation ax2 + bx + c = 0, is a heterogeneous equilibrium. x= (− b ± b2 − 4ac ) Heterogeneous equilibria often involve pure solids or liquids. We can simplify equilibrium 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 involved, then [X(s)] and [X(l)] are constant, x2 = (0.848 + 0.4118)/6.48 = 0.194 whatever the amount of ‘X’ is taken. Contrary 2022-23 204 CHEMISTRY 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 equilibrium with CaCO3(s) and CaO(s), is CaCO3 (s) CaO (s) + CO2 (g) (7.16) 2.0 ×105 Pa. Therefore, equilibrium constant at 1100K for the above reaction is: On the basis of the stoichiometric equation, we can write, Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00 CaO ( s) CO 2 ( g ) Similarly, in the equilibrium between 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), constant, therefore modified equilibrium the equilibrium constant is written as constant for the thermal decomposition of  Ni (CO)4  Kc =  calcium carbonate will be K´c = [CO2(g)] (7.17) [CO]4 or Kp = pCO 2 (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 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 K p partial pressures do not appear in the 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. For the reactions, Kc = [ AgNO ]3 2 H2(g) + I2(g) 2HI, Kc and Kp have no unit. [HNO ]3 2 N2O4(g) 2NO2 (g), Kc has unit mol/L and Kp has unit bar Problem 7.6 Equilibrium constants can also be The value of Kp for the reaction, expressed as dimensionless quantities if CO2 (g) + C (s) 2CO (g) the standard state of reactants and products are specified. For a pure gas, the is 3.0 at 1000 K. If initially PCO = 0.48 bar 2 standard state is 1bar. Therefore a pressure and PCO = 0 bar and pure graphite is of 4 bar in standard state can be expressed present, calculate the equilibrium partial as 4 bar/1 bar = 4, which is a pressures of CO and CO2. dimensionless number. Standard state (c0) Solution for a solute is 1 molar solution and all concentrations can be measured with For the reaction, respect to it. The numerical value of let ‘x’ be the decrease in pressure of CO2, equilibrium constant depends on the then standard state chosen. Thus, in this CO2(g) + C(s) 2CO(g) system both K p and Kc are dimensionless Initial quantities but have different numerical values due to different standard states. pressure: 0.48 bar 0 2022-23 EQUILIBRIUM 205 5. The equilibrium constant K for a reaction At equilibrium: is related to the equilibrium constant of the (0.48 – x)bar 2x bar corresponding reaction, whose equation is pC2 O obtained by multiplying or dividing the Kp = equation for the original reaction by a small pC O2 integer. Kp = (2x)2/(0.48 – x) = 3 Let us consider applications of equilibrium 4x2 = 3(0.48 – x) constant to: 4x2 = 1.44 – x predict the extent of a reaction on the basis 4x2 + 3x – 1.44 = 0 of its magnitude, a = 4, b = 3, c = –1.44 predict the direction of the reaction, and calculate equilibrium concentrations. x= (−b ± b2 − 4 ac ) 7.6.1 Predicting the Extent of a Reaction 2a The numerical value of the equilibrium = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 constant for a reaction indicates the extent of = (–3 ± 5.66)/8 the reaction. But it is important to note that = (–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 equilibrium is reached. The magnitude of Kc x = 2.66/8 = 0.33 or K p is directly proportional to the The equilibrium partial pressures are, concentrations of products (as these appear pCO = 2x = 2 × 0.33 = 0.66 bar in the numerator of equilibrium constant 2 expression) and inversely proportional to the pCO = 0.48 – x = 0.48 – 0.33 = 0.15 bar 2 concentrations of the reactants (these appear in the denominator). This implies that a high 7.6 APPLICATIONS OF EQUILIBRIUM value of K is suggestive of a high concentration CONSTANTS of products and vice-versa. Before considering the applications of We can make the following generalisations equilibrium constants, let us summarise the concerning the composition of important features of equilibrium constants as equilibrium mixtures: follows: If Kc > 103, products predominate over 1. Expression for equilibrium constant is reactants, i.e., if Kc is very large, the reaction applicable only when concentrations of the proceeds nearly to completion. Consider reactants and products have attained the following examples: 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 , independent of initial concentrations of the Kc = 2.4 × 1047. reactants and products. (b) H2(g) + Cl2(g) 2HCl(g) at 300K has 3. Equilibrium constant is temperature Kc = 4.0 × 1031. dependent having one unique value for a particular reaction represented by a (c) H 2(g) + Br 2(g) 2HBr (g) at 300 K, Kc = 5.4 × 1018 balanced equation at a given temperature. 