Equilibrium PDF

Unit 6 Equilibrium Chemical equilibria are important in numerous biological...

Unit 6 Equilibrium Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O2 molecules and the protein hemoglobin play After studying this unit you will be a crucial role in the transport and delivery of O2 from able to our 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 explain characteristics of the liquid surface into the vapour phase and number of equilibria involved in physical liquid molecules from the vapour phase strike the liquid and chemical processes; surface and are retained in the liquid phase. It gives rise write expressions for equilibrium to a constant vapour pressure because of an equilibrium in constants; which the number of molecules leaving the liquid equals the establish a relationship between Kp and Kc; number returning to liquid from the vapour. We say that explain various factors that the system has reached equilibrium state at this stage. affect the equilibrium state of a However, this is not static equilibrium and there is a lot of reaction; activity at the boundary between the liquid and the vapour. classify substances as acids or Thus, at equilibrium, the rate of evaporation is equal to the bases according to Arrhenius, rate of condensation. It may be represented by Bronsted-Lowry and Lewis concepts; H2O (l) H2O (vap) classify acids and bases as The double half arrows indicate that the processes weak or strong in terms of their ionization constants; in both the directions are going on simultaneously. The explain the dependence of degree mixture of reactants and products in the equilibrium state of ionization on concentration is called an equilibrium mixture. of the electrolyte and that of the common ion; Equilibrium can be established for both physical describe pH scale for representing processes and chemical reactions. The reaction may be hydrogen ion concentration; fast or slow depending on the experimental conditions and explain ionisation of water and the nature of the reactants. When the reactants in a closed its duel role as acid and base; vessel at a particular temperature react to give products, describe ionic product (Kw ) and the concentrations of the reactants keep on decreasing, pKw for water; while those of products keep on increasing for some time appreciate use of buffer after which there is no change in the concentrations solutions; calculate solubility product of either of the reactants or products. This stage of the constant. system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to 2024-25 Unit 6.indd 168 9/12/2022 11:58:20 AM  EQUILIBRIUM 169 this dynamic equilibrium stage that there is characteristic features. We observe that the no change in the concentrations of various mass of ice and water do not change with species in the reaction mixture. Based on the time and the temperature remains constant. extent to which the reactions proceed to reach However, the equilibrium is not static. the state of chemical equilibrium, these may The intense activity can be noticed at the be classified in three groups. boundary between ice and water. Molecules (i) The reactions that proceed nearly from the liquid water collide against ice and to completion and only negligible adhere to it and some molecules of ice escape concentrations of the reactants are into liquid phase. There is no change of mass left. In some cases, it may not be even of ice and water, as the rates of transfer of possible to detect these experimentally. molecules from ice into water and of reverse transfer from water into ice are equal at (ii) The reactions in which only small atmospheric pressure and 273 K. amounts of products are formed and most of the reactants remain unchanged It is obvious that ice and water are in at equilibrium stage. equilibrium only at particular temperature and pressure. For any pure substance at (iii) The reactions in which the concentrations atmospheric pressure, the temperature at of the reactants and products are which the solid and liquid phases are at comparable, when the system is in equilibrium is called the normal melting point equilibrium. or normal freezing point of the substance. The The extent of a reaction in equilibrium system here is in dynamic equilibrium and we varies with the experimental conditions such can infer the following: as concentrations of reactants, temperature, (i) Both the opposing processes occur etc. Optimisation of the operational conditions simultaneously. is very important in industry and laboratory (ii) Both the processes occur at the same so that equilibrium is favorable in the rate so that the amount of ice and water direction of the desired product. Some remains constant. important aspects of equilibrium involving physical and chemical processes are dealt in 6.1.2 Liquid-Vapour Equilibrium this unit along with the equilibrium involving This equilibrium can be better understood if ions in aqueous solutions which is called as we