Equilibrium

Introduction

Equilibrium is a fundamental concept in chemistry that describes a state in which the forward and reverse processes occur at equal rates, resulting in no net change in the system. This dynamic balance can be observed in both physical and chemical processes. In a system at equilibrium, although the macroscopic properties remain constant, the microscopic processes continue to occur. This chapter introduces the concept of equilibrium, emphasizing its dynamic nature and importance in understanding various chemical and physical phenomena. Equilibrium is essential for explaining how reactions proceed and how conditions such as temperature, pressure, and concentration affect the system. The chapter also highlights the relevance of equilibrium in biological and environmental processes, such as oxygen transport by hemoglobin and the toxicity of carbon monoxide due to its interaction with hemoglobin. **Table on page 1 (1×2)** | Chemical equilibria are important in numerous biologica and environmental processes. For example, equilibri involving O molecules and the protein hemoglobin pla After studying this unit you will be 2 a crucial role in the transport and delivery of O from able to 2 our lungs to our muscles. Similar equilibria involving C • 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 escap • explain characteristics of the liquid surface into the vapour phase and number o equilibria involved in physical liquid molecules from the vapour phase strike the liqui and chemical processes; surface and are retained in the liquid phase. It gives ris • write expressions for equilibrium to a constant vapour pressure because of an equilibrium i constants; which the number of molecules leaving the liquid equals th • establish a relationship between number returning to liquid from the vapour. We say tha K and K; p c the system has reached equilibrium state at this stage • explain various factors that affect the equilibrium state of a However, this is not static equilibrium and there is a lot o 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 th bases according to Arrhenius, rate of condensation. It may be represented by Bronsted-Lowry and Lewis concepts; H O (l) H O (vap) 2 2 • classify acids and bases as The double half arrows indicate that the processe weak or strong in terms of their in both the directions are going on simultaneously. Th ionization constants; mixture of reactants and products in the equilibrium stat • explain the dependence of degree of ionization on concentration is called an equilibrium mixture. of the electrolyte and that of the Equilibrium can be established for both physica common ion; processes and chemical reactions. The reaction may b • describe pH scale for representing hydrogen ion concentration; fast or slow depending on the experimental conditions an • explain ionisation of water and the nature of the reactants. When the reactants in a close its duel role as acid and base; vessel at a particular temperature react to give products • describe ionic product (K ) and the concentrations of the reactants keep on decreasing w pK for water; w while those of products keep on increasing for some tim • appreciate use of buffer | | | --- | --- | | | l a y O . , e f d e n e t . f . e s e e l e d d , , e | **Table on page 7 (1×2)** | | | | --- | --- | | | ) m e a e s s | **Table on page 8 (8×9)** | of the purple colour remained constant an are the products in the balanced chemical equilibrium was attained. Similarly, fo equation. On the basis of experimental studies of many reversible reactions, the Norwegian experiments 5 and 6, the equilibrium wa chemists Cato Maximillian Guldberg and attained from the opposite direction. Peter Waage proposed in 1864 that the Data obtained from all six sets o concentrations in an equilibrium mixture experiments are given in Table 6.2. are related by the following equilibrium It is evident from the experiments 1, 2 equation, 3 and 4 that number of moles of dihydroge C D K  (6.1) reacted = number of moles of iodine reacte c AB = ½ (number of moles of HI formed). Also (6.1) where K is the equilibrium constant and experiments 5 and 6 indicate that, c the expression on the right side is called the [H (g)] = [I (g)] equilibrium constant expression. 