4. The equilibrium constant for the reverse If Kc < 10–3, reactants predominate over reaction is equal to the inverse of the products, i.e., if Kc is very small, the reaction equilibrium constant for the forward proceeds rarely. Consider the following reaction. examples: 2022-23 206 CHEMISTRY (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, Kc = 4.1 × 10 –48 Consider the gaseous reaction of H 2 (b) N2(g) + O2(g) 2NO(g), with I2, at 298 K has Kc = 4.8 ×10 – 31. H2(g) + I2(g) 2HI(g); Kc = 57.0 at 700 K. If K c is in the range of 10 – 3 to 10 3 , Suppose we have molar concentrations appreciable concentrations of both [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M. reactants and products are present. (the subscript t on the concentration symbols Consider the following examples: means that the concentrations were measured (a) For reaction of H2 with I2 to give HI, at some arbitrary time t, not necessarily at equilibrium). Kc = 57.0 at 700K. Thus, the reaction quotient, Qc at this stage (b) Also, gas phase decomposition of N2O4 to of the reaction is given by, NO2 is another reaction with a value 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 too small nor too large. Hence, = 8.0 equilibrium mixtures contain appreciable Now, in this case, Qc (8.0) does not equal concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g) These generarlisations are illustrated in is not at equilibrium; that is, more H2(g) and Fig. 7.6 I2(g) will react to form more HI(g) and their concentrations will decrease till Qc = Kc. The reaction quotient, Q c is useful in predicting the direction of reaction by comparing the values of Qc and Kc. Thus, we can make the following 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 Kc except that the concentrations in If Qc < Kc, net reaction goes from left to right Qc are not necessarily equilibrium values. If Qc > Kc, net reaction goes from right to For a general reaction: left. aA+bB cC+dD (7.19) If Qc = Kc, no net reaction occurs. c d a b Qc = [C] [D] / [A] [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? 2022-23 EQUILIBRIUM 207 Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc found to be 9.15 bar. Calculate Kc, Kp and is given by, partial pressure at equilibrium. Qc = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10 –4M We know pV = nRT Qc = (3 ×10 –4)(3 × 10 –4) / (3 ×10 –4)2 = 1 Total volume (V ) = 1 L as Qc > Kc so the reaction will proceed in Molecular mass of N2O4 = 92 g the reverse direction. Number of moles = 13.8g/92 g = 0.15 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 pV = nRT the equilibrium concentrations, the following p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 three steps shall be followed: × 400 K Step 1. Write the balanced equation for the p = 4.98 bar reaction. N 2O4 2NO2 Step 2. Under the balanced equation, make a Initial pressure: 4.98 bar 0 table that lists for each substance involved in At equilibrium: (4.98 – x) bar 2x bar the reaction: Hence, (a) the initial concentration, ptotal at equilibrium = pN O + pNO 2 4 2 (b) the change in concentration on going to 9.15 = (4.98 – x) + 2x equilibrium, and 9.15 = 4.98 + x (c) the equilibrium concentration. 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 pN O = 4.98 – 4.17 = 0.81bar 2 4 the stoichiometry of the reaction to determine the concentrations of the other substances in pNO = 2x = 2 × 4.17 = 8.34 bar 2 terms of x. ( ) 2 K p = p NO2 / p N 2O4 Step 3. Substitute the equilibrium concentrations into the equilibrium equation = (8.34)2/0.81 = 85.87 ∆n for the reaction and solve for x. If you are to Kp = Kc(RT) solve a quadratic equation choose the 85.87 = Kc(0.083 × 400)1 mathematical solution that makes chemical sense. Kc = 2.586 = 2.6 Step 4. Calculate the equilibrium Problem 7.9 concentrations from the calculated value of x. 3.00 mol of PCl5 kept in 1L closed reaction Step 5. Check your results by substituting vessel was allowed to attain equilibrium them into the equilibrium equation. at 380K. Calculate composition of the Problem 7.8 mixture at equilibrium. Kc= 1.80 13.8g of N2O4 was placed in a 1L reaction Solution vessel at 400K and allowed to attain PCl5 PCl3 + Cl2 equilibrium Initial concentration: 3.0 0 0 N 2O4 (g) 2NO2 (g) 2022-23 208 CHEMISTRY  Let x mol per litre of PCl5 be dissociated, K = e–∆G /RT (7.23) At equilibrium: Hence, using the equation (7.23), the (3-x) x x reaction spontaneity can be interpreted in Kc = [PCl3][Cl2]/[PCl5] terms of the value of ∆G.   