consider the example of a transparent box ionic equilibrium. carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride 6.1 E Q U IL IBRI U M IN P H Y S I C A L (or phosphorus penta-oxide) is placed for PROCESSES a few hours in the box. After removing the The characteristics of system at equilibrium drying agent by tilting the box on one side, a are better understood if we examine some watch glass (or petri dish) containing water physical processes. The most familiar examples is quickly placed inside the box. It will be are phase transformation 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 solid gas pressure inside the box increases and reaches a constant value. Also the volume of water in 6.1.1 Solid-Liquid Equilibrium the watch glass decreases (Fig. 6.1). 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 its the box. As water evaporated the pressure in contents and the surroundings) at 273K and the box increased due to addition of water the atmospheric pressure are in equilibrium molecules into the gaseous phase inside state and the system shows interesting the box. The rate of evaporation is constant. 2024-25 Unit 6.indd 169 9/12/2022 11:58:20 AM  170 chemistry Fig. 6.1 Measuring equilibrium vapour pressure of water at a constant temperature However, the rate of increase in pressure vapour to liquid state is much less than the decreases with time due to condensation rate of evaporation. These are open systems of 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 Water and water vapour are in equilibrium of water molecules from the gaseous state into the liquid state also increases till the position at atmospheric pressure (1.013 bar) equilibrium is attained i.e., and at 100°C in a closed vessel. The boiling 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 equilibrium the pressure exerted by at which the liquid and vapours are at the 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, acetone and ether, it is observed that different 6.1.3 Solid – Vapour Equilibrium liquids have different equilibrium vapour Let us now consider the systems where solids pressures at the same temperature, and the sublime to vapour phase. If we place solid liquid which has a higher vapour pressure is iodine in a closed vessel, after sometime more volatile and has a lower boiling point. the vessel gets filled up with violet vapour If we expose three watch glasses containing and the intensity of colour increases with separately 1mL each of acetone, ethyl alcohol, time. After certain time the intensity of and water to atmosphere and repeat the colour becomes constant and at this stage experiment with different volumes of the equilibrium is attained. Hence solid iodine liquids in a warmer room, it is observed sublimes to give iodine vapour and the iodine that in all such cases the liquid eventually vapour condenses to give solid iodine. The disappears and the time taken for complete equilibrium can be represented as, evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the I2(solid) I2 (vapour) temperature. When the watch glass is open Other examples showing this kind of to the atmosphere, the rate of evaporation equilibrium are, remains constant but the molecules are dispersed into large volume of the room. As Camphor (solid) Camphor (vapour) a consequence the rate of condensation from NH4Cl (solid) NH4Cl (vapour) 2024-25 Unit 6.indd 170 9/12/2022 11:58:21 AM  EQUILIBRIUM 171 6.1.4 Equilibrium Involving Dissolution pressure of the gas above the solvent. of Solid or Gases in Liquids This 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 some of the dissolved carbon dioxide gas sugar 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 temperature, sugar crystals separate out if we how the soda water in bottle when left open cool the syrup to the room temperature. We call to the air for some time, turns ‘flat’. It can be it a saturated solution when no more of solute generalised that: can be dissolved in it at a given temperature. The concentration of the solute in a saturated (i) For solid liquid equilibrium, there is solution depends upon the temperature. In only one temperature (melting point) at a saturated solution, a dynamic equilibrium 1 atm (1.013 bar) at which the two exits between the solute molecules in the solid phases can coexist. If there is no state and in the solution: exchange of heat with the surroundings, the mass of the two phases remains Sugar (solution) Sugar (solid), and constant. the rate of dissolution of sugar = rate of crystallisation of sugar. (ii) For liquid vapour equilibrium, the vapour pressure is constant at a given Equality of the two rates and dynamic temperature. nature of equilibrium has been confirmed with the help of radioactive sugar. If we drop some (iii) For dissolution of solids in liquids, radioactive sugar into saturated solution of the solubility is constant at a given non-radioactive sugar, then after some time temperature. radioactivity is observed