2 eq 2 eq Knowing the above facts, in orde The equilibrium equation is also known as the law of mass action because in the early to establish a relationship betwee 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 u work better, let us consider reaction between consider the simple expression, gaseous H and I carried out in a sealed vessel 2 2 [HI(g)] / [H (g)] [I (g)] at 731K. eq 2 eq 2 eq H (g) + I (g)   2HI(g) It can be seen from Table 6.3 that if w 2 2 put the equilibrium concentrations of th 1 mol 1 mol 2 mol reactants and products, the above expressio Table 6.2 Initial and Equilibrium Concentrations of H , I and HI 2 2 | | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | **Table on page 14 (1×3)** | | At equilibrium: (0.48 – x)bar 2x bar p2 K = pCO p CO 2 K = (2x)2/(0.48 – x) = 3 p | 5. The equilibrium constant K for a reactio is related to the equilibrium constan of the corresponding reaction, whos equation is obtained by multiplying o dividing the equation for the origina reaction by a small integer. Let us consider applications of equilibriu constant to: | | --- | --- | --- | | 4x2 = 3(0.48 – x) • predict the extent of a reaction on th 4x2 = 1.44 – x basis of its magnitude, 4x2 + 3x – 1.44 = 0 • predict the direction of the reaction, an a = 4, b = 3, c = –1.44 • calculate equilibrium concentrations. b 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 equilibriu constant for a reaction indicates the exten = (–3 ± 5.66)/8 of the reaction. But it is important to not = (–3 + 5.66)/8 (as value of x cannot be that an equilibrium constant does not giv negative hence we neglect that value) any information about the rate at whic x = 2.66/8 = 0.33 the equilibrium is reached. The magnitud The equilibrium partial pressures are, of K or K is directly proportional to th c p concentrations of products (as these appea p = 2x = 2 × 0.33 = 0.66 bar CO2 in the numerator of equilibrium constan p = 0.48 – x = 0.48 – 0.33 = 0.15 bar expression) and inversely proportional to th CO2 concentrations of the reactants (these appea in the denominator). This implies that a hig 6.6 APPLICATIONS OF EQUILIBRIUM value of K is suggestive of a high concentratio CONSTANTS of products and vice-versa. Before considering the applications of We can make the following generalisation equilibrium constants, let us summarise the concerning the composition of equilibriu important features of equilibrium constants mixtures: as follows: 1. Expression for equilibrium constant is • If K > 103, products predominate ove c applicable only when concentrations of reactants, i.e., if K is very large, th c 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 with O at 500 2 2 independent of initial concentrations of has a very large equilibrium constan the reactants and products. K = 2.4 × 1047. c 3. Equilibrium constant is temperature (b) H (g) + Cl (g) 2HCl(g) at 300K ha 2 2 dependent having one unique value for K = 4.0 × 1031. c a particular reaction represented by a (c) H (g) + Br (g) 2HBr (g) at 300 K balanced equation at a given temperature. 2 2 K = 5.4 × 1018 c 4. The equilibrium constant for the reverse | 4x2 = 3(0.48 – x) 4x2 = 1.44 – x 4x2 + 3x – 1.44 = 0 a = 4, b = 3, c = –1.44 b b2 4ac x  2a = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 = (–3 ± 5.66)/8 = (–3 + 5.66)/8 (as value of x cannot be negative hence we neglect that value) x = 2.66/8 = 0.33 The equilibrium partial pressures are, p = 2x = 2 × 0.33 = 0.66 bar CO2 p = 0.48 – x = 0.48 – 0.33 = 0.15 bar CO2 | | **Table on page 16 (3×4)** | | Solution For the reaction the reaction quotient Q is c given by, Q = [B][C]/ [A]2 c as [A] = [B] = [C] = 3 × 10–4M Q = (3 ×10–4)(3 × 10–4) / (3 ×10–4)2 = 1 c as Q > K so the reaction will proceed in the c c reverse direction. | | The total pressure at equilbrium was found to be 9.15 bar. Calculate K, K and c p partial pressure at equilibrium. Solution We know pV = nRT Total volume (V ) = 1 L Molecular mass of N O = 92 g 2 4 Number of moles = 13.8g/92 g = 0.15 of the gas (n) Gas constant (R) = 0.083 bar L mol–1K–1 Temperature (T ) = 400 K pV = nRT