If ∆G < 0, then –∆G /RT is positive, and 1.8 = x2/ (3 – x) >1, making K >1, which implies x2 + 1.8x – 5.4 = 0 a spontaneous reaction or the reaction x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 which proceeds in the forward direction to x = [–1.8 ± √3.24 + 21.6]/2 such an extent that the products are present predominantly. x = [–1.8 ± 4.98]/2   If ∆G > 0, then –∆G /RT is negative, and x = [–1.8 + 4.98]/2 = 1.59 [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M < 1, that is , K < 1, which implies [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute 7.7 RELATIONSHIP BETWEEN quantity of product is formed. EQUILIBRIUM CONSTANT K, REACTION QUOTIENT Q AND GIBBS ENERGY G Problem 7.10  The value of Kc for a reaction does not depend The value of ∆G for the phosphorylation on the rate of the reaction. However, as you of glucose in glycolysis is 13.8 kJ/mol. have studied in Unit 6, it is directly related Find the value of Kc at 298 K. to the thermodynamics of the reaction and Solution in particular, to the change in Gibbs energy,  ∆G = 13.8 kJ/mol = 13.8 × 103J/mol ∆G. If,  Also, ∆G = – RT lnKc ∆G is negative, then the reaction is Hence, ln Kc = –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 non-spontaneous. Instead, as reverse Kc = e–5.569 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. ∆G is 0, reaction has achieved equilibrium; Hydrolysis of sucrose gives, at this point, there is no longer any free Sucrose + H2O Glucose + Fructose energy left to drive the reaction. Equilibrium constant Kc for the reaction  A mathematical expression of this is 2 ×1013 at 300K. Calculate ∆G at thermodynamic view of equilibrium can be 300K. described by the following equation: Solution  ∆G = ∆G + RT lnQ (7.21)   ∆G = – RT lnKc where, G is standard Gibbs energy.  ∆G = – 8.314J mol–1K–1× At equilibrium, when ∆G = 0 and Q = Kc, 300K × ln(2×1013) the equation (7.21) becomes,   ∆G = – 7.64 ×10 J mol–1 4 ∆G = ∆G + RT ln K = 0  ∆G = – RT lnK (7.22) 7.8 FACTORS AFFECTING EQUILIBRIA  lnK = – ∆G / RT One of the principal goals of chemical synthesis Taking antilog of both sides, we get, is to maximise the conversion of the reactants 2022-23 EQUILIBRIUM 209 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 H2(g) + I2(g) 2HI(g) economic importance. Annual world production of ammonia is about hundred If H2 is added to the reaction mixture at million tones, primarily for use as fertilizers. equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the Equilibrium constant, Kc is independent of reaction proceeds in a direction wherein H2 is initial concentrations. But if a system at consumed, i.e., more of H2 and I2 react to form equilibrium is subjected to a change in the HI and finally the equilibrium shifts in right concentration of one or more of the reacting (forward) direction (Fig.7.8). This is in substances, then the system is no longer at accordance with the Le Chatelier’s principle equilibrium; and net reaction takes place in which implies that in case of addition of a some direction until the system returns to reactant/product, a new equilibrium will be equilibrium once again. Similarly, a change in set up in which the concentration of the temperature or pressure of the system may 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. qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a 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/ 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 in the direction that consumes the added concentration for the reactants and substance. products in the reaction, H2(g) + I2 (g) 2HI(g) The concentration stress of a removed 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, Qc = [HI] / [H2][I2] 2022-23 210 CHEMISTRY Addition of hydrogen at equilibrium results replenish the Fe 3+ ions. Because the in value of Qc being less than Kc. 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 2– form stable complex ion [Hg(SCN)4]. 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 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 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 important building material) from CaCO3, A pressure change obtained by changing the 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 remembered that continuous removal of a of moles of gaseous reactants and total 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 ignored because the volume (and This can be demonstrated by the following concentration) of a solution/liquid is nearly reaction: – 2+ independent of pressure. Fe3+(aq)+ SCN (aq) [Fe(SCN)] (aq) (7.24) Consider the reaction, yellow colourless deep red CO(g) + 3H2(g) CH4(g) + H2O(g) Here, 4 mol of gaseous reactants (CO + 3H2) (7.25) become 2 mol of gaseous products (CH4 + 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 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 – or SCN ions. For