both in the solution (iv) For dissolution of gases in liquids, and in the solid sugar. Initially there were no the concentration of a gas in liquid radioactive sugar molecules in the solution is proportional to the pressure but due to dynamic nature of equilibrium, (concentration) of the gas over the liquid. there is exchange between the radioactive These observations are summarised in and non-radioactive sugar molecules between Table 6.1 the two phases. The ratio of the radioactive Table 6.1 Some Features of Physical Equilibria to non-radioactive molecules in the solution increases till it attains a constant value. Process Conclusion Liquid Vapour p H 2O constant at given Gases in liquids temperature H2O (l) H2O (g) When a soda water bottle is opened, some of Solid Liquid Melting point is fixed at the carbon dioxide gas dissolved in it fizzes constant pressure out rapidly. The phenomenon arises due H2O (s) H2O (l) to difference in solubility of carbon dioxide Solute(s) Solute Concentration of solute at 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 This equilibrium is governed by Henry’s temperature CO2(g) CO2(aq) law, which states that the mass of a gas [CO2(aq)]/[CO2(g)] is dissolved in a given mass of a solvent at constant at a given any temperature is proportional to the temperature 2024-25 Unit 6.indd 171 11/2/2022 4:10:42 PM  172 chemistry 6.1.5 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 Fig. 6.2 Attainment of chemical equilibrium. constant value of one of its parameters at a given temperature. Table 6.1 lists Eventually, the two reactions occur at the such 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 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, as the equilibrium can be reached from either 6.2 EQUILIBRIUM IN CHEMICAL direction. PROCESSES – DYNAMIC EQUILIBRIUM The dynamic nature of chemical equilibrium can be demonstrated in the Analogous to the physical systems chemical synthesis of ammonia by Haber’s process. reactions also attain a state of equilibrium. In a series of experiments, Haber started These reactions can occur both in forward with known amounts of dinitrogen and and backward directions. When the rates of dihydrogen maintained at high temperature the forward and reverse reactions become and pressure and at regular intervals equal, the concentrations of the reactants determined the amount of ammonia present. and the products remain constant. This He was successful in determining also the is the stage of chemical equilibrium. This concentration of unreacted dihydrogen and equilibrium is dynamic in nature as it consists dinitrogen. Fig. 6.4 (page 174) shows that after of a forward reaction in which the reactants a certain time the composition of the mixture give product(s) and reverse reaction in which remains the same even though some of the product(s) gives the original reactants. reactants are still present. This constancy in For a better comprehension, let us consider composition indicates that the reaction has 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 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. 6.2). in place of H2. The reaction mixtures starting This leads to a decrease in the rate of either with H2 or D2 reach equilibrium with forward reaction and an increase in the rate the same composition, except that D2 and of the reverse reaction, ND3 are present instead of H2 and NH3. After 2024-25 Unit 6.indd 172 11/2/2022 4:10:43 PM  EQUILIBRIUM 173 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 2nd 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. 1 2 1 2 (a) (b) Fig.6.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. 2024-25 Unit 6.indd 173 9/12/2022 11:58:21 AM  174 chemistry 2NH3(g) N2(g) + 3H2(g) Similarly let us consider the reaction, H 2 (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 constant at equilibrium (Fig. 6.5). 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. 6.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we Fig. 6.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product. N 2  g   3H 2  g  2NH3  g  equilibrium is attained, these two mixtures (H 2, N 2, NH 3 and D 2, N 2, ND 3) 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 Fig. 6.5 Chemical equilibrium in the reaction H2(g) when they reached equilibrium, then there + I2(g) 2HI(g) can be attained from would have been no mixing of isotopes in either direction this way. 6.3 LAW OF CHEMICAL EQUILIBRIUM Use of isotope (deuterium) in the formation AND EQUILIBRIUM CONSTANT of ammonia clearly indicates that chemical A mixture of reactants and products in the reactions reach a state of dynamic equilibrium state is called an equilibrium equilibrium in which the rates of forward mixture. In this section we shall address a and reverse reactions are equal and there number of important questions about the is no net change in composition. composition of equilibrium mixtures: What is Equilibrium can be attained from both the