p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 × 400 K p = 4.98 bar N O 2NO 2 4 2 Initial pressure: 4.98 bar 0 At equilibrium: (4.98 – x) bar 2x bar Hence, p at equilibrium = p + p total N2O4 NO2 9.15 = (4.98 – x) + 2x 9.15 = 4.98 + x x = 9.15 – 4.98 = 4.17 bar Partial pressures at equilibrium are, p = 4.98 – 4.17 = 0.81bar N2O4 p = 2x = 2 × 4.17 = 8.34 bar NO2 2 K = p  /p p NO N 2O 2 4 = (8.34)2/0.81 = 85.87 K = K (RT)∆n p c 85.87 = K(0.083 × 400)1 c K = 2.586 = 2.6 c Problem 6.9 3.00 mol of PCl kept in 1L closed reaction 5 vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. K= 1.80 c Solution PCl PCl + Cl 5 3 2 Initial concentration: 3.0 0 0 | | --- | --- | --- | --- | | 6.6.3 Calculating Equilibrium Concentrations In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed: Step 1. Write the balanced equation for the reaction. Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction: (a) the initial concentration, (b) the change in concentration on going to equilibrium, and (c) the equilibrium concentration. In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x. Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense. Step 4. Calculate the equilibrium concentrations from the calculated value of x. Step 5. Check your results by substituting them into the equilibrium equation. Problem 6.8 13.8g of N O was placed in a 1L reaction 2 4 | | | Molecular mass of N O = 92 g 2 4 Number of moles = 13.8g/92 g = 0.15 of the gas (n) Gas constant (R) = 0.083 bar L mol–1K–1 Temperature (T ) = 400 K pV = nRT p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 × 400 K p = 4.98 bar N O 2NO 2 4 2 Initial pressure: 4.98 bar 0 At equilibrium: (4.98 – x) bar 2x bar Hence, p at equilibrium = p + p total N2O4 NO2 9.15 = (4.98 – x) + 2x 9.15 = 4.98 + x x = 9.15 – 4.98 = 4.17 bar Partial pressures at equilibrium are, p = 4.98 – 4.17 = 0.81bar N2O4 p = 2x = 2 × 4.17 = 8.34 bar NO2 2 K = p  /p p NO N 2O 2 4 = (8.34)2/0.81 = 85.87 K = K (RT)∆n p c 85.87 = K(0.083 × 400)1 c K = 2.586 = 2.6 c Problem 6.9 3.00 mol of PCl kept in 1L closed reaction 5 vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. K= 1.80 c Solution | | | Problem 6.8 13.8g of N O was placed in a 1L reaction 2 4 | | | | | vessel at 400K and allowed to attain equilibrium N O (g) 2NO (g) 2 4 2 | | | **Table on page 17 (2×4)** | | Let x mol per litre of PCl be dissociated, 5 At equilibrium: (3-x) x x K = [PCl ][Cl ]/[PCl ] c 3 2 5 1.8 = x2/ (3 – x) x2 + 1.8x – 5.4 = 0 x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 | | | | --- | --- | --- | --- | | x = [–1.8 ± √3.24 + 21.6]/2 reaction which proceeds in the forwar direction to such an extent that th 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, an [PCl ] = 3.0 – x = 3 –1.59 = 1.41 M 5 e –∆G </RT 1, that is , K < 1, which implie [PCl ] = [Cl ] = x = 1.59 M a non-spontaneous reaction or a reactio 3 2 which proceeds in the forward directio 6.7 Relationship between to such a small degree that only a ver Equilibrium Constant K, minute quantity of product is formed. Reaction Quotient Q and Gibbs Energy G Problem 6.10 The value of K for a reaction does not depend The value of ∆G  for the phosphorylation of c on the rate of the reaction. However, as you glucose in glycolysis is 13.8 kJ/mol. Find have studied in Unit 5, it is directly related the value of K at 298 K. c to the thermodynamics of the reaction and Solution in particular, to the change in Gibbs energy, ∆G. If, ∆G  = 13.8 kJ/mol = 13.8 × 103J/mol • ∆G is negative, then the reaction is Also, ∆G  = – RT lnK spontaneous and proceeds in the forward c Hence, ln K = –13.8 × 103J/mol direction. c (8.314 J mol–1K–1 × 298 K) • ∆G is positive, then reaction is considered ln K = – 5.569 non-spontaneous. Instead, as reverse c reaction would have a negative ∆G, the K = e–5.569 c products of the forward reaction shall be K = 3.81 × 10–3 c converted to the reactants. Problem 6.11 • ∆G is 0, reaction has achieved equilibrium; Hydrolysis of sucrose gives, at this point, there is no longer any free energy left to drive the reaction. Sucrose + H O Glucose + Fructose 2 A mathematical expression of this Equilibrium constant K for the reaction is c thermodynamic