example, oxalic acid direction, a direction in which the number of (H2C2O4), reacts with Fe3+ ions to form the moles of the gas or pressure decreases (we stable complex ion [Fe(C 2 O 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 2 O] be the molar concentrations at relieved by dissociation of [Fe(SCN)] 2+ to equilibrium for methanation reaction. When 2022-23  EQUILIBRIUM 211 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 ∆H= – 92.38 kJ mol–1 double its value. is an exothermic process. According to CH4 ( g )  H2 O ( g ) Le Chatelier’s principle, raising the Qc =  temperature shifts the equilibrium to left and CO ( g )  H2 (g ) 3 decreases the equilibrium concentration of As Qc < Kc , the reaction proceeds in the ammonia. In other words, low temperature is forward direction. favourable for high yield of ammonia, but practically very low temperatures slow down In reaction C(s) + CO2(g) 2CO(g), when the reaction and thus a catalyst is used. pressure is increased, the reaction goes in the reverse direction because the number of moles Effect of Temperature – An experiment of gas increases in the forward direction. Effect of temperature on equilibrium can be demonstrated by taking NO2 gas (brown in 7.8.3 Effect of Inert Gas Addition colour) which dimerises into N 2 O 4 gas If the volume is kept constant and an inert gas (colourless). 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 NO 2 gas prepared by addition of Cu undisturbed. It is because the addition of an turnings to conc. HNO3 is collected in two inert gas at constant volume does not change 5 mL test tubes (ensuring same intensity of the partial pressures or the molar colour of gas in each tube) and stopper sealed concentrations of the substance involved in the  with araldite. Three 250 mL beakers 1, 2 and reaction. The reaction quotient changes only 3 containing freezing mixture, water at room if the added gas is a reactant or product temperature and hot water (363 K ), involved in the reaction. 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. After this one is placed in beaker 1 and the Whenever an equilibrium is disturbed by a other in beaker 3. The effect of temperature change in the concentration, pressure or on direction of reaction is depicted very well volume, the composition of the equilibrium in this experiment. At low temperatures in mixture changes because the reaction beaker 1, the forward reaction of formation of quotient, Qc no longer equals the equilibrium N2O4 is preferred, as reaction is exothermic, and constant, Kc. However, when a change in thus, intensity of brown colour due to NO2 temperature occurs, the value of equilibrium decreases. While in beaker 3, high constant, Kc is changed. temperature favours the reverse reaction of In general, the temperature dependence of the equilibrium constant depends on the sign of ∆H for the reaction. The equilibrium constant for an exothermic 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) N2O4 (g) 2022-23 212 CHEMISTRY 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 3+ – 2– [Co(H2O) 6] (aq) + 4Cl (aq) [CoCl4] (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 (V 2O5) is used as At room temperature, the equilibrium catalyst to increase the rate of the reaction. mixture is blue due to [CoCl4]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. 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 the direction of equilibrium, you have reaction by making available a new low energy incidently come across with the following pathway for the conversion of reactants to equilibrium which involves ions: products. It increases the rate of forward and reverse reactions that pass through the same Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) 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 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 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 two categories based on their ability to conduct exothermic reaction and proceeds with electricity. One category of substances decrease in total number of moles formed as conduct electricity in their aqueous solutions compared to the reactants. Equilibrium and are called electrolytes while the other do constant decreases with increase in not and are thus, referred to as non- temperature. At low temperature rate 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 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 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 2022-23 EQUILIBRIUM 213 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 Acids, bases and salts find widespread a very high dielectric constant of 80. Thus, occurrence in nature. Hydrochloric acid when sodium chloride is d