relationship between the concentrations sides, whether we start reaction by taking, of reactants and products in an equilibrium H2(g) and N2(g) and get NH3(g) or by taking mixture? How can we determine equilibrium NH3(g) and decomposing it into N2(g) and H2(g). concentrations from initial concentrations? N2(g) + 3H2(g) 2NH3(g) What factors can be exploited to alter the 2024-25 Unit 6.indd 174 9/12/2022 11:58:21 AM  EQUILIBRIUM 175 composition of an equilibrium mixture? The Six sets of experiments with varying initial last question in particular is important when conditions were performed, starting with only choosing conditions for synthesis of industrial gaseous H2 and I2 in a sealed reaction vessel chemicals such as H2, NH3, CaO etc. in first four experiments (1, 2, 3 and 4) and To answer these questions, let us consider only HI in other two experiments (5 and 6). a general reversible reaction: Experiment 1, 2, 3 and 4 were performed A+B C+D taking different concentrations of H2 and / or where A and B are the reactants, C and D I2, and with time it was observed that intensity are the products in the balanced chemical of the purple colour remained constant and equation. On the basis of experimental studies equilibrium was attained. Similarly, for of many reversible reactions, the Norwegian experiments 5 and 6, the equilibrium was chemists Cato Maximillian Guldberg and attained from the opposite direction. Peter Waage proposed in 1864 that the Data obtained from all six sets of concentrations in an equilibrium mixture experiments are given in Table 6.2. are related by the following equilibrium equation, It is evident from the experiments 1, 2, 3 and 4 that number of moles of dihydrogen Kc  CD         (6.1) reacted = number of moles of iodine reacted  A  B  = ½ (number of moles of HI formed). Also, (6.1) where Kc is the equilibrium constant and experiments 5 and 6 indicate that, the expression on the right side is called the [H2(g)]eq = [I2(g)]eq equilibrium constant expression. The equilibrium equation is also known as Knowing the above facts, in order the law of mass action because in the early to establish a relationship between days of chemistry, concentration was called concentrations of the reactants and products, “active mass”. In order to appreciate their several combinations can be tried. Let us work better, let us consider reaction between consider the simple expression, gaseous H2 and I2 carried out in a sealed vessel [HI(g)]eq / [H2(g)]eq [I2(g)]eq at 731K. H2(g) + I2(g)   2HI(g) It can be seen from Table 6.3 that if we 1 mol 1 mol 2 mol put the equilibrium concentrations of the reactants and products, the above expression Table 6.2 Initial and Equilibrium Concentrations of H2, I2 and HI 2024-25 Unit 6.indd 175 9/12/2022 11:58:21 AM  176 chemistry Table 6.3 Expression Involving the The equilibrium constant for a general Equilibrium Concentration of reaction, Reactants aA + bB cC + dD H2(g) + I2(g) 2HI(g) is expressed as, Kc = [C]c[D]d / [A]a[B]b (6.4) 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 Kc = [NO]4[H2O]6 / [NH3]4 [O2]5 Molar concentration of different species is is far from constant. However, if we consider indicated by enclosing these in square bracket the expression, and, as mentioned above, it is implied that [HI(g)]2eq / [H2(g)]eq [I2(g)]eq these are equilibrium concentrations. While writing expression for equilibrium constant, we find that this expression gives constant symbol for phases (s, l, g) are generally ignored. value (as shown in Table 6.3) in all the six cases. It can be seen that in this expression Let us write equilibrium constant for the the power of the concentration for reactants reaction, H2(g) + I2(g) 2HI(g) (6.5) and products are actually the stoichiometric as, Kc = [HI]2 / [H2] [I2] = x (6.6) coefficients in the equation for the chemical reaction. Thus, for the reaction H2(g) + I2(g) The equilibrium constant for the reverse 2HI(g), following equation 6.1, the equilibrium reaction, 2HI(g) H2(g) + I2(g), at the same constant Kc is written as, temperature is, Kc = [HI(g)]eq2 / [H2(g)]eq [I2(g)]eq (6.2) K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc (6.7) Generally the subscript ‘eq’ (used for Thus, K′c = 1 / Kc (6.8) 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. Kc = [HI(g)]2 / [H2(g)] [I2(g)] (6.3) 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 At a given temperature, the product of constant also reflects that change. For concentrations of the reaction products example, if the reaction (6.5) is written as, raised to the respective stoichiometric coefficient in the balanced chemical ½ H2(g) + ½ I2(g) HI(g) (6.9) equation divided by the product of the equilibrium constant for the above concentrations of the reactants raised to reaction is given by their individual stoichiometric coefficients K″c = [HI] / [H2]1/2[I2]1/2 = {[HI]2 / [H2][I2]}1/2 has a constant value. This is known as the Equilibrium Law or Law of Chemical = x1/2 = Kc1/2 (6.10) Equilibrium. On multiplying the equation (6.5) by n, we get 2024-25 Unit 6.indd 176 9/12/2022 11:58:22 AM EQUILIBRIUM 177 nH2(g) + nI2(g) D 2nHI(g) (6.11) Therefore, equilibrium constant for the N2(g) + O2(g) 2NO(g) reaction is equal to Kc n. These findings are Solution summarised in Table 6.4. It should be noted that because the equilibrium constants Kc For the reaction equilibrium constant, Kc and K ′c have different numerical values, it is can be written as, 2 important to specify the form of the balanced NO Kc = chemical equation when quoting the value of N 2 O2 an equilibrium constant. 