view of equilibrium can be 2 ×1013 at 300K. Calculate ∆G  at 300K. described by the following equation: Solution ∆G = ∆G + RT lnQ (6.21) ∆G  = – RT lnK where, G is standard Gibbs energy. ∆G  = – 8.314Jc mol–1K–1× At equilibrium, when ∆G = 0 and Q = K, 300K × ln(2×1013) c the equation (6.21) becomes, ∆G  = – 7.64 ×104 J mol–1 | x = [–1.8 ± √3.24 + 21.6]/2 x = [–1.8 ± 4.98]/2 x = [–1.8 + 4.98]/2 = 1.59 [PCl ] = 3.0 – x = 3 –1.59 = 1.41 M 5 [PCl ] = [Cl ] = x = 1.59 M 3 2 | | | | | | Problem 6.10 The value of ∆G  for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of K at 298 K. c Solution ∆G  = 13.8 kJ/mol = 13.8 × 103J/mol Also, ∆G  = – RT lnK c Hence, ln K = –13.8 × 103J/mol c (8.314 J mol–1K–1 × 298 K) ln K = – 5.569 c K = e–5.569 c K = 3.81 × 10–3 c Problem 6.11 Hydrolysis of sucrose gives, Sucrose + H O Glucose + Fructose 2 Equilibrium constant K for the reaction is c 2 ×1013 at 300K. Calculate ∆G  at 300K. Solution ∆G  = – RT lnK c ∆G  = – 8.314J mol–1K–1× 300K × ln(2×1013) ∆G  = – 7.64 ×104 J mol–1 | |

• Equilibrium occurs when forward and reverse processes happen at equal rates.
• It is a dynamic state with no net change in macroscopic properties.
• Equilibrium applies to both physical and chemical processes.
• Understanding equilibrium is crucial for predicting reaction behavior.
• Biological processes like oxygen transport involve chemical equilibria.
• Environmental phenomena also depend on equilibrium concepts.
• 📌 Equilibrium: A state where forward and reverse reaction rates are equal.
• 📌 Dynamic equilibrium: Continuous microscopic changes with no macroscopic change.

Equilibrium in Physical Processes

Physical processes such as evaporation, condensation, and dissolution often reach a state of dynamic equilibrium when conducted in a closed system. For example, when a liquid is placed in a closed container, molecules with relatively higher kinetic energy escape from the liquid surface into the vapor phase (evaporation). Simultaneously, vapor molecules strike the liquid surface and return to the liquid phase (condensation). Initially, evaporation dominates, increasing vapor pressure. However, as vapor concentration increases, the rate of condensation also increases until both rates become equal, establishing dynamic equilibrium. At this point, the vapor pressure becomes constant and is called the equilibrium vapor pressure. This equilibrium is temperature-dependent. Similar principles apply to the dissolution of solids in liquids, where the rate of dissolution equals the rate of crystallization at equilibrium. These equilibria are characterized by constant measurable properties such as vapor pressure or concentration at a given temperature. Importantly, equilibrium in physical processes can only be achieved in a closed system where no matter escapes or enters. **Table on page 3 (1×5)** | | | | | | | --- | --- | --- | --- | --- | | 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 th decreases with time due to condensation rate of evaporation. These are open system of vapour into water. Finally it leads to an and it is not possible to reach equilibrium i 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 position at atmospheric pressure (1.013 bar into the liquid state also increases till the and at 100°C in a closed vessel. The boilin 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 atmospheri H O(l) H O (vap) 2 2 pressure (1.013 bar), the temperatur At equilibrium the pressure exerted by at which the liquid and vapours are a the water molecules at a given temperature equilibrium is called normal boiling point o remains constant and is called the equilibrium the liquid. Boiling point of the liquid depend vapour pressure of water (or just vapour on the atmospheric pressure. It depends o pressure of water); vapour pressure of water the altitude of the place; at high altitude th 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 solid pressures at the same temperature, and the sublime to vapour phase. If we place soli liquid which has a higher vapour pressure is iodine in a closed vessel, after sometim more volatile and has a lower boiling point. the vessel gets filled up with violet vapou