2 2.8 10–3M Table 6.4 Relations between Equilibrium = Constants for a General Reaction 3.0 10 3 M 4.2 10 3 M and its Multiples. = 0.622 Chemical equation Equilibrium constant aA+bB c C + dD Kc 6.4 HOMOGENEOUS EQUILIBRIA cC+dD aA+bB K′c =(1/Kc ) In a homogeneous system, all the reactants and products are in the same phase. na A + nb B ncC + ndD K′″c = (Kcn ) For example, in the gaseous reaction, N 2(g) + 3H 2(g) 2NH 3(g), reactants and products are in the homogeneous phase. Problem 6.1 Similarly, for the reactions, The following concentrations were CH3COOC2H5 (aq) + H2O (l) CH3COOH (aq) obtained for the formation of NH3 from + C2H5OH (aq) N 2 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 consider equilibrium constant for some Solution homogeneous reactions. The equilibrium constant for the reaction, N2(g) + 3H2(g) 2NH3(g) can be written as, 6.4.1 Equilibrium Constant in Gaseous Systems  NH3  g  2 So far we have expressed equilibrium Kc = constant 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 = 1.2  10  2 2 involving gases, however, it is usually more 1.5  10  3.0  10  2 2 3 convenient to express the equilibrium constant in terms of partial pressure. = 0.355 × 103 = 3.55 × 102 The ideal gas equation is written as, pV = nRT Problem 6.2 n At equilibrium, the concentrations of ⇒ p = RT N 2=3.0 × 10 –3M, O 2 = 4.2 × 10 –3M and V NO= 2.8 × 10–3M in a sealed vessel at 800K. Here, p is the pressure in Pa, n is the number What will be Kc for the reaction of moles of the gas, V is the volume in m3 and T is the temperature in Kelvin 2024-25  178 chemistry Therefore, n/V is concentration expressed in mol/m3  NH3 ( g ) [RT ] 2 −2 =  = K c ( RT ) −2 If concentration c, is in mol/L or mol/dm3,  N 2 ( g ) H 2 ( g ) 3 and p is in bar then p = cRT, or K p = K c ( RT ) (6.14) −2 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 ( p )( p ) [ A ] [B] (RT )( a A b B a b a +b ) H2(g) + I2(g) 2HI(g) We can write either [ C ] [D] c d RT ) (c + d ) −(a +b ) b ( = HI ( g ) 2 [ A ] [B ] a Kc = H2 ( g ) I2 ( g ) [ C ] [D] c d RT ) ∆n = K c ( RT ) ∆n b ( = (6.15) or K c = ( p HI ) 2 (6.12) [ A ] [B ] a ( p )( p ) H2 I2 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 p H2 = H2 ( g ) RT because standard state for pressure is 1 bar. We know from Unit 1 that : Therefore, 1pascal, Pa=1Nm–2, and 1bar = 105 Pa HI ( g ) [RT ] Kp values for a few selected reactions at ( pHI )2 2 2 Kp = = different temperatures are given in Table 6.5 ( p )( p ) H2 I2 H2 ( g ) RT. I2 ( g ) RT Table 6.5 Equilibrium Constants, Kp for HI ( g ) 2 a Few Selected Reactions = = Kc (6.13) 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 6.3  NH3 ( g ) [RT ] 2 2 = PCl5, PCl3 and Cl2 are at equilibrium at 500  N 2 ( g ) RT. H 2 ( g ) (RT ) 3 3 K and having concentration 1.59M PCl3, 1.59M Cl2 and 1.41 M PCl5. 2024-25 Unit 6.indd 178 11/30/2022 15:46:47  EQUILIBRIUM 179 Calculate Kc for the reaction, the value 0.194 should be neglected because PCl5 PCl3 + Cl2 it will give concentration of the reactant Solution which is more than initial concentration. The equilibrium constant Kc for the above Hence the equilibrium concentrations are, reaction can be written as, [CO2] = [H2-] = x = 0.067 M Kc  PCl  Cl   1.59 3 2 2  1.79 [CO] = [H2O] = 0.1 – 0.067 = 0.033 M PCl 5 1.41 Problem 6.5 Problem 6.4 For the equilibrium, The value of K c = 4.24 at 800K for the 2NOCl(g) 2NO(g) + Cl2(g) reaction, the value of the equilibrium constant, Kc is CO (g) + H2O (g) CO2 (g) + H2 (g) 3.75 × 10 –6 at 1069 K. Calculate the Kp for Calculate equilibrium concentrations of CO2, the reaction at this temperature? H2, CO and H2O at 800 K, if only CO and H2O are present initially at concentrations Solution of 0.10M each. We know that, Solution Kp = Kc(RT)∆n For the reaction, For the above reaction, CO (g) + H2O (g) CO2 (g) + H2 (g) ∆n = (2+1) – 2 = 1 Kp = 3.75 ×10–6 (0.0831 × 1069) Initial concentration: Kp = 0.033 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. 6.5 HETEROGENEOUS EQUILIBRIA At equilibrium: Equilibrium in a system having more than one (0.1-x) M (0.1-x) M xM xM phase is called heterogeneous equilibrium. The equilibrium between water vapour and where x is the amount of CO2 and H2 at equilibrium. liquid water in a closed container is an example of heterogeneous equilibrium. Hence, equilibrium constant can be written as, H2O(l) H2O(g) Kc = x2/(0.1-x)2 = 4.24 In this example, there is a gas phase and x2 = 4.24(0.01 + x2-0.2x) a liquid phase. In the same way, equilibrium x2 = 0.0424 + 4.24x2-0.848x between a solid and its saturated solution, 3.24x2 – 0.848x + 0.0424 = 0 Ca(OH)2 (s) + (aq) Ca2+ (aq) + 2OH–(aq) a = 3.24, b = – 0.848, c = 0.0424 is a heterogeneous equilibrium. (for quadratic equation ax2 + bx + c = 0, Heterogeneous equilibria often involve x  b  b2  4ac  pure solids or liquids. We can simplify equilibrium expressions for the heterogeneous 2a equilibria involving a pure liquid or a pure x = 0.848±√(0.848)2– 4(3.24)(0.0424)/ solid, as the molar concentration of a pure solid or liquid is constant (i.e., independent (3.24×2) of the amount present). In other words if a x = (0.848 ± 0.4118)/ 6.48 substance ‘X’ is involved, then [X(s)] and [X(l)] x1 = (0.848 – 0.4118)/6.48 = 0.067 are constant, whatever the amount of ‘X’ is x2 = (0.848 + 0.4118)/6.48 = 0.194 taken. Contrary to this, [X(g)] and [X(aq)] will 2024-25 Unit 6.indd 179 9/12/2022 11:58:24 AM  180 chemistry vary as the amount of X in a given volume This shows that at a particular varies. Let us take thermal dissociation of temperature, there is a constant concentration calcium carbonate which is an interesting and or pressure of CO2 in equilibrium with CaO(s) important example of heterogeneous chemical and CaCO3(s). Experimentally it has been 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) (6.16) 2.0 ×105 Pa. Therefore, equilibrium constant On the basis of the stoichiometric equation, at 1100K for the above reaction is: we can write, Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00 CaO  s  CO2  g  Similarly, in the equilibrium between Kc  CaCO3  s  nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel), Since [CaCO3(s)] and [CaO(s)] are both constant, therefore modified equilibrium Ni (s) + 4 CO (g) Ni(CO)4 (g), constant for the thermal decomposition of the equilibrium constant is written as calcium carbonate will be  Ni CO 4  K´c = [CO2(g)] (6.17) Kc   or Kp = pCO (6.18) CO4 2 It must be remembered that for the existence of heterogeneous equilibrium pure Units of Equilibrium Constant solids or liquids must also be present (however The value of equilibrium constant Kc can be small the amount may be) at equilibrium, but calculated by substituting the concentration their concentrations or partial pressures do terms in mol/L and for Kp partial pressure not appear in the expression of the equilibrium is substituted in Pa, kPa, bar or atm. This constant. In the reaction, results in units of equilibrium constant based Ag2O(s) + 2HNO3(aq) 2AgNO3(aq) +H2O(l) on molarity or pressure, unless the exponents of both the numerator and denominator are  AgNO  2 3 same. Kc = HNO  2 For the reactions, 3 H2(g) + I2(g) 2HI, Kc and Kp have no unit. N2O4(g) 2NO2 (g), Kc has unit mol/L and Kp has unit bar Problem 6.6 Equilibrium constants can also be The value of Kp for the reaction, expressed as dimensionless quantities if the CO2 (g) + C (s) 2CO (g) standard state of reactants and products is 3.0 at 1000 K. If initially PCO = 0.48 2 are specified. For a pure gas, the standard bar and PCO = 0 bar and pure graphite is state is 1bar. Therefore a pressure of 4 bar present, calculate the equilibrium partial in standard state can be expressed as 4 pressures of CO and CO2. bar/1 bar = 4, which is a dimensionless number. Standard state (c0) for a solute is 1 Solution molar solution and all concentrations can be For the reaction, measured with 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 system then both Kp and Kc are dimensionless quantities CO2(g) + C(s) 2CO(g) but have different numerical values due to Initial different standard states. pressure: 0.48 bar 0 2024-25 Unit 6.indd 180 9/12/2022 11:58:24 AM  EQUILIBRIUM 181 5. The equilibrium constant K for a reaction At equilibrium: is related to the equilibrium constant (0.48 – x)bar 2x bar of the corresponding reaction, whose equation is obtained by multiplying or pC2 O dividing the equation for the original Kp = pC O2 reaction by a small integer. Kp = (2x)2/(0.48 – x) = 3 Let us consider applications of equilibrium constant to: 4x2 = 3(0.48 – x) predict the extent of a reaction on the 4x2 = 1.44 – x basis of its magnitude, 4x2 + 3x – 1.44 = 0 predict the direction of the reaction, and a = 4, b = 3, c = –1.44 b   calculate equilibrium concentrations. b2  4ac x 6.6.1 Predicting the Extent of a 2a Reaction = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 The numerical value of the equilibrium = (–3 ± 5.66)/8 constant for a reaction indicates the extent of the reaction. But it is important to note = (–3 + 5.66)/8 (as value of x cannot be that an equilibrium constant does not give negative hence we neglect that value) any information about the rate at which x = 2.66/8 = 0.33 the equilibrium is reached. The magnitude The equilibrium partial pressures are, of Kc or Kp is directly proportional to the concentrations of products (as these appear pCO = 