If we expose three watch glasses containing and the intensity of colour increases wit separately 1mL each of acetone, ethyl alcohol, time. After certain time the intensity o and water to atmosphere and repeat the colour becomes constant and at this stag experiment with different volumes of the equilibrium is attained. Hence solid iodin liquids in a warmer room, it is observed sublimes to give iodine vapour and the iodin that in all such cases the liquid eventually vapour condenses to give solid iodine. Th 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 I (solid) I (vapour) 2 2 temperature. When the watch glass is open Other examples showing this kind o | | | | | **Table on page 4 (6×4)** | partial pressure in the atmosphere. This i temperature, sugar crystals separate out if we how the soda water in bottle when left ope cool the syrup to the room temperature. We call to the air for some time, turns ‘flat’. It can b it a saturated solution when no more of solute generalised that: can be dissolved in it at a given temperature. (i) For solid liquid equilibrium, there i The concentration of the solute in a saturated only one temperature (melting point) a solution depends upon the temperature. In 1 atm (1.013 bar) at which the tw a saturated solution, a dynamic equilibrium phases can coexist. If there is n exits between the solute molecules in the solid exchange of heat with the surroundings state and in the solution: the mass of the two phases remain Sugar (solution) Sugar (solid), and constant. the rate of dissolution of sugar = rate of (ii) For liquid vapour equilibrium, th crystallisation of sugar. vapour pressure is constant at a give Equality of the two rates and dynamic temperature. nature of equilibrium has been confri med with (iii) For dissolution of solids in liquids the help of radioactive sugar. If we drop some the solubility is constant at a give radioactive sugar into saturated solution of temperature. non-radioactive sugar, then after some time 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 liqui radioactive sugar molecules in the solution is proportional to the pressur but due to dynamic nature of equilibrium, (concentration) of the gas over the liquid there is exchange between the radioactive These observations are summarised i and non-radioactive sugar molecules between Table 6.1 the two phases. The ratio of the radioactive Table 6.1 Some Features of Physical Equilibr to non-radioactive molecules in the solution increases till it attains a constant value. Process Conclusion Liquid Vapour p constant at given Gases in liquids H2O H O (l) H O (g) temperature When a soda water bottle is opened, some of 2 2 Solid Liquid Melting point is fixed at the carbon dioxide gas dissolved in it fizzes constant pressure out rapidly. The phenomenon arises due H O (s) H O (l) 2 2 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 CO (gas) CO (in solution) constant at a given 2 2 | | | s n e s t o o , s e n , n , d e . n ia | | --- | --- | --- | --- | | | Process | Conclusion | | | | Liquid Vapour H O (l) H O (g) 2 2 | p constant at given H2O temperature | | | | Solid Liquid H O (s) H O (l) 2 2 | Melting point is fixed at constant pressure | | | | Solute(s) Solute (solution) Sugar(s) Sugar (solution) | Concentration of solute in solution is constant at a given temperature | | | | Gas(g) Gas (aq) | [gas(aq)]/[gas(g)] is constant at a given | | | This equilibrium is governed by Henry’s law, which states that the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the | CO (g) CO (aq) 2 2 | | |

• Dynamic equilibrium occurs when evaporation and condensation rates are equal.
• Equilibrium vapor pressure is constant at a given temperature.
• Physical equilibria require a closed system to be maintained.
• Dissolution and crystallization rates equalize at equilibrium in saturated solutions.
• Measurable properties like vapor pressure remain constant at equilibrium.
• Temperature influences the equilibrium state in physical processes.
• 📌 Equilibrium vapor pressure: The pressure exerted by vapor in equilibrium with its liquid.
• 📌 Saturated solution: A solution where no more solute can dissolve at a given temperature.