2x = 2 × 0.33 = 0.66 bar 2 in the numerator of equilibrium constant pCO = 0.48 – x = 0.48 – 0.33 = 0.15 bar expression) and inversely proportional to the 2 concentrations of the reactants (these appear 6.6 APPLICATIONS OF EQUILIBRIUM in the denominator). This implies that a high CONSTANTS value of K is suggestive of a high concentration of products and vice-versa. Before considering the applications of equilibrium constants, let us summarise the We can make the following generalisations important features of equilibrium constants concerning the composition of equilibrium as follows: mixtures: 1. Expression for equilibrium constant is If Kc > 103, products predominate over applicable only when concentrations of reactants, i.e., if Kc is very large, the the reactants and products have attained reaction proceeds nearly to completion. constant value at equilibrium state. Consider the following examples: 2. The value of equilibrium constant is (a) The reaction of H 2 with O 2 at 500 K independent of initial concentrations of has a very large equilibrium constant, the reactants and products. Kc = 2.4 × 1047. 3. Equilibrium constant is temperature (b) H2(g) + Cl 2(g) 2HCl(g) at 300K has dependent having one unique value for Kc = 4.0 × 1031. a particular reaction represented by a (c) H 2(g) + Br 2(g) 2HBr (g) at 300 K, balanced equation at a given temperature. Kc = 5.4 × 1018 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 equilibrium constant for the forward reaction proceeds rarely. Consider the reaction. following examples: 2024-25 Unit 6.indd 181 9/12/2022 11:58:24 AM  182 chemistry (a) The decomposition of H2O into H2 and If Qc = Kc, the reaction mixture is already O2 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 (b) Also, gas phase decomposition of N2O4 stage of the reaction is given by, to NO2 is another reaction with a value Qc = [HI]t2 / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20) of K c = 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. 6.6 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 Fig.6.6 Dependence of extent of reaction on Kc the reaction (Fig. 6.7) : 6.6.2 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. 6.7 Predicting the direction of the reaction is defined in the same way as the equilibrium If Qc < Kc, net reaction goes from left to constant Kc except that the concentrations right in Qc are not necessarily equilibrium values. If Qc > Kc, net reaction goes from right to For a general reaction: left. aA+bB c C + d D (6.19) If Qc = Kc, no net reaction occurs. Qc = [C]c[D]d / [A]a[B]b (6.20) Problem 6.7 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 the reaction will proceed? direction of the products (forward reaction). 2024-25 Unit 6.indd 182 9/12/2022 11:58:25 AM  EQUILIBRIUM 183 Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc is found to be 9.15 bar. Calculate Kc, Kp and given by, partial pressure at equilibrium. Qc = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10–4M Qc = (3 ×10–4)(3 × 10 –4) / (3 ×10 –4)2 = 1 We know pV = nRT as Qc > Kc so the reaction will proceed in the Total volume (V ) = 1 L reverse direction. Molecular mass of N2O4 = 92 g 6.6.3 Calculating Equilibrium Number of moles = 13.8g/92 g = 0.15 Concentrations of the gas (n) In case of a problem in which we know the Gas constant (R) = 0.083 bar L mol–1K–1 initial concentrations but do not know any of Temperature (T ) = 400 K the equilibrium concentrations, the following three steps shall be followed: pV = nRT Step 1. Write the balanced equation for the p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 reaction. × 400 K Step 2. Under the balanced equation, make p = 4.98 bar a table that lists for each substance involved N2O4 2NO2 in the reaction: Initial pressure: 4.98 bar 0 (a) the initial concentration, At equilibrium: (4.98 – x) bar 2x bar (b) the change in concentration on going to Hence, equilibrium, and ptotal at equilibrium = pN + pNO 2O4 2 (c) the equilibrium concentration. 9.15 = (4.98 – x) + 2x In constructing the table, define x as the concentration (mol/L) of one of the substances 9.15 = 4.98 + x that reacts on going to equilibrium, then use x = 9.15 – 4.98 = 4.17 bar the stoichiometry of the reaction to determine Partial pressures at equilibrium are, the concentrations of the other substances in pN = 4.98 – 4.17 = 0.81bar terms of x. 2O4 pNO = 2x = 2 × 4.17 = 8.34 bar Step 3. Substitute the equilibrium 2   2 concentrations into the equilibrium equation K p = p NO2 / p N 2O4 for the reaction and solve for x. If you are = (8.34)2/0.81 = 85.87 to solve a quadratic equation choose the mathematical solution that makes chemical Kp = Kc(RT)∆n sense. 85.87 = Kc(0.083 × 400)1 Step 4. Calculate the equilibrium Kc = 2.586 = 2.6 concentrations from the calculated value of x. Step 5. Check your results by substituting Problem 6.9 them into the equilibrium equation. 3.00 mol of PCl5 kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the Problem 6.8 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 N 2O4 (g) 2NO2 (g) concentration: 3.0 0 0 2024-25 Unit 6.indd 183 9/12/2022 11:58:25 AM  184 chemistry Taking antilog of both sides, we get, Let x mol per litre of PCl5 be dissociated, At equilibrium: K = e–∆G/RT (6.23) (3-x) x x Hence, using the equation (6.23), the Kc = [PCl3][Cl2]/[PCl5] reaction spontaneity can be interpreted in terms of the value of ∆G . 1.8 = x2/ (3 – x) If ∆G  < 0, then –∆G /RT is positive, x2 + 1.8x – 5.4 = 0 and e –∆DG /RT>1, making K >1, which x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 implies a spontaneous reaction or the x = [–1.8 ± √3.24 + 21.6]/2 reaction which proceeds in the forward direction to such an extent that the x = [–1.8 ± 4.98]/2 products are present predominantly. x = [–1.8 + 4.98]/2 = 1.59 If ∆G  > 0, then –∆G /RT is negative, and [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M e –∆G [OH– ] Some substances like water are unique in their ability of acting both as an acid and a Neutral: [H3O+] = [OH– ] base. We have seen this in case of water in Basic : [H3O+] < [OH–] section 6.10.2. In presence of an acid, HA it accepts a proton and acts as the base while 6.11.2 The pH Scale in the presence of a base, B– it acts as an Hydronium ion concentration in molarity is acid by donating a proton. In pure water, one more conveniently expressed on a logarithmic H2O molecule donates proton and acts as an scale known as the pH scale. The pH of a acid and another water molecules accepts a solution is defined as the negative logarithm proton and acts as a base at the same time. The following equilibrium exists: to base 10 of the activity a H   of hydrogen 2024-25 Unit 6.indd 193 9/12/2022 11:58:26 AM  194 chemistry ion. In dilute solutions (< 0.01 M), activity when the hydrogen ion concentration, [H+] of hydrogen ion (H+) is equal in magnitude changes by a factor of 100, the value of pH to molarity represented by [H+]. It should changes by 2 units. Now you can realise why be noted that activity has no units and is the change in pH with temperature is often defined as: ignored. = [H+] / mol L–1 Measurement of pH of a solution is very From the definition of pH, the following essential as its value should be known when dealing with biological and cosmetic can be written, applications. The pH of a solution can be pH = – log aH+ = – log {[H+] / mol L–1} found roughly with the help of pH paper that Thus, an acidic solution of HCl (10–2 M) has different colour in solutions of different will have a pH = 2. Similarly, a basic solution pH. Now-a-days pH paper is available with of NaOH having [OH–] =10–4 M and [H3O+] = four strips on it. The different strips have 10–10 M will have a pH = 10. At 25 °C, pure different colours (Fig. 6.11) at the same pH. water has a concentration of hydrogen ions, The pH in the range of 1-14 can be determined [H + ] = 10–7 M. Hence, the pH of pure water is with an accuracy of ~0.5 using pH paper. given as: pH = –log(10–7) = 7 Acidic solutions possess a concentration of hydrogen ions, [H + ] > 10–7 M, while basic solutions possess a concentration of hydrogen ions, [H+] < 10–7 M. thus, we can summarise that Fig.6.11 pH-paper with four strips that may have Acidic solution has pH < 7 different colours at the same pH Basic solution has pH > 7 Neutral solution has pH = 7 For greater accuracy pH meters are used. pH meter is a device that measures the Now again, consider the equation (6.28) pH-dependent electrical potential of the test at 298 K solution within 0.001 precision. pH meters Kw = [H3O+] [OH–] = 10–14 of the size of a writing pen are now available Taking negative logarithm on both sides in the market. The pH of some very common of equation, we obtain substances are given in Table 6.5 (page 195). –log Kw = – log {[H3O+] [OH–]} Problem 6.16 = – log [H3O+] – log [OH–] The concentration of hydrogen ion in a = – log 10 –14 sample of soft drink is 3.8 × 10–3M. what pKw = pH + pOH = 14 (6.29) is its pH ? Note that although K w may change with temperature the variations in pH with Solution temperature are so small that we often pH = – log[3.8 × 10–3] ignore it. = – {log[3.8] + log[10–3]} pK w is a very important quantity for = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42 aqueous solutions and controls the relative Therefore, the pH of the soft drink is 2.42 concentrations of hydrogen and hydroxyl and it can be inferred that it is acidic. ions as their product is a constant. It should Problem 6.17 be noted that as the pH scale is logarithmic, a change in pH by just one unit also means Calculate pH of a 1.0 × 10 –8 M solution of change in [H+] by a factor of 10. Similarly, HCl. 2024-25 Unit 6.indd 194 9/12/2022 11:58:28 AM  EQUILIBRIUM 195 Table 6.5 The pH of Some Common Substances Name of the Fluid pH Name of the Fluid pH Saturated solution of NaOH ~15 Black Coffee 5.0 0.1 M NaOH solution 13 Tomato juice ~4.2 Lime water 10.5 Soft drinks and vinegar ~3.0 Milk of magnesia 10 Lemon juice